EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 2, Article Number 6119
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Inverse Boundary Value Problem for Pseudo
Hyperbolic Equation of the Fourth Order With
Nonlocal Integral Conditions of the Second Kind

Yashat T. Mehraliyev1,∗, Anar A. Mammadov2

1 Department of Differential and Integral Equations, Baku State University, 23, Z. Khalilov
Str., Baku, AZ1148, Azerbaijan
2 Department of Mathematics and Informatics, Baku Slavic University, 33, S. Rustam Str.,
Baku, AZ1014, Azerbaijan

Abstract. In this paper, we consider a nonlinear inverse boundary value problem for a fourth-
order pseudo hyperbolic equation with nonlocal conditions of the integral type. First, we introduce
the definition of a classical solution to the problem. The purpose of this paper is to determine the
unknown coefficient of the right-hand side and to solve the problem of interest. The problem is
considered in a rectangular domain. To study the solvability of the inverse problem, we perform a
transformation from the original problem to some auxiliary inverse problem with trivial boundary
conditions. Using the principle of contraction mappings, we prove the existence and uniqueness of
solutions to the auxiliary problem. Then we again perform a transformation to the problem and
as a result obtain the solvability of the inverse problem.

2020 Mathematics Subject Classifications: 35R30, 35L80, 35A01, 35A02, 35A09

Key Words and Phrases: Inverse boundary problem, Pseudo hyperbolic equation, Method
Fourier, Classic solution

1. Introduction

The foundations of the theory and practice of studying inverse problems were laid and
developed in the pioneering works [1], [2], [3], [4]. Subsequently, the applied significance
of inverse problems attracted the attention of many authors, and in recent decades nu-
merous articles and monographs devoted to inverse problems have been published (see,
for example, [5], [6], [7], [8], [9], [10], [11], [12], [13], [14], [15], [16], [17], [18], [19] and the
literature cited therein).

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.6119

Email addresses: yashar aze@mail.ru (Y. T. Mehraliyev), mammedov1@mail.ru (A. A. Mammadov)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



Y. T. Mehraliyev, A. A. Mammadov / Eur. J. Pure Appl. Math, 18 (2) (2025), 6119 2 of 9

2. Problem Statement on Determining the Unknown Coefficient and
the Constant Term

In the rectangle ΠT = {(x, t) : 0 ≤ x ≤ 1, 0 ≤ t ≤ T} we will consider the following
problem:

ωtt(x, t)− ωttxx(x, t) + ωxxxx(x, t)

= φ(t)ω(x, t) + ψ(t)r(x, t) + s(x, t) (x, t) ∈ ΠT , (1)

ω(x, 0) = ε(x) +

∫ T

0
σ1(t)ω(x, t)dt,

ωt(x, 0) = ν(x) +

∫ T

0
σ2(t)ω(x, t)dt (0 ≤ x ≤ 1), (2)

ω(0, t) = ωx(1, t) = ωxx(0, t) = ωxxx(1, t) = 0 (0 ≤ t ≤ T ), (3)

ω(xi, t) = ni(t) (xi ∈ (0, 1) i = 1, 2; x1 ̸= x2 0 ≤ t ≤ T ), (4)

where r(x, t), s(x, t), ε(x), ν(x), σi(t), ni(t) (i = 1, 2) given functions, and ω(x, t), φ(t),
ψ(t)-sought functions.

Let us designate

V 4,2(ΠT ) = {ω(x, t) : ω(x, t) ∈ C2(ΠT ), ωxxxx(x, t), ωttxx(x, t ∈ C(ΠT )}.

Definition 1. We call the triple of {ω(x, t), φ(t), ψ(t)} functions ω(x, t), φ(t) and ψ(t)
a classical solution of problem (1)-(4) if:

(i) ω(x, t) ∈ V 4,2(ΠT );

(ii) φ(t) ∈ C[0, T ] and ψ(t) ∈ C[0, T ];

(iii) functions ω(x, t), φ(t) and ψ(t) satisfy all conditions (1)-(4) in the usual sense.

