EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 10, No. 5, 2017, 981-994
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Inverse Problem for a Parabolic Equation in a
Rectangle Domain with Integral Conditions

Elvin Azizbayov1,∗, Yashar Mehraliyev2

1 Department of Computational Mathematics, Mechanics and Mathematics Faculty,
Baku State University, Baku, Azerbaijan
2 Department of Differential and Integral Equations, Mechanics and Mathematics Faculty,
Baku State University, Baku, Azerbaijan

Abstract. This paper is devoted to study of the nonlocal inverse boundary-value problem for a
second-order parabolic equation. The problem is considered in the rectangular domain. First, we
introduce a definition of a classical solution of the stated problem. Then, the initial problem is
reduced to an equivalent problem, for which using the method of contraction mappings principle
the theorem of the existence and uniqueness of solutions is proved. Moreover, using the equivalency,
we prove the existence and uniqueness of classical solution of the original problem.

2010 Mathematics Subject Classifications: 35A02, 35A09, 35R30, 35K10.
Key Words and Phrases: Inverse value problem, parabolic equation, classical solution, integral
condition.

1. Introduction

Inverse problems for differential equations are called the problem of finding the un-
known coefficients of differential equations, right-hand side, boundary or initial conditions,
the border of domain. The unknown elements of the initial-boundary value problems
defined for some additional information about solving equations. Such information are
different kinds of overdetermination condition [3], [9], [14], [16]. Inverse problems for dif-
ferential equations of mathematical physics are now playing an important role in the field
of natural sciences and their applications [1], [6], [7],[8], [18]. Coefficient inverse problems
are the problems in which, together with the solutions of differential equations is unknown
and is one (or more) of its coefficients. Many important applied problems relating to diffu-
sion processes, electromagnetic oscillations, elastic deformations, geophysics, seismology,
and computed tomography, scattering theory, acoustics, optics, theory of molecular oscil-
lations, radiolocation, gravity, and others, lead to the like inverse problems [2], [5], [11],
[4], [15],[13], [12], [17], [19]. In the submitted article the inverse boundary value problem
with nonlocal conditions for second order parabolic equation is studied.

∗Corresponding author.

Email addresses: eazizbayov@bsu.edu.az (E. Azizbayov), yashar aze@mail.ru (Y. Mehraliyev)

http://www.ejpam.com 981 c© 2017 EJPAM All rights reserved.



E. Azizbayov, Y. Mehraliyev / Eur. J. Pure Appl. Math, 10 (5) (2017), 981-994 982

2. Problem Statement

Denote by QT the domain {(x, t) : 0 ≤ x ≤ 1, 0 ≤ t ≤ T} and consider the equation

c(t)ut(x, t) = uxx(x, t) + a(t)u(x, t) + b(t)g(x, t) + f(x, t), (x, t) ∈ QT (1)

with nonlocal initial condition

u(x, 0) + δu(x, T ) = ϕ(x) (0 ≤ x ≤ 1), (2)

Neumann boundary condition

ux(0, t) = 0 (0 ≤ t ≤ T ), (3)

nonlocal integral condition

1∫
0

u(x, t)dx = 0 (0 ≤ t ≤ T ), (4)

and the additional conditions

u(0, t) = h1(t) (0 ≤ t ≤ T ), (5)

u(1, t) = h2(t) (0 ≤ t ≤ T ), (6)

where δ ≥ 0 is a fixed number, 0 < c(t), g(x, t), f(x, t), 0 ≤ p(t), hi(t) (i = 1, 2) are
given functions, u(x, t), a(t) and b(t) are unknown functions.

Definition 1. The triple {u(x, t), a(t), b(t)} is said to be a classical solution of problem
(1)-(6), if for the functions u(x, t), a(t) and b(t) satisfy the following conditions:

(i) The function u(x, t) and its derivatives ut(x, t), ux(x, t), uxx(x, t) are continuous in
the domain QT ;

(ii) the functions a(t) and b(t) are continuous on the interval [0, T ];

(iii) equation (1) and conditions (2)-(6) are satisfied in the usual sense.

