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On a final value problem for a nonlinear fractional pseudo-parabolic equation

  • Received: 01 May 2020 Revised: 01 July 2020 Published: 25 August 2020
  • 35K55, 35K70, 35K92, 47A52, 47J06

  • In this paper, we investigate a final boundary value problem for a class of fractional with parameter β pseudo-parabolic partial differential equations with nonlinear reaction term. For 0<β<1, the solution is regularity-loss, we establish the well-posedness of solutions. In the case that β>1, it has a feature of regularity-gain. Then, the instability of a mild solution is proved. We introduce two methods to regularize the problem. With the help of the modified Lavrentiev regularization method and Fourier truncated regularization method, we propose the regularized solutions in the cases of globally or locally Lipschitzian source term. Moreover, the error estimates is established.

    Citation: Vo Van Au, Hossein Jafari, Zakia Hammouch, Nguyen Huy Tuan. On a final value problem for a nonlinear fractional pseudo-parabolic equation[J]. Electronic Research Archive, 2021, 29(1): 1709-1734. doi: 10.3934/era.2020088

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  • In this paper, we investigate a final boundary value problem for a class of fractional with parameter β pseudo-parabolic partial differential equations with nonlinear reaction term. For 0<β<1, the solution is regularity-loss, we establish the well-posedness of solutions. In the case that β>1, it has a feature of regularity-gain. Then, the instability of a mild solution is proved. We introduce two methods to regularize the problem. With the help of the modified Lavrentiev regularization method and Fourier truncated regularization method, we propose the regularized solutions in the cases of globally or locally Lipschitzian source term. Moreover, the error estimates is established.



    We consider the final value problem:

    {utmΔut+(Δ)βu=f(t,x;u),in(0,T]×Ω,u(t,x)=0,on(0,T]×Ω,u(T,x)=φ(x),inΩ, (PT)

    where m>0, and ΩRd,(d1) is a bounded domain with smooth boundary Ω, the operator (Δ)β with is the fractional Laplace operator with 0<β1 and the final data φL2(Ω). Pseudo-parabolic equations have many applications in science and technology, especially in physical phenomena such as seepage of homogeneous fluids through a fissured rock, aggregation of populations, ... see e.g. Ting [24], R. Xu [17,31,32,33] and references therein.

    For β=1, the direct problem is

    utmΔutΔu=f(t,x;u),in(0,T]×Ω, (1.1)

    with conditions u(0,x)=u0(x),xΩ and u(t,x)=0,(t,x)(0,T]×Ω. Problem (1.1) has been studied by many authors. Specifically,

    1. f=0, see e.g. [11,22,24], the existence and uniqueness of solutions is established. Moreover, the asymptotic behavior and regularity are investigated.

    2. f(u)=up,p1, in [5], the authors investigate large time behavior of solutions. R. Xu et al. [32] proved the invariance of some sets, global existence, nonexistence and asymptotic behavior of solutions with initial energy J(u0)d and finite time blow-up with high initial energy J(u0)>d and some related works [18,34]. For the case of f(u)=|u|p2u, there are other results on the large time behavior of solutions of the pseudo-parabolic see [7,28,29,30,35,36] and their references.

    3. When the source term is a logarithmic nonlinearity f(u)=|u|pulog|u|, very recently, the work [10] focus on the initial conditions, which ensure the solutions to exist globally, blow up in finite time and blow up at infinite time. The asymptotic behaviour for the solutions has been considered in [4,6,8,12,34] and the references therein.

    4. For nonlocal source, f(u)=|u|pΩG(x,y)|u|p+1(y)dy,yΩ, the authors of [19] considered blow-up time for solutions, obtained a lower bound as well as an upper bound for the blow-up time under different conditions, respectively. Also, they investigated a nonblow-up criterion and compute an exact exponential decay, see also [9,23].

    For 0<β1, [14] considered the Cauchy problem

    utmΔut+(Δ)βu=up+1,inR+×Ω, (1.2)

    supplemented with initial condition u(0,x)=u0(x),xΩ and Dirichlet boundary condition u(t,x)=0,(t,x)R+×Ω. The paper established the global existence and time-decay rates for small-amplitude solutions.

    As mention above, initial value problems of nonlinear pseudo-parabolic equations have been considered in many papers see [1,4,5,6,7,8,9,10,11,12,13,14,15,19,21,22,23,24,32,35,36,37]. However, there are not many results devoted to Problem (PT). Our approach includes as special cases all previously on the reaction terms. In this work, we consider two cases; first, the source f is globally Lipschitz and in the second case, we consider f is general locally Lipschitz function (a coercive-type condition). At this point, we remark that there exists some locally Lipschitz functions, but we cannot determine its specific Lipschitz coefficient e.g. f=u(abu2),b>0 of the Ginzburg-Landau equation. Hence, we have to find another method to study the problem with the locally Lipschitz source which is similar to the Ginzburg-Landau equation, etc. (see Subsection 4.2.2 for more details).

    The solution of Problem (PT) is of the regularity-loss structure for 0<β<1, xΩ,t(0,T], we consider the existence and regularity of Problem (PT). In the case β>1, the regularity-gain type for xΩ,t(0,T] and the Problem (PT) is ill-posed. So the regularization methods are required. As we know, there are many regularization methods to suit each problem [2,3,16,20,25,26,27]. For problem (PT), we propose two methods to regularize solution: Modified Lavrentiev regularization (MLR) method and Fourier truncated regularization (FTR) method.

    The plan of the paper is as follows. Section 2 contains notations and formulation of a solution of Problem (PT) and the proof of its instability. In Section 3, the case 0<β<1, well-posedness of solutions of Problem (PT) is established. In Section 4, the case β>1, the proof that Problem (PT) is ill-posed and the well-posedness of the regularized problem are presented. We also propose two regularization methods: MLR method and FTR method for the globally Lipschitz or locally Lipschitz reaction terms, respectively.

    Let us recall that the spectral problem

    {(Δ)βej(x)=λβjej(x),inΩ,β>0,ej(x)=0,onΩ, (2.1)

    admits a family of eigenvalues

    0<λ1λ2λ3...λj....

    The notation B stands for the norm in the Banach space B. We denote by Lq(0,T;B),1q, the Banach space of real-valued measurable functions w:(0;T)B with norm

    wLq(0,T;B)=(T0w(t)qBdt)1q,for1q<,
    wL(0,T;B)=esssupt(0,T)w(t)B,forq=.

    The norm of the function space Ck([0,T];B),0k is denoted by

    wCk([0,T];B)=ki=0supt[0,T]w(i)(t)B<.

