# Well-Posedness and Numerical Study for Solutions of a Parabolic Equation with Variable-Exponent Nonlinearities

Authors: Jamal H. Al-Smail; Salim A. Messaoudi; Ala A. Talahmeh
Journal: International Journal of Differential Equations
 (2018)

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## Abstract

We consider the following nonlinear parabolic equation:ut-div(|∇u|p(x)-2∇u)=f(x,t), where f:Ω×(0,T)→R and the exponent of nonlinearity p(·) are given functions. By using a nonlinear operator theory, we prove the existence and uniqueness of weak solutions under suitable assumptions. We also give a two-dimensional numerical example to illustrate the decay of solutions.


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## Body

## 1. Introduction

LetΩ be a bounded domain in Rn with a smooth boundary ∂Ω. We consider the following initial and boundary value problem:Put-div∇upx-2∇u=fx,t,inΩ×0,Tux,t=0,on∂Ω×0,Tux,0=u0x,inΩ,where f:Ω×(0,T)→R and u0:Ω→R are given functions. The exponent p(·) is a given measurable function on Ω¯ such that(1)2<p1≤px≤p2<+∞,with(2)p1≔essinfx∈Ω⁡px,p2≔esssupx∈Ω⁡px.We also assume that p(·) satisfies the log-Hölder continuity condition:(3)px-py≤-Alog⁡x-y,∀x,y∈Ω,withx-y<δ,where A>0 and 0<δ<1 are constants. The term div(|∇u|p(·)-2∇u) is called the p(·)-Laplacian and denoted by Δp(·)u.The study of partial differential equations involving variable-exponent nonlinearities has attracted the attention of researchers in recent years. The interest in studying such problems is stimulated and motivated by their applications in elastic mechanics, fluid dynamics, nonlinear elasticity, electrorheological fluids, and so forth. In particular, parabolic equations involving thep(·)-Laplacian are related to the field of image restoration and electrorheological fluids which are characterized by their ability to change the mechanical properties under the influence of the exterior electromagnetic field. The rigorous study of these physical problems has been facilitated by the development of the Lebesgue and Sobolev spaces with variable exponents.Regarding parabolic problems with nonlinearities of variable-exponent type, many works have appeared. We note here that most of the results deal with blow-up and global nonexistence. Let us mention some of these works. For instance, Alaoui et al. [1] considered the following nonlinear heat equation:(4)utx,t-div∇umx-2∇u=uupx-2+f,in a bounded domain in Ω⊂Rn (n≥1) with a smooth boundary ∂Ω. Under appropriate conditions on the exponent functions m,p and for f=0, they showed that any solution with nontrivial initial datum blows up in finite time. They also gave a two-dimensional numerical example to illustrate their result. Pinasco [2] studied the following problem:(5)ut-Δu=fu,inΩ×0,Tux,t=0,on∂Ω×0,Tux,0=u0x,inΩ,where Ω⊂Rn is a bounded domain with a smooth boundary ∂Ω, and the source term is of the following form:(6)fu=axupxorfu=ax∫Ωuqyy,tdy,with p,q: Ω→(1,∞) and the continuous function a: Ω→R being given functions satisfying specific conditions. They established the local existence of positive solutions and proved that solutions with initial data sufficiently large blow up in finite time. Parabolic problems with sources like the ones in (5) appear in several branches of applied mathematics and they have been used to model chemical reactions, heat transfer, or population dynamics.Recently, Shangerganesh et al. [3] studied the following fourth-order degenerate parabolic equation:(7)ut+div∇Δupx-2∇Δu=f-divg,in a bounded domain Ω⊂Rn (n≥1) with a smooth boundary ∂Ω, and proved the existence and uniqueness of weak solutions of (7) by using the difference and variation methods under suitable assumptions on f,g and the exponents p.EquationP is a nonlinear diffusion equation which has been used to study image restoration and electrorheological fluids (see [4–11]). In particular, Bendahmane et al. [12] proved the well-posedness of a solution, for L1-data. Akagi and Matsuura [13] gave the well-posedness for L2 initial datum and discussed the long-time behaviour of the solution using the subdifferential calculus approach. In our paper, we give an alternative proof of the well-posedness of  P which is simpler than that in [13] using a theory of nonlinear evolution equations. In addition, we give a numerical example in 2D to illustrate the decay result obtained in [13].This paper consists of three sections in addition to the introduction. In Section2, we recall the definitions of the variable-exponent Lebesgue spaces, Lp(·)(Ω), the Sobolev spaces, W1,p(·)(Ω), as well as some of their properties. We also state, without proof, a proposition to be used in the proof of our main result. In Section 3, we state and prove the well-posedness of solution to our problem. In Section 4, we give a numerical verification of the decay result.


