INSTITUTE OF PHYSICS PUBLISHING
INVERSE PROBLEMS
Inverse Problems 18 (2002) 319–329
PII: S0266-5611(02)23955-6
Upper and lower estimates in determining point
sources in a wave equation
Vilmos Komornik1 and Masahiro Yamamoto2
1 Institut
de Recherche Mathématique Avancée, Université Louis Pasteur et CNRS, 7, Rue René
Descartes, 67084 Strasbourg Cedex, France
2 Department of Mathematical Sciences, The University of Tokyo, 3-8-1 Komaba Meguro Tokyo
153, Japan
E-mail: komornik@math.u-strasbg.fr and myama@ms.u-tokyo.ac.jp
Received 18 April 2001, in final form 3 October 2001
Published 24 January 2002
Online at stacks.iop.org/IP/18/319
Abstract
Applying a theorem of the graph theory, we obtain upper and lower estimates
between the Sobolev norms and the distance of finite combinations of Dirac
masses. These estimates are then applied to solve an inverse problem for the
wave equation.
1. Introduction and formulation of the main results
Given a positive number T and a function λ ∈ C 1 [0, T ], we consider the problem
u tt (x, t) = u xx (x, t) + p(x)u(x, t) + λ(t) f (x),
0 < x < 1,
u(x, 0) = u t (x, 0) = 0
u(0, t) = u(1, t) = 0
0 < x < 1,
0 < t < T,
(1)
0 < t < T.
The subscripts t and x represent partial differentiation. Throughout this paper, we fix
p ∈ C[0, 1] such that p 0 on [0, 1]. The wave equation in (1) is obtained from a more
general hyperbolic equation describing vibrations
∂ ũ
∂
γ (ξ ) (ξ, t) + λ(t) f˜(ξ ),
0 < ξ < , t > 0,
ρ(ξ )ũ tt (ξ, t) =
∂ξ
∂ξ
with smooth ρ > 0 and γ > 0 on [0, 1]. In fact, by means of the Liouville transformation
(see, for example, [1]), we can transform the latter hyperbolic equation with respect to ũ to the
wave equation in (1) for x ∈ [0, ] with some > 0.
It is well known (see, for example, [2]) that for given every f ∈ H −1 (0, 1), this problem
has a unique (weak) solution
u ∈ C([0, T ]; H01(0, 1)) ∩ C 1 ([0, T ]; L 2 (0, 1)) ∩ C 2 ([0, T ]; H −1(0, 1)).
0266-5611/02/020319+11$30.00
© 2002 IOP Publishing Ltd
Printed in the UK
319
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V Komornik and M Yamamoto
Now we consider a special case where f is the sum of finitely many Dirac masses
f = δa :=
n
δai ,
i=1
with a given integer n and with a = (a1 , . . . , an ) ∈ [0, 1]n . The resulting system
u tt (x, t) = u xx (x, t) + p(x)u(x, t) + λ(t)δa ,
0 < x < 1,
u(x, 0) = u t (x, 0) = 0
u(0, t) = u(1, t) = 0
0 < x < 1,
0 < t < T,
(2)
0<t <T
is a greatly simplified version of a model in seismology, where the determination of a is the first
step in identifying the location of the hypocentre of an earthquake (see, for example, [3]). In
this system, the force term concentrates at n points a1 , . . . , an (some of them may coincide) and
initiates the vibration from the equilibrium at t = 0. Throughout this paper, λ ∈ C 1 [0, T ] is
known and λ(0) = 0. The purpose of this paper is estimation of a by means of the observation
u x (0, t), 0 < t < T . The condition λ(0) = 0 means that the force acts at t = 0.
First we show an estimate of f H −1 (0,1) :
Proposition 1. If λ ∈ C 1 ([0, T ]), λ(0) = 0 and T 1, then two positive constants, c1 and
c2 , exist such that
T
|u x (0, t)|2 dt c2 f 2H −1 (0,1)
c1 f 2H −1 (0,1) 0
for all f ∈ H
−1
(0, 1).
In view of proposition 1, a natural question is now whether δa ∈ H −1 (0, 1) determines
continuously a = (a1 , . . . , an ). The answer is not completely affirmative even in the simplest
case n = 1. In this paper, we establish an upper and lower estimate between a = (a1 , . . . , an )
and u x (0, ·) L 2 (0,T ) . In order to distinguish two point sources a = (a1 , . . . , an ) and
b = (b1 , . . . , bn ), we introduce a semi-norm. We start with the case n = 1. In the following
we use the notation
x := dist (x, Z),
x ∈ R.
