# Qualitative Properties of Nonnegative Solutions for a Doubly Nonlinear Problem with Variable Exponents

Authors: Zakariya Chaouai; Abderrahmane El Hachimi
Journal: Abstract and Applied Analysis
 (2018)

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## Abstract

We consider the Dirichlet initial boundary value problem∂tum(x)-div⁡∇upx,t-2∇u=ax,tuq(x,t), where the exponents p(x,t)>1, q(x,t)>0, and m(x)>0 are given functions. We assume that a(x,t) is a bounded function. The aim of this paper is to deal with some qualitative properties of the solutions. Firstly, we prove that if ess⁡sup⁡p(x,t)-1<ess⁡inf⁡m(x), then any weak solution will be extinct in finite time when the initial data is small enough. Otherwise, when ess⁡sup⁡m(x)<ess⁡inf⁡p(x,t)-1, we get the positivity of solutions for large t. In the second part, we investigate the property of propagation from the initial data. For this purpose, we give a precise estimation of the support of the solution under the conditions that ess⁡sup⁡m(x)<ess⁡inf⁡p(x,t)-1 and either q(x,t)=m(x) or a(x,t)≤0 a.e. Finally, we give a uniform localization of the support of solutions for all t>0, in the case where a(x,t)<a1<0 a.e. and ess⁡sup⁡qx,t<ess⁡inf⁡p(x,t)-1.


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## Body

## 1. Introduction

This paper is devoted to studying qualitative properties of nonnegative weak solutions for the following doubly nonlinear parabolic problem with variable exponents(1)P:∂bx,u∂t-Δpx,tu=ax,tuqx,t-1uinQT=Ω×0,T,u=0on∂Ω×0,T,bx,ux,0=bx,u0xinΩ,whereΩ is a bounded domain of RN,N≥1, with smooth boundary ∂Ω, b(x,u)=um(x)-1u and Δp(x,t)u is defined as(2)Δpx,tu=div⁡∇upx,t-2∇u.The exponents p, q, m and the coefficient a are given measurable functions. It will be assumed throughout the paper that these functions satisfy some specific conditions.Problems of this form appear in various applications; for instance in models for gas or fluid flow in porous media ([1, 2]) and for the spread of certain biological populations ([3]). Our motivation to study problem P with variable exponents is the fact that it is considered as a model of an important class of non-Newtonian fluids which are well known as electrorheological fluids, see ([4]). It appears also as a model in image restoration ([5]) and in elasticity ([6]).It is well known that solutions of problems such asP exhibit various qualitative properties, which reflect natural phenomena, according to certain conditions on p(x,t), q(x,t), m(x), a(x,t), and u0, (see for example [7–13] and the references therein). Among the phenomena that interest us in this work is the finite speed of propagation, which means that if ρ0>0 is such that supp⁡(u0)⊂B(x0,ρ0), then supp⁡(u(x,t))⊂B(x0,ρ(t)), for any t∈(0,T), where ρ(t) is a positive function which depends on ρ0, (i.e., solutions with compact support). This property has various physical meanings; for instance, in the study of turbulent filtration of gas through porous media, a solution with compact support means that gas will remain confined to a bounded region of space, (see [14]).The phenomenon of finite speed of propagation was investigated by Kalashnikov in [15]. He considered, for N=1, the equation ∂b(u)/dt-Δu=0 in R×(0,∞) and, under specific conditions, proved that if the initial condition u0 has a compact support, then the condition ∫0+(1/b(s))ds<+∞ is necessary and sufficient for solutions to have compact support. This result was extended by Dìaz for N≥1, in [16]. Later, in [17] Dìaz and Hernández considered the doubly nonlinear problem with absorption term ∂b(u)/∂t-Δpu+uq-1u=0, in RN×(0,∞), where b(u)=um-1u. Under the assumption that u0 has a compact support and 0<q<p-1, they proved that any solution has a compact support for all t>0. This result was obtained by the construction of a local uniform super-solution. Let us recall that the finite speed of propagation phenomenon has been studied by many authors in the last decades, (see [18–21]).Besides, extinction and nonextinction are also important properties for solutions of evolution equations that have attracted many authors in the last few decades. Most of them focused on equations with constant exponents of nonlinearity, (see [22–26]). For example, Hong et al., dealt in [27] with the homogeneous equation ut-Δpum=0, in Ω×(0,∞), where p>1 and m>0. They proved that the condition 1<p<1+1/m is necessary and sufficient for extinction to occur. Moreover, Zhou and Mu ([28]) studied the extinction behavior of weak solutions for the equation with source term ut-Δpum=λuq, in Ω×(0,∞), where p>1, m,q,λ>0 and m(p-1)<1. They proved that q=m(p-1) is a critical extinction exponent.Otherwise, it is worth noting that problemP has been treated by Antontsev and Shamarev in several papers. In [29, 30], they proved the existence of weak and strong solutions. Moreover, under certain regularity hypotheses on m(x), p(x,t), and under the sign condition a(x,t)≤0 a.e, they studied properties of finite speed of propagation and extinction in finite time in [9, 10]. Their results were established by using the local energy method. Here, we shall use the so-called method of sub- and supersolutions to extend some of the results in [9, 10]. To the best of our knowledge, there are few results concerning the study of qualitative properties for parabolic equations with variable exponents by using this method. Furthermore, we shall also extend to the parabolic case some of the results by Zhang et al. in [31], where radial sub- and supersolutions for some elliptic problems with variable exponents are constructed, and some of the results by Chung and Park in [22] and by Yuan et al. in [27], to variable exponents case. In fact, we shall exploit their arguments in our parabolic problem setting with less conditions on the exponents p(x,t), q(x,t), and m(x) and the coefficient a(x,t).The present paper is organized as follows. In Section2, we introduce some basic facts about the variable exponents spaces. In Section 3, we give assumptions and general definitions; then, we establish a comparison principle which ensures the uniqueness of solutions. In Section 4, we investigate the extinction and nonextinction properties for the solution of P. Finally in Section 5, we study the property of finite speed of propagation.


