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Determine unknown source problem for time fractional pseudo-parabolic equation with Atangana-Baleanu Caputo fractional derivative

  • In this paper, we consider a pseudo-parabolic equation with the Atangana-Baleanu Caputo fractional derivative. Our main tool here is using fundamental tools, namely the Fractional Tikhonov method and the generalized Tikhonov method, the error estimate is shown. Finally, we provided numerical experiments to prove the correctness of our theory.

    Citation: Nguyen Duc Phuong, Le Dinh Long, Devender Kumar, Ho Duy Binh. Determine unknown source problem for time fractional pseudo-parabolic equation with Atangana-Baleanu Caputo fractional derivative[J]. AIMS Mathematics, 2022, 7(9): 16147-16170. doi: 10.3934/math.2022883

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  • In this paper, we consider a pseudo-parabolic equation with the Atangana-Baleanu Caputo fractional derivative. Our main tool here is using fundamental tools, namely the Fractional Tikhonov method and the generalized Tikhonov method, the error estimate is shown. Finally, we provided numerical experiments to prove the correctness of our theory.



    The fractional diffusion equation has been of significant interest for many decades and has many important applications in practice, where mathematical formulas are abstract and can be used to provide procedures for converting outside measurements into information about inside properties thus increasingly interested in the study of inverse problems and unknown sources for partial differential equations [1,2,3,4,5,6,7,8,9,10,11]. So, PDEs with fractional derivatives are a generalization of equations with integer-order partial derivatives, please see in the reference [12,13,14,15,16,17,18,19,20,21,22]. In this paper, we consider time-fractional PDE

    {ABC0Dαt(u(t,x)+kAu(t,x))+Au(t,x)=θ(t)f(x),in(0,T]×D,u(t,x)=0,on(0,T]×D,u(T,x)=g(x),inD, (1.1)

    Here the Atangana-Baleanu Caputo derivative ABC0Dαtu(t,x) is defined by

    ABC0Dαtu(t,x)=L(α)1αt0u(s,x)sEα,1(α(ts)α1α)ds, (1.2)

    where L(α) satisfies L(α)=1α+αΓ(α), and L(0)=L(1)=1, and Eα,1(α(ts)α1α) is the Mittag-Leffler function is introduced in Section 2. The problem (1.1) can be considered in the following two cases:

    The case α=1 : System (1.1) is called pseudo-parabolic. There are many works on the well-posedness of the pseudo-parabolic equation with classical derivative, see [23,24,25,26], here the condition u(T,x) is replaced by the initial condition u(0,x). An important goal in the scientific community is to investigate the existence, uniqueness, and stability of fractional differential equations (1.1).

    The case α1: System (1.1) is called the fractional pseudo-parabolic equation. From the given data θ and the measured data at the final time gL2(Ω), we recovered the source function f(x). The problem (1.1) is severely ill-posed. The problem (1.1) is called the classical pseudo-parabolic equation. The problem (1.1) is considered with the right hand side of function F(t,x)=θ(t)f(x). In this case, we would like to take a look at some related issues as follows:

    ● In case θ(t)=1,k=0 and Dαt is the Caputo derivative, then (1.1) is called the fractional diffusion equation, and this type has been surveyed a lot, see in [27,28,29].

    ● In case θ(t)1,k=0, and Dαt is the Caputo derivative, there are a number of studies on this case, see in [30,31].

    ● In case θ(t)1,k0, with Dαt is Atangana - Baleanu fractional derivative, until now, according to our understanding, we did not find papers to solve this case.

    Our problem recover of a space-dependent source function f(x) along with subject to the homogeneous initial condition. When exact data (θ,g) is given, observed data (θϵ,gϵ) will inevitably be noisy by

    θθϵL(0,T)+ggϵL2(D)ϵ. (1.3)

    Regarding the regularization methods, in [32], Ting Wei used a quasi boundary value method. In [33], Tuan-Long-Thinh used the Tikhonov regularization method. In [34], with a general filter method, the authors studied the problem of finding the source distribution for the linear bi-parabolic equation when we have the final observation. In [35], Zhang with the truncation regularization method. In [36], Yang and his group used fractional Tikhonov to identify the initial value problem for a time-fractional diffusion equation. In [37], Cheng and co-authors used iteration regularization to solve a time-fractional inverse diffusion problem in the frequency domain. The Landweber regularization method, see in [38,39]. In [40], Binh and co-authors studied an inverse source problem for the Rayleigh–Stokes problem using the Tikhonov method. In [41], Ma and his group identified the unknown space-dependent source term in a time-fractional diffusion equation by applying the generalized and revised generalized Tikhonov regularization methods.

