Hindawi Publishing Corporation
Boundary Value Problems
Volume 2009, Article ID 305291, 22 pages
doi:10.1155/2009/305291
Research Article
The Inverse Problem for Elliptic Equations from
Dirichlet to Neumann Map in
Multiply Connected Domains
Guochun Wen,1 Zuoliang Xu,2 and Fengmin Yang2
1
2
School of Mathematical Sciences, Peking University, Beijing 100871, China
School of Information, Renmin University of China, Beijing 100872, China
Correspondence should be addressed to Guochun Wen, wengc@math.pku.edu.cn
Received 10 July 2008; Accepted 5 January 2009
Recommended by Pavel Drabek
The present paper deals with the inverse problem for linear elliptic equations of second order
from Dirichlet to Neumann map in multiply connected domains. Firstly the formulation and
the complex form of the problem for the equations are given, and then the existence and global
uniqueness of solutions for the above problem are proved by the complex analytic method, where
we absorb the advantage of the methods in previous works and give some improvement and
development.
Copyright q 2009 Guochun Wen et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
1. Formulation of the Inverse Problem for Second-Order
Elliptic Equations from Dirichlet to Neumann Map
In 1–9, the authors posed and discussed the inverse problem of second-order elliptic
equations. In this paper, by using the complex analytic method, the corresponding problem
for linear elliptic complex equations of first-order in multiply connected domains is firstly
discussed, afterwards the existence and global uniqueness of solutions of the inverse problem
for the elliptic equations of second-order are obtained.
Let G be an N 1-connected domain bounded domain in the complex plane C with
2
the boundary ∂G L ∪N
j0 Lj ∈ Cμ 0 < μ < 1, where Lj j 1, . . . , N are inside of L0 .
Consider the linear elliptic equation of second-order:
uξξ uηη auξ buη 0 in G,
1.1
2
Boundary Value Problems
in which a aζ, b bζ are real functions of ζ ξ iη, and aζ, bζ ∈ Lp G, p> 2 is a
positive constant. Moreover let a b 0 in C \ G. The above condition is called Condition C.
In this paper the notations are the same as those in 10 or 11.
Denote
uξ − iuη
uζ ,
Wζ U iV 2
1.2
Wξ iWη
uξξ uηη
Wζ uζζ in G,
2
4
we can get
uζζ Wζ
1
Wξ iWη
2
1
− auξ buη
4
1 − a W W ib W − W
4
1
− a ibW a − ibW
4
1.3
−AζW − BζW
−2Re AζW
in G,
where A Aζ Bζ B a ib/4. We choose a conformal mapping z zζ from
the above general domain G onto the circular domain D with the boundary Γ ∪N
j0 Γj , Γ0 ΓN1 {|z| 1}, Γj {|z − zj | rj }, j 1, . . . , N, and z 0 ∈ D. In this case, the complex
equation 1.3 is reduced to the complex equation
Wz −ζ z A ζz W B ζz W ,
in D,
wz uzz −2Re A ζz ζ zuz −2Re A ζz Jzw
1.4
where uζζ uzz |z ζ|2 , Wζ uζ uz z ζ, wz uz , ζ ζz is the inverse function
of z zζ, and ζ z 1/z ζ Jz in D is a known Hölder continuously differentiable
function see 10, Section 2, Chapter I, hence the above requirement can be realized.
Introduce the Dirichlet boundary condition for 1.1 as follows:
u fζ on L ∂G,
u f ζz on Γ zL,
1.5
where fζ ∈ Cα1 L, fζz ∈ Cα1 Γ, α≤ p − 2/p is a positive constant, which is called
Problem D for 1.1 or 1.4. By 10, 11, Problem D has a unique solution u ∈ Wp2 G
or Wp2 D satisfying 1.1 or 1.4 and the Dirichlet boundary condition 1.5. From this
solution, we can define the Dirichlet to Neumann map Λ : Cα1 L → Cα L or Λ : Cα1 Γ →
Cα Γ by Λf ∂u/∂n.
Boundary Value Problems
3
Our inverse problem is to determine the coefficient a and b of 1.1 or Aζ in 1.3
from the map Λ. In the following, we will transform the Dirichlet to Neumann map Λ into a
equivalent boundary condition. In fact, if we find the derivative of positive tangent direction
with respect to the unit arc length parameter s arg zz ∈ Γ0 and s − argz − zj z ∈
Γj , j 1, . . . , N of the boundary Γ with s0 arg z arg1 0 0, then
∂f ζz
fs ∂s
⎧
uz zs uz zs uz iz − uz iz 2Re izuz ,
⎪
⎪
⎪
⎪
⎪
⎪
⎨uz z − zj uz z − zj
s
s
⎪
⎪
−uz i z − zj uz i z − zj
⎪
⎪
⎪
⎪
⎩
−2Re i z − zj uz ,
on Γ0 ,
1.6
on Γj , j 1, . . . , N.
