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Lakshmi - 11860 - An Introduction to Inverse Problems in Physics.indd 1
Lakshmi - 11860 - An Introduction to Inverse Problems in Physics.indd 1 14/5/2020 3:49:20 pm
14/5/2020 3:49:20 pm](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-9-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 2
2 Introduction
the other hand “inverse problems” are typically ill-posed problems, meaning that
a small error in the initial parameters can result in large errors in the final results.
Thus these problems are often unstable. For instance consider the simple case of
the matrix eigenvalue problem. Here the direct problem refers to the calculation of
the eigenvalues of a symmetric matrix, and the inverse eigenvalue problem is that
of the construction of a symmetric matrix when the set of eigenvalues are known.
Here the direct problem is well-posed, but not the inverse matrix eigenvalue prob-
lem.
We say a physical problem is well-posed if it has the following properties [1]:
(1) A solution for it is guaranteed to exist.
(2) The solution must be unique.
(3) The solution has to depend continuously on the data.
In physical sciences we find two distinct sets of inverse problems. First those
problems which can be precisely stated but the answer, in general, may not be
unique. For instance in classical dynamics, the inverse problem of vibration of a
linear system composed of masses and springs, Sec. 1.6, or in quantum theory the
inverse problem of bound states in a confining potential, Sec. 2.1 deal with examples
of this type. In the second group of inverse problems we can make a mathematically
well-defined formulation if and only if we make a number of simplifying assump-
tions so that the problem can be formulated without ambiguity. These assumptions
can be about the nature of the physical system, or an extension of the range of
validity of the observables or the input data beyond what is physically accessible or
measurable. For instance, in quantum scattering theory, consider the question of
the determination of a potential function from the energy dependence of the phase
shift. In this problem we have to make a number of assumptions, such as whether
the potential should be local or nonlocal and/or whether the potential should be
energy (or velocity) dependent or independent.
Inverse methods can be used to investigate different aspects of problems that
we encounter in theoretical physics. Among them we will consider the followings:
(a) For exactly solvable problems we can use these methods to study the uniqueness
of the results that we have found from the solution of the direct problem. Interest-
ing examples belonging to this group are Wigner’s derivation of the general form of
the commutation relation for the harmonic oscillator [2], [3] and the Bargmann con-
struction of phase equivalent potentials [4], and the fact that having the standard
quantum mechanical angular momentum eigenvalues, 2
( + 1), do not necessarily
imply the existence of a central force law [5].
(b) The elegant method of Gel’fand–Levitan [6], [7] and also the interesting method
of Marchenko [8] to solve the inverse problem of wave equation, i.e. the construc-
tion of a local energy independent short range potential for the Schrödinger equation
from the information known about the energy dependence of the phase shift, is the
central part of any discussion of the inverse scattering theory, particularly in nu-
clear and particle physics. A good review entitled “Inverse Problems in Classical
and Quantum Physics” by A.A. Almasy discusses a number of wide ranging prob-
lems including problems from particle physics and field theory.
(c) Except for a limited set of examples, we cannot find analytical solutions to the
inverse problems, therefore for practical applications we have to resort to](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-19-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 3
An Introduction to Inverse Problems in Physics 3
numerical methods. Unfortunately, unlike the direct problems, most inverse prob-
lems are ill-posed and are unstable. While many different ways of obtaining numer-
ical solutions have been proposed, some more accurate than the others, in general,
these solutions should be used as guides to the form of the output. In this book
we will try to sample a number of different inverse problems numerically. Nearly
all of the numerical solutions reported in this text have been done in 1980’s, with
the numerical techniques and computers available in that decade. Today solutions
of these problems may yield much better results.
This book begins with a review of a number of inverse problems of classical
dynamics including a brief discussion of construction of the Lagrangian and the
Hamiltonian from the equations of motion and also the conditions under which
such a construction is possible. Then the application of the inverse method in semi-
classical formulation of quantum mechanics for certain physical systems is stud-
ied. This is followed by a short account of the Heisenberg formulation. The next
chapter is devoted to the inverse problem for the Schrödinger equation using the
Gel’fand–Levitan formulation in one dimension as well as for the radial equation for
spherically symmetric potential in spherical polar coordinates. In the latter case it
is assumed that for a given partial wave one knows the phase shifts for all energies
and also the bound states energies. Marchenko’s approach is studied in the follow-
ing chapter, where it is applied to solve a number of physically interesting problems.
The Newton–Sabatier method for determining the potential from scattering
data given at fixed energy but for large number of partial waves is discussed next.
Then few applications such as finding solvable cases of the Fokker–Planck equation
are presented. Inverse problems for classical wave propagation are the subject of
the next few chapters. Here continued fractions are introduced as a powerful way of
getting approximate solutions for inverse problems of various types. This approach
combined with the input data given in the form of a rational fraction provides a reli-
able technique for solving many problems in atomic, nuclear and geophysical inverse
problems numerically. Among a number of applications, the solution of the inverse
problem of torsional vibration is discussed. The next two chapters are devoted to
the approximate determination of nuclear forces from the empirical knowledge of
the phase shifts and binding energies. The inverse problem of electrical conductiv-
ity which is of great interest in geophysics is the subject which is presented in the
following chapter. Finally, there is a brief review of the inverse problem of reflection
from a moving target.
I must also acknowledge the great books of Newton, [9], by Chadan and
Sabatier [8] and by Zakhariev and Suzko [11] that paved the way for my studies
of this subject over the years, and were very helpful in guiding me in different
ways. Only when I had completed the writing of the book that I came across the
very interesting dissertation entitled “Inverse Problems in Classical and Quantum
Physics” by A.A. Almasy who discusses a number of wide-ranging problems includ-
ing problems from particle physics and field theory [12].](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-20-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 4
4 References
References
[1] J. Hadamard, Sur les Problémes aux Dérivées Partielles et Leur Signification
Physique, Princeton University Bulletin, pp. 49-52.
[2] E.P. Wigner, Phys. Rev. 77, 711, (1950).
[3] M. Razavy, Heisenberg’s Quantum Mechanics, (World Scientific, Singapore,
2011) Chapter 5.
[4] V. Bagamann, Phys. Rev. 75, 301 (1949).
[5] M. Razavy, Phys. Letts. 88A, 215 (1982).
[6] I.M. Gel’fand and B.M. Levitan, Dokl. Akad. Nauk SSSR, 77, 557 (1951).
[7] I.M. Gel’fand and B.M. Levitan, Isvest. Akad. Nauk SSSR, 15, 309 (1951).
[8] V.A. Marchenko, Dokl. Akad. Nauk SSSR, 72, 457 (1950).
[9] R.G. Newton, Scattering Theory of Waves and Particles, (Second Edition,
1980).
[10] K. Chadan and P.C. Sabatier, Inverse Prblems in Quantum Scattering Theory,
Second Edition, (Springer-Verlag, 1989).
[11] B.N. Zakhariev and A.A. Suzko, Direct and Inverse Problems: Potentials in
Quantum Scattering, (Springer-Verlag, Berlin 1990).
[12] A.A. Almasy, Inverse Problems in Classical and Quantum Physics, Ph.D. Dis-
sertation, Johannes Gutenberg-Universität in Mainz (2007), arXiv.0912.0455
math-ph (2009).](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-21-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 5
Chapter 1
Inverse Problems in Classical
Dynamics
Some of the very interesting problems of classical mechanics belong to the category
of inverse problems. While these are elementary in nature, and are exactly solvable,
they provide a very good introduction to the physics of the inverse problems.
1.1 Inverse Problem for Trajectory
We will start our review of the inverse problems in classical mechanics by considering
a problem that goes back to I. Newton [1]. That is the derivation of the inverse-
square law of planetary motion from empirical laws of Kepler.
First we note that Kepler’s second law, i.e. the area swept out by the position
vector in unit time is constant, in polar coordinates, can be written as
mr2
θ̇ = = constant, (1.1)
where dot denotes the time derivative, and where m is the mass of the planet. In
addition we know that the orbit is an ellipse with the origin at one of the two foci.
Thus the orbit in polar coordinate is expressible as
r =
a
1 + cos θ
, 1. (1.2)
By differentiating r twice with respect to time and then using to eliminate θ̇, we
find
ṙ =
a sin θ
(1 + cos θ)2
, (1.3)
5](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-22-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 6
6 Shape of the Potential Energy
and
r̈ =
a
(cos θ)θ̇ =
2
mar2
cos θ, (1.4)
Eliminating cos θ between (1.2) and (1.4), we can write r̈ as a function of r only
r̈ =
2
mar2
a
r
− 1
. (1.5)
But we know that the radial component of the acceleration in polar coordinates ar
is given by
ar =
r̈ − r
θ̇
2
=
1
m2
2
r3
−
2
ar2
− r
2
r4
= −
2
m2a
1
r2
. (1.6)
Therefore the radial component of the force is given by Fr = − 2
ma r−2
.
1.2 Determination of the Shape of the Potential
Energy from the Period of Oscillation
Another classical problem which is of interest is to find the shape of a confining
potential (i.e. a potential V (x) that goes to +∞ as x → ±∞) from the energy
dependence of its period of oscillations [2].
For the one-dimensional motion the total energy, E, of the particle is given
by
E =
1
2
mẋ2
+ V (x), (1.7)
where x is the position of the particle and V (x) is a confining potential with the
condition V (0) = 0. Equation (1.7) is a first order differential equation for x and
can easily be solved for t as a function of x
t =
1
2
m
dx
E − V (x)
+ t0. (1.8)
The two constants of motion are E, which is the total energy and t0. The limits
of the motion are the two turning points where the potential energy is equal to the
total energy, E = V (x). We only consider confining potentials where the motion is
bounded by two such points. Now let us consider the coordinate x as function of V .
This function x(V ) is double-valued, i.e. for a given value of V we find two different
values of x. Therefore, the integrand in (1.8) can be divided into two parts, one
from x = x1 to x = 0 and the other from x = 0 to x = x2. We denote the function
x(V ) in these two regions by x = x1(V ) and x = x2(V ) respectively. We also note
that the limits of integration with respect to V are E and zero, therefore the period](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-23-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 7
An Introduction to Inverse Problems in Physics 7
is given by
T(E) =
√
2m
E
0
dx2(V )
dV
dV
√
E − V
+
√
2m
0
E
dx1(V )
dV
dV
√
E − V
=
√
2m
E
0
[dx2(V ) − dx1(V )]
dV
dV
√
E − V
. (1.9)
Now we divide both sides of (1.9) by
√
E − E, where E is a parameter, and then
integrate the result with respect to E from 0 to E. Thus we find Abel’s integral
equation [3]–[4]
E
0
T(E)dE
√
E − E
=
√
2m
E
0
E
0
dx2
dV
−
dx1
dV
dV dE
(E − E)(E − V )
. (1.10)
By changing the order of integration in Eq. (1.10) we have
E
0
T(E)dE
√
E − E
=
√
2m
E
0
dx2
dV
−
dx1
dV
dV ×
E
V
dE
(E − E)(E − V )
= π
√
2m [x2(E) − x1(E)] , (1.11)
where we have substituted the value of the last integral in Eq. (1.11) and the value
is π. Now by replacing E by V , we obtain
x2(V ) − x1(V ) =
1
π
√
2m
V
0
T(E)
√
V − E
dE. (1.12)
Thus by knowing T(E) we can find the difference x2(V ) − x1(V ), but not x2(V )
and x1(V ) separately. This means that there are infinitely many curves V = V (x)
which gives us the same T(E), and that these functions, V (x), differ from each
other in such a way that x2(V ) − x1(V ) is the same for every curve. By imposing
some other conditions on V (x), such as its symmetry about the V axis we find a
unique potential which is given by [2]–[6]
x(V ) =
1
2π
√
2m
V
0
T(E)dE
√
V − E
. (1.13)
As an example if T(E) ∝ E− 1
4 , then from (1.13) we find
x(V ) ∝
V
0
dE
E
1
4
√
V − E
= V
1
4
1
0
dx
x
1
4
√
1 − x
= V
1
4 β
3
4
,
1
4
, (1.14)
where β is the β function and is related to Γ function by [7],
β(z, w) = β(w, z) =
1
0
tz−1
(1 − t)w−1
dt =
Γ(z)Γ(w)
Γ(z + w)
. (1.15)](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-24-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 8
8 Action Equivalent Hamiltonoans
Thus taking the symmetry about the V axis we find that the potential is V (x) ∝
|x|4
, i.e. the motion is that of a classical anharmonic oscillator.
Abel’s Integral Equation — The integral equation (1.9) is an Abel integral
equation and is a special case of a class of the general integral equations of the form
[3], [6]
f(x) =
1
Γ(α)
x
0
η(ξ)dξ
(x − ξ)1−α
, 0 α 1, (1.16)
where Γ(α) is the Gamma function. In this integral equation f(x) is a known
function and we want to find η(ξ). The formal solution of (1.16) which is obtained
by inversion is given by
η(ξ) =
1
Γ(1 − α)
d
dξ
ξ
0
f(x)
(ξ − x)α
dx
=
1
Γ(1 − α)
f(0)
tα
+
ξ
0
f
(x)
(ξ − x)α
dx . (1.17)
Equation (1.16) has a continuous solution in the interval a ≤ ξ ≤ b subject to the
following conditions:
(a) f(x) must be continuous in this interval,
(b) f(0) = 0 and
(c)
ξ
0
f(x)(ξ − x)α−1
dx must have a continuous derivative in the interval [6], [5].
In a number of problems we find Abel’s integral equation of the form
f(x) =
b
x
η(ξ)dξ
ξ2 − x2
, (1.18)
where 0 ≤ x ≤ ξ ≤ b ≤ ∞. The solution of this integral equation is
η(ξ) = −
2
π
d
dξ
b
ξ
xf(x)dx
x2 − ξ2
, (1.19)
or [6]
η(ξ) =
2ξ
π
f(b)
b2 − ξ2
−
b
ξ
f
(x)dx
x2 − ξ2
. (1.20)
An elegant way of solving Abel’s integral equation is by utilizing the Laplace trans-
form method. This method is given, for example in Arfken’s book [8].
1.3 Action Equivalent Hamiltonians
Let us solve the inverse one-dimensional motion of particle for the simplest case
where the period T = 2π
ω is independent of energy. Then from (1.12) we conclude](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-25-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 9
An Introduction to Inverse Problems in Physics 9
that
x2(V ) − x1(V ) =
2V
m
1
ω
. (1.21)
This relation shows that V is a quadratic function of x. Let us write it as
V (x) =
⎧
⎪
⎨
⎪
⎩
1
2 mω2
+x2
for x 0
1
2 mω2
−x2
for x 0
, (1.22)
and this solution satisfies the condition V (0) = 0. From (1.21) and (1.22) it follows
that the angular frequency of the motion, ω, is related to ω+ and ω− by
ω =
2ω+ω−
ω+ + ω−
. (1.23)
Thus for all non-zero values of the parameters ω+ and ω− of the potential the
resulting angular frequency is ω, and therefore in an action-angle formulation of the
problem the Hamiltonian is given by [9], [10]
H = ωJx. (1.24)
A Two-Dimensional Motion with Closed Orbit — Let us study the direct
problem for the motion of a particle of unit mass in two dimensions x and y, where
the Hamiltonian is given by
H =
1
2
p2
x + ω2
x2
+
1
2
p2
y + ω2
y2
. (1.25)
If we transform this Hamiltonian to the action-angle variables, then we have [9]–[11].
H = ω(Jx + Jy). (1.26)
According to what we have found earlier for one-dimensional motion the simplest
two-dimensional motion for the Hamiltonian (1.26) given in terms of the canonical
variables is
H =
1
2
p2
x + p2
y + Ωx(x)x2
+ Ωy(y)y2
. (1.27)
where
Ωx(x) = ω
θ(x)
1 + λ1
+
θ(x)
1 − λ1
. (1.28)
and
Ωy(y) = ω
θ(y)
1 + λ2
+
θ(y)
1 − λ2
. (1.29)
In these relations θ(x) is the step function
θ(x) =
⎧
⎪
⎨
⎪
⎩
1 for x 0
0 for x 0
, (1.30)
and λ1 and λ2 are two arbitrary parameters |λ1,2| 1.](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-26-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 10
10 Abel’s Integral Equation
–1
–2
–1
1
2
–0.5 0.5 1
x
y
Figure 1.1: The closed orbit found from the Hamiltonian (1.27) for m = 1, ω = 1, λ1 = 0.4 and
λ2 = 0.2. Apart from the energy there are no apparent symmetries or conserved quantities for
this motion [11].
In conclusion we observe that the symmetry of the Hamiltonian (1.25) and
the symmetry of the Hamiltonian (1.27), which we found by solving the inverse
problem, are not the same, i.e. (1.25) remains invariant under the transformation
x y, but (1.27) does not, even though in both cases the orbits are closed. This
is particularly significant once we consider the question of the quantization.
We will come back to this important aspect of using the results of solving the
inverse problem later in connection with the quantum mechanical inverse problems.
1.4 Abel’s Original Inverse Problem
An inverse problem of historical interest has been called “determination of the shape
of a hill from the travel time measurement” [12]. If a particle slides up and down a
hill with no friction and with the initial energy E, assuming that we can measure
the travel time for up and down motion T(E) as a function of the initial velocity of
the particle, then under certain conditions we can find the shape of the hill. In the
direct problem, we know that the potential energy of the particle at the height y is
given by mgy(s), where s measures the arc length along the hill. The total energy
of the particle is the sum of kinetic and potential energies:
E =
1
2
m
ds
dt
2
+ V (s). (1.31)
At t = 0 we assume that the particle is at s(0) = 0, and has a velocity
ds
dt
t=0
= v0.
Solving (1.31) for ds
dt , choosing the positive square root since v0 0, and then](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-27-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 11
An Introduction to Inverse Problems in Physics 11
integrating (1.31) we obtain
t =
m
2
s
0
ds
E − V (s)
. (1.32)
Thus starting at s = 0, the particle moves up and reaches the classical turning point
s1, where E = V (s1). The corresponding value of t at this point is 1
2 T(E), i.e.
T(E) =
√
2m
s1(E)
0
ds
E − V (s)
. (1.33)
At the time t = 1
2 T(E), the velocity of the particle becomes zero and after that
the velocity becomes negative and the particle will start descending. Therefore the
solution for t ≥ 1
2 T(E), is obtained if we take the negative square root in (1.31).
The result in this case is
t =
1
2
T(E) +
m
2
s1(E)
0
ds
E − V (s)
,
1
2
T(E) ≤ t. (1.34)
Now if we set s1 = 0 in the above equation we find the time that takes the particle to
move back to the point s = 0 is just the same 1
2 T(E) given by (1.33). If V (s) is not
a monotonically increasing function of s, we can divide the upward (or downward)
motion of the particle into different segments and then find the total time [12].
