LAPLACE’S EQUATION IN THE EXTERIOR OF A CONVEX POLYGON.
THE EQUILATERAL TRIANGLE
A. Charalambopoulos
Department of Material Science and Engineering, University of Ioannina, Greece
G. Dassios(*) and A.S. Fokas
Department of Applied Mathematics and Theoretical Physics,
University of Cambridge, U.K.
ABSTRACT
A general method for studying boundary value problems for linear and for integrable
nonlinear partial differential equations in two dimensions was introduced in [3]. For
linear equations in a convex polygon [2,4,5], this method: (a) Expresses the solution
q(x,y) in the form of an integral (generalized inverse Fourier transform) in the
complex k-plane involving a certain function q̂(k) (generalized direct Fourier
transform) which is defined as an integral along the boundary of the polygon, (b)
Characterizes a generalized Dirichlet-to-Neumann map by analyzing the so-called
global relation. For simple polygons and simple boundary conditions, this
characterization is explicit. Here, we extend the above method to the case of elliptic
partial differential equations in the exterior of a convex polygon and we illustrate the
main ideas by studying the Laplace equation in the exterior of an equilateral triangle.
Regarding (a), we show that whereas q̂(k) is identical with that of the interior
problem, the contour of integration in the complex k-plane appearing in the formula
for q(x,y) depends on (x,y). Regarding (b), we show that the global relation is now
replaced by a set of appropriate relations which in addition to the boundary values
also involve certain additional unknown functions. In spite of this significant
complication we show that for certain simple boundary conditions the exterior
problem for the Laplace equation can be mapped to the solution of a Dirichlet
problem formulated in the interior of a convex polygon formed by three sides.
1. INTRODUCTION
(*)
On leave from the University of Patras and ICE-HT/FORTH Greece
Constructing analytic solutions for Laplace’s equation in the exterior of a polygon is
a long standing open problem in the theory of partial differential equations. On the
other hand, a method for solving Laplace’s equation in the interior of a convex
polygon was presented in [2]. This is a particular case of a general method for linear
boundary value problems for open or closed convex polygons introduced by the third
author in [3]. The basic elliptic equations in an equilateral triangle for a large class of
boundary conditions were studied in [1]. In the above works convexity played a
crucial role, and this is the reason for analyzing interior problems. In the present
work we investigate the solution of an exterior problem. In order to elucidate the
effects of non-convexity we restrict our analysis to the simple case of Laplace’s
equation.
We first show that the integral representation for the exterior solution, in addition to a
change of sign which reflects the change of direction of the normal to the boundary,
it differs from the analogous representation of the interior problem in the following
important way: For the interior problem for a convex n-polygon, the integral
representation involves the integrals of the n functions {q̂ j (k)} along n specific rays
n
j=1
in the complex k-plane. For the exterior problem, the integral representation involves
the same functions {q̂ j (k)} , but the rays of integration depend on (x,y). Actually the
n
j=1
extensions of the sides of the polygon divide the plane into certain domains, and one
of the rays of integration rotates by π as the point (x,y) moves into a neighboring
domain. In other words, the exterior problem has a constant spectral form but a
varying spectral representation.
For example, in the integral representation of the exterior solution for an equilateral
triangle, the three rays of integration rotate by π/3 as (x,y) moves successively in the
six convex domains generated by the three sides of the triangle. Thus, the rays return
to their original positions after (x,y) travels around the six domains.
For a convex polygon the so-called Dirichlet-to-Neumann map is characterized by
the global relation which is an equation coupling the Dirichlet and Neumann
boundary values [5]. We show that for problems formulated in the exterior of a
convex polygon the fundamental domain must be subdivided to an appropriate set of
convex subdomains, and in each of these subdomains there exists an appropriate
global relation. This construction introduces additional unknown functions which
2
makes the characterization of the Dirichlet-to-Neumann map more complicated. In
spite of this significant complication we will show that if one prescribes the same
function as Dirichlet boundary condition on each side of an equilateral triangle and if
this function is symmetric with respect to the midpoint of the side, then the exterior
problem can be mapped to the solution of a Dirichlet problem formulated in the
interior of a convex polygon formed by three sides. On the other hand, the general
Dirichlet problem gives rise to a matrix Riemann-Hilbert problem.
