Lecture Notes in Mathematics A collection of informal reports and seminars Edited by A. Dold, Heidelberg and B. Eckmann, Z0rich
167 Lavrentiev. Romanov Vasiliev Computing Center of the Academy of Sciences Novosibirsk / UdSSR
Multidimensional Inverse Problems for Differential Equations
$ Springer-Verla9 Berlin-Heidelberg. New York 1970
I S B N 3-540-05282-8 S p r i n g e r V e d a g B e r l i n • H e i d e l b e r g • N e w Y o r k I S B N 0-387-05282-8 S p r i n g e r Verlag N e w Y o r k • H e i d e l b e r g • Berlin
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CONTENTS
INTRODUCTION CHAPTER
I
................................................... Some
-
1. P r o b l e m
Problems
Ellipsoids
Problem
Integrals
along Center
the
4. On the
Problem
Values
CHAPTER
of D e t e r m i n i n g
2
over
Equation I. S t a t e m e n t
CHAPTER
Formulations
Problem
3 - Linearized
1. F o r m u l a t i o n
CHAPTER
4
-
of the
I.
Inverse
Heat
2. n - D i m e n s i o n a l 3. A p p l i c a t i o n
CHAPTER
5 - Inverse
Problem
and
Inverse
Linearlzed
Space
from
13
Its M e a n
for the
19
Telegraph
Its L i n e a r i z a t i o n
Problem
in
22 ..
Inverse
Problem
Equation
Inverse
Kinematic
Problem
Problem
of the
Heat
Its
Linearlzed
to G e o p h y s i c s
Conduction
Sources
for the
Linearization
Version
.......
of the
with
Inverse
Problems
Heat
Problems
Problems
for
Conduction
for S e c o n d - O r d e r
33 33
Inverse 36
Continuously
a Half-Plane
59
.......
59
..........
45
...............
49
Problems
to G e o p h y s i c s
31
Wave
.........................
Problems
28
for the
.....................................
Conduction
of the
and
24
in
................................. Differential
22
Two-
.........................................
Problem
Inverse
to R o t a t i o n
.......................................
of the
Active
from
...........................................
2. A n A p p l i c a t i o n Kinematic
Invariant
a Function
Problem
of a N o n l i n e a r
Equation
a Circle
..........................
Dynamic
Inverse
Three-Dimenslonal
Inverse
2 10
.....................................
Inverse
Space
4. D e r i v a t i o n
inside
1
over
.......................
of C u r v e s Circle
0ne-Dimenslonal
Dimensional
Integrals
...........................................
of the
2. L i n e a r i z e d
5. T w o
of the
of D e t e r m i n i n g
Linearized
-
Curves
a Function
a Family
Circles
.................
from
................................
to A n a l y t i c
about
Geometry
a Function
of R e v o l u t i o n
2. G e n e r a l i z a t i o n 3.
of I n t e g r a l
of D e t e r m i n i n g
IV
Elliptic
Equations
51
i.
Inverse
Problem
for E q u a t i o n
(i)
in a H a l f - P l a n e
........
52
2.
Inverse
Problem
for E q u a t i o n
(1)
in a H a l f - S p a c e
........
54
BIBLIOGRAPHY
.........................................................
57
INTRODUCTION An inverse problem for a differential
equation
the determination
or right-hand
tial equation
of the coefficients
from certain
Two significant
advances
inverse problems
functionals
[19],
[23],
operator
function of the operator. ving the determination equation
for STURM-LIOUVILLE
equations
the coefficient
(Ill,
in a second-
is required to be found from the spectral
In [19] and
[2], a number of problems
of the coefficients
are shown to be reducible
equations.
in the study of
equations.
[36]). In these problems,
order differential
side of a differen-
of its solution.
have been made previously
for differential
The first is in inverse problems [13],
is any problem involving
of a partial
to inverse problems
It is assumed there that the coefficients
invol-
differential for STURM-LIOUVILLE
are functions
of
a single variable. The second is in problems
of potential
[33]). In the inverse problems elliptic partial differential restrictive Thus,
additional
in [17] and
equation
theory
([17],
conditions
restrictions
is a star-shaped
dealing with the inverse problem of potential Until now, multidimensional ly little consideration. coefficients
rally arbitrary function space. Multidimensional M. BERZANSKII.
functions
inverse problems
some multidimensional
In [8] and
i
only,
side in other papers
problem,
the required
equations
are gene-
belonging to a certain
problems were first investigated
functions.
and
The same sort of
have been given comparative-
of several variables
in the papers of Ju.
theorem was proved for the solution
to the inverse problem for SCHR~DINGER'S wise analytic
0
sides of the differential
In [4] a uniqueness
side.
theory.
In a multidimensional
or right-hand
Highly
side in POISSON'S
domain.
are also imposed on the right-hand
[29],
side of an
are imposed on the right-hand
is assumed to be a function having values 1
[25],
equation has to be determined.
[20], the required right-hand
and the set where it is
[20],
of this type, the right-hand
equation in the class of piece-
[9] functions were constructed
inverse problems
of quantum scattering theory
that are similar to the GEL'FAND-LEVITAN
functions
occurring in the
for
-
V -
inverse problem for the STURM-LIOUVILLE This monograph
investigates
whose formulations
equation.
a number of multidimensional
inverse problems
differ from those of the papers mentioned above. A
portion of the results have been published
as short notes
([21S,
[22],
[30], Di]). A characteristic property
aspect of multidimensional
of not being well-posed
advantageous
to make use of the general notions
posed problems theoretical HADAMARD
developed
in [16],
[20],
of proofs of uniqueness
theorem.
theorems
is their
Thus,
and approaches
[34], and
it is
to ill-
[55]. Central to the
study of a problem that is not well-posed
is the proof of a uniqueness
primarily
inverse problems
in the sense of HADAMARD.
in the sense of
The monograph
consists
for the formulations
in
question. The inverse problems
for which uniqueness
theorems will be proved are
linear and their solution is reduced to the solution of first-order linear operator equations. them numerically [20] and
Thus algorithms
by application
of the general methods
the problems.
developed
the stability
compact
are uniformly
sets
of the various
(for example,
in
differential
formulations
the set of functions
on certain whose gradients
here to the study of inverse problems equations
over to higher-order
equations
The inverse problems
considered
to problems
for
of estimates
bounded).
We shall confine ourselves second-order
lead to special algorithms
They also make possible the derivation
characterizing
aspects
to solve
[35] for linear equations.
The methods used to prove uniqueness
specific
may be constructed
of integral
geometry
although some of the methods
for carry
~8]. for hyperbolic
equations
are reducible
and so Chapter i is devoted to some
of it. Chapter 2 establishes
uniqueness
to the inverse problem for the telegraph
theorems
for the solution
equation with the help of the
results of Chapter i while Chapter 3 does the same thing for the wave equation. Chapters 4 and 5 deal with inverse problems
for the heat equation and
for elliptic equations. They are reduced to the solution of certain integral equations of first kind. In Chapters
3 and 4, some applied problems
to corresponding
versions
are discussed that lead
of the inverse problem.
CHAPTER I Some Problems of Integral Geometry In accordance with the terminology used in
~2],
an Integral-geometrlc
problem is any problem involving the determination of a function defined in a domain through its integrals along a family of curves in the domain. One of the earliest and most familiar versions of such problems determination of a continuous
function of
is the
n-variables through its mean
values over spheres of arbitrary radius with centers lying on an (n-l)dimensional hyperplane.
The uniqueness
proved in R. COURANT'S book
of a solution to the problem is
[6]. At present,
the determination of a
function from its integrals over hyperplanes has been the problem dealt with most fully In monographs
[12]. The authors of the book, I.M. GEL'FAND,
M.I. GRAEV and N.Ja. VILENKIN were led to its consideration while working on problems In representation theory. The appearance of this book furthered to a considerable extent the development of Integral-geometrlc
problems.
and systematic elaboration
Some differently formulated problems of
integral geometry are contained in F. JOHN'S book As already indicated in the introduction,
we were led to problems
integral geometry while considering llnearized problems for the simplest
[18]. in
inverse
equations of mathematical physics.
Some of
them lead to the familiar problem of determining a function from its mean ~alues over spheres and others to new problems which it is the purpose of this chapter to consider.
The problem of finding a function
from its integrals is not well-posed in the sense of HADAMARD.
We shall
therefore proceed from A.N. TIHONOV'S notion and we shall prescribe the function space to which the solution of a problem is to belong. With applications
of integral geometry to the study of linearlzed problems
mind, the most natural one for our purposes is the space uous functions.
Throughout
C
in
of contin-
the following we shall assume the solutions
of corresponding problems in integral geometry to belong to the space C
or to some subset of it. We shall be interested in uniqueness
for theseproblems,
the determination of inversion formulas
theorems
for them
and the set of functions for which these formulas are valid. The latter is related to the question of existence of a solution. Section i considers the problem of reconstructing a two-dlmensional function from its integrals over a family of ellipses with one focus
-
2
-
fixed and the other running over the points
of a llne.
At the end of
the section it is shown how these results may be rephrased to encompass the p r o b l e m of r e c o n s t r u c t i n g grals over ellipsoids I are generalized
a function of
of revolution.
is also g e n e r a l i z e d
The formulation
by the introduction
grals along the curves.
Section
invariant to rotation
already,
uniqueness
for c o n s t r u c t i n g
of a weight
case
the d e t e r m i n a t i o n
of a funct-
over a family of In Section
for the p r o b l e m of d e t e r m i n i n g Although,
has been proved before function,
problem
factor in the inte-
from its integrals
over circles.
the required
of Section
a special
about the center of the circle.
formula is obtained
from its mean values
from its inte-
2 the results
of the integral-geometric
3 considers
ion inside a circle of unit radius an inversion
variables
to curves of a more general nature
being that of ellipses.
curves
n
In Section
a function
as we have indicated
and a basic a l g o r i t h m
the inversion
exists
formula has been
lacking for this problem. 1. Problem of D e t e r m i n i n 6
a Function
from Inte6rals
over E l l i p s o i d s
of Revolution Consider the following p r o b l e m
In
(n+l)-dimensional
grals of a function
u(x,s)
family of ellipsoids
with one focus
running over all points tained by r e v o l v i n g to be considered.
= u(xl,x2,...,Xn,S)
The inte-
are p r e s c r i b e d
on a
fixed at the origin and the other
of the hyperplane
an ellipse
space.
s = O. Only ellipsoids
ob-
around the llne Joining the two foci are
It is required to determine
u(x,s)
from the given
integrals. Denote the coordinates and by
S
of the second
the ellipsoid
focus by
of revolution
(x°,O)
= (x~,x~ .... ,x°,O).
defined by
x°,t (I) where
r(x,s,O,O) r(x,s,O,O)
and the foci Thus,
(0,0)
and
+ r(x,s,x°,O)
r(x,s,x°,O)
and
(x°,O),
= t
are the distances
between
(x,s)
respectively.
let the function
(2) be known.
v(x°,t) =/u(x,s)d~ Here
the origin.
~
Sxo,t is the solid angle
In accordance
in
(x,s)-space
with the above discussion,
4
with vertex at we shall assume
-3-
that
u(x,s)
apparently
belongs to the space of C-functions.
only m e a n i n g f u l
From
solution to this p r o b l e m in the class of even functions we shall do in the following.
In addition,
of generality
in a s s u m i n g that
computations,
we shall take
be played by ellipses At the c o n c l u s i o n
u(O,O)
n = I
of the section,
of a
s
is o b v i o u s l y
and this no loss
so that the role of ellipsoids w
will
will be the polar angle ~ .
we show how the c o r r e s p o n d i n g
results
n.
n = I , the following u n i q u e n e s s
T h e o r e m 1: If equation
there
of
: 0. To simplify the subsequent
and the solid angle
carry over to the case of a r b i t r a r y For
(2) it is
to pose the question of uniqueness
t h e o r e m holds.
(2) has a solution belonging to
C
satisfying
a H~Ider condition in a neighborhood of the origin,
then it
is unique. By solution here, we mean an even function of
s
v a n i s h i n g at the origin.
The idea of the proof of the t h e o r e m is to find all moments To this end,
it is convenient
related to the cartesian (3)
of an ellipse
(4)
and
eccentricity and
~os~
t
~
(x,s)
,
Sin ~
s : r
.
is
cos~)-1
are p a r a m e t e r s
of the ellipse;
u(x,s). (r,#)
by the formulas
in polar coordinates
r = p(1-~
p
where
coordinates
X : r
The equation
of
to go over to p o l a r coordinates
characterizing
the polar distance
they are e x p r e s s i b l e
in terms of
and
x°
by x° z : K- '
(4a) Formula
p :
t
(I - c 2)
.
