Theoretical Mathematics & Applications, vol. 4, no. 4, 2014, 43-53
ISSN: 1792- 9687 (print), 1792-9709 (online)
Scienpress Ltd, 2014
Noether Theorem of a Kind
of Complete Singular Integral Equation
with Hilbert Kernel on Open Arcs
Li-xia Cao 1, Ji-ying Liu 2 and Hui-jun Zheng 3
Abstract
We considered a kind of singular integral equation with Hilbert kernel on open
arcs lying in a period strip. By regularization method, we transformed the equation
into Fredholm equation and then obtained the solvable Noether theorem for this
kind of complete singular integral equations with Hilbert kernel on open arcs.
Mathematics Subject Classification: 30E25
1
2
3
School of Mathematics and Statistics, Northeast Petroleum University, Daqing,
Heilongjiang, 163318, P.R. China. E-mail: caolixia98237@163.com
School of Mathematics and Statistics, Northeast Petroleum University, Daqing,
Heilongjiang, 163318, P.R. China. E-mail: Liujiying216@126.com
School of Mathematics and Statistics, Northeast Petroleum University, Daqing,
Heilongjiang, 163318, P.R. China. E-mail: zhenghuijun02@163
The project of this thesis is supported by “Heilongjiang Province education department
natural science research item” (12541089).
Article Info: Received: October 9, 2014. Revised: November 19, 2014.
Published online : November 25, 2014.
44
Noether Theorem of a Kind of Complete Singular Integral Equation…
Keywords: Hilbert kernel; complete singular integral equation; general solutions;
solvable Noether theorem
1 Introduction
In [1,2,3,4,5], the authors discussed the singular integral equation with
Cauchy kernel on the real half-line on the real line, or on a complicated contour.
In [6], the authors discussed Numerical solution of a singular integral equation
with Cauchy kernel in the plane contact problem. While the solvable Noether
theorem for complete singular integral equations with Hilbert kernel on open arcs,
has few been discussed. Here we study a kind of complete singular integral
equation with Hilbert kernel on open arcs, by using the regularization method for
this kind of singular integral equation, we transform the equation into Fredholm
equation and give the solvable Noether theorem for the singular integral equation.
2
The complete singular integral equation with Hilbert
kernel on open arcs
Suppose that L = ∑ j L j is a finite set of nonintersecting open arcs, with
period aπ lying entirely in the in the same periodic region S :| Re z |< aπ 2 , and
being positively oriented. We consider the following the singular integral equation
with Hilbert kernel for a Hölder continuous function ϕ (t ) on L
Kϕ ≡ A(t0 )ϕ (t0 ) +
t − t0
1
K (t0 , t ) cot
ϕ (t )dt =
f (t0 ) , t 0 ∈ L .
∫
aπ i L
a
(1)
where A(t ), B(t ), f (t ) ∈ H ( L) , K (t0 , t ) ∈ H 0 ( L × L) are given functions with
