Solvability of nonlinear fractional integro-differential equation with nonlocal condition

Abstract

Purpose

This study describes the applicability of the a priori estimate method on a nonlocal nonlinear fractional differential equation for which the weak solution's existence and uniqueness are proved. The authors divide the proof into two sections for the linear associated problem; the authors derive the a priori bound and demonstrate the operator range density that is generated. The authors solve the nonlinear problem by introducing an iterative process depending on the preceding results.

Design/methodology/approach

The functional analysis method is the a priori estimate method or energy inequality method.

Findings

The results show the efficiency of a priori estimate method in the case of time-fractional order differential equations with nonlocal conditions. Our results also illustrate the existence and uniqueness of the continuous dependence of solutions on fractional order differential equations with nonlocal conditions.

Research limitations/implications

The authors’ work can be considered a contribution to the development of the functional analysis method that is used to prove well-positioned problems with fractional order.

Originality/value

The authors confirm that this work is original and has not been published elsewhere, nor is it currently under consideration for publication elsewhere.

Citation

Aicha, S. and Merad, A. (2023), "Solvability of nonlinear fractional integro-differential equation with nonlocal condition", Arab Journal of Mathematical Sciences, Vol. 29 No. 2, pp. 172-190. https://doi.org/10.1108/AJMS-05-2021-0109

1. Introduction

Fractional order partial differential equations have become one of the most popular areas of research in mathematical analysis. Their application has been utilized in various scientific fields, such as optimal control theory, chemistry, physics, mathematics, biology, finance and engineering [1–5].

Integro-differential equations are a combination of derivatives and integrals which are appealing to both researchers and scientists for their applications in many areas [6–9]. Numerous mathematical formulations of physical phenomena include integro-differential equations, which may arise in modelling biological fluid dynamics [10–15].

It is important to establish effective methods to solve fractional differential equations (FDEs). Recently, a great deal of attention was dedicated to FDE solutions utilizing different methods, including the Adomian decomposition method [16,17], the Laplace transform method [18], exponential differential operators [19], the F-expansion method [20], non-Nehari manifold method [21] and the reproducing kernel space method [22,23], in the search for exact or analytical solutions. The applicability of most techniques becomes difficult with the presence of the integral condition. The energy inequality method is a useful tool for studying nonlocal fractional and classical problems. Compared with other techniques, it has an essential role in establishing the solution’s existence and uniqueness proof and depends on density arguments and certain a priori bounds.

There have been few articles related to nonlinear fractional partial equations that employ the energy inequality method [24]. Furthermore, for partial differential equations with classical order, many results have utilized this method [25–28]. Motivated by the previous results, the authors studied a nonlocal nonlinear time-fractional order problem. Moreover, we demonstrate the solution’s uniqueness, existence and dependence on the given data.

This article is outlined in the following way: in Section 2, we present the main problem. The next section is focused on posing the associated linear problem and introducing some required preliminaries and functional spaces. Then, in Section 4, we develop the energy inequality method to demonstrate the linear problem’s strong solution’s uniqueness. In addition, we prove the strong solution’s existence in Section 5. Moreover, we derive a priori bound and demonstrate the generated operator range density in a Hilbert space. We solve the nonlinear problem in Section 6 by utilizing the results achieved in Sections 4 and 5, and an iteration process.

2. Statement of problem

In the region D=Ω×0,T, Ω=0,1, T < ∞, we pose the nonlinear fractional equation

(1)Lv=∂tβ+1Cv−∂∂xγx,t∂v∂x−∂2∂x∂tηx,t∂v∂x−∫0tξt−zvx,zdz=fx,t,v,∂v∂x

with 0 < β < 1.

Associated with initial condition

(2)ℓ1v=vx,0=φx,ℓ2v=∂vx,0∂t=ψx,x∈Ω,

and the boundary condition

(3)∫01vx,tdx=0,vx1,t=0,t∈0,T,

Such that the known functions γ, η and ξ verify Assumption 1, and data functions f, φ and ψ belong to suitable function spaces as mentioned in Section 3.

In the Caputo definition for a function v, the fractional derivatives of order β + 1 with 0 < β < 1 is defined as

(4)∂tβ+1Cv(x,t)=1Γ1−β∫01vττx,τt−ββdτ,

where Γ. is the gamma function

and the Riemann-Liouville integral of order 0 < β < 1, which is given by

(5)Dt−βv(x,t)=1Γβ∫01vx,τt−β1−βdτ.

3. Technical tools and associated linear problem

We define some function spaces and tools required to investigate the following linear problem associated with problems (1)–(3).

