Exercise 2.1.2 A Nonhyperbolic Cauchy Problem for the Wave Equation)
9.5 Discontinuous principal coefficient and recovery
Corollary 9.4.5. Letγ be any nonempty open part of∂and let the parabolic equation (9.0.1) have coefficients a0 =(deta∗)1/2,a =a0a∗, where a∗is a strictly positive symmetric C∞()-matrix function, b=0,c=0,∂∈C∞.
Then the local parabolic Dirichlet-to Neumann map uniquely determines a∗ modulo an isometry of the Riemannian manifold(,a).
In [KKL] there are also some partial results about recovery of more general hyperbolic equations modulo gauge transforms which similarly imply parallel results for parabolic inverse problems.
9.5 Discontinuous principal coefficient and recovery of a domain
The results and technique described above are applicable only to continuous coef- ficients of the principal part of an equation in divergent form, while in applications the discontinuous coefficient is quite important. As a typical example we will consider
(9.5.1) a =a0+kχ(Q•),
wherea0,kareC2-smooth functions andχ(Q•) is the characteristic function of the unknown domainQ•. In this section we describe uniqueness results for time- independent and time-dependent Q• with given one or many lateral boundary measurements. To prove the first result we will make use of the transform (9.2.1) and of the Laplace transform, which enable one to apply uniqueness theorems for hyperbolic and elliptic equations. To prove the second result we utilize the method of products of singular solutions developed for elliptic equations in Section 5.7.
In the next theoremγ is any open subset of the hyperplanexn =0 andis the half-space{xn <0}inRn.
Theorem 9.5.1. Let a0=1, and Q•=D×(0,T), where D is a bounded Lip- schitz xn-convex domain in the half-space{xn<0},2≤n, intersecting the strip γ×R−. Assume that k is known constant and
(9.5.2) −1<k<0.
Let the Dirichlet lateral data g0(x,t)be g0(x)ψ(t), whereψcorresponds via (9.2.1)to a functionψ∗(τ)that is positive as well as its first-order derivative on some interval(0, ε),|ψ0(τ)| ≤Cexp(Cτ)and g0=1on,g0∈C02(∂). Let u be a solution to the parabolic problem(9.0.1)–(9.0.3)with a given by(9.5.1)and with a0=1,b=0,c=0,u0=0.
Then∂νu onγ×(0,T)uniquely determines D.
PROOF. We consider the hyperbolic equation
∂τ2u∗−div((1+kχ(Q•))∇u∗)=0 on×R+
with zero initial data and with the lateral datau∗(x, τ)=g0(x)ψ∗(τ) on∂×R+. T∗ is to be chosen later. Then u(x,t) obtained from u∗ solves on ×R+ the parabolic problem (9.0.1)–(9.0.3) witha0=1,b=0,c=0, anda given by (9.5.1). As in Section 9.2, the Dirichlet lateral data for the parabolic problem are analytic with respect tot, due to our growth assumptions onψ∗. Therefore, the solution of the parabolic problem is analytic with respect totas well, and the data
∂νu(x,t) of the inverse problem given onγ ×(0,T) are uniquely determined on γ×R+. Let us choose anyx0∈γ such that the straight line through this point parallel to thexn-axis intersects∂D. Letx0 =(x0,xn0) be the intersection point with largestxn. Since∂Dis Lipschitz, there is a point of∂Dnearx0where there is a tangent (hyper-)plane to∂Dwhose normal is not perpendicular toγ. For brevity we can assume that this point isx0. By the uniqueness Theorem 8.5.1 for inverse hy- perbolic problems,∂D∩V is uniquely identified for some neighborhoodV ofx0.
The Laplace transform
U(x,s)= ∞
0
u(x,t)e−stdt
of the solution to the parabolic problem solves the elliptic problem with the pa- rameters,
−div((1+kχ(D))∇U)−sU =0 in,s>0.
The Dirichlet data of U(,s) on ∂ are g0(x)(s) and not zero at some s1. Moreover,∂νU(,s) on∂\Dis uniquely determined because the solution of the parabolic problem is uniquely determined. We have proved already that two possi- ble solutionsD1,D2of the inverse problem have a common piece of their bound- aries. From our assumptions aboutxn-convexity it follows that they arei-contact.
