BULLETIN of the
Bull. Malaysian Math. Sc. Soc. (Second Series) 26 (2003) 13−33
MALAYSIAN
MATHEMATICAL
SCIENCES
SOCIETY
Eigenproblem of the Generalized Neumann Kernel
ALI HASSAN MOHAMED MURID AND MOHAMED M.S. NASSER
Department of Mathematics, Faculty of Science, Universiti Teknologi Malaysia,
81310 UTM Skudai, Johor, Malaysia
e-mail: [email protected] and [email protected]
To the memory of Mohamad Rashidi Md. Razali,
a friend, a colleague and source of inspiration
Abstract. Recently, the Riemann problem in the interior domain of a smooth Jordan curve was
solved by transforming its boundary condition to a Fredholm integral equation of the second kind
with the generalized Neumann kernel. The eigenvalues λ = ±1 play an important role in the
solvability of these integral equations. In this paper, the necessary and sufficient conditions for
λ = ±1 to be eigenvalues of the generalized Neumann kernel are given and the corresponding
eigenfunctions are derived. Some examples are presented.
1. Introduction
Suppose that Γ : t = t ( s ), 0 ≤ s ≤ β be a smooth Jordan curve, Ω + and Ω − its interior
and exterior respectively such that the origin of the coordinate system belongs to Ω + and
∞ belongs to Ω − . The unit tangent to Γ at the point t will be denoted by
T (t ) = t ′( s ) / t ′( s ) . Let f + (t ) and f − (t ) denote the limiting values of the analytic
function f (z ) when the point z tends to the point t ∈ Γ from inside and outside of Γ
respectively. Assume that a, b and γ be three real functions of the point t ∈ Γ all
satisfying the Hölder condition and a 2 (t ) + b 2 (t ) ≠ 0 for all t ∈ Γ . The Riemann
Problem in Ω + consists of finding all functions f = u + iv that are analytic in Ω + ,
continuous on Ω + = Ω + ∪ Γ with limiting values of the real and imaginary parts on Γ
satisfying the linear relation
a (t )u + (t ) − b(t )v + (t ) = γ (t ) , t ∈ Γ .
(1.1)
14
A.H.M. Murid and M.M.S. Nasser
Let c(t ) ≡ a (t ) + ib(t ), t ∈ Γ, the boundary condition (1.1) may be rewritten as
[
]
Re c(t ) f + (t ) = γ (t ) , t ∈ Γ .
(1.2)
We assume that c(t ) = 1 on Γ which is no loss of generality as can seen by divided
(1.2) by c(t ) . When γ (t ) ≡ 0 we are faced with the homogeneous Riemann problem
[
]
Re c(t ) f + (t ) = 0 , t ∈ Γ .
(1.3)
Similarly, the Riemann problem for exterior domain Ω − consists of finding all
functions f = u + iv that are analytic in Ω − (including at ∞ ), continuous on
Ω − = Ω − ∪ Γ , and satisfy the boundary condition
a (t )u − (t ) − b(t )v − (t ) = γ (t ) , t ∈ Γ
(1.4)
which is equivalent to
[
]
Re c(t ) f − (t ) = γ (t ) , t ∈ Γ .
(1.5)
The homogeneous problem of the exterior domain Ω − is given by
[
]
Re c(t ) f − (t ) = 0 , t ∈ Γ .
(1.6)
Recently, when a(t ) and b(t ) have continuous first order derivatives, Murid et al.
[7] solved the Riemann problem using Fredholm integral equations of the second kind.
The kernel of these integral equations is a generalization to the familiar Neumann kernel
so it will be call called generalized Neumann kernel.
This kernel is very important in solving Dirichlet problem [3] and Riemann problem
[7] using Fredholm integral equations. The solvability of the Riemann problem (1.2) and
(1.5) depends on the index of the function c(t ) with respect to the curve Γ , for
definition of the index see [2, pp. 85−89]. However, the solvability of the related integral
equations depend on the eigenvalues λ = ±1. It is found that the possibility of λ = ±1
to be an eigenvalue of the generalized Neumann kernel depends on the index of c(t ).
The organization of this paper is as follow. In Section 2, we give a brief derivation
of the integral equations with generalized Neumann kernel related to the Riemann
problem in the interior and exterior domains. Section 3 contains the proof of the
continuity of the generalized Neumann kernel and some of its properties. In Section 4,
the necessary and sufficient condition for which λ = ±1 is an eigenvalue of the
generalized Neumann kernel is given. Section 5 contained two examples. The
conclusions are given in Section 6.
Eigenproblem of the Generalized Neumann Kernel
15
2. Fredholm integral equations related to Riemann problem
With the help of the Sokhotskyi formula [2, p.25] we are able to extend the results
obtained in [6] and [7] which will play a key role in deriving an integral equation related
to (1.5). Suppose that γ is a real function defined on Γ and satisfies the Hölder
condition. Suppose also that c(t ) is a complex valued function defined on Γ such that
c ′(t ) is continuous on Γ and c(t ) ≠ 0 for all t ∈ Γ and define the function L(z ) in
C \ Γ by
L( z ) =
1
2πi
2γ (t )
∫ Γ c(t )(t − z ) dt .
(2.1)
Theorem 2.1 derives integral equation related to Riemann problem in the interior
domain Ω + of a smooth Jordan curve Γ. The proof can be found in [7].
Theorem 2.1. [7] Let Γ be a smooth Jordan curve and Ω + be its interior. Suppose
that f (z ) is an analytic solution of the Riemann problem (1.2) in the interior domain
Ω + . Define g (t ) = c(t ) f + (t ) , t ∈ Γ , then g (t ) satisfies the integral equation
g (t ) − PV
∫ Γ N (c) (t , w) g (w)
dw = −c(t ) L− (t ) .
