Research article Special Issues

On one coefficient inverse boundary value problem for a linear pseudoparabolic equation of the fourth order

  • In the present work, we consider an inverse boundary value problem for a fourth order pseudo parabolic equation with periodic and integral condition. Using analytical and operator-theoretic methods, as well as the Fourier method, the existence and uniqueness of the classical solution of this problem is proved. By the contraction mapping principle is formulated as an auxiliary inverse problem which, in turn, is reduced to the operator equation in a specified Banach space using the method of spectral analysis.

    Citation: Yashar Mehraliyev, Seriye Allahverdiyeva, Aysel Ramazanova. On one coefficient inverse boundary value problem for a linear pseudoparabolic equation of the fourth order[J]. AIMS Mathematics, 2023, 8(2): 2622-2633. doi: 10.3934/math.2023136

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  • In the present work, we consider an inverse boundary value problem for a fourth order pseudo parabolic equation with periodic and integral condition. Using analytical and operator-theoretic methods, as well as the Fourier method, the existence and uniqueness of the classical solution of this problem is proved. By the contraction mapping principle is formulated as an auxiliary inverse problem which, in turn, is reduced to the operator equation in a specified Banach space using the method of spectral analysis.



    Modern problems of natural science lead to the need to generalize the classical problems of mathematical physics, as well as to the formulation of qualitatively new problems, which include non-local problems for differential equations. Among nonlocal problems, problems with integral conditions are of great interest. Integral conditions are encountered in the study of physical phenomena in the case when the boundary of the process flow region is inaccessible for direct measurements. 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. Various inverse problems for certain types of partial differential equations have been studied in many works. A more detailed bibliography and a classification of problems are found in [1,2,3,4,5]. Inverse problems for one-dimensional pseudo-parabolic equations of third-order were studied in [6]. The existence and uniqueness of the solution of the inverse problem for the third order pseudoparabolic equation with integral over-determination condition is studied in [7]. Khompysh [8] investigated the reconstruction of unknown coefficient in pseudo-parabolic inverse problem with the integral over determination condition and studied the uniqueness and existence of solution by means of method of successive approximations. Studies of wave propagation in cold plasma and magnetohydrodynamics also reduce to the partial differential equations of fourth-order. To the study of nonlocal boundary value problems (including integral conditions) for partial differential equations of the fourth-order are devoted large number of works, see, for example, [9,10]. It should be noted that boundary value problems with integral conditions are of particular interest. From physical considerations, the integral conditions are completely natural, and they arise in mathematical modelling in cases where it is impossible to obtain information about the process occurring at the boundary of the region of its flow using direct measurements or when it is possible to measure only some averaged (integral) characteristics of the desired quantity.

    In this article, we study the an inverse boundary value problem for a fourth order pseudo parabolic equation with periodic and integral condition to identify the time-dependent coefficients along with the solution function theoretically, i.e. existence and uniqueness.

    Statement of the problem and its reduction to an equivalent problem. In the domain DT={(x,t):0x1,0tT}, we consider an inverse boundary value problem of recovering the timewise dependent coefficients p(t) in the pseudo-parabolic equation of the fourth-order

    ut(x,t)butxx(x,t)+a(t)uxxxx(x,t)=p(t)u(x,t)+f(x,t) (1.1)

    with the initial condition

    u(x,0)+δu(x,T)=φ(x)(0x1), (1.2)

    boundary conditions

    u(0,t)=u(1,t),ux(0,t)=ux(1,t),uxx(0,t)=uxx(1,t)(0tT), (1.3)

    nonlocal integral condition

    10u(x,t)dx=0(0tT) (1.4)

    and with an additional condition

    u(0,t)=t0γ(τ)u(1,τ)dτ+h(t)(0tT), (1.5)

    where b>0, δ0-given numbers, a(t)>0,f(x,t),φ(x),γ(τ),h(t) -given functions, u(x,t) and p(t) - required functions.

    Denote

    ˉC4,1(DT)={u(x,t):u(x,t)C2,1(DT),utxx,uxxxxC(DT)}.

