An over-determined boundary value problem arising from neutrally coated inclusions in three dimensions

0.16 MB·English

by  Hyeonbae Kang

An over-determined boundary value problem arising from
neutrally coated inclusions in three dimensions∗
Hyeonbae Kang† Hyundae Lee† Shigeru Sakaguchi‡
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January 30, 2015
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n
a
J
9 Dedicated to the memory of Professor Kenjiro Okubo
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Abstract

]
P
We consider the neutral inclusion problem in three dimensions which is to prove if a
A
coated structure consisting of a core and a shell is neutral to all uniform fields, then the
h. core and the shell must be concentric balls if the matrix is isotropic and confocal ellipsoids
t if the matrix is anisotropic. We first derive an over-determined boundary value problem
a
in the shell of the neutral inclusion, and then prove in the isotropic case that if the over-
m
determined problem admits a solution, then the core and the shell must be concentric balls.
[
As a consequence it is provedthat the structure is neutralto all uniform fields if and only if

1 it consists of concentric balls provided that the coefficient of the core is larger than that of
v the shell.
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6
AMS subject classifications (MSC 2010). 35N25, 35Q60
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7 Key words. Neutral inclusion, over-determined problem, confocal ellipsoids, concentric balls.
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.
1
1 Introduction
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1 The purpose of this paper is to prove that the coated inclusions neutral to all uniform fields in
:
v the isotropic medium are concentric balls in three dimensions. The coated inclusion is depicted
Xi by (D,Ω) where D and Ω are bounded domains with Lipschitz boundaries in Rd (d = 2,3) such
that D ⊂ Ω. Here, D represents the core and Ω\D the shell. Theconductivity (or the dielectric
r
a
constant) is σ in the core and σ in the shell (σ 6= σ ). If the structure (D,Ω) is inserted into
c s c s
the free space Rd with conductivity σ where there is a uniform field −∇(a·x) = −a for some
m
constant vector a, then the field is perturbed in general. But for certain inclusions the field is
not perturbed, in other words, the field does not recognize the existence of the inclusion. For
example, the coated inclusion is made of concentric balls with specially chosen conductivities
(confocal ellipsoids if σ is anisotropic), one can see the uniform field is not perturbed. The
m
inclusion with this property is called a neutral inclusion (or neutrally coated inclusion) and the
neutralinclusionproblemistoshowthattheinclusionsofconcentricballs(orconfocalellipsoids)
are the only coated inclusions neutral to all uniform fields.
∗This work is supported by the Korean Ministry of Education, Sciences and Technology through NRFgrants
Nos. 2010-0017532 and2013R1A1A1A05009699, and byGrant-in-AidforScientificResearch (B)(♯ 26287020) of
Japan Society for thePromotion of Science.
†Department of Mathematics, InhaUniversity,Incheon 402-751, Korea (hbkang,[email protected]).
‡Research Center for Pure and Applied Mathematics, Graduate School of Information Sciences, Tohoku Uni-
versity,Sendai,980-8579, Japan ([email protected]).
1
Let σ denote the conductivity distribution of the medium so that
σ in D,
c
σ = σ in Ω\D, (1.1)
 s
σ in Rd\Ω.
m

Here we assume that σ and σ are constants (or isotropic matrices), but σ is allowed to be
c s m
anisotropic symmetric matrix. We consider the following problem:
∇·σ∇u = 0 in Rd,
(1.2)
(u(x)−a·x= O(|x|−2) as |x| → ∞,
where a is a constant vector. The term u(x)−a·x depicts the perturbation of the potential
due to insertion of the coated inclusion (D,Ω). If the potential is not perturbed, namely,
u(x)−a·x ≡ 0 in Rd\Ω, (1.3)
the coated inclusion (D,Ω) is said to be neutral to the field a. If (D,Ω) is neutral to the field
e for j = 1,...,d, where e is the standard basis of Rd, then (D,Ω) is neutral to all uniform
j j
fields.
