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Regularization of a final value problem for a linear and nonlinear biharmonic equation with observed data in Lq space

  • In this work, we focus on the final value problem of an inverse problem for both linear and nonlinear biharmonic equations. The aim of this study is to provide a regularized method for the bi-harmonic equation, once the observed data are obtained at a terminal time in Lq(Ω). We obtain an approximated solution using the Fourier series truncation method and the terminal input data in Lq(Ω) for q2. In comparision with previous studies, the most highlight of this study is the error between the exact and regularized solutions to be estimated in Lq(Ω); wherein an embedding between Lq(Ω) and Hilbert scale spaces Hρ(Ω) is applied.

    Citation: Anh Tuan Nguyen, Le Dinh Long, Devendra Kumar, Van Thinh Nguyen. Regularization of a final value problem for a linear and nonlinear biharmonic equation with observed data in Lq space[J]. AIMS Mathematics, 2022, 7(12): 20660-20683. doi: 10.3934/math.20221133

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  • In this work, we focus on the final value problem of an inverse problem for both linear and nonlinear biharmonic equations. The aim of this study is to provide a regularized method for the bi-harmonic equation, once the observed data are obtained at a terminal time in Lq(Ω). We obtain an approximated solution using the Fourier series truncation method and the terminal input data in Lq(Ω) for q2. In comparision with previous studies, the most highlight of this study is the error between the exact and regularized solutions to be estimated in Lq(Ω); wherein an embedding between Lq(Ω) and Hilbert scale spaces Hρ(Ω) is applied.



    Let Ω be a bounded domain in RN(N=1,2,3) a sufficiently smooth boundary Ω. In this study, we consider a nonlinear biharmonic equation, as follows:

    Δ2uutttt+2uttxx+uxxxx=G(x,t,u(x,t)),inΩ×(0,T], (1.1)

    satisfying the following boundary conditions:

    u(x,t)=Δu(x,t)=0,Ω×(0,T), (1.2)

    and

    {u(x,T)=f(x),ut(x,T)=0,inΩ,[0.4cm]Δu(x,T)=g(x),Δut(x,T)=0,inΩ, (1.3)

    wherein the condition (1.2) is the Navier boundary condition; and the condition (1.3) is the mixed Dirichlet-Neumann boundary condition. The function u=u(x,t) represents a concentration of contaminant at a position x and at time t. The data f,gLq(Ω) and GL(0,T;Lq(Ω)) are defined later on. However, in actual conditions, there are always included errors in the measurement methods of a physical process, so we have the following conditions:

    fδfLq(Ω)+gδgLq(Ω)+GδGL(0,T;Lq(Ω))δ. (1.4)

    The biharmonic equation plays an important role in engineering and physics. It arises in the deformation of thin plates, the motion of fluids, free boundary problems and nonlinear elasticity, see [1,2,3,4,5]. Therefore, the biharmonic equation has a long history of research. It has been studied by many authors at early time. The most highlighted studies on numerical methods for the biharmonic equation are described in [5,6,7,8,9,10]. In particular, Smith [6] presented a numerical method for solving the biharmonic difference equation using finite difference methods. Ehrlich [7] has improved the iteration scheme to lead the Smith's result to be a special case of his study. Recently, Tuan et al. [11] have studied an approximate solution for a nonlinear biharmonic equation with discrete random data. Especially, in applications to radar imaging, Matevossian et al. [12,13] have focused on the solution of the biharmonic equation with the Dirichlet, Neumann and Cauchy boundary value problems for the Poisson equation using the scattering model.

    Regarding the regularization for biharmonic equations, the authors in [14] considered a nonlinear biharmonic equation, and proved that problem (1.1) under the conditions (1.2) and (1.3) is ill-posed in the sense of Hadamard, and showed the error estimates. The corresponding regularized solutions in their study are strongly converged to the exact solution in L2(Ω) under some priori assumptions on the solution. Besides, there are many other studies on linear homogeneous biharmonic equations; however, most of previous studies are focused on the regularization for biharmonic equations in Lq(Ω) with q=2; and the convergent rate in Lq(Ω), with q2 is still not well implemented (Nam et al. [14]). Therefore, it can be stated that our study in this paper is one of the first results regarding the inverse problem for the biharmonic equation, once the observed data is obtained in the Lq(Ω) space with q2. The main objective of this study is to establish regularized solutions for problem (1.1) under the conditions (1.2) and (1.3) and showed the regularized solution is converged to the exact solution; in the linear case refered to (3.13), and in the nonlinear case referred to (3.73).

    For evaluation cases in Lq(Ω) spaces, the most obstacle is unable to use Parseval's equality; therefore, we applied the embedding between Lq(Ω) and Hilbert scales spaces Hρ(Ω) to overcome this limitation; and Lemma 2.1 will be used throughout this article. The manuscript is proceeding as follows:

    ● The first part deals with the inverse problem with a defined source function. In this subsection, we introduce the mild solution of problem (1.1) under the conditions (1.2) and (1.3) with the observed data fδ,gδLq(Ω), and GδL(0,T;Lq(Ω)). Then, applying the Fourier series truncation method, we estimate the error between the regular and exact solution in the L2NN4ρ(Ω).

    ● The second part of the manuscript investigates the inverse initial value problem for problem (1.1) under the conditions (1.2) and (1.3) with a nonlinear source function. In this section, the main results to be obtained are theorems: (ⅰ) The existence and the well-posedness of regularized solutions using Banach fixed point theorem; and (ⅱ) The convergent rate between the regularized solution and the exact solution through the estimation of ˜VδNδ(,t)u(,t)L2NN4ρ(Ω).