In what follows, we will assume existence; however it is possible to eliminate this
assumption. In fact, it turns out that the global relation is not only a necessary but
also a sufficient condition for existence. Indeed, combining the global relation with
the integral representation mentioned earlier one can prove that the question of
existence of solution of a given boundary value problem can be reduced to the
question of analyzing the global relation. This proof is not presented here, but it is
similar with the proof of the analogous result for the interior problem given in [7].
The paper is organized as follows. In Section 2 the integral representation for the
exterior of an n-polygon is derived. In Section 3 this general result is illustrated for
the particular case of the exterior of an equilateral triangle. Section 4 constructs the
global relation in the case that the same function is prescribed as Dirichlet boundary
condition on each side of the triangle. Furthermore, it is shown that if this function is
anti-symmetric with respect to the midpoint of the side, then the Dirichlet problem
can be mapped to the solution of a Dirichlet problem formulated in the interior of a
convex polygon formed by three sides.
2. THE INTEGRAL REPRESENTATION
Let z1 , z 2 ,..., z n , z n +1 = z1 be the n vertices of a bounded convex polygon Ω(i) in the
complex plane. Denote by Ω(e) the domain exterior to Ω(i) and by ∂Ω(i) the boundary
of Ω(i).
Proposition 2.1 Suppose that the real-valued function q(x,y) is a harmonic function
in Ω(e), which decays as x + y → ∞ and which is smooth on the boundary ∂Ω(i) .
Then the following relations are valid
3
⎧ 0
, z ∈ Ω(i)
1
∂q(z′) dz′ ⎪
= ⎨ ∂q(z)
2πi ∂Ω∫(i ) ∂z′ z′ − z ⎪ −
, z ∈ Ω(e) .
⎩ ∂z
1
∂q(z)
=
(2πi) 2
∂z
(2.1)
⎡
⎤ ∂q(z′)
eik1 (x − x ′) +ik 2 ( y − y′)
dk1dk 2 ⎥
dz′, z ∈ Ω(e)
⎢
∫(i ) ⎢ ∫∫2
′
k
ik
z
+
∂
1
2
∂Ω ⎣ R
⎦⎥
(2.2)
where z denotes the usual complex variable z = x + iy , and the integration is taken in
the positive direction.
z1 = z n +1
zn
z2
z n −1
Ω (e)
• z = x + iy
Ω(i)
zj
z j+ 2
z j+1
Figure 2.1
Proof. The equations
z = x + iy,
z = x − iy ,
(2.3)
imply
∂z =
1
1
(∂ x − i∂ y ), ∂ z = (∂ x + i∂ y ) .
2
2
(2.4)
Hence, Laplace’s equation becomes
∂ 2 q(z, z)
= 0,
∂ z∂z
(2.5)
which shows that the function qz is analytic. Hence, using the fact that q decays at
infinity, Cauchy’s theorem in the domain Ω(e) yields equation (2.1).
In the Appendix, we prove the following identity
1
1
e 1( ) 2 (
=
z − z′ 2π ∫∫
ik1 − k 2
R2
ik x − x ′ + ik
y − y′ )
dk1dk 2 .
(2.6)
4
Replacing in equation (2.1b) 1 ( z − z ′ ) by the RHS of equation (2.6) we find
equation (2.2).
QED.
It was shown in [6] that starting with representations of the type (2.2), it is possible to
obtain Ehrenpreis-Palamodov type representations by integrating with respect to kN,
where kN is the component of the vector k = (k1 , k 2 ) , normal to the boundary. Using
this approach, we will prove the following result.
