Identification of nonlinearity in conductivity equation via Dirichlet-to-Neumann map

15 Pages·0.16 MB·English

by  Hyeonbae Kang

Identification of Nonlinearity in Conductivity Equation
via Dirichlet-to-Neumann Map
∗
2
0
0 Hyeonbae Kang† Gen Nakamura‡
2
School of Mathematical Sciences Department of Mathematics

n Seoul National University Hokkaido University
a
J Seoul 151-747, Korea Sapporo 060-0810, Japan

9 [email protected] [email protected]

1 February 7, 2008
v
2
2
0
1
Abstract
0
2
We prove that the linear term and quadratic nonlinear term entering a
0
/ nonlinear elliptic equation of divergence type can be uniquely identified by
h
the Dirichlet to Neuman map. The unique identifiability is proved using the
p
complex geometrical optics solutions and singular solutions.
-
h
t Mathematics subject classification (MSC2000): 35R30
a
m Keywords: Inverseconductivityproblem,nonlinearequations,elasticitytensor,DirichlettoNeuman
: map, complex geometrical optics solutions, singular solutions
v
i
X
1 Introduction
r
a
Let Ω be a bounded domain in Rn (n 2) with a C2 boundary ∂Ω. We consider
≥
the following nonlinear boundary value problem:
C(x, u) = 0 in Ω,
(1.1) ∇· ∇
u =f.
∂Ω
(cid:26) |
The nonlinear function C takes the following form:
(1.2) C(x,q) = γq+Q(x,q) := γq+P(x,q)+R(x,q), x Ω, q Rn (or Cn).
∈ ∈
∗ThisworkwascompletedwhilebothauthorswerevisitingtheMathematicalSciencesResearch
Institute(MSRI).Wewouldliketothanktheinstituteforpartialsupportandprovidingstimulating
environment for theresearch.
†partly suppoted byKOSEF98-0701-03-5 and 2000-1-10300-001-1.
‡partly supported byGrant-in-Aid for Scientific Research (C)(No.12640153).
1
Here γ C2(Ω), and there exists C > 0 such that
0
∈
(1.3) γ(x) C x Ω.
0
≥ ∈
P(x,q) represents the quadratic nonlinearity:
(1.4) P(x,q) = c1 (x)q q , , cn(x)q q .
 kl k l ··· kl k l
1≤k≤l≤n 1≤k≤l≤n
X X
 
We suppose that ci C1(Ω) are real valued functions and
kl ∈
(1.5) γ , ci C (1 i n,1 k l n)
k kC1(Ω) k klkC1(Ω) ≤ 1 ≤ ≤ ≤ ≤ ≤
for some C > 0. R(x,q) C2(Ω H) with H := q Rn or Cn ; q h (h > 0)
1
∈ × { ∈ | |≤ }
represents the higher order nonlinearity, namely, it satisfies
∂ R(x,q) C q 3,
| xj | ≤ 2| |
∂ ∂ R(x,q) C q 2,
(1.6)  | xj qk | ≤ 2| |
∂ ∂ R(x,q) C q 1
 | qj qk | ≤ 2| |
(1 j,k n,x Ω,q H)
≤ ≤ ∈ ∈


