A high-order iterative scheme is established in order to get a convergent sequence at a rate of order
π
(
π
β₯
1
) to a local unique weak solution of a nonlinear Kirchhoff wave equation in the unit membrane. This extends a recent result in (EJDE, 2005, No. 138) where a recurrent sequence converges at a rate of order 2.
1. Introduction In this paper we consider the initial and boundary value problem
π’
π‘
π‘
ξ
β
β
π’
β
π΅
π
β
β
2
0
π’
ξ
ξ
π
π
+
1
π
π’
π
ξ
|
|
|
|
=
π
(
π
,
π‘
,
π’
)
,
0
<
π
<
1
,
0
<
π‘
<
π
,
l
i
m
π
β
0
+
β
π
π’
π
|
|
|
|
π’
(
π
,
π‘
)
<
β
,
π
(
1
,
π‘
)
+
β
π’
(
1
,
π‘
)
=
0
,
π’
(
π
,
0
)
=
Μ
π’
0
(
π
)
,
π’
π‘
(
π
,
0
)
=
Μ
π’
1
(
π
)
,
(
1
.
1
)
where
π΅
,
π
,
Μ
π’
0
,
Μ
π’
1
are given functions satisfying conditions specified later,
β
π’
π
β
2
0
=
β«
1
0
π
|
π’
π
(
π
,
π‘
)
|
2
π
π
, and
β
>
0
is a given constant.
Equation (1.1 )1 herein is the bidimensional nonlinear wave equation describing nonlinear vibrations of the unit membrane
Ξ©
1
=
{
(
π₯
,
π¦
)
βΆ
π₯
2
+
π¦
2
<
1
}
. In the vibration process, the area of the unit membrane and the tension at various points change in time. The condition on the boundary
π
Ξ©
1
describes elastic constraints, where the constant
β
1
has a mechanical signification. The boundary condition
|
l
i
m
π
β
0
+
β
π
π’
π
(
π
,
π‘
)
|
<
β
is satisfied automatically if
π’
is a classical solution of the problem (1.1 ), for example, with
π’
β
C
1
(
[
0
,
1
]
Γ
(
0
,
π
)
)
β©
πΆ
2
(
(
0
,
1
)
Γ
(
0
,
π
)
)
. This condition is also used in connection with Sobolev spaces with weight
π
(see [1 –3 ]).
Equation (1.1 )1 is related to the Kirchhoff equation
π
β
π’
π‘
π‘
=
ξ©
π
0
+
πΈ
β
ξ
2
πΏ
πΏ
0
|
|
|
|
π
π’
|
|
|
|
π
π¦
(
π¦
,
π‘
)
2
ξͺ
π’
π
π¦
π₯
π₯
(
1
.
2
)
presented by Kirchhoff in 1876 (see [4 ]). This equation is an extension of the classical D’Alembert wave equation which considers the effects of the changes in the length of the string during the vibrations. The parameters in (1.2 ) have the following meanings:
π’
is the lateral deflection,
πΏ
is the length of the string,
β
is the area of the cross-section,
πΈ
is the Young modulus of the material,
π
is the mass density, and
π
0
is the initial tension.
The Kirchhoff wave equation of the form (1.1 )1 received much attention. Many interesting results about the existence, stability, regularity in time variable, asymptotic behavior, and asymptotic expansion of solutions can be found, for example, in [2 , 3 , 5 –14 ] and references therein.
In [2 ], in a special case, sufficient conditions were established for a quadratic convergence to the solution of (1.1 ) with
π
(
π
,
π‘
,
π’
)
=
π
(
π
,
π’
)
and
π΅
(
β
π’
π
β
2
0
)
=
π
0
+
β
π’
π
β
2
0
,
π
0
>
0
. Based on the ideas about recurrence relations for a third-order method for solving the nonlinear operator equation
πΉ
(
π’
)
=
0
in [15 ], we extend the above result by the construction of a high-order iterative scheme for (1.1 )1 , where
π
and
π΅
are more generalized.
In this paper, we associate with (1.1 )1 a recurrent sequence
{
π’
π
}
defined by
π
2
π’
π
π
π‘
2
ξ
β
β
π’
β
π΅
π
π
β
β
2
0
ξ
ξ΅
π
2
π’
π
π
π
2
+
1
π
π
π’
π
ξΆ
=
π
π
π
β
1
ξ
π
=
0
1
π
π
!
π
π
π
π’
π
ξ·
π
,
π‘
,
π’
π
β
1
π’
ξΈ
ξ·
π
β
π’
π
β
1
ξΈ
π
,
(
1
.
3
)
0
<
π
<
1
,
0
<
π‘
<
π
, with
π’
π
satisfying (1.1 )2-3 . The first term
π’
0
is chosen as
π’
0
β‘
0
. If
π΅
β
πΆ
1
(
β
+
)
and
π
β
πΆ
π
(
[
0
,
1
]
Γ
β
+
Γ
β
)
, we prove that the sequence
{
π’
π
}
converges at a rate of order
π
to a unique weak solution of the problem (1.1 ). This result is a relative generalization of [2 , 3 , 8 , 9 , 14 , 16 ].
2. Preliminary Results, Notations, Function Spaces Put
Ξ©
=
(
0
,
1
)
. We omit the definitions of the usual function spaces
πΆ
π
(
Ξ©
)
,
πΏ
π
(
Ξ©
)
,
π»
π
(
Ξ©
)
, and
π
π
,
π
(
Ξ©
)
. For any function
π£
β
πΆ
0
(
Ξ©
)
we define
β
π£
β
0
as
β
π£
β
0
β«
=
(
1
0
π
π£
2
(
π
)
π
π
)
1
/
2
and define the space
π
0
as completion of the space
πΆ
0
(
Ξ©
)
with respect to the norm
β
β
β
0
. Similarly, for any function
π£
β
πΆ
1
(
Ξ©
)
we define
β
π£
β
1
as
β
π£
β
1
=
(
β
π£
β
2
0
+
β
π£
π
β
2
0
)
1
/
2
and define the space
π
1
as completion of the space
πΆ
1
(
Ξ©
)
with respect to the norm
β
β
β
1
. Note that the norms
β
β
β
0
and
β
β
β
1
can be defined, respectively, from the inner products
ξ
β¨
π’
,
π£
β©
=
1
0
π
π’
(
π
)
π£
(
π
)
π
π
,
β¨
π’
,
π£
β©
+
β¨
π’
π
,
π£
π
β©
.
(
2
.
1
)
Identifying
π
0
with its dual
π
ξ
0
we obtain the dense and continuous embedding
π
1
βͺ
π
0
β‘
π
ξ
0
βͺ
π
ξ
1
. The inner product notation will be reutilized to denote the duality pairing between
π
1
and
π
ξ
1
.
We then have the following lemmas, the proofs of which can be found in [1 ].
Lemma 2.1. There exist two constants
πΎ
1
>
0
and
πΎ
2
>
0
such that, for all
π£
β
πΆ
1
(
Ξ©
)
, we have (i)
β
π£
π
β
2
0
+
π£
2
(
1
)
β₯
β
π£
β
2
0
,
(ii)
|
π£
(
1
)
|
β€
πΎ
1
β
π£
β
1
,
(iii)
β
π
|
π£
(
π
)
|
β€
πΎ
2
β
π£
β
1
,
f
o
r
a
l
l
π
β
Ξ©
.
Lemma 2.2. The embedding
π
1
βͺ
π
0
is compact.