To investigate the problem (1)-(6), first consider the following problem:

c(t)y′(t) = a(t)y(t) (0 ≤ t ≤ T ), (7)

y(0) + δy(T ) = 0 (8)

where δ ≥ 0 is a given number c(t), a(t) ∈ C[0, T ] are given functions, and y = y(t)
is unknown function. Under the classic solution of problem (7),(8) we understand the
function y(t), continuous on the interval [0, T ] together with all its derivatives contained
in equation (7) satisfying both (7) and (8) in the classical sense.



E. Azizbayov, Y. Mehraliyev / Eur. J. Pure Appl. Math, 10 (5) (2017), 981-994 983

Lemma 1. Let δ ≥ 0, 0 < c(t) ∈ C[0, T ], and a(t) ∈ C[0, T ]. Then the problem (7), (8)
has a unique trivial solution.

Proof. It’s obvious that the general solution of equation (7) has the form:

y(t) = ce

t∫
0

a(τ)
c(τ)

dτ
. (9)

Using (7) we obtain

c

1 + δe

T∫
0

a(t)
c(t)

dt

 = 0.

By δ ≥ 0, from the latter relation we have c = 0. Putting the value of c = 0 in (9), we
get that the problem (7), (8) has only the trivial solution. The proof is complete.

Theorem 1. Suppose that

δ ≥ 0, 0 < c(t) ∈ C[0, T ], f(x, t) ∈ C(QT ),

1∫
0

f(x, t)dx = 0 (0 ≤ t ≤ T ),

g(x, t) ∈ C(QT ),

1∫
0

g(x, t)dx = 0 (0 ≤ t ≤ T ), hi(t) ∈ C1[0, T ], (i = 1, 2),

h(t) ≡ h1(t)g(1, t)− h2(t)g(0, t) 6= 0 (0 ≤ t ≤ T )

and the compatibility conditions
1∫

0

ϕ(x)dx = 0, (10)

h1(0) + δh1(T ) = ϕ(0), h2(0) + δh2(T ) = ϕ(1) (11)

hold. Then the problem of finding a classical solution of (1)-(6) is equivalent to the problem
of determining functions u(x, t) ∈ C2,1(QT ), a(t) ∈ C[0, T ], and b(t) ∈ C[0, T ], satisfying
equation (1), conditions (2) and (3), and the conditions

ux(1, t) = 0 (0 ≤ t ≤ T ), (12)

c(t)h′1(t) = uxx(0, t) + a(t)h1(t) + b(t)g(0, t) + f(0, t) (0 ≤ t ≤ T ), (13)

c(t)h′2(t) = uxx(1, t) + a(t)h2(t) + b(t)g(1, t) + f(1, t) (0 ≤ t ≤ T ). (14)



E. Azizbayov, Y. Mehraliyev / Eur. J. Pure Appl. Math, 10 (5) (2017), 981-994 984

Proof. Let {u(x, t), a(t), b(t)} be a classical solution of (1)-(6). Integrating both sides
of (1) with respect to x from 0 to 1 yields

c(t)
d

dt

1∫
0

u(x, t)dx = ux(1, t)− ux(0, t)

+a(t)

1∫
0

u(x, t)dx+ b(t)

1∫
0

g(x, t)dx+

1∫
0

f(x, t)dx (0 ≤ t ≤ T ). (15)

Under the assumptions
1∫
0

f(x, t)dx = 0,
1∫
0

g(x, t)dx = 0 (0 ≤ t ≤ T ), by virtue of (3)

we conclude that (12) is satisfied.
Setting x = 0, in (1) we obtain

c(t)ut(0, t) = uxx(0, t) + a(t)u(0, t) + b(t)g(0, t) + f(0, t) (0 ≤ t ≤ T ). (16)

