EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 2, 2023, 670-686 ISSN 1307-5543 – ejpam.com Published by New York Business Global Two-dimensional inverse boundary value problem for a third-order pseudo-hyperbolic equation with an additional integral condition Yashar T. Mehraliyev1, Sadikhzade R.Shafi1, Aysel T. Ramazanova2,∗ 1 Department of Differential and Integral Equations, Baku State University, Baku, Azerbaijan 2 Fakultät of Mathematik, Universität Duisburg-Essen, Essen, Germany Abstract. In this paper we study an inverse boundary value problem with an unknown time- dependent coefficient for a third-order pseudo-hyperbolic equation with an additional integral con- dition. The definition of the classical solution of the problem is given. The essence of the problem is that it is required together with the solution to determine the unknown coefficient. The problem is considered in a rectangular area. When solving the original inverse boundary value problem, the transition from the original inverse problem to some auxiliary inverse problem is carried out. The existence and uniqueness of a solution to an auxiliary problem are proved with the help of contracted mappings. Then the transition to the original inverse problem is again made, as a result, a conclusion is made about the solvability of the original inverse problem. 2020 Mathematics Subject Classifications: 31A25, 35L35 Key Words and Phrases: Inverse boundary value problem, third-order pseudo-hyperbolic equa- tion, Fourier method, classical solution 1. Introduction and Problem Statement It is known that the practical requirements often lead to the problem of determining the coefficients or the right hand side of the differential equations for some known data about their solutions. Such problems are called inverse problems in mathematical physics. Inverse problems arise in various fields of human activity, such as seismology, mineral exploration, biology, medical visualization, computed tomography, Earth remote sensing, spectral analysis, nondestructive control, etc. Fundamentals of the theory and practice of research of inverse problems were estab- lished and developed in the works published by Tikhonov [22], Lavrent’ev [16], Ivanov [10], Romanov [21], Isakov [6], and so on. