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Lecture_03_S08.pdf
LECTURE # 3:
ABSTRACT RITZ-GALERKIN METHOD
MATH610: NUMERICAL METHODS FOR PDES:
RAYTCHO LAZAROV
1. Variational Formulation

In the previous lecture we have introduced the following space
of functions
defined on (0, 1):
(1)
V =
v(x) is continuous function on (0, 1);
v′(x) exists in generalized sense and in L2(0, 1);
v(0) = v(1) = 0
and equipped it with the L2 and H1 norms
‖v‖ = (v,v)1/2 and ‖v‖V = (v,v)
1/2
V =
(∫ 1
0

(u′2 + u2)dx
)1
2
.
We also introduced the following variational and minimization
problems:
(V ) find u ∈ V such that a(u,v) = L(v), ∀ v ∈ V,
(M) find u ∈ V such that F(u) ≤ F(v), ∀ v ∈ V,
where a(u,v) is a bilinear form that is symmetric, coercive and
contin-
uous on V and L(v) is continuous on V and F(v) = 1
2
a(u,u) −L(v).
As an example we can take
a(u,v) ≡

∫ 1
0
(k(x)u′v′ + q(x)uv) dx and L(v) ≡
∫ 1
0
f(x)v dx.
Here we have assumed that there are positive constants k0, k1,
M such that
(2) k1 ≥ k(x) ≥ k0 > 0, M ≥ q(x) ≥ 0, f ∈ L2(0, 1).
These are sufficient for the symmetry, coercivity and continuity
of the
bilinear form a(., .) and the continuity of the linear form L(v).
The proof of these properties follows from the following
theorem:
Theorem 1. Let u ∈ V ≡ H10 (0, 1). Then the following
inequalities are
valid:

(3)
|u(x)|2 ≤ C1
∫ 1
0
(u′(x))2dx for any x ∈ (0, 1),∫ 1
0
u2(x)dx ≤ C0
∫ 1
0
(u′(x))2dx.
with constants C0 and C1 that are independent of u.
1
2 MATH610: NUMERICAL METHODS FOR PDES: RAYTCHO
LAZAROV
Proof: We give two proofs. The simple one proves the above
inequali-

ties with C0 = 1/2 and C1 = 1. The better proof establishes the
above
inequalities with C0 = 1/6 and C1 = 1/4.
Indeed, for any x ∈ (0, 1) we have:
u(x) = u(0) +
∫ x
0
u′(s)ds.
Since u ∈ H10 (0, 1) then u(0) = 0. We square this equality and
apply
Cauchy-Swartz inequality:
(4) |u(x)|2 =
∣ ∣ ∣ ∫ x
0
u′(s)ds
∣ ∣ ∣ 2 ≤ ∫ x
0

1ds
∫ x
0
(u′(s))2ds ≤ x
∫ x
0
(u′(s))2ds.
Taking the maximal value of x on the right hand side of this
inequality
we get the first inequality (3) with C1 = 1. Further, increasing
the r.h.s by
taking the integral in the whole interval and then integrating the
obtained
inequality for x ∈ (0, 1) we get the second inequality (3) with
C0 = 1/2.
This simple proof uses only one boundary condition u(0) = 0.
We can
improve the constants if we use also the second boundary
condition u(1) = 0.
Namely, we derive in the same manner the inequality

(5) |u(x)|2 = |
∫ 1
x
u′(s)ds |2 ≤
∫ 1
x
1ds
∫ 1
x
(u′(s))2 ≤ (1 −x)
∫ 1
x
(u′(s))2ds.
Now one multiplies (4) by 1 −x and (5) by x and adds the two
inequalities
to get the estimate:
|u(x)|2 ≤ x(1 −x)
∫ 1

0
(u′(s))2ds.
This will allows us to show (3) with C0 = 1/6 and C1 = 1/4,
correspondingly.
The appropriate inequalities for an arbitrary l are obtained by
change of
the variable.
Now the coercivity of the bilinear form follows easily from
these inequal-
ities and the assumptions (2). Indeed,
a(u,u) ≥ k0
∫ 1
0
u′2dx ≥
k0
2
∫ 1