In other words, x denotes the distance of x from the closest integer, so that for 0 < x < 1
we have
x = min{x, 1 − x}.
This is not a norm, but this notation is usual in number theory (see, for example, [4]).
Furthermore we set δ0 = δ1 := 0 (as an element of H −1(0, 1)) for brevity.
Proposition 2. Two positive constants, c1 and c2 , exist such that
c1 a − b1/2 δa − δb H −1 (0,1) c2 a − b1/2
for all a, b ∈ [0, 1].
Remarks.
• If a → 0 and b → 1, then a − b → 0, so that we can only identify a from δa
continuously if a is far from the boundary.
• Choosing b = 1, we obtain that δa → 0 in H −1 (0, 1) if a → 0 or if a → 1.
Upper and lower estimates in determining point sources in a wave equation
321
Turning to the case n 2, for any given a = (a1 , . . . , an ) and b = (b1 , . . . , bn ) in [0, 1]n ,
using the triangle inequality and then applying the preceding proposition, we obtain
n
δai − δbi H −1 (0,1) nc2 max ai − bi 1/2 .
(3)
δa − δb H −1 (0,1) 1in
i=1
But the converse inequality does not hold in general. Indeed, it suffices to take n = 2, a1 = a2
and b1 = a2 , b2 = a1 . Then δa − δb = 0, while
a1 − b1 = a2 − b2 = a1 − a2 > 0.
Hence we cannot hope for the inverse inequality to hold unless we minimize the right-hand
side of equation (3) for all possible permutations of the points b1 , . . . , bn .
If π is an arbitrary permutation of 1, 2, . . . , n, then, replacing bi by bπ(i) in the above
argument, we obtain that
δa − δb H −1 (0,1) nc2 max ai − bπ(i) 1/2 .
1in
Our key result is that the inverse inequality also holds true. We set
a − b := min max ai − bπ(i) π
1in
for brevity, where π runs over all permutations of the integers 1, 2, . . . , n.
Proposition 3. Given n 2, two positive constants, c1 and c2 , exist such that
c1 a − b1/2 δa − δb H −1 (0,1) c2 a − b1/2
for all a, b ∈ [0, 1]n .
For the proof of the inverse inequality, we will use the Hall–Rado theorem from the graph
theory in a crucial way.
Applying proposition 1 with f = δa − δb and using propositions 2 and 3, we obtain the
main result for an inverse wave source problem.
Theorem 4. Fix T 1 and a positive integer n. Two positive constants, c1 and c2 , exist such
that for every a = (a1 , . . . , an ) and b = (b1 , . . . , bn ) in [0, 1]n , the corresponding solutions
u a and u b of equation (2) satisfy the estimates
∂u a
∂u b
1/2
c1 a − b (0, ·) −
(0, ·)
c2 a − b1/2 .
2
∂x
∂x
L (0,T )
Remarks.
• We assume here that the number n of point sources is known. Otherwise we can still prove
the uniqueness, but according to the last remark following proposition 2, the stability no
longer holds without some a priori hypotheses concerning the location of the sources. As
for other estimates for our inverse problem, see [5].
• Usually, also in higher dimensions, the Hausdorff distance is used instead of a − b (see,
for example, Alessandrini et al [6–8]). Our example preceding proposition 3 shows that
the equivalence of theorem 4 would not hold true for the Hausdorff distance of the sets
{a1 , . . . , an } and {b1 , . . . , bn } instead of a − b. We consider an example: let n = 3 and
let
3 3 1 5
am = a1(m) , a2(m) , a3(m) =
, + ,
8 8 m 8
and
3 5 5
1
(m)
(m)
(m)
, , +
.
bm = b1 , b2 , b3
=
8 8 8 2m
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V Komornik and M Yamamoto
1
Then am − bm = 14 − 2m
and limm→∞ am − bm = 14 = 0. On the other hand,
the Hausdorff distance between am and bm is max{maxk=1,2,3 min=1,2,3 ak(m) − b(m) ,
maxk=1,2,3 min=1,2,3 a(m) − bk(m) } = m1 −→ 0 as m −→ ∞. As the limiting point
sources corresponding to limm→∞ am and limm→∞ bm , we have 2δ3/8 +δ5/8 and δ3/8 +2δ5/8
respectively, and so our adopted distance is better than the Hausdorff distance.