## 2. Preliminaries

In this section we give some elementary results for the generalized Lebesgue spacesLpxΩ and Sobolev spaces W1,pxΩ, where Ω is a bounded set of RN(N≥1) with smooth boundary. For more details, see ([11, 32, 33]). (3)C+Ω¯=px:Ω¯→p-,p+⊂1,∞;pisacontinousfunction,where(4)p+=supx∈Ω¯⁡px,p-=infx∈Ω¯⁡px.For any px∈C+Ω, we introduce the variable exponent Lebesgue space as follows: (5)LpxΩ=u:Ω→R,uisameasurablefunction,∫Ωuxpxdx<∞,endowed with the Luxemburg norm (6)upx,Ω=inf⁡λ:λ>0,∫Ωuxλpxdx≤1.Proposition 1 (see [11, 32, 33]).
⁡(i)
The spaceLp(x)Ω is a separable and reflexive Banach space, and its conjugate space is LqxΩ, where 1/px+1/qx=1. Moreover, for any u∈Lp(x)Ω and v∈Lq(x)Ω, we have (7)∫Ωuvdx≤1p-+1q-upx,Ωvqx,Ω.(ii)
Letp1,p2∈C+Ω¯ be given such that p1x≤p2x for any x∈Ω¯ then Lp2xΩ is continuously embedded into Lp1xΩ.Proposition 2 (see [11, 32, 33]).
Let(8)ρu=∫Ωupxdx,∀u∈LpxΩ.Then, we have(i)
upx,Ω<1(=1;>1)⇔ρu<1(=1;>1)(ii)
min⁡upx,Ωp+,upx,Ωp-≤ρu≤max⁡upx,Ωp+,upx,Ωp-(iii)
upx,Ω→0⇔ρu→0;upx,Ω→∞⇔ρu→∞.Now, we define the variable Sobolev spaceW1,pxΩ as follows: (9)W1,pxΩ=u∈LpxΩ:∇upx∈L1Ω,endowed with the norm (10)uW1,pxΩ=upx,Ω+∇upx,Ω.We say that p∈C+(Ω) satisfies the log-Hölder condition in Ω if(11)∀x,y∈Ω,x-y<1,px-py≤ω(x-y,where ω satisfies (12)lim⁡supτ→0+ωτln⁡1τ⁡<∞.Proposition 3 (see [11, 32, 33]).
⁡(i)
W1,p(x)(Ω) is a separable and reflexive Banach space.(ii)
Ifp(x) satisfies the log-Hölder condition (11), then the space C∞(Ω) is dense in W1,px(Ω). Moreover, we can define the Sobolev space with zero boundary values, W01,p(x)(Ω) as the completion of C0∞(Ω), with respect to the norm⁡.W1,pxΩ.Next, letm(x)>0 and p(x,t)>1 be given functions. For T>0 fixed, we denote QT=Ω×(0,T). Let m∈C0(Ω¯), we assume that p(x,t)∈C+(QT) satisfies the following log-Hölder condition in QT, (13)∀x,t,y,τ∈QT,suchthatx,t-y,τ=x-y2+t-τ2<1, we have(14)px,t-py,τ≤ω(x,t-y,τ,where ω satisfies (15)lim⁡supτ→0+ωτln⁡1τ<∞.For every fixed t∈[0,T], we introduce the following Banach. (16)VtΩ=u∈Lmx+1Ω∩W01,1Ω:∇u∈Lpx,tΩ,endowed with the norm (17)uVtΩ=umx+1,Ω+∇upx,t,Ω.We denote by W(QT) the following Banach space, (18)WQT=u∈Lmx+1QT:∇upx,t∈L1QT,u.,t∈VtΩa.e.t∈0,T,endowed with the norm (19)uWQT=umx+1,QT+∇upx,t,QT.We denote by W′(QT) the dual of W(QT).


## 3. Assumptions and Results

Throughout this paper we assume that the coefficients and the exponents of nonlinearity satisfy the following conditions,(20)thereexistpositiveconstantsp±,q±,m±anda1suchthat,foranyx,tinQT1<p-≤px,t≤p+,0<q-≤qx,t≤q+,0<m-≤mx≤m+,ax,t≤a1,and the initial data u0 satisfies(21)u0∈L∞Ω,u0≥0a.e.inΩ.Now, let us state the definition of weak solutions for the problem P.Definition 4.
We say thatu(x,t) is a super-(sub)solution of P on QT if (1)
u∈L∞(QT)∩W(QT) and (∂/∂t)b(x,u)∈W′(QT).(2)
for every nonnegative test functionϕ∈W(QT) and (∂/∂t)ϕ∈W′(QT), we have (22)∫QTϕ∂∂tbx,u+∇upx,t-2∇u∇ϕdxdt≥≤∫QTax,tuqx,t-1uϕdxdt.(3)
b(x,u.,0)≥(≤)b(x,u0)a.e.inΩ, andu≥(≤)0on∂Ω×(0,T).
A functionu is a weak solution of P if it is simultaneously a supersolution and a subsolution.The following result concerning the local existence of weak solutions of problemP is established in [29].Theorem 5.