Theorem 2.1. Under the assumptions of Proposition 2.1 the real-valued harmonic
function q admits the following integral representation:
∂q(z)
1 n
= − ∑ ∫ eikz qˆ j (k) dk,
2π j=1 lˆ
∂z
z ∈ Ω (e) ,
(2.7)
j
where the spectral functions {q̂ j (k)}
n
j=1
are defined by
zj
q̂ j (k) =
∫e
− ikz
q z (z) dz,
j = 1, 2,..., n,
(2.8)
z j+1
and the rays lˆj , j = 1, 2,..., n emanating from the origin are defined as follows:
⎧⎪l − = {k ∈ : arg k = π − arg(z j − z j+1 )} , x Nj < 0
lˆj = ⎨ j+
x Nj > 0
⎪⎩l j = {k ∈ : arg k = − arg(z j − z j+1 )} ,
(2.9)
where x Nj is the outward to Ω(e) normal component of the vector r − r′ , with r ∈ Ω (e)
and r′ on the j-th side. Finally,
z = x + iy
q(x, y) = 2 Re
∫
z0
∂q(z′)
dz′ .
∂z′
(2.10)
Proof. We introduce local coordinates on each side of the polygon in such a way that
the normal N̂ is exterior to Ω(e), i.e. it points towards Ω(i), see Figure 2.2. The Frenet
system on the j-th side is
ˆ = z j − z j+1 ,
T
z j − z j+1
ˆ = iT
ˆ.
N
(2.11)
z j+1
5
θ
Ω
(e)
N̂
T̂
r − r′
Ω(i)
r′
r
zj
O
Figure 2.2
We introduce the notations
k = (k1 , k 2 ) ,
r = (x, y) ,
r′ = (x′, y′)
(2.12)
and we denote by (k T , k N ) and (x T , x N ) the Frenet components of k and of r − r′ ,
respectively, i.e.
ˆ +k N
ˆ
k = k TT
N ,
ˆ +x N
ˆ .
r − r′ = x T T
N
(2.13)
Hence,
k1 (x − x ′) + k 2 (y − y′) = k T x T + k N x N ,
dk1dk 2 = dk T dk N .
(2.14)
Let θ denote the argument of z j − z j+1 , then
k1 + ik 2 = eiθ (k T + ik N ) .
(2.15)
Hence, in the local Frenet system equation (2.2) becomes
∂q(z)
1
=− 2
∂z
4π
n
∑ ∫∫
j=1
R2
⎡ z j ei(k T x T + k N x N )
∂q(z′) ⎤
⎢∫
e − iθ
dz′⎥ dk T dk Ν .
∂z′
⎢⎣ z j+1 k T + ik N
⎥⎦
(2.16)
ˆ .
x N = (r − r′) ⋅ N
(2.17)
Equation (2.13b) implies
In order to evaluate the integrals in the expression (2.l6) we need to determine the
sign of xn. Hence, for each side of the polygon we define the splitting
6
+
j
−
j
{
= {r ∈ Ω
}
< 0}
ˆ >0
= r ∈ Ω (e) (r − r′) ⋅ N
j
(e)
ˆ
(r − r′) ⋅ N
j
(2.18)
where r′ belongs to the j-side of the polygon. Note that since the polygon is convex,
+
j
−
j
is always a half-plane minus the polygon, while
r
is a full half-plane.
z j+1
r − r′
r′
θ−j
zj
Ω (e)
N̂ j
θ+j
Ω(i)
r − r′
−
j
+
j
r
Figure 2.3
Furthermore,
−
j
a) The case where r ∈
r∈
+
j
⇒ θ+j ∈ ( −π / 2, π / 2 ) ,
r∈
−
j
⇒ θ−j ∈ ( π / 2,3π / 2 ) .
(2.19)
.
In this case x N < 0 , and integrating in the lower kN-half plane yields
+∞
∫
−∞
⎧−2πe − k T x N ,
eik N x N
dk N = ⎨
k T + ik N
0,
⎩
kT < 0
k T > 0.
(2.20)
Therefore,
−
1
4π 2
∫∫
R
2
ei(k T x T + k N x N ) − iθ
1
e dk T dk Ν =
k T + ik N
2π
+∞
∫
e − ik T (x T +ix N ) e− iθ dk T .