0
(u′2 + u2)dx ≥
k0
2
‖u‖2V .
2. Abstract form of Ritz-Galerkin method
Instead of (V ), we shall consider its approximation. Namely, let
Vh be a
n-dimensional subspace of V . We consider the following
simpler problem:
(Vh) find uh ∈ Vh such that a(uh,v) = L(v), ∀ v ∈ Vh.
One can show that this problem is equivalent to the following
minimization
problem in Vh:
(Mh) find uh ∈ Vh such that F(uh) ≤ F(v), ∀ v ∈ Vh.
There some simple but important for the applications properties
that
are easily obtaind from the equivalence of the problems (Vh)
and (Mh).

Using the fact that a(u,u) = L(u) and a(uh,uh) = L(uh) we get
from the
inequality F(u) ≤ F(uh) that
1
2
a(u,u) −L(u) ≤
1
2
a(uh,uh) −L(uh) =⇒ −
1
2
a(u,u) ≤−
1
2
a(uh,uh)
LECTURE # 3: ABSTRACT RITZ-GALERKIN METHOD 3
which gives a(u,u) ≥ a(uh,uh). This inequality has a clear

physical inter-
pretation.
Since Vh is n-dimensional, we can assume that it is spanned by
n linearly
independent functions φj(x) ∈ V, j = 1, ...,n; i.e.
Vh =
n∑
j=1
cjφj(x), cj are arbitrary constants
We may relate the parameter h to n by h = 1/n.
Examples of Spaces Vh:
Example 1:

φj(x) = sin(jπx), j = 1, ...,n. This will produce the so-called
spectral
method.
Example 2:
φj(x) = xj(1 −x), j = 1, ...,p. The so-called p-version of FEM.
Example 3:
The construction of the space is done in the following manner:
split the
interval [0, 1] into n + 1 subintervals by introducing the points
xj = jh,
j = 0, ...,n + 1, where h = 1
n+1
. The space Vh consists of all continuous
functions on [0, 1] that are linear on each subinterval (element)
[xj−1,xj]
and vanish at x = 0 and x = 1. Obviously, the functions in the
space Vh are
determined by their values at the nodes xj, j = 1, ...,n. Solving

(Vh) with
such space Vh will lead to the finite element method.
The following set of functions can serve as a basis for Vh:
(6) φj(x) =
x−xj−1
h
, x ∈ [xj−1,xj];
xj+1 −x
h
, x ∈ [xj,xj+1];
0, elsewhere ;
j = 1, ...,n. The functions are constructed in such way that φj(x)
is 1 at
the node xj, 0 at all remaining nodes, and linear over the finite
elements.

This basis is called a nodal basis. Note, that this is just one
possible basis.
Another example is the so-called hierarchical basis.
Obviously, the solution uh of the problem (Vh) is in the form
uh(x) =
n∑
j=1
ξjφj(x), where ξj are unknown constants.
LAZAROV
0 1x
j
j
φ

Figure 1. A nodal basis function for linear finite elements
Then the method (Vh) can be written in the form
a(uh,v) = l(v), ∀ v ∈ Vh =⇒ a
j=1
ξjφj(x),φk
(7)
This produces a linear system called also
(Ritz or Galerkin system) for the unknown ξ ∈ Rn:
n∑
j=1
ξja(φj,φk) = L(φk), k = 1, ...,n, in matrix form Aξ = b,

where A ≡{ajk}nj,k=1 = {a(φj,φk)}
n
j,k=1, is a square n×n matrix
and b = {L(φj)}nj=1, and ξ = {ξj}
n
j=1 are vector-columns in R
n.
The matrix A is often called “stiffness” matrix while b is the
“load” vector
which is computed from the data. Since the bilinear form a(., .)
is coercive
the matrix A is nonsingular (show this) and therefore the system
Aξ = b
has unique solution for any b. However, the condition number
of A play an
importnat role in the numerical methods for solving the system
and there
is a necessity to discuss this in details.
3. Mixed boundary conditions
For the boundary value problem