Letm∈C0(Ω), p(x,t) satisfies the log-Hölder condition in QT (14), and let conditions (20) and (21) be fulfilled. Moreover, we assume that (23)∇1mx∈LβΩ,withsomeβ>1,and the exponents m, p satisfy one of the following conditions (1)
p is independent of t, and m(x)>0 in Ω,(2)
p(x,t)>1, m(x)>1, and ∇1/mx∈Lp(x,t)(QT),(3)
p(x,t)>1, m(x)>0, ∇1/mx∈Lp(x,t)(QT), and (24)1>1px,t+mxinQT.Then, the problem P has at least one nonnegative weak solution in QT∗, with (25)T∗=sup⁡θ:ut∞,Ω<∞,∀t∈0,θ.Moreover, for small τ the solution satisfies the estimate(26)u∞,Ω≤u0∞,ΩeAt,t∈0,τ.with a constant A depending only on the data.The following comparison principle is essential to prove uniqueness and qualitative properties of nonnegative solutions.Proposition 6.
Letu (respectively v) be a subsolution (respectively supersolution) of P, with the initial datum u0 (respectively v0), satisfying (21). We assume that (∂/∂t)b(x,u), (∂/∂t)b(x,v)∈L1(QT), and that conditions (20) are fulfilled. If either a(x,t)≤0 a.e. in QT, or m+≤q-, then we have u≤v a.e. in QT.Remark 7.
Note that the comparison principle is true for weak solutionsu with (∂/∂t)b(x,u)∈L1(QT)∩W′(QT) and recall that in the papers [29, 30], the authors gave some conditions on the data of problem P in order to ensure that this class of solutions is nonempty.Proof.
We consider the test functionϕη=signη⁡(u-v), where (27)signη⁡s=1,ifs>η,sη,ifs≤η,0,ifs<-η,and η>0 is small. It is easy to see that (28)ϕηs→sign+⁡sasη→0,where sign+⁡(s)=1, if s>0, and sign+⁡(s)=0, if s≤0. Moreover, we claim that for all u,v∈W(Qt) the function ϕη(u-v)∈W(QT). Indeed, we observe that for all s∈R, ϕηs≤1. Then, by Proposition 2(29)ϕηu-vmx+1,QT<CT.On the other hand, we have (30)∫QT∇ϕηu-vpx,tdxdt=∫0T∫u-v≤η∇ϕηu-vpx,tdxdt+∫0T∫u-v>η∇ϕηu-vpx,tdxdt+∫0T∫u-v<-η∇ϕηu-vpx,tdxdt=∫0T∫u-v≤η∇u-vηpx,tdxdt≤Cη∫QT∇upx,t+∇upx,tdx<∞.Hence, from Proposition 2 we get(31)∇ϕηu-vpx,t,QT<∞.Therefore, combining (29) and (31) we deduce the claim. On the other hand, from Definition 4, we obtain (32)∫QT∂∂tbx,u-bx,vϕηdxdt+∫QT∇upx,t-2∇u-∇vpx,t-2∇v∇ϕηdxdt≤∫QTax,tuqx,t-1u-vqx,t-1vϕηdxdt.Due to a monotonicity argument, we have (33)∫QT∇upx,t-2∇u-∇vpx,t-2∇v∇ϕηdxdt≥0,then(34)∫QT∂∂tbx,u-bx,vϕηdxdt≤∫QTax,tuqx,t-1u-vqx,t-1vϕηdxdt.By Lebesgue’s dominated convergence theorem, we have(35)limη→0⁡∫QT∂∂tbx,u-bx,vϕηdxdt=∫QT∂∂tbx,u-bx,v+dxdt.Now, we can write(36)∫QTax,tuqx,t-1u-vqx,t-1vϕηdxdt=∫QTax,t∫01∂∂ssu+1-svqx,t-1su+1-svϕηdsdxdt=∫QTax,t∫01qx,tsu+1-svqx,t-1u-vϕηdsdxdt.Then, from (34) and (35), by letting η→0, we obtain(37)∫QT∂∂tbx,u-bx,v+dxdt≤∫QTax,t∫01qx,tsu+1-svqx,t-1u-v+dsdxdt.Hence, if a(x,t)≤0 a.e. in QT, it follows that (38)∫QT∂∂tbx,u-bx,v+dxdt≤0.Then, by Gronwall’s lemma we deduce the desired result. Now, we continue the proof without any sign condition on a(x,t). From (37), by using m+≤q- and the Lebesgue’s dominated convergence theorem it follows that (39)∫QT∂∂tbx,u-bx,v+dxdt≤a1q+∫QT∫01su+1-svqx,t-mxsu+1-svmx-1u-v+dsdxdt≤∫QTu∞,QT+v∞,QTqx,t-mx∫01su+1-svmx-1u-v+dsdxdt≤Climη→0⁡∫QT∫01su+1-svmx-1u-vϕηdsdxdt≤Climη→0⁡∫QT1mx∫01∂∂ssu+1-svmx-1su+1-svdsϕηdxdt≤Cm-∫QTbx,u-bx,v+dxdt,where C is depending on the supnorms of u and v. Hence we deduce from Gronwall’s lemma that(40)∫Ωbx,u-bx,v+dx≤eCT∫Ωbx,ux,0-bx,vx,0+dx,which allows us to conclude the result.Definition 8.
We callu(x,t) a strong solution of P, if u is a weak solution and satisfies (41)∂∂tbx,u∈L1QT.