    Hence, this manuscript is organized as follows. In Section 2, some preliminaries such as definition and Lemmas are given. Section 3 introduces some results on regularization of problem (1.1) under the conditions (1.2) and (1.3) in the linear and non-linear cases. Numerical examples is described in Section 4 associate with observed data in Lq(Ω).

    We begin this section by introducing some preliminary definitions and basic lemmas that are needed for our analysis.

    Definition 2.1. Assume that Δ has the eigenvalues λk,kN:

    0<λ1λ2λ3, (2.1)

    and the corresponding eigenelements ek(x), which form an orthonormal basis in L2(Ω).

    Definition 2.2. Let , be an inner product in L2(Ω). The notation X stands for in the norm in the Banach space. We denote by Lq(0,T;X), 1q<, the Banach space of real-valued functions u:(0,T)X measurable, providing that

    uLq(0,T;X)=(T0u(t)qdt)1q<,for1q<, (2.2)

    while

    uL(0,T;X)=esssupt(0,T)u(t)X,forq=.

    Definition 2.3. (see [8]) For any σ0, we also define the space

    Hσ(Ω)={uL2(Ω):k=1λ2σk|u,ek|2<+}, (2.3)

    then Hσ(Ω) is a Hilbert space endowed with the norm

    uHσ(Ω)=(k=1λ2σk|u,ek|2)12. (2.4)

    Lemma 2.1. (see [15]) The following inclusions hold true:

    Lq(Ω)Hσ(Ω),ifd4<σ0,q2dd4σ,Hσ(Ω)Lq(Ω),if0<σd4,q2dd4σ. (2.5)

    First of all, we present the formula of a mild solution of problem (1.1) as follows.

    The solution of problem (1.1) can be written in the following Fourier series form:

    u(x,t)=k=1uk(t)ek(x),whereuk(t)=u(,t),ek (3.1)

    We have a particular solution of problem (1.1) in the form

    uk(t)=cosh(λk(Tt))f,ek+(Tt)sinh(λk(Tt))2λkg,ek+Tt(rt)cosh(λk(rt))2λkG(,r),ekdr+Ttsinh(λk(rt))2λkλkG(,r),ekdr. (3.2)

    Substituting the result into (3.1), we will have the formal solution of problem (1.1). The following steps, we are going to find the mild solution of problem (1.1) when the source function is linear and nonlinear.

    By applying the Fourier truncation method, we provide a regularization solution as follows:

    uNtrδ(x,t)=Ntrk=1cosh(λk(Tt))fδ,ekek(x)+Ntrk=1(Tt)sinh(λk(Tt))2λkgδ,ekek(x)+Ntrk=1(Tt(rt)cosh(λk(rt))2λkGδ(,t),ekdr)ek(x)+Ntrk=1(Ttsinh(λk(rt))2λkλkGδ(,t),ekdr)ek(x), (3.3)

    whereby Ntr is a parameter regularization which will be defined later. From (3.3), it allows us to deduce that the mild solution to problem (1.1) in the following form:

    u(x,t)=k=1cosh(λk(Tt))f,ekek(x)+(Tt)sinh(λk(Tt))2λkg,ekek(x)+k=1(Tt(rt)cosh(λk(rt))2λkG(,r),ekdr)ek(x)+k=1(Ttsinh(λk(rt))2λkλkG(,r),ekdr)ek(x). (3.4)

    Theorem 3.1. Let assume that fδ,gδ,GδLq(Ω)×Lq(Ω)×L(0,T;Lq(Ω)) are observed data such that

    fδfLq(Ω)+gδgLq(Ω)+GδGL(0,T;Lq(Ω))δ. (3.5)

    Let uL(0,T;Hσ+γ(Ω)) for any γ>0. By choosing Ntr=(1αTC1log(δ1))N for any 0<α<1, then we have uNtrδ(,t)u(,t)L2NN4σ(Ω) is of order

    max{δα(1αTC1log(δ1))2σ+N2(2q1),(1αTC1log(δ1))2γ}. (3.6)

    Proof. Because of the Sobolev embedding Lq(Ω)HN(q2)4q(Ω), we have

    fδfHN(q2)4q(Ω)+gδgHN(q2)4q(Ω)+GδGL(0,T;HN(q2)4q(Ω))C1fδfLq(Ω)+C1gδgLq(Ω)+C1GδGL(0,T;Lq(Ω))C1δ, (3.7)

    with C1 depends on N,p. Our goal in this theorem is to assess the convergence error of u(,t)uNtrδ(,t)Hσ(Ω). Next, we first introduce the following function:

    ˜UNtr(x,t)=Ntrk=1cosh(λk(Tt))f,ekek(x)+Ntrk=1(Tt)sinh(λk(Tt))2λkg,ekek(x)+Ntrk=1(Tt(rt)cosh(λk(rt))2λkG(,r),ekdr)ek(x)+Ntrk=1(Ttsinh(λk(rt))2λkλkG(,r),ekdr)ek(x). (3.8)

    Using the triangle inequality, we receive

    uNtrδ(,t)u(,t)Hσ(Ω)uNtrδ(,t)˜UNtr(,t)Hσ(Ω)+˜UNtr(,t)u(,t)Hσ(Ω). (3.9)

    Next, we evaluate (3.9) through two steps as follows:

