1. Introduction
The importance of problems with integral condition has been pointed out by Samarskiĭ [1]. Mathematical modelling by evolution problems with a nonlocal constraint of the form
(
1
/
(
1
-
α
)
)
∫
α
1
u
(
x
,
t
)
d
x
=
E
(
t
)
is encountered in heat transmission theory, thermoelasticity, chemical engineering, underground water flow, and plasma physics. For background information, we refer the reader to Benouar and Yurchuk [2], Bouziani and Benouar [3], Bouziani [4], Cannon and van der Hoek [5–7] Ionkin and Moiceev [8, 9], Kamynin [10], and Yurchuk [11, 12]. Mixed problems with nonlocal boundary conditions or with nonlocal initial conditions were studied in Bouziani [13], Byszewski [14–16], Gasymov [17], Ionkin [8, 9], Lazhar [18], and Said and Nadia [19]. The results and the method used here are a further elaboration of those in [2]. We should mention here that the presence of integral term in the boundary condition can greatly complicate the application of standard functional and numerical techniques. This work can be considered as a continuation of the results in [11, 20].
We consider the following mixed problem in the rectangle
Q
=
(
0
,
l
)
×
(
0
,
T
)
:
(1)
L
u
=
∂
u
∂
t
-
1
x
m
∂
∂
x
(
x
m
∂
u
∂
x
)
=
f
(
x
,
t
)
,
m
>
0
,
(2)
l
u
=
u
(
x
,
0
)
=
φ
(
x
)
,
|
u
(
0
,
t
)
|
<
∞
,
∫
α
l
x
m
u
(
x
,
t
)
d
x
=
0
,
α
>
0
.
2. A Priori Estimate
Let
E
be the Hilbert space of all sufficiently smooth functions
u
satisfying the second and third conditions in (2) and equipped with the norm
(3)
∥
u
∥
E
2
=
∫
Q
(
l
-
x
)
[
x
m
|
∂
u
∂
t
|
2
+
1
x
m
|
∂
∂
x
(
x
m
∂
u
∂
x
)
|
2
d
x
d
t
]
+
sup
0
≤
t
≤
T
{
∫
0
l
(
l
-
x
)
x
m
(
∂
x
∂
u
)
2
d
x
+
m
l
-
α
∫
α
l
x
m
-
1
u
2
d
x
}
.
The equality
(4)
x
m
+
1
∂
u
∂
x
=
∫
0
x
d
η
∬
η
x
∂
∂
ξ
(
ξ
m
∂
u
∂
ξ
)
d
ξ
+
∫
0
x
ξ
m
∂
u
∂
ξ
d
ξ
implies the following inequality:
(5)
x
m
|
∂
u
∂
x
|
≤
x
1
/
2
∫
0
x
(
∂
∂
ξ
(
ξ
m
∂
u
∂
ξ
)
)
2
d
ξ
+
∫
0
x
(
ξ
m
∂
u
∂
ξ
)
2
d
ξ
.
By (5) it follows that
(6)
lim
x
→
0
x
m
∂
u
∂
x
=
0
,
for any
u
∈
E
.
We will use (5) for the solutions
u
of the problem (1)-(2). For the right hand side
f
of (1) and initial condition of
u
from (2) we introduce the space
F
which is consisted of the vector function
ℱ
=
(
f
,
φ
)
with the norm
(7)
∥
ℱ
∥
F
2
=
∫
Q
x
m
|
f
(
x
,
t
)
|
2
d
x
+
∫
0
l
x
m
(
∂
φ
∂
x
)
2
d
x
+
m
l
-
α
∫
α
l
x
m
-
1
φ
2
d
x
.
Here it is assumed that
(8)
|
φ
(
0
)
|
<
∞
,
∫
α
l
x
m
φ
(
x
)
d
x
=
0
.
Theorem 1.
For any function
u
∈
E
such that
x
m
/
2
(
∂
u
/
∂
t
)
∈
L
2
(
Q
)
and
x
-
m
/
2
(
∂
/
∂
x
)
x
m
(
∂
u
/
∂
x
)
∈
L
2
(
Q
)
, the following inequality holds:
(9)
∥
u
∥
E
2
≤
c
∥
ℱ
∥
F
2
,
where
c
=
2
(
l
+
exp
(
T
/
2
α
2
)
)
.