(6)Lv=∂tβ+1Cv−∂∂xγx,t∂v∂x−∂2∂x∂tηx,t∂v∂x−∫0tξt−zvx,zdz=fx,t

(7)ℓ1v=vx,0=φx,ℓ2v=∂vx,0∂t=ψx,x∈Ω,

(8)∫01vx,tdx=0,vx1,t=0,t∈0,T,

First, we convert problems (1)–(3) into an equivalent operator form

(9)Lv=F=f,φ,ψ

where the unbounded operator L=L,ℓ1,ℓ2 with L: E → F is defined in DL such that

(10)DL=v∈L2D,∂tβ+1cv,∂v/∂t,∂v/∂x,∂2v/∂x2,∂3v/∂x2∂t∈L2D∫01vx,tdx=0,vx1,t=0,t∈0,T,

and v also verify the initial condition. Here E is Banach space containing elements having the finite norm

(11)∥v∥E2=Dtβ−1‖Ix∂v∂t‖L2(Ω)2+∥∂v∂t∥L2D2+∥v∥C0,T,L2Ω2,

and F is Hilbert space composed of functions normed with

(12)‖F‖F2=‖Ixf‖L2D2+‖φ‖L2Ω2+‖Ixψ‖L2Ω2.

Lemma 1.

[29] Let S(t) a nonnegative absolute continuous function verifying the inequality

(13) ∂tαCS(t)≤c1S(t)+c2(t),0<α<1,

for almost all t ∈ [0, T], where c₁ is a positive constant and c₂(t) is an integrable nonnegative function on [0, T]. Then,

(14) S(t)≤S(0)Eα(c1tα)+Γ(α)Eα,α(c1tα)Dt−αc2(t),

where

Eα(x)=∑n=0∞xnΓαn+1andEα,ν(x)=∑n=0∞xnΓαn+ν,

are Mittag-Leffler functions.

Lemma 2.

[29] On the interval [0, T], any absolute continuous function y(t) verifies the following estimate:

(15) StC∂tβS(t)≥12∂tβcS2t,0<β<1,

Lemma 3.

[30] For any n∈N, we have

(16) ∥Ix2nv∥L20,l2≤l22n∥v∥L20,l2,

where

Ix2nv=∫0x∫0ξ1…∫0ξ2n−1vη,tdηdξ2n−1…dξ1=∫0xx−ξ2n−12n−1!vξ,tdξ.

Cauchy ɛ-inequality [31]

(17)∣ab∣≤ε2a2+12εb2,

which holds for arbitrary a and b, and all ɛ > 0.

4. A priori estimate and consequences

Assumption 1.

For any x,t∈D, we suppose that

(18) c0≤γ(x,t)≤c1,∂γ(x,t)∂t≤c2,∂γ(x,t)∂x≤c3c4≤η(x,t),c5≤∂η(x,t)∂t≤c6,∂η(x,t)∂x≤c7∂2η(x,t)∂t2≤c8,∂2η(x,t)∂t∂x≤c9,ξ(x,t)≤c10

such that c_i(i = 0, …, 10) are positive constants.

Theorem 4.

Let Assumption 1 be fulfilled. Then, any function v ∈ D(L) verify the following estimate

(19) ∥v∥E≤C∥Lv∥F.

where C > 0 constant independent of v.

Proof.

We take the scalar product L2Dτ of equality (6) and the integro-differential operator Mv=−2Ix2∂v∂t, such that τ ∈ [0, T], we have

(20) 2∂tβ+1Cv,−Ix2∂v∂tL2Dτ−2∂∂xγx,t∂v∂x,−Ix2∂v∂tL2Dτ−2∂2∂x∂tηx,t∂v∂x,−Ix2∂v∂tL2Dτ−2∫0tξt−zvx,zdz,−Ix2∂v∂tL2Dτ=2fx,t,−Ix2∂v∂tL2Dτ.

The integration of the first three terms on the left-hand side (LHS) of Equation (20), taking into consideration initial and boundary conditions (2)−(3), gives

(21)−∂tβ+1Cv,−Ix2∂v∂tL2Dτ=∂tβC∂v∂t,Ix2∂v∂tL2Dτ=2∂tβCIx∂v∂t,Ix∂v∂tL2Dτ

(22)−2∂∂xγx,t∂v∂x,−Ix2vtL2Dτ=−∫Dτ∂γ∂tv2dxdt+2∫Dτ∂γ∂xvIx∂v∂tdxdt+∫01γx,τv2x,τdx−∫01γx,0φ2xdx