    For any ν0, we define the space

    Hν(Ω)={wL2(Ω):j=1λ2νjw,ej2L2(Ω)<},

    where ,L2(Ω) is the inner product in L2(Ω); Hν(Ω)) is a Hilbert space with the norm

    wHν(Ω)=(j=1λ2νjw,ej2L2(Ω))12.

    The Gevrey of order β class of functions with index η1,η2>0, defined by the spectrum of the Laplacian is denoted by

    Gβ(η1,η2)(Ω):={wL2(Ω):j=1λη1βjexp(η2λβj)w,ej2L2(Ω)<},

    and its norm given by

    wGβ(η1,η2)(Ω)=(j=1λη1βjexp(η2λβj)w,ej2L2(Ω))12.

    Next, we give the formulation of solution of the problem (PT).

    Now, assume that Problem (PT) has a unique solution, then we find its the form. Let u(t,x)=j=1uj(t)ej(x) be the decomposition of u in L2(Ω) with uj(t)=u(t,),ejL2(Ω). From (PT), taking the inner product of both sides of (PT) with ej(x), we obtain the ordinary differential equation

    (1+mλj)uj(t)+λβjuj(t)=fj(u)(t),

    where uj(t)=ddtu(t,),ejL2(Ω),fj(u)(t)=f(t,;u),ejL2(Ω), whose solution is

    uj(t)=exp((Tt)λβj1+mλj)φj11+mλjTtexp((τt)λβj1+mλj)fj(u)(τ)dτ,

    where φj=φ,ejL2(Ω).

    Definition 2.1 (Mild solution of Problem (PT)). A function u is a mild solution of (PT) if uC([0,T];L2(Ω)) and satisfies the following integral equation

    u(t,x)=j=1[exp((Tt)λβj1+mλj)φj11+mλjTtexp((τt)λβj1+mλj)fj(u)(τ)dτ]ej(x). (2.2)

    for all (t,x)(0,T)×Ω, and β>0.

    Now we introduce the main results in this paper.

    In this section, we prove that the Problem (PT) is well-posed. First prove that for the Problem (PT) exists a unique mild solution, then the regularity of the solution is established.

    We will make the following assumptions:

    (A1) Assume that f satisfy the global Lipschitz condition:

    f(t,;w1)f(t,;w2)L2(Ω)Kw1w2L2(Ω), (3.1)

    with K>0 independent of t,x,w1,w2, and (t,x)[0,T)×Ω,wiC([0,T];L2(Ω)),i=1,2.

    (A2) We set f(0):=f(t,x;0)=0,(t,x)[0,T)×Ω and

    f(t,;w)L2(Ω)KwL2(Ω),wC([0,T];L2(Ω)). (3.2)

    Theorem 3.1 (L-Existence). Let 0<β<1, assume that f satisfy the assumption (A1). Then, the integral equation (2.2) has a unique mild solution in L(0,T;L2(Ω)).

    Proof. We prove the existence of the solution uL(0,T;L2(Ω)) of the integral equation (2.2). For wL(0,T;L2(Ω)), we consider the following operator:

    H(w)(t,x)=j=1[exp((Tt)λβj1+mλj)φj11+mλjTtexp((τt)λβj1+mλj)fj(w)(τ)dτ]ej(x), (3.3)

    and we aim to show that Hk is a contraction mapping on the space L(0,T;L2(Ω)). In fact, we will prove that for every w1,w2L(0,T;L2(Ω)), it holds

    Hk(w1)(t,)Hk(w2)(t,)L2(Ω)(Kmλ1exp(Tmλ1)(Tt))kk!w1w2L(0,T;L2(Ω)). (3.4)

    We will prove (3.4) by induction. For k=1, using Parseval's relation and assumption (A1), one obtains

    H(w1)(t,)H(w2)(t,)L2(Ω)Ttj=111+mλjexp((τt)λβj1+mλj)(fj(w1)(τ)fj(w2)(τ))ejL2(Ω)dτTtj=11(1+mλj)2exp(2(τt)λβj1+mλj)|fj(w1)(τ)fj(w2)(τ)|2dτ1mλ1exp(Tmλ1)Ttf(τ,,w1)f(τ,,w2)L2(Ω)dτKmλ1exp(Tmλ1)Ttw1(τ,)w2(τ,)L2(Ω)dτKmλ1exp(Tmλ1)(Tt)w1w2L(0,T;L2(Ω)). (3.5)

    Assume now that (3.4) is satisfied for k=k0, let us prove that it is satisfied for k=k0+1. It holds

    Hk0+1(w1)(t,)Hk0+1(w2)(t,)L2(Ω)Kmλ1exp(Tmλ1)TtHk0(w1)(τ,)Hk0(w2)(τ,)L2(Ω)dτ1k0!(Kmλ1exp(Tmλ1))k0+1w1w2L(0,T;L2(Ω))Tt(Tτ)k0dτ1(k0+1)!(Kmλ1exp(Tmλ1))k0+1(Tt)k0+1w1w2L(0,T;L2(Ω))1(k0+1)!(KTmλ1exp(Tmλ1))k0+1w1w2L(0,T;L2(Ω)).

    Therefore, by the induction principle we get (3.4). Since the right hand side of (3.5) is independent of t, we deduce that

    Hk(w1)Hk(w2)L(0,T;L2(Ω))1k!(KTmλ1exp(Tmλ1))kw1w2L(0,T;L2(Ω)).

    When k is large enough, we have 1k1!(KTmλ1exp(Tmλ1))k10. Hence there exists k1 such that

    0<1k1!(KTmλ1exp(Tmλ1))k1<1.

    We claim that the mapping Hk1 is a contraction, i.e. Hk1(u)=u. We have

    Hk1(H(u))=H(Hk1(u))=H(u).

    Due to the uniqueness of the fixed point of Hk1, it holds H(u)=u. We conclude that the integral equation (3.3) has a unique solution in L(0,T;L2(Ω)).

    Theorem 3.2 (Regularity). Let 0<β<1, and f only satisfy the assumption (A1), we have the following:

    a) If φL2(Ω) and f(0)L1(0,T;L2(Ω)), then there exists C(T,m,λ1)>0 such that

    uL(0,T;L2(Ω))C(T,m,λ1)(φL2(Ω)+f(0)L1(0,T;L2(Ω))mλ1). (3.6)

    b) If φHβ(Ω), f(0)L(0,T;L2(Ω)) then there exists C(T,m,λ1,β)>0 such that

    uL(0,T;Hβ(Ω))C(T,m,λ1,β)(φHβ(Ω)+Tf(0)L(0,T;L2(Ω))mλ1β1). (3.7)

    Here, we recall f(0):=f(t,x;0),(t,x)[0,T)×Ω.