Since both domains produce the same Cauchy data on, they have to coincide by Theorem 4.3.2.
The proof is complete.
To formulate the result for domains changing in time, we define the lateral boundary∂xQ•of an open subset of the layerRn×(0,T) as the closure of∂Q•∩ {0<t <T}. We will say thatQ•isx-Lipschitz if∂xQ•is Lipschitz.
Theorem 9.5.2. Let n≥2. Suppose that Q• is an open x-Lipschitz subset of Q with∂xQ•∩∂xQ= ∅satisfying the conditions
(9.5.3) the sets(Q\Q•)∩ {t=τ}are connected when0< τ <T and that scalar matrix a0, a symmetric matrix k, and b,c satisfy the conditions
a0,k∈C2(),a=a0+χ(Q•)k,b=0,c=0,0<k, or −a0<k<0, on (9.5.4)
9.5. Discontinuous principal coefficient and recovery of a domain 281 Then the lateral boundary data
g1=∂νu onγ×(0,T)
of a solution u to(9.0.1)–(9.0.3)(with∂xQ•,u)=0) given for any g0∈C02(∂× (0,T))that are zero outsideγ×(0,T)uniquely determine Q•. If k is a scalar function, then it is uniquely determined on Q•.
Detailed proofs for scalarkare given in the paper of Elayyan and Isakov [ElI1].
They utilize the idea of using singular solutions of differential equations in in- verse problems introduced and used by Isakov [Is3]. Since these proofs are quite technical, we will outline only the basic steps.
Let us assume that there are two domainsQ•1,Q•2with twok1,k2producing the same boundary data. Let us denote byQ3θthe connected component of the open set
\(Q•1θ∪Q•2θ) whose boundary containsγ. HereQ•jθis defined asQ•j∩ {t =τ}.
LetQ3be the union ofQ3θover 0< θ <T and letQ4=Q\Q3.
Lemma 9.5.3. Under the conditions of Theorem9.5.2we have the following or- thogonality relations:
(9.5.5)
Q•1
k1∇v1ã ∇u∗2=
Q•2
k2∇v1ã ∇u∗2 for all solutions v1to the equation
∂tv1−div((a0+k2χ(Q1))∇v1)=0near Q4,v1=0when t <0, and all solutions u∗2to the adjoint equation
−∂tu∗2−div((a0+k2χ(Q2))∇u∗2)=0near Q4,u∗2=0when T <t.
PROOF. As we did several times before, we subtract two equations (9.0.1) with the coefficientsa2anda1and obtain for the differenceuof their solutionsu2andu1
the differential equation
(9.5.6) ∂tu−div((a0+k2χ(Q•2))∇u)=div((k2χ(Q•2)−k1χ(Q•1))∇u1) inQ.
Sinceu2,u1have the same lateral Cauchy data onγ×(0,T) by known uniqueness results (Theorem 3.3.10) they coincide onQ4, so their differenceuis zero there.
Writing the definition (9.0.4) of the generalized solution to equation (9.5.6) with the test functionv∗2, we get
Q
(∂tuv2∗+(1+k2χ(Q•2))∇uã ∇v2∗)= −
Q
(k2χ(Q•2)−k2χ(Q•1))∇u1ã ∇v∗2. Applying again the definition (9.0.4) of the generalized solution to the equation forv2∗with the test functionu, we complete the proof of the orthogonality relation (9.5.5) with u1 instead of v1. The final relation can be obtained by using the approximation ofv1byu1 exactly as was done for elliptic equations in Lemma
5.7.2.
Exercise 9.5.4. Prove that the relation (9.5.5) withv1replaced byu1implies the general (9.5.5).
{Hint: make use of uniqueness of the continuation and of parabolic potentials to prove the Runge property of solutions of parabolic equations.}
Returning to the proof of Theorem 9.5.2, we assume that
(9.5.7) Q•1=Q•2.
