Geophys. J. R. astr. Soc. (1970) 21, 337-357.
Inverse Problem for the One-dimensional Wave Equation
M. L. Gerver
(Received 1970 February 16)
Summary
The problems of determination of velocity-depth functions from traveltime curves or from dispersion curves show that the solution of an inverse
problem may not be unique.
We study here, as a preliminary analogy of such problems, the derivation of the unknown density function for an inhomogeneous string
capable of small transverse vibrations, with one end fixed and one free.
A unit impulse is applied at the free end, and the subsequent motion of
the free end is observed. We prove that the density as function of position
on the string is uniquely determined by these observations, under certain
conditions.
If a more general disturbance is applied, and similar observations are
made at an arbitrary point of the string, is the determination of density
still unique? We show that it is, provided all modes of free oscillation
of the string are excited when the string is symmetrical with free ends.
Further, we examine the stability of our solution. Could very large
variations of density correspond to small variations in the observed
motion? If so, a solution from actual data, liable to error, would be
useless. We show that the solution is stable for a wide class of strings
provided the observation point does not coincide with a node of one of the
first N modes, and that these modes are excited ' distinctly enough '. The
choice of N and the meaning of ' distinctly enough ' are fully explained.
The inverse problem of seismology is to determine the internal structure of the
Earth from observations of the surface motion resulting from earthquakes and
explosions. Some examples of inverse problems are the determination of velocitydepth functions from travel-time curves or from dispersion curves; and these examples
show that the solution of the inverse problem may not be unique. The basic question
arises: is it possible to determine the internal constitution of the Earth from observations on its surface?
A priori, it is not excluded that differing models of the Earth may be indistinguishable when examined from the surface. The complete mathematical solution of this
question is rather complicated. We will choose a simplified case: the inverse problem
for the one-dimensional wave equation-' equation of the string ', Many aspects
of this simplified problem will be important for the general problem. This simplified
problem is also of independent interest, it is a problem of interpretation of normal
incidence.
337
338
M. L. Gerver
It is convenient to start with the simplest variants of our simplified problem,
generalizing step by step.
1.8-source of oscillations. The questions of uniqueness
1 .1. Statement of the problem
Given is a string of finite length 1. Its left end is free, its right end fixed. The
tension p is constant along the string: p = 1. The density p depends on the distance
from the free end: p = p(x). At the initial moment there is no motion, and we
hit the free end, giving it a transverse unit impulse. The string will execute clastic
motion, with transverse displacement u(x, t).
The inverse problem is to find the density distribution, knowing the oscillation
of the left end. (Knowing A(t) = u(0, t), to find p(x)).
Our question now is: can we find p(x) uniquely? Or is it possible to find two
different strings, pl(x) # p 2 ( x ) with A,(t) = A,(t)?
The answer is p(x) can be determined uniquely from A(t).
1.2. Explanation of the result
I was surprised to learn that the structure of the whole string is determined by
the motion of a single point of it (see (5)). Now it seems natural to me, and I will
explain why. Let us consider A(t) in more detail. I shall start with a homogeneous
string.
Fig. 1
Inverse problem for the one-dimensional wave equation
339
t
Fig. 2
Remark. Instead of ‘ string ’ I should strictly speak of ‘ equation of string ’;
u(x, t ) is not a motion of a real string, but a fundamental solution of the equation:
pii-(pu’)’ = 6(x)6(t)
with
4 < 0
=0
1.2.1. The homogeneous string. At present we assume p = 1, if also p = 1, u(x, t )
can be reproduced as in Fig. 1; the corresponding A ( t ) = u(0, t ) is shown in Fig. 2.
At the initial moment the left end jumps to the level 1. It pushes the neighbouring
points; they also jump one after another to the level 1; the disturbance propagates
to the right with velocity equal in our units also to 1 (Fig. 3).
At the moment T the disturbance reaches the right end and the impulse is reflected
with coefficient ‘ - 1 ’. The points of the string, from the right to the left, begin to
return to their initial positions. At the moment 2T the whole string is in the zero
position. At this moment our impulse ‘ - 1 ’ is reflected from the free end with
coefficient ‘ + 1 ’ and the whole cycle repeats with the single difference that the
displacement is now downward, instead of upward.
The picture remains qualitatively the same if ,u and p are constants other than 1.
In that case the velocity of propagation of the impulse is V = J(p/p); the displacement is l/J(pp) (Fig. 4). If we hit the string with impulse i , the displacement
increases by a factor i (while the velocity remains the same) (Fig. 5).
1 .2.2. String consisting of two homogeneous segments; n-strings. Now let the string
consist of two homogeneous segments. Part of the impulse i is reflected from their
contact, part is transmitted through it. The law of conservation holds: the sum of
reflection and transmission coefficients is equal to 1 (Fig. 6).