It is clear that the equivalent boundary value problem is to find a solution Wζz, uζz
of the complex equation 1.4 with the boundary conditions
⎧
fs
⎪
⎪
z ∈ Γ0 ,
⎨Re izwz 2 ,
Re λzwz ⎪
⎪
⎩Re iz − zj wz − fs , z ∈ Γj , j 1, . . . , N,
2
u1 f ζ1 b0 ,
1.7
and the relation
z
uz 2Re wzdz b0
in D,
1.8
1
in which λz iz, z ∈ Γ0 and λz iz − zj , z ∈ Γj , j 1, . . . , N. It is easy to see that
2Re Wζdζ 2Re uz dz
Γj
Lj
Sj
−2Re i z − zj uz rj ds
1.9
0
Sj
fs rj ds 0,
0
where Sj 2πrj j 1, . . . , N is the arc length of Γj {|z − zj | rj } j 1, . . . , N and
applying the Green formula, we can see that the function uz determined by the integral in
1.8 in D is single-valued.
4
Boundary Value Problems
Under the above condition, the corresponding Neumann boundary condition is
∂u
uz zn uz zn
∂n
⎧
⎨uz zn uz zn uz zuz z 2 Im izuz
gz ⎩−u z − z − u z − z −2Imiz − z u z
j
z
j
j
z
un on Γ0 ,
1.10
on Γj , j 1, . . . , N,
where n is the unit outwards normal vector of Γ. The boundary value problem 1.1
or 1.4, 1.10 will be called Problem N. Taking into account the partial indexes of
K0 ΔΓ0 argλz ΔΓ0 arg iz and ΔΓ0 arg z are equal to −1 and Kj ΔΓj argλz ΔΓj arg iz − zj and ΔΓj arg z − zj j 1, . . . , N are equal to 1, thus the index of the above
boundary value problem is K K0 K1 · · · KN N − 1. In general the above Problem N
is not solvable, we need to give the modified boundary conditions as follows:
gz
1
un Re λzuz g0 , z ∈ Γj , j 0, 1, . . . , N,
2
2
u1 b0 on Γ,
1.11
where λz z, z ∈ Γ0 , and λz z − zj , z ∈ Γj , j 1, . . . , N, gz ∈ Cα Γ and g0 0 on
Γj j 1, . . . , N, g0 on Γ0 is an undetermined real constant see 11, Chapter VI. Hence,
the Dirichlet to Neumann map can be transformed into the boundary conditions as follows:
⎧
⎨2Re izuz 2iIm izuz 2izwz, z ∈ Γ0 ,
us iun ⎩−2iz − z wz,
z ∈ Γj , j 1, . . . , N,
j
⎧
us iun
⎪
⎪
,
z ∈ Γ0 ,
⎪
⎨
2iz
wz hz ⎪
us iun
⎪
⎪
, z ∈ Γj , j 1, . . . , N,
⎩− 2i z − zj
1.12
which will be called Problem DN for the complex equation 1.4 or 1.1 with the relation
1.8, where hz ∈ Cα Γ is a complex function satisfying the condition
Γj
Re i z − zj hz ds 0,
j 1, . . . , N.
1.13
For any function fζz ∈ Cα1 Γ in the Dirichlet boundary condition 1.5, there is a set
{gz} of the functions of Neumann boundary condition 1.10, where gz is corresponding
to the complex equation 1.4 one by one, namely if we know the boundary value fζz and
one complex equation in 1.4, then the boundary value gz can be determined. Inversely if
the gz in 1.10 is given, then one complex equation in 1.4 can be determined, which will
be verified later on. We denote the set of functions {eikz hz} by Rh , where k is a complex
number and hz is as stated in 1.12.
Boundary Value Problems
5
2. Some Relations of Inverse Problem for Second-Order
Elliptic Equations from Dirichlet to Neumann Map
According to 10, introduce the notations
T fz −
1
π
C
fζ
dσζ ,
ζ−z
2.1
in which fz ∈ Lp D, p > 2. Suppose that fz 0 in C \ D. Obviously T fz fz in C.
We consider the complex equation
gz JAg ek zBJg 0,
gz JAg ek zJAg 0
in C,
2.2
where gz eikz w, ek z eikzkz and k is a complex number. On the basis of the Pompeiu
formula see 10, Chapters I and III, the corresponding integral equation of the complex
equation 2.2 is as follows:
1
gz, k − T JAg ek JAg 2πi
gζ, k
dζ
Γ ζ−z
in D.
2.3
For simplicity we can only consider the following integral equation
gz, k − T JAg ek JAg 1
or i
2.4
in D
later on.
Lemma 2.1. If fz ∈ Lp D p > 2, then
lim max T ek f z 0.
2.5
k → ∞ z∈D
Proof. It suffices to prove that for any small positive number ε, there exists a sufficiently large
positive number N such that
T ek f z < ε
for z ∈ D, |k| ≥ N.