(2.21)
0
Setting k T = keiθ , and using
x T + ix N = (z − z′)e − iθ
(2.22)
we rewrite (2.21) as
7
∞e − iθ
1
2π
∫
0e
e
− ik (z − z′ )
− iθ
1
dk = −
2π
∞ei ( π−θ )
∫
0e
eik (z − z′) dk ,
(2.23)
i ( π−θ )
where the last equality is obtained by replacing k with keiπ .
Therefore,
−
1
4π 2
ei(k T x T + k N x N ) −iθ
1
e dk T dk Ν = −
eik (z − z′) dk ,
∫
k T + ik N
2π l −
∫∫
R2
(2.24)
j
with
l j− = {k ∈
+
j
b) The case where r ∈
arg k = π − arg(z j − z j+1 )} .
(2.25)
∩ Ω (e) .
In this case x N > 0 , and integrating in the upper kN-half plane yields
+∞
kT < 0
⎧ 0,
eik N x N
dk N = ⎨
− k T xN
,
k T + ik N
⎩2πe
∫
−∞
k T > 0.
(2.26)
Therefore
−
1
4π 2
∫∫
R2
ei(k T x T + k N x N ) −iθ
1
e dk T dk Ν = −
k T + ik N
2π
+∞
∫
eik T (x T +ix N ) e− iθ dk T .
(2.27)
0
Setting k T = keiθ , and using
x T + ix N = (z − z′)e− iθ ,
(2.28)
we arrive at
1
−
2π
+∞
∫
e
1
e dk T = −
2π
ik T (x T + ix N ) − iθ
0
∞e − iθ
∫
eik (z − z′) dk .
(2.29)
0e − iθ
Therefore,
1
− 2
4π
∫∫
R2
ei(k T x T + k N x N ) −iθ
1
e dk T dk Ν = −
eik (z − z′) dk ,
∫
k T + ik N
2π l +
(2.30)
j
with
l j+ = {k ∈
arg k = − arg(z j − z j+1 )} .
Hence, the proof of the Theorem is completed.
(2.31)
QED.
Remark: Expressions (2.7) and (2.8) provide the generalized Fourier transform pair
associated with Laplace’s equation in the exterior domain Ω(e). For a given point
z ∈ Ω(e) , lˆj can be fixed by studying the relative position of z with respect to the side:
8
The contours l j− (associated with the whole half-spaces), are exactly the same as the
corresponding contours for the interior problem [1], whereas the contours l j+
(associated with the half-spaces that contain the polygon), are the opposite of l j− .
3. THE EQUILATERAL TRIANGLE
Set
α=e
i
2π
3
1
3
.
= − +i
2
2
(3.1)
And let
z1 = −α
3
,
z1 = −α
3
z3 = −
,
3
,
(3.2)
be the vertices of the equilateral triangle of side , shown in Figure 3.1. The side S j
is located opposite to the vertex z j , j = 1, 2,3.
Ω (e)
z1 = −
3
α
S2
Ω(i)
N̂
T̂
T̂
N̂
z3 = −
3
N̂
S1
S3
T̂
z2 = −
3
α
Figure 3.1
On the sides S3 ,S2 ,S1 , the following parameterizations
9
⎛
⎞
+ is ⎟
S2 : z(s) = α ⎜
⎝2 3
⎠
⎛
⎞
S1 : z(s) = α ⎜
+ is ⎟
⎝2 3
⎠
⎫
⎪
⎪
⎪⎪
⎬
⎪
⎪
⎪
⎪⎭
1
S3 : ∂ z = − (∂ N + i∂ T )
2
α
S2 : ∂ z = − (∂ N + i∂ T )
2
α
S1 : ∂ z = − (∂ N + i∂ T )
2
⎫
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎭
S3 : z(s) =
2 3
+ is
− ≤s≤ ,
2
2
(3.3)
imply
(3.4)
where ∂ N and ∂ T denote normal and tangential differentiation along N̂ and T̂ .
Let q j (s) and ∂ N q j (s) denote the Dirichlet and Neumann values of the side S j ,
j=1,2,3, i.e.
q(x, y) ( x,y)∈S = q j (s),
j
q N (x, y) ( x,y)∈S = q N j (s), −
j
2
≤s≤
2
, j=1, 2, 3.