Remark 2.3. In Lemma 2.1 , the two constants
πΎ
1
and
πΎ
2
can be given explicitly as
πΎ
1
=
ξ
β
1
+
2
and
πΎ
2
=
ξ
β
1
+
5
. We also note that
l
i
m
π
β
0
+
β
π
π£
(
π
)
=
0
for all
π£
β
π
1
(see [17 , page 128/Lemma 5.40]). On the other hand, by
π»
1
(
π
,
1
)
βͺ
πΆ
0
(
[
π
,
1
]
)
,
0
<
π
<
1
and
β
π
β
π£
β
π»
1
(
π
,
1
)
β€
β
π£
β
1
for all
π£
β
π
1
, it follows that
π£
|
[
π
,
1
]
β
πΆ
0
(
[
π
,
1
]
)
. From both relations we deduce that
β
π
π£
β
πΆ
0
(
Ξ©
)
for all
π£
β
π
1
. Now, let the bilinear form
π
(
β
,
β
)
be defined by
ξ
π
(
π’
,
π£
)
=
β
π’
(
1
)
π£
(
1
)
+
1
0
π
π’
π
(
π
)
π£
π
(
π
)
π
π
,
π’
,
π£
β
π
1
,
(
2
.
2
)
where
β
is a positive constant. Then, there exists a unique bounded linear operator
π΄
βΆ
π
1
β
π
ξ
1
such that
π
(
π’
,
π£
)
=
β¨
π΄
π’
,
π£
β©
for all
π’
,
π£
β
π
1
. We then have the following lemma.
Lemma 2.4. The symmetric bilinear form
π
(
β
,
β
)
defined by (2.2 ) is continuous on
π
1
Γ
π
1
and coercive on
π
1
, that is, (i)
|
π
(
π’
,
π£
)
|
β€
πΆ
1
β
π’
β
1
β
π£
β
1
,
(ii)
π
(
π£
,
π£
)
β₯
πΆ
0
β
π£
β
2
1
,
for all
π’
,
π£
β
π
1
, where
πΆ
0
=
(
1
/
2
)
m
i
n
{
1
,
β
}
and
πΆ
1
β
=
1
+
(
1
+
2
)
β
.
The proof of Lemma 2.4 is straightforward and we omit it.
Lemma 2.5. There exists an orthonormal Hilbert basis
{
π€
π
}
of the space
π
0
consisting of eigenfunctions
π€
π
corresponding to eigenvalues
π
π
such that (i)
0
<
π
1
β€
π
2
β€
β―
β€
π
π
β
+
β
a
s
π
β
β
,
(ii)
π
(
π€
π
,
π£
)
=
π
π
β¨
π€
π
,
π£
β©
f
o
r
a
l
l
π£
β
π
1
a
n
d
π
β
β
. Note that it follows from (ii) that
{
π€
π
/
β
π
π
}
is automatically an orthonormal set in
π
1
with respect to
π
(
β
,
β
)
as inner product. The eigensolutions
π€
π
are indeed eigensolutions for the boundary value problem
π΄
π€
π
β‘
β
1
π
π
ξ΅
π
π
π
π
π€
π
ξΆ
π
π
=
π
π
π€
π
|
|
|
|
,
i
n
Ξ©
,
l
i
m
π
β
0
+
β
π
π
π€
π
|
|
|
|
π
r
(
π
)
<
+
β
,
π
π€
π
π
π
(
1
)
+
β
π€
π
(
1
)
=
0
.
(
2
.
3
)
The proof of Lemma 2.5 can be found in ([18 , page 87, Theorem 7.7]) with
π
=
π
1
,
π»
=
π
0
and
π
(
β
,
β
)
as defined by (2.2 ).
For any function
π£
β
πΆ
2
(
Ξ©
)
we define
β
π£
β
2
as
β
π£
β
2
=
ξ
β
π£
β
2
0
+
β
β
π£
π
β
β
2
0
+
β
π΄
π£
β
2
0
ξ
1
/
2
(
2
.
4
)
and define the space
π
2
as completion of
πΆ
2
(
Ξ©
)
with respect to the norm
β
β
β
2
. Note that
π
2
is also a Hilbert space with respect to the scalar product
β¨
π’
,
π£
β©
+
β¨
π’
π
,
π£
π
β©
+
β¨
π΄
π’
,
π΄
π£
β©
(
2
.
5
)
and that
π
2
can be defined also as
π
2
=
{
π£
β
π
1
βΆ
π΄
π£
β
π
0
}
.
We then have the following two lemmas the proof of which can be found in [1 ].
Lemma 2.6. The embedding
π
2
βͺ
π
1
is compact.
Lemma 2.7. For all
π£
β
π
2
we have
β
β
π£
(
i
)
π
β
β
πΏ
β
(
Ξ©
)
β€
1
β
2
β
π΄
π£
β
0
,
β
β
π£
(
i
i
)
π
π
β
β
0
β€
ξ
3
2
β
π΄
π£
β
0
,
(
i
i
i
)
β
π£
β
2
πΏ
β
(
Ξ©
)
β€
ξ©
2
β
π£
β
0
+
1
β
2
β
π΄
π£
β
0
ξͺ
β
π£
β
0
.
(
2
.
6
)
For a Banach space
π
, we denote by
β
β
β
π
its norm, by
π
ξ
its dual space, and by
πΏ
π
(
0
,
π
;
π
)
,
1
β€
π
β€
β
the Banach space of all real measurable functions
π’
βΆ
(
0
,
π
)
β
π
such that
β
π’
β
πΏ
π
(
0
,
π
;
π
)
=
ξ΅
ξ
π
0
β
π’
(
π‘
)
β
π
π
ξΆ
π
π‘
1
/
π
β
<
β
,
f
o
r
1
β€
π
<
β
,
π’
β
πΏ
β
(
0
,
π
;
π
)
=
e
s
s
s
u
p
0
<
π‘
<
π
β
π’
(
π‘
)
β
π
f
o
r
π
=
β
.
(
2
.
7
)
Let
π’
(
π‘
)
,
π’
ξ
(
π‘
)
=
π’
π‘
(
π‘
)
=
Μ
π’
(
π‘
)
,
π’
ξ
ξ
(
π‘
)
=
π’
π‘
π‘
(
π‘
)
=
Μ
π’
(
π‘
)
,
π’
π
(
π‘
)
=
β
π’
(
π‘
)
,
π’
π
π
(
π‘
)
(
2
.
8
)
denote
π’
(
π
,
π‘
)
,
π
π’
π
π
π‘
(
π
,
π‘
)
,
2
π’
π
π‘
2
(
π
,
π‘
)
,
π
π’
π
π
π
(
π
,
π‘
)
,
2
π’
π
π
2
(
π
,
π‘
)
,
(
2
.
9
)
respectively.
With
π
β
πΆ
π
(
Ξ©
Γ
β
+
Γ
β
)
,
π
=
π
(
π
,
π‘
,
π’
)
, we put
π·
1
π
=
π
π
/
π
π
,
π·
2
π
=
π
π
/
π
π‘
,
π·
3
π
=
π
π
/
π
π’
, and
π·
πΎ
π
=
π·
πΎ
1
1
π·
πΎ
2
2
π·
πΎ
3
3
π
,
πΎ
=
(
πΎ
1
,
πΎ
2
,
πΎ
3
)
β
β€
3
+
,
|
πΎ
|
=
πΎ
1
+
πΎ
2
+
πΎ
3
=
π
.