Much interest in neutrally coated inclusions was aroused by the work of Hashin and Shtrik-
man [7] and Hashin [6]. They showed that since insertion of neutral inclusions does not perturb
the outside uniform field, the effective conductivity of the assemblage filled with coated inclu-
sions of many different scales is σ . We refer to [14] for developments on neutral inclusions in
m
relation to the theory of composites. Another interest in neutral inclusions has been aroused
in relation to invisibility cloaking. The neutral inclusion is invisible from the probe by uniform
fields as observed in [12]. Recently, the idea of neutrally coated inclusions has been extended
to construct multi-coated circular structures which are neutral not only to uniform fields but
also to fields of higher order up to N for a given integer N [2]. It was proved there that the
multi-coated structurecombined with a transformation dramatically enhances the near cloaking
of [13]. Cloaking by transformation optics was proposed in [17] (and [5]).
As mentioned before, concentric balls (or disks) are neutral to all uniform fields by choosing
σ , σ and σ properly (σ is isotropic). Confocal ellipsoids (or ellipses) are also neutral to all
c s m m
uniform fields if σ is anisotropic [12] (see also section 3). Then a question arises naturally:
m
are there any other shapes which are neutral to all uniform fields? In two dimensions there are
no other shapes: if a coated inclusion (D,Ω) is neutral to all uniform fields in two dimensions,
then D and Ω are concentric disks (confocal ellipses if σ is anisotropic). This is proved when
m
σ =0 or ∞ in [15] and when σ is finite in [10]. In this paper we consider the neutral inclusion
c c
problem in three dimensions. We emphasize that the methods in [10, 15] use powerful tools
from complex analysis such as conformal mappings and harmonic conjugates, which cannot be
applied to three dimensions. It is worth mentioning that there are many different shapes of
coated inclusions neutral to a single uniform field as shown in two dimensions in [8, 15].
We firstshow thatif(D,Ω) isneutraltoall uniformfieldsinthreedimensionsandifσ > σ ,
c s
then the following problem admits a solution:
∆w = k in Ω\D,
∇w = 0 on ∂Ω, (1.4)

 ∇w(x) = Ax+d on ∂D,
where k(=6 0) is a constant, Ais a symmetric matrix, and d is a constant vector. We emphasize
that this is an over-determined problem because ∇w is prescribed on the boundaries. The
2
problem, which is of independent interest, is to prove that if (1.4) admits a solution in three
dimensions, then D and Ω are confocal ellipsoids. If D and Ω are confocal ellipsoids, then (1.4)
admits asolution andA shouldbeeither positive or negative-definite dependingon the sign of k
(see section 3). So a partof the problem is to show that A is either positive or negative-definite.
In two dimensions it is proved in [10] that if (1.4) admits a solution then D and Ω are confocal
ellipses (concentric disks if A is isotropic). However, the proof there is based on the powerful
result that there is a conformal mapping from Ω\D onto an annulus. So it cannot be extended
to three dimensions. The condition σ > σ , which is not natural, is required because of a
c s
technical reason for the derivation of (1.4) in subsection 2.2. Even though we do not know how
to do so, it is likely that the condition can be removed.
In this paper we solve the problem partially as the following theorem shows.
Theorem 1.1 Let D and Ω be bounded domains with Lipschitz boundaries in R3 with D ⊂ Ω.
Suppose that Ω\D is connected. If (1.4) admits a solution for A =cI for some constant c where
I is the identity matrix in three dimensions, then D and Ω are concentric balls.
As a consequence, we obtain the following theorem.
Theorem 1.2 Let D and Ω be bounded domains with Lipschitz boundaries in R3 with D ⊂ Ω.
Suppose that ∂D is connected and R3 \D is simply connected. If σ is isotropic, σ > σ and
m c s
(D,Ω) is neutral to all uniform fields, then D and Ω are concentric balls.