Then we may assume that there is a point (x0,t0)∈∂xQ•1\Q•2 that is contained as well in ∂Q3. By considering g0=0 when t <t0 and using the translation t →t−t0,x→x−x0, we can also assume thatt0=0,x0=0. Let us choose a ballB⊂Rnand a cylinderZ =B×(−τ, τ) such thatB⊂,Zis disjoint from Q2, and (∂xQ1)∩Z is a Lipschitz surface. By the Whitney extension theorem there is aC2(Q1∪Z)-functiona3coinciding witha0+k1onQ1. We will extend a3ontoQ\(Q1∪Z) asa0.
To complete the proof we need the following modification of the orthogonality relations (9.5.5):
(9.5.8)
Q•1
k1∇u3ã ∇u∗2=
Q•2
k2∇u3ã ∇u∗2 for any solutionu3to the parabolic equation
∂tu3−div(a3∇u3)=0 nearQ4, u3=0 whent<0
and for any solutionu∗2to the parabolic equation from Lemma 9.5.3. The derivation of (9.5.8) from (9.5.5) is quite similar to the derivation of the relation (5.7.4) from the relation (5.7.2) in the elliptic case. It suffices to approximateu3by solutions u3 of parabolic equations witha3replaced bya3,mand with the same initial and lateral boundary Dirichlet data as u3. The sequence of the coefficientsa3,m is chosen so that they are uniformly bounded from zero and from infinity, equal to a3in a neighborhood ofQ4depending onm, and pointwise convergent toa3. The orthogonality relation (9.5.5) is valid withv1=u3,m, and the relation (9.5.8) is obtained by passing to the limit whenm→ ∞.
To obtain a contradiction with the initial assumption, we will use asu3andu∗2the fundamental solutionsK+andK−of the forward and backward Cauchy problems for the parabolic equations with coefficientsa3 anda2 with poles at the points (y,0),(y, τ), whereyis outsideQ4and is close to the origin. To obtain bounds of integrals it is convenient to use new variables. We can assume that the direction enof thexn-axis coincides with the interior unit normal to∂xQ•1∩ {t =0}. This surface near the origin is the graph of a Lipschitz functionxn =q1(x1, . . . ,xn−1,t).
The substitution
xm=xm0,m=1, . . . ,n−1,xn=xn0+q1(0,t),t =t0
transforms the equations into similar ones (with additional first-order differential operators with respect to x0n). The domains Q•j are transformed into similar do- mains, with the additional property that the points (0,t),0<t <T, are in∂xQ•1.
9.5. Discontinuous principal coefficient and recovery of a domain 283 Sinceq1 is a Lipschiz function we may assume that (0, . . . ,0,1) is the interior normal to its graph at the origin. Further on, we will drop the index 0.
The singular (fundamental) solutions to be used have the following structure:
(9.5.9) K+=K1++K0+,K− =K1−+K0−, where the first terms are the so-called parametrices
K1+(x,t;y, τ)=C(deta−31(y)(t−τ))−n/2exp(−a−1(y)(x−y)
ã(x−y)(/(4(t−τ))),
K1−(x,t;y, τ)=C(a0(y))1/2((τ−t))−n/2exp(−|x−y|2/(4a0(y)(τ −t))), andK0+,K0−are remainders with weaker singularities satisfying the bounds
|∇xjK0+(x,y;t, τ)| ≤C|t−τ|(λ−n−j)/2ex p(−C|x−y|2/(t−τ)),
|∇xjK0−(x,y;t, τ)| ≤C|t−τ|(λ−n−j)/2ex p(−C|x−y|2/(τ −t)),j =0,1.
Letting in (9.5.8)u3=K+( ;y,0),u∗2=K−( ;y, τ), splittingK+,K−according to (9.5.9), and breaking the integration domainQ•1intoQ•1∩Zand its complement, we obtain
(9.5.10) I1=I2+I3,
where
I1=
Q•1∩Z
k1∇xK1+ã ∇xK1−, I2= −
Q•1\Z
k1∇xK+ã ∇xK−+
Q•2
k2∇xK+ã ∇xK−, and
I3=
Q•1∩Z
k1(∇xK1+ã ∇xK0−+ ∇xK0+ã ∇xK1−+ ∇xK0+ã ∇K0−).