    Proof. First, we set

     M1(φ)(t,x):=j=1exp((Tt)λβj1+mλj)φjej(x),M2(u)(t,x):=j=1[11+mλjTtexp((τt)λβj1+mλj)fj(u)(τ)dτ]ej(x).

    a) First, using the Parseval's relation, we infer that

    M1(φ)(t,)L2(Ω)=(j=1exp(2(Tt)λβj1+mλj)φ2j)1/2exp(Tmλ1β1)φL2(Ω), (3.8)

    where for β(0,1), we have exp((Tt)λβj1+mλj)exp(Tmλ1β1),t[0,T),jN. We also obtain

    M2(u)(t,)L2(Ω)Ttj=111+mλjexp((τt)λβj1+mλj)fj(u)(τ)ejL2(Ω)dτ
    =Tt(j=11(1+mλj)2exp(2(τt)λβj1+mλj)|fj(u)(τ)|2)1/2dτ1mλ1exp(Tmλ1β1)Ttf(τ,;u)L2(Ω)dτ, (3.9)

    where we have used 11+mλj1mλ1,jN and for β(0,1) we implies that

    exp(2(τt)λβj1+mλj)exp(Tmλ1β1),0tτT.

    Using (3.1), we obtain

    f(t,;u)f(t,;0)L2(Ω)Ku(t,)L2(Ω),

    then

    f(t,;u)L2(Ω)Ku(t,)L2(Ω)+f(t,;0)L2(Ω), (3.10)

    Combining (3.10) with (3.9), leads to

    M2(u)(t,)L2(Ω)1mλ1exp(Tmλ1β1)f(0)L1(0,T;L2(Ω))+Kmλ1exp(Tmλ1β1)Ttu(τ,)L2(Ω)dτ. (3.11)

    From (3.8) and (3.11) yields

    u(t,)L2(Ω)exp(Tmλ1β1)(φL2(Ω)+1mλ1f(0)L1(0,T;L2(Ω)))+Kmλ1exp(Tmλ1β1)Ttu(τ,)L2(Ω)dτ.

    Thanks to Grönall's inequality

    u(t,)L2(Ω)exp(Tmλ1β1)(φL2(Ω)+1mλ1f(0)L1(0,T;L2(Ω)))×exp(Kmλ1exp(Tmλ1β1)(Tt)),

    this imlpies (3.6).

    b) We observe that for β(0,1)

    M1(φ)(t,)Hβ(Ω)=(j=1λ2βjexp(2(Tt)λβj1+mλj)φ2j)1/2exp(Tmλ1β1)φHβ(Ω). (3.12)

    We conclude that

    M2(u)(t,)Hβ(Ω)Ttj=111+mλjexp((τt)λβj1+mλj)fj(u)(τ)ejHβ(Ω)dτ=Ttj=1λ2βj(1+mλj)2exp(2(τt)λβj1+mλj)|fj(u)(τ)|2dτ=Ttj=11m2λ22βjexp(2(τt)λβj1+mλj)|fj(u)(τ)|2dτ=1mλ1β1exp(Tmλ1β1)Ttj=1|fj(u)(τ)|2dτ1mλ1β1exp(Tmλ1β1)Ttf(τ,;u)L2(Ω)dτ. (3.13)

    Inequality (3.10) associated with (3.13) leads to

    M2(u)(t,)Hβ(Ω)Tt1mλ1β1exp(Tmλ1β1)(Ku(τ,)L2(Ω)+f(τ,;0)L2(Ω))dτTmλ1β1exp(Tmλ1β1)f(0)L(0,T;L2(Ω))+Kmλ1β1exp(Tmλ1β1)Ttu(τ,)L2(Ω)dτ. (3.14)

    Estimates (3.12) and (3.14) lead to

    u(t,)Hβ(Ω)j=1exp((Tt)λβj1+mλj)φjejHβ(Ω)+Ttj=111+mλjexp((τt)λβj1+mλj)fj(u)(τ)ejHβ(Ω)dτexp(Tmλ1β1)(φHβ(Ω)+Tmλ1β1f(0)L(0,T;L2(Ω)))+Kmλ12β1exp(Tmλ1β1)Ttu(τ,)Hβ(Ω)dτ, (3.15)

    where we have used that wHβ(Ω)λβ1wL2(Ω) for β(0,1). Grönwall's inequality allows to obtain

    u(t,)Hβ(Ω)(φHβ(Ω)+Tmλ1β1f(0)L(0,T;L2(Ω)))
    ×exp(KTmλ12β1exp(Tmλ1β1)+Tmλ1β1),

    we implies (3.7). The proof is complete.

    Definition 4.1 (Ill-posed). The well-posed problem in the sense of Hadamard is to satisfy the following properties:

    ⅰ) a solution exists;

    ⅱ) the solution is unique;

    ⅲ) the solution's behaviour changes continuously with the initial conditions.

    If at least one of the three properties above does not satisfy, the problem is ill-posed.

    Next, we give an example which shows that the solution ˜u(k)(t,x) (for any kN) of Problem (PT) is not stable (property ⅲ) is unsatisfied). For β>1, let us set

    ˜φ(k)(x)=λ1kek(x),kN, (4.1)
    ˜f(t,x;w)=j=1λjT1exp(Tλβj1+mλj)w(t,),ejL2(Ω)ej(x),m>1, (4.2)

    for (t,x)(0,T)×Ω and wC([0,T];L2(Ω)). The solution ˜u(k)(t,x) satisfies the integral equation

    ˜u(k)(t,x)=j=1[exp((Tt)λβj1+mλj)˜φ(k)j]ej(x)j=1[11+mλjTtexp((τt)λβj1+mλj)˜fj(u(k))(τ)dτ]ej(x), (4.3)

    where

    ˜φ(k)j=˜φ(k),ejL2(Ω),

    and

    ˜fj(˜u(k))(t)=˜f(t,;˜u(k)),ejL2(Ω).

    Step 1. We show that (4.3) has a unique solution ˜u(k)C([0,T];L2(Ω)).

    Indeed, we consider the function

    G(w)(t,x)=j=1[exp((Tt)λβj1+mλj)˜φ(k)j]ej(x)j=1[11+mλjTtexp((τt)λβj1+mλj)˜fj(w)(τ)dτ]ej(x).