3. The Hight Order Iterative Schemes Fix
π
β
>
0
, we make the following assumptions: (
π»
1
)
Μ
π’
0
β
π
2
a
n
d
Μ
π’
1
β
π
1
;
(
π»
2
)
π΅
β
πΆ
1
(
β
+
)
a
n
d
t
h
e
r
e
e
x
i
s
t
c
o
n
s
t
a
n
t
s
π
β
>
0
,
πΌ
>
1
,
π
0
,
π
1
>
0
s
u
c
h
t
h
a
t
(i)
π
β
β€
π΅
(
π
)
β€
π
0
(
1
+
π
πΌ
)
,
f
o
r
a
l
l
π
β₯
0
,
(ii)
|
π΅
ξ
(
π
)
|
β€
π
1
(
1
+
π
πΌ
β
1
)
,
f
o
r
a
l
l
π
β₯
0
;
(
π»
3
)
π
β
πΆ
π
(
Ξ©
Γ
[
0
,
π
β
]
Γ
β
)
a
n
d
s
a
t
i
s
ο¬
e
s
t
h
e
f
o
l
l
o
w
i
n
g
c
o
n
d
i
t
i
o
n
βΆ
f
o
r
a
l
l
π
>
0
,
ξ
πΎ
(
1
)
β
π
(
π
,
π
)
=
s
u
p
(
π
,
π‘
,
π’
)
β
π΄
π
|
|
|
|
ξ
β
π
ξ
β
π
π
π
π
π
π’
π
|
|
|
|
ξ
πΎ
(
π
,
π‘
,
π’
)
<
+
β
,
π
=
0
,
1
,
β¦
,
π
β
1
,
(
2
)
β
π
(
π
,
π
)
=
s
u
p
(
π
,
π‘
,
π’
)
β
π΄
π
|
|
|
|
ξ
β
π
ξ
β
π
π
π
π
π
π
π
π’
π
β
1
|
|
|
|
(
π
,
π‘
,
π’
)
<
+
β
,
π
=
1
,
β¦
,
π
β
1
,
(
3
.
1
)
where
π΄
π
=
{
(
π
,
π‘
,
π’
)
β
[
0
,
1
]
Γ
[
0
,
π
β
ξ
]
Γ
β
βΆ
|
π’
|
β€
π
β
2
+
1
/
2
}
. We put
ξ
πΎ
β
π
β§
βͺ
β¨
βͺ
β©
ξ
πΎ
(
π
,
π
)
=
(
1
)
β
0
ξ
ξ
πΎ
(
π
,
π
)
,
π
=
0
,
m
a
x
(
1
)
β
π
ξ
πΎ
(
π
,
π
)
,
(
2
)
β
π
ξ
(
π
,
π
)
,
π
=
1
,
β¦
,
π
β
1
.
(
3
.
2
)
With
π΅
and
π
satisfying assumptions
(
π»
2
)
and
(
π»
3
)
, respectively, for each
π
>
0
given, we introduce the following constants:
ξ
πΎ
π
(
π΅
)
=
s
u
p
0
β€
π
β€
π
2
ξ·
π΅
|
|
π΅
(
π
)
+
ξ
|
|
ξΈ
,
(
π
)
πΎ
0
(
π
,
π
)
=
s
u
p
(
π
,
π‘
,
π’
)
β
π΄
π
|
|
|
|
,
π
(
π
,
π‘
,
π’
)
πΎ
π
ξ
(
π
,
π
)
=
|
|
πΎ
|
|
β€
π
πΎ
0
(
π
,
π·
πΎ
π
)
.
(
3
.
3
)
For each
π
β
(
0
,
π
β
]
and
π
>
0
we get
ξ½
π
(
π
,
π
)
=
π£
β
πΏ
β
ξ·
0
,
π
;
π
2
ξΈ
βΆ
π£
ξ
β
πΏ
β
ξ·
0
,
π
;
π
1
ξΈ
,
π£
ξ
ξ
β
πΏ
2
ξ·
0
,
π
;
π
0
ξΈ
,
w
i
t
h
β
π£
β
πΏ
β
(
0
,
π
;
π
2
)
,
β
β
π£
ξ
β
β
πΏ
β
(
0
,
π
;
π
1
)
,
β
β
π£
ξ
ξ
β
β
πΏ
2
(
0
,
π
;
π
0
)
ξ
,
π
β€
π
1
ξ½
(
π
,
π
)
=
π£
β
π
(
π
,
π
)
βΆ
π£
ξ
ξ
β
πΏ
β
ξ·
0
,
π
;
π
0
.
ξΈ
ξΎ
(
3
.
4
)
We will choose as first initial term
π’
0
β‘
0
, suppose that
π’
π
β
1
β
π
1
(
π
,
π
)
,
(
3
.
5
)
and associate with the problem (1.1 ) the following variational problem.
Find
π’
π
β
π
1
(
π
,
π
)
(
π
β₯
1
)
so that
ξ«
π’
π
ξ
ξ
ξ¬
(
π‘
)
,
π£
+
π
π
ξ·
π’
(
π‘
)
π
π
ξΈ
(
π‘
)
,
π£
=
β¨
πΉ
π
(
π‘
)
,
π£
β©
,
β
π£
β
π
1
,
π’
π
(
0
)
=
Μ
π’
0
,
π’
ξ
π
(
0
)
=
Μ
π’
1
,
(
3
.
6
)
where
π
π
ξ
β
β
(
π‘
)
=
π΅
β
π’
π
β
β
(
π‘
)
2
0
ξ
,
πΉ
π
(
π
,
π‘
)
=
π
β
1
ξ
π
=
0
1
π·
π
!
π
3
π
ξ·
π
,
π‘
,
π’
π
β
1
π’
ξΈ
ξ·
π
β
π’
π
β
1
ξΈ
π
.
(
3
.
7
)
Then, we have the following theorem.
Theorem 3.1. Let assumptions
(
π»
1
)
-
(
π»
3
)
hold. Then there exist a constant
π
>
0
depending on
π
β
,
Μ
π’
0
,
Μ
π’
1
,
π΅
,
β
and a constant
π
>
0
depending on
π
β
,
Μ
π’
0
,
Μ
π’
1
,
π΅
,
β
,
π
such that, for
π’
0
β‘
0
, there exists a recurrent sequence
{
π’
π
}
β
π
1
(
π
,
π
)
defined by (3.6 ), (3.7 ).
Proof. The proof consists of several steps.Step 1. The Faedo-Galerkin approximation (introduced by Lions [19 ]). Consider as in Lemma 2.5 the basis
{
π€
π
}
for
π
1
and put
π’
π
(
π
)
(
π‘
)
=
π
ξ
π
=
1
π
(
π
)
π
π
(
π‘
)
π€
π
,
(
3
.
8
)
where the coefficients
π
(
π
)
π
π
satisfy the system of the following nonlinear differential equations:
ξ¬
Μ
π’
π
(
π
)
(
π‘
)
,
π€
π
ξ
+
π
π
(
π
)
ξ
π’
(
π‘
)
π
π
(
π
)
(
π‘
)
,
π€
π
ξ
=
ξ¬
πΉ
π
(
π
)
(
π‘
)
,
π€
π
ξ
π’
,
1
β€
π
β€
π
,
π
(
π
)
(
0
)
=
Μ
π’
0
π
,
Μ
π’
π
(
π
)
(
0
)
=
Μ
π’
1
π
,
(
3
.
9
)
where
Μ
π’
0
π
=
π
ξ
π
=
1
πΌ
π
(
π
)
π€
π
βΆ
Μ
π’
0
s
t
r
o
n
g
l
y
i
n
π
2
,
Μ
π’
1
π
=
π
ξ
π
=
1
π½
π
(
π
)
π€
π
βΆ
Μ
π’
1
s
t
r
o
n
g
l
y
i
n
π
1
.
π
(
3
.
1
0
)
π
(
π
)
ξ
β
β
(
π‘
)
=
π΅
β
π’
π
(
π
)
β
β
(
π‘
)
2
0
ξ
,
πΉ
π
(
π
)
(
π
,
π‘
)
=
π
β
1
ξ
π
=
0
1
π·
π
!
π
3
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
ξ
π’
π
(
π
)
β
π’
π
β
1
ξ
π
=
π
ξ
π
=
0
Ξ¨
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
ξ
π’
π
(
π
)
ξ
π
,
Ξ¨
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
=
π
β
1
ξ
π
=
π
1
π
!
(
π
β
π
)
!
(
β
1
)
π
β
π
π·
π
3
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
π’
π
β
π
π
β
1
,
0
β€
π
β€
π
β
1
.
(
3
.