The reflection coefficient d is determined by the formula:
1 +d
-
J ( P 1 Pl)
-
1 -d
J ( P 2 P2) *
This formula represents the following condition: after redistribution of the impulse
the displacement of the string in the vicinity of the contact is the same to the right
as to the left. The above formula leads to:
d = J ( P , PI)--J(PZ P J
J(P1
PI)+J(PZ
P2) ‘
340
M.L. Gerver
i
E
E
627-
€
Fig. 3
Hence the coefficient of reflection from the other side of the contact is (-d) (Fig. 7).
I f p = I,
We now introduce the definition. An n-string is a string consisting of n homogeneous segments, where the time of propagation of an impulse through each segment
is the same. If this time T / n is given, and if p = 1, the n-string S can be uniquely
represented by n velocities ( V , ... V,) or by V , and n-1 reflection coefficients
(d ...d , - where
dk
=
vk+ 1 - vk
vk+ 1
+ vk
'
Inverse problem for the one-dimensional wave equation
34 1
Fig. 4
Fig. 5
1.2.3. The proof of the theorem of uniqueness for n-strings. The case of p = 1.
The fundamental solution for the n-string is the piece-wise constant function u(x, t);
its discontinuities are the characteristics shown in Fig. 8. This solution can also be
represented by the identical Fig. 9; it shows how the original impulse, given at the
point x = 0 at t = 0, branches owing to reflection and transmission. It follows
from the above description of the case of the homogeneous string that:
(1) The impulse i, propagating along the segment with velocity V, generates the
increase of displacement i V ; and
342
M. L. Gerver
ld
-id
i(l+d)
Fig. 7
1 ig. 6
(2) At intervals T / n two impulses reach each contact from opposite directions.
If these impulses at some moment are i and j and the reflection coefficient is d,,
then the impulses are redistributed as follows:
I = idk + j ( l
i
+ dk),
Impulses j and J on one side of the contact can be uniquely determined from
impulses i , I on the other side and reflection coefficient dk.
If j = 0, i and I determine d,:
dk = I/i.
Suppose we know A(t) for an n-string S; i.e. we know the consecutive displacements A,, A,, ... . First of all we can determine V,:
v, = A , .
Then we can determine all impulses in the first segment. The first two of them
determine d , .
We proceed by induction. Suppose we know all impulses in the k-th segment.
Then we can determine dk from the first two of them. Knowing dk we can determine
all impulses in the (k+ 1)-th segment, and so on.
Determining ( V , , d , , . . . , d , - , ) we determine S . The theorem of uniqueness for
the n-string is thus proved.
1.3. W h a t i f p # l ?
p ( x ) and p(x) cannot both be determined from A(r). Consider two n-strings
S and S :
S = { P l y P I , * * * , P n , Pn>,
S={Ply
PI,
P n , Fn>*
Suppose the time of propagation T is the same for S and 3. We write
Ok =
d(Pk pk),
zk
= J(pk
Pk).
Then the following can be proved: A(t) is the same for S and 3 if and Only if 0, = zk
For all k, 1 < k < n. The proof is based on two facts: (1) impulse i produces
displacement i/a, in the k-th segment; and (2) dk = ( c k - e k + l ) / ( c k + c k + l ) *
343
Inverse problem for the one-dimensionalwave equation
____c
X
\i
I
Fig. 8
344
M. L. Gerver
-X 1
Fig. 9
Inverse problem for the one-dimeosioaal wave equation
345
1.4. Theorem of uniqueness for general strings
Assume that p(x), p(x) and p- ' ( x ) are positive measurable bounded functions'
x E (090.
u(x, t ) is the fundamental solution of
= s(x)s(t),
pii-(pu'Y
T is the travel-time of impulse along the string:
x, is the point, reached by the impulse at time t :
MoXis the mass of the part of the string,in (0, x):
1
X
MoX
=
P(0dC.
0
We introduce M ( t ) = MoXL
and, as before, u(0, t ) = A(t).
Theorem. The functions M(t), t E (0, T ) , and A(t), t E ( 0 , 2 T ) determine each
other uniquely.
To see that the above theorem of uniqueness for n-strings leads exactly to this
theorem, it is enough to notice that
since the velocity of the impulse at the point x, is
and in time dt the impulse propagates along the segment V (x,) dt, while the mass
of this segment is dM = p(x,) V(x,)dt = a(xr)dt.
1 .5. Some details: an observation time, boundary conditions, extrapolation
M ( t ) can be determined in the interval (0, to), to < T, from A(t) in the interval
(0,2tO).
Up to now the boundary condition u(t, l ) = 0 has not been used. If the right
end were not fixed but, say, free (u'(t, I ) = 0), A(t) would be the same in the interval
(0, 2 T ) . If A(t) is observed in the interval ( 0 , 2 T + E ) we can not only determine
M(z) in (0, T)but also: (a) discover that the impulse has returned from the right
end of the string (see (5)); and (b) find a boundary condition on this end, among the
set of homogeneous conditions
u(1, t) +hu'(l, t ) = 0
(see (6)). Different h correspond to different A(t) in (2T, 2T + E ) . It was not necessary to assume that the left end is free. We could introduce any other homogeneous
8
346
M. L. Gerver
boundary condition at the left end, except, of course, the fixed end condition
u(0, t ) = 0. For the case u(0, t ) = 0 we would have to apply the impulse and
observe the displacement at some other point.