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4743 Email addresses: aysel.ramazanova@uni-due.de (A.Ramazanova), yasharaze@mail.ru (Y. T. Mehraliyev) https://www.ejpam.com 670 © 2023 EJPAM All rights reserved. Y. T. Mehraliyev, S. R.Shafi, A. T. Ramazanova / Eur. J. Pure Appl. Math, 16 (2) (2023), 670-686 671 A more detailed bibliography and a classification of recent works connected with the investigation of inverse problems for partial differential equations can be found in mono- graphs and in articles [2, 3, 5, 7–9, 11, 13, 20] and references therein. It should be noted that pseudo-hyperbolic equations arise in the theory of unsteady flow of a viscous gas during the propagation of initial densifications in a viscous gas [23], in the theory of solutions [17] when describing the process of electron motion in the system “superconductor – dielectric with tunneling conductivity – superconductor”. The solvability of inverse problems in certain formulations, with certain overdetermination conditions for pseudohyperbolic equations, was the subject of study in [1, 4, 14, 15, 18, 19] and references therein. In this work we study a two-dimensional inverse boundary value problem for a third- order pseudo-hyperbolic equation with an additional integral condition. In the paper using the Fourier method and the contraction mappings principle, the existence and uniqueness of a classical solution to the considered nonlinear inverse boundary value problem is proved. Consider for the equation utt(x, y, t)− α∆ut(x, y, t)− β∆u(x, y, t) = a(t)u(x, y, t) + f(x, y, t) (x, y, t) ∈ DT , (1) in the domain DT = Qxy × {0 < t ≤ T}, where Qxy = {(x, y) : 0 < x < 1, 0 < y < 1} an inverse boundary problem with initial conditions u(x, y, 0) = ϕ(x, y), ut(x, y, 0) = ψ(x, y), 0 ≤ x, y ≤ 1, (2) with boundary conditions ux(0, y, t) = u(1, y, t) = 0, 0 ≤ y ≤ 1, 0 ≤ t ≤ T, (3) u(x, 0, t) = uy(x, 1, t) = 0, 0 ≤ y ≤ 1, 0 ≤ t ≤ T, (4) and with additional condition 1∫ 0 1∫ 0 ω(x, y)u(x, y, t)dxdy = h(t), 0 ≤ t ≤ T, (5) where α, β > 0 are given numbers, f(x, y, t), ϕ(x, y), ψ(x, y), ω(x, y), and h(t) (i = 1, 2) are given functions, u(x, y, t), a(t) are desired functions, and ∆ = ∂2 ∂x2 + ∂2 ∂y2 . Definition 1. The pair {u(x, y, t), a(t)} is said to be a classical solution of the inverse boundary value problem (1)-(5), if the following conditions are satisfied: the function u(x, y, t) ∈ C̃2,2,2(DT )∩C1,1,1(DT ), a(t) ∈ C[0, T ], satisfying equation (1) in DT , condition (2) in Q̄xy, condition (3) in [0, 1] × [0, T ] , condition (4) in [0, 1] × [0, T ] and condition (5) in [0, T ] . The following theorem holds: Y. T. Mehraliyev, S. R.Shafi, A. T. Ramazanova / Eur. J. Pure Appl. Math, 16 (2) (2023), 670-686 672 Theorem 1. Let ϕ(x, y) ∈ C(Q̄xy), ψ(x, y) ∈ C(Q̄xy), f(x, y, t) ∈ C(DT ), h(t) ∈ C2[0, T ], h(t) ̸= 0, 0 ≤ t ≤ T and the consistency condition 1∫ 0 1∫ 0 ω(x, y)ϕ(x, y)dxdy = h(0), 1∫ 0 1∫ 0 ω(x, y)ψ(x, y)dxdy = h′(0) (6) be satisfied. Then the problem of