# Global Existence and Extinction Singularity for a Fast Diffusive Polytropic Filtration Equation with Variable Coefficient

Authors: Dengming Liu; Changyu Liu
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

In this article, we deal with an inhomogeneous fast diffusive polytropic filtration equation. By using the energy estimate approach, Hardy–Littlewood–Sobolev inequality, and a series of ordinary differential inequalities, we prove the global existence result and obtain the conditions on the occurrence of the extinction phenomenon of the weak solution.


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## Body

## 1. Introduction

Our main objectives in this article are to deal with the global existence and the extinction phenomenon of the inhomogeneous fast diffusive polytropic filtration equation:(1)x−sut−div∇ump−2∇um=uq,x,t∈Ω×0,+∞,ux,t=0,x,t∈∂Ω×0,+∞,ux,0=u0x,x∈Ω,where Ω⊂ℝNN>p is a bounded domain with smooth boundary ∂Ω, x=x1,…,xN∈Ω, x=x12+⋯+xN2, u0x is a nonnegative and bounded function with u0m∈W01,pΩ, and the parameters m, s, p, and q satisfy(2)0<m≤1,0<mp−1<1,1−m2<q≤1,0≤s<Nqq+m.Inhomogeneous parabolic problems arise in a wide range of physical contexts (see for instance [1–3] and the references therein, where a more detailed physical background can be found). Problem (1) can be used to describe the compressible fluid flows in a homogeneous isotropic rigid porous medium with ux,t being the density of the fluid and αx=x−s acting as the volumetric moisture content. On the other hand parabolic models like (1), together with differential equation models, stochastic differential equations, and linear systems, are regarded as the powerful tools to solve lots of problems from control engineering, image processing, and other areas (see [4–8]). Because of the degeneracy and the singularity, problem (1) might not have classical solution in general, and hence, we introduce definition of the weak solution as follows.Definition 1.
By a local weak solution to problem (1), we understand a function u∈S=defu∈C0,T;L1Ω,u∈L2qΩ×0,T∩L2Ω×0,T,∇um∈LpΩ×0,T for some T>0, which moreover satisfies the following assumptions:(i)
For any0≤ϕ∈S˜=defϕ∈S,ϕ|∂Ω=0 and 0<t1<t2<T, one has(3)∫Ωx−sutϕdx+∫Ω∇ump−2∇um⋅∇ϕdx=∫Ωuqϕdx.(ii)
ux,t⟶u0x as t⟶0 with convergence in L1Ω.In the past few decades, many mathematicians have studied the global existence, blow-up, and extinction phenomena of the following parabolic equation:(4)x−sut−div∇ump−2∇um=fu,x,t∈Ω×0,+∞,subject to various assumptions (see [9–23] and the references therein). For the case s=0, the authors in [24–26] concerned with the global existence and blow-up properties of the solutions to problem (4) with m=1 and p≥2. Yuan et al. [27] considered problem (4) with m=1 and fu=0 and showed that the solution of problem (4) vanishes in finite time if and only if 1<p<2. Gu [28] studied problem (4) with m=1 and fu=−uq and claimed that the necessary and sufficient condition on the occurrence of extinction phenomenon is p∈1,2 or q∈0,1. Tian and Mu [29] and Jin and Yin [30] studied