The extrapolation of A(t) can be used to control the determination of M(t),
if A(t) is observed for 0 d t d to, to > T. We determine M(t), T and the boundary
conditions from part of the data, i.e. from A(t) in 0 < t < 2 T + e . Then we use
this M ( t ) and the boundary conditions for the solution of the direct problemthe determination of A(t) for t > 2T. Comparison of this A(r) with the observed
A(t) is the control of our solution. In any case A ( t ) in 0 d t < 2 T + E determines
uniquely A(t) in t > 2T.
2. More complicated source. The questions of dqueness
2 . 1 . Statement of the problem for an unknown source
Up to the present we assumed the source to be a &function. Suppose now the
string is excited by a more complicated source F ( x , t). Then
p U - ( p ~ ' ) ' = F ( X , t).
We assume at first that the source, as before, is concentrated at the left end of the
string:
F(x, 0 = f 0)w .
Then
U ( X ,r ) = U ( X t, ) *f (t).
and in particular
UP, 0 = A(t) * f (0,
or, after Fourier transformation (from t- to z-domain),
If the excitationf(t) is known, we can determine A(t) from U ( 0 , t); this reduces the
problem to the previous one.
We should not, however, assume that f ( t ) is known, since the problem for the
string is a model-problem for future inversion of an earthquake record, when the
source mechanism is unknown. So we have to determine both A(t) and f ( 2 ) from
U(0, t), i.e. to find two factors a(,) andf(z) from their product 8(O,z).
2 . 2 . Examples of non-uniqueness
The trivial example which comes directly to mind is:
2i (2) = 222 (2);
Then
3 2 (Z) = 2Jt (2).
2I (Z)31(Z) = a2(Z)32 (Z).
Physically this means that double the force, applied to a string with double mass,
produces the same effect as the original force applied to a string with the original
mass.
To avoid non-uniqueness of such a kind, we shall fix the mass of the string once
and for all. From now on all strings are considered to have the same known mass M.
This limitation does not lead to uniqueness. Consider two arbitrary strings
S, and S2. Denote by A,(t) and A2(t) the fundamental solutions for them at x = 0.
Assume fi(t) = A2(t) and f 2 ( t ) = Al(t). Then, since the convolution is commutative,
At *fl = A2 *fi.
Inverse problem for the one-dimensiona1wave equation
347
Observing that U,(O, t ) = A,(t) *f , ( t ) we have no means to decide whether we are
dealing with a string S1 under force f , ( t ) , or with a string S2 under force f2(t). Thus
we have succeeded in making two arbitrary strings indistinguishable. This is indeed
a strong non-uniqueness. How can we improve the situation? We still desire to
find something about the constitution of S by observing U(0, t).
Can we change the situation by introducing some limitations on the source f ?
2.3. Finite sources
Since f is related to an earthquake model we can assume, for example, that f ( t )
is a finite function, i.e. is different from zero only on some limited time-interval.
A(t) is different from zero on infinitely large t . Consequently, in the last example,
where f , = A , and f 2 = A,, neither f , and f 2 are finite functions.
Can we find two finite functions h,(t), h2(t), for which
A , * h i = A2 *h,?
The answer is yes (see Section 2 . 3 . 3 ) .
2 . 3 . 1 . Finitizable fwctions. Suppose a finite function 4 f 0 exists for which A I4
is finite. Let us call such functions as A finitizuble; it is possible to make them
finite by convolution with a proper finite function 4. This we shall call finitizing.
This possibility has the following physical interpretation: a string, after being hit,
oscillates incessantly and without attenuation, since we did not introduce attenuation
into the wave equation. However, we can determine the source 4 (t)6(x) in such a
way that after some time the string will return to the initial position and the oscillations will stop.
For a homogeneous string the function A(t) has a period 4T. If we give to the
string an impulse at t = 0, and the same impulse in the opposite direction at t = 4T,
the string will stop in its initial position:
+(t) = 8 ( t ) - s ( t - 4 T ) .
Another way to stop the string is to give it two identical impulses at t = 0 and t = 27':
4(t) = s ( t ) + s ( t + 2 T ) .
This can be generalized: such a generalized function #(t), $ ( t ) f 0 only on ( 0 , 2 T ) ,
exists for nearly any string S, so that A * 4 = 0 for t > 2T.
2.3.2. Minimal finitizing force (m-f-f). The condition $ ( t ) f 0 should hold
throughout an interval no less than 2T. Otherwise, as we shall see later, A * 4
will be not finite. That is why, if the finitking function is zero outside the interval
(0,2T),
we shall call it a minimal finitizing force (m-f-for m(t) for short).