(2) then becomes 2~
(5)
J u(r cosg,
r ~in~)d~
: v(p,~) ,
0 with
r
g i v e n by ( 4 ) .
We a p p l y t o b o t h s i d e s
of
defined by
(6)
Lv = p ~
v(z,¢)
dz
-
Z
0
•
(5) t h e o p e r a t o r
L
-4-
When equation (5) has a solution in C , both its legitimacy and the result of applying it are substantiated by the following sequence of equations : p 2~ ~c / LV : P T~
f
=PT[ (6a)
dz o ~ U ( r z C O S ~ , ~--
d~
0
: Py
rz s i n ~ ) d ~
u(r z e o s ~ ,
r z sin~)
d,.
~--
0 P
d~ ~-~ S
0 2~
u(r c o s ~
,
r sin~)
dr ~--
0
=/
U(rp c o s ~ ,
rp s i n ~ )
0
rp c o s ~ d f : / u ( x , s ) x d ~ p,c
The subscripts p and z on r indicate which of these two parameters is to be substituted in formula (4) when calculating r. Sp,¢ denotes an ellipse with parameters p and ¢ . Applying the operator
(7)
L
repeatedly to the resultant equation
Y
U(X,S)Xd~=
Lk
the k-fold iteration of
/
u(x,s)xkd~=
LV
S
and denoting by way
(8)
Sp,¢
L, we obtain in a similar
Lkv (k = 1,2,3,...)
If we define L°v m v(p,¢), formula (8) is also valid for k = 0 . Thus we have constructed in unique fashion a system of moments on each ellipse. Since u(x,s) is an even function of s, it is uniquely determined by these moments. exists, it is unique.
In other words,
We next consider how relations u(x,s)
explicitly in terms of
if a solution to equation
(4)
(8) may be used to express the function v(p,¢). At the same time, we shall study
-5
the properties to equation
needed by
v(p,c)
-
to assure the existence
(5). At this Juncture,
we shall slightly
of functions
for which we have proved the uniqueness
ted function
from its integrals
sider functions I °. each
u(r,~)
u(r,~)
disc
over ellipses.
is a continuous
positive
2 ° . In a n e i g h b o r h o o d
contract
the class
of the r e c o n s t r u c -
Namely,
we shall con-
satisfying the following conditions
r ~ r o , it is even in
arbitrary
of a solution
function of its arguments ~
and
u(O,~)
= O. Here
: in the r°
Is an
number. of the polar origin,
u(r,~)
satisfies
a HOLDER
condition,
(9)
lu(r,~)
where 3 ° . Each
A
and
,
u(r,~)
I ~ Ar~,
(~ > O)
are constants.
satisfies
(Io) k:O
the inequality
max l uk(r) r
]< ®
wherein 2~ ~(r)
: ~/u(r,~)
(Ii)
(k : 1,2,3,...)
cosk~d~,
o 2~
0 The functions class
for which conditions
I°-3 ° hold will be designated
as
U.
We shall also make a slight Consider
a circle of radius
of eccentricity to determine
0 ~ c < 1
a function
change in the statement r°
in the
of the problem.
(x,s)-plane
and all ellipses
falling inside the circle.
u(r,~)
~ U
It is required
from its integrals
over this
family of ellipses. Let
v(p,¢)
be a function
for which a solution to
parameter
c
radius
and the following relation
(12)
p
A
(P,~ ) c°sk~df=
0
(5) exists.
tend to zero in (8). Each ellipse becomes results
P-k-Lkv-z=O[] .
Let the
a circle of
in the limit
:
(k = 0,1,2,...)
-6-
For each fixed value of construct u G U
by virtue of condition
ellipses
having eccentricity
arbitrarily disc
r, these relations
a Fourier series for
u(r,~)
3 ° . Thus knowing the integrals ranging in the interval
small positive number,
r ~ ro
alone can now be used to
which is convergent
in an arbitrarily
small strip
u(r,~)
the function
0 ~ e i 6 , we determine
~ ~ e < i. This implies
~ is an
in the
along all ellipses
lying in this disc. This means that if we prescribe v(p,e)
along
0 i e ~ 6where
we can determine
and hence we can find its integrals
for any
it
completely
for
in turn that to any arbitrary
continuous
function there exists no solution to the stated problem.
This result relates to the fact that the given problem is not well-posed in the sense of HADAMARD.
Actually
belonging to the function
spaces
property
obtained above for
perties the functions sufficient
v(p,e).
u(r,~ )
for the existence
no closed linear manifold C k, H, Lp
of a
Mk v : I
of functions
Wp . (z) possesses the
Below we shall indicate what pro-
must possess u(r,~)
Consider a family of linear operators
(13)
or
Mk
that are necessary satisfying
and
equation
(5).
defined by the relations
[~] [~-J k ) (J~2))p2J-kLk-2J v ' ~ (-l)J( ~ (2(J+g) J=O ~=0 (k=i,2,...)
and I MoV : ~
(13a)
Using the system of relations to (5) results
(14)
1
It should
be
5u(~{[t+/t2_l]k+[t_
t2~_l]k }
cos~d~: t-
p,e
noted that for
Nevertheless,
e = O, equation
(8), we find that the application
of
Mk
in the formula / S
complex.
i v(p,~) L°v --~-~
Itl < i
(14) can be written
l-ecos~
the expressions
the entire expression
Mkv
in brackets
in braces
are
is real. For
as
2~ ~i f O
(14,)
Hence,
u(r,~)
cosk~d~
:
[MkV]~: o
if a solution to (5) exists which belongs to
in terms of
(i5)
v(p,e)
through the formula
u(p,~) =kiO[Mkv]e=O cosk~' .
U, it is expressible
-7-
The convergence of this series for any
u~U
follows from (14') and
condition 3 ° on the function u ( r , ~ ) . We now examine what properties are possessed by V, the image of the set U under the correspondence definable by (5). Theorem 2: The image
V
of
U
under (5) has the following properties:
i °. The functions MkV (k = 0,i,2,...) corresponding to a by (13) exist and are contlnuous t and [MkV]p=o = O. 2° . For any
v(p,a)
v(p,~)~V
the series
(16) k=O
maxlMkVlc:O p
: ~v
is convergent. 3 ° . The function u ( r , ~ ) constructed for satisfies the HGLDER condition, (17) 4 °. Each
lu(r,9')l v(p,¢)
v(p,¢)
from formula (i5)
~ Ar ~, (~ • 0) .
satisfies the inequality
(18)
Iv(p,~)i
~ 2~ ~v "
Properties 1°-3 ° follow in a trivial way from the corresponding properties of the functions u(r,~) and equations (14), (14') and (15). Only inequality (18) remains to be proved. To this end, note that the function u ( r , ~ ) given by (15) has to satisfy equation (5), i.e., the following identity must hold: (18a)
f 0
~ [MkV] e=0 c o s k ~ d ~ k=O p÷r
s v(p,¢)
.
[MkV] c :0 should be interpreted to mean that MkV Is first to be p÷r evaluated at ¢ = 0 and then p replaced by r as given by (4). From this Identlty we deduce that (18b)
Iv(p,c)l ~ 2~
and the theorem Is proved.
~ maxlMkVl¢:0 k=O p
= 2~ ~v
-8-
Theorem 3: A necessary and sufficient condition for equation
have a solution belonging to The necessity v(p,e)
to
follows
V
is that
v(p,¢)
belong to V.
from Theorem 2. Let us show that the belonging
is sufficient
consider for such a
U
(5) to
v(p,¢)
(i9)
for the existence
of a solution.
of
Indeed,
the series
[Mkv] :ocosk : k=O
By virtue of property continuous
function
coefficients
for
By properties even in
~
equation
2 ° , it is uniformly u(p, ~ )
u(p,~)
I°-3 ° of
and
of
Moreover, [MkV]~=O
the function u(p,~)
S
$ ~ V. Applying
Or by the linearity
of
(20b)
Mk
is in
is
the function
.
to equation
(20), we obtain
=
(k : 0,I,2,...)
V-V
V. Using inequality
(20c)
w(p,a)
or equivalently,
u
Mk ,
W
W
(clearly,
p,e
[MkW]e=O = O,
The function
the FOURIER
(k = 0,1,2,...).
u~U
construct
~(p,~) : f u ( r , ~ ) d ~
Evidently,
and defines a
= 0). It remains to show that it satisfies
(5). On the basis of
(2o)
produces
andS.
coincide with v(p,c),
u(O, ~ )
p
convergent
$(p,¢)
= v(p,e).
a solution to equation
(18), we find from this that
~ 0
This means that each function
(5) through
formula
v(p,e)
(i9). By Theorem I
it is unique. We return now to the case of arbitrary obtained Thus,
let
for e l l l p s ~ c a r r y Sxo t
n
and we outline how the results
over to the case of ellipsoids
be a family of ellipsoids
of revolution
of revolution. one of whose
foci is at the origin and the other at any point of hyperplane Knowing the integrals
of a function
u(x,s)
s = O.
over these ellipsoids,
one
-9
is required equation
to determine
-
it. In other words,
(2). The uniqueness
of a solution
strated Just as in the case of ellipses function.
To this end, we perform
formation
in
the origin
(x,s)-space
formation
the point
goes into
(y°,O),
(Yl,O,.. . o
,0,0).
parameters
(x,s) where
goes into
coordinate
amounting
(y,s)
and
transabout
(x°,O)
(y°,O)
depends
to be, in particular,
Yl : r~i
of the
to a rotation
and the point
= (yl,Y2,...Yn,S)
cosines ql,q2, .... qn of the radius Introduce spherical coordinates for
is demon-
that under this trans-
of the transformation
which may be taken
(21)
Q
s = 0 . Suppose
(y,s)
The matrix
to this equation
to solve
by finding the moments
an orthogonal
with matrix
in the hyperplane
it is necessary
=
on several
the direction
vector to the point (x°,O). y and s by the formulas
' s : r~n+ i ,
(i : 1,2,...,n) where
~i(i
vector
r
= 1,2,...,n+l) in
are the direction
(y,s)-space.
can then be written
The equation
of revolution
r : p(l - ~ i )-i
p
and
c
are given by o Yl ~ c2 E : K- ' p : (1 )
(22a) Equation
of the radius
in the form
(22) where
cosines
of an ellipsoid
(2) can now be rewritten
(23)
/
as
u(rQ.~)d~
= v(q,p,¢)
,
S q,p,a where ~ = (~1,~2,...,~n+I), q = (ql,q2,...,qn) and Sq,p, E is the surface of the ellipsoid of revolution with parameters q,p and c. Just as in the preceding,
by applying
keeping
L
q
fixed,
(24)
where
/
the operator
is the operator
(24a)
to equation
(k = 0,1,2,...)
of the orthogonality Yl Yl0 = x l
of the transformation,
Xl0 + X 2 X ~ + "°" + X nXn0
(23)
by (6), we obtain
u(rQ. )y d : Lkv
Sp,q,~ Note that by virtue
Lk
defined
"
-
Since
(x°,O)
is an arbitrary
taking the invarlance (24) for
c = 0
of
dm
10
point of the hyperplane into consideration,
SO, t
to Analytic
symmetric
section,
dimensional
uniquely.
an inversion
a function
These
Using the resultant
formula in a similar
from its integrals
section can easily be g e n e r a l i z e d
n = 1, in other words,
preceding
s
Curves
more general than ellipsoids. sider curves
(i = 1,2,...,n)
with center at the origin.
of
one can construct
The m e t h o d of determining
case
:
n = I.
2. G e n e r a l i z a t i o n
the p r e c e d i n g
t
our even function
system of moments, way to the case
u(x,s)
(Ai = 0,1,2,...),
is a sphere of radius
determine
for
AI ~2 ~n x I .x 2 ... x n dm,
u(x,s) So't
where
s = O, by
one can use equation
to derive the following moments
(24b)
moments
-
However,we plane
about the
the entire discussion
surfaces
Moreover,
to the
we shall only con-
(s = 0). However, easily carries
resulting by way of rotating
in
of r e v o l u t i o n
shall confine ourselves
curves. x-axls
discussed
to surfaces
as in the
over to
a plane
(n+l)-
curve around
an axis of symmetry.