To obtain the relation (9.5.10) we first takey=(0, . . . ,0,−δ) and then letδ →0.
The passage to the limit can be justified (whenτ >0 is fixed) by using the Lebesgue dominated convergence theorem and bounds onx-derivatives of the fundamental solutions, which follows from the given formulae for parametrices and from known results (see the book by Friedman [Fr]). After the passage to the limit, the integrals in (9.5.10) became functions ofτ only.
Lemma 9.5.5. We have
τ−n/2/C ≤ |I1|,
|I2| ≤C,
|I3| ≤Cτ−n/2+λ/2, where constant C does not depend onτ.
PROOF. From (9.5.9) we have
∇xK1+(x,t; 0,0)= −K1+(x,t; 0,0)/(2t)a1−1(0)x,
∇xK1−(x,t; 0,0)= −K1−(x,t; 0,0)/(2ta0(0))x.
Since a1=a0+k1 on Q1 anda0 is scalar, the matrices k1(0),a1(0),a−1/21 (0) commute, using in addition their symmetry we will have
|k1(x)(a1−1(0)x))ã(a−01(0)x)|
= |k1(0)a−1/21 (0)a−1/20 (0))xã(a−1/21 (0)a0−1/2(0)x) +(k1(x)−k1(0))(a1−1(0)xã(a0−1(0)x)|
≥ |x|2/C−C|x|3≥ |x|2/C, x∈ B
where we used the trangle inequality and condition (9.5.4) onk=k1and choose Bto have small radius. By our regularity assumptions there is an open coneCin Rnwith vertex at the origin and axis (0, . . . ,0,s),0<s< ε0that is contained in Q•1. Hence,
|I1| ≥1/C
C×(0,τ)
|x|2(t(τ −t))−n/2−1ex p(−|x|2/(mt)− |x|2/(m(t−τ)))dtd x
≥1/C
C
τ
0
|x|2(t(t−τ))−n/2−1ex p(−|x|2τ/(mt(τ −t)))dtd x wheremis the smallest of the eigenvalues ofa1,a0. Using the inequality
1/(tτ)≤1/(t(τ −t))≤2/(tτ) when 0<t < τ/2 and the previous bound we obtain
|I1| ≥1/C
C
τ/2
0
|x|2(tτ)−n/2−1ex p(−2|x|2/(mt))dtd x
=1/Cτ−n/2−1 ε
0
ρ2( τ/2
0
t−n/2−1ex p(−2ρ2/(mt))dt)ρn−1dρ where we have used the polar coordinatesρ= |x|inRn. Using the substitutiton w =2ρ2/(mt) in the last integral we yield
|I1| ≥1/Cτ−n/2−1 ε
0
ρ(
4ρ2/(mτ)wn/2−1e−w)dρ
≥1/Cτ−n/2−1
4ε2/(mτ)
0
(
√mτw/2
0
ρdρ)wn/2−1e−wdw
where we replaced the integration domain by the smaller one {0< ρ <
ε,4ρ2/(mτ)<w<4ε2/(mτ)}and interchanged order of integration. Calculat- ing the last inner integral we get
|I1| ≥1/Cτ−n/2
4ε2/(mτ)
0
wn/2e−wdw.
9.5. Discontinuous principal coefficient and recovery of a domain 285 The proof of the first bound of Lemma 9.5.5 is complete.
The second bound is obvious, since the fundamental solution is bounded away from singularities.
The third bound can be obtained similarly to the first one by using the bounds (9.5.9) onK0+,K0−. Of course, now we obtaining bounds from above. We will not give a complete argument, only some outlines. One of the integrals to be bounded looks like that one forI1, it is less than the integral of
|x|(τ −t)−n/2−1tλ/2−n/2−1/2e−|x|2/(mt)e−|x|2/(m(τ−t)
≤C|x|2+λ(t(τ −t))−n/2−1e−|x|2/(m1t)e−|x|2/(m1(τ−t)
when 0<m<m1. Here we used that |x|1+λt−(1+λ)/2ex p(−(m1− m)|x|2/(mm1t))<C. After these remarks upper bounding I3 repeats lower
bounding|I1|.