    However, to the best acknowledgment, the Tikhonov regularization method is very famous. Over the past decade, some modified Tikhonov methods have begun to be researched by mathematicians, such as the Fractional Tikhonov regularization method used in this article was considered by Smina Djennadi and his colleagues, see [42], and the generalized Tikhonov method can be found in reference [41]. For the reader's convenience, we would like to outline the main results:

    ● We give the ill-posedness of our inverse source problem.

    ● The first goal of this paper is to provide the fractional Tikhonov method to solve this inverse space-dependent source problem for the fractional pseudo-parabolic equation.

    ● The second goal of this paper is to provide the A generalized Tikhonov to solve the (1.1).

    ● And then finally, we provide an example of the illustration of the correctness of our theory and conclude this article.

    The layout of this article is as follows. We show the preliminaries and Lemma is used. In Section 3, we have the mild solution of Problem (1.1) and the condition stability for Problem (1.1) in Theorem 3.1. In Sections 4 and 5, we have the error estimate between the exact solution and its approximation by the Fractional Tikhonov and The generalized Tikhonov method. In Section 6, we have one example numerical experiment.

    Let us recall that the spectral problem

    {Δen(x)=λnen(x),inΩ,en(x)=0,onΩ, (2.1)

    admits a family of eigenvalues

    0<λ1λ2λ3...λn....

    For all r0, we define by Ar the following operator

    Arh=n=1h,φnλrnen,hH(Ar)={hL2(D): n=1|h,en|2λ2rn<}. (2.2)

    The domain H(Ar) is the Banach space equipped with the norm

    hH(Ar)=(n=1|h,en|2λ2rn)12,hH(Ar).

    Definition 2.1 ([43]). The definition of the Mittag-Leffler function as follows

    Eα,β(z)=n=1znΓ(αn+β),zC, (2.3)

    where α,β are arbitrary constants.

    Lemma 2.1. ([43]) Let 0<α<1. Then there exist positive constants A1,A2,A3 such that

    A11+yEα,1(y)A21+y,Eα,α(y)A31+y,forally0,αR. (2.4)

    Lemma 2.2. For λ>0,α>0 and mN, then

    dmdtmEα,1(λtα)=λtαmEα,αm+1(λtα), (2.5)
    ddt(tEα,2(λtα))=Eα,1(λtα), (2.6)
    ddt(tα1Eα,α(λtα))=tα2Eα,α1(λtα). (2.7)

    Lemma 2.3. [33] For t>0, and λ>0, and 0<α<1, we get

    αtEα,1(λtα)=λEα,1(λtα),

    Lemma 2.4. For α(0,1), we put Dα(λn,k)=ξnαL(α)(L(α)+λnξn(1α))2, it gives

    Dα(λn,k)1λnkαL(α)(L(α)λ1ξ1+(1α))2 (2.8)

    Proof. First of all, we notice that λnξn=λn1+kλn=λnλn(1λn+k)11λ1+k=1λ11+k, and we have

    Dα(λn,k)=ξnαL(α)(L(α)+λnξn(1α))2αL(α)λ2n1+kλn(L(α)λnξn+(1α))2=kαL(α)λn(L(α)λnξn+(1α))21λnkαL(α)(L(α)λ1ξ1+(1α))2 (2.9)

    Lemma 2.5. Let α(0,1), by denoting

    Hα(λn,k)=αλnξn(L(α)+λnξn(1α))1,Gα(λn,k,s)=Eα,α(Hα(λn,k)(ts)α)(ts)α1. (2.10)

    and ξn=(1+kλn)1, then we get the following estimate

    (1αα)(TA2Hα(λ1,k)T1α1α)T0Gα(λn,k,s)ds1Hα(λn,k). (2.11)

    Proof. For Eα,α(y)0 for 0<α<1 and y0, and using Lemma 2.1, we obtain

    a)T0Gα(λn,k,s)ds1Hα(λn,k)T0(1Eα,1(Hα(λn,k)tα))dt(1αα)(T0dtT0Eα,1(Hα(λn,k)tα))dt)(1αα)(TA2Hα(λ1,k)T0tαdt)(1αα)(TA2Hα(λ1,k)T1α1α), (2.12)
    b)T0Gα(λn,k,s)ds=1Hα(λn,k)T0dds(Eα,1(Hα(λn,k)(ts)α))ds=1Hα(λn,k)(Hα(λn,k))1(1Eα,1(Hα(λn,k)tα))1Hα(λn,k) (2.13)

    Lemma 2.6 ([42]). For constant N>0, sλ1, and 12σ<1, then we get

    sN2σ+[γ(ϵ)]s2σC2[γ(ϵ)]12σ, (2.14)

    where C2=C2(σ,N)>0 is independent on s and σ.