    Definition.By the classical solution of the inverse boundary value problem (1.1)-(1.5)we mean the pair {u(x,t),p(t)} functions u(x,t)ˉC4,1(DT), p(t)C[0,T] satisfying equation (1.1) in DT, condition (1.2) in [0, 1] and conditions (1.3)-(1.5) in [0, T].

    Theorem 1. Let be b>0,δ0,φ(x)C[0,1],f(x,t)C(DT), 10f(x,t)dx=0, 0<a(t)C[0,T], h(t)C1[0,T], h(t)0(0tT), γ(t)C[0,T],δγ(t)=0 (0tT) and

    10φ(x)dx=0,φ(0)=h(0)+δh(T).

    Then the problem of finding a solution to problem (1.1)-(1.5) is equivalent to the problem of determining the functions u(x,t)ˉC4,1(DT) and p(t)C[0,T], from (1.1)-(1.3) and

    uxxx(0,t)=uxxx(1,t)(0tT), (1.6)
    γ(t)u(1,t)+h(t)butxx(0,t)+a(t)uxxxx(0,t)=
    =p(t)(t0γ(τ)u(1,τ)dτ+h(t))+f(0,t)(0tT). (1.7)

    Proof. Let be {u(x,t),p(t)} is a classical solution to problem (1.1)-(1.5). Integrating equation (1.1) with respect to x from 0 to 1, we get:

    ddt10u(x,t)dxb(utx(1,t)utx(0,t))+a(t)(uxxx(1,t)uxxx(0,t))=
    =p(t)10u(x,t)dx+10f(x,t)dx(0tT). (1.8)

    Assuming that 10f(x,t)dx=0, taking into account (1.3) and (1.4), we arrive at the fulfillment of (1.6).

    Further, considering h(t)C1[0,T] and differentiating with respect to t (1.5), we get:

    ut(0,t)=γ(t)u(1,t)+h(t)(0tT) (1.9)

    Substituting x=0 into equation (1.1), we have:

    ut(0,t)butxx(0,t)+a(t)uxxxx(0,t)=p(t)u(0,t)+f(0,t)(0tT). (1.10)

    Now, suppose that {u(x,t),p(t)} is a solution to problem (1.1)-(1.3), (1.6), (1.7). Then from (1.8), taking into account (1.3) and (1.6), we find:

    ddt10u(x,t)dxp(t)10u(x,t)dx=0(0tT). (1.11)

    Due to (1.2) and 10φ(x)dx=0, it's obvious that

    10u(x,0)dx+δ10u(x,T)dx=10φ(x)dx=0. (1.12)

    Obviously, the general solution(1.11) has the form:

    10u(x,t)dx=cet0p(τ)dτ(0tT). (1.13)

    From here, taking into account (1.12), we obtain:

    10u(x,0)dx+δ10u(x,T)dx=c(1+δeT0p(τ)dτ)=0. (1.14)

    By virtue of δ0, from (1.14) we get that c=0, and substituting into (1.13) we conclude, that 10u(x,t)dx=0(0tT). Therefore, condition (1.4) is also satisfied.

    Further, from (1.7) and (1.10), we obtain:

    ddt[u(0,t)(t0γ(τ)u(1,τ)dτ+h(t))]=
    =p(t)[u(0,t)(t0γ(τ)u(1,τ)dτ+h(t))](0tT). (1.15)

    Let introduce the notation:

    y(t)u(0,t)(t0γ(τ)u(1,τ)dτ+h(t))(0tT) (1.16)

    and rewrite the last relation in the form:

    y(t)+p(t)y(t)=0(0tT). (1.17)

    From (1.16), taking into account (1.2), δγ(t)=0 (0tT) and φ(0)=h(0)+δh(T), it is easy to see that

    y(0)+δy(T)=u(0,0)h(0)+δ[u(0,T)(T0γ(τ)u(1,τ)dτ+h(T))]=u(0,0)+
    +δu(0,T)(h(0)+δh(T))δT0γ(τ)u(1,τ)dτ=φ(0)(h(0)+δh(T))=0. (1.18)

    Obviously, the general solution (1.17) has the form:

    y(t)=cet0p(τ)dτ(0tT). (1.19)

    From here, taking into account (1.18), we obtain:

    y(0)+δy(T)=c(1+δeT0a0(τ)a1(τ)dτ)=0. (1.20)

    By virtue of δ0, from (1.20) we get that c=0, and substituting into (1.19) we conclude that y(t)=0(0tT). Therefore, from (1.16) it is clear that the condition (1.5). The theorem has been proven.