Here, for the sake of simplicity, we assume that V (s) is a monotonic function of s.
With this condition the solution of the Abel integral equation is given by
ds(V )
dV
=
T(0)
π
√
2mV
+
1
π
√
2m
V
0
dT(E)
dE
dE
√
V − E
, (1.35)
where, because of the initial conditions, T(0) = 0. Finally we integrate (1.35) from
0 to V to get s1(V )
s1(V ) =
1
π
√
2m
V
0
T(E)dE
√
V − E
. (1.36)
Once we have s1(V ), by inverting, we find s(V ) and since V (s) = mgy(s) and
dx
ds
2
+
dy
ds
2
= 1, therefore we have a parametric equation for x(s) and y(s):
x(s) =
s
0
1 −
dy (s)
ds
2
ds
, and y(s) =
V (s)
mg
. (1.37)
Even when the profile of the hill cannot be described by a monotonic function
y = y(s), still there are cases that the problem can be solved [12].](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-28-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 13
An Introduction to Inverse Problems in Physics 13
b
b
db
dΘ
Θ
Figure 1.2: Scattering of a particle by a repulsive force.
Thus by integrating this equation and noting that the scattering angle, θ = π − 2φ,
we find
θ = θ(b) = π − 2
∞
r0
dr
r2
1 −
V (r)
E
1
b2
−
1
r2
− 1
2
. (1.44)
This is the solution of the direct problem, i.e. if we know the potential we can
calculate the scattering angle as a function of energy and of the impact parameter.
In the direct problem, the important concept is the cross section which is a
function of θ. Thus we consider a mono-energetic beam of particles incident in the
annular ring of area 2πbdb bounded by the impact parameters b and b + db, and
that the number of scattered particles through the solid angle dΩ = 2π sin θdθ is
σ(θ) (see Fig. 1.2). By equating the number of incoming and outgoing particles we
find [10]
σ(θ) = −
b
sin θ
db
dθ
. (1.45)
The Inverse Scattering Problem in Classical Mechanics — In order to solve
the inverse problem, we assume that θ(b) is known and we want to find V (r).
Following the work of Firsov and of Keller [13], [14], we introduce the new variables:
x =
1
b2
, u =
1
r
, θ(b) = θ̄(x), V (r) = V̄ (u), (1.46)
and
β(x) =
1
2
π − θ̄(x)
. (1.47)
With these changes, Eq. (1.44) can be written as
β(x) =
u0
u=0
x
1 −
V̄ (u)
E
− u2
− 1
2
du, (1.48)
where u0 = 1
r0
. Next we make the following substitution:
v(u) = 1 −
V̄ (u)
E
, w =
u2
v
and g(w) =
1
√
v
du
dw
. (1.49)](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-30-320.jpg)
![May 12, 2020 11:33 book-961x669 11860-main page 15
An Introduction to Inverse Problems in Physics 15
Inverse Problem of the Rutherford Scattering — We know that in the well-
known Rutherford scattering, the cross section is given by [10]
σ(θ) =
A
4 sin4 θ
2
, (1.58)
where A = e2
4E2 . Using Eq. (1.45) we can express b2
in terms of the scattering angle
θ
−
bdb
sin θdθ
=
A
4 sin4 θ
2
, (1.59)
or simplifying we find
1
b2
= x(θ) = A−1
tan2 θ
2
, θ = tan−1
√
Ax. (1.60)
Now we substitute x(θ) in (1.57) to find v:
v = exp
⎧
⎨
⎩
1
π
w
0
dw
√
w
2 tan−1
√
Aw
0
dθ
w − A−1 tan2 θ
2
1
2
⎫
⎬
⎭
. (1.61)
To evaluate the θ integral, we change the variable θ to φ, where
tan
θ
2
=
√
Aw sin φ. (1.62)
The last integral in (1.61) can be calculated [7]
2 tan−1
√
Aw
0
dθ
w − A−1 tan2 θ
2
1
2
= 2
√
A
π
2
0
dφ
1 + Aw sin2
φ
=
π
√
a
√
1 + Aw
. (1.63)
By substituting this integral in (1.52) we can carry out the integration over w
and
obtain v;
v = 1 + 2Aw + 2 Aw(2 + Aw). (1.64)
Since u = 1
r =
√
vw, from Eqs. (1.46) and (1.49), we find
V (r) = 2E
√
Au = eu =
e
r
, (1.65)
which is the Coulomb potential.
1.6 Inverse Problem of a Linear Chain of Masses
Coupled to Springs
A simple and much studied example of an inverse problem in classical dynamics
is that of the determination of masses and different spring constants for a finite](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-32-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 16
16 Masses Coupled to Strings
chain from the knowledge of the characteristic frequencies of the system [15], [16].
In order to state the questions relating to the input and output of this problem
clearly, we need to study certain aspects of the theory of matrices first. Therefore
we begin this section with the review of the mathematical tools needed for solving
this particular problem [16].
Rayleigh Quotient— Let x be a column n × 1 matrix (or a vector with n com-
ponents) and An be a symmetric n × n matrix, and let us consider the following
quotient called Rayleigh quotient [17]
R =
xT
· Anx
xT · x
. (1.66)
In this relation xT
is the transpose of x and dot denotes the scalar product of two
vectors. We note that by replacing x by bx, with b a constant R will not be affected,
since it is the ratio of two quadratic forms. Thus we can write R as the value of
xT
· Ax
subject to the normalization condition xT
· x = 1. This constraint on
x can be enforced by introducing a Lagrange multiplier λ, and then finding the
condition for R to be stationary. Now we consider the condition under which the
quadratic form
R = xT
· Anx − λxT
· x, (1.67)
becomes stationary. By differentiating R with respect to a matrix element xi and
setting it equal to zero we get
∂R
∂xi
= 2(a1ix1 + a2ix2 + · · · + anixn) − 2λxi = 0. (1.68)
If we collect all the terms ∂R
∂x1
, ∂R
∂x2
, · · · , ∂R
∂xn
, we find the equation
Anx − λx = 0, (1.69)
and this shows that λ is an eigenvalue of the matrix A.
From this direct problem we find a set of eigenvalues λ1, λ2, · · · , λn, but know-
ing this set is not sufficient to solve the inverse problem since in the latter we
need n masses, m1, m2, · · · , mn and n spring constants k1, k2, · · · , kn. The rest of
the needed information can be obtained from the eigenvalues μ1, μ2, · · · , μn−1 of
the matrix An−1 which will be defined below and in addition the total mass of the
system
n
i=1 mi. If we delete the row n and the column n from the matrix we
find (n − 1) × (n − 1) matrix An−1, with n − 1 eigenvalues. Whereas the matrix
(An − λ)x = 0 represents the motion of the system shown in Fig. 1.3 (the top fig-
ure) with all particles moving, the eigenvalue equation (An−1 − μ)x = 0 describes
the system shown in the same figure (below) where the nth mass is attached to
the wall and is not moving. Since x1 = 0 is equivalent to xT
· e1 = 0, where
e1 = [0, 0, · · · , 1], we can introduce two Lagrange multiplier λ and 2μ and consider
under what conditions
R = xT
· Anx − λxT
· x − 2μxT
· en, (1.70)](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-33-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 18
18 Masses Coupled to Strings
k1
m1
k2
m2
kn
mn
k1
m1
k2
m2
kn
mn
Figure 1.3: Linear chains of masses and springs. The top figure shows the case where all n
masses can move, and the bottom shows the case where the last mass mn is rigidly attached to
the wall [16].
where I is the unit matrix, we can evaluate both sides of this relation at (n, n)
position to obtain
det(λiIn−1) = x2
ni
n
k=1,k=i
(λi − λk). (1.80)
In this relation xni is the last entry of xi, and An−1 denotes the (n − 1)(n − 1)
matrix. Solving for x2
ni, we get
|x1i|2
=
n−1
k=1,k=i(μk − λi)
n
k=1,k=i(λk − λi)
, (1.81)
and this expression gives us the matrix elements of A. Thus we conclude that if
the eigenvalues μ1, μ2, · · · , μn−1 are known then we know the first entries in the
vectors x1, x2, · · · , xn. If we know λ1, λ2, · · · , λn and xn1, xn2, · · · , xnn we have
2n + 1 quantities. Because xn is normalized xn s are not independent.
The Jacobi Matrix — The Jacobi matrix is a square n×n matrix with just three
diagonal elements being nonzero. A typical Jacobi matrix has the following form
[18], [19]
Kn =
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣
a1 b1 0 0 · · · 0
b1 a2 b2 0 · · · 0
0 b2 a3 b3 · · · 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
· · · · · · · · · · · · · · · bn−1
· · · · · · · · · · · · bn−1 an
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
(1.82)
Since changing the sign of bj has no effect on the eigenvalues, we can assume that
b1, b2, · · · , bn−1 are all negative. We can collect the n column vector equations
(A − λiI)xi = 0 into one matrix equation
AX = XΛ, (1.83)](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-35-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 19
An Introduction to Inverse Problems in Physics 19
where
X = [x1, x2, · · · , xn]. (1.84)
Let us now denote the transpose of X by U, i.e. XT
= U, then AUT
= UT
Λ and
on transposing we have
UA = ΛU, (1.85)
or in particular if A = J is a Jacobi matrix we have
[u1, u2, · · · , un]J = Λ[u1, u2, · · · , un], (1.86)
where U = [u1, u2, · · · , un]. Now let us consider a method for determining the
elements aj, bj of this J matrix from the set of eigenvalues λj and μj. There is a
wealth of literature on the inversion of symmetric Jacobi matrices and the reader
is referred to the following papers where in these works, further references can be
found to other works [16]–[25]. Let us consider a simple way of this inversion.
The first column of Eq. (1.86) is
a1u1 + b1u2 = Λu1. (1.87)
Since u1, u2, · · · , un form an orthogonal set, from (1.87) we find
a1 = uT
1 · Λu1. (1.88)
To find b1 we write (1.87) as
b1u2 = (Λ − a1)u1 = z2. (1.89)
The vector z2 is known from its definition and u2 is a unit vector. Thus
b2
1 = z2
2
, b1 = z2 , (1.90)
and
u2 =
1
b1
z2. (1.91)
Now let us consider the second column of (1.86);
b1u1 + a2u2 + b2u3 = Λu2. (1.92)
Again we multiply this equation by uT
2 , using the fact that uT
2 · u1 = uT
2 · u3 = 0
and get
a2 = uT
2 Λu2, (1.93)
and
b2u3 = z3 = (Λ − a2)u2 − b1u1. (1.94)
Since every term in the right-hand side of this equation is known therefore z3 can
be found. From the last relation we deduce that
b2 = z3 , u3 =
z3
b2
. (1.95)](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-36-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 20
20 Masses Coupled to Strings
We continue this process to find all ai s and bi s.
Lanczos Algorithm — The procedure of finding ai and bi, i = 1, 2 · · · n is called
Lanczos algorithm [20]. The original Lanczos algorithm concerns with the solution
of the following problem:
Given a symmetric matrix A and a normalized vector x1 (i.e. xT
1 ·x1 = 1), compute
a Jacobian matrix J and an orthogonal matrix X = [x1, x2, · · · , xn] such that
A = XJXT
. Variants of Lanczos’s algorithm, and different ways of finding the
elements in a Jacobi matrix from the eigenvalues are given in Refs. [21]–[28].
Eigenfrequencies of the Mass-Spring Chain — Now let us consider the prob-
lem of n masses m1, m2, · · · , mn connected by springs with spring constants k1,
k2, · · · , kn to each other. For the system composed of masses and springs, Fig. 1.3
(top), we can write this eigenvalue equation as
⎡
⎢
⎢
⎢
⎢
⎣
(k1 + k2) −k2 . . . 0
−k2 (k2 + k3) . . . 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . −kn
. . . . . . −kn kn
⎤
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎣
x1
x2
. . .
xn−1
xn
⎤
⎥
⎥
⎥
⎥
⎦
= λ
⎡
⎢
⎢
⎢
⎢
⎣
m1 0 · · · 0
0 m2 · · · 0
. . . . . . . . . . . . . . . .
· · · · · · mn−1 0
· · · · · · · · · mn
⎤
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎣
x1
x2
. . .
xn−1
xn
⎤
⎥
⎥
⎥
⎥
⎦
, (1.96)
where λ = ω2
, and ω1, ω2 · · · ωn are the natural frequencies of the system.
Here we want to find the characteristic frequencies ωi s of such a system under
two different boundary conditions: In the first one the mass m1 is attached with a
spring k1 to a rigid wall, but mn is free to move, as shown in Fig. 1.3 at the top.
The second is when the last mass mn is fixed by attaching it to the wall, as shown
in the lower figure Fig. 1.3. The equations of motion for the system at the top can
be written as
(K − λM)u = 0, λ = ω2
, (1.97)
where K is the Jacobi matrix with the elements related to the spring constants ki
as shown in (1.97). If we want to write (1.97) in the standard form of an eigenvalue
equation we introduce the matrix D;
M = D2
, D−1
KD−1
= J, Du = x, (1.98)
and then (1.97) can be written as
(J − λI)x = 0, (1.99)
where I is a unit n × n matrix. Using Lanczos algorithm we find the elements of
J, and then we try to find matrices for K and M from J [16], [28]. As we can see](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-37-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 21
An Introduction to Inverse Problems in Physics 21
from (1.98) the matrices K and A are related to each other K = DJD.
By inspection we see that
K[1, 1, · · · , 1] = k1e1, (1.100)
where e1 = [1, 0, · · · , 0]. Thus
DAD[1, 1, · · · , 1] = k1e1, (1.101)
and therefore
A[d1, d2, · · · , dn] = D−1
k1e1 = d−1
1 k1e1. (1.102)
Since the off diagonal elements of the K matrix are negative, Eq. (1.96), when we
solve
K[x1, x2, · · · , xn] = e1, (1.103)
then the solution x1, x2, · · · , xn will have positive components. The solution of the
last equation in (1.98) is d = cx, where c can be found from the total mass of the
system
m =
n
i=1
mi =
n
i=1
d2
i = d 2
= c2
x 2
. (1.104)
Therefore, knowing m we can calculate x 2
then find c and d, and hence the
masses mi = d2
i . Having found di s we can compute K = DAD and find the spring
constants k1, k2, · · · , kn [15], [16].
An Analytically Solvable Spring-Mass Problem — Let us consider the special
case of Eq. (1.96) when n = 2. Then the eigenvalues are the roots of the quadratic
equation
k1 + k2 − λm1 −k2
−k2 k2 − λm2
= 0. (1.105)
In the direct problem we know k1, k2, m1 and m2, and the eigenvalue equation
(1.105) can be solved for λ1 and λ2, the two roots of this equation. In the inverse
problem for this case we are given the eigenvalues plus m1 and m2 and from (1.105)
we obtain k1 in terms of k2:
k1 =
1
k2
(m1m2λ1λ2), (1.106)
where k2 is the root of the quadratic equation
m1 + m2
m1m2
k2
2 − (λ1 + λ2)k2 + m2λ1λ2 = 0. (1.107)
This equation has a real positive root if and only if
(λ1 − λ2)2
4m2
m1
λ1λ2. (1.108)
This result indicates that, in general, for the accuracy of the numerical solution of
the inverse problem the eigenvalues should not be close to each other [16].](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-38-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 22
22 Non-exponential Decay in Linear Chain
Some Other Solvable Mass-Spring Systems — There are a number of other
interesting problems closely related to the mass-spring system that we have studied.
Among them is the question of recovery of masses and the spring constants of a
finite system when we have the natural frequencies of N masses connected with
springs with different spring constants plus the frequencies of the system when it
is perturbed by modifying one mass and adding one spring to the system [29].
Another inverse problem of interest is the determination of a finite element model
of longitudinally vibrating rod with one end fixed and the other end supported on
a spring. This case has been studied by Wei and Dai [30].
Let us note that as long as the number n of masses (and springs) are finite
the motion is reversible (see e.g. Ref. [31]).
1.7 Direct Problem of Non-exponential and of
Exponential Decays in a Linear Chain
One of the natural generalizations of the mass-spring system is to study mechanical
models where both the number of particles and springs go to infinity. The exact
solution of such a problem is if interest, since the motion becomes irreversible i.e.
once the system starts from a given configuration it will not come back to its original
state.
First let us study a very special type of inverse problem for this case. Consider
a linear chain of particles each of mass Mj, attached to each other by means of
springs with spring constants Kj. Assume that initially all of the particles are at
rest in their equilibrium position except one particle which is displaced and released
with zero initial velocity. How we should choose different masses and different spring
constants so that the motion of this particle decays exponentially?
The Decay of Motion in the Schrödiger Chain — Before discussing this ques-
tion, let us investigate the solution of the direct problem for the case where masses
are all equal and the connecting springs all have the the same spring constant, as-
suming that there are infinite number of masses and springs. This problem was first
studied and solved by Schrödinger [32]. If we denote the mass of each particle by
m and the spring constant by k, then the motion of the jth particle in the chain is
given by
mẍj = k(ξj+1 + ξj−1 − 2ξj), j = 0, ±1, ±2, · · · , (1.109)
where ξj is the displacement of the jth particle from the equilibrium and k is the
spring constant which we write in terms of another constant ν with the dimension
of wave number,
k =
1
4
mν2
. (1.110)
Let assume that initially the central particle j = 0 is displaced by a distance A, and
then released with zero initial velocity, while all of the other particles are initially](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-39-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 23
An Introduction to Inverse Problems in Physics 23
at rest in their equilibrium position. Thus at t = 0 we have the initial conditions
ξ0(t = 0) = A, ξj(t = 0) = 0, j = 0, (1.111)
and
˙
ξ0(t = 0) = 0, j = 0, ±1, ±2 · · · . (1.112)
A method for solving the system of equations (1.109) is to construct the generating
function G(z, t) which we define by [33], [34]
G(z, t) =
+∞
j=−∞
ξj(t)z2j
. (1.113)
We multiply (1.109) by zj
, we sum over all j s, and use (1.113) to obtain the
following equation for G(z, t):
d2
G(z, t)
dt2
=
1
4
ν2
1
z
− z
2
G(z, t). (1.114)
We can find the initial conditions for this differential equation from (1.111)–(1.113).
