Exercise 2.1.2 A Nonhyperbolic Cauchy Problem for the Wave Equation)
4.3 The inverse conductivity problem
Of special interest is the so-called conductivity equation div(a∇u)=0, in , a∈ L∞().
(4.3.1)
First we give a simple global uniqueness result due to Alessandrini [Al2]. Let u1,u2 be solutions to the Dirichlet problem (4.3.1), (4.0.2) with scalara=a1, a=a2, which are equal to 1 near∂and nonconstantg. If a1≤a2 inand
∂νu1 =∂νu2 on , then a1=a2 on . Here is any nonvoid open part of
∂.
In fact,u1,u2 are harmonic near∂, and they have the same Cauchy data on . Therefore, they coincide near∂. By the definition (4.0.3) of a weak solution,
a1|∇u1|2=
∂u1∂νu1=
∂u2∂νu2=
a2|∇u2|2≥
a1|∇u2|2. By the Dirichlet principle, u1 is the unique minimizer of the first integral (the energy integral), so we conclude thatu1=u2on, and thereforea1 =a2.
Now we consider
a=1+kχ(D), D¯ ⊂ (4.3.2)
whereDis a subdomain ofto be found withC1-piecewise smooth boundary.
Unless specified, in this section we assume thatkis a given nonzero constant.
We discuss the direct problem in more detail. By Theorem 4.1, ifg0∈C2+λ(∂) and∂∈C2+λ, then a solutionuto the Dirichlet problem (4.0.1), (4.0.2) belongs toC2+λ(∩V), where V is a neighborhood of∂. Denote u on \D byue andu on Dbyui. When∂D is only Lipschitz, thenu is continuous in ¯and
∇ue(∇ui) has nontangential limits at almost every point of∂D. Moreover, these limits belong to L2(∂D), and for any sequence ofν-approximating D domains Dm+1⊂Dm, we have∇ue2(∂Dm)<C (property B). These results are con- tained or directly follow from the paper of Escauriaza, Fabes, and Verchota [EsFV].
ν-approximation means that in a local coordinate system whereDis given by the in- equality{xn<d(x)}with a Lipschitzd, domainsDmare defined as{xn < fm(x)}
wherefmsatisfy the conditions:d+(m+1)−1 < fm≤d+m−1,∇fm→ ∇dal- most everywhere and|∇fm| ≤C. The estimate on the boundaries of approximating
4.3. The inverse conductivity problem 103 domains follows from the estimate of the maximal functions in the paper [EsFV].
When∂D∈C1+λ, the classical method of simple layer potentials described in the book of Miranda [Mi] (compare with the proof of Lemma 4.3.8) gives solu- tions with∇ue(∇ui) continuous up to∂D. When∂D∈C2+λ, then by Theorem 4.1ue∈C2+λ(\D) andui ∈C2+λ(D). Equation (4.3.1) is then equivalent to the following equations and refraction conditions on∂D:
ue=0 on\D, ui =0 onD, ue=ui, ∂νue=(1+k)∂νui on∂D.
(4.3.3)
Exercise 4.3.1. Let u∈ H(1)(), ∇ue have nontangential limits almost every- where on∂Dand possess the propertyB, and so are∇ui. Prove that the differential equations (4.3.3) and the refraction conditions are equivalent to the differential equation (4.3.1) with the coefficientagiven by (4.3.2).
{Hint: In the definition (4.0.3) of a solution to equation (4.3.1) first consider test functionsφ inC0∞(D) and inC∞0 (\D) and conclude that the differential equations are satisfied; then integrate by parts in this definition in Dand\D, transferring derivatives ontoue,ui, and use the density of the test functions in L2(∂D). When integrating by parts in LipschitzDuseν-approximations, integrate in the approximating domains, and pass to the limit with respect to these domains.}
Domains D1,D2 are called i-contact domains if the sets \Dj, \(D1∪ D2),D1∩D2are connected; the sets∂D1∩∂D2,int(D1∩D2) are disjoint, and there is a nonemptyC2-surface that belongs to the boundaries of both\(D1∪D2) andD1∩D2. In other words, these domains have a common piece of the bound- aries. They are located on “one side” of this piece, and they satisfy some natural topological restrictions. In particular, two domains are i-contact when they are xn-convex, their boundaries have a common part, and these domains overlap.
Theorem 4.3.2. Let D1, D2 be i -contact Lipschitz solutions to the inverse con- ductivity problem(4.3.1),(4.3.2),(4.0.2),(4.0.6)with nonconstant Dirichlet data.
Then D1=D2.