## 2. Preliminaries

We present some preliminary facts about the Lebesgue and Sobolev spaces with variable exponents (see [1, 14–16]). Let p:Ω→[1,∞] be a measurable function, where Ω is a domain of Rn. We define the Lebesgue space with a variable-exponent p(·) by(8)Lp·Ω≔u:Ω⟶R;measurableinΩ:ϱLp·Ωλu<∞,forsomeλ>0,where(9)ϱLp·Ωu=∫Ωuxpxdxis called a modular. Equipped with the Luxembourg-type norm,(10)uLp·Ω≔infλ>0:ϱLp·Ωuλ≤1,Lp(·)(Ω) is a Banach space (see [10]).Lemma 1 (Hölder’s inequality [10]).
Letp,q,s≥1 be measurable functions defined on Ω such that(11)1sy=1py+1qy, for a.e. y∈Ω. If f∈Lp(·)(Ω) and g∈Lq(·)(Ω), then fg∈Ls(·)(Ω) and(12)fgs·≤2fp·gq·.Lemma 2 (see [10]).
Letp be a measurable function on Ω. Then,(a)
fp(·)≤1 if and only if ϱp(·)(f)≤1;(b)
forf∈Lp(·)(Ω), if fp(·)≤1, then ϱp(·)(f)≤fp(·); and if fp(·)≥1, then fp(·)≤ϱp(·)(f);(c)
fp(·)≤1+ϱp(·)(f).Lemma 3 (see [10]).
If (1) holds, then(13)minup·p1,up·p2≤ϱp·u≤maxup·p1,up·p2, for any u∈Lp(·)(Ω).We next define the variable-exponent Sobolev spaceW1,p(·)(Ω) as follows:(14)W1,p·Ω=u∈Lp·Ωsuchthat∇uexists,∇u∈Lp·Ω. This space is a Banach space with respect to the norm uW1,p(·)(Ω)=up(·)+∇up(·). Furthermore, W01,p(·)(Ω) is the closure of C0∞(Ω) in W1,p(·)(Ω). The dual of W01,p(·)(Ω) is defined as W-1,p′(·)(Ω), by the same way as the usual Sobolev spaces where 1/p(·)+1/p′(·)=1.Lemma 4 (see [10]).
LetΩ be a bounded domain of Rn and p(·) satisfies (1) and (3), and then(15)up·≤C∇up·,∀u∈W01,p·Ω,where the positive constant C depends on p(·) and Ω. In particular, the space W01,p(·)(Ω) has an equivalent norm given by uW1,p(·)(Ω)=∇up(·).Lemma 5 (see [10]).
Ifp(·)∈C(Ω¯), q:Ω→[1,∞) is a continuous function and(16)essinfx∈Ω⁡p∗x-qx>0withp∗x=npxesssupx∈Ω⁡n-pxifp2<n∞ifp2≥n. Then the embedding W01,p(·)(Ω)↪Lq(·)(Ω) is continuous and compact.Definition 6 (see [17]).
LetV be a separable Banach space and H be a Hilbert space such that V⊂H⊂V′ with continuous embedding and V is dense in H. Let A:V→V′ be a nonlinear operator.(1)
A is said to be monotone if A(u)-A(v),u-vV′×V≥0. If, in addition, we have(17)Au-Av,u-vV′×V≠0,∀u≠v,then A is said to be strictly monotone.(2)
A is said to be bounded, if A(S) is bounded in V′, whenever S is bounded in V.(3)
A is said to be hemicontinuous, if the real function(18)λ⟼Au+λv,wis continuous from R to R, for any fixed u,v,w∈V.We end this section with a proposition which is exactly like Theorem7.1 [17].Proposition 7.
Letu0∈H and f∈Lp′((0,T),V′). Suppose that A: V→V′ is a bounded monotone and hemicontinuous (nonlinear) operator satisfying, for some α,β>0 and for some p>1,(19)Av,v≥αvVp-β,∀v∈V.Then the following problem(20)ut+Au=fu·,0=u0,has a unique weak solution:(21)u∈Lp0,T,Vwithut∈Lp′0,T,V′, where 1/p+1/p′=1.