2 . 3 . 3 . An example of non-uniqueness for finite sources. Let S , and S2 be two
arbitrary strings. Let us choose m l ( t ) and m2(t) in such a way that A , * m , and
A2 * m, are finite.
Write
hl=m,*A2*m2,
h, = m2 * A , * m , .
Then,evidently,
A , * h , = A2 * h 2
while hl and h, are finite. Thus we see that the determination of S from U(0, t)
is not unique when the source is finite.
348
M. L. Gerver
2.4. Special and non-special sources
There is another way to seek a unique solution. The source 6 ( t ) 6 ( x ) has the
following important quality: it excites all modes of free oscillations of the string:
the expansion of the fundamental solution into a series of modes of free oscillations
contains all modes with non-zero coefficients.
We shall call the source a special one if it fails to excite the oscillations of at
least one eigen frequency. Later on we shall rigorously explain the following statement: if the string S and the source f ( t ) 6 ( x ) (with a finite f ( t ) ) are chosen independently, then with probability 100 per cent the source will be a non-special one
for S.
2 . 5 . The theorem on finding two factors from their product
Let us impose on S additional limitations: S is a string with both ends free:
for t > 0
~ ' ( 0 ,t ) = ~ ' ( 1 t, ) = 0.
Furthermore, S is symmetrical about its centre:
p ( x ) = p(1-x);
p ( x ) = p(1-x)*.
Then we will prove the following theorem:
If the source f (t)6(x) is non-special, we can determine uniquely both A(t) and
f ( t ) from U(0, t ) = A(t) * f ( t ) ; or, in terms of the Fourier transformation (in the
z-domain), the two factors z(z) and f ( 2 ) can be uniquely determined from
m t z ) = J<Z,
m.
One consequence is that M(t) or, if p = 1, p(x) can be determined uniquely from
8 (0, t ) if the source is non-special when the string is symmetrical with free ends.
3. Questions of stability
The last theorem looks surprising. Before giving the proof let us raise a critical
question: is the determination of p(x) from U(0, t), assuming for simplicity p ( x ) = 1,
really stable? If not, i.e. if two quite different pl(x), p z ( x ) may correspond to close
Ul(O, t), U,(O,t ) then p1 and p, cannot in practice be distinguished and our theorem
of uniqueness, without stability, will be a useless mathematical curiosity. The
question of stability is not trivial even for the fundamental solution. Let us investigate the question: if A(t) are close, are the corresponding p ( x ) necessarily close too?
3.1.1. Examples of instability for strings with very heavy or very light parts. The
answer to the above question is no, if the strings can have very heavy or very light
parts. Some examples follow.
Consider two n-strings, S and S with n = 3, such that their first two pairs of
corresponding segments are identical and the third pair are different:
v, = vl, v, = v,,
v3
#
v3.
* It is suffcient to demand that the function o ( t ) = z/{p(x,) p(x,)} is symmetrical: u(t) = o ( T - t )
(compare with 1.4).
349
Inverse problem for the one-dimensional wave equation
Then d , = a, but d , #
a,,
since
If V,, V,, V3 are arbitrary, but V, is very large or very small, d 2 and 22, will be very
close to - 1 or + I respectively. Then A , = A,, A, = A, and A, is very close to A,.
At the same time V, can differ from V3 by as much as we wish.
3.1.2. Additional limitations. This example does not mean that the determination
of p(x) from A ( t ) is always unstable. It means only that some additional limitations
on S are necessary.
Suppose that a positive constant K exists, such that
K-' < p(x) < K .
Unfortunately even in this case one can find S , and S , with significantly different
pl(x), p,(x) but practically coincident A,(t), A2(t). The example of such instability
is based on the following theorem.
3.2.1. The theorem on asymptotics of the fundamental solution for strings with
periodic density. Suppose S is an infinite string, p ( x ) is periodic and piecewise
constant, and for each homogeneous segment the time of propagation of impulse is 1.
Then the fundamental solution u(x, t ) has, for t + 00, a limit J ( l / p o ) which is
independent of x. (Here po is the average density.) S can also be considered as
semi-infinite, extending over the region 0 < x < 00, if p ( x ) is symmetrical with
p(x) = p(-x).
A,(t) and A2(t) for two such semi-infinite strings, S, and S,, are
asymptotically equal for large t if the average density is the same: i.e. for t > to(&)
IA,(t)-A,(t)l < 8.
3.2.2. An example of instability for strings with rapidly oscillating density. Changing
the time-scale, we can make the interval [0, to(&)]arbitrarily small. Let us fix beforehand any number T, say T = n in this new scale. A,(t) and A,(t) will be arbitrarily
close in L,[O, 2x1. Let us cut off from S, and S , the segments [0, x,] and [0, x,],
for which the propagation time is n. This gives us the example of instability
because pl(x), X E [0, xl] and pz(x), X E [0, x,] are not necessarily close.
