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Chapter 2
Mechanics Problems from the Functional
Analysis Viewpoint




In the past, an engineer could calculate mechanical stresses and strains using a pen-
cil and a logarithmic slide rule. Modern mechanical models, on the other hand, are
nonlinear, and even the linear models are complicated. Numerical methods in struc-
tural dynamics cannot be applied without computers running specialized programs.
However, a researcher should have a solid grasp of the equations that underlie a
numerical model and the types of results that can be expected. New models ap-
pear in mechanics on a regular basis. Some of these, when written out in detail, can
span multiple pages and are clearly beyond pencil-and-paper approaches. Although
functional analysis does not provide a detailed picture of the results to be expected
from a complicated model, it can answer questions regarding whether the problem
is mathematically well-posed (e.g., whether a solution exists and is unique). It may
also indicate whether the results can be obtained by a general computer program or
whether a special program, based on a knowledge of the general properties of the
model, is required. In other words, functional analysis can provide valuable insight
even to those who rely heavily on numerical approaches.



2.1 Introduction to Sobolev Spaces

In his famous book [33], S.L. Sobolev (1908–1989) introduced some normed spaces
that now bear his name; they are denoted by W m,p (Ω). The norm in W m,p (Ω) is
                                   ⎛                       ⎞1/p
                                   ⎜
                                   ⎜
                                   ⎜                       ⎟
                                                           ⎟
                                                           ⎟
                                u =⎜
                                   ⎜
                                   ⎜
                                   ⎝               |D u| dΩ⎟ ,
                                                    α   p  ⎟
                                                           ⎟
                                                           ⎠                                 (2.1.1)
                                         Ω |α|≤m


where m is an integer, p ≥ 1, and Ω is compact in Rn . (The Dα notation was in-
troduced on p. 19.) Indeed this is a norm on the set of functions that possess con-
tinuous derivatives up to order m on Ω, with satisfaction of axiom N3 ensured by
Minkowski’s inequality (1.3.6). The completion of the resulting normed space is the
Banach space W m,p (Ω).


L.P. Lebedev et al., Functional Analysis in Mechanics, Springer Monographs in Mathematics,      115
DOI 10.1007/978-1-4614-5868-5_2, © Springer Science+Business Media, LLC 2013
116                              2 Mechanics Problems from the Functional Analysis Viewpoint

   For Ω a segment [a, b], the spaces W m,p (a, b) were introduced by Stefan Banach
(1892–1945) in his dissertation. Our interest in Sobolev spaces is clear, as the ele-
ments of each of our energy spaces will belong to W m,2 (Ω) for some m.
Problem 2.1.1. (a) Demonstrate that if f (x) ∈ W 1,2 (a, b) and g(x) ∈ C (1) (a, b), then
f (x)g(x) ∈ W 1,2 (a, b). (b) Show that A f (x) = f (x) defines a bounded linear operator
acting from W 1,2 (0, 1) to L2 (0, 1).
Generalized Notions of Derivative. For u ∈ L p (Ω), K.O. Friedrichs (1901–1982)
[12] introduced the notion of strong derivative, calling v ∈ L p (Ω) a strong derivative
Dα (u) if there is a sequence {ϕn } ⊂ C (∞) (Ω) such that

          |u(x) − ϕn (x)| p dΩ → 0 and           |v(x) − Dα ϕn (x)| p dΩ → 0 as n → ∞ .
      Ω                                      Ω

Since C (∞) (Ω) is dense in any C (k) (Ω), an element of W m,p (Ω) has all strong deriva-
tives up to the order m lying in L p (Ω).
   Another version of generalized derivative was proposed by Sobolev. He used,
along with the classical integration by parts formula, an idea from the classical cal-
culus of variations: if
                                       u(x) ϕ(x) dΩ = 0
                                      Ω
for all infinitely differentiable functions ϕ(x) having compact support in an open
domain Ω, then u(x) = 0 almost everywhere (everywhere if u(x) is to be continuous).
We say that a function ϕ(x), infinitely differentiable in Ω, has compact support in
Ω ⊂ Rn if the closure of the set

                                  M = {x ∈ Ω : ϕ(x)         0}

is a compact set in Ω. In the sense introduced by Sobolev, v ∈ L p (Ω) is called a weak
derivative Dα u of u ∈ L p (Ω) if for every infinitely differentiable function ϕ(x) with
compact support in Ω we have

                           u(x) Dα ϕ(x) dΩ = (−1)|α|        v(x) ϕ(x) dΩ .          (2.1.2)
                       Ω                                Ω

   The two definitions of generalized derivative are equivalent [33]. The proof
would exceed the scope of our presentation, and the same is true for some other
facts of this section.
Imbedding Theorems. The results presented next are particular cases of Sobolev’s
imbedding theorem. Later we will give proofs of some particular results when study-
ing certain energy spaces for elastic models which turn out to be subspaces of
Sobolev spaces. The theorems are formulated in terms of imbedding operators, a
notion elucidated further in Sect. 2.2.
   We assume that the compact set Ω ⊂ Rn satisfies the cone condition. This means
there is a finite circular cone in Rn such that any point of the boundary of Ω can
be touched by the vertex of the cone while the cone lies fully inside Ω. This is the
2.1 Introduction to Sobolev Spaces                                                     117

condition under which Sobolev’s imbedding theorem is proved. We denote by Ωr an
r-dimensional piecewise smooth hypersurface in Ω. (This means that, at any point
of smoothness, in a local coordinate system, it is described by functions having all
derivatives continuous up to order m locally, if we consider W m,p (Ω).)
   The theory of Sobolev spaces is a substantial branch of mathematics (see Adams
[1], Lions and Magenes [25], etc.). We formulate only what is needed for our pur-
poses. This is Sobolev’s imbedding theorem with some extensions:

Theorem 2.1.1 (Sobolev–Kondrashov). The imbedding operator of the Sobolev
space W m,p (Ω) to Lq (Ωr ) is continuous if one of the following conditions holds:
(i)      n > mp, r > n − mp, q ≤ pr/(n − mp);
(ii)     n = mp, q is finite with q ≥ 1.
If n < mp, then the space W m,p (Ω) is imbedded into the H¨ lder space H α (Ω) when
                                                           o
α ≤ (mp − n)/p, and the imbedding operator is continuous.
    The imbedding operator of W m,p (Ω) to Lq (Ωr ) is compact (i.e., takes every
bounded set of W m,p (Ω) into a precompact set of the corresponding space1 ) if
(i)      n > mp, r > n − mp, q < pr/(n − mp) or
(ii)     n = mp and q is finite with q ≥ 1.
If n < mp then the imbedding operator is compact to H α (Ω) when α < (mp − n)/p.

   Note that this theorem provides imbedding properties not only for functions but
also for their derivatives:

                     u ∈ W m,p (Ω) =⇒ Dα u ∈ W m−k,p (Ω) for |α| = k .

Also available are stricter results on the imbedding of Sobolev spaces on Ω into
function spaces given on manifolds Ωr of dimension less than n; however, such
trace theorems require an extended notion of Sobolev spaces.
   Let us formulate some useful consequences of Theorem 2.1.1.

Theorem 2.1.2. Let γ be a piecewise differentiable curve in a compact set Ω ⊂ R2 .
For any finite q ≥ 1, the imbedding operator of W 1,2 (Ω) to the spaces Lq (Ω) and
Lq (γ) is continuous (and compact), i.e.,

                             max{ u    Lq (Ω) ,   u   Lq (γ) }   ≤m u   W 1,2 (Ω)   (2.1.3)

with a constant m which does not depend on u(x).

Theorem 2.1.3. Let Ω ⊂ R2 be compact. If α ≤ 1, the imbedding operator of
W 2,2 (Ω) to H α (Ω) is continuous; if α < 1, it is compact. For the first derivatives,
the imbedding operator to Lq (Ω) and Lq (γ) is continuous (and compact) for any
finite q ≥ 1.


1
    The notion of compact operator will be explored in Chap. 3.
118                                2 Mechanics Problems from the Functional Analysis Viewpoint

Theorem 2.1.4. Let γ be a piecewise smooth surface in a compact set Ω ⊂ R3 .
The imbedding operator of W 1,2 (Ω) to Lq (Ω) when 1 ≤ q ≤ 6, and to L p (γ) when
1 ≤ p ≤ 4, is continuous; if 1 ≤ q < 6 or 1 ≤ p < 4, respectively, then it is compact.

   We merely indicate how such theorems are proved. We will establish similar re-
sults for the beam problem (see (2.3.5)) and for the clamped plate problem (see
(2.3.21)) via integral representations of functions from certain classes. In like man-
ner, Sobolev’s original proof is given for Ω a union of bounded star-shaped domains.
A domain is called star-shaped with respect to a ball B if any ray with origin in B
intersects the boundary of the domain only once. For a domain Ω which is bounded
and star-shaped with respect to a ball B, a function u(x) ∈ C (m) (Ω) can be repre-
sented in the form

                                         α          α
                  u(x) =                x1 1 · · · xn n         Kα (y) u(y) dΩ
                            |α|≤m−1                         B

                                           1
                            +                                     Kα (x, y) Dα u(y) dΩy         (2.1.4)
                                   Ω   |x − y|n−m         |α|=m

where Kα (y) and Kα (x, y) are continuous functions. Investigating properties of the
integral terms on the right-hand side of the representation (2.1.4), Sobolev formu-
lated his results; later they were extended to more general domains.
   Another method is connected with the Fourier transformation of functions. In the
case of W m,2 (Ω), it is necessary to extend functions of C (m) (Ω) outside Ω in such a
way that they belong to C m (Rn ) and W m,2 (Rn ). Then using the Fourier transforma-
tion
                         u(y) = (2π)−n/2
                         ˆ                  e−ix·y u(x) dx1 · · · dxn
                                                     Rn
along with the facts that

                                       u(x)   L2 (Rn )    = u(y)
                                                            ˆ         L2 (Rn )

and
                                Dα u(x) = (iy1 )α1 · · · (iyn )αn u(y)
                                                                  ˆ
for u ∈ L2 (Rn ), we can present the norm in W m,2 (Ω) in the form

                      u(x)      2
                                W m,2 (Rn )   =            yα1 · · · yαn u(y)
                                                            1         n ˆ
                                                                                 2
                                                                                 L2 (Rn )   .
                                                  |α|≤m


We can then study the properties of the weighted space Lw (Rn ); this transformed
                                                               2

problem is simpler, as many of the problems involved are algebraic estimates of
Fourier images.
    Moreover, we can consider W p,2 (Rn ) with fractional indices p. These lead to nec-
essary and sufficient conditions for the trace problem: given W m,2 (Ω), find the space
W p,2 (∂Ω) in which W m,2 (Ω) is continuously imbedded. The inverse trace problem
is, given W p,2 (Ω), find the maximal index m such that every element u ∈ W p,2 (∂Ω)
2.2 Operator of Imbedding                                                           119

can be extended to Ω, u∗ ∈ W m,2 (Ω), in such a way that

                               u∗    W m,2 (Ω)   ≤c u     W p,2 (∂Ω)   .

In this way, many results from the contemporary theory of elliptic (and other types
of) equations and systems are obtained. We should mention that the trace theorems
are formulated mostly for smooth manifolds, hence are not applicable to practical
problems involving domains with corners.
    As stated in Lemma 1.16.3, all Sobolev spaces are separable. The same holds for
all the energy spaces we introduce.



2.2 Operator of Imbedding

Let Ω be a Jordan measurable compact set in Rn . Any element of C (k) (Ω) also be-
longs to C(Ω). This correspondence — between an element of the space C (k) (Ω) and
the same element in the space C(Ω) — is a linear operator. Although it resembles an
identity operator, its domain and range differ and it is called the imbedding operator
from C (k) (Ω) to C(Ω). Clearly, C (k) (Ω) is a proper subset of C(Ω); this situation is
typical for an imbedding operator that gives the correspondence between the same
elements but considered in different spaces. For a function u that belongs to both
spaces, we have
                                  u C(Ω) ≤ u C (k) (Ω) ,
which shows that the imbedding operator is bounded or continuous (what can be said
about its norm?). Moreover, by a consequence of Arzel` ’s theorem, it is compact.
                                                       a
   We get a similar imbedding operator when considering the spaces p and q for
1 ≤ p < q. As (1.4.3) states, any element x of p belongs to q and

                                       x   q    ≤ c1 x     p   .

Here and below, the constants ck are independent of the element of the spaces, so
the imbedding operator from p to q is bounded.
   We encounter another imbedding operator when considering the correspondence
between the same equivalence classes in the spaces L p (Ω) and Lq (Ω): if 1 ≤ p < q,
then any element u of Lq (Ω) belongs to L p (Ω) and

                                 u    L p (Ω)    ≤ c2 u   Lq (Ω)   .

So the imbedding operator from Lq (Ω) to L p (Ω) is also bounded. (Note that the
relations between p and q are different in the spaces r and Lr (Ω).)
    The situation is slightly different when we consider the relation between the
spaces C(Ω) and L p (Ω). It is clear that any function u from C(Ω) has a finite norm in
L p (Ω). To use the imbedding idea, we identify the function u with the equivalence
class from L p (Ω) that contains the stationary sequence (u, u, . . .), and in this way
120                             2 Mechanics Problems from the Functional Analysis Viewpoint

consider u as an element of L p (Ω). Because

                                         u dΩ ≤ c3 u             C(Ω)   ,
                                     Ω

the imbedding operator from C(Ω) to L p (Ω) is bounded for any 1 ≤ p < ∞.
    This practice of identification must be extended when we consider imbedding
operators involving Sobolev spaces as treated in this book. In the imbedding from
W 1,2 (Ω) to L p (Ω), the elements of W 1,2 (Ω) — the equivalence classes of functions
from C (1) (Ω) — are different from the equivalence classes from C(Ω) that constitute
L p (Ω); however, each class from C (1) (Ω) is contained by some class from C(Ω), and
we identify these classes. In this sense, we say that W 1,2 (Ω) is imbedded into L p (Ω)
and the imbedding operator is continuous (or even compact).
    The imbedding of a Sobolev space into the space of continuous functions is even
more complicated, although the general ideas are the same. We will consider this in
more detail when dealing with the energy space for a plate. In this case we cannot
directly identify an element of a Sobolev space, which is an equivalence class, with
a continuous function. However, we observe that for a class of equivalent sequences
in the norm of a Sobolev space, for any sequence that enters into the class, there is
a limit element, a continuous function that does not differ for other representative
sequences, and we identify the whole class with this function. The inequality relat-
ing the Sobolev norm of the class and the continuous norm of this function states
that this identification or correspondence is a bounded operator whose linearity is
obvious. Some other inequalities state that the operator is compact.



2.3 Some Energy Spaces

A Beam. Earlier we noted that the set S of all real-valued continuous functions y(x)
having continuous first and second derivatives on [0, l] and satisfying the boundary
conditions
                          y(0) = y (0) = y(l) = y (l) = 0                    (2.3.1)
is a metric space under the metric
                                             l                                    1/2
                                 1
                    d(y, z) =                    B(x)[y (x) − z (x)]2 dx                .   (2.3.2)
                                 2       0

We called this an energy space for the clamped beam. We can introduce an inner
product
                                 1 l
                        (y, z) =      B(x)y (x)z (x) dx                (2.3.3)
                                 2 0
and norm
                                                     l                      1/2
                                     1
                            y =                          B(x)[y (x)]2 dx
                                     2           0
2.3 Some Energy Spaces                                                                                                          121

such that d(y, z) = y − z . But this space is not complete (it is clear that there
are Cauchy sequences whose limits do not belong to C (2) (0, l); the reader should
construct an example). To have a complete space, we must apply the completion
theorem. The actual energy space denoted by E B is the completion of S in the metric
(2.3.2) (or, what amounts to the same thing, in the inner product (2.3.3)).
    Let us consider some properties of the elements of E B . An element y(x) ∈ E B is a
set of Cauchy sequences equivalent in the metric (2.3.2). Let {yn (x)} be a represen-
tative sequence of y(x). Then, according to (2.3.2),
                            l                                                     1/2
                1
                                B(x)[yn (x) − ym (x)]2 dx                               → 0 as n, m → ∞ .
                2       0

If we assume that
                                            0 < m1 ≤ B(x) ≤ m2 ,
then the sequence of second derivatives {yn (x)} is a Cauchy sequence in the norm of
L2 (0, l) because
                        l                                                     l
             m1             [yn (x) − ym (x)]2 dx ≤                               B(x)[yn (x) − ym (x)]2 dx .
                    0                                                     0

Hence {yn (x)} is a representative sequence of some element of L2 (0, l), and we can
regard y ∈ Eb as having a second derivative that belongs to L2 (0, l).
   Now consider {yn (x)}. For any y(x) ∈ S we get
                                                                      x
                                               y (x) =                    y (t) dt .
                                                                  0

So for a representative {yn (x)} of a class y(x) ∈ E B we have
                                           x                                                  l
          |yn (x) − ym (x)| ≤                  |yn (t) − ym (t)| dt ≤                             1 · |yn (t) − ym (t)| dt
                                       0                                                  0
                                                      l                                             1/2
                                   ≤ l1/2                 [yn (x) − ym (x)]2 dx
                                                  0
                                                                  l                                              1/2
                                   ≤ (l/m1 )1/2                       B(x)[yn (x) − ym (x)]2 dx
                                                              0

                                   → 0 as n, m → ∞ .                                                                         (2.3.4)

It follows that {yn (x)} converges uniformly on [0, l]; hence there exists a limit func-
tion z(x) which is continuous on [0, l]. This function does not depend on the choice
of representative sequence (Problem 2.3.1 below). The same holds for a sequence
of functions {yn (x)}: its limit is a function y(x) continuous on [0, l]. Moreover,

                                                      y (x) = z(x) .
122                           2 Mechanics Problems from the Functional Analysis Viewpoint

To prove this, it is necessary to repeat the arguments of Sect. 1.11 on the differen-
tiability of the elements of C (k) (Ω), with due regard for (2.3.4). From (2.3.4) and the
similar inequality for {yn (x)} we get
                                                            l                     1/2
                                                    1
                 max |y(x)| + |y (x)| ≤ m                       B(x)[y (x)]2 dx         (2.3.5)
                 x∈[0,l]                            2   0

for some constant m independent of y(x) ∈ E B . So each element y(x) ∈ E B can be
identified with an element y(x) ∈ C (1) (0, l) in such a way that (2.3.5) holds. This
correspondence is an imbedding operator, and we interpret (2.3.5) as a statement
that the imbedding operator from E B to C (1) (0, l) is continuous (cf., Sect. 1.21).
Henceforth we refer to the elements of E B as if they were continuously differentiable
functions, attaching the properties of the uniquely determined limit functions to the
corresponding elements of E B themselves.

Problem 2.3.1. Prove that the function z(x), discussed above, is independent of the
choice of representative sequence.

   We are interested in analyzing all terms that appear in the statement of the equi-
librium problem for a body as a minimum potential energy problem. So we will
consider the functional giving the work of external forces. For the beam, it is
                                             l
                                    A=           F(x)y(x) dx .
                                         0

This is well-defined on E B if F(x) ∈ L(Ω); moreover, (2.3.5) shows that the work of
external forces can contain terms of the form

                                    [Fk y(xk ) + Mk y (xk )] ,
                                k

which can be interpreted as the work of point forces and point moments, respec-
tively. This is a consequence of the continuity of the imbedding from E B to C (1) (0, l).

Remark 2.3.1. Alternatively we can define E B on a base set S 1 of smoother func-
tions, in C (4) (0, l) say, satisfying (2.3.1). The result is the same, since S 1 is dense
in S with respect to the norm of C (2) (0, l). However, sometimes such a definition is
convenient.

Remark 2.3.2. Readers familiar with the contemporary literature in this area may
have noticed that Western authors usually deal with Sobolev spaces, studying the
properties of operators corresponding to problems under consideration; we prefer to
deal with energy spaces, studying first their properties and then those of the corre-
sponding operators. Although these approaches lead to the same results, in our view
the physics of a particular problem should play a principal role in the analysis —
in this way the methodology seems simpler, clearer, and more natural. In papers de-
voted to the study of elastic bodies, we mainly find interest in the case of a clamped
boundary. Sometimes this is done on the principle that it is better to deal solely with
2.3 Some Energy Spaces                                                                            123

homogeneous Dirichlet boundary conditions, but often it is an unfortunate conse-
quence of the use of the Sobolev spaces H k (Ω). The theory of these spaces is well
developed but is not amenable to the study of other boundary conditions. Success
in the investigation of mechanics problems can be much more difficult without the
benefit of the physical ideas brought out by the energy spaces.

Remark 2.3.3. In defining the energy space of the beam, we left aside the question
of smoothness of the stiffness function B(x). From a mathematical standpoint this
is risky since, in principle, B(x) can be nonintegrable. But in the case of an actual
physical beam, B(x) can have no more than a finite number of discontinuities and
must be differentiable everywhere else. For simplicity, we shall continue to make
realistic assumptions concerning physical parameters such as stiffness and elastic
constants; in particular, we shall suppose whatever degree of smoothness is required
for our purposes.

A Membrane (Clamped Edge). The subset of C (1) (Ω) consisting of all functions
satisfying
                           u(x, y) ∂Ω = 0                              (2.3.6)
with the metric
                                                                                       1/2
                                      ∂u ∂v         2        ∂u ∂v         2
               d(u, v) =                −               +      −               dx dy           (2.3.7)
                                  Ω   ∂x ∂x                  ∂y ∂y

is an incomplete metric space. If we introduce an inner product

                                             ∂u ∂v ∂u ∂v
                           (u, v) =               +      dx dy                                 (2.3.8)
                                         Ω   ∂x ∂x ∂y ∂y
consistent with (2.3.7), we get an inner product space. Its completion in the metric
(2.3.7) is the energy space E MC for the clamped membrane, a real Hilbert space.
   What can we say about the elements U(x, y) ∈ E MC ? If {un (x, y)} is a representa-
tive of U(x, y), then
                                                                     1/2
                  ∂un ∂um     2       ∂un ∂um            2
                     −            +      −                   dx dy         → 0 as n, m → ∞ .
           Ω      ∂x   ∂x             ∂y   ∂y

and we see that the sequences of first derivatives {∂un /∂x}, {∂un /∂y} are Cauchy
sequences in the norm of L2 (Ω). What about U(x, y) itself? If we extend each un (x, y)
by zero outside Ω, we can write
                                                    x
                                                        ∂un (s, y)
                                  un (x, y) =                      ds
                                                0          ∂s
(assuming, without loss of generality, that Ω is confined to the band 0 ≤ x ≤ a).
Squaring both sides and integrating over Ω, we get
124                                2 Mechanics Problems from the Functional Analysis Viewpoint
                                                                                            2
                                                                 x
                                                                     ∂un (s, y)
                       u2 (x, y) dx dy =                                        ds dx dy
                   Ω
                        n
                                                  Ω          0          ∂s
                                                                                                     2
                                                                 a
                                                                           ∂un (s, y)
                                         ≤                           1·               ds dx dy
                                                  Ω          0                ∂s
                                                                     a
                                                                          ∂un (s, y) 2
                                         ≤               a                             ds dx dy
                                                  Ω              0           ∂s
                                                                 ∂un (x, y) 2
                                         ≤ a2                                 dx dy .
                                                         Ω          ∂x

This means that if {∂un /∂x} is a Cauchy sequence in the norm of L2 (Ω), then so is
{un }. Hence we can consider elements U(x, y) ∈ E MC to be such that U(x, y), ∂U/∂x,
and ∂U/∂y belong to L2 (Ω).
   As a consequence of the last chain of inequalities, we get Friedrichs’ inequality

                                                                         ∂U   2        ∂U        2
                           U 2 (x, y) dx dy ≤ m                                    +                 dx dy
                       Ω                                     Ω           ∂x            ∂y

which holds for any U(x, y) ∈ E MC and a constant m independent of U(x, y).
Membrane (Free Edge). Although it is natural to introduce the energy space using
the energy metric (2.3.7), we cannot distinguish between two states u1 (x, y) and
u2 (x, y) of the membrane with free edge if

                               u2 (x, y) − u1 (x, y) = c = constant .

This is the only form of “rigid” displacement possible for a membrane. We first show
that no other rigid displacements (i.e., displacements associated with zero strain
energy) are possible. The proof is a consequence of Poincar´ ’s inequality
                                                              e
                                                     2                        ∂u   2         ∂u      2
            u2 dx dy ≤ m                u dx dy +                                      +                 dx dy    (2.3.9)
        Ω                           Ω                                Ω        ∂x             ∂y

for a function u(x, y) ∈ C (1) (Ω). The constant m does not depend on u(x, y).
Proof of Poincar´ ’s inequality [9]. We first assume that Ω is the square [0, a] × [0, a]
                e
and write down the identity
                                                x2
                                                      ∂u(s, y1 )                        y2
                                                                                                ∂u(x2 , t)
             u(x2 , y2 ) − u(x1 , y1 ) =                         ds +                                      dt .
                                             x1          ∂s                            y1          ∂t

Squaring both sides and then integrating over the square, first with respect to the
variables x1 and y1 and then with respect to x2 and y2 , we get
2.3 Some Energy Spaces                                                                                                                                125


                          u2 (x2 , y2 ) − 2u(x2 , y2 )u(x1 , y1 ) + u2 (x1 , y1 ) dx1 dy1 dx2 dy2
         Ω        Ω
                                                                                                                2
                                          x2
                                               ∂u(s, y1 )                        y2
                                                                                      ∂u(x2 , t)
     =                                                    ds +                                   dt dx1 dy1 dx2 dy2
              Ω           Ω           x1          ∂s                            y1       ∂t
                                                                                                                                 2
                                          a
                                                       ∂u(s, y1 )                         a
                                                                                                      ∂u(x2 , t)
     ≤                                         1·                 ds +                        1·                 dt dx1 dy1 dx2 dy2
              Ω           Ω           0                   ∂s                          0                  ∂t
                                                   a
                                                       ∂u(s, y1 )       2                     a
                                                                                                  ∂u(x2 , t) 2
     ≤ 2a                                                         ds +                                         dt dx1 dy1 dx2 dy2
                  Ω               Ω            0          ∂s                              0          ∂t
                      a           a
                                              ∂u       2       ∂u   2
     ≤ 2a4                                                 +            dx dy .
                  0           0               ∂x               ∂y

Note that, along with the Schwarz inequality for integrals, we have used the elemen-
tary inequality (a + b)2 ≤ 2(a2 + b2 ) which follows from the fact that (a − b)2 ≥ 0.
The beginning of this chain of inequalities is
                                                                                                        2
         a2           u2 (x, y) dx dy − 2                                       u(x, y) dx dy + a2                               u2 (x, y) dx dy
                  Ω                                                     Ω                                                    Ω

so
                                                                                      2                                 ∂u   2       ∂u   2
      2a2                 u2 dx dy ≤ 2                              u dx dy + 2a4                                                +            dx dy
                  Ω                                             Ω                                           Ω           ∂x           ∂y

and we get (2.3.9) with m = max(a2 , 1/a2 ).
    It can be shown that Poincar´ ’s inequality holds on more general domains. A
                                   e
modification of (2.3.9) will appear in Sect. 2.12.
    Let us return to the free membrane problem. Provided we consider only the mem-
brane’s state of stress, any two states are identical if they are described by functions
u1 (x, y) and u2 (x, y) whose difference is constant. We gather all functions (such that
the difference between any two is a constant) into a class denoted by u∗ (x, y). There
is a unique representative of u∗ (x, y) denoted by ub (x, y) such that

                                                                    ub (x, y) dx dy = 0 .                                                        (2.3.10)
                                                                Ω

For this balanced representative (or balanced function), Poincar´ ’s inequality be-
                                                                e
comes
                                                                                     ∂ub      2        ∂ub          2
                          u2 (x, y) dx dy ≤ m                                                     +                     dx dy .                  (2.3.11)
                      Ω
                           b
                                                                            Ω        ∂x                ∂y

As the right-hand side is zero for a “rigid” displacement, so is the left-hand side and
it follows that the balanced representative associated with a rigid displacement must
be zero. Hence u(x, y) = c is the only permissible form for a rigid displacement of a
membrane.
126                            2 Mechanics Problems from the Functional Analysis Viewpoint

    Because (2.3.11) has the same form as Friedrichs’ inequality, we can repeat our
former arguments to construct the energy space E MF for a free membrane using the
balanced representatives of the classes u∗ (x, y). In what follows we shall use this
space E MF , remembering that its elements satisfy (2.3.10).
    The condition (2.3.10) is a geometrical constraint resulting from our mathemat-
ical technique. Solving the static free membrane problem, we must remember that
the formulation of the equilibrium problem does not impose this constraint — the
membrane can move as a “rigid body” in the direction normal to its own surface.
But if we consider only deformations and the strain energy defined by the first par-
tial derivatives of u(x, y), the results must be independent of such motions. Consider
then the functional of the work of external forces

                           A=              F(x, y)U(x, y) dx dy .
                                       Ω

If we use the space E MF , then A makes sense if F(x, y) ∈ L2 (Ω) (guaranteed by
(2.3.11) together with the Schwarz inequality in L2 (Ω)). This is the only restriction
on external forces for a clamped membrane. However, in the case of equilibrium for
a free membrane, the functional A must be invariant under transformations of the
form u(x, y) → u(x, y) + c with any constant c. This requires

                                          F(x, y) dx dy = 0 .                    (2.3.12)
                                      Ω

Again, we consider the equilibrium problem where rigid motion, however, is possi-
ble. Since we did not introduce inertia forces, we have formally equated the mass of
the membrane to zero. In this situation of zero mass, any forces with nonzero resul-
tant would make the membrane as a whole move with infinite acceleration. Thus,
(2.3.12) also precludes such physical nonsense.
   Meanwhile, Sobolev’s imbedding theorem permits us to incorporate forces ψ
acting on the boundary S of Ω into the functional describing the work of external
forces:
                             F(x, y)u(x, y) dx dy +   ψu ds .                (2.3.13)
                           Ω                                S
In this case, the condition of invariance of the functional under constant motions
yields the condition
                                   F(x, y) dx dy +        ψ ds = 0 .             (2.3.14)
                               Ω                      S
Note that in the formulation of the Neumann problem, this quantity ψ appears in the
boundary condition:
                                     ∂u
                                          =ψ.
                                     ∂n S
   The reader may ask why, if the membrane can move as a rigid body, it cannot
rotate freely in the manner of an ordinary free rigid body. The answer is that in this
model, unlike the other models for elastic bodies in this book, rotation of the mem-
2.3 Some Energy Spaces                                                                              127

brane as a rigid body alters the membrane strain energy. In the membrane model,
the deflection u(x, y) is the deflection that is imposed on the prestressed state of the
membrane. The membrane equilibrium problem is an example of a problem for a
prestressed body.
   There is another way to formulate the equilibrium problem for a free mem-
brane, based on a different method of introducing the energy space. Let us return
to Poincar´ ’s inequality (2.3.9). Denote
          e
                                                                   2                 1/2
                               u    1   =             u dx dy + D(u)
                                                  Ω

where
                                                      ∂u   2       ∂u      2
                              D(u) =                           +               dx dy ,
                                              Ω       ∂x           ∂y
and recall that the norm in W 1,2 (Ω) is
                                                                       2                1/2
                          u   W 1,2 (Ω)   =               u2 dx dy + D(u)                     .
                                                      Ω

By (2.3.9), the norms · 1 and · W 1,2 (Ω) are equivalent on the set of continuously
differentiable functions, and hence on W 1,2 (Ω). Let us use the norm · 1 on W 1,2 (Ω).
Now the space W 1,2 (Ω) is the completion, with respect to the norm · 1 , of the
functions continuously differentiable on Ω. Let us take an element U(x, y) ∈ W 1,2 (Ω)
and select from U(x, y) an arbitrary representative Cauchy sequence {uk (x, y)}. A
smooth function uk (x, y) can be uniquely written in the form

                  uk (x, y) = uk (x, y) + ak
                              ˜                           where                    uk dx dy = 0 .
                                                                                   ˜
                                                                               Ω

Since
                                                           2
        uk − um   2
                  1   =            (ak − am ) dx dy + D(˜ k − um ) → 0 as k, m → ∞ ,
                                                        u     ˜
                               Ω

we see that {ak } is a numerical Cauchy sequence that has a limit corresponding to
U. This means that {ak } belongs to c, the space of numerical sequences each of
which has a limit. It is easy to show that this limit does not depend on the choice
of representative sequence {uk (x, y)}, and we can regard it as a rigid displacement
of the membrane. Now, because the membrane is not geometrically fixed and can
be moved through any uniform displacement with no change in energy, we place
all the elements of W 1,2 (Ω) that characterize a strained state of the membrane into
the same class such that for any two elements U (x, y) and U (x, y) of the class and
any representative sequences {uk (x, y)} and {uk (x, y)} taken from them, {˜ k (x, y)} and
                                                                           u
{˜ k (x, y)} are equivalent in the norm (i.e., D(˜ k − uk ) → 0) and the difference se-
 u                                               u     ˜
quence {ak − ak } is in c. This construction of the classes is equivalent to the con-
struction of the factor space W 1,2 (Ω)/c, which can also be called the energy space
128                          2 Mechanics Problems from the Functional Analysis Viewpoint

for the free membrane. The zero of this energy factor space is the set of the elements
of W 1,2 (Ω) each containing as a representative an element from c. Clearly if we wish
to have the energy functional defined in this space, the necessary condition for its
continuity is that its linear part — the work of external forces — must be zero over
any constant displacement. The latter involves the self-balance condition (2.3.12). It
is seen that the principal part of any class-element of the new energy space, defined
by the sequences {˜ k (x, y)}, coincides with the corresponding sequences for the el-
                     u
ements of the space E MF ; moreover, this correspondence between the two energy
spaces maintains equal norms in both spaces. So we can repeat the procedure to
establish the existence-uniqueness theorem in the new energy space, using the proof
in E MF .
    The restriction (2.3.12) (or (2.3.14)) is necessary for the functional of external
forces to be uniquely defined for an element U∗ (x, y). We shall use the same nota-
tion E MF for this type of energy space since there is a one-to-one correspondence,
preserving distances and inner products, between the two types of energy space for
the free membrane. Moreover, we shall always make clear which version we mean.
    Those familiar with the theory of the Neumann problem for Laplace’s equation
should note that the necessary condition for solvability that arises in mathematical
physics as a mathematical consequence, i.e.,