## 4. Finite Time Extinction and Nonextinction

This section is devoted to studying extinction and positivity properties for nonnegative solutions of problemP, without any sign condition on the coefficient a(x,t), and according to the ranges of p(x,t), q(x,t), and m(x). The proof of the results is based on the construction of suitable sub- and supersolutions and on the use of the preceding comparison principle given in Proposition 6.


### 4.1. Finite Time Extinction

We state and prove our main extinction result.Theorem 9.
Letu be a strong solution of P. Assume that m+<q-, p+-1<m-, supQT⁡∇px,t<∞ and u0∞ is small enough. Then, there exists a finite time T1 such that for all t≥T1(42)ux,t=0,a.e.x∈Ω.Proof.
We consider the following function(43)vx,t=kT1-t+αln⁡l+x1+⋯+xnwhere(44)l=supx1,…,xn∈Ω⁡x1+⋯+xn+2,(45)T1=u0∞kln⁡21/α,and(46)α=1m--p++1,where k>0 will be specified later. Our goal is to prove that v is a supersolution of P and by comparison principle, we can thus deduce the result. Firstly, we shall show that (47)v∈L∞QT∩WQT,∂∂tvmx∈W′QT,∀T>0.For all x∈Ω and t>0, we have (48)kT1-tαln⁡l+x1+⋯+xn≤kT1αln⁡2l<∞,and(49)∇u=kT1-tαN1/2l+x1+⋯+xn≤KT1αN1/22<∞,which implies that v∈L∞(QT)∩W(QT). Moreover, we have(50)∂∂tvmx=-αmxkln⁡l+x1+⋯+xnmxT1-t+αmx-1,then(51)∂∂tvmx≤αm+km-max⁡T1αm+-1,T1αm--1ln⁡2lm+,and hence (∂/∂t)vm(x)∈L∞(QT). Due to the embedding (52)L∞QT=L1QT′↪W′QT,we get that (∂/∂t)vm(x)∈W′(QT).
On the other hand, it is clear thatv(x,0)≥u∞≥u0(x), for a.e. x∈Ω, and v(x,t)≥0, for all x∈∂Ω, t≥0. Next, we prove that (53)∂∂tvmx-Δpx,tv≥ax,tvqx,t,inQT.Since ax,t≤a1, it suffices to prove that(54)∂∂tvmx-Δpx,tv≥a1vqx,t,inQT.By simple calculations, we obtain(55)-Δpx,tv=1N1/2kT1-t+αN1/2l+x1+⋯+xnpx,t-1px,t-1Nl+x1+⋯+xn-∑i=1N∂p∂xiln⁡kT1-t+αN1/2l+x1+⋯+xn≥1N1/2kT1-t+αN1/2l+x1+⋯+xnpx,t-1p--1Nl+x1+⋯+xn-supQT⁡∇pkT1-t+αNl+x1+⋯+xn≥kT1-t+αpx,t-1Npx,t/2l+x1+⋯+xnpx,tp--1-supQT⁡∇pkT1α=kT1-t+αpx,t-1Npx,t/2l+x1+⋯+xnpx,tp--1-supQT⁡∇pu0∞ln⁡2.We set(56)S=p--1-supQT⁡∇pu0∞ln⁡2.If supQT⁡∇p=0, then S=p--1>0. Otherwise, since u0∞ is small enough, then we can assume that u0∞<p--1ln⁡(2)/supQT⁡∇p, to deduce that S>0.
Now, we are looking for conditions onk to get (54). Thanks to (50) and (55), it is sufficient to have(57)αmxkln⁡l+x1+⋯+xnmxT1-t+αmx-1≥a1kT1-t+αln⁡l+x1+⋯+xnqx,t,and(58)kT1-t+αpx,t-1Npx,t/2l+x1+⋯+xnpx,tS≥2αmxkln⁡l+x1+⋯+xnmxT1-t+αmx-1.As u0∞ is small enough, we can assume also that u0∞≤kln⁡(2). Then it yields T1≤1, which implies that T1-t+≤1. Since α=1/(m--p++1) and q->m+, thus (57) and (58) reduce to(59)T1-t+αmx-1αmxkln⁡l+x1+⋯+xnmx-a1kln⁡l+x1+⋯+xnqx,t≥0,and(60)T1-t+αpx,t-1kpx,t-1Npx,t/2l+x1+⋯+xnpx,tS-2αmxkln⁡l+x1+⋯+xnmx≥0,which are satisfied if (61)kqx,t-mx≤αm-ln⁡2m+a1ln⁡2lq+,and(62)kmx-px,t+1≤SNp-/22αm+2lp+ln⁡2lm+.By setting E=SNp-/2/2αm+2lp+ln⁡2lm+ and F=αm-ln⁡2m+/a1ln⁡2lq+, we can choose k such that (63)k=min⁡E1/m--p++1,E1/m+-p-+1,F1/q--m+,F1/q+-m-.Therefore, we get the desired result.Next, we will mention an extinction result where there is no condition between the ranges ofp(x,t) and m(x).Proposition 10.
Letu be a strong solution of P. Assume that m->q+ and a(x,t)≤-c<0, supQT⁡∇px,t<∞ and u0∞ is small enough. Then, there exists a finite time T1 such that for all t≥T1(64)ux,t=0,a.e.x∈Ω.Proof.
We consider the same supersolutionv(x,t) as in the proof of Theorem 9 but we choose here k=1, which means (65)vx,t=T1-t+αln⁡l+x1+⋯+xnwhere(66)l=supx1,…,xn∈Ω⁡x1+⋯+xn+2,(67)T1=u0∞ln⁡21/α,and(68)α=1m--q+.We have already shown in Theorem 9 that -Δp(x,t)v≥0. We claim that (69)∂∂tvmx≥ax,tuqx,t,by using the same lines as in the proof of Theorem 9. Since q+<m-, it is therefore sufficient to have (70)T1-t+αqx,tcln⁡l+x1+⋯+xnqx,t-αmxln⁡l+x1+⋯+xnmx≥0,which is satisfied if we choose (71)c=αm+ln⁡2lm+ln⁡2q-.Consequently, by the comparison principle we deduce the extinction of solution in finite time.