    Step 1: Estimate of uNtrδ(,t)˜UNtr(,t)Hσ(Ω), we find that

    uNtrδ(x,t)˜UNtr(x,t)=Ntrk=1cosh(λk(Tt))ffδ,ekek(x)+Ntrk=1(Tt)sinh(λk(Tt))2λkggδ,ekek(x)+Ntrk=1(Tt(rt)cosh(λk(rt))2λkG(,r)Gδ(,r),ekdr)ek(x)+Ntrk=1(Ttsinh(λk(rt))2λkλkG(,r)Gδ(,r),ekdr)ek(x)=A1(x,t)+A2(x,t), (3.10)

    whereby

    A1(x,t)=Ntrk=1cosh(λk(Tt))ffδ,ekek(x)+Ntrk=1(Tt)sinh(λk(Tt))2λkggδ,ekek(x),A2(x,t)=Ntrk=1(Tt(rt)cosh(λk(rt))2λkG(,r)Gδ(,r),ekdr)ek(x)+Ntrk=1(Ttsinh(λk(rt))2λkλkG(,r)Gδ(,r),ekdr)ek(x). (3.11)

    First of all, estimating the A1(x,t), it is easy to check that cosh(x)exp(x) and sinh(x)exp(x),x>0, this implies that

    cosh(λk(Tt))exp(λk(Tt)),sinh(λk(Tt))exp(λk(Tt)). (3.12)

    Therefore, it gives

    A1(,t)2Hσ(Ω)2Ntrk=1λ2σN(q2)2qkλN(q2)2qkexp(2λkT)|ffδ,ek|2+2Ntrk=1λ2σN(q2)2qkλN(q2)2qkT2exp(2λkT)4λk|ggδ,ek|2. (3.13)

    In the fact that λkC2k2N, and noting that σN4N2q, we can verify that for kNtr,

    2λ2σN(q2)2qkexp(2λkT)+T22λkλ2σN(q2)2qkexp(2λkT)=2exp(2λkT)λ2σN(q2)2qk+T22λ1exp(2λkT)λ2σN(q2)2qkC3exp(2C2k1NT)k4σNq2q+C4exp(2C2k1NT)k4σNq2q, (3.14)

    in which C3=2C2σN(q2)2q2 and C4=T22λ1C2σN(q2)2q2.

    Combining (3.13) and (3.14), we have

    A1(,t)2Hσ(Ω)C5exp(2TC2k1N)k4σNq2q(Ntrk=1λN(q2)2qk|ffδ,ek|2+Ntrk=1λN(q2)2qk|ggδ,ek|2)C5exp(2TC2k1N)k4σNq2q(ffδ2HN(q2)4q(Ω)+ggδ2HN(q2)4q(Ω)), (3.15)

    with C5=2max{C3,C4}. It follows from (3.7) that

    A1(,t)2Hσ(Ω)2C5exp(2C2(Ntr)1NT)(Ntr)4σNq2qδ2. (3.16)

    Next, considering the term A2(.,t)Hσ(Ω), applying the Parseval's equality, we can see that

    A2(.,t)2Hσ(Ω)2Ntrk=1λ2σk[Tt(rt)exp(λk(rt))2λkG(,r)Gδ(,r),ekdr]2+2Ntrk=1λ2σk[Ttexp(λk(rt))2λkλkG(,r)Gδ(,r),ekdr]2. (3.17)

    From (3.17), using the Hölder's inequality, we receive

    A2(.,t)2Hσ(Ω)2Ntrk=1λ2σkTt(rt)2exp(2λk(rt))4λ2k|G(,r)Gδ(,r),ek|2dr+2Ntrk=1λ2σkTtexp(2λk(rt))4λ3k|G(.,r)Gδ(.,r),ek|2dr(2T2+1)Ntrk=1exp(2λkT)λ2σN(q2)2qkλN(q2)2qkTt|G(.,r)Gδ(.,r),ek|2dr. (3.18)

    Because of λkC2k2N, and noting that σN4N2q, we can verify that for kNtr,

    λ2σN(q2)2qkexp(2λk(rt))C3exp(2C2k1NT)k4σNq2qC3exp(2C2(Ntr)1NT)(Ntr)4σNq2q. (3.19)

    From (3.18), we noticed that

    A2(.,t)2Hσ(Ω)C3exp(2C2(Ntr)1NT)(Ntr)4σNq2q(2T2+1)Ntrk=1λN(q2)2qkTt|G(.,r)Gδ(.,r),ek|2drC3exp(2C2(Ntr)1NT)(Ntr)4σNq2q(2T2+1)TtG(,r)Gδ(,r)2HN(q2)4q(Ω)dr. (3.20)

    Due to condition (3.7) and from the estimation in (3.20), one has

    A2(.,t)2Hσ(Ω)C3exp(2C2(Ntr)1NT)(Ntr)4σNq2q(2T3+T)δ2. (3.21)

    From all the estimation above, we received

    uNtrδ(,t)˜UNtr(,t)2Hσ(Ω)2A1(.,t)2Hσ(Ω)+2A2(.,t)2Hσ(Ω)2C5exp(2C2(Ntr)1NT)(Ntr)4σNq2qδ2+2C3C1exp(2C2(Ntr)1NT)(Ntr)4σNq2q(2T3+T)δ2. (3.22)

    Step 2: Estimate of ˜UNtr(,t)u(,t)Hσ(Ω), from (3.4) and (3.8), and using the Parseval's inequality, we deduce that