This paper is organized as follows. In section 2 we show that if (D,Ω) is neutral to all
uniform fields then (1.4) admits a solution. In section 3 we construct a solution to (1.4) when D
and Ω are confocal ellipsoids. Section 4 is to prove Theorem 1.1. In section 5 we formulate the
problem (1.4) using Newtonian potentials and relate the problem with a known characterization
of ellipsoids.
2 Derivation of the over-determined problem
In this section we derive (1.4) out of the neutral inclusion problem. We will do so only in three
dimensionssince(1.4)hasbeenderivedintwo dimensions[10]. Weassumethat∂D isconnected
and R3\D is simply connected.
Suppose, after diagonalization, that
σm = diag[σm,1,σm,2,σm,3]. (2.1)
Let u , j =1,2,3, be the solution to
j
∇·σ∇u = 0 in R3,
j
(2.2)
u (x)−x = O(|x|−2) as |x|→ ∞.
j j
(cid:26)
The structure being neutral to all three fields means that u (x)−x = 0 in R3\Ω for j = 1,2,3.
j j
Let
1
w = u (2.3)
j j
β
j
where
σ
m,j
β := −1, j = 1,2,3,
j
σ
s
and w = (w1,w2,w3)T (T for transpose). Set also
B= diag[1/β1,1/β2,1/β3]. (2.4)
We will show the following:
3
(i) ∇w is symmetric and divw is constant, and hence there is a function ψ in Ω\D such that
w = ∇ψ and ∆ψ = TrB+1 in Ω\D. (2.5)
(ii) w(x) = c0x+d, x∈ D for some constant c0 and constant vector d (under the assumption
that σ > σ ).
c s
We emphasize that it is in (ii) where the condition σ > σ is required.
c s
Once we have (i) and (ii), then we can show that (1.4) has a solution. In fact, since u = x
j j
on ∂Ω, we have
∇ψ(x) = Bx on ∂Ω.
Note that ∇ψ(x) = c0x+d on ∂D. Now define
1
w(x) := ψ(x)− x·Bx. (2.6)
2
Then w satisfies (1.4) with k = 1 and A = c0I−B. We emphasize that if σm is isotropic, so are
B and A.
2.1 Proof of (i)
Let us first deal with the case when 0 < σc < ∞. Denote by ν = (n1,n2,n3)T the outward unit
normal vector field to ∂Ω or ∂D. Note that the solution u (j = 1,2,3) to (2.2) satisfies the
j
following transmission conditions on two interfaces :
∂u ∂u
j j
uj|+−uj|− = 0, σm,j −σs = 0 on ∂Ω (2.7)
∂ν + ∂ν −
(cid:12) (cid:12)
(cid:12) (cid:12)
and
(cid:12) (cid:12)
∂u ∂u
j j
uj|+−uj|− = 0, σs −σc = 0 on ∂D (2.8)
∂ν + ∂ν −
(cid:12) (cid:12)
where + denotes the limit from outside and −(cid:12)that from i(cid:12)nside of Ω or D. If (D,Ω) is neutral
(cid:12) (cid:12)
to x , then u (x)−x = 0 in R3\Ω, so we see from (2.7) that
j j j
∂u
j
u | =x , σ = σ n on ∂Ω. (2.9)
j − j s m,j j
∂ν −
(cid:12)
(cid:12)
In other words, u is the solution to the following over-determined problem:
j (cid:12)
∇·σ∇u = 0 in Ω,
j
∂u σ (2.10)
j m,j
 u = x , = n on ∂Ω.
j j j
 ∂ν σs
Let v ∈ C2(Ω). Then we see from the divergence theorem and (2.8) that
j
∂u ∂v ∂u ∂v
j j j j
v −u = − u ∆v + v −u
j j j j j j
Z∂Ω ∂ν (cid:12)− ∂ν ZΩ\D Z∂D ∂ν (cid:12)+ ∂ν
(cid:12) σ (cid:12) ∂u ∂u ∂v
(cid:12) = − u ∆v + c −1(cid:12) j v + j v −u j
j j j j j
ZΩ\D (cid:18)σs (cid:19)Z∂D ∂ν (cid:12)− Z∂D ∂ν (cid:12)− ∂ν
σ (cid:12) (cid:12)
= − u ∆v + c −1 ∇u ·(cid:12)∇v − u ∆v(cid:12)
j j j j j j
σ
ZΩ\D (cid:18) s (cid:19)ZD ZD
σ
c
= − u ∆v + −1 ∇u ·∇v .