### 4.2. Nonextinction of Solutions

The following theorem deals with the positivity of weak solutions.Theorem 11.
Letu be a strong solution of P and u0 not identically zero. Assume that m+<q-, m+<p--1, and supQT⁡∇px,t<∞. Then, there exists a finite time Tp such that for all t≥Tp(72)ux,t>0,a.e.x∈Ω.The method of proof is inspired from [27], where the constant exponents case is treated. However, some difficulties arise in the construction of subsolutions due to the fact that the exponents are variable. The proof of this theorem is divided into two lemmas. In the first lemma, we show by using a comparison function that the support of weak solution is nondecreasing with respect to time. In the second lemma, we show that the solution is positive locally in Ω; then, by a finite covering argument, we deduce the result.Lemma 12.
Letu be a strong solution of P. Assume that m+<q- and the initial condition u0 is nontrivial. Then(73)supp⁡u.,s⊂supp⁡u.,t,for all 0<s<t.The proof of Lemma12 follows the same lines as that of lemma 4.2 in [22], where the constant exponents case is studied. For completeness, we shall give it here.Proof.
The argument used here is based on a comparison function with which we show that the support of solution is increasing. For that we considerω an arbitrary set which is a nonzero measure subset of Ω such that infx∈ωu0≠0. We divide the proof in two cases, firstly we treat the case where m-≥1 and then the case where m+<1. If m-≥1, we consider the following function:(74)v1x,t=δm+-q-+q--m+a1t1/m+-q-,inω×0,∞,0on∂ω×0;∞,where δ=min⁡infx∈ωu0,1. It is clear that v1∈L∞(QT)∩W(QT), and it is easy to verify that (75)∂∂tv1mx∈L∞QT↪W′QT.On the other hand, by direct calculations we get (76)∂∂tv1m+=-a1m+v1q-.Hence (77)v1m+-1∂v1∂t=-a1v1q-.Since m(x)≥1 and v1(x,t)≤1 for all x∈ω, t>0, it follows that (78)∂∂tv1mx-Δpx,tv1-ax,tv1qx,t≤0,x∈ω,t>0.Moreover, from the definition of v1, we have v1(x,0)≤u0, almost everywhere in ω, and v1(x,t)=0 for all, x∈∂ω, t>0. Thus, by comparison principle we conclude that for any arbitrary ω where infx∈ωu0>0, the weak solution of P satisfies (79)ux,t>0a.e.x∈ω,andallt>0;and the result follows in this case.
Ifm+<1, we consider the following function:(80)v2x,t=δm+-q-+q--m+m-a1t1/m+-q-,inω×0,∞0,on∂ω×0;∞where δ=min⁡infx∈ωu0,1. By the same argument used previously we obtain that v2∈L∞(QT)∩W(QT), and (∂/∂t)v2m(x)∈W′(QT). Moreover, by direct calculations we get (81)∂∂tv2m+=-a1m+m-v2q-,since m+<1 and v2(x,t)≤1 for all x∈ω, t>0. Hence (82)∂∂tv2mx+a1v2qx,t≤0,Therefore, we have (83)∂∂tv2mx-Δpx,tv2-ax,tv2qx,t≤0,x∈ω,t>0.Thus, by the same argument used previously we deduce our result.Lemma 13.
Under the same assumptions of Theorem11, let the initial condition satisfies infx∈Br(x0)⁡u0≠0, for some 0<r<R≤1/2. Then, there exists Tp>0, such that for any t≥Tp, (84)ux,t>0,a.e.x∈BRx0,where BR(x0)=x∈Ω:x-x0<R.Proof.
We consider the following function:(85)ψx,t=krS-α/λtHp--1/γx,t,where(86)Hx,t=1-x-x0S1/λtp-/p--1+,St=rλ/α+λσkβrγt,0≤t≤Tp,and(87)Tp=Rλ-rλ/αλσkβrγ,where k and α are positive constants small and large enough, respectively, γ is a positive constant such that(88)γ≤p--1p+-1p--1-m+,and λ, γ, β are positive constants and will be determined later. By direct calculations, we get(89)∇ψ=krp-γS-α+1/λtHp--1/γ-1x,tx-x0S1/λt1/p--1,and(90)∂∂tψmx=-mxσαkβ+mxrγ+mxS-αmx/λ-1tHmxp--1/γx,t+mxσp-γkβ+mxrγ+mxS-αmx/λ-1tHmxp--1/γ-1x,tx-x0S1/λtp-/p--1.We can see easily that H(x,t)≤1 a.e. in BR(x0)×(0,Tp). Then, by using (88) we obtain (91)ψ∈L∞BRx0×0,Tp∩WBRx0×0,Tp,and(92)∂∂tψmx∈L∞BRx0×0,Tp↪W′BRx0×0,Tp.Moreover, from the definition of ψ we have (93)ψx,0=kHp--1/γx,0≤k≤u0xa.e.x∈BRx0.Since for all t∈[0,Tp], (94)S1/λt≤S1/λTp=R,then H(x,t)=0 on ∂BR(x0)×[0,Tp], which implies that ψ(x,t)=0 on ∂BR(x0)×[0,Tp]. Now, let us show that (95)∂∂tψmx-Δpx,tψ-ax,tψqx,t≤0,inBRx0×0,Tp.To do so, it suffices to show that(96)∂∂tψmx-Δpx,tψ+a1ψqx,t≤0,inBRx0×0,Tp.Using (89), we obtain by simple calculations(97)-Δpx,tψ=∇ψpx,t-1x-x0N-1+px,t-1p--1+∇ψpx,t-1ln⁡∇ψ∑i=1N∂px,t∂xixi-x0,ix-x0-∇ψpx,t-1p--1γ-1p-p--1H-1x,tx-x0S1/λt1/p--1px,t-1S1/λt.We set C1=(N-1+(p+-1)/(p--1)), μ=supQTp∇px,t, and (98)Gx,t=mxσαkβ+mxrγ+mxS-αmx/λ-1tHmxp--1/γx,t.To get (96), by (90) and (97) it suffices to have(99)Gx,tα-3p-γx-x0S1/λtp-/p--1H-1x,t≥0,(100)Gx,t≥3C1x-x0krp-γS-α+1/λtHp--1/γ-1x,tx-x0S1/λt1/p--1px,t-1,(101)Gx,t≥3a1krS-α/λtHp--1/γx,tqx,tand(102)p--1γ-1p-S1/λtH-1x,tx-x0S1/λt1/p--1≥μkrp-γS-α+1/λtHp--1/γ-1x,tx-x0S1/λt1/p--1Now, our goal is to choose the constants γ, λ, and β in order to verify each of the inequalities above. Firstly, let us show the inequality (99). Since α is large enough and R<1/2, we have (103)α>ln⁡rln⁡2p--1/p-R,from which we get(104)12p--1/p->Rr1/α>x-x0S1/λ0.Then, it yields (105)Hx,t≥x-x0S1/λtp-/p--1,inBRx0×0,Tp.Therefore, we have(106)3α-p-γx-x0S1/λtp-/p--1H-1x,t≥3α-p-γ≥0,so (99) is satisfied. Next, to get (100) and (101), we use the fact that s(t)≤1 and H(x,t)≤1 a.e. in BR(x0)×(0,Tp). By setting (107)γ=min⁡p--1p+-1p--1-m+,q--m+,β=min⁡p--1-m+,q--m+λ=max⁡α+p-p--1p+-1-αm-,αq+-m-,and using α large enough, so that (108)α>max⁡Z3C1σm-,3a1m-σ,where(109)Z=max⁡p-γp+-1,p-γp--1,we obtain the inequalities (100) and (101). Finally, to get the inequality (102), it suffices to have (110)p-p--1γ-1Sα/λ0≥μkrp-γ,which reduces to (111)p--1γ-1≥μkγ.Since k is small enough, the last inequality holds, which means that the inequality (102) is satisfied and allows us to deduce inequality (96). Then, we obtain by comparison principle, for each t∈[0,Tp], (112)ux,t≥ψx,ta.e.x∈BRx0.Therefore, the result follows from Lemma 12.Proof of Theorem11.
The proof is similar to that ofTheorem 1.2, in [27], and we omit the details here.