G(z, 0) = A,
dG(z, t)
dt
t=0
= 0. (1.115)
With these conditions we find the solution of (1.113) to be
G(z, t) = A cosh
1
2
νt
z −
1
z
. (1.116)
This G(z, t) is the generating function for the Bessel function of even order [35];
cosh
1
2
νt
z −
1
z
=
+∞
j=−∞
J2j(νt)z2j
, (1.117)
and therefore
ξj(t) = AJ2j(νt). (1.118)
Thus in this model the energy of the central particle is lost to the other particles in
the system, and the rate of energy loss is
dE0
dt
t→∞
→
mν2
A2
2πt
cos(2νt), (1.119)
i.e. the energy transfer between this particle and the rest of the system has an
oscillatory behaviour and goes to zero as t−1
.](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-40-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 24
24 Non-uniform Chain
1.8 Inverse Problem of Dynamics for a
Non-uniform Chain
Let us first consider the propagation of a disturbance along a non-uniform chain.
If Mj denotes the mass of the jth particle in the chain and the spring constant
between the particle j and j + 1 is Kj, then we have the set of equations of motion
Mjẍj = Kj(xj+1 − xj) + Kj−1(xj−1 − xj), j = 1, 2 · · · N, (1.120)
where xj is the displacement of the jth particle from its position of equilibrium.
Following the work of Dyson we introduce the variable ξj s by [36]
ξj = M
1
2
j xj, (1.121)
where all Mj s are greater than zero. Also let us introduce the set of constants
ω1, ω2, · · · , ω2N−2 by the following relations
ω2
2j−1 =
Kj
Mj
, ω2
2j =
Kj
Mj+1
, (1.122)
then the equations of motion take the form
¨
ξj = (ω2j−1ω2j)ξj+1 + (ω2j−3ω2j−2)ξj−1 −
ω2
2j−1 + ω2
2j−2
ξj. (1.123)
Now we define the variables z1, z2, · · · , zN−1 by
ż = ω2jξj+1 − ω2j−1ξj, (1.124)
and then we can rewrite (1.123) as
˙
ξj = ω2j−1zj − ω2j−2zj−1. (1.125)
By combining (1.124) and (1.125) we arrive at a single set of first order equations:
Ẋj = ωjXj+1 − ωj−1Xj−1, (1.126)
where X1, X2, · · · , X2N−1 are defined by
X2j−1 = ξj, X2j = zj. (1.127)
The characteristic frequencies νj of the chain are the eigenvalues of a (2N − 1) ×
(2N − 1) matrix Λ whose elements are given by
Λj+1,j = −Λj,j+1 = iωj. (1.128)
Here we want to choose the parameters Mj s and Kj s in such a way that in the
limit of N → ∞ we can find an exact solution for the motion of the particles, i.e. a
solvable many-body problem. Let us consider a model where
ωj = jλ, (1.129)](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-41-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 25
An Introduction to Inverse Problems in Physics 25
where λ is a positive constant. For this choice of ωj, the ratio of the masses of two
neighbouring particles according to Eq. (1.122) is
Mj+1
Mj
=
(2j − 1)2
4j2
, j = 1, 2, · · · . (1.130)
By substituting (1.121), (1.129) and (1.130) in (1.120), we find the equation of
motion for xj;
ẍj = λ2
(2j − 1)2
(xj+1 − xj) + 4λ2
(j − 1)2
(xj−1 − xj). (1.131)
Assuming that initially only the particle j = 1 is displaced by one unit of length
and with no initial velocity, while all other particles in the chain are at rest with
no initial displacement, i.e.
x1(0) = 1, ẋj(0) = 0, j = 1, 2, · · · , (1.132)
and
xj(0) = 0, j = 1, (1.133)
then we can write (1.131) with the initial conditions (1.132), (1.133) as a set of
equations
Ẋj = λ[jXj+1 − (j − 1)Xj−1], (1.134)
subject to the initial conditions
X1(0) = M
1
2
1 , Xj(0) = 0, j = 1. (1.135)
We can solve this problem by the same method of constructing the generating
function that we discussed earlier. Let us assume that the generating function is of
the form
G(z, t) =
∞
j=1
Xj(t)zj−1
, (1.136)
then by differentiating G(z, t) we find
z
∂
∂z
(zG(z, t)) =
∞
j=1
(j − 1)Xj−1zj−1
, (1.137)
and
∂
∂z
G(z, t) =
∞
j=1
jXj+1zj−1
. (1.138)
Now if we multiply (1.134) by zj−1
and sum over all j values we find the following
first order partial differential equation
∂G(z, t)
∂t
= λ
∂G(z, t)
∂z
− z
∂
∂z
(zG(z, t))
. (1.139)](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-42-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 27
An Introduction to Inverse Problems in Physics 27
which shows that for times t 1
λ the displacement decays exponentially in time [37].
Since we have found the exact solution of motion according to the second law of
dynamics, the time-reversed motion is also a solution of the problem. The damped
motion of the first particle results from its coupling to a many-particle system.
1.9 Direct and Inverse Problems of Analytical
Dynamics
In classical mechanics the Newton second law of motion establishes a relation be-
tween the motion of the bodies and the forces acting between them. Thus at first
it seems that the direct problem should be the derivation (or construction) of the
Lagrangian and the Hamiltonian from the second and the third laws of motion.
Historically this was the direction of the development of analytical mechanics taken
by Lagrange, Hamilton, Poisson and Jacobi. They showed how useful the analytical
formulation of the classical mechanics (i.e. Lagrangian and Hamiltonian formula-
tions) can be, both as a concise way of writing different mechanical problems and
also as a very powerful method of finding approximate and exact solutions [38]–
[41].
It is also possible to start with a Hamiltonian (or Lagrangian) and obtain
the equations of motion using the method of calculus of variations. That is, if we
choose the classical Hamiltonian function, H(pj, qj, t), as the basic mathematical
way of expressing a given motion, then the direct method shows that we can obtain
a unique set of equations from the canonical equations of motion,
dqj(t)
dt
=
∂H
∂pj
, and
dpj(t)
dt
= −
∂H
∂qj
, j = 1, 2 · · · N. (1.145)
In these relations qj(t) is the coordinate of the jth particle at the time t, pj(t) is its
canonical momentum conjugate to qj(t), and N is the number of degrees of freedom
of the system. By eliminating pj(t) s between the two equations in (1.145) we find
the equations of motion for q1(t), q2(t), · · · , qn(t). Here we expect to get a unique
set of differential equations for qj(t) s. Thus the direct problem is well-defined for
those problems for which the Hamiltonians (or Lagrangians) are known.
For the direct problem we start with Eqs. (1.145) and determine the motion in
2N dimensional phase space p1(t) · · · pN (t); q1(t) · · · qN (t), or we find the equations
of motion in the coordinate space:
H(pj, qj, t) →
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
mjq̈j = Fj(q1, · · · , qN ; p1, · · · , pN )
mjq̇j = Qj(q1, · · · , qN ; p1, · · · , pN )
mjṗj = Pj(q1, · · · , qN ; p1, · · · , pN )
. (1.146)
Thus if we start with H(pj, qj, t), find the equations of motion either in the co-
ordinate space or in the phase space, and then use these equations to obtain the](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-44-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 28
28 Direct and Inverse Problems of Analytical Dynamics
first integrals, i.e. to solve the inverse problem. We will not always recover just
H(pj, qj, t), but among the pq-equivalent Hamiltonian, there will be one Hamilto-
nian that we started with. That is, we have a set of Hamiltonians giving us the
correct equations of motion in coordinate space;
mjq̈j = Fj(q1, · · · , qN ; p1, · · · , pN ; t) → K(pj, qj; t), (q-equivalent Hamiltonians),
(1.147)
and another, smaller set generating the motion in phase space;
mjq̇j = Qj(q1, · · · , qN ; p1, · · · , pN ; t)
mjṗj = Pj(q1, · · · , qN ; p1, · · · , pN ; t),
→ H1(pj, qj), (1.148)
the latter is pq equivalent Hamiltonian. As a very simple example, let us consider
the Hamiltonian for the damped motion of a single particle moving along the x axis:
H =
p2
2m
e−λt
. (1.149)
From this H we obtain the equation of motion in the phase space
dp
dt
=
∂H
∂q
= 0,
dq
dt
=
p
m
e−λt
, (1.150)
or [43]
d2
q
dt2
= −λ
dq
dt
. (1.151)
This equation of motion can also be found from the more complicated Hamiltonian
H1, where
H1 =
1
3mp0
(2p0p)
3
2 e−λt
, (1.152)
and p0 is a constant.
Nonuniqueness of the Hamiltonian for Conservative Many-particle Sys-
tem — Starting with the equations of motion in the coordinate space for a con-
servative system we find nonuniqueness in determination of the Lagrangian and
the Hamiltonian functions. For instance let us consider a system of N interacting
particles with the equations of motion
mj
d2
rj
dt2
= −∇jV (r1, · · · , rN ), j = 1, 2, · · · , N. (1.153)
If we sum (1.153) over all j s we find that
d
dt
N
j=1
mjvj = 0, (1.154)](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-45-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 30
30 Equations of Motion and the Lagrangian
But first let us examine the conditions for the existence of a Lagrangian function
and how it can be constructed [38]–[41].
The laws governing the motion of a dynamical system can be formulated ac-
cording to Hamilton’s principle and this principle states that there is a function
L (x1, · · · xN ; ẋ1 · · · ẋN , t) and that this L satisfies the following condition [2]:
If the system at the instant t1 has the coordinates x
(1)
i and at t2 has the co-
ordinates x
(2)
i then this condition implies that the functional differential δS defined
by
δS = δ
t2
t1
L (xi, ẋi, t) dt =
t2
t1
N
j=1
δL (xi, ẋi, t)
δxj
δxj(t)dt
=
t2
t1
j,k
μjk (xi, ẋi, t)
ẍk −
1
m
fk (xi, ẋi, t)
δxj(t)dt,
j = 1, 2, · · · N, (1.160)
must vanish identically for any continuous variation of the path consistent with the
requirement
δxj (t1) = δxj (t2) = 0. (1.161)
In Eq. (1.160) μjk (xi, ẋi, t) is a positive definite matrix which may be viewed
as an integrating factor [38]–[42]. Thus the requirement of the extremum of action
S implies the vanishing of the integrand between any two arbitrary instances t1 and
t2. Since we have imposed the condition of positive definiteness on the integrating
factor μjk and since the variation δxj(t) is arbitrary (apart from the continuity
condition and the variational conditions of (1.161)), therefore we conclude that
ẍk(t) =
1
m
fk (xi, ẋi, t) , k = 1, 2, · · · N, (1.162)
which is Newton’s second law of motion.
The Lagrangian function determined from the requirement of δS ≡ 0 is not
unique. We can add the total time derivative of any function Φ (x, ẋ, t) to the
Lagrangian without affecting the resulting equations of motion [43]. For the special
case of μjk = δjk we find the Lagrangian to be the same as that given by (1.156).
The second variation of S with respect to δxj(t) and δxk(t
) must satisfy the
identity:
δ2
S
δxj (t) δxk (t)
=
δ2
S
δxk (t) δxj (t)
. (1.163)
By imposing this condition on (1.160) we find the following set of equations which
μjk s have to satisfy:
μjk = μkj, (1.164)
∂μkj
∂ẋl
=
∂μjl
∂ẋk
=
∂μlk
∂ẋj
, (1.165)
D̂μjk +
1
2m
l
μjl
∂fl
∂ẋk
+ μkl
∂fl
∂ẋj
= 0, (1.166)](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-47-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 31
An Introduction to Inverse Problems in Physics 31
and
D̂
k
μik
∂fk
∂ẋj
− μjk
∂fk
∂ẋi
− 2
k
μik
∂fk
∂ẋj
− μjk
∂fk
∂ẋi
= 0, (1.167)
where in Eqs. (1.166) and (1.167) D̂ denotes the following differential operator
D̂ =
∂
∂t
+
k
ẋk
∂
∂xk
+
1
m
fk (xi, ẋi, t)
∂
∂ẋk
. (1.168)
These are the Helmholtz conditions [38], [39].
We can construct the Lagrangian from the equations of motion
mẍk = fk (xi, ẋi, t) , k = 1, 2, · · · N, (1.169)
by solving a linear partial differential equation. To this end we first write the
following identity:
δL
δxk
=
∂L
∂xk
−
d
dt
∂L
∂ẋk
≡
∂L
∂xk
−
∂2
L
∂t∂ẋk
−
j
ẋj
∂2
L
∂xj∂ẋk
−
j
ẍj
∂2
L
∂ẋj∂ẋk
= 0, (1.170)
and then we substitute for ẍ from the equation of motion to obtain
∂L
∂xk
−
∂2
L
∂t∂ẋk
−
j
ẋj
∂2
L
∂xj∂ẋk
−
1
m
j
∂2
L
∂ẋj∂ẋk
fj (xi, ẋi, t) = 0, k = 1 2, · · · N.
(1.171)
This equation together with the set of equations (1.164)–(1.167) determine both
L(x, ẋ, t) and μ(x, ẋ, t).
For another method of constructing Lagrangians and Hamiltonians for non-
linear systems the reader is referred to a paper by Musielak et al. [44].
In the case of a one-dimensional motion Eq. (1.171) simplifies and later we
will use it to find the Lagrangian for simple damped systems. But there are other
ways of constructing the Lagrangian for motions confined to one dimension [45],
[46]. For instance we can start with the equation of motion
mẍ = f (x, ẋ) , (1.172)
and write the Lagrangian for this motion as
L (x, ẋ) = ẋ
ẋ
1
v2
G(v, x)dv, (1.173)
where G(v, x) is a function to be determined later. By substituting L (x, ẋ) in the
Euler–Lagrange equation we find
d
dt
∂L
∂ẋ
−
∂L
∂x
=
ẍ
ẋ
∂G (ẋ, x)
∂ẋ
+
∂G (ẋ, x)
∂x
≡ mẍ − f (ẋ, x) . (1.174)](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-48-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 32
32 Equations of Motion and the Lagrangian
From Eq. (1.174) it follows that G (ẋ, x) is the solution of the following differential
equation
1
mẋ
f (x, ẋ)
∂
∂ẋ
G (ẋ, x) = −
∂
∂x
G (ẋ, x) , (1.175)
provided that
1
ẋ
∂
∂ẋ
G (ẋ, x) = 0. (1.176)
Now let us consider two examples using this method of construction of the
Lagrangian. When the equation of motion for a damped system is of the form
ẍ +
ẋ
dg(ẋ)
dẋ
dV (x)
dx
= 0, (1.177)
then the Lagrangian which is found by solving Eq. (1.175) is
L (x, ẋ) = ẋ
ẋ
1
v2
F [z(x, v)] dv, (1.178)
where F is an arbitrary function of its argument and
z(v, x) = g(v) + V (x). (1.179)
However the condition
1
v
dg(v)
dv
dF(z)
dz
= 0, (1.180)
must be satisfied. Thus for the very simple case of ẍ + λẋ = 0 and for F(z) = z we
find
L (x, ẋ) = ẋ ln ẋ − λx. (1.181)
As a second example let us consider the damped harmonic oscillator
ẍ + λẋ + ω2
0x = 0. (1.182)
By calculating G (ẋ, x) we find that G (ẋ, x) = F(z), where now
z =
1
2
x +
λ
2 − iω
ẋ
ν∗
x +
λ
2 + iω
ẋ
ν = z∗
, (1.183)
with z∗
being the complex conjugate of z and
ν2
=
λ
2 + iω
λ
2 − iω
, (1.184)
with ω2
= ω2
0 − λ2
4 .
These q-equivalent Hamiltonians are acceptable as generators of motion in an-
alytical dynamics, and depending on the integrating factors used, there are infinitely
many Hamiltonians. However as we will see in Sec. 3.3 they are not admissible for](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-49-320.jpg)
![May 12, 2020 11:33 book-961x669 11860-main page 33
An Introduction to Inverse Problems in Physics 33
quantizing a particular system. Since their classical phase space trajectories are very
different from those of the standard Hamiltonians they may not be also admissible
even in classical statistical mechanics.
The Inverse Problem for Liouville Equation — The inverse problem of the
classical Liouville equation can be described as the problem of finding a time-
dependent Hamiltonian H which generates a given phase space density ρ, and both
the Hamiltonian H and the density ρ satisfy the same Liouville equation [6]. Frisch
and collabortors have shown that, subject to some physically well-defined subsidiary
conditions, the solution exists and is unique, when ρ is chosen to be the solution of
the standard Fokker–Planck equation when the particles are free, are in a constant
external field, or are harmonically bound.
Let us consider a system of particles and denote the coordinate and the canon-
ical momentum of the jth particle by qj and pj respectively. As we have seen ear-
lier we can derive the equation of motion of these particles from a time-dependent
Hamiltonian H(q, p, t) ≡ H(qi, pi, t), Eq. (1.145). We assume the following initial
conditions for the first order equations of motion:
qj(0) = qj, pj(0) = pj, H(0) = H0. (1.185)
In classical statistical mechanics one wants to find the time-dependent density
ρ(q, p, t) ≡ ρ (1.186)
corresponding to a given H = H(q, p, t) which satisfies Liouville’s equation
∂ρ
∂t
+ {ρ, H} =
∂ρ
∂t
+
j
∂ρ
∂qj
∂H
∂pj
−
∂ρ
∂pj
∂H
∂qj
= 0, (1.187)
subject to the initial condition
ρ(0) = ρ0(q, p). (1.188)
Now we define the dynamical inverse problem for the Liouville equation by the fol-
lowing procedure [6]:
(1) We assume a marginally consistent, time-dependent point process given by
ρ(q, p, t) and its marginal distributions.
(2) We try to find a “Hamiltonian” function H(q, p, t) corresponding to the ρ given
in (1.186) which is a solution of Eq. (1.187).
By considering (1.187) as a linear partial differential equation for H = H(q, p, t),
we have the characteristic equations for H
dpj
− ∂ρ
∂qj
=
dqj
∂ρ
∂pj
=
dH
∂ρ
∂t
. (1.189)
These equations are not identical with Hamilton’s canonical equations (1.145) for
H except when ρ is judiciously chosen [6]. Solution of (1.189) only determines the
general solution of (1.187). If one specifies the initial condition on H and in addition
provides information about H on a characteristic strip or singular region of phase
space, then one can show the existence of the Cauchy problem for H with very weak
conditions on the continuously differentiable density ρ [48].](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-50-320.jpg)
![May 14, 2020 8:55 book-961x669 11860-main page 34
34 References
1.11 Langevin and Fokker–Planck Equations
If a particle of mass m is immersed in a fluid with the friction coefficient γ, its
equation of motion (in one dimension) is given by
dv
dt
+ γv = −K(x) + A(t). (1.190)
In this equation v is the speed of the particle, x is its position, K(x) is the acceler-
ation due to mechanical forces and A(t) is the acceleration due to random forces.