PROOF. Let0 be the common piece of∂D1and∂D2. The functionsue1 andue2 are harmonic on \(D1∪D2), and they have the same Cauchy data on . By uniqueness in the Cauchy problem they coincide there. Using (4.3.3) we conclude thatui1=ui2and∂νui1=∂νui2on0. Again by uniqueness in the Cauchy problem, ui1 =ui2onD1∩D2.
Let us assume thatD1\D2is not empty. The functionsui1andue2are harmonic in D1\D2. Due to our assumptions onD1,D2, the boundary ofD1\D2consists only of points of∂(D1∪D2) or of∂(D1∩D2). In the first case, by using the refraction conditions (4.3.3), we haveui1=ue1=ue2, and in the second case,ui1=ui2 =ue2. So the functionsui1andue2are harmonic inD1\D2and coincide on its boundary.
By the maximum principles they are equal in D1\D2. Therefore, they have the same normal derivatives on∂(D1\D2).
We consider∂D1\D2. On this set we have∂νui1=∂νue2=∂νue1becauseue1=ue2 near this set. On the other hand, the refraction conditions give (1+k)∂νui1=∂νue1 on the same set, which implies that∂νui1 =0 on∂D1\D2. Similarly,∂νui1=0 on the remaining part of∂(D1\D2). By uniqueness in the Neumann problem,ui1is constant, and thereforeue2 is constant, which contradicts our assumptions about the Dirichlet data foru.
This contradiction shows thatD1⊂D2. Similarly,D2⊂D1.
The proof is complete.
Exercise 4.3.3. Prove that if∂νui1=0 on∂(D1\D2), thenui1is constant onD1\D2. {Hint: make use of integration by parts and ofν-approximation of D1\D2by their open subsets with smooth boundaries.}
Exercise 4.3.4. Letbe the cylinderG×(−H,0), whereGis a boundedC2+λ- domain inRn−1. LetD be its subdomain{−H <xn <d(x),x∈G}, whered is a Lipschitz function,−H <d <0. Letbe the union of two nonempty open parts ofG× {−H}and ofG× {0}.
Show that Dis uniquely identified by the data of the inverse problem (4.3.1), (4.3.2), (4.0.2), (4.0.6), provided thatg0=0 on∂G×(−H,0).
We observe that these two exercises are interesting forDwith piecewise smooth and even withC2+λ-boundaries, and proofs for general Lipschitz domains are only slightly more difficult, because one has to use some approximation.
Now we turn to noncontact domains. Not as much is known here. We are able to give only the following global uniqueness result.
Theorem 4.3.5. Let us assume that D1, D2are either (i) two convex polyhedra or (ii) two bounded cylinders with strictly convex bases. Let
diamDj <dist(Dj, ∂).
(4.3.4)
If the solutions uj of(4.3.1), (4.3.2), (4.0.2) with nonconstant g0 and with D=Dj satisfy the condition∂νu1=∂νu2on, then D1=D2.
It is observed in the paper of Friedman and Isakov [FrI] that condition (4.3.4) can be dropped whenis a half-space. Theorems 4.3.2, 4.3.5 are from [FrI].
We mention that global uniqueness is known for unions of a finite numbers of discs in the plane and for ballsDin the three-dimensional space (see the papers of Isakov and Powell [IsP] and of Kang and Seo [KS1], [KS2]), but it is not known for ellipses or ellipsoids.
Outline of a proof for n=2. First we prove the following Lemma.
Lemma 4.3.6. Let the origin be a vertex of a convex polygon D and let uehave a harmonic continuation onto a ball B(0;ε). Then there is a rotation of the plane such that ueon this ball is invariant with respect to this rotation.
4.3. The inverse conductivity problem 105 PROOF. We can assume that Dε=D∩B(0;ε) is {0< σ < θ,0<r< ε}. Ac- cording to relations (4.3.3), the functionui solves the following Cauchy prob- lem:ui =0 inDε,ui =ue, ∂νui =(1+k)−1∂νueon{0<x1< ε,x2=0}. We can consider the same extended Cauchy problem with the data on the interval {−ε <x1< ε}. Since its dataue are analytic near the origin, for someε there is a solutionui that is analytic on B(0;ε) (e.g., see the book of John [Jo4]). A solution is unique onDε, souihas an analytic continuation ontoB(0;ε) for some positiveε.