## 4.1. Finite Time Extinction

We state and prove our main extinction result.Theorem 9.
Letu be a strong solution of P. Assume that m+<q-, p+-1<m-, supQT⁡∇px,t<∞ and u0∞ is small enough. Then, there exists a finite time T1 such that for all t≥T1(42)ux,t=0,a.e.x∈Ω.Proof.
We consider the following function(43)vx,t=kT1-t+αln⁡l+x1+⋯+xnwhere(44)l=supx1,…,xn∈Ω⁡x1+⋯+xn+2,(45)T1=u0∞kln⁡21/α,and(46)α=1m--p++1,where k>0 will be specified later. Our goal is to prove that v is a supersolution of P and by comparison principle, we can thus deduce the result. Firstly, we shall show that (47)v∈L∞QT∩WQT,∂∂tvmx∈W′QT,∀T>0.For all x∈Ω and t>0, we have (48)kT1-tαln⁡l+x1+⋯+xn≤kT1αln⁡2l<∞,and(49)∇u=kT1-tαN1/2l+x1+⋯+xn≤KT1αN1/22<∞,which implies that v∈L∞(QT)∩W(QT). Moreover, we have(50)∂∂tvmx=-αmxkln⁡l+x1+⋯+xnmxT1-t+αmx-1,then(51)∂∂tvmx≤αm+km-max⁡T1αm+-1,T1αm--1ln⁡2lm+,and hence (∂/∂t)vm(x)∈L∞(QT). Due to the embedding (52)L∞QT=L1QT′↪W′QT,we get that (∂/∂t)vm(x)∈W′(QT).
On the other hand, it is clear thatv(x,0)≥u∞≥u0(x), for a.e. x∈Ω, and v(x,t)≥0, for all x∈∂Ω, t≥0. Next, we prove that (53)∂∂tvmx-Δpx,tv≥ax,tvqx,t,inQT.Since ax,t≤a1, it suffices to prove that(54)∂∂tvmx-Δpx,tv≥a1vqx,t,inQT.By simple calculations, we obtain(55)-Δpx,tv=1N1/2kT1-t+αN1/2l+x1+⋯+xnpx,t-1px,t-1Nl+x1+⋯+xn-∑i=1N∂p∂xiln⁡kT1-t+αN1/2l+x1+⋯+xn≥1N1/2kT1-t+αN1/2l+x1+⋯+xnpx,t-1p--1Nl+x1+⋯+xn-supQT⁡∇pkT1-t+αNl+x1+⋯+xn≥kT1-t+αpx,t-1Npx,t/2l+x1+⋯+xnpx,tp--1-supQT⁡∇pkT1α=kT1-t+αpx,t-1Npx,t/2l+x1+⋯+xnpx,tp--1-supQT⁡∇pu0∞ln⁡2.We set(56)S=p--1-supQT⁡∇pu0∞ln⁡2.If supQT⁡∇p=0, then S=p--1>0. Otherwise, since u0∞ is small enough, then we can assume that u0∞<p--1ln⁡(2)/supQT⁡∇p, to deduce that S>0.
Now, we are looking for conditions onk to get (54). Thanks to (50) and (55), it is sufficient to have(57)αmxkln⁡l+x1+⋯+xnmxT1-t+αmx-1≥a1kT1-t+αln⁡l+x1+⋯+xnqx,t,and(58)kT1-t+αpx,t-1Npx,t/2l+x1+⋯+xnpx,tS≥2αmxkln⁡l+x1+⋯+xnmxT1-t+αmx-1.As u0∞ is small enough, we can assume also that u0∞≤kln⁡(2). Then it yields T1≤1, which implies that T1-t+≤1. Since α=1/(m--p++1) and q->m+, thus (57) and (58) reduce to(59)T1-t+αmx-1αmxkln⁡l+x1+⋯+xnmx-a1kln⁡l+x1+⋯+xnqx,t≥0,and(60)T1-t+αpx,t-1kpx,t-1Npx,t/2l+x1+⋯+xnpx,tS-2αmxkln⁡l+x1+⋯+xnmx≥0,which are satisfied if (61)kqx,t-mx≤αm-ln⁡2m+a1ln⁡2lq+,and(62)kmx-px,t+1≤SNp-/22αm+2lp+ln⁡2lm+.By setting E=SNp-/2/2αm+2lp+ln⁡2lm+ and F=αm-ln⁡2m+/a1ln⁡2lq+, we can choose k such that (63)k=min⁡E1/m--p++1,E1/m+-p-+1,F1/q--m+,F1/q+-m-.Therefore, we get the desired result.Next, we will mention an extinction result where there is no condition between the ranges ofp(x,t) and m(x).Proposition 10.
Letu be a strong solution of P. Assume that m->q+ and a(x,t)≤-c<0, supQT⁡∇px,t<∞ and u0∞ is small enough. Then, there exists a finite time T1 such that for all t≥T1(64)ux,t=0,a.e.x∈Ω.Proof.
We consider the same supersolutionv(x,t) as in the proof of Theorem 9 but we choose here k=1, which means (65)vx,t=T1-t+αln⁡l+x1+⋯+xnwhere(66)l=supx1,…,xn∈Ω⁡x1+⋯+xn+2,(67)T1=u0∞ln⁡21/α,and(68)α=1m--q+.We have already shown in Theorem 9 that -Δp(x,t)v≥0. We claim that (69)∂∂tvmx≥ax,tuqx,t,by using the same lines as in the proof of Theorem 9. Since q+<m-, it is therefore sufficient to have (70)T1-t+αqx,tcln⁡l+x1+⋯+xnqx,t-αmxln⁡l+x1+⋯+xnmx≥0,which is satisfied if we choose (71)c=αm+ln⁡2lm+ln⁡2q-.Consequently, by the comparison principle we deduce the extinction of solution in finite time.