(2.2)
where
N (c)(t , w) =
⎡ c(t ) T ( w) ⎤
Im ⎢
⎥ , t , w ∈ Γ and t ≠ w .
π
⎣ c( w) w − t ⎦
1
(2.3)
Similarly we can obtain an integral equation related to Riemann problem (1.5) in the
exterior domain Ω − with an additional condition f (∞) = 0 .
Theorem 2.2.
Let Γ be a smooth Jordan curve and Ω − be its exterior. Suppose that
f (z ) is an analytic solution of the Riemann problem (1.4) in the exterior domain Ω −
with the condition f (∞) = 0 . Define g (t ) = c(t ) f − (t ) , then g (t ) satisfies the integral
equation
g (t ) + PV
∫ Γ N (c)(t , w) g (w)
dw = c(t ) L+ (t ) .
(2.4)
16
A.H.M. Murid and M.M.S. Nasser
Proof:
Let f (z ) be a solution of (1.4) in Ω − then f (z ) is analytic in Ω − and
continuous
Ω− ,
on
f − (t ) = f (t ) , t ∈ Γ .
hence
From
(1.5)
we
have
c(t ) f (t ) + c(t ) f (t ) = 2γ (t ) which leads to
f (t ) = −
2γ (t )
c(t )
, t ∈ Γ.
f (t ) +
c(t )
c(t )
(2.5)
According to [2, p. 2], f (z ) satisfies
1
2πi
f ( w)
∫ Γ w − z = 0,
z ∈ Ω+ .
(2.6)
Taking the limit Ω + ∋ z → t ∈ Γ and applying Sokhotskyi formula to (2.6), we get
1
1
f (t ) + PV
2
2πi
f ( w)
∫Γ w − t
dw = 0 .
(2.7)
Conjugating both side of (2.7) the using (2.5), we get
−
1 c(t ) ⎡
2γ (t ) ⎤
1
f (t ) −
+ PV
2 c(t ) ⎢⎣
2πi
c(t ) ⎥⎦
c( w)
∫ Γ c(w)(w − t )
⎡
2γ ( w) ⎤
⎢ f ( w) − c( w) ⎥ dw = 0
⎣
⎦
which leads to
−
1 c(t )
1
f (t ) + PV
2 c(t )
2πi
c( w) f ( w)
dw
w −t
∫ Γ c(w)
⎛ 1 2γ (t )
1
= −⎜⎜
+ PV
πi
2
c
(
t
)
2
⎝
−
∫Γ
⎞
2γ ( w)
dw ⎟⎟ .
c( w)( w − t )
⎠
(2.8)
Using the fact that T ( w) dw = dw and from the definition of L( z ) , (2.8) becomes
1 c(t )
1
f (t ) − PV
2 c(t )
2πi
c( w) T ( w)
f ( w) dw = L+ (t )
w −t
∫ Γ c(w)
(2.9)
Equation (2.6) may be written in the form
1
1
f (t ) + PV
2
2πi
T ( w)
∫Γ w − t
f ( w) dw = 0 , t ∈ Γ
(2.10)
Eigenproblem of the Generalized Neumann Kernel
17
Multiply (2.9) by c(t ) and add the result to (2.10) multiplied by c(t ) , and using the
definitions of g (t ) and N (c) we get (2.4).
3. Generalized Neumann kernel
The kernel N (c) (t , w) defined by (2.3) is continuous at all points (t , w) ∈ Γ × Γ except
for t = w where it is undefined. In Theorem 3.2, it will be shown that if Γ is
sufficiently smooth then N (c) (t , w) is continuous even at t = w where
N (c) ( w, w) = lim N (c) (t , w) . Then, the symbol PV appears in (2.2) and (2.4) will be
t→w
dropped and the equations (2.2) and (2.4) are Fredholm integral equations of the second
kind.
Theorem 3.1. Let the smooth Jordan curve Γ : t = t ( s) , 0 ≤ s ≤ β , be such that
c ′(t ( s )) and t ′′(s) exist and are continuous on [ 0, β ] . Then the limit of N (c) (τ , w) as
τ → w exists for every w = t ( s) ∈ Γ , and
lim N (c) ( w, w) = κ ( w)
t→w
(3.1)
where
κ ( w) =
⎡ t ′′( s ) 2c ′(t ( s ))t ′( s ) ⎤
1
.
Im ⎢
−
2π t ′( s )
c(t ( s )) ⎥⎦
⎣ t ′( s )
(3.2)
Moreover this limit exists uniformly.
Proof:
Let τ = t (σ ) , w = t ( s) . Then
N (c)(τ , w) =
1
π
⎡ c(τ ) T ( w) ⎤
⎡ c(t (σ ))
t ′( s ) ⎤
1
Im ⎢
⎥ = π t ′( s ) Im ⎢ c(t ( s )) t ( s ) − t (σ ) ⎥
c
w
w
τ
−
(
)
⎣
⎦
⎣
⎦
Using
t (σ ) − t ( s ) = t ′( s)(σ − s) +
1
2
t ′′( s) (σ − s) 2 + ο ((σ − s) 2 ) ,
c(t (σ )) = c(t ( s)) + c ′(t ( s ))t ′( s)(σ − s) + ο ((σ − s))
and
1
= 1 − θ + ο (θ )
1+θ
(3.3)
18
A.H.M. Murid and M.M.S. Nasser
for θ near to 0, we have for s near enough to σ ,
⎫
1 ⎧
1 t ′′( s )
t ′( s)
(σ − s) + ο (σ − s)⎬
= −
⎨1 +
σ −s ⎩
2 2t ′( s )
t ( s ) − t (σ )
⎭
=−
⎫
1 ⎧
1 t ′′( s)
(σ − s) + ο (σ − s)⎬
⎨1 −
2 t ′( s )
σ −s⎩
⎭
= −
1
1 t ′′( s )
+
+ ο (1)
σ − s 2 t ′( s )
−1
and
c(t (σ ))
c ′(t ( s ))t ′( s )
(σ − s ) + ο (σ − s )
= 1+
c(t ( s ))
c(t ( s ))
implies
c(t (σ ))
t ′( s)
1
1 t ′′( s ) c ′(t ( s ))t ′( s)
=
+
−
+ ο (1) .