Similarly, from the equation (1), we get

c(t)ut(1, t) = uxx(1, t) + a(t)u(1, t) + b(t)g(1, t) + f(1, t) (0 ≤ t ≤ T ). (17)

Further, assuming hi(t) ∈ C1[0, T ] (i = 1, 2) and differentiating (5) and (6), we have

ut(0, t) = h1(t) (0 ≤ t ≤ T ), (18)

ut(1, t) = h2(t) (0 ≤ t ≤ T ) (19)

respectively.
From (16), by (5) and (18), we conclude that the relation (13) is fulfilled.
Analogously, from (17), by using the relations (6) and (19), we arrive at the satisfying

of (14).
Now, suppose that {u(x, t), a(t), b(t)} is the solution of (1)-(3), (12)-(14). Then from

(15), by means of (3) and (12), we find

c(t)
d

dt

1∫
0

u(x, t)dx = a(t)

1∫
0

u(x, t)dx (0 ≤ t ≤ T ). (20)

By virtue of (2) and (10), it is not hard to see that

1∫
0

u(x, 0)dx+ δ

1∫
0

u(x, T )dx =

1∫
0

(u(x, 0) + δu(x, T )) dx =

1∫
0

ϕ(x)dx = 0. (21)

Since, by Lemma 1, the problems (20) and (21) has only a trivial solution, it follows
that

1∫
0

u(x, t)dx = 0 (0 ≤ t ≤ T ),



E. Azizbayov, Y. Mehraliyev / Eur. J. Pure Appl. Math, 10 (5) (2017), 981-994 985

i.e. the condition (4) holds.
Moreover, from (13), (14), (16) and (17) we find

c(t)
d

dt
(u(0, t)− h1(t)) = a(t)(u(0, t)− h1(t)) (0 ≤ t ≤ T ), (22)

c(t)
d

dt
(u(1, t)− h2(t)) = a(t)(u(1, t)− h2(t)) (0 ≤ t ≤ T ). (23)

Using (2) and the compatibility conditions (11) we have

u(0, 0)− h1(0) + δ(u(0, T )− h1(0)) = (u(0, 0) + δu(0, T ))− (h1(0) + δh1(T ))

= ϕ(0)− (h1(0) + δh1(T )) = 0, (24)

u(1, 0)− h1(0) + δ(u(1, T )− h1(0)) = (u(1, 0) + δu(1, T ))− (h2(0) + δh2(T ))

= ϕ(1)− (h2(0) + δh2(T )) = 0. (25)

From (22), (24), and (23),(25), by Lemma 1, we conclude that conditions (5) and (6)
are satisfied. The lemma is thus proved.

3. Solvability of inverse boundary-value problem

Since for λk = kπ (k = 0, 1, . . .), the system {cosλkx}∞k=0 form an orthogonal system
in L2(0, 1). We’ll seek the first component u(x, t) of classical solution u(x, t), a(t), b(t) of
the problem (1)-(3), (12)-(14) in the form

u(x, t) =
∞∑
k=0

uk(t) cosλkx (λk = kπ), (26)

where

uk(t) = mk

1∫
0

u(x, t) cosλkxdx (k = 0, 1, 2, ...),

and

mk =

{
1, k = 0,
2, k = 1, 2, ....

Then applying the formal scheme of the Fourier method, from (1) and (2) we obtain

c(t)u′0(t) = F0(t;u, a, b) (0 ≤ t ≤ T ), (27)

c(t)u′k(t) + λ2kuk(t) = Fk(t;u, a, b) (k = 1, 2 . . . ; 0 ≤ t ≤ T ), (28)

uk(0) + δuk(T ) = ϕk (k = 0, 1, 2 . . .) (29)

where
Fk(t;u, a, b) = fk(t) + b(t)gk(t) + a(t)uk(t) (k = 0, 1, 2 . . .),



E. Azizbayov, Y. Mehraliyev / Eur. J. Pure Appl. Math, 10 (5) (2017), 981-994 986

gk(t) = mk

1∫
0

g(x, t) cosλkxdx (k = 0, 1, 2, . . .),

fk(t) = mk

1∫
0

f(x, t) cosλkxdx (k = 0, 1, 2, . . .),

ϕk = mk

1∫
0

ϕ(x) cosλkxdx (k = 0, 1, 2, . . .).