The functions ui,ue are harmonic near the origin, so they admit the expan- sionsu(x)=
(amcosmσ +bmsinmσ)rm(m=0,1, . . .) with the coefficients aim,aem, . . . for ui,ue. From the refraction conditions (4.3.3) at σ =0 we ob- tain ame =aim and bem=(1+k)bim. The same conditions at σ =θ then give (bem−bmi) sinmθ =0 and (ame −(1+k)ami) sinmθ=0. Summing up, we obtain bemsinmθ=ame sinmθ=0.
Ifθ/πis irrational, thenbme =aem=0 for allm=1, . . .. Therefore, the function ueis constant, which contradicts the assumption that the Dirichlet datagare not constant. Ifθ/πisp/q,q ≥1 and (p,q)=1, then for all nonzero coefficients we must have sinmpπ/q =0, which means thatmp/qis an integer; som=lq,l = . . . ,−1,0,1, . . .. Returning to the representation ofueby the series, we conclude that this function is invariant with respect to rotation by 2π/q.
The proof is complete.
We return to the proof of Theorem 4.3.5. Let us assume thatD1is notD2. Then we may assume that the origin is a vertex of D1 which is not contained in the convex hull of D1∪D2. Sinceue1,ue2are harmonic on the domain\(D1∪D2) and have the same Cauchy data on, they are equal on this domain; and sinceue2 has a harmonic continuation onto a neighborhood of the origin,ue1has a harmonic continuation as well. By condition (4.3.4) there is a sectorSof the diskB(0;R), whereR>diamD1, with angle greater thanπthat is contained in\D1. Rotating this sectorq times by the angle 2π/q, one can obtain the harmonic continuation ofue1onto the whole disk that contains D1 and therefore onto. Indeed, letO be this rotation. Thenu(O(x))=u(x) when|x|< εby Lemma 4.3.6. Since both functions are harmonic, they agree also on the intersection of S and O(S), so u(O(x)) is the harmonic continuation ofuonto O(S), and we can proceed with rotations. The functionsue1 andui1 are harmonic inD1, and they coincide on its boundary. Therefore, they are equal inD1. Using the second refraction condition (4.3.3), we conclude that∂νue1=0 on∂D1. By uniqueness in the Neumann problem for harmonic functions,ue1is constant onD1, and by uniqueness of the harmonic continuation, it is constant on, which contradicts the assumption that the Dirichlet datag0are not constant. So the initial assumption thatD1is notD2is wrong, and the proof is complete.
The proof for three-dimensional space and whenis a half-space is given for polyhedra in the paper of Friedman and Isakov [FrI] and for cylindrical Din the paper of Isakov and Powell [IsP]. Recently, Seo in the paper [Se] showed that two boundary measurements uniquely determine a (not necessarily convex) polygonD
andk. We observe that these proofs heavily use properties of harmonic functions, and so far they are not extended to solutions of the Helmholtz equation. This extension is quite desirable since many inverse problems use prospecting by time harmonic waves.
While there are no global uniqueness results for general convexDin the absence of interior sources, we can prove uniqueness of such Dwhen a field is generated by a source insideD.
We consider the conductivity equation
div(−a∇u)= f inR3, lim
|x|→∞u(x)=0. (4.3.5)
LetDbe a subdomain of a bounded domain, andaC1-surface inR3\. We assume that f andkare given. For the inverse conductivity problem we consider the additional data
u=g0, ∂νu =g1 on. (4.3.6)
Theorem 4.3.7. Let us assume that f ≥0, f ∈ L1(); f =0outside D, f ≡0.
Let D be a convex domain and a=kχ(D)where k>0 on. (4.3.7)
Then D is uniquely determined by the data(4.3.6).
In the proof we make use of the following results.
Lemma 4.3.8. If two domains D1and D2produce the same data(4.3.6), then
∂D1
v∂νui1=
∂D2
v∂νui2 for all functions v harmonic near D1∪D2.
PROOF. From the definition (4.0.3) of the equality div(a1∇u1)=div(a2∇u2) inR3 we get
∂0
∂νu1v−
0
a1∇u1ã ∇v=
∂0
∂νu2v−
0
a2∇u2ã ∇v
for any functionv∈C1(0), where0is a smooth domain containingD1∪D2. As above, the equality (4.3.6) implies thatu1=u2on\(D1∪D2). Let us choose 0 to be contained a neighborhood ofD1∪D2 wherev is harmonic. Then the boundary integrals are equal, and using (4.3.2), we conclude that
0
∇u1∇v+k
D1
∇u1ã ∇v=
0
∇u2ã ∇v+k
D2
∇u2ã ∇v.
Integrating by parts in the integrals overand using the harmonicity ofv and the equalityu1=u2outsideD1∪D2, we conclude that these integrals are equal.