A 2 (t ) − B 2 (t ) ≠ 0 .We assume the different points c1 , c 2 , , c n are the all nodes
Li-xia Cao and Ji-ying Liu
45
on L (including the end-points of L j and the discontinuity points of A(t ), B(t )
and f (t ) ). Without loss of generality, no special condition is needed except for the
requirement that ϕ ∈ h =
h(c1 , c2 , , cq ) (q ≤ n) , that is ϕ is bounded near
c1 , c2 , and cq , while ϕ has a singularity of order less than 1 at any other nodes.
For L , we permit it to arrive at the boundary of the region S , and in this case we
regarded c and c ± aπ as the same one.
If we let B(t ) = K (t , t ) , then (1) can be rewritten as followed
Kϕ ≡ A(t0 )ϕ (t0 ) +
B(t0 )
t − t0
ϕ (t )dt + ∫ k (t0 , t )ϕ (t )dt =
f (t0 ), t0 ∈ L ,
cot
∫
L
aπ i L
a
(2)
where
k (t0 , t ) =
[ K (t0 , t ) − K (t , t )]cot [ (t − t0 ) a ] ,
(3)
Since K (t 0 , t ) ∈ H α ( 0 < α ≤ 1 ), we have
λ
k (t0 , t ) ≤ A t − t0 , 0 ≤ λ = 1 − α < 1 .
(4)
When k (t 0 , t ) satisfies (4), kϕ ≡ ∫ k (t0 , t )ϕ (t )dt becomes an F—I operator, and
L
K 0ϕ ≡ A(t0 )ϕ (t0 ) +
B(t0 )
t − t0
ϕ (t )dt ,
cot
∫
L
aπ i
a
(5)
is actually the characteristic operator of operator K . Therefore, Eq.(1) can be
rewritten as
( K 0 + k )ϕ = f ,
K ′ψ ≡ A(t0 )ψ (t0 ) −
(6)
t − t0
1
K (t , t0 )ψ (t ) cot
dt =
g (t0 ), t0 ∈ L
∫
L
aπ i
a
(7)
or
K ′ψ ≡ A(t0 )ψ (t0 ) −
t − t0
1
(7)′
B(t )ψ (t ) cot
dt ± ∫ k (t , t0 )ψ (t )dt =
g (t0 ),
∫
L
aπ i L
a
is the associative equation of Eq. (1). Setting
S (t ) = A(t ) + B (t ) , D(t ) = A(t ) − B(t ) ,
κj =
(8)
1
A(t ) − B (t )
1
D
[log
]L =
[arg ] L ,
2πi
A(t ) + B (t )
2π
S
j
j
(9)
46
Noether Theorem of a Kind of Complete Singular Integral Equation…
p
thus the index of Eq. (1) is κ = ∑ κ j .
j =1
3
Regularization for the singular integral equation with
Hilbert kernel on open arcs
We may rewrite (2) as follows
K 0ϕ = f (t 0 ) − kϕ ,
(10)
when t 0 and t1 are on the same smooth arc ( including endpoints), we have
k (t 0 , t ) = [ K (t 0 , t ) − K (t 0 , t 0 )] cot
∞
t − t0
a
α
≤ A t − t0
+ ∑ bn (t − t 0 ) n ≤ M λ ( 0 ≤ λ =1 − α < 1) ,
a
t − t 0 n =1
t − t0
so ∫ k (t0 , t )ϕ (t )dt is F—I operator, and k (t 0 , t ) ∈ H 0 ( L) , therefore f − kϕ ∈ H 0 .
L
Let
p
X ( z ) = ∏ (tan
j =1
cj λ
z
− tan ) j e Γ ( z ) ,
a
a
(11)
∗
K ∗ f ≡ A∗ (t 0 ) f (t 0 ) − B (t0 ) Z (t0 ) ∫ f (t ) cot t − t0 dt .
L Z (t )
aπ i
a
(12)
If κ ≥ 0 , then ϕ must satisfies
t
ϕ (t 0 ) = K ∗ ( f − kϕ ) + B ∗ (t 0 ) Z (t 0 ) Pκ (tan 0 ) ,
a
(13)
that is
(13)′
ϕ (t 0 ) + K ∗ kϕ = f ∗ (t 0 ) ,
where
f ∗ (t 0 ) = K ∗ f + B ∗ (t 0 ) Z (t 0 ) Pκ (tan
t0
).
a
(14)
Obviously, f ∗ (t 0 ) ∈ h . On the contrary, if ϕ ∈ h (for some Pκ ) satisfies (13)′,
then it must satisfy (13), that is ϕ is the solution for (2).
Li-xia Cao and Ji-ying Liu
47
If κ < 0 , if and only if
 f (t ) − kϕ
j −1 t
j +1 t
= 0, =
j 1, , −κ − 1,
 ∫L Z (t ) (tan a + tan a )dt