    Then for any w1,w2C([0,T];L2(Ω)), we obtain

    G(w1)(t,)G(w2)(t,)L2(Ω)Ttj=1[11+mλjexp((τt)λβj1+mλj)(˜fj(w1)(τ)˜fj(w2)(τ))]ejL2(Ω)dτ=Tt[j=1(11+mλjexp((τt)λβj1+mλj))2λ2jT2exp(Tλβj1+mλj)|w1,j(τ)w2,j(τ)|2]1/2dτ=Tt[j=11m2T2exp((τtT)λβj1+mλj)|w1,j(τ)w2,j(τ)|2]1/2dτTt1mTw1(τ,)w2(τ,)L2(Ω)dτ1mw1w2C([0,T];L2(Ω)),

    where we denote wi,j(t):=wi(t,),ejL2(Ω),i=1,2.

    This implies that

    G(w1)G(w2)C([0,T];L2(Ω))1mw1w2C([0,T];L2(Ω)).

    Hence G is a contraction because m>1. Using the Banach fixed-point theorem, we conclude that G(w)=w has a unique solution ˜u(k)C([0,T];L2(Ω)).

    Step 2. The solution of Problem (4.3) is instable. We have

    ˜u(k)(t,)L2(Ω)j=1exp((Tt)λβj1+mλj)˜φ(k)jejL2(Ω)=:M3(˜φ(k))(t)j=1[11+mλjTtexp((τt)λβj1+mλj)˜fj(˜u(k))(τ)dτ]ejL2(Ω)=:M4(˜u(k))(t).

    It is easy to see that (here, noting that from (4.2), we have ˜fj(0)=0)

    |M4(˜u(k))(t)|=G(˜u(k))(t)G(0)(t)L2(Ω)1m˜u(k)C([0,T];L2(Ω)).

    Hence

    ˜u(k)(t,)L2(Ω)|M3(˜φ(k))(t)|1m˜u(k)C([0,T];L2(Ω)).

    This leads to

    ˜u(k)C([0,T];L2(Ω))mm+1sup0tT|M3(˜φ(k))(t)|. (4.4)

    We continue to estimate the right hand side of the latter inequality. Indeed, since {ej(x)}j1 is a basis of L2(Ω), i.e.,

    {ek,ejL2(Ω)=1,k=j,ek,ejL2(Ω)=0,kj,

    we have

    |M3(˜φ(k))(t)|=j=1exp(2(Tt)λβj1+mλj)˜φ(k)j,ej2L2(Ω)=j=1exp(2(Tt)λβj1+mλj)λ1kek,ej2L2(Ω)=1λkexp((Tt)λβk1+mλk).

    Since the function χ(t):=exp((Tt)λβk1+mλk) is a decreasing function with respect to variable t[0,T] and β>1, we deduce that

    sup0tT|M3(˜φ(k))(t)|sup0tT(1λkexp((Tt)λβk1+mλk))=exp(Tλβk1+mλk)1λk. (4.5)

    Combining (4.4) and (4.5) yields

    ˜u(k)C([0,T];L2(Ω))mm+1exp(Tλβk1+mλk)1λk.

    As k, we see that

    limk˜φ(k)L2(Ω)=limkekL2(Ω)λk=limk1λk=0,

    but

    limk˜u(k)C([0,T];L2(Ω))limkmm+1exp(Tλβk1+mλk)1λk=.

    Thus, it is shown that Problem (PT)is ill-posed in the Hadamard sense in L2-norm for β>1.

    In order to obtain the stable numerical solutions, we propose two regularization methods to solve the Problem (PT) in two cases of f:

    f is globally Lipschitz: MLR method.

    f is locally Lipschitz: FTR method.

    In this subsection, the functions f(t,x;u) is satisfy the globally Lipschitz (A1). To approximate u, we introduce the regularized solutions Uεα given by MLR method as follows

    Uεα(t,x)=j=1[Lj(α,T)exp((Tt)λβj1+mλj)φεj]ej(x)j=111+mλj[TtLj(α,T)exp((τt)λβj1+mλj)fj(Uεα)(τ)dτ]ej(x), (4.6)

    where we set

    Lj(α,T)=exp(Tλβj1+mλj)αλβj+exp(Tλβj1+mλj), (4.7)

    and the coefficient α:=α(ε) satifies limε0+α=0; it plays the role of a regularization parameter.

    The following technical lemma plays the key role in our analysis.

    Lemma 4.2. For 0tτT, we have

    (a)|Lj(α,T)exp((Tt)λβj1+mλj)|αtT1(Tlog(α1T))1tT, (4.8)
    (b)|Lj(α,T)exp((τt)λβj1+mλj)|αtτT(Tlog(α1T))τtT. (4.9)

    Proof. a). First, from (4.7), it holds

    |Lj(α,T)exp((Tt)λβj1+mλj)|=exp(tλβj1+mλj)αλβj+exp(Tλβj1+mλj)=exp(tλβj1+mλj)(αλβj+exp(Tλβj1+mλj))tT(αλβj+exp(Tλβj1+mλj))1tT1(αλβj+exp(Tλβj1+mλj))1tT1(αλβj+exp(Tλβj))1tT. (4.10)

    Using the inequality

    1ν1ζ+exp(ζν2)ν2ν1log(ν2ν1), (4.11)

    for νi>0,i=1,2 and 0<ζ<eν2, if α<eT, we obtain

    1(αλβj+exp(Tλβj))1tT(α1Tlog(α1T))1tT.

    Whereupon

    |Lj(α,T)exp((Tt)λβj1+mλj)|(α1Tlog(α1T))1tT. (4.12)

    Using (4.12) into (4.10), we obtain (4.8).

    With the same argument as in the proof of (4.8), we obtain (4.9). This concludes the proof of the lemma.

    A. The well-posedness of the regularized solution (4.6). In this subsection, we will obtain the existence and regularity results for the regularized solution (4.6).

    Theorem 4.3 (Existence-uniqueness). Suppose that f satisfying the assumption (A1) then the regularized solution (4.6) has unique weak solution UεαC([0,T];L2(Ω)).

    Proof. For any FC([0,T];L2(Ω)), we define the function

    F:C([0,T];L2(Ω))C([0,T];L2(Ω))

    by

    F(ϑ)(t,x):=j=1[Lj(α,T)exp((Tt)λβj1+mλj)φεj]ej(x)j=1[11+mλjTtLj(α,T)exp((τt)λβj1+mλj)fj(ϑ)(τ)dτ]ej(x). (4.13)

    We also define Fk as follows

    Fk(ϑ)=FF(F(ϑ))ktimes.