    It is known [5] that the system

    1,cosλ1x,sinλ1x,...,cosλkx,sinλkx,... (2.1)

    forms the basis of L2(0,1), where λk=2kπ(k=0,1,...).

    Since system (2.1) forms a basis in L2(0,1), it is obvious that for each solution {u(x,t),a(t)} problem (1.1)–(1.3), (1.6), (1.7):

    u(x,t)=k=0u1k(t)cosλkx+k=1u2k(t)sinλkx(λk=2πk), (2.2)

    where

    u10(t)=10u(x,t)dx,u1k(t)=210u(x,t)cosλkxdx(k=1,2,...),
    u2k(t)=210u(x,t)sinλkxdx(k=1,2,...).

    Applying the formal scheme of the Fourier method, to determine the desired coefficients u1k(t)(k=0,1,...) and u2k(t)(k=1,2,...) functions u(x,t) from (1.1) and (1.2) we get:

    u10(t)=F10(t;u,p)(0tT), (2.3)
    (1+bλ2k)uik(t)+a(t)λ4kuik(t)=Fik(t;u,p)(i=1,2;0tT;k=1,2,...), (2.4)
    u10(0)+δu10(T)=φ10, (2.5)
    uik(0)+δuik(T)=φik(i=1,2;k=1,2,...), (2.6)

    where

    F1k(t;u,a,b)=p(t)u1k(t)+f1k(t)(k=0,1,...),
    f10(t)=10f(x,t)dx,f1k(t)=210f(x,t)cosλkxdx(k=1,2,...),
    φ10=10φ(x)dx,φ1k=210φ(x)cosλkxdx(k=1,2,...),
    F2k(t;u,a,b)=p(t)u2k(t)+f2k(t),
    f2k(t)=210f(x,t)sinλkxdx(k=1,2,...),φ2k=210φ(x)sinλkxdx(k=1,2,...).

    Solving problem (2.3)-(2.6), we find:

    u10(t)=(1+δ)1(φ10δT0F0(τ;u,p)dτ)+t0F10(τ;u,p)dτ(0tT), (2.7)
    uik(t)=et0a(s)λ4k1+bλ2kds1+δeT0a(s)λ4k1+bλ2kdssφik+11+bλ2kt0Fik(τ;u,p)etτa(s)λ4k1+bλ2kdsdτ
    δeT0a(s)λ4k1+bλ2kds1+δeT0a(s)λ4k1+bλ2kds11+bλ2kT0Fik(τ;u,p)etτa(s)λ4k1+bλ2kdsdτ(i=1,2;0tT;k=1,2,...). (2.8)

    After substituting the expression u1k(t)(k=0,1,...), u2k(t)(k=1,2,...) in (2.2), to define a component u(x,t) solution of problem (1.1)-(1.3), (1.6), (1.7), we obtain:

    u(x,t)=(1+δ)1(φ0δT0F0(τ;u,p)dτ)+t0F0(τ;u,p)dτ+
    +k=1{et0a(s)λ4k1+bλ2kds1+δeT0a(s)λ4k1+bλ2kdssφ1kk+11+bλ2kt0F1k(τ;u,p)etτa(s)λ4k1+bλ2kdsdτ
    δeT0a(s)λ4k1+bλ2kds1+δeT0a(s)λ4k1+bλ2kds11+bλ2kT0F1k(τ;u,p)etτa(s)λ4k1+bλ2kdsdτ}cosλkx+
    +k=1{et0a(s)λ4k1+bλ2kds1+δeT0a(s)λ4k1+bλ2kdssφ2kk+11+bλ2kt0F2k(τ;u,p)etτa(s)λ4k1+bλ2kdsdτ
    δeT0a(s)λ4k1+bλ2kds1+δeT0a(s)λ4k1+bλ2kds11+bλ2kT0F2k(τ;u,p)etτa(s)λ4k1+bλ2kdsdτ}sinλkx. (2.9)