(23)−2∂2∂x∂tηx,t∂v∂x,−Ix2∂v∂tL2Dτ=2∫Dτη∂v∂t2dxdt+∫01∂ηx,τ∂tv2x,τdx−∫Dτ∂2η∂t2v2dxdt+2∫Dτ∂η∂x∂v∂tIx∂v∂tdxdt+2∫Dτ∂2η∂x∂tvIx∂v∂tdxdt−∫01∂ηx,0∂tφ2xdx

Substituting (21)−(23) into (20) yields

(24)2∂tβCIx∂v∂t,Ix∂v∂tL2Dτ+2∫Dτη∂v∂t2dxdt+∫01γx,τ+∂ηx,τ∂tv2x,τdx=2fx,t,−Ix2∂v∂tL2Dτ+2∫0tξt−zvx,zdz,−Ix2∂v∂tL2Dτ+∫Dτ∂γ∂t+∂2η∂t2v2dxdt+∫01γx,0+∂ηx,0∂tφ2xdx−2∫Dτ∂η∂x∂v∂tIx∂v∂tdxdt−2∫Dτ∂γ∂x+∂2η∂x∂tvIx∂v∂tdxdt

By applying inequality (16), we estimate the first and the last two terms on the right-hand side (RHS) of (24); as such it follows that

(25)−2∫DτfIx2∂v∂tdxdt≤∫DτIxf2dxdt+∫DτIx∂v∂t2dxdt,

(26)2∫Dτ∫0tξt−zvx,zdzIx2∂v∂tdxdt≤c10T2∥v∥LDτ2+12∫DτIx∂v∂t2dxdt

(27)−2∫Dτ∂γ∂x+∂2η∂x∂tvIx∂v∂tdxdt≤2∫Dτ∂γ∂x2+∂2η∂x∂t2v2dxdt+∫DτIx∂v∂t2dxdt,

(28)−2∫Dτ∂η∂x∂v∂tIx∂v∂tdxdt≤c4∫Dτ∂v∂t2dxdt+1c4∫Dτ∂η∂x2Ix∂v∂t2dxdt.

By Lemma 2, the first term on the LHS of (20) becomes

(29)2∂tβCIx∂v∂t,Ix∂v∂tL2Dτ≥∫Dτ∂tβCIx∂v∂t2dxdt,

Hence, by Formulas (25)–(29) and Assumption (1), we obtain

∫Dτ∂tβCIx∂v∂t2dxdt+∫0τ‖∂v∂t.,t‖L2Ω2dt+‖v.,τ‖L2Ω2≤δ1∫0τ‖Ixf.,t‖LΩ2dt+‖φ‖L2Ω2+∫0τ‖Ix∂v∂t.,t‖LΩ2dt+δ2∫0τ‖v.,t‖L2Ω2dt,

where

δ1=max1,c1+c6,52+c72c4minc4,c0+c5,1,

δ2=c2+c8+c32+c92+c10T2minc4,c6+c5,1.

Now, since

(30)∫0τ∂tβC∥Ix∂v∂t2∥L2Ωdt=Dτβ−1‖Ix∂v∂t‖L2(Ω)2−τ1−β(1−β)Γ(1−β)‖Ixψ‖L2(Ω)2,

then

(31)Dτβ−1∥Ix∂v∂t‖L2Ω2+∫0τ∥∂v∂t.,t∥L2Ω2dt+∥v.,τ∥L2Ω2≤δ3∫0τ∥Ixf.,t∥LΩ2dt+∥φ∥L2Ω2+‖Ixψ‖L2(Ω)2+∫0τ∥Ix∂v∂t.,t∥LΩ2dt+δ2∫0τ∥v.,t∥L2Ω2dt,

where

δ3=maxδ1,T1−β1−βΓ1−β,

We need to drop the last term on the RHS of (31). Therefore, we use Gronwall’s lemma, which yields

(32)Dτβ−1‖Ix∂v∂t‖L2(Ω)2+∫0τ∥∂v∂t.,t∥L2Ω2dt+∥v.,τ∥L2Ω2≤δ4∫0τ∥Ixf.,t∥LΩ2d+∥φ∥L2Ω2t+‖Ixψ‖L2(Ω)2+∫0τ‖Ix∂v∂t.,t∥LΩ2dt,

where

δ4=expδ2Tδ3,

Now, by discarding the last two terms on the LHS of (32) then posing Sτ=∫0τ‖Ix∂v∂t.,t‖L2(Ω)2dt, ∂tβ+1CSτ=Dτβ−1‖Ix∂v∂t‖L2(Ω)2,with S0=0, in Lemma (1), we obtain