In fact, we prove the following more general theorem containing both propositions 2 and 3
as its special case corresponding to q = 2.
Theorem 5. Given a positive integer n and a real number 1 < q ∞, two positive constants,
c1 and c2 , exist such that
c1 a − b1/q δa − δb W −1,q (0,1) c2 a − b1/q
(4)
for all a = (a1 , . . . , an ) and b = (b1 , . . . , bn ) in [0, 1]n .
Remarks.
• The proof will show that we can take c1 = (2n)−1/ p and c2 = n.
• Our result shows in particular that the W −1,∞ norm is too weak for the continuous
determination of a, even inside (0, 1). Indeed we have
δa − δb W −1,∞ () c1
for all a = b. However they can be arbitrarily close to each other.
2. Proof of theorem 5 for n = 1
The following proposition is more precise than the case n = 1 of theorem 5. We fix two
conjugate exponents 1 p < ∞ and 1 < q ∞, i.e. p−1 + q −1 = 1. We recall that
δ0 = δ1 := 0.
Proposition 6. We have
−1/ p
δa − δb W −1,q (0,1) = |a − b|1− p + (1 − |a − b|)1− p
(5)
for all a, b ∈ [0, 1] such that |a − b| < 1. It follows that
2−1/ p a − b1/q δa − δb W −1,q (0,1) a − b1/q
(6)
for all a, b ∈ [0, 1].
Proof. It suffices to establish equation (5) because it easily implies equation (6). The case
a = b being obvious, we assume in the sequel by symmetry that a < b.
1, p
If ϕ ∈ W0 (0, 1), then
b
ϕ (x) dx (b − a)1/q ϕ L p (a,b) .
|(δa − δb )(ϕ)| = |ϕ(a) − ϕ(b)| = a
Here ϕ denotes the differentiation in the variable under consideration. For example,
2
(x), ϕ (x) = ddxϕ2 (x), ψ (t) = dψ
(t). Furthermore, extending ϕ to a 1-periodic
ϕ (x) = dϕ
dx
dt
function, we have
a+1
|(δa − δb )(ϕ)| = |ϕ(a + 1) − ϕ(b)| = ϕ (x) dx (a + 1 − b)1/q ϕ L p (b,a+1) .
b
Upper and lower estimates in determining point sources in a wave equation
Hence
p
ϕ 1, p
W0 (0,1)
and therefore
323
|ϕ (x)| p dx |(δa − δb )(ϕ)| p (b − a)− p/q + (a + 1 − b)− p/q
a
= |(δa − δb )(ϕ)| p (b − a)1− p + (a + 1 − b)1− p
a+1
=
−1/ p
δa − δb W −1,q (0,1) (b − a)1− p + (a + 1 − b)1− p
.
1, p
W0 (0, 1)
On the other hand, we consider the function ϕ ∈


 (b − a)x
ϕ(x) := (b − a)a − (a + 1 − b)(x − a)


(b − a)(x − 1)
(7)
given by the formula
if 0 x a,
if a x b,
if b x 1.
We have
|(δa − δb )(ϕ)| = |ϕ(a) − ϕ(b)| = (b − a)(a + 1 − b)
and
ϕ
p
W01, p (0,1)
= a(b − a) p + (b − a)(a + 1 − b) p + (1 − b)(b − a) p
= (a + 1 − b)(b − a) p + (b − a)(a + 1 − b) p ,
so that
p
δa − δb W −1,q (0,1) and therefore
(b − a) p (a + 1 − b) p
(a + 1 − b)(b − a) p + (b − a)(a + 1 − b) p
−1/ p
δa − δb W −1,q (0,1) (b − a)1− p + (a + 1 − b)1− p
.
(8)
The lemma follows from (7) and (8).
3. Proof of theorem 5 for n 2
We need the following important theorem from the graph theory.
Theorem 7 (Hall–Rado). Consider an even graph having 2n points a1 , . . . , an , b1 , . . . , bn .