    Lemma 2.7 ([42]). For constant sλ1, with 12σ<1, we obtain

    [γ(ϵ)]s(2σr)N2σ+s2σ[γ(ϵ)]{C4(σ,r,N)[γ(ϵ)]r2σ,0<r2σ,C5(σ,r,λ1)[γ(ϵ)]r>2σ. (2.15)

    where C4=C4(σ,r,N) and C5=C5(σ,r,N,λ1).

    Lemma 2.8 ([42]). For constant sλ1>0, and 12σ<1, we get

    [γ(ϵ)]s(2σ1r)N2σ+s2σ[γ(ϵ)]{C6(σ,r,N)[γ(ϵ)]r+12σ,0<r2σ1,C7(σ,r,N,λ1)[γ(ϵ)]r>2σ1. (2.16)

    Lemma 2.9. Let θ0,θ1 are positive constants such that θ0<θ<θ1. By choosing ϵ(0,θ14), and B(θ0,θ1)=θ1+θ04, we obtain

    θ04|θϵ(t)|B(θ0,θ1). (2.17)

    Proof. This provision was detailed in [5]. We omit it here.

    In this section, we consider a linear problem as follows

    {ABC0Dα(u(t,x)+kAu(t,x))+Au(t,x)=F(t,x),in(0,T]×D,u(t,x)=0,on(0,T]×D,u(0,x)=u0(x),inD, (3.1)

    where u0 and F are given functions. Let u(t,x)=n=1un(t)en(x) be the Fourier series in L2(Ω) with un(t)=Du(t,x)en(x)dx, then we have the fractional integro-differential equation involving the Atangana-Baleanu fractional derivative in the form

    ABC0Dαt(1+kλn)un(t)+λnun(t)=Fn(t), (3.2)

    along with the following condition un(0)=Du(0,x)en(x)dx, and Fn(0)=λnun(0), using Lemma 3.4 in [43], the solution (3.1) is written by Fourier series as follows

    un(t)=L(α)L(α)+λn1+kλn(1α)Eα,1(αλn1+kλntαL(α)+λn1+kλn(1α))u0,n+(11+kλn)1αL(α)+λn1+kλn(1α)Fn(t)+(11+kλn)αL(α)(L(α)+λn1+kλn(1α))2t0Eα,α(αλn1+kλn(ts)αL(α)+λn1+kλn(1α))(ts)α1Fn(s)ds. (3.3)

    From now on, for a shorter, we put ξn=(1+kλn)1, this implies that

    u(t,x)=n=1L(α)L(α)+λnξn(1α)Eα,1(αλnξntαL(α)+λnξn(1α))(Du(0,x)en(x)dx)en(x),+n=1ξn(1α)L(α)+λnξn(1α)(DFn(t)en(x)dx)en(x)+n=1ξnαL(α)(L(α)+λnξn(1α))2t0Eα,α(αλnξn(ts)αL(α)+λnξn(1α))(ts)α1(DFn(s)en(x)dx)dsen(x). (3.4)

    Let t=T and noting that u(0,x)=0, F(t,x)=θ(t)f(x), and θ(T)=0, we find that

    Dg(x)en(x)dx=(Df(x)en(x)dx)ξnαL(α)(L(α)+λnξn(1α))2T0Eα,α(αλnξn(Ts)αL(α)+λnξn(1α))(Ts)α1θ(s)ds. (3.5)

    Let us denote and (2.10), we have

    Dα(λn,k)=ξnαL(α)(L(α)+λnξn(1α))2,Hα(λn,k)=αλnξn(L(α)+λnξn(1α))1. (3.6)

    From (3.5) and (3.6), this leads to

    Dg(x)en(x)dx=(Df(x)en(x)dx)Dα(n,k)T0Gα(λn,k,s)θ(s)ds, (3.7)

    and we can rewrite the formula (3.8) as follows:

    (Df(x)en(x)dx)=(Dg(x)en(x)dx)(Dα(λn,k)T0Gα(λn,k,s)θ(s)ds)1 (3.8)

    and

    f(x)=n=1(Dα(λn,k)T0Gα(λn,k,s)θ(s)ds)1(Dg(x)en(x)dx)en(x). (3.9)

    with Dα(λn,k) are defined as (3.6).

    Theorem 3.1. The inverse problem (1.1) is ill-posed.

    Proof. A linear operator K:L2(D)L2(D) as follows.

    Kf(x)=n=1(Dα(λn,k)T0Gα(λn,k,s)θ(s)ds)(Df(x)en(x)dx)en(x)=Dp(x,ξ)f(ξ)dξ. (3.10)

    where

    p(x,ξ)=n=1(Dα(λn,k)T0Gα(λn,k,s)θ(s)ds)en(x)en(ξ).