problem (4) with m=1 and fu=λuq with λ>0 and showed that q=p−1 is the critical extinction exponent of the solutions. When s=0 and fu=λuq with λ>0, Jin et al. [31] and Zhou and Mu [32] concluded that the critical extinction exponent of the solution to problem (1) is q=mp−1. Compared with s=0, there are few literatures for the case s>0. By Hardy inequality and potential well method, Tan [33] obtained the global existence and blow-up results of problem (4) with s=2 and m=1. Wang [34] generated the results in [33] to the case 0<s≤2. Zhou generated the results in [33] to the case s≥2 and gave the global existence and blow-up results for (4) with p=2 and m=1 in [35, 36], respectively. To the best knowledge of us, there is little work on the global existence and extinction behavior of problem (1). In a recent paper, Deng and Zhou [37] considered the special case m=1 and analysed the effect of the singular potential on the global existence and extinction behavior of the solutions.In order to state well our results, we first introduce some definitions, fundamental facts, and useful symbols. SinceΩ is a bounded domain in ℝN, then there is a ball B0,R⊂ℝN centered at 0 with radius(5)R=supx∈Ωx12+⋯+xN2,such that Ω⊆B0,R.We denote the norm ofLrΩ by ⋅r and the norm of W1,rΩ by ⋅W1,rΩ; that is, for any ϕ∈LrΩ,(6)ϕr=∫Ωϕxrdx1/rif1≤r<+∞,esssupx∈Ωϕx,ifr=+∞,and for any ϕ∈W1,rΩ, ϕW1,rΩ=rϕrr+∇ϕrr. According to Poincaré’s inequality, one can see that ∇⋅r is equivalent to ⋅W1,rΩ in W01,rΩ, and hence, we equip W01,rΩ with the norm ϕ=def∇ϕr.Letux,t be a weak solution of problem (1). Define an energy functional as the following form:(7)Eu=1p∫Ω∇umpdx−mm+q∫Ωum+qdx.Then, by (3), one can easily show that(8)∂Eu∂t=−m∫Ωum−1ut2dx≤0,which tells us that Eu is nonincreasing with respect to t.We state our main results as follows.Theorem 1.
Suppose that the parametersm, p, q, and s satisfy (2), and the initial data u0x are a nonnegative and bounded function with u0m∈W01,pΩ. Let ux,t be a solution of problem (1). Then, the maximal existence time of ux,t is T=+∞; that is, ux,t is a global solution. Moreover,(i)
If0<maxmp−1,1−m/2<q≤1 and 0≤s≤minp,Nq/m+q and there is a constant a>maxm+1−αm/mα−mp,−1/p with α=pN−s/N−p such that(9)∫Ωx−su0mpa+1+1dxq−mp−1/mpa+1+1≤pa+12a+1pR−smpa+1+q/mpa+1+1|Ω|q−1/mpa+1+1κ1N−p/N−s−mpa+1/mpa+1+1κ2−N−p/N−s,where R, κ1, and κ2 are given by (5), (12), and Lemma 2, respectively, then the solution ux,t of problem (1) vanishes in finite time.(ii)
If0<1−m/2<q≤mp−1<1 and(10)∫Ωx−su0m+1dx>0andEu0≤0,whenq=mp−1,∫Ωx−su0m+1dx>0andEu0<0,whenq<mp−1,where(11)Eu0=1p∫Ω∇u0mpdx−mm+q∫Ωu0m+qdx,
then the solutionux,t of problem (1) does not possess extinction phenomenon.The rest of this article is organized as follows. In Section2, we collect some useful auxiliary lemmas. The last section is mainly focused on the global existence and the conditions on the occurrence of the extinction phenomenon of the solution. By Hardy–Littlewood–Sobolev inequality and some ordinary differential inequalities, the proof of Theorem 1 will be given in Section 3.