(33)∫0τ‖Ix∂v∂t.,t‖L2(Ω)2dt≤δ5Dτ−1−β∥Ixf∥LΩ2+∥φ∥L2Ω2+∥Ixψ‖L2(Ω)2,

where

δ5=Γ(β)Eβ,β(c17Tβ)max1,Tβ+11+βΓ(1+β)

Combining (32)−(33) yields

(34)Dτβ−1‖Ix∂v∂t‖L2(Ω)2+∫0τ‖∂v∂t.,t‖L2Ω2dt+‖v.,τ‖L2Ω2≤δ6Dτ−β−1‖Ixf‖L2(Ω)2+∫0τ‖Ixf.,t‖LΩ2dt+‖φ‖L2Ω2+‖Ixψ‖L2(Ω)2,

where

δ6=maxδ4δ5,δ5,

From given inequality

(35)Dτ−1−β‖Ixf∥LΩ2≤TβΓ(1+β)∫0τ∥Ixf∥LΩ2dt,

we reduce inequality (34) as follows

(36)Dτβ−1‖Ix∂v∂t‖L2(Ω)2+∫0τ∥∂v∂t.,t∥L2Ω2dt+∥v.,τ∥L2Ω2≤δ7∫0T∥Ixf.,t∥LΩ2dt+∥φ∥L2Ω2+‖Ixψ‖L2(Ω)2.

δ7=δ61+TβΓ(1+β)

Since the RHS of estimate (36) is independent of τ, we can take the supremum on the LHS with respect to τ over [0, T]. Thus, we get the desired inequality (19). Theorem (4) proof is complete. ■

5. Existence of the linear problem solution

The current section’s aim is to prove the existence of the strong solution of problems (6)–(8). It remains to demonstrate the density of the range R(L).

Proposition 5.

[32] The operator L engendered by problems (1)–(3) has a closure.

Defining the operator equation solution

L̅v=F=f,φ,ψ,

as a strong solution of problems (6)–(8). The inequality (19) can be extended into

(37)∥v∥E≤∥L̅v∥F,∀v∈DL̅.

the inequality demonstrated above assures the strong solution uniqueness.

Corollary 6.

The range of the operator L̅ is closed in F and R(L̅)=R(L)̅ and L̅−1=L−1̅.

Theorem 7.

Let Theorem (4) conditions be verified. Then, for any F=(f,g,h)∈ F, the problems (6)–(8) have a unique solution v such that v=L̅−1F=L−1̅F.

Proposition 8.

Let Assumption (1) be fulfilled. If for a certain function g ∈ L²(Q), and every v ∈ D(L) verifying homogenous initial conditions, we have

(38) (Lv,g)L2(D)=0,

then g vanishes almost everywhere in D are as follows:

Proof.

Introducing a new function σ(x, t) verifies conditions (2) and (3), and σ, σ_x,Itσx, Itσ and ∂tβ+1Cσ∈L2D, then we pose

vx,t=It2σ,

where

Itσ=∫0tσ(x,s)ds,It2σ=∫0t∫0sσ(x,z)dzds.

Equation (38) then becomes

(39) ∂tβ+1cIt2σ−∂∂xγx,tIt2∂σ∂x−∂2∂x∂tηx,tIt2∂σ∂x−∫0tξt−zIt2σx,zdz,gL2(D)=0.

Now, we consider the function

(40) gx,t=−ItIx2σ.

Obviously, the function g included in L²(D). Equations (39)–(40) lead to

(41) −∂tβ+1CIt2σ,ItIx2σL2(D)+∂∂xγx,tIt2∂σ∂x,ItIx2σL2(D)+∂2∂x∂tηx,tIt2∂σ∂x,ItIx2σL2(D)+∫0tξt−zIt2σx,zdz,ItIx2σL2(D)=0

Note that the function σ verifies conditions (2)−(3), then we have

(42) −∂tβ+1CIt2σ,ItIx2σL2(D)=−∂tβCIxItσ,IxItσL2(D)

(43) ∂∂xγx,tIt2∂σ∂x,ItIx2σL2(D)=12∫01γIt2σ2dx−12∫D∂γ∂tIt2σ2dxdt+∂γx,t∂xIt2σ,ItIxσL2(D)

(44) ∂2∂x∂tηx,tIt2∂σ∂x,ItIx2σL2(D)=∂2η∂x∂tIt2σ,ItIxσL2(D)+∂η∂tIt2σ,ItσL2(D)+∫D∂2η∂x2ItIxσ2dxdt+∫DηItσ2dxdt