Consider
a two-parameter
coordinates
family of curves
given in polar
r = pf(E,E COS~), f(E,,)
is an analytic
small n e i g h b o r h o o d ~--~n f(O,O)
/~(~,~
and
f(O,O)
,
in an arbitrary
~ 0
and
cosy) u(r,~)d~: v(p,c),
p,e
be given along these curves, of
e
and
n
~ O. From formula
sider d e t e r m i n i n g Therefore
¢
of the origin such that
S
function
function of
~ O. Let the integrals
(2)
~(0,0)
pjc
by
(I) where
S
~(E,,)
from
v(p,¢)
is a known analytic
of the origin such that
(2), it is apparently
uniquely
throughout
wherein
in a n e i g h b o r h o o d
only m e a n i n g f u l
a function
u(r,~)
to coneven in ~.
the following when we speak of a solution to
we shall mean an even function of ~.
(2)
-
il
-
The following theorem holds. Theorem 4: If equation (2) has a solution that is bounded everywhere, belongs to C and satisfies a HULDER condition at the origin, then it is unique.
The method of proving this theorem is similar to that of Theorem I and so we shall find the moments of Expanding
@(~.n)
u(r. ~ )
on each circle
in a series with respect to
,
Sp. o.
and substituting It
in formula (2). we obtain V(p,¢) :
(3)
~ ckak(¢) Vk(P,¢), k=O
where
f (4)
vk(P,¢) : 9 u ( r , ~ ) Sp.~
(5)
ak(e)
cosk~d ~,
1
~k$
kl
~n k In=O
=
I
If we are able to determine the functions Vk(P.O) uniquely in terms of v(p.e), thls wlll prove that the reconstruction of u ( r . ~ ) is unique. From (3). we find that
(5a)
Vo(p,O) :
Introduce the operator
L
aotU)
defined by
P dz Lv : ~-~ .r * V(Z,¢) ~-0 and apply it to (47. Its validity is Justified by the H~LDER condition (6)
for the function
u(r,~).
Indeed,
P LV k
~.~
Z-" O
(6a)
Sp,,
S
SE. c
L f(''' oos ) j
CO sk~d~ Sp. a
pf(~,c COS~) U (r,~) ~-O
-
12
-
Expanding the expression in brackets g c o s ~ , we obtain
in the integrand in powers of
LVk = b°(g) vk(P'E) +n i i [Cn(g ) + Ebn(E) ] cn-ivk+n(p,a),
(7)
wherein bn(g) derivatives of
and Cn(g) are expressible in terms of the partial f(¢,n). By virtue of the conditions imposed on the
latter function,
ci(O) ~ O. The last relation allows us to now derive
an equation for This results in
vl(P,O). Apply
L
to (3) and use formula (7) for Lv k.
Lv = do(z) Vo(P,g) + ~ ck-ldk(g) Vk(P,e) k:i
(8)
+ ~o(g )
Vo(Z,g ) z-- + O
gk-
dz
k=i
O
in which dl(O) = ao(O)Cl(O)M 0. Letting ~ ÷ O in equation (8) and dividing by di(O) , we arrive at the following VOLTERRA equation for vl(P,O)
: P
(9) X
vl(P,O) + X
d~ Vl(~,O) ~-- = fl(p)
0 being some numerical parameter. The equation is meaningful
in that
vi(P,O) satisfies a HOLDER condition at p=O . Equation (9) does not have a unique solution in general since it has an elgenfunction of the form Cp -x if X < O, C being an arbitrary constant. The equation has no other elgenfunctions.
Note however that the boundedness
of
u(r,~) together with (4) implies the boundedness of all vk(P,a) and so the solution we seek for (9) has to be bounded. But since for X < O, Cp -x tends to infinity as p ~ = if C ~ O, we can assert that any bounded solution to (9) is unique. If we apply L to equation (3) k times in succession, we let g tend to zero and we divide by the non-zero coefficient ao(0)c~(O) , _ obtain for the function equation of the form (9a)
z
Wk(Z) +
J[ko(Z
Wk(Z) ~ vk(ex p z, O)
we
a VOLTERRA integral
- ~)k-l+Xl(z - ~)k-2+...+Xk_2(z
- ~)+Xk_i] Wk(~) : fk(z),
where ~o' XI' ..., Xk_ 2 and ~k-1 are certain numerical coefficients and fk(z) is a contlnuous function expressible in terms of the functions [LJv]g=O (J : O,l,2,...,k). It is easy to show by reducing
-
the V O L T E R R A efficients
13
-
equation to a differential
that there are at most
k
equation with constant
linearly
independent
co-
elgenfunctions
each of w h i c h tends to infinity as
z ÷ ~. Hence any b o u n d e d
of the integral
equation is unique.
Thus all of the functions
(k = O,1,2,...)
can be found in a unique way. This implies the unique-
ness of a solution to equation in the class of functions
(2). An inversion
having convergent
solution Vk(P,O)
formula can be derived
FOURIER
series by p r o c e e d i n g
in the same way as in Sec.i. The following
example
shows that n o t h i n g
significant
if the hypothesis
of the t h e o r e m
could be relaxed.
Even for our earlier considered
a solution
is unbounded,
certain situations uniquely Indeed,
it is possible
boundedness
determined
of
u(r,~)
case of ellipses,
to choose a weight
in such a way that the required
function
if in
function will not be
by its integrals.
consider the function
(i0) where
concerning
could be gained
u(r, ~ )
T
is a positive
are arbitrary numerical
number,
= rY N
N X Ak cosk~, k=l
is a positive
coefficients.
integer and the
Choose as weight
Ak
function
¢ ( E , n ) ~ (1 -n) Y. Then if
Sp,a
is the ellipse with parameters
equation
(4) of Sec.1,
and
E
given by
we have
f
(lOa)
p
_~
(1-s c o s ~ P ) T u ( r , ~ ) d ~
= 0
Sp,s for any
p
and
0 ~ ¢ < 1. But this means that equation
ponding to the function also the nontrivial
v(p,E)
solution
3. Problem of D e t e r m l n i n ~ a Family
of Curves
~ 0 not only the trivial
given by (iO).
a Function
Invariant
inside a Circle
to Rotation
from Integrals on
around the Center of the CIEcle
In this section we shall consider the following problem: function
is defined
two-parameter of interest.
inside the unit circle
family of curves The first
(2) has corressolution but
are given.
A continuous
and its integrals As before
is w h e t h e r the function
along a
two questions
determined
are
through these
-
integrals
is unique
and second
14
-
if the function
is unique how to con-
struct it. Introduce polar coordinates
(r,~)
with the pole situated at the
center of the circle. Consider a t w o - p a r a m e t e r
family of curves having the following proper-
ties: 1 ° . The family
is invariant
2 ° . Each curve begins
to rotation about the center of the circle.
and ends on the unit circle.
3 ° , Each curve consists
of two branches whose
equations
are expressible
in the form ~j
(I)
= ~ - (-l)Jgj
(r,p)
r~-p , J = 1,2
Here
p
is the distance
(O
from the center of the circle to the point of
the curve closest to the center of the circle
(we shall h e n c e f o r t h
refer to this point as the certex of the curve) angle of the point. by the coordinates
~I(P'P)
= ~2 (p'p)
Concerning
by an equation
conditions
twice d i f f e r e n t i a b l e the curves
and ~p
~
9j(r,p)
for
~ r~
0 < p ~ r ~ I
and (r/-~-6)3 i ~
r = const, ~
function of a r c l e n g t h
in at most two points
from the center of the circle can
such as (I) only if the center of the
(r-p) ~ (0 < ~ < ~)
on the curve. H o w e v e r functions
assume values greater than
(r-p) ~
center of
multiplying
under certain additional
if one considers
for which the vertices
so that there exist t w o - s i d e d
tangents,
piecewise
are corners
then
of
may also ½. We shall not touch upon these cases al-
though from our considerations with them.
the functions
In the latter case, the factor
may be replaced by
smoothness
is the polar
circle at the vertex does not coincide withAthe
the unit circle. Cj(r,p)
r
differentiable
each circle
and whose vertex is at a distance osculating
~
in this same region.
Observe that a twice c o n t i n u o u s l y whose graph intersects
in
~ O" The functions
will also be assumed continuous
be r e p r e s e n t e d
and
Thus each curve of the given family is c h a r a c t e r i z e d of its vertex.
we shall assume them to be continuous and
.
~
below it will be clear how to p r o c e e d
-
15
-
f---vv The continuity of the functions
~j(r,p)
and
#£r-p)
BrBp
with
respect to p are additional restrictions on the density with which the curves of the family cover the unit circle. Thus suppose that the integrals of u(r,~) with respect to arclength are known on the family of curves satisfying conditions I°-3 ° :
Z .1--u(r, SPj)
(2)
J=l
~r
P
~r
: v(o,~)
The following theorem holds : Theorem 5: If uCr,~) is continuous in the unit diec~ then it can be uniquely determined from the function vCp,~). The theorem is proved by finding the FOURIER coefficients of u(r,~). Introduce the notation 2~ (r) : ~ u ( r , ~ ) exp ( i k ~ ) d ~ (3) o 2~ vk(r) ~ / 0 v(r,~) exp(ik~) d ~ , (k=O,*i,±2,...).
Multiplying the left and rlght-hand sides of (2) by (2~) -i exp(ik~) and integrating with respect to ~ from 0 to 2~, we obtain 2
2~
J=l
: ~ (3a)
1
0
+(r
~j
~
p
)2
u(r.~j) exp(ik.) djdr 0
J=i
= 2/~Z +(r~)2' ~ ~ /d u(r.~j ~ )j exp~[ik~j d+ik(_,)J,, r 2 (r".o).V~Z~] J~l p =J!1 ~ uk(r) p
0 +(r
exp[Ik(-1)J,j(r,p)~
dr
-
Thus the function (4)
uk(r)
16
-
satisfies the equation Rk(r,P) --dr : Vk(P) ,
uk(r) D
where
(5)
2~ /i+(r ~~r )2 exp[ik(-i)J~j(r,0) J:1
Rk(r,p) = ~
The functions uk(r) of a real variable.
and
Using the representation derive the formula
Vk(P)
rW~-p].
in (4) are complex-valued functions
(1) for the functions
~I' we can easily
2 (6)
L
Rk(r,p) :
÷ (r-0)
2
J:1 • exp[ik(-1)J~j(r,o) It is apparent from this that Rk(r,p) D{O < p ~ r ~ I} together with ~ (6a)
~
] •
is continuous in the domain Rk(r,p). In addition
(O
Rk(p,p) = p~l(p,p) ~ O,
.
These properties can be used to easily reduce the integral equation (4) to a VOLTERRA equation of second kind. Indeed, apply to (4) the operator L defined by (7)
Lv = fJ ( p - s ~
v(p)dp
8
and then change the order of integration• This yields i (8)
/uk(r)Qk(r,s)dr
: LVk,
S
where r
(9)
Qk(r, s) : / S
Rk(r,P) dp ¢(r-p)(o-s)'
Making the change of variables (9a)
r÷s p = -~-
+
cos e
-
in the last integral,
we arrive
17
-
at
w (10)
Qk (r's)
F
=
r+s
r-s
Rk(r' T
+ T
c°se)
de .
0 Formula s
in
(10) implies D
together
that
wlth
(lOa)
Qk(r,s)
qk(s,s)
Differentiating
equation
sides of the equation
is a continuous
function
of
r
and
r~-s ~s-~Qk(r , s ). In addition,
WS¢l(S,S)
:
(8) wlth respect
by
Qk(S,S),
to
and then dividing
s
both
we obtain
I
y uk(r)
(11)
-Uk(S) +
Tk(r,s)dr = ~
~-~ Lv k ,
S
where (12)
Tk(r,s ) :
The kernel
of equation
time,
Tk(r,s)
~
has a unique
(r's)
solution
coefficients
of
for
0 < s ~ r ~ 1
Uk(S).
Recalling
u(r, ~ ) ,
At the same
and hence that the
we conclude
that
(11)
uk(r) a solution
(2) Is also unique.
Since the kernels in question, structed
~Qk
(11) has a weak polar singularity.
is continuous
continuous
are the FOURIER to equation
I
of equation
namely,
(11) are known
they are given by
for each such equation.
for the family
(12), a resolvent
As the result
of
curves
may be con-
of inverting
(11), we
obtain the set of formulas,
(13)
uk(r) : MkV, (k = 0,11,'2,...)
wherein
the
Mk
We now consider images
are fixed operators what properties
of the functions
before,
Namely,
we consider
convergent
FOURIER
series,
functions
v(0,~)
under the mapping
we make a slight contraction
functions.