                                         ψ ds = 0 ,
                                     S

is a particular case of (2.3.14) when F(x, y) = 0. This means that for solvability
of the problem, the external forces acting on the membrane edge should be self-
balanced.
    Finally, we note that Poisson’s equation governs not only membranes, but also
situations in electricity, magnetism, hydrodynamics, mathematical biology, and
other fields. So we can consider spaces such as E M in various other sciences. It
is clear that the results will be the same.
    We will proceed to introduce other energy spaces in a similar manner: they will
be completions of corresponding metric (inner product) spaces consisting of smooth
functions satisfying certain boundary conditions. The problem is to determine prop-
erties of the elements of those completions. As a rule, metrics must contain all the
strain energy terms (we now discuss only linear systems). For example, we can con-
sider a membrane whose edge is elastically supported; then we must include the
energy of elastic support in the expression for the energy metric.
Bending a Plate. Here we begin with the work of internal forces on variations of
displacements

                  −(w1 , w2 ) = −        Dαβγδ ργδ (w1 ) ραβ (w2 ) dx dy       (2.3.15)
                                     Ω

where w1 (x, y) is the normal displacement of the plate midsurface Ω, w2 (x, y) can be
considered as its variation, ραβ (u) are components of the change-of-curvature tensor,
2.3 Some Energy Spaces                                                                                       129

                                  ∂2 u                    ∂2 u                          ∂2 u
                   ρ11 (u) =           ,        ρ12 =          ,            ρ22 =            ,
                                  ∂x2                     ∂x∂y                          ∂y2

Dαβγδ are components of the tensor of elastic constants of the plate such that

                                      Dαβγδ = Dγδαβ = D βαγδ                                             (2.3.16)

and, for any symmetric tensor ραβ there exists a constant m0 > 0 such that
                                                                 2
                                   Dαβγδ ργδ ραβ ≥ m0                   ρ2 .
                                                                         αβ                              (2.3.17)
                                                                α,β=1


We suppose the Dαβγδ are constants, but piecewise continuity of these parameters
would be sufficient.
   For the theory of shells and plates, Greek indices will assume values from the
set {1, 2} while Latin indices will assume values from the set {1, 2, 3}. The repeated
index convention for summation is also in force. For example, we have
                                                      2
                                       aαβ bαβ ≡            aαβ bαβ .
                                                    α,β=1

   We first consider a plate with clamped edge ∂Ω:

                                                    ∂w
                                           w ∂Ω =               =0.                                      (2.3.18)
                                                    ∂n    ∂Ω

(Of course, the variation of w must satisfy (2.3.18) as well.) Let us show that on S 4 ,
the subset of C (4) (Ω) consisting of those functions which satisfy (2.3.18), the form
(w1 , w2 ) given in (2.3.15) is an inner product. We begin with the axiom P1:
                                                                                    2
        (w, w) =             Dαβγδ ραβ (w) ργδ (w) dx dy ≥ m0                             ρ2 (w) dx dy
                                                                                           αβ
                         Ω                                                      Ω α,β=1

                                   ∂2 w    2        ∂2 w    2        ∂2 w   2
                  = m0                         +2               +                dx dy ≥ 0 .
                              Ω    ∂x2              ∂x∂y             ∂y2

If w = 0 then (w, w) = 0. If (w, w) = 0 then, on Ω,

                         ∂2 w                   ∂2 w                     ∂2 w
                              =0,                    =0,                      =0.
                         ∂x2                    ∂x∂y                     ∂y2
It follows that
                                     w(x, y) = a1 + a2 x + a3 y ,
where the ai are constants. By (2.3.18) then, w(x, y) = 0. Hence P1 is satisfied.
Satisfaction of P2 follows from (2.3.16), and it is evident that P3 is also satisfied.
130                                       2 Mechanics Problems from the Functional Analysis Viewpoint

Thus S 4 with inner product (2.3.15) is an inner product space; its completion in the
corresponding metric is the energy space E PC for a clamped plate.
   Let us consider some properties of the elements of E PC . It was shown that

                             ∂2 w     2              ∂2 w          2           ∂2 w      2 2
             m0                           +2                           +                       dx dy
                    Ω        ∂x2                     ∂x∂y                      ∂y2

                             ≤                Dαβγδ ργδ (w) ραβ (w) dx dy ≡ (w, w) .                                       (2.3.19)
                                          Ω

From this and the Friedrichs inequality, written first for w and then for the first
derivatives of w ∈ S 4 as well, we get

                                                    ∂w     2               ∂w    2
             w2 dx dy ≤ m1                                     +                         dx dy
         Ω                                    Ω     ∂x                     ∂y
                                                    ∂2 w       2                ∂2 w       2        ∂2 w   2
                         ≤ m2                                      +2                          +               dx dy
                                              Ω     ∂x2                         ∂x∂y                ∂y2

                         ≤ m3                     Dαβγδ ργδ (w) ραβ (w) dx dy ≡ m3 (w, w) .                                (2.3.20)
                                              Ω

Hence if {wn } ⊂ S 4 is a Cauchy sequence in E PC , then the sequences

                     ∂wn                      ∂wn                      ∂2 wn                   ∂2 wn             ∂2 wn
          {wn },         ,                        ,                          ,                       ,                 ,
                      ∂x                       ∂y                       ∂x2                    ∂x∂y               ∂y2

are Cauchy sequences in L2 (Ω). So we can say that an element W of the completion
E PC is such that W(x, y) and all its derivatives up to order two are in L2 (Ω).
   We now investigate W(x, y) further. Let w ∈ S 4 and w(x, y) ≡ 0 outside Ω. Sup-
pose Ω lies in the domain {(x, y) : x > 0, y > 0}. Then the representation
                                                               x           y
                                                                               ∂2 w(s, t)
                                      w(x, y) =                                           ds dt
                                                           0           0         ∂s∂t
holds. Using H¨ lder’s inequality and (2.3.20), we get
              o
                         x        y
                                      ∂2 w(s, t)                                               ∂2 w(s, t)
      |w(x, y)| ≤                                ds dt ≤                                 1·               ds dt
                     0        0         ∂s∂t                                         Ω           ∂s∂t
                                                        ∂2 w(s, t) 2                       1/2
                  ≤ (mes Ω)1/2                                       ds dt                         ≤ m4 (w, w)1/2 .        (2.3.21)
                                                    Ω     ∂s∂t
This means that if {wn } ⊂ S 4 is a Cauchy sequence in the metric of E PC , then it
converges uniformly on Ω. Hence there exists a limit function

                                                  w0 (x, y) = lim wn (x, y)
                                                                       n→∞
2.3 Some Energy Spaces                                                                                        131

which is continuous on Ω; this function is identified, as above, with the correspond-
ing element of E PC and we shall say that E PC is continuously imbedded into C(Ω).
   The functional describing the work of external forces

                             A=                    F(x, y) W(x, y) dx dy
                                              Ω

now makes sense if F(x, y) ∈ L(Ω); moreover, it can contain the work of point forces

                                              F(xk , yk ) w0 (xk , yk )
                                      k

and line forces
                                              F(x, y) w0 (x, y) ds
                                          γ

where γ is a line in Ω and w0 (x, y) is the corresponding limit function for W(x, y).
Remark 2.3.4. Modern books on partial differential equations often require that
F(x, y) ∈ H −2 (Ω). This is a complete characterization of external forces — how-
ever, it is difficult for an engineer to verify this property.
   Now let us consider a plate with free edge. In this case, we also wish to use the
inner product (2.3.15) to create an energy space. As in the case of a membrane with
free edge, the axiom P1 is not fulfilled: we saw that from (w, w) = 0 it follows that

                                  w(x, y) = a1 + a2 x + a3 y .                                            (2.3.22)

This admissible motion of the plate as a rigid whole is called a rigid motion, but still
differs from real “rigid” motions of the plate as a three-dimensional body.
    Poincar´ ’s inequality (2.3.9) implies that the zero element of the correspond-
            e
ing completion consists of functions of the form (2.3.22). Indeed, taking w(x, y) ∈
C (4) (Ω) we write down Poincar´ ’s inequality for ∂w/∂x:
                                 e

          ∂w 2                            ∂w      2                       ∂2 w   2       ∂2 w       2
               dx dy ≤ m                     dx dy +                                 +                  dx dy ,
      Ω   ∂x                          Ω   ∂x                        Ω     ∂x2            ∂x∂y

and then the same inequality for ∂w/∂y with the roles of x and y interchanged. From
these and (2.3.9) we get

                             ∂w   2           ∂w     2                                          2
                      w2 +            +                  dx dy ≤ m1                  w dx dy
                  Ω          ∂x               ∂y                                 Ω

                                  ∂w      2                         ∂w               2
                       +             dx dy +                           dx dy
                              Ω   ∂x                            Ω   ∂y
                                   ∂w 2        2         ∂w2    2         ∂2 w   2
                       +                           +2               +                dx dy .
                             Ω     ∂x2                   ∂x∂y             ∂y2

From (2.3.19) it follows that
132                                  2 Mechanics Problems from the Functional Analysis Viewpoint

                                 ∂w       2        ∂w   2                                     2
                         w2 +                 +             dx dy ≤ m2              w dx dy
                     Ω           ∂x                ∂y                           Ω

                                         ∂w      2                   ∂w          2
                          +                 dx dy +                     dx dy
                                     Ω   ∂x                      Ω   ∂y

                          +          Dαβγδ ργδ (w) ραβ (w) dx dy .                                  (2.3.23)
                                 Ω

For any function w(x, y) ∈ C (4) (Ω), we can take suitable constants ai and find a
function wb (x, y) of the form

                                wb (x, y) = w(x, y) + a1 + a2 x + a3 y                              (2.3.24)

such that
                                              ∂wb                            ∂wb
              wb dx dy = 0 ,                      dx dy = 0 ,                    dx dy = 0 .        (2.3.25)
          Ω                              Ω     ∂x                        Ω    ∂y

As for the membrane with free edge, we can now consider a subset S 4b of C (4) (Ω)
consisting of balanced functions satisfying (2.3.25). We construct an energy space
E PF for a plate with free edge as the completion of S 4b in the metric induced by the
inner product (2.3.15).
   From (2.3.25), (2.3.23), and (2.3.19), we see that an element W(x, y) ∈ E PF is
such that W(x, y) and all its “derivatives” up to order two are in L2 (Ω). We could
show the existence of a limit function

                                         w0 (x, y) = lim wn ∈ C(Ω)
                                                        n→∞

for any Cauchy sequence {wn }, but in this case the technique is more complicated
and, in what follows, we have this result as a particular case of the Sobolev imbed-
ding theorem.
   Note that (2.3.25) can be replaced by

          w(x, y) dx dy = 0 ,                     x w(x, y) dx dy = 0 ,              y w(x, y) dx dy = 0 ,
      Ω                                       Ω                                 Ω

since these also uniquely determine the ai for a class of functions of the form
(2.3.24). (This possibility follows from Sobolev’s general result [33] on equivalent
norms in Sobolev spaces.)
   The system (2.3.25) represents constraints that are absent in nature. For a static
problem, there must be a certain invariance of some objects under transformations
of the form (2.3.24) with arbitrary constants ak . In particular, the work of external
forces should not depend on the ak if the problem is stated correctly. This leads to
the necessary conditions
2.3 Some Energy Spaces                                                                                      133


          F(x, y) dx dy = 0 ,              x F(x, y) dx dy = 0 ,                        y F(x, y) dx dy = 0 .
      Ω                                Ω                                        Ω
                                                                        (2.3.26)
The mechanical sense of (2.3.26) is clear: the resultant force and moments must
vanish. This is the condition for a self-balanced force system.
Problem 2.3.2. What is the form of (2.3.26) if the external forces contain point and
line forces?
    An energy space for a free plate, as for the membrane with free edge, can be
introduced in another way: namely, we begin with an element of W 2,2 (Ω), selecting
a representative sequence {wk (x, y)+ak +bk x+ck y}, where w satisfies (2.3.25) and ak ,
                                ˜                              ˜
bk , ck are constants. It is easy to show that the numerical sequences {ak }, {bk }, {ck } are
Cauchy and therefore belong to the space c. Next we combine into a class-element
all the elements of W 2,2 (Ω) whose differences are a linear polynomial a + bx + cy,
and state that the energy space is the factor space of W 2,2 (Ω) by the space which is
the completion of the space of linear polynomials with respect to the norm

                                a + bx + cy = (a2 + b2 + c2 )1/2 .

In the factor space, the zero is the class of all Cauchy sequences whose differ-
ences are equivalent (in the norm of W 2,2 (Ω)) to a sequence {ak + bk x + ck y} with
{ak }, {bk }, {ck } ∈ c. The norm of W, an element in the factor space, is
                                                                          1/2
                                     Dαβγδ ργδ (W) ραβ (W) dx dy                .
                                 Ω

The elements of the factor space are uniquely identified with the elements of the
energy space E PF whose elements satisfy (2.3.25). Moreover, the identification is
isometric so, as for the membrane, we can use this factor space as the energy space
and repeat the proof of the existence-uniqueness theorem in terms of the elements of
the energy factor space. The self-balance condition for the external forces is a nec-
essary condition to have the functional of the work of external forces be meaningful.
The reader may also consider mixed boundary conditions: how must the treatment
be modified if the plate is clamped only along a segment AB ⊂ Ω so that

                                           w(x, y) AB = 0 ,

with the rest of the boundary free of geometrical constraints?
Linear Elasticity. We return to the problem of linear elasticity, considered in
Sect. 1.5. Let us introduce a functional describing the work of internal forces on
variations v(x) of the displacement field u(x):

                         −(u, v) = −                ci jkl   kl (u) i j (v) dΩ      .                  (2.3.27)
                                                Ω

For the notation, see (1.5.7)–(1.5.9). We recall that the strain energy E4 (u) of an
elastic body occupying volume Ω is related to the introduced inner product as fol-
134                                          2 Mechanics Problems from the Functional Analysis Viewpoint

lows:
                                                 (u, u) = u             2
                                                                            = 2E4 (u) .
                             i jkl
The elastic moduli c may be piecewise continuous functions satisfying (1.5.8)
and (1.5.9), which guarantee that all inner product axioms are satisfied by (u, v) for
vector functions u, v continuously differentiable on Ω, except P1: from (u, u) = 0 it
follows that u = a + b × x. Note that (u, v) is consistent with the metric (1.5.10).
   Let us consider boundary conditions prescribed by

                                                              u(x) ∂Ω = 0 .                                                  (2.3.28)

If we use the form (2.3.27) on the set S 3 of vector-functions u(x) satisfying (2.3.28)
and such that each of their components is of class C (2) (Ω), then (u, v) is an inner
product and S 3 with this inner product becomes an inner product space. Its com-
pletion E EC in the corresponding metric (or norm) is the energy space of an elastic
body with clamped boundary. To describe the properties of the elements of E EC , we
establish Korn’s inequality.

Lemma 2.3.1 (Korn). For a vector function u(x) ∈ S 3 , we have
                                         3
                                                     ∂ui      2
                     |u|2 +                                           dΩ ≤ m                  ci jkl     kl (u) i j (u) dΩ
                 Ω                   i, j=1
                                                     ∂x j                                 Ω


for some constant m which does not depend on u(x).

Proof. By (1.5.9) and Friedrichs’ inequality, it is sufficient to show that
                                     3                                                        3
                                                 ∂ui      2
                                                              dΩ ≤ m1                                  2
                                                                                                       i j (u) dΩ   .
                               Ω i, j=1          ∂x j                                    Ω i, j=1
                                                                                            i≤ j

Consider the term on the right:
                         3                                                   3
                                                            1                        ∂ui ∂u j 2
        A≡                       2
                                 i j (u) dΩ           =                                  +      dΩ
                 Ω i, j=1                                   4           Ω i, j=1     ∂x j ∂xi
                    i≤ j                                                   i≤ j
                  ⎧                                                                                                 ⎫
                  ⎪
                  ⎪ 3
                  ⎪                                                                                                 ⎪
                                                                                                                    ⎪
                                                                                                                    ⎪
                  ⎪
                  ⎪
                  ⎨     ∂ui                      2       1
                                                                  3
                                                                        ∂ui      2         ∂u j     2               ⎪
                                                                                                           ∂ui ∂u j ⎪
                                                                                                                    ⎬
          =       ⎪
                  ⎪                                  +                               +                  +2          ⎪ dΩ .
                  ⎪
                  ⎪ i=1 ∂xi
                 Ω⎪                                      4 i, j=1       ∂x j               ∂xi             ∂x j ∂xi ⎪
                                                                                                                    ⎪
                                                                                                                    ⎪
                                                                                                                    ⎪
                  ⎩                                                                                                 ⎭
                                                              i< j

Integrating by parts (twice) the term
                                             3                                                      3
                     1                        ∂ui ∂u j      1                                          ∂ui ∂u j
              B≡                                       dΩ =                                                     dΩ
                     2               Ω i, j=1 ∂x j ∂xi      2                                 Ω i, j=1 ∂xi ∂x j
                                         i< j                                                     i< j
2.4 Generalized Solutions in Mechanics                                                               135

and using the elementary inequality |ab| ≤ (a2 + b2 )/2, we get
                          3                                                     3
               1                 ∂ui   2     ∂u j       2        1                  ∂ui 2
         |B| ≤                             +                dΩ =                          dΩ .
               4      Ω i, j=1   ∂xi         ∂x j                2         Ω i=1    ∂xi
                         i< j

Therefore
                                                    3
                                       1                    ∂ui 2
                                 A≥                               dΩ ,
                                       4      Ω i, j=1      ∂x j

which completes the proof.

    By Korn’s inequality, each component of an element U ∈ E EC belongs to E MC ,
i.e., the ui and their first derivatives belong to L2 (Ω).
    Note that the construction of an energy space is the same if the boundary condi-
tion (2.3.28) is given only on some part ∂Ω1 of the boundary of Ω:

                                           u(x) ∂Ω = 0 .
                                                    1


Korn’s inequality also holds, but its proof is more complicated (see, for example,
[26, 11]).
    If we consider an elastic body with free boundary, we encounter issues similar
to those for a membrane or plate with free edge: we must circumvent the difficulty
with the zero element of the energy space. The restrictions

                             u dΩ = 0 ,                         x × u(x) dΩ = 0 ,                (2.3.29)
                         Ω                                  Ω

provide that the rigid motion u = a + b × x becomes zero, and that Korn’s inequality
remains valid for smooth vector functions satisfying (2.3.29). So by completion, we
get an energy space E EF with known properties: all Cartesian components of vectors
pertain to the space W 1,2 (Ω).
    As for a free membrane, we can also organize an energy space of classes — a
factor space — in which the zero element is the set of all elements whose differences
between any representative sequences in the norm of (W 1,2 (Ω))3 are equivalent to a
sequence of the form {ak + bk × x} such that the Cartesian components of the vectors
{ak } and {bk } constitute some elements of the space c.



2.4 Generalized Solutions in Mechanics

We now discuss how to introduce generalized solutions in mechanics. We begin
with Poisson’s equation

                         −Δu(x, y) = F(x, y) ,                   (x, y) ∈ Ω ,                     (2.4.1)
136                              2 Mechanics Problems from the Functional Analysis Viewpoint

where Ω is a bounded open domain in R2 . The Dirichlet problem consists of this
equation supplemented by the boundary condition

                                          u ∂Ω = 0 .                                   (2.4.2)

Let u(x, y) be its classical solution; i.e., let u ∈ C (2) (Ω) satisfy (2.4.1) and (2.4.2). Let
ϕ(x, y) be a function with compact support in Ω. Again, this means that ϕ ∈ C (∞) (Ω)
and the closure of the set M = {(x, y) ∈ Ω : ϕ(x, y) 0} lies in Ω.
   Multiplying both sides of (2.4.1) by ϕ(x, y) and integrating over Ω, we get

               −         ϕ(x, y) Δu(x, y) dx dy =            F(x, y) ϕ(x, y) dx dy .   (2.4.3)
                     Ω                                   Ω

If this equality holds for every infinitely differentiable function ϕ(x, y) with compact
support in Ω, and if u ∈ C (2) (Ω) and satisfies (2.4.2), then, as is well known from
the classical calculus of variations, u(x, y) is the unique classical solution to the
Dirichlet problem.
    But using (2.4.3), we can pose this Dirichlet problem directly without using the
differential equation (2.4.1); namely, u(x, y) is a solution to the Dirichlet problem
if, obeying (2.4.2), it satisfies (2.4.3) for every ϕ(x, y) that is infinitely differentiable
with compact support in Ω. If F(x, y) belongs to L p (Ω) then we can take, as it seems,
u(x, y) having second derivatives in the space L p (Ω); such a u(x, y) is not a classical
solution, and it is natural to call it a generalized solution.
    We can go further by applying integration by parts to the left-hand side of (2.4.3)
as follows:
                   ∂u ∂ϕ ∂u ∂ϕ
                        +      dx dy =                  F(x, y) ϕ(x, y) dx dy .        (2.4.4)
              Ω    ∂x ∂x ∂y ∂y                      Ω

In such a case we may impose weaker restrictions on a solution u(x, y) and call it the
generalized solution if it belongs to E MC , the energy space for a clamped membrane.
Equation (2.4.4) defines this solution if it holds for every ϕ(x, y) that has a compact
support in Ω. Note the disparity in requirements on u(x, y) and ϕ(x, y).
   Further integration by parts on the left-hand side of (2.4.4) yields

               −         u(x, y) Δϕ(x, y) dx dy =            F(x, y) ϕ(x, y) dx dy .   (2.4.5)
                     Ω                                   Ω

Now we can formally consider solutions from the space L(Ω) and this is a new class
of generalized solutions.
   This way leads to the theory of distributions originated by Schwartz [32]. He
extended the notion of generalized solution to a class of linear continuous function-
als, or distributions, defined on the set D(Ω) of all functions infinitely differentiable
in Ω and with compact support in Ω. For this it is necessary to introduce the con-
vergence and other structures of continuity in D(Ω). Unfortunately D(Ω) is not a
normed space (see, for example, Yosida [44] — it is a locally convex topological
space) and its presentation would exceed our scope. This theory justifies, in particu-
2.4 Generalized Solutions in Mechanics                                                                 137

lar, the use of the so-called δ-function, which was introduced in quantum mechanics
via the equality
                                  ∞
                                      δ(x − a) f (x) dx = f (a) ,                                   (2.4.6)
                                 −∞
valid for every continuous function f (x). Physicists considered δ(x) to be a function
vanishing everywhere except at x = 0, where its value is infinity. Any known theory
of integration gave zero for the value of the integral on the left-hand side of (2.4.6),
and the theory of distributions explained how to understand such strange functions.
It is interesting to note that the δ-function was well known in classical mechanics,
too; if we consider δ(x − a) as a unit point force applied at x = a, then the integral
on the left-hand side of (2.4.6) is the work of this force on the displacement f (a),
which is indeed f (a).
    So we have several generalized statements of the Dirichlet problem, but which
one is most natural from the viewpoint of mechanics?
    From mechanics it is known that a solution to the problem is a minimizer of the
total potential energy functional

                                 ∂u   2       ∂u    2
                I(u) =                    +             dx dy − 2               Fu dx dy .          (2.4.7)
                           Ω     ∂x           ∂y                           Ω


According to the calculus of variations, a minimizer of I(u) on the subset of C (2) (Ω)
consisting of all functions satisfying (2.4.2), if it exists, is a classical solution to
the Dirichlet problem. But we also can consider I(u) on the energy space E MC if
F(x, y) ∈ L p (Ω) for p > 1. Indeed, the first term in I(u) is well-defined in E MC and
can be written in the form u 2 ; the second,

                          Φ(u) = −                F(x, y) u(x, y) dx dy ,
                                              Ω

is a linear functional with respect to u(x, y). It is also continuous in E MC ; by virtue
of H¨ lder’s inequality with exponents p and q = p/(p − 1), we have
     o
                                                               1/p                            1/q
                      Fu dx dy ≤              |F| p dx dy                      |u|q dx dy
                  Ω                       Ω                                Ω

                                 ≤ m1 F       L p (Ω)    u   W 1,2 (Ω)   ≤ m2 u    E MC   .

To show this, we have used the imbedding Theorem 2.1.2 and the Friedrichs in-
equality. By Theorem 1.21.1, Φ(u) is continuous in E MC , and therefore so is I(u).
   Thus I(u) is of the form

                                   I(u) = u         2
                                                        + 2Φ(u) .                                   (2.4.8)

Let u0 ∈ E MC be a minimizer of I(u), i.e.,

                               I(u0 ) ≤ I(u) for all u ∈ E MC .                                     (2.4.9)
138                               2 Mechanics Problems from the Functional Analysis Viewpoint

We try a method from the classical calculus of variations. Take u = u0 + v where v
is an arbitrary element of E MC . Then

                  I(u) = I(u0 + v) = u0 + v            2
                                                           + 2Φ(u0 + v)
                      = (u0 + v, u0 + v) + 2Φ(u0 + v)
                      = u0    2
                                  + 2 (u0 , v) +   2
                                                       v   2
                                                               + 2Φ(u0 ) + 2 Φ(v)
                      = u0    2
                                  + 2Φ(u0 ) + 2 [(u0 , v) + Φ(v)] +          2
                                                                                 v   2
                                                                                         .

From (2.4.9), we get

                             2 [(u0 , v) + Φ(v)] +         2
                                                               v   2
                                                                       ≥0.

Since is an arbitrary real number (in particular it can take either sign, and the first
term, if nonzero, dominates for small in magnitude), it follows that

                                      (u0 , v) + Φ(v) = 0 .                                  (2.4.10)

In other words,
              ∂u0 ∂v ∂u0 ∂v
                    +       dx dy −                    F(x, y) v(x, y) dx dy = 0 .           (2.4.11)
          Ω   ∂x ∂x   ∂y ∂y                        Ω

This equality holds for every v ∈ E MC and defines the minimizer u0 ∈ E MC . Note
that (2.4.11) has the same form as (2.4.4).
   So we have introduced a notion of generalized solution that has an explicit me-
chanical background.

Definition 2.4.1. An element u ∈ E MC is called the generalized solution to the
Dirichlet problem if u satisfies (2.4.11) for any v ∈ E MC .

   We can also obtain (2.4.11) from the principle of virtual displacements (work).
This asserts that in the state of equilibrium, the sum of the work of internal forces
(which is now the variation of the strain energy with a negative sign) and the work
of external forces is zero on all virtual (admissible) displacements.
   In the case under consideration, both approaches to introducing generalized en-
ergy solutions are equivalent. In general, however, this is not so, and the virtual work
principle has wider applicability. If F(x, y) is a nonconservative load depending on
u(x, y), we cannot use the principle of minimum total energy; however, (2.4.11) re-
mains valid since it has the mathematical form of the virtual work principle. In what
follows, we often use this principle to pose problems in equation form.
   Since the part of the presentation from (2.4.8) up to (2.4.10) is general and does
not depend on the specific form (2.4.11) of the functional I(u), we can formulate

Theorem 2.4.1. Let u0 be a minimizer of a functional I(u) = u 2 +2Φ(u) given in an
inner product (Hilbert) space H, where the functional Φ(u) is linear and continuous.
Then u0 satisfies (2.4.10) for every v ∈ H.
2.4 Generalized Solutions in Mechanics                                                                  139

   Equation (2.4.10) is a necessary condition for minimization of the functional
I(u), analogous to the condition that the first derivative of an ordinary function must
vanish at a point of minimum.
   We can obtain (2.4.10) formally by evaluating
                                       d
                                         I(u0 + v)             =0.                                  (2.4.12)
                                       d                  =0

This is valid for the following reason. Given u0 and v, the functional I(u0 + v) is
an ordinary function of the numerical variable , and assumes a minimum value at
   = 0. The left-hand side of (2.4.12) can be interpreted as a partial derivative at
u = u0 in the direction v, and is called the Gˆ teaux derivative of I(u) at u = u0 in
                                                 a
the direction of v. We shall return to this issue later.
   The Dirichlet problem for a clamped membrane is a touchstone for all static prob-
lems in continuum mechanics. In a similar way we can introduce a natural notion
of generalized solution for other problems under consideration. As we said, each
of them can be represented as a minimization problem for a total potential energy
functional of the form (2.4.10) in an energy space. For example, equation (2.4.11),
a particular form of (2.4.10) for a clamped membrane, is the same for a free mem-
brane — we need only replace E MC by E MF . The quantity Φ(u), to be a continuous
linear functional in E MF , must be supplemented with self-balance condition (2.3.12)
for the load.
   Let us concretize equation (2.4.10) for each of the other problems we have under
consideration.
Plate. The definition of generalized solution w0 ∈ E P is given by the equation

                      Dαβγδ ργδ (w0 ) ραβ (w) dx dy −                  F(x, y) w(x, y) dx dy
                  Ω                                                Ω
                                m
                            −         Fk w(xk , yk ) −        f (s) w(x, y) ds = 0                  (2.4.13)
                                k=1                       γ


(see the notation of Sect. 2.3) which must hold for every w ∈ E P . The equation is
the same for any kind of homogeneous boundary conditions (i.e., for usual ones)
but the energy space will change from one set of boundary conditions to another. If
a plate can move as a rigid whole, the requirement that

                              F(x, y) ∈ L(Ω) ,                f (s) ∈ L(γ) ,

for Φ to be a continuous linear functional, must be supplemented with self-balance
conditions for the load:
                                         m
           F(x, y) wi (x, y) dx dy +          Fk wi (xk , yk ) +           f (s) wi (x, y) ds = 0   (2.4.14)
       Ω                                k=1                            γ
140                                 2 Mechanics Problems from the Functional Analysis Viewpoint

for i = 1, 2, 3, where w1 (x, y) = 1, w2 (x, y) = x, and w3 (x, y) = y. This condition
is necessary if we use the space where the set of rigid plate motions is the zero of
the space. If from each element of the energy space we select a representative us-
ing (2.3.25), then on the energy space of all representatives with norm · P we can
prove existence and uniqueness of the energy solution without self-balance condi-
tion (2.4.14). But for solvability of the initial problem for a free plate, we still should
require (2.4.14) to hold as constraints (2.3.25) are absent in the problem statement.
    Note that for each concrete problem we must specify the energy space. The same
is true for the following problem.
Linear Elasticity. The generalized solution u is defined by the integro-differential
equation

                       ci jkl   kl (u) i j (v) dΩ   −          F(x, y, z) · v(x, y, z) dΩ
                   Ω                                       Ω

                                −         f(x, y, z) · v(x, y, z) dS = 0 ,                     (2.4.15)
                                      Γ

where F and f are forces distributed over Ω and over some surface Γ ⊂ Ω, respec-
tively. If we consider the Dirichlet (or first) problem of elasticity, which is

                                               u(x) ∂Ω = 0

where ∂Ω is the boundary of Ω, the solution u should belong to the space E EC and
equation (2.4.15) must hold for every virtual displacement v ∈ E EC . Note that in
this case,
                                          f(x, y, z) · v(x, y, z) dS = 0 .
                                    ∂Ω
   The load, thanks to Theorem 2.1.4 and Korn’s inequality, is of the class

            Fi (x, y, z) ∈ L6/5 (Ω) ,          fi (x, y, z) ∈ L4/3 (Γ)       (i = 1, 2, 3) ,

where Fi (x, y, z) and fi are Cartesian components of F and f, respectively, and Γ is
a piecewise smooth surface in Ω. This provides continuity of Φ(w).
   In the second problem of elasticity there are given forces distributed over the
boundary:
                               ci jkl kl (x) n j (x) ∂Ω = fi (x) ,
where the n j are Cartesian components of the unit exterior normal to ∂Ω. This can
be written in tensor notation as

                                            σ(x) · n ∂Ω = f(x)

where the components of stress tensor σ are related to the components of the strain
tensor by σi j = ci jkl kl .
   As for equilibrium problems for free membranes and plates, in the case of a free
elastic body we must require that the load be self-balanced:
2.5 Existence of Energy Solutions to Some Mechanics Problems                        141


                               F(x) dΩ +         f(x) dS = 0 ,
                           Ω                 Γ

                               x × F(x) dΩ +           x × f(x) dS = 0 .       (2.4.16)
                           Ω                       Γ

The solution is required to be in E EF .
   We have argued that it is legitimate to introduce the generalized solution in such
a way. Of course, full legitimacy will be assured when we prove that this solution
exists and is unique in the corresponding space.
   We emphasize once more that the definition of generalized solution arose in a
natural way from the variational principle of mechanics.



2.5 Existence of Energy Solutions to Some Mechanics Problems

In Sect. 2.4 we introduced generalized solutions for several mechanics problems
and reduced those problems to a solution of the abstract equation

                                    (u, v) + Φ(v) = 0                           (2.5.1)

in an energy (Hilbert) space. We obtained some restrictions on the forces to provide
continuity of the linear functional Φ(v) in the energy space. The following theorem
guarantees solvability of those mechanics problems in a generalized sense.
Theorem 2.5.1. Assume Φ(v) is a continuous linear functional given on a Hilbert
space H. Then there is a unique element u ∈ H that satisfies (2.5.1) for every v ∈ H.
Proof. By the Riesz representation theorem there is a unique u0 ∈ H such that the
continuous linear functional Φ(v) is represented in the form Φ(v) = (v, u0 ) ≡ (u0 , v).
Hence (2.5.1) takes the form

                                   (u, v) + (u0 , v) = 0 .                      (2.5.2)

We need to find u ∈ H that satisfies (2.5.2) for every v ∈ H. Rewriting it in the form

                                     (u + u0 , v) = 0 ,

we see that its unique solution is u = −u0 .
   This theorem answers the question of solvability, in the generalized sense, of
the problems treated in Sect. 2.4. To demonstrate this, we rewrite Theorem 2.5.1 in
concrete terms for a pair of problems.
Theorem 2.5.2. Assume F(x, y) ∈ L(Ω) and f (x, y) ∈ L(γ) where Ω ⊂ R2 is com-
pact and γ is a piecewise smooth curve in Ω. The equilibrium problem for a plate
with clamped edge has a unique generalized solution: there is a unique w0 ∈ E PC
which satisfies (2.4.13) for all w ∈ E PC .
142                           2 Mechanics Problems from the Functional Analysis Viewpoint

   Changes for a plate which is free of clamping are evident: we must add the self-
balance condition (2.4.14) for forces and replace the space E PC by E PF .

Theorem 2.5.3. Assume all Cartesian components of the volume forces F(x, y, z)
are in L6/5 (Ω) and those of the surface forces f(x, y, z) are in L4/3 (S ), where Ω is
compact in R3 and S is a piecewise smooth surface in Ω. Then the problem of
equilibrium of an elastic body occupying Ω, with clamped boundary, has a unique
generalized solution u ∈ E EC ; namely, u(x, y, z) satisfies (2.4.15) for every v ∈ E EC .

   In both theorems, the load restrictions provide continuity of the corresponding
functionals Φ, the work of external forces.

Problem 2.5.1. Formulate existence theorems for the other mechanics problems dis-
cussed in Sect. 2.4.