    Now from (1.7), taking into account (2.2), we have:

    p(t)=[h(t)]1{h(t)f(0,t)+γ(t)u10(t)p(t)t0γ(τ)u10(τ)dτ+
    +k=1(bλ2ku1k(t)+a(t)λ4ku1k(t)+γ(t)u1k(t)p(t)t0γ(τ)u1k(τ)dτ). (2.10)

    Further, from (2.4), taking into account (2.8), we obtain:

    bλ2ku1k(t)+a(t)λ4ku1k(t)+γ(t)u1k(t)=F1k(t;u,p)u1k(t)+γ(t)u1k(t)=
    =bλ2k1+bλ2kF1k(t;u,p)+(a(t)λ4k1+bλ2k+γ(t))u1k(t)=
    =bλ2k1+bλ2kFk(t;u,p)+(a(t)λ4k1+bλ2k+γ(t))[et0a(s)λ4k1+bλ2kds1+δeT0a(s)λ4k1+bλ2kdssφ1k+
    +11+bλ2kt0F1k(τ;u,p)etτa(s)λ4k1+bλ2kdsdτ
    δeT0a(s)λ4k1+bλ2kds1+δeT0a(s)λ4k1+bλ2kds11+bλ2kT0F1k(τ;u,p)etτa(s)λ4k1+bλ2kdsdτ](0tT;k=1,2,...). (2.11)
    p(t)=[h(t)]1{h(t)f(0,t)+
    +γ(t))[(1+δ)1(φ10δT0F0(τ;u,p)dτ)+t0F10(τ;u,p)dτ]
    p(t)t0γ(τ)u10(τ)dτ+k=1[bλ2k1+bλ2kF1k(t;u,p)+
    +(a(t)λ4k1+bλ2k+γ(t))[et0a(s)λ4k1+bλ2kds1+δeT0a(s)λ4k1+bλ2kdssφ1k+11+bλ2kt0F1k(τ;u,p)etτa(s)λ4k1+bλ2kdsdτ
    +11+bλ2kt0F1k(τ;u,p)etτa(s)λ4k1+bλ2kdsdτ
    δeT0a(s)λ4k1+bλ2kds1+δeT0a(s)λ4k1+bλ2kds11+bλ2kT0F1k(τ;u,p)etτa(s)λ4k1+bλ2kdsdτ]+
    +p(t)t0γ(τ)u1k(τ)dτ]}. (2.12)

    Thus, the solution of problem (1.1)–(1.3), (1.6), (1.7)is reduced to the solution of system (2.9), (2.12) with respect to unknown functions u(x,t) and p(t).

    To study the question of the uniqueness of the solution of problem (1.1)–(1.3), (1.6), (1.7) the following plays an important role.

    Lemma 1. If {u(x,t),p(t)}-any solution of problem (1.1)–(1.3), (1.6), (1.7), then the functions

    u10(t)=10u(x,t)dx,u1k(t)=210u(x,t)cosλkxdx(k=1,2,...),
    u2k(t)=210u(x,t)sinλkxdx(k=1,2,...)

    satisfy the system consisting of equations (27), (28) on [0,T].

    It is obvious that if u10(t)=10u(x,t)dx, u1k(t)=210u(x,t)cosλkxdx(k=1,2,...), u2k(t)=210u(x,t)sinλkxdx(k=1,2,...) is a solution to system (2.7), (2.8), then the pair {u(x,t),p(t)} functions u(x,t)=k=0u1k(t)cosλkx+k=1u2k(t)sinλkx(λk=2πk) and p(t) is a solution to system (2.9), (2.12).

    Consequence. Let system (29), (32) have a unique solution. Then problem (1.1)–(1.3), (1.6), (1.7) cannot have more than one solution, i.e. if problem (1.1)-(1.3), (1.6), (1.7) has a solution, then it is unique.

    In order to study the problem (1.1)–(1.3), (1.6), (1.7) consider the following spaces.