4.3. The inverse conductivity problem 107 Then similarly, integrating by parts in the integrals overD1,D2, we arrive at the
conclusion of this lemma.
Lemma 4.3.9. Under the conditions of Theorem4.3.7we have∂νui<0on∂D.
PROOF. Let
h(x)=(1+k)
D
f(y)G(x,y)d y,
where G(x,y)=1/(4π|x−y|) is the standard fundamental solution of the Laplace operator. We have div(−a∇h)= f inD, andh is harmonic outsideD.
SinceDis convex, we have
∂ν(x)G(x,y)= −(4π|x−y|3)−1ν(x)ã(x−y)≤0
whenx∈∂Dandy∈D, so∂νh<0 on∂D. Whenusolves the direct problem, then the functionw =u−h is harmonic inD, and outside D, it goes to zero at infinity, is continuous inR3, and satisfies the refraction condition
∂νwe−(1+k)∂νwi =k∂νh on∂D.
(4.3.8)
Moreover, any suchwgenerates a solution to the original problem foru. To find wone can make use of the classical single layer potential
w(x)=U1(x;ρ)=
∂D
ρ(y)G(x,y)d(y),
with densityρto be found. The well-known jump relations for first order derivatives of this potential ∂νwe= −ρ/2+∂νU1, ∂νwi =ρ/2+∂νU1 on ∂D (e.g., [Is4], section 1.6, [Mi], p. 35, for smooth Dand [EsFV] for Lipschitz D) where∂νU1
denotes the potential on∂Dwith the kernel∂νGtransform the refraction condition (4.3.8) into the integral equation
−ρ/2−(1+k)ρ/2−k∂νU1=k∂νh on∂D or
(I−K)ρ= −k/(1+k/2)∂νh (4.3.9)
whereIis the identity operator andKρ= −k/(1+k/2)∂νU1.
The right side of the equation (4.3.9) is positive on∂D. Since the kernel∂νG of the operator∂νU1 is non positive the operatorK maps nonnegative functions into nonnegative functions. According to the results of Escauriaza, Fabes, and Verchota the eigenvalues of the operatorρ →∂νU1are contained in (−1/2,1/2].
Due to the convexity assumptions, this operator has non-positive kernel, so its spectral radius inL2does not exceed 1/2, the spectral radius of K inL2is less than 1, and a solution ρ to equation (4.3.9) is the sum of the Neumann series (I+K +K2+. . .)h1, whereh1is the right side of (4.3.9). Sinceh1is positive, so isρ.
The proof will be complete when we write
∂νue=∂νh+∂νwe=∂νh−ρ/2+∂νU1
and express∂νU1from (4.3.9) to conclude that∂νue= −(1/k+1)ρ <0.
PROOF OFTHEOREM4.3.7. We will modify the orthogonality method used to prove Theorem 4.1.1.
Assume that there are two different domainsD1,D2producing the same exterior data. If one of them is contained in another, then we obtain a contradiction by using the argument from the beginning of Section 4.3. So we can assume that both D1\D2 and D2\D1 are nonempty. We denote by 1e, 1i, 2e, 2i the exterior and interior parts∂D1\D2, ∂D2∩D1, ∂D2\D1, ∂D1∩D2of their boundaries. By using harmonic approximation and stability of the Dirichlet problem with respect to a domain (see, e.g., [Is4]) we can extend the orthogonality relation of Lemma 4.3.8 onto functionsvharmonic inD1∪D2and continuous on the closure of this star-shaped domain. Letg0be any continuous function on∂(D1∪D2),0≤g0 ≤1, andva solution of the Dirichlet problem inD1∪D2with the datag0. We choose g0=0 on some open part of2e. By using the comparison principle one can show then thatv<1−ε(F) for any compact setFinD1∪D2(see, e.g., [Is4], Lemma 1.7.4). By maximum principles 0≤v.
From Lemma 4.3.8 we have
1e
g0∂νui1+
2i
v∂νui1=
2e
g0∂νui2+
1i
v∂νui2.
By using the inequalities forvand Lemma 3.4.8 we bound the left side from above and the right side from below to obtain
1e
g0∂νui1≥
2e
g0∂νui2+
1i
∂νui2+ε,
whereεdoes not depend ong0. We can approximate inL1(2e∪1e) by suchg0 the function that is 1 on1eand 0 on2e, which give the inequality
1e
∂νui1−
1i
∂νui2>0.