2.6 Operator Formulation of an Eigenvalue Problem

We have seen how to use the Riesz representation theorem to prove the existence
and uniqueness of a generalized solution. Now let us consider another application
of the Riesz representation theorem: how to cast a problem as an operator equation.
   The eigenvalue equation for a membrane has the form

                                  ∂ 2 u ∂2 u
                                       +     = −λu .                             (2.6.1)
                                  ∂x2 ∂y2
Similar to the equilibrium problem for a membrane, we can introduce a general-
ized solution to the eigenvalue problem for a clamped membrane by the integro-
differential equation

                          ∂u ∂v ∂u ∂v
                               +      dx dy = λ                uv dx dy .        (2.6.2)
                      Ω   ∂x ∂x ∂y ∂y                      Ω

The eigenvalue problem is to find a nontrivial element u ∈ E MC and a corresponding
number λ such that u satisfies (2.6.2) for every v ∈ E MC .
  First we reformulate this problem as an operator equation

                                       u = λKu                                   (2.6.3)

in the space E MC . For this, consider the term

                                 F(v) =         uv dx dy
                                            Ω

as a functional in E MC , with respect to v, when u is a fixed element of E MC . It is
seen that F(v) is a linear functional. By the Schwarz inequality
2.6 Operator Formulation of an Eigenvalue Problem                                                     143
                                                                  1/2                       1/2
           |F(v)| =              uv dx dy ≤            u2 dx dy                  v2 dx dy
                             Ω                     Ω                         Ω

hence by the Friedrichs inequality

                                    |F(v)| ≤ m u v = m1 v                                          (2.6.4)

(hereafter the norm · and the inner product (·, ·) are taken in E MC ). So F(v) is a
continuous linear functional acting in the Hilbert space E MC . By the Riesz represen-
tation theorem, F(v) has the unique representation

                             F(v) ≡          uv dx dy = (v, f ) = ( f, v) .                        (2.6.5)
                                         Ω

What have we shown? For every u ∈ E MC , by this representation, there is a unique
element f ∈ E MC . Hence the correspondence

                                                 u→ f

is an operator f = K(u) from E MC to E MC .
    Let us display some properties of this operator. First we show that it is linear. Let

                                  f1 = K(u1 )    and       f2 = K(u2 ) .

Then
                             (λ1 u1 + λ2 u2 )v dx dy = (K(λ1 u1 + λ2 u2 ), v)
                         Ω
while on the other hand,

                 (λ1 u1 + λ2 u2 )v dx dy = λ1              u1 v dx dy + λ2            u2 v dx dy
             Ω                                         Ω                          Ω

                                             = λ1 (K(u1 ), v) + λ2 (K(u2 ), v)
                                             = (λ1 K(u1 ) + λ2 K(u2 ), v) .

Combining these we have

                      (K(λ1 u1 + λ2 u2 ), v) = (λ1 K(u1 ) + λ2 K(u2 ), v) ,

hence
                             K(λ1 u1 + λ2 u2 ) = λ1 K(u1 ) + λ2 K(u2 )
because v ∈ E MC is arbitrary. Therefore linearity is proven.
   Now let us rewrite (2.6.4) in terms of this representation:

                                      |(K(u), v)| ≤ m u v .

Take v = K(u); then
144                          2 Mechanics Problems from the Functional Analysis Viewpoint

                                K(u)   2
                                           ≤ m u K(u)
and it follows that
                                   K(u) ≤ m u .                                 (2.6.6)
Hence K is a continuous operator in E MC .
  Equation (2.6.2) can now be written in the form

                                 (u, v) = λ(K(u), v) .

Since v is an arbitrary element of E MC , this equation is equivalent to the operator
equation
                                    u = λK(u)
with a continuous linear operator K.
   By (2.6.6), we get

                  λK(u) − λK(v) = |λ| K(u − v) ≤ m |λ| u − v .

If
                                       m |λ| < 1 ,
then λK is a contraction operator in E MC and, by the contraction mapping principle,
there is a unique fixed point of λK which clearly is u = 0. So the set |λ| < 1/m does
not contain real eigenvalues of the problem. Further, we shall see (and this is well
known in mechanics) that eigenvalues in this problem must be real. The fact that
the set |λ| < 1/m does not contain real eigenvalues, and so any eigenvalues of the
problem, has a clear mechanical sense: the lowest eigenfrequency of oscillation of a
bounded clamped membrane is strictly positive. Note that from (2.6.2), when v = u
it follows that an eigenvalue must be positive.
    In a similar way, we can introduce eigenvalue problems for plates and elastic
bodies. Here we can obtain corresponding equations of the form (2.6.3) with con-
tinuous linear operators and can also show that the corresponding lowest eigenvalues
are strictly positive. All this we leave to the reader; later we shall consider eigen-
value problems in more detail.
    In what follows, we shall see that, using the Riesz representation theorem, one
can also introduce operators and operator equations for nonlinear problems of me-
chanics. One of them is presented in the next section.



2.7 Problem of Elastico-Plasticity; Small Deformations

Following the lines of a paper by I.I. Vorovich and Yu.P. Krasovskij [40] that was
published in a sketchy form, we consider a variant of the theory of elastico-plasticity
(Il’yushin [16]), and justify the method of elastic solutions for corresponding bound-
ary value problems.
2.7 Problem of Elastico-Plasticity; Small Deformations                                                                                    145

   The system of partial differential equations describing the behavior of an elastic-
plastic body occupying a bounded volume Ω is

           ν    ω ∂θ
              −      + (1 − ω)Δuk
         ν − 2 3 ∂xk
                                      3              3
                   2 dω                        ∗             ∗     ∂2 ul     Fk
                  − eI                                                     +    = 0 (k = 1, 2, 3) ,                                   (2.7.1)
                   3 deI            s,t=1
                                               ks
                                                    l=1
                                                             lt
                                                                  ∂x s ∂xt   G

where ν is Poisson’s ratio, G is the shear modulus, F = (F1 , F2 , F3 ) are the volume
forces, and ω(ei ) is a function of the variable eI , the intensity of the strain tensor
which defines plastic properties of the material with hardening:
             √
                                                                                                                              2 1/2
      eI =    3
               2
                      (   11   −   22 )
                                          2
                                              +(    11   −    33 )
                                                                     2
                                                                         +(      22   −   33 )
                                                                                                 2
                                                                                                     + 6(   2
                                                                                                            12   +   2
                                                                                                                     13   +   23 )    .

The function ω(eI ) must satisfy
                                                                                 dω(eI )
                                   0 ≤ ω(eI ) ≤ ω(eI ) + eI                              ≤λ<1.                                        (2.7.2)
                                                                                  deI
Other bits of notation are

                                     θ ≡ θ(u) =               11 (u)         +   22 (u)   +      33 (u)     ,

and                                                                       ⎧          √
                                                                          ⎪ ∂uk θ
                                                                          ⎪
                                                                          ⎪            2
                                                                          ⎪
                                                                          ⎪ ∂x − 3 e ,
                                                                          ⎪                                           k=s,
                      1 ∂ui ∂u j                                   ∗
                                                                          ⎪
                                                                          ⎨    s      I
                  =         +    ,                                       =⎪
                                                                          ⎪
             ij
                      2 ∂x j ∂xi                                   ks     ⎪ ∂uk ∂u s
                                                                          ⎪
                                                                          ⎪              1
                                                                          ⎪
                                                                          ⎪
                                                                          ⎩      +      √    ,                        k       s.
                                                                            ∂x s ∂xk     2eI
   If ω(eI ) ≡ 0 we get the equations of linear elasticity for an isotropic homoge-
neous body. By analogy with elasticity problems, to pose a boundary value problem
for (2.7.1) we must supplement the equations with boundary conditions. We con-
sider a mixed boundary value problem: a part S 0 of the boundary ∂Ω of a body
occupying the domain Ω is fixed,

                                                                  uS =0,                                                              (2.7.3)
                                                                         0


and the remainder S 1 = ∂Ω  S 0 is subjected to surface forces f(x) (see [16]):

                                                             σ·nS =f ,                                                                (2.7.4)
                                                                             1


where σ is the stress tensor and n is the external unit normal to S 1 .
   When ω(eI ) is small (as it is if eI is small) we have a nonlinear boundary value
problem which is, in a certain way, a perturbation of a corresponding boundary value
problem of linear elasticity. It leads to the idea of using an iterative procedure, the
method of elastic solutions, to solve the former. This procedure looks like that of
146                                  2 Mechanics Problems from the Functional Analysis Viewpoint

the contraction mapping principle if we can make the problem take the correspond-
ing operator form. Then it remains to show that the operator of the problem is a
contraction. Now we begin to carry out the program.
   Let us introduce the notation

              u, v = 2 {[
                     9        11 (u)     −    22 (u)][ 11 (v)         −      22 (v)]

                         +[   11 (u)     −    33 (u)][ 11 (v)         −      33 (v)]

                         +[   22 (u)     −    33 (u)][ 22 (v)         −      33 (v)]

                         + 6[       12 (u) 12 (v)      +     13 (u) 13 (v)     +       23 (u) 23 (v)]}   .            (2.7.5)

If we consider the terms on the right-hand side of (2.7.5) as coordinates of vectors
a = (a1 , . . . , a6 ), b = (b1 , . . . , b6 ),

                      ai = ci (u) ,              bi = ci (v)              (i = 1, . . . , 6) ,

                  √                                                                √
       c1 (w) =     2
                  3 [ 11 (w)        −   22 (w)]    ,               c2 (w) =         2
                                                                                   3 [ 11 (w)       −   33 (w)]   ,
                  √
       c3 (w) =     2
                  3 [ 22 (w)        −   33 (w)]    ,               c4 (w) =        2
                                                                                   √ 12 (w)
                                                                                    3
                                                                                                ,

       c5 (w) =   2
                  √ 13 (w)
                   3
                                ,                                  c6 (w) =        2
                                                                                   √ 23 (w)
                                                                                    3
                                                                                                ,

then the form u, v is a scalar product between a and b in R6 :
                                                              6
                                             u, v =                ai bi .
                                                             i=1

Besides,
                                                       6
                                        u, u =               c2 (u) = e2 (u)
                                                              i        I                                              (2.7.6)
                                                       i=1

and by the Schwarz inequality we get
                                             6
                          | u, v | =               ci (u)ci (v) ≤ eI (u)eI (v) .                                      (2.7.7)
                                             i=1

On the set C2 of vector functions satisfying the boundary condition (2.7.3) and such
that each of their components is of class C (2) (Ω), we introduce an inner product

                      (u, v) =                3
                                              2G        u, v + 1 K θ(u) θ(v) dΩ .
                                                               2                                                      (2.7.8)
                                         Ω

This coincides with a special case of the inner product (2.3.27) in the linear the-
ory of elasticity. So the completion of C2 in the metric corresponding to (2.7.8) is
the energy space of linear elasticity E EM (M for “mixed”) if we suppose that the
2.7 Problem of Elastico-Plasticity; Small Deformations                                                            147

condition (2.7.3) provides u = 0 if

                         u   2
                                 =         3   2
                                           2 GeI (u)   + 1 Kθ2 (u) dΩ = 0 .
                                                         2
                                       Ω

The norm of E EM is equivalent to one of W 1,2 (Ω) × W 1,2 (Ω) × W 1,2 (Ω) (see Sect. 2.3
and Fichera [11]). (By H1 × H2 we denote the Cartesian product of Hilbert spaces
H1 and H2 , the elements of which are pairs (x, y) for x ∈ H1 and y ∈ H2 . The scalar
product in H1 × H2 is defined by the expression (x1 , x2 )1 +(y1 , y2 )2 where x1 , x2 ∈ H1
and y1 , y2 ∈ H2 .)
   By the principle of virtual displacements, the integro-differential equation of
equilibrium of an elastico-plastic body is
                                                  3                             3
 (u, v) − 3 G
          2             ω(eI (u)) u, v dΩ −                      Fi vi dΩ −                      fi vi dS = 0 .
                    Ω                            i=1         Ω                  i=1         S1
                                                                              (2.7.9)
This equation can be obtained using the equations (2.7.1) and the boundary con-
ditions (2.7.3)–(2.7.4). Conversely, using the technique of the classical calculus of
variations we can get (2.7.1) and the natural boundary conditions (2.7.4). Thus, in a
certain way, (2.7.9) is equivalent to the above statement of the problem. So we can
introduce

Definition 2.7.1. A vector function u ∈ E EM is called the generalized solution of the
problem of elastico-plasticity with boundary conditions (2.7.3)–(2.7.4) if it satisfies
(2.7.9) for every v ∈ E EM .

   For correctness of this definition we must impose some restrictions on external
forces. It is evident that they coincide with those for linear elasticity. So we assume
that
            Fi (x1 , x2 , x3 ) ∈ L6/5 (Ω) , fi (x1 , x2 , x3 ) ∈ L4/3 (S 1 ) .  (2.7.10)
Consider the form
                                                        3                             3
   B[u, v] = 3 G
             2                ω(eI (u)) u, v dΩ +                        Fi vi dΩ +                    fi vi dS
                          Ω                            i=1           Ω                i=1         S1


as a functional in E EM with respect to v(x1 , x2 , x3 ) when u(x1 , x2 , x3 ) ∈ E EM is
fixed. As in linear elasticity, the load terms, thanks to (2.7.10), are continuous linear
functionals with respect to v ∈ E EM . In accordance with (2.7.5) and (2.7.2), we get

          3
          2G            ω(eI (u)) u, v dΩ ≤ λ 3 G
                                              2                      | u, v | dΩ ≤ λ u v ,
                   Ω                                             Ω

so this part of the functional is also continuous.
   Therefore we can apply the Riesz representation theorem to B[u, v] and obtain

                                     B[u, v] = (v, g) ≡ (g, v) .
148                                   2 Mechanics Problems from the Functional Analysis Viewpoint

This representation uniquely defines a correspondence

                                                      u→g

where u, g ∈ E EM . We obtain an operator A acting in E EM by the equality

                                                    g = A(u) .

Equation (2.7.9) is now equivalent to

                                            (u, v) − (A(u), v) = 0                               (2.7.11)

or, since v ∈ E EM is arbitrary,
                                                    u = A(u) .                                   (2.7.12)
   The operator A is nonlinear. We shall show that it is a contraction operator. For
this, take arbitrary elements u, v, w ∈ E EM and consider

      (A(u) − A(v), w) = 3 G
                         2                       [ω(eI (u)) u, w − ω(eI (v)) v, w ] dΩ .         (2.7.13)
                                             Ω

First, let u, v, w be in C2 . At every point of Ω, by (2.7.7), we can estimate the inte-
grand from (2.7.13) as follows. We have

                  Int ≡ ω(eI (u)) u, w − ω(eI (v)) v, w
                                        6                                6
                     = ω(eI (u))             ci (u) ci (w) − ω(eI (v))         ci (v) ci (w) .
                                       i=1                               i=1

Let us introduce a real-valued function f (t) of a real variable t by the relation
                              6
                    f (t) =         ω(eI (tu + (1 − t)v))ci (tu + (1 − t)v)ci (w) .
                              i=1

It is seen that
                                             Int = | f (1) − f (0)| .
As f (t) is continuously differentiable, the classical mean value theorem gives

                   f (1) − f (0) = f (z)(1 − 0) = f (z) for some z ∈ [0, 1] ,

or, in the above terms, we get
2.7 Problem of Elastico-Plasticity; Small Deformations                                                                              149

                                            6
                                  d
                 Int =                              ω(eI (tu + (1 − t)v)) ci (tu + (1 − t)v) ci (w)
                                  dt       i=1
                                                                                                                       t=z

                                   dω(eI (tu + (1 − t)v)) deI (tu + (1 − t)v)
                      =                                                       ·
                                            deI                    dt
                                   6                                                      6
                              ·         ci (tu + (1 − t)v) ci (w) + ω                          ci (u − v) ci (w)             .
                                                                                                                      t=z
                                  i=1                                                    i=1

(Here we have used the linearity of ci (u) in u and, thus, in t.) Let us consider the
term
                          6
                                  deI (tu + (1 − t)v)
             T=                                       ci (tu + (1 − t)v) ci (w)
                       i=1
                                           dt
                          6                     6
                                  d                                             1/2
                  =                                     c2 (tu + (1 − t)v)
                                                         j                            ci (tu + (1 − t)v) ci (w)
                       i=1
                                  dt        j=1

                                        6
                                  2             c j (tu + (1 − t)v) c j (u − v)
                          6
                                       j=1
                  =                                                                        ci (tu + (1 − t)v) ci (w) .
                                                    6                             1/2
                       i=1
                                       2                 c2 (tu
                                                          j       + (1 − t)v)
                                                j=1

Applying the Schwarz inequality, we obtain
                      6                                            1/2    6                     1/2
                              c2 (tu + (1 − t)v)
                               j                                               c2 (u − v)
                                                                                j
           6
                      j=1                                                j=1
  |T | ≤                                                                                              · |ci (tu + (1 − t)v)| |ci (w)|
                                                6                               1/2
           i=1
                                                        c2 (tu + (1 − t)v)
                                                         j
                                            j=1

so that by (2.7.6)
                                            6                            6
                                                                   1/2
                      |T | ≤                        c2 (u − v)
                                                     j                         |ci (tu + (1 − t)v)| |ci (w)|
                                        j=1                              i=1

                                                              6                                        6
                                                                                               1/2                   1/2
                              ≤ eI (u − v)                         c2 (tu + (1 − t)v)
                                                                    i                                       c2 (w)
                                                                                                             i
                                                             i=1                                      i=1

                              = eI (u − v) eI (tu + (1 − t)v) eI (w) .

Similarly,
150                               2 Mechanics Problems from the Functional Analysis Viewpoint

       6                           6                           6
                                                      1/2                    1/2
            ci (u − v) ci (w) ≤         c2 (u − v)
                                         i                          c2 (w)
                                                                     i              = eI (u − v) eI (w) .
      i=1                         i=1                         i=1

Combining all these, we get

                      dω(eI (tu + (1 − t)v))
             Int ≤                           eI (tu + (1 − t)v)) eI (u − v) eI (w)
                               deI

                      + ω(eI (tu + (1 − t)v)) eI (u − v) eI (w)
                                                                              t=z

                                                 dω(eI (tu + (1 − t)v))
                 = ω(eI (tu + (1 − t)v)) +                              ·
                                                          deI

                      · eI (tu + (1 − t)v))           eI (u − v) eI (w) .
                                                t=z

By the condition (2.7.2), we have

                                       Int ≤ λeI (u − v) eI (w)                                      (2.7.14)

at every point of Ω.
    Returning to (2.7.13) we have, using (2.7.14),

                  |(A(u) − A(v), w)| ≤ λ                   3
                                                           2 G eI (u   − v) eI (w) dΩ .
                                                       Ω

In accordance with the norm of E EM it follows that

                             |(A(u) − A(v), w)| ≤ λ u − v w

or, putting w = A(u) − A(v), we get

                     A(u) − A(v) ≤ λ u − v ,                       λ = constant < 1 .                (2.7.15)

Being obtained for u, v, w ∈ C2 , this inequality holds for all u, v, w ∈ E EM since in
this inequality we can pass to the limit for corresponding Cauchy sequences in E EM .
   Inequality (2.7.15) states that A is a contraction operator in E EM ; hence, by the
contraction mapping principle, (2.7.12) has a unique solution that can be found us-
ing the iterative procedure

                            uk+1 = A(uk )             (k = 0, 1, 2, . . .) .

This procedure begins with an arbitrary element u0 ∈ E EM ; when u0 = 0, it is
called the method of elastic solutions since at each step we must solve a problem of
linear elasticity with some given load terms. From a practical standpoint, the method
works best when the constant λ is small.
   So we can formulate
2.8 Bases and Complete Systems; Fourier Series                                         151

Theorem 2.7.1. Assume S 0 is a piecewise smooth surface of nonzero area and that
conditions (2.7.2) and (2.7.10) hold. Then a mixed boundary value problem of
elastico-plasticity has a unique generalized solution in the sense of Definition 2.7.1;
the iterative procedure (2.7.15) defines a sequence of successive approximations
uk ∈ E EM that converges to the solution u ∈ E EM and

                                            λk
                               uk − u ≤        u0 − u1 .                          (2.7.16)
                                           1−λ
   It is clear that we cannot apply this theorem when, say, S 1 = ∂Ω. In such a
case, we must add the self-balance conditions (2.4.16). These guarantee that we can
repeat the above method for a free elastic-plastic body, and so we can formulate

Theorem 2.7.2. Assume that all the requirements of Theorem 2.7.1 and the self-
balance conditions (2.4.16) are met. Then there is a unique generalized solution of
the boundary value problem for a bounded elastic-plastic body, and it can be found
by an iterative procedure of the form (2.7.15).

Problem 2.7.1. Is an estimate of the type (2.7.16) valid in Theorem 2.7.2?

   We recommend that the reader prove Theorem 2.7.2 in detail, in order to gain
experience with the technique.

Remark 2.7.1. We should call attention to the way in which we obtained the main
inequality of this section: it was proved for smooth functions and then extended to
the general case. This is a standard technique in the treatment of nonlinear problems
of mechanics.



2.8 Bases and Complete Systems; Fourier Series

If a linear space Y has finite dimension n, then there is a set {g1 , . . . , gn } of n lin-
early independent elements, called a basis of Y, such that every y ∈ Y has a unique
representation
                                            n
                                      y=         αk gk
                                           k=1

where the αk are scalars. We now consider an infinite dimensional normed space X.

Definition 2.8.1. A system of elements {ek } is a (countable) basis of X if any ele-
ment x ∈ X has a unique representation
                                            ∞
                                      x=         αk ek
                                           k=1

where the αk are scalars.
152                           2 Mechanics Problems from the Functional Analysis Viewpoint

   It is clear that a basis {ek } is linearly independent since the equation
                                               ∞
                                      0=            αk ek
                                              k=1

has the unique solution αk = 0 for each k.
   A normed space with a countable basis is separable: a countable set of all linear
combinations n ck ek (n arbitrary) with rational coefficients ck is dense in the
                k=1
space.

Problem 2.8.1. Prove this.

   We are familiar with some systems of functions which could be bases in certain
spaces: for example,
                                   {gk } =      √1
                                                 2π
                                                      eikx                       (2.8.1)

in L2 (0, 2π). Later, we confirm this example.
   Now we consider the system of monomials {xk } (k = 1, 2, . . .) in C(0, 1). If it is a
basis, then any function f (x) ∈ C(0, 1) could be represented in the form
                                               ∞
                                    f (x) =          αk xk ,
                                               k=0

where the series converges uniformly on [0, 1]. This means the function is analytic,
but we know there are continuous functions on [0, 1] that are not analytic. Hence the
system {xk } is not a basis. On the other hand, the Weierstrass theorem states that this
system possesses properties similar to those of a basis. To generalize this similarity,
we introduce

Definition 2.8.2. A countable system {gk } of elements in a normed space X is com-
plete (or total) in X if for any x ∈ X and any positive number ε there is a finite linear
combination n(ε) αi gi such that
                 i=1

                                        n(ε)
                                   x−          αi gi < ε .
                                        i=1

    By Definition 2.8.2 and the Weierstrass theorem, the system of monomials {xk }
is complete in C(0, 1). Because C(0, 1) is dense in L p (0, 1) for p ≥ 1, this system is
also complete in L p (0, 1).

Problem 2.8.2. Which systems are complete in L p (Ω) or W k,p (Ω)?

  If a normed space has a countable complete system, then the space is separable.
The reader should be able to name a countable dense set to verify this.

Problem 2.8.3. Name such a set.
2.8 Bases and Complete Systems; Fourier Series                                                153

   The problem of existence of a basis in a certain normed space is difficult, but
there is a special case where it is fully solved: a separable Hilbert space. The reader
will find here the theory of Fourier series largely repeated in abstract terms. We
begin with
Definition 2.8.3. A system {xk } of elements of a Hilbert space H is orthonormal if
for all integers m, n,
                                  (xm , xn ) = δmn
where δmn is the Kronecker delta symbol.
   We know that, at least for Rn , there are some advantages in using an orthonormal
system of vectors as a basis.
   Suppose we have an arbitrary linearly independent system of elements of a
Hilbert space H, say { f1 , . . . , fn }, and let Hn be the subspace of H spanned by
this system. We would like to use the system to construct an orthonormal system
{g1 , . . . , gn } that is also a basis of Hn . This can be accomplished by the Gram–
Schmidt procedure:
(1)   The first element of the new system is g1 = f1 / f1 , g1 = 1.
(2)   Take e2 = f2 − ( f2 , g1 )g1 ; then (e2 , g1 ) = ( f2 , g1 ) − ( f2 , g1 ) g1 2 = 0, so the
      second element is g2 = e2 / e2 .
(3)   Take e3 = f3 − ( f3 , g1 )g1 − ( f3 , g2 )g2 ; then (e3 , g1 ) = 0 and (e3 , g2 ) = 0. Since
      e3 0, we get the third element as g3 = e3 / e3 .
 .
 .
 .
(i)   Let ei = fi − ( fi , g1 )g1 − · · · − ( fi , gi−1 )gi−1 . It is seen that (ei , gk ) = 0 for
      k = 1, . . . , i − 1, hence we set gi = ei / ei .
This process can be continued ad infinitum since all ek 0 (why?). So we obtain
an orthonormalized system {g1 , g2 , g3 , . . .}. The process is, however, found to be
unstable for numerical computation.
   As is known from linear algebra, a system { f1 , . . . , fn } is linearly independent in
an inner product space if and only if the Gram determinant

                                     ( f1 , f 1 ) · · · ( f 1 , f n )
                                           .
                                           .       ..         .
                                                              .
                                           .          .       .
                                     ( f n , f1 ) · · · ( fn , fn )

is nonzero. For an orthonormal system of elements the Gram determinant, being
the determinant of the identity matrix, equals +1; hence an orthonormal system is
linearly independent.
Problem 2.8.4. Provide a more direct proof that an orthonormal system is linearly
independent.
   Let {gk } (k = 1, 2, . . .) be an orthonormal system in a complex Hilbert space H.
For an element f ∈ H, the numbers αk defined by αk = ( f, gk ) are called the Fourier
coefficients of f . Now we prove
154                                           2 Mechanics Problems from the Functional Analysis Viewpoint

Theorem 2.8.1. A complete orthonormal system {gk } in a Hilbert space H is a basis
of H; any f ∈ H has the unique Fourier series representation
                                                                      ∞
                                                               f =          αk gk                                                         (2.8.2)
                                                                      k=1

where αk = ( f, gk ) are the Fourier coefficients of f .
Proof. First we consider the problem of the best approximation of an element f ∈
H by elements of a subspace Hn spanned by g1 , . . . , gn . In Sect. 1.19 we showed
that this problem has a unique solution. Now we show that it is n αk gk . Indeed,
                                                                     k=1
consider an arbitrary linear combination n ck gk . Then
                                          k=1

                   n                2                 n                        n
          f−            ck gk           = f−               ck gk , f −              ck gk
               k=1                                   k=1                      k=1
                                                                 n                          n                            n            2
                                        = f     2
                                                     − f,             ck gk −                   ck gk , f +                   ck gk
                                                                k=1                     k=1                             k=1
                                                           n                   n                        n
                                        = f     2
                                                     −          ck αk −             ck αk +                     ck ck
                                                          k=1                 k=1                   k=1
                                                           n                   n
                                        = f     2
                                                     −          |αk |2 +            |ck − αk |2 .
                                                          k=1                 k=1

Because the right-hand side takes its minimum value when ck = αk , we have
               n                2                                n            2                             n
        f−             αk gk        = min            f−               ck gk        = f          2
                                                                                                    −            |αk |2 ≥ 0 ;             (2.8.3)
                                        c1 ,...,cn
             k=1                                                k=1                                     k=1

moreover, we obtain Bessel’s inequality
                                                      n
                                                           |( f, gk )|2 ≤ f             2
                                                                                            .                                             (2.8.4)
                                                     k=1

   Denote by
                                                                       n
                                                               fn =         αk gk                                                         (2.8.5)
                                                                      k=1

the nth partial sum of the Fourier series for f . Let us show that { fn } is a Cauchy
sequence in H. By Bessel’s inequality,
                                                           n
                                                                |αk |2 ≤ f          2
                                                                                        ;
                                                          k=1

hence
2.8 Bases and Complete Systems; Fourier Series                                                               155
                                   n+m              2       n+m
              fn − fn+m   2
                              =           αk gk         =           |αk |2 → 0 as n → ∞ .
                                  k=n+1                     k=n+1

    Now we show that { fn } converges to f . Indeed, by completeness of the system
{gk } in H, for any ε > 0 we can find a number N and coefficients ck (ε) such that
                                              N                2
                                     f−            ck (ε)gk         <ε.
                                             k=1

By (2.8.3),
                                         N              2                   N                   2
                 f − fN   2
                              = f−           αk gk          ≤ f−                    ck (ε)gk        <ε,
                                      k=1                               k=1

so the sequence { fN } converges to f and thus

                                              f = lim fn .                                                (2.8.6)
                                                    n→∞

This completes the proof.

   From (2.8.6) we can obtain Parseval’s equality
                                      ∞
                                             |( f, gk )|2 = f       2
                                                                            ,                             (2.8.7)
                                     k=1

which holds whenever {gk } is a complete orthonormal system in H. Indeed, by
(2.8.3),
                                     n              2                                    n
                  0 = lim f −             αk gk         = lim           f       2
                                                                                    −         |αk |2 .
                      n→∞                                   n→∞
                                    k=1                                                 k=1

   Now we introduce

Definition 2.8.4. A system {ek } (k = 1, 2, . . .) in a Hilbert space H is closed in H if
from the system of equations

                              ( f, ek ) = 0 for all k = 1, 2, 3, . . .

it follows that f = 0.

   It is clear that a complete orthonormal system of elements is closed in H.

Problem 2.8.5. Provide a detailed proof.

   The converse statement holds as well. We formulate

Theorem 2.8.2. Let {gk } be an orthonormal system of elements in a Hilbert space
H. This system is complete in H if and only if it is closed in H.
156                           2 Mechanics Problems from the Functional Analysis Viewpoint

Proof. We need to demonstrate only that a closed orthonormal system in H is com-
plete. Proving Theorem 2.8.1, we established that for any element f ∈ H the se-
quence of partial Fourier sums (2.8.5) is a Cauchy sequence. By completeness of
H, there exists f ∗ = limn→∞ fn that belongs to H. To complete the proof we need to
show that f = f ∗ . We have
                                             n
               ( f − f ∗ , gm ) = lim f −         αk gk , gm = αm − αm = 0 .
                               n→∞
                                            k=1

By Definition 2.8.4, it follows that f = f ∗ , hence {gk } is complete.
    It is normally simpler to check whether a system is closed than to check whether
it is complete. At the beginning of this section we established that any system of
linearly independent elements in H can be transformed into an orthonormal sys-
tem equivalent to the original system in a certain way. So we draw the following
conclusion.
Theorem 2.8.3. A complete system {gk } in H is closed in H; conversely, a system
closed in H is complete in H.
Problem 2.8.6. Write out a detailed proof.
   As stated above, the existence of a countable basis in a Hilbert space provides its
separability. The converse statement is also valid. We formulate that as
Theorem 2.8.4. A Hilbert space H has a countable orthonormal basis if and only if
H is separable.
    The proof follows immediately from the previous theorem. Indeed, in H select a
countable set of elements that is dense everywhere in H. Using the Gram–Schmidt
procedure, produce an orthonormal system of elements from this set (removing any
linearly dependent elements). Since the initial system is dense it is complete and
thus, as a result of the Gram–Schmidt procedure, we get an orthonormal basis of the
space.
    Remember that all of the energy spaces we introduced above are separable.
Hence each of them has a countable orthonormal basis (nonunique, of course). If a
Hilbert space is not separable, by Bessel’s inequality and Lemma 1.16.5, it follows
that for any element x of a nonseparable space the set of nonzero coefficients of
Fourier αk is countable. Repeating the above considerations we can get that (2.8.2)
is valid in this case as well.
    In conclusion, we consider whether the system (2.8.1) is a basis of the complex
space L2 (0, 2π). From standard calculus it is known that the system is orthonormal
in L2 (0, 2π) (the reader, however, can check this). Weierstrass’s theorem on the ap-
proximation of a function continuous on [0, 2π] can be formulated as the statement
that the set of trigonometric polynomials, i.e., finite sums of the form k αk eikx , is
dense in the complex space C(0, 2π). But the set of functions C(0, 2π) is the base
for construction of L2 (0, 2π), hence the finite sums k αk eikx are dense in L2 (0, 2π).
This shows that (2.8.1) is an orthonormal basis of L2 (0, 2π).
2.9 Weak Convergence in a Hilbert Space                                          157

2.9 Weak Convergence in a Hilbert Space

We know that in Rn , the convergence of a sequence of vectors is equivalent to
coordinate-wise convergence.
   In a Hilbert space H, the Fourier coefficients ( f, gk ) of an element f ∈ H play
the role of the coordinates of f . Suppose {gk } is an orthonormal basis of H. What
can we say about convergence of a sequence { fn } if, for every fixed k, the numerical
sequence {( fn , gk )} is convergent?
   Let us consider {gn } as a sequence. It is seen that for every k,

                                    lim (gn , gk ) = 0 ,
                                   n→∞

hence we have coordinate-wise convergence of {gn } to zero. But the sequence {gn }
is not convergent, since
                                     √
                          gn − gm = 2 for n m .

Therefore, coordinate-wise convergence in a Hilbert space is not equivalent to the
usual form of convergence in the space. We define a new type of convergence in a
Hilbert space.
Definition 2.9.1. Let H be a Hilbert space. A sequence {xk } ⊂ H is weakly conver-
gent to x0 ∈ H if for every continuous linear functional F in H,

                                  lim F(xk ) = F(x0 ) .
                                  k→∞

If every numerical sequence {F(xk )} is a Cauchy sequence, then {xk } is a weak
Cauchy sequence.
   To distinguish between weak convergence and convergence as defined on p. 29,
we shall refer to the latter as strong convergence. We retain the notation xk → x for
strong convergence and adopt xk        x for weak convergence.
   Definition 2.9.1 is given in a form which (with suitable modifications) is valid in
a metric space. But in a Hilbert space any continuous linear functional, by the Riesz
representation theorem, takes the form F(x) = (x, f ) where f is an element of H. So
Definition 2.9.1 may be rewritten as follows:
Definition 2.9.2. Let H be a Hilbert space. A sequence {xn } ⊂ H is weakly conver-
gent to x0 ∈ H if for every element f ∈ H we have

                                 lim (xn , f ) = (x0 , f ) .
                                 n→∞

If every numerical sequence {(xn , f )} is a Cauchy sequence, then {xk } is a weak
Cauchy sequence.
   We have seen that some weak Cauchy sequences in H are not strong Cauchy
sequences. But a strong Cauchy sequence is always a weak Cauchy sequence, by
virtue of the continuity of the linear functionals in the definition.
158                                2 Mechanics Problems from the Functional Analysis Viewpoint

   We formulate a simple sufficient condition for strong convergence of a weakly
convergent sequence:

Theorem 2.9.1. Suppose that xk                x0 , where xk , x0 belong to a Hilbert space H.
Then xk → x0 if xk → x0 .