(3.5)
Then, the spectral functions q̂ j (k) are defined by the following
z1
q̂ 3 (k) = ∫ e− ikz q z (z) dz =
z2
z3
q̂ 2 (k) = ∫ e− ikz q z (z) dz =
z1
z2
q̂1 (k) = ∫ e− ikz q z (z) dz =
z3
1
[−iE(−ik)Ψ 3 (k) + E(−ik)Φ 3 (k)],
2
(3.6)
1
[−iE(−iαk)Ψ 2 (αk) + E(−iαk)Φ 2 (αk)],
2
(3.7)
1
[−iE(−iαk)Ψ1 (αk) + E(−iαk)Φ1 (αk)],
2
(3.8)
where
E(k) = e
{Φ } ,{Ψ }
j
3
3
1
j 1
k
2 3
,
k ∈ C,
(3.9)
denote the Dirichlet and Neumann integrals,
/2
Φ j (k) =
∫
i
e ks q j (s)ds,
k ∈ C, j = 1, 2,3,
eks q N j (s)ds,
k ∈ C, j = 1, 2,3,
− /2
(3.10)
/2
Ψ j (k) =
∫
(3.11)
− /2
10
and dot denotes differentiation.
Indeed, introducing local coordinates it follows that
2qˆ 3 (k) =
z1
∫
e− ikz (∂ x − i∂ y )q(x, y)dz = −iE(−ik)Ψ 3 (k) + E(−ik)Φ 3 (k) .
(3.12)
z2
Similarly, the functions q̂ 2 (k) and q̂1 (k) can be obtained through rotations by
and
2π
3
4π
respectively.
3
The extensions of the three sides of the triangle partition the exterior of the triangle in
the six regions denoted by R11 , R 22 , R 33 , R 21 , R 32 , R13 , as in the Figure 3.2.
l1-
l2+
l2+
R11
l3-
3
l
z1
l1+
R 13
l2-
l3-
l1R 33
R 21
z3
l3+
l1+
l2+
R 32
l3+
z2
2
l
l1+
R 22
l1-
l3+
l2-
Figure 3.2
11
In summary, the integral representations for the exterior of the equilateral triangle
depicted in Figure 3.1 is given by equation (2.7), where the spectral functions are
defined by equations (3.6)-(3.11) and the appropriate rays (defined in (2.9)) in the
particular domains are chosen as follows
⎧
⎪
⎪
⎪⎪
(lˆ1 , lˆ2 , lˆ3 ) = ⎨
⎪
⎪
⎪
⎪⎩
(l1− , l2+ , l3− ) , z ∈ R11
(l1+ , l2+ , l3− ) , z ∈ R13
(l1+ , l2− , l3− ) , z ∈ R 33
(3.13)
(l1+ , l2− , l3+ ) , z ∈ R 32
(l1− , l2− , l3+ ) , z ∈ R 22
(l1− , l2+ , l3+ ) , z ∈ R 21 .
Remarks.
1. As z moves from a region R ij to the counterclockwise neighboring region, two of
the rays lˆj remain the same and the third one changes direction. As a consequence,
the set of the three rays (lˆ1 ,lˆ2 ,lˆ3 ) advances by an angle π / 3 clockwise. Thus after z
moves through all six regions, the three rays advance by an angle 6π / 3 = 2π , and
they return to their starting position.
2. Comparing with the interior problem, where the three rays l j are symmetrically
located, making an angle 2π / 3 between each other, for the exterior problem the
angles between the rays are always π / 3, π / 3 and 4π / 3 .
4. THE GLOBAL RELATION
Before deriving the set of global relations needed for the exterior of the triangle we
first establish some symmetry relations.
Proposition 4.1. Assume that there exists a unique function q(x, y) which satisfies
the Laplace equation in the exterior of an equilateral triangle, which decays to zero as
| x | + | y | → ∞ , and which satisfies
q j (s) = f (s) , j = 1, 2,3 , −
2
≤s≤
2
.
(4.1)
Then the function Q(z) = q z (z) satisfies the symmetry relation
12
Q(z) = αQ(αz), z ∈ Ω(e) .