(14)
the
u(r, ~ )
for the family of curves must
defined
in question.
have as by (2). As
in the class of continuous
functions
u(r, ~ )
or more precisely,
~ maxluk(r) I < k:-® r
having
for which
absolutely
-
The set of functions labeled
u(r,~)
U. The set of images
be denoted by
18
for which inequality v(p,~)
V
U
(14) holds will be
under the mapping (2) wlll
possesses the following properties:
I°. For each v(p,~), the functions continuous.
MkV (k = O, *i, ±2,...)
are
v(p,a),
(15) 3°. Each
of
V.
Theorem 6: The set
2° . For each
-
X maxlMkvl k=-~ r veV
(16)
: ~v < "
admits the estimate
Iv(p,~)l
_< :v "j~1
+ ( r T#-~) ~ dr p
The fulfillment of the first and second conditions follows from (13) and (14). To prove the validity of inequality
(16), we consider the
inversion formula for the given problem. After the FOURIER coefficients of u(r, ~ ) have been found using (15), can be determined through the formula (17)
u(r,~)
= ~
(MkV)e i k ~
u(r,~),
being a member of
U,
'
The substitution of thls expression Into formula (2) must reduce It to an identity and so we have (18)
[ / i (MkV)elk~J J:l k -P
/i+(r
This identity leads to inequality
dr.
v(p,~)
.
(16) In a trivial way and so all three
conditions have been proved. The fact that
v(p,~)
belongs to
V
Is
also a sufficient condition for a solution to exist for equation (2). Indeed, suppose
vgV.
(18a) By conditions
From it, construct the series ~ (MkV)elk~ k=-~
I°
and
2°
the series converges to a continuous
function u ( r , ~ ) for which the functions MkV efficients. Hence by condition 2° It belongs to
are its FOURIER coU.
-
On the basis 2 ~ J=i
(19)
Since
of
-
we can compute
i
fki-®
function
~(~,a)
defined by
~r~ )2' dr = $(0,a)
(Mkv)eik~J~+(r
O
u(r,~)
By formula
u(r,~),
19
~ U, it follows
that
$ e V, and we can apply
Mk
to (i9).
(13) we then obtain
(19a)
MkV =
M
k v~
p (k=O,±i
,
±2,
, ,
.)
W=v-~
Or by the linearlty
of the operator
(19b)
MkW = O,
The element
w
This
that
implies
But thls means to equation Collecting Theorem
belongs
to
w(p,~)
family
and the unit
we arrive
(2) belonging
D
$(p,~)
at the following
condition to
conditlons
of the unit
circle,
we can construct
inequality = v(~,~)
we have constructed
(i6). .
a solution
U
is that
v(o,~)
4. On the Problem of Determining
to
unit
disc
a Function
by a curve
convergence
(2) in
D
to exist
belong to V.
i°-3 °. If we consider
disc bounded
then by the local
a solution
theorem:
for a solution
the case where the entire
satisfying
In a subdomain
satisfies
5, it is unique.
and sufficient
We have considered
curves
and it therefore v(p,a) ~ V
the above results,
7: A necessary
(k=O,±i,±2,...)
~ O, or in other words,
(2). By Theorem
covered by curves
series,
V
that to each
to equation Remark.
Mk,
is only of the
of FOURIER
In a similar way.
from its Mean Values
over Circles In COURANT'S s,
book
[6], the question
u(xi,x2,...Xn,S)
radius
wlth centers
from its mean values at points
It is shown that
such a function
At the same time,
the function
approximately.
has been developed
is unique
s = 0
function
of
of arbitrary is considered.
within the class of conti-
an algorithm
However,
for the problem
an even
over spheres
of the hyperplane
nuous
functions.
of determining
is given
up until now,
and the question
for constructing
no inversion of existence
formula of a
-
solution has gone unclarified.
20
-
In what follows, we shall in a certain
sense fill in this gap. But in order to simplify the computations, we shall consider a planar version of the problem in terms of circles even though it has no basic significance. Thus, suppose that the integrals of centered on the line s = O, i.e. 2~ (I)
/u(x+r.cos~, 0
Applying to (1) the operator
u(x,s)
= v(x,r)
r.sln~)d~ L
are known along circles
defined by r
(2)
Lv = xv(x,r) + r ~
~v(x,p)d~ 0
we obtain Lv
:
x _~_2~ u (x+r- cos~, r •sin~) d~+r
/f u(~,s)d~ds (x_~)2+s2
_/
0 -~ 0 (2a)
9'
_/r~-C~-x) -
x-r 2~
= x / ©
"9
x+r
-Jr~- (-;;-x)
2~ u(x+r.cos~,r-sin~)d~+/u(x+r.oosq, O
r.sln~)r.cos~d~
2w :
Let
Lkv
/ 0
u(x+r.cos~,r.sln~)(x+r.cos~) d~.
denote the k-fold application of
L
to
v(x,r). Then in a
similar way, we have (3)
Lkv =
f
u(x+r.cos~,
r.sln~)(x+r,cos~)kd~
0
In this connection,
L°v m v(x,r).
From this it will follow in particular that the even function (of s) u(x,s) can be determined from v(x,r) in unique fashion. In analogy with Sec. I, we let
Mk k
[3] (4)
be an operator defined by k -J
MkV = ; Z (-l)J( [ (oC~+O~)(JO.) J:O £ : 0 ~'~ ~"
r J
v,
(k : 1,2,...) MoV : ~-V v(x,r).
- 2i -
Its application to equation
(I) results in the following set of
relations : 2~ (5)
MkV = ~ - V f
u(x+r.cos~,
r. s i n ~ ) ~ [ t + J ~ - l ] k + [ t - 4 ~ - i ] k ~
0 d ~ (t: x + cos ~ ) For
(k = 1,2,...)
x = O, they may be written in the form 2~ (6) l J u(r.cos~ r.sin~)cosk~d~ [V]Mk_x= 0 : ~
0
.
(k = 1,2,3,...)
From thls we obtain an inversion formula for continuous functions belonging to U, namely, (?)
u(r.cos~,
r.sln~)
= ~ [MkV]x=o.cosk ~ k=O
.
Formula (I) now easily leads to an equality for the function v(x,r) corresponding to which the solution to (i) Is given by (7). It amounts to the following:
(8)
Finally, MkV
[v(x,r)]
_< 21, ~ max[MkV]x=O k=O r
It is possible to formulate a theorem In terms of the functions
as was done previously.
Theorem 8: A necessary and sufficient condition for a solution to exist for equation (I) belonging to the set U is that v(z,r) belong to the set conditions I. For each
V
of functions
satisfying
the following
: v(x,r)~V,
the functions
Mkv
(k = 0, I .... )
are continuous. ~.
~ maXIMkVlx= 0 < ® . k=O r
~. Each
v(x,r)~ V
satisfies
inequality
C8A.
The proof of the theorem is carried out along the same lines as before and so we shall not stop to give it.
CHAPTER 2 Llnearlzed
Inverse Dynamic
Problem for the
Telegraph Equation Consider the problem of determining telegraph
the coefficient
a(x,y,z)
in the
equation S2u S2u + -+- S +2 -B2h u~t 2 = Sx---2 ~y2 Sz 2
In the domain
z ~ 0
a(x,y,z)u
from the value of the solution to the CAUCHY
problem for the equation at
z = 0. The CAUCHY data may In principle
depend on several parameters. The first one to examine thls sort of problem was Ju.M. BEREZANSKI~ He showed that plecewlse xo
and
on values
a(x,y,z)
analytic Yo
functions
are parameters
In thls chapter, to the problem.
function,
from the function
u(x,y,O,xo,Yo,t).
Here
region.
it Is necessary
Thus, to reconstruct
a
to know a function of flve
Each llnearized problem Is then reduced to an integral-
we need to know information coefficient
a(x,y,z).
I. Statement
of the Inverse
Consider the telegraph
for the function
[4].
in the class of
we shall examine two kinds of linear approximations
problem considered
(I)
uniquely
connected with the CAUCHY data that take
In a certain two-dlmenslonal
three-dimenslonal variables.
geometric
can be determined
In Chapter I. To determine of the same dlmensionallty
the coefficient, as that of the
Problem and Its Linearlzat~on
equation
~2u
.. = au + a(M)u + f(M,Mo,t) ~t 2 u(M,Mo,t )
In the domain
z ~ 0
initial and boundary condltlons:
(2)
u(M,Mo,O)
(3)
~
= 8
u(M1,Mo,t)
u(M,Mo,O) = O,
= O,
t • 0
under the following
-
23
-
Here M(x,y,z), Mo(xo,Yo,O) and Mi(x,y,O) are points in threedimensional space, a is the Laplacian in x,y and z, and a(M) is a continuous function in the domain z ~ O. Throughout the following we shall assume that f(M,Mo,t) is a function of the form (4) where
f(M,Mo,t) 6(M-Mo,t)
=
~(M-Mo,t)
is the DIHAC delta-function
(see
[11]).
By a solution
u(M,Mo,t) to equation (i) under conditions (2) and (3), we shall mean a generalized solution of the equation regular at Inflnity. If a(M) is glven, then determining the function u(M,Mo,t) is called the direct problem for equation (i). We now state the corresponding inverse problem: it is required to find continuous function a(M) knowing the valueSat z = O of the solution to equation (i) under conditions (2) and (3). Thus suppose that the function (5)
~(MI,Mo,t)
=
u(MI,Mo,t)
has been prescribed. It Is convenient to reduce the stated problem by use of the boundary condition
(3) to an equivalent problem for the whole space. This we do
by continuing space z < O.
a(M)
and
f(M,Mo,t)
We represent the solution (6) Uo(M,Mo,t)
u(M,Mo,t)
as even functions into the half-
u(M,Mo,t) = Uo(M,Mo,t)
in the form + Ul(M,Mo,t)
being a function satisfying the equation
(7)
~2u St 2 = Au + 6(M-Mo,t)
and boundary conditions the equation
(2) and (3). We then obtain for
ui(M,Mo,t)
S2u 1
(8)
~
= Am I ÷ a(M) Eu° + ul]
and conditions similar to (2) and (3). If a(M) is allowed to decrease indefinitely in this equation, its solution Ul(M,Mo,t) will obviously
-
24
-
tend to zero. Taking this Into consideration, we shall regard
a(M)
to be so small that the second order term in (8) may be neglected, namely, the product linearized equation
a(M)ul(M,Mo,t).
~2u i ~ St 2
(9)
=
As a result, we obtain the
Au I + a(M)Uo(M,Mo,t)
and it is this equation for which we shall consider the problem of reconstructing
a(M)
from the known value of its solution for
(10)
Ullz: 0
under conditions on
:
ul(M,Mo,t)
z = O,
9~1(MI,Mo,t)
analogous to (2) and (3).
2. Linearized One-Dimensional Inverse Problem in Two-Dimensional Space In this section, we shall deal with the linearlzed problem of determining
a(M)
a function of
in the half-plane
y ~ O
on the assumption it is Just
y. Carrying over the discussion of Sec. I to the two-
dimensional case, we arrive at the problem of determining the solution to the equation ~2u I (i)
V~t
in the domain
y & 0
~2u I
a(y)
from
~2u I
: ~~x
+
~y~
+
a(y)Uo(X,y,t)
under the initial and boundary conditions
(2)
ul(x,y,O)
=
(3)
S-~ ul(x,O,t)
~-~ u1(x,y,O) :
O,
:
O,
(t > O)
the solution being known at a single point: (4) The point
Ul(Xl,O,t) Mo
: ~(t)
.
will be regarded as the origin. The solution
uo(x,y,t)
of the two-dlmensional analogue of equation (7) of Sec. I in the halfplane y ~ 0 is given by [32]
2 _×2_y2' 0
,
t 2
<
x2
+
y2
-
25
-
while the solution to (i) under conditions as
(6,
Ul(X,y,t):
(2) and (3) can be expressed
~-7 ~ [ f f
a(n)u°(''n'~)d~dn
0
4 t _ , ) 2_r2'
r_
] d~
wherein r = /('(x _~)2 + (y-n) 2 '
(6a)
We now let the point (x,y) tend to (xi,O) in formula (6). This results in the £ollowing integral equation of the first kind for the function a(y) : (7)
t (#(t) = ~ f [ / /
a(n)Uo(~,~,T)d~dn
] d~
ri:t-
o
=
The region of integration in (7) is the interior of the
n
)
cone
T = t - A ~ - x l ) 2 + n 21
(8)
and as is apparent from (5), the function inside the cone (9)
= ~
T
uo(x,y,t)
is nonvanishing
.
