Exercise 2.1.2 A Nonhyperbolic Cauchy Problem for the Wave Equation)
8.5 Recovery of discontinuity of the speed of propagation
ψ=0 on the remaining part of20.
Observe that the assumptiona∈C∞() is needed only for the formula (8.4.6) (in factCN with large N is sufficient). All other parts of the proof require only a ∈C1().
By extending this method Belishev and Kurylev [Be3], [BeK], [KKL] obtained a complete solution of uniqueness problem for an anisotropic equation with com- plete spectral data or with the lateral hyperbolic Dirichlet-to-Neumann map. They proved the following result
Theorem 8.4.6. Let a be a C∞()positive symmetric matrix. Let a•be another matrix of the coefficients of the operator A which produces the same complete spec- tral data (eigenvalues and boundary values of normalized Neumann eigenfunction) or the same lateral Neumann-to-Dirichlet map as a. Then there is an isometry of the Riemannian manifold(,a)onto the Riemannian manifold(,a•)which is identical on∂
There are other of recovery of the speed of the propagation which are not using the formula (8.4.6). In particular, in [Be3] a so-called mark function (a harmonic function with singularity insideand with zero Dirichlet data on ∂is used).
Also, Gaussian beams can be incorporated in the reconstruction process [KKL].
The anisotropic inverse hyperbolic problem has been considered by Sylvester and Uhlmann [SyU4], who obtained uniqueness results forain a linearized (around the Euclidean metric) version of the inverse problem. They showed that by using the progressive wave expansion for the Cauchy problem one can find Riemannian distances between all boundary points of. Then from known results [Rom] one concludes that a conformal metric (ais scalar) is unique, and by using uniqueness in the Radon transform and a harmonic map, one can prove the linearized version in the general case.
8.5 Recovery of discontinuity of the speed of propagation
An important applied problem concerns recovery of a discontinuity surface of the speed of propagation from boundary observations.
We consider the hyperbolic equation (8.0.1), where
Au= −div((1+kχ(D×(−∞,T)))∇u), a0=1
in the half-spaceR3−×(−∞,T) with zero initial datau=0 whent<0. Given R>0, we let γ0=∂∩B(0;R), so the Neumann data g1 are supported in B(0;R)×[0,T].
With respect to domainsDj, we assume that they are subgraphs{(x,t) :x3 <
dj(x1,x2)}of Lipschitz functionsdj. We introduce the uniqueness set U= {−T/2<x3,|x|<R}.
We will make use of the special Neumann data. To describe them we consider any functionφ∈ H(2)(R) such that it is zero on (−∞,0) andφ, φare positive on the interval (0, τ) for someτ. Then the functionφ(t+x3) is a solution to the wave equation in the half-space with the Neumann datag1,φ(x,t)=φ(t).
Theorem 8.5.1(Uniqueness of Discontinuity Surface). Assume that (8.5.1) −1<kj<0, dj <0.
If solutions uj to the initial boundary value problem(8.0.1)–(8.0.3)with k= kj,D=Dj, and the Neumann data g1=g1,φ onγ0× {0, τ}for someτ satisfy the condition
(8.5.2) u1=u2onγ0×(0,T), then D1∩U=D2∩U .
PROOF. Our first claim is that
(8.5.3) u1 =u2onQT\(Q1∪Q2),
whereQTis{0<t+x3,t−x3<T,|x|<R,x3<0}. To prove this, we denote by Tr(x0,t0) the triangle{t0+x30<t+x3,t−x3<t0−x30,x3<0}, which is a translated and scaled triangle Tr from Lemma 3.4.6. Observe that (x0,t0) is a vertex of this triangle. Let (x0,t0) be a point inQT\(Q1∪Q2). Then there is positive such that the set{x:|x−x0|< ε} ×Tr(x0,t0) is contained inQT\(Q1∪Q2).
On this set the functionu=u2−u1satisfies the wave equation, and it has zero Cauchy data on the part of the boundary of this set contained in {x3=0}. By Lemma 3.4.7 we haveu=0 on this set and hence at the point (x0,t0).
Let us assume that the claim of Theorem 8.5.1 is wrong. After (possible) rela- beling we can then assume that there is a pointy(0)∈(∂D2\D1)∩U. Since∂D2 is a Lipschitz surface, it has an (exterior) normal almost everywhere, so we can in addition assume that there is an exterior normalν aty(0). Lety(δ)=y(0)+ δν.
Let us introduce the functions
v(x,t)=φ(t+x3),v∗(x,t;y, τ)= |x−y|−1φ(−A0(t−τ)− |x−y|), whereφis the function from the beginning of this section andA0=(1+k2)1/2is the speed of propagation insideD2. We have
v =0 whent+x3 <0 v∗ =0 outside conA0(y, τ).
(8.5.4)
8.5. Recovery of discontinuity of the speed of propagation 251 Moreover, one can check that
(∂t2−)v=0,
(∂t2−(1+k2))v∗ =0 whenx=y.
(8.5.5)
We have the following orthogonality relation.