Assume that some pairs (ai , b j ) are connected such that the following two conditions hold:
• for every k = 1, . . . , n and for every subsequence a = (ai1 , . . . , aik ) of a = (a1 , . . . , an ),
at least k elements b j of the sequence b are connected to one of them;
• for every k = 1, . . . , n and for every subsequence b = (bi1 , . . . , bik ) of b = (b1 , . . . , bn ),
at least k elements a j of the sequence a are connected to one of them.
Then there is a permutation π of the integers 1, . . . , n such that ai is connected to bπ(i) for
every i .
It is clear that the conditions of the theorem are also necessary. See, for example, [9]
or [10] for the proof of this theorem and [11] for other applications.
Proof of theorem 5 for n 2. The second inequality of equation (4) follows easily, by
applying the triangle inequality, then the second inequality of equation (6) in proposition 6,
and finally the Hölder inequality. We obtain for every permutation π the inequalities
n
n
δai − δbπ(i) W −1,q (0,1) ai − bπ(i) 1/q
δa − δb W −1,q (0,1) i=1
i=1
n max ai − bπ(i) 1/q
1in
.
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V Komornik and M Yamamoto
Hence
δa − δb W −1,q (0,1) na − b1/q .
Turning to the proof of the inverse estimate, we write
r := a − b
for brevity. There is nothing to be proven if r = 0, and so we assume in the following that r > 0.
Let us consider the even graph with points ai , bi , i = 1, . . . , n, and let us connect ai to
b j if ai − b j < r . If this graph satisfies the two conditions of the Hall–Rado theorem, then
there is a permutation π of the integers 1, . . . , n such that ai − bπ(i) < r for every i . Then
max ai − bπ(i) < r = a − b,
1in
contradicting the definition of a − b. So at least one of the two conditions of the preceding
theorem is violated.
Let us assume by symmetry that the first condition is not satisfied. Then there is a
subsequence (ai1 , . . . , aik ) of a such that at most k − 1 points b j belong to the union of the
open ‘balls’
m = 1, . . . , k.
Br (aim ) := {x ∈ (0, 1) : x − aim < r },
Let us introduce the functions ϕ0 , ϕ : [0, 1] → R by the formulae
ϕ0 (x) := min{x − ai1 , . . . , x − aik , r }
and
ϕ(x) := ϕ0 (0) − ϕ0 (x).
1, p
Then ϕ is continuous, piecewise linear and ϕ(0) = ϕ(1) = 0, so that ϕ ∈ W0 (0, 1).
Furthermore ϕ0 and ϕ have the following additional properties:
ϕ0 (ai1 ) = · · · = ϕ0 (aik ) = 0,
ϕ0 (b j ) = r
0 ϕ0 (x) r
for all but at most k − 1 indices j,
for all x ∈ ,
and
for almost all x ∈ (0, 1),
|ϕ (x)| 1
ϕ = 0 outside a set of measure 2kr in (0, 1).
The last property follows from the fact that the measure of each ball Br (aim ) is equal to 2r .
(We may have an inequality if the balls overlap.)
It follows from these properties that
1
p
ϕ 1, p
=
|ϕ (x)| p dx 2kr 2nr
W0 (0,1)
0
and
n
n
n
n
ϕ(ai ) −
ϕ(bi )
= ϕ0 (ai ) −
ϕ0 (bi )
|(δa − δb )(ϕ)| = i=1
n
i=1
ϕ0 (bi ) −
i=1
n
i=1
i=1
ϕ0 (ai ) (n − k + 1)r − (n − k)r = r.
i=1
Therefore
p
δa − δb W −1,q (0,1) rp
= (2n)−1 r p−1
2nr
and
δa − δb W −1,q (0,1) (2n)−1/ p r ( p−1)/ p = (2n)−1/ p a − b1/q
as desired.
Upper and lower estimates in determining point sources in a wave equation
325
4. Proof of proposition 1
It suffices to consider the limiting case T = 1: the case T > 1 then follows easily by the
triangle inequality.
First step
We reformulate our problem for the solutions of a homogeneous equation with a
nonhomogeneous initial condition. For this, we introduce the integral operators S and S ∗
by the formulae
t
(Sϕ)(t) =
λ(t − s)ϕ(s) ds,
0<t<1
0
and
(S ∗ ψ)(t) =
1
λ(η − t)ψ(η) dη,
0 < t < 1.
t
Writing H := L 2 (0, 1) for brevity, we have
(Sϕ, ψ) H = (ϕ, S ∗ ψ) H ,
ϕ, ψ ∈ H.