To complete the proof of Theorem 9.5.2 we first show that domains coincide.
From (9.5.10) and Lemma 9.5.5 we have
τ−n/2≤C(1+τ−n/2+λ/2)
and lettingτ →0 we will have a contradiction. The contradiction shows that the initial assumption is wrong, soQ•1=Q•2.
When the domains coincide, one can prove equality of scalarkjonQ•1by using the methods of sections 5.1,9.4.
The first claim is that
(9.5.11) k1 =k2, ∇k1= ∇k2on∂xQ•.
As in the proof for domains we assume the opposite. Then we can assume that the origin 0∈∂xQ•1andε <k2−k1on some ballBcentered at the origin. As above there is functiona4∈C2(Rn) conciding witha2onQ•1. By repeating the proof of (9.5.8) we conclude that
Q•1
(k2−k1)∇u3ã ∇u∗4=0
for all solutionsu3to the equation∂tu3−div(a3∇u3)=0 nearQ•1which are zero whent <0 and for all solutionsu∗4 to the equation∂tu∗4+div(a4∇u∗4)=0 near Q•1which are zero whenT <t. Repeating the proof of uniqueness for domains we will obtain a contradiction. Hence we have the first equality (9.5.11). Repeating this argument with use of normal derivatives of fundamental solutions we prove the second equality (9.5.11).
To conclude it suffices to show that k1=k2 on the intersection 0 of all Q•1∪ {t =θ}over allθ∈(0,T). LetQ0 =0×(0,T). By using local represen- tations of∂xQ•1and using thatmi nof a family ofLi pschi t zfunctions is Lipschitz one can show that (at least for small T)0 is Lipschitz. Due to (9.5.11) and to time independence,k1=k2 onQ\0×(0,T). We can assume thatT is small, since increasing T shrinks Q0. Letting a5=a0+χ(0)k1,a6=a0+χ(0)k2
and adjusting the derivation of (5.7.3) from (5.7.2) to the parabolic case we obtain from Lemma 9.7.4 that
(9.5.12)
Q0
(k2−k1)∇xu5ã ∇xu∗6=0,
for all solutionsu5to the equation∂tu5−div(a5∇xu5)=0 nearQ0which are zero whent <0 and for all solutionsu∗6to the equation∂tu∗6+div(a6∇xu∗6)=0 near Q0which are zero whenT <t.
We claim that (9.5.12) implies equality of the lateral Dirichlet-to Neumann maps for parabolic equations with coefficientsa5,a6.
Indeed, letu5andu6be solutions of these equations inQwith zero initial data and with the same (smooth) lateral Dirichlet datag0supported inγ×(0,T). By subtracting these equations and lettingu=u6−u5we yield
∂tu−div(a6∇xu)=div((a5−a6)∇xu5) on Q.
From the definition of a generalized solution ( before Theorem 9.1) we have
−
∂×(0,T)
a0∂νuv+
Q
(a6∇xuã ∇xv−u∂tv)=
Q
(a6−a5)∇xu5ã ∇xv (9.5.13)
for any functionv∈ L2(0,T;H(1)()) with∂tv∈L2(Q),v=0 on× {T}. Since coefficientsa5,a6it follows ( by forming finite differences in time) from Theorem 9.1 that ∂tu5, ∂tu6∈ L2(Q), and ∂tu∗6∈ L2(Q) for any solution to the adjoint equation∂tu∗6+div(a6∇xu∗)=0 inQwith zero data att =T and smooth lateral Dirichlet data on∂xQ. Using again the definition of a(weak) solution to the adjoint equation with the test functionuwe obtain
Q
(a6∇u∗6ãu−u∗6∂tu)=0,
because u =0 on∂xQ. Letting v=u∗6 in (9.5.13) and using the last integral equality as well as (9.5.12) we yield
∂×(0,T)
a0∂νuu∗6 =0.
Sinceu∗6on∂×(0,T) is an arbitrary smooth function,∂νu=0 on∂×(0,T).