2.6
In fact, noting that ek z e2iRekz e2|kz|i cosφarg z , φ arg k, |ek z| 1, and using the
Hölder inequality, we have
T ek f z ≤ Lp f
≤ M1
1/q
ek ζ q
dσζ
D ζ−z
2qi|kζ| cosφarg ζ e
D
|ζ − z|q
2.7
1/q
dσζ
,
6
Boundary Value Problems
where M1 1 Lp f, 1 < q p/p − 1 < 2. Now we estimate the integral
1/q
e2qi|kζ| cosφarg ζ J0 dσ
.
ζ
D
|ζ − z|q
2.8
We choose two sufficiently small positive constants δ and η, and divide the domain D into
three parts: D1 {|ζ| ≤ δ}, D2 {D \ D1 } ∩ {| arg ζ φ| ≤ η} ∪ {| arg ζ φ − π| ≤ η}, and
D3 D \ {D1 ∪ D2 }, such that for the above positive number ε, we can get
e2qi|kζ| cosφarg ζ q J dσ
ζ
1
D1
|ζ − z|q
<
ε
3M1
q
,
J1 <
ε
,
3M1
e2qi|kζ| cosφarg ζ q J dσ
ζ
2
D2
|ζ − z|q
2.9
≤
|ζ − z|1−q d|ζ − z|dθ
D2
≤
J2 ≤
ε
3M1
q
,
ε
,
3M1
where θ arg ζ. Moreover noting that |dθ φ| |d cosθ φ/ sinθ φ| ≤ |d cosθ φ/ sin η|, if ζ ∈ D3 , and then
e2qi|kζ| cosφarg ζ q J dσζ 3
q
D3
|ζ − z|
e2qi|kζ| cosφarg ζ 1
cosφ
arg
ζ
≤ ζ
−
z|d2q|kζ
d
|ζ − z|q−1
2qk minD3 ζ| sin η| D3
1
|de2qi|kζ| cosφarg ζ |
≤
d|ζ
−
z|
q−1
2q|kδ sin η| D3
|ζ − z|
≤
ε
3M1
q
,
J3 ≤
ε
3M1
for |k| ≥ N.
2.10
Boundary Value Problems
7
Thus we obtain
T ek f z ≤ Lp fJ1 Lp fJ2 Lp fJ3 ≤ M1 J1 J2 J3 < ε
for z ∈ D, |k| ≥ N.
2.11
This shows that the formula 2.6 is true.
Lemma 2.2. If Lp A, D ≤ k0 , p > 2, where k0 is a positive constant, then the solution gz, k of
2.2 satisfies the estimate
Cα gz, k, D ≤ M2 M2 p, α, k0 , D ,
2.12
in which M2 is a positive constant.
Proof. First of all, we verify that any solution gz, k of 2.2 satisfies the boundedness
estimate
C gz, k, D ≤ M3 M3 p, α, k0 , D ,
2.13
where M3 is a positive constant. Suppose that 2.13 is not true, then there exists a sequence
of coefficients {Am z}, which satisfy the same condition of coefficient Az and weakly
converges to A0 z, and the corresponding integral equations
gmz JAm gm ek JAm gm 0
in D, m 1, 2, . . .
2.14
possess the solutions gm z, k m 1, 2, . . ., but Cgm z, k, D m 1, 2, . . . are unbounded.
Hence we can choose a subsequence of {gm z, k} denoted by {gm z, k} again, such that
hm Cgm z, k, D → ∞ as m → ∞, and can assume hm ≥ 1. Obviously gm z, k gm z, k/hm m 1, 2, . . . are solutions of the integral equations
gmz JAm gm ek JAm gm 0 in D, m 1, 2, . . . .
2.15
Noting that Lp Am gm ≤ k0 , Lp ek Am gm , D ≤ k0 , we can derive the estimate
Cα T JAm gm T ek JAm gm , D ≤ M4 M4 p, α, k0 , D ,
2.16
Cα gm , D ≤ M5 M5 p, α, k0 , D .
2.17
see 10, 11, thus
gm z, k} again, which
Hence from {
gm z, k}, we can choose a subsequence denoted by {
uniformly converges to g0 z in D, it is clear that g0 z is a solution of the equation
g0z JA0 g0 0,
or g0 z T JA0 g0 0
in D.
2.18
8
Boundary Value Problems
On the basis of the result in 10, Section 5, Chapter III, the solution g0 z 0 in D, however,
g0 z∗ , D 1, which is
from C
gm z, k, D 1, there exists a point z∗ ∈ D, such that C
impossible. This shows that 2.13 and then the estimate 2.12 are true.
Lemma 2.3. Under the above conditions, one has
lim gz, k g0 z
k→∞
in D,
2.19
where g0 z is a unique solution of the equation
g0z JAg0 0 in D.