j j j j
σ
ZΩ (cid:18) s (cid:19)ZD
4
On the other hand, we see from (2.9) that
∂u ∂v σ ∂v
j j m,j j
v −u = n v −y
j j j j j
Z∂Ω ∂ν (cid:12)− ∂ν Z∂Ω σs ∂ν
(cid:12) σm,j ∂vj
(cid:12) = −1 − y ∆v .
j j
σ ∂y
s Ω j Ω
(cid:18) (cid:19)Z Z
Equating two identities above we obtain
∂v
j
(y −u )∆v +α ∇u ·∇v −β = 0, j = 1,2,3 (2.11)
j j j j j j
∂y
ZΩ ZD ZΩ j
for v ∈C2(Ω), where α and β are defined for ease of notation to be
j j
σ σ
c m,j
α = −1 and β = −1. (2.12)
j
σ σ
s s
1
Let w be defined by w := u as in (2.3). Then (2.11) can be rephrased as
j j βj j
1 ∂v
j
( y −w )∆v +α ∇w ·∇v − = 0, j = 1,2,3. (2.13)
j j j j j
β ∂y
ZΩ j ZD ZΩ j
Summing (2.13) over j = 1,2,3 we have
3 3 3
1 ∂v
j
( y −w )∆v +α ∇w ·∇v − = 0
j j j j j
β ∂y
ZΩj=1 j ZD j=1 ZΩj=1 j
X X X
for vj ∈ C2(Ω). If we use vector notation w = (w1,w2,w3)T and v = (v1,v2,v3)T (T for
transpose), then the above identity can be rewritten as
(By−w)·∆v+α ∇w :∇v− divv = 0. (2.14)
ZΩ ZD ZΩ
Here and afterwards A : B denote the contraction of two matrices A and B, i.e., A : B =
a b = Tr(ATB).
ij ij
Let Γ be the fundamental solution of the Laplace operator in R3, i.e.,
P
1
Γ(x) := − , x6= 0. (2.15)
4π|x|
Let v (y) = Γ(x−y) for a fixed x ∈ Ω. Since ∆v (y) = δ(x−y), by applying the divergence
j j
theorem over Ω\B (x) for sufficiently small ǫ (B (x) is the ball of radius ǫ centered at x) we
ǫ ǫ
see from (2.13) that
1 ∂
wj(x) = xj +α ∇wj(y)·∇yΓ(x−y)dy+ NΩ(x), x∈ Ω, j = 1,2,3, (2.16)
β ∂x
j ZD j
where NΩ is the Newtonian potential on a domain Ω, i.e.,
NΩ(x) := Γ(x−y)dy, x∈ R3. (2.17)
Ω
Z
Let
f (x) := ∇w (y)·∇ Γ(x−y)dy, j = 1,2,3,
j j y
ZD
5
and let f = (f1,f2,f3)T. Note that fj is harmonic in R3\D, and (2.16) can be rewritten as
1
w(x) = αf(x)+∇ x·Bx+NΩ(x) , x ∈ Ω, j =1,2,3. (2.18)
2
(cid:18) (cid:19)
For any fixed x ∈ R3\Ω, let
∂
v (y) = Γ(x−y), j = 1,2,3.
j
∂x
j
Then divv(y) = −∆ Γ(x−y) = 0 and ∆v(y) = 0 for y ∈Ω. So we see from (2.14) that
y
∇w :∇v = 0,
ZD
and hence
∂
divf(x) = ∇w (y)·∇ Γ(x−y)dy = ∇w : ∇v = 0, x ∈R3\Ω. (2.19)
j
∂x
ZD j j ZD
X
Since f is harmonic in R3\D, (2.19) holds for all x ∈ R3\D.