    ˜UNtr(x,t)u(x,t)=k=Ntr+1cosh(λk(Tt))f,ekek(x)+Ntrk=1(Tt)sinh(λk(Tt))2λkg,ekek(x)+k=Ntr+1(Tt(rt)cosh(λk(rt))2λkG(,r),ekdr)ek(x)+k=Ntr+1(Ttsinh(λk(rt))2λkλkG(,r),ekdr)ek(x). (3.23)

    From (3.23), for any γ>0, we received

    ˜UNtr(,t)u(,t)2Hσ(Ω)=k=Ntr+1λ2σk|u(,t),ek|2=k=Ntr+1λ2γkλ2σ+2γk|u(,t),ek|2. (3.24)

    In case k>Ntr, there exists a postive constant C6>0 such that λ2γkC6k4γN, we have

    ˜UNtr(,t)u(,t)2Hσ(Ω)C6(Ntr)4γNu(,t)2Hσ+γ(Ω)C6(Ntr)4γNu2L(0,T;Hσ+γ(Ω)). (3.25)

    Combining (3.22) to (3.25), we conclude that

    uNtrδ(,t)u(,t)2Hσ(Ω)2uNtrδ(,t)˜UNtr(,t)2Hσ(Ω)+2˜UNtr(,t)u(,t)2Hσ(Ω)8C6exp(2TC2(Ntr)1N)(Ntr)4σNq2qδ2+4C3C1exp(2TC2(Ntr)1N)(Ntr)4σNq2q(2T3+T)δ2+2C6(Ntr)4γNu2L(0,T;Hσ+γ(Ω)). (3.26)

    By choosing Ntr=(1αTC2log(1δ))N, we need the following results:

    exp(2TC2(Ntr)1N)(Ntr)4σNq2qδ2=δ2α(1αTC2log(1δ))4σ+N(2q1). (3.27)

    The provision of this theorem is completed.

    In this section, we will study the initial inverse problem for nonlinear of source term.

    {Δ2uutttt+2uttxx+uxxxx=G(x,t,u(x,t)),inΩ×(0,T],u(x,t)=Δu(x,t)=0,Ω×(0,T), (3.28)

    with the final condition

    {u(x,T)=f(x),ut(x,T)=0,inΩ,Δu(x,T)=g(x),Δut(x,T)=0,inΩ. (3.29)

    We assume that νL2(Ω), let Pi(zt),i={1,2,3,4} is an operator defined as follows:

    P1(zt)ν=k=1cosh(λk(zt))ν,ekek(x),P2(zt)ν=k=1(zt)sinh(λk(zt))2λkν,ekek(x),P3(zt)ν=k=1(zt)cosh(λk(zt))2λkν,ekek(x),P4(zt)ν=k=1sinh(λk(zt))2λkλkν,ekek(x). (3.30)

    From the way to set the operator (3.30), the mild solution of the problem (3.28) under the condition (3.29) is as follows:

    u(t)=P1(Tt)f+P2(Tt)g+TtP3(rt)G(u(r))dr+TtP4(rt)G(u(r))dr. (3.31)

    We regularized the mild solution (3.31) by Fourier method. Assume that we have L2(Ω), then we define δ,Ntr as follows:

    TNδ=Nδk=1,ekek(x), (3.32)

    without loss of generality, we can completely assume that fδ,gδLq(Ω) such that ffδLq(Ω)+ggδLq(Ω)δ. Therefore, we build the structure of regularized solution and symbols it is ˜VδNδ.

    ˜VδNδ(t)=P1(Tt)TNδfδ+P2(Tt)TNδgδ+TtP3(rt)TNδG(˜VδNδ(r))dr+TtP4(rt)TNδG(˜VδNδ(r))dr. (3.33)

    The next theorem will provide details about the existence and the well-posedness of regularized solutions.

    Theorem 3.2. Let the terminal data Lq(Ω), then the nonlinear integral equation (3.33) has a unique solution ˜VδNδ(x,t)L(0,T;L2NN4ρ(Ω)), then we have the following estimate:

    ˜VδNδ(,t)L2NN4ρ(Ω)2(Tt)(Nδ)ρN(2q)4qexp((Nδ+m)(Tt))gδLq(Ω), (3.34)

    where mNδ+2Lf(Nδ)ρmax{T,1}Nδ.

    Proof. Let any Hσ1(Ω), suppose that σ1σ, we get

    P1(Tt)TNδ+P2(Tt)TNδ2Hσ1(Ω)2P1(Tt)TNδ2Hσ1(Ω)+2P2(Tt)TNδ2Hσ1(Ω)2Nδk=1λ2σ12σkcosh2(λk(Tt))λ2σk|,ek|2+2Nδk=1λ2σ12σk(Tt)2sinh2(λk(Tt))4λkλ2σk|,ek|2. (3.35)

    From (3.35) and in view of (3.12), we receive

    P1(Tt)TNδ+P2(Tt)TNδ2Hσ1(Ω)2(Nδ)2σ12σexp(2Nδ(Tt))2Hσ(Ω)+2(Tt)2(Nδ)2σ12σexp(2Nδ(Tt))2Hσ(Ω). (3.36)

    By a similar argument above (3.36), we can find also that

    P3(rt)TNδ2Hσ1(Ω)=Nδk=1λ2σ12σk(rt)2cosh2(λk(rt))4λ2kλ2σk|,ek|2(Nδ)2σ12σ(rt)2exp(2Nδ(rt))Hσ(Ω), (3.37)

    and we can see that

    P4(rt)TNδ2Hσ1(Ω)=Nδk=1λ2σ12σksinh2(λk(rt))4λ3kλ2σk|,ek|2(Nδ)2σ12σexp(2Nδ(rt))Hσ(Ω). (3.38)