Observe that according to the refraction conditions (4.3.3) we have
∂νui1=(1+k)−1∂νue1 =(1+k)−1∂νue2on1e
becauseue1=ue2 outside D1∪D2, and similarly,∂νui2=(1+k)−1∂νue2 on1i, so we can replaceui1 andui2byue2in the above integral. Sinceue2is harmonic in D1\D2andνis the interior normal on1i, the integral must be zero.
We have a contradiction, which shows that the initial assumptionD1=D2was wrong.
The proof is complete.
The orthogonality relations can also be used to estimate some functionals of the conductivity coefficienta. Indeed, from the definition of a weak solution (4.0.3)
4.3. The inverse conductivity problem 109 we have
a∇uã ∇φ=
∂g1φ,
and the right side is known from the data (4.0.6). In particular, whena has the form (4.3.2),
∇uã ∇φ+k
D
∇uã ∇φ=
∂g1φ
∇u0ã ∇φ=
∂∂νu0φ, (4.3.10)
ifu0is a harmonic function.
Exercise 4.3.10. Prove that the data (4.0.2), (4.0.6) of the inverse conductivity problem with a of the form (4.3.2) in a unique and stable way determine the integrals
k
∂D
u∂νv for any harmonic functionvin.
Now we will show how to use the orthogonality relations (4.3.10) to evaluate size ofD. The most recent and advanced paper on this subject is written of Alessandrini, Rosset, and Seo [AlRS] where one can find more general and detailed results as well as references to previous work. The idea is to compare solutions of the problems witha =1+kχ(D) and witha =1. We will assume thatu =u0=g0 on∂. Choosing in (4.3.10) the test functionφ=u−u0, subtracting the integral equalities (4.3.10) and utilizing thatφ=0 on∂we yield
|∇u− ∇u0|2+k
D
∇uã ∇(u−u0)=0. From the first equality (4.3.10) withφ=u0,
k
D
∇uã ∇u0=
∂∂νuu−
∇uã ∇u0=
∂(∂νu−∂νu0)g0
where we also used the second equality (4.3.10) withφ=u0. With the aid of the last integral identity we finally arive at the relation
|∇u− ∇u0|2+k
D
∇uã ∇u=
∂(∂νu−∂νu0)g0. (4.3.11)
If 0<kthen the both terms on the left side are nonnegative and therefore bounded by the known integral over∂. By using the elementary inequalityb2 ≤2(|b− c|2+c2) we can eliminate unknown term withuin the integral (4.3.11) over:
k
D
|∇u0|2 ≤2k(
D
|∇(u−u0)|2+
D
|∇u|2)≤2(k+1)
∂(∂νu−∂νu0)g0.
This inequality can be used in the estimation of size ofD. For example, letting u0(x)=xjwe will have
k/2(k+1)vol D ≤
∂(∂νu(;j)−νj)xj
whereu(;j) solves the Dirichlet problem with the datau(x;j)=xj,x∈∂.
Similarly one obtains the integral identity
(1+kχ(D))|∇(u−u0)|2−k
D
|∇u0|2=
∂(∂νu0−∂νu)g0, which can be used to evaluatevol Dwhen−1<k<0 and also for bounding the right side of (4.3.11) by the integral overDto conclude that
1/k
(∂νu−∂νu0)g0≤
D
|∇u0|2≤2(1+1/k)
(∂νu−∂νu0)g0
(4.3.12)
In [AlRS] they considered more general elliptic equations and by using so-called doubling inequalitites to bound the inner integral in (4.3.12) by vol D obtained two-sided estimates ofvol Dfor arbitrary nonzero Neumann boundary data with constants depending on this data. A good review of available results on size estima- tion (including evaluation of elastic cavities) is given in the paper of Alessandrini, Morassi, and Rosset in [I3].
We observe that for nonlinear conductivity equation (4.3.1) witha=a(u) there is a global uniqueness and certain existence results due to Pilant and Rundell [PiR3]. We demonstrate their method in a slightly more general many-dimensional situation. We consider a domain⊂Rnwith theC2+λ-boundary. Letbe aC1- curve on∂. We prescribe the following boundary value data:
a(u)∂νu =g1 on∂, g1∈C1+λ,
∂g1=0 and the additional boundary data for determination ofa,
u =g0on. Letbbe a point of∂and
v(x)= u(x)
u(b)
a(s)ds.
Then equation (4.3.1) and the Neumann boundary conditions (4.0.6) are trans- formed into the following equation and boundary conditions:
−v=0 in,
∂νv=g1on∂; (4.3.13)
and the additional boundary data give g0(x)
g0(b)
a=v(x) whenx∈. (4.3.14)