The propagation time for each period became very small after we changed the
time-scale in arranging this example. That is why the parts [0, xi]and [0, x,] which
we cut off contain many periods: p(x) changes rapidly, its form is manyfolded (i.e.
oscillating) and the impulse, passing through the folds, attenuates quickly, so that
A(t) represents only the average density.
3.2.3. The generalization of the theorem and of the example
The last theorem can be generalized as follows:
This formula leads to a more complicated example of instability. For any
we can find two strings S, and S , such that
E
>0
IA,(t)--Az(t)l < E for t E [O, 2x1,
but p l ( x ) and p,(x) are not close, and do not become close after any smoothing.
350
M. L. Gerver
3 . 3 . The class FKV
Thus an additional restriction is necessary for stable determination of p ( x ) from
A(t). We have to eliminate the rapidly oscillating p(x) by the condition: a constant
V exists, such that
varp(x) < V.
(This means that, for an arbitrary set of points x,, 0 < xo < x1 < ... < x,
< I,
Let FKV be the set of functions f ( x ) , x E [0,1] for which
0 < K-' < f ( x ) < K ;
From now on we assume
varf ( x ) < V.
p(x) E
with some fixed K, V. If p f 1 we shall assume also p ( x ) E FKv. The strings with
p(x) E FKV and p ( x ) E FKv form a natural and reasonably wide class, for which the
solution of the inverse problem is stable.* The exact statements follow.
3.4. The formulation of theorems of stability
Let (PI, pl) and (p2, p2) characterize two strings from the class just determined;
let A,(t), Ml(t) and A2(t), M,(t) be the corresponding functions A, M. M ( t ) is
investigated on the interval [0, TI, and A(t) on [0,2T]. Let us fix T, say T = n.
As already mentioned, M , ( t ) = M2(t), t E [0,n] if and only if
A , ( t ) = A,(t),
tE
[O, 2x1.
Theorem: If p i , p1 and p2, p 2 belong to Fgv, then for any E > 0 there exists S > 0
such that
j.
[A1(t)-A2(t)]2dt < E,
0
if
IM1(0-M2(t)l
< 6.
This theorem means uniform stability in the direct problem: if M ( t ) are close in
C[O, n], then A(t) are close in L2[0,2n].
The stability criterion for the inverse problem can be derived from this in a
rather general form. The functions M ( t ) are uniformly bounded and have uniformly
bounded derivatives. Consequently they are compact in C[O, n]. Their transformation into A(t) is continuous and one-to-one. Consequently it is mutually
continuous. This means that M ( t ) are close in C[O, n] if A(t) are close in L2[0,2n].
* Two remarks: (1) It is essential that our limitations on strings are not comedod with any
smoothness of p and p. The functions p, p E F ~may
, have discontinuities, the number of which is
irrelevant in the theorems of stability (3.5); and (2) The condition that each function p(x) and p(x)
belongs to Fxyis not necessary; we can demand instead that
where
(compare with the analogous remark in 2.5); the solution of the inverse problem is stable also in
the corresponding wider class of strings.
Inverse problem for the onedimensional wave equation
351
3.5. A general scheme for investigation of stability in inverse problems
Note that basically this scheme is rather common in investigations of inverse
problems: if a compact 92 is transformed into R:
F
rn+R
and the transformation F is mutually unique, then to check the continuity of F is
often simpler than to prove the continuity of the inverse transformation F - I . And
continuity of F-' follows from continuity and mutual uniqueness of F.
3.6. The theorem of stability for the direct problem. Plan of proof
Our theorem on continuity of the transformation of M(r) into A(r) can be proved
as follows. Denote by Ak the eigenvalues (i.e. the squares of the eigenfrequencies 0,)
and by yk the eigenfunctions of a string (p, p):
Assume
(pyk')'+Ak pj'k = 0; y,'(o) = yk(0 = 0;
& = ek2.
Then
and consequently
A(t) =
m
k=l
ck-sin0,r
-
ek
The proof of the theorem of stability can be reduced to the verification of three
lemmas:
Lemma 1. A C > 0 exists, depending only on the constants K, V in definition of
F R Y , such that for any k
lek-k( < c.
Ickl < c,
Lemma 2. A number N o can be found for any E > 0, t > 0, such that for any N,,
N, (subject to No < N , < N2)
Lemma 3. A 6 > 0 can be found for any E > 0, k > 0 such that
if
<
and I ~ k l - ~ k <
Z ~e
IM,(t)-M,(t)l < 6 for t e [0, n].
lekl-ek21
Here
c k i , i = 1, 2, are &, c
k for the string (pi,
pi).
We have represented A(t) through the spectral parameters of the string; i.e.
through the eigenfrequencies 0, and the normalizing factors C,. It follows from
the assumption p E F K V , p E FKv that ck are finite and
assuming T = n we have the statement of Lemma 1.