    For m>0, we denote by Lm(0,T;L2NN4ρ(Ω)) the function space L(0,T;L2NN4ρ(Ω)) with the following norm:

    νm:=esssup0tTexp(m(tT))ν(,t)L2NN4ρ(Ω),νL2NN4ρ(Ω). (3.39)

    Next, we define a nonlinear map M:Lm(0,T;L2NN4ρ(Ω))Lm(0,T;L2NN4ρ(Ω)) by

    Mu(t)=P1(Tt)TNδfδ+P2(Tt)TNδgδ+TtP3(rt)TNδG(u(r))dr+TtP4(rt)TNδG(u(r))dr. (3.40)

    ● {Case 1:} u=0, we have M(u=0)=P1(Tt)TNδfδ+P2(Tt)TNδgδ. From Lemma 2.1 and 0ρN4, with the Sobolev embedding Hρ(Ω)L2NN4ρ(Ω), there exists a constant C1 depends on N,ρ such that

    P1(Tt)TNδfδ+P2(Tt)TNδgδL2NN4ρ(Ω)C7(N,ρ)P1(Tt)TNδfδ+P2(Tt)TNδHρ(Ω). (3.41)

    From (3.41), use evaluation results in (3.36), taking square root on the both side, we choose σ1=ρ and σ=N(2q)4q, it gives

    P1(Tt)TNδfδ+P2(Tt)TNδgδHρ(Ω)2(Nδ)ρN(2q)4qexp(Nδ(Tt))fδHN(2q)4q(Ω)+2(Tt)(Nδ)ρN(2q)4qexp(Nδ(Tt))gδHN(2q)4q(Ω). (3.42)

    For 1<q<2, using Lemma 2.1, we find that Lq(Ω)HN(2q)4q(Ω). Therefore, there exist a constant C8(N,ρ) such that

    P1(Tt)TNδfδ+P2(Tt)TNδgδHN(2q)4q(Ω)C8(N,ρ)(2(Nδ)ρN(2q)4qexp(Nδ(Tt))fδLq(Ω)+2(Tt)(Nδ)ρN(2q)4qexp(Nδ(Tt))gδLq(Ω)). (3.43)

    Combining (3.41) to (3.43), this leads to

    P1(Tt)TNδfδ+P2(Tt)TNδgδL2NN4ρ(Ω)C9(N,ρ)(2(Nδ)ρN(2q)4qexp(Nδ(Tt))fδLq(Ω)+2(Tt)(Nδ)ρN(2q)4qexp(Nδ(Tt))gδLq(Ω)), (3.44)

    whereby C9(N,ρ)=C8(N,ρ)C7(N,ρ). Combining with above arguments, we deduce that

    P1(Tt)TNδfδ+P2(Tt)TNδgδLm(0,T;L2NN4ρ(Ω)). (3.45)

    ● {Case 2:} In this case, we take two function u1,u2Lm(0,T;L2NN4ρ(Ω)), from (3.40), it is easy to see that

    Mu1(t)Mu2(t)=TtP3(rt)(TNδG(u1(r))TNδG(u2(r)))dr+TtP4(rt)(TNδG(u1(r))TNδG(u2(r)))dr. (3.46)

    Taking any m>0, this implies that

    exp(m(tT))(Mu1(t)Mu2(t))Hρ(Ω)Ttexp(m(tT))P3(rt)(TNδG(u1(r))TNδG(u2(r)))Hρ(Ω)D1(r,t)dr+Ttexp(m(tT))P4(rt)(TNδG(u1(r))TNδG(u2(r)))Hρ(Ω)D2(r,t)dr. (3.47)

    If we reuse estimates of (3.37) and (3.38) with σ1=ρ and σ=0, and in the two reviews below, we have used the Sobolev embedding L2NN4ρ(Ω)L2(Ω). So, D1(r,t) and D2(r,t) can be bounded as follows:

    D1(r,t)(Nδ)ρ(rt)exp(Nδ(rt))G(u1(r))G(u2(r))L2(Ω)Lf(Nδ)ρ(rt)exp(Nδ(rt))u1(r)u2(r)L2(Ω)Lf(Nδ)ρ(rt)exp(Nδ(rt))u1(r)u2(r)L2NN4ρ(Ω), (3.48)

    and by prove similarly, we obtain

    D2(r,t)(Nδ)ρexp(Nδ(rt))G(u1(r))G(u2(r))L2(Ω)Lf(Nδ)ρexp(Nδ(rt))u1(r)u2(r)L2NN4ρ(Ω). (3.49)

    Combining (3.40) to (3.49), we have

    exp(m(tT))(Mu1(t)Mu2(t))Hρ(Ω)Lf(Nδ)ρTtexp(m(tT))(rt)2exp(2Nδ(rt))u1(r)u2(r)L2NN4ρ(Ω)dr+Lf(Nδ)ρTtexp(m(tT))exp(2Nδ(rt))u1(r)u2(r)L2NN4ρ(Ω). (3.50)

    From (3.50), make a simple transformation, we get

    exp(m(tT))(Mu1(t)Mu2(t))Hρ(Ω)Lf(Nδ)ρTtexp(m(tr))(rt)exp(Nδ(rt))exp(m(rT))u1(r)u2(r)L2NN4ρ(Ω)dr+Lf(Nδ)ρTtexp(m(tr))exp(Nδ(rt))exp(m(rT))u1(r)u2(r)L2NN4ρ(Ω). (3.51)