j
Again fix x∈ R3\Ω. Let {i,j,k} be a permutation of {1,2,3} and let
∂ ∂
v (y) = Γ(x−y), v (y) =− Γ(x−y), v = 0, y ∈ Ω.
i j k
∂x ∂x
j i
Then, ∆v = 0 and divv = 0 in Ω. So we have from (2.14)
∂ ∂
∇w (y)·∇ Γ(x−y)dy− ∇w (y)·∇ Γ(x−y)dy = 0,
i j
∂x ∂x
ZD j ZD i
which implies that
∂ f (x) = ∂ f (x) (2.20)
i j j i
for all x ∈ R3\Ω and hence for all x∈ R3\D. Moreover, since R3\D is simply connected, by
the Stokes theorem there is ϕ such that
f(x) = ∇ϕ(x), x∈ R3\D. (2.21)
Because of (2.19), we have
∆ϕ(x) = 0, x∈ R3\D. (2.22)
Let
1
ψ(x) = αϕ(x)+ x·Bx+NΩ(x), x ∈ Ω\D. (2.23)
2
Then, we have from (2.18) and (2.21)
w(x) = ∇ψ(x), x∈ Ω\D, j = 1,2,3. (2.24)
Since ∆NΩ(x) = 1 for x∈ Ω, we have from (2.22) that
∆ψ(x) = TrB+1, x ∈Ω\D. (2.25)
Sofar we have shown that ∇w is symmetric, divw is constant, and (2.5) holds whenσ is finite.
c
6
We now assume that σ = 0. In this case the problem (2.10) becomes
c
∆u = 0 in Ω\D,
j
∂u
 j = 0 on ∂D,
(2.26)
 ∂ν


 ∂u σ
 j m,j
u = x , = n on ∂Ω.
j j j
∂ν σ
s




So, we see in a way similarto (2.11) that
∂v ∂v
j j
y ∆v − u ∆v − u −β = 0 (2.27)
j j j j j j
∂ν ∂y
ZΩ ZΩ\D Z∂D ZΩ j
for all v ∈ C2(Ω). So we obtain a representation of the solution similar to (2.16):
j
1 ∂ ∂
wj(x) = xj − wj(y) Γ(x−y)dσ(y)+ NΩ(x), x ∈Ω\D. (2.28)
β ∂ν ∂x
j Z∂D j
So, we infer in theexactly same way as in the previous sections that ∇w is symmetric and divw
is constant, and there is a function ψ such that (2.5) holds.
Suppose that σ = ∞. In this case the problem (2.10) becomes
c
∆u = 0 in Ω\D,
j
u = γ (constant) on ∂D,
j j
 (2.29)
 uj = xj, ∂uj = σm,jnj on ∂Ω.
∂ν σ
s


The constant γ is determined by the condition
j
∂u
j
= 0.
∂ν +
Z∂D (cid:12)
(cid:12)
We then obtain similarly to (2.11) (cid:12)
∂u ∂v
j j
y ∆v − u ∆v + v −γ ∆v −β = 0 (2.30)
j j j j j j j j
∂ν ∂y
ZΩ ZΩ\D Z∂D ZD ZΩ j
for all v ∈ C2(Ω). We then obtain a representation of the solution similar to (2.16):
j
1 ∂w ∂
j
wj(x) = xj + (y)Γ(x−y)dσ(y)+ NΩ(x), x∈ Ω\D. (2.31)
β ∂ν ∂x
j Z∂D j
So, we infer that ∇w is symmetric, divw is constant, and there is a function ψ such that (2.5)
holds.