Solving problem (27)-(29) we get

u0(t) = (1 + δ)−1

ϕ0 − δ
T∫
0

1

c(t)
F0(t;u, a, b)dt

+

t∫
0

1

c(τ)
F0(τ ;u, a, b)dτ, (30)

uk(t) =
e
−

t∫
0

λ2k
c(s)

ds

1 + δe
−
T∫
0

λ2
k

c(s)
ds

ϕk −
δe
−
T∫
0

λ2k
c(s)

ds

1 + δe
−
T∫
0

λ2
k

c(s)
ds

T∫
0

1

c(τ)
Fk(τ ;u, a, b)e

−
t∫
τ

λ2k
c(s)

ds
dτ

+

t∫
0

1

c(τ)
Fk(τ ;u, a, b)e

−
t∫
τ

λ2k
c(s)

ds
dτ (k = 1, 2, . . .). (31)

After substituting expressions uk(t) (k = 0, 1, ...) in (26), we have

u(x, t) = (1 + δ)−1

ϕ0 − δ
T∫
0

1

c(t)
F0(t;u, a, b)dt

+

t∫
0

1

c(τ)
F0(τ ;u, a, b)dτ

+
∞∑
k=1


e
−

t∫
0

λ2k
c(s)

ds

1 + δe
−
T∫
0

λ2
k

c(s)
ds

ϕk −
δe
−
T∫
0

λ2k
c(s)

ds

1 + δe
−
T∫
0

λ2
k

c(s)
ds

T∫
0

1

c(τ)
Fk(τ ;u, a, b)e

−
t∫
τ

λ2k
c(s)

ds
dτ

+

t∫
0

1

c(τ)
Fk(τ ;u, a, b)e

−
t∫
τ

λ2k
c(s)

ds
dτ

 cosλkx. (32)

Now, using (26), from (13) and (14) we find

a(t) = [h(t)]−1{(c(t)h′1(t)− f(0, t))g(1, t)− (c(t)h′2(t)− f(1, t))g(0, t)

+
∞∑
k=1

λ2kuk(t)(g(1, t)− (−1)kg(0, t))

}
, (33)



E. Azizbayov, Y. Mehraliyev / Eur. J. Pure Appl. Math, 10 (5) (2017), 981-994 987

b(t) = [h(t)]−1{h1(t)(c(t)h′2(t)− f(1, t))− h2(t)(c(t)h′1(t)− f(0, t))

+

∞∑
k=1

λ2kuk(t)((−1)kh1(t)− h2(t))

}
, (34)

where
h(t) ≡ h1(t)g(1, t)− h2(t)g(0, t) 6= 0 (0 ≤ t ≤ T ). (35)

Putting the expression of (31) in (33) and (34) we obtain

a(t) = [h(t)]−1{(c(t)h′1(t)− f(0, t))g(1, t)− (c(t)h′2(t)− f(1, t))g(0, t)

+

∞∑
k=1

λ2k

 e
−

t∫
0

λ2k
c(s)

ds

1 + δe
−
T∫
0

λ2
k

c(s)
ds

ϕk −
δe
−
T∫
0

λ2k
c(s)

ds

1 + δe
−
T∫
0

λ2
k

c(s)
ds

T∫
0

1

c(τ)
Fk(τ ;u, a, b)e

−
t∫
τ

λ2k
c(s)

ds
dτ

+

t∫
0

1

c(τ)
Fk(τ ;u, a, b)e

−
t∫
τ

λ2k
c(s)

ds
dτ

 (g(1, t)− (−1)kg(0, t)), (36)

b(t) = [h(t)]−1{h1(t)(c(t)h′2(t)− f(1, t))− h2(t)(c(t)h′1(t)− f(0, t))