for some C > 0 independent of x and q.
2
Thefollowingtheorem isprobablywell-known. However, sincewearenotableto
finda proper reference, we give a short proof of it based on the contraction mapping
principle in the Appendix at the end of this paper. For the elasticity equation, a
proof using the implicit function theorem can be found in [4].
Theorem 1.1 Let n < p < . There exist ǫ and δ < h/2 such that for any
∞
f W2−1/p,p(∂Ω) satisfying f < ǫ, (1.1) admits a unique solution u
∈ k kW2−1/p,p(∂Ω)
such that u < δ. Moreover there exists C > 0 independent of f such that
W2,p(Ω)
k k
(1.7) u C f .
k kW2,p(Ω) ≤ k kW2−1/p,p(∂Ω)
We definetheDirichlet-to-Neumann (DN) mapΛ (f)forf with f
C k kW2−1/p,p(∂Ω)
< ǫ to be
(1.8) Λ (f) := C(x, u) ν W1−1/p,p(∂Ω)
C ∂Ω
∇ | · ∈
where u is the unique solution of (1.1) such that u < δ.
W2,p(Ω)
k k
In this paper we consider the inverse boundary value problem to identify C by
means of the DN map Λ . We are particularly interested in finding the linear term
C
γ and the second order nonlinearity P(x,q).
Thisinverse problem has interest in it’s own right as the oneto findconductivity
distribution when the conductivity varies depending on the currents. It may also
2
considered as a simplified model for the nonlinear elasticity equation. In elasticity
γ(x) corresponds to the Lam´e moduli and cl (x) can be thought as the third order
kl
elasticity tensor. In acousto-elasticity, this higher elasticity tensor is important [5].
We obtain the following unique identifiability theorem for γ and P.
Theorem 1.2 Suppose that n 3. Let C(j)(x,q) = γ q + Q(j), j = 1,2, where
j
≥
Q(j) = P(j)(x,q)+R(j)(x,q),
P(j)(x,q) = c(j)1(x)q q , , c(j)n(x)q q ,
 kl k l ··· kl k l
1≤k≤l≤n 1≤k≤l≤n
X X
 