c(t ( s)) t (σ ) − t ( s ) σ − s 2 t ′( s )
c(t ( s))
(3.4)
⎡ c(t (σ ))
⎡ 1 t ′′( s) c′(t ( s))t ′( s) ⎤
t ′( s ) ⎤
= Im ⎢
−
lim Im ⎢
⎥
σ →s
c(t ( s )) ⎥⎦
⎣ c(t ( s)) t (σ ) − t ( s) ⎦
⎣ 2 t ′( s)
(3.5)
Hence
Thus from (3.3) and (3.5), we have
lim N (c) (t (σ ), t ( s )) =
σ →s
1
π t ′( s )
⎡ 1 t ′′( s ) c′(t ( s )) t ′( s ) ⎤
−
Im ⎢
c(t ( s )) ⎥⎦
⎣ 2 t ′( s )
implying (3.1) and (3.2). To show that the limit (3.5), hence also (3.1), exists uniformly,
for any τ , w ∈ Γ , τ = t (σ ) and w = t (s) , let ε be any given positive real number, we
must find δ (ε ) > 0 such that σ − s < 0 implies
⎡ c(t (ς ))
⎡ 1 t ′′( s) c′(t ( s)) t ′( s) ⎤
t ′( s) ⎤
− Im ⎢
−
Im ⎢
<ε.
⎥
c(t ( s)) ⎥⎦
⎣ c(t ( s)) t (σ ) − t ( s) ⎦
⎣ 2 t ′( s)
(3.6)
19
Eigenproblem of the Generalized Neumann Kernel
From (3.4) and (3.6), we have
⎡ c(t (σ ))
t ′( s ) ⎤
Im ⎢
− Im
(
(
))
(
) − t ( s ) ⎥⎦
σ
c
t
s
t
⎣
⎡ 1 t ′′( s ) c ′(t ( s ))t ′( s ) ⎤
1
⎡
⎤
⎢ 2 t ′( s ) − c(t ( s )) ⎥ = Im ⎢− σ − s + ο (1)⎥
⎣
⎦
⎣
⎦
= Im [ο (1) ] = ο (1)
(3.7)
Since (3.4) holds for all τ , w ∈ Γ , τ = t (σ ) and w = t (s ) , such that σ near enough to
s, thus from (3.7) there exists δ (ε ) > 0 such that for all τ , w ∈ Γ , τ = t (σ ) and
w = t (s) , σ − s < 0 implies (3.6).
Theorem 3.2. Under the hypotheses of Theorem 3.1 the kernel N (c) (t , w) defined by
⎧ 1
⎡ c(t ) T ( w) ⎤
Im ⎢
, w ≠ t , w, t ∈ Γ ,
⎪
c( w) w − t ⎥⎦
⎪π
⎣
N (c ) (t , w) = ⎨
⎡ t ′′( s ) 2c ′(t ( s )) t ′( s ) ⎤
1
⎪
Im ⎢
−
, w = t ∈ Γ,
⎪⎩ 2π t ′( s )
c(t ( s )) ⎥⎦
⎣ t ′( s )
(3.8)
is continuous on Γ × Γ .
Proof: To prove the continuity of the kernel N (c), (t , w) at a point (t0 , t0 ) ∈ Γ × Γ , it
must be shown that for any given ε > 0 , we must find δ > 0 such that
max { t − t 0 , w − t 0
}
< δ
(3.9)
implies
N (c) (t , w) − N (c) (t0 , t 0 ) < ε .
(3.10)
By triangle inequality (3.10) will hold if both
N (c) (t , w) − N (c) (t , t ) <
ε
2
(3.11)
and
N (c) (t , t ) − N (c ) (t 0 , t 0 ) <
ε
2
.
(3.12)
20
A.H.M. Murid and M.M.S. Nasser
From Theorem 3.1 the limit lim N (c) (t , w) = N (c) (t , t ) exists uniformly for all t ∈ Γ
w→t
which implies that there exists δ 1 > 0 such that w − t < δ 1 implies that the inequality
(3.11) is held. The function κ (t ) defined by (3.2) is continuous on the compact set Γ ,
hence uniform continuous. Therefore there exists δ 2 > 0 such that for all t ∈ Γ
satisfies t − t 0 < δ 1 implies κ (t ) − κ (t 0 ) <
ε
2
, then the inequality (3.12) holds. Let
1
min {δ 1 , δ 2 } , therefore for all t , w ∈ Γ , if max { t − t 0 , w − t 0 } < δ then
2
the inequality (3.10) holds, hence N (c)(t , w) is continuous at (t 0 , t 0 ) ∈ Γ × Γ.
δ =
The Neumann kernel arises frequently in the integral equation of potential theory and
conformal mapping. If c(t ) ≡ 1 the kernel (3.8) is identical with the Neumann kernel
[3, pp. 282−286] so the kernel (3.8) will be called the generalized Neumann kernel and it
will be denoted by N (c) (t , w) or only N (c) when there is no confusion. When
c(t ) ≡ 1 , we write N (t , w) or only N.
Some of the properties of the generalized Neumann kernel are listed in the
following remarks.
Remark 3.1.