    Denote by Bα2,T [6] the set of all functions of the form

    u(x,t)=k=0u1k(t)cosλkx+k=1u2k(t)sinλkx(λk=2πk),

    considered in DT, where each of the functions u1k(t)(k=0,1,...), u2k(t)(k=1,2,...) continuous on [0,T] and

    J(u)=u10(t)C[0,T]+{k=1(λαku1k(t)C[0,T])2}12+{k=1(λαku2k(t)C[0,T])2}12<+,

    α0. We define the norm in this set as follows:

    u(x,t)Bα2,T=J(u).

    Through EαT denote the space Bα2,T×C[0,T] vector - functions z(x,t)={u(x,t),p(t)} with norm

    z(x,t)EαT=u(x,t)Bα2,T+p(t)C[0,T].

    It is known that Bα2,T and EαT are Banach spaces.

    Now consider in space E5T operator

    Φ(u,p)={Φ1(u,p),Φ2(u,p)},

    operator

    Φ1(u,p))=˜u(x,t)k=0˜u1k(t)cosλkx+k=1˜u2k(t)sinλkx,Φ2(u,p)=˜p(t),

    ˜u10(t),˜uik(t)(i=1,2;k=1,2,...),˜p(t) are equal to the right-hand sides of (2.7), (2.8) and (2.12), respectively.

    It is easy to see that

    1+bλ2k>bλ2k,1+δ1,.1+δeT0a(s)λ4k1+bλ2kds1.

    Then, we have:

    ˜u0(t)C[0,T]|φ10|+(1+δ)T(T0|f10(τ)|2dτ)12+(1+δ)Tp(t)C[0,T]u10(t)C[0,T], (2.13)
    (k=1(λ5k˜uik(t)C[0,T])2)123(k=1(λ5k|φik|)2)12+3(1+δ)bT(T0k=1(λ3k|fik(τ)|)2dτ)12+
    +3(1+δ)bTp(t)C[0,T](k=1(λ5kuik(t)C[0,T])2)12(i=1,2), (2.14)
    ˜p(t)C[0,T][h(t)]1C[0,T]{h(t)f(0,t)C[0,T]+
    +γ(t)C[0,T][|φ0|+(1+δ)T(T0|f0(τ)|2dτ)12+(1+δ)Tp(t)C[0,T]u0(t)C[0,T]]+
    +Tγ(t)C[0,T]p(t)C[0,T]u10(t)C[0,T]+
    +(k=1λ2k)12[(k=1(λkf1k(t))2C[0,T])12+p(t)C[0,T](k=1(λ3ku1k(t)C[0,T])2)12+
    +(γ(t)C[0,T]+1ba(t)C[0,T])[(k=1(λ3k|φ1k|)2)12+T(1+δ)b(T0k=1(λk|f1k(τ)|)2dτ)12+
    +T(1+δ)bp(t)C[0,T](k=1(λ5ku1k(t)C[0,T])2)12]++Tγ(t)C[0,T]p(t)C[0,T](k=1(λ5ku1k(t)C[0,T])2)12]}. (2.15)

    Let us assume that the data of problem (1.1)–(1.3), (1.6), (1.7) satisfy the following conditions:

    1.φ(x)W2(5)(0,1),φ(0)=φ(1),φ(0)=φ(1),
    φ(0)=φ(1),φ(0)=φ(1),φ(4)(0)=φ(4)(1);
    2.f(x,t),fx(x,t),fxx(x,t)C(DT),fxxx(x,t)L2(DT),
    f(0,t)=f(1,t),fx(0,t)=fx(1,t),fxx(0,t)=fxx(1,t)(0tT);
    3.b>0,δ0,γ(t),a(t)C[0,T],h(t)C1[0,T],h(t)0(0tT).