Proof. Consider xk − x0 2 . We get

         xk − x0     2
                         = (xk − x0 , xk − x0 ) = xk     2
                                                             − (x0 , xk ) − (xk , x0 ) + x0   2
                                                                                                  .

By Definition 2.9.2 we have

                               lim [(x0 , xk ) + (xk , x0 )] = 2 x0        2
                                                                               ,
                               k→∞

hence xk − x0    2
                     → 0.

    We shall see later that for some numerical methods it is easier to first establish
weak convergence of approximate solutions and then strong convergence, than to es-
tablish strong convergence directly. The last theorem allows us to justify a method
successively, beginning with a simple approximate result and then passing to the
needed one. That is why weak convergence is a major preoccupation in this presen-
tation.

Theorem 2.9.2. In a Hilbert space, every weak Cauchy sequence {xn } is bounded.

Proof. We will prove the theorem for a complex Hilbert space; the proof is valid for
a real space as well. Suppose to the contrary that there is a weak Cauchy sequence
{xn } which is not bounded in H. So let xn → ∞ as n → ∞. We will show that this
yields a contradiction.
    First we consider the set U of all numbers of the form (xn , y), where y belongs to
a closed ball B(y0 , ε) with arbitrary ε > 0 and center y0 ∈ H, which are momentarily
fixed. We claim that U is unbounded from above. Indeed, elements of the form
yn = y0 + εxn /(2 xn ) belong to B(y0 , ε) since
                                                     εxn             ε
                                     yn − y0 =                   =     .
                                                    2 xn             2
As {xn } is a weak Cauchy sequence, the numerical sequence {(xn , y0 )} converges and
therefore is bounded. Since xn → ∞ we get
                                           ε                            ε
           |(xn , yn )| = (xn , y0 ) +        (xn , xn ) = (xn , y0 ) +   xn → ∞
                                         2 xn                           2
as n → ∞. We see that U is unbounded for any fixed y0 .
   Now we show that unboundedness of any set U for any y0 yields a contradiction.
Take the ball B(y0 , ε1 ) with ε1 = 1 and y0 = 0. Because U is unbounded from above,
we can take any y1 ∈ B(y0 , ε1 ) and then find xn1 such that

                                           |(xn1 , y1 )| > 1 .                                        (2.9.1)
2.9 Weak Convergence in a Hilbert Space                                                159

By continuity of the inner product in both its variables, we can find a closed ball
B(y1 , ε2 ) such that B(y1 , ε2 ) ⊂ B(y0 , ε1 ) and such that (2.9.1) holds not only for y1
but for all y ∈ B(y1 , ε2 ):

                             |(xn1 , y)| > 1 for all y ∈ B(y1 , ε2 ) .

Then, in the ball B(y1 , ε2 ), we similarly take an interior point y2 and find xn2 , with
n2 > n1 , such that
                                     |(xn2 , y2 )| > 2 ,
and, after this, a closed ball B(y2 , ε3 ) such that B(y2 , ε3 ) ⊂ B(y1 , ε2 ) and

                             |(xn2 , y)| > 2 for all y ∈ B(y2 , ε3 ) .

Repeating this procedure ad infinitum, we produce a sequence of closed balls
B(yk , εk+1 ) such that B(y0 , ε1 ) ⊃ B(y1 , ε2 ) ⊃ B(y2 , ε3 ) ⊃ · · · , and corresponding
terms xnk , nk+1 > nk , of the sequence {xn } such that

                            |(xnk , y)| > k for all y ∈ B(yk , εk+1 ) .

Since H is a Hilbert space there is at least one element y∗ which belongs to every
B(yk , εk+1 ), so
                                   |(xnk , y∗ )| > k .
Thus we find a continuous linear functional F ∗ (x) = (x, y∗ ) for which the numerical
sequence {F ∗ (xnk )} is not a Cauchy sequence. This contradicts the definition of weak
convergence of {xk }.

   This proof yields another important result:

Lemma 2.9.1. Assume {xk } is an unbounded sequence in H, i.e., xk → ∞. Then
there exists y∗ ∈ H and a subsequence {xnk } such that (xnk , y∗ ) → ∞ as k → ∞.

Proof. Let zn = xn / xn . For any y with unit norm, the numerical sequence (zn , y)
is bounded and thus we can select a convergent subsequence from it. If there exists
such a unit element y∗ and a subsequence {znk } for which (znk , y∗ ) → a 0, then the
statement of the lemma is valid for the subsequence {xnk } and y∗ if a > 0; if a < 0,
then y∗ must be changed to −y∗ . Indeed, if a > 0 then (xnk , y∗ ) = (znk , y∗ ) xnk → ∞.
    Now we suppose that we cannot find such an element y∗ and a subsequence
{znk } for which (znk , y∗ ) → a   0. So (zn , y) → 0 for any y ∈ H. By the Riesz
representation theorem, this means that {zn } converges weakly to zero. We will prove
that the statement of Lemma 2.9.1 holds for the latter class of sequences as well. For
this we repeat two steps of the proof of Theorem 2.9.2.
    First we show that for any center y0 and radius ε, the numerical set (xn , y)
with y running over B(y0 , ε) is unbounded. Indeed, taking the sequence yn =
y0 + ε/(2 xn ) xn we get an element from B(y0 , ε). Next,
                                             ε                            ε
               (xn , yn ) = (xn , y0 ) +        (xn , xn ) = (zn , y0 ) +     xn .
                                           2 xn                           2
160                             2 Mechanics Problems from the Functional Analysis Viewpoint

Since ε is finite and (zn , y0 ) → 0 as n → ∞, we have (xn , yn ) → ∞.
   Another step of the proof of Theorem 2.9.2, establishing the existence of a subse-
quence {xnk } and an element y∗ such that (xnk , y∗ ) → ∞, requires only that xn → ∞
and that for any ε > 0 the set (xn , y) is unbounded when y runs over B(y0 , ε), which
was just proved. Thus we immediately state the validity of Lemma 2.9.1 for all the
unbounded sequences.

   This is used in proving the principle of uniform boundedness, which we have
established in a more general form (Theorem 1.23.2).

Theorem 2.9.3. Let {Fk (x)} (k = 1, 2, . . .) be a family of continuous linear function-
als defined on a Hilbert space H. If supk |Fk (x)| < ∞, then supk Fk < ∞.

Proof. By the Riesz representation theorem, each of the functionals Fk (x) has the
form
               Fk (x) = (x, fk ) , where fk ∈ H ,     fk = F k .
So the condition of the theorem can be rewritten as

                                    sup |(x, fk )| < ∞ .                           (2.9.2)
                                      k

By Lemma 2.9.1, the assumption that supk fk = ∞ implies the existence of x0 ∈ H
and { fkn } such that
                         |(x0 , fkn )| → ∞ as k → ∞ .
This contradicts (2.9.2).

Corollary 2.9.1. Let {Fk (x)} be a sequence of continuous linear functionals given on
H, such that for every x ∈ H the numerical sequence {Fk (x)} is a Cauchy sequence.
Then there is a continuous linear functional F(x) on H such that

                             F(x) = lim Fk (x) for all x ∈ H                       (2.9.3)
                                     k→∞

and
                                  F ≤ lim inf Fk < ∞ .                             (2.9.4)
                                          k→∞

Proof. The limit on the right-hand side of (2.9.3), existing by the condition, defines
a functional F(x) which is clearly linear. Since the condition of Theorem 2.9.3 is
met, we have supk Fk < ∞; from

                            |F(x)| = lim |Fk (x)| ≤ sup Fk    x
                                    k→∞              k

it follows that F(x) is continuous. Moreover (recall Problem 1.23.1),

                        |F(x)| = lim |Fk (x)| ≤ lim inf Fk     x ,
                                  k→∞             k→∞

i.e., (2.9.4) is proved also.
2.9 Weak Convergence in a Hilbert Space                                               161

  The following theorem gives an equivalent but more convenient definition of
weak convergence.
Theorem 2.9.4. A sequence {xn } is weakly Cauchy in a Hilbert space H if and only
if the following pair of conditions holds:
(i)     {xn } is bounded in H, i.e., there is a constant M such that xn ≤ M;
(ii)    for any fα ∈ H from a system { fα } which is complete in H, the numerical
        sequence (xn , fα ) is a Cauchy sequence.
Proof. Necessity of the conditions follows from the definition of weak convergence
and Theorem 2.9.2.
   Now we prove sufficiency. Suppose the conditions (i) and (ii) hold. Take an arbi-
trary continuous linear functional defined, by the Riesz representation theorem, by
an element f ∈ H and consider the numerical sequence

                                     dnm = (xn , f ) − (xm , f ) .

As the system { fα } is complete, there is a linear combination
                                                   N
                                            fε =         ck fk
                                                   k=1

such that
                                         f − fε < ε/3M .
Then

                |dnm | = |(xn − xm , f )|
                      = |(xn − xm , fε + f − fε )|
                      ≤ |(xn − xm , fε )| + |(xn − xm , f − fε )|
                            N
                      ≤          |ck | |(xn − xm , fk )| + ( xn + xm ) f − fε .
                           k=1

Since, by (ii), the sequences {(xn , fk )} (k = 1, . . . , N) are Cauchy sequences, we can
find a number R such that
                       N
                            |ck | |(xn − xm , fk )| < ε/3 for all m, n > R
                      k=1

hence
                      |dnm | ≤ ε/3 + 2Mε/(3M) = ε for m, n > R .
This means that {(xn , f )} is a weak Cauchy sequence.
Problem 2.9.1. Show that a sequence {xn } is weakly convergent to x0 in H if and
only if the following pair of conditions holds:
162                           2 Mechanics Problems from the Functional Analysis Viewpoint

(i)    {xn } is bounded in H;
(ii)   for any fα from a system { fα }, fα ∈ H, which is complete in H, we have
       limn→∞ (xn , fα ) = (x0 , fα ).
   Because weak convergence differs from strong convergence, we are led to con-
sider weak completeness of a Hilbert space.
Theorem 2.9.5. Any weak Cauchy sequence {xn } in a Hilbert space H converges
weakly to an element of this space.
   In other words, a Hilbert space H is also weakly complete.
Proof. For any fixed y ∈ H we define F(y) = limn→∞ (y, xn ). The functional F(y),
whose linearity is evident, is defined on the whole of H. From the inequality

                                      |(y, xn )| ≤ M y

where M is a constant such that xn ≤ M, it follows that

                          |F(y)| ≤ M y        and        F ≤M.

Therefore F(y) is a continuous linear functional which, by the Riesz representation
theorem, can be written in the form

                   F(y) = (y, f ) ,       f ∈H,          f = F ≤M.

But this means that f is a weak limit of {xn }.
   From this proof also follows
Lemma 2.9.2. If a sequence {xn } ⊂ H converges weakly to x0 in H and xn ≤ M
for all n, then x0 ≤ M.
Problem 2.9.2. Provide the details.
   This states that a closed ball about zero is weakly closed. Any closed subspace
of a Hilbert space is also weakly closed. We also formulate Mazur’s theorem that
any closed convex set in a Hilbert space is weakly closed. The interested reader can
find a proof in Yosida [44].
Theorem 2.9.6 (Mazur). Assume that a sequence {xn } in a Hilbert space H con-
verges weakly to x0 ∈ H. Then there is a subsequence {xnk } of {xn } such that the
                             1  N
sequence of arithmetic means N k=1 xnk converges strongly to x0 .
Problem 2.9.3. Show that a weakly closed set is closed.
   Let us consider the problem of weak compactness of a set in a Hilbert space. We
have seen that a ball in an infinite dimensional Hilbert space is not strongly compact.
But for weak compactness, an analog of the Bolzano–Weierstrass theorem holds as
follows:
2.9 Weak Convergence in a Hilbert Space                                            163

Theorem 2.9.7. A bounded sequence {xn } in a separable Hilbert space contains a
weak Cauchy subsequence.

   In other words, a bounded set in a Hilbert space is weakly precompact.

Proof. In a separable Hilbert space there is an orthonormal basis {gn }. By Theo-
rem 2.9.4 it suffices to show that there is a subsequence {xnk } such that, for each
fixed gm , the numerical sequence {(xnk , gm )} is a Cauchy sequence.
   The bounded numerical sequence {(xn , g1 )} contains a convergent subsequence
{(xn1 , g1 )}. Considering the numerical sequence {(xn1 , g2 )}, for the same reason we
can choose a convergent sequence {(xn2 , g2 )}. Continuing this process, on the kth
step we obtain a convergent numerical subsequence {(xnk , gk )}.
   Choosing now the elements xnn , we obtain a sequence {xnn } such that for any
fixed gm the numerical sequence {(xnn , gm )} is a Cauchy sequence. That is, {xnn } is a
weak Cauchy sequence.

   This theorem has important applications. In justifying certain numerical meth-
ods we can sometimes prove boundedness of the set of approximate solutions in
a Hilbert (as a rule, energy) space, and hence obtain a subsequence of approxima-
tions that converges weakly to an element; then we can show that this element is a
solution.
   Let us apply this procedure to the approximation problem, namely, we want to
minimize a functional
                                  F(x) = x − x0 2                          (2.9.5)
over a real Hilbert space when x0 is a fixed element of H, x0          M, and x is an
arbitrary element of a closed subspace M ⊂ H.
   In Sect. 1.19 we established the existence of a minimizer of F(x). We now treat
this problem once more, as though this existence were unknown to us.
   This very simple problem (at least in theory) exhibits the following typical steps,
which are common for the justification of approximate solutions to many boundary
value problems:
1. the formulation of an approximation problem and the demonstration of its solv-
   ability;
2. a global a priori estimate of the approximate solutions that does not depend on
   the step of approximation;
3. the demonstration of convergence of the approximate solutions to a solution of
   the initial problem, and a study of the nature of convergence.
Thus we begin to study our problem with Step 1, the formulation of the approxima-
tion problem.
   We try to solve the problem approximately, using the Ritz method. Assume {gk } is
a complete system in M such that any of its finite subsystems is linearly independent.
Consider Mn spanned by (g1 , . . . , gn ) and find an element which minimizes F(x) on
Mn . A solution of this problem, denoted by xn , is the nth Ritz approximation of the
solution.
164                                  2 Mechanics Problems from the Functional Analysis Viewpoint

   A real-valued function f (t) = F(xn + tgk ) of the real variable t takes its minimal
value at t = 0 and, thanks to differentiability of f (t),
                                               d f (t)
                                                               =0.
                                                 dt      t=0

This yields
      d                                  d
 0=      xn − x0 + tgk     2
                               t=0
                                     =      (xn − x0 + tgk , xn − x0 + tgk ) t=0 = 2(xn − x0 , gk ) ,
      dt                                 dt
so xn − x0 is orthogonal to each gk (k = 1, . . . , n).
   Using the representation
                                                         n
                                               xn =           ckn gk ,
                                                        k=1

we get a linear system of algebraic equations called the Ritz system of nth approxi-
mation:
                      n
                           ckn (gk , gm ) = (x0 , gm )                (m = 1, . . . , n) .    (2.9.6)
                     k=1

The determinant of this system is the Gram determinant of a linearly independent
system (g1 , . . . , gm ) that is not equal to zero. So the system (2.9.6) has a unique
solution (ˆ 1n , . . . , ckn ).
          c              ˆ
   Step 2. Now we will find a global estimate of the approximate solutions that does
not depend on n. Although, in this case, we know that the approximate solution
exists, we can get the estimate without this knowledge. Hence it is called an a priori
estimate.
   We begin with the definition of xn :

                           xn − x0        2
                                              ≤ x − x0         2
                                                                   for all x ∈ Mn .

As x = 0 ∈ Mn , it follows that

                                              xn − x0    2
                                                              ≤ x0    2
                                                                          ,

from which
                                              xn   2
                                                       ≤ 2 xn       x0 ,
hence
                                                   xn ≤ 2 x0 .                                (2.9.7)
This is the required estimate.
Remark 2.9.1. It is possible to get a sharper estimate than (2.9.7); however, for this
problem it is only necessary to establish the existence of a bound.
   Step 3. Our last goal is to show that the sequence of approximations converges to
a solution of the problem. First we demonstrate that this convergence is weak, and
then that it is strong.
2.9 Weak Convergence in a Hilbert Space                                                      165

   By (2.9.7), the sequence {xn } is bounded and, thanks to Theorem 2.9.7, contains a
weakly convergent subsequence {xnk } whose weak limit x∗ belongs to M (remember
that a closed subspace is weakly closed).
   For any fixed gm , we can pass to the limit as k → ∞ in the equality

                                     (xnk − x0 , gm ) = 0

and get
                                     (x∗ − x0 , gm ) = 0
because (x, gm ) is a continuous linear functional in x ∈ H.
   Now consider (x∗ − x0 , h) where h ∈ M is arbitrary but fixed. By completeness of
the system g1 , g2 , g3 , . . . in M, given ε > 0 we can find a finite linear combination
                                                   N
                                       hε =              ck gk
                                                   k=1

such that
                                   h − hε ≤ ε/(3 x0 ) .
Then

                   |(x∗ − x0 , h)| = |(x∗ − x0 , h − hε + hε )|
                                  ≤ |(x∗ − x0 , h − hε )| + |(x∗ − x0 , hε )|
                                  = |(x∗ − x0 , h − hε )|
                                  ≤ x∗ − x0              h − hε
                                         ∗
                                  ≤( x           + x0 ) h − hε
                                  ≤ (2 x0 + x0 ) ε/(3 x0 ) = ε .

Therefore, for any h ∈ M we get

                                     (x∗ − x0 , h) = 0 .                                  (2.9.8)

Finally, considering values of (2.9.5) on elements of the form x = x∗ + h when
h ∈ M, we obtain, by (2.9.8),

                F(x∗ + h) = (x∗ − x0 + h, x∗ − x0 + h)
                            = x∗ − x0        2
                                                 + 2(x∗ − x0 , h) + h        2


                            = x∗ − x0        2
                                                 + h     2
                                                             ≥ x∗ − x0   2
                                                                             = F(x∗ ) .

It follows that x∗ is a solution of the problem, and existence of solution has been
proved.
    Now we can show that the approximation sequence converges strongly to a solu-
tion of the problem. By Theorem 1.19.3, a minimizer of F(x) is unique; this gives
166                           2 Mechanics Problems from the Functional Analysis Viewpoint

us weak convergence of the sequence {xn } on the whole. Indeed, suppose to the con-
trary that {xn } does not converge weakly to x∗ . Then there is an element f ∈ H such
that
                                   (xn , f ) (x∗ , f ) .                       (2.9.9)
By boundedness of the numerical set {(xn , f )}, the statement (2.9.9) implies that
there is a subsequence {xnk } such that there exists

                                 lim (xnk , f )    (x∗ , f ) .                  (2.9.10)
                                 k→∞

Problem 2.9.4. Prove (2.9.10).

   But, for the subsequence {xnk }, we can repeat the above considerations and find
that {xnk } contains a subsequence which converges weakly to a solution of the prob-
lem. Since the solution is unique, this contradicts (2.9.10). Finally, multiplying both
sides of (2.9.6) by the Ritz coefficient cmn and summing over m, we get
                                         ˆ

                                   (xn , xn ) = (x0 , xn ) .

We can pass to the limit as n → ∞, obtaining

                        lim (xn , xn ) = lim (x0 , xn ) = (x0 , x∗ ) .
                        n→∞              n→∞

By (2.9.8) with h = x∗ we have

                                   (x0 , x∗ ) = (x∗ , x∗ ) ,

so
                                   lim xn     2
                                                  = x∗   2
                                                             .
                                  n→∞

Therefore, by Theorem 2.9.1, the sequence {xn } converges strongly to x∗ .
   So we have demonstrated, via the Ritz method, a general way of justifying the
solution of a minimal problem and the Ritz method itself. The method is common
to a wide variety of problems, some nonlinear. In the latter case, many difficulties
center on Steps 2 or 3, depending on the problem. The problem under discussion can
also be interpreted another way, and this is of so much importance that we devote a
separate section to it.



2.10 The Ritz and Bubnov–Galerkin Methods in Linear
     Problems

We reconsider the problem of minimizing the quadratic functional (2.4.8) in a
Hilbert space, namely,

                            I(x) = x     2
                                             + 2Φ(x) → min .                    (2.10.1)
                                                               x∈H
2.10 Ritz and Bubnov–Galerkin Methods                                                          167

Assuming Φ(x) is a continuous linear functional, the Riesz representation theorem
yields
                                Φ(x) = (x, −x0 )
where x0 ∈ H is uniquely defined by Φ(x). Then

                       I(x) = x     2
                                        − 2(x, x0 ) = x − x0            2
                                                                            − x0   2
                                                                                       .
           2
Since x0       is fixed, the problem (2.4.1) is equivalent to

                                  F(x) = x − x0         2
                                                            → min .
                                                                  x∈H

This problem has the unique (and obvious) solution x = x0 . Of much interest is
the fact that it coincides with the problem of the previous section if M = H. So
application of the Ritz method in this problem is justified. Let us recall those results
in terms of the new problem.
    Let {gk } be a complete system in H, every finite subsystem of which is linearly
independent, and let the nth Ritz approximation to a minimizer be
                                                  n
                                          xn =         ckn gk .
                                                 k=1

The system giving the nth approximation of the Ritz method is
                       n
                            ckn (gk , gm ) = −Φ(gm )          (m = 1, . . . , n) .         (2.10.2)
                      k=1

Let us collect the results in

Theorem 2.10.1. The following statements hold.
(i)    For each n ≥ 1, the system (2.10.2) of nth approximation of the Ritz method
       has the unique solution c1n , . . . , cnn .
(ii)   The sequence {xn } of Ritz approximations defined by (2.10.2) converges
       strongly to the minimizer of the quadratic functional x 2 + 2Φ(x), where
       Φ(x) is a continuous linear functional on H.

   It is interesting to note that if {gk } is an orthonormal basis of H, then (2.10.2)
gives the Fourier coefficients of the solution in H.
   Concerning Bubnov’s method, we only mention that it appeared when A.S. Bub-
nov, reviewing an article by S. Timoshenko, noted that the Ritz equations can be
obtained by multiplying by gm , a function of a complete system, the differential
equation of equilibrium in which u was replaced by
                                                  n
                                         un =          ckn gk ,
                                                 k=1
168                               2 Mechanics Problems from the Functional Analysis Viewpoint

integrating the latter over the region, and then integrating by parts. In our terms this
is
                        (un , gm ) = −Φ(gm )    (m = 1, . . . , n) .
Since this system indeed coincides with (2.10.2), Theorem 2.10.1 also justifies Bub-
nov’s method.
   Galerkin was the first to propose multiplying by fm , a function of another sys-
tem, for better approximation of the residual. The corresponding system is, in our
notation,
                      (un , fm ) = −Φ( fm )  (m = 1, . . . , n) .
Discussion of this modification of the method can be found in Mikhlin [27].
   Finally, we note that the finite element method for solution of mechanics prob-
lems is a particular case of the Bubnov–Galerkin method, hence it is also justified
for the problems we consider.



2.11 Curvilinear Coordinates, Nonhomogeneous Boundary
     Conditions

We have considered some problems of mechanics using the Cartesian coordinate
system. Almost all of the textbooks present the theory of the same problems in
Cartesian frames; the few exceptions are the textbooks on the theory of shells and
curvilinear beams, where it is impossible to consider the problems in Cartesian
frames. However, in practice other coordinate systems occur frequently. The ques-
tion arises whether it is necessary to investigate the boundary value problems for
other coordinates, or whether it is enough to reformulate the results for Cartesian
systems. For the generalized statements of mechanics problems in energy spaces,
the answer is simple: it is possible to reformulate the results, and a key tool is a sim-
ple change of the coordinates. This change allows us to reformulate the imbedding
theorems in energy spaces, to establish the requirements for admitting classes of
loads, etc. We note that it is a hard problem to obtain similar results independently,
without the use of coordinate transformations, if the coordinate frame has singular
points.
    Let us consider a simple example of a circular membrane with fixed edge (Dirich-
let problem). In Cartesian coordinates we have the Sobolev imbedding theorem
                                                                                1/2
                                 1/p              ∂u   2       ∂u   2
                |u(x)| p dx dy         ≤m                  +            dx dy         (2.11.1)
            Ω                                Ω    ∂x           ∂y

for p ≥ 1, which is valid for any u ∈ W 1,2 (Ω) ≡ E MC satisfying the boundary
                                       ˙
condition
                                  u ∂Ω = 0 .                           (2.11.2)
2.11 Curvilinear Coordinates, Nonhomogeneous Boundary Conditions                                                                       169

Taking a function u ∈ C (1) (Ω) satisfying (2.11.2), in both integrals of (2.11.1) we
pass to the polar coordinate system:
                                                                                                                             1/2
        R       2π                   1/p                               R        2π
                                                                                        ∂u   2       1 ∂u     2
                     |u| p r dφ dr           ≤m                                                  +                 r dφ dr         (2.11.3)
    0       0                                                      0        0           ∂r           r2 ∂φ

where (r, φ) are the polar coordinates in a disk of radius R. Passing to the limit along
a Cauchy sequence of E MC in the inequality (2.11.1), which is valid in Cartesian
coordinates, shows us that it remains valid in the form (2.11.3) in polar coordinates.
Inequality (2.11.3) is an imbedding theorem in the energy space of the circular mem-
brane in terms of polar coordinates. The expression
                                                                                                                   1/2
                                                 R            2π
                                                                        ∂u      2       1 ∂u         2
                            u =                                                     +                    r dφ dr                   (2.11.4)
                                             0            0             ∂r              r2 ∂φ

is the norm in this coordinate system and
                                                     R             2π
                                                                           ∂u ∂v 1 ∂u ∂v
                             (u, v) =                                           +         r dφ dr
                                                 0            0            ∂r ∂r r2 ∂φ ∂φ

is the corresponding inner product.
    The requirement imposed on forces for existence of a generalized solution has
the form
                                         R           2π
                                                          |F|q r dφ dr < ∞                           (q > 1) .
                                     0           0
We have a natural form of the norm in the energy space (which is determined by
the energy itself) using curvilinear coordinates, as well as a form of the imbedding
theorem (i.e., properties of elements of the energy space and natural requirements
on forces for the problem to be uniquely solvable).
    Then we note that we can replace formally the Cartesian system by any other
system of coordinates which is admissible for smooth functions, and also change
formally any variables in any expression which makes sense in the energy space
considered in Cartesian coordinates.
    Finally, note that a norm like (2.11.4) is usually called a weighted norm because
of the presence of weight factors, here connected with powers of r. There is an
abstract theory of such weighted Sobolev spaces, not being so elementary as in the
space we have considered.
    For more complicated problems such as elasticity problems, we can use the same
method of introducing curvilinear coordinates; here we can change not only the in-
dependent variables (x1 , x2 , x3 ), but also unknown components of vectors of dis-
placements and prescribed forces, to the new coordinate system. We leave it to the
reader to write down an equation determining a generalized solution, the forms of
norm and scalar product, and restrictions for forces as well as imbedding inequali-
ties, in other curvilinear coordinate systems such as cylindrical and spherical.
170                           2 Mechanics Problems from the Functional Analysis Viewpoint

   Now let us consider two questions connected with nonhomogeneous boundary
value problems in mechanics. The first is to identify the whole class of admissible
external forces for which an energy solution exists. We know that the condition for
existence of a solution is that the functional of external forces

                                           F(x)v(x) dΩ                          (2.11.5)
                                       Ω

(say, in the membrane problem) is continuous and linear with respect to v(x) on
an energy space. We shall show how this condition can be expressed in terms of
so-called spaces with negative norms, a notion due to P.D. Lax [17].
    The functional (2.11.5) can be considered as the scalar product of F(x) by v(x) in
L2 (Ω). But v(x) belongs to an energy space E whose norm, for simplicity, is assumed
to be such that v E = 0 implies v = 0. We know that v ∈ L2 (Ω) if v ∈ E; moreover,
E is dense in L2 (Ω). For any F(x) ∈ L2 (Ω), we can introduce a new norm

                            F   E   = sup           F(x)v(x) dΩ .
                                      v E ≤1    Ω


It is clear that L2 (Ω) with this norm is not complete (since all v ∈ L p (Ω) for any
p > 2, p < ∞). The completion of L2 (Ω) in the norm · E is called the space
with negative norm, denoted E − . In Lax [17] (and in other books, for example,
Yosida [44]) it is shown that the set of all continuous linear functionals on E can be
identified with E − since E is dense in L2 (Ω).
    So the condition F(x) ∈ E − is necessary and sufficient for the work functional
(2.11.5) to be continuous with respect to v(x) on E.
                                                                           ˙
    In Lax [17], such a construction was introduced for a Sobolev space W k,2 (Ω); the
                                                                −k,2
corresponding space with negative norm was denoted by W (Ω). An equivalent
approach to the introduction of W −k,2 (Ω) involves use of the Fourier transformation
in Sobolev spaces (cf., Yosida [44]).
    The notion of the space with negative norm is useful for studying problems, but it
is not too informative when we want to know whether certain forces are of a needed
class; here sufficient conditions are more convenient.
    Secondly, we discuss how to handle nonhomogeneous boundary conditions (of
Dirichlet type). Consider, for example, the problem

                                        −Δv = F ,                               (2.11.6)
                                           v ∂Ω = ϕ .                           (2.11.7)

We can try the classical approach, finding a function v0 (x) that satisfies (2.11.7), i.e.,

                                        v0 ∂Ω = ϕ .

Now we are seeking v(x) in the form v = u + v0 , where u(x) satisfies the homoge-
neous boundary condition
                                   u ∂Ω = 0 .                             (2.11.8)
2.12 Bramble–Hilbert Lemma and Its Applications                                               171

An integro-differential equation of equilibrium of the membrane is

         ∂u ∂ψ ∂u ∂ψ                       ∂v0 ∂ψ ∂v0 ∂ψ
              +      dΩ +                        +       dΩ =                      Fψ dΩ (2.11.9)
     Ω   ∂x ∂x ∂y ∂y                   Ω   ∂x ∂x   ∂y ∂y                       Ω

wherein virtual displacements must also satisfy (2.11.8):

                                           ψ|∂Ω = 0 .

We recognize the term
                                       ∂v0 ∂ψ ∂v0 ∂ψ
                                             +       dΩ
                                   Ω   ∂x ∂x   ∂y ∂y
as a continuous linear functional on E MC if ∂v0 /∂x and ∂v0 /∂y belong to L2 (Ω). In
that case there is a generalized solution to the problem, i.e., u ∈ E MC satisfying
(2.11.9) for any ψ ∈ E MC .
    We have supposed that there exists an element of W 1,2 (Ω) satisfying (2.11.7).
In more detailed textbooks on the theory of partial differential equations, one may
find the conditions for a function ϕ given on the boundary that are sufficient for the
existence of v0 . Corresponding theorems for v0 from Sobolev spaces are called trace
theorems. The trace theorems assume the boundary is sufficiently smooth. The case
of a piecewise smooth boundary, frequently encountered in practice, has not been
completely studied yet. The problem of the traces of functions is beyond the scope
of this book.
    A final remark is in order. In mathematics we normally deal with dimensionless
quantities, and we have followed that practice here. However, variables with dimen-
sional units can be used without difficulty, provided we check carefully for units in
all inequalities and equations, and introduce additional factors as needed. In partic-
ular, the constants in imbedding theorems normally carry dimensional units, hence
these constants change if the units are changed.



2.12 Bramble–Hilbert Lemma and Its Applications

This lemma is widely used to establish the convergence rate for the finite element
method (see, for example, Ciarlet [7]). It gives a bound for a functional with special
properties in a Sobolev space. We remark that the lemma can be viewed as a simple
consequence of the theorem on equivalent norming of W l,p (Ω) in Sobolev [33].
   Recall Poincar´ ’s inequality (2.3.9),
                  e
                                           2            ∂u   2       ∂u   2
               u2 dS ≤ m           u dS        +                 +            dS ,        (2.12.1)
           S                   S                   S    ∂x           ∂y

which was derived when S was the square [0, a] × [0, a].
  The proof of (2.3.9) is easily extended to the case of an n-dimensional cube. We
now discuss how to extend it to a compact set Ω that is star-shaped with respect to a
172                                         2 Mechanics Problems from the Functional Analysis Viewpoint

square S ; that is, any ray starting in S intersects the boundary of Ω exactly once. We
shall establish the following estimate, which is also called Poincar´ ’s inequality:
                                                                       e
                                                          2                                  ∂u      2        ∂u      2
                      u2 dΩ ≤ m1               u dS           + m2                                       +                dΩ .              (2.12.2)
                 Ω                         S                                      Ω          ∂x               ∂y

Let us rewrite this in a system of polar coordinates (r, φ) having origin at the center
of S . Let ∂Ω be given by the equation r = R(φ) ≥ a/2, R(φ) < R0 . Then (2.12.2) has
the form
     2π        R(φ)                                                2                   2π           R(φ)
                                                                                                             ∂u       2       1 ∂u      2
                      u2 r dr dφ ≤ m1               u dS +m2                                                              +                 r dr dφ .
 0         0                                    S                                 0             0            ∂r               r2 ∂φ

Because of (2.12.1), it follows that it is sufficient to get the estimate
          2π      R(φ)                              2π         a/2
                                                                          ∂u 2                                   2π       R(φ)
                         u2 r dr dφ ≤ m3                                u2 r dr dφ + m4
                                                                               dr dφ                                             r
    0     a/2                    0    a/4                     0   a/4     ∂r
                                                                               (2.12.3)
with constants independent of u ∈ C (1) (Ω) (C (1) (Ω) is introduced in Cartesian coor-
dinates!). We now proceed to prove this.
   The starting point is the representation
                                                         r2
                                                              ∂u(r, φ)                                   a/4 ≤ r1 ≤ a/2 ,
                      u(r2 , φ) = u(r1 , φ) +                          dr ,
                                                     r1         ∂r                                       a/4 ≤ r2 ≤ R0 ,

from which, by squaring both sides and applying elementary transformations, we
get

                                                                          1 √ ∂u
                                                                   r2                                        2
                 u2 (r2 , φ) ≤ 2u2 (r1 , φ) + 2                           √   r    dr
                                                               r1           r   ∂r
                                                               r2
                                                                        dr            r2
                                                                                                ∂u 2
                              ≤ 2u2 (r1 , φ) + 2                                           r         dr
                                                              r1         r        r1            ∂r
                                                                       R(φ)
                                                                                   ∂u 2                                          4R0
                              ≤ 2u2 (r1 , φ) + m5                             r         dr ,                 m5 = 2 ln               .
                                                                   a/4             ∂r                                             a

Multiplying this chain of inequalities by r1 r2 and then integrating it first with respect
to r2 from a/2 to R(φ) and then with respect to r1 from a/4 to a/2, we have
                 a/2         R(φ)                                                      a/2                                     R(φ)
                       r1           u2 (r2 , φ)r2 dr2 dr1 ≤ 2                                  u2 (r1 , φ)r1 dr1                      r2 dr2
               a/4          a/2                                                   a/4                                         a/2
                                                    a/2            R(φ)                                      R(φ)
                                                                                                                          ∂u 2
                                       + m5                               r1 r2 dr1 dr2                               r        dr
                                                a/4           a/2                                          a/4            ∂r
or
2.12 Bramble–Hilbert Lemma and Its Applications                                                                           173
                R(φ)                                 a/2                                           R(φ)
                                                                                                              ∂u 2
     3 2
    32 a               u2 (r, φ)r dr ≤ R2                  u2 (r, φ)r dr +         3 2 2
                                                                                  64 a R0 m5              r        dr .
               a/2
                                        0
                                                    a/4                                          a/4          ∂r

Finally, integrating this with respect to φ over [0, 2π] and multiplying it by 32/(3a2 ),
we establish (2.12.3) and hence (2.12.2).
   We can similarly extend Poincar´ ’s inequality to the case of a multiconnected do-
                                     e
main Ω which is a union of star-shaped domains, and to the case of an n-dimensional
domain Ω with n > 2. The latter extension is
                                                                             n
                                                               2                          ∂u 2
                               u2 dΩ ≤ m1               u dΩ + m2                              dΩ ,               (2.12.4)
                           Ω                        C                       i=1       Ω   ∂xi

where C ⊂ Ω is a hypercube in Rn .
   We can apply the inequality (2.12.4) to any derivative Dα u, |α| < k. Combining
these estimates successively, we derive the inequality needed to prove the Bramble–
Hilbert lemma:
                                                               2
           u   2
               W k,2 (Ω)
                           ≤ m3                     Dα u dΩ        + m4                   |Dα u|2 dΩ .            (2.12.5)
                                  0≤|α|<k       C                           |α|=k     Ω


This estimate permits us to introduce another form of equivalent norm in W k,2 (Ω).
(Question to the reader: Which one?) Note that the estimate was obtained for func-
tions of C (k) (Ω), but the now standard procedure of completion provides that it is
valid for any u ∈ W k,2 (Ω).

