Exercise 2.1.2 A Nonhyperbolic Cauchy Problem for the Wave Equation)
4.4 Methods of the theory of one complex variable
One can prescribe the Dirichlet datag0∈C2+λin the direct problem and the additional Neumann data on that are transformed into the boundary relations (4.3.14) on∂and (4.3.13) on. Uniqueness ofa on the range ofg0when the maximum and minimum ofg0are achieved onstill holds, as shown below.
Leta1,a2be two conductivity coefficients andv1,v2the corresponding functions v. Leta =a2−a1,w =v2−v1. Then subtracting the relations (4.3.13), (4.3.14), we obtain
w =0 in,
∂νw =0 on,
g0(x)
g0(b)
a =w(x) on∂.
By Giraud’s extremum principle, a nonconstantw cannot attain its maximum or minimum overat a point of∪, so they are attained on∂\. Sinceais positive, this means that the maximum and minimum ofg0are on∂\as well, which contradicts the choice ofg0. Therefore,wis constant on, which must be zero becausew(b)=0. Thena1=a2on the range ofg0.
The choice of the Dirichlet data for the direct problem has advantages, because then the range ofg0is under control.
4.4 Methods of the theory of one complex variable
In the plane case one can enjoy additional opportunities of the theory of functions of one complex variable. In particular, conformal mappings and Riemann-Hilbert type boundary value problems are very useful when studying the free boundary problems where a domainDis to be found from certain exterior data.
We introduce the complex variablesz=x1+i x2,z=x1−i x2and the related complex differentiations
∂=2−1(∂1−i∂2), ∂¯ =2−1(∂1+i∂2).
(4.4.1)
The function ¯∂u then is one-half of the gradient of u. It is easy to show that =4∂∂. For references on theory of one complex variable we refer to the book¯ of Ahlfors [Ah].
Lemma 4.4.1. Let D be a bounded simply connected domain in R2 with rec- tifiable boundary. Let f ∈ L∞(D). Let u be a function that is harmonic in R2\D, is Cln|x| +u0, where u0 goes to zero at infinity and u0∈C1(R2\D).
Let U(; fχ(D))be the logarithmic potential of D with density f . Then the equality
U(; fχ(D))=u outside D (4.4.2)
is equivalent to the equality
∂u =F+φ+ on∂D, (4.4.3)
where F is a solution to the equation−4 ¯∂F = f in D,F∈C(D), andφ+∈C(D) is a complex analytic function on D.
PROOF. The potential U(; fχ(D)) is contained in C1(R2), and −4∂∂U¯ (; fχ(D))= fχ(D). Since u and the potential are equal outside D, their anti- gradients∂are equal on∂D. Letφ+=∂U(; fχ(D))−F. Then ¯∂φ+=0 onD, and we have the equality (4.4.3).
Now let us start with (4.4.3). Define the functionV as∂U(; fχ(D))−∂u on R2\Dand as∂U(;fχD))−F−φ+inD. Both functions outside and inside of Dare analytic functions, continuous up to∂Dand equal on∂D. By the continuity principle,Vis complex analytic inC. In addition, gradients ofuand of the potential go to zero at infinity, soV is bounded. By Liouville’s theorem, it is constant, which must be zero due to the behavior at infinity. Since the gradient of the difference of uand the potential is zero, this difference is constant outside D, which turns out to be zero if we remember the behavior at infinity.
The proof is complete.
Letz(t) be a conformal mapping of the unit disk B(0; 1) onto D. Since Dis Jordan,z(t) is continuous on the closure of the unit disk. LettingU =∂u and transplanting the relation (4.4.3) onto the boundary of this disk, we obtain
U(z(t))=F(z(t),z(t))+φ+(t) when|t| =1,
whereφ+is a function that is complex analytic in the unit disk and continuous on its closure. In particular, when f is constant, we may choose F(z,z)= −f z/4, which gives the useful relation
U(z(t))= −f z(1/t)/4+φ+(t) whentt= |t|2=1, (4.4.4)
which helps, say, to analyze the connection between regular points of the exterior potential∂u and the conformal mapz(t). The relation (4.4.4) has been obtained by V. Ivanov [Iv].
Indeed, letD•be a domain containingC\D, the part of∂Dinside D•, and γ,G•be the inverse images of,D∩D• under the mappingz(t). The exterior potentialU(; fχ(D)) has a harmonic continuation fromC\DontoD•if and only ifz(t) has an analytic continuation onto the interior of B(0; 1)∪γ∪G∗•, where G∗•is the image ofG•under the inversiont →1/t.
If the exterior potentialu has a continuation, then the left side of (4.4.4) is complex analytic inG•.z(1/t) is complex analytic outside of the unit disk and continuous up to its boundary. Since it is equal onγ to the function 4/f(φ+(t)− U(z(t)), which is complex analytic onG•and continuous up toγ, by the continuity principle we conclude thatz(1/t) has an analytic continuation ontoG•. Applying inversions, we obtain the claim forz(t).
4.4. Methods of the theory of one complex variable 113 If the conformal mapping has a continuation, then we can use the same argument to show thatUhas an analytic continuation ontoD•, and thereforeuhas a harmonic continuation.