    Using the fact that, we get

    u1u2m:=esssup0rTexp(m(rT))u1(r)u2(r)L2NN4ρ(Ω). (3.52)

    From (3.47) and (3.52), we follow that

    exp(m(tT))(Mu1(t)Mu2(t))Hρ(Ω)Lf(Nδ)ρ(Ttexp(m(tr))(rt)exp(Nδ(rt))dr)u1u2m+Lf(Nδ)ρ(Ttexp(m(tr))exp(Nδ(rt))dr)u1u2m. (3.53)

    Form (3.53), we can see that

    (Ttexp(m(tr))(rt)exp(Nδ(rt))dr)TTtexp((Nδm)(rt))drTmNδ,(Ttexp(m(tr))exp(Nδ(rt))dr)1mNδ. (3.54)

    We combine estimation from (3.53) to (3.54), and Sobolev embedding as in Hρ(Ω)L2NN4ρ(Ω), one obtains

    exp(m(tT))(Mu1(t)Mu2(t))L2NN4ρ(Ω)exp(m(tT))(Mu1(t)Mu2(t))Hρ(Ω)max{T,1}Lf(Nδ)ρmNδu1u2m. (3.55)

    Therefore, we have

    esssup0tTexp(m(tT))(Mu1(t)Mu2(t))L2NN4ρ(Ω)max{T,1}Lf(Nδ)ρmNδu1u2m. (3.56)

    For any u1,u2Lm(0,T;L2NN4ρ(Ω)), we conclude that

    Mu1Mu2mmax{T,1}Lf(Nδ)ρmNδu1u2m. (3.57)

    From (3.57), if we choose mNδ+2Lf(Nδ)ρmax{T,1}Nδ, we can see that M is a contraction mapping from Lm(0,T;L2NN4ρ(Ω))Lm(0,T;L2NN4ρ(Ω)), we conclude that M has a fixed point ˜VδNδLm(0,T;L2NN4ρ(Ω)) throught the Banach fixed point. For the estimation (3.57), let assume that u1=˜VδNδ and u2=0, and we denote ˜VδNδ=M(˜VδNδ), and Mu2=P1(Tt)TNδfδ+P2(Tt)TNδgδ, we get

    ˜VδNδm=M(˜VδNδ)mM(˜VδNδ)P1(Tt)TNδfδP2(Tt)TNδgδm+P1(Tt)TNδfδ+P2(Tt)TNδgδm12˜VδNδm+(2(Nδ)ρN(2q)4qexp(Nδ(Tt))fδLq(Ω)+2(Tt)(Nδ)ρN(2q)4qexp(Nδ(Tt))gδLq(Ω)). (3.58)

    Using the estimation (3.44), thus, we obtain that

    ˜VδNδm2(2(Nδ)ρN(2q)4qexp(Nδ(Tt))fδLq(Ω)+2(Tt)(Nδ)ρN(2q)4qexp(Nδ(Tt))gδLq(Ω)). (3.59)

    In the theory below, we are going to show the error estimate between the regularized and exact solutions in the space of Lq(Ω) type.

    Theorem 3.3. Assume that problem (1.1) under the condition (1.2) has a unique solution uL(0,T;Hn+ρ(Ω)) for any n>0 and 0<ρ<N4. In addition, we assume that there exists a positive constant M such that

    M=esssup0tT(exp(2tλk)λ2ςk|u(t),ek|2)12, (3.60)

    where ς>0, taking the noisy data fδ,gδLq(Ω) such that

    fδfLq(Ω)+gδgLq(Ω)δ,1<q<. (3.61)

    By choosing Nδ such that

    limδ0Nδ=,limϵ0(Nδ)ρN(2q)4qexp(NδT)δ=0. (3.62)

    Specifically in this section, we choose Nδ as follows:

    Nδ=(T1γ)2[log(δ1)]2for any0<γ<1, (3.63)

    then we have

    ˜VδNδ(,t)u(,t)L2NN4ρ(Ω) is of order. (3.64)
    max{(log(δ1))2n,(log(δ1))2ρ2ς,(log(δ)1)2ρN(2q)2qδγ}. (3.65)

    Proof. In this provision, we provide the upper bound for the term ˜VδNδ(,t)u(,)Hρ(Ω), using the triangle inequality, then we get

    ˜VδNδ(,t)u(,t)L2(Ω)˜VδNδ(,t)˜VNδ(,t)L2(Ω)+˜VNδ(,t)u(,t)L2(Ω). (3.66)

    From definition (3.33), we can know that

    ˜VNδ(t)=P1(Tt)TNδf+P2(Tt)TNδg+TtP3(rt)TNδG(˜VNδ(r))dr+TtP4(rt)TNδG(˜VNδ(r))dr. (3.67)

    From (3.33) and (3.67), we see that

    ˜VNδ(t)˜VδNδ(t)=P1(Tt)TNδ(fδf)+P2(Tt)TNδ(gδg)+TtP3(rt)[TNδG(˜VδNδ(r))TNδG(˜VNδ(r))]dr+TtP4(rt)[TNδG(˜VδNδ(r))TNδG(˜VNδ(r))]dr. (3.68)