## 4.2. Nonextinction of Solutions

The following theorem deals with the positivity of weak solutions.Theorem 11.
Letu be a strong solution of P and u0 not identically zero. Assume that m+<q-, m+<p--1, and supQT⁡∇px,t<∞. Then, there exists a finite time Tp such that for all t≥Tp(72)ux,t>0,a.e.x∈Ω.The method of proof is inspired from [27], where the constant exponents case is treated. However, some difficulties arise in the construction of subsolutions due to the fact that the exponents are variable. The proof of this theorem is divided into two lemmas. In the first lemma, we show by using a comparison function that the support of weak solution is nondecreasing with respect to time. In the second lemma, we show that the solution is positive locally in Ω; then, by a finite covering argument, we deduce the result.Lemma 12.
Letu be a strong solution of P. Assume that m+<q- and the initial condition u0 is nontrivial. Then(73)supp⁡u.,s⊂supp⁡u.,t,for all 0<s<t.The proof of Lemma12 follows the same lines as that of lemma 4.2 in [22], where the constant exponents case is studied. For completeness, we shall give it here.Proof.
The argument used here is based on a comparison function with which we show that the support of solution is increasing. For that we considerω an arbitrary set which is a nonzero measure subset of Ω such that infx∈ωu0≠0. We divide the proof in two cases, firstly we treat the case where m-≥1 and then the case where m+<1. If m-≥1, we consider the following function:(74)v1x,t=δm+-q-+q--m+a1t1/m+-q-,inω×0,∞,0on∂ω×0;∞,where δ=min⁡infx∈ωu0,1. It is clear that v1∈L∞(QT)∩W(QT), and it is easy to verify that (75)∂∂tv1mx∈L∞QT↪W′QT.On the other hand, by direct calculations we get (76)∂∂tv1m+=-a1m+v1q-.Hence (77)v1m+-1∂v1∂t=-a1v1q-.Since m(x)≥1 and v1(x,t)≤1 for all x∈ω, t>0, it follows that (78)∂∂tv1mx-Δpx,tv1-ax,tv1qx,t≤0,x∈ω,t>0.Moreover, from the definition of v1, we have v1(x,0)≤u0, almost everywhere in ω, and v1(x,t)=0 for all, x∈∂ω, t>0. Thus, by comparison principle we conclude that for any arbitrary ω where infx∈ωu0>0, the weak solution of P satisfies (79)ux,t>0a.e.x∈ω,andallt>0;and the result follows in this case.
Ifm+<1, we consider the following function:(80)v2x,t=δm+-q-+q--m+m-a1t1/m+-q-,inω×0,∞0,on∂ω×0;∞where δ=min⁡infx∈ωu0,1. By the same argument used previously we obtain that v2∈L∞(QT)∩W(QT), and (∂/∂t)v2m(x)∈W′(QT). Moreover, by direct calculations we get (81)∂∂tv2m+=-a1m+m-v2q-,since m+<1 and v2(x,t)≤1 for all x∈ω, t>0. Hence (82)∂∂tv2mx+a1v2qx,t≤0,Therefore, we have (83)∂∂tv2mx-Δpx,tv2-ax,tv2qx,t≤0,x∈ω,t>0.Thus, by the same argument used previously we deduce our result.Lemma 13.
Under the same assumptions of Theorem11, let the initial condition satisfies infx∈Br(x0)⁡u0≠0, for some 0<r<R≤1/2. Then, there exists Tp>0, such that for any t≥Tp, (84)ux,t>0,a.e.x∈BRx0,where BR(x0)=x∈Ω:x-x0<R.Proof.
We consider the following function:(85)ψx,t=krS-α/λtHp--1/γx,t,where(86)Hx,t=1-x-x0S1/λtp-/p--1+,St=rλ/α+λσkβrγt,0≤t≤Tp,and(87)Tp=Rλ-rλ/αλσkβrγ,where k and α are positive constants small and large enough, respectively, γ is a positive constant such that(88)γ≤p--1p+-1p--1-m+,and λ, γ, β are positive constants and will be determined later. By direct calculations, we get(89)∇ψ=krp-γS-α+1/λtHp--1/γ-1x,tx-x0S1/λt1/p--1,and(90)∂∂tψmx=-mxσαkβ+mxrγ+mxS-αmx/λ-1tHmxp--1/γx,t+mxσp-γkβ+mxrγ+mxS-αmx/λ-1tHmxp--1/γ-1x,tx-x0S1/λtp-/p--1.We can see easily that H(x,t)≤1 a.e. in BR(x0)×(0,Tp). Then, by using (88) we obtain (91)ψ∈L∞BRx0×0,Tp∩WBRx0×0,Tp,and(92)∂∂tψmx∈L∞BRx0×0,Tp↪W′BRx0×0,Tp.Moreover, from the definition of ψ we have (93)ψx,0=kHp--1/γx,0≤k≤u0xa.e.x∈BRx0.Since for all t∈[0,Tp], (94)S1/λt≤S1/λTp=R,then H(x,t)=0 on ∂BR(x0)×[0,Tp], which implies that ψ(x,t)=0 on ∂BR(x0)×[0,Tp]. Now, let us show that (95)∂∂tψmx-Δpx,tψ-ax,tψqx,t≤0,inBRx0×0,Tp.To do so, it suffices to show that(96)∂∂tψmx-Δpx,tψ+a1ψqx,t≤0,inBRx0×0,Tp.Using (89), we obtain by simple calculations(97)-Δpx,tψ=∇ψpx,t-1x-x0N-1+px,t-1p--1+∇ψpx,t-1ln⁡∇ψ∑i=1N∂px,t∂xixi-x0,ix-x0-∇ψpx,t-1p--1γ-1p-p--1H-1x,tx-x0S1/λt1/p--1px,t-1S1/λt.We set C1=(N-1+(p+-1)/(p--1)), μ=supQTp∇px,t, and (98)Gx,t=mxσαkβ+mxrγ+mxS-αmx/λ-1tHmxp--1/γx,t.To get (96), by (90) and (97) it suffices to have(99)Gx,tα-3p-γx-x0S1/λtp-/p--1H-1x,t≥0,(100)Gx,t≥3C1x-x0krp-γS-α+1/λtHp--1/γ-1x,tx-x0S1/λt1/p--1px,t-1,(101)Gx,t≥3a1krS-α/λtHp--1/γx,tqx,tand(102)p--1γ-1p-S1/λtH-1x,tx-x0S1/λt1/p--1≥μkrp-γS-α+1/λtHp--1/γ-1x,tx-x0S1/λt1/p--1Now, our goal is to choose the constants γ, λ, and β in order to verify each of the inequalities above. Firstly, let us show the inequality (99). Since α is large enough and R<1/2, we have (103)α>ln⁡rln⁡2p--1/p-R,from which we get(104)12p--1/p->Rr1/α>x-x0S1/λ0.Then, it yields (105)Hx,t≥x-x0S1/λtp-/p--1,inBRx0×0,Tp.Therefore, we have(106)3α-p-γx-x0S1/λtp-/p--1H-1x,t≥3α-p-γ≥0,so (99) is satisfied. Next, to get (100) and (101), we use the fact that s(t)≤1 and H(x,t)≤1 a.e. in BR(x0)×(0,Tp). By setting (107)γ=min⁡p--1p+-1p--1-m+,q--m+,β=min⁡p--1-m+,q--m+λ=max⁡α+p-p--1p+-1-αm-,αq+-m-,and using α large enough, so that (108)α>max⁡Z3C1σm-,3a1m-σ,where(109)Z=max⁡p-γp+-1,p-γp--1,we obtain the inequalities (100) and (101). Finally, to get the inequality (102), it suffices to have (110)p-p--1γ-1Sα/λ0≥μkrp-γ,which reduces to (111)p--1γ-1≥μkγ.Since k is small enough, the last inequality holds, which means that the inequality (102) is satisfied and allows us to deduce inequality (96). Then, we obtain by comparison principle, for each t∈[0,Tp], (112)ux,t≥ψx,ta.e.x∈BRx0.Therefore, the result follows from Lemma 12.Proof of Theorem11.
The proof is similar to that ofTheorem 1.2, in [27], and we omit the details here.