352
M. L. Gerver
The statement of Lemma 2 is that the expansion of A(t), for strings from F,,
converges uniformly in the mean in any finite interval. Consequently not the series,
but finite sums, can be compared instead of A,, Az. These sums are close (not only
in the mean but even in C), if M,(t), M z ( t ) are close in C; this follows from Lemma 3.
4. The spectral properties of strings
4.1. The eigenji-equencies and normalizing factors
We have mentioned all these details in order to analyse the important relation
between M ( t ) , A ( t ) and the spectral parameters e k , ck of a string (see (2), (6)). A(?)
and consequently M ( t ) can be determined uniquely from the two sets (0,) and {Ck}.
It follows from the proof, given above, that we need only a finite set of numbers
el, ..., O N and C1,..., C, to determine M ( t ) with a given limit E for the error; N
increases when E decreases. ‘ To find M ( t ) with an error e ’ means: ‘ to find Mo(t)
such that for t E [0, 711
IM,(t)-M(t)l < E ’ .
Conversely the sets {Ok}, {ck}are determined uniquely by M ( t ) : if, for two strings
( p , , pl) and (p2, p2), M , ( t ) = M2(t), then the spectral parameters 8ky ck are also
identical. Consequently, any one of the following determines the other two
uniquely: (1) M ( t ) on [0, R ] , (2) A(?) on [0,2n];and (3) a pair of sequences { e k } , {C,}.
4.2. Unique determination of the structure of astring by two spectra of eigenfrequencies.
Zeros and poles of &z)
There is also a fourth datum, equivalent to these three: a couple of sets { e k } ,
{a,} together with normalizing factor
1
*
T
dx
0
dt
0
Here ek, as before, are eigenfrequencies in the problem with the left end free and
are eigenfrequencies in a second problem: that of the string
right end fixed.
with both ends fixed. It is more convenient to prove this statement after Fourier
transformation of the fundamental solution u(x, t ) into G(x, z). We have for G(x, z)
the equation
A =22,
(pG’)’+ApC = 0,
and boundary conditions:
Denote by y ( x , A) the solution of the same equation, for which
For the ratio
(py‘)1 = - 1 7
(Y)l
= 0.
we have, evidently,
( 4 0
i.e.
= 1,
( 4 l
= 0,
nverse problem for the one-dimensional wave equation
353
In particular
Using this formula, we can prove that
Both y(0, A) and y’(0, A) are entire analytic functions of order 3. Denote by z k
and s, the zeros of these functions, where k = 1,2, ...; the numbers k increase with
the moduli of the zeros.
Then
y‘(0, A) = y’(0,O)
k==l
Evidently y(0, A) = 0 if and only if A is an eigenvalue in the problem with both ends
fixed, i.e. if 1 = a,’. It is comparatively easy to verify that all these zeros are simple
ones: i.e.
Consequently
zk
=
a,’.
Quite analogously s k = 8,’ are the simple zeros of y’(0, A). For d = 0 the equation
for y(x, A) is:
(PY‘) = 0
py’ = const.
or
This together with the initial conditions for y(x, A) leads to
P(0) Y‘(0,O) =
-1 ,
Finally
{a,}and Z is indeed equivalent to the knowledge
of &) or A(t). If I is unknown, but both spectra (03, {ak}are given, A(& and
consequently M(t), can be determined to within an arbitrary constant multiplier.
Using in addition the total mass M of the string, we determine M(t) uniquely. So,
if M is known, M ( t ) is uniquely determined by (0,) and (a,}(compare with [l]and
[3] [4n. Let us denote by (1-1)
the problem with both ends fixed and by (0-1)
the problem with left end free, and right end fixed. It is unimportant that 0, relates
We could use instead any other two homogeneous
to (1-1).
to (0-1) and
boundary problems, if their boundary conditions are the same at one end and
different at the other (for example (0-1)and (0-0)).
We see that the knowledge of {O,},
3 54
M.L. Gcrvcr
4.3. The conditions of symmetry
If the string (p, p) is symmetrical, i.e.
A x ) = P(l-X),
A4 = P ( l - X h
and the boundary conditions are symmetrid too (e.g. 1-1 or 0-0 or url0 = u(,,,
urJl= -uIl), then the single spectrum of eigenfrequences determines M(r) uniquely.
This is due to the fact that, at the centre of a symmetrical string with symmetrical
conditions at the ends, an eigenfunction and its derivative are equal to zero alterfor a symmetrical string (p, p,
nately. Hence, for example, the spectrum (1-1)
0 < x < r) is the sum of two alternating spectra, (1-0) and (/-I), for the left half
of the string (p, p, 0 < x < l/2). If the boundary conditions are not symmetrical
(for example of (0-1) type) even a symmetrical string cannot be determined uniquely
by a single spectrum.