andR(j) satisfy(1.3), (1.5), and(1.6), respectively, forthesameconstants C ,C ,C
0 1 2
and h. If
(1.9) Λ (f)= Λ (f) for all complex-valued small f W2−1/p,p(∂Ω),
C(1) C(2) ∈
then
(1)j (2)j
(1.10) γ (x) = γ (x) and c (x) = c (x), x Ω
1 2 kl kl ∈
for 1 j n and 1 k l n.
≤ ≤ ≤ ≤ ≤
The dimensional restriction n 3 is imposed since we are using the complex
≥
geometrical optics solutions of Sylvester-Uhlmann [13] to prove the Theorem.
In most models of nonlinear elasticity, the higher order tensors do not depend
j
on x. So we consider the same inverse problem when the coefficients c are con-
kl
stants. In this case we obtain the following uniqueness theorem including the two
dimensions. One thing to be noted is the condition (1.9) (and (1.11) below). Since
the equation considered in this paper is nonlinear, Λ (f) = Λ (f) for all real-
C(1) C(2)
valued small f W2−1/p,p(∂Ω) does not imply the same for complex-valued small
∈
f W2−1/p,p(∂Ω). Therefore, in the situation where the data Λ (f) for only real-
C
∈
valuedsmallf areavailable, Theorem1.2maynotbeapplied. Thefollowingtheorem
uses only real-valued Dirichlet data.
Theorem 1.3 Suppose that n 2. Let C(j)(x,q) as in Theorem 1.2 except that the
≥
(j)i
coefficients c are constants. If
kl
(1.11) Λ (f) = Λ (f) for all real-valued small f W2−1/p,p(∂Ω),
C(1) C(2) ∈
then
(1)j (2)j
(1.12) γ (x)= γ (x) and c = c
1 2 kl kl
for 1 j n and 1 k l n.
≤ ≤ ≤ ≤ ≤
3
There have been some related works on the identification of nonlinear terms en-
tering partial differential equations. Sun proved the global uniqueness for the non-
linear conductivity γ(x,u) when the Dirichlet data are complex valued and small
[11]. Nakamura and Sun considered the similar problem for nonlinear elasticity
model of St.Venant-Kirchhoff [10]. They proved that even in the presence of non-
linearity, the linear term can be identified uniquely by means of DN map. Sun and
Uhlmannprovedtheglobaluniquenessuptodiffeomorphismfixingtheboundaryfor
the two dimensional nonlinear anisotropic conductivity A(x,u), and also the same
resultfor the3dimensionalanalytic conductivity [12]. Quiterecently, Isakov proved
the unique identifiability of the nonlinearity c(u,p) entering the quasilinear elliptic
equation ∆u+ c(u, u) = 0 (and corresponding parabolic equation) by means of
∇
DN map on ∂Ω [7]. G. Uhlmann informed us that Hervas and Sun proved a result
related to ours in the two dimensional case [6]. They proved that the constant co-
efficients nonlinear terms with extra symmetry can be identified from the DN map
defined for complex valued small Dirichlet data.
Theorem 1.2 and Theorem 1.3 are proved by investigating the first and second
terms in the asymptotic expansion of Λ near 0. The first term is Λ , the DN map
C γ
corresponding to the conductivity γ. This fact was also observed in [10] and [7].
We derive certain cubic relation from the second term. This is included in Section
2. Then using the complex geometrical optics solutions of Sylvester-Uhlmann [13],