## 2. Preliminaries

In this section, as preliminaries, we collect some well-known results, which play an important role in our proof of Theorem1.Lemma 1 (see [37]).
SupposeN>s and Ω⊂ℝN is a bounded domain. Then, we have(12)κ1=def∫Ωx−sdx≤∫B0,Rx−sdx=∫0R∫∂B0,rx−sdSxdr=ωN∫0Rr−srN−1dr=ωNN−sRN−s<+∞,where B0,R is the ball in ℝN centered at 0 with radius R=supx∈Ωx12+⋯+xN2 satisfying Ω⊂B0,R and(13)ωN=NπN/2ΓN/2+1denotes the surface area of the unit sphere ∂B0,1, and Γ is the usual Gamma function.Lemma 2 (see [38]).
SupposeN≥2, 1<μ<N, 0≤ϑ≤μ, and σ=μN−ϑ/N−μ. Then, there is a positive constant κ2=κ2μ,ϑ,N such that(14)∫Ωuxσxϑdx≤κ2∫Ω∇uμdxN−ϑ/N−μholds for any u∈W01,μΩ, where Ω⊂ℝN is a bounded domain.Lemma 3 (see [39]).
Assumeθ, δ, and β are positive constants. Let yt be a nonnegative absolutely continuous function satisfying(15)dydt+δyθt≥β,t>0.Then, we have(16)yt≥miny0,βδ1/θ.Lemma 4 (see [40]).
Suppose0<k<r≤1. Let yt be the solution of the ordinary differential inequality:(17)dydt+Cyk≤γyr,t>0,y0=y0>0,where C>0 and 0<γ<Cy0k−r/2. Then, there are two positive constants η and ξ such that, for t≥0,(18)0≤yt≤ξe−ηt.