2.2 Proof of (ii)
The transmission conditions (2.8) on ∂D can be rephrased as
w|+ = w|−, σs∇w|+ν = σc∇w|−ν. (2.32)
Let t1 and t2 be two orthonormal tangent vector fields to ∂D. Then, we have
(divw)− = h(∇w)−ν,νi+h(∇w)−t1,t1i+h(∇w)−t2,t2i,
7
and
(divw)+ = h(∇w)+ν,νi+h(∇w)+t1,t1i+h(∇w)+t2,t2i.
Here (divw)− denotes the limit of divw to ∂D from inside D, and (divw)+ denotes that from
outside D. Since
h(∇w)−tj,tji= h(∇w)+tj,tji, j = 1,2,
we have
(divw)−−(divw)+ = h(∇w)−ν,νi−h(∇w)+ν,νi.
It then follows from the second identity in (2.32) that
σ
(∇w)T− − σc(∇w)− ν,ν = (divw)−−(divw)+. (2.33)
s
(cid:28) (cid:29)
(cid:0) (cid:1)
On the other hand, since (∇w)+ is symmetric, we obtain
σ σ
(∇w)T − c(∇w) ν,t = hν,(∇w) t i− ch(∇w) ν,t i
− σ − j − j σ − j
s s
(cid:28) (cid:29)
(cid:0) (cid:1) = hν,(∇w)+tji−h(∇w)+ν,tji = 0. (2.34)
We then infer from (2.33) and (2.34) that
σ
(∇w)T− − σc(∇w)− ν = (divw)−ν −(divw)+ν. (2.35)
s
(cid:18) (cid:19)
Recall that divw is constant in Ω\D. Let
(divw)+
v(x) = w(x)− x, x∈ D. (2.36)
2+ σc
σs
Then one can see from (2.35) that
σ
(∇v)T − c(∇v) ν −(divv)ν = 0 on ∂D. (2.37)
σ
s
(cid:18) (cid:19)
Let g be a smooth vector field on D. It follows from (2.37) and the divergence theorem that
σ
0 = ν · (∇v)g− c(∇v)Tg−(divv)g dσ
σ
Z∂D (cid:18) s (cid:19)
σ
= div (∇v)g− c(∇v)Tg−(divv)g dx.
σ
ZD (cid:18) s (cid:19)
One can easily show that
σ σ
div (∇v)g− c(∇v)Tg−(divv)g = ∇vT : ∇g− c∇v :∇g−(divv)(divg),
σ σ
s s
(cid:18) (cid:19)
and so we obtain
σ
∇vT : ∇g− c∇v :∇g−(divv)(divg) = 0. (2.38)
σ
ZD s
Using notation
1 1
∇v := (∇v+∇vT) and ∇˘v := (∇v−∇vT),
2 2
b
8
it can be rewritten as
σ σ
(1− c) ∇v : ∇g−(1+ c) ∇˘v :∇˘g− (divv)(divg) = 0. (2.39)
σ σ
s ZD s ZD ZD
If σ > σ , then we takbe g =bv so that
c s
σ σ
(1− c) |∇v|2−(1+ c) |∇˘v|2 − (divv)2 = 0. (2.40)
σ σ
s ZD s ZD ZD
Thus, we infer that v is constantbin D and hence
(divw)+
w(x) = x+a constant vector, x∈ D. (2.41)
2+ σc
σs
If σ = ∞, then u is constant on ∂D, and hence (∇w)t = 0 on ∂D for any tangential vector t
c
to ∂D. Since ∇w is symmetric and divw is constant, it implies that
(∇w)ν = c0ν on ∂D
for some constant c0. So, we can see that (ii) holds.
3 Existence of solutions on confocal ellipsoids
We first mention that the solution w to (1.4) is unique in the sense that if w1 and w2 are two
solutions (with different k, A’s, and d’s), then w1 = Cw2 + E for some constants C and E.
In fact, if w is a solution to (1.4) with k = k 6= 0, A = A and d = d (j = 1,2), then
j j j j
w = w1 − kk12w2 satisfies ∆w = 0 in Ω\D and ∇w = 0 on ∂Ω, so we have that w must be a
constant.
