# A Nonlinear Fractional Problem with a Second Kind Integral Condition for Time-Fractional Partial Differential Equation

Authors: Benbrahim Abdelouahab; Taki-Eddine Oussaeif; Adel Ouannas; Khaled M. Saad; Hadi Jahanshahi; Ahmed Diar; Awad M. Aljuaid; Ayman A. Aly
Journal: Journal of Function Spaces
 (2022)

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## Abstract

The aim of this research is to demonstrate the existence and the uniqueness of the weak solution for a semilinear fractional parabolic problem with the special case of the second integral boundary condition. For this aim, we split the proof into two parts; to study the main linear problem part, we used the variable separation method, and concerning the semilinear problem part, we apply an iterative method and a priori estimate for the study of the weak solution.


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## Body

## 1. Introduction

Fractional calculus is a mathematical analysis branch that explores the various different possibilities of describing the differentiation operator power of real or complex numbers. Fractional differential equations are extraordinary differential equations [1]. Many natural phenomena and modern problems of physics, mechanics, biology and technology, chemistry, engineering, etc. can be modeled by fractional partial differential equations (FPDEs). For details, see ([2–12]) and the references therein.The first who drew attention to these problems with an integral one-space variables condition is Cannon [13], which gold of the study of heat conduction in a bar heated thin is demonstrated by using the potential method, and the importance of the problems with integral conditions has been pointed out by Samarskii [14]. The basic physical meaning of integral conditions (total energy, average temperature, the total mass of impurities, total flux, moments, etc.) has served as the main reason for the growing interest in this kind of problem.In modern physics and technology, many problems are found using nonlocal conditions for partial differential equations, which are defined using integral conditions. So, the first type of integral condition is given by the following:(1)∫01kx,tux,tdx=Et,where k is a given function.Or second type(2)ul,t=∫01kx,tux,tdx,∀t∈0,T,∂u∂xl,t=∫01kx,tux,tdx,∀t∈0,T,can be used when it is impossible to directly measure the quantity sought on the border. Its total value or its average is known. We find a lot of research studied this kind of integral like [15–17].Many researchers have widely studied the solvability “existence and uniqueness” of solutions of fractional and integer partial differential equation problem by many method; see, for example, [18–23].There are a few works that study the nonlinear fractional partial differential equation with initial and boundary conditions, not to mention the nonlocal problem like [24].In this work, we developed the study of fractional problems to partial differential equations on the one hand and beyond this in the nonlinear direction, which simulates heat diffusion in the complex phenomena. We also relied on the modeling of Neumann’s condition, i.e., heat flow by an integral condition, which is closer to reality in this situation.Motivated by this, in our article, we treat for the existence and the uniqueness of the weak solution of the mixed semilinear problem for a fractional partial differential equation with an integral condition of the second type, where we start by studying the nonlocal linear problem by the variable separation method, to demonstrate the existence and the uniqueness of the weak solution of the semilinear problem; we then apply an iterative method based on the results obtained by the linear problem.