 f (t ) − kϕ (sin v − cos v tan t )dt =
0,
 ∫L Z (t )
2
2
a
(15)
Eq.(13) has solution in class h , and its unique solution is ϕ (t 0 ) = K ∗ ( f − kϕ ) ,
that is (13)′ with Pκ ≡ 0 .
Since k is a Fredholm operator, we have

∫ kϕ  tan
j −1
L
t
t
+ tan j +1  Z (t )dt
=
a
a

∫ ϕ (t )k ′  tan
j −1
L
t
t
+ tan j +1  Z (t ) dt ,=
j 1, , −κ − 1,
a
a
thus

j −1 t
j +1 t
 tan a + tan a
= ∫ ϕ (t ) Pj (t )dt , =
f (t )dt
j 1, , −κ − 1,
∫
Z (t )
L
L

 sin v − cos v tan t

2
2
a f (t )dt = ϕ (t ) P (t )dt ,
0
∫L
∫
(
)
Z
t
L
(15)′
where
{
}
j −1
j +1
Pj (t ) =
k ′  tan j −1 at + tan j +1 at  Z (t ) =
∫ [tan a1 + tan a1 ] Z (t )dt1 , j = 1,,−κ − 1 ,
t
t
L
k (t1 , t )
t
v
v
 v
v
t