(9)
Introducing the Hilbert space
(g, h)V := (g , h ) H
V := {g ∈ H 1(0, 1); g(1) = 0},
and identifying H with its dual, we have the dense and continuous imbeddings
V ⊂ H = H ⊂ V
as usual, and we deduce from equation (9) the identity
(Sϕ, ψ) H = ϕ, S ∗ ψV ,V ,
ϕ ∈ H,
ψ ∈ V.
(10)
Here we note that H01 (0, 1) and V are proper subspaces of V and H −1 (0, 1) respectively,
which require special care in the following argument.
Since
1
λ (η − t)ϕ(η) dη,
0<t <1
(S ∗ ψ) (t) = λ(0)ψ(t) −
t
is a Volterra integral equation of the second kind, we see that
S∗
is an isomorphism of
H
onto
V.
V
onto
H.
Using equation (10), it follows that
S
is an isomorphism of
Now, for a given f ∈ H
−1
(0, 1), the problem
u tt (x, t) = u xx (x, t) + p(x)u(x, t) + λ(t) f (x),
0 < x < 1,
u(x, 0) = u t (x, 0) = 0,
u(0, t) = u(1, t) = 0,
(11)
0 < x < 1,
0 < t < 1,
0<t<1
has a unique solution u = u( f ) satisfying
u ∈ C([0, 1]; H01(0, 1)) ∩ C 1 ([0, 1]; L 2 (0, 1)) ∩ C 2 ([0, 1]; H −1(0, 1)).
Using Duhamel’s principle as in [5] for example, it follows that the function
w := S −1 u ∈ C([0, 1]; L 2 (0, 1)) ∩ C 1 ([0, 1]; H −1(0, 1))
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V Komornik and M Yamamoto
solves the problem
wtt (x, t) = wxx (x, t) + p(x)w(x, t),
w(x, 0) = 0,
wt (x, 0) = f,
w(0, t) = w(1, t) = 0,
0 < x < 1,
0 < x < 1,
0 < t < 1,
0 < t < 1.
In particular, we have
wx (0, ·) = S −1 u x (0, ·)
and therefore, using equation (11)
c1 u x (0, ·) H wx (0, ·)V c2 u x (0, ·) H ,
where two positive constants, c1 and c2 , are independent of the particular choice of f .
Second step
We prove that the positive constants, c1 and c2 , exist such that
c1 wx (0, ·)V f H −1 (0,1) c2 wx (0, ·)V (12)
−1
for all f ∈ H (0, 1).
Firstly, functions ϕn ∈ H 2 (0, 1) ∩ H01 (0, 1) and λn > 0, n ∈ N exist such that
−ϕn (x) − p(x)ϕn (x) = λn ϕn (x),
0<x <1
and that {ϕn }n∈N is an orthonormal basis in L (0, 1). Here we note that p 0 on [0, 1]. We
set ρn = ϕn (0), n ∈ N. Then we see that ρn = 0, n ∈ N. Moreover, from chapter 1, section 2
in [1] for example, a constant c3 > 0 exists such that
(13)
λn = nπ + O n1
|ρn | = c3 n + O(1),
2
as n → ∞. Also, it is known that positive constants c4 and c5 exist, which are independent of
f and g, such that
1/2
∞
2
c4 g H01(0,1) λk βk
c5 g H01 (0,1)
(14)
if g =
∞
k=1
k=1
βk ϕk in L 2 (0, 1), and that
1/2
∞
2
αk
c5 f H −1 (0,1)
c4 f H −1 (0,1) k=1
if f =
∞ √
λk αk ϕk in H −1 (0, 1). For simplicity, we set
k=1
(Av)(x) = v (x) + p(x)v(x),
0<x <1
for v ∈ H 2 (0, 1), and we write equations (14) and (15), respectively, as
1/2
∞
2
g H01 (0,1) ∼
λk βk
k=1
and
f H −1 (0,1) ∼
∞
k=1
1/2
αk2
.
(15)
Upper and lower estimates in determining point sources in a wave equation
327
Applying the Fourier method, we obtain the expansion
∞
w(x, t) =
αk sin λk t ϕk (x)
k=1
with suitable real coefficients αk (see, for example, [2, 12]). Then
f (x) =
∞
αk λk ϕk (x).
