Copyright © 2006 SBMAC ISSN 0101-8205
www.scielo.br/cam
On an inverse boundary value problem*
ALBERTO P. CALDERÓN
Department of Mathematics, The University of Chicago
Abstract. This paper is a reprint of the original work by A. P. Calderón published by the Brazilian Mathematical Society (SBM) in ATAS of SBM (Rio de Janeiro), pp. 65-73, 1980. The original paper had no abstract, so this reprint to be truthful to the original work is published with no abstract.
Mathematical subject classification: 35J25, 35Q60, 47F05, 86A20, 86A22.
Key words: inverse problems, boundary value problems, identification problem, elliptic equa-tions.
In this note we shall discuss the following problem. Let D be a bounded do-main inRn, n ≥ 2, with Lipschitzian boundaryd D, andγ be a real bounded
measurable function inDwith a positive lower bound. Consider the differential operator
Lγ(w)= ∇ ·(γ∇w)
acting on function ofH1(D)and the quadratic formQγ(φ) ,where the functions
inH1(Rn) ,defined by
Qγ(φ) =
D
γ (∇w)2d x, w∈ H1 Rn
, w|d D = 0
Lγw = ∇ ·(γ∇w) = 0 in D.
#693/06. Received: 14/XI/06.
The problem is then to decide whetherγ is uniquely determined byQγ and to
calculateγ in terms Qγ,ifγ is indeed determined byQγ.
This problem originates in the following problem of electrical prospection. If D represents an inhomogeneous conducting body with electrical conductivity
γ, determineγ by means of direct current steady state electrical measurements
carried out on the surface ofD, that is without penetratingD. In this physical sit-uationQγ(φ)represents the power necessary to maintain an electrical potential
γ in∂D.
In principleQγ can be determined through measurements effected ond Dand
contains all information aboutγ which can be thus obtained.
But let us return to our mathematical problem. Let us introduce the following norms in the space of functionsγ on d D and in the space of quadratic forms Q(φ)
φ2 =
D
|∇u|2d x ; u|d D=φ , u=0 in D.
Q = sup
φ≤1 |Q(φ)|
Then the mapping
:γ → Qγ
is bounded and analytic in the subset of L∞(D)consisting of functions which
are real and have a positive lower bound. Our goal is then to determine whether
is injective, and invertif this is the case. This we are yet unable to do and
is, as far as we know, an open problem. However we shall show thatd|γ=const is indeed injective, that is, the linearized problem has an affirmative answer. If d|γ=const, which is a linear operator, had a closer range, one could conclude thatitself is injective in a sufficiently small neighborhood ofγ =const. But the range of dis not closed and the desired conclusion cannot be obtained in
this fashion. Nevertheless, as we shall see below, ifγ is sufficiently close to a
constant, it is nearly determined by Qγ and in some cases it can be calculated
with an error much smaller thanγ −constL∞.
To show this let us first obtain an expression for the solution of the equation
LetW =u+v, whereu =L1u=0,u|d D=φ. Then
Lγ(W)= L1+γ(u+v)= L1v+Lγv+Lγu=0
Sinceu|d D =W|d Dwe havev|d D=0 andv∈ H01(D), the closure in H1(Rn) of functions ofC∞with support in D. ButL1, as an operator fromH01(D)into H−1(D), has a bounded inverseG, and from the last expression we obtain
v+G Lδv = −G Lδu.
and
v= −
∞
0
(−1)j(G Lδ)j
(G Lδu) (1)
Since for W ∈ H01(D), LδWH−1(D) ≤ δL∞W
H01(D) if A denotes the
norm ofG, the series above will converge forδL∞A<1 and
vH−1(D)≤ A
δL∞φ
1−AδL∞
(2)
From (1) it follows thatφ is analytic atγ =1. The same argument would show thatφis analytic at any other pointγ.