P0 (t ) =
k ′  sin − cos tan  Z (t )  =
[sin − cos tan 1 ]dt1 ,
∫
L
a
2
Z (t )
a
2
2
 2

where=
Pj (t )( j 0,1, , κ − 1) are the known functions in class H 0 , and when κ ≤ 0 the
Eq. (2) in class h is equivalent to Eq. (13) ′ in class h (with Pκ ≡ 0 in (14))
and the additive condition (15) or (15)′ by the same reason to the case that
when κ ≥ 0 .
Now, Let us demonstrate K ∗ k is a weak Fredholm operator. Because
K ∗ f ≡ A∗ (t 0 ) f (t 0 ) −
B ∗ (t 0 ) Z (t 0 )
aπi
∫
L
t − t0 ,
f (t )
dt
cot
Z (t )
a
we have
K ∗kϕ = A∗ (t 0 )(kϕ )(t 0 ) −
B ∗ (t 0 ) Z (t 0 )
aπi
∫ cot
L
t − t dt
0
⋅ k (t , t1 )ϕ (t1 )dt1 .
a Z (t ) ∫L
48
Noether Theorem of a Kind of Complete Singular Integral Equation…
Since the order of Hilbert integral and common integral can be exchanged, we
obtain
K ∗ kϕ = A∗ (t 0 )(kϕ )(t 0 ) − B ∗ (t 0 ) Z (t 0 )
∫k
∗
L
(t0 , t1 )ϕ (t1 )dt1 ,
(16)
where
k ∗ (t0 , t ) =
t − t0
k (t , t1 )
1
cot
dt ,
∫
aπ i L Z (t )
a
(17)
has at the most less than one order singularity at the node t 0 and so that it is a
weak Fredholm kernel.
Applying Fredholm Theorem, we obtain the following results.
Theorem 3.1 Equation
ϕ (t 0 ) + K ∗ kϕ = f ∗ (t 0 ) , t 0 ∈ L
(18)
is solvable in class h if and only if
∫
L
f ∗ (t ) w j (t )dt = 0 ,
j = 1,2,  , v ,
(19)
where {w j (t )}1v is the set of linearly independent solutions of the homogeneous
associative equation
w(t 0 ) + k ′K ∗′ w = 0
(20)
in class h ′ .
When the solvable condition (19) is satisfied, the solution of (18) in class h can be
written as
v
ϕ (t 0 ) = Γf ∗ + ∑ C j x j (t 0 ) ,
(21)
j =1
where C1 , , Cv are arbitrary complex constants, {χ j (t )}1v is the set of linearly
independent solutions of the homogeneous equation of Eq.(18), while
Γf ∗ ≡ f ∗ (t0 ) + ∫ Γ(t0 , t ) f ∗ (t )dt ,
L
and Γ(t 0 , t ) is the generalized resolvent kernel.
Li-xia Cao and Ji-ying Liu
49
3.1 Noether theorem of a kind of singular integral equation with
Hilbert kernel on open arcs
Applying the results above and following [1], we may obtain the following
result.
Theorem 4.1 (Noether Theorem)
(a) The number of the linearly independent solutions of the homogeneous
equation for (2) ( f ≡ 0 ) in class h = h(c1 ,  , c q ) is finite, denoted by l .
(b) Eq.(2) is solvable in class h = h(c1 ,  , c q ) if and only if
∫
L
fψ j dt = 0 , j = 1, , l ′
is satisfied, where {ψ j }l ′ is the set of linearly independent solutions of the
homogeneous associative equation
K 0′ψ ≡ A(t0 )ψ (t0 ) −
t − t0
1
=
B(t )ψ (t ) cot
0
∫
L
aπi
a
in class h ′ .
(c) If κ is the index of Eq.(2) in class h = h(c1 , c2 , , cq ) , then l − l ′ = κ .
Proof.
(a) is true because that the Fredholm Eq. regularized from Kϕ = 0 has only a
finite number of linearly independent solutions and the solutions of the latter must
be the solutions of Eq. Kϕ = 0 .And (c) is clearly true.Now we aim to prove (b).
Case 1. κ ≥ 0 . Eq. (2) in class h is equivalent to Eq. (13)′, where
t
f ∗=
(t0 ) K ∗ f + B∗ (t0 ) Z (t0 ) Pκ (tan 0 )
a
with the coefficients of the polynomial
(
)
t
t
t
t0
Pκ tan
=
A0 tan κ + A1 tan κ −1 +  + Aκ −1 tan + Aκ
a
a
a
a
satisfying
50
Noether Theorem of a Kind of Complete Singular Integral Equation…
− cos
v
v
v 1
f (t )
sin
dt.
( Aκ − Aκ −2 + ) + sin ( Aκ −1 − Aκ −3 + ) =
∫
L
2
2
2 aπ i Z (t )
(22)
Without loss of generality, we let A1 , A2 ,..., Aκ are arbitrary constants, then A0
is determined by (22). On the other hand, the solvable conditions for Eq.(13 )′ are
given by (19), we put (14) for f ∗ (t 0 ) into (19), and introduce the note
δ j = ∫ w j K ∗ fdt =
, j 1, 2, …, v ,
(23)
L
then (19) can be rewritten as
κ
∑γ
k =1
jk
Ak = δ j ,
j = 1, …, v ,
(24)
where (γ jk ) is a given constant matrix independent to f (t ) .
It is always true for the operator K that
∫
L
fKgdt = ∫ gK ′fdt as long as
L
both f and g have no integrable singularity at the same knot. Thus, we may rewrite
δ j as
′
δ j = ∫ fK ∗ w j dt = ∫ fw ∗j dt ,
(25)
w∗j (t ) = K ∗′ w j =
, j 1, 2, …, v
(26)
L
L
where
are linearly independent and satisfies w ∗j ∈ h ′ .
Set rank (δ j k ) = ρ ( ρ ≤ v, ρ ≤ κ ) . Without loss of generality, we suppose that the
ρ -order determinant of (δ j k ) in the top left is not 0. Then the compatibile
conditions are given by
γ 11