Hence, the integration in (7) actually extends over the region common to the two cones. Introduce the notation /r2(~,~,)
nl = g e xl
-- ¢ / ~ n
t
,
/,~
2'
(2 : 7- + ~-~nl cn I - n v~ We change the order of integration in (7) and take into account the evenness of the integrand in
n. Equation
(7) can thus be rewritten as
-
(11)
~(t)
~
:
26
-
t)
a(n) K ( t , n ) d n
,
O where ~2
~-rl
K(t,n) : i
(12)
n
dr d~
~:
2'] d(t_~)2_~][2_r2
r~
We make a change of variables in the iterated integral (12) setting 2T-(t-rl+r 2) t-r l-r 2
z :
(12a)
The formula for the kernel (13)
i
nI
2~-x I
u :t
K(t,n)
then becomes
i --
K(t,n) = ~n--~
f
-I -I ,~l_z2)[4p l+(t_pl_p2)(l_z)] [4o2÷(t_P1_.2)(1+z)]
Here we have introduced the notation
(14)
~i 2
=
Xl r1(~--
+
u
Xl r2(~-- + u
=
~ ,
t
~ ,
Equation (II) is a VOLTERRA equation of first kind. To answer the question as to whether it is reducible to an equation of second kind, we must ascertain the behavior of its kernel as n + nl(t). We shall now do this. Using (I0) and (14), we can easily deduce the following relations for n + nl:
(14a)
01
5
t :
~
nl-n )(u2-i) ÷ Ol((nl-n) ~)
~
o2 : ~
+ Xl
~
3
I
~
K(t,n)
3
- 2n~
are i n f i n i t e s i m a l s of order (nl-n) g.
-1
: ~ as
n ÷ q1"
3
2
3
where Ol((nl-n) g) and 02((nl-n) g) Hence we conclude that the kernel (15)
2
~(~1~ )(u21) + °2((~i-n)~)
-1
~
[ ~
+ 03((nl-nl)
+ o(~I-,) ]
dzdu
-
27
-
This formula remains valid as n ~ O. A more detailed analysis shows that the origin is a singular point for K(t,n), namely, its limit the~e does not exist. However K(t,n) remains bounded in a neighborhood of the origin. Formula (i5) leads to
(15a)
K(t,n) = ~
The function Ki(t,n) except at the origin.
Kl(t,n) .
is continuous together with its derivatives
We now replace the variable t in (11) by its value in terms of As a result, equation (II) becomes nI = J a(n) K ( ~ , n)dn . 0 Apply to the left-hand and right-hand sides of (16) the operator defined by (16)
n I.
~ ( ~ )
(17)
d
I
0
L
.
s-nl dni
We thus obtain S~ JO
(17a)
d
=~
~ 0
a(n)
s-nl
a(n)l#
s-nl
Ln
: a(s) R(s,s)
~
anl
~ ~/
K ( ~ n ) d
K(
n)dn
dn I
dn
s + S
where
a(n) ~~
R(s,n) dn
O s
(i8)
R(s,n) : f s / ~ l S _ u
Sn i K ( ~ ,
n)dn I
From (i8) it is easy to deduce that (i8a)
R(s,s)
= i
.
As a result, we arrive at the following VOLTERRA equation of second kind (19)
a(s) + J a(n)%-j ~ R(s,n)dn 0
= L~
.
-
The kernel
of this equation
of the point s ~-- R(s,n) ~s such that
remains
continuous
is unique.
~(t)
~ )l
R(s,
On the other hand,
3. Two Formulations
with the exception that the function
and that there
~
exists
an
s~
< I
L~
exists
of
a(s)
a solution
from the function
will exist
if and only
and is continuous.
of the Linearized
Inverse
Problem
in Three-Dimen-
Uo(M,Mo,t)
that satisfies
Space
For three-dimensional equation
s = 0
that the determination
is such that
sional
everywhere
to show further
{0 < s < s , n <= s}.
From this it follows ~(t)
at
Is ~
in the domain
-
is continuous
s = O. It is possible
(19a)
if
28
space,
(7) of Sec.
such as (2) and
the function
i and homogeneous
initial
and boundary
conditions
(3) Is given by (t-r(M,M o))
(i) In this
Uo(M,Mo,t) formula
The solution boundary
r(M,M o)
to equation
conditions
(2>
=
2~ r(M,M o )
is the distance (9) of Sec.
radius With
t
centered
(2) as point
the linearized
First Formulation: ul(Mo,Mo,t) approach
(5)
Mo
runs over the points
of a ball of
M. one may consider
Mo(Xo,Yo,0)
two formulations
be a variable
be a given function.
In formula
(2) and using
~(~o,2t) :
and
:
problem.
Let
= ~(Mo,t)
initial
dVp
at the point
inverse
and
r(P,M)
P(~,o,~)
of departure,
formula
M
a(P) Uo (P,Zo,t-r(P,M))
r(P,M)~t variable
the points
I under corresponding
is given by EIRCHHOFF'S
u1(M,Mo,t)= ~f~
The integration
between
i
8w2t 2
Letting the point
expression
fS SMo,t
point and let
a(P)dS
M
(I), we obtain
of
M o.
-
29
-
where SM ,t is a sphere of radius is spherlc~l surface element.
t
wlth center at
MO
and
dS
Thus in this version of the llnearized inverse problem we arrive at the integral-geometrlc problem of determining a function from its mean values over spheres of arbitrary radius with centers at points of the plane
z = O. Thls problem, as we have already pointed out, has a
unique solution and thus we have the following theorem. Theorem 9: The coefficient
a(M) in the Zinearized ~nverse problem can be determined f~om uI(Mo, Mo, t) in a unique manner.
We mention that the results in COURANT'S book for the center
Mo
[6] show that it suffices
to belong to a point set in the plane
z = 0
lying
In an c-nelghborhood of some fixed point of this plane. In the case where
a(M) = a(z), it suffices to know the function
U(Mo,Mo,t) at a fixed point M o. The function a(z) can then be found explicitly. To this end, introduce spherical coordinates at the point Mo, letting
@
be the angle between the z-axls and the radius vector
of the variable point of integration. Formula (3) can then be written in the form (4)
~(Mo,2t)
: ~Sa(t
cos@)sln@d@
.
0 Equation (4) has the explicit solution a(z) : 2~ d
(5) Second Formulation:
The point
ml(MI,Mo,t) = (M1,t) of the plane z = O.
[z~(Mo,2Z) ] . Mo(Xo,Yo,O)
is given, where
Substituting the expression for
Uo(M,Mo,t)
find In this case that the function : (6)
ui(M'M°'t)
I 8-~
###
is fixed and the function
Mi(x,y,O)
Is a variable point
given by (1) Into (2), we
ul(M,Mo,t)
Is given by
a(P)~(t-r(P'M)-r(P'Mo)) r(P,M)r(P,M o)
dVp
r(P,M)~t It is apparent from this formula that the Integrand is nonvanlshing only on a surface whose equation is
-
(7)
30
-
r(P,M o) + r(P,M) : t
.
This is nothing more than the equation of an ellipsoid of revolution with loci at the points
Mo
and
M. Since
Mo
is fixed, it may be
regarded as the origin of our fixed cartesian coordinate system. We also introduce a moving cartesian coordinate system
~',n',~'
~'-axis coinciding with the line passing through the loci
M
with and
Mo
of the ellipsoid. To simplify the integral in (6), we consider in addition to the ellipsoid (7) a family of confocal ellipsoids of revolution given by (8)
r(P,M o) + r(P,M) = T
We now pass from cartesian coordinates r,@,~
~',n',~'
to spherical coordinates
by means of
(8a)
i I = r sin~ C O S ? , ' r sin@ s i n ~ , t
The ellipsoids
r =
~
r Cos~
(8) may then be written in the form
(9)
where
.
is the distance between
2 2 • -p T-p COS@
I M
O
and
'
M.
It is not hard to show that 4Vp (9a) r(P,M)r(P,M o) and so expression
(iO) where d~
2r 2 = --~ • -0
sin@ d@ d ~ d T
(6) can be represented in the form
1 u1(M,Mo,t) = 4 2[t2_r2(M,Mo)] SM, t
JS SM,t
a(P)r2(p'Mo )d~ '
denotes the ellipsoid of revolution defined by (7)
is the element of solid angle with vertex at point
Letting the point
M
tend to
Ml(x,y,O)
an integral equation of first kind for
and
M o.
in relation (I0), we obtain a(M), namely,
-
31
-
// (II)
~
aCP)r2CP,Mo)dw = 4,2[t2-r2(MI,Mo )] ~(Ml,t) .
SMI,t Since the factor r2(p,Mo ) can always be grouped with the function a(P) and their product satisfies a HOLDER condition at Mo, we can apply the results of Sec. i of Chapt. I to arrive at the following theorem. Theorem I0: The coefficient a(M) in the linearized telegraph equation can be determined from the function u1(MI,Mc, t) in a ~nique m~nner.
Finally, the algorithm of Sec. I, Chapt. I may be used to construct a(M).
4. Derivation of a Nonlinear Differential Equation for the Inverse Problem We next consider how the coefficient a(M) in the is related to the solution itself. To this end, we equation (8) of the first section of this chapter. the term a(M)u i in this equation and use similar the preceding section, we can derive the formula 1
(I~
~S
telegraph equation shall make use of If we do not neglect reasoning to that of
a(P)r2(p'Mo)d~
ul(M'M°'t> : 4~2[t2~r2(M'M°>] SM,t fJf + ~
a(P)uI(P'M°'t-r(P'M)) r(P,M) dvp
.
DM,t DM, t here denotes the domain having the ellipsoid of revolution SM, t as boundary. It is apparent from this relation that the solution ul(M,Mo,t) is bounded. Therefore, if we pass to the limit in (I) letting t * r(M,Mo) , the integral over DM, t will vanish. Moreover, the ellipsoid of revolution degenerates to the line segment Joining M and M o. Denoting by r°(Mo,M) the unit vector in the direction of the line from M o to M and letting the parameter t tend to r(M,M o) in (I), we obtain in the limit the expression I r~ M'MO) -~ (2) ui(M,Mo,r(M,Mo) ) = 4 w r ( M , M o ) j a(r.r°(Mo,M))dr .
-
32
-
Computing the dlrec~lonal derivative of both members of this relation in the direction (3)
r°(Mo,M), we conclude that
a(M) = 4,
3
[rCM,Mo)~I(M,Mo,r(M,Mo)) ]
•
37(Mo,M) Formula (3) which we have obtained for the coefficient in the telegraph equation has the same type of structure as the formula arising in the spectral version of the one-dimensional inverse STU~M-LIOUVILLE problem (see [1] and [13]). Substituting this expression for a(M) in formula (8) of Sec. I, we arrive at the following nonlinear differential equation wit~ shifted argument : (4)
~
= AU 1
+
4,(uo+U 1)
_. [r(Mo,M)uI(M,Mo,r(M,Mo))] ~r ° (M o ,M)
If this equation should be solvable under the initial conditions
(5)
It=O
=
~-ff ~ ullt= 0 =
0
and boundary conditions
(6)
then we could afterwards find
a(M)
using (3). The questions of
existence and uniqueness of a solution to the above problem for (4) require further investigation.
.
CHAPTER 3 Linearized Inverse Kinematic Problem for the Wave Equation
i. Formulation of the Problem and Its Linearization Consider the wave equation
(i) where
n n
2 S2u ~t 2
is a function of
x
=
Au
and
y
and
A
is the LAPLACIAN with
respect to the same variables. Let u(x,xo,Y,t) be the generalized solution to equation (I) in the half-plane y • 0 satisfying the following initial and boundary conditions:
(2)
S u(X,Xo,Y,t ) It:O = O , u(X,Xo,Y,t) It=O = ~-~ ~
u(X,Xo,Y,t )ly=O = ~(X-Xo,t)
,
~(X-Xo,t) being the DIRAC delta-function. Further, let T(xl,x o) be the maximum of the numbers T such that for given x o and xi, the support of
u(xl,xo,O,t)
is a subset of the interval
We pose the following inverse problem for equation
T ~ t < ~.
o
(1): Given the
function T(xl,Xo), to determine n(x,y). This problem is called the inverse kinematic problem. It was studied in the papers [i5] and [37] for the case where the function n is independent of x. Some results for the problem are to be found in [3]. The function
X(Xl,X o)
(3)
is the minimum of the functional
J(r) = J
n(x,y)ds
r(xl,x o) where
r(xl,x O) is a curve Joining the points (xi,0) and (Xo,O) lying in the upper half-plane and ds is the element of arclength along
Win the process characterized by (I), T(xl,x o) disturbance produced at
(Xo,O)
reaches
is the time at which a
(Xl,O).