1
1
)
Let us suppose that
π’
π
β
1
satisfies (3.5 ). Then we have the following lemma.Lemma 3.2. Let assumptions
(
π»
1
)
-
(
π»
3
)
hold. For fixed
π
>
0
and
π
>
0
, then, the system (3.8 )–(3.11 ) has a unique solution
π’
π
(
π
)
(
π‘
)
on an interval
[
0
,
π
π
(
π
)
]
β
[
0
,
π
]
. Proof of Lemma 3.2 . The system of (3.8 )–(3.11 ) is rewritten in the form
Μ
π
(
π
)
π
π
(
π‘
)
=
β
π
π
π
π
(
π
)
(
π‘
)
π
(
π
)
π
π
ξ¬
πΉ
(
π‘
)
+
π
(
π
)
(
π‘
)
,
π€
π
ξ
π
,
1
β€
π
β€
π
,
(
π
)
π
π
(
0
)
=
πΌ
π
(
π
)
,
Μ
π
(
π
)
π
π
(
0
)
=
π½
π
(
π
)
,
(
3
.
1
2
)
and it is equivalent to the system of integral equations
π
(
π
)
π
π
(
π‘
)
=
πΌ
π
(
π
)
+
π½
π
(
π
)
π‘
β
π
π
ξ
π‘
0
ξ
π
π
π
0
π
π
(
π
)
(
π
)
π
(
π
)
π
π
(
ξ
π
)
π
π
+
π‘
0
ξ
π
π
π
0
ξ¬
πΉ
π
(
π
)
(
π
)
,
π€
π
ξ
π
π
,
(
3
.
1
3
)
for
1
β€
π
β€
π
. Omitting the index
π
, it is written as follows:
[
π
]
,
π
=
β±
(
3
.
1
4
)
where
β±
[
π
]
=
(
β±
1
[
π
]
,
β¦
,
β±
π
[
π
]
)
,
π
=
(
π
1
,
β¦
,
π
π
)
,
β±
π
[
π
]
(
π‘
)
=
π
π
(
π‘
)
β
π
π
ξ
π‘
0
ξ
π
π
π
0
Μ
π
(
π
)
π
π
(
+
π
)
π
π
π
β
1
ξ
π
=
1
ξ
π‘
0
ξ
π
π
π
0
ξ«
Ξ¨
π
ξ·
β
,
π
,
π’
π
β
1
ξΈ
π’
π
(
π
)
,
π€
π
ξ¬
π
π
π
,
1
β€
π
β€
π
,
π
(
π‘
)
=
πΌ
π
(
π
)
+
π½
π
(
π
)
ξ
π‘
+
π‘
0
ξ
π
π
π
0
ξ«
Ξ¨
0
ξ·
β
,
π
,
π’
π
β
1
ξΈ
,
π€
π
ξ¬
Μ
Μ
π
[
π
]
ξ·
π
π
,
1
β€
π
β€
π
,
π
(
π‘
)
=
(
π‘
)
=
π΅
β
β
π’
(
π‘
)
β
2
0
ξΈ
,
π’
(
π‘
)
=
π
ξ
π
=
1
π
π
(
π‘
)
π€
π
.
(
3
.
1
5
)
For every
π
π
(
π
)
β
(
0
,
π
]
and
π
>
0
that will be chosen later, we put
π
=
πΆ
0
(
[
0
,
π
π
(
π
)
]
;
β
π
)
,
π
=
{
π
β
π
βΆ
β
π
β
π
β€
π
}
, where
β
π
β
π
=
s
u
p
0
β€
π‘
β€
π
π
(
π
)
|
π
(
π‘
)
|
1
,
|
π
(
π‘
)
|
1
=
β
π
π
=
1
|
π
π
(
π‘
)
|
, for each
π
=
(
π
1
,
β¦
,
π
π
)
β
π
. Clearly
π
is a closed nonempty subset in
π
, and we have the operator
β±
βΆ
π
β
π
. In what follows, we will choose
π
>
0
and
π
π
(
π
)
>
0
such that(i)
π
i
s
m
a
p
p
e
d
i
n
t
o
i
t
s
e
l
f
b
y
β±
,
(ii)
β±
βΆ
π
β
π
i
s
c
o
n
t
r
a
c
t
i
v
e
. Proof (i). First we note that, for all
π
=
(
π
1
,
β¦
,
π
π
)
β
π
,
π’
(
π‘
)
=
π
ξ
π
=
1
π
π
(
π‘
)
π€
π
,
β
π’
(
π‘
)
β
0
=
ξΆ
ξ΅
ξ΅
ξ΅
β·
π
ξ
π
=
1
π
2
π
|
|
|
|
(
π‘
)
β€
π
(
π‘
)
1
,
β
β
β
π’
(
π‘
)
2
0
β€
π
(
π’
(
π‘
)
,
π’
(
π‘
)
)
=
π
ξ
π
,
π
=
1
π
π
(
π‘
)
π
π
ξ·
π€
(
π‘
)
π
π
,
π€
π
ξΈ
=
π
ξ
π
=
1
π
π
π
2
π
(
π‘
)
β€
π
π
β
β
π’
(
π‘
)
2
0
,
β
π’
(
π‘
)
β
0
β€
|
|
|
|
π
(
π‘
)
1
β€
β
π
β
π
β€
π
,
β
β
π’
(
π‘
)
β
0
β€
β
π
π
|
|
|
|
π
(
π‘
)
1
β€
β
π
π
π
,
β
π’
(
π‘
)
β
1
β€
ξ
1
πΆ
0
β
ξ
π
(
π’
(
π‘
)
,
π’
(
π‘
)
)
β€
1
πΆ
0
β
π
π
β
π’
(
π‘
)
β
0
β€
ξ
π
π
πΆ
0
|
|
|
|
π
(
π‘
)
1
β€
ξ
π
π
πΆ
0
π
,
(
3
.
1
6
)
so
Μ
π
ξ·
β
(
π‘
)
=
π΅
β
β
π’
(
π‘
)
2
0
ξΈ
β€
π
0
ξ·
β
1
+
β
β
π’
(
π‘
)
0
2
πΌ
ξΈ
β€
π
0
ξ·
1
+
π
π
2
πΌ
π
2
πΌ
ξΈ
.
(
3
.
1
7
)
On the other hand, by
|
|
π’
π
β
1
|
|
ξ
β€
π
1
2
+
β
2
|
|
Ξ¨
β‘
π
,
0
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
|
|
=
|
|
|
|
|
π
β
1
ξ
π
=
0
1
π
!
(
β
1
)
π
π·
π
3
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
π’
π
π
β
1
|
|
|
|
|
β€
π
β
1
ξ
π
=
0
|
|
|
1
π·
π
!
π
3
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
π’
π
π
β
1
|
|
|
β€
πΎ
π
(
π
,
π
)
π
β
1
ξ
π
=
0
1
ξ©
π
ξ
π
!
1
2
+
β
2
ξͺ
π
=
πΎ
π
(
π
,
π
)
π
β
1
ξ
π
=
0
π
π
,
π
!
(
3
.
1
8
)
we have
|
|
ξ«
Ξ¨
0
ξ·
π‘
,
π’
π
β
1
ξΈ
,
π€
π
ξ¬
|
|
β€
β
β
Ξ¨
0
ξ·
π‘
,
π’
π
β
1
ξΈ
β
β
0
β
β
π€
π
β
β
0
=
β
β
Ξ¨
0
ξ·
π‘
,
π’
π
β
1
ξΈ
β
β
0
β€
1
β
2
πΎ
π
(
π
,
π
)
π
β
1
ξ
π
=
0
π
π
.
π
!
(
3
.
1
9
)
By Lemma 2.1 , (iii), and the assumption (
π»
3
), we deduce from (3.16 ) that
|
|
ξ«
Ξ¨
π
ξ·
π
,
π’
π
β
1
ξΈ
π’
π
(
π
)
,
π€
π
ξ¬
|
|
=
|
|
|
|
|
π
β
1
ξ
π
=
π
1
π
!
(
π
β
π
)
!
(
β
1
)
π
β
π
ξ«
π·
π
3
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
π’
π
β
π
π
β
1
π’
π
(
π
)
,
π€
π
ξ¬
|
|
|
|
|
=
|
|
|
|
|
π
β
1
ξ
π
=
π
1
π
!