finding a classical solution to problem (1)-(5) is equivalent to the problem of determining the functions u(x, y, t) ∈ C̃2,2,2(D̄T ), a(t) ∈ C[0, T ] from (1)-(4) and h′′(t)− α 1∫ 0 1∫ 0 ω(x, y)∆ut(x, y, t)dxdt− β 1∫ 0 1∫ 0 ω(x, y)∆u(x, y, t)dxdt = = a(t)h(t) + 1∫ 0 1∫ 0 ω(x, t)f(x, y, t)dxdy (0 ≤ t ≤ T ). (7) Proof. Let {u(x, y, t), a(t)} be a classical solution to problem (1)-(5), and u(x, y, t) ∈ C̃2,2,2(D̄T ). Assuming h(t) ∈ C2[0, T ] and differentiating two times (5), we get: ut(0, 1, t) = h′(t) , utt(0, 1, t) = h′′(t) (0 ≤ t ≤ T ). (8) Further, multiplying Eq. (1) by the function ω(x, y), integrating the equation over x from 0 to 1, we have: d2 dt2 1∫ 0 1∫ 0 ω(x, y)u(x, y, t)dxdy − α 1∫ 0 1∫ 0 ω(x, y)∆ut(x, y, t)dxdy− −β 1∫ 0 1∫ 0 ω(x, y)∆u(x, y, t)dxdy = = a(t) 1∫ 0 1∫ 0 ω(x, y)u(x, y, t)dxdy + 1∫ 0 1∫ 0 ω(x, y)f(x, y, t)dxdy (0 ≤ t ≤ T ). (9) From (9), taking into account (5) and (8), the fulfillment of (7) follows. Now, suppose that {u(x, y, t), a(t)} is a solution to the problem (1)-(4), (7). Then from (7) and (9) we find: d2 dt2  1∫ 0 1∫ 0 ω(x, y)u(x, y, t)dxdy − h(t)  = Y. T. Mehraliyev, S. R.Shafi, A. T. Ramazanova / Eur. J. Pure Appl. Math, 16 (2) (2023), 670-686 673 = a(t)  1∫ 0 1∫ 0 ω(x, y)u(x, y, t)dxdy − h(t)  (0 ≤ t ≤ T ). (10) Due to (2) and (6), we have : 1∫ 0 1∫ 0 ω(x, y)u(x, y, 0)dxdy − h(0) = 1∫ 0 1∫ 0 ω(x, y)ϕ(x, y)dxdy − h(0) = 0, 1∫ 0 1∫ 0 ω(x, y)ut(x, y, 0)dxdy − h′(0) = 1∫ 0 1∫ 0 ω(x, y)ψ(x, y)dxdy − h′(0) = 0. (11) From (10), (11) we conclude that 1∫ 0 1∫ 0 ω(x, y)u(x, y, t)dxdy − h(t) = 0 (0 ≤ t ≤ T ), i.e. condition (5) is satisfied. 2. Solvability of the existence and uniqueness of the classical solution of the inverse boundary value problem The first component of the solution {u(x, y, t), a(t) } of the problem (1)-( 4), (7) will be sought in the form: u(x, y, t) = ∞∑ n=1 ∞∑ k=1 uk,n(t) cosλkx sin γny, (12) where λk = π 2 (2k − 1), k = 1, 2, ..., γn = π 2 (2n− 1), n = 1, 2, ... , uk,n(t) = 4 1∫ 0 1∫ 0 u(x, y, t) cosλkx sin γnydxdy, k, n = 1, 2, .... Applying the method of separation of variables to determine the desired coefficients uk,n(t) (k = 1, 2, ...;n = 1, 2, ...), of the function u(x, y, t) from (1), (2), we get: u′′k,n(t) + αµ2k,nu ′ k.n(t) + βµ2k,nuk,n(t) = Fk,n(t;u, a), k, n = 1, 2, ..., 0 ≤ t ≤ T, (13) uk,n(0) = ϕk,n, u ′ k,n(0) = ψk,n, k, n = 1, 2, ..., (14) Y. T. Mehraliyev, S. R.Shafi, A. T. Ramazanova / Eur. J. Pure Appl. Math, 16 (2) (2023), 670-686 674 where µ2k,n = λ2k + γ2n, k, n = 1, 2, ..., Fk,n(t;u, a) = fk,n(t) + a(t)uk,n(t), k, n = 1, 2, ..., fk,n(t) = 4 1∫ 0 1∫ 0 f(x, y, t) cosλkx sin γnydxdy, k, n = 1, 2, ..., ϕk,n = 4 1∫ 0 1∫ 