Let us consider the following problem: Find a triple of {ω(x, t), φ(t), ψ(t)} functions
ω(x, t) ∈ V 4,2(ΠT ), φ(t) ∈ C[0, T ] and ψ(t) ∈ C[0, T ] from relations (1)-(3), and

h′′i (t)− ωttxx(xi, t) + ωxxxx(xi, t) =

= φ(t)ni(t) + ψ(t)r(xi, t) + s(xi, t) (i = 1, 2; 0 ≤ t ≤ T ). (5)

Similarly ([1]) the following is proved

Theorem 1. Let s(x, t), r(x, t) ∈ C(DT ), ε(x), v(x) ∈ C[0, 1], n(t) ≡ n1(t)r(x2, t) −
n2(t)r(x1, t) ̸= 0 (0 ≤ t ≤ T ), ni(t) ∈ C2[0, T ] (i = 1, 2), σi(t) ∈ C[0, T ], and the matching
conditions

ni(0) =

∫ T

0
σ1(t)ni(t)dt+ ε(xi), n′i(0) =

∫ T

0
σ2(t)ni(t)dt+ v(xi), i = 1, 2,



Y. T. Mehraliyev, A. A. Mammadov / Eur. J. Pure Appl. Math, 18 (2) (2025), 6119 3 of 9

are satisfied.
Then the following assertions are valid:
i) each classical solution {ω(x, t), φ(t), ψ(t)} of the problem (1)-(4) is a solution of

problem (1)-(3), (5), as well;
ii) each solution {ω(x, t), φ(t), ψ(t)} of the problem (1)-(3), (5), if(

T ∥σ2(t)∥C[0,T ] + ∥σ1(t) ∥C[0,T ] +
T

2
∥φ(t)∥C[0,T ]

)
T < 1,

is a classical solution of problem (1)-(4).

3. Study of Solvability of the Inverse Problem

It is obvious that component ω(x, t) of the solution {ω(x, t), φ(t), ψ(t)} of (1)-(3), (5)
has the form:

ω(x, t) =

∞∑
k=1

ωk(t) sinλkx
(
λk =

π

2
(2k − 1)

)
, (6)

where ωk(t) = 2
∫ 1
0 ω(x, t) sinλkxdx (k = 1, 2, ...)-Fourier coefficients of component ω(x, t)

in the field in the L2(0, 1) system
(
λk = π

2 (2k − 1)
)∞
k=1

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

(1 + λ2k)ω
′′
k(t) + λ4kωk(t) = Qk(t;ω, φ, ψ) (0 ≤ t ≤ T ; k = 1, 2, ...), (7)

ωk(0) = εk +

∫ T

0
σ1(t)ωk(t)dt, ω′

k(0) = vk +

∫ T

0
σ2(t)ωk(t)dt (k = 1, 2, ...), (8)

where
Qk(t;ω, φ, ψ) = φ(t)ωk(t) + ψ(t)rk(t) + sk(t),

sk(t) = 2

∫ 1

0
s(x, t) sinλkx dx, rk(t) = 2

∫ 1

0
r(x, t) sinλkx dx,

εk = 2

∫ 1

0
ε(x) sinλkxdx, vk = 2

∫ 1

0
v(x) sinλkxdx (k = 1, 2, ...).

It is easy to see that the solution to problem (7)-(8) has the form:

ωk(t) =

(
εk +

∫ T

0
σ1(t)ωk(t)dt

)
cosβkt +

1

βk

(
vk +

∫ T

0
σ2(t)ωk(t)dt

)
sinβkt+

+
1

βk(1 + λ2k)

∫ t

0
Qk(τ ;ω, φ, ψ) sinβk(t− τ)dτ (k = 1, 2, ...), (9)

where

β2k =
λ4k

1 + λ2k
(k = 1, 2, ...).