2.20
Proof. Denote by gz, k the solution of 2.2 in D. From Lemma 2.2, we know that the
solution gz, k satisfies the estimate 2.12. Moreover by using 2.5, that is,
lim max T ek JAg z 0,
k → ∞ z∈D
2.21
we can choose subsequences {kn } and {gz, kn }, where kn → ∞ as n → ∞, such that
{gz, kn } in D uniformly converges to g0 z as n → ∞, which is a solution of 2.20 in D
see 11. The uniqueness of solutions of 2.20 can be seen from the proof of Lemma 2.4
below.
Lemma 2.4. The solution g0 z of 2.20 can be expressed as
g0 z Φze−T JA
in D,
2.22
where Φz 1 in D.
Proof. On the basis of the results as in 10, Section 5, Chapter III, we know that the integral
equations
g0 z − T JAg0 1
1
in D,
in C
2.23
have the unique solutions g0 z in D and C respectively, this shows that the function g0 z in
D can be extended in C. Moreover by the result in 10, 11, the solution g0 z can be expressed
as Wz g0 z Φze−T JA in C. Note that T JA → 0 as z → ∞, and the entire function
Φz in C satisfies the condition Φz → 1 as z → ∞, hence Φz 1 in C, and then
g0 z e−T JA in D.
Theorem 2.5. For the inverse problem of the equation
g0 z z JAg0 0
in D,
2.24
Boundary Value Problems
9
with the boundary condition
g0 z≡
/ 0
on Γ,
2.25
one can obtain
T JA − ln g0 z
on Γ,
2.26
which is a known function.
Proof. From the expression 2.22 of the solution g0 z in D and Φz 1 in D, it follows that
2.26 is true.
3. The Inverse Scattering Method for Second-Order
Elliptic Equations from Dirichlet to Neumann Map
For the complex equation 1.4, through the transformation Wz wzeT JA , we can obtain
that the function Wz satisfies the complex equation
Wz CzW 0 in C,
3.1
where C Cz BζzJzeT JA−T JA AζzJzeT JA−T JA and C Cz 0 in C \ D, in
this case every function hzeikz in RA is reduced to hzeikzT JA , hence later on it suffices to
discuss the complex equation 3.1 and system of complex equations
φjz −1j−1 Czek zφj 0 in C, j 1, 2.
3.2
where ek z eikzkz . In the following we will find two solutions φ1 z and iφ2 z of
complex equation φz Czek zφz 0 with the conditions φ1 z → 1 and iφ2 z → i as
z → ∞.
Now we find two solutions W1 z and W2 z in C of 3.1 with the conditions W1 z ∼
−ikz
and W2 z ∼ ie−ikz for sufficiently large |z|. In other words, there exist two solutions
e
φ1 z eikz W1 z and φ2 z −ieikz W2 z in C of 3.2 with the conditions φ1 z → 1 and
φ2 z → 1 as z → ∞. Denote
m1 z, k φ1 z φ2 z
,
2
m2 z, k ek z
φ2 z − φ1 z
,
2
3.3
obviously m1 z, k, m2 z, k satisfy the system of first-order complex equations
m1 z Cm2 ,
m2
z
− ikm2 Cm1 ,
ek e−k m2 z Cm1 ,
φ2 − φ1
ikm2 Cm1 ikm2 ,
m2 z ek
2
z
3.4
10
Boundary Value Problems
such that m1 z, k → 1 and m2 z, k → 0 |m2 z, k| |ek zm2 z, k| → 0 as z → ∞.
According to the way in 8, we can obtain the following two lemmas.
Lemma 3.1. Under the above conditions, there exist two functions m1 z, k, m2 z, k satisfying the
system of complex equations:
m1 z, k k T kek zm2 z, k 0,
m2 z, k k T kek zm1 z, k 0,
3.5
where
i
T k −
π
i
e−k ζCζm1 ζ, kdσζ −
2π
C
C
e−ikz Cζ W1 − iW2 dσζ .
3.6
Proof. In the following we verify the 3.5. From 3.4, we have
Cζm2
1
dσζ ,
m1 1 π C z−ζ
Cζ m2 k
1
dσζ
m1 k π C z−ζ
CT kek ζm1 ζ, k
1
dσζ
−
π C
z−ζ
−T kek zm2 z, k,
e−k Cm1
1
dσζ ,
e−k m2 π C z−ζ
ek Cm1
1
e k m2 dσζ ,
π C z−ζ
ek e−k m1 k m1k − i ζ − z z m1
m1k − i ζ − z m1 − izm1 ,
m2k ek e−k m2 k izm2
C e−k m1 k
ek z
dσζ izm2
π
C
z−ζ
Cζe−k ζm1k
1
i
dσζ ek
e−k ζCζm1 ζ, kdσζ
π C
π C
z−ζ
1
T kCζm2 ζ, k
dσζ T k
−ek
π C
z−ζ
−T kek zm1 .
3.7
Boundary Value Problems
11
In addition, from 3.5 it follows that
m1 z, k m2 z, k
m1 z, k − m2 z, k
k
−T kek z m1 z, k m2 z, k ,
k
T kek z m1 z, k − m2 z, k .