(4.2)
Furthermore, if f(s) is symmetric (even) or antisymmetric (odd) with respect to the
midpoint of the side, then Q(z) satisfies the following additional relations
f (−s) = f (s) ⇒ Q(z) = Q(z),
and
f (−s) = −f (s) ⇒ Q(z) = −Q(z) .
(4.3)
Proof. Cauchy’s theorem implies
z3
z
z2
⎤ Q(ζ )
1 ⎡1
Q(z) = −
dζ
⎢∫ + ∫ + ∫ ⎥
2πi ⎣⎢ z2
⎥ ζ−z
z1
z3 ⎦
(4.4)
where z1 , z 2 , z3 are shown in Figure 3.1. On the side S3 , ζ (s) is given by the first of
equations (3.3), thus dζ = ids . On the side S2 , ζ (s) is given by the second of
equations (3.3), thus dζ = iαds . Similarly for the side S1 and hence equation (4.4)
becomes
/2
1
Q(z) = −
2π − ∫/ 2
⎛ Q ( τ ( s ) ) αQ ( ατ ( s ) ) αQ ( ατ ( s ) ) ⎞
+
+
⎜
⎟ ds
⎜ τ(s) − z
ατ(s) − z
ατ(s) − z ⎟⎠
⎝
(4.5)
with
τ(s) =
2 3
+ is,
− ≤s≤
2
2
(4.6)
Replacing in equation (4.5) z by α z and multiplying the resulting equation by α we
find
/2
1
αQ(αz) = −
2π − ∫/ 2
⎛ Q ( τ ( s ) ) αQ ( ατ ( s ) ) αQ ( ατ ( s ) ) ⎞
+
+
⎜
⎟ ds .
⎜ ατ(s) − z
τ(s) − z
ατ(s) − z ⎟⎠
⎝
(4.7)
Equations (4.5) and (4.7) are consistent with equation (4.2), as well as with equation
Q ( z ) = αQ ( αz ) which follows from equation (4.2) by replacing z with αz .
If f (s) = f (−s) (the even symmetric case), then the reflection of the triangle with
respect to the x-axis leads to the same geometry and boundary data and hence, due to
uniqueness of the exterior Dirichlet problem solution, it follows that
q(z, z) = q(z, z) .
(4.8)
Differentiating with respect to z we obtain
q z (z, z) = q z (z, z) = q z (z, z) = q z (z, z)
(4.9)
where for the last equation we have used the fact that q is real. The first and the last
equation imply
13
Q(z) = Q(z) .
(4.10)
If f (s) = −f (−s) (the odd symmetric case), reflection with respect to the x-axis yields
a Dirichlet problem with opposite data. Then uniqueness of the solution and linearity
imply
q(z, z) = −q(z, z) .
(4.11)
Q(z) = − Q(z) .
(4.12)
and hence
QED.
Proposition 4.1. Let the real function q(x,y) satisfy Laplace’s equation in the domain
Ω(e) exterior to the equilateral triangle depicted in Figure 3.1, and let the smooth
function f(s) be the prescribed Dirichlet data on each side of the triangle. Assume that
there exists a unique solution q(x,y) which has sufficient smoothness all the way to
the boundary and which vanishes at infinity. Then, the unknown functions
/2
∫
Ψ (k) =
k∈C ,
eks q N (s)ds ,
(4.13)
0
+∞
∫
G(k) =
0
+∞
∫
H(k) =
0
⎛
⎞
eks g ⎜ s +
⎟ ds ,
3⎠
⎝
Re k ≤ 0 ,
(4.14)
⎛
⎞
eks h ⎜ s +
⎟ ds ,
⎝ 2 3⎠
Re k ≤ 0 ,
(4.15)
are coupled with the known function
/2
F(k) =
∫
i
k∈C ,
e ks f (s)ds ,
(4.16)
0
by the global relation
k
2
iΨ (k) + e G(iαk) + H(−ik) = F(k) ,
−π ≤ arg k ≤ −
π
3
(4.17)
where q N is the outward normal derivative of q on the boundary of the triangle and
g, h are the following functions
h(s) = 2Q ( s ) ,
g ( s ) = 2Q ( seiπ ) ,
0<s<∞.