## 3. Well-Posedness

In this section, we state and prove the well-posedness of our problem.Theorem 8.
Letu0∈L2(Ω),f∈Lp′(·)((0,T),W-1,p′(·)(Ω)). Assume that (1) and (3) hold. Then P has a unique weak solution:(22)u∈Lp10,T,W01,p·Ω∩L∞0,T,L2Ω,ut∈Lp2′0,T,W-1,p′·Ω,where 1/p(·)+1/p′(·)=1.Proof.
We verify the conditions of Proposition7. Let V=W01,p(·)(Ω), and equip it with the norm,(23)uW1,p·Ω=∇up·. So, V′=W-1,p′(·)(Ω). Define A:V→V′ by(24)Au=-div∇upx-2∇u.Boundedness ofA. For all u,v∈V,(25)Au,v=∫Ω∇upx-2∇u·∇v≤∫Ω∇upx-1∇v.Hölder inequality gives(26)∫Ω∇upx-1∇v≤2∇up·-1p′·vV.Combining (25) and (26), we obtain(27)AuV′≤2∇up·-1p′·.Then, Lemma 2 implies(28)∇up·-1p′·≤1+ϱp·∇u.Combining (27) and (28), we arrive at(29)AuV′≤21+ϱp·∇u.Let S⊂V such that uV≤M, for all u∈S. That is, ∇up(·)≤M.
IfM≤1, then Lemma 2 implies ϱp(·)(∇u)≤1 and (29) gives A(u)V′≤4<+∞.
IfM>1, then ϱp(·)(∇u)≤Mp2. Thus, (29) implies A(u)V′≤2(1+Mp2)<+∞.
HenceA is bounded.
Monotonicity of
A
. Let u,v∈V. Then(30)Au-Av,u-vV′×V=-∫div∇upx-2∇u-∇vpx-2∇vu-vdx=∫∇upx-2∇u-∇vpx-2∇v∇u-∇vdx. By using the inequality,(31)apx-2a-bpx-2b·a-b≥0,for all a,b∈Rn and a.e. x∈Ω. Thus we obtain A(u)-A(v),u-v≥0.
Hence,A is monotone.
To verify (19), we note that, for all u∈V, we have(32)Au,u=∫Ω∇upxdx=ϱp·∇u.If uV=∇up(·)>1, then by Lemma 3, we get(33)ϱp·∇u≥∇up·p1.Combining (32) and (33), we easily see that(34)Au,u≥uVp1.
IfuV=∇up(·)≤1, then by Lemma 2, we obtain(35)Au,u=ϱp·∇u≥uV-1≥uVp1-1. Therefore, we have(36)Au,u≥uVp1-1,∀u∈V.Hemicontinuity ofA. Let u,v,w∈V be fixed. Let(37)gλ=Au+λv,w=∫Ω∇u+λ∇vpx-2∇u+λ∇v·∇wdx.Let λk→λ (real) and consider(38)gλk=∫Ω∇u+λk∇vpx-2∇u+λk∇v·∇wdx. Since(39)∇u+λk∇vpx-2∇u+λk∇v·∇w⟶∇u+λ∇vpx-2∇u+λ∇v·∇w, for a.e. x∈Ω and (40)∇u+λk∇vpx-1∇w≤C∇upx-1∇w+∇vpx-1∇w∈L1Ω, where C=max{2p2-2,2p2-2(1+|M|p2-1)}>0, then, by the classical dominated convergence theorem,(41)gλk⟶gλask⟶∞. Hence, A is hemicontinuous.
Therefore, conditions of Proposition7 are satisfied and problem P has a unique solution.