(D)
−(k(x)u′)′ + q(x)u = f(x), in (0, 1)
u(0) = 0,
u(1) + k(1)u′(1) = β1
we introduced the space V :
(8) V =
v(x) is continuous function on (0, 1);
v′(x) exists in a generalized sense and is in L2(0, 1);
v(0) = 0
and the variational formulation (V ) with
a(u,v) ≡
∫ 1

0
(k(x)u′v′ + q(x)uv) dx + u(1)v(1)
and
L(v) ≡
∫ 1
0
f(x)v dx + β1v(1).
Examples of Spaces Vh for the problem (D):
Example 1:
φj(x) = sin((j − 0.5)πx), j = 1, ...,n. This will produce the so-
called
spectral method.
Example 2:

φj(x) = xj, j = 1, ...,p. This will produce the so-called p-version
of
Galerkin method.
Example 3: (the finite element method)
The construction of the space is done in the following manner:
split the
interval [0, 1] into n subintervals by introducing the points xj =
jh, j =
0, ...,n, where h = 1
n
. The space Vh consists of all continuous functions on
[0, 1] that are linear on each subinterval (element) [xj−1,xj] and
vanish at
x = 0. Obviously, the space Vh is determined by the values of a
function at
the nodes xj, j = 1, ...,n. Solving (Vh) with such space Vh will
lead to the
finite element method.
In this case, a nodal basis will consist of all functions from the

previous
example corresponding to internal nodes as well as one more
function that
will involve the value at end xn = 1: The following set of
functions can serve
as a basis for Vh:
(9) φn(x) =
x−xn−1
h
, x ∈ [xn−1,xn];
0, elsewhere.
Another possible basis in Vh is the so-called hierarchical basis
which utilizes
hierarchy of grids.
4. Neumann boundary conditions

Now we shall consider the following simple model problem for
the un-
known function u(x):
(D)
−u′′ + u = f(x), in (0, 1)
u′(0) = 0.
u′(1) = 0.
In the previous lecture we have introduced the set of functions
defined on
(0, 1) that are is continuous function, have piece-wise
continuous deriva-
tive.This set has been equipped with the norms
||v||2 = (v,v) and ||v||2V = (v,v)V = (v,v) + (v
′,v′).
LAZAROV
After completing the set V in the norm || · ||V we get the

Sobolev space
H1(0, 1) of functions having generalized first derivatives in
L2(0, 1). Note,
that the functions in V do not satisfy any boundary conditions.
Therefore,
V ≡ H1(0, 1).
We also introduced the following variational problem:
(V ) find u ∈ V such that a(u,v) = L(v), ∀ v ∈ V,
where
a(u,v) ≡
∫ 1
0
(u′v′ + uv) dx and L(v) ≡
∫ 1
0
f(x)v dx.
We shall study the Ritz system for this particular BVP. It is
obvious, that
a(u,v) = (u,v)V so this form is trivially coercive.

We have introduced the following finite dimensional space:
split the in-
terval [0, 1] into n− 1 subintervals by introducing the points xj
= (j − 1)h,
j = 1, ...,n, where h = 1
n−1 ; the space Vh consists of all continuous on [0, 1]
functions that are linear on each subinterval (element) [xj−1,xj].
Obviously,
the functions in the space Vh can be determined by their values
at the nodes
xj, j = 1, ...,n. The approximate problem (Vh) for such space Vh
will lead
to the finite element method with linear elements.
The following set of functions can serve as a basis for Vh:
(10) φj(x) =
x−xj−1
h

, x ∈ [xj−1,xj];
xj+1 −x
h
, x ∈ [xj,xj+1];
0, elsewhere ;
j = 2, ...,n− 1 and two additional functions defined at the end-
points:
(11)
φ1(x) =
x2 −x
h
, x ∈ [x1,x2];
0, elsewhere ;
φn(x) =

x−xn−1
h
, x ∈ [xn−1,xn];
0, elsewhere .
The functions are constructed in such way that φj(x) is 1 at the
node xj,
0 at all remaining nodes and linear over the finite elements.
This basis is
called nodal basis.
The solution uh of the problem (Vh) is in the form
uh(x) =
n∑
j=1
ξjφj(x), where ξj are unknown constants.