Insertion of Equations (42)–(44) into (41), yields

(45) −2∂tβCIxItσ,IxItσL2(D)+∫01γIt2σ2dx=∫D∂γ∂tIt2σ2dxdt−2∫D∂2η∂x2ItIxσ2dxdt−2∫DηItσ2dxdt−2∂2η∂x∂t+∂γx,t∂xIt2σ,ItIxσL2(D)−2∂η∂tIt2σ,ItσL2(D)−2∫0tξt−zIs2σx,zdz,ItIx2σL2(D)

According to Lemma 1, we bound the first term on the LHS of (45); we have

(46) 2∂tβCIxItσ,IxItσL2(D)≥∂tβC∥IxItσ∥L2(D)2,

Also, we bound the last three terms on the RHS of (45) utilizing inequality 17, and we then get

(47) 2∂γx,t∂x+∂2η∂x∂tIt2σ,ItIxσL2(D)≤2c32+c92∥It2σ∥L2(D)2+∥IxItσ∥L2(D)2

(48) 2∂η∂tIt2σ,ItσL2(D)≤2c4∥It2σ∥L2(D)2+c622c42∥Itσ∥L2(D)2

(49) ∫0tξt−zIz2σx,zdz,Ix2ItσL2(D)≤c10T2∥It2σ∥LD2+12∥IxItσ∥LD2,

The insertion of estimates (46)–(48) in Equation (45) gives

(50) Dτβ−1‖IxItσ‖L2(D)2+∫01It2σ2dx≤δ8∫DIt2σ2dxdt+‖IxItσ‖L2(D)2

with

(51) δ8=maxc2+2c32+c92+c10T2+c622c4,2c10+32min1,c0

Eliminating the first term on the LHS of (50), using Lemma 2, with

(52) Sτ=∫0τ∫01It2σ2dxdt

observe that S0=0, then we get

(53) Sτ≤T⁡expTδ8‖IxItσ‖L2(Dτ)2,

Similarly, by discarding the second integral on the LHS of (50) and applying (53), we obtain

(54) Dτβ−1∥IxItσ∥L2(D)2≤δ8T⁡expTδ8+1∥IxItσ∥L2(D)2

by Lemma 2, with

Sτ=∫0τ∫01IxItσ2dxdt

and

∂τβCSτ=Dτβ−1‖IxItσ‖L2(D)2,S0=0,

it follows that

(55) ‖IxItσ‖L2(Dτ)2≤S(0)Eβ(δ8T⁡expTδ8+1τβ)+Γ(β)Eβ,β(δ8T⁡expTδ8+1τβ)Dτ−β0=0,

for any τ ∈ [0, T]. Hence inequality (55) shows that g = 0 ae in D. Continuing Theorem 7 proof, we assume that for a certain function G=g,g0,g1∈RL⊥, we have

(56) Lv,gL2D+l1v,g0L2Ω+l2v,g1L2Ω=0,

then we should show that g₀ = 0, g₁ = 0. Putting v ∈ D(L), verifying homogenous initial conditions into (56), yields

(57) Lv,gL2D=0,∀v∈DL,

By applying Proposition (8) to (57), we see that g = 0. Consequently, (56) becomes

(58) l1v,g0L2Ω+l2v,g1L2Ω=0∀v∈DL,

Since l₁v and l₂v are independent and their ranges l₁ and l₂ are everywhere dense in L2Ω, we conclude that g₀ = g₁ = 0, this complete the proof of Theorem 7. ■

6. The study of the nonlinear problem

This section is devoted to solving the main problems (1)–(3). Consider now the auxiliary problem with the homogenous equation:

(59)LV=∂tβ+1CV−∂∂xγx,t∂V∂x−∂2∂x∂tηx,t∂V∂x−∫0tξt−zVx,zdz=0

(60)ℓ1V=Vx,0=0,ℓ2V=∂Vx,0∂t=0,x∈Ω,

(61)∫01Vx,tdx=0,Vx1,t=0,t∈0,T,

If V and v are solutions of problems (8)-(6),(1)-(3), respectively, then h = v − V satisfies

(62)Lh=∂tβ+1Ch−∂∂xγx,t∂h∂x−∂2∂x∂tηx,t∂h∂x−∫0tξt−zhx,zdz=Gx,t,h,∂h∂x

(63)ℓ1h=hx,0=0,ℓ2h=∂h∂tx,0=0,x∈Ω,

(64)∫01hx,tdx=0,hx1,t=0,t∈0,T,

such that the function G(x,t,h,∂h∂x)=f(x,t,h+V,∂h∂x+∂V∂x), verifies the following condition

(65)|G(x,t,w1,y1)−G(x,t,w2,y2)|≤L|w1−w2|+|y1−y2|forall(x,t)∈D.