Proof.
We set
M
u
=
x
m
u
with
0
≤
x
≤
α
and
(10)
M
u
=
l
-
x
l
-
α
x
m
u
(
x
,
t
)
+
1
l
-
α
J
(
x
m
u
)
,
J
ϑ
(
x
)
=
∫
α
x
ϑ
(
ξ
)
d
ξ
,
with
α
≤
x
≤
l
.
Consider the following equality:
(11)
∫
0
τ
∫
0
l
L
u
M
∂
u
∂
t
d
x
d
t
=
∫
0
τ
∫
0
l
∂
u
∂
t
M
∂
u
∂
t
d
x
d
t
-
∫
0
τ
∫
0
l
1
x
m
∂
∂
x
(
x
m
∂
u
∂
x
)
M
∂
u
∂
t
d
x
d
t
.
It can be seen that the following equalities hold:
(12)
∫
0
τ
∫
0
l
∂
u
∂
t
M
∂
u
∂
t
d
x
d
t
=
∫
0
τ
∫
0
α
x
m
|
∂
u
∂
t
|
2
d
x
+
∫
0
τ
∫
α
l
[
(
l
-
x
l
-
α
)
x
m
|
∂
u
∂
t
|
2
+
1
l
-
α
∂
u
∂
t
∫
α
x
ξ
∂
u
(
ξ
,
t
)
d
ξ
∂
t
d
x
d
t
(
l
-
x
l
-
α
)
x
m
|
∂
u
∂
t
|
2
]
,
∫
α
l
∂
u
∂
t
∫
α
x
ξ
m
∂
u
(
ξ
,
t
)
∂
t
d
ξ
d
x
=
m
2
∫
α
l
∂
u
∂
t
|
∫
α
x
ξ
m
∂
u
(
ξ
,
t
)
∂
t
d
ξ
|
2
d
x
.
By (12), we obtain the following equality:
(13)
∫
0
τ
∫
0
l
∂
u
∂
t
M
∂
u
∂
t
d
x
d
t
=
∫
0
τ
∫
0
α
x
m
|
∂
u
∂
t
|
2
d
x
d
t
+
∫
0
τ
∫
α
l
l
-
x
l
-
α
x
m
|
∂
u
∂
t
|
2
d
x
d
t
+
m
l
-
α
∫
0
τ
∫
α
l
1
x
m
+
1
|
∫
α
x
ξ
m
∂
u
(
ξ
,
t
)
∂
t
d
ξ
|
2
d
x
d
t
.
Integrating by parts (and using (6)), we get
(14)
-
∫
0
τ
∫
0
α
1
x
m
∂
∂
x
(
x
m
∂
u
∂
x
)
x
m
∂
u
∂
t
d
x
d
t
=
-
∫
0
τ
x
m
∂
u
∂
x
∂
u
∂
t
d
t
|
x
=
0
x
=
α
+
∫
0
τ
∫
0
α
x
m
∂
u
∂
x
∂
2
u
∂
x
∂
t
d
x
d
t
=
-
∫
0
τ
α
m
∂
u
(
α
,
t
)
∂
α
∂
u
(
α
,
t
)
∂
t
d
t
+
1
2
∫
0
α
x
m
(
∂
u
(
x
,
τ
)
∂
x
)
2
d
x
-
1
2
∫
0
α
x
m
(
d
φ
(
x
)
d
x
)
2
d
x
,
-
∫
0
τ
∫
α
l
l
-
x
l
-
α
1
x
m
∂
∂
x
(
x
m
∂
u
∂
x
)
x
m
∂
u
∂
t
d
x
d
t
=
-
∫
0
τ
l
-
x
l
-
α
x
m
∂
u
∂
x
∂
u
∂
t
d
t
|
x
=
α
x
=
l
+
∫
0
τ
∫
α
l
l
-
x
l
-
α
x
m
∂
u
∂
x
∂
2
u
∂
x
∂
t
d
x
d
t
-
1
l
-
α
∫
0
τ
∫
α
l
x
m
∂
u
∂
x
∂
u
∂
t
d
x
d
t
=
∫
0
τ
α
∂
u
(
α
,
t
)
∂
α
∂
u
(