we prove Theorem 1.2 in Section 3. We prove Theorem 1.3 using the singular
solutions of Alessandrini [2]. Since the coefficients to be determined are constants,
boundary determination is sufficient. This is the reason why the singular solutions
are effectively used. The proof is included in Section 4. Appendix is to prove
Theorem 1.1.
2 Asymptotics of DN map
Throughout this paper denotes the Wk,p-norm on Ω, and = . Also,
k,p p 0,p
k k k k k k
C > 0 denotes the general constant in estimate independent of the functions being
estimated. We will use the following estimates repeatedly: if k 1, p > n, and
≥
n = 2,3, then
(2.1) uv C u v .
k,p k,p k,p
k k ≤ k k k k
This inequality holds by the Sobolev embedding theorem (see [1]). We also use the
chain rule for the derivative of the composition R(x, u(x)) for u W2,p(Ω) and its
∇ ∈
estimate given in Chapter II, section 3 of [14].
Suppose that f W2−1/p,p(∂Ω), p > n, and let t R1 be a small parameter.
∈ ∈
Let u(t) be the solution of
C(x, u) = 0 in Ω,
∇· ∇
(u∂Ω = tf
|
4
such that u(t) δ. Then there exists a constant C > 0 such that
2,p
k k ≤
(2.2) u(t) Ct f .
k k2,p ≤ k kW2−1/p,p(∂Ω)
Let u = u (f) W2,p(Ω) be the solution of
1 1
∈
(γ u ) = 0 in Ω,
1
∇· ∇
(u1 ∂Ω = f.
|
Then by the regularity of the Dirichlet problem for elliptic equations, there exist a
constant C > 0 such that we have
(2.3) u (f) C f
k 1 k2,p ≤ k kW2−1/p,p(∂Ω)
for any f W2−1/p,p(∂Ω). Let u be the solution of
2
∈
(γ u ) = P(x, u ) in Ω,
2 1
(2.4) ∇· ∇ −∇· ∇
(u2 ∂Ω = 0.
|
It then follows from the regularity of elliptic equations, (2.1), and (2.3) that
u C P(x, u )
2 2,p 1 p
k k ≤ k∇· ∇ k
C u 2
≤ k∇ 1k1,p
C f 2 .
≤ k kW2−1/p,p(∂Ω)
We now define v(t) by
(2.5) u(t) = t(u +tv(t)).
1
Then, we have from (2.2) and (2.3)
t2 v(t) u(t) +t u
2,p 2,p 1 2,p
k k ≤ k k k k
Ct f .
≤ k kW2−1/p,p(∂Ω)
Lemma 2.1 There existt and C depending on f, not ont, suchthat forall t < t ,
0 f 0
(2.6) v(t) u C t.
2 2,p f
k − k ≤
In particular, we have
(2.7) v(t) C .
2,p f
k k ≤
5
Proof. Put w(t) := v(t) u . Then a straight-forward computation shows that
2
−
γ w(t) = tg(t) in Ω,
∇· ∇ −
(w(t) ∂Ω = 0
|
where
g(t) = t−1 P(x, u +t v(t)) P(x, u ) +t−3 R(x,t u +t2 v(t)).
1 1 1
∇· ∇ ∇ − ∇ ∇· ∇ ∇
h i
It follows from (1.4), (1.6), and (2.1) that
g(t) C( u v(t) +t v(t) 2 + u 3 +t3 v(t) 3 ).
k kp ≤ k 1k2,pk k2,p k k2,p k 1k2,p k k2,p
It then follows from (2) that
v(t) u = w(t) Ct g(t)
2 2,p 2,p p
k − k k k ≤ k k
C(t v(t) +t).
2,p
≤ k k
Thus, if t is so small that Ct < 1, then by (2) we obtain
0 0 2
v(t) 2( u +1) C.
2,p 2 2,p
k k ≤ k k ≤
Then (2.6) follows from (2). This completes the proof. (cid:3)
Lemma 2.2 For a given f W2−1/p,p(∂Ω), we have
∈
1
(1) lim [Λ (tf) tΛ (f)]= 0,
C γ
t→0 t −
1
(2) lim [Λ (tf) tΛ (f)]= ν [γ u +P(x, u )].
t→0 t2 C − γ · ∇ 2 ∇ 1
The convergence is in the topology of W1−1/p,p(∂Ω).
Proof. From (2.5) we have
Λ (tf) tΛ (f) = ν [C(x, u(t)) tγ u ]
C γ 1
− · ∇ − ∇
= ν [t2γ v(t) +Q(x, u(t))].
· ∇ ∇
It follows from the trace theorem [1], (2.2) and (2.7) that
t−1(Λ (tf) tΛ (f)) tγ v(t) +t−1Q(x, u(t))
k C − γ kW1−1/p,p(∂Ω) ≤ k ∇ ∇ k1,p