## 2. Preliminaries

Definition 1 (see [25, 26]).
For any0<α<1, the Caputo and Riemann Liouville derivatives are defined, respectively, as follows:
(1)
The left Caputo derivatives:(3)0CDtαux,t≔1Γα−1∫0t∂ux,t∂Γ1t−ταdτ(2)
The left Riemann-Liouville derivatives:(4)0RDtαux,t≔1Γα−1∂∂t∫0tux,t∂τ1τ−tαdτ(3)
The right Caputo derivatives:(5)0RDtαux,t≔−1Γα−1∫0t∂ux,t∂τ1τ−tαdτ(4)
The right Riemann-Liouville derivatives:(6)tRDtαux,t≔1Γα−1∂∂t∫0tux,tτ−tαdτProposition 2 (see [25, 26]).
Forn=1, we have
(7)0RDtαux,t=0cDtαux,t+u0Γ1−αtα,0RDtαux,t=0cDtαux,t+uTΓ1−αT−tα.
Ifux,0=0, then we have
(8)0RDtαux,t=0cDtαux,t.Definition 3 (see [27]).
For any realσ>0 and finite interval a,b of the real axis R, we define the seminorm
(9)ul2HαΩ≔RDtαu2L2Ω,and norm
(10)ulHαΩ≔uL2Ω2+ul2H0αΩ1/2.We then define lH0αΩ as the closure of C0∞Ω with respect to the norm ⋅lH0αΩ.Definition 4 (see [27]).
For any realσ>0, we define the seminorm
(11)ur2H0αΩ≔tRDTαuL22Ω,and norm
(12)urH0σΩ≔uL22Ω+ur2H0σΩ1/2.
We then defineRH0αΩ as the closure of C0∞Ω with respect to the norm ⋅rH0αΩ.Definition 5.
For any realσ>0, we define the seminorm
(13)ucHαΩ=DRtαu,DtRαuL2Ωcosαπ1/2,and norm
(14)ucHαΩ=uL2Ω2+ucHαΩ21/2.Lemma 6 (see [27, 28]).
For any realσ∈ℝ+ if u∈lHαΩ and v∈C0∞Ω, then
(15)RDtαut,vtL2Ω=ut,TRDαvtL2Ω.Lemma 7 (see [27, 28]).
For0<σ<2,σ≠1,u∈H0σ/2Ω on a
(16)RDtσut=RDtσ/2RDtσ/tut.Lemma 8 see ([27, 28]).
Forσ∈ℝ+,σ≠n+1/2, the seminorms ⋅lHσΩ,⋅rHσΩ and ⋅cHσΩ are equivalent. Then, we pose
(17)⋅lHσΩ≅⋅rHσΩ.Lemma 9 see ([27, 28]).
For any realσ>0, the space RH0σΩ with respect to the norm (12) is complete.


## 3. Formulation of the Nonlinear Problem (18)

LetQ=x,t∈ℝ2with0<x<1and0<t<T.We would like to deal with the following nonlinear problem:(18)0cDtαux,t−a∂2ux,t∂x2+bux,t=fx,t,u,∂u∂x,∀x,t∈Q,ux,0=φx,∀x∈0,1,u0,t=0,∀t∈0,1,∂u∂x1,t=∫01ux,tdx,∀t∈0,T,where 0<α<1 and a,b∈ℝ+∗.In the remainder of this section, we assumef∈L2Q to be Lipschitz; i.e., there is a constant k>0 as
(19)fx,t,u,ux−fx,t,v,vxL2Q≤ku−vL2Q+ux−vxL2Q,∀u,ux,v,x∈L2Q.


## 4. Study of the Associated Linear Problem

### 4.1. Position of the Associated Linear Problem (20)

InQ=0,1×0,T, with T<∞, we examine the following associated linear problem:
(20)0cDtαux,t−a∂2ux,t∂x2+bux,t=fx,t,∀x,t∈Q,ux,0=φx,∀x∈0,1,u0,t=0,∀t∈0,T,∂u∂x1,t=∫01ux,tdx,∀t∈0,T,which can be written in this operational form:
(21)Lu=0cDtαux,t−a∂2ux,t∂x2+bux,t=fx,t.With the initial condition(22)ℓu=ux,0=φx,x∈0,1,Dirichlet boundary condition(23)u0,t=0,t∈0,T,and the integral condition of the second kind
(24)∂u1,t∂x=∫01ux,tdx,t∈0,T.