4.4. The general formula for
A(z)
If u(x, t) is a fundamental solution of a problem (x-e) with boundary conditions
( x ) au(O,O
+ B A O ) ~’(0,t) = 0,
) t > 0,
(*I Y N , t)+Sp(l)u’(l, t ) = 0,
then A(z) = C(0,z) is a meromorphic function which, except for a constant factor,
is equal to
are frequencies of the problem
where 6, are frequencies of the problem (x-*) and
(I - *) (for fixed left end and given initial condition (I) for the right end).
The last formula will be slighay changed if there is a zero frequency among
Ok(O0 = 0). In this case an additional term zz will appear in the denominator.
Such a situation will occur for the problem (0-0).
4.5. The proof of the theorem of section 2.5
We can now prove the theorem stated in section‘ 2 . 5 on the determination of
both factors from their product
m 4 z ) = A ( 4 J(4.
d(z) is proportional to the product
Here 1-0, and G O k are frequencies for the problems (1-0)and (0-0).
Smce a source 6(x) f(r) is non-special, f ( z ) # 0 when z = f(r0,. Hence the
poles of O(0,z) coincide with the spectrum of (0-0);in fact they are the points
k0 - 4 .
Inverse problem for the one-dimmsional wave Cqo.tlan
355
Since the string @, p) is symmetrical and its mass M is supposed to be known
it is sufficient to know &ok to determine M(r).
Thus the condition of non-speciality for the source provides the possibility of
determining all poles of A(z) from b(0,z). Due to the symmetry of the string and
of the boundary conditions, and to knowledge of M ywe have sufficient information
to determine M(r) (or A(t)). As we know the product b(0,z) and one of its factors
A(z) we can find the other onef(z). The theorem is now proved.
4.6. The meaning of terms special source’ and ‘ m-f-f.’
The condition of speciality of a source 6(x) f (r) occurs when f(z) is equal to
zero at least at one pole of A(z). An oscillation with the corresponding eigenfrequency
is not excited and we cannot determine it from b(0,z).
Both the numerator and denominator of a@)are Fourier transforms of finite
functions: a = b(z)/rFi(z). The function m(t) has been mentioned above as a
minimal finitizing force (m---f ). The zeros of its Fourier transform coincide exactly
with the poles of a(z). When f ( t ) = m(t) then D(r) = A(?)* m(t), and the zeros of
b(z) coincide with the zeros of a(z). The length of the segment outside which the
finite function +(r) is equal to zero is related to the rate of increase of I &z)l at infinity,
and through the latter with the density of the set of zeros of &). The fact that for
eigenfrequencies 0,
x
Ok
-+k k+m T
means that m(r) cannot be concentrated on too short a segment. This is the explanation of the result stated above: for were m(t) concentrated on a segment of length
< 2T, zeros of m(z) would be s aced less close than they actually are. Then m(z)
K
could not be zero at all poles of (z) and the function D(t) would not be a finite one.
We can say that a source 6(x) m(r) is ‘ the most special ’ among all possible sources,
if fi(z) is equal to zero at all poles of &). To construct such a source one needs to
know the structure of the string. If one does not, there are poor chances for even
one zero to coincide with a pole. The zeros off(z) may be distributed on the whole
z-plane. The probability that one of them will coincide with a member of the discrete
sequence of eigenfrequencies is zero.
If we still want to take into account such a possibility we should observe U‘(0, r)
besides U(0, r).
If we are dealing with a finite point source concentrated at the free end 0, the
possibility of determining uniquely A(r) and f ( t ) from U(0, t ) and U’(0, r) is now
quite evident: indeed in this case U,l(O, r) = -f (r)/p(O).
5. The theorems of uniqueness and stability for an arbitrary finite source
But it is possible to consider a concentrated source acting at an arbitrary point
5 E [0,I], where the displacement V satisfies
PV - (PV’)’= S(x-0
f(0,
or even a distributed source, where the displacement W satisfies
p W - ( p W ’ Y = F(x, 2).
The function F(t) is assumed to be 0 outside an interval 0 < t < Tp, F(x,t) = 0 out;p(x),
side a rectangle IIF : a < x < by 0 < t < Tp ; f(t) EL^ [0,TP],F(x,t)EL~(II,,)
p ( x ) E FRY. It is unnecessary to put the observation point at 0. Let this point be
at x, i.e. let the data be
V ( x , r) and V’(x, t )
(point source),
or
W ( x , t ) and W’(x, r)
(distributed source),
where x is fixed.
356
M. L. Cerver
5.1. Thefunctions A,,(t) and B,,(t)
Let A&<) describe the displacement of the point x excited by a &function
concentrated at <,i.e. A,,(t) = V ( x , t ) under the condition that f (t) = S(t). With
the same condition the derivative V' is given by
0.