+
∞∑
k=1

λ2k

 e
−

t∫
0

λ2k
c(s)

ds

1 + δe
−
T∫
0

λ2
k

c(s)
ds

ϕk −
δe
−
T∫
0

λ2k
c(s)

ds

1 + δe
−
T∫
0

λ2
k

c(s)
ds

T∫
0

1

c(τ)
Fk(τ ;u, a, b)e

−
t∫
τ

λ2k
c(s)

ds
dτ

+

t∫
0

1

c(τ)
Fk(τ ;u, a, b)e

−
t∫
τ

λ2k
c(s)

ds
dτ

 ((−1)kh1(t)− h2(t)). (37)

Analogously [10], the following lemma was proved.

Lemma 2. Let {u(x, t), a(t), b(t)} be an arbitrary solution of (1)-(3), (12)-(14), then the
functions

uk(t) = mk

1∫
0

u(x, t) cosλkxdx (k = 0, 1, 2, . . .)

satisfy system (30) and (31) on the interval [0, T ].

Remark 1. From Lemma 2 it follows that to prove the uniqueness of the solution of
problem (1)-(3),(12)-(14), it is suffices to prove the uniqueness of the solution of system
(32),(36) and (37).



E. Azizbayov, Y. Mehraliyev / Eur. J. Pure Appl. Math, 10 (5) (2017), 981-994 988

Now, consider the following space.
Denote by B3

2,T [10] the set of all functions of the form

u(x, t) =
∞∑
k=0

uk(t) cosλkx (λ = kπ),

considered in domain QT , where the functions uk(t) (k = 0, 1, 2, ...) are continuous on the
interval [0, T ] and satisfies the following condition

‖u0(t)‖C[0,T ] +

( ∞∑
k=1

(
λ3k ‖uk(t)‖C[0,T ]

)2) 1
2

< +∞.

In the space B3
2,T the operations addition and multiplication by a scalar, we define in

the usual way, and the norm defined by the following formula

‖u(x, t)‖B3
2,T

= ‖u0(t)‖C[0,T ] +

( ∞∑
k=1

(
λ3k ‖uk(t)‖C[0,T ]

)2) 1
2

.

We denote by E3
T , the Banach space B3

2,T×C[0, T ]×C[0, T ] of vector functions z(x, t) =
{u(x, t), a(t), b(t)} with norm

‖z(x, t)‖B3
2,T

= ‖u(x, t)‖B3
2,T

+ ‖a(t)‖C[0,T ] + ‖b(t)‖C[0,T ] .

It is known that B3
2,T and E3

T are Banach spaces.
Now consider the operator

Φ(u, a, b) = {Φ1(u, a, b),Φ2(u, a, b),Φ3(u, a, b)}

in the space E3
T , where

Φ1(u, a, b) = ũ(x, t) ≡
∞∑
k=0

ũk(t) cosλkx, Φ2(u, a, b) = ã(t), Φ3(u, a, b) = b̃(t),

and the functions ũ0(t), ũk(t) (k = 1, 2, ...), ã(t) and b̃(t) are equal to the right-hand
sides of (30),(31),(36) and (37) respectively.

Using simple transformations from (30),(31),(36) and (37) we obtain

‖ũ0(t)‖C[0,T ] ≤ (1 + δ)−1

|ϕ1|+ δ

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

√T
 T∫

0

|f0(τ)|2 dτ


1
2

+T ‖a(t)‖C[0,T ] ‖u0(t)‖C[0,T ] +
√
T ‖b(t)‖C[0,T ]

 T∫
0

|g0(τ)|2 dτ


1
2






E. Azizbayov, Y. Mehraliyev / Eur. J. Pure Appl. Math, 10 (5) (2017), 981-994 989

+

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

√T
 T∫

0

|f0(τ)|2 dτ


1
2

+ T ‖a(t)‖C[0,T ] ‖u0(t)‖C[0,T ]