Then the method (Vh) can be written in the form (7).
This basis of the space Vh will produce a tridiagonal matrix A.
Indeed,
a(φj,φk) = 0, for |j −k| > 1. Also, for j = k we get
a(φj,φj) =
∫ xj+1
xj−1
(
1
h2
+ φ2j (x)
)
dx =
2
h
+
2h
3
, for 1 < j < n,

a(φ1,φ1) =
∫ x2
x1
(
1
h2
+ φ21(x)
)
dx =
1
h
+
h
3
, for j = 1,
a(φn,φn) =
∫ xn
xn−1

(
1
h2
+ φ2n(x)
)
dx =
1
h
+
h
3
, for j = n.
Similarly, for k = j + 1 we get
a(φj,φj+1) =

∫ xj+1
xj
(
−1
h2
+ φj(x)φj+1(x)
)
dx =
−1
h
+
h
6
.
The coefficients below the main diagonal are recover from the
symmetry of
the matrix A.
Thus, the matrix A of the Ritz-system has the form A = A0 +

A1, where:
(12) A1 =
1
h
1 −1 0 . . . 0
−1 2 −1 . . . 0
0 −1 2 . . . 0
. . . . . . .
0 0 0 . . . 1
= h6
2 1 0 . . . 0
1 4 1 . . . 0
0 1 4 . . . 0

. . . . . . .
0 0 0 . . . 2
Matrix A1 is called “stiffness” matrix, while the matrix A0 is
called “mass”
matrix. Both matrices are symmetric and A0 is positive definite
while A1 is
semi-definite.
5. Issues to be addressed
In the genral case we are facing the following issues:
• to assemble the matrix A and to solve the system Aξ = b;
• alternatively, if an iterative method is used that requires only
the
matrix-vector multiplication Aξ, then one should prepare a pro-
cedure of matrix vector multiplication (possibly without
explicitly
forming the matrix A);
• estimate the condition number of the matrix A for a prticular
chice

of the basis of the space Vh;
• estimate the error e = u−uh;
• to develope an algorithm that adaptively choses the mesh so
that
the error is uniformly distributed in the domain and is dreven
below
a desired level.
We need to develop the mathematical tools for studying these
problems.
This includes: estimate for the condition number of A,
deriving/finding fast
methods for solving the system, proving various integral
inequalities, deriv-
ing the approximation error with piece-wise polynomial
functions, estimates
in various Sobolev norms, etc.
6. An estimate of the condition number of the global matrix A
for Neumann BC
Further, we shall use the following definition of a condition
number of a

symmetric and positive definite matrix:
cond(A) =
max λ(A)
min λ(A)
,
where λ(A) is an eigenvalue of A, i.e. Aξ = λξ, for some ξ a
nonzero vector
in Rn.
Often it is not possible to compute the eigenvalues and the
condition
number, but for practical purposes it is enough to have an upper
bound for
cond(A). For this we need upper and lower bounds for the
eigenvalues of A.
LAZAROV
Simple calculations show that

h
6
≤ λ(A0) ≤ h and 0 ≤ λ(A1) ≤
4
h
.
So we produce the following bound from above for the
condition number of
the matrix A
(13) cond(A) ≤
max λ(A0) + max λ(A1)
min λ(A0) + min λ(A1)
≤
4/h + h
h/6
=
24
h2

+ 6 = O(h−2).
Remark 1. Note, that A0 and A1 are square matrices of size n
and one
finds that cond(A0) ≤ 6 i.e. the condition number of A0 does
not depend
on the size of the matrix. Such matrices are called well-
conditioned. In
contrast, A1 has condition number O(h−2) which increases
quadratically,
when h → 0. Such matrices are called ill-conditioned.
7. Exercises
The following matrices play essential role in the finite element,
finite vol-
ume and finite difference methods for two-point boundary value
problems
and the solution of the corresponding linear systems. The
spectral properties
of these matrices are used very often in the computational
practice.
(1) Find the exact eigenvalues of the matrices B1,B0 ∈ Rn×n
given by

(14) B1 =
2 −1 0 . . . 0 0
−1 2 −1 . . . 0 0
0 −1 2 . . . 0 0
. . . . . . . .
0 0 0 . . . −1 2
and
(15) B0 =
4 1 0 . . . 0 0
1 4 1 . . . 0 0