(16)
k=1
Introducing the function
g(x) := −
∞
αk
√ ϕk (x),
λk
k=1
we have g ∈ H01 (0, 1) and
f 2H −1 (0,1) ∼ g2H 1 (0,1) ∼
0
∞
αk2
k=1
by equations (14) and (15). It remains to prove the relation
wx (0, ·)2V ∼
∞
αk2 .
(17)
k=1
Let us consider the set F of functions f of the form (16) with only finitely many nonzero
coefficients αk such that
∞
ρk
√ αk = 0.
λk
k=1
This set is dense in H −1 (0, 1). Indeed, it suffices to show that if an element
v(x) =
∞
βk ϕk (x)
k=1
of H01 (0, 1) is orthogonal to F, then v = 0. Considering the functions
fm (x) :=
λm+1
λm
ϕm+1 (x) −
ϕm (x),
ρm+1
ρm
m ∈ N,
in F, we obtain that
0 = fm , vV ,V =
λm+1
λm
βm+1 −
βm ,
ρm+1
ρm
so that
βk =
ρk λ1
β1 ,
λk ρ1
k ∈ N.
2
Since v ∈ H01 (0, 1) implies that ∞
k=1 λk |βk | converges, we conclude by equation (13)
that β1 = 0; that is, βk = 0 for k ∈ N, namely, v = 0. Thus we have proven that F is dense
in H −1(0, 1).
Consequently it suffices to prove equation (17) for f ∈ F: the general case will then
follow by continuity.
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V Komornik and M Yamamoto
Given ψ ∈ V arbitrarily, by integrating by parts, we obtain that
1
1
∞
wx (0, t)ψ(t) dt =
αk ρk sin λk tψ(t) dt
wx (0, ·), ψV ,V =
0
0
k=1
∞
1
=
0
ρk d
(1 − cos λk t)ψ(t) dt
αk √
λk dt
k=1
1
∞
1 ρk ρk
αk √ (1 − cos λk t)ψ(t) 0 −
αk √
ψ (t) dt
λ
λ
k
k 0
k=1
k=1
1
∞
ρk
+
αk √ (cos λk t)ψ (t) dt
λk
0 k=1
1
∞
ρk
αk √ (cos λk t)ψ (t) dt.
=
λk
0 k=1
∞ ρ k
Here we used ψ(1) = 0 and k=1 √λ αk = 0, and note that ∞
k=1 · · · is a finite sum by f ∈ F.
k
Applying equation (13) and the Cauchy–Schwarz inequality, we see that
1
∞
2 1 2
∞ αk ρk
wx (0, ·), ψV ,V 2 √
cos
λ
t
dt
|ψ
(t)|
dt
c
αk2 ψ2V .
k 6
λ
=
∞
0
k=1
k
0
k=1
Here c6 > 0 is independent of f and ψ. Hence
∞
αk2 .
wx (0, ·)2V c6
k=1
Choosing
ψ(t) =
∞
αk
√ sin λk t
λk
k=1
in the above computation, again in terms of equation (13) and the Cauchy–Schwarz inequality,
we can obtain the reverse inequality, so that equation (17) follows.
Remarks.
• In the case of p = 0, the proof of the inequalities
u x (0, ·) L 2 (0,T ) ∼ f (H 1 (0,1))
can be simpler. However we need the norm f H −1 (0,1) , not f (H 1 (0,1)) , for applying
theorem 5.
• Relation (12) is the H −1 version of the usual observability inequalities employing L 2
norms. See [12] and [2] for the derivation observability inequalities employing L 2
norms, and pp 73–5 in Lions [2] for a general argument for transferring L 2 -observability
inequalities to H −1-versions.
• For general initial data w(·, 0) = 0 and wt (·, 0) = 0, we should take T = 2 instead of
T = 1 in equation (12). The reason for the sufficiency to take T = 1 here is that now we
have w(·, 0) = 0. We refer to the section contrôlabilité exacte élargie in [2] for a more
general treatment of this question.
Acknowledgment
Part of this work was done during the visit of the first author to the Department of Mathematical
Sciences of the University of Tokyo in February 1999. He thanks the department for its
hospitality.
Upper and lower estimates in determining point sources in a wave equation
329
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