(4.18)
Proof. Let D denote the domain depicted in Figure 4.1, where the boundary ∂D is
determined by the following contour
14
⎧
⎫ ⎧
∂D = ⎨z = s
≤ s ≤ R ⎬ ∪ ⎨z = Reiθ 0 ≤ θ ≤
2 3
⎩
⎭ ⎩
π
i
⎧
∪ ⎨z = se 3
⎩
π⎫
⎬
3⎭
⎫ ⎧
⎫
≤ s ≤ R ⎬ ∪ ⎨z =
+ is 0 ≤ s ≤ ⎬ .
2⎭
3
2 3
⎭ ⎩
(4.19)
y
∂D
z1
2
D
π
3
z3
−
3
R
2 3
x
z2
Figure 4.1
Since q satisfies Laplace’s equation in D it follows that
∫
e − ikz Q(z)dz = 0,
Q = qz .
(4.20)
∂D
For R finite, equation (4.20) is valid for every k ∈ . However, as R → ∞ , using the
fact that
0 ≤ arg z ≤
π
,
3
(4.21)
it follows that equation (4.20) is valid for k in the sector
−π ≤ arg k ≤ −
π
.
3
(4.22)
Equation (4.20) yields
R
∫
e
− iks
π/3
Q(s)ds +
+∫ e
/2
e
iθ
⎛
⎞
− ik ⎜
+ is ⎟
⎝2 3
⎠
iθ
Q(Re )i Re dθ +
0
/2 3
0
∫
− ik Reiθ
⎛
⎞
Q⎜
+ is ⎟ ids = 0 .
⎝2 3
⎠
/ 3
∫
e
i
π
− ikse 3
i
π
3
i
π
3
Q(se )e ds
R
(4.23)
15
As R → ∞ Jordan’s Lemma implies that the second integral in (4.23) vanishes.
Using
π
i ⎞
⎛ iπ ⎞
⎛
Q ⎜ se 3 ⎟ = αQ ⎜ sαe 3 ⎟ = αQ(seiπ )
⎝
⎠
⎝
⎠
(4.24)
and
i
i
1
⎛
⎞ 1
Q⎜
+ is ⎟ = q x (s) − q y (s) = − (q N (s) + i f (s))
2
2
⎝2 3
⎠ 2
(4.25)
equation (4.23) becomes
∞
∫
e
− iks
Q(s)ds + αe
0
∫
π / 3
3
∫
e
− iks e
i
π
3
Q(s eiπ )ds
∞
/2 3
1
−
2
i
e
⎛
⎞
− ik ⎜
+ is ⎟
⎝2 3
⎠
i
(q N (s) + i f (s))ids = 0 .
(4.26)
/2
Let
2Q(s) = h(s) ,
iπ
2Q(se ) = g(s),
⎫
2 3 ⎪⎪
⎬
⎪
s>
3 ⎪⎭
s>
(4.27)
Then equation (4.26) becomes
ie
− ik
/2
2 3
∫
0
i
e ks (q N (s) + i f (s))ds +
+∞
∫
i
π
3
e− ikse g(s)ds +
/ 3
+∞
∫
e− iks h(s)ds = 0 . (4.28)
/2 3
Inserting (4.13) - (4.16) in (4.28) we arrive at the global relation (4.17).
QED.
Remark : According to the second of equations (4.27) the functions g, h assume real
values in the even symmetric cases and imaginary values in the odd symmetric case.
It turns out that the case of odd Dirichlet data leads to a problem formulated in the
interior of a convex polygon formed by three sides.
16
Theorem 4.1. Let the real function q(x,y) satisfy Laplace’s equation in the domain
Ω (e) exterior to the equilateral triangle depicted in Figure 3.1, and let the odd smooth
function f(s) be the prescribed Dirichlet data on each side of the triangle. Then the
Neumann values q Ν (s) on each side of the triangle can be obtained by solving the
Laplace equation in the domain Ω depicted in Figure 4.2 with the contour
w1 → ∞, z1, z2 , w2 → ∞ and with the following Dirichlet boundary conditions:
i
q T = f (s) ,
qT = 0
− 2 ≤s ≤
(z
on the sides
on S3
2
1,
w1 ) and
(4.29)
(z
2,
w2 ) .