0 ϕ(x, y) cosλkx sin γnydxdy, k, n = 1, 2, ..., ψk,n = 4 1∫ 0 1∫ 0 ψ(x, y) cosλkx sin γnydxdy, k, n = 1, 2, .... Let’s assume that α2π2 4 − β > 0 . Then solving problem (13), (14), we find: uk,n(t) = 1 γk,n [( µ2,k,ne µ1,k,nt − µ1,k,ne µ2,k,nt ) ϕk,n+ ( eµ2,k,nt − eµ1,k,n t ) ψk,n+ + t∫ 0 Fk,n(τ ;u, a) ( eµ2,k,n(t−τ) − eµ1,k,n(t−τ) ) dτ  , (15) where µi,k,n = − αµ2k,n 2 + (−1)iµk,n √ α2µ2k,n 4 − β (i = 1, 2) , γk,n = µ2,k,n − µ1,k,n = 2µk,n √ α2µ2k,n 4 − β . After substituting the expression from (15) into (12), to determine the component of the solution to problem (1)-(3), (7), we obtain: u(x, y, t) = ∞∑ k=1 ∞∑ n=1 { 1 γk,n [( µ2,k,ne µ 1,k,n t − µ1,k,ne µ 2,k,n t ) ϕk,n+ ( eµ2,k,n t − eµ1,k,n t ) ψk,n+ + t∫ 0 Fk,n(τ ;u, a) ( eµ2,k,n (t−τ) − eµ2,k,n (t−τ) ) dτ  cosλkx sin γny. (16) Y. T. Mehraliyev, S. R.Shafi, A. T. Ramazanova / Eur. J. Pure Appl. Math, 16 (2) (2023), 670-686 675 From (6),(7), we have: a(t)h(t) = h′′(t)− 1∫ 0 1∫ 0 ω(x, y)f(x, y, t)dxdy+ + ∞∑ k=1 ∞∑ n=1 pk,n ( αu′k,n (t) + βuk,n (t) ) (0 ≤ t ≤ T ), (17) where pk,n = 1∫ 0 1∫ 0 ω(x, y) cosλkx sin γnydxdy. (18) Differentiating (15) two times, we get: u′k,n(t) = 1 γk,n [ µ1,k,nµ2,k,n ( eµ1,k,nt − eµ2k,nt ) ϕk,n+ + ( µ2,k,ne µ2,k,nt − µ1,k,ne µ1,k,nt ) ψk,n+ + t∫ 0 Fk,n(τ ;u, a) ( µ 2,k,n eµ2,k,n(t−τ) − µ1,k,ne µ1,k,n(t−τ) ) dτ  , (19) u′′k,n(t) = 1 γ k,n [ µ1,k,nµ2,k,n ( µ1,k,ne µ1,k,nt − µ2,k,ne µ2,k,nt ) ϕk,n+ , + ( µ22,k,ne µ2,k,nt − µ21,k,ne µ1,k,nt ) ψk,n+ + t∫ 0 Fk,n(τ ;u, a) ( µ22,k,ne µ2,k,n(t−τ) − µ21,k,ne µ1,k,n(t−τ) ) dτ + Fk,n(t;u, a). (20) Due to (13) and (20) we have : αµ2k,nu ′ k,n(t) + βµ2k,nuk,n(t) = Fk,n(t;u, a) − u′′k,n(t) = = − 1 γ k,n [ µ1,k,nµ2,k,n ( µ1,k,ne µ1,k,nt − µ2,k,ne µ2,k,nt ) ϕk,n+ + ( µ22,k,ne µ2,k,nt − µ21,k,ne µ1,k,nt ) ψk,n+ + t∫ 0 Fk,n(τ ;u, a) ( µ22,k,ne µ2,k,n(t−τ) − µ21,k,ne µ1,k,n(t−τ) ) dτ  . (21) Y. T. Mehraliyev, S. R.Shafi, A. T. Ramazanova / Eur. J. Pure Appl. Math, 16 (2) (2023), 670-686 676 In order to obtain an equation for the second component a(t) of the solution {u(x, y, t), a(t) } of problem (1)-(4), (7), we substitute expression (21) into (17): a(t) = [h (t)]−1 h′′(t)− 1∫ 0 1∫ 0 ω(x, y)f(x, y, t)dxdy + + ∞∑ k=1 ∞∑ n=1 pk,n µ2k,nγk,n [ µ1,k,nµ2,k,n ( µ1,k,ne µ1,k,nt − µ2,k,ne µ2,k,nt ) ϕk,n+ + ( µ22,k,ne µ2,k,nt − µ21,k,ne µ1,k,nt ) ψk,n+ + t∫ 0 Fk,n(τ ;u, a) ( µ22,k,ne µ2,k,n(t−τ) − µ21,k,ne µ1,k,n(t−τ) ) dτ  . (22) Thus, the solution of problem (1)-(4),(7) is reduced to the solution of system (15), (22) with respect to unknown functions u(x, y, t) and a(t). To study the question of the uniqueness of the solution of problem (1)-(4), (7), the following lemma plays an