B,,(t) = K'(x,
It is clear that
1
V(x,
0 = 4,(0
*f(t),
W(x,0 =
J A,&)
* F(C, o d e ,
a
1
V'b,
0 = B,,(t) *fW,
W'(x, 0 =
J Bx&) * F(C, Odt.
a
Hence
1
RXY
4 = A&) f(4Y
m x ,4 =
J L,(4m , z >dC,
a
t < x ,4 = llx,(4J ( 4 , it,l(x,z) =
j
J&) F ( W d < .
a
<
Let us for example consider the problem (0-0) and let 0 c x < < I. Then it
may be shown that
Here 1-42' is the k-th eigenfrequency for a string {(p, p), 5 < x < I}, with similar
is the mass of the segment [0, x].
notation for other frequencies; "@i
Several conclusions can be drawn from these formulae, but we limit ourselves to
the following.
5 . 2 . The theorems of uniqueness
1. Let (p, p) be a symmetrical string with known mass My and let 1/2 < x <
if a source is concentrated at the point (, or 112 < x < u < b if a source is distributed.
Then M ( t ) can be determined uniquely by V ( x , t ) and K'(x, t ) (or by W ( x , t ) and
W;(x, 2)). For the point sourcef(t) also may be found.
2. Again let (p, p) be a symmetrical string with known mass M and in addition
let the source be non-special, i.e. let all free oscillations be excited and let the observation point x be not at a node of any eigenfunction for the problem (0-4). Then
to determine M ( t ) it is sufficient to know V ( x , t) or W ( x , t). In the first casef(t)
can be determined uniquely from V ( x , t). For both theorems observations are
<
Inverse problem for the one-dimensional wave equation
357
+
made for the time interval 0 c t c 4 T TF,which is equal to the duration of the
source function plus quadrupal the travel time of the impulse along the string.
In this way the theorems of uniqueness for an arbitrary finite source are formulated.
5.3. The theorems of stability
Let {f } o be a compact subset of L2[O,TF]which does not include the function
f ( t ) = 0 and let {F}, be a compact subset of &(nF)which does not include F(x,t) = 0.
If f ( t )~ { f }F(x,t)
~ , E { F } , the following stability theorems are true.
1‘. The determination of M ( t ) is stable under conditions of theorem 1.
2’. Let us consider how the theorem of stability 2’ is related to theorem of
uniqueness 2. To find M ( t ) with given error E one needs to know only a finite number
N ( E )of lower eigenfrequencies. It is sufficient to know about higher frequencies
that they are like &/n. So it is sufficient to assume that the observation point does
not coincide with any of the nodes of the lower eigenfunctions and that the source
excitates all free oscillations with orders k Q N(e). But besides these weaker assumptions a stronger limitation is necessary. We require that first N(e) free oscillations
should be excited ‘distinctly enough’. One possible variant of the exact formulation
of this condition is as follows. Let us call a triplet of string, source and observation
point a special triplet if at least one of the first N(E)frequencies is not observed. Then
all triplets considered should belong to some compact which does not include any
special triplet.
If these limitations are obeyed M(t) can be determined from W ( x , t),
t E (0,TF 4T) or from V (x, t), t E (0, TF 4T ) with an :error E, provided W (x, t ) or
V ( x , t ) were given accurately, i.e. with differences less than B(E) from the exact
values in L,. In addition the sourcef ( t ) can be determined by V ( x , 2) with stability
in L,.
+
+
This work was done at the Institute of Physics of the Earth, Moscow, during
1968 and 1969. Some of the results were obtained earlier by G. Borg, I. M. Gelfand,
M. G. Krein, B. M. Levitan, and V. A. Marchenko, while some of the new results were
obtained by the author together with several students and post-graduate students of
Moscow University, among whom I. N. Bernstein should be especially mentioned.
Institute of Physics of the Earth,
Moscow.
References
(1) Borg, G., 1945. Eine Umkehrung der Sturm-Liouvillschen Eigenwertaufgabe,
Bestimmung der Differentialgleichung fur die Eigenwerte, Acta Math., 78, 1-96.
(2) Gelfand, I. M. & Levitan, B. M., 1951. On determination of differential equation
from its spectral function, Izvestia Akademii Nauk SSSR, Ser. mathem., 15,
309-360.
(3) Krein, M.G., 1951. The solution of Stunn-Liouville inverse problem, Doklady
Akad. Nauk SSSR, 76,21-24.
(4) Krein, M. G., 1951. Determination of density of inhomogeneous symmetrical
string from its eigenfrequencies, Doklady Akad. Nauk SSSR, 76, 345-348.
(5) Krein, M. G.,1954. On a method of efficient solution of inverse boundary
problem, Doklady Akad. Nauk SSSR, 94, 13-16.
(6) Marchenko, V.A., 1950. Some questions of the theory of the differential operator
of the second order, Doklady Akad. Nauk SSSR,72,457-460.
(7) Marchenko, V. A., 1952. Some questions of the theory of one-dimensional
differential operators of the second order, I, Trudy Mosk. Mathem. obtshestva,
1.327-420.