(
π
β
π
)
!
(
β
1
)
π
β
π
ξ‘
ξ
β
π
ξ
β
π
π·
π
3
π
ξ·
π
,
π
,
π’
π
β
1
ξΈ
ξ
β
π
ξ
π
β
π
π’
π
β
π
π
β
1
ξ
β
π
ξ
π
π’
π
(
π
)
,
π€
π
ξ’
|
|
|
|
|
β€
π
β
1
ξ
π
=
π
1
ξ
πΎ
π
!
(
π
β
π
)
!
β
π
(
π
,
π
)
π
π
β
π
πΎ
π
2
β
π’
(
π
)
β
π
1
ξ‘
ξ
β
π
ξ
π
β
π
,
|
|
π€
π
|
|
ξ’
β€
π
β
1
ξ
π
=
π
1
ξ
πΎ
π
!
(
π
β
π
)
!
β
π
(
π
,
π
)
π
π
β
π
πΎ
π
2
ξ©
ξ
π
π
πΆ
0
π
ξͺ
π
1
β
β€
2
+
π
β
π
π
β
1
ξ
π
=
π
1
1
π
!
(
π
β
π
)
!
β
ξ
πΎ
2
+
π
β
π
β
π
(
π
,
π
)
π
π
β
π
πΎ
π
2
ξ©
ξ
π
π
πΆ
0
π
ξͺ
π
,
1
β€
π
β€
π
β
1
.
(
3
.
2
0
)
It follows that
|
|
β±
π
[
π
]
(
|
|
β€
|
|
π
π‘
)
π
(
|
|
π‘
)
+
π
π
π
0
ξ·
1
+
π
π
2
πΌ
π
2
πΌ
ξΈ
ξ
π‘
0
ξ
π
π
π
0
|
|
π
π
(
|
|
+
1
π
)
π
π
2
π‘
2
π
β
1
ξ
π
=
1
π
β
1
ξ
π
=
π
1
1
π
!
(
π
β
π
)
!
β
ξ
πΎ
2
+
π
β
π
β
π
(
π
,
π
)
π
π
β
π
πΎ
π
2
ξ©
ξ
π
π
πΆ
0
π
ξͺ
π
.
(
3
.
2
1
)
Thus
|
|
β±
[
π
]
(
|
|
π‘
)
1
β€
|
|
|
|
π
(
π‘
)
1
+
π
π
π
0
ξ·
1
+
π
π
2
πΌ
π
2
πΌ
ξΈ
ξ
π‘
0
ξ
π
π
π
0
|
|
|
|
π
(
π
)
1
+
1
π
π
2
π
π‘
2
π
β
1
ξ
π
=
1
π
β
1
ξ
π
=
π
1
1
π
!
(
π
β
π
)
!
β
ξ
πΎ
2
+
π
β
π
β
π
(
π
,
π
)
π
π
β
π
πΎ
π
2
ξ©
ξ
π
π
πΆ
0
π
ξͺ
π
β€
β
π
β
π
+
1
2
π·
π
ξ
π
π
(
π
)
ξ
2
,
(
3
.
2
2
)
where
β
π
β
π
=
s
u
p
0
β€
π‘
β€
π
|
π
(
π‘
)
|
1
and
π·
π
=
π·
π
(
π
,
π
,
π
,
π
,
π
,
π
)
=
π
π
π
0
ξ·
1
+
π
π
2
πΌ
π
2
πΌ
ξΈ
π
+
π
π
β
1
ξ
π
=
1
π
β
1
ξ
π
=
π
1
1
π
!
(
π
β
π
)
!
β
ξ
πΎ
2
+
π
β
π
β
π
(
π
,
π
)
π
π
β
π
πΎ
π
2
ξ©
ξ
π
π
πΆ
0
π
ξͺ
π
.
(
3
.
2
3
)
Hence, we obtain
[
π
]
β
β
β±
π
β€
β
π
β
π
+
1
2
π·
π
ξ
π
π
(
π
)
ξ
2
,
(
3
.
2
4
)
choosing
π
>
β
π
β
π
and
π
π
(
π
)
β
(
0
,
π
]
, such that
1
2
π·
π
ξ
π
π
(
π
)
ξ
2
β€
π
β
β
π
β
π
,
1
2
ξ
π·
π
ξ
π
π
(
π
)
ξ
2
<
1
,
(
3
.
2
5
)
where
ξ
π·
π
=
ξ
π·
π
(
π
,
π
,
π
,
π
,
π
,
π
,
π
)
β‘
π
0
π
π
ξ·
1
+
π
π
2
πΌ
π
2
πΌ
ξΈ
+
2
π
1
π
2
π
π
2
ξ·
1
+
π
π
πΌ
β
1
π
2
πΌ
β
2
ξΈ
+
π
π
β
1
ξ
π
=
1
π
ξ©
ξ
π
π
πΆ
0
πΎ
2
ξͺ
π
π
π
β
1
π
β
1
ξ
π
=
π
1
1
π
!
(
π
β
π
)
!
β
ξ
πΎ
2
+
π
β
π
β
π
(
π
,
π
)
π
π
β
π
.
(
3
.
2
6
)
Then
[
π
]
β
β
β±
π
β€
β
π
β
π
+
1
2
π·
π
ξ
π
π
(
π
)
ξ
2
β€
π
,
β
π
β
π
,
(
3
.
2
7
)
which means that
β±
maps
π
into itself.Proof (ii). We now prove that, for all
π
,
π
β
π
, for all
π‘
β
[
0
,
π
π
(
π
)
]
,
|
|
β±
[
π
]
[
π
]
|
|
(
π‘
)
β
β±
(
π‘
)
1
β€
1
2
ξ
π·
π
π‘
2
β
π
β
π
β
π
,
β
π
β
β
,
(
3
.
2
8
)
where
ξ
π·
π
is defined as (3.26 ). Proof of (3.28 ) is as follows. For all
π
=
1
,
2
,
β¦
,
π
, for all
π‘
β
[
0
,
π
π
(
π
)
]
, we have
|
|
β±
π
[
π
]
(
π‘
)
β
β±
π
[
π
]
(
|
|
π‘
)
β€
π
π
ξ
π‘
0
ξ
π
π
π
0
|
|
Μ
π
[
π
]
(
ξ·
π
π
)
π
(
π
)
β
π
π
(
ξΈ
|
|
π
)
π
π
+
π
π
ξ
π‘
0
ξ
π
π
π
0
|
|
ξ·
Μ
π
[
π
]
Μ
π
[
π
]
ξΈ
π
(
π
)
β
(
π
)
π
|
|
+
(
π
)
π
π
π
β
1
ξ
π
=
1
ξ
π‘
0
ξ
π
π
π
0
|
|
ξ«
Ξ¨
π
ξ·
π
,
π’
π
β
1
π’
ξΈ
ξ·
π
(
π
)
β
π£
π
ξΈ
(
π
)
,
π€
π
ξ¬
|
|
π
π
,
(
3
.
2
9
)
where
Μ
π
[
π
]
ξ·
β
(
π‘
)
=
π΅
β
β
π’
(
π‘
)
2
0
ξΈ
,
Μ
π
[
π
]
ξ·
β
(
π‘
)
=
π΅
β
β
π£
(
π‘
)
2
0
ξΈ
,
π’
(
π‘
)
=
π
ξ
π
=
1
π
π
(
π‘
)
π€
π
,
π£
(
π‘
)
=
π
ξ
π
=
1
π
π
(
π‘
)
π€
π
,
(
3
.