+
√
T ‖b(t)‖C[0,T ]

 T∫
0

|g0(τ)|2 dτ


1
2

 , (38)

( ∞∑
k=1

(
λ3k ‖ũk(t)‖C[0,T ]

)2) 1
2

≤
√

5

( ∞∑
k=1

(
λ3k |ϕk|

)2) 1
2

+
√

5(1 + δ)

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

√
T

 T∫
0

∞∑
k=1

(
λ3k |fk(τ)|

)2
dτ


1
2

+
√

5(1 + δ)

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

T ‖a(t)‖C[0,T ]

( ∞∑
k=1

(
λ3k ‖uk(t)‖C[0,T ]

)2) 1
2

+2
√

2T (1 + δ)

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

‖b(t)‖C[0,T ]

 T∫
0

∞∑
k=1

(
λ3k |gk(τ)|

)2
dτ


1
2

, (39)

‖ã(t)‖C[0,T ] ≤
∥∥[h(t)]−1

∥∥
C[0,T ]

×{
∥∥(c(t)h′1(t)− f(0, t))g(1, t)− (c(t)h′2(t)− f(1, t))g(0, t))

∥∥
C[0,T ]

+

( ∞∑
k=1

λ−2k

) 1
2

‖|g(0, t)|+ |g(1, t)|‖C[0,T ]

( ∞∑
k=1

(
λ3k |ϕk|

)2) 1
2

+(1 + δ)

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

√
T

 T∫
0

∞∑
k=1

(
λ3k |fk(τ)|

)2
dτ


1
2

+(1 + δ)

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

T ‖a(t)‖C[0,T ]

( ∞∑
k=1

(
λ3k ‖uk(t)‖C[0,T ]

)2) 1
2

+(1 + δ)

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

√
T ‖b(t)‖C[0,T ]

 T∫
0

∞∑
k=1

(
λ3k |gk(τ)dτ |

)2
1
2


 , (40)

∥∥∥b̃(t)∥∥∥
C[0,T ]

≤
∥∥[h(t)]−1

∥∥
C[0,T ]



E. Azizbayov, Y. Mehraliyev / Eur. J. Pure Appl. Math, 10 (5) (2017), 981-994 990

×{
∥∥h1(t)(c(t)h′2(t)− f(1, t))− h2(t)(c(t)h′1(t)− f(0, t))

∥∥
C[0,T ]

+

( ∞∑
k=1

λ−2k

) 1
2

‖|h1(t)|+ |h2(t)|‖C[0,T ]

( ∞∑
k=1

(
λ3k |ϕk|

)2) 1
2

+(1 + δ)

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

√
T

2∑
i=1

 T∫
0

∞∑
k=1

(
λ3k |fk(τ)|

)2
dτ


1
2

+(1 + δ)

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

T ‖a(t)‖C[0,T ]

( ∞∑
k=1

(
λ3k ‖uk(t)‖C[0,T ]

)2) 1
2

+(1 + δ)

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

√
T ‖b(t)‖C[0,T ]

 T∫
0

∞∑
k=1

(
λ3k |gk(τ)dτ |

)2
1
2


 . (41)

Suppose that the data for problem (1)-(3), and (12)-(14) satisfy the following conditions

A1) ϕ(x) ∈ C2[0, 1], ϕ′′′(x) ∈ L2(0, 1), ϕ′(0) = ϕ′(1) = 0;

A2) f(x, t), fx(x, t), fxx(x, t) ∈ C2[0, 1], fxxx(x, t) ∈ L2(QT ),
fx(0, t) = fx(1, t) = 0 (0 ≤ t ≤ T );

A3) g(x, t), gx(x, t), gxx(x, t) ∈ C2[0, 1], gxxx(x, t) ∈ L2(QT ),
gx(0, t) = gx(1, t) = 0 (0 ≤ t ≤ T );

A4) δ ≥ 0, hi(t) ∈ C1[0, T ] (i = 1, 2), h(t) = h1(t)g(1, t)−h2(t)g(0, t) 6= 0 (0 ≤ t ≤ T ).

