Exercise 2.1.2 A Nonhyperbolic Cauchy Problem for the Wave Equation)
8.3 Many measurements: use of beam solutions
We consider the operator
Au= −u+b0∂tu+bã ∇u+cu.
LetPbe a half-space inRnand0=∩P,γ0=∂∩P. We define the local lateral Neumann-to-Dirichlet mapε(b,c;T) :g1→g) onγ0×(0,T), whereu is a solution to the mixed problem (8.0.1), (8.0.2), (8.0.3) with f =0, zero initial data, and suppg1⊂0×(0, ε). In this section we assume that b0,b,cdo not depend ont. Byj,εwe denote(bj,cj;T).
8.3. Many measurements: use of beam solutions. 237 Theorem 8.3.1. Suppose that0is simply connected, b0,b∈C2(), c∈L∞(),
(8.3.1) diam0<T,
and b is given onγ0.
Then the local Neumann-to-Dirichlet map ε(b,c;T) uniquely determines b0,curl b,4c+bãb−2 divb in0.
The proof is based on the following results.
Lemma 8.3.2. εuniquely determines the integral (8.3.2)
Q
(−u(b0+b0∗)∂tv−u(b+b∗)ã ∇v+(c−divb)uv)
for any solution u to the problem(8.0.1)–(8.0.3)with given∂νu on∂×(0,T), (8.3.3) ∂νu =0on(∂\γ0)×(0,T)and in C02(∂×(0,T)),
and any (given) solution v ∈H(2)(Q)to the backward initial value problem for the hyperbolic equation
(+b0∗∂t+b∗ã ∇)v=0in Q, (8.3.4)
v=∂tv=0on× {T}, (8.3.5)
∂νv=0on(∂\γ0)×(0,T), (8.3.6)
where=∂t2−and b∗0,b∗ are some given functions with the same properties as b0,b.
PROOF. First observe that ε uniquely determines T. Indeed, let us pick any g1 supported inγ0ì(0,T). Let represent it as the sum g1,1+ ã ã ã +g1,m with suppg1,k inγ0×[tk,tk+ε]. Denote by uk a solution to the mixed hyperbolic problem (8.0.1)–(8.0.3) with largeT and with Neumann datag1,k. Since the coef- ficients of the differential equation do not depend ont, the functionuk(x,t−tk) solves the mixed hyperbolic problem with the translated datag1,k(x,t−tk) that are supported in [0, ε]. Thenεuniquely determinesuk(x,t−tk) whenx∈γ0, 0<t<T, and thereforeukis uniquely determined onγ0×(0,T). Thus we obtain uonγ0×(0,T). So we are givenT.
Multiplying the equation forubyvand integrating by parts gives 0=
Q
((+b0∂t+bã ∇ +c)u)v
=
∂×(0,T)
(∂νvu−∂νuv+bãνuv) +
Q
((−b0∂t−bã ∇ −divb+c)v)u,
where we used thatb0does not depend ontand the Cauchy data foruare zero at t =0 and forvatt =T, so the integrals over× {0}and over× {T}are zero.
From conditions (8.3.3), (8.3.6), the integral over∂×(0,T) is reduced to the integral overγ0×(0,T), where∂νu,v, ∂νvare given. The mapT then uniquely determinesuon this over part of the lateral boundary, and hence we are given the boundary integral. Using the equation forv, we complete the proof.
Lemma 8.3.3. For any x∈Rn, any directionσ, and any functionφ∈C0∞(Rn) there is a solution u to equation(8.0.1)of the form
u(x,t)=φ(x+tσ)B(x,t) exp(iτ(xãσ+t))+r(x,t), B(x,t)=exp(−1/2
t
0
(b0+bãσ)(x+sω)ds) (8.3.7)
with
(8.3.8) r =∂tr=0on× {0}, ∂νr =0on∂×(0,T), and
(8.3.9) τr2(Q)+ r(1)(Q)≤C(φ). PROOF. Since
(vw)=vw+2∂tv∂tw−2∇vã ∇w+wv
and (exp(iτ(xãω+t))=0, equation (8.0.1) foruis equivalent to the following equation forr:
(8.3.10) (+b0∂t+bã ∇ +c)r=F, where
F =(−−b0∂t−bã ∇ −c)(exp(iτ(xãσ+t))
= −((iτ(2∂t−2∇ãσ+(b0+bãσ))
+(+b0∂t+bã ∇ +c)) exp(iτ(xãσ+t)), =φ(x+tσ)B.