    It is obvious that p(x,ξ)=p(ξ,x), we know that K is self-adjoint operator. Next, its compactness is considered and the finite rank operators as follows:

    KNf(x)=nN(Dα(λn,k)T0Gα(λn,k,s)θ(s)ds)(Df(x)en(x)dx)en(x). (3.11)

    From (3.8), we find that the problem of seeking f(x) can be transformed into the following operator equation:

    (Kf)(x)=g(x)=nN(Dα(λn,k)T0Gα(λn,k,s)θ(s)ds)(Df(x)en(x)dx)en(x). (3.12)

    where KN:fg is a linear operator. To prove that KN is a compact operator, defining KN the finite rank operator as

    (KNf)(x)=nN(Dα(λn,k)T0Gα(λn,k,s)θ(s)ds)(Df(x)en(x)dx)en(x). (3.13)

    Using the estimate of Lemma 2.5, we notice that Dα,β(λn,k)T0Gα,β(λn,k,s)θ(s)dsθ1λn

    So, the KNfKfL2(D) can be bounded as follows:

    KNfKfL2(D)nNθ1λn|Df(x)en(x)dx|2θ21N2f2L2(D) (3.14)

    Hence, KNfKfL2(D)0 in the sense of L(L2(D),L2(D))as N and K is a linear operator. Next, we show one example to illustrate the non-well posed of problem (1.1). Considering the input final data gm(x)=em(x)(λm)12 and gm=0 and the corresponding source solution fm(x)=em(x)(λmDα(λm,k)T0Gα(λm,k,s)θ(s)ds)1 and f(x)=0. The error estimate in L2 between gm and g is

    gmgL2(D)=1λm0,asm. (3.15)

    Applying Lemma 2.5, then the corresponding error between source fm and f is

    fmfL2(D)λmθ1,asm. (3.16)

    From the condition (3.15) and (3.16), we conclude that the problem (3.13) is non-well posed.

    Theorem 3.2. Let r>0, and fH(Ar(D)) such that

    fH(Ar(D))E,E>0. (3.17)

    Then we have

    fL2(D)|Vα(λ1,k,T,A2,θ0)|rr+1E1r+1grr+1. (3.18)

    where

    Vα(λ1,k,T,A2,θ0)=θ04kL(α)(L(α)λ1ξ1+(1α))2(T(1α)A2T1αHα(λ1,k)) (3.19)

    Proof. This theorem can refer similarly in [43]. We omit here.

    To get the regularized solution with the given data gϵ(x)L2(D), minimizing the quantity Jσ[γ(ϵ)](f)L2(D) such that

    Jσγ(ϵ)(f)=Kfg2σ+[γ(ϵ)]f2L2(D), (4.1)

    whereby [γ(ϵ)] is a regularization parameter, and .σ is a weighted seminorm defined as ξσ=W12ξL2(D) for any ξ with W=(KK)σ1 some of parameter 12σ<1 called the fractional parameter. In here, we have some reviews as follows:

    ● In case σ=12, it is the quasi-boundary value method, interested readers can find a view in the document [42].

    ● In case σ=1, it is classical Tikhonov regularization method, this method is fully described in the following documents [33].

    ● For 12<σ<1, it is the new fractional Tikhonov regularization.

    The minimizer f satisfy the norm equation

    ((KK)[γ(ϵ)]+σI)fσ=(KK)σ1Kg. (4.2)

    Using the singular decomposition for the compact self-adjoint operator, we get

    fσγ(ϵ)(x)=n=1|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|2σ1|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|2σ+[γ(ϵ)](Dg(x)en(x)dx)en(x),12σ<1. (4.3)

    For the observed data, we have

    fσϵ,γ(ϵ)(x)=n=1|Dα(λn,k)T0Gα(λn,k,s)θϵ(s)ds|2σ1|Dα(λn,k)T0Gα(λn,k,s)θϵ(s)ds|2σ+[γ(ϵ)](Dgϵ(x)en(x)dx)en(x),12σ<1. (4.4)

    Next, with two formulas of regularized in (4.3) and (4.4), we consider the convergent rate of ffσε,γ(ε)L2(D), by discussing how to select the regularization parameter, with a-priori strategies.

    In this subpart, under a suitable choice for the regularization parameter [γ(ϵ)], we will give an error estimate for ffσϵ,γ(ϵ)L2(D).