important role. Lemma 1. If {u(x, y, t), a(t)} is any classical solution to the problem (1)-(4),(7), then the functions uk,n(t) = 4 1∫ 0 1∫ 0 u(x, y, t) cosλkx sin γnydxdy, k, n = 1, 2, .... satisfy of the system (15). Proof. Let {u(x, t), a(t)} be any solution to the problem (1)-(4), (7). Then multiplying both sides of equation (1), by the function 4 cosλkx sin γny , (k = 1, 2, ...; n = 1, 2, ...), integrating the resulting equality over x and y from 0 to 1, and using the relations 4 1∫ 0 1∫ 0 utt(x, y, t) cosλkx sin γnydxdy = = d2 dt2 4 1∫ 0 1∫ 0 u(x, y, t) cosλkx sin γnydxdy  = u′′k,n(t) (k = 1, 2, ...; n = 1, 2, ...), Y. T. Mehraliyev, S. R.Shafi, A. T. Ramazanova / Eur. J. Pure Appl. Math, 16 (2) (2023), 670-686 677 4 1∫ 0 1∫ 0 uxx(x, y, t) cosλkx sin γnydxdy = = −λ2k 4 1∫ 0 1∫ 0 u(x, y, t) cosλkx sin γnydxdy  = −λ2kuk,n(t) (k = 1, 2, ...; n = 1, 2, ...), 4 1∫ 0 1∫ 0 uyy(x, y, t) cosλkx sin γnydxdy = −γ2n 4 1∫ 0 1∫ 0 u(x, y, t) cosλkx sin γnydxdy  = −γ2nuk,n(t) (k = 1, 2, ...; n = 1, 2, ...), 4 1∫ 0 1∫ 0 utxx(x, y, t) cosλkx sin γnydxdy = = −λ2k 4 1∫ 0 1∫ 0 ut(x, y, t) cosλkx sin γnydxdy  = −λ2ku′k,n(t) (k = 1, 2, ...; n = 1, 2, ...), 1∫ 0 1∫ 0 utyy(x, y, t) cosλkx sin γnydxdy = = −γ2n 4 1∫ 0 1∫ 0 ut(x, y, t) cosλkx sin γnydxdy  = −γ2nu′′k,n(t) (k = 1, 2, ...; n = 1, 2, ...), we obtain that equation (13) is satisfied. Similarly, from (2) we conclude at condition (14). Thus, uk,n(t) (k = 1, 2, ...; n = 1, 2, ...) are a solution to problem (13), (14). Then from this, it directly follows that the functions uk,n(t) (k = 1, 2, ...; n = 1, 2, ...) satisfy on [0, T ] of the system (15). Y. T. Mehraliyev, S. R.Shafi, A. T. Ramazanova / Eur. J. Pure Appl. Math, 16 (2) (2023), 670-686 678 It is obvious that if uk,n(t) = 4 1∫ 0 1∫ 0 u(x, y, t) cosλkx sin γnydxdy, (k = 1, 2, ...; n = 1, 2, ...) are a solution of system (15), then the pair {u(x, t), a(t)} of the functions u(x, y, t) = ∞∑ n=1 ∞∑ k=1 uk,n(t) cosλkx sin γny and a(t) is a solution of system (15), (21). From Lemma 1 it follows that: Remark 1. Let system (15), (22) have a unique solution. Then problem (1)-(4), (7) cannot have more than one solution, i.e. if problem (1)-(4), (7) has a solution, then it is unique. 1. We denote by B3 2,T [12], a consisting of all functions u(x, y, t) of the form u(x, y, t) = ∞∑ n=1 ∞∑ k=1 uk,n(t) cosλkx sin γny, considered in DT , where uk,n(t) (k = 1, 2, ...; n = 1, .2, ..) is continuous on [0, T ] and{ ∞∑ n=1 ∞∑ k=1 ( µ3k,n∥uk,n(t)∥C[0,T ] )2 } 1 2 < +∞, where µk,n = √ λ2k + γ2n (k = 1, 2, ...;n = 1, 2, ...). The norm in this set is defined as follows: ∥u(x, y, t)∥B3 2,T = { ∞∑ n=1 ∞∑ k=1 ( µ3k,n∥uk,n(t)∥C[0,T ] )2 } 1 2 . 