    Since (3.66) and (3.68), we receive

    ˜VδNδ(,t)u(,t)L2(Ω)u(,t)˜VNδ(,t)L2(Ω)B1+P1(Tt)TNδ(fδf)L2(Ω)+P2(Tt)TNδ(gδg)L2(Ω)B2+TtP3(rt)[TNδG(˜VδNδ(r))TNδG(˜VNδ(r))]drL2(Ω)B3+TtP4(rt)[TNδG(˜VδNδ(r))TNδG(˜VNδ(r))]drL2(Ω)B4. (3.69)

    We rated ˜VδNδ(,t)u(,t)L2(Ω) through four steps as follows:

    Step 1: Estimate of B1, we have

    B1u(,t)˜VNδ(,t)L2(Ω)=(k=Nδ+1exp(2tλk)λ2ςkexp(2tλk)λ2ςk|u(t),ek|2)12(Nδ)ςexp(tNδ)M. (3.70)

    Step 2: Estimate of B2, for 1<q<2, we follow Lemma 2.1 in combination wiith the Sobolev embedding, one has

    Lq(Ω)HN(2q)4q(Ω). (3.71)

    This implies that

    B2P1(Tt)TNδ(fδf)L2(Ω)+P2(Tt)TNδ(gδg)L2(Ω)2(Nδ)N(2q)4qexp(Nδ(Tt))fδfHN(2q)4q(Ω)+2(Tt)(Nδ)N(2q)4qexp(Nδ(Tt))gδgHN(2q)4q(Ω)2C10(N,q)(Nδ)N(2q)4qexp(Nδ(Tt))fδfLq(Ω)+2C10(N,q)(Tt)(Nδ)N(2q)4qexp(Nδ(Tt))gδgLq(Ω). (3.72)

    Step 3: Estimate of B3, by using the similar argument as in (3.48), we can find that

    B3LfTt(rt)exp(Nδ(rt))VδNδ(r)VNδ(r)L2(Ω)dr. (3.73)

    Step 4: Estimate of B4, by using the similar argument as in (3.49), one obtains

    B4LfTtexp(Nδ(rt))VδNδ(r)VNδ(r)L2(Ω). (3.74)

    Combining (3.66), estimation of Steps 1–4, we conclude that

    ˜VδNδ(,t)u(,t)L2(Ω)(Nδ)ςexp(tNδ)M+22C10(N,q)(Nδ)N(2q)4qexp(Nδ(Tt))δ+LfTt(rt)exp(Nδ(rt))VδNδ(r)VNδ(r)L2(Ω)dr+LfTtexp(Nδ(rt))VδNδ(r)VNδ(r)L2(Ω). (3.75)

    Multiplying both sides by exp(tNδ), we have

    exp(tNδ)˜VδNδ(,t)u(,t)L2(Ω)(Nδ)ςM+22C10(N,q)(Nδ)N(2q)4qexp(NδT)δ+2Lfmax{T,1}Ttexp(Nδr)VδNδ(r)VNδ(r)L2(Ω)dr. (3.76)

    Applying the Grönwall's inequality, we get

    exp(tNδ)˜VδNδ(,t)u(,t)L2(Ω)((Nδ)ςM+22C10(N,q)(Nδ)N(2q)4qexp(NδT)δ)exp(2Lfmax{T,1}(Tt)). (3.77)

    This implies that

    ˜VδNδ(,t)u(,t)L2(Ω)exp(tNδ)((Nδ)ςM+22C10(N,q)(Nδ)N(2q)4qexp(NδT)δ)×exp(2Lfmax{T,1}(Tt)). (3.78)

    Next, we have the following inequality:

    ˜VδNδ(,t)u(,t)L2NN4ρ(Ω)C11(N,ρ)˜VδNδ(,t)u(,t)Hρ(Ω). (3.79)

    It is easy to see that

    ˜VδNδ(,t)u(,t)Hρ(Ω)˜VδNδ(,t)˜VNδ(,t)Hρ(Ω)E1+˜VNδ(,t)u(,t)Hρ(Ω)E2. (3.80)

    Now, for 0<ς<ρ, we immediately have a rating the first term of (3.80) as follows:

    E1(Nδk=1λ2ρk|˜VδNδ(,t)˜VNδ(,t),ek|2)12(Nδ)ρ˜VδNδ(,t)˜VNδ(,t)L2(Ω)exp(tNδ)((Nδ)ρςM+22C10(N,q)(Nδ)ρN(2q)4qexp(NδT)δ)×exp(2Lfmax{T,1}(Tt)). (3.81)

    E2 can be bounded as follows:

    E2(k=Nδλ2nkλ2n+2ρk|u(,t),ek|2)12(Nδ)nuL(0,T;Hn+ρ(Ω)). (3.82)

    From the observation above, we have

    ˜VδNδ(,t)u(,t)L2NN4ρ(Ω)C11(N,ρ)˜VδNδ(,t)u(,t)Hρ(Ω)C11(N,ρ)[(Nδ)nuL(0,T;Hn+ρ(Ω))+exp(tNδ)((Nδ)ρςM+22C10(N,q)(Nδ)ρN(2q)4qexp(NδT)δ)exp(2Lfmax{T,1}(Tt))]. (3.83)

    In this section, we carry out a numerical example in order to verify our proposed theory. In other words, we consider the stable property of the regularized solution based on the Fourier truncation method. First of all, we introduce some definitions to support the numerical implementation as follows. By choosing Ω×(0,T):=(0,π)×(0,1), we have the eigenvalues λk and the corresponding eigenvector ek which are the complete orthonormal system of eigenfunctions forming an orthogonal basis such that Δek=λkek and ek|Ω=0 for kN. Here, we choose λk=k2π2,ek(x)=2sin(kπx). We are going to find a function X satisfied