    Lemma 4.1. Assume that the condition (1.3) holds, then we have

    fσγ(ϵ)fσϵ,γ(ϵ)L2(D)C2[γ(ϵ)]12σϵ+max{ϵ,[γ(ϵ)]112σ}(32θ20+2θ21λ2β1C22)12fL2(D). (4.5)

    Proof. To prove this Lemma, we have used two formulas (4.3), (4.4), for 12σ<1, we have

    fσγ(ϵ)(x)fσϵ,γ(ϵ)(x)=n=1|Dα(λn,k)T0Gα(λn,k,s)θϵ(s)ds|2σ1|Dα(λn,k)T0Gα(λn,k,s)θε(s)ds|2σ+[γ(ϵ)](D(gϵ(x)g(x))en(x)dx)en(x)E1+n=1[|Dα(λn,k)T0Gα(λn,k,s)θϵ(s)ds|2σ1|Dα(λn,k)T0Gα(λn,k,s)θϵ(s)ds|2σ+[γ(ϵ)]|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|2σ1|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|2σ+[γ(ϵ)]](Dg(x)en(x)dx)en(x).E2 (4.6)

    Now, we need establish the upper bound of fσγ(ϵ)fσε,γ(ε)L2(D). For convenience, we consider the following step.

    Step1:_ Estimate of E1, with V depends on α,λ1,k,T,A2 and gϵgL2(D)ϵ, applying Lemma 2.6 with s=λn, we get

    E21supn1(|B(θ0,θ1)|2σ1λn|Vα|2σ+λ2σn[γ(ϵ)])2ϵ2C22[γ(ϵ)]1σϵ2. (4.7)

    whereby Vα is defined as (3.19) and C2 depends on V,σ and B.

    Step2:_ Estimate of E2, applying the inequality (a+b)22a2+2b2, after simple transformation steps, we have

    E222([Dα(λn,k)T0Gα(λn,k,s)(θε(s)θ(s))ds|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds||Dα(λn,k)T0Gα(λn,k,s)θϵ(s)ds|])2(Dg(x)en(x)dx)2+2([γ(ϵ)]|Dα(λn,k)T0Gα(λn,k,s)θε(s)ds|2σ1|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|(|Dα(λn,k,s)T0Gα(λn,k,s)θε(s)ds|2σ+[γ(ϵ)]))2(Df(x)en(x)dx)2(32ϵ2θ20+2[γ(ϵ)]2θ21λ2β1(|Dα(λn,k)T0Gα(λn,k,s)θε(s)ds|2σ1|Dα(λn,k)T0Gα(λn,k,s)θε(s)ds|2σ+[γ(ϵ)])2)(Df(x)en(x)dx)2. (4.8)

    From (4.8), the right hand side can be bounded as follow:

    E22(32ε2θ20+2θ21λ2β1C22[γ(ϵ)]21σ)f2L2(D). (4.9)

    Combining (4.6) to (4.9), we have

    fσγ(ϵ)fσϵ,γ(ϵ)L2(D)C2[γ(ϵ)]12σϵ+(32ε2θ20+2θ21λ2β1C22[γ(ϵ)]21σ)12fL2(D). (4.10)

    Lemma 4.2. Suppose that fHr(D), then it gives

    fϵγ(ϵ)fL2(D){C4(σ,r,Vα)[γ(ϵ)]r2σE,0<r2σ,C5(σ,r,λ1)[γ(ϵ)]E,r>2σ. (4.11)

    Proof. From (3.8) and (4.3), we received

    fσγ(ϵ)f2L2(D)n=1([γ(ϵ)]|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|2σ+[γ(ϵ)])(Dg(x)en(x)dx)|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|2L2(D)n=1([γ(ϵ)]λrn|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|2σ+[γ(ε)])λrn(Df(x)en(x)dx)2L2(D)supn1([γ(ϵ)]λ2σrn|4Vα|2σ+λ2σn[γ(ϵ)])2fHr(D). (4.12)

    Due to Lemma 4.2, the upper bound of the right hand side of (4.12) as follows:

    ● If r(0,2σ), then we have

    fσγ(ϵ)fL2(D)C4(σ,r,V)[γ(ϵ)]r2σE. (4.13)

    ● If r2σ, then we have

    fσγ(ϵ)fL2(D)C5(σ,r,λ1)[γ(ϵ)]E. (4.14)

    Theorem 4.1. Assume that the noise assumption (1.3) are hold, and the a-priori bound condition (3.17). Then we have the following estimate:

    ● If r(0,2σ), by choosing [γ(ϵ)]=(ϵE)2σr+2, then we have

    ffσϵ,γ(ϵ)L2(D)max{ϵr+1r+2,ϵ,ϵ2σ1r+2,εrr+2}P1. (4.15)

    where P1 depends on C2,E,θ0,θ1,Vα,C4.

    ● If r2σ, by choosing [γ(ϵ)]=(ϵE)σσ+1, then we have

    ffσϵ,γ(ϵ)L2(D)max{ϵσσ+1,ϵ,ϵσ1/2σ+1,ϵσσ+1}P2. (4.16)

    and P2 depends on C2,E,θ0,θ1,λ1,C5.