0 1 4 . . . 0 0
. . . . . . . .
0 0 0 . . . 1 4
Hint: Show that λj(B1) = 4 sin2
πj
2(n+1)
, j = 1, . . . ,n and then use
the fact that B1 +B0 = I, where I is the identity matrix in Rn.
From
these calculations follow that both B1 and B0 are positive
definite.
(2) Estimate that eigenvalies of the scaled “stiffness” matrix B1
∈ Rn×n
(16) B1 =

1 −1 0 . . . 0 0
−1 2 −1 . . . 0 0
0 −1 2 . . . 0 0
. . . . . . . .
0 0 0 . . . −1 1
and the scaled “mass” matrix B0 ∈ Rn×n
(17) B0 =
2 1 0 . . . 0 0
1 4 1 . . . 0 0
0 1 4 . . . 0 0
. . . . . . . .
0 0 0 . . . −1 2

Remark 2. Using the technique applied above we can show that
in this case
the eigenvalues are λj(B1) = 4 sin2
πj
2(n−1) , j = 0, . . . ,n− 1.
Remark 3. The eigenvalues and eigenvectors of these algebraic
problems
and problems obtained by approximation of the same
differential operator
with third type boundary conditions could be found in the
monograph of
Samarskii [6, pp. 104–109].
References
[1] L. C. Evans, Partial Differential Equations, Graduate
Studies in Mathematics, vol.

19, AMS, 1998.
[2] Ch. Grossmann, H.-O. Ross, and M. Stynes, Numerical
Treatment of Partial Differ-
ential Equations, Springer, Berlin, 2005.
[3] M. Renardy and R. Rogers, An Introduction to Partial
Differential Equations, Texts
in Applied Mathematics, Springer-Verlag, 1993.
[4] P. Knabner and L. Angermann, Numerical Methods for
Elliptic and Parabolic PDEs,
Springer-Verlag, New Yrok Inc, 2003.
[5] S. Larsen and V. Thomee, Partial Differential Equations
with Numerical Methods,
Springer, 2003.
[6] A.A. Samarskii, The Theory of Difference Schemes,
Monographs and Textbooks in
Pure and Appled Mathematics, Marcel Dekker, Inc, New York,
2001.
Lecture_08_S08.pdf

LECTURE # 8:
MULTIDIMENSIONAL SECOND ORDER ELLIPTIC
PROBLEMS
MATH610: NUMERICAL METHODS FOR PDES – R.
LAZAROV
1. Introduction and preliminaries
First, we introduce some notations that will be used further.
Here Ω will
denote a polygonal bounded domain in Rd, d = 2, 3 with
boundary ∂Ω.
Further, for the vector q = (q1, . . . , qd) and for a scalar
function v we define
the divergence ∇ · q and the gradient ∇ v, correspondingly, by
∇ · q = ∂q1
∂x1
+ · · · + ∂qd
∂xd

and ∇ v =
(
∂v
∂x1
, . . . ,
∂
∂xd
)
.
The Stokes theorem will be used in the following form:
∫
∂Ω
q · n ds =
∫
Ω

∇ · q dx.
Here, n is the outward unit vector to ∂Ω and q·n denotes the
inner product
of two vectors on Rd.
We shall use the Hilbert space H1(Ω) of functions defined on Ω
and having
their generalized derivatives in L2(Ω). The subspace of those
functions in
H1(Ω) that vanish on the boundary ∂Ω will be denoted by H10
(Ω). The
L2 and H1-inner products of these spaces and the corresponding
norms are
defined as follows:
(u, v) =
∫
Ω
uv dx, (u, v)1 = (u, v) + (∇ u, ∇ v),(1)
‖u‖ = (u, u)1/2, ‖u‖1 = (u, u)1/21 .(2)
For the elements in the space H1(Ω) we shall use the following
Poincare

inequality:
(3)
∫
Ω
u2 dx ≤ M0
∫
Ω
|∇ u|2 dx
where the constant M0 > 0 does not depend on u.
We shall give proof of this inequality for d = 2.Without loss of
generality,
we can assume that Ω is contained in the unit square Π, i.e. Ω ⊂
Π :=
(0, 1) × (0, 1). Then we can extend a function u ∈ H1(Ω) to Π
by zero
1