IfBis given by formula (8.3.7), then the factor ofiτin the above formula is zero.
This is easy to derive by using the substitutions=θ+tin the integral defining B. So
(8.3.11) F=F1exp(iτ(xãσ +t)), F12(Q)+ ∂tF12(Q)≤C.
Since the coefficientsb0,b,cdo not depend ont, the function R(x,t)=
t
0
r(x,s)ds
solves the mixed hyperbolic problem (8.3.10), (8.3.8) with the right side F2=
t
0
F= t
0
F1(,s)(iτ)−1∂s(exp(iτ(xãσ+s))ds.
8.3. Many measurements: use of beam solutions. 239 Integrating by parts with respect tosand using the bounds (8.3.11), we conclude thatF22(Q)≤C/τ. The standard energy estimates for mixed hyperbolic prob- lems given by Theorem 8.1 imply then that∂tR2(Q)≤C/τ, so we have the bound (8.3.9) forr=∂tR.
SinceF2(Q)≤C, applying again the energy estimates we obtain the bound
(8.3.9) forr(1)and thereby complete the proof.
Lemma 8.3.4. Assume that b0∗,b∗do not depend on t and have the same regularity properties as b0,b. For x, σ, φin Lemma8.3.3there is a solution v to the hyperbolic equation(+b0∗∂t+b∗ã ∇)v=0in Q of the form
v(x,t)=φ(x+tσ)B∗(x,t) exp(−iτ(xãσ+t))+r∗(x,t), B∗(x,t)=exp(−1/2
t
0
(b∗0+b∗ãσ)(x+sσ)ds) (8.3.7∗)
with
(8.3.8∗) r∗ =∂tr∗=0on× {T}, ∂νr∗ =0on∂×(0,T), and
(8.3.9∗) τr∗|2(Q)+ r∗(1)(Q)≤C(φ).
The proof is similar to that of Lemma 8.3.3.
PROOF OFTHEOREM8.3.1.
LetL be any straight line such that it intersection L0with0 is contained in the half-spaceP. Letσ be a direction of this line. We will show that our data determine the integral ofb0+bãσ overL.
From our assumptions it is possible to find an interval [y,z] inRn containing L0such that|y−z|<T, and both yandz do not belong to0. Let us choose δ >0 so small that theδ-neighborhoods ofyandzdo not intersect0as well, and theδ-neighborhood of [y,z] is contained inP. Letφ∈C0∞in theδ-neighborhood of yand zero elsewhere. We choose the directionσ = −|z−y|−1(z−y). Then the functionφ(x+tσ) is zero in (Rn\P×(0,T) near×[0,T]. The boundary conditions (8.3.8) forrguarantee thatuconstructed in Lemma 8.3.3 satisfies all the conditions of Lemma 8.3.2. So does the functionv, where we takeb∗0 =0,b∗=0.
According to Lemma 8.3.2, we are given the integrals (8.3.2). Using the formulae (8.3.7), (8.3.7∗) and the bounds (8.3.9), (8.3.9∗), we conclude that we are given
(8.3.12) iτ
Q
(b0+bãσ)(x)φ2(x+tσ)B(x,t)d xdt+ ã ã ã, where. . .denotes the terms bounded with respect toτ.
Dividing byτand lettingτ → +∞yields the integral (8.3.12). Extendingb0,b as zero ontoRn\0and substitutingX =x+tσ,θ=tin the integrals, we obtain
the integral
φ2(X) T
0
(b0+bãσ)(X−θσ)B(X−θσ, θ)dθ
d X.