α
,
t
)
∂
t
d
t
+
1
2
∫
α
l
l
-
x
l
-
α
x
m
(
∂
u
(
x
,
τ
)
∂
x
)
2
d
x
-
1
2
∫
α
l
l
-
x
l
-
α
x
m
|
∂
φ
(
x
)
∂
x
|
2
d
x
-
1
l
-
α
∫
0
τ
∫
α
l
x
m
∂
u
∂
x
∂
u
∂
t
d
x
d
t
,
-
∫
0
τ
∫
α
l
1
l
-
α
1
x
m
∂
∂
x
(
x
m
∂
u
∂
x
)
∫
α
x
ξ
m
∂
u
(
ξ
,
t
)
∂
t
d
ξ
d
x
d
t
=
1
l
-
α
∫
0
τ
∫
α
l
x
m
∂
u
∂
x
∂
u
∂
t
d
x
d
t
-
m
l
-
α
∫
0
τ
∫
α
l
1
x
∂
u
∂
x
∫
α
x
ξ
m
∂
u
(
ξ
,
t
)
∂
t
d
ξ
d
x
d
t
=
1
l
-
α
∫
0
τ
∫
α
l
x
m
∂
u
∂
x
∂
u
∂
t
d
x
d
t
-
m
l
-
α
∫
0
τ
∫
α
l
u
x
2
∫
α
x
ξ
m
∂
u
(
ξ
,
t
)
∂
t
d
ξ
d
x
d
t
+
m
2
(
l
-
x
)
∫
α
l
u
2
(
x
,
τ
)
x
m
-
1
d
x
-
m
2
(
l
-
x
)
∫
α
l
x
m
-
1
φ
2
(
x
)
d
x
.
The formulas (14) imply the following:
(15)
-
∫
0
τ
∫
0
α
1
x
m
∂
∂
x
(
x
m
∂
u
∂
x
)
M
∂
u
∂
t
d
x
d
t
=
1
2
{
1
2
∫
0
α
x
m
(
∂
u
(
x
,
τ
)
∂
x
)
2
d
x
+
∫
α
l
[
l
-
x
l
-
α
x
m
(
∂
u
(
x
,
τ
)
∂
x
)
2
+
m
·
x
m
(
l
-
α
)
u
2
(
x
,
τ
)
]
d
x
-
∫
0
α
x
m
|
∂
φ
(
x
)
∂
x
|
2
d
x
-
∫
α
l
[
l
-
x
l
-
α
x
m
|
∂
φ
(
x
)
∂
x
|
2
-
m
l
-
α
x
m
-
1
φ
2
(
x
)
1
2
∫
0
α
x
m
(
∂
u
(
x
,
τ
)
∂
x
)
2
]
d
x
}
-
m
l
-
α
∫
0
τ
∫
0
α
u
(
x
,
t
)
x
2
∫
α
x
ξ
m
∂
u
(
ξ
,
t
)
∂
t
d
ξ
d
t
d
x
.
Adding (13) and (15), we get
(16)
∫
0
τ
∫
0
l
L
u
M
∂
u
∂
t
d
x
d
t
=
∫
0
τ
∫
0
l
ψ
(
x
)
x
m
|
∂
u
∂
t
|
2
d
x
d
t
+
1
2
{
∫
0
l
ψ
(
x
)
x
m
(
∂
u
(
x
,
τ
)
∂
x
)
2
d
x
+
m
(
l
-
α
)
∫
α
l
x
m
-
1
u
2
(
x
,
τ
)
d
x
-
∫
0
l
ψ
(
x
)
x
m
|
d
φ
d
x
|
2
d
x
-
m
(
l
-
α
)
∫
α
l
x
m
-
1
φ
2
(
x
)
d
x
}
+
m
(
l
-
α
)
∫
0
τ
∫
α
l
1
x
m
+
1
|
∫
α
x
ξ
m
∂
u
(
ξ
,
t
)
∂
t
|
2
d
x
d
t
-
m
(
l
-
α
)
∫
0
τ
∫
α
l
u
x
2
∫
α
x
ξ
m
∂
u
(
ξ
,
t
)
∂
t
d
x
d
t
,
where the function
(17)
ψ
(
x
)
=
{
1
,
0
≤
x
≤
α
l
-
x
l
-
α
,
α
≤
x
≤
l
.