It is clear that N (c) (t , w) is a real kernel thus its adjoint kernel
N (c) (t , w) is given by N * (c) (t , w) = N (c)( w, t ) = N (c) ( w, t ) . Moreover, for all
t ≠ w and t , w ∈ Γ , we have
*
N (c) ( w, t ) =
⎡ c( w) T (t ) ⎤
⎡ T (t ) / c(t ) T ( w) ⎤
1
1
Im ⎢
⎥ = − π Im ⎢ T ( w) / c( w) w − t ⎥ = − N (cˆ) (t , w) . (3.13)
−
π
(
)
c
t
t
w
⎣
⎦
⎣
⎦
Therefore N * (c) = − N (cˆ) where cˆ (t ) = T (t ) / (t ) = T (t ) c(t ) , t ∈ Γ. Since Γ is a
smooth Jordan curve, T (t ) is the unit tangent vector at t ∈ Γ, therefore T (t ) = e iθ (t )
where θ (t ) is the angle between the tangent to the contour Γ and the real axis. In going
round the contour Γ in anticlockwise direction θ (t ) acquires the increment 2π .
Therefore ind Γ (T ) = 1 and hence
ind Γ (c ) = ind Γ (T ) − ind Γ (c) = 1 − ind Γ (c) .
(3.14)
21
Eigenproblem of the Generalized Neumann Kernel
If Γ is the unit circle then T ( w) = iw and
Remark 3.2.
N (c) (t , w) =
⎡ c(t ) − c( w) + c( w) T ( w) ⎤
⎡ c( w) − c(t ) iw ⎤ 1
1
1
⎡ iw ⎤
Im ⎢
⎥ = − π Im ⎢ w − t
⎥ + π Im⎢ w − t ⎥
−
π
c
w
w
t
c
w
(
)
(
)
⎣
⎦
⎣
⎦
⎣
⎦
⎡ c( w) − c(t ) iw
c( w) − c(t ) − i w ⎤
1
−
⎥ +
⎢
πi
w
−
t
c
(
w
)
w
−
t
2
c
(
w
)
⎦
⎣
1
1 c( w) − c(t ) ⎡ w
t ⎤
=
−
⎢ c( w) + c(t ) ⎥
2π 2π
w−t
⎣
⎦
=−
1
2πi
⎡ iw
− iw ⎤
−
⎥
⎢
⎣w − t w − t ⎦
which implies that N (c) (t , w) is symmetric when Γ is the unit circle.
4. Eigenvalue of the generalized Neumann kernel N (c ) ( t, w )
In this section we give the main results of this paper. First we need the following
theorems from [2, pp. 221 & 226] with slight modifications. The first theorem discusses
the solvability of the Riemann problem in the interior domain Ω + and the second
theorem discusses the solvability in the exterior domain Ω − .
Theorem 4.1. [2, p. 222] Let x = ind Γ (c) , in the case x ≤ 0 the homogeneous
Riemann problem (1.3) has −2 x + 1 linearly independent solutions and the
non-homogeneous problem (1.2) is absolutely soluble and its solution depends linearly
on −2 x + 1 arbitrary real constants. In the case x > 0 the homogeneous Riemann
problem (1.3) has only the trivial solution and the non-homogeneous problem (1.2) is
soluble only if 2 x − 1 conditions are satisfied. If the latter conditions are satisfied the
non-homogeneous problem has a unique solution.
Theorem 4.2. [2, p. 226]
Let x = ind Γ (c) , in the case x ≥ 0 the homogeneous
Riemann problem (1.6) in the exterior domain Ω − has 2 x + 1 linearly independent
solutions and the non-homogeneous problem (1.4) is absolutely soluble and its solution
depends linearly on 2 x + 1 arbitrary real constants. In the case x < 0 the homogeneous
Riemann problem (1.6) in the exterior domain Ω − has only the trivial solution and the
non-homogeneous problem is soluble only if −2 x − 1 conditions are satisfied. If the
latter conditions are satisfied the non-homogeneous problem has a unique solution.
Remark 4.1.
If f (z ) is any analytic function in Ω − and vanishes at infinity then its
Taylor series in the vicinity of z = ∞ is given by f ( z ) =
a1
z
+
a2
z2
+
a3
z3
+
. Suppose
that the function f 1 ( z ) is defined by f 1 ( z ) = z f ( z ) then its Taylor series in the vicinity
22
A.H.M. Murid and M.M.S. Nasser
of z = ∞ is given f1 ( z ) = a1 +
a2
z
+
a3
z2
+
and hence f 1 ( z ) is analytic in Ω − and
bounded at infinity.
Corollary 4.1.
Let x = ind Γ (c) , if x > 0 then the homogeneous Riemann problem
(1.6) in the exterior domain Ω − with the condition f (∞) = 0 has 2 x − 1 linearly
independent solutions. If x ≤ 0 then the homogeneous Riemann problem (1.6) in the
exterior domain Ω − with the condition f (∞) = 0 has only the trivial solution.
Proof: To prove this Corollary, we shall prove that the homogeneous Riemann problem
(1.6) with the condition f (∞) = 0 is equivalent to the homogeneous Riemann problem
[
]
Re c1 (t ) f1− (t ) = 0
(4.1)
where c1 (t ) = c(t ) / t , t ∈ Γ , and f 1 is merely analytic at infinity. Suppose that f (z )
is a solution to (1.6) with f (∞) = 0 and define f 1 ( z ) = z f ( z ) , z ∈ Ω − . According to
the Remark 3.1
f1 ( z)
is analytic in
Ω−
and bounded at infinity. Since
f 1− (t ) = t f − (t ) , t ∈ Γ we get f 1 ( z ) is a solution to (4.1). Similarly let f 1 ( z ) is a
solution to (4.1) and define f ( z ) = f1 ( z ) / z , z ∈ Ω − . Then f (z ) is analytic in Ω − ,
vanishes at infinity and f − (t ) = f 1− (t ) / t , t ∈ Γ. Thus f (z ) is a solution to (1.6)
with f (∞) = 0. Therefore the homogeneous Riemann problem (1.6) with the condition
f (∞) = 0 is equivalent to the homogeneous Riemann problem (4.1) in Ω − . Let
x1 = ind Γ (c1 ) then x1 = ind Γ (c) − ind Γ (t ) = x − 1 . If x > 0 then x1 ≥ 0 , from
Theorem 4.2 the homogeneous Riemann problem (4.1) and hence (1.6) with the condition
f (∞) = 0 has exactly 2 x − 1 real linearly independent solution. If x ≤ 0 then
x1 < 0 , according to Theorem 4.2 the homogeneous Riemann problem (4.1) and hence
(1.6) with the condition f (∞) = 0 has only the trivial solution.