Lemma 2.12.1 (Bramble–Hilbert [5]). Assume F(u) is a continuous linear func-
tional on W k,2 (Ω) such that for any polynomial Pr (x) of order less than k,

                                                     F(Pr (x)) = 0 .                                              (2.12.6)

Then there is a constant m∗ depending only on Ω such that
                                                      ⎛                             ⎞1/2
                                                      ⎜
                                                      ⎜
                                                      ⎜                             ⎟
                                                                                    ⎟
                                                                                    ⎟
                               |F(u)| ≤ m F W k,2 (Ω) ⎜
                                            ∗         ⎜
                                                      ⎜
                                                      ⎝                     |D u| dΩ⎟ .
                                                                                  α ⎟
                                                                                    ⎟
                                                                                    ⎠
                                                                                      2
                                                                                                                  (2.12.7)
                                                               |α|=k    Ω


Proof. From (2.12.5) and continuity of F(u) on W k,2 (Ω), it follows that
                           ⎡                                                                        ⎤1/2
                           ⎢
                           ⎢
                           ⎢                                        2                               ⎥
                                                                                                    ⎥
                                                                                                    ⎥
                           ⎢
    |F(u)| ≤ m F W k,2 (Ω) ⎢
                           ⎢                              Dα u dΩ +                                 ⎥
                                                                                          |Dα u|2 dΩ⎥ .
                                                                                                    ⎥             (2.12.8)
                           ⎣                                                                        ⎦
                                      0≤|α|<k        C                      |α|=k     Ω


By (2.12.6),
                                        F(u(x) + Pk−1 (x)) = F(u(x))
where Pk−1 (x) is an arbitrary polynomial of order k − 1. Fixing u(x) ∈ W k,2 (Ω), we
can always choose a polynomial P∗ (x) such that
                                   k−1
174                                              2 Mechanics Problems from the Functional Analysis Viewpoint


                         Dα (u(x) + P∗ (x)) dΩ = 0 for all 0 ≤ |α| ≤ k − 1 .
                                     k−1
                     C

Substituting u(x) + P∗ (x) into (2.12.8), we get (2.12.7) since
                     k−1

                                             Dα P∗ (x) = 0 for |α| = k .
                                                 k−1

This completes the proof.

  Let us consider some simple applications of this lemma. Assume that we find
numerically, by Simpson’s rule,
                                             1
                                                 u(x) dx for u(x) ∈ W 2,2 (0, 1) .
                                         0

What is a bound on the error? First we find the error in one step of the trapezoidal
rule:
                             xk +h
                                            h
                 Fk (u) =          u(x) dx − [u(xk + h) + u(xk )] .
                           xk               2
It is clear that Fk (u) is a linear and continuous functional in W 2,2 (0, 1). Making the
change of variable x = xk + hz in the integral, we get
                         1
      |Fk (u)| = h           u(xk + hz) dz − 1 [u(xk ) + u(xk + h)] ≤ 2h max |u(xk + zh)| .
                                             2
                     0                                                                                 z∈[0,1]
                                                                                                                          (2.12.9)
By the elementary inequality (Problem 2.12.1 below)

                         √            1                                   1/2
                                                                                     √
       max | f (x)| ≤        2                   f 2 (x) + [ f (x)]2 dx         ≤     2 f              W 2,2 (0,1)   ,   (2.12.10)
      x∈[0,1]                     0

relation (2.12.9) gives
                                             √
                                 |Fk (u)| ≤ 2 2h u(xk + hz)                     W 2,2 (0,1)   .

Since Fk (a+bx) = 0 for any constants a, b we can apply the Bramble–Hilbert lemma
and obtain
                  √                  1                            1/2                              xk +h                   1/2
      |Fk (u)| ≤ 2 2hm                   [u (xk + hz)]2 dz              = m1 h5/2                          [u (x)]2 dx           .
                                 0                                                                xk

This is the needed error bound for one step of integration.
   Consider now the bound on total error when [0, 1] is subdivided into N equal
parts
                        1             N−1
                                    h
             F(u) =       u(x) dx −       [u(xk ) + u(xk+1 )] , xk = kh .
                      0             2 k=0

This is linear and continuous in W 2,2 (0, 1), and
2.12 Bramble–Hilbert Lemma and Its Applications                                                                        175

                                                       N−1
                                            f (u) =             Fk (u) .
                                                       k=0

We get
                    N−1              N−1                               N−1                 xk +h                 1/2
         |F(u)| =         Fk (u) ≤          |Fk (u)| ≤ m1 h5/2                                     [u (x)]2 dx
                    k=0              k=0                               k=0             xk

                        √         N−1         xk +h                        1/2
               ≤ m1 h5/2 N                            [u (x)]2 dx                .
                                     k=0    xk


Thus the needed bound on the error of the trapezoidal rule is
                                                            1                         1/2
                             |F(u)| ≤ m1 h2                     [u (x)]2 dx                    .
                                                        0

No improvements in the order of the error result if we take functions smoother than
those from W 2,2 (0, 1). But if v ∈ W 1,2 (0, 1), the bound is worse:
                                                            1                        1/2
                              |F(v)| ≤ m2 h                     [v (x)]2 dx                .
                                                        0

Problem 2.12.1. Prove (2.12.10).

   Another example of the application of Lemma 2.12.1 is given by

Problem 2.12.2. Show that the local error of approximation of the first derivatives
of a function u(x1 , x2 ) ∈ W 3,2 (Ω), Ω ⊂ R2 , by symmetric differences, is

              ∂u(0, 0) u(h1 , 0) − u(−h1 , 0)   ∂u(0, 0) u(0, h2 ) − u(0, −h2 )
     l(u) =           −                       +         −
                ∂x1              2h1              ∂x2              2h2
              M(h2 + h2 )
          ≤    √
                 1     2
                          u     W 3,2 (Ω)
                 h1 h2
if 0 < c1 < h1 /h2 < c2 < ∞.
http://www.springer.com/978-1-4614-5867-8

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Functional analysis in mechanics 2e