Ct.
≤
This proves (1).
Since
Q(x, u(t)) = t2P(x, u )+t3O( u v(t) + v(t) 2+ u 3+ v(t) 3),
1 1 1
∇ ∇ |∇ ||∇ | |∇ | |∇ | |∇ |
(2) follows from (2.6). This completes the proof. (cid:3)
6
3 Proof of Theorem 1.2
Lemma 2.2 says that we can recover Λ and ν [γ u +P(x, u )] on ∂Ω from Λ .
γ 2 1 C
· ∇ ∇
If Λ = Λ , then Λ = Λ . It then follows from well-known results ([3],
C(1) C(2) γ1 γ2
[8, 9], [13]) that γ = γ in Ω. Put γ = γ = γ . By Lemma 2.2 (2), we have
1 2 1 2
(3.1) ν [γ u(1) +P(1)(x, u )] = ν [γ u(2) +P(2)(x, u )] on ∂Ω
· ∇ 2 ∇ 1 · ∇ 2 ∇ 1
where u W2,p(Ω) is a solution of (γ u ) = 0 in Ω, and u(j) is the solution of
1 ∈ ∇· ∇ 1 2
(2.4) when P = P(j).
Let v W2,p(Ω) be a solution of (γ v) = 0 in Ω with v = g. Since
∂Ω
∈ ∇ · ∇ |
(j)
u = 0, it follows from the divergence theorem that
2 |∂Ω
ν [γ u(j)+P(j)(x, u )]gdσ = [γ u(j)+P(j)(x, u )] vdx
· ∇ 2 ∇ 1 ∇ 2 ∇ 1 ·∇
Z∂Ω ZΩ
= P(j)(x, u ) vdx.
1
∇ ·∇
ZΩ
We thus have
(3.2) P(1)(x, u ) vdx = P(2)(x, u ) vdx,
1 1
∇ ·∇ ∇ ·∇
ZΩ ZΩ
in other words,
n n
∂u ∂u ∂v ∂u ∂u ∂v
(1)i 1 1 (2)i 1 1
(3.3) c (x) dx = c (x) dx
jl ∂x ∂x ∂x jl ∂x ∂x ∂x
i=11≤j≤l≤nZΩ j l i i=11≤j≤l≤nZΩ j l i
X X X X
for all u ,v W2,p(Ω), solutions of γ u= 0 in Ω. Therefore, it suffices to prove
1
∈ ∇· ∇
that if
n
∂u ∂u ∂v
(3.4) ci (x) 1 1 dx = 0
jl ∂x ∂x ∂x
i=11≤j≤l≤nZΩ j l i
X X
for all u ,v W2,p(Ω), solutions of γ u= 0, then
1
∈ ∇· ∇
(3.5) ci 0, i= 1, ,n, 1 j l n.
jl ≡ ··· ≤ ≤ ≤
Let u W2,p(Ω) be another solution of γ u = 0. Then u +u is also a
2 1 2
∈ ∇· ∇
solution of the equation, and hence we obtain by polarizing (3.4) that
n
∂u ∂u ∂u ∂u ∂v
(3.6) ci (x) 1 2 + 1 2 dx = 0
jl ∂x ∂x ∂x ∂x ∂x
i=11≤j≤l≤nZΩ (cid:18) j l l j(cid:19) i
X X
7
for all u ,u ,v W2,p(Ω), solutions of γ u= 0. The arguments we made so far
1 2
∈ ∇· ∇
are true for n 2.
≥
Let n = 3 and use complex geometrical optics solutions. Let k R3 and choose
∈
unit vectors ξ and η in R3 so that
(3.7) k ξ = k η = ξ η = 0.
· · ·
Then choose t,s > 0 so that
k 2
(3.8) t2 = | | +s2.
4
Define ρ(1),ρ(2) C3 by
∈
k k
(3.9) ρ(1) = tη+i( +sξ), ρ(2) = tη+i( sξ)
2 − 2 −
Then by the fundamental work of Sylvester-Uhlmann [13] there exist solutions u ,
j
j = 1,2, of γ u= 0 of the form
∇· ∇
(3.10) u (x) = γ−1/2eρ(j)·x(1+ψ (x,ρ(j))), x Ω
j j
∈
where ψ satisfies
j
C
(3.11) ψ and ψ C
k jkL∞(Ω) ≤ s k∇ jkL∞(Ω) ≤
| |
for some C independent of t. We apply these solutions u to (3.6) and obtain
j
3
∂v
γ(x)−1ci (x) (x)eik·x
jl ∂x ×
i=11≤j≤l≤3ZΩ i
X X
(γ1/2∂ γ−1/2(1+ψ )+ρ(1)(1+ψ )+∂ ψ )
j 1 j 1 j 1 ×
h
(3.12) (γ1/2∂ γ−1/2(1+ψ )+ρ(2)(1+ψ )+∂ ψ )
l 2 l 2 l 2
+(γ1/2∂ γ−1/2(1+ψ )+ρ(1)(1+ψ )+∂ ψ )
l 1 l 1 l 1 ×
(γ1/2∂ γ−1/2(1+ψ +ρ(2)(1+ψ )+∂ ψ ) dx
j 2 j 2 j 2
= 0. i
Here ∂ = ∂/∂x . Set ζ := η+iξ and note that
j j
ρ(1) ρ(2)
lim = ζ, lim = ζ.
s→∞ s s→∞ s −
8
Therefore, by dividing the both sides of (3.12) by t, taking the limit s , and
→ ∞
using (3.11), we get
3
∂v
(3.13) ζ ζ ci (x)γ(x)−1 (x)eik·xdx = 0.
j l jl ∂x
ZΩ i=11≤j≤l≤3 i
X X
Since (3.13) holds for all k R3, we have
∈
3
∂v
(3.14) ζ ζ ci (x) (x) = 0, x Ω.
 j l jl  ∂x ∈
i
i=1 1≤j≤l≤3
X X
 