## 3. Proof of Theorem1

In this section, we will give the proof of the global existence result and the conditions on the occurrence of the extinction phenomenon of the solutionux,t.Case 1.
If1−m/2<q<1. Taking the test function ϕ=umx,t in (3), and using Hölder’s inequality, one has(19)1m+1ddt∫Ωx−sum+1dx+ump=∫Ωuq+mdx=∫Ωx−sq+m/1+muq+mxsq+m/1+mdx≤∫Ωx−sum+1dxm+q/m+1∫Ωxsm+q/1−qdx1−q/m+1≤κ3∫Ωx−sum+1dxm+q/m+1,which implies that(20)ddt∫Ωx−sum+1dx≤κ3m+1∫Ωx−sum+1dxm+q/m+1,where(21)κ3=∫B0,Rxsq+m/1−qdx1−q/1+m=ωN1−qsq+m+N1−qRsq+m+N1−q/1−q1−q/1+m.
Integrating (20) from 0 to t, one gets(22)∫Ωx−sum+1dx≤κ31−qt+∫Ωx−su0m+1dx1−q/m+1m+1/1−q.
From (19) and (22), it follows that(23)uq+mq+m≤κ3κ31−qt+∫Ωx−su0m+1dx1−q/m+1m+q/1−q.
On the other hand, taking the test functionϕ=umt in (3), then by using Cauchy’s inequality with ε and Hölder’s inequality, one can obtain(24)4mm+12∫Ωx−sum+1/2t2dx+1pddtump=2mm+1∫Ωxs/2um+2q−1/2x−s/2um+1/2tdx≤2mεm+1∫Ωx−sum+1/2t2dx+m2εm+1∫Ωxsum+2q−1dx≤2mεm+1∫Ωx−sum+1/2t2dx+m2εm+1∫Ωxsq+m/1−qdx1−q/q+m∫Ωuq+mdxm+2q−1/q+m.
Letε be sufficiently small to ensure that 4m/m+12−2mε/m+1≥0, then by (21), (23), and (24), one has(25)ddtump≤mpκ322εm+1κ31−qt+∫Ωx−su0m+1dx1−q/m+1m+2q−1/1−q.
Integrating (25) from 0 to t yields that(26)ump≤u0mp+mpκ322εm+1∫0tκ31−qτ+∫Ωx−su0m+1dx1−q/m+1m+2q−1/1−qdτ≤u0mp−mpκ32εm+1m+q∫Ωx−su0m+1dxm+q/m+1+mpκ32εm+1m+qκ31−qt+∫Ωx−su0m+1dx1−q/m+1m+q/1−q,which means that the solution ux,t of the problem (1) is global.Case 2.
Ifq=1, taking the test function ϕ=umx,t in (3), then we can see that(27)1m+1ddt∫Ωx−sum+1dx+ump=∫Ωx−sum+1xsdx≤Rs∫Ωx−sum+1dx,which tells us that(28)∫Ωx−sum+1dx≤em+1Rst∫Ωx−su0m+1dx,um+1m+1≤Rsem+1Rst∫Ωx−su0m+1dx.
On the other hand, taking the test functionϕ=umt in (3), then Cauchy’s inequality with ε leads to(29)4mm+12∫Ωx−sum+1/2t2dx+1pddtump=2mm+1∫Ωxs/2um+1/2x−s/2um+1/2tdx≤2mεm+1∫Ωx−sum+1/2t2dx+m2εm+1∫Ωxsum+1dx≤2mεm+1∫Ωx−sum+1/2t2dx+mRs2εm+1∫Ωum+1dx.
Choosingε∈0,2/m+1 to guarantee that 4m/m+12−2λmε/m+1≥0, then by (29), one has(30)ddtump≤mpR2s2εm+1em+1Rst∫Ωx−su0m+1dx,which implies that(31)ump≤u0mp+mpRs2εm+12em+1Rst−1∫Ωx−su0m+1dx.
Then, the proof of the global existence result is complete.
Now, we take our attention to the extinction singularity of the solutionux,t to problem (1). We denote α=pN−s/N−p. Noticing that 0≤s<p, we can verify that α>p. Let a be a constant satisfying(32)a>maxm+1−αmmα−mp,−1p.
From (32), it follows that(33)0<mpa+1+1αma+1<1.
Selecting the test functionϕ=umpa+1x,t in (3), one has(34)1mpa+1+1ddt∫Ωx−sumpa+1+1dx+pa+1a+1p∫Ω∇uma+1pdx=∫Ωumpa+1+qdx.
Making use of Hölder’s inequality, one can find that(35)∫Ωx−sumpa+1+1dx=∫Ωx−smpa+1+1/αma+1uma+1⋅mpa+1+1/ma+1x−s1−mpa+1+1/αma+1dx≤∫Ωx−suαma+1dxmpa+1+1/αma+1∫Ωx−sdx1−mpa+1+1/αma+1=κ11−mpa+1+1/αma+1∫Ωx−suαma+1dxmpa+1+1/αma+1,where κ1 is the same as that in (12). Since 0<mp−1<1 and 0<m≤1, one can deduce that 1<p<1+1/m. This together with the assumption N>p one has 1<p<minN,1+1/m. Meanwhile, recalling that 0≤s<minp,Nq/q+m, then it follows from Lemma 2 that(36)∫Ωx−suαma+1dx=∫Ωx−suma+1⋅pN−s/N−pdx≤κ2∫Ω∇uma+1pdxN−s/N−p,where κ2 is given in Lemma 2. Combining (35) with (32), one sees(37)∫Ωx−sumpa+1+1dx≤κ11−mpa+1+1/αma+1κ2mpa+1+1/αma+1∫Ω∇uma+1pdxN−smpa+1+1/αma+1N−p=κ11−mpa+1+1/αma+1κ2mpa+1+1/αma+1∫Ω∇uma+1pdxmpa+1+1/mpa+1.