### 4.2. Resolution of Problem (20) by the Variable Separation Method

Let the following associated homogeneous linear problem:(25)0cDtσuu,t−a∂2ux,t∂x2+bux,t=0,∀x,t∈Q,ux,0=φx,∀x∈0,1,u0,t=0,∀t∈0,T,∂u1,t∂x=∫01ux,tdx,∀t∈0,T.We can certainly show that problem (25) admits a unique solution, for that we put
(26)ux,t=XxYt.Substituting (26) into (25), we obtain
(27)0cDtαY.X−aX"Y+bXY=0,XxY0=φx,X0Yt,X′1Yt=∫01XxYt.Therefore, we get forλ>0(28)0cDtαYY=aX"X−b=−λ.We begin by showing a Sturm-Liouville problem:(29)X"x+μaXx=0,X0=0,X′1=∫01Xxdx.Withμ=b−λ, it has a solution which is given by
(30)Xx=Acosμax+Bsinμax.OrA and B are two real arbitrary.Using the Dirichlet condition, we find(31)A=0.It follows that(32)Xx=Bsinμax.Now, we use the integral condition, and we get(33)X′1=Bμacosμa=∫01Xxdx=∫01Bsinμaxdx−Bμ/acosμax10=−Bμ/acosμa+Bμ/a.Hence, we can assert that the eigenvalue is given by the following equation:(34)μ=aacrossaμ+a.Our next goal is to determine the explicit form ofYt. According to the superposition theorem, we pose
(35)ux,t=∑n≥1Xnx⋅Ynt.Replacing (35) by (20) leads to
(36)∑n≥1ocDtaYn+b−μnYn⋅sinμnax=∑n≥1sinμnax⋅fnt,∀n∈ℕ,which implies
(37)0cDtαYnt+b−μnYnt=fnt.As(38)ux,0=∑n≥1sinμnaxYn0=φx=∑n≥0φn⋅sinμnax,then
(39)φn∫01φx⋅sinμnaxdx,hence
(40)Yn0=φn.We can then solve in a simple way the fractional ordinary differential problem (37)−(40) using the Laplace transform method; so it comes
(41)L0cDtαYnt=sFs−Yn0s1−α,then
(42)φn=sFs−Yn0s1−α+b−μnFs=Lfnt.Therefore,(43)Fs=1sα+b−μnsα−1φn+Lfnt.Finally, we obtain(44)Ynt=L−11sα+b−μnsα−1φn+Lfnt.Hence, we obtain the solution of (25) in the following explicit form:
(45)∑n≥0Bnsinμnax⋅L−11sα+b−μnsα−1φn+Lfnt.


## 4.1. Position of the Associated Linear Problem (20)

InQ=0,1×0,T, with T<∞, we examine the following associated linear problem:
(20)0cDtαux,t−a∂2ux,t∂x2+bux,t=fx,t,∀x,t∈Q,ux,0=φx,∀x∈0,1,u0,t=0,∀t∈0,T,∂u∂x1,t=∫01ux,tdx,∀t∈0,T,which can be written in this operational form:
(21)Lu=0cDtαux,t−a∂2ux,t∂x2+bux,t=fx,t.With the initial condition(22)ℓu=ux,0=φx,x∈0,1,Dirichlet boundary condition(23)u0,t=0,t∈0,T,and the integral condition of the second kind
(24)∂u1,t∂x=∫01ux,tdx,t∈0,T.