Then from (33)-(35) we find that

‖ũ(x, t)‖B3
2,T
≤ A1(T ) +B1(T ) ‖a(t)‖C[0,T ] ‖u(x, t)‖B3

2,T
+D1(T ) ‖b(t)‖C[0,T ] , (42)

‖ã(t)‖C[0,T ] ≤ A2(T ) +B2(T ) ‖a(t)‖C[0,T ] ‖u(x, t)‖B3
2,T

+D2(T ) ‖b(t)‖C[0,T ] , (43)∥∥∥b̃(t)∥∥∥
C[0,T ]

≤ A3(T ) +B3(T ) ‖a(t)‖C[0,T ] ‖u(x, t)‖B3
2,T

+D3(T ) ‖b(t)‖C[0,T ] , (44)

where

A1(T ) = (1 + δ)−1

(
2 ‖ϕ(x)‖L2(0,1)

+ 2δ

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

‖f(x, t)‖L2(QT )

)

+

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

‖f(x, t)‖L2(QT )
+
√

5
∥∥ϕ′′′(x)

∥∥
L2(0,1)

+
√

5T

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

‖fxxx(x, t)‖L2(QT )
,



E. Azizbayov, Y. Mehraliyev / Eur. J. Pure Appl. Math, 10 (5) (2017), 981-994 991

B1(T ) =
(
δ(1 + δ)−1 + 1

)
T

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

,

D1(T ) = (δ(1 + δ)−1 +
√

5)
√
T

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

‖gxxx(x, t)‖L2(QT )
,

A2(T ) =
∥∥[h(t)]−1

∥∥
C[0,T ]

×{
∥∥(c(t)h′1(t)− f(0, t))g(1, t)− (c(t)h′2(t)− f(1, t))g(0, t))

∥∥
C[0,T ]

+

( ∞∑
k=1

λ−2k

) 1
2

‖|g(x1, t)|+ |g(x2, t)|‖C[0,T ]

×

[∥∥ϕ′′′(x)
∥∥
L2(0,1)

+ (1 + δ)

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

√
T ‖fxxx(x, t)‖L2(QT )

]}
,

B2(T ) =
∥∥[h(t)]−1

∥∥
C[0,T ]

( ∞∑
k=1

(λ−2k )

) 1
2

‖|g(x1, t)|+ |g(x2, t)|‖C[0,T ] (1 + δ)T

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

,

D2(T ) =
∥∥[h(t)]−1

∥∥
C[0,T ]

( ∞∑
k=1

(λ−2k )

) 1
2

‖|g(1, t)| |g(0, t)|‖C[0,T ]

×(1 + δ)

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

√
T ‖fxxx(x, t)‖L2(QT )

,

A3(T ) =
∥∥[h(t)]−1

∥∥
C[0,T ]

{
∥∥h1(t)(c(t)h′2(t)− f(1, t))− h2(t)(c(t)h′1(t)− f(0, t))

∥∥
C[0,T ]

+

( ∞∑
k=1

λ−2k

) 1
2

‖|h1(t)|+ |h2(t)|‖C[0,T ]

×

[
2
∥∥ϕ′′′(x)

∥∥
L2(0,1)

+ 2(1 + δ)

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

√
T ‖fxxx(x, t)‖L2(QT )

]}
,

B3(T ) =
∥∥[h(t)]−1

∥∥
C[0,T ]

( ∞∑
k=1

(λ−2k )

) 1
2

‖|h1(t)|+ |h2(t)|‖C[0,T ] (1 + δ)T

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

,

D3(T ) =
∥∥[h(t)]−1

∥∥
C[0,T ]

( ∞∑
k=1

(λ−2k )

) 1
2

‖|h1(t)|+ |h2(t)|‖C[0,T ]

×(1 + δ)

∥∥∥∥ 1

c(t)

∥∥∥∥
C[0,T ]

√
T ‖fxxx(x, t)‖L2(QT )

.