Y. T. Mehraliyev, A. A. Mammadov / Eur. J. Pure Appl. Math, 18 (2) (2025), 6119 4 of 9

Now, after substituting expression ωk(t) (k = 1, 2, ...) to determine ω(x, t), we have:

ω(x, t) =

∞∑
k=1

(
εk +

∫ T

0
σ1(t)ωk(t)dt

)
cosβkt+

1

βk

(
vk +

∫ T

0
σ2(t)ωk(t)dt

)
sinβkt+

+
1

βk(1 + λ2k)

∫ t

0
Qk(τ ;ω, φ, ψ) sinβk(t− τ)dτ

}
sinλkx. (10)

Next, using equation (9), from (5) and (6) we obtain:

φ(t) = [n(t)]−1{(n′′1(t)− s(x1, t))r(x2, t)− (n′′2(t)− s(x2, t))r(x1, t)+

+

∞∑
k=1

β2k

[(
εk +

∫ T

0
σ1(t)ωk(t)dt

)
cosβkt +

1

βk

(
vk +

∫ T

0
σ2(t)ωk(t)dt

)
sinβ

+
1

βk(1 + λ2k)

∫ t

0
Qk(τ ;ω, φ, ψ) sinβk(t − τ)dτ+

+
1

λ2k
Qk(τ ;ω, φ, ψ)

 (r(x2, t) sinλkx1−r(x1, t) sinλkx2)

 , (11)

ψ(t) = [n(t)]−1{(n′′2(t)− s(x2, t))n1(t)− (n′′1(t)− s(x1, t))n2(t)+

+
∞∑
k=1

β2k

[(
εk +

∫ T

0
σ1(t)ωk(t)dt

)
cosβkt +

1

βk

(
vk +

∫ T

0
σ2(t)ωk(t)dt

)
sinβ +

+
1

βk(1 + λ2k)

∫ t

0
Qk(τ ;ω, φ, ψ) sinβk(t − τ)dτ+

+
1

λ2k
Qk(τ ;ω, φ, ψ)

 (r(x2, t) sinλkx1−r(x1, t) sinλkx2)

+

+
1

λ2k
Qk(τ ;ω, φ, ψ)

 (n1(t) sinλkx2−n2(t) sinλkx1)

 . (12)

To study the problem of the uniqueness of the solution of problem (1)-(3), (5), the
following lemma plays an important role.

Lemma 1. If - {ω(x, t), φ(t), ψ(t)} any solution of problem (1)-(3), (5), then the function
ωk(t) = 2

∫ 1
0 ω(x, t) sinλkxdx (k = 1, 2, ...), i.e. the Fourier coefficients ω(x, t) in the

system
(
λk = π

2 (2k − 1)
)∞
k=1

satisfy the [0, T ]system (9).

This lemma implies the validity of the following



Y. T. Mehraliyev, A. A. Mammadov / Eur. J. Pure Appl. Math, 18 (2) (2025), 6119 5 of 9

Corollary 1. Let system (10), (11), (12) have a unique solution. Then problem (1)-(3),
(5) cannot have more than one solution, i.e. if problem (1)-(3), (5) has a solution, then
it is unique.

In order to study the problem (1)-(3), (5), we define the following spaces.
Denote by B5

2,T [20], [21] the set of all functions ω(x, t) of the form

ω(x, t) =

∞∑
k=1

ωk(t) sinλkx
(
λk =

π

2
(2k − 1)

)
,

defined on ΠT , where each of the functions ωk(t) ∈ C[0, T ] (k = 1, 2, . . .) and

JT (ω) ≡

( ∞∑
k=1

(λ5k ∥ωk(t)∥C[0,T ])
2

) 1
2

< +∞.

The norm in this space is defined as

∥ω(x, t)∥B5
2,T

= J(ω).