Soa5,a6generate the same lateral Dirichlet-to Neumann map. Now by using sta- bilization of solutions of parabolic problems for largetas in the proof of Theorem 9.4.1 we conclude that the elliptic equations
−div(aj∇vj)+svj=0, in, j=5,6,
have the same Dirichlet-to Neumann maps. By Theorem 5.7.1,a5=a6andk1 = k2.
Finally we prove similar results for impenetrable domains for single boundary measurements which in particular guarantee that under natural assumptions voids are uniquely identified by their exterior thermal field.
9.5. Discontinuous principal coefficient and recovery of a domain 287 Theorem 9.5.6. Let u solve the parabolic equation (9.0.1) with f =0in Q\Q• where Q• is a subdomain of Q satisfying the conditions of Theorem 9.5.2. Let u have the initial condition u0=0 on× {0}\Q• and the Dirichlet condition (9.0.3) with nonnegative g0 ∈C2(∂×[0,T])which is not identically zero on
∂×(0, θ)for anyθ >0. Let (9.5.14)
u =0on∂xQ•(D) or∂ν(a)u =0on∂xQ• and Q•=•×(0,T) (N).
Then the additional Neumann data (9.0.6) for any open subsetγof∂uniquely determine Q•.
We will prove this result by the Schiffer’s method which is used in section 6.3 to show uniqueness of a soft obstacle.
PROOF. Let us assume that there are two different domains Q•1,Q•2 producing the same lateral boundary data. We will introduce the domains Q3,Q4as in the proof of Theorem 9.5.2. We may assume that Q•2 is not contained in Q•1, hence as in the proof of Theorem 6.3.1 there is a connected componentQ0of Q4\Q•1. The lateral boundary∂xQ0=1∪2where1⊂∂xQ•1and2⊂∂xQ•1∩∂xQ3. Sinceu1,u2satisfy the same parabolic equation inQ3by uniqueness in the lateral Cauchy problem (Theorem 3.3.10)u1=u2onQ3. Henceu1=u2 =0 on2in case (D) and∂ν(a)u1=0 on2in case (N).
Replacingu1byeτtv1we will have forv1the parabolic equation (9.0.1) withc replaced byc+τa0. To show thatv1=0 onQ0we will make use of the definition of weak solution inQ0:
Q0
(a0∂tv1φ+(a∇v1)ã ∇φ+(bã ∇v1+(c0+τa0)v1)φ)
−
1
(a∇v1)ãνφ−
2
(a∇v1)ãνφ=0
for any test function φ∈ L2(Q\Q•1) with∇φ∈ L2(Q\Q•1),φ∈ H(1/2)(Q\Q•1).
Lettingϕ=v1, observing that (∂tv1)v1=1/2∂t(v12), and integrating by parts in the first integral with use of zero boundary conditions on1, 2and att =0 we obtain
∂Q0∩{t=0}1/2a0v12+
Q0
(a∇v1ã ∇v1+bã ∇v1v1+(a0τ +c−∂ta0/2)v21)=0.
Using the elementary inequality−ε|∇v1|2−C(ε)v12≤bã ∇v1v1and positivity of the matrixawe yield
Q0
(ε0|∇v1|2−ε|∇v1|2+(a0τ+c−∂ta0/2−C(ε))v21≤0.
Choosingε < ε0and thenτto be large, we obtain thatv1=0 inQ0. By uniqueness of the continutation fromQ0(Theorem 3.3.10) we havev1=0,u1=0 onQ3∩ {T1<t<T2}whereT1=i n f t andT2=supt over (x,t)∈ Q0. Hence g0=0
on∂×(T1,T2). In case (N)T1=0 and we have a contradiction with conditions of Theorem 9.5.6 ong0. In case (D)u1is nonnegative by Theorem 9.2 and hence by the second part of Theorem 9.2 we haveu1 =0 on (Q\Q•1)∩ {0<t <T2}, g0=0 on∂×(0,T2) and we have a contradiction again.
The proof is complete.
Instead of smooth functiong0one can use the Dirac delta-function concentrated at a point of∂× {0}. The proof is valid in this case with minor technical ad- justments regarding definition of a weak soltion with less regular Dirichlet data.
Vessella [Ve] obtained a logarithmic stability estimate for this problem.