2 MATH610: NUMERICAL METHODS FOR PDES – R.
LAZAROV
outside Ω. The extended function is denoted by ū. It belongs to
H10 (Π) and
obviously,
∫
Π
ū2 dx =
∫
Ω
u2 dx.
Next, we write the equality
2
∫
Π

ū2 dx =
∫
Π
{(∫ x1
0
∂
∂x1
ū(ξ, x2) dξ
)2
dx(4)
+
(∫ x2
0
∂
∂x2

ū(x1, ξ) dξ
)2 }
dx
and apply Cauchy-Schwarz inequality to each of the line
integrals:
2
∫
Π
ū2 dx ≤
∫
Π
{
x1
∫ 1
0
(

∂
∂x1
ū(ξ, x2) dξ
)2
(5)
+x2
∫ 1
0
(
∂
∂x2
ū(x1, ξ) dξ
)2 }
dx.
Using Fubini theorem,we get finally:
2

∫
Ω
u2 dx = 2
∫
Π
ū2 dx ≤ 1
2
∫
Π
|∇ū|2 dx(6)
=
1
2
∫
Ω
|∇ u|2 dx,

which is the required inequality with M0 = 1/4. If the domain Ω
is contained
in a rectangle (0, l1)×(0, l2) the required inequality follows by
change of the
variables.
Further, we shall need the following two inequalities valid for
functions in
H1(Ω):
(7)
∫
∂Ω
u2 ds ≤ C‖u‖21,
and
(8)
∫
Ω
u2 dx ≤ C

{∫
Ω
|∇ u|2 ds +
∫
∂Ω
u2 ds
}
.
Here the constant C does not depend on u but depend on the
domain Ω.
One can prove these inequalities for rectangular domains simply
by using
the corresponding estimates from the one-dimensional case. The
proofs are
left as an exercise for this part of the class (see, e.g. [3, 7]).
MULTIDIMENSIONAL ELLIPTIC PROBLEMS 3

2. Problem formulation
In this lecture we shall consider the following Dirichlet
boundary-value
problem: find u(x) such that:
(D)
Lu := ∇ ·
(
−K(x)∇ u + b(x)u
)
+ q(x)u = f (x), x ∈ Ω
u(x) = 0, x ∈ ∂Ω.
where the coefficients K(x), b, q and f are given functions on Ω.
We
assume that Ω is a bounded domain with Lipschitz boundary
∂Ω, K(x) is a
symmetric and uniformly in Ω positive definite matrix and the
coefficients
K(x), b(x), q(x) are measurable and bounded function in Ω.

This is the divergent form of the problem. Quite often second
order
problems are given in the following non-divergent form:
(9)
Lu := ∇ · (−K(x)∇u) + b̃(x)∇ u + q(x)u = f (x), x ∈ Ω
u(x) = 0, x ∈ ∂Ω.
If the vector field b̃(x) is differentiable then these two forms are
equivalent.
In case when b ≡ b̃ and ∇ · b = 0, then these two form coinside.
In some applications this equation describes: (1) deflection of
an elastic
membrane under transverse load f (then K = I, b ≡ 0, q ≡ 0); (2)
the
pressure distribution in a porous media (K is the permeability
tensor, b ≡
0, q ≡ 0); (3) concentration distribution of a chemical in a flow
with velocity
b and absorption coefficient q. The quantity
q(x) = −K(x)∇ u + b(x)u
is often called total flux (mass, thermal, etc) with −K(x)∇ u the

diffusive
part and b(x)u convective part of the flux.
For deriving the variational formulation of this problem we
follow the
standard approach used in the 1-dimensional problems. We
multiply the
differential equation (D) by a test function v ∈ H10 (Ω) and
integrate over Ω:
∫
Ω
(
∇ · (−K(x)∇ u + b(x)u) + q(x)u
)
v dx =
∫
Ω
f (x)v dx.