Theorem 7 shows that the solution of problems (6)–(8) is unique and depends continuously on the initial data. It remains to establish a similar proof for problems (62)–(64). We introduce the space

(66)C̃1D=w∈C1Dsuchthat,∂w2∂t∂x∈CD

Suppose that h and u∈C̃1(D) verify homogenous initial and boundary conditions h(x,T)=0,h(x,0)=0,∫01w(x,t)dx=0. For u∈C̃1(D), we have

(67)Lh,IxuL2(D)=∂tβ+1Ch,IxuL2(D)−∂∂xγx,t∂h∂x,IxuL2(D)−∂2∂x∂tηx,t∂h∂x,IxuL2(D)−∫0tξt−zhx,zdz,IxuL2(D)

Computation of all terms of Equation (67), using conditions on h and u, gives

(68)∂tβ+1Ch,IxuL2(D)=−∂tβ+1CIxh,uL2(D)

(69)−∂∂xγ∂h∂x,IxuL2(D)=γ∂h∂x,uL2(D)

(70)−∂2∂x∂tη∂h∂x,IxuL2(D)=∂∂tη∂h∂x,uL2(D)

(71)−∫0tξt−zhx,zdz,IxuL2(D)=∫0tξt−zIxhx,zdz,uL2(D)

Insertion of (68)–(71) into (67) yields

(72)R(h,u)=u,IxGL2(D)

such that

(73)R(h,u)=−∂tβ+1CIxh,uL2(D)+γ∂h∂x,uL2(D)+∂∂tη∂h∂x,uL2(D)+∫0tξt−zIxhx,zdz,uL2(D)

Definition 9.

A function h ∈ L²(0, T, H¹(Ω)) is considered as the problems (62)–(64) weak solution if it satisfies (64) and (72) holds.

Constructing an iteration sequence as follows: let h(0) = 0, and let defining the sequence (h(n))n∈N as follows: if h⁽ⁿ⁻¹⁾ is given, then for n∈N solve the following problem:

(74)Lh=∂tβ+1Ch(n)−∂∂xγ∂h(n)∂x−∂2∂x∂tη∂h(n)∂x−∫0tξt−zh(n)x,zdz=Gx,t,h(n−1),∂h(n−1)∂x

(75)ℓ1h(n)=h(n)x,0=0,ℓ2h(n)=∂h(n)∂tx,0=0,x∈Ω,

(76)∫01h(n)x,tdx=0,hx(n)1,t=0,t∈0,T,

Theorem 10.

For each fixed n, assume that the solution of problems (74)–(76) h⁽ⁿ⁾(x, t) is unique. If we put H⁽ⁿ⁾(x, t) = h⁽ⁿ⁺¹⁾(x, t) − h⁽ⁿ⁾(x, t), then we obtain

(77) LH(n)=∂tβ+1CH(n)−∂∂xγ∂H(n)∂x−∂2∂x∂tη∂H(n)∂x−∫0tξt−zH(n)x,zdz=Ψn−1x,t

(78) ℓ1H(n)=H(n)x,0=0,ℓ2H(n)=∂H(n)x,0∂t=0,x∈Ω,

(79) ∫01H(n)x,tdx=0,Hx(n)1,t=0,t∈0,T,

with

Ψn−1x,t=Gx,t,hn,∂hn∂x−Gx,t,hn−1,∂hn−1∂x

Lemma 11.

Under Assumptions (1), and supposing that the condition (65) holds, then for the linearized problems (77)–(79), the following estimate holds

(80) ‖H(n)‖L2(0,T,H1Ω)≤K‖H(n−1)‖L2(0,T,H1Ω)

where K > 0 is constant given by

K=expδ10T1+Γ(β)Eβ,β(δ9⁡expδ10Ttβ)TβΓ1+β

Proof.

We take the scalar product in L²(D^τ), τ ∈ [0, T] of (77) and the integro-differential operator MH(n)=−Ix2H(n), we get

(81) 2∂tβ+1CH(n),−Ix2∂H(n)∂tL2Dτ−2∂∂xγx,t∂H(n)∂x,−Ix2∂H(n)∂tL2Dτ−2∂2∂x∂tηx,t∂H(n)∂x,−Ix2∂H(n)∂tL2Dτ−2∫0tξt−zH(n)x,zdz,−Ix2∂H(n)∂tL2Dτ=2Ψn−1x,t,−Ix2∂H(n)∂tL2Dτ.