We need the following lemma the proof of which will be given at the end of this
section.
Lemma 3.1 Suppose n 2. There exist solutions v W2,p(Ω), j = 1, ,n, of
j
≥ ∈ ···
γ v = 0 in Ω such that
j
∇· ∇
∂v
j
(3.15) det (x) = 0, for almost all x Ω.
∂x 6 ∈
(cid:18) i (cid:19)
Lemma 3.1 and (3.14) yield
(3.16) ζ ζ ci (x) = 0, x Ω, i= 1,2,3.
j l jl ∈
1≤j≤l≤n
X
Note that (3.16) holds for all ζ in the set := ζ C3 ζ ζ = 0, ζ = √2 . If
V { ∈ | · | | }
ζ = (0,z,√ 1z) for z C with z = 1, then ζ . With this ζ, (3.16) becomes
− ∈ | | ∈ V
(3.17) ci (x) ci (x)+√ 1ci (x) = 0, x Ω.
22 − 33 − 23 ∈
Since ci are real, we have ci (x) = ci (x) and ci = 0. By substituting ζ =
jl 22 33 23
(z,0,√ 1z) and ζ = (z,√ 1z,0) in order into (3.16), we obtain that
− −
(3.18) ci = ci = ci and ci = 0 (j = l).
11 22 33 jl 6
Because of (3.18), (3.6) now takes the form
3 3
∂u ∂u ∂v
(3.19) ci (x) 1 2 dx = 0
jj ∂x ∂x ∂x
i=1 j=1ZΩ j j i
XX
for all uu ,v. By the same argument as above (using u and v), we can conclude
, 2 2
that
3 3
∂u
(3.20) ζ ζ ci (x) 1(x) = 0
j i jj ∂x
" # j
j=1 i=1
X X
9
for all u and ζ . It then follows from Lemma 3.1 that
1
∈V
3
(3.21) ζ ζ ci (x) = 0, j = 1,2,3,
j i jj
i=1
X
and we conclude that ci = 0, i,j = 1,2,3. The proof is complete. (cid:3)
jj
Proof of Lemma 3.1. We assume that n = 3. The same proof works for the two
dimensional case since complex geometrical optics solutions exist in two dimensions.
Choose ζ(1),ζ(2),ζ(3) so that they are linearly independent over C, e.g., ζ(1) =
∈ V
(1,√ 1,0), ζ(2) = (1,0,√ 1), ζ(3) = (0,1,√ 1). Let
− − −
v (x) = γ−1/2etζ(j)·x(1+ψ (x,tζ(j))), x Ω, j = 1,2,3
j j
∈
as before. Then, we have
v ζ(1)(1+ψ )+O(t−1)
1 1
det ∇v = γ−3/2et(ζ(1)+ζ(2)+ζ(3))·xt3det ζ(2)(1+ψ )+O(t−1) .
2 2
∇   
v ζ(3)(1+ψ )+O(t−1)
3 3
∇
   
It then follows from (3.11) that if t is sufficiently large, then
v (x)
1
∇
det v (x) = 0.
2
∇  6
v (x)
3
∇
 
This finishes the proof. (cid:3)
4 Proof of Theorem 1.3
We need to prove that if
n
∂u ∂u ∂v
(4.1) ci dx = 0
kl ∂x ∂x ∂x
i=11≤k≤l≤n ZΩ k l i
X X
for all u,v W2,p(Ω) real solutions of γ u= 0, then
∈ ∇· ∇
(4.2) ci = 0, i = 1, ,n, 1 k l n.
kl ··· ≤ ≤ ≤
Since (4.1) holds only for real-valued solutions u,v, we can not use the complex
geometrical optics solutions. However, since the coefficients ci are assumed to be
kl
constants, we can use instead the singular solutions of Alessandrini. We only deal
with the three dimensional case. Two dimensional case can be proved in the same
way.
10