    We shall prove by induction, for any couple ϑ1,ϑ2C([0,T];L2(Ω)), that

    Fk(ϑ1)(t,)Fk(ϑ2)(t,)H2(Ω)(Kα1Tmlog(α1T)(Tt))kk!ϑ1ϑ2C([0,T];H2(Ω)). (4.14)

    For k=1, by using (A1) and Lemma 4.2 and noting that α1Tlog(α1T)1, it holds

    αtτT(Tlog(α1T))τtTα1Tlog(α1T),for all0tτT,

    then we have

    F(ϑ1)(t,)F(ϑ2)(t,)L2(Ω)Ttj=1[11+mλjLj(α,T)exp((τt)λβj1+mλj)(fj(ϑ1)(τ)fj(ϑ2)(τ))]ejL2(Ω)dτTt[j=11(1+mλj)2|Lj(α,T)exp((τt)λβj1+mλj)|2|fj(ϑ1)(τ)fj(ϑ2)(τ)|2]1/2dτKα1Tmλ1log(α1T)Tt[j=1|ϑ1,j(τ)ϑ2,j(τ)|2]1/2dτKα1Tmλ1log(α1T)Ttϑ1(τ,)ϑ2(τ,)L2(Ω)dτKα1T(Tt)mλ1log(α1T)ϑ1ϑ2C([0,T];L2(Ω)),ϑi,j=ϑi,ejL2(Ω),i=1,2.

    Hence, (4.14) holds for k=1.

    Assume that (4.14) holds for k=N. We show now that (4.14) holds for k=N+1. In fact, we have

    FN+1(ϑ1)(t,)FN+1(ϑ2)(t,)Hν(Ω)Kα1Tmlog(α1T)TtFN(ϑ1)(τ,)FN(ϑ2)(τ,)L2(Ω)dτ(Kα1Tmlog(α1T))N+1Tt(Tτ)NN!dτϑ1ϑ2C([0,T];L2(Ω))1(N+1)!(Kα1Tmlog(α1T)(Tt))N+1ϑ1ϑ2C([0,T];L2(Ω)).

    By the induction principle, we deduce that (4.14) holds for all kN. Notice that, as α is fixed, then 1k!(Kα1Tmlog(α1T)(Tt))k tends to 0 when k, so there exists a positive integer k0 such that

    1k0!(Kα1Tmlog(α1T)(Tt))k0<1.

    It means that Fk0 is a contraction. Finally, it follows the desired conclusion that the problem (4.6) has a unique solution UεαC([0,T];L2(Ω)). The proof is complete.

    Given a constant α(0,1) (which will be assumed from now on) and a function wC([0,T];L2(Ω)), we denote the scaling with α as follows:

    wα,=sup0tT(αtTw(t,)L2(Ω)). (4.15)

    Theorem 4.4 (Regularity). Assume that f satisfy the assumptions (A1) and (A2). We have the following results:

    a) If φεL2(Ω), then Uεα given by (4.6) satisfies

    Uεαα,α1Tlog(T)exp(KT2mλ1log(T))φεL2(Ω). (4.16)

    b) If φεH1(Ω), then

    UεαL(0,T;H1(Ω))α1Tlog(T)exp(KT2α1mλ1log(T))φεH1(Ω). (4.17)

    Proof. We rewrite (4.6) as

    Uεα(t,x)=j=1[Lj(α,T)exp((Tt)λβj1+mλj)φεj]ej(x)=:M5(φε)(t,x)j=1[11+mλjTtLj(α,T)exp((τt)λβj1+mλj)fj(Uεα)(τ)dτ]ej(x)=:M6(Uεα)(t,x). (4.18)

    a) Using (3.10) (noting that f(t,x;0)=0, for all (t,x)(0,T)×Ω) and Lemma 4.2b), we first observe that

    M6(Uεα)(t,)L2(Ω)Ttj=1[11+mλjLj(α,T)exp((τt)λβj1+mλj)fj(Uεα)(τ)]2dτ1mλ1TtαtτT(Tlog(α1T))τtTf(τ,;Uεα)L2(Ω)dτKTmλ1log(T)TtαtτTUεα(τ,)L2(Ω)dτ, (4.19)

    where we have used (Tlog(α1T))τtTTlog(T), for α(0,1),0tτT.

    The next observation, from Lemma 4.2a), is that

    M5(φε)(t,)L2(Ω)j=1[Lj(α,T)exp((Tt)λβj1+mλj)φεj]2αtT1(Tlog(α1T))1tTφεL2(Ω)αtT1Tlog(T)φεL2(Ω), (4.20)

    noting that (Tlog(α1T))τtTTlog(T), for all 0tT,0<α<1.

    Combining (4.18)-(4.20) and multiplying the two sides of of the result obtained by αtT, we deduce that

    αtTUεα(t,)L2(Ω)M5(φε)(t,)L2(Ω)+M6(Uεα)(t,)L2(Ω)α1Tlog(T)φεL2(Ω)+KTmλ1log(T)TtατTUεα(τ,)L2(Ω)dτ.

    Using Grönwall's inequality, we get

    αtTUεα(t,)L2(Ω)α1Tlog(T)φεL2(Ω)exp(KT2mλ1log(T)). (4.21)

    Since the right side of (4.21) does not depend on t, we have

    Uεαα,α1Tlog(T)exp(KT2mλ1log(T))φεL2(Ω).

    This leads to (4.16).

    b) From Lemma 4.2a), we first observe that

    M5(φε)(t,)H1(Ω)j=1[λjLj(α,T)exp((Tt)λβj1+mλj)φεj]2αtT1(Tlog(α1T))1tTφεH1(Ω)α1Tlog(T)φεH1(Ω). (4.22)

    The next observation, using Lemma 4.2b), is that

    M6(Uεα)(t,)H1(Ω)Ttj=1[λj1+mλjLj(α,T)exp((τt)λβj1+mλj)fj(Uεα)(τ)]2dτ1mTtαtτT(Tlog(α1T))τtTf(τ,;Uεα)L2(Ω)dτKTα1mlog(T)TtUεα(τ,)L2(Ω)dτ, (4.23)

    where we have used αtτT(Tlog(α1T))τtTα1Tlog(T),0tτT for 0<α<1.

    From (4.22) and (4.23), we deduce that

    Uεα(t,)H1(Ω)α1Tlog(T)φεH1(Ω)+KTα1mλ1log(T)TtUεα(τ,)H1(Ω)dτ,

    where wL2(Ω)wH1(Ω)λ1. Thanks to Grönwall's inequality, we get

    Uεα(t,)H1(Ω)α1Tlog(T)φεH1(Ω)exp(KTα1mλ1log(T)(Tt)),

    which implies (4.17). This concludes the proof.

    B. Error estimate In this subsection, by using MLR method, the error between the exact solution and the regularized solution is obtained. Now, we can formulate the main theorem.