By E5
T we denote the space of the vector functions {ω(x, t), φ(t), ψ(t)} such that

ω(x, t) ∈ B5
2,T , φ(t), ψ(t) ∈ C[0, T ], and equip this space by the norm

∥η∥E5
T
= ∥ω(x, t)∥B5

2,T
+ ∥φ(t)∥C[0,T ] + ∥ψ(t)∥C[0,T ] .

Clearly, B5
2,T and E5

T are Banach spaces.

Now we consider in E5
T the operator

H(ω, φ, ψ) = {H1(ω, φ, ψ), H2(ω, φ, ψ), H3(ω, φ, ψ)},

where H1(ω, φ, ψ) = ω̃(x, t) ≡
∞∑
k=1

ω̃k(t) sinλkx, H2(ω, φ, ψ) = φ̃(t),

H3(ω, φ, ψ) = ψ̃(t),ω̃k(t ( k = 1, 2, ...), φ̃(t) and ψ̃(t) are the right hand sides of (9) and
(13), (14) correspondingly.

Now, let the given problems satisfy the following conditions:
1.ε(x) ∈ C4[0, 1], ε(5)(x) ∈ L2(0, 1),

ε(0) = ε′(1) = ε′′(0) = ε′′′(1) = ε(4)(0) = 0;
2.v(x) ∈ C2[0, 1], v(4)(x) ∈ L2(0, 1), v(0) = v′(1) = v′′′(0) = v′′′(1) = 0;
3.s(x, t), sx(x, t) ∈ C(DT ), sxx(x, t) ∈ L2(DT ), s(0, t) = sx(1, t) = 0 (0 ≤ t ≤ T );
4.r(x, t), rx(x, t) ∈ C(DT ), rxx(x, t) ∈ L2(DT ), r(0, t) = rx(1, t) = 0 (0 ≤ t ≤ T );
5.σi(t) ∈ C[0, T ], ni(t) ∈ C2[0, T ] (i = 1, 2),

n(t) ≡ n1(t)r(x2, t)− n2(t)r(x1, t) ̸= 0 (0 ≤ t ≤ T ).
Now, from (15)-(17) we find:

∥ω̃(x, t)∥B5
2,T

≤ A1(T ) +B1(T ) ∥φ(t)∥C[0,T ] ∥ω(x, t)∥B5
2,T

+



Y. T. Mehraliyev, A. A. Mammadov / Eur. J. Pure Appl. Math, 18 (2) (2025), 6119 6 of 9

+C1(T ) ∥ω(x, t)∥B5
2,T

+D1(T ) ∥ψ(t)∥C[0,T ] , (13)

∥φ̃(t)∥C[0,T ] ≤ A2(T ) +B2(T ) ∥φ(t)∥C[0,T ] ∥ω(x, t)∥B5
2,T

+

+C2(T ) ∥ω(x, t)∥B5
2,T

+D2(T ) ∥ψ(t)∥C[0,T ] , (14)∥∥∥ψ̃(t)∥∥∥
C[0,T ]

≤ A3(T ) +B3(T ) ∥φ(t)∥C[0,T ] ∥ω(x, t)∥B5
2,T

+

+C3(T ) ∥ω(x, t)∥B5
2,T

+D3(T ) ∥ψ(t)∥C[0,T ] , (15)

where
A1(T ) =

√
7
∥∥∥ε(5)(x)∥∥∥

L2(0,1)
+
√
14
∥∥∥v(4)(x)∥∥∥

L2(0,1)
+

+
√
10T ∥sxx(x, t)∥L2(DT ) , B1(T ) =

√
14T,

C1(T ) =
√
14T

(
∥σ1(t)∥C[0,T ]+ ∥σ2(t)∥C[0,T ]

)
, D1(T ) =

√
10T ∥rxx(x, t)∥L2(DT ) ,

A2(T ) =
∥∥∥[n(t)]−1

∥∥∥
C[0,T ]