The Fokker–Planck equation for the distribution in phase space of the single
particle, P(x, v, t) is given by
∂P
∂t
+ v
∂P
∂x
− K(x)
∂P
∂v
= γ
∂(vP)
∂v
+ Dγ2 ∂2
P
∂v2
. (1.191)
The solution of (1.191) subject to the boundary condition
P(x, v, 0) = δ(x − x0)δ(v − v0), (1.192)
can be obtained for the following forms of K(x):
K(x) =
⎧
⎪
⎨
⎪
⎩
0 free particle
F0 a constant force
ω2
x harmononically bound particle
. (1.193)
Solutions of the Fokker–Planck equation for these cases are given in [6], and the gen-
eral Hamiltonian for a the motion of a particle in a viscous medium with frictional
force βv is discussed in detail in [49]–[51].
References
[1] I. Newton, The Principia: Mathematical Principles of Natural Philosophy,
(University of California Press, 1999).
[2] L.D. Landau and E.M. Lifshitz, Mechanics, (Pergamon Press. Oxford, 1960),
Chap. III.
[3] N.H. Abel, J. Reine Angew. Math. 1, 13 (1826).
[4] A.H. Carter, Am. J. Phys. 68, 698 (2000).
[5] M. Bocher, An Introduction to the Study of Integral Equations, Cambridge
Tracts in Mathematics and Mathematical Physics, No. 10 (Hefner, New York,
1960).](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-51-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 35
An Introduction to Inverse Problems in Physics 35
[6] R. Gorenflo and S. Vessella, Abel Integral Equations, Analysis and Applications,
(Springer-Verlag, Berlin, 1991).
[7] I.S. Gradshteyn and I.M. Ryzhik, Tables of Integrals, Series, and the Products,
(Academic Press, New York, 1965) p. 875.
[8] G. Arfken, Mathematical Methods for Physicists, Third Edition, (Academic
Press, New York, 1985) p. 875.
[9] E.T. Whittaker, A Treatise on the Analytical Dynamics of Particles and Rigid
Bodies, Fourth Edition (Cambridge University Press, London 1960).
[10] H. Goldstein, Classical Mechanics, (Addison-Wesley, Reading, 1956).
[11] J.F. Marko and M. Razavy, Lettere Al Nuovo Cimento, 40, 533, 1984.
[12] J.B. Keller, Am. Math. Monthly, 83, 107 (1976).
[13] O.B. Firsov, Zh. Eksp. Teor. Fiz. 24, 279 (1953). Firsov’s solution is describled
in L.D. Landau and E.M. Lifshitz, Mechanics, (Pergamon Press, Oxford, 1960)
p. 52.
[14] J.B. Keller, I. Kay and J. Shmoys, Phys. Rev. 557, 102 (1956).
[15] M.T. Chu and G.H. Golub, Inverse Eigenvalue Problems: Theory, Algorithm
and Applications, (Oxford University Press, London, 2005).
[16] G.M.L. Gladwell, Inverse Problems in Vibration, Second Edition, Volume 119
of Solid Mechanics and Its Applications, (Dordrecht, Kluwer, 2004).
[17] See for instance The Symmetric Eigenvalue Problem, (SIAM, Classics in Ap-
plied Mathematics, 1988).
[18] D. Belkić, in Advances in Quantum Chemistry V. 145, Chapter 4 (2011).
[19] I.M. Gel’fand and S.V. Fornin, Calculus of Variation, translated by R.A. Sil-
verman, (Dover, New York, 1991).
[20] C. Lanczos, J. Res. Natl. Bur. Stand. 45, 255 (1950), ibid. 49, 33 (1958).
[21] For other methods of inversion of a Jacobi matrix see D.N. Ghosh Roy. Methods
of Inverse Problems in Physics, (CRC Press Boca Raton, 1991) Chapter 11.
[22] G. Meurant, SIAM J. Matrix Anal. Appl. 13, 707 (1992).
[23] X. Wu and E. Jiang, J. Shanghai University, 11, 27 (2007).
[24] G.M.I. Gladwell and N.B. Williams, Inverse Problems, 4, 1013 (1988).
[25] R.A. Usmani, Computer Mat. Appl. 27, 59 (1994).
[26] J.H. Wilkinson, The Algebraic Eigenvalue Problem, (Oxford University Press,
Oxford, 1965) p. 388.](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-52-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 36
36 References
[27] A.S. Housholder, The Theory of Matrices in Mathematical Analysis, (Blaisdell,
New York, 1964) p. 254.
[28] G.H. Golub and G. Meurant, Matrices, Moments and Quadrature with Appli-
cations, (Princeton University Press, Princeton, 2010), Chapter 4.
[29] R. del Rio and M. Kudryavtsev, Inverse Problems, 28, 055007 (2012).
[30] Y. Wei and H. Dai, J. Comp. Appl. Math, 300, 172 (2016).
[31] H.J. Kreuzer, Nonequilibrium Thermodynamics and Its Statistical Foundations,
(Oxford University Press, Oxford, 1981).
[32] E. Schrödinger, Ann. Phys. (Leipzig), 44, 916 (1914).
[33] R. Bellman and K.L. Cook, Differential-Difference Equtions, (Academic Press,
New York, 1963).
[34] M. Razavy, Can. J. Phys. 57, 1731 (1979).
[35] I.S. Gradshtetyn and I.M. Ryzhik, Table of Integrals, Series and Products,
(Academic Press, New York, 1965) p. 973.
[36] F.J. Dyson, Phys. Rev. 92, 1331 (1953).
[37] M. Razavy, Can. J. Phys. 58, 1019 (1980).
[38] P. Havas, Nuovo Cimento Supp. 5, 363 (1957).
[39] H. Helmholtz, J. Reine Angew. Math. 100, 137 (1887).
[40] R.M. Santilli, Foundations of Theoretical Mechanics: The Inverse Problem in
Newtonian Mechanics, (Springer-Verlag, New York, 1978).
[41] V. Dodonov, V.I. Man’ko and V.D. Skarzhinsky, Had. J. 4, 1734 (1981).
[42] C. Lanczos, The Variational Principles of Mechanics, Fourth Edition, (Dover,
New York, 1986).
[43] For a number of similar examples see M. Razavy Classical and Quantum Dis-
sipative Systems, Second Edition (World Scientific, Singapore, 2017).
[44] Z.E. Musielak, D. Roy and L.D. Swift, Chaos, Solitons Fractals, 38, 894
(2008).
[45] J.A. Kobussen, Acta Phys. Austriaca, 59, 293 (1979).
[46] S. Okubo, Phys. Rev. A 23, 2776 (1981).
[47] H.L. Frisch, G. Forgacs and S.T. Chui, Phys. Rev. A 20, 561 (1979).
[48] H. Bateman, Differential Equations, (Chelsea, New York,1966).](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-53-320.jpg)
![May 5, 2020 14:13 book-961x669 11860-main page 37
An Introduction to Inverse Problems in Physics 37
[49] M. Razavy, Can. J. Phys. 56, 311 (1978).
[50] M. Razavy, Can. J. Phys. 56. 1372 (1978).
[51] M. Razavy, Classical and Quantum Disiipative Motion, Second Edition, (World
Scientific, Singapore, 2017).](https://image.slidesharecdn.com/17405257-250601100122-277cb68e/85/An-Introduction-To-Inverse-Problems-In-Physics-Mohsen-Razavy-54-320.jpg)
An Introduction To Inverse Problems In Physics Mohsen Razavy
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- 6. An Introduction to in Inverse Problems Physics
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- 8. NEW JERSEY • LONDON • SINGAPORE • BEIJING • SHANGHAI • HONG KONG • TAIPEI • CHENNAI • TOKYO World Scientific An Introduction to in Inverse Problems Physics M Razavy University of Alberta, Canada
- 9. Published by :RUOG6FLHQWL¿F3XEOLVKLQJR3WH/WG 7RK7XFN/LQN6LQJDSRUH 86$R৽FH:DUUHQ6WUHHW6XLWH+DFNHQVDFN1- 8.R৽FH6KHOWRQ6WUHHWRYHQW*DUGHQ/RQGRQ:++( Library of Congress Cataloging-in-Publication Data 1DPHV5D]DY0RKVHQDXWKRU 7LWOH$QLQWURGXFWLRQWRLQYHUVHSUREOHPVLQSKVLFV05D]DY8QLYHUVLWRI$OEHUWDDQDGD 'HVFULSWLRQ1HZ-HUVH:RUOG6FLHQWL¿F@_,QFOXGHVELEOLRJUDSKLFDOUHIHUHQFHVDQGLQGH[ ,GHQWL¿HUV/1 SULQW _/1 HERRN _,6%1 KDUGFRYHU _ ,6%1 HERRN _,6%1 HERRNRWKHU 6XEMHFWV/6+,QYHUVHSUREOHPV 'L൵HUHQWLDOHTXDWLRQV _'L൵HUHQWLDOHTXDWLRQV1XPHULFDOVROXWLRQV_ 0DWKHPDWLFDOSKVLFV ODVVL¿FDWLRQ/4'5 SULQW _/4' HERRN _''GF /UHFRUGDYDLODEOHDWKWWSVOFFQORFJRY /HERRNUHFRUGDYDLODEOHDWKWWSVOFFQORFJRY British Library Cataloguing-in-Publication Data $FDWDORJXHUHFRUGIRUWKLVERRNLVDYDLODEOHIURPWKH%ULWLVK/LEUDU RSULJKW‹E:RUOG6FLHQWL¿F3XEOLVKLQJR3WH/WG $OOULJKWVUHVHUYHG7KLVERRNRUSDUWVWKHUHRIPDQRWEHUHSURGXFHGLQDQIRUPRUEDQPHDQVHOHFWURQLFRU PHFKDQLFDOLQFOXGLQJSKRWRFRSLQJUHFRUGLQJRUDQLQIRUPDWLRQVWRUDJHDQGUHWULHYDOVVWHPQRZNQRZQRUWR EHLQYHQWHGZLWKRXWZULWWHQSHUPLVVLRQIURPWKHSXEOLVKHU )RUSKRWRFRSLQJRIPDWHULDOLQWKLVYROXPHSOHDVHSDDFRSLQJIHHWKURXJKWKHRSULJKWOHDUDQFHHQWHU ,QF5RVHZRRG'ULYH'DQYHUV0$86$,QWKLVFDVHSHUPLVVLRQWRSKRWRFRSLVQRWUHTXLUHGIURP WKHSXEOLVKHU )RUDQDYDLODEOHVXSSOHPHQWDUPDWHULDOSOHDVHYLVLW KWWSVZZZZRUOGVFLHQWL¿FFRPZRUOGVFLERRNVW VXSSO 3ULQWHGLQ6LQJDSRUH Lakshmi - 11860 - An Introduction to Inverse Problems in Physics.indd 1 Lakshmi - 11860 - An Introduction to Inverse Problems in Physics.indd 1 14/5/2020 3:49:20 pm 14/5/2020 3:49:20 pm
- 10. May 5, 2020 14:13 book-961x669 11860-main page v To Haide, Maryam and Rabeé v
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- 12. May 6, 2020 14:51 book-961x669 11860-main page vii Preface In this book we consider a number of inverse problems from classical dynamics of vibrating linear chains of particles connected with springs to the quantum scat- tering of particles, and from inverse scattering applied to the construction of local nuclear potentials to the determination of forces in atom-atom collisions. While different techniques for inversion are reviewed, the main emphasis is on the well- known methods developed by Gel’fand–Levitan, and by Marchenko for cases where the input information is for fixed partial waves and those advanced by Newton and Sabatier, and by Lipperheide and Fiedeldey where the available information is given at fixed energy. In all these cases when the spectral function is assumed to be a rational function of the variable of the problem, the complexity of the solution is greatly reduced. A large section of the book is devoted to the method based on the continued fraction expansion method, where it is applied to a number of problems of classical wave propagation, nuclear scattering, torsional vibration and inverse problem of geomagnetic induction. For more than thirty years, my colleagues and I have been working on different inverse problems with mixed results. There were times that we were successful in finding solutions for certain problems and there were times when the results were not promising. In the latter group, the problems were either analytically too complicated, or different numerical techniques were unstable mainly due to the accumulation of the round-off errors. In writing this book, I am indebted to a number of friends and collaborators who worked with me on and off on inverse problems, among them Dr. Marie Hron, Professor W. van Dijk and in particular, Professor M.A. Hooshyar. In addition, I was fortunate to have the assistance of a number of graduate and undergraduate physics students. The support of my wife, Ghodssi, was essential in writing this book, and I am grateful to her for her encouragement. Edmonton, Alberta, Canada. January 2020 vii
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- 14. May 5, 2020 14:13 book-961x669 11860-main page ix Contents Preface vii Introduction 1 1 Inverse Problems in Classical Dynamics 5 1.1 Inverse Problem for Trajectory . . . . . . . . . . . . . . . . . . . . 5 1.2 Determination of the Shape of the Potential Energy from the Period of Oscillation . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.3 Action Equivalent Hamiltonians . . . . . . . . . . . . . . . . . . . 8 1.4 Abel’s Original Inverse Problem . . . . . . . . . . . . . . . . . . . 10 1.5 Inverse Scattering Problem in Classical Mechanics . . . . . . . . . 12 1.6 Inverse Problem of a Linear Chain of Masses Coupled to Springs . 15 1.7 Direct Problem of Non-exponential and of Exponential Decays in a Linear Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 1.8 Inverse Problem of Dynamics for a Non-uniform Chain . . . . . . 24 1.9 Direct and Inverse Problems of Analytical Dynamics . . . . . . . 27 1.10 From the Classical Equations of Motion to the Lagrangian and Hamiltonian Formulations . . . . . . . . . . . . . . . . . . . . . . 29 1.11 Langevin and Fokker–Planck Equations . . . . . . . . . . . . . . . 34 2 Inverse Problems in Semiclassical Formulation of Quantum Mechanics 39 2.1 Quantum Mechanical Bound States for Confining Potentials . . . 39 2.2 Semiclassical Formulation of the Inverse Scattering Problem . . . 41 3 Inverse Problems and the Heisenberg Equations of Motion 47 3.1 Equations of Motion Derived from the Hamiltonian Operator . . . 48 3.2 Determination of the Commutation Relations From the Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 49 3.3 Construction of the Hamiltonian Operator as an Inverse Problem 52 4 Inverse Scattering Problem for the Schrödinger Equation and the Gel’fand–Levitan Formulation 55 4.1 The Jost Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 4.2 The Jost Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 4.3 The Levinson Theorem . . . . . . . . . . . . . . . . . . . . . . . . 61 ix
- 15. May 5, 2020 14:13 book-961x669 11860-main page x x Contents 4.4 The Gel’fand–Levitan Equation . . . . . . . . . . . . . . . . . . . 63 4.5 Inverse Problem for One-dimensional Schrödinger Equation . . . . 68 4.6 Bargmann Potentials . . . . . . . . . . . . . . . . . . . . . . . . . 73 4.7 The Jost and Kohn Method of Inversion . . . . . . . . . . . . . . 77 5 Marchenko’s Formulation of the Inverse Scattering Problem 83 5.1 Mathematical Preliminaries . . . . . . . . . . . . . . . . . . . . . . 83 5.2 Bound States Embedded in Continuum . . . . . . . . . . . . . . . 91 5.3 More Solvable Potentials Found from Inverse Scattering . . . . . . 92 5.4 The Inverse Problem for Reflection and Transmission from a Barrier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 5.5 A Special Problem in Electromagnetic Inverse Scattering . . . . . 98 5.6 Construction of Reflectionless Potentials . . . . . . . . . . . . . . 104 5.7 Symmetric Reflectionless Potentials Supporting a Given Set of Bound States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 6 Newton–Sabatier Approach to the Inverse Problem at Fixed Energy 115 6.1 Construction of the Potential at Fixed Energy . . . . . . . . . . . 115 6.2 Criticism of the Newton–Sabatier Method of Inversion at a Fixed Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 6.3 On the Results of the Numerical Solution of Inverse Problems . . 123 6.4 Modified Form of the Gel’fand–Levitan for Fixed Energy Problems and the Langer Transform . . . . . . . . . . . . . . . . . 124 6.5 Lipperheide and Fiedeldey Approach to the Inverse Problem at Fixed Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 6.6 Completeness of the Set of Jost Solutions f(λ, k, r) . . . . . . . . 136 6.7 Generalized Gel’fand–Levitan Approach to Inversion . . . . . . . 141 6.8 The Method of Schnizer and Leeb . . . . . . . . . . . . . . . . . . 145 6.9 Analysis of Atom-Atom Scattering Using Complex Angular Momentum Formulation . . . . . . . . . . . . . . . . . . . . . . . 147 7 Discrete Forms of the Schrödinger Equation and the Inverse Problem 153 7.1 Zakhariev’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . 154 7.2 The Method of Case and Kac for Discrete Form of Inverse Scattering Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 155 7.3 Discrete Form of the Spectral Density for Solving the Inverse Problem on Semi-axis 0 ≤ r ∞ . . . . . . . . . . . . . . . . . . 163 8 R Matrix Theory and Inverse Problems 173 8.1 Inverse Problem for R Matrix Formulation of Scattering . . . . . 178 8.2 The Finite-difference Analogue of the R Matrix Theory of Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 8.3 Shell-model Hamiltonian in Tri-diagonal Form . . . . . . . . . . . 182 8.4 Continued Fraction Expansion of the R Matrix . . . . . . . . . . 183
- 16. May 5, 2020 14:13 book-961x669 11860-main page xi An Introduction to Inverse Problems in Physics xi 9 Solvable Models of Fokker–Planck Equation Obtained Using the Gel’fand–Levitan Method 187 9.1 Solution of the Fokker–Planck Equation for Symmetric and Asymmetric Double-Well Potentials . . . . . . . . . . . . . . . . . 191 10 The Eikonal Approximation 195 10.1 Finding the Impact Parameter Phase Shifts from the Cross Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 11 Inverse Methods Applied to Study Symmetries and Conservation Laws 207 11.1 Classical Degeneracy and Its Quantum Counterpart . . . . . . . . 208 11.2 Inverse Problem for Angular Momentum Eigenvalues . . . . . . . 209 11.3 Quantum Potentials Proportional to . . . . . . . . . . . . . . . 213 12 Inverse Problems in Quantum Tunneling 217 12.1 Nonlinear Equation for Variable Reflection Amplitude . . . . . . . 217 12.2 Inverse One-dimensional Tunneling Problem . . . . . . . . . . . . 219 12.3 A Method for Finding the Potential from the Reflection Amplitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 12.4 Finding the Shape of the Potential Barrier in One-Dimensional Tunneling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 12.5 Construction of a Symmetric Double-Well Potential from the Known Energy Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . 228 12.6 The Inverse Problem of Molecular Spectra . . . . . . . . . . . . . 230 12.7 The Inverse Problem of Tunneling for Gamow States . . . . . . . 233 12.8 Inverse Problem of Survival Probability . . . . . . . . . . . . . . . 236 13 Inverse Problems Related to the Classical Wave Propagation 241 13.1 Determination of the Wave Velocity in an Inhomogeneous Medium from the Reflection Coefficient . . . . . . . . . . . . . . . . . . . . 241 13.2 Solvable Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 246 13.3 Extension of the Inverse Method to Reflection from a Layered Medium where the Asymptotic Values of c(t) at t → ±∞ are Different . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248 13.4 Direct and Inverse Problems of Wave Propagation Using Travel Time Coordinate . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 13.5 R Matrix and the Inverse Problems of Wave Propagation . . . . . 262 13.6 Inverse Problem for Acoustic Waves: Determination of the Wave Velocity and Density Profiles . . . . . . . . . . . . . . . . . . . . . 265 13.7 Inversion of Travel Time Data in the Geometrical Acoustic Limit 265 13.8 Riccati Equation for Solving the Direct Problem for Variable Velocity and Density . . . . . . . . . . . . . . . . . . . . . . . . . 267 13.9 Finite Difference Equation for Acoustic Pressure in an Inhomogeneous Medium: Direct and Inverse Problems . . . . . . 268
- 17. May 5, 2020 14:13 book-961x669 11860-main page xii xii Contents 13.10 Determination of the Wave Velocity and the Density of the Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270 13.11 Rational Representation of the Input Data . . . . . . . . . . . . . 271 13.12 Direct and Inverse Methods Based on Continued Fraction Expansion Applied to Two Simple Models . . . . . . . . . . . . . 271 13.13 Inverse Problem of Wave Propagation Using Schwinger’s Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 14 The Inverse Problem of Torsional Vibration 285 15 Local Nucleon-Nucleon Potentials Found from the Inverse Scattering Problem at Fixed Energy 293 15.1 Constructing the S Matrix from Empirical Data . . . . . . . . . . 294 15.2 A Method for the Numerical Calculation of the Local Potential Using the Gel’fand–Levitan Formulation . . . . . . . . . . . . . . 299 15.3 Direct and Inverse Problems for Nucleon-Nucleon Scattering Using Continued Fraction Formulation . . . . . . . . . . . . . . . 302 15.4 Inverse Problem of Scattering in the Presence of the Tensor Force 305 15.5 Potential Model for Generating the Input Data for Testing the Inversion Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 15.6 Inverse Method of Nucleon-Nucleon Phase Shift and the Calculation of Nuclear Structure . . . . . . . . . . . . . . . . . . . 312 16 The Inverse Problem of Nucleon-Nucleus Scattering 317 16.1 Solving the Inverse Nucleon-Nucleus Problem . . . . . . . . . . . 320 16.2 Inverse Scattering Theory Incorporating Both Coulomb and Nuclear Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 16.3 Inverse Scattering Method for Two Identical Nuclei at Fixed Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 17 Two Inverse Problems of Electrical Conductivity in Geophysics 333 17.1 Inverse Problem of Electrical Conductivity in One-Dimension . . 333 17.2 The Inverse Problem of Geomagnetic Induction at a Fixed Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 18 Determination of the Mass Density Distribution Inside or on the Surface of a Body from the Measurement of the External Potential 349 19 The Inverse Problem of Reflection from a Moving Object 355 A Expansion Algorithm for Continued J-fractions 361 B Reciprocal Differences of a Quotient and Thiele’s Theorem 367 Index 371