  • 1. Chapter 2 Mechanics Problems from the Functional Analysis Viewpoint In the past, an engineer could calculate mechanical stresses and strains using a pen- cil and a logarithmic slide rule. Modern mechanical models, on the other hand, are nonlinear, and even the linear models are complicated. Numerical methods in struc- tural dynamics cannot be applied without computers running specialized programs. However, a researcher should have a solid grasp of the equations that underlie a numerical model and the types of results that can be expected. New models ap- pear in mechanics on a regular basis. Some of these, when written out in detail, can span multiple pages and are clearly beyond pencil-and-paper approaches. Although functional analysis does not provide a detailed picture of the results to be expected from a complicated model, it can answer questions regarding whether the problem is mathematically well-posed (e.g., whether a solution exists and is unique). It may also indicate whether the results can be obtained by a general computer program or whether a special program, based on a knowledge of the general properties of the model, is required. In other words, functional analysis can provide valuable insight even to those who rely heavily on numerical approaches. 2.1 Introduction to Sobolev Spaces In his famous book [33], S.L. Sobolev (1908–1989) introduced some normed spaces that now bear his name; they are denoted by W m,p (Ω). The norm in W m,p (Ω) is ⎛ ⎞1/p ⎜ ⎜ ⎜ ⎟ ⎟ ⎟ u =⎜ ⎜ ⎜ ⎝ |D u| dΩ⎟ , α p ⎟ ⎟ ⎠ (2.1.1) Ω |α|≤m where m is an integer, p ≥ 1, and Ω is compact in Rn . (The Dα notation was in- troduced on p. 19.) Indeed this is a norm on the set of functions that possess con- tinuous derivatives up to order m on Ω, with satisfaction of axiom N3 ensured by Minkowski’s inequality (1.3.6). The completion of the resulting normed space is the Banach space W m,p (Ω). L.P. Lebedev et al., Functional Analysis in Mechanics, Springer Monographs in Mathematics, 115 DOI 10.1007/978-1-4614-5868-5_2, © Springer Science+Business Media, LLC 2013
  • 2. 116 2 Mechanics Problems from the Functional Analysis Viewpoint For Ω a segment [a, b], the spaces W m,p (a, b) were introduced by Stefan Banach (1892–1945) in his dissertation. Our interest in Sobolev spaces is clear, as the ele- ments of each of our energy spaces will belong to W m,2 (Ω) for some m. Problem 2.1.1. (a) Demonstrate that if f (x) ∈ W 1,2 (a, b) and g(x) ∈ C (1) (a, b), then f (x)g(x) ∈ W 1,2 (a, b). (b) Show that A f (x) = f (x) defines a bounded linear operator acting from W 1,2 (0, 1) to L2 (0, 1). Generalized Notions of Derivative. For u ∈ L p (Ω), K.O. Friedrichs (1901–1982) [12] introduced the notion of strong derivative, calling v ∈ L p (Ω) a strong derivative Dα (u) if there is a sequence {ϕn } ⊂ C (∞) (Ω) such that |u(x) − ϕn (x)| p dΩ → 0 and |v(x) − Dα ϕn (x)| p dΩ → 0 as n → ∞ . Ω Ω Since C (∞) (Ω) is dense in any C (k) (Ω), an element of W m,p (Ω) has all strong deriva- tives up to the order m lying in L p (Ω). Another version of generalized derivative was proposed by Sobolev. He used, along with the classical integration by parts formula, an idea from the classical cal- culus of variations: if u(x) ϕ(x) dΩ = 0 Ω for all infinitely differentiable functions ϕ(x) having compact support in an open domain Ω, then u(x) = 0 almost everywhere (everywhere if u(x) is to be continuous). We say that a function ϕ(x), infinitely differentiable in Ω, has compact support in Ω ⊂ Rn if the closure of the set M = {x ∈ Ω : ϕ(x) 0} is a compact set in Ω. In the sense introduced by Sobolev, v ∈ L p (Ω) is called a weak derivative Dα u of u ∈ L p (Ω) if for every infinitely differentiable function ϕ(x) with compact support in Ω we have u(x) Dα ϕ(x) dΩ = (−1)|α| v(x) ϕ(x) dΩ . (2.1.2) Ω Ω The two definitions of generalized derivative are equivalent [33]. The proof would exceed the scope of our presentation, and the same is true for some other facts of this section. Imbedding Theorems. The results presented next are particular cases of Sobolev’s imbedding theorem. Later we will give proofs of some particular results when study- ing certain energy spaces for elastic models which turn out to be subspaces of Sobolev spaces. The theorems are formulated in terms of imbedding operators, a notion elucidated further in Sect. 2.2. We assume that the compact set Ω ⊂ Rn satisfies the cone condition. This means there is a finite circular cone in Rn such that any point of the boundary of Ω can be touched by the vertex of the cone while the cone lies fully inside Ω. This is the
  • 3. 2.1 Introduction to Sobolev Spaces 117 condition under which Sobolev’s imbedding theorem is proved. We denote by Ωr an r-dimensional piecewise smooth hypersurface in Ω. (This means that, at any point of smoothness, in a local coordinate system, it is described by functions having all derivatives continuous up to order m locally, if we consider W m,p (Ω).) The theory of Sobolev spaces is a substantial branch of mathematics (see Adams [1], Lions and Magenes [25], etc.). We formulate only what is needed for our pur- poses. This is Sobolev’s imbedding theorem with some extensions: Theorem 2.1.1 (Sobolev–Kondrashov). The imbedding operator of the Sobolev space W m,p (Ω) to Lq (Ωr ) is continuous if one of the following conditions holds: (i) n > mp, r > n − mp, q ≤ pr/(n − mp); (ii) n = mp, q is finite with q ≥ 1. If n < mp, then the space W m,p (Ω) is imbedded into the H¨ lder space H α (Ω) when o α ≤ (mp − n)/p, and the imbedding operator is continuous. The imbedding operator of W m,p (Ω) to Lq (Ωr ) is compact (i.e., takes every bounded set of W m,p (Ω) into a precompact set of the corresponding space1 ) if (i) n > mp, r > n − mp, q < pr/(n − mp) or (ii) n = mp and q is finite with q ≥ 1. If n < mp then the imbedding operator is compact to H α (Ω) when α < (mp − n)/p. Note that this theorem provides imbedding properties not only for functions but also for their derivatives: u ∈ W m,p (Ω) =⇒ Dα u ∈ W m−k,p (Ω) for |α| = k . Also available are stricter results on the imbedding of Sobolev spaces on Ω into function spaces given on manifolds Ωr of dimension less than n; however, such trace theorems require an extended notion of Sobolev spaces. Let us formulate some useful consequences of Theorem 2.1.1. Theorem 2.1.2. Let γ be a piecewise differentiable curve in a compact set Ω ⊂ R2 . For any finite q ≥ 1, the imbedding operator of W 1,2 (Ω) to the spaces Lq (Ω) and Lq (γ) is continuous (and compact), i.e., max{ u Lq (Ω) , u Lq (γ) } ≤m u W 1,2 (Ω) (2.1.3) with a constant m which does not depend on u(x). Theorem 2.1.3. Let Ω ⊂ R2 be compact. If α ≤ 1, the imbedding operator of W 2,2 (Ω) to H α (Ω) is continuous; if α < 1, it is compact. For the first derivatives, the imbedding operator to Lq (Ω) and Lq (γ) is continuous (and compact) for any finite q ≥ 1. 1 The notion of compact operator will be explored in Chap. 3.
  • 4. 118 2 Mechanics Problems from the Functional Analysis Viewpoint Theorem 2.1.4. Let γ be a piecewise smooth surface in a compact set Ω ⊂ R3 . The imbedding operator of W 1,2 (Ω) to Lq (Ω) when 1 ≤ q ≤ 6, and to L p (γ) when 1 ≤ p ≤ 4, is continuous; if 1 ≤ q < 6 or 1 ≤ p < 4, respectively, then it is compact. We merely indicate how such theorems are proved. We will establish similar re- sults for the beam problem (see (2.3.5)) and for the clamped plate problem (see (2.3.21)) via integral representations of functions from certain classes. In like man- ner, Sobolev’s original proof is given for Ω a union of bounded star-shaped domains. A domain is called star-shaped with respect to a ball B if any ray with origin in B intersects the boundary of the domain only once. For a domain Ω which is bounded and star-shaped with respect to a ball B, a function u(x) ∈ C (m) (Ω) can be repre- sented in the form α α u(x) = x1 1 · · · xn n Kα (y) u(y) dΩ |α|≤m−1 B 1 + Kα (x, y) Dα u(y) dΩy (2.1.4) Ω |x − y|n−m |α|=m where Kα (y) and Kα (x, y) are continuous functions. Investigating properties of the integral terms on the right-hand side of the representation (2.1.4), Sobolev formu- lated his results; later they were extended to more general domains. Another method is connected with the Fourier transformation of functions. In the case of W m,2 (Ω), it is necessary to extend functions of C (m) (Ω) outside Ω in such a way that they belong to C m (Rn ) and W m,2 (Rn ). Then using the Fourier transforma- tion u(y) = (2π)−n/2 ˆ e−ix·y u(x) dx1 · · · dxn Rn along with the facts that u(x) L2 (Rn ) = u(y) ˆ L2 (Rn ) and Dα u(x) = (iy1 )α1 · · · (iyn )αn u(y) ˆ for u ∈ L2 (Rn ), we can present the norm in W m,2 (Ω) in the form u(x) 2 W m,2 (Rn ) = yα1 · · · yαn u(y) 1 n ˆ 2 L2 (Rn ) . |α|≤m We can then study the properties of the weighted space Lw (Rn ); this transformed 2 problem is simpler, as many of the problems involved are algebraic estimates of Fourier images. Moreover, we can consider W p,2 (Rn ) with fractional indices p. These lead to nec- essary and sufficient conditions for the trace problem: given W m,2 (Ω), find the space W p,2 (∂Ω) in which W m,2 (Ω) is continuously imbedded. The inverse trace problem is, given W p,2 (Ω), find the maximal index m such that every element u ∈ W p,2 (∂Ω)
  • 5. 2.2 Operator of Imbedding 119 can be extended to Ω, u∗ ∈ W m,2 (Ω), in such a way that u∗ W m,2 (Ω) ≤c u W p,2 (∂Ω) . In this way, many results from the contemporary theory of elliptic (and other types of) equations and systems are obtained. We should mention that the trace theorems are formulated mostly for smooth manifolds, hence are not applicable to practical problems involving domains with corners. As stated in Lemma 1.16.3, all Sobolev spaces are separable. The same holds for all the energy spaces we introduce. 2.2 Operator of Imbedding Let Ω be a Jordan measurable compact set in Rn . Any element of C (k) (Ω) also be- longs to C(Ω). This correspondence — between an element of the space C (k) (Ω) and the same element in the space C(Ω) — is a linear operator. Although it resembles an identity operator, its domain and range differ and it is called the imbedding operator from C (k) (Ω) to C(Ω). Clearly, C (k) (Ω) is a proper subset of C(Ω); this situation is typical for an imbedding operator that gives the correspondence between the same elements but considered in different spaces. For a function u that belongs to both spaces, we have u C(Ω) ≤ u C (k) (Ω) , which shows that the imbedding operator is bounded or continuous (what can be said about its norm?). Moreover, by a consequence of Arzel` ’s theorem, it is compact. a We get a similar imbedding operator when considering the spaces p and q for 1 ≤ p < q. As (1.4.3) states, any element x of p belongs to q and x q ≤ c1 x p . Here and below, the constants ck are independent of the element of the spaces, so the imbedding operator from p to q is bounded. We encounter another imbedding operator when considering the correspondence between the same equivalence classes in the spaces L p (Ω) and Lq (Ω): if 1 ≤ p < q, then any element u of Lq (Ω) belongs to L p (Ω) and u L p (Ω) ≤ c2 u Lq (Ω) . So the imbedding operator from Lq (Ω) to L p (Ω) is also bounded. (Note that the relations between p and q are different in the spaces r and Lr (Ω).) The situation is slightly different when we consider the relation between the spaces C(Ω) and L p (Ω). It is clear that any function u from C(Ω) has a finite norm in L p (Ω). To use the imbedding idea, we identify the function u with the equivalence class from L p (Ω) that contains the stationary sequence (u, u, . . .), and in this way
  • 6. 120 2 Mechanics Problems from the Functional Analysis Viewpoint consider u as an element of L p (Ω). Because u dΩ ≤ c3 u C(Ω) , Ω the imbedding operator from C(Ω) to L p (Ω) is bounded for any 1 ≤ p < ∞. This practice of identification must be extended when we consider imbedding operators involving Sobolev spaces as treated in this book. In the imbedding from W 1,2 (Ω) to L p (Ω), the elements of W 1,2 (Ω) — the equivalence classes of functions from C (1) (Ω) — are different from the equivalence classes from C(Ω) that constitute L p (Ω); however, each class from C (1) (Ω) is contained by some class from C(Ω), and we identify these classes. In this sense, we say that W 1,2 (Ω) is imbedded into L p (Ω) and the imbedding operator is continuous (or even compact). The imbedding of a Sobolev space into the space of continuous functions is even more complicated, although the general ideas are the same. We will consider this in more detail when dealing with the energy space for a plate. In this case we cannot directly identify an element of a Sobolev space, which is an equivalence class, with a continuous function. However, we observe that for a class of equivalent sequences in the norm of a Sobolev space, for any sequence that enters into the class, there is a limit element, a continuous function that does not differ for other representative sequences, and we identify the whole class with this function. The inequality relat- ing the Sobolev norm of the class and the continuous norm of this function states that this identification or correspondence is a bounded operator whose linearity is obvious. Some other inequalities state that the operator is compact. 2.3 Some Energy Spaces A Beam. Earlier we noted that the set S of all real-valued continuous functions y(x) having continuous first and second derivatives on [0, l] and satisfying the boundary conditions y(0) = y (0) = y(l) = y (l) = 0 (2.3.1) is a metric space under the metric l 1/2 1 d(y, z) = B(x)[y (x) − z (x)]2 dx . (2.3.2) 2 0 We called this an energy space for the clamped beam. We can introduce an inner product 1 l (y, z) = B(x)y (x)z (x) dx (2.3.3) 2 0 and norm l 1/2 1 y = B(x)[y (x)]2 dx 2 0
  • 7. 2.3 Some Energy Spaces 121 such that d(y, z) = y − z . But this space is not complete (it is clear that there are Cauchy sequences whose limits do not belong to C (2) (0, l); the reader should construct an example). To have a complete space, we must apply the completion theorem. The actual energy space denoted by E B is the completion of S in the metric (2.3.2) (or, what amounts to the same thing, in the inner product (2.3.3)). Let us consider some properties of the elements of E B . An element y(x) ∈ E B is a set of Cauchy sequences equivalent in the metric (2.3.2). Let {yn (x)} be a represen- tative sequence of y(x). Then, according to (2.3.2), l 1/2 1 B(x)[yn (x) − ym (x)]2 dx → 0 as n, m → ∞ . 2 0 If we assume that 0 < m1 ≤ B(x) ≤ m2 , then the sequence of second derivatives {yn (x)} is a Cauchy sequence in the norm of L2 (0, l) because l l m1 [yn (x) − ym (x)]2 dx ≤ B(x)[yn (x) − ym (x)]2 dx . 0 0 Hence {yn (x)} is a representative sequence of some element of L2 (0, l), and we can regard y ∈ Eb as having a second derivative that belongs to L2 (0, l). Now consider {yn (x)}. For any y(x) ∈ S we get x y (x) = y (t) dt . 0 So for a representative {yn (x)} of a class y(x) ∈ E B we have x l |yn (x) − ym (x)| ≤ |yn (t) − ym (t)| dt ≤ 1 · |yn (t) − ym (t)| dt 0 0 l 1/2 ≤ l1/2 [yn (x) − ym (x)]2 dx 0 l 1/2 ≤ (l/m1 )1/2 B(x)[yn (x) − ym (x)]2 dx 0 → 0 as n, m → ∞ . (2.3.4) It follows that {yn (x)} converges uniformly on [0, l]; hence there exists a limit func- tion z(x) which is continuous on [0, l]. This function does not depend on the choice of representative sequence (Problem 2.3.1 below). The same holds for a sequence of functions {yn (x)}: its limit is a function y(x) continuous on [0, l]. Moreover, y (x) = z(x) .
  • 8. 122 2 Mechanics Problems from the Functional Analysis Viewpoint To prove this, it is necessary to repeat the arguments of Sect. 1.11 on the differen- tiability of the elements of C (k) (Ω), with due regard for (2.3.4). From (2.3.4) and the similar inequality for {yn (x)} we get l 1/2 1 max |y(x)| + |y (x)| ≤ m B(x)[y (x)]2 dx (2.3.5) x∈[0,l] 2 0 for some constant m independent of y(x) ∈ E B . So each element y(x) ∈ E B can be identified with an element y(x) ∈ C (1) (0, l) in such a way that (2.3.5) holds. This correspondence is an imbedding operator, and we interpret (2.3.5) as a statement that the imbedding operator from E B to C (1) (0, l) is continuous (cf., Sect. 1.21). Henceforth we refer to the elements of E B as if they were continuously differentiable functions, attaching the properties of the uniquely determined limit functions to the corresponding elements of E B themselves. Problem 2.3.1. Prove that the function z(x), discussed above, is independent of the choice of representative sequence. We are interested in analyzing all terms that appear in the statement of the equi- librium problem for a body as a minimum potential energy problem. So we will consider the functional giving the work of external forces. For the beam, it is l A= F(x)y(x) dx . 0 This is well-defined on E B if F(x) ∈ L(Ω); moreover, (2.3.5) shows that the work of external forces can contain terms of the form [Fk y(xk ) + Mk y (xk )] , k which can be interpreted as the work of point forces and point moments, respec- tively. This is a consequence of the continuity of the imbedding from E B to C (1) (0, l). Remark 2.3.1. Alternatively we can define E B on a base set S 1 of smoother func- tions, in C (4) (0, l) say, satisfying (2.3.1). The result is the same, since S 1 is dense in S with respect to the norm of C (2) (0, l). However, sometimes such a definition is convenient. Remark 2.3.2. Readers familiar with the contemporary literature in this area may have noticed that Western authors usually deal with Sobolev spaces, studying the properties of operators corresponding to problems under consideration; we prefer to deal with energy spaces, studying first their properties and then those of the corre- sponding operators. Although these approaches lead to the same results, in our view the physics of a particular problem should play a principal role in the analysis — in this way the methodology seems simpler, clearer, and more natural. In papers de- voted to the study of elastic bodies, we mainly find interest in the case of a clamped boundary. Sometimes this is done on the principle that it is better to deal solely with
  • 9. 2.3 Some Energy Spaces 123 homogeneous Dirichlet boundary conditions, but often it is an unfortunate conse- quence of the use of the Sobolev spaces H k (Ω). The theory of these spaces is well developed but is not amenable to the study of other boundary conditions. Success in the investigation of mechanics problems can be much more difficult without the benefit of the physical ideas brought out by the energy spaces. Remark 2.3.3. In defining the energy space of the beam, we left aside the question of smoothness of the stiffness function B(x). From a mathematical standpoint this is risky since, in principle, B(x) can be nonintegrable. But in the case of an actual physical beam, B(x) can have no more than a finite number of discontinuities and must be differentiable everywhere else. For simplicity, we shall continue to make realistic assumptions concerning physical parameters such as stiffness and elastic constants; in particular, we shall suppose whatever degree of smoothness is required for our purposes. A Membrane (Clamped Edge). The subset of C (1) (Ω) consisting of all functions satisfying u(x, y) ∂Ω = 0 (2.3.6) with the metric 1/2 ∂u ∂v 2 ∂u ∂v 2 d(u, v) = − + − dx dy (2.3.7) Ω ∂x ∂x ∂y ∂y is an incomplete metric space. If we introduce an inner product ∂u ∂v ∂u ∂v (u, v) = + dx dy (2.3.8) Ω ∂x ∂x ∂y ∂y consistent with (2.3.7), we get an inner product space. Its completion in the metric (2.3.7) is the energy space E MC for the clamped membrane, a real Hilbert space. What can we say about the elements U(x, y) ∈ E MC ? If {un (x, y)} is a representa- tive of U(x, y), then 1/2 ∂un ∂um 2 ∂un ∂um 2 − + − dx dy → 0 as n, m → ∞ . Ω ∂x ∂x ∂y ∂y and we see that the sequences of first derivatives {∂un /∂x}, {∂un /∂y} are Cauchy sequences in the norm of L2 (Ω). What about U(x, y) itself? If we extend each un (x, y) by zero outside Ω, we can write x ∂un (s, y) un (x, y) = ds 0 ∂s (assuming, without loss of generality, that Ω is confined to the band 0 ≤ x ≤ a). Squaring both sides and integrating over Ω, we get
  • 10. 124 2 Mechanics Problems from the Functional Analysis Viewpoint 2 x ∂un (s, y) u2 (x, y) dx dy = ds dx dy Ω n Ω 0 ∂s 2 a ∂un (s, y) ≤ 1· ds dx dy Ω 0 ∂s a ∂un (s, y) 2 ≤ a ds dx dy Ω 0 ∂s ∂un (x, y) 2 ≤ a2 dx dy . Ω ∂x This means that if {∂un /∂x} is a Cauchy sequence in the norm of L2 (Ω), then so is {un }. Hence we can consider elements U(x, y) ∈ E MC to be such that U(x, y), ∂U/∂x, and ∂U/∂y belong to L2 (Ω). As a consequence of the last chain of inequalities, we get Friedrichs’ inequality ∂U 2 ∂U 2 U 2 (x, y) dx dy ≤ m + dx dy Ω Ω ∂x ∂y which holds for any U(x, y) ∈ E MC and a constant m independent of U(x, y). Membrane (Free Edge). Although it is natural to introduce the energy space using the energy metric (2.3.7), we cannot distinguish between two states u1 (x, y) and u2 (x, y) of the membrane with free edge if u2 (x, y) − u1 (x, y) = c = constant . This is the only form of “rigid” displacement possible for a membrane. We first show that no other rigid displacements (i.e., displacements associated with zero strain energy) are possible. The proof is a consequence of Poincar´ ’s inequality e 2 ∂u 2 ∂u 2 u2 dx dy ≤ m u dx dy + + dx dy (2.3.9) Ω Ω Ω ∂x ∂y for a function u(x, y) ∈ C (1) (Ω). The constant m does not depend on u(x, y). Proof of Poincar´ ’s inequality [9]. We first assume that Ω is the square [0, a] × [0, a] e and write down the identity x2 ∂u(s, y1 ) y2 ∂u(x2 , t) u(x2 , y2 ) − u(x1 , y1 ) = ds + dt . x1 ∂s y1 ∂t Squaring both sides and then integrating over the square, first with respect to the variables x1 and y1 and then with respect to x2 and y2 , we get
  • 11. 2.3 Some Energy Spaces 125 u2 (x2 , y2 ) − 2u(x2 , y2 )u(x1 , y1 ) + u2 (x1 , y1 ) dx1 dy1 dx2 dy2 Ω Ω 2 x2 ∂u(s, y1 ) y2 ∂u(x2 , t) = ds + dt dx1 dy1 dx2 dy2 Ω Ω x1 ∂s y1 ∂t 2 a ∂u(s, y1 ) a ∂u(x2 , t) ≤ 1· ds + 1· dt dx1 dy1 dx2 dy2 Ω Ω 0 ∂s 0 ∂t a ∂u(s, y1 ) 2 a ∂u(x2 , t) 2 ≤ 2a ds + dt dx1 dy1 dx2 dy2 Ω Ω 0 ∂s 0 ∂t a a ∂u 2 ∂u 2 ≤ 2a4 + dx dy . 0 0 ∂x ∂y Note that, along with the Schwarz inequality for integrals, we have used the elemen- tary inequality (a + b)2 ≤ 2(a2 + b2 ) which follows from the fact that (a − b)2 ≥ 0. The beginning of this chain of inequalities is 2 a2 u2 (x, y) dx dy − 2 u(x, y) dx dy + a2 u2 (x, y) dx dy Ω Ω Ω so 2 ∂u 2 ∂u 2 2a2 u2 dx dy ≤ 2 u dx dy + 2a4 + dx dy Ω Ω Ω ∂x ∂y and we get (2.3.9) with m = max(a2 , 1/a2 ). It can be shown that Poincar´ ’s inequality holds on more general domains. A e modification of (2.3.9) will appear in Sect. 2.12. Let us return to the free membrane problem. Provided we consider only the mem- brane’s state of stress, any two states are identical if they are described by functions u1 (x, y) and u2 (x, y) whose difference is constant. We gather all functions (such that the difference between any two is a constant) into a class denoted by u∗ (x, y). There is a unique representative of u∗ (x, y) denoted by ub (x, y) such that ub (x, y) dx dy = 0 . (2.3.10) Ω For this balanced representative (or balanced function), Poincar´ ’s inequality be- e comes ∂ub 2 ∂ub 2 u2 (x, y) dx dy ≤ m + dx dy . (2.3.11) Ω b Ω ∂x ∂y As the right-hand side is zero for a “rigid” displacement, so is the left-hand side and it follows that the balanced representative associated with a rigid displacement must be zero. Hence u(x, y) = c is the only permissible form for a rigid displacement of a membrane.
  • 12. 126 2 Mechanics Problems from the Functional Analysis Viewpoint Because (2.3.11) has the same form as Friedrichs’ inequality, we can repeat our former arguments to construct the energy space E MF for a free membrane using the balanced representatives of the classes u∗ (x, y). In what follows we shall use this space E MF , remembering that its elements satisfy (2.3.10). The condition (2.3.10) is a geometrical constraint resulting from our mathemat- ical technique. Solving the static free membrane problem, we must remember that the formulation of the equilibrium problem does not impose this constraint — the membrane can move as a “rigid body” in the direction normal to its own surface. But if we consider only deformations and the strain energy defined by the first par- tial derivatives of u(x, y), the results must be independent of such motions. Consider then the functional of the work of external forces A= F(x, y)U(x, y) dx dy . Ω If we use the space E MF , then A makes sense if F(x, y) ∈ L2 (Ω) (guaranteed by (2.3.11) together with the Schwarz inequality in L2 (Ω)). This is the only restriction on external forces for a clamped membrane. However, in the case of equilibrium for a free membrane, the functional A must be invariant under transformations of the form u(x, y) → u(x, y) + c with any constant c. This requires F(x, y) dx dy = 0 . (2.3.12) Ω Again, we consider the equilibrium problem where rigid motion, however, is possi- ble. Since we did not introduce inertia forces, we have formally equated the mass of the membrane to zero. In this situation of zero mass, any forces with nonzero resul- tant would make the membrane as a whole move with infinite acceleration. Thus, (2.3.12) also precludes such physical nonsense. Meanwhile, Sobolev’s imbedding theorem permits us to incorporate forces ψ acting on the boundary S of Ω into the functional describing the work of external forces: F(x, y)u(x, y) dx dy + ψu ds . (2.3.13) Ω S In this case, the condition of invariance of the functional under constant motions yields the condition F(x, y) dx dy + ψ ds = 0 . (2.3.14) Ω S Note that in the formulation of the Neumann problem, this quantity ψ appears in the boundary condition: ∂u =ψ. ∂n S The reader may ask why, if the membrane can move as a rigid body, it cannot rotate freely in the manner of an ordinary free rigid body. The answer is that in this model, unlike the other models for elastic bodies in this book, rotation of the mem-
  • 13. 2.3 Some Energy Spaces 127 brane as a rigid body alters the membrane strain energy. In the membrane model, the deflection u(x, y) is the deflection that is imposed on the prestressed state of the membrane. The membrane equilibrium problem is an example of a problem for a prestressed body. There is another way to formulate the equilibrium problem for a free mem- brane, based on a different method of introducing the energy space. Let us return to Poincar´ ’s inequality (2.3.9). Denote e 2 1/2 u 1 = u dx dy + D(u) Ω where ∂u 2 ∂u 2 D(u) = + dx dy , Ω ∂x ∂y and recall that the norm in W 1,2 (Ω) is 2 1/2 u W 1,2 (Ω) = u2 dx dy + D(u) . Ω By (2.3.9), the norms · 1 and · W 1,2 (Ω) are equivalent on the set of continuously differentiable functions, and hence on W 1,2 (Ω). Let us use the norm · 1 on W 1,2 (Ω). Now the space W 1,2 (Ω) is the completion, with respect to the norm · 1 , of the functions continuously differentiable on Ω. Let us take an element U(x, y) ∈ W 1,2 (Ω) and select from U(x, y) an arbitrary representative Cauchy sequence {uk (x, y)}. A smooth function uk (x, y) can be uniquely written in the form uk (x, y) = uk (x, y) + ak ˜ where uk dx dy = 0 . ˜ Ω Since 2 uk − um 2 1 = (ak − am ) dx dy + D(˜ k − um ) → 0 as k, m → ∞ , u ˜ Ω we see that {ak } is a numerical Cauchy sequence that has a limit corresponding to U. This means that {ak } belongs to c, the space of numerical sequences each of which has a limit. It is easy to show that this limit does not depend on the choice of representative sequence {uk (x, y)}, and we can regard it as a rigid displacement of the membrane. Now, because the membrane is not geometrically fixed and can be moved through any uniform displacement with no change in energy, we place all the elements of W 1,2 (Ω) that characterize a strained state of the membrane into the same class such that for any two elements U (x, y) and U (x, y) of the class and any representative sequences {uk (x, y)} and {uk (x, y)} taken from them, {˜ k (x, y)} and u {˜ k (x, y)} are equivalent in the norm (i.e., D(˜ k − uk ) → 0) and the difference se- u u ˜ quence {ak − ak } is in c. This construction of the classes is equivalent to the con- struction of the factor space W 1,2 (Ω)/c, which can also be called the energy space
  • 14. 128 2 Mechanics Problems from the Functional Analysis Viewpoint for the free membrane. The zero of this energy factor space is the set of the elements of W 1,2 (Ω) each containing as a representative an element from c. Clearly if we wish to have the energy functional defined in this space, the necessary condition for its continuity is that its linear part — the work of external forces — must be zero over any constant displacement. The latter involves the self-balance condition (2.3.12). It is seen that the principal part of any class-element of the new energy space, defined by the sequences {˜ k (x, y)}, coincides with the corresponding sequences for the el- u ements of the space E MF ; moreover, this correspondence between the two energy spaces maintains equal norms in both spaces. So we can repeat the procedure to establish the existence-uniqueness theorem in the new energy space, using the proof in E MF . The restriction (2.3.12) (or (2.3.14)) is necessary for the functional of external forces to be uniquely defined for an element U∗ (x, y). We shall use the same nota- tion E MF for this type of energy space since there is a one-to-one correspondence, preserving distances and inner products, between the two types of energy space for the free membrane. Moreover, we shall always make clear which version we mean. Those familiar with the theory of the Neumann problem for Laplace’s equation should note that the necessary condition for solvability that arises in mathematical physics as a mathematical consequence, i.e., ψ ds = 0 , S is a particular case of (2.3.14) when F(x, y) = 0. This means that for solvability of the problem, the external forces acting on the membrane edge should be self- balanced. Finally, we note that Poisson’s equation governs not only membranes, but also situations in electricity, magnetism, hydrodynamics, mathematical biology, and other fields. So we can consider spaces such as E M in various other sciences. It is clear that the results will be the same. We will proceed to introduce other energy spaces in a similar manner: they will be completions of corresponding metric (inner product) spaces consisting of smooth functions satisfying certain boundary conditions. The problem is to determine prop- erties of the elements of those completions. As a rule, metrics must contain all the strain energy terms (we now discuss only linear systems). For example, we can con- sider a membrane whose edge is elastically supported; then we must include the energy of elastic support in the expression for the energy metric. Bending a Plate. Here we begin with the work of internal forces on variations of displacements −(w1 , w2 ) = − Dαβγδ ργδ (w1 ) ραβ (w2 ) dx dy (2.3.15) Ω where w1 (x, y) is the normal displacement of the plate midsurface Ω, w2 (x, y) can be considered as its variation, ραβ (u) are components of the change-of-curvature tensor,
  • 15. 2.3 Some Energy Spaces 129 ∂2 u ∂2 u ∂2 u ρ11 (u) = , ρ12 = , ρ22 = , ∂x2 ∂x∂y ∂y2 Dαβγδ are components of the tensor of elastic constants of the plate such that Dαβγδ = Dγδαβ = D βαγδ (2.3.16) and, for any symmetric tensor ραβ there exists a constant m0 > 0 such that 2 Dαβγδ ργδ ραβ ≥ m0 ρ2 . αβ (2.3.17) α,β=1 We suppose the Dαβγδ are constants, but piecewise continuity of these parameters would be sufficient. For the theory of shells and plates, Greek indices will assume values from the set {1, 2} while Latin indices will assume values from the set {1, 2, 3}. The repeated index convention for summation is also in force. For example, we have 2 aαβ bαβ ≡ aαβ bαβ . α,β=1 We first consider a plate with clamped edge ∂Ω: ∂w w ∂Ω = =0. (2.3.18) ∂n ∂Ω (Of course, the variation of w must satisfy (2.3.18) as well.) Let us show that on S 4 , the subset of C (4) (Ω) consisting of those functions which satisfy (2.3.18), the form (w1 , w2 ) given in (2.3.15) is an inner product. We begin with the axiom P1: 2 (w, w) = Dαβγδ ραβ (w) ργδ (w) dx dy ≥ m0 ρ2 (w) dx dy αβ Ω Ω α,β=1 ∂2 w 2 ∂2 w 2 ∂2 w 2 = m0 +2 + dx dy ≥ 0 . Ω ∂x2 ∂x∂y ∂y2 If w = 0 then (w, w) = 0. If (w, w) = 0 then, on Ω, ∂2 w ∂2 w ∂2 w =0, =0, =0. ∂x2 ∂x∂y ∂y2 It follows that w(x, y) = a1 + a2 x + a3 y , where the ai are constants. By (2.3.18) then, w(x, y) = 0. Hence P1 is satisfied. Satisfaction of P2 follows from (2.3.16), and it is evident that P3 is also satisfied.
  • 16. 130 2 Mechanics Problems from the Functional Analysis Viewpoint Thus S 4 with inner product (2.3.15) is an inner product space; its completion in the corresponding metric is the energy space E PC for a clamped plate. Let us consider some properties of the elements of E PC . It was shown that ∂2 w 2 ∂2 w 2 ∂2 w 2 2 m0 +2 + dx dy Ω ∂x2 ∂x∂y ∂y2 ≤ Dαβγδ ργδ (w) ραβ (w) dx dy ≡ (w, w) . (2.3.19) Ω From this and the Friedrichs inequality, written first for w and then for the first derivatives of w ∈ S 4 as well, we get ∂w 2 ∂w 2 w2 dx dy ≤ m1 + dx dy Ω Ω ∂x ∂y ∂2 w 2 ∂2 w 2 ∂2 w 2 ≤ m2 +2 + dx dy Ω ∂x2 ∂x∂y ∂y2 ≤ m3 Dαβγδ ργδ (w) ραβ (w) dx dy ≡ m3 (w, w) . (2.3.20) Ω Hence if {wn } ⊂ S 4 is a Cauchy sequence in E PC , then the sequences ∂wn ∂wn ∂2 wn ∂2 wn ∂2 wn {wn }, , , , , , ∂x ∂y ∂x2 ∂x∂y ∂y2 are Cauchy sequences in L2 (Ω). So we can say that an element W of the completion E PC is such that W(x, y) and all its derivatives up to order two are in L2 (Ω). We now investigate W(x, y) further. Let w ∈ S 4 and w(x, y) ≡ 0 outside Ω. Sup- pose Ω lies in the domain {(x, y) : x > 0, y > 0}. Then the representation x y ∂2 w(s, t) w(x, y) = ds dt 0 0 ∂s∂t holds. Using H¨ lder’s inequality and (2.3.20), we get o x y ∂2 w(s, t) ∂2 w(s, t) |w(x, y)| ≤ ds dt ≤ 1· ds dt 0 0 ∂s∂t Ω ∂s∂t ∂2 w(s, t) 2 1/2 ≤ (mes Ω)1/2 ds dt ≤ m4 (w, w)1/2 . (2.3.21) Ω ∂s∂t This means that if {wn } ⊂ S 4 is a Cauchy sequence in the metric of E PC , then it converges uniformly on Ω. Hence there exists a limit function w0 (x, y) = lim wn (x, y) n→∞
  • 17. 2.3 Some Energy Spaces 131 which is continuous on Ω; this function is identified, as above, with the correspond- ing element of E PC and we shall say that E PC is continuously imbedded into C(Ω). The functional describing the work of external forces A= F(x, y) W(x, y) dx dy Ω now makes sense if F(x, y) ∈ L(Ω); moreover, it can contain the work of point forces F(xk , yk ) w0 (xk , yk ) k and line forces F(x, y) w0 (x, y) ds γ where γ is a line in Ω and w0 (x, y) is the corresponding limit function for W(x, y). Remark 2.3.4. Modern books on partial differential equations often require that F(x, y) ∈ H −2 (Ω). This is a complete characterization of external forces — how- ever, it is difficult for an engineer to verify this property. Now let us consider a plate with free edge. In this case, we also wish to use the inner product (2.3.15) to create an energy space. As in the case of a membrane with free edge, the axiom P1 is not fulfilled: we saw that from (w, w) = 0 it follows that w(x, y) = a1 + a2 x + a3 y . (2.3.22) This admissible motion of the plate as a rigid whole is called a rigid motion, but still differs from real “rigid” motions of the plate as a three-dimensional body. Poincar´ ’s inequality (2.3.9) implies that the zero element of the correspond- e ing completion consists of functions of the form (2.3.22). Indeed, taking w(x, y) ∈ C (4) (Ω) we write down Poincar´ ’s inequality for ∂w/∂x: e ∂w 2 ∂w 2 ∂2 w 2 ∂2 w 2 dx dy ≤ m dx dy + + dx dy , Ω ∂x Ω ∂x Ω ∂x2 ∂x∂y and then the same inequality for ∂w/∂y with the roles of x and y interchanged. From these and (2.3.9) we get ∂w 2 ∂w 2 2 w2 + + dx dy ≤ m1 w dx dy Ω ∂x ∂y Ω ∂w 2 ∂w 2 + dx dy + dx dy Ω ∂x Ω ∂y ∂w 2 2 ∂w2 2 ∂2 w 2 + +2 + dx dy . Ω ∂x2 ∂x∂y ∂y2 From (2.3.19) it follows that
  • 18. 132 2 Mechanics Problems from the Functional Analysis Viewpoint ∂w 2 ∂w 2 2 w2 + + dx dy ≤ m2 w dx dy Ω ∂x ∂y Ω ∂w 2 ∂w 2 + dx dy + dx dy Ω ∂x Ω ∂y + Dαβγδ ργδ (w) ραβ (w) dx dy . (2.3.23) Ω For any function w(x, y) ∈ C (4) (Ω), we can take suitable constants ai and find a function wb (x, y) of the form wb (x, y) = w(x, y) + a1 + a2 x + a3 y (2.3.24) such that ∂wb ∂wb wb dx dy = 0 , dx dy = 0 , dx dy = 0 . (2.3.25) Ω Ω ∂x Ω ∂y As for the membrane with free edge, we can now consider a subset S 4b of C (4) (Ω) consisting of balanced functions satisfying (2.3.25). We construct an energy space E PF for a plate with free edge as the completion of S 4b in the metric induced by the inner product (2.3.15). From (2.3.25), (2.3.23), and (2.3.19), we see that an element W(x, y) ∈ E PF is such that W(x, y) and all its “derivatives” up to order two are in L2 (Ω). We could show the existence of a limit function w0 (x, y) = lim wn ∈ C(Ω) n→∞ for any Cauchy sequence {wn }, but in this case the technique is more complicated and, in what follows, we have this result as a particular case of the Sobolev imbed- ding theorem. Note that (2.3.25) can be replaced by w(x, y) dx dy = 0 , x w(x, y) dx dy = 0 , y w(x, y) dx dy = 0 , Ω Ω Ω since these also uniquely determine the ai for a class of functions of the form (2.3.24). (This possibility follows from Sobolev’s general result [33] on equivalent norms in Sobolev spaces.) The system (2.3.25) represents constraints that are absent in nature. For a static problem, there must be a certain invariance of some objects under transformations of the form (2.3.24) with arbitrary constants ak . In particular, the work of external forces should not depend on the ak if the problem is stated correctly. This leads to the necessary conditions
  • 19. 2.3 Some Energy Spaces 133 F(x, y) dx dy = 0 , x F(x, y) dx dy = 0 , y F(x, y) dx dy = 0 . Ω Ω Ω (2.3.26) The mechanical sense of (2.3.26) is clear: the resultant force and moments must vanish. This is the condition for a self-balanced force system. Problem 2.3.2. What is the form of (2.3.26) if the external forces contain point and line forces? An energy space for a free plate, as for the membrane with free edge, can be introduced in another way: namely, we begin with an element of W 2,2 (Ω), selecting a representative sequence {wk (x, y)+ak +bk x+ck y}, where w satisfies (2.3.25) and ak , ˜ ˜ bk , ck are constants. It is easy to show that the numerical sequences {ak }, {bk }, {ck } are Cauchy and therefore belong to the space c. Next we combine into a class-element all the elements of W 2,2 (Ω) whose differences are a linear polynomial a + bx + cy, and state that the energy space is the factor space of W 2,2 (Ω) by the space which is the completion of the space of linear polynomials with respect to the norm a + bx + cy = (a2 + b2 + c2 )1/2 . In the factor space, the zero is the class of all Cauchy sequences whose differ- ences are equivalent (in the norm of W 2,2 (Ω)) to a sequence {ak + bk x + ck y} with {ak }, {bk }, {ck } ∈ c. The norm of W, an element in the factor space, is 1/2 Dαβγδ ργδ (W) ραβ (W) dx dy . Ω The elements of the factor space are uniquely identified with the elements of the energy space E PF whose elements satisfy (2.3.25). Moreover, the identification is isometric so, as for the membrane, we can use this factor space as the energy space and repeat the proof of the existence-uniqueness theorem in terms of the elements of the energy factor space. The self-balance condition for the external forces is a nec- essary condition to have the functional of the work of external forces be meaningful. The reader may also consider mixed boundary conditions: how must the treatment be modified if the plate is clamped only along a segment AB ⊂ Ω so that w(x, y) AB = 0 , with the rest of the boundary free of geometrical constraints? Linear Elasticity. We return to the problem of linear elasticity, considered in Sect. 1.5. Let us introduce a functional describing the work of internal forces on variations v(x) of the displacement field u(x): −(u, v) = − ci jkl kl (u) i j (v) dΩ . (2.3.27) Ω For the notation, see (1.5.7)–(1.5.9). We recall that the strain energy E4 (u) of an elastic body occupying volume Ω is related to the introduced inner product as fol-