2. The spaces E3 T denote the space consisting of the topological product B3 2,T×C[0, T ] . The norm of element z = {u, a} is determined by the formula ∥z∥E3 T = ∥u(x, y, t)∥B3 2,T + ∥a(t)∥C[0,T ]. It is obvious that B3 2,T and E3 T are Banach spaces. Y. T. Mehraliyev, S. R.Shafi, A. T. Ramazanova / Eur. J. Pure Appl. Math, 16 (2) (2023), 670-686 679 Now consider in the space E3 T the operator Φ(u, a) = {Φ1(u, a),Φ2(u, a)}, where Φ1(u, a) = ũ(x, y, t) ≡ ∞∑ n=1 ∞∑ k=1 ũk,n(t) cosλkx sin γny, Φ2(u, a) = ã(t), and ũk,n(t) (k = 1, 2, ...; n = 1, 2, ...) and ã(t) are equal to, respectively, the right sides of (15), and (21). It is easy to get that µ3k,n ≤ (λ2k + γ2k)(λk + γn) = λ3k + λ2k γn + γ2nλk + γ3n, |γk,n| > α√ 2 µ2k,n, |µi,k,n| ≤ αµ2k,n (i = 1, 2), |µ1,k,nµ2,k,n| = βµ2k,n (i = 1, 2), |pk,n| ≤ ∥ω(x, y)∥C(Q̄xy) . Taking into account this ratio, we have:{ ∞∑ n=1 ∞∑ k=1 ( µ3k,n∥ũk,n(t)∥C[0,T ] )2 } 1 2 ≤ 4 ( ∞∑ n=1 ∞∑ k=1 ( λ3k |ϕk,n| )2) 1 2 + +4 ( ∞∑ n=1 ∞∑ k=1 ( λ2k γn |ϕk,n| )2) 1 2 + 4 ( ∞∑ n=1 ∞∑ k=1 ( λkγ 2 n |ϕk,n| )2) 1 2 + +4 ( ∞∑ n=1 ∞∑ k=1 ( γ3n |ϕk,n| )2) 1 2 + + 4 α ( ∞∑ n=1 ∞∑ k=1 (λk |ψk,n|)2 ) 1 2 + 4 α ( ∞∑ n=1 ∞∑ k=1 ( γn |ψk,n|)2 ) 1 2 + + 4 √ T α   T∫ 0 ∞∑ n=1 ∞∑ k=1 (λk |fk,n(τ)|)2 dτ  1 2 +  T∫ 0 ∞∑ n=1 ∞∑ k=1 (γn |fk,n(τ)|)2 dτ  1 2 + + 4T α ∥a(t)∥C[0,T ] ( ∞∑ n=1 ∞∑ k=1 ( µ3k,n ∥uk,n(t)∥C[0,T ] )2) 1 2 , (23) Y. T. Mehraliyev, S. R.Shafi, A. T. Ramazanova / Eur. J. Pure Appl. Math, 16 (2) (2023), 670-686 680 ∥ã(t)∥C[0,T ] ≤ ∥∥∥[h(t)]−1 ∥∥∥ C[0,T ]  ∥∥∥∥∥∥h′′(t)− 1∫ 0 1∫ 0 ω(x, y)f(x, y, t)dxdy ∥∥∥∥∥∥ C[0,T ] + +2 √ 2 ( ∞∑ n=1 ∞∑ k=1 µ−2 k,n ) 1 2 ∥ω(x, y)∥C[Q̄xy ] β( ∞∑ n=1 ∞∑ k=1 ( λ3k |ϕk,n| )2) 1 2 +β ( ∞∑ n=1 ∞∑ k=1 ( λ2k γn |ϕk,n| )2) 1 2 + β ( ∞∑ n=1 ∞∑ k=1 ( λkγ 2 n |ϕk,n| )2) 1 2 + +β ( ∞∑ n=1 ∞∑ k=1 ( γ3n |ϕk,n| )2) 1 2 + +α ( ∞∑ n=1 ∞∑ k=1 (λk |ψk,n|)2 ) 1 2 + α ( ∞∑ n=1 ∞∑ k=1 ( γn |ψk,n|)2 ) 1 2 + +α √ T   T∫ 0 ∞∑ n=1 ∞∑ k=1 (λk |fk,n(τ)|)2 dτ  1 2 +  T∫ 0 ∞∑ n=1 ∞∑ k=1 (γn |fk,n(τ)|)2 dτ  1 2 + +αT∥a(t)∥C[0,T ] ( ∞∑ n=1 ∞∑ k=1 ( µ3k,n ∥uk,n(t)∥C[0,T ] )2) 1 2  . (24) Let us assume that the data of problem (1)-(4), (7) satisfy the following conditions: 1. α > 0, β > 0, α2 8 − β > 0; 2. ϕ(x, y), ϕx(x, y), ϕxx(x, y), ϕy(x, y), ϕxy(x, y), ϕyy(x, y) ∈ C(Q̄xy), ϕxxy(x, y), ϕxyy(x, y), ϕxxx(x, y), ϕyyy(x, y) ∈ L2(Qxy), ϕx(0, y) = ϕ(1, y) = ϕxx(1, y) = 0, 0 ≤ y ≤ 1, ϕ(x, 0) = ϕy(x, 1) = ϕyy(x, 0) = 0, 0 ≤ x ≤ 1; 3. ψ(x, y), ψx(x, y), ψy(x, y), ψxx(x, y), ψxy(x, y), ψyy(x, y) ∈ C(Q̄xy), ψxxy(x, y), ψxyy(x, y), ψxxx(x, y), ψyyy(x, y) ∈ L2(Qxy), ψx(0, y) = ψ(1, y) = ψxx(1, y) = 0, 0 ≤ y ≤ 1, ψ(x, 0) = ψy(x, 1) = ψyy(x, 1) = 0, 0 ≤ x ≤ 1; 4. f(x, y, t), fx(x, y, t), fy(x, y, t), fxx(x, y, t), fxy(x, y, t), fyy(x, y, t) ∈ C(DT ), fxxx(x, y, t), fxxy(x, y, t), fxyy(x, y, t), fyyy(x, y, t) ∈ L2(DT ), fx (0, y, t) = f (1, y, t) = fxx (0, y, t) = 0, 0 ≤ y ≤ 1, 0 ≤ t ≤ T, f (x, 0, t) = fy (x, 1, t) = fyy (x, 1, t) = 0, 0 ≤ x ≤ 1, 0 ≤ t ≤ T ; Y. T. Mehraliyev, S. R.Shafi, A. T. Ramazanova / Eur. J. Pure Appl. Math, 16 (2) (2023), 670-686 681 5. h(t) ∈ C2[0, T ], h(t) ̸= 0 (0 ≤ t ≤ T ). Then from (26) - (28), respectively, we obtain: ∥u(x, y, t)∥B3 2,T ≤ A1(T ) +B1(T )∥a(t)∥C[0,T ]∥u(x, y, t)∥B3 2,T + C1(T )∥b(t)∥C[0,T ], (25) ∥ã (t)∥C[0,T ] ≤ A2 (T ) +B2 (T ) ∥a (t)∥C[0,T ]∥u (x, y, t)∥B3 2,T + C2 (T ) ∥b (t)∥C[0,T ], (26) where A1(T ) = 4∥ϕxxx(x, y)∥L2(Qxy) + 4∥ϕxxy(x, y)∥L2(Qxy) + 4 ∥ϕxyy(x, y)∥ L2(Qxy)+ +4∥ϕyyy(x, y)∥L2(Qxy) + 4 α ∥ψx(x, y)∥L2(Qxy) + 4 α ∥ψy(x, y)∥L2(Qxy) + + 4 √ T α ( ∥fx(x, y, t)∥L2(DT ) + ∥fy(x, y, t)∥L2(DT ) ) , A2(T ) = ∥∥∥[h(t)]−1 ∥∥∥ C[0,T ]  ∥∥∥∥∥∥h′′(t)− 1∫ 0 1∫ 0 ω(x, y)f(x, y, t)dxdy ∥∥∥∥∥∥ C[0,T ] + +2 √ 2 ( ∞∑ n=1 ∞∑ k=1 µ−2 k,n ) 1 2 ∥ω(x, y)∥C[Q̄xy ]β∥ϕxxx(x, y)∥L2(Qxy)+ +β∥ϕxxy(x, y)∥L2(Qxy) + β ∥ϕxyy(x, y)∥ L2(Qxy) + β∥ϕyyy(x, y)∥L2(Qxy) + +α∥ψx(x, y)∥L2(Qxy) + α∥ϕy(x, y)∥L2(Qxy) + +α √ T ( ∥fx(x, y, t)∥L2(DT ) + ∥fy(x, y, t)∥L2(DT ) )]} , B2 (T ) = 2 √ 2α ∥∥∥[h(t)]−1 ∥∥∥ C[0,T ] ( ∞∑ n=1 ∞∑ k=1 µ−2 k,n ) 1 2 ∥ω(x, y)∥C[Q̄xy ] T. From inequalities (25)-(26), we conclude: ∥u(x, y, t)∥B3 2,T + ∥ã(t)∥C[0,T ] ≤ A(T ) +B(T )∥a(t)∥C[0,T ]∥u(x, y, t)∥B3 2,T , (27) where A(T ) = A1(T ) +A2(T ), B(T ) = B1(T ) +B2(T ). So, we can prove the following theorem: Y. T. Mehraliyev, S. R.Shafi, A. T. Ramazanova / Eur. J. Pure Appl. Math, 16 (2) (2023), 670-686 682 Theorem 2. Let conditions 1-5 be satisfied and B(T )(A(T ) + 2)2 < 1. Then problem (1)-(4), (7) has unique solution in the ball K = KR(∥z∥E3 T ≤ R = A(T )+2) of the spaces E3 T . Proof. In the space E3 T consider the equation z = Φz, where z = {u, a}, the components Φi(u, a)(i = 1, 2) of the operator (u, a) are defined by the right-hand sides of equations (15), (22). Consider the operator (u, a) in the sphere K = KR(∥z∥E3 T ≤ R = A(T ) + 2) from E3 T . Similarly to (27), we obtain that for any z, z1 ∈ KR fair estimates: ∥Φz∥E3 T ≤ A(T ) +B(T )∥a(t)∥C[0,T ] ∥u(x, y, t)∥B3 2,T ≤ A(T ) +B(T ) (A(T ) + 2)2, (28) ∥Φz1 − Φz2∥E3 T ≤ B(T )R ( ∥a1(t)− a2(t)∥C[0,T ] + ∥u1(x, y, t)− u2(x, y, t)∥B3 2,T ) . (29) Then estimates (30) and (31), taking into account (28), it follows that the operator Φ acts in the sphere K = KR and is contractive. Therefore, in the sphere K = KR the operator Φ has a unique fixed point {u, a} , that is a solution of