- 18. May 5, 2020 14:13 book-961x669 11860-main page 1 Introduction In a vast body of problems, called inverse problems of physics, we can find two distinct categories. In the first one, we have a well-defined physical system, and the solution of the direct problem is unambiguous, stable and in general unique. Now supposing that the final solution or the result of a theory is known, can we reverse the process and find the initial question or find out the law(s) governing the physical system from the known results? The second category which contains the majority of interesting problems in physics consists of a group of problems where we have a set of laws or rules and some input data, and from these we can calculate certain results. These results may or may not be empirically observables. Now within the confines of these laws or set of rules, can we interchange the role of the input and the output, and thus replace the direct problem with the inverse one? In other words we want to know whether it is possible to begin with the empirically determined results and find the underlying basic law? Let us illustrate the first category by the following examples: (1) Consider a system composed of identical masses connected linearly to each other by springs with different spring constants. In the direct problem we can calculate the characteristic frequencies of such a system. This is achieved by diagonalizing a tri-diagonal matrix. For the inverse problem the question is that if the the total mass and all characteristic frequencies are known, is it possible determine each one of the spring constants? (2) A very important set of inverse problems in mathematical physics can be stated in the following way: Suppose that we are given a second order linear differential equation such as the Schrödinger equation for a fixed partial wave and subject to specific boundary conditions. We can solve the direct problem when the local poten- tial V (r) is known and we can calculate the bound state energies and the scattering phase shift, the latter as a function of energy. The inverse problem is to find a way to determine a local, energy independent V (r) from the phase shifts and the binding energies. For most of the important physical applications that we encounter, a precise statement of the mathematical problem cannot be given, and we have to resort to approximate models and methods to describe the system, particularly when we are trying to solve inverse problems. The direct problem is, in general, well-posed, and the results are not very sensitive to the small variations of the input data. On 1
- 19. May 5, 2020 14:13 book-961x669 11860-main page 2 2 Introduction the other hand “inverse problems” are typically ill-posed problems, meaning that a small error in the initial parameters can result in large errors in the final results. Thus these problems are often unstable. For instance consider the simple case of the matrix eigenvalue problem. Here the direct problem refers to the calculation of the eigenvalues of a symmetric matrix, and the inverse eigenvalue problem is that of the construction of a symmetric matrix when the set of eigenvalues are known. Here the direct problem is well-posed, but not the inverse matrix eigenvalue prob- lem. We say a physical problem is well-posed if it has the following properties [1]: (1) A solution for it is guaranteed to exist. (2) The solution must be unique. (3) The solution has to depend continuously on the data. In physical sciences we find two distinct sets of inverse problems. First those problems which can be precisely stated but the answer, in general, may not be unique. For instance in classical dynamics, the inverse problem of vibration of a linear system composed of masses and springs, Sec. 1.6, or in quantum theory the inverse problem of bound states in a confining potential, Sec. 2.1 deal with examples of this type. In the second group of inverse problems we can make a mathematically well-defined formulation if and only if we make a number of simplifying assump- tions so that the problem can be formulated without ambiguity. These assumptions can be about the nature of the physical system, or an extension of the range of validity of the observables or the input data beyond what is physically accessible or measurable. For instance, in quantum scattering theory, consider the question of the determination of a potential function from the energy dependence of the phase shift. In this problem we have to make a number of assumptions, such as whether the potential should be local or nonlocal and/or whether the potential should be energy (or velocity) dependent or independent. Inverse methods can be used to investigate different aspects of problems that we encounter in theoretical physics. Among them we will consider the followings: (a) For exactly solvable problems we can use these methods to study the uniqueness of the results that we have found from the solution of the direct problem. Interest- ing examples belonging to this group are Wigner’s derivation of the general form of the commutation relation for the harmonic oscillator [2], [3] and the Bargmann con- struction of phase equivalent potentials [4], and the fact that having the standard quantum mechanical angular momentum eigenvalues, 2 ( + 1), do not necessarily imply the existence of a central force law [5]. (b) The elegant method of Gel’fand–Levitan [6], [7] and also the interesting method of Marchenko [8] to solve the inverse problem of wave equation, i.e. the construc- tion of a local energy independent short range potential for the Schrödinger equation from the information known about the energy dependence of the phase shift, is the central part of any discussion of the inverse scattering theory, particularly in nu- clear and particle physics. A good review entitled “Inverse Problems in Classical and Quantum Physics” by A.A. Almasy discusses a number of wide ranging prob- lems including problems from particle physics and field theory. (c) Except for a limited set of examples, we cannot find analytical solutions to the inverse problems, therefore for practical applications we have to resort to
- 20. May 5, 2020 14:13 book-961x669 11860-main page 3 An Introduction to Inverse Problems in Physics 3 numerical methods. Unfortunately, unlike the direct problems, most inverse prob- lems are ill-posed and are unstable. While many different ways of obtaining numer- ical solutions have been proposed, some more accurate than the others, in general, these solutions should be used as guides to the form of the output. In this book we will try to sample a number of different inverse problems numerically. Nearly all of the numerical solutions reported in this text have been done in 1980’s, with the numerical techniques and computers available in that decade. Today solutions of these problems may yield much better results. This book begins with a review of a number of inverse problems of classical dynamics including a brief discussion of construction of the Lagrangian and the Hamiltonian from the equations of motion and also the conditions under which such a construction is possible. Then the application of the inverse method in semi- classical formulation of quantum mechanics for certain physical systems is stud- ied. This is followed by a short account of the Heisenberg formulation. The next chapter is devoted to the inverse problem for the Schrödinger equation using the Gel’fand–Levitan formulation in one dimension as well as for the radial equation for spherically symmetric potential in spherical polar coordinates. In the latter case it is assumed that for a given partial wave one knows the phase shifts for all energies and also the bound states energies. Marchenko’s approach is studied in the follow- ing chapter, where it is applied to solve a number of physically interesting problems. The Newton–Sabatier method for determining the potential from scattering data given at fixed energy but for large number of partial waves is discussed next. Then few applications such as finding solvable cases of the Fokker–Planck equation are presented. Inverse problems for classical wave propagation are the subject of the next few chapters. Here continued fractions are introduced as a powerful way of getting approximate solutions for inverse problems of various types. This approach combined with the input data given in the form of a rational fraction provides a reli- able technique for solving many problems in atomic, nuclear and geophysical inverse problems numerically. Among a number of applications, the solution of the inverse problem of torsional vibration is discussed. The next two chapters are devoted to the approximate determination of nuclear forces from the empirical knowledge of the phase shifts and binding energies. The inverse problem of electrical conductiv- ity which is of great interest in geophysics is the subject which is presented in the following chapter. Finally, there is a brief review of the inverse problem of reflection from a moving target. I must also acknowledge the great books of Newton, [9], by Chadan and Sabatier [8] and by Zakhariev and Suzko [11] that paved the way for my studies of this subject over the years, and were very helpful in guiding me in different ways. Only when I had completed the writing of the book that I came across the very interesting dissertation entitled “Inverse Problems in Classical and Quantum Physics” by A.A. Almasy who discusses a number of wide-ranging problems includ- ing problems from particle physics and field theory [12].
- 21. May 5, 2020 14:13 book-961x669 11860-main page 4 4 References References [1] J. Hadamard, Sur les Problémes aux Dérivées Partielles et Leur Signification Physique, Princeton University Bulletin, pp. 49-52. [2] E.P. Wigner, Phys. Rev. 77, 711, (1950). [3] M. Razavy, Heisenberg’s Quantum Mechanics, (World Scientific, Singapore, 2011) Chapter 5. [4] V. Bagamann, Phys. Rev. 75, 301 (1949). [5] M. Razavy, Phys. Letts. 88A, 215 (1982). [6] I.M. Gel’fand and B.M. Levitan, Dokl. Akad. Nauk SSSR, 77, 557 (1951). [7] I.M. Gel’fand and B.M. Levitan, Isvest. Akad. Nauk SSSR, 15, 309 (1951). [8] V.A. Marchenko, Dokl. Akad. Nauk SSSR, 72, 457 (1950). [9] R.G. Newton, Scattering Theory of Waves and Particles, (Second Edition, 1980). [10] K. Chadan and P.C. Sabatier, Inverse Prblems in Quantum Scattering Theory, Second Edition, (Springer-Verlag, 1989). [11] B.N. Zakhariev and A.A. Suzko, Direct and Inverse Problems: Potentials in Quantum Scattering, (Springer-Verlag, Berlin 1990). [12] A.A. Almasy, Inverse Problems in Classical and Quantum Physics, Ph.D. Dis- sertation, Johannes Gutenberg-Universität in Mainz (2007), arXiv.0912.0455 math-ph (2009).
- 22. May 5, 2020 14:13 book-961x669 11860-main page 5 Chapter 1 Inverse Problems in Classical Dynamics Some of the very interesting problems of classical mechanics belong to the category of inverse problems. While these are elementary in nature, and are exactly solvable, they provide a very good introduction to the physics of the inverse problems. 1.1 Inverse Problem for Trajectory We will start our review of the inverse problems in classical mechanics by considering a problem that goes back to I. Newton [1]. That is the derivation of the inverse- square law of planetary motion from empirical laws of Kepler. First we note that Kepler’s second law, i.e. the area swept out by the position vector in unit time is constant, in polar coordinates, can be written as mr2 θ̇ = = constant, (1.1) where dot denotes the time derivative, and where m is the mass of the planet. In addition we know that the orbit is an ellipse with the origin at one of the two foci. Thus the orbit in polar coordinate is expressible as r = a 1 + cos θ , 1. (1.2) By differentiating r twice with respect to time and then using to eliminate θ̇, we find ṙ = a sin θ (1 + cos θ)2 , (1.3) 5
- 23. May 5, 2020 14:13 book-961x669 11860-main page 6 6 Shape of the Potential Energy and r̈ = a (cos θ)θ̇ = 2 mar2 cos θ, (1.4) Eliminating cos θ between (1.2) and (1.4), we can write r̈ as a function of r only r̈ = 2 mar2 a r − 1 . (1.5) But we know that the radial component of the acceleration in polar coordinates ar is given by ar = r̈ − r θ̇ 2 = 1 m2 2 r3 − 2 ar2 − r 2 r4 = − 2 m2a 1 r2 . (1.6) Therefore the radial component of the force is given by Fr = − 2 ma r−2 . 1.2 Determination of the Shape of the Potential Energy from the Period of Oscillation Another classical problem which is of interest is to find the shape of a confining potential (i.e. a potential V (x) that goes to +∞ as x → ±∞) from the energy dependence of its period of oscillations [2]. For the one-dimensional motion the total energy, E, of the particle is given by E = 1 2 mẋ2 + V (x), (1.7) where x is the position of the particle and V (x) is a confining potential with the condition V (0) = 0. Equation (1.7) is a first order differential equation for x and can easily be solved for t as a function of x t = 1 2 m dx E − V (x) + t0. (1.8) The two constants of motion are E, which is the total energy and t0. The limits of the motion are the two turning points where the potential energy is equal to the total energy, E = V (x). We only consider confining potentials where the motion is bounded by two such points. Now let us consider the coordinate x as function of V . This function x(V ) is double-valued, i.e. for a given value of V we find two different values of x. Therefore, the integrand in (1.8) can be divided into two parts, one from x = x1 to x = 0 and the other from x = 0 to x = x2. We denote the function x(V ) in these two regions by x = x1(V ) and x = x2(V ) respectively. We also note that the limits of integration with respect to V are E and zero, therefore the period
- 24. May 5, 2020 14:13 book-961x669 11860-main page 7 An Introduction to Inverse Problems in Physics 7 is given by T(E) = √ 2m E 0 dx2(V ) dV dV √ E − V + √ 2m 0 E dx1(V ) dV dV √ E − V = √ 2m E 0 [dx2(V ) − dx1(V )] dV dV √ E − V . (1.9) Now we divide both sides of (1.9) by √ E − E, where E is a parameter, and then integrate the result with respect to E from 0 to E. Thus we find Abel’s integral equation [3]–[4] E 0 T(E)dE √ E − E = √ 2m E 0 E 0 dx2 dV − dx1 dV dV dE (E − E)(E − V ) . (1.10) By changing the order of integration in Eq. (1.10) we have E 0 T(E)dE √ E − E = √ 2m E 0 dx2 dV − dx1 dV dV × E V dE (E − E)(E − V ) = π √ 2m [x2(E) − x1(E)] , (1.11) where we have substituted the value of the last integral in Eq. (1.11) and the value is π. Now by replacing E by V , we obtain x2(V ) − x1(V ) = 1 π √ 2m V 0 T(E) √ V − E dE. (1.12) Thus by knowing T(E) we can find the difference x2(V ) − x1(V ), but not x2(V ) and x1(V ) separately. This means that there are infinitely many curves V = V (x) which gives us the same T(E), and that these functions, V (x), differ from each other in such a way that x2(V ) − x1(V ) is the same for every curve. By imposing some other conditions on V (x), such as its symmetry about the V axis we find a unique potential which is given by [2]–[6] x(V ) = 1 2π √ 2m V 0 T(E)dE √ V − E . (1.13) As an example if T(E) ∝ E− 1 4 , then from (1.13) we find x(V ) ∝ V 0 dE E 1 4 √ V − E = V 1 4 1 0 dx x 1 4 √ 1 − x = V 1 4 β 3 4 , 1 4 , (1.14) where β is the β function and is related to Γ function by [7], β(z, w) = β(w, z) = 1 0 tz−1 (1 − t)w−1 dt = Γ(z)Γ(w) Γ(z + w) . (1.15)
- 25. May 5, 2020 14:13 book-961x669 11860-main page 8 8 Action Equivalent Hamiltonoans Thus taking the symmetry about the V axis we find that the potential is V (x) ∝ |x|4 , i.e. the motion is that of a classical anharmonic oscillator. Abel’s Integral Equation — The integral equation (1.9) is an Abel integral equation and is a special case of a class of the general integral equations of the form [3], [6] f(x) = 1 Γ(α) x 0 η(ξ)dξ (x − ξ)1−α , 0 α 1, (1.16) where Γ(α) is the Gamma function. In this integral equation f(x) is a known function and we want to find η(ξ). The formal solution of (1.16) which is obtained by inversion is given by η(ξ) = 1 Γ(1 − α) d dξ ξ 0 f(x) (ξ − x)α dx = 1 Γ(1 − α) f(0) tα + ξ 0 f (x) (ξ − x)α dx . (1.17) Equation (1.16) has a continuous solution in the interval a ≤ ξ ≤ b subject to the following conditions: (a) f(x) must be continuous in this interval, (b) f(0) = 0 and (c) ξ 0 f(x)(ξ − x)α−1 dx must have a continuous derivative in the interval [6], [5]. In a number of problems we find Abel’s integral equation of the form f(x) = b x η(ξ)dξ ξ2 − x2 , (1.18) where 0 ≤ x ≤ ξ ≤ b ≤ ∞. The solution of this integral equation is η(ξ) = − 2 π d dξ b ξ xf(x)dx x2 − ξ2 , (1.19) or [6] η(ξ) = 2ξ π f(b) b2 − ξ2 − b ξ f (x)dx x2 − ξ2 . (1.20) An elegant way of solving Abel’s integral equation is by utilizing the Laplace trans- form method. This method is given, for example in Arfken’s book [8]. 1.3 Action Equivalent Hamiltonians Let us solve the inverse one-dimensional motion of particle for the simplest case where the period T = 2π ω is independent of energy. Then from (1.12) we conclude
- 26. May 5, 2020 14:13 book-961x669 11860-main page 9 An Introduction to Inverse Problems in Physics 9 that x2(V ) − x1(V ) = 2V m 1 ω . (1.21) This relation shows that V is a quadratic function of x. Let us write it as V (x) = ⎧ ⎪ ⎨ ⎪ ⎩ 1 2 mω2 +x2 for x 0 1 2 mω2 −x2 for x 0 , (1.22) and this solution satisfies the condition V (0) = 0. From (1.21) and (1.22) it follows that the angular frequency of the motion, ω, is related to ω+ and ω− by ω = 2ω+ω− ω+ + ω− . (1.23) Thus for all non-zero values of the parameters ω+ and ω− of the potential the resulting angular frequency is ω, and therefore in an action-angle formulation of the problem the Hamiltonian is given by [9], [10] H = ωJx. (1.24) A Two-Dimensional Motion with Closed Orbit — Let us study the direct problem for the motion of a particle of unit mass in two dimensions x and y, where the Hamiltonian is given by H = 1 2 p2 x + ω2 x2 + 1 2 p2 y + ω2 y2 . (1.25) If we transform this Hamiltonian to the action-angle variables, then we have [9]–[11]. H = ω(Jx + Jy). (1.26) According to what we have found earlier for one-dimensional motion the simplest two-dimensional motion for the Hamiltonian (1.26) given in terms of the canonical variables is H = 1 2 p2 x + p2 y + Ωx(x)x2 + Ωy(y)y2 . (1.27) where Ωx(x) = ω θ(x) 1 + λ1 + θ(x) 1 − λ1 . (1.28) and Ωy(y) = ω θ(y) 1 + λ2 + θ(y) 1 − λ2 . (1.29) In these relations θ(x) is the step function θ(x) = ⎧ ⎪ ⎨ ⎪ ⎩ 1 for x 0 0 for x 0 , (1.30) and λ1 and λ2 are two arbitrary parameters |λ1,2| 1.