The following theorems discuss the eigen problem of the generalized Neumann
kernel. We shall discuss only the cases λ = ±1 which are very important in the
discussion of the solvability of the Fredholm integral related to Riemann problem.
Theorem 4.3.
If x = ind Γ (c) > 0 , then λ = −1 is an eigenvalue of the generalized
Neumann kernel N (c) (t , w) .
Since x > 0 , according to Corollary 4.1, the homogeneous Riemann problem
(1.6) with the condition f (∞) = 0 has 2 x − 1 linearly independent solutions in
Proof:
23
Eigenproblem of the Generalized Neumann Kernel
Ω − , f j ( z ) , j = 1, 2,
, 2x − 1 .
From
Theorem
2.2
c (t ) f j− (t )
satisfy
the
homogeneous Fredholm integral equation
c(t ) f j− (t ) +
∫ Γ N (c) (t , w) c( w)
f j− ( w) dw = 0 .
(4.2)
Hence λ = −1 is an eigenvalue of N (c) .
Theorem 4.4. If x = ind Γ (c) > 0 and φ (t ) is a real eigenfunction of the generalized
Neumann kernel N (c) (t , w) corresponding to the eigenvalue λ = −1 then the function
Φ( z ) defined by
Φ( z ) =
1
2π i
iφ ( w)
∫ Γ c(w) (w − z ) dw
(4.3)
is a solution to the homogeneous Riemann problem (1.6) in Ω − .
The function Φ(z ) defined by (4.3) is analytic in (C \ Γ) ∪ {∞} . Taking the
Proof:
+
limit Ω z → t ∈ Γ and using the Sokhotskyi formula to (4.3), we get
2c(t ) Φ + (t ) = iφ (t ) +
1
π
c(t ) T ( w)
∫ Γ c(w) (w − t ) φ ( w)
dw
(4.4)
Taking the imaginary part of both sides of (4.4) we have
[
2 Im c(t )Φ + (t )
]
= φ (t ) +
∫ Γ N (t , w) φ ( w)
dw .
(4.5)
Since φ is an eigenfunction of N (c) (t , w) corresponding to the eigenvalue λ = −1 ,
(4.5) reduces to the following homogeneous Riemann problem in Ω +
[
]
Im c(t ) Φ + (t ) = 0 .
(4.6)
Since x > 0 , according to Theorem 4.1 the problem (4.6) has only the trivial solution.
Therefore Φ ( z ) = 0 for all z ∈ Ω + . Consequently, according to [2, p. 25] the function
Φ( z ) defined by (4.3) satisfies
c (t )Φ − (t ) = −iφ (t ) .
(4.7)
24
A.H.M. Murid and M.M.S. Nasser
Taking the real part of both sides of
homogeneous Riemann problem (1.6).
(4.7), we find that Φ(z ) is a solution of
Corollary 4.2. If x = ind Γ (c) > 0 then the eigenfunctions of N (c) corresponding
to the eigenvalue λ = −1 are
φ j (t ) = ic(t ) f j− (t ) , t ∈ Γ , j = 1, 2,
, 2x − 1
(4.8)
where f j are the linearly independent solution of the homogeneous Riemann problem
(1.6) in the exterior domain Ω − with the condition f (∞) = 0 .
Proof:
From Theorem 4.3, λ = −1 is an eigenvalue of N (c) and the functions
φ j (t ) ≡ ic (t ) f j− (t ) , j = 1, 2,
, 2 x + 1 are linearly independent solutions of the
homogeneous Fredholm integral equation
φ (t ) +
∫ Γ N (c) (t , w) φ (w)
dw = 0.
(4.9)
Since Re [ c (t ) f j− (t ) ] = 0 , hence φ j (t ) are real eigenfunctions of N (c) (t , w)
corresponding to the eigenvalue λ = −1 . We next show that these are the only
independent eigenfunctions of N (c) (t , w) corresponding to the eigenvalue λ = −1 .
Let φ be any real eigenfunction of N (c) (t , w) corresponding to the eigenvalue λ = −1 .
From Theorem 4.4 the function G ( z ) defined by
G( z) =
1
2π i
iφ (t )
∫ Γ c(t ) (t − z )
(4.10)
dt .
is a solution to the homogeneous Riemann problem (1.6) with G (∞) = 0 and
c(t ) G − (t ) = −iφ (t ) . Since
f j , j = 1, 2,
, 2 x − 1 are the linearly independent
solutions of the homogeneous Riemann problem (1.6) with
G( z) =
∑
2 x −1
k = 1 bk
f (∞) = 0 , hence
f k ( z ) . Consequently,
φ (t ) = ic (t )G − (t ) =
2 x −1
∑k =1
ibk c(t ) f k− (t ) =
2 x −1
∑k =1
bk φ k (t ) .
Therefore φ1 , φ 2 , , φ 2 x − 1 are the only linearly independent real eigenfunctions of
N (c) (t , w) corresponding to the eigenvalue λ = −1.
25
Eigenproblem of the Generalized Neumann Kernel
Theorem 4.5.
If x = ind Γ (c) ≤ 0 , then λ = 1 is an eigenvalue of the generalized
Neumann kernel N (c) .
Proof: Since x ≤ 0 , according to Theorem 4.1 the homogeneous Riemann problem
(1.3) has −2 x + 1 linearly independent solution f j ( z ) , j = 1, 2, ,−2 x + 1 , z ∈ Ω + .