Exploiting (34) and (37), one can arrive at(38)1mpa+1+1ddt∫Ωx−sumpa+1+1dx+pa+1a+1pκ1N−p/N−s−mpa+1/mpa+1+1κ2−N−p/N−s∫Ωx−sumpa+1+1dxmpa+1/mpa+1+1≤∫Ωuq+mpa+1dx.
In what follows, for the sake of simplicity, we denoteyt=∫Ωx−sumpa+1+1dx, and C1=mpa+1+1pa+1/a+1pκ1N−p/N−s−mpa+1/mpa+1+1κ2−N−p/N−s.
Ifmax1−m/2,mp−1<q=1, then from (38), one can immediately know that(39)dydt+C1ympa+1/mpa+1+1t≤C2yt,where C2=mpa+1+1Rs. Remembering that 0<mp−1<1 and (32), one can check that 0<mpa+1/mpa+1+1<1. If 2C2<C1ymp−1−1/mpa+1+10, then Lemma 4 tells us that there are two positive constants η1 and ξ1 satisfying(40)0≤yt≤ξ1e−η1t.
PuttingT0=max0,mpa+1+1/η11−mp−1ln2C2/C1ξ11−mp−1/mpa+1+1, then for any t>T0, (40) leads to(41)2C2y1−mp−1/mpa+1+1t<C1,which together with (39) yields(42)dydt+C12ympa+1/mpa+1+1≤0,t≥T0.
Integrating above inequality fromT0 to t leads to(43)y1−mp−1/mpa+1+1t≤y1−mp−1/mpa+1+1T0−C11−mp−12mpa+1+1t−T0,t≥T0.
The above inequality means that(44)limt⟶T1yt=limt⟶T1∫Ωx−sumpa+1+1dx=0,where(45)T1=T0+2mpa+1+1C11−mp−1y1−mp−1mpa+1+1T0.
Ifmax1−m/2,mp−1<q<1. In view of Hölder’s inequality, one has(46)∫Ωumpa+1+qdx≤Ω1−q/mpa+1+1∫Ωumpa+1+1dxmpa+1+q/mpa+1+1≤Ω1−q/mpa+1+1Rsmpa+1+q/mpa+1+1∫Ωx−sumpa+1+1dxmpa+1+q/mpa+1+1.
Combining (38) with (46), one can conclude that(47)dydt+C1ympa+1/mpa+1+1t≤C3ympa+1+q/mpa+1+1t,where(48)C3=λmpa+1+1Ω1−q/mpa+1+1Rsmpa+1+q/mpa+1+1.
Recalling that0<mp−1<q<1 and (32), one can check that(49)0<mpa+1mpa+1+1<mpa+1+qmpa+1+1<1.
Then, by (47) and Lemma 4, one knows that there are two positive constants η2 and ξ2 satisfying(50)0≤yt≤ξ2e−η2t,provided that 2C3yq−mp−1/mpa+1+10<C1. Setting T2=max0,mpa+1+1/η2q−mp−1ln2C3/C1ξ2q−mp−1/mpa+1+1, then for any t>T1, (50) leads to(51)2C3yq−mp−1/mpa+1+1t<C1,which together with (47) yields(52)dydt+C12ympa+1/mpa+1+1≤0,t≥T2.
The remainder proof is the same as the previous one in the caseq=1, and we omit it here. Up to now, the proof of the extinction phenomenon of the solution ux,t to problem (1) is complete.
Now, we begin to prove the nonextinction result. Denoting(53)Mt=1m+1∫Ωx−sum+1dx,and taking the derivative of Mt with respect to t, one has(54)M′t=∫Ωx−sumutdx=∫Ωum+qdx−∫Ω∇umpdx=q−mp−1m+q∫Ωum+qdx−pEu.
Ifq=mp−1. From (8), one knows that Eu is nonincreasing with respect to t. Then, for any t≥0, one has(55)M′t≥−pEu0.
Integrating, one obtains(56)Mt≥M0−pEu0t,fort≥0,which tells us that Mt>0 since M0>0 and Eu0≤0; that is, the solution ux,t of problem (1) does not possess extinction phenomenon.
Ifq<mp−1, with the help of Hölder’s inequality, one gets(57)q−mp−1m+q∫Ωum+qdx≥q−mp−1m+qMm+q/m+1t∫Ωxsm+q/1−qdx1−q/m+1.
Exploiting (54) and (57), one can claim that(58)M′t−C4Mm+q/m+1t≥−pEu0,where(59)0>C4=q−mp−1m+q∫Ωxsm+q/1−qdx1−q/m+1>q−mp−1m+qRsm+q/m+1Ω1−q/m+1>−∞.
SinceM0>0 and Eu0<0, then from (58) and Lemma 3, it follows that(60)Mt≥minM0,pEu0C4m+1/m+q>0,which means that the solution ux,t of problem (1) does not possess extinction phenomenon.



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