## 4.2. Resolution of Problem (20) by the Variable Separation Method

Let the following associated homogeneous linear problem:(25)0cDtσuu,t−a∂2ux,t∂x2+bux,t=0,∀x,t∈Q,ux,0=φx,∀x∈0,1,u0,t=0,∀t∈0,T,∂u1,t∂x=∫01ux,tdx,∀t∈0,T.We can certainly show that problem (25) admits a unique solution, for that we put
(26)ux,t=XxYt.Substituting (26) into (25), we obtain
(27)0cDtαY.X−aX"Y+bXY=0,XxY0=φx,X0Yt,X′1Yt=∫01XxYt.Therefore, we get forλ>0(28)0cDtαYY=aX"X−b=−λ.We begin by showing a Sturm-Liouville problem:(29)X"x+μaXx=0,X0=0,X′1=∫01Xxdx.Withμ=b−λ, it has a solution which is given by
(30)Xx=Acosμax+Bsinμax.OrA and B are two real arbitrary.Using the Dirichlet condition, we find(31)A=0.It follows that(32)Xx=Bsinμax.Now, we use the integral condition, and we get(33)X′1=Bμacosμa=∫01Xxdx=∫01Bsinμaxdx−Bμ/acosμax10=−Bμ/acosμa+Bμ/a.Hence, we can assert that the eigenvalue is given by the following equation:(34)μ=aacrossaμ+a.Our next goal is to determine the explicit form ofYt. According to the superposition theorem, we pose
(35)ux,t=∑n≥1Xnx⋅Ynt.Replacing (35) by (20) leads to
(36)∑n≥1ocDtaYn+b−μnYn⋅sinμnax=∑n≥1sinμnax⋅fnt,∀n∈ℕ,which implies
(37)0cDtαYnt+b−μnYnt=fnt.As(38)ux,0=∑n≥1sinμnaxYn0=φx=∑n≥0φn⋅sinμnax,then
(39)φn∫01φx⋅sinμnaxdx,hence
(40)Yn0=φn.We can then solve in a simple way the fractional ordinary differential problem (37)−(40) using the Laplace transform method; so it comes
(41)L0cDtαYnt=sFs−Yn0s1−α,then
(42)φn=sFs−Yn0s1−α+b−μnFs=Lfnt.Therefore,(43)Fs=1sα+b−μnsα−1φn+Lfnt.Finally, we obtain(44)Ynt=L−11sα+b−μnsα−1φn+Lfnt.Hence, we obtain the solution of (25) in the following explicit form:
(45)∑n≥0Bnsinμnax⋅L−11sα+b−μnsα−1φn+Lfnt.