Integrations by parts all terms of (81), by using conditions (78)−(79), proceeding as in the establishment of Theorem 4, yields

(82) Dβ−1‖Ix∂H(n)∂t‖L2(Ω)2+c0+c5∥H(n).,τ∥L2Ω2≤∫0τ∥IxΨn−1x,t∥LΩ2dt+52+c722c4∫0τ∥Ix∂H(n)∂t.,t∥LΩ2dt+2c32+c92+c2+c8+c10T2∫0τ∥H(n).,t∥L2Ω2dt

On the other hand, applying to Equation (77) the operator Ix, and taking into consideration condition (79), multiplying the resulting equation with ∂T(n)∂x and integrating over D^τ, we get

(83) ∫Dτ∂tβ+1CIxH(n)∂H(n)∂xdxdt−∫Dτγx,t∂H(n)∂x2dxdt−∫Dτ∂∂tηx,t∂H(n)∂x∂H(n)∂xdxdt−∫Dτ∫0tξt−zIxH(n)x,z∂H(n)∂xdzdxdt=∫DτIxΨn−1x,t∂H(n)∂xdxdt

After integration by parts of all the terms of (83) and taking into consideration conditions (78), (79) and using inequality (17), we have

(84) ∫Dτ∂tβ+1CH(n)H(n)dxdt+c0∫0τ∥∂H(n)∂x.,t∥L2Ωdt+12c4∥∂H(n)∂x.,τ∥L2Ω2≤12∫0τ∥Ψn−1∥L2Ωdt+c10T2+1∫0τ∥H(n).,t∥L2Ωdt

Combination of inequalities (83)−(84) gives

(85) Dτβ−1‖Ix∂H(n)∂t‖L2Ω+∫0τ∂tβ+1H(n)H(n)dxdt+c0∫0τ∥∂H(n)∂x.,t∥L2Ωdt+12c4∥∂H(n)∂x.,τ∥L2Ω2+c0+c5∥H(n).,τ∥L2Ω≤∫0τ∥Ψn−1∥L2Ωdt+52+C722C4∫0τ∥Ix∂H(n)∂t.,t∥L2Ωdt+2c32+c92+c2+c8+c10T2+1∫0τ∥H(n).,t∥L2Ω2dt

Eliminating the last term on the RHS of (85), by using Gronwall’s lemma, it comes

(86) Dτβ−1‖Ix∂H(n)∂t‖L2Ω+∫0τ∂tβ+1H(n)H(n)dxdt+c0+c5∥H(n).,τ∥L2Ω+12c4∥∂H(n)∂x.,τ∥L2Ω2+c0∫0τ∥∂H(n)∂x.,t∥L2Ωdt≤expδ10∫0τ∥Ψn−1∥LΩ2dt+δ9∫0τ∥Ix∂H(n)∂t.,t∥LΩ2dt

where

δ9=52+C722C4δ10=2c32+c92+c2+c8+c10T2+1

To discard the last integral on the RHS of inequality 86, we drop the three first elements then use the Gronwall’s lemma, it follows

(87) ∫0τ‖Ix∂H(n)∂t‖L2Ω2dt≤Γ(β)Eβ,β(δ9⁡expδ10Ttβ)expδ9tDt−β∥Ψn−1∥LΩ2

On the other hand, via the condition (65), we get

(88) ∫0τ∥Ψn−1∥LΩ2dt≤2L2∫0τ∥H(n−1).,t∥L2Ω2+∥∂H(n−1).,t∂x∥L2Ω2dt

Combining (86)–(88) and by using (35), we get

(89) Dβ−1‖Ix∂H(n)∂t‖L2(Ω)2+∫0τ∂tβ+1H(n)H(n)dxdt+∫0τ∥∂H(n)∂x.,t∥L2Ω2dt+∥∂H(n)∂x.,τ∥L2Ω2+∥H(n).,τ∥L2Ω2≤δ11L2∫0T‖H(n−1).,t‖L2Ω2+‖∂H(n−1)∂x.,t‖L2Ω2dt

where

δ11=expδ10T1+Γ(β)Eβ,β(δ9⁡expδ10Ttβ)TβΓ1+β

After discarding the first two terms on the LHS of inequality (89), we get

(90) ∫0τ∥∂H(n)∂x.,t∥L2Ω2dt+∥H(n).,τ∥L2Ω2+∥∂H(n)∂x.,τ∥L2Ω2≤δ11L2∫0T‖H(n−1).,t‖L2Ω2+‖∂H(n−1)∂x.,t‖L2Ω2dt

Here, the RHS doesn’t depend on τ so, we can replace the LHS by upper bounds with respect to τ, we obtain

(91) ∫0T∥∂H(n)∂x.,t∥L2Ω2dt+∥H(n).,τ∥L2Ω2+∥∂H(n)∂x.,τ∥L2Ω2≤δ11L2∫0T‖H(n−1).,t‖L2Ω2+‖∂H(n−1)∂x.,t‖L2Ω2dt

Now, we integrate over 0,T, we get

(92) ∫0T∥H(n).,τ∥L2Ω2dt+∫0T∥∂H(n)∂x.,t∥L2Ω2dt≤δ12L2∫0T‖H(n−1).,t‖L2Ω2+‖∂H(n−1)∂x.,t‖L2Ω2dt

δ12=δ11L2Tmin1,T

We get then the desired inequality (80).