Sinceφis an arbitrary smooth function supported neary, we are given the interior integral whenX =y, i.e.,
(8.3.13)
T
0
(b0+bãσ)(y−θσ)B(y−θσ, θ)dθ.
Using the substitutions=θ+s1in the integral (8.3.7) definingBand differenti- ating with respect toθ, we obtain
d/dθB(y−θσ, θ)=d/dθexp
−1 2
0
−θ(b0+bãσ)(y+s1ω)ds1
= −1
2(b0+bãω)(y−θσ)B(y−θσ, θ).
Therefore, the integral (8.3.13) is the difference of the value of the function
−2B(y−θσ, θ) at the points θ=T andθ =0. The value at 0 is −2, so we are given the value atθ=T. Taking the logarithm, we obtain
L
(b0+bãσ).
Since the direction ofLis−σas well, we are given the integral ofb0−bãσ, and therefore the integrals ofb0and ofbãσ over all suchL.
The next step is to show that these integrals determineb0and curlbin0. First, we reduce then-dimensional case to the two-dimensional one by intersecting with two-dimensional planes and considering only lines L in these planes. By Corollary 7.1.2 (whereK is the closure of the convex hull of\P) the integrals ofb0overLnot crossing\Puniquely determineb0in0.
The recovery of curlbis based on the equality
p(σ)curlb=
L
bãσ,
where p(σ) is a half-plane with exterior normal orthogonal toσ that does not intersect\P. To prove this equality, introduce orthonormal coordinatesx1,x2
such that thex1-direction coincides with the exterior unit normal top(σ). Then the x2-direction is parallel toσ. We have curlb=∂1b2−∂2b1. Integrating by parts, we obtain
p(σ)
curlb=
L
b2+
p(σ)∩∂(ν1b2−ν2b1)=
L
bãσ+I.
Since the direction (−ν2, ν1) is tangential to ∂, the integral I is given by the conditions of Theorem 8.3.2. Given the integrals of curlbover all such p(ω), we can find the integrals of curlboverLby considering parallel translationsp(σ) by
8.3. Many measurements: use of beam solutions. 241 θand then differentiating the integrals over the translated p(σ) with respect toθ.
As above, the integrals overLuniquely determine curlbin0.
As in the proof of Theorem 5.4.1, curlbuniquely identifiesb•=b− ∇φ, where φis aC1(0)-function. We can assume that this function is zero at a point of0. Since bis given there by the conditions of Theorem 8.3.1, bothφ and∂νφare given there as well. As in the proof of Theorem 5.5.1, the substitutionu=eφ/2v transforms the equation foruinto the equation
v+b0∂tv+b•ã ∇v+
c+1
2divb•−1
4|b•|2−1
2divb+1 4|b|2
v=0.
Sinceφand∂νφare given onγ0, the Neumann-to-Dirichlet map for the original equation uniquely determines the map for the new equation. To complete the proof, it suffices to show thatc−12divb+14|b|2is uniquely determined whenb0,b•are given. We will again make use of integrals (8.3.2), where nowb0,b=b•are given.
We will use in the integrals (8.3.2) new functionsu• given in Lemma 8.3.3 for the•-equation and solutionsvto the hyperbolic equation (8.3.4) withb∗0 = −b0, b∗= −b. Multiplying in the integrand, we will have
φ2(x+tω)(c(x)+divb•−1
4|b•|2−1
2divb+1
4|b|2)+O(τ−1).
Therefore, we are given integrals of the first term overQ. Repeating the argument from the beginning of the proof of Theorem 8.3.1, we conclude that we know (nonweighted) integrals ofcover all straight lines L joining boundary points in γ0. So, as above, uniqueness in the Radon transform guarantees thatc−12divb+
1
4|b|2is unique in0.
The proof is complete.