## 5. Finite Speed of Propagation Property

In this section we shall give precise estimates for the of support of the solution ofP, depending on the size of the support of u0. Let us emphasize that each estimation is obtained under a sign condition on a(x,t) and depending on the range of the exponents p(x,t), q(x,t), and m(x). As in [21], the proof is based on the construction of local supersolutions and on the use of the comparison principle.Concerning the construction of supersolutions, we shall proceed as in [31].Note that under some conditions on the data, ifa(x,t) is positive, then the solutions will blow up in finite time (see [10]). For that it requires to construct a supersolution defined locally in time, which means in (0,T) for any T≤T∗, where T∗ is the maximal existence time. We denote ΩR=Ω¯∩B(0,R).Theorem 14.
Letu be a strong solution of P. Let 0<R<+∞ be such that supp⁡u0⊂ΩR. We assume m+<p--1, q(x,t)=m(x), and supQT∇px,t<∞. Then, for any t∈(0,T), there exists a unique compactly supported solution of P such that (113)supp⁡ut,.⊂x∈Ω¯:x≤ρt,where ρ(t) will be specified in the proof below.Proof.
The idea of the proof is to construct a suitable supersolutionu¯ with compact support which is not necessarily defined in the whole Ω. Then, by the comparison principle, we deduce directly that x,t∈QT:u¯=0⊂x,t∈QT:u=0, and the result follows. For all t∈(0,T), we define u¯x,t as follows(114)u¯r,t=KTem-+1λtct+kR-rα,R≤r<R+ctk=ρt0,r≥R+ctkwhere r=x, (115)α=p-p--1-m+,M=max⁡1,u0∞,(116)λ=a1m-2,(117)ct=MKT1/αeλtp+-1-m-/α,and(118)KT=m-λe-m-+1λTp+-1-m-kα-1p+-1+ϵn-1kp--1αp+-11/p+-1-m-,with sufficiently small ϵ>0.
Firstly, we denoteQTR=(Ω∖ΩR)×(0,T). It is clear that u¯∈C1(QTR) and by direct calculations, we have(119)∂∂tu¯mx=m-+1mxλu¯mx+αmxKTem-+1λtmxct+kR-rαmx-1c′t.Thus, it follows that (120)∂∂tu¯mx∈L∞QTR⊂W′QTR.Moreover, from the definition of u¯ we have (121)u¯R,0=M≥u0∞≥u0.Next, we will show that (122)∂∂tu¯mx-Δpx,tu¯≥ax,tu¯mxonQTR.Since ax,t≤a1, it suffices to show that(123)∂u¯mx∂t-Δpx,tu¯≥a1u¯mxonQTR.Using the hypothesis p+>m-+1, we have c′t>0 for all t∈(0,T). Then, from (119) and (116) we obtain(124)∂∂tu¯mx≥m-+1mxλu¯mx≥a1u¯mx+mxλu¯mx.Due to the last inequality, it remains to prove that(125)-Δpx,tu¯+mxλu¯mx≥0.onQTR.We have(126)∂u¯∂xi=-αkKTem-+1λtct+kR-rα-1xix,and(127)∇u¯=αkKTem-+1λtct+kR-rα-1.We set(128)νTt=αkKTem-+1λt.Then(129)Δpx,tu¯=ct+kR-rα-1px,t-1-1νTtpx,t-1kα-1px,t-1-hx,t,where(130)hx,t=ln⁡ct+kR-rct+kR-rα-1∑i=1N∂p∂xixix+N-1x+∑i=1N∂p∂xixixln⁡νTtct+kR-r=n∑i=1N∂p∂xixixln⁡ct+kR-r1/nct+kR-r1/nct+kR-r1-1/nα-1+N-1x+∑i=1N∂p∂xixixln⁡νTct+kR-r1/nct+kR-r1-1/n,for all n≥3. By straightforward considerations, (boundedness of different functions), there exist positive constants A,B≥1 such that, for all t∈(0,T) and R≤r<ρ(t), (131)n∑i=1N∂p∂xixixln⁡ct+kR-r1/nct+kR-r1/nα-1≤AN-1x+∑i=1N∂p∂xixixln⁡νTtct+kR-r1/n≤B.Thus (132)hx,t≤A+Bct+kR-r1-1/n≤A+Bct+kR-r1/nn-1.In this case we have kr-R≤ct, and from (118), we can choose k small enough, so that ct is also small for all t∈(0,T), whence we have (133)A+Bct+kR-r1/n≤ϵ,for sufficiently small ϵ>0, which implies that(134)Δpx,tu¯≤ct+kR-rα-1px-1-1νTtpx,t-1kα-1px,t-1+ϵn-1.Set W(x,t)=k(α-1)(p(x,t)-1)+ϵn-1. Now, to get (125) it suffices to have (135)ct+kR-rα-1pr,t-1-1νTtpx,t-1Wr,t≤ct+kR-rαmxmxλKmxem-+1λmxt.Since ct+kR-r is small enough then, from the value of α and K(T), we obtain that (136)ct+kR-rα-1px,t-1-1≤ct+kR-rαmx,and(137)αkKTem-+1λtpx,t-1Wx,t≤mxλKTmxem-+1λmxt,which implies (125). Therefore we deduce the desired result.Theorem 15.
Letu be a strong solution of P. Let 0<R<+∞ be such that supp⁡u0⊂ΩR. We assume m+<p--1, supQT∇px,t<∞, and a(x,t)≤0. Then, for any t∈(0,T), there exists a unique compactly supported solution of P such that(138)supp⁡ut,.⊂x∈Ω¯:x≤ρt,where ρ(t) is specified in the proof above.Proof.
We consider the same supersolutionu¯r,t as in the proof of Theorem 14 and we just need to prove that (139)∂u¯mx∂t-Δpx,tu¯≥ax,tu¯qx,tonQTR,where QTR=(Ω∖ΩR)×(0,T). Since a(x,t)≤0, it is sufficient to prove that (140)∂u¯mx∂t-Δpx,tu¯≥0onQTR.Combining the same lines as in the previous theorem and the comparison principle in Proposition 6, we conclude the result.Finally, we state the following result on uniform localization of the support of solution.Theorem 16.
Letu be a strong solution of P. Let 0<R<+∞ be such that supp⁡u0⊂ΩR. We assume q+<p--1, supQT∇px,t<∞, and a(x,t)≤a2<0. Then for any t∈(0,∞), there exists a unique compactly supported solution of P such that(141)supp⁡ut,.⊂x∈Ω¯:x≤R1where R1 is positive constants determined in the proof below.Proof.
In order to get the desired estimation of the support of the solution, we define a suitable local supersolution associated with the stationary problem related toP. Let M=maxu0L∞Ω,1. We define u^ as follows:(142)u^r=KR1-rαR≤r<R10r≥R1,where r=x, and(143)K=MR1-Rα,where R1-R≤1 and α>1 is a constant that will be determined hereafter. It is easy to verify that u^∈C1(QTR) for any T>0, where QTR=(Ω∖ΩR)×(0,T). Concerning initial condition, we have u^(R,0)=u^(R)≥M≥u0. Now, let us show that(144)∂bx,u^∂t-Δpx,tu^≥ax,tu^qx,t,x,t∈QTR.We have (145)∂u^∂xi=-αKR1-rα-1xir,and(146)∇u^=αKR1-rα-1.Then(147)Δpx,tu^=αKpx,t-1R1-rα-1px,t-1-1α-1px,t-1-Hx,t,with(148)Hx,t=α-1R1-rln⁡R1-r∑i=1N∂p∂xixix-N-1r+∑i=1N∂p∂xixixln⁡KαR1-r.By straightforward considerations, there exists a constant A≥1 such that, for every r with R≤r<R1, we have (149)α-1px,t-1-Hx,t≤A.Hence, we get (150)Δpx,tu^≤AαKpx,t-1R1-rα-1px,t-1-1.Now, in order to obtain the inequality (144), we need to show that(151)Δpx,tu^≤-ax,tu^rqx,t.To do so, we just need to get(152)AαKpx,t-1R1-rα-1px,t-1-1≤-a2Kqx,tR1-rαqx,t,Due to the fact that R1-r≤R1-R≤1, (152) holds if (153)α-1px,t-1-1-αqx,t≥0,Aαpx,t-1Kpx,t-1-qx,t≤-a2.Then, it is suffices to take (154)α=p-p--1-q+,a2=-Aαp+-1Kp+-1-q-,to complete the proof.Remark 17.
As in [20, 21], we can assume in this section that Ω is a set of RN, not necessarily bounded.



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