- 27. May 5, 2020 14:13 book-961x669 11860-main page 10 10 Abel’s Integral Equation –1 –2 –1 1 2 –0.5 0.5 1 x y Figure 1.1: The closed orbit found from the Hamiltonian (1.27) for m = 1, ω = 1, λ1 = 0.4 and λ2 = 0.2. Apart from the energy there are no apparent symmetries or conserved quantities for this motion [11]. In conclusion we observe that the symmetry of the Hamiltonian (1.25) and the symmetry of the Hamiltonian (1.27), which we found by solving the inverse problem, are not the same, i.e. (1.25) remains invariant under the transformation x y, but (1.27) does not, even though in both cases the orbits are closed. This is particularly significant once we consider the question of the quantization. We will come back to this important aspect of using the results of solving the inverse problem later in connection with the quantum mechanical inverse problems. 1.4 Abel’s Original Inverse Problem An inverse problem of historical interest has been called “determination of the shape of a hill from the travel time measurement” [12]. If a particle slides up and down a hill with no friction and with the initial energy E, assuming that we can measure the travel time for up and down motion T(E) as a function of the initial velocity of the particle, then under certain conditions we can find the shape of the hill. In the direct problem, we know that the potential energy of the particle at the height y is given by mgy(s), where s measures the arc length along the hill. The total energy of the particle is the sum of kinetic and potential energies: E = 1 2 m ds dt 2 + V (s). (1.31) At t = 0 we assume that the particle is at s(0) = 0, and has a velocity ds dt t=0 = v0. Solving (1.31) for ds dt , choosing the positive square root since v0 0, and then
- 28. May 5, 2020 14:13 book-961x669 11860-main page 11 An Introduction to Inverse Problems in Physics 11 integrating (1.31) we obtain t = m 2 s 0 ds E − V (s) . (1.32) Thus starting at s = 0, the particle moves up and reaches the classical turning point s1, where E = V (s1). The corresponding value of t at this point is 1 2 T(E), i.e. T(E) = √ 2m s1(E) 0 ds E − V (s) . (1.33) At the time t = 1 2 T(E), the velocity of the particle becomes zero and after that the velocity becomes negative and the particle will start descending. Therefore the solution for t ≥ 1 2 T(E), is obtained if we take the negative square root in (1.31). The result in this case is t = 1 2 T(E) + m 2 s1(E) 0 ds E − V (s) , 1 2 T(E) ≤ t. (1.34) Now if we set s1 = 0 in the above equation we find the time that takes the particle to move back to the point s = 0 is just the same 1 2 T(E) given by (1.33). If V (s) is not a monotonically increasing function of s, we can divide the upward (or downward) motion of the particle into different segments and then find the total time [12]. Here, for the sake of simplicity, we assume that V (s) is a monotonic function of s. With this condition the solution of the Abel integral equation is given by ds(V ) dV = T(0) π √ 2mV + 1 π √ 2m V 0 dT(E) dE dE √ V − E , (1.35) where, because of the initial conditions, T(0) = 0. Finally we integrate (1.35) from 0 to V to get s1(V ) s1(V ) = 1 π √ 2m V 0 T(E)dE √ V − E . (1.36) Once we have s1(V ), by inverting, we find s(V ) and since V (s) = mgy(s) and dx ds 2 + dy ds 2 = 1, therefore we have a parametric equation for x(s) and y(s): x(s) = s 0 1 − dy (s) ds 2 ds , and y(s) = V (s) mg . (1.37) Even when the profile of the hill cannot be described by a monotonic function y = y(s), still there are cases that the problem can be solved [12].
- 29. May 5, 2020 14:13 book-961x669 11860-main page 12 12 Inverse Scattering Problem in Classical Dynamics 1.5 Inverse Scattering Problem in Classical Mechanics As another example of application of inverse problems in classical mechanics we want to consider the inverse scattering problem of a projectile of mass m from a fixed (or a massive) target. The projectile is coming from infinity, and has a total energy E = 1 2 mv2 ∞ and we assume that the interaction potential is repulsive and goes to zero as r → ∞. The trajectory of the particle is a hyperbola and its shape is determined by the impact parameter b which is equal to the closest distance to the center that the particle will reach if the potential were zero everywhere. However because of the repulsive nature of the potential the nearest point that the projectile will reach the target is r = r0. The hyperbolic path of the particle has two asymptotes corresponding to t = −∞ and t = +∞, and the interaction causes a defection of the asymptotes by an angle θ which is the scattering angle. We use the polar coordinates (r, φ) to describe the position of the particle as a function of time. Since the force is central, from the laws of motion we find two conserved quantities, the angular momentum and the energy E: = −bmv∞, (1.38) and E = 1 2 mṙ2 (t) + 2 2mr2 + V (r). (1.39) In these relations dot means derivative with respect to time, is the angular mo- mentum and b is the impact parameter. Since the angular momentum is conserved we have φ̇(t) = mr2 = − b r2 2E m 0, (1.40) so that φ(t) is a monotonically decreasing function of t. From Eq. (1.39) we find ṙ(t) to be: ṙ(t) = ± 2 m 1 − b2 r2 E − V (r) 1 2 . (1.41) Now if at the time t = t0 the projectile reaches the nearest distance from the target, i.e. r0 = r(t0), then we choose the minus sign in (1.41) for −∞ t t0, and the plus sign for t0 t +∞. Thus the value of r0 is determined from E 1 − b2 r2 0 − V (r0) = 0. (1.42) For finding the asymptotes, we have to know the trajectory. This is obtained by eliminating t between (1.40) and (1.41): dφ = − 2E m b r2 dt = − 1 r2 1 − V (r) E 1 b2 − 1 r2 − 1 2 dr. (1.43)
- 30. May 5, 2020 14:13 book-961x669 11860-main page 13 An Introduction to Inverse Problems in Physics 13 b b db dΘ Θ Figure 1.2: Scattering of a particle by a repulsive force. Thus by integrating this equation and noting that the scattering angle, θ = π − 2φ, we find θ = θ(b) = π − 2 ∞ r0 dr r2 1 − V (r) E 1 b2 − 1 r2 − 1 2 . (1.44) This is the solution of the direct problem, i.e. if we know the potential we can calculate the scattering angle as a function of energy and of the impact parameter. In the direct problem, the important concept is the cross section which is a function of θ. Thus we consider a mono-energetic beam of particles incident in the annular ring of area 2πbdb bounded by the impact parameters b and b + db, and that the number of scattered particles through the solid angle dΩ = 2π sin θdθ is σ(θ) (see Fig. 1.2). By equating the number of incoming and outgoing particles we find [10] σ(θ) = − b sin θ db dθ . (1.45) The Inverse Scattering Problem in Classical Mechanics — In order to solve the inverse problem, we assume that θ(b) is known and we want to find V (r). Following the work of Firsov and of Keller [13], [14], we introduce the new variables: x = 1 b2 , u = 1 r , θ(b) = θ̄(x), V (r) = V̄ (u), (1.46) and β(x) = 1 2 π − θ̄(x) . (1.47) With these changes, Eq. (1.44) can be written as β(x) = u0 u=0 x 1 − V̄ (u) E − u2 − 1 2 du, (1.48) where u0 = 1 r0 . Next we make the following substitution: v(u) = 1 − V̄ (u) E , w = u2 v and g(w) = 1 √ v du dw . (1.49)
- 31. May 5, 2020 14:13 book-961x669 11860-main page 14 14 Inverse Scattering Problem in Classical Dynamics With these changes (1.48) becomes the standard Abel integral equation β(x) = x w=0 g(w)dw √ x − w , 0 ≤ x ≤ xmax, (1.50) where xmax = b−2 min. In Eq. (1.50) when the square root is zero, i.e. for x = w, then u = u0. Now Eq. (1.48) is the Abel integral equation for which we know what the solution is (see Eq. (1.17)) g(w) = 1 π d dw w 0 β(x)dx √ w − x . (1.51) Having obtained g(w) we want to determine V (r). Now from (1.49), we have u = √ vw, therefore g(w) = √ w 2v dv dw + 1 2 √ w . (1.52) We solve this equation for v as a function of w; v = v̄(w) = exp w 0 2g (w̄) w̄− 1 2 − w̄−1 dw̄. (1.53) Let us note that u = 0 implies w = 0, and that V̄ (0) = V (∞) = 0. In this way we obtain a parametric representation of V (r), viz, from (1.48) and (1.49) we find V = E (1 − v̄(w)) , r = 1 wv̄(w) , (1.54) for 0 ≤ w ≤ xmax = b−2 min. Another way to simplify our formulation is to integrate (1.51) by parts, and the integration yields the following result: g(w) = d dw √ w − 1 π w 0 √ w − x dθ dx dx = 1 2 √ w + 1 2π w 0 dθ dx dx √ w − x . (1.55) Now by changing the integration variable to θ, and noting that θ(0) = 0, we find g(w) = 1 2 √ w + 1 2π θ(w) 0 dθ w − x(θ) . (1.56) Using this result Eq. (1.53) becomes v = exp 1 π w 0 dw √ w θ(w ) 0 dθ w − x(θ) . (1.57) Now we will use these results to solve the inverse problem of Rutherford scattering.
- 32. May 12, 2020 11:33 book-961x669 11860-main page 15 An Introduction to Inverse Problems in Physics 15 Inverse Problem of the Rutherford Scattering — We know that in the well- known Rutherford scattering, the cross section is given by [10] σ(θ) = A 4 sin4 θ 2 , (1.58) where A = e2 4E2 . Using Eq. (1.45) we can express b2 in terms of the scattering angle θ − bdb sin θdθ = A 4 sin4 θ 2 , (1.59) or simplifying we find 1 b2 = x(θ) = A−1 tan2 θ 2 , θ = tan−1 √ Ax. (1.60) Now we substitute x(θ) in (1.57) to find v: v = exp ⎧ ⎨ ⎩ 1 π w 0 dw √ w 2 tan−1 √ Aw 0 dθ w − A−1 tan2 θ 2 1 2 ⎫ ⎬ ⎭ . (1.61) To evaluate the θ integral, we change the variable θ to φ, where tan θ 2 = √ Aw sin φ. (1.62) The last integral in (1.61) can be calculated [7] 2 tan−1 √ Aw 0 dθ w − A−1 tan2 θ 2 1 2 = 2 √ A π 2 0 dφ 1 + Aw sin2 φ = π √ a √ 1 + Aw . (1.63) By substituting this integral in (1.52) we can carry out the integration over w and obtain v; v = 1 + 2Aw + 2 Aw(2 + Aw). (1.64) Since u = 1 r = √ vw, from Eqs. (1.46) and (1.49), we find V (r) = 2E √ Au = eu = e r , (1.65) which is the Coulomb potential. 1.6 Inverse Problem of a Linear Chain of Masses Coupled to Springs A simple and much studied example of an inverse problem in classical dynamics is that of the determination of masses and different spring constants for a finite
- 33. May 5, 2020 14:13 book-961x669 11860-main page 16 16 Masses Coupled to Strings chain from the knowledge of the characteristic frequencies of the system [15], [16]. In order to state the questions relating to the input and output of this problem clearly, we need to study certain aspects of the theory of matrices first. Therefore we begin this section with the review of the mathematical tools needed for solving this particular problem [16]. Rayleigh Quotient— Let x be a column n × 1 matrix (or a vector with n com- ponents) and An be a symmetric n × n matrix, and let us consider the following quotient called Rayleigh quotient [17] R = xT · Anx xT · x . (1.66) In this relation xT is the transpose of x and dot denotes the scalar product of two vectors. We note that by replacing x by bx, with b a constant R will not be affected, since it is the ratio of two quadratic forms. Thus we can write R as the value of xT · Ax subject to the normalization condition xT · x = 1. This constraint on x can be enforced by introducing a Lagrange multiplier λ, and then finding the condition for R to be stationary. Now we consider the condition under which the quadratic form R = xT · Anx − λxT · x, (1.67) becomes stationary. By differentiating R with respect to a matrix element xi and setting it equal to zero we get ∂R ∂xi = 2(a1ix1 + a2ix2 + · · · + anixn) − 2λxi = 0. (1.68) If we collect all the terms ∂R ∂x1 , ∂R ∂x2 , · · · , ∂R ∂xn , we find the equation Anx − λx = 0, (1.69) and this shows that λ is an eigenvalue of the matrix A. From this direct problem we find a set of eigenvalues λ1, λ2, · · · , λn, but know- ing this set is not sufficient to solve the inverse problem since in the latter we need n masses, m1, m2, · · · , mn and n spring constants k1, k2, · · · , kn. The rest of the needed information can be obtained from the eigenvalues μ1, μ2, · · · , μn−1 of the matrix An−1 which will be defined below and in addition the total mass of the system n i=1 mi. If we delete the row n and the column n from the matrix we find (n − 1) × (n − 1) matrix An−1, with n − 1 eigenvalues. Whereas the matrix (An − λ)x = 0 represents the motion of the system shown in Fig. 1.3 (the top fig- ure) with all particles moving, the eigenvalue equation (An−1 − μ)x = 0 describes the system shown in the same figure (below) where the nth mass is attached to the wall and is not moving. Since x1 = 0 is equivalent to xT · e1 = 0, where e1 = [0, 0, · · · , 1], we can introduce two Lagrange multiplier λ and 2μ and consider under what conditions R = xT · Anx − λxT · x − 2μxT · en, (1.70)
- 34. May 5, 2020 14:13 book-961x669 11860-main page 17 An Introduction to Inverse Problems in Physics 17 will become stationary. The condition on R is ∂R ∂xi = 2 n j=1 aijxj − 2λxj − 2μδi1 = 0 (1.71) which can be written as Anx − λx − μe1 = 0. (1.72) Since x1, x2, · · · , xn span Rn, we can write x = n i=1 βixi, (1.73) and by substituting this expression in (1.72) we find n i=1 (λi − λ)βixi − μe1 = 0. (1.74) From the orthogonality of the eigenvectors xi, i.e. xT j · xi = δij we obtain (λj − λ)βj = μxT · e1 = μx1j, (1.75) where x1j denotes the first component of xj. Thus we have βj = μx1j (λj − λ) . (1.76) Now by imposing the constraint condition x1 = 0 we get f(λ) ≡ n i=1 |x1i|2 λi − λ = 0. (1.77) This relation shows that the eigenvalues μ1, μ2, · · · , μn−1 are the roots of f(λ), and that these eigenvalues interlace λ1, λ2, · · · , λn, i.e. λ1 μ1 λ2 μ2, · · · , μn−1 λn. (1.78) If xni is normalized i.e. n i=1 |xni|2 = 1, and if we know all the elements xn1, xn2, · · · , xnn then from Eq. (1.77) we can find μ1, μ2, · · · , μn−1. Thus in the direct problem knowing the matrix An enables us to determine the set of μ s. But this way of looking at the problem is useful since we can also solve the inverse problem. That is if we are given (n−1) μ s, and we know they interlace λ s, then we can find the elements xn1, xn2, · · · , xnn. Using the relation for the spectral decomposition of a given symmetric matrix A, adj (λiI − A) = n k=1,k=i (λi − λk)xixT i , (1.79)
- 35. May 5, 2020 14:13 book-961x669 11860-main page 18 18 Masses Coupled to Strings k1 m1 k2 m2 kn mn k1 m1 k2 m2 kn mn Figure 1.3: Linear chains of masses and springs. The top figure shows the case where all n masses can move, and the bottom shows the case where the last mass mn is rigidly attached to the wall [16]. where I is the unit matrix, we can evaluate both sides of this relation at (n, n) position to obtain det(λiIn−1) = x2 ni n k=1,k=i (λi − λk). (1.80) In this relation xni is the last entry of xi, and An−1 denotes the (n − 1)(n − 1) matrix. Solving for x2 ni, we get |x1i|2 = n−1 k=1,k=i(μk − λi) n k=1,k=i(λk − λi) , (1.81) and this expression gives us the matrix elements of A. Thus we conclude that if the eigenvalues μ1, μ2, · · · , μn−1 are known then we know the first entries in the vectors x1, x2, · · · , xn. If we know λ1, λ2, · · · , λn and xn1, xn2, · · · , xnn we have 2n + 1 quantities. Because xn is normalized xn s are not independent. The Jacobi Matrix — The Jacobi matrix is a square n×n matrix with just three diagonal elements being nonzero. A typical Jacobi matrix has the following form [18], [19] Kn = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ a1 b1 0 0 · · · 0 b1 a2 b2 0 · · · 0 0 b2 a3 b3 · · · 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . · · · · · · · · · · · · · · · bn−1 · · · · · · · · · · · · bn−1 an ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ (1.82) Since changing the sign of bj has no effect on the eigenvalues, we can assume that b1, b2, · · · , bn−1 are all negative. We can collect the n column vector equations (A − λiI)xi = 0 into one matrix equation AX = XΛ, (1.83)
- 36. May 5, 2020 14:13 book-961x669 11860-main page 19 An Introduction to Inverse Problems in Physics 19 where X = [x1, x2, · · · , xn]. (1.84) Let us now denote the transpose of X by U, i.e. XT = U, then AUT = UT Λ and on transposing we have UA = ΛU, (1.85) or in particular if A = J is a Jacobi matrix we have [u1, u2, · · · , un]J = Λ[u1, u2, · · · , un], (1.86) where U = [u1, u2, · · · , un]. Now let us consider a method for determining the elements aj, bj of this J matrix from the set of eigenvalues λj and μj. There is a wealth of literature on the inversion of symmetric Jacobi matrices and the reader is referred to the following papers where in these works, further references can be found to other works [16]–[25]. Let us consider a simple way of this inversion. The first column of Eq. (1.86) is a1u1 + b1u2 = Λu1. (1.87) Since u1, u2, · · · , un form an orthogonal set, from (1.87) we find a1 = uT 1 · Λu1. (1.88) To find b1 we write (1.87) as b1u2 = (Λ − a1)u1 = z2. (1.89) The vector z2 is known from its definition and u2 is a unit vector. Thus b2 1 = z2 2 , b1 = z2 , (1.90) and u2 = 1 b1 z2. (1.91) Now let us consider the second column of (1.86); b1u1 + a2u2 + b2u3 = Λu2. (1.92) Again we multiply this equation by uT 2 , using the fact that uT 2 · u1 = uT 2 · u3 = 0 and get a2 = uT 2 Λu2, (1.93) and b2u3 = z3 = (Λ − a2)u2 − b1u1. (1.94) Since every term in the right-hand side of this equation is known therefore z3 can be found. From the last relation we deduce that b2 = z3 , u3 = z3 b2 . (1.95)