From theorem 2.1 c (t ) f j+ (t ) satisfy the homogeneous Fredholm integral equation
c(t ) f j+ (t ) −
∫ Γ N (c) (t , w) c (w)
f j+ ( w) dw = 0 ,
(4.11)
and hence λ = 1 is an eigenvalue of N (c) .
If x = ind Γ (c) ≤ 0 and φ (t ) is a real eigenfunction of the generalized
Neumann kernel N (c) (t , w) corresponding to the eigenvalue λ = 1 then the function
Φ(z ) defined by
Theorem 4.6.
Φ( z ) =
1
2π i
iφ ( w)
∫ Γ c (w) (w − z ) dw
(4.12)
is a solution to the homogeneous Riemann problem (1.3) in Ω + .
Proof:
The function Φ( z ) defined by (4.12) is analytic in (C \ Γ) ∪ {∞} with
Φ (∞) = 0 . Taking the limit Ω − z → t ∈ Γ and using the Sokhotskyi formula to
(4.12), we get
2c(t ) Φ − (t ) = −iφ (t ) +
1
π
c(t )T ( w)
∫ Γ c( w)(w − t ) φ (w)
dw
(4.13)
Taking the imaginary part of both sides of (4.13) we have
[
]
2 Im c(t )Φ − (t ) = −φ (t ) +
Since φ is an eigenfunction of
∫ Γ N (t , w) φ ( w)
dw .
(4.14)
N (c) (t , w) corresponding to the eigenvalue λ = 1 ,
(4.14) reduces to the following homogeneous Riemann problem in Ω − ,
[
Im c(t ) Φ − (t )
]
= 0.
(4.15)
26
A.H.M. Murid and M.M.S. Nasser
with the condition Φ (∞ ) = 0 . Since x ≤ 0 , according to Corollary 4.1 the problem
(4.15) has only the trivial solution. Therefore Φ( z ) = 0 for all z ∈ Ω − . Consequently,
according to [2, p. 25] the function Φ(z ) defined by (4.12) satisfies
c(t ) Φ + (t ) = iφ (t ) .
(4.16)
Taking the real part of both sides of (4.16), we find that Φ ( z ) is a solution of (1.3).
Corollary 4.3. If x = ind Γ (c) ≤ 0 then the eigenfunctions of N (c ) corresponding to
the eigenvalue λ = 1 are
φ j (t ) = ic (t ) f j+ (t ) , t ∈ Γ , j = 1, 2,
, − 2x + 1
(4.17)
where f j are the linearly independent solution of the homogeneous Riemann problem
(1.3) in the interior domain Ω + .
Proof:
From Theorem 4.5, λ = 1 is an eigenvalue of N (c) and the functions
φ j (t ) ≡ ic (t ) f j+ (t ) , j = 1, 2,
, − 2 x + 1 are linearly independent solutions of the
homogeneous Fredholm integral equation
ϕ (t ) − ∫ Γ N (c) (t , w) ϕ ( w) dw = 0 .
Since Re [c (t ) f j+ (t )] = 0 ,
hence φ j (t )
are real eigenfunctions of the
(4.18)
N (c)
corresponding to the eigenvalue λ = 1. We next show that these are the only
independent eigenfunctions of N (c) corresponding to the eigenvalue λ = 1. Let φ be
any real eigenfunction of N (c) corresponding to the eigenvalue λ = 1. From Theorem
4.6 the function G (z ) defined by
G( z) =
1
2π i
iφ (t )
∫ Γ c(t ) (t − z )
(4.19)
dt .
is a solution to the homogeneous Riemann problem (1.3) and c(t )G + (t ) = iφ (t ) .
Therefore G ( z ) =
∑−k 2=x1+ 1 bk f k ( z ) and hence
−2 x + 1
−2 x + 1
ϕ (t ) = −ic(t )G − (t ) = −∑k = 1 bk ic(t ) f k (t ) = −∑k = 1 bk φ k (t ) .
(4.20)
27
Eigenproblem of the Generalized Neumann Kernel
Therefore φ1 , φ 2 q ,
, φ − 2 x + 1 are the only linearly independent real eigenfunctions of
N (c) corresponding to the eigenvalue λ = 1.
Corollary 4.4.
Suppose
x = ind Γ (c) ≤ 0 .
that
If
φ j (t ) , t ∈ Γ, j = 1, 2,
,
−2 x + 1 are linearly independent real eigenfunctions of N (c) corresponding to the
eigenvalue λ = 1 then the general solution of the homogeneous Riemann problem (1.3)
is given by
f h ( z) =
where a j , j = 1, 2,
Proof.
− 2 x +1
∑ j =1
aj
1
2π i
iφ j (t )
∫ Γ c(t ) (t − z ) dt
(4.21)
, − 2 x + 1 are arbitrary real constants.
From Theorem 4.6 the function f h (z ) is a solution to (1.3). Since φ j are
linearly independent, then so are the integrals in (4.21). According to Theorem 4.1 and
since f h (z ) contains −2 x + 1 arbitrary real constants, the functions f h (z ) is the general
solution to(1.3).
The following two Theorems are proved only for the unit disk where the kernel
N (c) (t , w) in this case is symmetric.
Theorem 4.7. Suppose that Γ is the unit disk. If x = ind Γ (c) ≤ 0 then λ = −1 is not
an eigenvalue of the generalized Neumann kernel N (c) .
Proof: In accordance with the Fredholm’s Alternative to prove that λ = −1 is not an
eigenvalue to N (c) it is sufficient to prove that the homogeneous equation
ρ (t ) + ∫ Γ N (c) (t , w) ρ ( w) dw = 0
(4.22)
has only the trivial solution. Let ρ (t ) be any solution of (4.22) and let us set
H ( z) =
1
2π i
ρ ( w)
∫ Γ c( w) ( w − z ) dw .