The relation (4.4.4) can be considered as a nonlinear boundary value problem for analytic functionsz(t), φ+(t) in the unit disk. This problem can be solved at least locally, which gives also some constructive method of finding Dgiven the exterior potentialu.
We refer for the corresponding results and details to the book of Cherednichenko [Cher] and of Isakov [Is4] and the references given there.
A similar technique is even more useful when studying the inverse conductivity problem.
LetDbe a simply connected subdomain ofwith the Lipschitz boundary. Let ube a solution to the conductivity problem (4.3.3) with the Dirichlet datau =gon
∂. By using the refraction condition on∂Dit is easy to observe that the integral of∂νueover∂Dis (1+k) times the integral of∂νui, which is zero because the functionui is harmonic insideD. Therefore, we can find the harmonic conjugate veofuein\D. SinceDis simply connected, there is the harmonic conjugatevi touiinD. We can assume that bothveandviare equal to zero at some point of∂D.
We assume that the pair (τ, ν), whereτ is a unit tangent to∂D, is oriented as the coordinate vectorse1,e2, and we recall the following form of the Cauchy-Riemann equation foru,v:
∂τu=∂νv, ∂τv= −∂νu, (4.4.5)
whereushould be replaced byueoutside Dand byui insideDand so isv. We introduce the complex analytic functionsUe=ue+i veandUi =ui+i vi. Since we are given the Cauchy data forueon, this function can be considered as given.
Letz(t) be the conformal mapping of the unit disk onto Dnormalized in the standard way:z(0)=0,z(0)>0.
Lemma 4.4.2. A domain D is a solution to the inverse conductivity problem with the exterior data ueif and only if
ψ+(t)=(2+k)Ue(z(t))+kUe(z(t))when|t| =1 (4.4.6)
for some functionψ+ that is complex analytic in B(0; 1)and whose first-order derivatives are in L2(∂B(0; 1)).
PROOF. LetDbe a solution to the inverse conductivity problem. From the Cauchy- Riemann conditions (4.4.5) and from the refraction conditions (4.3.3) we conclude that∂τve=(1+k)∂τvion∂D. Integrating along∂Dand using the special choice ofve,vi, we conclude thatve=(1+k)vion∂D. This relation and the continuity of the functionuyield
Ue+Ue=Ui+Ui, (Ue−Ue)=(1+k)(Ui−Ui) on∂D.
By substitutingUifrom the first relation into the second one we obtain the equality (4.4.6) withψ+(t)=(2+2k)Ui(z(t)).
Now let the relation (4.4.6) be valid, wherez(t) is the normalized conformal map of the unit disk ontoD. LetUDe be the exterior analytic function constructed for the exterior part of the solution of the direct conductivity problem. Then we have the relation (4.4.6) forUD withψ+ replaced by (2+2k)UDi(z(t)). LetUi(z)= ψ+(t(z))/(2+2k). Subtracting the relations (4.4.6) forU andUD and letting U•be their difference, we obtain the equality (2+2k)U•i =(2+k)U•e+kUe•on
∂D. SinceRUeandRUDe have the same Dirichlet data on∂, we conclude that RU•e=0 on∂. Subtracting the boundary relation forU•and its complex adjoint as well as adding these relations, we arrive at the equalities
U•e−Ue•=(1+k)(U•i−Ui•), U•e+Ue•=U•i+Ui•, or
ue•=ui•,v•e=(1+k)ve• on∂D.
Differentiating the second relation in the tangential direction and using once more the Cauchy-Riemann conditions (4.4.5), we conclude that∂νue•=(1+k)∂νui•on
∂D. Sou•is a solution to the direct conductivity problem with zero Dirichlet data on∂, and it follows thatue•=0 outsideD. Therefore,ue=ueD, and the proof is
complete.
The relation (4.4.6) can be considered as a nonlinear boundary value problem for analytic functionsψ+,zin the unit disk whose solution (with givenUe) produces a solution Dto the inverse conductivity problem. In particular, one can obtain analyticity results similar to those for inverse gravimetry, provided that∂Ue=0 on∂D(or∇ue=0, which is the same). This condition is controlled by the index (winding number) of the vector field∇ue, and it plays a crucial role in Theorem 4.4.3.
In proving analyticity and even uniqueness results similar to Theorem 4.3.5 one can use the Schwarz function and complete regularity results for domains having such a function that were obtained by Sakai [Sa]. Letγ be a portion of
∂Dfor an open set D⊂C. Let the origin be a point of. A functionSdefined on D∪is called a Schwarz function of D∪ifS is complex analytic in D, continuous in D∪, andS(z)=zon. Sakai showed that loosely speaking, near the origin is either a regular analytic curve or the union of two such curves or a regular analytic cusp directed intoD. If the exterior potentialUhas a complex analytic continuation across, then lettingt =t(z) in (4.4.4), we obtain the relation z=4(φ+(z)−U(z)) on, so the right side is a Schwarz function, and we obtain analyticity of. Similarly, if we can resolve equation (4.4.6) with respect toz, we obtain the same claim. This equation can be locally resolved when∂U =0 on, which can be guaranteed by index theory, as we will show later.