Observe that using L2(Q)-bounds of solutions of hyperbolic initial boundary value problems with source terms in negative Sobolev spaces H(−1)(Q) (see the book of Lions and Magenes [LiM], Vol. 1, Theorem 9.3, p. 288) one can obtain uniqueness ofb0,b∈C1(0). One has only to make minor changes in the proofs of Lemmas 8.3.3, 8.3.4. Using the structure of fundamental solutions of hyperbolic equations and integral geometry, Romanov [Rom] obtained uniqueness of curlb andc−12divb+14|b|2in 1974. Then Rakesh and Symes showed uniqueness of c(b=0,b0=0) in the paper [RS], where they first used beam solutions in inverse hyperbolic problems. In these papers0 =; i.e., they consider the data on the whole lateral boundary. Beam solutions seem to be a very useful tool in inverse problems that has not yet been completely utilized. For review of the subject we can refer to the books of Arnaud [Ar], Katchalov, Kurylev, and Lassas [KKL]
and to the paper of Ralston [Ra]. The idea to use rapidly oscillating solutions in the inverse hyperbolic problems with given results of all boundary measurements originated from the paper of Sylvester and Uhlmann [SyU2] on elliptic equations, which has been discussed in Sections 5.2, 5.3.
To formulate a stability estimate, we impose the following constraints onb0j,cj :
|b0j|2()+ |cj|1()<M,j =1,2.
We viewj,εas an operator fromL2(0×(0, ε)) intoH(1/6)(0×(0,T)) and denote byδthe operator norm of2,ε−1,ε.
Theorem 8.3.5. There are C, λdepending only on P, , ε,M such that (a) When n=3, we have|b02−b01|0(0)+ |c2−c1|0(0)≤Cδλ. (b) When n=2and b01=b02, we have|c2−c1|0(0)≤C(−lnδ)−λ. This result is proven in the paper of Isakov and Sun [IsS1] using the scheme of the proof of Theorem 8.3.1. Observe that the (H¨older) stability estimates in the three-dimensional case are much better than in the two-dimensional one and in the inverse conductivity problem (Theorem 5.2.3). We think that by modifying the method of the proof one can obtain H¨older-type stability in the two-dimensional case as well.
When one is given the complete Dirichlet-to-Neumann map for any initial data and the lateral Neumann data, and one measures the Dirichlet data and the Cauchy data at the final moment of time, then there are results about unique- ness of the coefficient c(x,t)∈ L∞(Q) (see, e.g., [Is7]) that are similar to Lemma 9.6.3.
The beam solutions can be constructed for more general hyperbolic equations. In principle, this construction shows that for scalarathe lateral Neumann-to-Dirichlet map (whenTis sufficiently large) uniquely determines geodesic distances between point ofγ0, which is not enough to obtain uniqueness ofain0without additional assumptions.
Whenγ0=∂, there is another way to show thatl(whenT is large) uniquely determines geodesic distances between points of∂. A simple argument based on the structure of the fundamental solution of the Cauchy problem for second-order hyperbolic equations (progressive wave expansion) is given by Uhlmann [U1]. We will give a minor part of it as the following exercise.
Exercise 8.3.6. Consider the hyperbolic equation (8.0.1) with the given lateral Neumann-to-Dirichlet map. Show that a solution of the Cauchy problem for this equation in Rn×(0,T) with the (given) datau0,u1, f supported in Rn\is uniquely determined in (Rn\)×(0,T).
Given all geodesic distances on∂, a (conformal, or scalar) Riemannian metric a(x)|d x|2is uniquely determined, provided that its geodesics are regular (see, e.g., [Rom, p. 93]). So if0=∂,T is large, andagenerates a regular geodesic, the lateral Neumann-to-Dirichlet map uniquely determinesa. Recently, Eskin [Es2]
showed that one also can recover unknown inclusions and established a connection with Aharonov-Bohm effect from theoretical physics.
For a discussion of the uniqueness problem for a matrixawe refer to the paper of Sylvester and Uhlmann [SyU4].