- 37. May 5, 2020 14:13 book-961x669 11860-main page 20 20 Masses Coupled to Strings We continue this process to find all ai s and bi s. Lanczos Algorithm — The procedure of finding ai and bi, i = 1, 2 · · · n is called Lanczos algorithm [20]. The original Lanczos algorithm concerns with the solution of the following problem: Given a symmetric matrix A and a normalized vector x1 (i.e. xT 1 ·x1 = 1), compute a Jacobian matrix J and an orthogonal matrix X = [x1, x2, · · · , xn] such that A = XJXT . Variants of Lanczos’s algorithm, and different ways of finding the elements in a Jacobi matrix from the eigenvalues are given in Refs. [21]–[28]. Eigenfrequencies of the Mass-Spring Chain — Now let us consider the prob- lem of n masses m1, m2, · · · , mn connected by springs with spring constants k1, k2, · · · , kn to each other. For the system composed of masses and springs, Fig. 1.3 (top), we can write this eigenvalue equation as ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ (k1 + k2) −k2 . . . 0 −k2 (k2 + k3) . . . 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . −kn . . . . . . −kn kn ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ x1 x2 . . . xn−1 xn ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ = λ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ m1 0 · · · 0 0 m2 · · · 0 . . . . . . . . . . . . . . . . · · · · · · mn−1 0 · · · · · · · · · mn ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ x1 x2 . . . xn−1 xn ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ , (1.96) where λ = ω2 , and ω1, ω2 · · · ωn are the natural frequencies of the system. Here we want to find the characteristic frequencies ωi s of such a system under two different boundary conditions: In the first one the mass m1 is attached with a spring k1 to a rigid wall, but mn is free to move, as shown in Fig. 1.3 at the top. The second is when the last mass mn is fixed by attaching it to the wall, as shown in the lower figure Fig. 1.3. The equations of motion for the system at the top can be written as (K − λM)u = 0, λ = ω2 , (1.97) where K is the Jacobi matrix with the elements related to the spring constants ki as shown in (1.97). If we want to write (1.97) in the standard form of an eigenvalue equation we introduce the matrix D; M = D2 , D−1 KD−1 = J, Du = x, (1.98) and then (1.97) can be written as (J − λI)x = 0, (1.99) where I is a unit n × n matrix. Using Lanczos algorithm we find the elements of J, and then we try to find matrices for K and M from J [16], [28]. As we can see
- 38. May 5, 2020 14:13 book-961x669 11860-main page 21 An Introduction to Inverse Problems in Physics 21 from (1.98) the matrices K and A are related to each other K = DJD. By inspection we see that K[1, 1, · · · , 1] = k1e1, (1.100) where e1 = [1, 0, · · · , 0]. Thus DAD[1, 1, · · · , 1] = k1e1, (1.101) and therefore A[d1, d2, · · · , dn] = D−1 k1e1 = d−1 1 k1e1. (1.102) Since the off diagonal elements of the K matrix are negative, Eq. (1.96), when we solve K[x1, x2, · · · , xn] = e1, (1.103) then the solution x1, x2, · · · , xn will have positive components. The solution of the last equation in (1.98) is d = cx, where c can be found from the total mass of the system m = n i=1 mi = n i=1 d2 i = d 2 = c2 x 2 . (1.104) Therefore, knowing m we can calculate x 2 then find c and d, and hence the masses mi = d2 i . Having found di s we can compute K = DAD and find the spring constants k1, k2, · · · , kn [15], [16]. An Analytically Solvable Spring-Mass Problem — Let us consider the special case of Eq. (1.96) when n = 2. Then the eigenvalues are the roots of the quadratic equation k1 + k2 − λm1 −k2 −k2 k2 − λm2 = 0. (1.105) In the direct problem we know k1, k2, m1 and m2, and the eigenvalue equation (1.105) can be solved for λ1 and λ2, the two roots of this equation. In the inverse problem for this case we are given the eigenvalues plus m1 and m2 and from (1.105) we obtain k1 in terms of k2: k1 = 1 k2 (m1m2λ1λ2), (1.106) where k2 is the root of the quadratic equation m1 + m2 m1m2 k2 2 − (λ1 + λ2)k2 + m2λ1λ2 = 0. (1.107) This equation has a real positive root if and only if (λ1 − λ2)2 4m2 m1 λ1λ2. (1.108) This result indicates that, in general, for the accuracy of the numerical solution of the inverse problem the eigenvalues should not be close to each other [16].
- 39. May 5, 2020 14:13 book-961x669 11860-main page 22 22 Non-exponential Decay in Linear Chain Some Other Solvable Mass-Spring Systems — There are a number of other interesting problems closely related to the mass-spring system that we have studied. Among them is the question of recovery of masses and the spring constants of a finite system when we have the natural frequencies of N masses connected with springs with different spring constants plus the frequencies of the system when it is perturbed by modifying one mass and adding one spring to the system [29]. Another inverse problem of interest is the determination of a finite element model of longitudinally vibrating rod with one end fixed and the other end supported on a spring. This case has been studied by Wei and Dai [30]. Let us note that as long as the number n of masses (and springs) are finite the motion is reversible (see e.g. Ref. [31]). 1.7 Direct Problem of Non-exponential and of Exponential Decays in a Linear Chain One of the natural generalizations of the mass-spring system is to study mechanical models where both the number of particles and springs go to infinity. The exact solution of such a problem is if interest, since the motion becomes irreversible i.e. once the system starts from a given configuration it will not come back to its original state. First let us study a very special type of inverse problem for this case. Consider a linear chain of particles each of mass Mj, attached to each other by means of springs with spring constants Kj. Assume that initially all of the particles are at rest in their equilibrium position except one particle which is displaced and released with zero initial velocity. How we should choose different masses and different spring constants so that the motion of this particle decays exponentially? The Decay of Motion in the Schrödiger Chain — Before discussing this ques- tion, let us investigate the solution of the direct problem for the case where masses are all equal and the connecting springs all have the the same spring constant, as- suming that there are infinite number of masses and springs. This problem was first studied and solved by Schrödinger [32]. If we denote the mass of each particle by m and the spring constant by k, then the motion of the jth particle in the chain is given by mẍj = k(ξj+1 + ξj−1 − 2ξj), j = 0, ±1, ±2, · · · , (1.109) where ξj is the displacement of the jth particle from the equilibrium and k is the spring constant which we write in terms of another constant ν with the dimension of wave number, k = 1 4 mν2 . (1.110) Let assume that initially the central particle j = 0 is displaced by a distance A, and then released with zero initial velocity, while all of the other particles are initially
- 40. May 5, 2020 14:13 book-961x669 11860-main page 23 An Introduction to Inverse Problems in Physics 23 at rest in their equilibrium position. Thus at t = 0 we have the initial conditions ξ0(t = 0) = A, ξj(t = 0) = 0, j = 0, (1.111) and ˙ ξ0(t = 0) = 0, j = 0, ±1, ±2 · · · . (1.112) A method for solving the system of equations (1.109) is to construct the generating function G(z, t) which we define by [33], [34] G(z, t) = +∞ j=−∞ ξj(t)z2j . (1.113) We multiply (1.109) by zj , we sum over all j s, and use (1.113) to obtain the following equation for G(z, t): d2 G(z, t) dt2 = 1 4 ν2 1 z − z 2 G(z, t). (1.114) We can find the initial conditions for this differential equation from (1.111)–(1.113). G(z, 0) = A, dG(z, t) dt t=0 = 0. (1.115) With these conditions we find the solution of (1.113) to be G(z, t) = A cosh 1 2 νt z − 1 z . (1.116) This G(z, t) is the generating function for the Bessel function of even order [35]; cosh 1 2 νt z − 1 z = +∞ j=−∞ J2j(νt)z2j , (1.117) and therefore ξj(t) = AJ2j(νt). (1.118) Thus in this model the energy of the central particle is lost to the other particles in the system, and the rate of energy loss is dE0 dt t→∞ → mν2 A2 2πt cos(2νt), (1.119) i.e. the energy transfer between this particle and the rest of the system has an oscillatory behaviour and goes to zero as t−1 .
- 41. May 5, 2020 14:13 book-961x669 11860-main page 24 24 Non-uniform Chain 1.8 Inverse Problem of Dynamics for a Non-uniform Chain Let us first consider the propagation of a disturbance along a non-uniform chain. If Mj denotes the mass of the jth particle in the chain and the spring constant between the particle j and j + 1 is Kj, then we have the set of equations of motion Mjẍj = Kj(xj+1 − xj) + Kj−1(xj−1 − xj), j = 1, 2 · · · N, (1.120) where xj is the displacement of the jth particle from its position of equilibrium. Following the work of Dyson we introduce the variable ξj s by [36] ξj = M 1 2 j xj, (1.121) where all Mj s are greater than zero. Also let us introduce the set of constants ω1, ω2, · · · , ω2N−2 by the following relations ω2 2j−1 = Kj Mj , ω2 2j = Kj Mj+1 , (1.122) then the equations of motion take the form ¨ ξj = (ω2j−1ω2j)ξj+1 + (ω2j−3ω2j−2)ξj−1 − ω2 2j−1 + ω2 2j−2 ξj. (1.123) Now we define the variables z1, z2, · · · , zN−1 by ż = ω2jξj+1 − ω2j−1ξj, (1.124) and then we can rewrite (1.123) as ˙ ξj = ω2j−1zj − ω2j−2zj−1. (1.125) By combining (1.124) and (1.125) we arrive at a single set of first order equations: Ẋj = ωjXj+1 − ωj−1Xj−1, (1.126) where X1, X2, · · · , X2N−1 are defined by X2j−1 = ξj, X2j = zj. (1.127) The characteristic frequencies νj of the chain are the eigenvalues of a (2N − 1) × (2N − 1) matrix Λ whose elements are given by Λj+1,j = −Λj,j+1 = iωj. (1.128) Here we want to choose the parameters Mj s and Kj s in such a way that in the limit of N → ∞ we can find an exact solution for the motion of the particles, i.e. a solvable many-body problem. Let us consider a model where ωj = jλ, (1.129)
- 42. May 5, 2020 14:13 book-961x669 11860-main page 25 An Introduction to Inverse Problems in Physics 25 where λ is a positive constant. For this choice of ωj, the ratio of the masses of two neighbouring particles according to Eq. (1.122) is Mj+1 Mj = (2j − 1)2 4j2 , j = 1, 2, · · · . (1.130) By substituting (1.121), (1.129) and (1.130) in (1.120), we find the equation of motion for xj; ẍj = λ2 (2j − 1)2 (xj+1 − xj) + 4λ2 (j − 1)2 (xj−1 − xj). (1.131) Assuming that initially only the particle j = 1 is displaced by one unit of length and with no initial velocity, while all other particles in the chain are at rest with no initial displacement, i.e. x1(0) = 1, ẋj(0) = 0, j = 1, 2, · · · , (1.132) and xj(0) = 0, j = 1, (1.133) then we can write (1.131) with the initial conditions (1.132), (1.133) as a set of equations Ẋj = λ[jXj+1 − (j − 1)Xj−1], (1.134) subject to the initial conditions X1(0) = M 1 2 1 , Xj(0) = 0, j = 1. (1.135) We can solve this problem by the same method of constructing the generating function that we discussed earlier. Let us assume that the generating function is of the form G(z, t) = ∞ j=1 Xj(t)zj−1 , (1.136) then by differentiating G(z, t) we find z ∂ ∂z (zG(z, t)) = ∞ j=1 (j − 1)Xj−1zj−1 , (1.137) and ∂ ∂z G(z, t) = ∞ j=1 jXj+1zj−1 . (1.138) Now if we multiply (1.134) by zj−1 and sum over all j values we find the following first order partial differential equation ∂G(z, t) ∂t = λ ∂G(z, t) ∂z − z ∂ ∂z (zG(z, t)) . (1.139)
- 43. May 5, 2020 14:13 book-961x669 11860-main page 26 26 Non-uniform Chain 0.1 0.2 0.3 0.4 10 5 15 20 25 30 t j=100 j=200 j=6 j=2 xj Figure 1.4: The displacement of the jth particle in the non-uniform chain as a function of time, Eq. (1.143), for j = 2, 6, 20, 100. The initial condition on G(z, t) can be found from Eqs. (1.135) and (1.136), i.e. G(z, t = 0) = M 1 2 1 . (1.140) The solution of (1.134) satisfying the boundary condition (1.140) can be expressed as G(z, t) = M 1 2 1 cosh λt (1 + z tanh λt)−1 . (1.141) Now to find Xj(t), we expand G(z, t) in powers of z and compare the results with (1.136). Thus we find Xj(t) to be Xj(t) = M 1 2 1 (−1)j−1 cosh λt (tanh λt)j−1 . (1.142) From Xj(t) we can find the displacement of the jth particle in this model; M 1 2 j xj(t) = M 1 2 1 cosh λt (tanh λt)2j−2 . (1.143) In particular the motion of the first particle is given by x1(t) = 1 cosh λt , (1.144)
- 44. May 5, 2020 14:13 book-961x669 11860-main page 27 An Introduction to Inverse Problems in Physics 27 which shows that for times t 1 λ the displacement decays exponentially in time [37]. Since we have found the exact solution of motion according to the second law of dynamics, the time-reversed motion is also a solution of the problem. The damped motion of the first particle results from its coupling to a many-particle system. 1.9 Direct and Inverse Problems of Analytical Dynamics In classical mechanics the Newton second law of motion establishes a relation be- tween the motion of the bodies and the forces acting between them. Thus at first it seems that the direct problem should be the derivation (or construction) of the Lagrangian and the Hamiltonian from the second and the third laws of motion. Historically this was the direction of the development of analytical mechanics taken by Lagrange, Hamilton, Poisson and Jacobi. They showed how useful the analytical formulation of the classical mechanics (i.e. Lagrangian and Hamiltonian formula- tions) can be, both as a concise way of writing different mechanical problems and also as a very powerful method of finding approximate and exact solutions [38]– [41]. It is also possible to start with a Hamiltonian (or Lagrangian) and obtain the equations of motion using the method of calculus of variations. That is, if we choose the classical Hamiltonian function, H(pj, qj, t), as the basic mathematical way of expressing a given motion, then the direct method shows that we can obtain a unique set of equations from the canonical equations of motion, dqj(t) dt = ∂H ∂pj , and dpj(t) dt = − ∂H ∂qj , j = 1, 2 · · · N. (1.145) In these relations qj(t) is the coordinate of the jth particle at the time t, pj(t) is its canonical momentum conjugate to qj(t), and N is the number of degrees of freedom of the system. By eliminating pj(t) s between the two equations in (1.145) we find the equations of motion for q1(t), q2(t), · · · , qn(t). Here we expect to get a unique set of differential equations for qj(t) s. Thus the direct problem is well-defined for those problems for which the Hamiltonians (or Lagrangians) are known. For the direct problem we start with Eqs. (1.145) and determine the motion in 2N dimensional phase space p1(t) · · · pN (t); q1(t) · · · qN (t), or we find the equations of motion in the coordinate space: H(pj, qj, t) → ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ mjq̈j = Fj(q1, · · · , qN ; p1, · · · , pN ) mjq̇j = Qj(q1, · · · , qN ; p1, · · · , pN ) mjṗj = Pj(q1, · · · , qN ; p1, · · · , pN ) . (1.146) Thus if we start with H(pj, qj, t), find the equations of motion either in the co- ordinate space or in the phase space, and then use these equations to obtain the
- 45. May 5, 2020 14:13 book-961x669 11860-main page 28 28 Direct and Inverse Problems of Analytical Dynamics first integrals, i.e. to solve the inverse problem. We will not always recover just H(pj, qj, t), but among the pq-equivalent Hamiltonian, there will be one Hamilto- nian that we started with. That is, we have a set of Hamiltonians giving us the correct equations of motion in coordinate space; mjq̈j = Fj(q1, · · · , qN ; p1, · · · , pN ; t) → K(pj, qj; t), (q-equivalent Hamiltonians), (1.147) and another, smaller set generating the motion in phase space; mjq̇j = Qj(q1, · · · , qN ; p1, · · · , pN ; t) mjṗj = Pj(q1, · · · , qN ; p1, · · · , pN ; t), → H1(pj, qj), (1.148) the latter is pq equivalent Hamiltonian. As a very simple example, let us consider the Hamiltonian for the damped motion of a single particle moving along the x axis: H = p2 2m e−λt . (1.149) From this H we obtain the equation of motion in the phase space dp dt = ∂H ∂q = 0, dq dt = p m e−λt , (1.150) or [43] d2 q dt2 = −λ dq dt . (1.151) This equation of motion can also be found from the more complicated Hamiltonian H1, where H1 = 1 3mp0 (2p0p) 3 2 e−λt , (1.152) and p0 is a constant. Nonuniqueness of the Hamiltonian for Conservative Many-particle Sys- tem — Starting with the equations of motion in the coordinate space for a con- servative system we find nonuniqueness in determination of the Lagrangian and the Hamiltonian functions. For instance let us consider a system of N interacting particles with the equations of motion mj d2 rj dt2 = −∇jV (r1, · · · , rN ), j = 1, 2, · · · , N. (1.153) If we sum (1.153) over all j s we find that d dt N j=1 mjvj = 0, (1.154)
- 46. May 5, 2020 14:13 book-961x669 11860-main page 29 An Introduction to Inverse Problems in Physics 29 which implies the conservation of total momentum of the system. We can construct a Galilean invariant Lagrangian by adding a term 1 2μ N j=1 (mjvj) 2 , (1.155) to the standard Lagrangian which is: L (r, ṙ) = N j=1 mjv2 j − V. (1.156) Thus if μ is a constant having the dimension of mass, then we write L(μ) as Lμ (r, ṙ) = N j=1 mjv2 j + 1 2μ N j=1 (mjvj)2 − V. (1.157) The equations of motion derived from (1.157) are given by (1.153), since the time derivative of the total momentum is zero, Eq. (1.154). However the relation between the canonical momentum pk and the velocity vk is now more complicated, i.e. pk = ∂Lμ (r, ṙ) ∂vk = mkvk + mk μ N j=1 mjvj. (1.158) From this Lagrangian we find a Hamiltonian which also depends on the parameter μ; Hμ (r, p) = N j=1 p2 k 2m − ( N j=1 pj)2 2μ + N j=1 mj + V, (1.159) where pk s are given by (1.158). 1.10 From the Classical Equations of Motion to the Lagrangian and Hamiltonian Formulations For most conservative systems it is easy to write down the standard Lagrangian (or the Hamiltonian) describing the motion. However if the forces acting on particles in a conservative system are known, we can ask whether it is possible to construct a unique Lagrangian (or Hamiltonian) for the motion or not? But when the system is dissipative, then even the question of existence of any Lagrangian (or Hamiltonian) is questionable. Helmholtz’s Conditions for Existence of a Lagrangian — The non-uniqueness of the Lagrangians function for dynamical systems will be studied in detail later.