    Proof. Using the triangle inequality, we get

    ffσϵ,γ(ϵ)L2(D)ffσγ(ϵ)L2(D)+fσγ(ϵ)fσϵ,γ(ϵ)L2(D) (4.17)

    From (4.3), (4.4), results from Lemma 4.1 and Lemma 4.2, Theorem 4.1 is proven.

    From [41], we have a stable solution to problem (1.1) with the observed data (gϵ,θϵ) by generalized Tihkhonov regularization method which minimizes the quantity

    J(fϵ,[γ(ϵ)])=Kfgϵ2L2(D)+[γ(ϵ)]f2Hr(D), (5.1)

    Let fϵ,[γ(ϵ)] be a solution of the problem (5.1) which satisfies

    KKfϵ,[γ(ϵ)]+[γ(ϵ)]λ2rnfϵ,[γ(ϵ)]=Kgϵ. (5.2)

    Using the SVD for compact self-adjoint operator, we get

    fϵ,[γ(ϵ)](x)=n=1|Dα(λn,k)T0Gα(λn,k,s)θϵ(s)ds||Dα(λn,k)T0Gα(λn,k,s)θϵ(s)ds|2+[γ(ϵ)]λ2rn(Dgϵ(x)en(x)dx)en(x), (5.3)

    and

    f[γ(ϵ)](x)=n=1|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds||Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|2+[γ(ϵ)]λ2rn(Dg(x)en(x)dx)en(x). (5.4)

    Lemma 5.1. Assume that fHr(D) holds, assume that the couple (g,θ) is noised by (gϵ,θϵ) such that the condition (1.3) satisfies, then we have

    f[γ(ϵ)]fϵ,[γ(ϵ)]L2(D)ϵ2[γ(ϵ)]λ2r1+5ϵθ0fL2(D). (5.5)

    Proof. From (5.3) and (5.4), we have

    fγ(ϵ)(x)fϵ,γ(ϵ)(x)Z1+Z2+Z3. (5.6)

    whereby Z1,Z2 and Z3 are defined in (5.7) as follows:

    (5.7)

    The proof falls naturally into three parts.

    Step1:_ Estimate of Z1, we have

    Z12L2(D)14[γ(ϵ)]n=11λ2rn(D(gϵ(x)g(x))en(x)dx)2ϵ24[γ(ϵ)]λ2r1 (5.8)

    Step2:_ Estimate of Z2, it gives

    (5.9)

    Step3:_ Estimate of Z3, we have

    Z32L2(D)n=1|Dα(λn,k)T0Gα(λn,k,s)(θ(s)θε(s))ds|2|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|2|Dα(λn,k)T0Gα(λn,k,s)θε(s)ds|2(Dg(x)en(x)dx)2n=1θθϵL(0,T)|Dα(λn,k)T0Gα(λn,k,s)ds|2|Dα(λn,k)T0Gα(λn,k,s)θε(s)ds|2(Dg(x)en(x)dx)2|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|216ϵ2θ20f2L2(D). (5.10)

    Combining (5.8)–(5.10), we get

    fϵfϵ,[γ(ϵ)]L2(D)ϵ2[γ(ϵ)]12λr1+5ϵθ0fL2(D). (5.11)

    Lemma 5.2. Suppose that fHr(D), then we have

    ● If r>1, then we have

    ffγ(ϵ)L2(D)R2(λ1,r,θ0,Vα,E)[γ(ϵ)]12. (5.12)

    ● If r(0,1), then we have

    ffγ(ϵ)L2(D)(R1(θ0,Vα,E))12[γ(ϵ)]rr+1. (5.13)

    whereby R1(θ0,Vα,E)=((2θ0Vα)2E2+E2) and R2(λ1,r,θ0,Vα,E)=λ(1r)12θ0VαE.

    Proof. From (5.4) and (3.8), we get:

    ff[γ(ϵ)]L2(D)=n=1[γ(ϵ)]λ2rn(Dg(x)en(x)dx)|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|[[γ(ϵ)]λ2rn+|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|2]=n=1[γ(ϵ)]λ2rn(Dg(x)en(x)dx)|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|[[γ(ϵ)]λ2rn+|Dα(λn,k)T0Gα(k,β,λn,s)θ(s)ds|2]supnN|V(n)|n=1λ2rn(Dg(x)en(x)dx)|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|=supnN|W(n)|fHr(D), (5.14)

    Hence, W(n) is given by

    V(n)=[γ(ε)][γ(ε)]λ2rn+|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|2 (5.15)

    Next, Estimate of V(n), we have

    W(n)[γ(ϵ)]2[γ(ϵ)]12λrn|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|[γ(ϵ)]12λ1rn2θ0Vα (5.16)

    We divide into two following cases

    ● If r1, then we have

    ff[γ(ϵ)]L2(D)[γ(ϵ)]12λ(1r)12θ0VαE. (5.17)