## 4. Numerical Study

In this section, we present some numerical results and applications of the problem:P∗ut-div∇upx,y-2∇u=0,inQ=Ω×0,Tu=0,on∂Q=∂Ω×0,Tux,y,0=u0x,y,inΩ,which is a well-posed problem due to Theorem 8. Our objective is to provide a numerical verification of the following decay result:Proposition 9 (see [13]).
Assume that (1) and (3) hold. Then the solution of P∗ satisfies the following:(i)
Ifp2=2, then there exists a constant c2>0 such that (42)ut2≤u02e-c2t,∀t≥0.(ii)
Ifp2>2, there exists a constant c1>0 and t1≥0 such that(43)ut2≤u021+c1t-1/p2-2,∀t≥t1.We consider two applications to illustrate numerically an exponential decay for the casep(x,y)=2 and a polynomial decay for an exponent function p(·) satisfying conditions (1)–(3).For this purpose, we introduce a numerical scheme forP∗, prove its convergence in Section 4.1, and show the decay results in Section 4.2.


### 4.1. Numerical Method

In this part, we present a linearized numerical scheme to obtain the numerical results of the systemP∗ and confirm the decay results. The system is fully discretized through a finite difference method for the time variable and a finite element Galerkin method for the space variable. Useful background about the numerical and error analysis of these methods is found in [18]. More interestingly in [19], Li and Wang introduced a numerical scheme to solve strongly nonlinear parabolic systems and proved unconditional error estimates of the scheme. Our problem P∗ is highly nonlinear due to the presence of the gradient and nonlinear exponent in the diffusivity coefficient, which can be zero inside the spatial domain. Below, we introduce our numerical scheme for the purpose of confirming the decay results.The parabolic equation(44)ut-div∇upx,y-2∇u=0,inQ=Ω×0,Tis discretized using finite differences for the time derivative and a finite element method for the p(·)-Laplacian term. For this, we divide the time interval [0,T] into N equal subintervals by(45)tn=nτ,τ=TN and denote by(46)unx,y≔ux,y,tn,n=0,1,…,N.The term ut is approximated using the first-order forward finite difference formula:(47)utn+1≔un+1-unτ.Semidiscrete Problem. A linear semidiscrete formulation of P∗ takes the following form: given p and u0, find {u1,u2,…,un+1} such that(48)utn+1-div∇unpx,y-2∇un+1=0,inΩun+1=0,on∂Ωu0=u0x,y,inΩ. This problem is elliptic and admits a unique solution [20], for every n=0,1,…,N. Also, the Rothe approximation u(n) to the exact solution u, given by(49)unx,y,t≔ui-1x,y+t-ti-1ui-ui-1τ,t∈ti-1,ti,i=1,2,…,n,is well defined and u(n)→u in L2(Ω) as τ→0, see [1].Full-Discrete Problem. The variable un+1 is discretized in space by a finite element method. For this, let Ωh be a triangulation of Ω with a maximal element size h. Let also vh be a test function in the linear Lagrangian space P1(Ωh) such that vh=0 on ∂Ωh.The semidiscrete problem is then written in a weak form to define the full-discrete problem: givenph, un∈P1(Ωh), find uhn+1∈P1(Ωh) such that(50)∫Ωhuhn+1-uhnτvh+∇uhnphx,y-2∇uhn+1·∇vhdΩh=0,∀vh∈P1Ωh.For ph≥2, the above problem has a unique solution uhn+1∈H01(Ωh) for every nontrivial uhn∈H1(Ωh). This follows from the Lax-Milgram Lemma, and the Galerkin approximation uhn+1 converges to un+1 in H01(Ωh) as h→0; see [18].