## 5. Solvability of the Weak Solution of the Nonlinear Problem (18)

The key point to discuss in this section is to study the existence and uniqueness of the problem’s weak solution (18). The key idea is to apply an iterative method and estimate a priori.In this way, we define an auxiliary problem with a homogeneous equation:(46)OcDtαux,t−a∂2ux,tαx2+byx,t=Gx,t,y,yx,∀x,t∈Q,ux,0=φx,∀x∈0,1,u0,t=0,∀t∈0,T,∂u1,t∂x=∫01ux,tdx,∀t∈0,T.Ifu is a solution of (18) and v is a solution to (46), then y=u−v satisfies
(47)Ly=0cDtαyx,t−a∂2yx,t∂x2+byx,t=Gx,t,y,yx,(48)yx,0=0,∀x∈0,1,(49)y0,t=0,∀t∈0,t,(50)∂y1,t∂=0,∀t∈0,t,or Gx,t,y,yx=fx,t,u+v,ux+vx. From now on, we make the assumption: the function G is Lipschitz, that is, there is a positive constant k as
(51)Gx,t,y1,y1x−Gx,t,y2,y2xL2Q≤ky1−y2L2Q+y1x−y2xL2Q,∀y1,y1x,y2,y2x∈L2Q.In order to get the desired results, it is necessary to propose the concept of the studied solution.Letϑ=ϑx,t arbitrary function of V as
(52)V=ϑ∈C1Q,ϑ0,t=∂∂xϑ1,t=0,t∈0,T.Multiply 9 by ϑ and integrate it on Qτ; we find
(53)∫Qτ0cDtαyx,t·ϑx,tdxdt−∫Qτa∂2y∂x2x,t·ϑx,tdxdt+∫Qτbyx,t·ϑx,tdxdt=∫QτGx,t,y,yx·ϑx,tdxdt.By the use of integration by parts and the conditions on y,ϑ, we get
(54)∫Qτ0RDtαyx,t·ϑx,tdxdt+∫Qτa∂y∂xx,t·∂ϑ∂xx,tdxdt+∫Qτbyx,t·ϑx,tdxdt=∫QτGx,t,y,yx·ϑx,tdxdt.Then, (54) establishes the following formula:
(55)Ay,ϑ=∫QτGx,t,y,yx·ϑx,tdxdt,or
(56)Ay,ϑ=∫Qτ0RDtαyx,t·ϑx,tdxdt+∫Qτa∂y∂xx,t·∂ϑ∂xx,tdxdt+∫Qτbyx,t·ϑx,tdxdt.Definition 10.
We say the solution to problems (47)–(50) is weak; any function y∈L20,T;H10,1 such as (33), (49), and (50) are achieved.Now, we build a recurring sequence defined as follows: fromyn−1 we can define ynn∈ℕ, where the first element is given by y0=0; then, for n=1,2,3,⋯, the following problem is solved:
(57)0RDtαynx,t−a∂2ynx,t∂x2+bynx,t=Gx,t,yn−1,yxn−1,ynx,0=0,yn0,t=0,∂yn1,t∂x=0.For anyn fixed, thanks to the study of (20) which we gave the solution explicitly using the variable separation method; then, problem (57) has a unique solution ynx,t.Now, supposeznx,t=yn+1x,t−ynx,t; so we get a new problem:
(58)0RDtαznx,t−a∂2znx,t∂x2+bznx,t=pn−1x,t,znx.0=0,zn0,t=0,∂zn1,t∂x=0,or
(59)pn−1x,t=Gx,t,yn,yxn−Gx,t,yn−1,yxn−1.Lemma 11.
We assume that (51) be satisfied. So we obtain for problem (58) the following a priori estimate:
(60)znL20,T,H10,1≤czn−1L20,T,H10,1,or
(61)c=k22εmina,b−ε/2.Proof.
We multiply byzn this equation
(62)0RDtαznx,t−a∂2znx,t∂x2+bznx,t=pn−1x,t,and we integrate it on Qτ; we get
(63)∫Qτ0RDtαznx,t·znx,,tdxdt−∫Qτa∂2znx,t∂x2·znx,tdxdt+∫Qτbznx,t2dxdt=∫Qτpn−1x,t·znx,tdxdt.
Performing integration by part and using lemma 2, 3, 4, and 5 by the same way in the articles [15–24], we obtain
(64)∫010RDtα/2znx,τ2dx+∫Qτa∂zn∂xx,t2dxdt+∫Qτbznx,t2dxdt=∫Qτpn−1x,t·znx,tdxdt.
Using Cauchy withε− inequality, we get
(65)∫01D0Rtα/2znx,τ2dx+∫Qta∂zn∂xx,t2dxdt+∫Qtbznx,t2dxdt≤12ε∫Qtapn−12dxdt+∫Qtznx,t2dxdt.
Exploiting the fact that(66)pn−1x,t2=Gx,t,yn,yxn−Gx,t,yn−1,yxn−12,we conclude from (51) that
(67)∫Qτpn−1x,t2≤∫Qτk2yn−yn−1+yxn−yxn−12=∫Qτk2zn−12+zxn−12≤k2zn−1x,tL20,T;H10,12.
Eliminating the first term from (64), we get