3.8
It is easy to see that
ψ1 z, k m1 z, k m2 z, k,
3.9
ψ2 z, k m1 z, k − m2 z, k
satisfy the system of complex equations
ψ1k T kek zψ1 0,
ψ2k − T kek zψ2 0
3.10
with the conditions ψ1 eikz Ψ1 ∼ 1 and ψ2 −ieikz Ψ2 ∼ 1 for sufficient large |k|, and Ψ1 e−ikz ψ1 , Ψ2 ie−ikz ψ2 are the solutions of the complex equation
Ψk T kΨ 0
for k ∈ C.
3.11
Later on we will verify T k ∈ L∞ C.
Similarly to the way from 3.2 to 3.6, we can obtain the following result.
Lemma 3.2. Under the above conditions, there exist two functions W1 z, k, W2 z, k satisfying the
system of complex equations:
W1 z, k z CzW1 z, k 0,
W2 z, k
z
CzW2 z, k 0
in C,
3.12
where
i
Cz −
π
i
−
2π
C
e−k zT k m1 z, k dσk
C
e
−ik z
T k Ψ1 z, k − iΨ2 z, k dσk .
3.13
Proof. Now we verify that 3.12 and 3.13 are true. Denote
ψ1 z, k ψ2 z, k
,
n1 2
ψ2 z, k − ψ1 z, k
n2 ek z
,
2
3.14
12
Boundary Value Problems
we see that n1 z, k, n2 z, k satisfy the system of first-order complex equations
n1 k T kn2 ,
n2 k − izn2 T kn1 ,
ek e−k n2 k T kn1 ,
ψ2 − ψ1
izn2 T kn1 izn2 ,
n2 k ek
2
3.15
k
such that n1 z, k → 1 and n2 z, k → 0 |n2 z, k| |ek zn2 z, k| → 0 as k → ∞. Thus
we have
T k n2
1
n1 1 dσk ,
π C k − k
T k n2 z
1
n1 z dσk
π C k − k
CT k ek n1 z, k
1
−
dσk
π C
k − k
−Czek zn2 z, k,
e−k T k n1
1
e−k n2 dσk ,
π C k − k
ek T k n1
1
ek n2 dσk ,
π C k − k
ek e−k n1 z n1z − i k − k k n1
n1z − i k − k n1 − ikn1 ,
n2z ek e−k n2 z ikn2
T k e−k n1 z
ek z
dσk ikn2
π
C
k − k
T k e−k n1z
1
i
ek
dσk e−k zT k n1 z, k dσk
π C
π C
k − k
CT k n2 z, k
1
− ek
dσk Cz
π C
k − k
3.16
−Czek zn1 .
In addition, from 3.12 it follows that
n1 z, k n2 z, k z Czek z n1 z, k n2 z, k 0,
n1 z, k − n2 z, k z − Czek z n1 z, k − n2 z, k 0.
3.17
Boundary Value Problems
13
It is obvious that
φ1 z, k n1 z, k n2 z, k,
φ2 z, k n1 z, k − n2 z, k
3.18
φ2z − Czek zφ2 0
3.19
satisfy the system of complex equations
φ1z Czek zφ1 0,
with the conditions φ1 eikz W1 ∼ 1 and φ2 −ieikz W2 ∼ 1 for sufficient large |z|, and
W1 e−ikz φ1 , W2 ie−ikz φ2 are the solutions of the complex equation
Wz CzWz 0 for z ∈ C.
3.20
From 3.6 and Lemma 3.3 below, the functions H1 z, k W1 z, k, H2 z, k W2 z, k on Γ can be obtained, then
i
T k −
e−ikζ C W1 − iW2 dσζ
2π C
i
e−ikζ W1 − iW2 ζ dσζ
2π D
1
−
e−ikζ W1 − iW2 dζ
4π Γ
i
νe−ikz W1 − iW2 dS.
4π Γ
3.21
Here we use the Green formula
vz dx dy −
D
1
2i
Γ
v dz 1
2
Γ
νv dS,
3.22
and for Γ0 {|z| 1}, ν z e−iθ e−i arg z , and Γj {|z − zj | rj }, ν −z − zj /rj −e−iθ −e−i argz−zj , j 1, . . . , N, dz −iν dS, S θ, z ∈ Γ0 , − dz −dz − zj iν dS, S rj θ, z ∈ Γj , j 1, . . . , N. This shows that the function T k for k ∈ C is known, and then
we can solve the solutions m1 , m2 of equations in 3.5. On the basis of Lemma 3.2, we can
obtain the system of complex equations in 3.12 and the coefficient Cz BzJzeT JA−T JA
of 3.1. This is just the so-called inverse scattering method. We mention that sometimes
W1 z, k, W2 z, k are written as W1 z, W2 z.