3
0
)
so
|
|
β±
[
π
]
(
[
π
]
(
|
|
π‘
)
β
β±
π‘
)
1
β€
π
π
ξ
π‘
0
ξ
π
π
π
0
Μ
π
[
π
]
(
|
|
|
|
π
)
π
(
π
)
β
π
(
π
)
1
π
π
+
π
π
ξ
π‘
0
ξ
π
π
π
0
|
|
Μ
π
[
π
]
Μ
π
[
π
]
|
|
|
|
|
|
(
π
)
β
(
π
)
π
(
π
)
1
+
π
π
π
ξ
π
=
1
π
β
1
ξ
π
=
1
ξ
π‘
0
ξ
π
π
π
0
|
|
ξ«
Ξ¨
π
ξ·
π
,
π’
π
β
1
π’
ξΈ
ξ·
π
(
π
)
β
π£
π
ξΈ
(
π
)
,
π€
π
ξ¬
|
|
π
π
β€
π
π
ξ
π‘
0
ξ
π
π
π
0
Μ
π
[
π
]
|
|
|
|
(
π
)
π
(
π
)
β
π
(
π
)
1
π
π
+
π
π
π
ξ
π‘
0
ξ
π
π
π
0
|
|
Μ
π
[
π
]
Μ
π
[
π
]
|
|
+
(
π
)
β
(
π
)
π
π
π
ξ
π
=
1
π
β
1
ξ
π
=
1
ξ
π‘
0
ξ
π
π
π
0
|
|
ξ«
Ξ¨
π
ξ·
π
,
π’
π
β
1
π’
ξΈ
ξ·
π
(
π
)
β
π£
π
ξΈ
(
π
)
,
π€
π
ξ¬
|
|
π
π
β‘
π½
1
+
π½
2
+
π½
3
,
(
3
.
3
1
)
in which
π½
1
=
π
π
ξ
π‘
0
ξ
π
π
π
0
Μ
π
[
π
]
(
|
|
|
|
π
)
π
(
π
)
β
π
(
π
)
1
π
π
β€
π
π
π
0
ξ·
1
+
π
π
2
πΌ
π
2
πΌ
ξΈ
ξ
π‘
0
ξ
π
π
π
0
|
|
|
|
π
(
π
)
β
π
(
π
)
1
π
π
β‘
π
1
ξ
π‘
0
ξ
π
π
π
0
|
|
|
|
π
(
π
)
β
π
(
π
)
1
π
π
.
(
3
.
3
2
)
In order to consider
π½
2
, we also note that
Μ
π
[
π
]
Μ
π
[
π
]
(
π
)
β
(
π
)
=
π΅
ξ
ξ·
β
(
π
)
β
β
π’
(
π
)
2
0
β
β
β
β
π£
(
π
)
2
0
ξΈ
,
(
3
.
3
3
)
where
β
π
=
π
β
β
π’
(
π
)
2
0
+
β
(
1
β
π
)
β
β
π£
(
π
)
2
0
,
0
β€
π
β€
π
π
π
2
,
0
<
π
<
1
,
(
3
.
3
4
)
and
π΅
ξ
(
π
)
satisfy the following inequality:
|
|
π΅
ξ
|
|
(
π
)
β€
π
1
ξ·
1
+
π
πΌ
β
1
ξΈ
β€
π
1
ξ·
1
+
π
π
πΌ
β
1
π
2
πΌ
β
2
ξΈ
.
(
3
.
3
5
)
It implies that
|
|
Μ
π
[
π
]
Μ
π
[
π
]
|
|
=
|
|
π΅
(
π
)
β
(
π
)
ξ
ξ·
β
(
π
)
β
β
π’
(
π
)
2
0
β
β
β
β
π£
(
π
)
2
0
ξΈ
|
|
β€
π
1
ξ·
1
+
π
π
πΌ
β
1
π
2
πΌ
β
2
ξΈ
2
π
π
π
|
|
|
|
π
(
π
)
β
π
(
π
)
1
=
2
π
π
π
π
1
ξ·
1
+
π
π
πΌ
β
1
π
2
πΌ
β
2
ξΈ
|
|
|
|
π
(
π
)
β
π
(
π
)
1
,
(
3
.
3
6
)
and then
π½
2
=
π
π
π
ξ
π‘
0
ξ
π
π
π
0
|
|
Μ
π
[
π
]
(
Μ
π
[
π
]
(
|
|
π
)
β
π
)
π
π
β€
2
π
2
π
π
2
π
1
ξ·
1
+
π
π
πΌ
β
1
π
2
πΌ
β
2
ξΈ
ξ
π‘
0
ξ
π
π
π
0
|
|
|
|
π
(
π
)
β
π
(
π
)
1
π
π
β‘
π
2
ξ
π‘
0
ξ
π
π
π
0
|
|
|
|
π
(
π
)
β
π
(
π
)
1
π
π
.
(
3
.
3
7
)
It remains to estimate
π½
3
. By
ξ
β
ξ
π
π’
(
π
)
π
β
ξ
β
ξ
π
π£
(
π
)
π
=
π
β
1
ξ
π
=
0
ξ
β
ξ
π
π’
(
π
)
π
ξ
β
ξ
π
π£
(
π
)
π
β
π
β
1
β
π
(
π’
(
π
)
β
π£
(
π
)
)
,
(
3
.
3
8
)
we obtain
|
|
|
ξ
β
ξ
π
π’
(
π
)
π
β
ξ
β
ξ
π
π£
(
π
)
π
|
|
|
β€
πΎ
π
2
π
β
1
ξ
π
=
0
β
π’
(
π
)
β
π
1
β
π£
(
π
)
β
1
π
β
π
β
1
β
π’
(
π
)
β
π£
(
π
)
β
1
β€
πΎ
π
2
π
β
1
ξ
π
=
0
β
β
π’
(
π
)
π
1
β
β
π£
(
π
)
1
π
β
π
β
1
β
β
π’
(
π
)
β
π£
(
π
)
1
β€
πΎ
π
2
π
β
1
ξ
π
=
0
ξ©
ξ
π
π
πΆ
0
π
ξͺ
π
ξ©
ξ
π
π
πΆ
0
π
ξͺ
π
β
π
β
1
ξ
π
π
πΆ
0
|
|
|
|
π
(
π
)
β
π
(
π
)
1
=
πΎ
π
2
π
β
1
ξ
π
=
0
ξ©
ξ
π
π
πΆ
0
ξͺ
π
π
π
β
1
|
|
|
|
π
(
π
)
β
π
(
π
)
1
ξ©
ξ
=
π
π
π
πΆ
0
πΎ
2
ξͺ
π
π
π
β
1
|
|
|
|
π
(
π
)
β
π
(
π
)
1
.
(
3
.
3
9
)
On the other hand,
Ξ¨
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
=
π
β
1
ξ
π
=
π
1
π
!
(
π
β
π
)
!
(
β
1
)
π
β
π
π·
π
3
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
π’
π
β
π
π
β
1
,
ξ
β
π
ξ
β
π
|
|
Ξ¨
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
|
|
=
|
|
|
|
|
π
β
1
ξ
π
=
π
1
(
π
!
(
π
β
π
)
!
β
1
)
π
β
π
ξ
β
π
ξ
β
π
π·
π
3
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
ξ
β
π
ξ
π
β
π
π’
π
β
π
π
β
1
|
|
|
|
|
β€
|
|
|
|
|
π
β
1
ξ
π
=
π
1
ξ
β
π
!
(
π
β
π
)
!
π
ξ
β
π
|
|
π·
π
3
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
|
|
ξ
β
π
ξ
π
β
π
|
|
π’
π
β
π
π
β
1
|
|
|
|
|
|
|
β€
π
β
1
ξ
π
=
π
1
ξ
πΎ
π
!
(
π
β
π
)
!
β
π
(
π
,
π
)
π
π
β
π
ξ
β
π
ξ
π
β
π
.
(
3
.