    Theorem 4.5 (Error estimate). Let α:=α(ε) satisfy

    {limε0+α=0,limε0+εα1=M0<. (4.24)

    Assume that f satisfy the assumptions (A1)(A2) and Problem (PT) has a unique solution u satisfying

    uL(0,T;Gβ(η1,η2)(Ω))anduL(0,T;Gβ(η1,η2)(Ω))P0, (4.25)

    for some known positive constants P0 and η12,η22Tmλ1. Then (for all t[0,T])

    Uεα(t,)u(t,)L2(Ω)2(M0+P0)exp(K2Tm2λ21(Tt))αtT(Tlog(α1T))1tT. (4.26)

    Remark 1. The error estimate in (4.26) is of order O(ε)=αtT(Tlog(α1T))1tT,t[0,T].

    If tT, then O(ε)=α tends to zero according to (4.24).

    If t0, then O(ε)=Tlog1(α1T) tends to zero as ε0+.

    Remark 2. Choose α=εrε0+0, for some 0<r<1, then the estimate in (4.26) is of order

    εrtT[T/log(T/εr)]tTT.

    Proof. For all t(0,T), the triangle inequality gives

    Uεα(t,)u(t,)L2(Ω)Uεα(t,)Uα(t,)L2(Ω)=:Mα,ε7(t,x)+Uα(t,)u(t,)L2(Ω)=:Mα,ε8(t,x), (4.27)

    where Uα(t,x) is the solution of Problem (4.6) corresponding to exact datum φ(x). In order to estimate Mα,ε7(t,x) and Mα,ε8(t,x), we need to divide the proof in two steps.

    Step 1. Estimate Mα,ε7(t,x).

    Thanks to Parseval's relation and basic inequality (a+b)22a2+2b2,a,b0, we obtain

    Uεα(t,)Uα(t,)2L2(Ω)2j=1[Lj(α,T)exp((Tt)λβj1+mλj)(φεjφj)]2=:Mα,ε7a(t)+2j=1[11+mλjTtLj(α,T)exp((τt)λβj1+mλj)(fj(Uεα)(τ)fj(Uα)(τ))dτ]2=:Mα,ε7b(t). (4.28)

    To estimate Mα,ε7a(t), let us use Lemma 4.2a) yields

    |Mα,ε7a(t)|2j=1α2tT2(Tlog(α1T))22tT|φεjφj|22α2tT2(Tlog(α1T))22tTφεφ2L2(Ω)2α2tT2(Tlog(α1T))22tTε2. (4.29)

    Next, we estimate Mα,ε7b(t). Using Hölder's inequality, (A1) and Lemma 4.2b), one obtains

    |Mα,ε7b(t)|=2j=1[11+mλjTtLj(α,T)exp((τt)λβj1+mλj)(fj(Uεα)(τ)fj(Uα)(τ))dτ]22Tj=11(1+mλj)2Tt|Lj(α,T)exp((τt)λβj1+mλj)|2|fj(Uεα)(τ)fj(Uα)(τ)|2dτ2Tm2λ21Ttα2t2τT(Tlog(α1T))2τ2tTj=1|fj(Uεα)(τ)fj(Uα)(τ)|2dτ2Tm2λ21Ttα2t2τT(Tlog(α1T))2τ2tTf(Uεα)(τ)f(Uα)(τ)2L2(Ω)dτ
    2K2Tm2λ21Ttα2t2τT(Tlog(α1T))2τ2tTUεα(τ,)Uα(τ,)2L2(Ω)dτ. (4.30)

    Combining the results in (4.28)-(4.30), we get

    Uεα(t,)Uα(t,)2L2(Ω)2α2tT2(Tlog(α1T))22tTε2+2K2Tm2λ21Ttα2t2τT(Tlog(α1T))2τ2tTUεα(τ,)Uα(τ,)2L2(Ω)dτ.

    Multiplying by α2tT(Tlog(α1T))2tT both sides, we deduce that

    α2tT(Tlog(α1T))2tTUεα(t,)Uα(t,)2L2(Ω)2α2(Tlog(α1T))2ε2+2K2Tm2λ21Ttα2τT(Tlog(α1T))2τTUεα(τ,)Uα(τ,)2L2(Ω)dτ.

    Grönwall's inequality allows to obtain

    α2tT(Tlog(α1T))2tTUεα(t,)Uα(t,)2L2(Ω)2α2(Tlog(α1T))2ε2exp(2K2Tm2λ21(Tt)).

    Similar calculations yield

    Uεα(t,)Uα(t,)L2(Ω)2α1εexp(K2Tm2λ21(Tt))αtT(Tlog(α1T))1tT. (4.31)

    Step 2. Estimate Mα,ε8(t). Let us define an operator

    Θw(t,x)=j=1[Lj(α,T)w(t,),ejL2(Ω)]ej(x),forwC([0,T];L2(Ω)).

    It clearly follows that

    Θu(t,x)=j=1[Lj(α,T)exp((Tt)λβj1+mλj)φj]ej(x)j=1[11+mλjTtLj(α,T)exp((τt)λβj1+mλj)fj(u)(τ)dτ]ej(x).

    The triangle inequality allows to write

    Uα(t,)u(t,)2L2(Ω)2Uα(t,)Θu(t,)2L2(Ω)+2Θu(t,)u(t,)2L2(Ω)=:Mα8a(t)+Mα8b(t).

    We estimate for Mα8a(t) as follows (similarly as in (4.30)):

    |Mα8a(t)|=2j=1[Tt11+mλjLj(α,T)exp((τt)λβj1+mλj)(fj(Uα)(τ)fj(u)(τ))dτ]22K2Tm2λ21Ttα2t2τT(Tlog(α1T))2τ2tTUα(τ,)u(τ,)2L2(Ω)dτ.

    The term Mα8b(t) is estimated by using (4.8) as

    |Mα8b(t)|=j=1(Lj(α,T)1)2|uj(t)|2=2j=1(exp(Tλβj1+mλj)αλβj+exp(Tλβj1+mλj)1)2|uj(t)|2=2j=1(αλβjαλβj+exp(Tλβj1+mλj))2|uj(t)|22α2j=1(exp(tλβj1+mλj)αλβj+exp(Tλβj1+mλj))2λ2βjexp(2Tλβj1+mλj)|uj(t)|22α2j=1α2tT2(Tlog(α1T))22tTλ2βjexp(2Tmλ1λβj)|uj(t)|22α2tT(Tlog(α1T))22tTu(t,)2Gβ(η1,η2)(Ω),

    where we have used λβj1+mλjλβjmλ1, for all jN and η12,η22Tmλ1.