{
∥∥(n′′1(t)− s(x1, t))r(x2, t)− (n′′2(t)− s(x2, t))r(x1, t) )

∥∥
C[0,T ]

+

+2 ∥|r(x2, t)|+ |r(x1, t)|∥C[0,T ]

( ∞∑
k=1

λ−2
k

) 1
2 [∥∥∥ε(5)(x)∥∥∥

L2(0,1)
+

+
√
2
∥∥∥v(4)(x)∥∥∥

L2(0,1)
+
√
2T ∥sxx(x, t)∥L2(DT ) +

∥∥∥∥sxx(x, t)∥C[0,T ]

∥∥∥
L2(0.1)

]}
,

B2(T ) = 2
∥∥∥[n(t)]−1

∥∥∥
C[0,T ]

∥|r(x2, t)|+ |r(x1, t)|∥C[0,T ]

( ∞∑
k=1

λ−2
k

) 1
2

(T + 1),

C2(T ) = 2
∥∥∥[n(t)]−1

∥∥∥
C[0,T ]

∥|r(x2, t)|+ |r(x1, t)|∥C[0,T ]

( ∞∑
k=1

λ−2
k

) 1
2

×T
(
∥σ1(t)∥C[0,T ]+ ∥σ2(t)∥C[0,T ]

)
,

D2(T ) = 2
∥∥∥[n(t)]−1

∥∥∥
C[0,T ]

∥|r(x2, t)|+ |r(x1, t)|∥C[0,T ]

( ∞∑
k=1

λ−2
k

) 1
2

×
(√

2T ∥rxx(x, t)∥L2(DT ) +
∥∥∥∥rxx(x, t)∥C[0,T ]

∥∥∥
L2(0.1)

)
,

A3(T ) =
∥∥∥[n(t)]−1

∥∥∥
C[0,T ]

{
∥∥(n′′2(t)− s(x2, t))n1(t)− (n′′1(t)− s(x1, t))n2(t)

∥∥
C[0,T ]

+

+2 ∥|n2(t)|+ |n1(t)|∥C[0,T ]

( ∞∑
k=1

λ−2
k

) 1
2 [∥∥∥ε(5)(x)∥∥∥

L2(0,1)
+
√
2
∥∥∥v(4)(x)∥∥∥

L2(0,1)
+



Y. T. Mehraliyev, A. A. Mammadov / Eur. J. Pure Appl. Math, 18 (2) (2025), 6119 7 of 9

+
√
3T ∥sxx(x, t)∥L2(DT ) +

∥∥∥∥sxx(x, t)∥C[0,T ]

∥∥∥
L2(0.1)

]}
,

B3(T ) = 2
∥∥∥[n(t)]−1

∥∥∥
C[0,T ]

∥|n2(t)|+ |n1(t)|∥C[0,T ]

( ∞∑
k=1

λ−2
k

) 1
2

(T + 1),

C2(T ) = 2
∥∥∥[n(t)]−1

∥∥∥
C[0,T ]

∥|n2(t)|+ |n1(t)|∥C[0,T ]

( ∞∑
k=1

λ−2
k

) 1
2

×

×T
(
∥σ1(t)∥C[0,T ]+ ∥σ2(t)∥C[0,T ]

)
,

D3(T ) = 2
∥∥∥[n(t)]−1

∥∥∥
C[0,T ]

∥|n2(t)|+ |n1(t)|∥C[0,T ]

( ∞∑
k=1

λ−2
k

) 1
2

×

×
(√

2T ∥rxx(x, t)∥L2(DT ) +
∥∥∥∥rxx(x, t)∥C[0,T ]

∥∥∥
L2(0.1)

)
.