4
0
)
Hence, we deduce from (3.39 ), (3.40 ) that
π½
3
=
π
ξ
π
=
1
π
β
1
ξ
π
=
1
ξ
π‘
0
ξ
π
π
π
0
|
|
|
|
ξ‘
ξ
β
π
ξ
β
π
Ξ¨
π
ξ·
π
,
π’
π
β
1
ξΈ
ξ΅
ξ
β
ξ
π
π’
(
π
)
π
β
ξ
β
ξ
π
π£
(
π
)
π
ξΆ
,
π€
π
ξ’
|
|
|
|
β€
π
π
π
ξ
π
=
1
π
β
1
ξ
π
=
1
π
β
1
ξ
π
=
π
1
ξ
πΎ
π
!
(
π
β
π
)
!
β
π
(
π
,
π
)
π
π
β
π
π
ξ©
ξ
π
π
πΆ
0
πΎ
2
ξͺ
π
π
π
β
1
Γ
ξ
π‘
0
ξ
π
π
π
0
|
|
|
|
ξ‘
ξ
β
π
ξ
π
β
π
,
|
|
π€
π
|
|
ξ’
|
|
|
|
|
|
|
|
π
(
π
)
β
π
(
π
)
1
=
π
π
π
ξ
π
=
1
π
β
1
ξ
π
=
1
π
β
1
ξ
π
=
π
1
ξ
πΎ
π
!
(
π
β
π
)
!
β
π
(
π
,
π
)
π
π
β
π
π
ξ©
ξ
π
π
πΆ
0
πΎ
2
ξͺ
π
π
π
β
1
1
β
Γ
ξ
2
+
π
β
π
π‘
0
ξ
π
π
π
0
|
|
|
|
π
(
π
)
β
π
(
π
)
1
π
π
=
π
π
β
1
ξ
π
=
1
π
ξ©
ξ
π
π
πΆ
0
πΎ
2
ξͺ
π
π
π
β
1
π
β
1
ξ
π
=
π
1
ξ
πΎ
π
!
(
π
β
π
)
!
β
π
(
π
,
π
)
π
π
β
π
1
β
Γ
ξ
2
+
π
β
π
π‘
0
ξ
π
π
π
0
|
|
|
|
π
(
π
)
β
π
(
π
)
1
π
π
β‘
π
3
ξ
π‘
0
ξ
π
π
π
0
|
|
|
|
π
(
π
)
β
π
(
π
)
1
π
π
.
(
3
.
4
1
)
We deduce that
|
|
β±
[
π
]
(
[
π
]
(
|
|
π‘
)
β
β±
π‘
)
1
β€
ξ·
π
1
+
π
2
+
π
3
ξΈ
ξ
π‘
0
ξ
π
π
π
0
|
|
|
|
π
(
π
)
β
π
(
π
)
1
1
π
π
β€
2
ξ
π·
π
π‘
2
β
π
β
π
β
π
.
(
3
.
4
2
)
We note that
π
1
+
π
2
+
π
3
=
ξ
π·
π
ξ
π·
(
π
,
π
,
π
,
π
,
π
,
π
,
π
)
=
π
.
(
3
.
4
3
)
It follows from (3.28 ) that
[
π
]
[
π
]
β
β
β±
β
β±
π
β€
1
2
ξ
π·
π
ξ
π
π
(
π
)
ξ
2
β
π
β
π
β
π
,
β
π
,
π
β
π
.
(
3
.
4
4
)
By (3.25 ), it follows that
β±
βΆ
π
β
π
is contractive. We deduce that
β±
has a unique fixed point in
π
; that is, the system (3.8 )–(3.11 ) has a unique solution
π’
π
(
π
)
(
π‘
)
on an interval
[
0
,
π
π
(
π
)
]
. The proof of Lemma 3.2 is complete. The following estimates allow one to take constant
π
π
(
π
)
=
π
for all
π
and
π
.Step 2. A priori estimates. Put
π
π
(
π
)
(
π‘
)
=
π
π
(
π
)
(
π‘
)
+
π
π
(
π
)
(
ξ
π‘
)
+
π‘
0
β
β
Μ
π’
π
(
π
)
(
β
β
π
)
2
0
π
π
,
(
3
.
4
5
)
where
π
π
(
π
)
β
β
(
π‘
)
=
Μ
π’
π
(
π
)
β
β
(
π‘
)
2
0
+
π
π
(
π
)
ξ
π’
(
π‘
)
π
π
(
π
)
(
π‘
)
,
π’
π
(
π
)
ξ
,
π
(
π‘
)
π
(
π
)
ξ
(
π‘
)
=
π
Μ
π’
π
(
π
)
(
π‘
)
,
Μ
π’
π
(
π
)
ξ
(
π‘
)
+
π
π
(
π
)
β
β
(
π‘
)
π΄
π’
π
(
π
)
β
β
(
π‘
)
2
0
,
(
3
.
4
6
)
with
π΄
is defined by (2.2 ). Then it follows that
π
π
(
π
)
(
π‘
)
=
π
π
(
π
)
(
ξ
0
)
+
π‘
0
Μ
π
π
(
π
)
(
ξ
π
ξ
π’
π
)
π
(
π
)
(
π
)
,
π’
π
(
π
)
(
ξ
+
β
β
π
)
π΄
π’
π
(
π
)
(
β
β
π
)
2
0
ξ
ξ
π
π
+
2
π‘
0
ξ¬
πΉ
π
(
π
)
(
π
)
,
Μ
π’
π
(
π
)
ξ
ξ
(
π
)
π
π
+
2
π‘
0
π
ξ
πΉ
π
(
π
)
(
π
)
,
Μ
π’
π
(
π
)
ξ
+
ξ
(
π
)
π
π
π‘
0
ξ¬
πΉ
π
(
π
)
(
π
)
,
Μ
π’
π
(
π
)
ξ
ξ
(
π
)
π
π
β
π‘
0
π
π
(
π
)
ξ¬
(
π
)
π΄
π’
π
(
π
)
(
π
)
,
Μ
π’
π
(
π
)
ξ
(
π
)
π
π
β‘
π
π
(
π
)
(
0
)
+
5
ξ
π
=
1
πΌ
π
.
(
3
.
4
7
)
We will now require the following lemma.Lemma 3.3. We have
(
i
)
0
<
π
β
β€
π
π
(
π
)
(
π‘
)
β€
π
0
ξ
β
β
1
+
β
π’
π
(
π
)
β
β
(
π‘
)
0
2
πΌ
ξ
,
|
|
Μ
π
(
i
i
)
π
(
π
)
|
|
β€
(
π‘
)
2
π
1
β
π
β
ξ
π
π
(
π
)
(
π‘
)
+
π
β
1
β
πΌ
ξ
π
π
(
π
)
ξ
(
π‘
)
πΌ
ξ
,
β
β
πΉ
(
i
i
i
)
π
(
π
)
β
β
(
π‘
)
0
β€
π
β
1
ξ
π
=
0
Μ
π
π
(
0
)
ξ΅
ξ
π
π
(
π
)
ξΆ
(
π‘
)
π
,
(
β
β
β
π
i
v
)
πΉ
π
π
π
(
π
)
(
β
β
β
π‘
)
0
β€
π
β
1
ξ
π
=
0
Μ
π
π
(
1
)
ξ΅
ξ
π
π
(
π
)
(
ξΆ
π‘
)
π
,
(
3
.
4
8
)
where
Μ
π
π
(
0
)
,
Μ
π
π
(
1
)
,
π
=
0
,
1
,
β¦
,
π
β
1
are defined as follows:
Μ
π
π
(
0
)
=
β§
βͺ
βͺ
βͺ
β¨
βͺ
βͺ
βͺ
β©
Μ
π
0
(
0
)
=
1
β
2
πΎ
π
(
π
,
π
)
π
β
1
ξ
π
=
0
π
π
π
!
,
π
=
0
,
Μ
π
π
(
0
)
=
πΎ
π
2
ξ
β
π
β
πΆ
0
ξ
π
π
β
1
ξ
π
=
π
π
π
β
π
1
π
!
(
π
β
π
)
!