(68)b∫Qτznx,τ2dx+a∫Qτ∂zn∂xx,t2dxdt≤∫010RDtα/2znx,τ2dx+∫Qτa∂zn∂xx,t2dxdt+∫Qτbznx,t2dxdt≤12ε∫Qτpn−12dxdt,which enables us to deduce
(69)znx,tL20,T;H10,12≤k22εmina,b−ε/2%zn−1x,tL20,T;H10,12.Our next concern will be the convergence of the sequenceynn, so we are thus looking for the convergence of the series ∑n=1∞zn.For this, we use the criterion of convergence of series which gives us(70)k22εmina,b−ε/2<1,Then,(71)k<2εmina,b−ε/2.Asznx,t=yn+1x,t−ynx,t and y0x,t=0, we have
(72)∑i=0n−1zi=∑i=0n−1yi+1x,t−yix,t=y1−y0+y2−y1+⋯+yn−yn−1=yn.Note that we have actually proved that the sequenceynn defined by
(73)ynx,t=∑i=0n−1zi,is convergent to an element y∈L20,T,H10,1. Having disposed of this preliminary step, the task is now to show that limn⟶∞ynx,t=yx,t is a solution of problems (47)–(50).In order to get this result, it will be necessary to check thaty satisfied
(74)Ay,ϑ=∫QτGx,t,y,yx·ϑx,tdxdt.From (57), we get
(75)Ayn,ϑ=∫Qτ0RDtαynx,t·ϑx,tdxdt+∫Qτa∂yn∂xx,t·∂ϑ∂xx,tdxdt+∫Qτbynx,t·ϑx,tdxdt.AsA is linear, we get
(76)Ayn,ϑ=Ayn−y,ϑ+Ay,ϑ=∫Qτ0RDtαyn−yx,t·ϑx,tdxdt+∫Qτa∂yn−y∂xx,t·∂ϑ∂xx,tdxdt∫Qτbyn−yx,t·ϑx,tdxdt+∫Qτ0RDtαyx,t·ϑx,tdxdt+∫Qτa∂y∂xx,t·∂ϑ∂xx,tdxdt+∫Qτbyx,t·ϑx,tdxdt.Applying the Cauchy Schwartz inequality onAyn−y,ϑ, we get
(77)∫Qτ0RDtαyn−yx,t·ϑx,tdxdt+∫Qτa∂yn−y∂xx,t·∂ϑ∂xx,tdxdt+∫Qτbyn−yx,t·ϑx,tdxdt≤max1,aγ,bϑxL2Q0RDtαyn−yL20,T,H10,1+yn−yxL20,T,H10,1,where γ is Poincare’s inequality constant. On the other hand, as
(78)yn⟶ydansL20,T,H10,1≊H1Q.So(79)yn⟶ydansL2Q,yxn⟶yxinL2Q.Lettingn⟶+∞, we find
(80)limn⟶+∞Ayn−y,ϑ=0.From (80) and going to the limit in (76), we obtain
(81)limn⟶+∞Ayn,ϑ=Ay,ϑ.Theorem 12.
If the condition (51) is satisfied and
(82)k<2εmina,b−ε/2.
So problems (47)–(50) admit a weak solution which belongs to L20,T;H10,1.It remains to be proven that problems (47)−(50) admit a unique solution.Theorem 13.
According to (51), the solution of problems (47)−(50) is unique.Proof.
Suppose thaty1 and y2 in L20,T;%H10,1 are two solutions of (47)−(50), then Z=y1−y2 satisfied Z∈L20,T;H10,1 and
(83)LZ=0RDtαZx,t−a∂2Zx,t∂x2+bZx,t=ψx,t,∀x,t∈Q¯,Zx,0=0,∀x∈0,1,Z0,t=0∀t∈0,t,∂Z1,t∂X=0∀t∈0,t,
Or(84)ψx,t=Gx,t,y1,y1x−Gx,t,y2,y2x.
Similar analysis to that in the proof of Lemma11, it shows that
(85)ZL20,T;H10,1≤ZL20,T;H10,1.
Orc is the same constant of Lemma 11. As c<1, so according to (85), it comes that
(86)1−cZL20,T;H10,1≤0.
We conclude thaty1=y2 in L20,T;H10,1 is the desired conclusion.


## 6. Conclusion

The study of fractional nonlinear parabolic problems is of great theoretical interest, because of new modeling with fractional derivatives which plays an important role in the description in the part of modeling in mathematics. Also, the use of the condition of Neumann in terms of integral condition which means the flux of the solution is an average gives a very good model of boundary condition. Then, there remains the numerical part as perspective in the fractional problem especially when there is a nonlinear term and integral condition of the second type.



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