    Then from (2.10)–(2.12), we have:

    ˜u(x,t)B52,TA1(T)+B1(T)p(t)C[0,T]u(x,t)B52,T, (2.16)
    ˜p(t)C[0,T]A2(T)+B2(T)p(t)C[0,T]u(x,t)B52,T, (2.17)

    where

    A1(T)=φ(x)L2(0,1)+(1+δ)Tf(x,t)L2(DT)+23φ(5)(x)L2(0,1)+
    +23b(1+δ)Tfxxx(x,t)L2(DT),B1(T)=(1+δ)(1+3b)T,
    A2(T)=[h(t)]1C[0,T]{h(t)f(0,t)C[0,T]+
    +γ(t)C[0,T](φ(x)L2(0,1)+(1+δ)Tf(x,t)L2(DT))+
    +(k=1λ2k)12[fx(x,t)C[0,T]L2(0,1)+
    +(γ(t)C[0,T]+1ba(t)C[0,T])(φ(3)(x)L2(0,1)+T(1+δ)bfx(x,t)L2(DT))]},
    B2(T)=[h(t)]1C[0,T](k=1λ2k)12[(γ(t)C[0,T]+1ba(t)C[0,T])T(2+δ)b+
    +Tγ(t)C[0,T]+1].

    From inequalities (2.16), (2.17) we conclude:

    u(x,t)B52,T+˜p(t)C[0,T]A(T)+B(T)p(t)C[0,T]u(x,t)B52,T, (2.18)
    A(T)=A1(T)+A2(T),B(T)=B1(T)+B2(T).

    We can prove the following theorem.

    Theorem 2. Let conditions 1-3 be satisfied and

    (A(T)+2)2B(T)<1. (2.19)

    Then problem (1.1)–(1.3), (1.6), (1.7) has in K=KR(zE5TR=A(T)+2) in the space E5T only one solution.

    Proof. In space E5T consider the equation

    z=Φz, (2.20)

    where z={u,p}, components P Φ1(u,p),Φ2(u,p) of operators Φ(u,p) are defined by the right-hand sides of equations (2.9) and (2.12).

    Consider the operator Φ(u,p) in a ball K=KR from E5T. Similarly to (2.18) we obtain that for any z={u,p}, z1={u1,p1}, z2={u2,p2}KR :

    ΦzE5TA(T)+B(T)p(t)C[0,T]u(x,t)B52,T, (2.21)
    Φz1Φz2E5TB(T)R(p1(t)p2(t)C[0,T]+u1(x,t)u2(x,t)B52,T). (2.22)

    Then from estimates (2.21), (2.22), taking into account (2.19), it follows that the operator Φ acts in a ball K=KR and is contractive. Therefore, in the ball K=KR operator Φ has a single fixed point {u,p}, which is the only one in the ball K=KR solution of equation (2.20), i.e. is the only one solution in the ball K=KR of system (2.9), (2.12) in the ball.

    Functions u(x,t), as an element of space B52,T is continuous and has continuous derivatives ux(x,t),uxx(x,t), uxxx(x,t),uxxxx(x,t) in DT.

    From (2.4), it is easy to see that

    (k=1(λkuik(t)C[0,T])2)122ba(t)C[0,T](k=1(λ5kuik(t)C[0,T])2)12+
    +2bfx(x,t)+p(t)ux(x,t)C[0,T]L2(0,1)(i=1,2).

    Hence it follows that ut(x,t) and utxx continuous in DT.

    It is easy to check that equation (1.1) and conditions (1.2), (1.3), (1.6), (1.7) are satisfied in the usual sense. Consequently, {u(x,t),p(t)} is a solution to problem (1.1)–(1.3), (1.6), (1.7). By the corollary of Lemma 1, it is unique in the ball K=KR. The theorem has been proven.

    With the help of Theorem 1, the unique solvability of the original problem (1.1)–(1.5) immediately follows from the last theorem.

    Theorem 3. Let all the conditions of Theorem 1 be satisfied, 10f(x,t)dx=0(0tT), δγ(t)=0 (0tT) and the matching condition is met:

    10φ(x)dx=0,φ(0)=h(0)+δh(T).

    Then problem (1.1)–(1.5) has in the ball K=KR(zE5TR=A(T)+2) from E5T the only classical solution.

    The article considered an inverse boundary value problem with a periodic and integral condition, when the unknown coefficient depends on time for a linear pseudoparabolic equation of the fourth order. An existence and uniqueness theorem for the classical solution of the problem is proved.

    The authors have declared no conflict of interest.



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