    ● If r(0,1), then we consider λ1rn satisfies the following two cases :

    λ1rn[γ(ε)]andλ1rn>[γ(ε)]. (5.18)

    then we rewrite N by N=Q1Q2 where

    Q1={nN,λ1rn[γ(ε)]},Q2={nN,λ1rn>[γ(ϵ)]}. (5.19)

    From (5.15) and (5.18), we have:

    ff[γ(ϵ)]2L2(D)=supnQ1[[γ(ϵ)]12λ1rn2θ0Vα]2nQ1λ2rn(Df(x)en(x)dx)2+nQ2[[γ(ε)][γ(ε)]λ2rn+|Dα(λn,k)T0Gα(λn,k,s)θ(s)ds|2]2λ2rn(Df(x)en(x)dx)2(2θ0Vα)2[γ(ϵ)]12f2Hr(D)+supnQ2λ4rnnQ2λ2rn(Df(x)en(x)dx)2(2θ0Vα)2[γ(ϵ)]12f2Hr(D)+[γ(ϵ)]4r1rf2Hr(D). (5.20)

    Choose =1r2(1+r), we have

    ff[γ(ϵ)]2L2(D)[γ(ϵ)]2rr+1((2θ0Vα)2E2+E2). (5.21)

    Theorem 5.1. Let θϵ,θL(0,T) such that θϵθL(0,T)ϵ and (θ,θε) satisfy the condition in Lemma (2.9), and the noise assumption (3.17) hold. By choosing the regularization parameter [γ(ϵ)]=(ϵE), we receive

    ● If r>1, then we have

    ffϵ,γ(ϵ)L2(D)ϵ12(R2(λ1,r,θ0,Vα)E12+E1212λr1+5ε12fL2(D)θ0). (5.22)

    ● If r(0,1), then we have

    ffϵ,γ(ϵ)L2(D)ϵrr+1(R1(θ0,Vα,E))12+E1r+112λr1+5ε1r+1fL2(D)θ0) (5.23)

    whereby R1(θ0,Vα,E)=((2θ0Vα)2E2+E2) and R2(λ1,r,θ0,Vα)=λ(1r)12θ0Vα.

    Proof. By the triangle inequality, we have

    ffϵ,γ(ϵ)L2(D)ffγ(ϵ)L2(D)+fγ(ϵ)fϵ,γ(ϵ)L2(D). (5.24)

    Combining Lemma 5.1 and Lemma 5.2, we have

    ● If r>1, then we have

    ffϵ,γ(ϵ)L2(D)is of orderε12. (5.25)

    ● If r(0,1), then we have

    ffϵ,γ(ϵ)L2(D)is of orderεrr+1. (5.26)

    This theorem is proven.

    In this section, we present one numerical example. By choosing Ω=(0,π), T=1, α=0.95, and σ=0.45, and r=0.25 k=0.01 are shown in this section, respectively. For computing the generalized Mittag-Leffler function and the accuracy control in computing is 1010, we have the Matlab codes given by Podlubny [44]. To do this, we consider the problem as follows:

    ABC0Dαt(u(t,x)+kΔu(t,x))+Δu(t,x)=θ(t)f(x),(x,t)(0,π)×(0,1). (6.1)

    where ABC0Dαtu(x,t) is the Atangana-Baleanu fractional derivative is given by (1.2). In this calculation, we chose the operator Δu=2x2u, we have chosen λn=n2,n=1,2,... and en(x) = 2πsin(nx), respectively. We have the function

    g(x)=2πsin(4x),θ(t)=((116k)Γ(4)t3αΓ(4α)16t3). (6.2)

    In general, the whole numerical procedure is summarized in the following steps:

    Step1:_ Finite difference to discretize the time and spatial variable for x(0,π) as follows:

    xk=kΔx,0kN,Δx=πN

    Step2:_ The input data (g,θ) is noised by observation data (gε,θε) such that

    θϵ=θ+1πϵ(2rand()1),gϵ=g+1πϵ(2rand()1). (6.3)

    Step3:_ The relative error estimation is defined by

    Error1=(Nn=1fσϵ,γ(ϵ)(xn)f(xn)2L2(0,π)Nn=1f(xn)2L2(0,π))12,Error2=(Nn=1fϵ,γ(ϵ)(xn)f(xn)2L2(0,π)Nn=1f(xn)2L2(0,π))12, (6.4)

    where Error1 is the error estimate between the exact solution and the regularized solution by the Fractional Tikhonov method, and Error2 is the error estimate between the exact solution and the regularized solution by generalized Tikhonov method. In addition, taking for the priori parameter choice rule of E is large enough. By [43], we get