We now construct asolution to (1.4) whenD andΩ areconfocal ellipsoids. To doso, assume
that ∂D is given by
x2 x2 x2
1 2 3
+ + = 1. (3.1)
c2 c2 c2
1 2 3
We then use the confocal ellipsoidal coordinates ρ, µ, ξ such that
x2 x2 x2
1 2 3
+ + = 1,
c2+ρ c2+ρ c2+ρ
1 2 3
x2 x2 x2
1 2 3
+ + = 1,
c2+µ c2+µ c2+µ
1 2 3
x2 x2 x2
1 2 3
+ + = 1,
c2+ξ c2+ξ c2+ξ
1 2 3
subject to the conditions −c2 < ξ < −c2 < µ < −c2 < ρ. Then the confocal ellipsoid ∂Ω is
3 2 1
given by ρ =ρ0 for some ρ0 > 0.
Let
2 2 2
g(ρ) = (c +ρ)(c +ρ)(c +ρ), (3.2)
1 2 3
and define
∞ 1
ϕ (ρ) = ds, j = 1,2,3. (3.3)
j Zρ (c2j +s) g(s)
p
9
Then the function w defined by
3 3
1 ∞ 1 1 1
w(x) = 2Zρ g(s)ds− 2 j=1ϕj(ρ)x2j + 2 j=1ϕj(ρ0)x2j (3.4)
X X
p
is a solution of (1.4). In fact, we can see that
3 3
∂ 1 ∞ 1 1 1 ∂ρ
ds− ϕ (ρ)x2 = − − ϕ′(ρ)x2 −ϕ (ρ)x .
∂xi 2 Zρ g(s) 2 j=1 j j  g(ρ) j=1 j j∂xi i i
X X
 p   p 
Since
3 3 x2 1
ϕ′(ρ)x2 = − j = − ,
j j (c2 +ρ) g(ρ) g(ρ)
j=1 j=1 j
X X
we have p p
3
∂ 1 ∞ 1 1
2
ds− ϕ (ρ)x = −ϕ (ρ)x ,
∂xi 2Zρ g(s) 2 j=1 j j i i
X
from which we see that  p 
∇w(x) = −(ϕ1(ρ)x1,ϕ2(ρ)x2,ϕ3(ρ)x3)+(ϕ1(ρ0)x1,ϕ2(ρ0)x2,ϕ3(ρ0)x3). (3.5)
Using the relation
∂ρ 2x x2 x2 x2 −1
i 1 2 3
= + + , (3.6)
∂x c2+ρ (c2+ρ)2 (c2+ρ)2 (c2+ρ)2
i i (cid:20) 1 2 3 (cid:21)
we see that ∆w is constant. Note that ∇w = 0 on ∂Ω (ρ= ρ0) and ∇w = Ax on ∂D where
A = diag[ϕ1(ρ0)−ϕ1(0),ϕ2(ρ0)−ϕ2(0),ϕ3(ρ0)−ϕ3(0)]. (3.7)
We emphasize that A is negative-definite.
4 Proof of Theorem 1.1
Let w be the solution to (1.4) with A = cI. We notice that c 6= 0. Indeed, if c = 0, then we have
∂w ∂w
0 6= k|Ω\D|= ∆w dx = dσ− dσ = 0− ν ·d dσ = 0,
∂ν ∂ν
ZΩ\D Z∂Ω Z∂D Z∂D
which is a contradiction. Since c6= 0, by introducing new variables
1
y = x+ d,
c
we may assume that d = 0. Set
∂ ∂
A = x −x , i6= j. (4.1)
ij j i
∂x ∂x
i j
It is worth mentioning that A is the angular derivative. Observe that A commutes with ∆,
ij ij
namely, A ∆ = ∆A . So, we have ∆A w = 0 in Ω\D. Note that A w = 0 on ∂Ω. Since
ij ij ij ij
∇w(x) = cx on ∂D, we see that A w = 0 on ∂D. Then the maximum principle yields that
ij
A w = 0 in Ω\D. (4.2)
ij
10