equation (15),(22). The function u(x, y, t), as an element of space B3 2,T , is continuous and has contin- uous derivatives ux(x, y, t), uxx(x, y, t), uy(x, y, t), uxy(x, y, t), uyy(x, y, t), uxxx(x, y, t), uyyy(x, y, t) in DT . Further, from (19), we find:{ ∞∑ n=1 ∞∑ k=1 ( µ3k,n ∥∥u′k,n(t)∥∥C[0,T ] )2 } 1 2 ≤ 4 √ 2β α ( ∞∑ n=1 ∞∑ k=1 ( λ3k |ϕk,n| )2) 1 2 + + 4 √ 2β α ( ∞∑ n=1 ∞∑ k=1 ( λ2k γn |ϕk,n| )2) 1 2 + 4 √ 2β α ( ∞∑ n=1 ∞∑ k=1 ( λkγ 2 n |ϕk,n| )2) 1 2 + + 4 √ 2β α ( ∞∑ n=1 ∞∑ k=1 ( γ3n |ϕk,n| )2) 1 2 + +4 √ 2 ( ∞∑ n=1 ∞∑ k=1 ( λ3k |ψk,n| )2) 1 2 + 4 √ 2 ( ∞∑ n=1 ∞∑ k=1 ( λ2k γn |ψk,n| )2) 1 2 + +4 √ 2 ( ∞∑ n=1 ∞∑ k=1 ( λkγ 2 n |ψk,n| )2) 1 2 + 4 √ 2 ( ∞∑ n=1 ∞∑ k=1 ( γ3n |ψk,n| )2) 1 2 + Y. T. Mehraliyev, S. R.Shafi, A. T. Ramazanova / Eur. J. Pure Appl. Math, 16 (2) (2023), 670-686 683 +4 √ 2T   T∫ 0 ∞∑ n=1 ∞∑ k=1 ( λ3k |fk,n(τ)| )2 dτ  1 2 +  T∫ 0 ∞∑ n=1 ∞∑ k=1 ( λ2kγn |fk,n(τ)| )2 dτ  1 2 + +  T∫ 0 ∞∑ n=1 ∞∑ k=1 ( λkγ 2 n |fk,n(τ)| )2 dτ  1 2 +  T∫ 0 ∞∑ n=1 ∞∑ k=1 ( γ3n |fk,n(τ)| )2 dτ  1 2 + +4 √ 2T∥a(t)∥C[0,T ] ( ∞∑ n=1 ∞∑ k=1 ( µ2k,n ∥uk,n(t)∥C[0,T ] )2) 1 2 , or { ∞∑ n=1 ∞∑ k=1 ( µ3k,n ∥∥u′k,n(t)∥∥C[0,T ] )2 } 1 2 ≤ 4 √ 2β α ∥ϕxxx(x, y)∥L2(Qxy) + 4∥ϕxxx(x, y)∥L2(Qxy) + 4∥ϕxxy(x, y)∥L2(Qxy) + 4 ∥ϕxyy(x, y)∥ L2(Qxy)+ + 4 √ 2β α ∥ϕxxy(x, y)∥L2(Qxy) + 4 √ 2β α ∥ϕxyy(x, y)∥L2(Qxy) + + 4 √ 2β α ∥ϕyyy(x, y)∥L2(Qxy) + +4 √ 2∥ψxxx(x, y)∥L2(Qxy) + 4 √ 2∥ψxxy(x, y)∥L2(Qxy) + +4 √ 2∥ψxyy(x, y)∥L2(Qxy) + 4 √ 2∥ψyyy(x, y)∥L2(Qxy) + +4 √ 2T ( ∥fxxx(x, y, t)∥L2(DT ) + ∥fxxy(x, y, t)∥L2(DT )+ +∥fxyy(x, y, t)∥L2(DT ) + ∥fyyy(x, y, t)∥L2(DT ) ) + +4 √ 2T∥a(t)∥C[0,T ] ( ∞∑ n=1 ∞∑ k=1 ( µ3k,n ∥uk,n(t)∥C[0,T ] )2) 1 2 . From the last relation, it is clear that ut(x, y, t), utx(x, y, t), uty(x, y, t), utxx(x, y, t), utyy(x, y, t) are continuous in DT . REFERENCES 684 Now, from (13) it is easy to see that{ ∞∑ n=1 ∞∑ k=1 ( µk,n ∥∥u′′k,n(t)∥∥C[0,T ] )2 } 1 2 ≤ 2 α{ ∞∑ n=1 ∞∑ k=1 ( µ3k,n ∥∥u′k,n(t)∥∥C[0,T ] )2 } 1 2 + +β { ∞∑ n=1 ∞∑ k=1 ( µ3k,n∥u k,n(t)∥C[0,T ] )2 } 1 2 + ∥∥∥∥fx(x, y, t) + fy(x, y, t)∥C[0,T ] ∥∥∥ L2(Qxy) + + ∥∥∥∥a(t) (ux(x, y, t) + uy(x, y, t))∥C[0,T ] ∥∥∥ L2(Qxy) ] . It is easy to verify that utt(x, y, t) is continuous in DT . Obvious that equation (1) and conditions (2)–(4), (7) are satisfied in the usual sense. Thus, the solution to problem (1)-(4), (7) is a triple of functions {u(x, t), a(t)} and by the corollary of Lemma 1, it is unique in the ball K = KR. Using Theorems 1 and 2, we obtain the unique solvability of problem (1)–(5). Theorem 3. Let all the conditions of Theorem 2 and the consistency condition (6) be satisfied. 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