Now, it can be shown that the following inequalities hold:
(18)
∫
0
τ
∫
α
l
u
x
2
∫
α
x
ξ
m
∂
u
(
ξ
,
t
)
∂
t
d
x
d
t
≤
∫
0
τ
∫
α
l
1
x
m
+
1
|
∫
α
x
ξ
m
∂
u
(
ξ
,
t
)
∂
t
|
2
d
x
d
t
+
1
4
α
2
∫
0
τ
∫
α
l
x
m
-
1
u
2
(
x
,
t
)
d
x
d
t
,
∫
0
τ
∫
0
l
L
u
M
∂
u
∂
t
d
x
d
t
≤
1
2
∫
0
τ
∫
0
l
x
m
|
L
u
|
2
d
x
d
t
+
1
2
∫
0
τ
∫
0
l
ψ
(
x
)
x
m
|
∂
u
∂
t
|
2
d
x
d
t
,
l
-
x
≤
ψ
(
x
)
≤
1
.
The equality (16) and the inequalities (18) imply the following inequality:
(19)
∫
0
τ
∫
0
l
(
l
-
x
)
x
m
|
∂
u
∂
t
|
2
d
x
d
t
+
∫
0
l
(
l
-
x
)
x
m
|
∂
u
(
x
,
τ
)
∂
x
|
2
d
x
+
m
l
-
α
∫
α
l
x
m
-
1
u
2
(
x
,
τ
)
d
x
≤
1
2
α
2
m
l
-
α
∫
0
τ
∫
α
l
x
m
-
1
u
2
d
x
d
t
+
∫
0
l
x
m
|
∂
φ
∂
x
|
2
d
x
+
m
l
-
α
∫
α
l
x
m
-
1
φ
2
(
x
)
d
x
+
∫
0
τ
∫
0
l
x
m
(
L
u
)
2
d
x
d
t
.
Lemma 2.
Let on
[
0
,
T
]
continuous nonnegative functions
g
1
,
g
, and
h
be given, where
h
is nondecreasing. Then from the inequality
(20)
g
1
(
τ
)
+
g
(
τ
)
≤
c
∫
0
τ
g
(
t
)
d
t
+
h
(
τ
)
implies the inequality
(21)
g
1
(
τ
)
+
g
(
τ
)
≤
e
c
τ
h
(
τ
)
.
The above lemma can be proved by iteration method. We omit the details.
In order to apply the lemma, we set
(22)
g
(
τ
)
=
m
l
-
α
∫
α
l
x
m
-
1
u
2
(
x
,
τ
)
d
x
.
Let
g
1
(
τ
)
be the first two terms in the left hand side of (19),
h
(
τ
)
the last three terms in the right hand side of (19), and
c
=
1
/
2
α
2
.
As a consequence of (19), we obtain the following inequality:
(23)
∫
0
τ
∫
0
l
(
l
-
x
)
x
m
|
∂
u
∂
t
|
2
d
x
d
t
+
∫
0
l
(
l
-
x
)
x
m
|
∂
u
(
x
,
τ
)
∂
x
|
2
d
x
+
m
l
-
α
∫
α
l
x
m
-
1
u
2
(
x
,
τ
)
d
x
≤
e
T
/
2
α
2
∥
ℱ
∥
F
2
.
Here we have used the notation (7). The right hand side of inequality (23) does not depend on
ℱ
. Consequently, in the right hand side of (23) the supremum can be taken. Thus, we get the following inequality:
(24)
∫
0
T
∫
0
l
(
l
-
x
)
x
m
|
∂
u
∂
t
|
2
d
x
d
t
+
sup
0
≤
t
≤
T
[
∫
0
l
(
l
-
x
)
x
m
(
∂
u
∂
x
)
2
d
x
+
m
l
-
α
∫
α
l
x
m
-
1
u
2
(
x
,
t
)
d
x
]
≤
e
T
/
2
α
2
∥
ℱ
∥
F
2
.
By (1), we get the following inequality:
(25)
∫
0
T
∫
0
l
(
l
-
x
)
1
x
m
(
∂
∂
x
(
x
m
∂
u
∂
x
)
)
2
d
x
d
t
≤
2
∫
0
T
∫
0
l
(
l
-
x
)
x
m
|
∂
u
∂
t
|
2
d
x
d
t
+
2
l
∫
0
T
∫
0
l
x
m
|
f
|
2
d
x
d
t
.
The inequalities (24) and (25) imply inequality (9).