14
Boundary Value Problems
Lemma 3.3. Under the above conditions, the functions h1 z, h2 z as stated in 1.12 are the
solutions of the system of integral equations
1
1 − iSk h1 e−ikz ,
2
1 h1 ζeikζ−z
dζ,
Sk h1 π Γ
ζ−z
1
1 − iSk h2 ie−ikz ,
2
1 h2 ζeikζ−z
Sk h2 dζ.
π Γ
ζ−z
3.23
We first prove one lemma see 7.
Lemma 3.4. The function gz, k eikz hj z ∈ Rh , j 1, 2 is a solution of the integral equations
gz, k T JAg T ek JAg eikz h1 z
gz, k eikz h2 z
1
i
in D,
3.24
on Γ,
if and only if it is a solution of the integral equation
1
1
gz, k 2
2πi
gζ, k
dζ Γ ζ−z
h1 z
1
2
2πi
h2 z
1
2
2πi
1,
i,
gζ, k ⎧
⎨eikζ h1 ζ,
⎩eikζ h2 ζ,
h1 ζeikζ−z
dζ e−ikz ,
ζ−z
Γ
h2 ζeikζ−z
dζ ie−ikz
ζ−z
Γ
3.25
on Γ.
Proof. It is clear that we can only discuss the case of h1 . If gz, k is a solution of 3.24, then
gz −JAg − ek JAg. On the basis of the Pompeiu formula
1
gz, k 2πi
1
2πi
gζ, k
dζ − T gζ, k ζ
Γ ζ−z
gζ, k
dζ − T JAg ek JAg
ζ
−
z
Γ
3.26
in D
see 10, Chapters I and III, we have
1
gz, k T JAg T ek JAg 1 2πi
gζ, k
dζ
Γ ζ−z
in D,
3.27
Boundary Value Problems
15
where gζ, k eikζ h1 ζ on Γ. Moreover by using the Plemelj-Sokhotzki formula for Cauchy
type integral see 12, 13
1
1
2πi
gζ, k
1
dζ gz, k,
ζ
−
z
2
Γ
gζ, k eikζ h1 ζ on Γ,
3.28
which is the formula 3.25.
On the contrary if 3.25 is true, then there exists a solution of equation gz −AJg −
ek JAg in D with the boundary values gζ, k eikζ h1 ζ on Γ, thus we have 3.26, where the
integral 1/2πi Γ gζ, k/ζ − zdζ in D is analytic, whose boundary value on Γ is
1
lim
z ∈D → z∈Γ 2πi
gζ, k
1
1
dζ gz, k 2
2πi
Γ ζ−z
gζ, k
dζ 1,
Γ ζ−z
3.29
hence
1
2πi
gζ, k
dζ 1
Γ ζ−z
3.30
in D,
and the formula 3.24 is true.
Proof of Lemma 3.3. On the basis of the theory of integral equations see 12, 13, we can
obtain the solutions h1 z and h2 z of 3.23. From Lemma 3.4, we define the functions
W1 z, k W2 z, k ⎧
h1 ζeikζ−z
1
⎪
−ikz
⎪
dζ,
−
e
⎪
⎪
⎨
2πi Γ
ζ−z
z ∈ C \ D,
⎪
⎪
⎪
CζW1 ζ, k
1
⎪
−ikz
⎩e
dσζ , z ∈ D,
π C
ζ−z
3.31
⎧
h2 ζeikζ−z
1
⎪
−ikz
⎪
dζ,
−
ie
⎪
⎪
⎨
2πi Γ
ζ−z
z ∈ C \ D,
⎪
⎪
⎪
CζW2 ζ, k
1
⎪
−ikz
⎩ie
dσζ ,
π C
ζ−z
z ∈ D,
which are analytic in C \ D with the boundary values h1 z, h2 z on Γ respectively, and
satisfy the complex equation 3.1.
Moreover according to 6, 7, we can obtain the following two lemmas.
Lemma 3.5. Under the above conditions, one has
ikz
e W1 z, k − 1 1
W
p,2 Cz
≤ M1 ,
− ieikz W2 z, k − 1
1
Wp,2
Cz
≤ M1 ,
for |k| ≥ R,
3.32
16
Boundary Value Problems
where p > 2, the positive constant M1 M1 k, p, R is only dependent on k, p and R, and R is a
sufficiently large positive number. Moreover the function T k in 3.6 satisfies T k ∈ L∞ Ck. In
particular, T k ∈ Lp1 ,2 Ck, where p1 0 < p1 < ∞ is a non-negative number.