(4.23)
Taking the limit Ω + z → t ∈ Γ and using the Sokhotskyi formula to (4.23), we get
2c(t ) H + (t ) = ρ (t ) + PV
1
πi
c(t ) ρ ( w)
dw
w−t
∫ Γ c(w)
(4.24)
28
A.H.M. Murid and M.M.S. Nasser
Taking the real parts of both sides of (4.24) and using (4.22) we conclude that H (z ) is a
solution to the homogeneous Riemann problem (1.3). Since x ≤ 0 , according to
Corollary 4.3 the kernel N (c) has an eigenvalue λ = 1 with the corresponding linearly
independent real eigenfunctions φ1 , φ 2 , , φ − 2 x + 1 . Since Γ is the unit circle then
N (c) (t , w) is symmetric and the eigenfunctions φ j may be assumed to be orthonormal
[9, p. 129], i.e.
∫ Γ φ i (t ) φ j (t )
dt = δ ij , i, j = 1, 2,
, − 2x + 1 ,
(4.25)
where δ ij = 1 if i = j and δ ij = 0 if i ≠ j . From Corollary 4.4 the general solution
of the homogeneous Riemann problem (1.3) is given by
f h ( z) =
where ai , i = 1, 2,
− 2 x +1
∑ i =1
ai
1
2π i
iφi (t )
∫ Γ c(t )(t − z )
dt , z ∈ Ω + ,
(4.26)
, − 2 x + 1 are arbitrary real constants. Thus there exists −2 x + 1
, a − 2 x +1 such that
certain real constants a1 , a 2 ,
H ( z) =
1
2πi
− 2 x +1
ρ ( w)
∫ Γ c(w) (w − z )
dw =
1
2πi
∫Γ
i ∑ i =1
aiφi ( w)
c( w) ( w − z )
dw , z ∈ Ω + .
(4.27)
Letting
φ (t ) =
−2 x + 1
∑ i =1
aiφ i (t ) , t ∈ Γ ,
(4.28)
then we get from (4.27)
1
2πi
∫Γ
ρ ( w) − iφ ( w)
c ( w) ( w − z )
dw = 0 , z ∈ Ω +
(4.29)
According to [2, p. 25] the function G (z ) defined in Ω − by
G( z) = −
is analytic in
1
2πi
∫Γ
ρ ( w) − i φ ( w)
c( w) ( w − z )
dw = 0 , z ∈ Ω −
(4.30)
Ω− , vanishes at infinity and satisfies
c(t )G − (t ) = ρ (t ) − iφ (t ) , t ∈ Γ .
(4.31)
29
Eigenproblem of the Generalized Neumann Kernel
Letting G ( z ) ≡ iG ( z ) , then from (4.31) include that G (z ) is a solution to the Riemann
problem
[
]
Re c(t ) G − (t ) = φ (t )
(4.32)
in Ω − with the condition G (∞ ) = 0 . Since x ≤ 0 , according to [2, p. 300], there exists
a real function η (t ) , t ∈ Γ depends on −2 x + 1 real arbitrary constants such that
G( z) =
1
2πi
η (t )
∫ Γ c(t )(t − z ) dt .
(4.33)
Taking the limit Ω − z → t ∈ Γ and using the Sokhotskyi formula to (4.33), we get
2c(t )G − (t ) = −η (t ) + PV
1
πi
c(t ) η ( w)
∫ Γ c(w) w − t dw
(4.34)
Taking the real parts of both sides of (4.24) and using (4.32) we conclude that η is a
solution to Fredholm integral equation
η (t ) − ∫ Γ N (c) (t , w) η ( w) dw = −2φ (t ) .
(4.35)
Since λ = 1 is an eigenvalue, in accordance with Fredholm alternative [4, p. 45] φ (t )
must be orthogonal with all eigenfunctions of the adjoint kernel N (c) ( w, t ) . Since
N (c) (t , w) is symmetric, hence φ (t ) must be orthogonal with φ1 , φ 2 ,
Therefore for all j = 1, 2,
, − 2x + 1
we must have
, φ − 2 x +1 .
∫ Γ φ (t ) φ j (t ) | dt |= 0 .
Using
(4.28) we get
1
2π i
− 2 x +1
∑ i =1
ai
∫ Γ φi (t ) φ j (t )
dt = 0 , j = 1, 2,
, − 2 x + 1.
Using (4.25) in (4.36) we conclude that a j = 0 for all
Consequently
φ (t ) = 0 .
Substituting
φ (t ) = 0
in
j = 1, 2,
(4.32)
(4.36)
, − 2x + 1 .
we
have
−
Re [ c(t )G (t ) ] = 0 , t ∈ Γ with G (∞) = 0 . Since x ≤ 0 , then from Corollary 4.1 we
get G ( z ) = 0
for all z ∈ Ω − and hence G − (t ) = 0 , t ∈ Γ which implies that
G − (t ) = 0 , t ∈ Γ . Substituting G − (t ) = 0 and φ (t ) = 0 in (4.31) we get ρ (t ) = 0 .
Our assertion is thereby proved.
30
A.H.M. Murid and M.M.S. Nasser
Theorem 4.8. Suppose that Γ is the unit circle. If x = ind Γ (c) > 0 then λ = 1 is not
an eigenvalue of the generalized Neumann kernel N (c) (t , w) .
Proof. Since Γ is the unit circle, from Remark 3.2 the kernel N (c) (t , w) is symmetric.