    Combining all these inequalities, we deduce

    Uα(t,)u(t,)2L2(Ω)2α2tT(Tlog(α1T))22tTu(t,)2Gβ(η1,η2)(Ω)+2K2Tm2λ21Ttα2t2τT(Tlog(α1T))2τ2tTUα(τ,)u(τ,)2L2(Ω)dτ.

    Multiplying by α2tT(Tlog(α1T))2tT, we obtain

    α2tT(Tlog(α1T))2tTUα(t,)u(t,)2L2(Ω)
    2(Tlog(α1T))2u(t,)2Gβ(η1,η2)(Ω)+2K2Tm2λ21Ttα2τT(Tlog(α1T))2τTUα(τ,)u(τ,)2L2(Ω)dτ.

    Using Grönwall's inequality, we thus obtain

    Uα(t,)u(t,)L2(Ω)2αtT(Tlog(α1T))1tTu(t,)Gβ(η1,η2)(Ω)exp(K2Tm2λ21(Tt)). (4.32)

    Combining (4.27), (4.31) and (4.32), we complete the proof of Theorem 4.5.

    In Subsection 4.2.1 has addressed the Problem (PT) in which f is a globally Lipschitz function, in the rest of this paper, we extend the analysis to a locally Lipschitz function f. Up to the present, the results of Problem (PT) for the locally Lipschitz cases are still very scarce. Hence, we have to find another regularization method to study the problem with the locally Lipschitz source. Thus, the FTR method is a very effective method for this case.

    We begin by establishing the locally Lipschitz properties of f by the following assumption:

    (A3) Assume that for each ϱ>0, there exists Kϱ>0 such that

    |f(t,x;u)f(t,x;v)|Kϱ|uv|,ifmax{|u|,|v|}ϱ, (4.33)

    and

    Kϱ=sup{|f(t,x;u)f(t,x;v)uv|,|u|,|v|ϱ,uv,(t,x)[0,T)×Ω},

    Kϱ is a non-decreasing function of ϱ. We assume that limϱKϱ=.

    Example 1. Let f1(u)=u|u|2. Easy calculations show that

    |f1(u)f1(v)|=|u|u|2v|v|2|=|(uv)|u|2+v(|u|2|v|2)|=(|u|2+|v||u|+|v|2)|uv|.

    Clearly, f1 is not globally Lipschitz. For ϱmax{|u|,|v|}, from (4.33), one can compute explicitely the coefficient Kϱ=3ϱ2.

    Example 2. Let f2(u)=u(abu2) (Ginzburg-Landau type), with aR,b>0. It can be easily seen that f2 is locally Lipschitz source. Moreover, we are possible to verify computationally the coefficient Kϱ=3ϱ2max{|a|,b}. In order to solve the problem with the locally Lipschitz sources as above (and some other types), we outline our ideas to construct an approximation of the function f. For all ϱ>0, we define

    fϱ(t,x;u):=f(t,x;˜u),where˜u:={ϱ,ifu(,ϱ),u,ifu[ϱ,ϱ],ϱ,ifu(ϱ,). (4.34)

    With this definition, we claim that fϱ is global Lipschitz function. Before stating the main theorem, we first consider the following lemma.

    Lemma 4.6. Let fϱεL([0,T]×Ω×R) given as (4.34). Then we have

    |fϱε(t,x;u)fϱε(t,x;v)|Kϱε|uv|,(t,x)[0,T)×Ω,u,vR, (4.35)

    where limε0+ϱε=limε0+Kϱε=.

    Proof. The proof can be found in [2].

    Remark 3. For ϱε>0 satisfying limε0+ϱε=, we implies that

    ˜uu,almost everywhere in[0,T)×Ω,fϱε(t,x;u)f(t,x;˜u),almost everywhere in[0,T)×Ω.

    Based on the above analysis, we propose the regularized solution by using FTR method as follows

    VεΛε(t,x)=[Λε]j=1[exp((Tt)λβj1+mλj)φεj11+mλjTtexp((τt)λβj1+mλj)fϱε,j(VεΛε)(τ)dτ]ej(x), (4.36)

    where fϱε,j(VεΛε)(τ)=fϱε(τ,;VεΛε),ejL2(Ω).

    The regularity estimates of the solution VεΠα given by (4.36) is possible that but we will not develop this point here because it is an argument analogous to the previous one. We continue with the error estimate result.

    Theorem 4.7 (Error estimate). Suppose that we can choose ϱε,Λε>0 such that limε0+Λε=limε0+ϱε= and satisfying

    limε0+exp((Λε)βmλ1T)ε=N0<, (4.37)

    and let us choose

    Kϱε=|log(log(εr))mλ1T|1/2,asεtends to0+.

    Let fϱε satisfy Lemma 4.6. Then nonlinear integral system (4.36) has a unique solution VεΛεC([0,T];L2(Ω)). Assuming further that equation (4.36) has a unique exact solution u satisfying

    uL(0,T;Gβ(η1,η2)(Ω)),withη12+2δ,η22Tmλ1,

    and

    uL(0,T;Gβ(η1,η2)(Ω))~P0,for some known constant~P00.

    Then the following stability estimate holds for any t[0,T],δ>0,

    VεΛε(t,)u(t,)L2(Ω)2(~P0(Λε)β(1+δ)+N0)(log(1εr))TtTexp(t(Λε)βmλ1). (4.38)

    Remark 4. We can choose Λε=m1/βλ1/β1βlog(εr)T1/βε0+, for some r(0,1), then the condition in (4.37) is fulfiled. Then the estimate in (4.38) is of order

    εrtT(log(εr))1tT0asε0+,for allt[0,T).

    Remark 5. Obviously, from (4.26), one can raise the question about I0 in (4.25) is large, then the estimate in (4.26) is not exist naturally. By using FTR method, this problem is improved as in (4.38) by the term I0(Λε)β+δ0 as ε goes to 0+.

    Proof. We divide the proof into two parts. In Part a, we prove that the integral equation (4.36) has a unique solution VεΛεC([0,T];L2(Ω)). In Part b, the error between the exact solution and regularized solution is obtained.

    Part a. The existence and uniqueness of a solution of the integral equation (4.36).

    For ζC([0,T];L2(Ω)), we consider the following function

    L(ζ)(t,x)=[Λε]j=1[exp((Tt)λβj1+mλj)φεj11+mλjTtexp((τt)λβj1+mλj)fϱ,j(ζ)(τ)dτ]ej(x). (4.39)

    Let us define Lk as follows

    Lk(ζ)=LL(L(ζ))ktimes.