### 4.2. Numerical Results

In this subsection, we present the following numerical applications ofP∗:(1)
Exponential decay: forph(x,y)=2, we show, for some c>0, that(51)g1t≔uhnuh0≤e-ct,∀t≥0.(2)
Polynomial decay: forph(x,y)=1/5x2+2.5, we show, for some c>0 and t0>0, that(52)g2t≔uhnuh0≤1+ct-1/p2-2,∀t≥t0.Here, ⌈·⌉ denote the greatest integer function.In both applications, we set the following parameters:(53)T=100,τ=0.1,h=0.1,Ωh=-5,5×-5,5.Figure1 shows the mesh used for Ωh, which involves 23702 triangles and 12052 vertices.Figure 1
Mesh of the domain[-5,5]×[-5,5].The initial condition is taken to beuh0(x,y)=e-0.5(x2+y2) and projected into P1(Ωh); see Figure 2.Figure 2
Initial condition:uh0(x,y)=e-0.5(x2+y2).The numerical results are obtained using the noncommercial software, FreeFem++ [21].Application 1.
p
h
(
x
,
y
)
=
2 satisfies the required conditions (1)–(3). Figure 3 shows the numerical solutions for t=5, t=10, t=20, and t=50.Figure 3
Numerical solutions for Application1.
(a)
u
h
5
(b)
u
h
10
(c)
u
h
20
(d)
u
h
50Withc=0.1, Figure 4(a) shows that uhn decays exponentially as(54)g1t=uhnuh0≤e-0.1t,0≤t≤T.Figure 4
Exponential decay.
(a)
Solid line:y=e-0.1t; dashed: y=g1(t)
(b)
y
=
g
1
(
t
)
/
e
-
0.1
tThis is also confirmed by Figure4(b) that shows the ratio y=g1(t)/e-0.1t is less than one and remains decreasing for a large value of T.Application 2.
The exponent functionph(x,y)=1/5x2+2.5, in Figure 5, satisfies the required conditions (1)–(3) as(i)
p1=2.5, p2=7.5;(ii)
|p(x,y)-p(x0,y0)|=1/5x2-x02≤-202log(1/δ)/log|x-y| for |x-y|<δ with 0<δ<1.Figure 5
p
h
(
x
,
y
)
=
1
/
5
x
2
+
2.5.Figure6 shows the numerical solutions for t=5, t=10, t=20, and t=50.Figure 6
Numerical solutions for Application2.
(a)
u
h
5
(b)
u
h
10
(c)
u
h
20
(d)
u
h
50In this case, the solutionuhn has a polynomial decay. With c=1, Figure 7(a) shows that(55)g2t=uhnuh0≤1+t-2/11,0≤t≤T. This is also confirmed by Figure 7(b), which shows that the ratio y=g2(t)/(1+t)-2/11 remains less than one and decreasing until T.Figure 7
Polynomial decay.
(a)
Solid line:y=(1+t)-2/11; dashed: y=g2(t)
(b)
y
=
g
2
(
t
)
/
(
1
+
t
)
-
2
/
11We conclude that the numerical results in above applications verify Proposition9.


## 4.1. Numerical Method

In this part, we present a linearized numerical scheme to obtain the numerical results of the systemP∗ and confirm the decay results. The system is fully discretized through a finite difference method for the time variable and a finite element Galerkin method for the space variable. Useful background about the numerical and error analysis of these methods is found in [18]. More interestingly in [19], Li and Wang introduced a numerical scheme to solve strongly nonlinear parabolic systems and proved unconditional error estimates of the scheme. Our problem P∗ is highly nonlinear due to the presence of the gradient and nonlinear exponent in the diffusivity coefficient, which can be zero inside the spatial domain. Below, we introduce our numerical scheme for the purpose of confirming the decay results.The parabolic equation(44)ut-div∇upx,y-2∇u=0,inQ=Ω×0,Tis discretized using finite differences for the time derivative and a finite element method for the p(·)-Laplacian term. For this, we divide the time interval [0,T] into N equal subintervals by(45)tn=nτ,τ=TN and denote by(46)unx,y≔ux,y,tn,n=0,1,…,N.The term ut is approximated using the first-order forward finite difference formula:(47)utn+1≔un+1-unτ.Semidiscrete Problem. A linear semidiscrete formulation of P∗ takes the following form: given p and u0, find {u1,u2,…,un+1} such that(48)utn+1-div∇unpx,y-2∇un+1=0,inΩun+1=0,on∂Ωu0=u0x,y,inΩ. This problem is elliptic and admits a unique solution [20], for every n=0,1,…,N. Also, the Rothe approximation u(n) to the exact solution u, given by(49)unx,y,t≔ui-1x,y+t-ti-1ui-ui-1τ,t∈ti-1,ti,i=1,2,…,n,is well defined and u(n)→u in L2(Ω) as τ→0, see [1].Full-Discrete Problem. The variable un+1 is discretized in space by a finite element method. For this, let Ωh be a triangulation of Ω with a maximal element size h. Let also vh be a test function in the linear Lagrangian space P1(Ωh) such that vh=0 on ∂Ωh.The semidiscrete problem is then written in a weak form to define the full-discrete problem: givenph, un∈P1(Ωh), find uhn+1∈P1(Ωh) such that(50)∫Ωhuhn+1-uhnτvh+∇uhnphx,y-2∇uhn+1·∇vhdΩh=0,∀vh∈P1Ωh.For ph≥2, the above problem has a unique solution uhn+1∈H01(Ωh) for every nontrivial uhn∈H1(Ωh). This follows from the Lax-Milgram Lemma, and the Galerkin approximation uhn+1 converges to un+1 in H01(Ωh) as h→0; see [18].