Next let us calculatedφ|γ=1. We have
Q1+δ(φ) =
D
(1+δ)|∇W|2d x =
D
[(1+δ)|∇u|2+2(∇u· ∇u)
+2δ(∇u· ∇v)+(1+δ)|∇v|2]d x
(3)
The contribution of the second term in the integrand of the last integral vanishes on account of the fact thatu=0. Furthermore, from (1) one sees readily that the parts linear inδ of the last two terms in the integrand vanish. Thus setting
dγ =δwe obtain
d Qγ(φ)|γ=1=
D
δ|∇u|2d x, u=0, u|d D =φ
To show thatd Qγ(φ)|γ=1 is injective, we merely have to show that if the last integral vanishes for alluwithu =0 thenδ=0. But if the integral vanishes for all suchu, then we also have
D
wheneveru1=u2=0 in D. Now letZ be any vector inRn andaanother vector such that|a| = |Z|, (a.Z)=0. Then the functions
u1(x) = eπi(Z.x)+π(a.x), u2 = ei(Z.x)−(a.x). (5) are harmonic, and substituting in (3) we obtain
2π|Z|2
D
δ(x)e2πi(Z.x)d x = 0, ∀Z whence it follows thatδ=0.
Now let us return toQγ(W). We set againγ =1+δand introduce the bilinear
form
B(φ1, φ2) = 1 2
Qγ(W1+W2)−Qγ(W1)−Qγ(W2) and settingWj =uj +vj, j =1,2, uj =0, uj|d D=φj we obtain
B(φ1, φ2) =
D
(1+δ)(∇u1.∇u2)+δ(∇u1.∇v2)+(∇u2.∇v1) + (1+δ)(∇v1.∇v2)d x
Now, substitution of the exponentials in (5) foru1 and u2 in the preceding expression (takingato be a function of Zsuch that|a| = |Z|, (a.Z)=0) yields
ˆ
γ (Z) = ˆF(Z)+R(Z) (6)
where γ (ˆ Z) is a Fourier transform of γ extended to be zero outside D, the function
ˆ
F(Z) = 1 2π2|Z|2 B
eiπ(Z.x)+π(a.x),eiπ(Z.x)−π(a.x)
is known and, as follows readily from (2),
|R(Z)| ≤ Cδ2L∞e2πr|Z| (7)
provided thatAδL∞ ≤1−ε, whereCdepends only on Dandε, andr is the
radius of the smallest sphere containing D. Now R(Z) is too large to permit
estimatingγ (x). However, under favorable circumstances it is still possible to
obtain satisfactory information aboutγ. Chooseα, 1< α <2, then for
|Z| ≤ 2−α 2πr log
1 δL∞
we have|R(Z)| ≤ Cδα
L∞. Letηbe a function such thatηˆ ∈ C
∞, suppˆ
η ⊂
{|Z| ≤1} ˆη(0)=1, and letησ =σnη(σZ). Then we have
ˆ
γ (Z)ηˆ Z
σ
= ˆF(Z)ηˆ Z
σ
+ R(Z)ηˆ Z
σ
and
(γ ∗ησ)(x) = (F∗ησ)(x) + ρ(x) (9)
where∗denotes convolution and
|ρ(Z)| ≤ Cδα
L∞
η(ˆ Z
σ)
d Z = C1δαL∞
log 1 δL∞
n
whereC1depends only on D,αeε.
Thus ifδL∞is sufficiently small, (9) gives an approximation forγ∗ησ with
an error which is much smaller thanδL∞. Clearly, ifδL∞ is small thenσ is
large andγ ∗ησ is itself in some sense, a good approximation toγ.
Approximations to the functionγ itself may be obtained if one assumes that γ, extended to be equal to 1 outsideD, is inCm,m >n. In this case one obtains
ˆ
δ (Z) = ˆF1(Z) + ˆR(Z)
whereF1is known and R(Z)is the same as in (6). One then calculatesδ (x)by integrating over|Z| ≤ σ withσ as in (8) and estimates the error by using the
decay ofδˆat∞. Thus one obtains
γ (x) = F2(x) + σ (x) whereF2(x)is known and
|ρ (x)| ≤CδαL∞
log 1 δL∞
n
+C M
log 1 δL∞
m+n
whereMis a bound for the derivatives of ordermof γ.
REFERENCES
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[2] J.R. Cannon, Determination of an unknown coefficient in a parabolic differential equation.
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[3] J.R. Cannon, Determination of the unknown coefficientk(u)in the equationV·k(u)∇u=0 from overspecified boundary data.J. Math. Anal. Appl.,18(1967), 112–114.
[4] J.R.Cannon and Paul Du Chateau, Determining unknown coefficients in a non-linear heat conduction problem.SIAM J. Appl. Math.,24(3) (1973), 298–314.
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