    To explore the proof, we only need to prove that there exists k0N such that the operator Lk0 which also maps C([0,T];L2(Ω)) into itself, is a contraction. In fact, we prove by mathematical induction that for any couples ζ1,ζ2C([0,T];L2(Ω)) then

    Lk(ζ1)Lk(ζ2)C([0,T];L2(Ω))
    [Kϱεexp(T(Λε)βmλ1)(Tt)]kk!ζ1ζ2C([0,T];L2(Ω)),

    for ζ1,ζ2C([0,T];L2(Ω)). The proof of the latter inequality is similar to that of Theorem 4.3 and thus it is omitted.

    Part b. The error estimate between the exact solution u and the regularized solution VεΛε. Using the triangle inequality, we have

    VεΛε(t,)u(t,)[L2(Ω)]2VεΛε(t,)WΛε(t,)L2(Ω)=:MΛε,ε9(t)+WΛε(t,)u(t,)L2(Ω)=:MΛε10(t), (4.40)

    where

    WΛε(t,x)=[Λε]j=1[exp((Tt)λβj1+mλj)φj11+mλjTtexp((τt)λβj1+mλj)fϱε,j(WΛε)(τ)dτ]ej(x).

    To estimate (4.40), we divide the right-hand side of (4.40) into two steps

    Step 1. Estimate MΛε,ε9(t). One has

    VεΛε(t,x)WΛε(t,x)=[Λε]j=1[exp((Tt)λβj1+mλj)(φεjφj)]ej(x)
    +[Λε]j=1[11+mλjTtexp((τt)λβj1+mλj)(fϱε,j(WΛε)(τ)fϱε,j(VεΛε)(τ))dτ]ej(x). (4.41)

    Using the Hölder's inequality with Parseval's relation, we obtain

    |MΛε,ε9(t)|2=2[Λε]j=1[exp((Tt)λβj1+mλj)(φεjφj)]2+2[Λε]j=1[11+mλjTtexp((τt)λβj1+mλj)(fϱε,j(WΛε)(τ)fϱε,j(VεΛε)(τ))dτ]22exp(2(Tt)(Λε)βmλ1)[Λε]j=1|φεjφj|2+2Tm2λ21Ttexp(2(τt)(Λε)βmλ1)[Λε]j=1|fϱε,j(WΛε)(τ)fϱε,j(VεΛε)(τ)|2dτ2exp(2(Tt)(Λε)βmλ1)ε2+2TK2ϱεm2λ21Ttexp(2(τt)(Λε)βmλ1)WΛε(τ,)VεΛε(τ,)2L2(Ω)dτ. (4.42)

    Multiplying both sides (4.42) by exp(2(Λε)βmλ1t), we get

    exp(2t(Λε)βmλ1)VεΛε(t,)WΛε(t,)2L2(Ω)2exp(2T(Λε)βmλ1)ε2+2TK2ϱεm2λ21Ttexp(2τ(Λε)βmλ1)WΛε(τ,)VεΛε(τ,)2L2(Ω)dτ.

    Applying Grönwall's inequality yields

    exp(2t(Λε)βmλ1)VεΛε(t,)WΛε(t,)2L2(Ω)2exp(2(Λε)βmλ1T)ε2exp(2TK2ϱεm2λ21(Tt)).

    Consequently,

    VεΛε(t,)WΛε(t,)L2(Ω)2exp((Λε)βmλ1(Tt))εexp(2TK2ϱεm2λ21(Tt)). (4.43)

    Step 2. Estimate of MΛε10(t). One has

    |MΛε10(t)|2=2j=[Λε]+1[exp((Tt)λβj1+mλj)φj11+mλjTtexp((τt)λβj1+mλj)fϱε,j(WΛε)(τ)dτ]2
    +2[Λε]j=1[11+mλjTtexp((τt)λβj1+mλj)(fj(u)(τ)fϱε,j(WΛε)(τ))dτ]2=:MΛε10a(t)+MΛε10b(t). (4.44)

    To estimate MΛε10a(t), we proceed as follows for δ>0 be arbitrary

    |MΛε10a(t)|=2j=[Λε]+1exp(2tλβj1+mλj)[exp(Tλβj1+mλj)φj11+mλjTtexp(τλβj1+mλj)fϱε,j(WΛε)(τ)dτ]22(Λε)2β2βδexp(2t(Λε)βmλ1)j=[Λε]+1λ2(β+βδ)jexp(2Tλβjmλ1)u(t,),ej()2L2(Ω)2(Λε)2β2βδexp(2t(Λε)βmλ1)u(t,)2Gβ(η1,η2)(Ω),for2η1,η22Tmλ1. (4.45)

    Since limε0+ϱε=, for a sufficiently small ε>0, there is an ϱε>0 such that ϱεuL(0,T;L2(Ω)). For this value of ϱε (from (3.54)) we have fϱε(t,x;u)f(t,x;u). We estimate of MΛε10b(t) as follows

    |MΛε10b(t)|=2[Λε]j=1[11+mλjTtexp((τt)λβj1+mλj)(fϱε,j(u)(τ)fϱε,j(WΛε)(τ))dτ]22Tm2λ21Ttexp(2(τt)(Λε)βmλ1)[Λε]j=1|fϱε,j(u)(τ)fϱε,j(WΛε)(τ)|2dτ2TK2ϱεm2λ21Ttexp(2(τt)(Λε)βmλ1)WΛε(τ,)u(τ,)2L2(Ω)dτ. (4.46)

    Combining (4.45) and (4.46), we get

    WΛε(t,)u(t,)2L2(Ω)2(Λε)2β2βδexp(2t(Λε)βmλ1)u(t,)2Gβ(η1,η2)(Ω)+2TK2ϱεm2λ21Ttexp(2(τt)(Λε)βmλ1)WΛε(τ,)u(τ,)2L2(Ω)dτ. (4.47)

    Multiplying by exp(2t(Λε)βmλ1) both sides and using Grönwall's inequality, we obtain

    WΛε(t,)u(t,)L2(Ω)2exp(t(Λε)βmλ1)(Λε)β+βδu(t,)Gβ(η1,η2)(Ω)exp(TK2ϱεm2λ21(Tt)). (4.48)

    Combining (4.40), (4.43) and (4.48), we obtain (4.38). This completes the proof of Theorem 4.7.

    The paper investigate a final boundary value problem for a class of pseudo-parabolic partial differential equations with nonlinear reaction term. For 0<β<1, the well-posedness of solution is established. For β>1, the problem is ill-posed. Thus, we propose two methods to regularize the problem. The error estimates are given in the cases of globally or locally Lipschitzian source term.



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