  • 20. 134 2 Mechanics Problems from the Functional Analysis Viewpoint lows: (u, u) = u 2 = 2E4 (u) . i jkl The elastic moduli c may be piecewise continuous functions satisfying (1.5.8) and (1.5.9), which guarantee that all inner product axioms are satisfied by (u, v) for vector functions u, v continuously differentiable on Ω, except P1: from (u, u) = 0 it follows that u = a + b × x. Note that (u, v) is consistent with the metric (1.5.10). Let us consider boundary conditions prescribed by u(x) ∂Ω = 0 . (2.3.28) If we use the form (2.3.27) on the set S 3 of vector-functions u(x) satisfying (2.3.28) and such that each of their components is of class C (2) (Ω), then (u, v) is an inner product and S 3 with this inner product becomes an inner product space. Its com- pletion E EC in the corresponding metric (or norm) is the energy space of an elastic body with clamped boundary. To describe the properties of the elements of E EC , we establish Korn’s inequality. Lemma 2.3.1 (Korn). For a vector function u(x) ∈ S 3 , we have 3 ∂ui 2 |u|2 + dΩ ≤ m ci jkl kl (u) i j (u) dΩ Ω i, j=1 ∂x j Ω for some constant m which does not depend on u(x). Proof. By (1.5.9) and Friedrichs’ inequality, it is sufficient to show that 3 3 ∂ui 2 dΩ ≤ m1 2 i j (u) dΩ . Ω i, j=1 ∂x j Ω i, j=1 i≤ j Consider the term on the right: 3 3 1 ∂ui ∂u j 2 A≡ 2 i j (u) dΩ = + dΩ Ω i, j=1 4 Ω i, j=1 ∂x j ∂xi i≤ j i≤ j ⎧ ⎫ ⎪ ⎪ 3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ∂ui 2 1 3 ∂ui 2 ∂u j 2 ⎪ ∂ui ∂u j ⎪ ⎬ = ⎪ ⎪ + + +2 ⎪ dΩ . ⎪ ⎪ i=1 ∂xi Ω⎪ 4 i, j=1 ∂x j ∂xi ∂x j ∂xi ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ i< j Integrating by parts (twice) the term 3 3 1 ∂ui ∂u j 1 ∂ui ∂u j B≡ dΩ = dΩ 2 Ω i, j=1 ∂x j ∂xi 2 Ω i, j=1 ∂xi ∂x j i< j i< j
  • 21. 2.4 Generalized Solutions in Mechanics 135 and using the elementary inequality |ab| ≤ (a2 + b2 )/2, we get 3 3 1 ∂ui 2 ∂u j 2 1 ∂ui 2 |B| ≤ + dΩ = dΩ . 4 Ω i, j=1 ∂xi ∂x j 2 Ω i=1 ∂xi i< j Therefore 3 1 ∂ui 2 A≥ dΩ , 4 Ω i, j=1 ∂x j which completes the proof. By Korn’s inequality, each component of an element U ∈ E EC belongs to E MC , i.e., the ui and their first derivatives belong to L2 (Ω). Note that the construction of an energy space is the same if the boundary condi- tion (2.3.28) is given only on some part ∂Ω1 of the boundary of Ω: u(x) ∂Ω = 0 . 1 Korn’s inequality also holds, but its proof is more complicated (see, for example, [26, 11]). If we consider an elastic body with free boundary, we encounter issues similar to those for a membrane or plate with free edge: we must circumvent the difficulty with the zero element of the energy space. The restrictions u dΩ = 0 , x × u(x) dΩ = 0 , (2.3.29) Ω Ω provide that the rigid motion u = a + b × x becomes zero, and that Korn’s inequality remains valid for smooth vector functions satisfying (2.3.29). So by completion, we get an energy space E EF with known properties: all Cartesian components of vectors pertain to the space W 1,2 (Ω). As for a free membrane, we can also organize an energy space of classes — a factor space — in which the zero element is the set of all elements whose differences between any representative sequences in the norm of (W 1,2 (Ω))3 are equivalent to a sequence of the form {ak + bk × x} such that the Cartesian components of the vectors {ak } and {bk } constitute some elements of the space c. 2.4 Generalized Solutions in Mechanics We now discuss how to introduce generalized solutions in mechanics. We begin with Poisson’s equation −Δu(x, y) = F(x, y) , (x, y) ∈ Ω , (2.4.1)
  • 22. 136 2 Mechanics Problems from the Functional Analysis Viewpoint where Ω is a bounded open domain in R2 . The Dirichlet problem consists of this equation supplemented by the boundary condition u ∂Ω = 0 . (2.4.2) Let u(x, y) be its classical solution; i.e., let u ∈ C (2) (Ω) satisfy (2.4.1) and (2.4.2). Let ϕ(x, y) be a function with compact support in Ω. Again, this means that ϕ ∈ C (∞) (Ω) and the closure of the set M = {(x, y) ∈ Ω : ϕ(x, y) 0} lies in Ω. Multiplying both sides of (2.4.1) by ϕ(x, y) and integrating over Ω, we get − ϕ(x, y) Δu(x, y) dx dy = F(x, y) ϕ(x, y) dx dy . (2.4.3) Ω Ω If this equality holds for every infinitely differentiable function ϕ(x, y) with compact support in Ω, and if u ∈ C (2) (Ω) and satisfies (2.4.2), then, as is well known from the classical calculus of variations, u(x, y) is the unique classical solution to the Dirichlet problem. But using (2.4.3), we can pose this Dirichlet problem directly without using the differential equation (2.4.1); namely, u(x, y) is a solution to the Dirichlet problem if, obeying (2.4.2), it satisfies (2.4.3) for every ϕ(x, y) that is infinitely differentiable with compact support in Ω. If F(x, y) belongs to L p (Ω) then we can take, as it seems, u(x, y) having second derivatives in the space L p (Ω); such a u(x, y) is not a classical solution, and it is natural to call it a generalized solution. We can go further by applying integration by parts to the left-hand side of (2.4.3) as follows: ∂u ∂ϕ ∂u ∂ϕ + dx dy = F(x, y) ϕ(x, y) dx dy . (2.4.4) Ω ∂x ∂x ∂y ∂y Ω In such a case we may impose weaker restrictions on a solution u(x, y) and call it the generalized solution if it belongs to E MC , the energy space for a clamped membrane. Equation (2.4.4) defines this solution if it holds for every ϕ(x, y) that has a compact support in Ω. Note the disparity in requirements on u(x, y) and ϕ(x, y). Further integration by parts on the left-hand side of (2.4.4) yields − u(x, y) Δϕ(x, y) dx dy = F(x, y) ϕ(x, y) dx dy . (2.4.5) Ω Ω Now we can formally consider solutions from the space L(Ω) and this is a new class of generalized solutions. This way leads to the theory of distributions originated by Schwartz [32]. He extended the notion of generalized solution to a class of linear continuous function- als, or distributions, defined on the set D(Ω) of all functions infinitely differentiable in Ω and with compact support in Ω. For this it is necessary to introduce the con- vergence and other structures of continuity in D(Ω). Unfortunately D(Ω) is not a normed space (see, for example, Yosida [44] — it is a locally convex topological space) and its presentation would exceed our scope. This theory justifies, in particu-
  • 23. 2.4 Generalized Solutions in Mechanics 137 lar, the use of the so-called δ-function, which was introduced in quantum mechanics via the equality ∞ δ(x − a) f (x) dx = f (a) , (2.4.6) −∞ valid for every continuous function f (x). Physicists considered δ(x) to be a function vanishing everywhere except at x = 0, where its value is infinity. Any known theory of integration gave zero for the value of the integral on the left-hand side of (2.4.6), and the theory of distributions explained how to understand such strange functions. It is interesting to note that the δ-function was well known in classical mechanics, too; if we consider δ(x − a) as a unit point force applied at x = a, then the integral on the left-hand side of (2.4.6) is the work of this force on the displacement f (a), which is indeed f (a). So we have several generalized statements of the Dirichlet problem, but which one is most natural from the viewpoint of mechanics? From mechanics it is known that a solution to the problem is a minimizer of the total potential energy functional ∂u 2 ∂u 2 I(u) = + dx dy − 2 Fu dx dy . (2.4.7) Ω ∂x ∂y Ω According to the calculus of variations, a minimizer of I(u) on the subset of C (2) (Ω) consisting of all functions satisfying (2.4.2), if it exists, is a classical solution to the Dirichlet problem. But we also can consider I(u) on the energy space E MC if F(x, y) ∈ L p (Ω) for p > 1. Indeed, the first term in I(u) is well-defined in E MC and can be written in the form u 2 ; the second, Φ(u) = − F(x, y) u(x, y) dx dy , Ω is a linear functional with respect to u(x, y). It is also continuous in E MC ; by virtue of H¨ lder’s inequality with exponents p and q = p/(p − 1), we have o 1/p 1/q Fu dx dy ≤ |F| p dx dy |u|q dx dy Ω Ω Ω ≤ m1 F L p (Ω) u W 1,2 (Ω) ≤ m2 u E MC . To show this, we have used the imbedding Theorem 2.1.2 and the Friedrichs in- equality. By Theorem 1.21.1, Φ(u) is continuous in E MC , and therefore so is I(u). Thus I(u) is of the form I(u) = u 2 + 2Φ(u) . (2.4.8) Let u0 ∈ E MC be a minimizer of I(u), i.e., I(u0 ) ≤ I(u) for all u ∈ E MC . (2.4.9)
  • 24. 138 2 Mechanics Problems from the Functional Analysis Viewpoint We try a method from the classical calculus of variations. Take u = u0 + v where v is an arbitrary element of E MC . Then I(u) = I(u0 + v) = u0 + v 2 + 2Φ(u0 + v) = (u0 + v, u0 + v) + 2Φ(u0 + v) = u0 2 + 2 (u0 , v) + 2 v 2 + 2Φ(u0 ) + 2 Φ(v) = u0 2 + 2Φ(u0 ) + 2 [(u0 , v) + Φ(v)] + 2 v 2 . From (2.4.9), we get 2 [(u0 , v) + Φ(v)] + 2 v 2 ≥0. Since is an arbitrary real number (in particular it can take either sign, and the first term, if nonzero, dominates for small in magnitude), it follows that (u0 , v) + Φ(v) = 0 . (2.4.10) In other words, ∂u0 ∂v ∂u0 ∂v + dx dy − F(x, y) v(x, y) dx dy = 0 . (2.4.11) Ω ∂x ∂x ∂y ∂y Ω This equality holds for every v ∈ E MC and defines the minimizer u0 ∈ E MC . Note that (2.4.11) has the same form as (2.4.4). So we have introduced a notion of generalized solution that has an explicit me- chanical background. Definition 2.4.1. An element u ∈ E MC is called the generalized solution to the Dirichlet problem if u satisfies (2.4.11) for any v ∈ E MC . We can also obtain (2.4.11) from the principle of virtual displacements (work). This asserts that in the state of equilibrium, the sum of the work of internal forces (which is now the variation of the strain energy with a negative sign) and the work of external forces is zero on all virtual (admissible) displacements. In the case under consideration, both approaches to introducing generalized en- ergy solutions are equivalent. In general, however, this is not so, and the virtual work principle has wider applicability. If F(x, y) is a nonconservative load depending on u(x, y), we cannot use the principle of minimum total energy; however, (2.4.11) re- mains valid since it has the mathematical form of the virtual work principle. In what follows, we often use this principle to pose problems in equation form. Since the part of the presentation from (2.4.8) up to (2.4.10) is general and does not depend on the specific form (2.4.11) of the functional I(u), we can formulate Theorem 2.4.1. Let u0 be a minimizer of a functional I(u) = u 2 +2Φ(u) given in an inner product (Hilbert) space H, where the functional Φ(u) is linear and continuous. Then u0 satisfies (2.4.10) for every v ∈ H.
  • 25. 2.4 Generalized Solutions in Mechanics 139 Equation (2.4.10) is a necessary condition for minimization of the functional I(u), analogous to the condition that the first derivative of an ordinary function must vanish at a point of minimum. We can obtain (2.4.10) formally by evaluating d I(u0 + v) =0. (2.4.12) d =0 This is valid for the following reason. Given u0 and v, the functional I(u0 + v) is an ordinary function of the numerical variable , and assumes a minimum value at = 0. The left-hand side of (2.4.12) can be interpreted as a partial derivative at u = u0 in the direction v, and is called the Gˆ teaux derivative of I(u) at u = u0 in a the direction of v. We shall return to this issue later. The Dirichlet problem for a clamped membrane is a touchstone for all static prob- lems in continuum mechanics. In a similar way we can introduce a natural notion of generalized solution for other problems under consideration. As we said, each of them can be represented as a minimization problem for a total potential energy functional of the form (2.4.10) in an energy space. For example, equation (2.4.11), a particular form of (2.4.10) for a clamped membrane, is the same for a free mem- brane — we need only replace E MC by E MF . The quantity Φ(u), to be a continuous linear functional in E MF , must be supplemented with self-balance condition (2.3.12) for the load. Let us concretize equation (2.4.10) for each of the other problems we have under consideration. Plate. The definition of generalized solution w0 ∈ E P is given by the equation Dαβγδ ργδ (w0 ) ραβ (w) dx dy − F(x, y) w(x, y) dx dy Ω Ω m − Fk w(xk , yk ) − f (s) w(x, y) ds = 0 (2.4.13) k=1 γ (see the notation of Sect. 2.3) which must hold for every w ∈ E P . The equation is the same for any kind of homogeneous boundary conditions (i.e., for usual ones) but the energy space will change from one set of boundary conditions to another. If a plate can move as a rigid whole, the requirement that F(x, y) ∈ L(Ω) , f (s) ∈ L(γ) , for Φ to be a continuous linear functional, must be supplemented with self-balance conditions for the load: m F(x, y) wi (x, y) dx dy + Fk wi (xk , yk ) + f (s) wi (x, y) ds = 0 (2.4.14) Ω k=1 γ
  • 26. 140 2 Mechanics Problems from the Functional Analysis Viewpoint for i = 1, 2, 3, where w1 (x, y) = 1, w2 (x, y) = x, and w3 (x, y) = y. This condition is necessary if we use the space where the set of rigid plate motions is the zero of the space. If from each element of the energy space we select a representative us- ing (2.3.25), then on the energy space of all representatives with norm · P we can prove existence and uniqueness of the energy solution without self-balance condi- tion (2.4.14). But for solvability of the initial problem for a free plate, we still should require (2.4.14) to hold as constraints (2.3.25) are absent in the problem statement. Note that for each concrete problem we must specify the energy space. The same is true for the following problem. Linear Elasticity. The generalized solution u is defined by the integro-differential equation ci jkl kl (u) i j (v) dΩ − F(x, y, z) · v(x, y, z) dΩ Ω Ω − f(x, y, z) · v(x, y, z) dS = 0 , (2.4.15) Γ where F and f are forces distributed over Ω and over some surface Γ ⊂ Ω, respec- tively. If we consider the Dirichlet (or first) problem of elasticity, which is u(x) ∂Ω = 0 where ∂Ω is the boundary of Ω, the solution u should belong to the space E EC and equation (2.4.15) must hold for every virtual displacement v ∈ E EC . Note that in this case, f(x, y, z) · v(x, y, z) dS = 0 . ∂Ω The load, thanks to Theorem 2.1.4 and Korn’s inequality, is of the class Fi (x, y, z) ∈ L6/5 (Ω) , fi (x, y, z) ∈ L4/3 (Γ) (i = 1, 2, 3) , where Fi (x, y, z) and fi are Cartesian components of F and f, respectively, and Γ is a piecewise smooth surface in Ω. This provides continuity of Φ(w). In the second problem of elasticity there are given forces distributed over the boundary: ci jkl kl (x) n j (x) ∂Ω = fi (x) , where the n j are Cartesian components of the unit exterior normal to ∂Ω. This can be written in tensor notation as σ(x) · n ∂Ω = f(x) where the components of stress tensor σ are related to the components of the strain tensor by σi j = ci jkl kl . As for equilibrium problems for free membranes and plates, in the case of a free elastic body we must require that the load be self-balanced:
  • 27. 2.5 Existence of Energy Solutions to Some Mechanics Problems 141 F(x) dΩ + f(x) dS = 0 , Ω Γ x × F(x) dΩ + x × f(x) dS = 0 . (2.4.16) Ω Γ The solution is required to be in E EF . We have argued that it is legitimate to introduce the generalized solution in such a way. Of course, full legitimacy will be assured when we prove that this solution exists and is unique in the corresponding space. We emphasize once more that the definition of generalized solution arose in a natural way from the variational principle of mechanics. 2.5 Existence of Energy Solutions to Some Mechanics Problems In Sect. 2.4 we introduced generalized solutions for several mechanics problems and reduced those problems to a solution of the abstract equation (u, v) + Φ(v) = 0 (2.5.1) in an energy (Hilbert) space. We obtained some restrictions on the forces to provide continuity of the linear functional Φ(v) in the energy space. The following theorem guarantees solvability of those mechanics problems in a generalized sense. Theorem 2.5.1. Assume Φ(v) is a continuous linear functional given on a Hilbert space H. Then there is a unique element u ∈ H that satisfies (2.5.1) for every v ∈ H. Proof. By the Riesz representation theorem there is a unique u0 ∈ H such that the continuous linear functional Φ(v) is represented in the form Φ(v) = (v, u0 ) ≡ (u0 , v). Hence (2.5.1) takes the form (u, v) + (u0 , v) = 0 . (2.5.2) We need to find u ∈ H that satisfies (2.5.2) for every v ∈ H. Rewriting it in the form (u + u0 , v) = 0 , we see that its unique solution is u = −u0 . This theorem answers the question of solvability, in the generalized sense, of the problems treated in Sect. 2.4. To demonstrate this, we rewrite Theorem 2.5.1 in concrete terms for a pair of problems. Theorem 2.5.2. Assume F(x, y) ∈ L(Ω) and f (x, y) ∈ L(γ) where Ω ⊂ R2 is com- pact and γ is a piecewise smooth curve in Ω. The equilibrium problem for a plate with clamped edge has a unique generalized solution: there is a unique w0 ∈ E PC which satisfies (2.4.13) for all w ∈ E PC .
  • 28. 142 2 Mechanics Problems from the Functional Analysis Viewpoint Changes for a plate which is free of clamping are evident: we must add the self- balance condition (2.4.14) for forces and replace the space E PC by E PF . Theorem 2.5.3. Assume all Cartesian components of the volume forces F(x, y, z) are in L6/5 (Ω) and those of the surface forces f(x, y, z) are in L4/3 (S ), where Ω is compact in R3 and S is a piecewise smooth surface in Ω. Then the problem of equilibrium of an elastic body occupying Ω, with clamped boundary, has a unique generalized solution u ∈ E EC ; namely, u(x, y, z) satisfies (2.4.15) for every v ∈ E EC . In both theorems, the load restrictions provide continuity of the corresponding functionals Φ, the work of external forces. Problem 2.5.1. Formulate existence theorems for the other mechanics problems dis- cussed in Sect. 2.4. 2.6 Operator Formulation of an Eigenvalue Problem We have seen how to use the Riesz representation theorem to prove the existence and uniqueness of a generalized solution. Now let us consider another application of the Riesz representation theorem: how to cast a problem as an operator equation. The eigenvalue equation for a membrane has the form ∂ 2 u ∂2 u + = −λu . (2.6.1) ∂x2 ∂y2 Similar to the equilibrium problem for a membrane, we can introduce a general- ized solution to the eigenvalue problem for a clamped membrane by the integro- differential equation ∂u ∂v ∂u ∂v + dx dy = λ uv dx dy . (2.6.2) Ω ∂x ∂x ∂y ∂y Ω The eigenvalue problem is to find a nontrivial element u ∈ E MC and a corresponding number λ such that u satisfies (2.6.2) for every v ∈ E MC . First we reformulate this problem as an operator equation u = λKu (2.6.3) in the space E MC . For this, consider the term F(v) = uv dx dy Ω as a functional in E MC , with respect to v, when u is a fixed element of E MC . It is seen that F(v) is a linear functional. By the Schwarz inequality
  • 29. 2.6 Operator Formulation of an Eigenvalue Problem 143 1/2 1/2 |F(v)| = uv dx dy ≤ u2 dx dy v2 dx dy Ω Ω Ω hence by the Friedrichs inequality |F(v)| ≤ m u v = m1 v (2.6.4) (hereafter the norm · and the inner product (·, ·) are taken in E MC ). So F(v) is a continuous linear functional acting in the Hilbert space E MC . By the Riesz represen- tation theorem, F(v) has the unique representation F(v) ≡ uv dx dy = (v, f ) = ( f, v) . (2.6.5) Ω What have we shown? For every u ∈ E MC , by this representation, there is a unique element f ∈ E MC . Hence the correspondence u→ f is an operator f = K(u) from E MC to E MC . Let us display some properties of this operator. First we show that it is linear. Let f1 = K(u1 ) and f2 = K(u2 ) . Then (λ1 u1 + λ2 u2 )v dx dy = (K(λ1 u1 + λ2 u2 ), v) Ω while on the other hand, (λ1 u1 + λ2 u2 )v dx dy = λ1 u1 v dx dy + λ2 u2 v dx dy Ω Ω Ω = λ1 (K(u1 ), v) + λ2 (K(u2 ), v) = (λ1 K(u1 ) + λ2 K(u2 ), v) . Combining these we have (K(λ1 u1 + λ2 u2 ), v) = (λ1 K(u1 ) + λ2 K(u2 ), v) , hence K(λ1 u1 + λ2 u2 ) = λ1 K(u1 ) + λ2 K(u2 ) because v ∈ E MC is arbitrary. Therefore linearity is proven. Now let us rewrite (2.6.4) in terms of this representation: |(K(u), v)| ≤ m u v . Take v = K(u); then
  • 30. 144 2 Mechanics Problems from the Functional Analysis Viewpoint K(u) 2 ≤ m u K(u) and it follows that K(u) ≤ m u . (2.6.6) Hence K is a continuous operator in E MC . Equation (2.6.2) can now be written in the form (u, v) = λ(K(u), v) . Since v is an arbitrary element of E MC , this equation is equivalent to the operator equation u = λK(u) with a continuous linear operator K. By (2.6.6), we get λK(u) − λK(v) = |λ| K(u − v) ≤ m |λ| u − v . If m |λ| < 1 , then λK is a contraction operator in E MC and, by the contraction mapping principle, there is a unique fixed point of λK which clearly is u = 0. So the set |λ| < 1/m does not contain real eigenvalues of the problem. Further, we shall see (and this is well known in mechanics) that eigenvalues in this problem must be real. The fact that the set |λ| < 1/m does not contain real eigenvalues, and so any eigenvalues of the problem, has a clear mechanical sense: the lowest eigenfrequency of oscillation of a bounded clamped membrane is strictly positive. Note that from (2.6.2), when v = u it follows that an eigenvalue must be positive. In a similar way, we can introduce eigenvalue problems for plates and elastic bodies. Here we can obtain corresponding equations of the form (2.6.3) with con- tinuous linear operators and can also show that the corresponding lowest eigenvalues are strictly positive. All this we leave to the reader; later we shall consider eigen- value problems in more detail. In what follows, we shall see that, using the Riesz representation theorem, one can also introduce operators and operator equations for nonlinear problems of me- chanics. One of them is presented in the next section. 2.7 Problem of Elastico-Plasticity; Small Deformations Following the lines of a paper by I.I. Vorovich and Yu.P. Krasovskij [40] that was published in a sketchy form, we consider a variant of the theory of elastico-plasticity (Il’yushin [16]), and justify the method of elastic solutions for corresponding bound- ary value problems.
  • 31. 2.7 Problem of Elastico-Plasticity; Small Deformations 145 The system of partial differential equations describing the behavior of an elastic- plastic body occupying a bounded volume Ω is ν ω ∂θ − + (1 − ω)Δuk ν − 2 3 ∂xk 3 3 2 dω ∗ ∗ ∂2 ul Fk − eI + = 0 (k = 1, 2, 3) , (2.7.1) 3 deI s,t=1 ks l=1 lt ∂x s ∂xt G where ν is Poisson’s ratio, G is the shear modulus, F = (F1 , F2 , F3 ) are the volume forces, and ω(ei ) is a function of the variable eI , the intensity of the strain tensor which defines plastic properties of the material with hardening: √ 2 1/2 eI = 3 2 ( 11 − 22 ) 2 +( 11 − 33 ) 2 +( 22 − 33 ) 2 + 6( 2 12 + 2 13 + 23 ) . The function ω(eI ) must satisfy dω(eI ) 0 ≤ ω(eI ) ≤ ω(eI ) + eI ≤λ<1. (2.7.2) deI Other bits of notation are θ ≡ θ(u) = 11 (u) + 22 (u) + 33 (u) , and ⎧ √ ⎪ ∂uk θ ⎪ ⎪ 2 ⎪ ⎪ ∂x − 3 e , ⎪ k=s, 1 ∂ui ∂u j ∗ ⎪ ⎨ s I = + , =⎪ ⎪ ij 2 ∂x j ∂xi ks ⎪ ∂uk ∂u s ⎪ ⎪ 1 ⎪ ⎪ ⎩ + √ , k s. ∂x s ∂xk 2eI If ω(eI ) ≡ 0 we get the equations of linear elasticity for an isotropic homoge- neous body. By analogy with elasticity problems, to pose a boundary value problem for (2.7.1) we must supplement the equations with boundary conditions. We con- sider a mixed boundary value problem: a part S 0 of the boundary ∂Ω of a body occupying the domain Ω is fixed, uS =0, (2.7.3) 0 and the remainder S 1 = ∂Ω S 0 is subjected to surface forces f(x) (see [16]): σ·nS =f , (2.7.4) 1 where σ is the stress tensor and n is the external unit normal to S 1 . When ω(eI ) is small (as it is if eI is small) we have a nonlinear boundary value problem which is, in a certain way, a perturbation of a corresponding boundary value problem of linear elasticity. It leads to the idea of using an iterative procedure, the method of elastic solutions, to solve the former. This procedure looks like that of
  • 32. 146 2 Mechanics Problems from the Functional Analysis Viewpoint the contraction mapping principle if we can make the problem take the correspond- ing operator form. Then it remains to show that the operator of the problem is a contraction. Now we begin to carry out the program. Let us introduce the notation u, v = 2 {[ 9 11 (u) − 22 (u)][ 11 (v) − 22 (v)] +[ 11 (u) − 33 (u)][ 11 (v) − 33 (v)] +[ 22 (u) − 33 (u)][ 22 (v) − 33 (v)] + 6[ 12 (u) 12 (v) + 13 (u) 13 (v) + 23 (u) 23 (v)]} . (2.7.5) If we consider the terms on the right-hand side of (2.7.5) as coordinates of vectors a = (a1 , . . . , a6 ), b = (b1 , . . . , b6 ), ai = ci (u) , bi = ci (v) (i = 1, . . . , 6) , √ √ c1 (w) = 2 3 [ 11 (w) − 22 (w)] , c2 (w) = 2 3 [ 11 (w) − 33 (w)] , √ c3 (w) = 2 3 [ 22 (w) − 33 (w)] , c4 (w) = 2 √ 12 (w) 3 , c5 (w) = 2 √ 13 (w) 3 , c6 (w) = 2 √ 23 (w) 3 , then the form u, v is a scalar product between a and b in R6 : 6 u, v = ai bi . i=1 Besides, 6 u, u = c2 (u) = e2 (u) i I (2.7.6) i=1 and by the Schwarz inequality we get 6 | u, v | = ci (u)ci (v) ≤ eI (u)eI (v) . (2.7.7) i=1 On the set C2 of vector functions satisfying the boundary condition (2.7.3) and such that each of their components is of class C (2) (Ω), we introduce an inner product (u, v) = 3 2G u, v + 1 K θ(u) θ(v) dΩ . 2 (2.7.8) Ω This coincides with a special case of the inner product (2.3.27) in the linear the- ory of elasticity. So the completion of C2 in the metric corresponding to (2.7.8) is the energy space of linear elasticity E EM (M for “mixed”) if we suppose that the
  • 33. 2.7 Problem of Elastico-Plasticity; Small Deformations 147 condition (2.7.3) provides u = 0 if u 2 = 3 2 2 GeI (u) + 1 Kθ2 (u) dΩ = 0 . 2 Ω The norm of E EM is equivalent to one of W 1,2 (Ω) × W 1,2 (Ω) × W 1,2 (Ω) (see Sect. 2.3 and Fichera [11]). (By H1 × H2 we denote the Cartesian product of Hilbert spaces H1 and H2 , the elements of which are pairs (x, y) for x ∈ H1 and y ∈ H2 . The scalar product in H1 × H2 is defined by the expression (x1 , x2 )1 +(y1 , y2 )2 where x1 , x2 ∈ H1 and y1 , y2 ∈ H2 .) By the principle of virtual displacements, the integro-differential equation of equilibrium of an elastico-plastic body is 3 3 (u, v) − 3 G 2 ω(eI (u)) u, v dΩ − Fi vi dΩ − fi vi dS = 0 . Ω i=1 Ω i=1 S1 (2.7.9) This equation can be obtained using the equations (2.7.1) and the boundary con- ditions (2.7.3)–(2.7.4). Conversely, using the technique of the classical calculus of variations we can get (2.7.1) and the natural boundary conditions (2.7.4). Thus, in a certain way, (2.7.9) is equivalent to the above statement of the problem. So we can introduce Definition 2.7.1. A vector function u ∈ E EM is called the generalized solution of the problem of elastico-plasticity with boundary conditions (2.7.3)–(2.7.4) if it satisfies (2.7.9) for every v ∈ E EM . For correctness of this definition we must impose some restrictions on external forces. It is evident that they coincide with those for linear elasticity. So we assume that Fi (x1 , x2 , x3 ) ∈ L6/5 (Ω) , fi (x1 , x2 , x3 ) ∈ L4/3 (S 1 ) . (2.7.10) Consider the form 3 3 B[u, v] = 3 G 2 ω(eI (u)) u, v dΩ + Fi vi dΩ + fi vi dS Ω i=1 Ω i=1 S1 as a functional in E EM with respect to v(x1 , x2 , x3 ) when u(x1 , x2 , x3 ) ∈ E EM is fixed. As in linear elasticity, the load terms, thanks to (2.7.10), are continuous linear functionals with respect to v ∈ E EM . In accordance with (2.7.5) and (2.7.2), we get 3 2G ω(eI (u)) u, v dΩ ≤ λ 3 G 2 | u, v | dΩ ≤ λ u v , Ω Ω so this part of the functional is also continuous. Therefore we can apply the Riesz representation theorem to B[u, v] and obtain B[u, v] = (v, g) ≡ (g, v) .
  • 34. 148 2 Mechanics Problems from the Functional Analysis Viewpoint This representation uniquely defines a correspondence u→g where u, g ∈ E EM . We obtain an operator A acting in E EM by the equality g = A(u) . Equation (2.7.9) is now equivalent to (u, v) − (A(u), v) = 0 (2.7.11) or, since v ∈ E EM is arbitrary, u = A(u) . (2.7.12) The operator A is nonlinear. We shall show that it is a contraction operator. For this, take arbitrary elements u, v, w ∈ E EM and consider (A(u) − A(v), w) = 3 G 2 [ω(eI (u)) u, w − ω(eI (v)) v, w ] dΩ . (2.7.13) Ω First, let u, v, w be in C2 . At every point of Ω, by (2.7.7), we can estimate the inte- grand from (2.7.13) as follows. We have Int ≡ ω(eI (u)) u, w − ω(eI (v)) v, w 6 6 = ω(eI (u)) ci (u) ci (w) − ω(eI (v)) ci (v) ci (w) . i=1 i=1 Let us introduce a real-valued function f (t) of a real variable t by the relation 6 f (t) = ω(eI (tu + (1 − t)v))ci (tu + (1 − t)v)ci (w) . i=1 It is seen that Int = | f (1) − f (0)| . As f (t) is continuously differentiable, the classical mean value theorem gives f (1) − f (0) = f (z)(1 − 0) = f (z) for some z ∈ [0, 1] , or, in the above terms, we get
  • 35. 2.7 Problem of Elastico-Plasticity; Small Deformations 149 6 d Int = ω(eI (tu + (1 − t)v)) ci (tu + (1 − t)v) ci (w) dt i=1 t=z dω(eI (tu + (1 − t)v)) deI (tu + (1 − t)v) = · deI dt 6 6 · ci (tu + (1 − t)v) ci (w) + ω ci (u − v) ci (w) . t=z i=1 i=1 (Here we have used the linearity of ci (u) in u and, thus, in t.) Let us consider the term 6 deI (tu + (1 − t)v) T= ci (tu + (1 − t)v) ci (w) i=1 dt 6 6 d 1/2 = c2 (tu + (1 − t)v) j ci (tu + (1 − t)v) ci (w) i=1 dt j=1 6 2 c j (tu + (1 − t)v) c j (u − v) 6 j=1 = ci (tu + (1 − t)v) ci (w) . 6 1/2 i=1 2 c2 (tu j + (1 − t)v) j=1 Applying the Schwarz inequality, we obtain 6 1/2 6 1/2 c2 (tu + (1 − t)v) j c2 (u − v) j 6 j=1 j=1 |T | ≤ · |ci (tu + (1 − t)v)| |ci (w)| 6 1/2 i=1 c2 (tu + (1 − t)v) j j=1 so that by (2.7.6) 6 6 1/2 |T | ≤ c2 (u − v) j |ci (tu + (1 − t)v)| |ci (w)| j=1 i=1 6 6 1/2 1/2 ≤ eI (u − v) c2 (tu + (1 − t)v) i c2 (w) i i=1 i=1 = eI (u − v) eI (tu + (1 − t)v) eI (w) . Similarly,
  • 36. 150 2 Mechanics Problems from the Functional Analysis Viewpoint 6 6 6 1/2 1/2 ci (u − v) ci (w) ≤ c2 (u − v) i c2 (w) i = eI (u − v) eI (w) . i=1 i=1 i=1 Combining all these, we get dω(eI (tu + (1 − t)v)) Int ≤ eI (tu + (1 − t)v)) eI (u − v) eI (w) deI + ω(eI (tu + (1 − t)v)) eI (u − v) eI (w) t=z dω(eI (tu + (1 − t)v)) = ω(eI (tu + (1 − t)v)) + · deI · eI (tu + (1 − t)v)) eI (u − v) eI (w) . t=z By the condition (2.7.2), we have Int ≤ λeI (u − v) eI (w) (2.7.14) at every point of Ω. Returning to (2.7.13) we have, using (2.7.14), |(A(u) − A(v), w)| ≤ λ 3 2 G eI (u − v) eI (w) dΩ . Ω In accordance with the norm of E EM it follows that |(A(u) − A(v), w)| ≤ λ u − v w or, putting w = A(u) − A(v), we get A(u) − A(v) ≤ λ u − v , λ = constant < 1 . (2.7.15) Being obtained for u, v, w ∈ C2 , this inequality holds for all u, v, w ∈ E EM since in this inequality we can pass to the limit for corresponding Cauchy sequences in E EM . Inequality (2.7.15) states that A is a contraction operator in E EM ; hence, by the contraction mapping principle, (2.7.12) has a unique solution that can be found us- ing the iterative procedure uk+1 = A(uk ) (k = 0, 1, 2, . . .) . This procedure begins with an arbitrary element u0 ∈ E EM ; when u0 = 0, it is called the method of elastic solutions since at each step we must solve a problem of linear elasticity with some given load terms. From a practical standpoint, the method works best when the constant λ is small. So we can formulate
  • 37. 2.8 Bases and Complete Systems; Fourier Series 151 Theorem 2.7.1. Assume S 0 is a piecewise smooth surface of nonzero area and that conditions (2.7.2) and (2.7.10) hold. Then a mixed boundary value problem of elastico-plasticity has a unique generalized solution in the sense of Definition 2.7.1; the iterative procedure (2.7.15) defines a sequence of successive approximations uk ∈ E EM that converges to the solution u ∈ E EM and λk uk − u ≤ u0 − u1 . (2.7.16) 1−λ It is clear that we cannot apply this theorem when, say, S 1 = ∂Ω. In such a case, we must add the self-balance conditions (2.4.16). These guarantee that we can repeat the above method for a free elastic-plastic body, and so we can formulate Theorem 2.7.2. Assume that all the requirements of Theorem 2.7.1 and the self- balance conditions (2.4.16) are met. Then there is a unique generalized solution of the boundary value problem for a bounded elastic-plastic body, and it can be found by an iterative procedure of the form (2.7.15). Problem 2.7.1. Is an estimate of the type (2.7.16) valid in Theorem 2.7.2? We recommend that the reader prove Theorem 2.7.2 in detail, in order to gain experience with the technique. Remark 2.7.1. We should call attention to the way in which we obtained the main inequality of this section: it was proved for smooth functions and then extended to the general case. This is a standard technique in the treatment of nonlinear problems of mechanics. 2.8 Bases and Complete Systems; Fourier Series If a linear space Y has finite dimension n, then there is a set {g1 , . . . , gn } of n lin- early independent elements, called a basis of Y, such that every y ∈ Y has a unique representation n y= αk gk k=1 where the αk are scalars. We now consider an infinite dimensional normed space X. Definition 2.8.1. A system of elements {ek } is a (countable) basis of X if any ele- ment x ∈ X has a unique representation ∞ x= αk ek k=1 where the αk are scalars.
  • 38. 152 2 Mechanics Problems from the Functional Analysis Viewpoint It is clear that a basis {ek } is linearly independent since the equation ∞ 0= αk ek k=1 has the unique solution αk = 0 for each k. A normed space with a countable basis is separable: a countable set of all linear combinations n ck ek (n arbitrary) with rational coefficients ck is dense in the k=1 space. Problem 2.8.1. Prove this. We are familiar with some systems of functions which could be bases in certain spaces: for example, {gk } = √1 2π eikx (2.8.1) in L2 (0, 2π). Later, we confirm this example. Now we consider the system of monomials {xk } (k = 1, 2, . . .) in C(0, 1). If it is a basis, then any function f (x) ∈ C(0, 1) could be represented in the form ∞ f (x) = αk xk , k=0 where the series converges uniformly on [0, 1]. This means the function is analytic, but we know there are continuous functions on [0, 1] that are not analytic. Hence the system {xk } is not a basis. On the other hand, the Weierstrass theorem states that this system possesses properties similar to those of a basis. To generalize this similarity, we introduce Definition 2.8.2. A countable system {gk } of elements in a normed space X is com- plete (or total) in X if for any x ∈ X and any positive number ε there is a finite linear combination n(ε) αi gi such that i=1 n(ε) x− αi gi < ε . i=1 By Definition 2.8.2 and the Weierstrass theorem, the system of monomials {xk } is complete in C(0, 1). Because C(0, 1) is dense in L p (0, 1) for p ≥ 1, this system is also complete in L p (0, 1). Problem 2.8.2. Which systems are complete in L p (Ω) or W k,p (Ω)? If a normed space has a countable complete system, then the space is separable. The reader should be able to name a countable dense set to verify this. Problem 2.8.3. Name such a set.
  • 39. 2.8 Bases and Complete Systems; Fourier Series 153 The problem of existence of a basis in a certain normed space is difficult, but there is a special case where it is fully solved: a separable Hilbert space. The reader will find here the theory of Fourier series largely repeated in abstract terms. We begin with Definition 2.8.3. A system {xk } of elements of a Hilbert space H is orthonormal if for all integers m, n, (xm , xn ) = δmn where δmn is the Kronecker delta symbol. We know that, at least for Rn , there are some advantages in using an orthonormal system of vectors as a basis. Suppose we have an arbitrary linearly independent system of elements of a Hilbert space H, say { f1 , . . . , fn }, and let Hn be the subspace of H spanned by this system. We would like to use the system to construct an orthonormal system {g1 , . . . , gn } that is also a basis of Hn . This can be accomplished by the Gram– Schmidt procedure: (1) The first element of the new system is g1 = f1 / f1 , g1 = 1. (2) Take e2 = f2 − ( f2 , g1 )g1 ; then (e2 , g1 ) = ( f2 , g1 ) − ( f2 , g1 ) g1 2 = 0, so the second element is g2 = e2 / e2 . (3) Take e3 = f3 − ( f3 , g1 )g1 − ( f3 , g2 )g2 ; then (e3 , g1 ) = 0 and (e3 , g2 ) = 0. Since e3 0, we get the third element as g3 = e3 / e3 . . . . (i) Let ei = fi − ( fi , g1 )g1 − · · · − ( fi , gi−1 )gi−1 . It is seen that (ei , gk ) = 0 for k = 1, . . . , i − 1, hence we set gi = ei / ei . This process can be continued ad infinitum since all ek 0 (why?). So we obtain an orthonormalized system {g1 , g2 , g3 , . . .}. The process is, however, found to be unstable for numerical computation. As is known from linear algebra, a system { f1 , . . . , fn } is linearly independent in an inner product space if and only if the Gram determinant ( f1 , f 1 ) · · · ( f 1 , f n ) . . .. . . . . . ( f n , f1 ) · · · ( fn , fn ) is nonzero. For an orthonormal system of elements the Gram determinant, being the determinant of the identity matrix, equals +1; hence an orthonormal system is linearly independent. Problem 2.8.4. Provide a more direct proof that an orthonormal system is linearly independent. Let {gk } (k = 1, 2, . . .) be an orthonormal system in a complex Hilbert space H. For an element f ∈ H, the numbers αk defined by αk = ( f, gk ) are called the Fourier coefficients of f . Now we prove
  • 40. 154 2 Mechanics Problems from the Functional Analysis Viewpoint Theorem 2.8.1. A complete orthonormal system {gk } in a Hilbert space H is a basis of H; any f ∈ H has the unique Fourier series representation ∞ f = αk gk (2.8.2) k=1 where αk = ( f, gk ) are the Fourier coefficients of f . Proof. First we consider the problem of the best approximation of an element f ∈ H by elements of a subspace Hn spanned by g1 , . . . , gn . In Sect. 1.19 we showed that this problem has a unique solution. Now we show that it is n αk gk . Indeed, k=1 consider an arbitrary linear combination n ck gk . Then k=1 n 2 n n f− ck gk = f− ck gk , f − ck gk k=1 k=1 k=1 n n n 2 = f 2 − f, ck gk − ck gk , f + ck gk k=1 k=1 k=1 n n n = f 2 − ck αk − ck αk + ck ck k=1 k=1 k=1 n n = f 2 − |αk |2 + |ck − αk |2 . k=1 k=1 Because the right-hand side takes its minimum value when ck = αk , we have n 2 n 2 n f− αk gk = min f− ck gk = f 2 − |αk |2 ≥ 0 ; (2.8.3) c1 ,...,cn k=1 k=1 k=1 moreover, we obtain Bessel’s inequality n |( f, gk )|2 ≤ f 2 . (2.8.4) k=1 Denote by n fn = αk gk (2.8.5) k=1 the nth partial sum of the Fourier series for f . Let us show that { fn } is a Cauchy sequence in H. By Bessel’s inequality, n |αk |2 ≤ f 2 ; k=1 hence