(93) ‖H(n)‖L2(0,T,H1Ω)≤δ12‖H(n−1)‖L(0,T,H1Ω)

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Using the convergence of series criteria we conclude that ∑n=1∞Hn converges if δ11L2Tmin1,T<1, namely if L<min1,Tδ11T. Since Hnx,t=hn+1x,t−hnx,t, then the sequence hnn∈N given by hnx,t=∑i=0n−1Hi+h0x,t, i∈N converges to a function h∈L20,T,H10,1.

In order to show that this limit is the solution of problems (77)–(79), it is sufficient to demonstrate that h verifies (64) and (72).

We have, from problems (74)–(76), that

(94) R(hn,u)=u,IxGx,t,hn−1,∂hn−1∂xL2(D)

Precisely

(95) R(hn−h,u)+R(h,u)=u,IxGx,t,hn−1,∂hn−1∂x−IxGx,t,h,∂h∂xL2(D)+u,IxGx,t,h,∂h∂xL2(D)

using Equation (74), then (95) becomes

(96) R(hn−h,u)=−∂tβ+1cIxhn−h,uL2(D)+γ∂hn−h∂x,uL2(D)+∂∂tη∂hn−h∂x,uL2(D)+∫0tξt−zIxhn−hx,zdz,uL2(D)

By integrating the parts on all terms on the LHS, and taking into consideration conditions on v and w, (96) transforms into

(97) R(hn−h,u)=−∂tβ+1Chn−h,IxuL2(D)+γ∂hn−h∂x,uL2(D)+η∂hn−h∂x,∂u∂tL2(D)+∫0tξt−zIxhn−hx,zdz,uL2(D)

Applying Cauchy-Schwartz inequality yields

(98) R(hn−h,u)≤δ13∥hn−h∥L20,T,H1Ω∥u∥L2D+∥∂u∂t∥L2D

where

δ13=maxc1+Tc02,c6

and from (95) we have the following estimate

(99) u,IxGx,t,hn−1,∂hn−1∂x−IxGx,t,h,∂h∂xL2(DT)≤L2∥hn−h∥L20,T,H1Ω∥u∥L2(D

Passing to the limit n → ∞ in (97), and taking into consideration (98)-(99), we obtain

(100) R(h,u)=u,IxGx,t,h,∂h∂xL2D

To conclude that problems (77)–(79) admit a weak solution, we prove that (64) holds. Since limn→∞∥hn−h∥L20,T;H1Ω=0 then, we deduce that ∫01hdx=0 and ∂h∂x1,t=0.

Therefore, we have established this result:

Theorem 12.

Suppose that conditions of Lemma (11) hold, and that L<min1,Tδ11T, then the problems (62)–(64) admit a weak solution in L²(0, T, H¹(Ω)).

Now, we prove the uniqueness of problems (62)–(64).

Theorem 13.

Under conditions of Lemma (11), the problems (62)–(64) admits unique solutions.

Proof.

Suppose that the problems (62)– (64) admit v₁ and v₂ as solutions in L²(0, T, H¹(Ω)), then H = v₁ − v₂ belongs to L²(0, T, H¹(Ω)) and verifies

(101) LH=∂tβ+1CH−∂∂xγx,t∂H∂x−∂2∂x∂tηx,t∂H∂x−∫0tξt−zHx,zdz=Ψx,t

(102) ℓ1H=Hx,0=0,ℓ2H=Htx,0=0,x∈Ω,

(103) ∫01Hx,tdx=0,Hx1,t=0,t∈0,T,

where Ψx,t=Gx,t,v1,∂v1∂x−Gx,t,v2,∂v2∂x.

This will be done by establishing the same proof of Lemma 11; we obtain

(104)‖H‖L2(0,T,H1Ω)≤K‖H‖L2(0,T,H1Ω)

Since K < 1, then from (80) we have (1−K)‖H‖L2(0,T,H1Ω)≤0, from which we deduce that v₁ = v₂ in L20,T,H1Ω. ■