β
ξ
πΎ
2
+
π
β
π
β
π
(
π
,
π
)
,
1
β€
π
β€
π
β
1
,
Μ
π
π
(
1
)
=
β§
βͺ
βͺ
βͺ
βͺ
β¨
βͺ
βͺ
βͺ
βͺ
β©
ξ©
1
β
2
ξͺ
+
2
π
πΎ
π
(
π
,
π
)
π
β
1
ξ
π
=
0
π
π
πΎ
π
!
,
π
=
0
,
π
2
ξ
β
π
β
πΆ
0
ξ
π
π
β
1
ξ
π
=
π
π
π
β
π
ξ¬
π
!
(
π
β
π
)
!
π
πΎ
2
β
1
ξ
πΎ
β
π
+
ξ©
1
(
π
,
π
)
β
ξͺ
ξ
πΎ
π
β
π
+
3
+
2
π
β
π
+
1
ξ
ξ
(
π
,
π
)
,
1
β€
π
β€
π
β
1
,
π
=
π
1
2
+
β
2
.
(
3
.
4
9
)
Proof of Lemma 3.3 . Proof (i), (ii). Note that
π
π
(
π
)
(
π‘
)
β₯
π
π
(
π
)
(
π‘
)
β₯
π
β
π
ξ
π’
π
(
π
)
(
π‘
)
,
π’
π
(
π
)
ξ
(
π‘
)
β₯
π
β
β
β
β
π’
π
(
π
)
β
β
(
π‘
)
2
0
,
π
π
(
π
)
(
π‘
)
β₯
π
π
(
π
)
ξ
(
π‘
)
β₯
π
Μ
π’
π
(
π
)
(
π‘
)
,
Μ
π’
π
(
π
)
ξ
β₯
β
β
(
π‘
)
β
Μ
π’
π
(
π
)
β
β
(
π‘
)
2
0
.
(
3
.
5
0
)
We deduce that
π
π
(
π
)
ξ
β
β
(
π‘
)
=
π΅
β
π’
π
(
π
)
β
β
(
π‘
)
2
0
ξ
β€
π
0
ξ
β
β
1
+
β
π’
π
(
π
)
β
β
(
π‘
)
0
2
πΌ
ξ
,
|
|
Μ
π
π
(
π
)
|
|
|
|
|
π΅
(
π‘
)
=
2
ξ
ξ
β
β
β
π’
π
(
π
)
β
β
(
π‘
)
2
0
ξ
|
|
|
|
|
|
ξ¬
β
π’
π
(
π
)
(
π‘
)
,
β
Μ
π’
π
(
π
)
ξ
|
|
|
(
π‘
)
β€
2
π
1
ξ
β
β
1
+
β
π’
π
(
π
)
β
β
(
π‘
)
0
2
πΌ
β
2
ξ
β
β
β
π’
π
(
π
)
β
β
(
π‘
)
0
β
β
β
Μ
π’
π
(
π
)
β
β
(
π‘
)
0
β€
2
π
1
ξ΅
1
+
π
β
1
β
πΌ
ξ
π
π
(
π
)
ξ
(
π‘
)
πΌ
β
1
ξΆ
1
β
π
β
π
π
(
π
)
=
(
π‘
)
2
π
1
β
π
β
ξ
π
π
(
π
)
(
π‘
)
+
π
β
1
β
πΌ
ξ
π
π
(
π
)
ξ
(
π‘
)
πΌ
ξ
.
(
3
.
5
1
)
Proof (iii). We have
β
β
πΉ
π
(
π
)
β
β
(
π‘
)
0
β€
β
β
Ξ¨
0
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
β
β
0
+
π
β
1
ξ
π
=
1
β
β
β
Ξ¨
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
ξ
π’
π
(
π
)
ξ
π
β
β
β
0
.
(
3
.
5
2
)
By (3.18 )3 , we have
β
β
Ξ¨
0
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
β
β
0
β€
1
β
2
πΎ
π
(
π
,
π
)
π
β
1
ξ
π
=
0
π
π
π
!
β‘
Μ
π
0
(
0
)
.
(
3
.
5
3
)
On the other hand, it follows from (3.49 ) and
π’
π
β
1
β
π
1
(
π
,
π
)
that
β
β
β
Ξ¨
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
ξ
π’
π
(
π
)
ξ
π
β
β
β
0
β€
π
β
1
ξ
π
=
π
1
β
β
β
π·
π
!
(
π
β
π
)
!
π
3
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
π’
π
β
π
π
β
1
ξ
π’
π
(
π
)
ξ
π
β
β
β
0
=
π
β
1
ξ
π
=
π
1
β
β
β
ξ
β
π
!
(
π
β
π
)
!
π
ξ
π
β
π
ξ
β
π
ξ
β
π
π·
π
3
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
π’
π
β
π
π
β
1
ξ
β
π
ξ
π
ξ
π’
π
(
π
)
ξ
π
β
β
β
0
β€
π
β
1
ξ
π
=
π
1
β
β
β
ξ
β
π
!
(
π
β
π
)
!
π
ξ
π
β
π
β
β
β
0
ξ
πΎ
β
π
(
π
,
π
)
π
π
β
π
πΎ
π
2
β
β
π’
π
(
π
)
β
β
(
π‘
)
π
1
=
π
β
1
ξ
π
=
π
1
1
π
!
(
π
β
π
)
!
β
ξ
πΎ
2
+
π
β
π
β
π
(
π
,
π
)
π
π
β
π
πΎ
π
2
β
β
π’
π
(
π
)
β
β
(
π‘
)
π
1
β€
πΎ
π
2
ξ
β
π
β
πΆ
0
ξ
π
π
β
1
ξ
π
=
π
π
π
β
π
1
π
!
(
π
β
π
)
!
β
ξ
πΎ
2
+
π
β
π
β
π
ξ΅
ξ
(
π
,
π
)
π
π
(
π
)
ξΆ
(
π‘
)
π
β‘
Μ
π
π
(
0
)
ξ΅
ξ
π
π
(
π
)
ξΆ
(
π‘
)
π
.
(
3
.
5
4
)
It follows from (3.52 )–(3.54 ) that
β
β
πΉ
π
(
π
)
β
β
(
π‘
)
0
β€
π
β
1
ξ
π
=
0
Μ
π
π
(
0
)
ξ΅
ξ
π
π
(
π
)
ξΆ
(
π‘
)
π
,
(
3
.
5
5
)
where
Μ
π
π
(
0
)
,
0
β€
π
β€
π
β
1
are defined by (3.49 )1 .Proof (iv). We have
π
πΉ
π
π
π
(
π
)
π
(
π
,
π‘
)
=
Ξ¨
π
π
0
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
+
π
β
1
ξ
π
=
1
π
Ξ¨
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
ξ
π’
π
(
π
)
ξ
π
β
1
β
π’
π
(
π
)
+
π
β
1
ξ
π
=
1
ξ
π
Ξ¨
π
π
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
π’
ξ
ξ
π
(
π
)
ξ
π
.
(
3
.
5
6
)
Hence
β
β
β
π
πΉ
π
π
π
(
π
)
β
β
β
(
π‘
)
0
β€
β
β
β
π
Ξ¨
π
π
0
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
β
β
β
0
+
π
β
1
ξ
π
=
1
π
β
β
β
Ξ¨
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
ξ
π’
π
(
π
)
ξ
π
β
1
β
π’
π
(
π
)
β
β
β
0
+
π
β
1
ξ
π
=
1
β
β
β
ξ
π
Ξ¨
π
π
π
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
π’
ξ
ξ
π
(
π
)
ξ
π
β
β
β
0
β‘
πΏ
1
+
πΏ
2
+
πΏ
3
.
(
3
.
5
7
)
We shall estimate step by step the terms on the right-hand side of (3.57 ) as follows.(iv.1) Estimating
πΏ
1
=
β
(
π
/
π
π
)
Ξ¨
0
(
π
,
π‘
,
π’
π
β
1
)
β
0
. We have
π
Ξ¨
π
π
0
ξ·
π
,
π‘
,
π’
π
β
1
ξΈ
=
π
β
1
ξ
π
=
1