-
34
-
the curve. Now let n(x,y) = no(X,y) + nl(x,y)
(4)
where no(X,y) is a given function and smooth small function. Correspondingly, as the sum
(5)
ni(x,y) is a sufficiently T(xi,xo) may be represented
T(xl,x o) = To(Xl,X o) + ~l(xl,Xo)
Here ~o(Xi,Xo) corresponds to the function is a minimum of the functional
•
no(X,y) , or in other words,
F
(6)
Jo(r) =
~
no(X,y)ds .
r(xl,x o) The function holds :
Ti(xi,Xo)
is then also small. The following theorem
Theorem li: The function
x1(Zl,Zo) can be represented to within infinitesimals of order n I2 by
(7)
Ti(xi,x o) =
~
ni(x,y)ds
r°(xl,x o) where r°(xl,x o) is the curve of the family r(xl,x o) which the minimum of the functional (6) is attained.
for
Denote by rl(Xl,Xo ) the curve of the family r(xl,x o) for which the minimum of the functional (3) is attained. We can then write down the following relation : r(xl,Xo)-
(8)
~
n(x,y)ds = [ S
r°(xl,Xo)
+r s
no(X,y)ds- f
krl(xl,Xo )
ni(x,y)ds-
Lrl (x 1,x o)
f
no(X,y)ds]+
rO(xl,Xo )
ni(x,y)ds] •
r ° (x 1,x o)
The extremum principle clearly implies the following inequalities: (9)
x(xi,x o) -
J
n(x,y)ds <_ 0 ,
r°(x1,x o)
f no(x,)d I rl(xl,xo)
no(X,,)dO
r°(xl,Xo )
.A
-
35
-
Therefore in order that relation (8) hold, it is necessary that the expressions in brackets in (8) have the same order of smallness. We determine
Xl(Xl,Xo)
zl(xi'x°) : /
from (8) in the form n!(×'Y)ds ÷ I
rC(x!,xo)
(ga)
r1(xl,x o)
,×o)
nO(x'y)dsl
l,Xo)
r°(×l,x o)
Noting that for smooth ni(x,y) differentiable), the expression (9b)
(~~:i no<x,y)ds -r°~(i 1
(for instance, twice continuously
f n1(x,y)ds - J ni(x,y)ds ri(xl,Xo ~ r°(xl,Xo )
is of second order as compared to the function (9c)
J
ni(x,y)ds
r°(xl,xo) and using the result obtained above, we arrive at formula (7) on neglecting small terms of higher order. Formula (7) may also be derived through linearization of the eiconal equation (9d)
Igradxl T(Xl,Xo) I = n(x)
If we substitute the expression for n in this equation and neglect the terms we obtain
(9e)
and T(xi,x o) from (4) and (5) n~ and Igradxl~l(Xl,Xo)12,
(grad To, grad Ti) = n o n i
or in other words, (9f)
dT I ds
=
nl "
This is precisely equivalent to formula (7). Thus formula (7) holds in linear approximation. Since the function no(X,y) is known, by solving a geometric optics problem, we can construct the two-parameter family of curves r°(xl,Xo). The determination of n(x,y) is then reduced to the construction of function nl(x,y)
-
from
Ti(xl,Xo).
problem.
no(X,y)
a ~ O
the point
and
b ~ O,
no(X,y)
is given by
: (ay + b) -I
r°(xl,x o)
((xi+Xo)/2 , -b/a)
(see
always be considered known for letting
-
In other words, we are led to an integral-geometric
In the special case where
(9g) where
36
is a circular arc with center at
[32]). The function
~ = O
x i ÷ Xo, we can determine
ni(x,y)
may
(if we pass to the limit in (7)
n~(xo,O)).
If we extend it in a
continuous llne
fashion into the strip -~ ~ y ~ 0 and then evenly about the y = - b ~, we wind up with a problem of determining function
nl(x,y)
from its mean values over circles of arbitrary radius with b centers on the llne y = - ~, which was considered in Chapt. i.
Another less trivial example of the use of the linearized the inverse kinematic
2. Application
problem will be considered
version of
in the next section.
of the Linearized Version of the Inverse Kinematic
Problem to Geophysics The problem treated above may be used to give a more precise ization of the global velocity verse waves propagating earth is an elastic
distribution
in the earth.
of longitudinal
Under the assumption
sphere for which the velocity
function of radius only, the data accumulated of earthquakes
has been used to construct
for (longitudinal earth's
radius
and transverse)
(see, for example,
and
and transthat the
of a disturbance
is a
from a fairly large number
velocity
distribution
seismic waves propagating ~]
character-
curves
along the
Ki4]). A theoretical
basis
for the ~roblem and a method for solving it were given in the previously mentioned
articles
[14] and
[37]. In principle,
to solve the problem,
one should according to the method know the travel-tlmes on the earth's Actually, stations
to any point
surface for the seismic waves generated by an earthquake.
earthquake situated
tremors
are recorded at a network of seismological
fairly distant
from one another.
struct more reliable velocity propagation
curves,
Therefore,
to con-
one ought to average
these curves over data obtained
from many earthquakes.
been done by many geophysicists
and there now exists a whole collection
Such work has
of velocity propagation
curves for longitudinal
The variations
curves range from iO to 15%. This is explained
in these
and transverse
waves.
by the fact that the various authors have in general made use of
-
different
earthquakes
data. The velocity
recorded
37
-
at different
distribution
stations to obtain their
curves for longitudinal
and transverse
waves most widely accepted today correspond to the JEFFREYS-BULLEN hodograph. Meanwhile,
geophysicists
information
now have at their disposal rather reliable
that the structure
respect to geographic of disturbances
of the earth is inhomogeneous
coordinates.
should therefore
And so the propagation
hodographs
have been observed
deviations
from the JEFFREYS-BULLEN
for travel-times
the propagation
velocities
the distribution Let
n(r,@,~)
r,@,~
This provides
of the
= no(r)
is a known function, coordinates.
time of a seismic wave from point coordinates
But the are small.
the deviations
a basis for assuming that
(longitudinal
speed. We can thus represent
are spherical
ordinates
of waves.
for example,
differ little from those corresponding
n(r,@,~) no(r)
hodograph,
This
from the average
to
depending Just on the radius.
(A) where
for the travel-times
be the reciprocal
wave propagation
deviations
of the order of 15-20 minutes,
do not exceed 5 - 6 seconds.
velocities
also depend on these coordinates.
is confirmed by the fact that systematic
Namely,
with
(9o,~)~ (@,~)
+ nl(r,@,~) nl(r,@,~)
Let M
We can represent
is a small function
on the earth's M
and
denote the travelsurface with co-
on the earth's
(in this connection,
to be a unit sphere).
,
T(@O,~O,@,~)
o
to another point
or transverse)
it in the form
surface with
we are considering the earth
it in the form
(B) where
TO(90,~O,@,~)
designates
subject to a velocity and TI(@O,~O,@,~)__ function.
in accordance
Observe that the curve at
Mo
arrives
at
M
the travel-tlme
distribution
F°(M,Mo )
corresponding
from point
Mo
to the function
with the above discussion
the plane of the great circle passing through
M
n = no(r) and
M o.
M
n(r),
is a small
along which the disturbance
in the shortest time under
to
produced lies in
-
38
-
If we carry over the results of Sec. I to the three-dlmensional we can obtain
for
Tl(@o,~o,@,~)
the following approximate
case,
formula
:
F°(M,M o ) We now let the points unlt circle. considered items i ° Condition
In Sec. and
2°
3°
and
Mo
vary over the circumference
3 of Chapt.
I. Indeed, the curves
of the requirements
monotone
function
no(r) of
at least In the earth's mantle.
In each great circular cross-section form a two-parameter
velocity
In
no(r)
know the travel-tlmes
problem,
In making practical
of waves,
n1(r,@,~)
available
in
if we merely
in cross-sectlons
passing
overdetermlnancy
can
the results obtained. use of the above procedure
are situated
of
one must not take very
ni(r,@,~)
because
seismological
fairly sparsely and also the resulting
distorted by observational allow for inhomogeneltles
errors.
In so doing,
large magnitude.
I, we can also point out
that the method may be used to clarify the local structure for those regions where there are a sufficiently stations and earthquakes.
data is
one wlll be able to
In the earth of sufficiently
On the basis of our remark in Sec. 3 of Chapt.
mological
can
Since these circles
can be recovered
for example,
many terms in the FOURIER expansion stations
requirement
we determine
of the earth,
nl(r,@,~)
curves,
In the norm of the space
through the earth's polar axis. The forementloned be used to average
that It be
distribution
family, we have an overdetermlnancy
The function
satisfy
Is a twice continuously
The dlfferentlablllty
C. Thus solving the Integral-geometric
our problem.
F°(M,M o)
r. The requirement
by making small changes
geometry
imposed on the family of curves.
follows from the one-dimensional
be satisfied
of the
we arrive at the problem of integral
will also hold If
dlfferentlable monotone
M
As a result,
of the earth
large number of seis-
CHAPTER 4 Inverse Heat Conduction
Problems with Continuously Active Sources
This chapter will deal with two inverse problems 2Su a ~-~ = au + f,
(I) in a half-plane,
(la) ~(t)
where
f
a = const,
is a function of the form
f(xl,x2,...,Xn,t)
being a known function.
n-dimenslonal
for the heat equation
= ~(t)fl(Xl,X2,...,Xn)
,
They are then extended to the case of
space.
I. Inverse Heat Conduction
Problems
I °- First inverse problem:
It is required to determine the function
f(x,y)
for a Half-Plane
from the equation 2 ~u
(2)
a
in the half-plane
y ~ 0
~ t
~
(3)
.
t
= Au + v t t ~ f ~ x , y ~
,
under the following conditions u(x,y,O)
= O,
u(x,O,t)
: h(x,t)
u(x,yl,t)=
r(x,t)
:
.
Let
(3a)
v(x,y,~)
=~e
-x2t u(x,y,t)dt
.
0 By virtue of (2) and (3), the function differential
v(x,y,k)
satisfies the
equation Av-
a 2 ~ 2 v = -¢(~)f(x,y)
,
(4)
¢(k) =
~(t)dt
.
0 The last two relations
in (3) go over into the following
:
-
40
-
v(x,yl,k) : ~ - 1 2 t r ( x , t ) d t
(4a)
= ri(x,~)
-O v(x,O,k)
=
~_~
2t
h(x,t)dt = hi(x,~)
O A fundamental solution for equation (4) is the HANKEL function
(4b)
~ Ho(1) (lair)
1 K o (a~r), : ~-~
r :
It satisfies this equation everywhere except at
x2~+y2
.
r = O.
The GREEN'S function of first kind is expressed in terms of the fundamental solution by
(4c)
G(x,y;~,n)
: ~
[Ko(~kR I) - Ko(akR2)l
R i = [(x-~)2+(y-n)2] ~ ,
,
i
R 2 = [(x-~)2+(y+n)2] ~
Therefore the solution to (4) may be represented in the form (5)
:
KI(axR o)
+¢(k) O ~ _ J G ( x , y ; ~ , n )
f(~,n)d~dn
,
i
RO = [(x_~)2 + y212 Setting Y = Yl in this last equation, we obtain an integral equation of first kind for the unknown function f(~,n), namely /f[Ko(a~R3) 0
--~
- Ko(a~R4)]f(~,n)d~dn
1 R 3 : [(x-~)2+(n-Yl)2]~;
: g(x,~),,
I R 4 : [(x-~)2+(n+yl)2]~;
(6) g(x,~) : ~ 2
[~rl(x,X ) _ a X y l j
~K 1 1 (aXR)h i(~,x)d~],
R :
The right-hand side g(x,~) of equation (6) is subject to the sole condition that the equation have as solution a function f(x,y) such that:
-
41
-
i. f(x,y) : O for Y < YI' Yl > 0 ; 2. f(x,y) ~ LI(D) , D = {-- < x < ®; O < y < ®}. Taking FOURIER transforms in equation (6) with respect to
x
and
using the relation (see [7]) (7)
/ K o LFb ( c 2 + t 2 )]I ~cos
e-lcl
ut dt
0 we obtain
(Ta)
/F(¢, n)(e- I r'-Yll ~ e 0
= ~
I "+Yl I ~1"1
/g(x,X)
e -i~x dx ,
with
(Tb)
-
F(~,n)
Taking into consideration that
I
/f(~,n)
f(x,y) = 0
e -i~
for
d~ .