From inequalities (18)-(20) we conclude

∥ω̃(x, t)∥
B5,3

2,T
+ ∥φ̃(t)∥C[0,T ] +

∥∥∥ψ̃(t)∥∥∥
C[0,T ]

+

+B(T ) ∥φ(t)∥C[0,T ] ∥ω(x, t)∥B5
2,T

+ C(T ) ∥ω(x, t)∥B5
2,T

+D(T ) ∥ψ(t)∥C[0,T ] , (16)

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

C(T ) = C1(T ) + C2(T ) + C3(T ), D(T ) = D1(T ) +D2(T ) +D3(T ).

So, we can prove the following theorem.

Theorem 2. Let conditions 1-5 be satisfied and

(A(T ) + 2)(B(T )(A(T ) + 2) + C(T ) +D(T )) < 1. (17)

The problem (1)-(3), (5) has a unique solution in the ball K = KR(||η ||E5
T

≤ R =

A(T ) + 2) of the space E5
T .

Proof. In the space E5
T consider the equation

η = Hη, (18)

where η = {ω, φ, ψ}, the components Hi(ω, φ, ψ) (i = 1, 2, 3) of the operator H(ω, φ, ψ)
are defined by the right hand sides of equations (10), (11) and (12).

Consider the operator H(ω, φ, ψ) in the ball K = KR from E5
T . Similarly to (18) we

obtain that the estimations

∥Hη∥E5
T
≤ A(T ) +B(T ) ∥φ(t)∥C[0,T ] ∥ω(x, t)∥B5

2,T
+ C(T ) ∥ω(x, t)∥B5

2,T
+



Y. T. Mehraliyev, A. A. Mammadov / Eur. J. Pure Appl. Math, 18 (2) (2025), 6119 8 of 9

+D(T ) ∥ψ(t)∥C[0,T ] ≤ A(T ) + (A(T ) + 2)(B(T )(A(T ) + 2) + C(T ) +D(T )), (19)

∥Hη1 −Hη2∥E5
T
≤ B(T )(A(T ) + 2)

(
∥ω1(x, t)− ω2(x, t)∥B5

2,T
+ ∥φ1(t)− φ2(t)∥C[0,T ]

)
+C(T ) ∥ω1(x, t)− ω2(x, t)∥B5

2,T
+D(T ) ∥ψ1(t)− ψ2(t)∥C[0,T ] , (20)

for the arbitrary η, η1, η2 ∈ KR. Taking into account (17), from estimates (19), (20) it
follows that the operator H acts in the ball K = KR and is contracting. Therefore in the
ball K = KR the operator H has a single fixed point {ω, φ, ψ} which is a unique solution
to equation (18) in the ball K = KR, i.e. {ω, φ, ψ} is a unique solution to system (10),
(11) and (12) in the ball K = KR.

The function ω(x, t) as an element of the space B5
2,T has continuous derivatives ωx(x, t),

ωxx(x, t), ωxxx(x, t), ωxxxx(x, t), in ΠT .
Similarly, [17] it can be shown that ωtt(x, t), ωttx(x, t), ωttxx(x, t), are continuous in

ΠT .
It is easy to verify that equation (1) and conditions (2), (3) and (5) are satisfied in

the usual sense. Therefore, {ω(x, t), φ(t), ψ(t)} is a solution to problem (1)-(3), (5), and,
by virtue of the corollary of Lemma 1, it is unique in the ball K = KR. The theorem is
proved.

Using Theorem 1, we prove the following

Theorem 3. Let all conditions of Theorem 2 be satisfied and

ni(0) =

∫ T

0
σ1(t)ni(t)dt+ ε(xi), n′i(0) =

∫ T

0
σ2(t)ni(t)dt+ v(xi), i = 1, 2,

(
T ∥σ2(t)∥C[0,T ] + ∥σ1(t) ∥C[0,T ] +

T

2
(A(T ) + 2)

)
T < 1.

Then problem (1)-(4) has unique classical solution in the ball K = KR(||η ||E5
T
≤ R =

A(T ) + 2) from E5
T .

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