Proof. From Lemma 3.1, noting that φj z, k → 1, z → ∞, j 1, 2, we have
φ1 z, k e
ikz
1
W1 z, k 1 π
φ2 z, k −ieikz W2 z, k 1 −
1
π
Cζek ζφ1 ζ, k
dσζ ,
z−ζ
C
3.33
Cζek ζφ2 ζ, k
dσζ .
z−ζ
C
On the basis of the result in 10, we can get
φ1 z, k − 1
1
Wp,2
Cz
1
Cek ζφ1 ζ, k
dσζ π C
z−ζ
1
Wp,2
Cz
≤ M2 Cek ζφ1 ζ, kLp,2 Cz
3.34
≤ M3 CLp,2 Cz
≤ M1 k, p, R
in which |k| ≥ R and Mj Mj k, p, R j 2, 3 are positive constants only dependent on
k, p and R. Similarly, we can obtain the second estimate in 3.32.
In addition, for
T k −
i
2π
i
e−ikζ C W1 − iW2 dσζ −
2π
C
C
e−k ζC φ1 φ2 dσζ ,
3.35
we have
i
T k
≤
e
ζCφ
φ
dσ
−k
1
2
ζ
L∞ C
2π C
L∞ Ck
i
≤
e−k ζCm1 ζ, kdσζ π C
3.36
L∞ Ck
≤
1
CLp,2 Cz m1 ζ, kLq,2 Cz L∞ Ck ,
π
in which q p/p − 1, 1 < q < 2. It is not difficult to see that T k ∈ Lp1 ,2 Ck, where
p1 0 < p1 < ∞ is a non-negative constant.
Boundary Value Problems
17
Lemma 3.6. Under the above conditions, one can find the coefficients Q Qz of the complex system
of first-order equations Dk m2 m2z − ikm2 Qm1 Q C in D as follows
1
Qz lim
k0 → ∞ πr 2
1
lim
k0 → ∞ πr 2
|k−k0 |≤r
Dk m2 z, kdσk
3.37
|k−k0 |≤r
Qm1 z, kdσk ,
in which dσk dRe k dIm k.
Proof. From the formula 3.4, we can get
lim
k0 → ∞
|k−k0 |≤r
Dk m2 z, kdσk Q lim
k0 → ∞
|k−k0 |≤r
m1 z, kdσk
3.38
πr 2 Qz,
where m1 z, k → 1 as k → ∞, hence the the formula 3.37 is true.
Theorem 3.7. For the inverse problem of Problem DN for 1.3 with Condition C, one can
reconstruct the coefficients aζ and bζ.
Proof. Similarly to 9, we will use the generalized Cauchy formula
Fz 1
2πi
Γ
Ω1 z, ζFdζ −
1
2πi
Γ
Ω2 z, ζFdζ
in D,
3.39
for the complex equation
Fz e
−T AJ
z
e
−T AJ
−AJ −C
J
J
e−T AJ −CF
J
J
3.40
to find the function F Fz e−T AJ in D, in which Ω1 z, ζ, Ω2 z, ζ are the standard kernels
of equation 3.40 see 10, Chapter III. In fact, denote F e−T AJ in D, and F e−T AJ on
Γ is known from Theorem 2.5, then according to 3.39, we can find the function Fz in D.
Moreover from
− ln F
z
CFJ/FJ C
J
J
eT AJ−T AJ AJ
thus the coefficient A aζ ibζ/4 in G is obtained.
in D,
3.41
18
Boundary Value Problems
4. The Global Uniqueness Result for Inverse Problem of First-Order
Elliptic Complex Equations from Dirichlet to Neumann Map
For the elliptic equation of second-order
ujξξ ujηη aj ujξ bj ujη 0 in G, j 1, 2,
4.1
in which aj aj ζ, bj bj ζ are real functions of ζ ξ iη ∈ G, j 1, 2, and aj , bj ∈
Lp G, j 1, 2, p> 2 is a positive constant. Moreover define aj bj 0 j 1, 2 in C \ G.
Denote
ujξ − iujη
ujζ ujζ
Wj ζ Uj iVj 2
in G, j 1, 2,
4.2
and we can get
Wjξ iWjη
Wjζ 2
−
1
aj ujξ bj ujη
4
4.3
−Aj ζWj − Bj ζWj
−2Re Aj Wj
in G, j 1, 2,
where Aj Aj ζ Bj ζ aj ibj /4, j 1, 2. As stated in Section 1, suppose that the
above equations satisfy Condition C, and through a conformal mapping z zζ, the complex
equations in 4.3 can be reduced to the following form
Wjz −ζ z Aj ζz Wj Bj ζz Wj ,
wjz −2Re Aj ζz ζ zujz
−2Re Aj ζz Jzwj
in D, j 1, 2,
4.4
where D is a circular domain, and Jz ζ z.
If wj z ujz j 1, 2 are the corresponding solutions of 4.4 from the Dirichlet
to Neumann maps Λj j 1, 2, and Λ1 Λ2 Λ, then the boundary conditions of the
inverse boundary value problem for second-order elliptic equations in 4.1 from Dirichlet to
Neumann map can be reduced to
wj z ujz hz
on Γ, j 1, 2,
4.5
where hz ∈ Cα Γ, 0 < α ≤ p − 2/p is a known complex function. In the following we
will prove the uniqueness theorem as follows.