Y < Yi' we can write
this last equation in the form
/a2~2+2 ' F(~,n) e -n
dn :
Q(m,~)
,
Yl
(8)
Q(~,~) =
/a2X2+~2'
G(~,X) ,
2w sinh yl ~ G(~,k) -
1 /
2~
g(x,k)
e-i~x
dx .
Let us show that we have the right to take FOURIER transforms in equation (6) with respect to
x
under the above assumptions on
f(x,y).
From (7) it follows that the kernel of the integral equation (6) is an absolutely integrable function. Then by the theorem convolution for two absolutely integrable functions, we infer that absolutely integrable in the argument is not hard to derive for
G(~,k)
2~sinhy1~-~
x
and hence
the estimate 22 2' ~ e -Yl/a ~ +~ / Yl
g(x,k) is also G(~,X)
exists. It
-
The fact that
f(~,n)g LI(D)
42
-
implies @o
(9a)
lF(~,n)l
~
-
~
/.F(~.n)[dn
i [lf(¢,n)Id¢
<
-
I
,
//,f(~..)'d~dn
~-~
Yl
: ~ • ®
Yi-"
Substituting the last inequality into (9), we finally obtain (io)
Is(~,~)l
•
""
Introduce the notation t = n - Yl ' (iOa)
p2 = a2~2 + 2
,
Fl(~,t) = F(~,t+y i) : Fil(~,t) + iFi2(m,t) Ql(~,p) = e
Q(w,
,
= Qli(W,p) + iQl2(u,p)
.
Then equation (8) can be split into two independent integral equations for the unknown functions FII and FI2 , namely, (il)
/Fii(~,t) 0
e "pt dt = Qii(~,p)
,
(i=1,2).
The functions Fll and Fi2 are clearly continuous and bounded since f(~,n)~Ll(D). Hence it follows that each of the equations in (li) has Just one Solution and it may be expressed by the formula K26~ : (_l)n(~)n+ i (i2)
Fli(~,z) = lim n÷®
~n Qii(~'~ ) ~pn n|
,
(i=1,2).
Further, it is known that if the FOURIER transform (i3)
F(~,n) -
i
~f(~,n) 3
e -i~
d~
of a summable function f(~,n) (for fixed n) is equal to zero for all ~, then f(~,n) = 0 almost everywhere. Therefore, the unique solution to equation (13) is given by (13a)
f(~,n) = L
~F(~,n)
e i~d~
"
-
43
-
Thus the inverse problem formulated above for equation one solution.
(2) has at most
2°- Second inverse problem: We reduced the inverse problem (2), (3) to integral equation (8) with the help of formula (5) which was the solution to the DIRICHLET problem for equation (4). An integral equation analogous to (8) may be derived by considering the NEUMANN problem for the same equation. Let u(x,y,t) be the solution to equation y ~ 0 such that
(14)
u(x,y,O)
= 0 ,
u(x,O,t)
: h(x,t)
~-- u(x,O,t) SY
(2) in the half-plane
,
= r(x,t)
Taking LAPLACE transforms with respect to up with equation (4) for v.
t
. in equation
The boundary conditions in (14) go over into the following V(x,O,~)
=~
(14a)
-~2th(x,t)dt
= ~i(x,~)
(2) we wind
:
,
0 ~--Sy V(X,O,A)
=~
-k2tr(x,t)dt
= ~2(x,A)
0 It is easy to show that the function (14b)
N(x,y;~,n)
= ~-~ [Ko(a~R i) + Ko(a;~R2)], I 1 R 1 = [(x-~)m+(y-~)2]~; R 2 = [(x-~)2+(y+n)2] ~
satisfies the differential equation (4) everywhere except at x = ~, Y = n where it has a logarithmic singularity. The normal derivative of this function vanishes along the boundary of the half-plane. Hence, N(x,y;~,n) is the GREEN'S function of second kind for the half-plane. The solution to equation (4) can thus be represented in the following form :
-
v(x,y,k)
I/
= ~
44
-
Ko(akR o) ~2(~,k)d5
--C@
(15) 0 -®
I R°
Settlng
y = 0
In (15), we obtaln the integral equatlon Ko(a~
(16)
[(x_~12+y2]~
--
(x-5)2+n2 ~}f(~,n)d~dn
= g(x,X)
,
0 -®
g(x,~) =
1
['Vl(X,
~)
/ -
Ko(ak Jx-~ J) ~2(~,n)d~ ]
for the unknown functlon f(~,n). If we take FOURIER transforms wlth respect to x, we wlnd up wlth (17)
Je -n~
/
(~,n)dn = ~
o
G(~,k) -
1
G(~,X)
g(x,X) e -l~Xdx .
~_® Introduce the notation p2 = a2X2+ 2, (17a)
,(p,~) = ~1 pa(~, } / ~ - ~ ) = ,1(p,~),i ,2(p,~) F(w,n) = Fl(~,n)+iF2(~,n)
Separating real
.
and imaginary parts In (17), we obtain two indepen-
dent integral equations for the unknown functions
FI
~ -PhFl(~,n)d n = ¢i(~,p) . 0 To Justify taking FOURIER transforms wlth respect to
and
F2 :
(18)
we may subject
g(x,~)
x
In (16~
to the condition that the solution to (16)
be a certain function f(x,y)~ LI(D) , D = {-® • x • ®; 0 < y < ®}. We can then represent the unique solution to equation (17) by means of formula (12). On inverting the FOURIER transform (13), we arrive at the unique solution f(x,y) to the inverse problem (2), (14).
-
45
-
We point out that the solution to the integral equation expressed differently. If we suppose that 1. f ( x , y ) e L l ( - 2. F ( m , z ) 6 L 2 ( O
(18) may be
< x < ®), y 6 ( 0 , - ) , < z < ~), ~ ( - - , ® ) ,
then the solution to (18) has the following form [26]: (18a)
where
F(x)
Fi(m,z)
= l.l.m. ~ A+®
I
t dt ¢I(*'~) ~t 0
Is the gamma-functlon.
2. n-Dlmenslonal
Inverse Heat Conduction Problems
We shall consider the same problems now as in the preceding section but for n-dlmenslonal space. Since the reasoning does not involve any essential changes, our presentation wlll be as condensed as possible. I °- First inverse problem:
It is required to find the function
f(X,Xn) , x = (xl,x2,... , Xn_l) , from the equation (1)
a 2 Su = AU + ~(t)
In the n-dimenslonal half-space are given for u(X,Xn,t) :
(2)
u(X,Xn,O)
Direct verification (3)
f(x,x n)
D= {x n ~ O} providing the following
=
O,
u(x,O,t)
= h(x,t),
u(x,a,t)
= r(x,t)
,
= const.
shows that the function v(X,Xn,~)
:~
-~2tu(X,Xn,t)dt
0 satisfies the differential equation
(4)
~2 ~v-a2~2v = -¢(X)f(X,Xn) , A =
~2 +
.+
"" ¢(~) = ~ 0
-k2t~(t)dt
~X-~n,
- 46 -
and relations v(x,O,k) = / 0
(5)
"k2th(x,t)dt = hl(x,k) ,
v(x,~,~) = O ~ "~2tr(x,t)dt = rl(x,~)
Let
Kn_2(x) T The function
.
be the cylindrical HANKEL function of imaginary argument.
(6)
Q~(X,Xn;~,~ n) = ~
ak n-2 (~-~)-'~-Kn_ 2 (aAR) ,
1--/R = [(x1-~I)2+ • .. +(Xn-~n)2] ~ , with singularity at
(~,~n)
~ = (~l,~2,...,~n_l)
is a fundamental solution of equation (4).
The GREEN'S function of first kind or fundamental solution for (4), being a function vanishing along the boundary of the half-space, can be expressed by the formula (6a)
Gk(X,Xn;~,~ n) = Qk(X,Xn;~,~ n) - Qk(X,Xn;~,~ n)
.
The solution to equation (4) is then expressible as V(X*Xn'k) :
(7)
where Setting
(8)
Sn
-•han
l(~'k) ~ n
Qk(X'Xn;~'0)d~
+ ¢(k)/f(~,~n ) Gk(X,Xn;~,~n)d~d~ n D is the surface of the half-space D. xn = ~
,
in relation (7), we arrive at the integral equation
ff(~,~n ) G~(x,~;~,~n)d~d~ n = g(x,~) , O g(x,k) = ~ i
[ri(x,~)
+/h
Sn for the unknown function
f(X,Xn).
I(5 ,~) -~n Qk (x, e; ~,O)dS]
-
47
-
Suppose that for given g(x,~), the solution to equation (8) Is a function f(x,x n) satisfying the conditions I. f(x,x n) = O
for
x n < a, a > O,
2. f(X,Xn)E LI(D) . Let us show that equation (8) can have at most one solution under these conditions. We first evaluate the integral (8a)
Since
Jn = (
1
)n-1
n-1 Q~(x,~I0,0) e
:
dXl,dX2...dXn_ 1.
Qk Is even In all variables Xl, the integral may be written as 1-2n n-2 ~ ~_ n-2 Jn = 2 n - l ( 2 ~ ) - - - ~ ( a x ) T f . . ~ R - - - g - K n _ 2 (akR) o --2-
(9)
n-1 x ~ = COSekXkdXldX2..dXn_l . .
; .R = .Ix2 + x2 +
1 + x n_1 2 + ~ 21~ J •
Applying the formula [7] n+2 ( / ~ - ~ ) - - ~ - Kn-2 [ b / ~ ] o -F" I 3-n n-3
=
cos ut dt
(blc I
Kn_3 [I c I/
VJ]
--Eto (9), we obtain the £ollowing representation: l-n 11/_2 - ' a -"zi (I0) Jn = (2~) e • , . "l'.LdD...
2
I
" " "+~n-1
1
The integral equation (8) is obviously of convolutlon-type in the variables Xl,X2,...,Xn_ I. From (I0) it follows that the kernel of the equation Is an absolutely Integrable function. Since f(x,x n) Is likewise absolutely integrable In domain D, applying to equation (8) the convolution theorem for two absolutely Integrable functions and making use of (I0), we obtain ; F(~n,~) (11)
{e-'~n-~I/a2~2+'~'2' -'~n+a'/a2~2+' 12~ - e ~ d~n
o
=
2n-1/a2x2+I~ 12' G(~,~)
,
-
48
-
where
~ (~1,~2,.
• . ,mn_l)
1~12
,
=
2 2
2
~1+~2+...+~n_
1
,
n-I (11a) G(~,k) = (
)n-i f
1
... f
-® F(Xn,~ ) = (
1
)n-1
// ...
~ ~kXk dx , g(x,k) e -i k=l n-I f(x,x n) e
k=l
dx .
By the first assumption, f(X,Xn) vanishes for x n ~ ~ and so F(~n,m) vanishes for ~n < ~" Therefore equation (il) has the simpler form (12)
S F(~n,~)
e
_ n/a2 2+l ld~z n = 2n-2/a22+llZ
G(~,X)
.
sinh( ~/a2~ 2+ I~ I z) We have obtained exactly the same integral equation as in the twodimensional case. Its solution may be found using formula (12) of Sec.1 of this chapter. Repeating the reasoning of Sec. I, we arrive at the conclusion that the solution to the inverse problem (I), (2) is unique in the class of summable functions. 2 °- Second inverse problem: but with the condition (12a)
Consider the inverse problem (I) and (2)
u(x,~,t)
= r(x,t)
replaced by (12b)
~xn U(x,O,t)
Eliminating the variable transformation relations
t
from equation
(3), we obtain for
v(x,O,X) (12c)
= m(x,t)
~ v(x,O,X) 8x n
=
(I) by application of the
V(X,Xn,~)
-X 0
.
equation
(4) and the
h(x,t)dt = h1(x,X)
f.
e-X
0
tm(x,t)dt
= m1(x,X )
.