Lemma 8.5.2. Under the conditions of Theorem8.5.1, we have (8.5.6)
Q2
k2∇vã ∇v∗=0
for any v∗ =v∗(;y(δ),(|y3(0)| +δ)/A0+ε)whenδ, ε(δ)are positive and small.
PROOF. Subtracting equations (8.0.1) foru2 andu1 and lettingu=u2−u1, we obtain
(∂t2u−div((1+k2χ(Q2))∇u)=div((k2χ(Q2)−k1χ(Q1))∇u1).
From (8.5.3) we have u=0 on QT\(Q1∪Q2). In addition, when δ, ε are small we have∇vã ∇v∗=0 outside a small neighborhood of the pointY(0)= (y(0),|y3(0)|) onQ2. To convince yourself of this observe that the intersection of the backward cone with the half-space{t+x3>0}will be in a small neighborhood ofY(0) for smallδ, since the surface of this cone is “steeper” than the boundary of the half-space (because the speed of propagation insideQ2corresponding to this cone is less than outside). Using the definition ofy(δ) and a continuity argument, we can find smallto guarantee that the intersection of the cone, the half-space, andQ2is simultaneously not empty and in a small neighborhood ofY(0).
According to the definition (8.0.5) of the generalized solution we have (8.5.7)
V
(∂tu∂tψ−(1+k2χ(Q2))∇uã ∇ψ)= −
V
k2∇u1ã ∇ψ for any test function ψ∈C0∞(V). Since u=0 on V\Q2, we can replace 1+ k2χ(Q2) by 1+k2and use asψonC∞(V)-function that is zero near∂V ∩Q2; in particular,ψ=v∗with smallδ, ε. Integrating by parts on the left side of (8.5.7) to apply all partial differentiations to ψ and exploiting thatv∗ solves the wave equation (8.5.5) on V ∩Q2, we conclude that the left side of (8.5.7) is zero.
Therefore, we have obtained (8.5.6) withu1instead ofv. Our choice of the lateral Neumann data guarantees thatv1has the same Neumann data asvnear× {0}. Observing that by condition (8.5.1) the speed of propagation outsideQ1is greater than that inside, we conclude thatv=u1inV.
The proof is complete.
We return to the proof of Theorem 8.5.1.
From the definition ofv,v∗and the equality
∂3|x−y(δ)| = |x−y(δ)|−1(x3−y3(δ))
we have
−∇vã ∇v∗= −φ(t+x3)∂3v∗
=φ(t+x3)(|x−y(δ)|−3φ(−A0(t−τ)− |x−y(δ)|)| + |x−y(δ)|−2φ(−A0(t−τ)− |x−y(δ)|))(x3−y3(δ)), due to the assumptionφ>0. Besides,x3−y3(δ)>0 when x∈Q2∩V, pro- vided thatδandεare small. Hence
(8.5.8) −
Q2
k2∇vã ∇v∗>0 whenδ, εare small.
For smallδandε, by Lemma 8.5.2 we have the equality (8.5.6), which contradicts the inequality (8.5.8). This contradiction shows that the assumption Q1∩U = Q2∩Uis wrong.
The proof is complete.
In geophysics one is also interested in spherical wavesφ(t+ |x−a|) instead of plane waves. A minor modification of Theorem 8.5.1 gives a sharp uniqueness result in this case. Let a functionφsatisfy the same conditions as in Theorem 8.5.1.
Thenφ(t+ |x|) is a solution to the wave equation in the half-space{x3<0}with the Neumann datahφs(x,t)=φ(t+ |x|)|x|−1x3.
Let us introduce the spherical uniqueness setUs= {|x|<T/2,|x|<R}. Theorem 8.5.3s. Let condition(8.5.1)be satisfied. If for solutions ujof the initial boundary value problem(8.0.1)–(8.0.3)with k=kj,D=Dj, and g1 =hφs on 0×(0, τ)for someτ we have the equality(8.5.2), then D1∩Us =D2∩Us. Exercise 8.5.3. Give a proof of Theorem 8.5.1s.
We expect that the proof of Theorem 8.5.1 can be repeated with the natural changes when applying sharp uniqueness of the continuation result, considering the intersection of two cones instead of the cone and the half-plane, and proving the inequality (8.5.8).
We think that the scheme of the proof of Theorem 8.5.1 can be transformed into an efficient algorithm of numerical reconstruction. To do so one considers one (unknown) equation (8.0.1),a=1+kχ(D), and makes use of solutionsv∗of the wave equation to form a functional similar to the integral (8.5.8). This functional can be found from the boundary measurements and the continuation of the wave field. For some simple geometries ofDthis continuation is not needed. Then one can changeT and calculate this functional. The first time it is positive, the wave strikes∂D. By using spherical waves from different sourcesx0, one can calculate distances fromx0to∂Dand recoverD.
In the paper [Is11] there is a similar result for the classical elasticity system.
Hansen [H] considered a linearized variant of this problem. Rakesh [Ra] made use