Boundary Value Problems
19
Theorem 4.1. For the inverse problem of Problem DN for 1.1 (or 1.3) with Condition C, one can
uniquely determine the coefficients a, b. In other words, if Λ1 Λ2 for 4.1, then a1 a2 , b1 b2 .
We first prove the Carleman estimate see 7.
Lemma 4.2. If the complex function uz ∈ Wp1 D with the condition uz 0 on Γ, and the real
function φz ∈ Wp2 D p > 2 then one has the Carleman estimate
2 φ
uz e dσz .
Δφ|u| e dσz ≤ 4
2 φ
D
4.6
D
Proof. It is sufficient to prove the equality
2 φ
uz uφz 2 eφ dσz 4
uz e dσz ,
Δφ|u|2 eφ dσz 4
D
D
4.7
D
in which φz ∈ Wp2 D, and uz ∈ Wp1 D with the condition uz 0 on Γ. We first consider
the complex form of the Green formula about v uz
1
4
D
1
1 uzz dx dy uz dz,
uxx uyy dx dy ux dy − uy dx 4
2i Γ
D
D
1
1
vz dx dy v dz, or
vz dx dy −
v dz,
2i
2i
D
Γ
D
Γ
4.8
with u ∈ C2 D.
If uz, φz are the above functions, by using the Green formula, we have
1
uz uz e dx dy u uz e z dx dy uuz eφ dz 0,
2i Γ
D
D
D
2 φ
φ
uz e dx dy uz uz e dx dy −
u uz eφ z dx dy
D
D
D
φ
u uzz uz φz e dx dy,
−
D
uz uφz 2 eφ dx dy uz uφz uz uφz eφ dx dy
D
D
φ
φ
u uz e uφz e z dx dy uφz uz uφz eφ dx dy
−
D
D
φ
u uzz uz φz uφzz e dx dy −
u uz uφz φz eφ dx dy
−
D
D
φ
uφz uz uφz e dx dy
D
u uzz uz φz uφzz eφ dx dy,
−
uuz eφ z dx dy φ
D
φ
4.9
20
Boundary Value Problems
thus
2 φ
uz e dx dy −
D
uz uφz 2 eφ dx dy D
uuφzz eφ dx dy
D
1 2
|u| Δφeφ dx dy.
D4
4.10
This is just the formula 4.7 for uz ∈ C2 D. Due to the density of C2 D in Wp1 Dp > 2,
it is known that 4.7 is also true for uz ∈ Wp1 D with the condition uz 0 on Γ.
Lemma 4.3. Under the above conditions, one can derive
C1 C2 ,
z ∈ D.
4.11
Proof. On the basis of h1 z h2 z on Γ, and the results of Lemmas 3.1 and 3.2, it follows
that the corresponding coefficients T1 k T2 k, and then C1 z C2 z in D. This shows
that the formula 4.11 is true.
Proof of Theorem 4.1. Similarly to 7, we can prove
A 1 A2
4.12
in D.
From 4.11, we have
C1 A1 JeT JA1 −T JA1 A2 JeT JA2 −T JA2 C2 ,
z ∈ D.
4.13
If we define Aj z 0, z ∈ C \ D, then C1 C2 , z ∈ C, and denote
Ez T J A2 − A1 ,
Ez J A2 − A1 ,
Fz eT JA2 −A1 −T JA2 −A1 ,
4.14
one gets
JA1 FzJA2 ,
Ez T JA2 − A1 T 1 − FzJA2
in C.
4.15
Setting that iθz T JA2 − A1 − T JA2 − A1 , obviously θz is a real function, and
1 − Fz eiθ − 1 eiθ/2 − e−iθ/2 2 sin θ ≤ |θ|
2 T J A2 − A1 − T J A2 − A1 ≤ 2T J A2 − A1 2Ez,
Ez z J A2 − A1 ,
Ez ≤ JA2 1 − Fz ≤ 2JA2 Ez,
z
where Ez T JA2 − A1 0 on Γ is derived from Theorem 2.5.
4.16
Boundary Value Problems
21
Finally we use the Carleman estimate for uz Ez and 4.16, and can get
2
2 φ
Ez e dσz
ΔφEz eφ dσz ≤ 4
D
D
JA2 2 Ez2 eφ dσz .
≤ 16
4.17
D
Taking into account A2 ∈ Lp D, p > 2, and choosing
9
φz π
|JA|2 ln |ζ − z|dσζ
for Az max 1, A2 ,
4.18
D
it is easy to see that Δφ 4φzz 18|JA|2 in D, and then
JAz2 Ez2 eφ dσz ≤ 0,
2
Ez 0 in D.
4.19
D
Consequently
J A2 − A1 0,
A2 − A1 0
in D,
4.20
this shows the coefficients a1 a2 , b1 b2 of equations in 4.1 in G.
Acknowledgment
The research was supported by the National Natural Science Foundation of China
10671207.
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