## 4.2. Numerical Results

In this subsection, we present the following numerical applications ofP∗:(1)
Exponential decay: forph(x,y)=2, we show, for some c>0, that(51)g1t≔uhnuh0≤e-ct,∀t≥0.(2)
Polynomial decay: forph(x,y)=1/5x2+2.5, we show, for some c>0 and t0>0, that(52)g2t≔uhnuh0≤1+ct-1/p2-2,∀t≥t0.Here, ⌈·⌉ denote the greatest integer function.In both applications, we set the following parameters:(53)T=100,τ=0.1,h=0.1,Ωh=-5,5×-5,5.Figure1 shows the mesh used for Ωh, which involves 23702 triangles and 12052 vertices.Figure 1
Mesh of the domain[-5,5]×[-5,5].The initial condition is taken to beuh0(x,y)=e-0.5(x2+y2) and projected into P1(Ωh); see Figure 2.Figure 2
Initial condition:uh0(x,y)=e-0.5(x2+y2).The numerical results are obtained using the noncommercial software, FreeFem++ [21].Application 1.
p
h
(
x
,
y
)
=
2 satisfies the required conditions (1)–(3). Figure 3 shows the numerical solutions for t=5, t=10, t=20, and t=50.Figure 3
Numerical solutions for Application1.
(a)
u
h
5
(b)
u
h
10
(c)
u
h
20
(d)
u
h
50Withc=0.1, Figure 4(a) shows that uhn decays exponentially as(54)g1t=uhnuh0≤e-0.1t,0≤t≤T.Figure 4
Exponential decay.
(a)
Solid line:y=e-0.1t; dashed: y=g1(t)
(b)
y
=
g
1
(
t
)
/
e
-
0.1
tThis is also confirmed by Figure4(b) that shows the ratio y=g1(t)/e-0.1t is less than one and remains decreasing for a large value of T.Application 2.
The exponent functionph(x,y)=1/5x2+2.5, in Figure 5, satisfies the required conditions (1)–(3) as(i)
p1=2.5, p2=7.5;(ii)
|p(x,y)-p(x0,y0)|=1/5x2-x02≤-202log(1/δ)/log|x-y| for |x-y|<δ with 0<δ<1.Figure 5
p
h
(
x
,
y
)
=
1
/
5
x
2
+
2.5.Figure6 shows the numerical solutions for t=5, t=10, t=20, and t=50.Figure 6
Numerical solutions for Application2.
(a)
u
h
5
(b)
u
h
10
(c)
u
h
20
(d)
u
h
50In this case, the solutionuhn has a polynomial decay. With c=1, Figure 7(a) shows that(55)g2t=uhnuh0≤1+t-2/11,0≤t≤T. This is also confirmed by Figure 7(b), which shows that the ratio y=g2(t)/(1+t)-2/11 remains less than one and decreasing until T.Figure 7
Polynomial decay.
(a)
Solid line:y=(1+t)-2/11; dashed: y=g2(t)
(b)
y
=
g
2
(
t
)
/
(
1
+
t
)
-
2
/
11We conclude that the numerical results in above applications verify Proposition9.



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