Proof. The second of equations (4.3) applied on z = x, - ∞ < x < Figure 4.2) yields
Q(x) + Q(x) = 0 ,
-∞ < x < 3.
(4.30)
3 (see
(4.31)
Hence,
q x ( x,0 ) = 0,
-∞ < x < -
3 .
(4.32)
w1 → ∞
qT ≡ 0
z1
i
π
3
w3 → ∞
q T = f (s)
z3
qT ≡ 0
z2
qT ≡ 0
w2 → ∞
Figure 4.2
Therefore, the symmetry relations (4.2) and Q ( z ) = αQ ( αz ) imply that q T = 0 on
the rays ( z1, w1 → ∞ ) and ( z2 , w2 → ∞ ) . That completes the proof of the theorem.
17
Remark: A technique for the numerical evaluation of the Dirichlet to Neumann map
for the Laplace equation in the interior of a convex polygon is presented [8, 9]. Using
this technique it is straightforward to evaluate the Neumann values q Ν (s) on the
i
side z1z2 in terms of f ( s ) .
APPENDIX
We will show that
1
eik1x +ik 2 y
1
.
dk1dk 2 =
∫∫
2π 2 ik1 − k 2
x + iy
(A.1)
We introduce the polar system ρˆ , φˆ (see Figure A.1),
x = ρ cos φ, y = ρ sin φ .
(A.2)
(k, k)
(x, y) = ρ(cos φ,sin φ)
φ̂
(k1 , k 2 )
φ
(x, y)
Figure A.1
The vector k̂ has the coordinates (k1,k2) in the system xˆ , yˆ and the coordinates
(k I , k II ) in the system ρˆ , φˆ . The polar base vectors are
ρˆ = (cos φ,sin φ) ,
φˆ = (− sin φ, cos φ) .
(A.3)
18
Since ρˆ , φˆ are obtained via a rotation of the Cartesian system by an angle +φ we
have that
⎡ k1 ⎤ ⎡ cos φ − sin φ ⎤ ⎡ k Ι ⎤
⎢ k ⎥ = ⎢ sin φ cos φ ⎥ ⎢ k ⎥ .
⎦ ⎣ ΙΙ ⎦
⎣ 2⎦ ⎣
(A.4)
Also,
ik1x + ik 2 y = iρk I
(A.5)
ik1 − ik 2 = (k I + ik II )ieiφ
(A.6)
which imply that
dk1dk 2 = dk1 ∧ dk 2 = (cos φdk I − sin φdk II ) ∧ (sin φdk I + cos φdk II )
= cos 2 φdk I ∧ dk II − sin 2 φdk II ∧ dk I = dk I ∧ dk II = dk I dk II .
(A.7)
Formula (A.7) is a consequence of the fact that rotations leave the Lebesque measure
invariant.
Using (A.5), (A.6) and (A.7) we obtain
1
eik1x +ik 2 y
e − iφ
eiρk I
=
dk
dk
dk I dk II .
1
2
2π ∫∫2 ik1 − k 2
2πi ∫∫2 k I + ik II
(A.8)
Since ρ > 0 , complex integration in the lower half plane implies
+∞
∫
−∞
⎧2πieρk II ,
e
⎪
dk I = ⎨ πi,
k I + ik II
⎪ 0,
⎩
iρk I
k II < 0
k II = 0 .
k II > 0
(A.9)
Hence, from (A.8) we obtain
e − iφ
2π
∫∫
2
eiρk Ι
e − iφ
dk Ι dk ΙΙ =
k I + ik II
ρ
+∞
∫
0
e − t dt =
e − iφ
1
=
.
x + iy
ρ
(A.10)
Acknowledgement. The present work was performed under the Marie Curie Chair
of Excellence Project BRAIN, granted to the second and the third author by the
European Commission under code number EXC 023928.
19
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