- 47. May 5, 2020 14:13 book-961x669 11860-main page 30 30 Equations of Motion and the Lagrangian But first let us examine the conditions for the existence of a Lagrangian function and how it can be constructed [38]–[41]. The laws governing the motion of a dynamical system can be formulated ac- cording to Hamilton’s principle and this principle states that there is a function L (x1, · · · xN ; ẋ1 · · · ẋN , t) and that this L satisfies the following condition [2]: If the system at the instant t1 has the coordinates x (1) i and at t2 has the co- ordinates x (2) i then this condition implies that the functional differential δS defined by δS = δ t2 t1 L (xi, ẋi, t) dt = t2 t1 N j=1 δL (xi, ẋi, t) δxj δxj(t)dt = t2 t1 j,k μjk (xi, ẋi, t) ẍk − 1 m fk (xi, ẋi, t) δxj(t)dt, j = 1, 2, · · · N, (1.160) must vanish identically for any continuous variation of the path consistent with the requirement δxj (t1) = δxj (t2) = 0. (1.161) In Eq. (1.160) μjk (xi, ẋi, t) is a positive definite matrix which may be viewed as an integrating factor [38]–[42]. Thus the requirement of the extremum of action S implies the vanishing of the integrand between any two arbitrary instances t1 and t2. Since we have imposed the condition of positive definiteness on the integrating factor μjk and since the variation δxj(t) is arbitrary (apart from the continuity condition and the variational conditions of (1.161)), therefore we conclude that ẍk(t) = 1 m fk (xi, ẋi, t) , k = 1, 2, · · · N, (1.162) which is Newton’s second law of motion. The Lagrangian function determined from the requirement of δS ≡ 0 is not unique. We can add the total time derivative of any function Φ (x, ẋ, t) to the Lagrangian without affecting the resulting equations of motion [43]. For the special case of μjk = δjk we find the Lagrangian to be the same as that given by (1.156). The second variation of S with respect to δxj(t) and δxk(t ) must satisfy the identity: δ2 S δxj (t) δxk (t) = δ2 S δxk (t) δxj (t) . (1.163) By imposing this condition on (1.160) we find the following set of equations which μjk s have to satisfy: μjk = μkj, (1.164) ∂μkj ∂ẋl = ∂μjl ∂ẋk = ∂μlk ∂ẋj , (1.165) D̂μjk + 1 2m l μjl ∂fl ∂ẋk + μkl ∂fl ∂ẋj = 0, (1.166)
- 48. May 5, 2020 14:13 book-961x669 11860-main page 31 An Introduction to Inverse Problems in Physics 31 and D̂ k μik ∂fk ∂ẋj − μjk ∂fk ∂ẋi − 2 k μik ∂fk ∂ẋj − μjk ∂fk ∂ẋi = 0, (1.167) where in Eqs. (1.166) and (1.167) D̂ denotes the following differential operator D̂ = ∂ ∂t + k ẋk ∂ ∂xk + 1 m fk (xi, ẋi, t) ∂ ∂ẋk . (1.168) These are the Helmholtz conditions [38], [39]. We can construct the Lagrangian from the equations of motion mẍk = fk (xi, ẋi, t) , k = 1, 2, · · · N, (1.169) by solving a linear partial differential equation. To this end we first write the following identity: δL δxk = ∂L ∂xk − d dt ∂L ∂ẋk ≡ ∂L ∂xk − ∂2 L ∂t∂ẋk − j ẋj ∂2 L ∂xj∂ẋk − j ẍj ∂2 L ∂ẋj∂ẋk = 0, (1.170) and then we substitute for ẍ from the equation of motion to obtain ∂L ∂xk − ∂2 L ∂t∂ẋk − j ẋj ∂2 L ∂xj∂ẋk − 1 m j ∂2 L ∂ẋj∂ẋk fj (xi, ẋi, t) = 0, k = 1 2, · · · N. (1.171) This equation together with the set of equations (1.164)–(1.167) determine both L(x, ẋ, t) and μ(x, ẋ, t). For another method of constructing Lagrangians and Hamiltonians for non- linear systems the reader is referred to a paper by Musielak et al. [44]. In the case of a one-dimensional motion Eq. (1.171) simplifies and later we will use it to find the Lagrangian for simple damped systems. But there are other ways of constructing the Lagrangian for motions confined to one dimension [45], [46]. For instance we can start with the equation of motion mẍ = f (x, ẋ) , (1.172) and write the Lagrangian for this motion as L (x, ẋ) = ẋ ẋ 1 v2 G(v, x)dv, (1.173) where G(v, x) is a function to be determined later. By substituting L (x, ẋ) in the Euler–Lagrange equation we find d dt ∂L ∂ẋ − ∂L ∂x = ẍ ẋ ∂G (ẋ, x) ∂ẋ + ∂G (ẋ, x) ∂x ≡ mẍ − f (ẋ, x) . (1.174)
- 49. May 5, 2020 14:13 book-961x669 11860-main page 32 32 Equations of Motion and the Lagrangian From Eq. (1.174) it follows that G (ẋ, x) is the solution of the following differential equation 1 mẋ f (x, ẋ) ∂ ∂ẋ G (ẋ, x) = − ∂ ∂x G (ẋ, x) , (1.175) provided that 1 ẋ ∂ ∂ẋ G (ẋ, x) = 0. (1.176) Now let us consider two examples using this method of construction of the Lagrangian. When the equation of motion for a damped system is of the form ẍ + ẋ dg(ẋ) dẋ dV (x) dx = 0, (1.177) then the Lagrangian which is found by solving Eq. (1.175) is L (x, ẋ) = ẋ ẋ 1 v2 F [z(x, v)] dv, (1.178) where F is an arbitrary function of its argument and z(v, x) = g(v) + V (x). (1.179) However the condition 1 v dg(v) dv dF(z) dz = 0, (1.180) must be satisfied. Thus for the very simple case of ẍ + λẋ = 0 and for F(z) = z we find L (x, ẋ) = ẋ ln ẋ − λx. (1.181) As a second example let us consider the damped harmonic oscillator ẍ + λẋ + ω2 0x = 0. (1.182) By calculating G (ẋ, x) we find that G (ẋ, x) = F(z), where now z = 1 2 x + λ 2 − iω ẋ ν∗ x + λ 2 + iω ẋ ν = z∗ , (1.183) with z∗ being the complex conjugate of z and ν2 = λ 2 + iω λ 2 − iω , (1.184) with ω2 = ω2 0 − λ2 4 . These q-equivalent Hamiltonians are acceptable as generators of motion in an- alytical dynamics, and depending on the integrating factors used, there are infinitely many Hamiltonians. However as we will see in Sec. 3.3 they are not admissible for
- 50. May 12, 2020 11:33 book-961x669 11860-main page 33 An Introduction to Inverse Problems in Physics 33 quantizing a particular system. Since their classical phase space trajectories are very different from those of the standard Hamiltonians they may not be also admissible even in classical statistical mechanics. The Inverse Problem for Liouville Equation — The inverse problem of the classical Liouville equation can be described as the problem of finding a time- dependent Hamiltonian H which generates a given phase space density ρ, and both the Hamiltonian H and the density ρ satisfy the same Liouville equation [6]. Frisch and collabortors have shown that, subject to some physically well-defined subsidiary conditions, the solution exists and is unique, when ρ is chosen to be the solution of the standard Fokker–Planck equation when the particles are free, are in a constant external field, or are harmonically bound. Let us consider a system of particles and denote the coordinate and the canon- ical momentum of the jth particle by qj and pj respectively. As we have seen ear- lier we can derive the equation of motion of these particles from a time-dependent Hamiltonian H(q, p, t) ≡ H(qi, pi, t), Eq. (1.145). We assume the following initial conditions for the first order equations of motion: qj(0) = qj, pj(0) = pj, H(0) = H0. (1.185) In classical statistical mechanics one wants to find the time-dependent density ρ(q, p, t) ≡ ρ (1.186) corresponding to a given H = H(q, p, t) which satisfies Liouville’s equation ∂ρ ∂t + {ρ, H} = ∂ρ ∂t + j ∂ρ ∂qj ∂H ∂pj − ∂ρ ∂pj ∂H ∂qj = 0, (1.187) subject to the initial condition ρ(0) = ρ0(q, p). (1.188) Now we define the dynamical inverse problem for the Liouville equation by the fol- lowing procedure [6]: (1) We assume a marginally consistent, time-dependent point process given by ρ(q, p, t) and its marginal distributions. (2) We try to find a “Hamiltonian” function H(q, p, t) corresponding to the ρ given in (1.186) which is a solution of Eq. (1.187). By considering (1.187) as a linear partial differential equation for H = H(q, p, t), we have the characteristic equations for H dpj − ∂ρ ∂qj = dqj ∂ρ ∂pj = dH ∂ρ ∂t . (1.189) These equations are not identical with Hamilton’s canonical equations (1.145) for H except when ρ is judiciously chosen [6]. Solution of (1.189) only determines the general solution of (1.187). If one specifies the initial condition on H and in addition provides information about H on a characteristic strip or singular region of phase space, then one can show the existence of the Cauchy problem for H with very weak conditions on the continuously differentiable density ρ [48].
- 51. May 14, 2020 8:55 book-961x669 11860-main page 34 34 References 1.11 Langevin and Fokker–Planck Equations If a particle of mass m is immersed in a fluid with the friction coefficient γ, its equation of motion (in one dimension) is given by dv dt + γv = −K(x) + A(t). (1.190) In this equation v is the speed of the particle, x is its position, K(x) is the acceler- ation due to mechanical forces and A(t) is the acceleration due to random forces. The Fokker–Planck equation for the distribution in phase space of the single particle, P(x, v, t) is given by ∂P ∂t + v ∂P ∂x − K(x) ∂P ∂v = γ ∂(vP) ∂v + Dγ2 ∂2 P ∂v2 . (1.191) The solution of (1.191) subject to the boundary condition P(x, v, 0) = δ(x − x0)δ(v − v0), (1.192) can be obtained for the following forms of K(x): K(x) = ⎧ ⎪ ⎨ ⎪ ⎩ 0 free particle F0 a constant force ω2 x harmononically bound particle . (1.193) Solutions of the Fokker–Planck equation for these cases are given in [6], and the gen- eral Hamiltonian for a the motion of a particle in a viscous medium with frictional force βv is discussed in detail in [49]–[51]. References [1] I. Newton, The Principia: Mathematical Principles of Natural Philosophy, (University of California Press, 1999). [2] L.D. Landau and E.M. Lifshitz, Mechanics, (Pergamon Press. Oxford, 1960), Chap. III. [3] N.H. Abel, J. Reine Angew. Math. 1, 13 (1826). [4] A.H. Carter, Am. J. Phys. 68, 698 (2000). [5] M. Bocher, An Introduction to the Study of Integral Equations, Cambridge Tracts in Mathematics and Mathematical Physics, No. 10 (Hefner, New York, 1960).
- 52. May 5, 2020 14:13 book-961x669 11860-main page 35 An Introduction to Inverse Problems in Physics 35 [6] R. Gorenflo and S. Vessella, Abel Integral Equations, Analysis and Applications, (Springer-Verlag, Berlin, 1991). [7] I.S. Gradshteyn and I.M. Ryzhik, Tables of Integrals, Series, and the Products, (Academic Press, New York, 1965) p. 875. [8] G. Arfken, Mathematical Methods for Physicists, Third Edition, (Academic Press, New York, 1985) p. 875. [9] E.T. Whittaker, A Treatise on the Analytical Dynamics of Particles and Rigid Bodies, Fourth Edition (Cambridge University Press, London 1960). [10] H. Goldstein, Classical Mechanics, (Addison-Wesley, Reading, 1956). [11] J.F. Marko and M. Razavy, Lettere Al Nuovo Cimento, 40, 533, 1984. [12] J.B. Keller, Am. Math. Monthly, 83, 107 (1976). [13] O.B. Firsov, Zh. Eksp. Teor. Fiz. 24, 279 (1953). Firsov’s solution is describled in L.D. Landau and E.M. Lifshitz, Mechanics, (Pergamon Press, Oxford, 1960) p. 52. [14] J.B. Keller, I. Kay and J. Shmoys, Phys. Rev. 557, 102 (1956). [15] M.T. Chu and G.H. Golub, Inverse Eigenvalue Problems: Theory, Algorithm and Applications, (Oxford University Press, London, 2005). [16] G.M.L. Gladwell, Inverse Problems in Vibration, Second Edition, Volume 119 of Solid Mechanics and Its Applications, (Dordrecht, Kluwer, 2004). [17] See for instance The Symmetric Eigenvalue Problem, (SIAM, Classics in Ap- plied Mathematics, 1988). [18] D. Belkić, in Advances in Quantum Chemistry V. 145, Chapter 4 (2011). [19] I.M. Gel’fand and S.V. Fornin, Calculus of Variation, translated by R.A. Sil- verman, (Dover, New York, 1991). [20] C. Lanczos, J. Res. Natl. Bur. Stand. 45, 255 (1950), ibid. 49, 33 (1958). [21] For other methods of inversion of a Jacobi matrix see D.N. Ghosh Roy. Methods of Inverse Problems in Physics, (CRC Press Boca Raton, 1991) Chapter 11. [22] G. Meurant, SIAM J. Matrix Anal. Appl. 13, 707 (1992). [23] X. Wu and E. Jiang, J. Shanghai University, 11, 27 (2007). [24] G.M.I. Gladwell and N.B. Williams, Inverse Problems, 4, 1013 (1988). [25] R.A. Usmani, Computer Mat. Appl. 27, 59 (1994). [26] J.H. Wilkinson, The Algebraic Eigenvalue Problem, (Oxford University Press, Oxford, 1965) p. 388.
- 53. May 5, 2020 14:13 book-961x669 11860-main page 36 36 References [27] A.S. Housholder, The Theory of Matrices in Mathematical Analysis, (Blaisdell, New York, 1964) p. 254. [28] G.H. Golub and G. Meurant, Matrices, Moments and Quadrature with Appli- cations, (Princeton University Press, Princeton, 2010), Chapter 4. [29] R. del Rio and M. Kudryavtsev, Inverse Problems, 28, 055007 (2012). [30] Y. Wei and H. Dai, J. Comp. Appl. Math, 300, 172 (2016). [31] H.J. Kreuzer, Nonequilibrium Thermodynamics and Its Statistical Foundations, (Oxford University Press, Oxford, 1981). [32] E. Schrödinger, Ann. Phys. (Leipzig), 44, 916 (1914). [33] R. Bellman and K.L. Cook, Differential-Difference Equtions, (Academic Press, New York, 1963). [34] M. Razavy, Can. J. Phys. 57, 1731 (1979). [35] I.S. Gradshtetyn and I.M. Ryzhik, Table of Integrals, Series and Products, (Academic Press, New York, 1965) p. 973. [36] F.J. Dyson, Phys. Rev. 92, 1331 (1953). [37] M. Razavy, Can. J. Phys. 58, 1019 (1980). [38] P. Havas, Nuovo Cimento Supp. 5, 363 (1957). [39] H. Helmholtz, J. Reine Angew. Math. 100, 137 (1887). [40] R.M. Santilli, Foundations of Theoretical Mechanics: The Inverse Problem in Newtonian Mechanics, (Springer-Verlag, New York, 1978). [41] V. Dodonov, V.I. Man’ko and V.D. Skarzhinsky, Had. J. 4, 1734 (1981). [42] C. Lanczos, The Variational Principles of Mechanics, Fourth Edition, (Dover, New York, 1986). [43] For a number of similar examples see M. Razavy Classical and Quantum Dis- sipative Systems, Second Edition (World Scientific, Singapore, 2017). [44] Z.E. Musielak, D. Roy and L.D. Swift, Chaos, Solitons Fractals, 38, 894 (2008). [45] J.A. Kobussen, Acta Phys. Austriaca, 59, 293 (1979). [46] S. Okubo, Phys. Rev. A 23, 2776 (1981). [47] H.L. Frisch, G. Forgacs and S.T. Chui, Phys. Rev. A 20, 561 (1979). [48] H. Bateman, Differential Equations, (Chelsea, New York,1966).
- 54. May 5, 2020 14:13 book-961x669 11860-main page 37 An Introduction to Inverse Problems in Physics 37 [49] M. Razavy, Can. J. Phys. 56, 311 (1978). [50] M. Razavy, Can. J. Phys. 56. 1372 (1978). [51] M. Razavy, Classical and Quantum Disiipative Motion, Second Edition, (World Scientific, Singapore, 2017).
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