E. Azizbayov, Y. Mehraliyev / Eur. J. Pure Appl. Math, 10 (5) (2017), 981-994 992

From the inequalities (42)-(44) we conclude that

‖ũ(x, t)‖B3
2,T

+ ‖ã(t)‖C[0,T ] +
∥∥∥b̃(t)∥∥∥

C[0,T ]

≤ A(T ) +B(T ) ‖a(t)‖C[0,T ] ‖u(x, t)‖B3
2,T

+D(T ) ‖b(t)‖C[0,T ] , (45)

where
A(T ) = A1(T ) +A2(T ) +A3(T ), B(T ) = B1(T ) +B2(T ) +B3(T ),

D(T ) = D1(T ) +D2(T ) +D3(T ).

Theorem 2. Suppose that the conditions A1)−A4) and the inequality

[B(T )(A(T ) + 2) +D(T )](A(T ) + 2) < 1 (46)

hold, then problem (1)-(3), (12)-(14) has a unique solution in the ball K = KR(‖z‖E3
T
≤

R ≤ A(T ) + 2) of the space E3
T .

Proof. In the space E3
T , we consider the equation

z = Φz, (47)

where z = {u, a, b}, and components Φi(u, a, b) (i = 1, 2, 3), of operator Φ(u, a, b), defined
by the right side of equations (32), (36) and (37) respectively.

Consider the operator Φ(u, a, b), in the ball K = KR of the space E3
T . Similarly, with

the aid of (45) we get that for any z1, z2, z3 ∈ KR the following inequalities hold

‖Φz‖E3
T
≤ A(T ) +B(T ) ‖a(t)‖C[0,T ] ‖u(x, t)‖B3

2,T
+D(T ) ‖b(t)‖C[0,T ]

≤ A(T ) +B(T )(A(T ) + 2)2 +D(T )(A(T ) + 2) < A(T ) + 2, (48)

‖Φz1 − Φz2‖E3
T
≤ B(T )R

(
‖a1(t)− a2(t)‖C[0,T ] + ‖u1(x, t)− u2(x, t)‖B3

2,T

)
+D(T ) ‖b1(t)− b2(t)‖C[0,T ] . (49)

Then by (46), from (48) and (49) it is clear that the operator Φ on the set K = KR

satisfy the conditions of the contraction mapping principle. Therefore the operator Φ has
a unique fixed point {z} = {u, a, b}, in the ball K = KR, which is a solution of equation
(47); i.e. in the ball K = KR is the unique solution of the systems (32), (36) and (37).
Then the function u(x, t), as an element of space B3

2,T , is continuous and has continuous
derivatives ux(x, t) and uxx(x, t) in QT .

From the equation (28) we obtain( ∞∑
k=1

(λk
∥∥u′k(t)∥∥C[0,T ]

)2

) 1
2

< +∞.



REFERENCES 993

Hence it follows that the function u(x, t) is continuous in QT . Further, it is possible to
verify that equation (1), and conditions (2), (3), (12), (13), (14), are satisfied in the usual
sense. Consequently, {u(x, t), a(t), b(t)} is a solution of (1)-(3), (11)-(12), and due to the
Lemma 2, it is unique in the ball K = KR. The theorem is thus proved.

From Theorem 2 and Theorem 1, it follows directly that the following assertion is
valid.

Theorem 3. Suppose that all assumptions of Theorem 2, and the compatibility conditions
(10) and (11) hold. If

1∫
0

f(x, t)dx = 0,

1∫
0

g(x, t)dx = 0 (0 ≤ t ≤ T ),

then problem (1)-(6) has a unique classical solution in the ball K = KR.

Acknowledgements

The authors would like to express their deep gratitude to the editorial team and the
anonymous referees of European Journal of Pure and Applied Mathematics, for the careful
reading of the manuscript as well as their valuable comments and suggestions which helped
to improve the present paper.

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