γ ρ1
γ ρ + j ,1
or

γ 1ρ
δ1


 γ ρρ
 γ ρ + j,ρ
δρ
δρ+ j
= 0,
j = 1, 2, …, v − ρ ,
(27)
Li-xia Cao and Ji-ying Liu
51
ρ
δ ρ + j + ∑ a jk δ k =
0, j = 1, 2, …, v − ρ ,
(2 7)′
k =1
where (a jk ) is constant matrix independent to f (t ) .
Put (25) for δ j into (2 7)′ , we obtain the solvable conditions for Eq. (13 )′ as
follows
∫ λ (t ) f (t )dt = 0,
L
j
j = 1, 2, …, v − ρ ,
(28)
where
ρ
λ j (t ) = wρ∗ + j (t ) + ∑ a jk wk∗ (t ), =
j 1, 2, …, v − ρ ,
(29)
k =1
are definite, linearly independent functions, belonging to h ′ .
Suppose that the solvable conditions are satisfied, then Eq.(13) is solvable and
(27) is also satisfied, thus we can express { A j }1ρ using arbitrary constants { Ak }vρ +1 .
Put the expression for { A j }1ρ into (24), we get
Aj
=
κ −ρ
∑B
κ
A
+ ∑ Γ jk δ k ,
jk ρ + k
k 1=
k 1
=
j = 1, 2, …, ρ
(30)
where ( B jk ) and (Γ jk ) are all given constant matrix, independent to f (t). Put the
expression (30) for { A j }1ρ into the right side of Eq.(13), we see Eq.(13) is always
solvable for arbitrary constants { Ak }vρ +1 . From (25) , we obtain by (14) and (23) the
general solutions for Eq.(13)
ϕ (t 0 ) = Γ ∗ K ∗ f + C1 χ 1 (t 0 ) + C 2 χ 2 (t 0 ) +  Cκ + v − ρ χ κ + v − ρ (t 0 ),
(31)
where χ 2 , χ 2 , , χ v and C1 , C2 , …, Cv are same to (21), χ V +1 , χ v + 2 ,  , χ κ +v − ρ
are some definite functions, independent to f and belonging to the class h , and
for the sake of consistency, we have rewritten Aρ +1 , Aρ + 2 … Aκ as
Cv +1 , Cv + 2 , …, Cκ +v − ρ respectively, and have put the terms relating to f into the
expression Γ ∗ K ∗ f with Γ ∗ F ≡ F (t 0 ) + ∫ Γ ∗ (t 0 , t )F (t )dt , where Γ ∗ (t 0 , t ) and
L
Γ(t 0 , t ) are same except some definite terms.
52
Noether Theorem of a Kind of Complete Singular Integral Equation…
In particular, the homogeneous Eq.(2) ( f ≡ 0) is equivalent to
ϕ (t 0 ) + K ∗ kϕ = b ∗ (t 0 ) Z (t 0 ) Pκ (t 0 )
(32)
and the general solution for it is
ϕ (t0 ) = C1 χ1 (t0 ) + C2 χ 2 (t0 ) + …Cκ + v − ρ χ κ + v − ρ (t0 ) ,
where {C j }1κ +v − ρ are arbitrary constants, and
(33)
{ χ j (t )}1κ +v − ρ
is a full set of linear
independent solutions for Kϕ = 0 .
Case 2. κ < 0 . At this time, Pκ ≡ 0 and the solvable conditions for Eq.(13 )′ is
δ j = 0 ( j = 1,2, ,v) , i.e.
∫ λ (t ) f (t )dt = 0, j = 1,2, , v,
L
(34)
j
where λ j ∈ h′( j = 1,2, , n ) are definite and linearly independent. Assume (34) is
valid, then Eq.(13 )′ is solvable in class h if and only if the solvable condition
∫ λ (t ) f (t )dt = 0, j
L
j = v + 1, v + 2, …, v + σ . ( σ ≤ −κ , λ j ∈ h ′ )
(35)
Combining (34) and (35), we obtain the following solvable condition for Eq.(2) in
class h .
Now we come to the result (b). ∀ϕ ∈ h,ψ ∈ h ′ , we see ∫ ψKϕdt = ∫ ϕK ′ψdt and
L
L
the solvable necessary condition for Kϕ = f in class h is
∫
L
where
{ψ }
l′
j
1
f ψ j dt = 0, =
j 1, 2 …l ′ ,
(36)
is the is the set of linearly independent solutions of the
homogeneous associative equation K ′ψ = 0 in class h ′ .
By the preceding discussion, we see Kϕ = f is solvable in class h if and
only if both (34) and (35), i.e.
∫ λ (t ) f (t )dt = 0,
L
j
j = 1, …, m
(m ∈ 
+
, λ j ∈ h′ )
(37)
are satisfied. Assume condition (36) is valid, then for any g (t ) ∈ H with
g (c j ) = 0 =
( j 1, 2, …, n) , we have Kg ∈ H 0 . Equation Kϕ = Kg in class H and
Li-xia Cao and Ji-ying Liu
53
then in class h must be solvable. So we have
0 = ∫ λ j Kgdt = ∫ gK ′λ j dt , j = 1, …, m .
L
L
By arbitrariness of g (t ) , we see K ′λ j = 0 , that is, λ j ∈ h ′ is a solution for
K ′ψ = 0 . So every λ j must be a linear combination of {ψ k }1l ′ . Now by (36) and
(37), we see the result (b) is true.
□
References
[1] Lu J.K., Boundary Value Problems for Analytic Functions, World Sci Publ,
Singapore, 2004.
[2] D. Pylak, R. Smarzewski and M.A. Sheshko, A Singular Integral Equation
with a Cauchy Kernel on the Real Half-Line, J. Differential Equations, 41,
(2005), 1775-1788.
[3] B.G. Gabdulkhaevt and I. N. Tikhonov, Methods for Solving a Singular
Integral Equation with Cauchy Kernel on the Real Line, J. Differential
equations, 44, (2008), 980-990.
[4] M.A. Sheshko and S.M. Sheshko, Singular integral equation with Cauchy
kernel on the real axis, J. Differential Equations, 46, (2010), 568-585.
[5] M.A. Sheshko and S.M. Sheshko, Singular integral equation with Cauchy
kernel on a complicated contour, Differential Equations, 47, (2011),
1344-1356.
[6] M.R. Capobianco and G. Criscuolo, Numerical solution of a singular integral
equation with Cauchy kernel in the plane contact problem, J. Quarterly of
Applied Mathematics, 69, (2011), 79-89.