In accordance with the Fredholm’s Alternative to prove λ = 1 is not an eigenvalue to
N (c) it is sufficient to prove that the homogeneous equation
μ (t ) − ∫ Γ N (c)(t , w) μ ( w) dw = 0
(4.37)
or, since N (c) (t , w) is symmetric, the associated homogenous equation
μ (t ) − ∫ Γ N (c)( w, t ) μ ( w) dw = 0
only the trivial solutions. Let μ (t )
and c1 (t ) = T (t ) / c(t ) , t ∈ Γ . From Remark
x1 = ind Γ (c1 ) = 1 − x ≤ 0 and (4.38) becomes
has
μ (t ) +
(4.38)
be
any solution of (4.38)
3.1, N (c) ( w, t ) = − N (c1 ) (t , w) ,
∫ Γ N (c1 )(t , w) μ (w) dw
= 0.
(4.39)
In view of Theorem 4.7 λ = −1 is not an eigenvalue of N (c1 ) and hence μ (t ) = 0 .
Consequently λ = 1 is not an eigenvalue of N (c) .
5. Examples
Example 1.
Suppose that c(t ) ≡ 1 and Γ is any smooth Jordan curve. Since
x = Ind Γ c = 0 , according to Corollary 4.3 N (c) has a simple eigenvalue λ = 1 .
Since the homogeneous Riemann problem Re [ f (t )] = 0 has only one independent
solution f ( z ) = iω , ω is arbitrary real number. Therefore the eigenfunction of N (c)
corresponding to the eigenvalue λ = 1 is θ (t ) = 1 .
Example 2. Consider the case when c(t ) = t n , n is an integer and Γ is the unit
circle. First we consider the case n ≤ 0 . From Corollary 4.3 λ = 1 is an eigenvalue to
N (c) with −2n + 1 linearly independent eigenfunctions. According to [2, p. 221] the
homogeneous Riemann problem Re [c(t ) f (t )] = 0 has the general solution
f H ( z) =
−2 n + 1
∑ k =1
ck f k ( z )
(4.40)
31
Eigenproblem of the Generalized Neumann Kernel
where
, c− 2 n +1 are arbitrary real constants and for k = 1, 2,
c1 , c2 ,
,− n,
(
)
f 1 ( z ) = iz − n , f 2 k ( z ) = z − n − k − z − n − k , f 2 k + 1 ( z ) = i z − n + k + z − n − k .
(4.41)
Therefore
(
)
f H ( z ) = c1 iz − n +
∑k =1 c 2 k ( z − n + k
−n
)
∑ k =1 c 2 k i ( z − n + k
−n
− z −n − k +
+ z −n − k
)
(4.42)
and
ic (t ) f H+ (t ) = it n f H+ (t )
{
−n
(
)
−n
(
= it n c1 (it − n ) + ∑k =1 c 2 k t − n + k − t − n − k + ∑k =1 c 2 k + 1 i t − n + k + t − n − k
)}
= −c1 + 2i ∑k =1 ( c 2 k cos ks − c 2 k + 1 sin ks )
−n
where c1 , c2 ,
, c− 2 n +1 are arbitrary real number.
n ≤ 0 , the eigenfunctions of N (c) are given by
Therefore from theorem 4.2 if
θ1 ( s) = 1, θ 2k ( s ) = cos ks, θ 2k +1 ( s) = sin ks, k = 1, 2,
,− n
(4.43)
Next we consider the case n > 0 . From Corollary 4.2 λ = −1 is an eigenvalue to N (c)
Let c1 (t ) = t n −1 , according to
with 2n − 1 linearly independent eigenfunctions.
[2, p. 225] the homogeneous Riemann problem Re [c1 (t ) f − (t )] = 0 has the general
solution
f H ( z) =
where c1 , c2 ,
2 n −1
∑ k =1
(4.44)
ck f k ( z )
, c2 n −1 are arbitrary real constants and for k = 1, 2,
, n − 1,
(
)
f1 ( z ) = i z − n +1 , f 2 k ( z ) = z − n +1+ k − z − n +1− k , f 2 k +1 ( z ) = i z − n +1+ k + z − n +1− k . (4.45)
Therefore
(
f H ( z ) = c1 iz − n + 1
)+
∑k =1
+
∑k =1
(
)
n −1
c 2k z − n +1+ k − z − n +1− k
n −1
c2k i z − n +1+ k + z − n +1− k
(
)
(4.46)
32
A.H.M. Murid and M.M.S. Nasser
and
ic1 (t ) f H (t ) = it n −1 f H (t )
{ (
)
n −1
(
)
n −1
(
= it n −1 c1 it − n + 1 + ∑k =1 c 2 k t − n + 1 + k − t − n + 1 − k + ∑k =1 c 2 k i t − n + 1 + k + t − n + 1 − k
)}
= −c1 + 2i ∑ k =1 (c2 k cos ks − c2 k +1 sin ks )
n −1
, c2 n −1 are arbitrary real number. Therefore from theorem 4.2 if n > 0 ,
the eigenfunctions of N (c) are given by
where c1 , c2 ,
θ1 ( s) = 1, θ 2k ( s) = cos ks, θ 2k +1 ( s) = sin ks, k = 1, 2,
,n −1
(4.47)
6. Conclusions
The solvability of the Riemann problem (1.2) depends on the index of the function c(t )
on Γ and the solvability of the Fredholm integral equation (2.2) related to the Riemann
problem (1.2) and the integral equation (2.4) related to the Riemann problem (1.5)
depend on whether λ = 1 or λ = −1 is an eigenvalue of the generalized Neumann
kernel. In this paper we presented the relation between the index of c(t ) and the
eigenvalues λ = ±1 of the generalized Neumann kernel N (c) . Theorems 4.7 and 4.8
are proved only for the unit circle and these proofs need to be extended for an arbitrary
smooth Jordan curve.
Acknowledgement. This work was supported in part by the Research Management
Centre, UTM, Project Vote: 74049. This support is gratefully acknowledged. We also
would like to acknowledge the support and contribution of our late colleague Professor
Dr. Mohamad Rashidi Md. Razali. Thanks are also due to an unknown referee for
suggesting a number of improvements.
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Keywords: Riemann problem, Fredholm integral equation, eigenvalue, index of functions.