    10xγ1(1x)β1Eα,β(z(1x)α)dx=Γ(β)Eα,α+β(z). (6.5)

    From (6.5), by replacing β=α, and z=αn21+kn2L(α)+n21+kn2(1α), we can find that

    10xγ1(1x)α1Eα,α(αn21+kn2L(α)+n21+kn2(1α))dx=Γ(γ)Eα,α+γ(αn21+kn2L(α)+n21+kn2(1α)) (6.6)

    From (3.9), we have the exact solution

    f(x)=2πNn=1(Dα(n2,k)T0Gα(n2,k,s)θ(s)ds)1(π0g(x)sin(nx))sin(nx). (6.7)

    From (4.4), by choosing the regularization parameter γ(ϵ)=(ϵE)12, with N is a truncation number large enough, we have the regularized solution with Fractional Tikhonov as follows

    fσϵ,γ(ϵ)(x)=2πNn=1|Dα(n2,k)10Gα(n2,k,s)θϵ(s)ds|2σ1|Dα(n2,k)10Gα(n2,k,s)θϵ(s)ds|2σ+(ϵE)12(π0gϵ(x)sin(nx)dx)sin(nx), (6.8)

    From (5.3), with the Generalized Tikhonov method, and γ(ϵ)=ϵE, we receive

    fϵ,[γ(ϵ)](x)=2πNn=1|Dα(n2,k)10Gα(n2,k,s)θϵ(s)ds||Dα(n2,k)10Gα(n2,k,s)θϵ(s)ds|2+(ϵE)n4r(π0gϵ(x)sin(nx)dx)sin(nx), (6.9)

    and Dα(n2,k) is defined in Lemma 2.4. From (6.6), the integral 10Gα(n2,k,s)θϵ(s)ds calculated

    10Gα(n2,k,s)θϵ(s)ds=n21+kn2αL(α)(L(α)+n21+kn2(1α))210Eα,α(αn21+kn2L(α)+n21+kn2(1α)(1s)α)(1s)α1θϵ(s)ds=(116k)Γ(4)Eα,4(αn21+kn2L(α)+n21+kn2(1α))16Γ(4)Eα,α+4(αn21+kn2L(α)+n21+kn2(1α))+1πϵ(2rand()1)L(α)+n21+kn2(1α)αn21+kn2(1Eα,1(αn21+kn2L(α)+n21+kn2(1α))). (6.10)

    In these calculations, we choose N=100 and ϵ=0.5, ϵ=0.25 and ϵ=0.125. Figure 1(a) shows the 2D graphs of the source function with the exact solution and its approximation for the case by the Fractional Tikhonov method. Figure 1(b) shows the error estimate between the exact solution and regularized solution for the case by the Fractional Tikhonov method. Figure 2(a) shows the 2D graphs of the source function with the exact solution and its approximation for the case by the generalized Tikhonov method. Figure 2(b) shows the error estimate between the exact solution and regularized solution for the case by the generalized Tikhonov method, respectively. Figure 3 shows the 2D graphs of the source function and its approximation, by two methods with ϵ=0.5, ϵ=0.25 and ϵ=0.125, respectively. The error estimates between the source function with the exact data and the measurement data are presented in Table 1. From the observations above, the approximation results are acceptable.

    Figure 1.  The exact solution and the regularized solutions are shown by the left graph and the graph of error with ϵ=0.5,ϵ=0.25 and ϵ=0.125 for the Fractional Tikhonov method and the error estimation.
    Figure 2.  The exact solution and the regularized solutions are shown by the left graph and the graph of error with ϵ=0.5,ϵ=0.25 and ϵ=0.125 for the Generalized Tikhonov method and the error estimation.
    Figure 3.  The exact solution and the regularized solution for two case : Fractional Tikhonov method and generalized Tikhonov method and the error estimation with ϵ=0.5,ϵ=0.25 and ϵ=0.125.
    Table 1.  The error estimation.
    α=0.95
    Err1 Err2
    ϵ=0.5 0.292922331 0.292928835
    ϵ=0.25 0.159947533 0.152469901
    ϵ=0.125 0.101037573 0.095171607

     | Show Table
    DownLoad: CSV

    In this study, we applied the Fractional Tikhonov method and the Generalized Tikhonov method to regularize the inverse problem to identify an unknown source term in fractional diffusion equations involving fractional derivative with Atangana-Baleanu Caputo derivative. This problem (1.1) is ill-posed in the sense of Hadamard. So, Under a priori parameter choice rule, we show the result for the convergent estimate between the sought solution and the regularized solution. Finally, we show an example to illustrate our proposed regularization.

    This research is supported by Industrial University of Ho Chi Minh City (IUH) under grant number 130/HD-DHCN. Le Dinh Long is thankful to Van Lang University.

    The authors declare no conflict of interest.



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