  • 41. 2.8 Bases and Complete Systems; Fourier Series 155 n+m 2 n+m fn − fn+m 2 = αk gk = |αk |2 → 0 as n → ∞ . k=n+1 k=n+1 Now we show that { fn } converges to f . Indeed, by completeness of the system {gk } in H, for any ε > 0 we can find a number N and coefficients ck (ε) such that N 2 f− ck (ε)gk <ε. k=1 By (2.8.3), N 2 N 2 f − fN 2 = f− αk gk ≤ f− ck (ε)gk <ε, k=1 k=1 so the sequence { fN } converges to f and thus f = lim fn . (2.8.6) n→∞ This completes the proof. From (2.8.6) we can obtain Parseval’s equality ∞ |( f, gk )|2 = f 2 , (2.8.7) k=1 which holds whenever {gk } is a complete orthonormal system in H. Indeed, by (2.8.3), n 2 n 0 = lim f − αk gk = lim f 2 − |αk |2 . n→∞ n→∞ k=1 k=1 Now we introduce Definition 2.8.4. A system {ek } (k = 1, 2, . . .) in a Hilbert space H is closed in H if from the system of equations ( f, ek ) = 0 for all k = 1, 2, 3, . . . it follows that f = 0. It is clear that a complete orthonormal system of elements is closed in H. Problem 2.8.5. Provide a detailed proof. The converse statement holds as well. We formulate Theorem 2.8.2. Let {gk } be an orthonormal system of elements in a Hilbert space H. This system is complete in H if and only if it is closed in H.
  • 42. 156 2 Mechanics Problems from the Functional Analysis Viewpoint Proof. We need to demonstrate only that a closed orthonormal system in H is com- plete. Proving Theorem 2.8.1, we established that for any element f ∈ H the se- quence of partial Fourier sums (2.8.5) is a Cauchy sequence. By completeness of H, there exists f ∗ = limn→∞ fn that belongs to H. To complete the proof we need to show that f = f ∗ . We have n ( f − f ∗ , gm ) = lim f − αk gk , gm = αm − αm = 0 . n→∞ k=1 By Definition 2.8.4, it follows that f = f ∗ , hence {gk } is complete. It is normally simpler to check whether a system is closed than to check whether it is complete. At the beginning of this section we established that any system of linearly independent elements in H can be transformed into an orthonormal sys- tem equivalent to the original system in a certain way. So we draw the following conclusion. Theorem 2.8.3. A complete system {gk } in H is closed in H; conversely, a system closed in H is complete in H. Problem 2.8.6. Write out a detailed proof. As stated above, the existence of a countable basis in a Hilbert space provides its separability. The converse statement is also valid. We formulate that as Theorem 2.8.4. A Hilbert space H has a countable orthonormal basis if and only if H is separable. The proof follows immediately from the previous theorem. Indeed, in H select a countable set of elements that is dense everywhere in H. Using the Gram–Schmidt procedure, produce an orthonormal system of elements from this set (removing any linearly dependent elements). Since the initial system is dense it is complete and thus, as a result of the Gram–Schmidt procedure, we get an orthonormal basis of the space. Remember that all of the energy spaces we introduced above are separable. Hence each of them has a countable orthonormal basis (nonunique, of course). If a Hilbert space is not separable, by Bessel’s inequality and Lemma 1.16.5, it follows that for any element x of a nonseparable space the set of nonzero coefficients of Fourier αk is countable. Repeating the above considerations we can get that (2.8.2) is valid in this case as well. In conclusion, we consider whether the system (2.8.1) is a basis of the complex space L2 (0, 2π). From standard calculus it is known that the system is orthonormal in L2 (0, 2π) (the reader, however, can check this). Weierstrass’s theorem on the ap- proximation of a function continuous on [0, 2π] can be formulated as the statement that the set of trigonometric polynomials, i.e., finite sums of the form k αk eikx , is dense in the complex space C(0, 2π). But the set of functions C(0, 2π) is the base for construction of L2 (0, 2π), hence the finite sums k αk eikx are dense in L2 (0, 2π). This shows that (2.8.1) is an orthonormal basis of L2 (0, 2π).
  • 43. 2.9 Weak Convergence in a Hilbert Space 157 2.9 Weak Convergence in a Hilbert Space We know that in Rn , the convergence of a sequence of vectors is equivalent to coordinate-wise convergence. In a Hilbert space H, the Fourier coefficients ( f, gk ) of an element f ∈ H play the role of the coordinates of f . Suppose {gk } is an orthonormal basis of H. What can we say about convergence of a sequence { fn } if, for every fixed k, the numerical sequence {( fn , gk )} is convergent? Let us consider {gn } as a sequence. It is seen that for every k, lim (gn , gk ) = 0 , n→∞ hence we have coordinate-wise convergence of {gn } to zero. But the sequence {gn } is not convergent, since √ gn − gm = 2 for n m . Therefore, coordinate-wise convergence in a Hilbert space is not equivalent to the usual form of convergence in the space. We define a new type of convergence in a Hilbert space. Definition 2.9.1. Let H be a Hilbert space. A sequence {xk } ⊂ H is weakly conver- gent to x0 ∈ H if for every continuous linear functional F in H, lim F(xk ) = F(x0 ) . k→∞ If every numerical sequence {F(xk )} is a Cauchy sequence, then {xk } is a weak Cauchy sequence. To distinguish between weak convergence and convergence as defined on p. 29, we shall refer to the latter as strong convergence. We retain the notation xk → x for strong convergence and adopt xk x for weak convergence. Definition 2.9.1 is given in a form which (with suitable modifications) is valid in a metric space. But in a Hilbert space any continuous linear functional, by the Riesz representation theorem, takes the form F(x) = (x, f ) where f is an element of H. So Definition 2.9.1 may be rewritten as follows: Definition 2.9.2. Let H be a Hilbert space. A sequence {xn } ⊂ H is weakly conver- gent to x0 ∈ H if for every element f ∈ H we have lim (xn , f ) = (x0 , f ) . n→∞ If every numerical sequence {(xn , f )} is a Cauchy sequence, then {xk } is a weak Cauchy sequence. We have seen that some weak Cauchy sequences in H are not strong Cauchy sequences. But a strong Cauchy sequence is always a weak Cauchy sequence, by virtue of the continuity of the linear functionals in the definition.
  • 44. 158 2 Mechanics Problems from the Functional Analysis Viewpoint We formulate a simple sufficient condition for strong convergence of a weakly convergent sequence: Theorem 2.9.1. Suppose that xk x0 , where xk , x0 belong to a Hilbert space H. Then xk → x0 if xk → x0 . Proof. Consider xk − x0 2 . We get xk − x0 2 = (xk − x0 , xk − x0 ) = xk 2 − (x0 , xk ) − (xk , x0 ) + x0 2 . By Definition 2.9.2 we have lim [(x0 , xk ) + (xk , x0 )] = 2 x0 2 , k→∞ hence xk − x0 2 → 0. We shall see later that for some numerical methods it is easier to first establish weak convergence of approximate solutions and then strong convergence, than to es- tablish strong convergence directly. The last theorem allows us to justify a method successively, beginning with a simple approximate result and then passing to the needed one. That is why weak convergence is a major preoccupation in this presen- tation. Theorem 2.9.2. In a Hilbert space, every weak Cauchy sequence {xn } is bounded. Proof. We will prove the theorem for a complex Hilbert space; the proof is valid for a real space as well. Suppose to the contrary that there is a weak Cauchy sequence {xn } which is not bounded in H. So let xn → ∞ as n → ∞. We will show that this yields a contradiction. First we consider the set U of all numbers of the form (xn , y), where y belongs to a closed ball B(y0 , ε) with arbitrary ε > 0 and center y0 ∈ H, which are momentarily fixed. We claim that U is unbounded from above. Indeed, elements of the form yn = y0 + εxn /(2 xn ) belong to B(y0 , ε) since εxn ε yn − y0 = = . 2 xn 2 As {xn } is a weak Cauchy sequence, the numerical sequence {(xn , y0 )} converges and therefore is bounded. Since xn → ∞ we get ε ε |(xn , yn )| = (xn , y0 ) + (xn , xn ) = (xn , y0 ) + xn → ∞ 2 xn 2 as n → ∞. We see that U is unbounded for any fixed y0 . Now we show that unboundedness of any set U for any y0 yields a contradiction. Take the ball B(y0 , ε1 ) with ε1 = 1 and y0 = 0. Because U is unbounded from above, we can take any y1 ∈ B(y0 , ε1 ) and then find xn1 such that |(xn1 , y1 )| > 1 . (2.9.1)
  • 45. 2.9 Weak Convergence in a Hilbert Space 159 By continuity of the inner product in both its variables, we can find a closed ball B(y1 , ε2 ) such that B(y1 , ε2 ) ⊂ B(y0 , ε1 ) and such that (2.9.1) holds not only for y1 but for all y ∈ B(y1 , ε2 ): |(xn1 , y)| > 1 for all y ∈ B(y1 , ε2 ) . Then, in the ball B(y1 , ε2 ), we similarly take an interior point y2 and find xn2 , with n2 > n1 , such that |(xn2 , y2 )| > 2 , and, after this, a closed ball B(y2 , ε3 ) such that B(y2 , ε3 ) ⊂ B(y1 , ε2 ) and |(xn2 , y)| > 2 for all y ∈ B(y2 , ε3 ) . Repeating this procedure ad infinitum, we produce a sequence of closed balls B(yk , εk+1 ) such that B(y0 , ε1 ) ⊃ B(y1 , ε2 ) ⊃ B(y2 , ε3 ) ⊃ · · · , and corresponding terms xnk , nk+1 > nk , of the sequence {xn } such that |(xnk , y)| > k for all y ∈ B(yk , εk+1 ) . Since H is a Hilbert space there is at least one element y∗ which belongs to every B(yk , εk+1 ), so |(xnk , y∗ )| > k . Thus we find a continuous linear functional F ∗ (x) = (x, y∗ ) for which the numerical sequence {F ∗ (xnk )} is not a Cauchy sequence. This contradicts the definition of weak convergence of {xk }. This proof yields another important result: Lemma 2.9.1. Assume {xk } is an unbounded sequence in H, i.e., xk → ∞. Then there exists y∗ ∈ H and a subsequence {xnk } such that (xnk , y∗ ) → ∞ as k → ∞. Proof. Let zn = xn / xn . For any y with unit norm, the numerical sequence (zn , y) is bounded and thus we can select a convergent subsequence from it. If there exists such a unit element y∗ and a subsequence {znk } for which (znk , y∗ ) → a 0, then the statement of the lemma is valid for the subsequence {xnk } and y∗ if a > 0; if a < 0, then y∗ must be changed to −y∗ . Indeed, if a > 0 then (xnk , y∗ ) = (znk , y∗ ) xnk → ∞. Now we suppose that we cannot find such an element y∗ and a subsequence {znk } for which (znk , y∗ ) → a 0. So (zn , y) → 0 for any y ∈ H. By the Riesz representation theorem, this means that {zn } converges weakly to zero. We will prove that the statement of Lemma 2.9.1 holds for the latter class of sequences as well. For this we repeat two steps of the proof of Theorem 2.9.2. First we show that for any center y0 and radius ε, the numerical set (xn , y) with y running over B(y0 , ε) is unbounded. Indeed, taking the sequence yn = y0 + ε/(2 xn ) xn we get an element from B(y0 , ε). Next, ε ε (xn , yn ) = (xn , y0 ) + (xn , xn ) = (zn , y0 ) + xn . 2 xn 2
  • 46. 160 2 Mechanics Problems from the Functional Analysis Viewpoint Since ε is finite and (zn , y0 ) → 0 as n → ∞, we have (xn , yn ) → ∞. Another step of the proof of Theorem 2.9.2, establishing the existence of a subse- quence {xnk } and an element y∗ such that (xnk , y∗ ) → ∞, requires only that xn → ∞ and that for any ε > 0 the set (xn , y) is unbounded when y runs over B(y0 , ε), which was just proved. Thus we immediately state the validity of Lemma 2.9.1 for all the unbounded sequences. This is used in proving the principle of uniform boundedness, which we have established in a more general form (Theorem 1.23.2). Theorem 2.9.3. Let {Fk (x)} (k = 1, 2, . . .) be a family of continuous linear function- als defined on a Hilbert space H. If supk |Fk (x)| < ∞, then supk Fk < ∞. Proof. By the Riesz representation theorem, each of the functionals Fk (x) has the form Fk (x) = (x, fk ) , where fk ∈ H , fk = F k . So the condition of the theorem can be rewritten as sup |(x, fk )| < ∞ . (2.9.2) k By Lemma 2.9.1, the assumption that supk fk = ∞ implies the existence of x0 ∈ H and { fkn } such that |(x0 , fkn )| → ∞ as k → ∞ . This contradicts (2.9.2). Corollary 2.9.1. Let {Fk (x)} be a sequence of continuous linear functionals given on H, such that for every x ∈ H the numerical sequence {Fk (x)} is a Cauchy sequence. Then there is a continuous linear functional F(x) on H such that F(x) = lim Fk (x) for all x ∈ H (2.9.3) k→∞ and F ≤ lim inf Fk < ∞ . (2.9.4) k→∞ Proof. The limit on the right-hand side of (2.9.3), existing by the condition, defines a functional F(x) which is clearly linear. Since the condition of Theorem 2.9.3 is met, we have supk Fk < ∞; from |F(x)| = lim |Fk (x)| ≤ sup Fk x k→∞ k it follows that F(x) is continuous. Moreover (recall Problem 1.23.1), |F(x)| = lim |Fk (x)| ≤ lim inf Fk x , k→∞ k→∞ i.e., (2.9.4) is proved also.
  • 47. 2.9 Weak Convergence in a Hilbert Space 161 The following theorem gives an equivalent but more convenient definition of weak convergence. Theorem 2.9.4. A sequence {xn } is weakly Cauchy in a Hilbert space H if and only if the following pair of conditions holds: (i) {xn } is bounded in H, i.e., there is a constant M such that xn ≤ M; (ii) for any fα ∈ H from a system { fα } which is complete in H, the numerical sequence (xn , fα ) is a Cauchy sequence. Proof. Necessity of the conditions follows from the definition of weak convergence and Theorem 2.9.2. Now we prove sufficiency. Suppose the conditions (i) and (ii) hold. Take an arbi- trary continuous linear functional defined, by the Riesz representation theorem, by an element f ∈ H and consider the numerical sequence dnm = (xn , f ) − (xm , f ) . As the system { fα } is complete, there is a linear combination N fε = ck fk k=1 such that f − fε < ε/3M . Then |dnm | = |(xn − xm , f )| = |(xn − xm , fε + f − fε )| ≤ |(xn − xm , fε )| + |(xn − xm , f − fε )| N ≤ |ck | |(xn − xm , fk )| + ( xn + xm ) f − fε . k=1 Since, by (ii), the sequences {(xn , fk )} (k = 1, . . . , N) are Cauchy sequences, we can find a number R such that N |ck | |(xn − xm , fk )| < ε/3 for all m, n > R k=1 hence |dnm | ≤ ε/3 + 2Mε/(3M) = ε for m, n > R . This means that {(xn , f )} is a weak Cauchy sequence. Problem 2.9.1. Show that a sequence {xn } is weakly convergent to x0 in H if and only if the following pair of conditions holds:
  • 48. 162 2 Mechanics Problems from the Functional Analysis Viewpoint (i) {xn } is bounded in H; (ii) for any fα from a system { fα }, fα ∈ H, which is complete in H, we have limn→∞ (xn , fα ) = (x0 , fα ). Because weak convergence differs from strong convergence, we are led to con- sider weak completeness of a Hilbert space. Theorem 2.9.5. Any weak Cauchy sequence {xn } in a Hilbert space H converges weakly to an element of this space. In other words, a Hilbert space H is also weakly complete. Proof. For any fixed y ∈ H we define F(y) = limn→∞ (y, xn ). The functional F(y), whose linearity is evident, is defined on the whole of H. From the inequality |(y, xn )| ≤ M y where M is a constant such that xn ≤ M, it follows that |F(y)| ≤ M y and F ≤M. Therefore F(y) is a continuous linear functional which, by the Riesz representation theorem, can be written in the form F(y) = (y, f ) , f ∈H, f = F ≤M. But this means that f is a weak limit of {xn }. From this proof also follows Lemma 2.9.2. If a sequence {xn } ⊂ H converges weakly to x0 in H and xn ≤ M for all n, then x0 ≤ M. Problem 2.9.2. Provide the details. This states that a closed ball about zero is weakly closed. Any closed subspace of a Hilbert space is also weakly closed. We also formulate Mazur’s theorem that any closed convex set in a Hilbert space is weakly closed. The interested reader can find a proof in Yosida [44]. Theorem 2.9.6 (Mazur). Assume that a sequence {xn } in a Hilbert space H con- verges weakly to x0 ∈ H. Then there is a subsequence {xnk } of {xn } such that the 1 N sequence of arithmetic means N k=1 xnk converges strongly to x0 . Problem 2.9.3. Show that a weakly closed set is closed. Let us consider the problem of weak compactness of a set in a Hilbert space. We have seen that a ball in an infinite dimensional Hilbert space is not strongly compact. But for weak compactness, an analog of the Bolzano–Weierstrass theorem holds as follows:
  • 49. 2.9 Weak Convergence in a Hilbert Space 163 Theorem 2.9.7. A bounded sequence {xn } in a separable Hilbert space contains a weak Cauchy subsequence. In other words, a bounded set in a Hilbert space is weakly precompact. Proof. In a separable Hilbert space there is an orthonormal basis {gn }. By Theo- rem 2.9.4 it suffices to show that there is a subsequence {xnk } such that, for each fixed gm , the numerical sequence {(xnk , gm )} is a Cauchy sequence. The bounded numerical sequence {(xn , g1 )} contains a convergent subsequence {(xn1 , g1 )}. Considering the numerical sequence {(xn1 , g2 )}, for the same reason we can choose a convergent sequence {(xn2 , g2 )}. Continuing this process, on the kth step we obtain a convergent numerical subsequence {(xnk , gk )}. Choosing now the elements xnn , we obtain a sequence {xnn } such that for any fixed gm the numerical sequence {(xnn , gm )} is a Cauchy sequence. That is, {xnn } is a weak Cauchy sequence. This theorem has important applications. In justifying certain numerical meth- ods we can sometimes prove boundedness of the set of approximate solutions in a Hilbert (as a rule, energy) space, and hence obtain a subsequence of approxima- tions that converges weakly to an element; then we can show that this element is a solution. Let us apply this procedure to the approximation problem, namely, we want to minimize a functional F(x) = x − x0 2 (2.9.5) over a real Hilbert space when x0 is a fixed element of H, x0 M, and x is an arbitrary element of a closed subspace M ⊂ H. In Sect. 1.19 we established the existence of a minimizer of F(x). We now treat this problem once more, as though this existence were unknown to us. This very simple problem (at least in theory) exhibits the following typical steps, which are common for the justification of approximate solutions to many boundary value problems: 1. the formulation of an approximation problem and the demonstration of its solv- ability; 2. a global a priori estimate of the approximate solutions that does not depend on the step of approximation; 3. the demonstration of convergence of the approximate solutions to a solution of the initial problem, and a study of the nature of convergence. Thus we begin to study our problem with Step 1, the formulation of the approxima- tion problem. We try to solve the problem approximately, using the Ritz method. Assume {gk } is a complete system in M such that any of its finite subsystems is linearly independent. Consider Mn spanned by (g1 , . . . , gn ) and find an element which minimizes F(x) on Mn . A solution of this problem, denoted by xn , is the nth Ritz approximation of the solution.
  • 50. 164 2 Mechanics Problems from the Functional Analysis Viewpoint A real-valued function f (t) = F(xn + tgk ) of the real variable t takes its minimal value at t = 0 and, thanks to differentiability of f (t), d f (t) =0. dt t=0 This yields d d 0= xn − x0 + tgk 2 t=0 = (xn − x0 + tgk , xn − x0 + tgk ) t=0 = 2(xn − x0 , gk ) , dt dt so xn − x0 is orthogonal to each gk (k = 1, . . . , n). Using the representation n xn = ckn gk , k=1 we get a linear system of algebraic equations called the Ritz system of nth approxi- mation: n ckn (gk , gm ) = (x0 , gm ) (m = 1, . . . , n) . (2.9.6) k=1 The determinant of this system is the Gram determinant of a linearly independent system (g1 , . . . , gm ) that is not equal to zero. So the system (2.9.6) has a unique solution (ˆ 1n , . . . , ckn ). c ˆ Step 2. Now we will find a global estimate of the approximate solutions that does not depend on n. Although, in this case, we know that the approximate solution exists, we can get the estimate without this knowledge. Hence it is called an a priori estimate. We begin with the definition of xn : xn − x0 2 ≤ x − x0 2 for all x ∈ Mn . As x = 0 ∈ Mn , it follows that xn − x0 2 ≤ x0 2 , from which xn 2 ≤ 2 xn x0 , hence xn ≤ 2 x0 . (2.9.7) This is the required estimate. Remark 2.9.1. It is possible to get a sharper estimate than (2.9.7); however, for this problem it is only necessary to establish the existence of a bound. Step 3. Our last goal is to show that the sequence of approximations converges to a solution of the problem. First we demonstrate that this convergence is weak, and then that it is strong.
  • 51. 2.9 Weak Convergence in a Hilbert Space 165 By (2.9.7), the sequence {xn } is bounded and, thanks to Theorem 2.9.7, contains a weakly convergent subsequence {xnk } whose weak limit x∗ belongs to M (remember that a closed subspace is weakly closed). For any fixed gm , we can pass to the limit as k → ∞ in the equality (xnk − x0 , gm ) = 0 and get (x∗ − x0 , gm ) = 0 because (x, gm ) is a continuous linear functional in x ∈ H. Now consider (x∗ − x0 , h) where h ∈ M is arbitrary but fixed. By completeness of the system g1 , g2 , g3 , . . . in M, given ε > 0 we can find a finite linear combination N hε = ck gk k=1 such that h − hε ≤ ε/(3 x0 ) . Then |(x∗ − x0 , h)| = |(x∗ − x0 , h − hε + hε )| ≤ |(x∗ − x0 , h − hε )| + |(x∗ − x0 , hε )| = |(x∗ − x0 , h − hε )| ≤ x∗ − x0 h − hε ∗ ≤( x + x0 ) h − hε ≤ (2 x0 + x0 ) ε/(3 x0 ) = ε . Therefore, for any h ∈ M we get (x∗ − x0 , h) = 0 . (2.9.8) Finally, considering values of (2.9.5) on elements of the form x = x∗ + h when h ∈ M, we obtain, by (2.9.8), F(x∗ + h) = (x∗ − x0 + h, x∗ − x0 + h) = x∗ − x0 2 + 2(x∗ − x0 , h) + h 2 = x∗ − x0 2 + h 2 ≥ x∗ − x0 2 = F(x∗ ) . It follows that x∗ is a solution of the problem, and existence of solution has been proved. Now we can show that the approximation sequence converges strongly to a solu- tion of the problem. By Theorem 1.19.3, a minimizer of F(x) is unique; this gives
  • 52. 166 2 Mechanics Problems from the Functional Analysis Viewpoint us weak convergence of the sequence {xn } on the whole. Indeed, suppose to the con- trary that {xn } does not converge weakly to x∗ . Then there is an element f ∈ H such that (xn , f ) (x∗ , f ) . (2.9.9) By boundedness of the numerical set {(xn , f )}, the statement (2.9.9) implies that there is a subsequence {xnk } such that there exists lim (xnk , f ) (x∗ , f ) . (2.9.10) k→∞ Problem 2.9.4. Prove (2.9.10). But, for the subsequence {xnk }, we can repeat the above considerations and find that {xnk } contains a subsequence which converges weakly to a solution of the prob- lem. Since the solution is unique, this contradicts (2.9.10). Finally, multiplying both sides of (2.9.6) by the Ritz coefficient cmn and summing over m, we get ˆ (xn , xn ) = (x0 , xn ) . We can pass to the limit as n → ∞, obtaining lim (xn , xn ) = lim (x0 , xn ) = (x0 , x∗ ) . n→∞ n→∞ By (2.9.8) with h = x∗ we have (x0 , x∗ ) = (x∗ , x∗ ) , so lim xn 2 = x∗ 2 . n→∞ Therefore, by Theorem 2.9.1, the sequence {xn } converges strongly to x∗ . So we have demonstrated, via the Ritz method, a general way of justifying the solution of a minimal problem and the Ritz method itself. The method is common to a wide variety of problems, some nonlinear. In the latter case, many difficulties center on Steps 2 or 3, depending on the problem. The problem under discussion can also be interpreted another way, and this is of so much importance that we devote a separate section to it. 2.10 The Ritz and Bubnov–Galerkin Methods in Linear Problems We reconsider the problem of minimizing the quadratic functional (2.4.8) in a Hilbert space, namely, I(x) = x 2 + 2Φ(x) → min . (2.10.1) x∈H
  • 53. 2.10 Ritz and Bubnov–Galerkin Methods 167 Assuming Φ(x) is a continuous linear functional, the Riesz representation theorem yields Φ(x) = (x, −x0 ) where x0 ∈ H is uniquely defined by Φ(x). Then I(x) = x 2 − 2(x, x0 ) = x − x0 2 − x0 2 . 2 Since x0 is fixed, the problem (2.4.1) is equivalent to F(x) = x − x0 2 → min . x∈H This problem has the unique (and obvious) solution x = x0 . Of much interest is the fact that it coincides with the problem of the previous section if M = H. So application of the Ritz method in this problem is justified. Let us recall those results in terms of the new problem. Let {gk } be a complete system in H, every finite subsystem of which is linearly independent, and let the nth Ritz approximation to a minimizer be n xn = ckn gk . k=1 The system giving the nth approximation of the Ritz method is n ckn (gk , gm ) = −Φ(gm ) (m = 1, . . . , n) . (2.10.2) k=1 Let us collect the results in Theorem 2.10.1. The following statements hold. (i) For each n ≥ 1, the system (2.10.2) of nth approximation of the Ritz method has the unique solution c1n , . . . , cnn . (ii) The sequence {xn } of Ritz approximations defined by (2.10.2) converges strongly to the minimizer of the quadratic functional x 2 + 2Φ(x), where Φ(x) is a continuous linear functional on H. It is interesting to note that if {gk } is an orthonormal basis of H, then (2.10.2) gives the Fourier coefficients of the solution in H. Concerning Bubnov’s method, we only mention that it appeared when A.S. Bub- nov, reviewing an article by S. Timoshenko, noted that the Ritz equations can be obtained by multiplying by gm , a function of a complete system, the differential equation of equilibrium in which u was replaced by n un = ckn gk , k=1
  • 54. 168 2 Mechanics Problems from the Functional Analysis Viewpoint integrating the latter over the region, and then integrating by parts. In our terms this is (un , gm ) = −Φ(gm ) (m = 1, . . . , n) . Since this system indeed coincides with (2.10.2), Theorem 2.10.1 also justifies Bub- nov’s method. Galerkin was the first to propose multiplying by fm , a function of another sys- tem, for better approximation of the residual. The corresponding system is, in our notation, (un , fm ) = −Φ( fm ) (m = 1, . . . , n) . Discussion of this modification of the method can be found in Mikhlin [27]. Finally, we note that the finite element method for solution of mechanics prob- lems is a particular case of the Bubnov–Galerkin method, hence it is also justified for the problems we consider. 2.11 Curvilinear Coordinates, Nonhomogeneous Boundary Conditions We have considered some problems of mechanics using the Cartesian coordinate system. Almost all of the textbooks present the theory of the same problems in Cartesian frames; the few exceptions are the textbooks on the theory of shells and curvilinear beams, where it is impossible to consider the problems in Cartesian frames. However, in practice other coordinate systems occur frequently. The ques- tion arises whether it is necessary to investigate the boundary value problems for other coordinates, or whether it is enough to reformulate the results for Cartesian systems. For the generalized statements of mechanics problems in energy spaces, the answer is simple: it is possible to reformulate the results, and a key tool is a sim- ple change of the coordinates. This change allows us to reformulate the imbedding theorems in energy spaces, to establish the requirements for admitting classes of loads, etc. We note that it is a hard problem to obtain similar results independently, without the use of coordinate transformations, if the coordinate frame has singular points. Let us consider a simple example of a circular membrane with fixed edge (Dirich- let problem). In Cartesian coordinates we have the Sobolev imbedding theorem 1/2 1/p ∂u 2 ∂u 2 |u(x)| p dx dy ≤m + dx dy (2.11.1) Ω Ω ∂x ∂y for p ≥ 1, which is valid for any u ∈ W 1,2 (Ω) ≡ E MC satisfying the boundary ˙ condition u ∂Ω = 0 . (2.11.2)
  • 55. 2.11 Curvilinear Coordinates, Nonhomogeneous Boundary Conditions 169 Taking a function u ∈ C (1) (Ω) satisfying (2.11.2), in both integrals of (2.11.1) we pass to the polar coordinate system: 1/2 R 2π 1/p R 2π ∂u 2 1 ∂u 2 |u| p r dφ dr ≤m + r dφ dr (2.11.3) 0 0 0 0 ∂r r2 ∂φ where (r, φ) are the polar coordinates in a disk of radius R. Passing to the limit along a Cauchy sequence of E MC in the inequality (2.11.1), which is valid in Cartesian coordinates, shows us that it remains valid in the form (2.11.3) in polar coordinates. Inequality (2.11.3) is an imbedding theorem in the energy space of the circular mem- brane in terms of polar coordinates. The expression 1/2 R 2π ∂u 2 1 ∂u 2 u = + r dφ dr (2.11.4) 0 0 ∂r r2 ∂φ is the norm in this coordinate system and R 2π ∂u ∂v 1 ∂u ∂v (u, v) = + r dφ dr 0 0 ∂r ∂r r2 ∂φ ∂φ is the corresponding inner product. The requirement imposed on forces for existence of a generalized solution has the form R 2π |F|q r dφ dr < ∞ (q > 1) . 0 0 We have a natural form of the norm in the energy space (which is determined by the energy itself) using curvilinear coordinates, as well as a form of the imbedding theorem (i.e., properties of elements of the energy space and natural requirements on forces for the problem to be uniquely solvable). Then we note that we can replace formally the Cartesian system by any other system of coordinates which is admissible for smooth functions, and also change formally any variables in any expression which makes sense in the energy space considered in Cartesian coordinates. Finally, note that a norm like (2.11.4) is usually called a weighted norm because of the presence of weight factors, here connected with powers of r. There is an abstract theory of such weighted Sobolev spaces, not being so elementary as in the space we have considered. For more complicated problems such as elasticity problems, we can use the same method of introducing curvilinear coordinates; here we can change not only the in- dependent variables (x1 , x2 , x3 ), but also unknown components of vectors of dis- placements and prescribed forces, to the new coordinate system. We leave it to the reader to write down an equation determining a generalized solution, the forms of norm and scalar product, and restrictions for forces as well as imbedding inequali- ties, in other curvilinear coordinate systems such as cylindrical and spherical.
  • 56. 170 2 Mechanics Problems from the Functional Analysis Viewpoint Now let us consider two questions connected with nonhomogeneous boundary value problems in mechanics. The first is to identify the whole class of admissible external forces for which an energy solution exists. We know that the condition for existence of a solution is that the functional of external forces F(x)v(x) dΩ (2.11.5) Ω (say, in the membrane problem) is continuous and linear with respect to v(x) on an energy space. We shall show how this condition can be expressed in terms of so-called spaces with negative norms, a notion due to P.D. Lax [17]. The functional (2.11.5) can be considered as the scalar product of F(x) by v(x) in L2 (Ω). But v(x) belongs to an energy space E whose norm, for simplicity, is assumed to be such that v E = 0 implies v = 0. We know that v ∈ L2 (Ω) if v ∈ E; moreover, E is dense in L2 (Ω). For any F(x) ∈ L2 (Ω), we can introduce a new norm F E = sup F(x)v(x) dΩ . v E ≤1 Ω It is clear that L2 (Ω) with this norm is not complete (since all v ∈ L p (Ω) for any p > 2, p < ∞). The completion of L2 (Ω) in the norm · E is called the space with negative norm, denoted E − . In Lax [17] (and in other books, for example, Yosida [44]) it is shown that the set of all continuous linear functionals on E can be identified with E − since E is dense in L2 (Ω). So the condition F(x) ∈ E − is necessary and sufficient for the work functional (2.11.5) to be continuous with respect to v(x) on E. ˙ In Lax [17], such a construction was introduced for a Sobolev space W k,2 (Ω); the −k,2 corresponding space with negative norm was denoted by W (Ω). An equivalent approach to the introduction of W −k,2 (Ω) involves use of the Fourier transformation in Sobolev spaces (cf., Yosida [44]). The notion of the space with negative norm is useful for studying problems, but it is not too informative when we want to know whether certain forces are of a needed class; here sufficient conditions are more convenient. Secondly, we discuss how to handle nonhomogeneous boundary conditions (of Dirichlet type). Consider, for example, the problem −Δv = F , (2.11.6) v ∂Ω = ϕ . (2.11.7) We can try the classical approach, finding a function v0 (x) that satisfies (2.11.7), i.e., v0 ∂Ω = ϕ . Now we are seeking v(x) in the form v = u + v0 , where u(x) satisfies the homoge- neous boundary condition u ∂Ω = 0 . (2.11.8)
  • 57. 2.12 Bramble–Hilbert Lemma and Its Applications 171 An integro-differential equation of equilibrium of the membrane is ∂u ∂ψ ∂u ∂ψ ∂v0 ∂ψ ∂v0 ∂ψ + dΩ + + dΩ = Fψ dΩ (2.11.9) Ω ∂x ∂x ∂y ∂y Ω ∂x ∂x ∂y ∂y Ω wherein virtual displacements must also satisfy (2.11.8): ψ|∂Ω = 0 . We recognize the term ∂v0 ∂ψ ∂v0 ∂ψ + dΩ Ω ∂x ∂x ∂y ∂y as a continuous linear functional on E MC if ∂v0 /∂x and ∂v0 /∂y belong to L2 (Ω). In that case there is a generalized solution to the problem, i.e., u ∈ E MC satisfying (2.11.9) for any ψ ∈ E MC . We have supposed that there exists an element of W 1,2 (Ω) satisfying (2.11.7). In more detailed textbooks on the theory of partial differential equations, one may find the conditions for a function ϕ given on the boundary that are sufficient for the existence of v0 . Corresponding theorems for v0 from Sobolev spaces are called trace theorems. The trace theorems assume the boundary is sufficiently smooth. The case of a piecewise smooth boundary, frequently encountered in practice, has not been completely studied yet. The problem of the traces of functions is beyond the scope of this book. A final remark is in order. In mathematics we normally deal with dimensionless quantities, and we have followed that practice here. However, variables with dimen- sional units can be used without difficulty, provided we check carefully for units in all inequalities and equations, and introduce additional factors as needed. In partic- ular, the constants in imbedding theorems normally carry dimensional units, hence these constants change if the units are changed. 2.12 Bramble–Hilbert Lemma and Its Applications This lemma is widely used to establish the convergence rate for the finite element method (see, for example, Ciarlet [7]). It gives a bound for a functional with special properties in a Sobolev space. We remark that the lemma can be viewed as a simple consequence of the theorem on equivalent norming of W l,p (Ω) in Sobolev [33]. Recall Poincar´ ’s inequality (2.3.9), e 2 ∂u 2 ∂u 2 u2 dS ≤ m u dS + + dS , (2.12.1) S S S ∂x ∂y which was derived when S was the square [0, a] × [0, a]. The proof of (2.3.9) is easily extended to the case of an n-dimensional cube. We now discuss how to extend it to a compact set Ω that is star-shaped with respect to a
  • 58. 172 2 Mechanics Problems from the Functional Analysis Viewpoint square S ; that is, any ray starting in S intersects the boundary of Ω exactly once. We shall establish the following estimate, which is also called Poincar´ ’s inequality: e 2 ∂u 2 ∂u 2 u2 dΩ ≤ m1 u dS + m2 + dΩ . (2.12.2) Ω S Ω ∂x ∂y Let us rewrite this in a system of polar coordinates (r, φ) having origin at the center of S . Let ∂Ω be given by the equation r = R(φ) ≥ a/2, R(φ) < R0 . Then (2.12.2) has the form 2π R(φ) 2 2π R(φ) ∂u 2 1 ∂u 2 u2 r dr dφ ≤ m1 u dS +m2 + r dr dφ . 0 0 S 0 0 ∂r r2 ∂φ Because of (2.12.1), it follows that it is sufficient to get the estimate 2π R(φ) 2π a/2 ∂u 2 2π R(φ) u2 r dr dφ ≤ m3 u2 r dr dφ + m4 dr dφ r 0 a/2 0 a/4 0 a/4 ∂r (2.12.3) with constants independent of u ∈ C (1) (Ω) (C (1) (Ω) is introduced in Cartesian coor- dinates!). We now proceed to prove this. The starting point is the representation r2 ∂u(r, φ) a/4 ≤ r1 ≤ a/2 , u(r2 , φ) = u(r1 , φ) + dr , r1 ∂r a/4 ≤ r2 ≤ R0 , from which, by squaring both sides and applying elementary transformations, we get 1 √ ∂u r2 2 u2 (r2 , φ) ≤ 2u2 (r1 , φ) + 2 √ r dr r1 r ∂r r2 dr r2 ∂u 2 ≤ 2u2 (r1 , φ) + 2 r dr r1 r r1 ∂r R(φ) ∂u 2 4R0 ≤ 2u2 (r1 , φ) + m5 r dr , m5 = 2 ln . a/4 ∂r a Multiplying this chain of inequalities by r1 r2 and then integrating it first with respect to r2 from a/2 to R(φ) and then with respect to r1 from a/4 to a/2, we have a/2 R(φ) a/2 R(φ) r1 u2 (r2 , φ)r2 dr2 dr1 ≤ 2 u2 (r1 , φ)r1 dr1 r2 dr2 a/4 a/2 a/4 a/2 a/2 R(φ) R(φ) ∂u 2 + m5 r1 r2 dr1 dr2 r dr a/4 a/2 a/4 ∂r or
  • 59. 2.12 Bramble–Hilbert Lemma and Its Applications 173 R(φ) a/2 R(φ) ∂u 2 3 2 32 a u2 (r, φ)r dr ≤ R2 u2 (r, φ)r dr + 3 2 2 64 a R0 m5 r dr . a/2 0 a/4 a/4 ∂r Finally, integrating this with respect to φ over [0, 2π] and multiplying it by 32/(3a2 ), we establish (2.12.3) and hence (2.12.2). We can similarly extend Poincar´ ’s inequality to the case of a multiconnected do- e main Ω which is a union of star-shaped domains, and to the case of an n-dimensional domain Ω with n > 2. The latter extension is n 2 ∂u 2 u2 dΩ ≤ m1 u dΩ + m2 dΩ , (2.12.4) Ω C i=1 Ω ∂xi where C ⊂ Ω is a hypercube in Rn . We can apply the inequality (2.12.4) to any derivative Dα u, |α| < k. Combining these estimates successively, we derive the inequality needed to prove the Bramble– Hilbert lemma: 2 u 2 W k,2 (Ω) ≤ m3 Dα u dΩ + m4 |Dα u|2 dΩ . (2.12.5) 0≤|α|<k C |α|=k Ω This estimate permits us to introduce another form of equivalent norm in W k,2 (Ω). (Question to the reader: Which one?) Note that the estimate was obtained for func- tions of C (k) (Ω), but the now standard procedure of completion provides that it is valid for any u ∈ W k,2 (Ω). Lemma 2.12.1 (Bramble–Hilbert [5]). Assume F(u) is a continuous linear func- tional on W k,2 (Ω) such that for any polynomial Pr (x) of order less than k, F(Pr (x)) = 0 . (2.12.6) Then there is a constant m∗ depending only on Ω such that ⎛ ⎞1/2 ⎜ ⎜ ⎜ ⎟ ⎟ ⎟ |F(u)| ≤ m F W k,2 (Ω) ⎜ ∗ ⎜ ⎜ ⎝ |D u| dΩ⎟ . α ⎟ ⎟ ⎠ 2 (2.12.7) |α|=k Ω Proof. From (2.12.5) and continuity of F(u) on W k,2 (Ω), it follows that ⎡ ⎤1/2 ⎢ ⎢ ⎢ 2 ⎥ ⎥ ⎥ ⎢ |F(u)| ≤ m F W k,2 (Ω) ⎢ ⎢ Dα u dΩ + ⎥ |Dα u|2 dΩ⎥ . ⎥ (2.12.8) ⎣ ⎦ 0≤|α|<k C |α|=k Ω By (2.12.6), F(u(x) + Pk−1 (x)) = F(u(x)) where Pk−1 (x) is an arbitrary polynomial of order k − 1. Fixing u(x) ∈ W k,2 (Ω), we can always choose a polynomial P∗ (x) such that k−1
  • 60. 174 2 Mechanics Problems from the Functional Analysis Viewpoint Dα (u(x) + P∗ (x)) dΩ = 0 for all 0 ≤ |α| ≤ k − 1 . k−1 C Substituting u(x) + P∗ (x) into (2.12.8), we get (2.12.7) since k−1 Dα P∗ (x) = 0 for |α| = k . k−1 This completes the proof. Let us consider some simple applications of this lemma. Assume that we find numerically, by Simpson’s rule, 1 u(x) dx for u(x) ∈ W 2,2 (0, 1) . 0 What is a bound on the error? First we find the error in one step of the trapezoidal rule: xk +h h Fk (u) = u(x) dx − [u(xk + h) + u(xk )] . xk 2 It is clear that Fk (u) is a linear and continuous functional in W 2,2 (0, 1). Making the change of variable x = xk + hz in the integral, we get 1 |Fk (u)| = h u(xk + hz) dz − 1 [u(xk ) + u(xk + h)] ≤ 2h max |u(xk + zh)| . 2 0 z∈[0,1] (2.12.9) By the elementary inequality (Problem 2.12.1 below) √ 1 1/2 √ max | f (x)| ≤ 2 f 2 (x) + [ f (x)]2 dx ≤ 2 f W 2,2 (0,1) , (2.12.10) x∈[0,1] 0 relation (2.12.9) gives √ |Fk (u)| ≤ 2 2h u(xk + hz) W 2,2 (0,1) . Since Fk (a+bx) = 0 for any constants a, b we can apply the Bramble–Hilbert lemma and obtain √ 1 1/2 xk +h 1/2 |Fk (u)| ≤ 2 2hm [u (xk + hz)]2 dz = m1 h5/2 [u (x)]2 dx . 0 xk This is the needed error bound for one step of integration. Consider now the bound on total error when [0, 1] is subdivided into N equal parts 1 N−1 h F(u) = u(x) dx − [u(xk ) + u(xk+1 )] , xk = kh . 0 2 k=0 This is linear and continuous in W 2,2 (0, 1), and
  • 61. 2.12 Bramble–Hilbert Lemma and Its Applications 175 N−1 f (u) = Fk (u) . k=0 We get N−1 N−1 N−1 xk +h 1/2 |F(u)| = Fk (u) ≤ |Fk (u)| ≤ m1 h5/2 [u (x)]2 dx k=0 k=0 k=0 xk √ N−1 xk +h 1/2 ≤ m1 h5/2 N [u (x)]2 dx . k=0 xk Thus the needed bound on the error of the trapezoidal rule is 1 1/2 |F(u)| ≤ m1 h2 [u (x)]2 dx . 0 No improvements in the order of the error result if we take functions smoother than those from W 2,2 (0, 1). But if v ∈ W 1,2 (0, 1), the bound is worse: 1 1/2 |F(v)| ≤ m2 h [v (x)]2 dx . 0 Problem 2.12.1. Prove (2.12.10). Another example of the application of Lemma 2.12.1 is given by Problem 2.12.2. Show that the local error of approximation of the first derivatives of a function u(x1 , x2 ) ∈ W 3,2 (Ω), Ω ⊂ R2 , by symmetric differences, is ∂u(0, 0) u(h1 , 0) − u(−h1 , 0) ∂u(0, 0) u(0, h2 ) − u(0, −h2 ) l(u) = − + − ∂x1 2h1 ∂x2 2h2 M(h2 + h2 ) ≤ √ 1 2 u W 3,2 (Ω) h1 h2 if 0 < c1 < h1 /h2 < c2 < ∞.