We will assume that D=Dσ is close to a C1+λ-domain D0 in the following sense. Let be a family of complex analytic functions σ on B(0; 1),|σ|1+λ(B(0; 1))≤ M, such thatz(t)+σ(t) is the normalized conformal mapping ofB(0; 1) onto some simply connected domainDσ. A sufficient condition
4.4. Methods of the theory of one complex variable 115 is thatσ(0)=0,Iσ(0)=0, andM is small. We will assume thatis closed in C1+λ(B(0; 1)).
Theorem 4.4.3. Let∂D0 be C1+λ. Let us assume that the boundary data g0∈ C2(∂)have a unique local maximum and unique local minimum on∂.
For the familythere is a numberεsuch that if Dσis a solution to the inverse conductivity problem and|σ|0(B(0; 1))< ε, then Dσ =D0.
A complete proof of this result is given in the papers of Alessandrini, Isakov, and Powell [AlIP] and of Powell [Pow]. It is based on linearization of the boundary condition (4.4.6), the study of the corresponding linear problem A(ψ, σ)=ψ1, and the contracting remainderB. The linearization is represented by the boundary value problem for complex analytic functionsψ, σ in the unit disk B(0; 1) with the nonlinear boundary condition
A(ψ, σ)(t)=Bσ(t) when|t| =1, (4.4.7)
where
A(ψ, σ)=ψ−(2+k)aσ−kaσ, a(t)=∂u(z(t)) and
Bσ =(2+k)B1σ +B1σ, B1σ =Ue(z+σ)−Ue(z)−∂ue(z)σ.
HereAis considered as an operator from×intoCλ(∂B(0; 1)). We define as the space of functions complex analytic inB(0; 1) and H¨older in its closure, and as the space of functionsσ ∈ such thatσ(0)=0,Iσ(0)=0. The operator Ais continuous from × onto its rangeR⊂Cλ(B(0; 1)), which is known to be finite-codimensional. The main claim about Ais that under the conditions of Theorem 4.4.3 its kernel is zero, and therefore this Fredholm operator is in- vertible fromRinto×. A proof of this claim given in [AlIP] is based on Muskhelishvili’s [Mus] theory of one-dimensional singular integral equations (as suggested and proven in the paper of Bellout, Friedman, and Isakov [BelFI] for analytic∂D0). The crucial step is to show that the index (winding number) of the vector fieldais well-defined on∂B(0; 1) and is zero. We observe that the boundary value problem for the conformal map was derived and used by Cherednichenko [Cher].
First, we observe that a solution to the elliptic differential equation (4.3.1) inR2 cannot have a zero of infinite order unless this solution is constant (see Section 3.3). This observation implies that all zeros of the gradient of a nonconstant solutionuto the equation div((1+kχ(D))∇u)=0 inare isolated, andunear a zero admits the following representation:u(x)−u(x0)=HN(x−x0)+O(|x− x0|N+λ), where HN is a homogeneous function of degree N ([AlIP], Theorem 4). the geometric interpretations of the index of∇u around zero gradient given in the papers of Alessandrini and Magnanini [AIM] and of Alessandrini, Isakov, and Powell [AlIP] as the multiplicityN of a zero of the gradient and homotopic invariance of the index show that the sum of indices of all critical points ofuinside
is equal to the index of∇uover∂. From the conditions of Theorem 4.4.3 on the boundary data we obtain that the index of∇uover∂is 0. So there is no zero of∇uein\D0and of∇ui inD. Hence the index ofa=∂u over∂Dis zero.
Another way to understand why the conditions ong0imply the absence of zeros of∇uinsideis to use level curves ofu. If there is a zero inside, then more than two level curves are entering∂, which contradicts the condition ong.
It is interesting that the linearized problem for analytic functions can be trans- formed into the oblique derivative problem for harmonic functionsvinD0 with the boundary condition ∇ueã ∇v=0 on ∂D0, as observed by Bellout, Fried- man, Isakov [BelFI]. Our conditions guarantee that this problem is elliptic in the two-dimensional case, but it is always nonelliptic in higher dimensions. This is a partial explanation why even local uniqueness results are not available in the three-dimensional case. There are expectations of obtaining local uniqueness re- sults from three special boundary measurements.
One can anticipate conditional logarithmic (local) stability in the situation of Theorem 4.4.3 which probably can be obtained by using sharp stability results for analytic/harmonic continuation up to the boundary of D (like in [AlBRV], [Ron] for different problems) and standard elliptic theory which implies that the inverse of the linearized operatorAin (4.4.7) is continuos fromC1+λinto itself.
Logarithmic stability should be optimal as suggested by the general theory in Di Christo, Rondi [DR].
Exercise 4.4.4. Prove that ifg0is chosen as in Theorem 4.4.3, then Theorem 4.3.5 about convex polygons is valid without the assumption (4.3.4).
{Hint: Use that the index of∇uover∂is zero and that from the proof of Lemma 4.3.6 one can conclude that ifuehas a harmonic continuation onto a neighborhood of a vertex, then this vertex must be a zero of∇ue.}