    Δ2uutttt+2uttxx+uxxxx=G(x,t,u(x,t)),(x,t)(0,1)×(0,1], (4.1)

    under the boundary conditions

    u(x,t)=Δu(x,t)=0,(x,t){0,1}×(0,1), (4.2)

    and the boundary conditions at T=1 as

    {u(x,1)=f(x),ut(x,1)=0,x(0,1),[0.4cm]Δu(x,1)=g(x),Δut(x,1)=0,x(0,1). (4.3)

    Finally, we use the finite difference method with the following partitions of temporal and spatial variable. For x[0,1] and t[0,1], let us consider the partitions DΩ×DT as follows:

    DΩ:={x1=0,x2=1NΩ,x3=2NΩ,,xi=i1NΩ,xNΩ=1, for i=1,2,...,NΩ,NΩ+1},DT:={t1=0,t2=1NT,t3=2Nt,,tj=j1NT,tNT=1, for j=1,2,...,NT,NT+1}.

    In Python software, the solutions can be re-write by the following form matrix:

    [u(x1,t1)u(x1,t2)u(x1,t3)u(x1,tNT+1)u(x2,t1)u(x2,t2)u(x2,t3)u(x2,tNT+1)u(x3,t1)u(x3,t2)u(x3,t3)u(x3,tNT+1)u(xNΩ+1,t1)u(xNΩ+1,t1)u(xNΩ+1,t2)u(xNΩ+1,tNT+1)](NΩ+1)×(NT+1)

    In this example, by choosing the solution u(x,t)=cosh(1t)sin(πx) to test the proposed results with the following input data:

    {G(x,t)=(π42π2+1)cosh(1t)sin(πx),(x,t)(0,1)×(0,1),f(x)=sin(πx),x(0,1),g(x)=π2sin(πx),x(0,1). (4.4)

    During the measurement of electromagnetic fields in applications of environmental and geophysical imaging, the exact data is approximated by the function fδ,gδ,Gδ as follows:

    fδfLq(Ω)+gδgLq(Ω)+GδGL(0,T;Lq(Ω))δ, (4.5)

    where

    {Gδ()=G()+δrand()/3,fδ()=f()+δrand()/3,gδ()=g()+δrand()/3. (4.6)

    By applying the Fourier truncation method, we provide a regularization solution as follows:

    uNtrδ(x,t)=Ntrk=1cosh(λk(Tt))fδ,ekek(x)+Ntrk=1(Tt)sinh(λk(Tt))2λkgδ,ekek(x)+Ntrk=1(Tt(rt)cosh(λk(rt))2λkGδ(,t),ekdr)ek(x)+Ntrk=1(Ttsinh(λk(rt))2λkλkGδ(,t),ekdr)ek(x), (4.7)

    where Ntr is a parameter regularization.

    We use the following estimation to evaluate the error between the regularized and exact solution at a certain time t:

    ENtrδ(t)=NΩ+1i=1|uNtrδ(xi,t)u(xi,t)|2NΩ+1.

    Table 1 and Figures 15 show the error estimate between the exact and regularized solutions at five observation times t{0.1,0.3,0.5,0.7,0.9} with δ{0.5,0.05,0.005}, respectively. Figure 6 presents the 3D graphs of the exact and regularized solutions, it shows they are quite similar. Overall, it shows that the error becomes smaller as the noise δ is decreased. It also shows the regularized solution is converged to the exact solution in this example.

    Table 1.  The error estimation between the regularized and exact solutions with NΩ=100,NT=100.
    t δ1=0.5 δ2=0.05 δ3=0.005
    0.1 0.09600230092017747 0.06839424680602106 0.035043593672148474
    0.3 0.09297773184568403 0.06548087722374361 0.030044924035854536
    0.5 0.08631247486802868 0.05829101885421206 0.031096834115623428
    0.7 0.08256109054607924 0.04950471042879639 0.028282438899535974
    0.9 0.08226034798718103 0.03610621656184547 0.024246403928042365

     | Show Table
    DownLoad: CSV
    Figure 1.  The solutions at t=0.1 and the error between exact and regularized solutions.
    Figure 2.  The solutions at t=0.3 and the error between exact and regularized solutions.
    Figure 3.  The solutions at t=0.5 and the error between exact and regularized solutions.
    Figure 4.  The solutions at t=0.7 and the error between exact and regularized solutions.
    Figure 5.  The solutions at t=0.9 and the error between exact and regularized solutions.
    Figure 6.  The 3D solutions on DΩ×DT(0,1)×(0,1) for δ=0.005.

    In this study, we focused on the final value problem of an inverse problem for both linear and nonlinear bi-harmonic equations. The regularized method for the biharmonic equation is proposed using the Fourier series truncation method and the terminal input data in Lq(Ω) for q2. The error between the exact and regularized solutions is estimated in Lq(Ω) using the embedding between Lq(Ω) and Hilbert scale spaces Hρ(Ω). The proposoed method has been verified by a numerical example; wherein, the regularized solution is well converged to the exact solution. It shows that our proposal method is capable of solving the final value problem of an inverse problem for both linear and nonlinear biharmonic equations.

    The authors would like to thank for the support from the National Research Foundation of Korea under grant number NRF-2020K1A3A1A05101625, and from the Institute of Construction and Environmental Engineering at Seoul National University.

    The authors declare no conflicts of interest.



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