We use the identity
(
∇ · (−K(x)∇ u + b(x)u)} v = ∇ · {(−K(x)∇ u + b(x)u) v
)
(10)
−
(
− K(x)∇ u + b(x)u
)
· ∇ v,
4 MATH610: NUMERICAL METHODS FOR PDES – R.
LAZAROV
so that after applying the Stokes theorem we transform the right
hand side
of the above identity to the form:
∫

∂Ω
(
K(x)∇ u − b(x)u
)
· n v ds +
∫
Ω
(
K(x)∇ u − b(x)u
)
· ∇ v dx.
Now we use the fact that v vanishes on ∂Ω to get
∫
Ω
(
(K(x)∇ u − b(x)u) · ∇ v + q(x)uv

)
dx =
∫
Ω
f (x)v dx.
We rewrite this integral identity in the abstract form
a(u, v) = L(v) ∀ v ∈ H10 (Ω),
where
a(u, v) =
∫
Ω
(
K(x)∇ u · ∇ v − ub(x) · ∇ v + q(x)uv
)
dx

and
L(v) =
∫
Ω
f (x)v dx.
Thus, we have shown that the solution of the problem (D)
satisfies the
following variational problem:
(V ) find u ∈ H10 (Ω) such that a(u, v) = L(v), ∀ v ∈ H10 (Ω) .
So we have reformulated the differential problem (D) in terms
of integral
identity involving the bilinear form a(·, ·) and the linear form
L(·). Again,
we can use the general theoretical framework and Lax-Milgram
theorem to
show the existence and the uniqueness of the solution u ∈ H10
(Ω). We shall
prove that under reasonable conditions on the coefficients the
bilinear form
a(·, ·) is coercive and continuous in V = H10 (Ω) so we can

apply the general
theoretical framework for such problems.
Now we give conditions on the coefficients of the differential
equation (D)
that are sufficient for the coercivity and the continuity of the
bilinear form
a(·, ·):
(C)
ξT K(x)ξ ≥ k0ξT ξ, ∀ ξ ∈ Rd, k0 = const > 0,
q(x) + 1
2
∇ · b(x) ≥ 0, ∀ x ∈ Ω.
Theorem 1. Assume that the conditions (C) are satisfied. Then
the bilinear
form a(·, ·) is coercive and continuous in V , i.e. there are
positive constants
α and C such that
(11)
a(u, u) ≥ α‖u‖21, (coercivity)
a(u, v) ≤ C0‖u‖1 ‖v‖1. (continuity)

MULTIDIMENSIONAL ELLIPTIC PROBLEMS 5
Proof: First, we note that
(12) −u b · ∇ v = −1
2
∇ · (bu2) + 1
2
u2 ∇ · b.
Then applying the Stokes theorem and the condition (C) and the
fact that
v vanishes on ∂Ω we get the following for for a(u, u)
a(u, u) =
∫
Ω
(K∇ u · ∇ u + (q + 0.5∇ · b)u2) dx ≥ k0||∇ u||2.
Using Poincare inequality (3) we get the desired result

regarding the coer-
civity. The continuity of the bilinear from is a simple
consequence of the
boundness of the coefficients.
Let Vh be a finite dimensional subspace H10 (Ω). The Ritz-
Galerkin method
can be formulated in the already discussed abstract form:
(Vh) find uh ∈ Vh ⊂ H10 (Ω) such that a(uh, v) = L(v), ∀ v ∈
Vh.
Our goal now is to construct the space Vh and to show how the
Ritz-system
derived from (Vh) is computed and solved.
3. Other types of boundary conditions
Instead of Dirichlet boundary conditions one can put various
other types
of boundary conditions on ∂Ω. Below we give two natural
boundary condi-
tions that are widely used in the applications.
Case b ≡ 0; Then we have diffusion-reaction equation and the

following
Robin condition can be prescribed on the whole boundary ∂Ω or
on part of
it:
(13) K(x)∇ u · n + σ(x)u = g(x) ∀ x ∈ ∂Ω.
Here σ(x) ≥ 0 and g(x) are given functions on ∂Ω. If σ(x) ≡ 0
then this is
the classical Neumann boundary condition. The meaning of this
…

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