- 49 -
Direct verification shows that the expression (12d)
Nx(X,Xn;~,~ n) = Qk(X,Xn;~,~ n) + Q~(X,Xn;~,-~ n)
Is the GREEN'S function of second klnd for the half-space it we can represent the solution to (4) by ~X,Xn,k) (13)
D. By using
= -2 ~Sml(x,k) Qk(X,Xn;~,O)d~ n
+ ¢(~)#f(~,~n ) Nx(X,Xn;~,~n)d~ D
•
we set x n = 0 in this last expression, we wlnd up with an integral equation for f(~,~n ), namely If
ff(~,~n ) Q~(x,O;~,~n)d~d~ n = g(x,~), D g(x,~) = I ~ ~½ 3 (x'~) h+ ~ m I (x'~) l Q~(x'O;~'O)d~ "
(14)
Suppose the rlght-hand slde g(x,k) equation has as solution a function
of equation (14) is such that the f(X,Xn)E LI(D).
Taking FOURIER transforms in (14) with respect to Xl,X2,... Xn_ i making use of (IO), we obtain the integral equation
and
L
-~n/a2x2+l~l 2
(14a)
(~n,m) e
d~ n : 2n-1/a2k2+Iml2G(~,k)
.
0 Hence the inverse problem with CAUCHY data on the boundary of the halfspace
D
can have at most one solution.
3. Application of the Problems to Geophysics As we know, there is a large range of mathematical physics preblems dealing wlth heating or cooling of bodies containing internal seurces of heat. We point out, for example, the problem of the effect of radioactive decay on the temperature of the earth's crust [363. The gist ef thls problem is as follows.
-
50
-
Radioactive
decay of elements
temperature
s a t i s f y i n g the heat equation
causes the earth's
crust to heat up, its
2~u a ~-~ = Au + f ,
(i4b) f = ~(t) The function and
~(t)
fl
characterizes
~
.
the volumetric
thermal
source strength,
is given by rl) t
(14c) where
fl(x,y,z)
~
T~tJ is the half-life
Thus knowing,
= ~e
-~t
of the c o r r e s p o n d i n g
for example,
radioactive
element.
the functions
u(x,y,O,t)
= n(x,y,t)
,
(14d) ~--~z u(x,y,O,t) we can determine
the volumetric
scattered in the earth's
= m(x,y,t)
,
strength of radioactive
crust u n d e r the conditions
elements
specified
above.
CHAPTER 5 Inverse Let
v
Problems
be a function
(1)
for S e c o n d - 0 r d e r Elliptic
satisfying the differential
av = (a+~b)v,
a(P)
equation
> O, a(P)+~b(P) p = (xl,x2,
in a domain bounded
D
u n d e r certain boundary
continuous
functions
and
k
v
takes on p r e s c r i b e d
...
conditions.
conditions
vl
, x n)
Here
a
and
are usually
values on the boundary
(la)
> 0,
b
are
is a parameter.
The f o l l o w i n g three types of boundary i)
Equations
S
of
considered:
D :
: f
IS
2)
the normal derivative
of
(Ib)
v
is p r e s c r i b e d
on
S :
~sSVl : ~ l
5)
v
satisfies
on
S
(Ic)
[SV ~-~ + hV]s : 4,
where Problems
the condition
h
and the function
(1) in
S
of domain
is the fundamental S. Finally,
(Id)
are prescribed.
D. The GREEN'S
defined as the fundamental
solution
~ 0
I) - 3) may be solved with the help of the GREEN'S
for equatlon boundary
?
h : const
Rh
function G(P,Q)
solution of equation
D. The GREEN'S
function
(I) v a n i s h i n g N(P,Q)
on the
of second kind
solution of (I) whose normal derivative
the GREEN'S
functions
of first kind is
vanishes
on
function of third kind is the fundamental
of the equation
for which
[[~-B-Rh + hR hi
: 0
.
s
Apart
from the direct problems
for (I) involving the d e t e r m i n a t i o n
a solution under one of the p a r t i c u l a r
boundary
conditions,
ef
of interest
-
52
-
In a certain sense are the inverse problems dealing wlth the determination of the function
b(P)
from certain properties of the solutions
to the equation. In this connection,
one can set up various inverse
problems depending on the nature of the information known about the solutions to equation (I). In thls chapter, we shall stop to consider one such formulation. Let
GI(P,Q)
and
G2(P,Q')
be the GREEN'S functions of first kind for
equation (I) in D corresponding to ~=~I and ~=~2" We cut out of D two infinitely small spheres described around the points Q and Q'. Denote the resultant domain by and G2(P,Q') in DI, we have (2)
GI(Q,Q') - G2(Q,Q')
D i. Applying GREENS'S theorem to GI(P,Q)
: (k2-kl)fb(P)GI(P,Q)G2(P,Q')dP D
Analogous relations also hold for the other GREEN'S functions. ticular, when klnd for G : (3)
k2=k
and
kl=O
G(Q,Q') - Gk(Q,Q')
. In par-
we obtain a FREDHOLM equation of second
: kfb(P)G(P,Q)Gk(P,Q')dP
.
D For sufficiently small
~
Therefore differentiating have (4)
Its solution is an analytic function of (3) wlth respect to
k
and setting
~G~(Q'Q') I = - f b (P)G(P,Q)G(P,Q')dP ~ ~=0 D
k.
k=O, we
.
Thls may be regarded as an integral equation of first klnd for the function b(P). Below we shall consider some specific inverse problems for equation (I). I. Inverse Problem for Equation (I) In a Half-Plan e Let
a(P) = a 2 = const., P = (~,~), b(P) = 0 for
domain D be the half-plane (4) becomes
n < Yl' and let the
o > O. Under these conditions, equation
(5) Sfb(~,~)[Ko(arl)-Ko(ar2)]~Ko(ar3)-Ke(ar4)]d~do 0
-~
= f(xl,x 2)
-
where
Ko(ar)
(5a)
53
-
is the HANKEL function of imaginary argument and I I r I : [((-xl)2+(n-yl)2]~ , r 2 = [(6-xl)2+(n+yl)2] ~, I 1 r 3 = [(~-Xm)2+(n-yl)2]~ , r 4 = [(~-x2)2+(n+yl)2]~ , Yl : const • 0 .
We impose on the rlght-hand side f(xl,x2) of integral equation (5) the single requirement that the solution b(~,n) of the equation belong to LI(D). We take FOURIER transforms In (5) wlth respect to x I In thls connection equation (7) of Sec.1, Chapt.4.
and
x2
using
Equation (5) then assumes the form f ;
b(~,n) e -I(~1+~2)~ e - n ( ~ +
~~dn=F
i(ml'm2 )'
Yl -~
=
F(~I,~2) , 2 n slnhy I / ~ i s Inhy I / ~ 2
:
~
f(xl,x 2) e -cm
dXldX 2 .
--c@
It Is not hard to show that (6a)
IF(~1,~2) [<
-
2/'~a2+~2) (a2+~2) '
_®
Ib(~,n ) Idea n
which implies that it is possible to take FOURIER transforms in (5). In what follows we shall regard
~I positive and
U : ~1+~2 ,
~2
negative. Let
U E (-~,~)
(7) V:
/~i
V g (2a,®) .
+ /~J~ 2 ,
The JACOBIAN
(7a)
"
~ (L~'I,~ 2 )
-
,.~
-
-
~
0
(~I,~2) ÷ (u,v)
ls one-to-one and so has an inverse.
Substituting
(7) In (6), we obtain
(8)
fe
-nVr(u,n)dn
= F2(u,v)
,
Yl (9)
r(u,n) = f b ( ~ , n )
e -lU~d~ .
Since b(~,n) is by hypothesis an absolutely Integrable function, r(u,n) Is continuous and bounded. Therefore, the unique solution to (8) may be expressed In the form [26] (_l)n(~)n+ 1
~n ~v n
r(u, Yl+t) = llm n÷® Yl v F3(u,v) = e F2(u,v)
(ga)
F3(u, ~) I
nl .
It Is known that equation (9) has a unique solution (almost everywhere) In the class of absolutely l~tegrable functions. The solution may be represented by # -
(10)
b(~,n)
= ~fr(u,n)
e lU~du .
Thus we have proved the following uniqueness theorem for equation
(I)
In a half-plane. Theorem:
The inverse problem for equation (I) has at most one solution in the class of absolutely integrable functions.
The entire above discussion Is clearly valid for the GREEN'S function of second kind, which may be expressed in the form (10a)
N(P,Q)
= 2~ [ ~ (arl)+Ko (ar2)]
2. Inverse Problem for Equation
"
(I) In a Half-Space
We shall assume a(P) = a 2 = const., P = (~,n,~), b(P) = 0 and the domain D to be the half-space ~ O.
for ~< z I
-
55
-
Equation (4) thus assumes the form .
.
~,~J _
(1)
.
.
aR 1
b(~,n,~)(~
-aR 2
_- e
R1
_
-aR 3
-aR 4
)(e
e
R2
R3
)dgdnd~= f(xl,x2,Yl,Y2)
R4
I I RI: [(~-Xl) 2+ (n-y1) 2+ (~-Zl) 2] ~; R2: [(~-Xl)2+ (n-Y1)2+ (~+Zl) 2] ~ ; 1 1 R3= [(~-x2)2+ (n-Y2) 2+ (~-zl) 2] ~ ; R4: [(~-x2)2+ (n-Y2) 2+ (~+zl)2] ~ ; z I = const > O;
Let the function f(xl,x2,yl,y2) be such that the solution of equation (I) belongs to LI(D). As before, on taking FOURIER transforms wlth respect to Y2 In (I), we obtain fi/b(~'n'~)e-i(~l+~2)~
b(~,n,~)
Xl,X2,Y I
and
- I(~2+~4)n
z1-®-®
•e
(2)
/2 21 2 '2 2 2 -~(va +~IT~2 + /a +~3+~4)
d~dnd~
a'2+ 2~ 2 /'2+ 2. 2 ~ 1 ~ 2 ~a ~3T~4 F(~I '~2' ~3 '~4 ) F(~I,~2,~3,~ 4)
/ 2 + 2+ 2 / 2 + 2+ 2 ~slnh zl~a '"I ~2 slnh zlfa '~3 ~4 ~e
=
I
f
oo
f(xl,x2,yl,y2) e
-I(~1xl+~2Yl+,.,3x2+~4Y2 )
( 2~ ) 2_~_®
dXldX2dYldY 2
It is not hard to show that the FOURIER transform of f(xl,x2,Yl,Y 2) exists and therefore one may consider equation (2) instead of (1). Let
w I > O, ~3 < O, m4 = 0
(3)
u
-- ~ I
+
v
=
'
~2
and
w 2 6 (-®,®)
~
'
2
2'+ /a2+ 2'
and introduce the notation
u6(-®,')
,
v~(-®,-)
,
I
w
/a 2 :
+wl+~
2
w~(2a,®)
)
The JACOBIAN (3a)
~ (u,v,w)
=
~3
.
~i
$ 0
,
-
56
-
is continuous for the considered values of (~I,m2,~3) + (u,v,w) Substituting
the mapping
(3) into (2), we have ;
(4)
~i " Therefore,
is one-to-one and has an inverse.
/;b(~,n,~)
e "lu~-iVne -~Wd~dnd~
= F2(u,v,w)
,
zI .... F2(u,v,w)
: Fl[~l(u,v,w) , ~2(u,v,w),
~3(u,v,w)]
Let
j
/e-i(uE+vn).b(E,n,;)dEdn
= h(u,v,E),
(5)
/h(u,v,~) zI
e-~Wd~ = F 2 ( u , v , w )
Repeating the reasoning of the first section, we again arrive at the conclusion that the inverse problem for equation (I) in a half-space has at most one solution for which the following representation is valid:
b(~,n,~)
i / jh(u
= (2~) 2
~) el(U~+Vn)dudv
,v,
,
(_:t~ntn~n÷l an F 3 ( u , v , ~ ) (5a)
h(u,v,t+z i) = lim
i t~i
n÷~
aw n
n!
zlw F3(u,v,w)
= e
F2(u,v,w)
.
Thus the only difference between the inverse problem for a half-space and the corresponding problem for the half-plane is that we have made use of excess information. Namely, to determine a function of three variables a function of four variables
b(~,n,~), we employ
f(xi,x2,Yl,Y2).
The case where the given function is gated further.
f(xl,x2,Y i)
has to be investi-
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-
58
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[12] GEL'FAND, I.M., GRAEV, M.I. and VILENKIN, N.Ja., Generalized Functions, Vol.5: Integral Geometry and Representation Theory, Academic Press, New York 1965. ~13]
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-
59
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RAPPOPORT,I.M.,On a two-dlmenslonal inverse problem in potential theory, Dokl. Akad. Nauk SSSR, 28, 1940.
~0]
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