# On the Solutions of a Porous Medium Equation with Exponent Variable

Authors: Huashui Zhan
Journal: Discrete Dynamics in Nature and Society
 (2019)

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## Abstract

The paper studies the initial-boundary value problem of a porous medium equation with exponent variable. How to deal with nonlinear term with the exponent variable is the main dedication of this paper. The existence of the weak solution is proved by the monotone convergent method. Moreover, according to the different boundary value conditions, the stability of weak solutions is studied. In some special cases, the stability of weak solutions can be proved without any boundary value condition.


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## Body

## 1. Introduction

Letρ be the density, let V be the velocity, and let p be the pressure of the ideal barotropic gas through a porous medium. The motion is governed by the mass conservation law (1)ρt+div⁡ρV=0,the Darcy law (2)V=-kx∇p,and the equation of stage (3)p=Pρ,where k(x) is a given matrix. One of the most common cases is P(s)=μsα with μ,α=const. Then we obtain a semilinear parabolic equation on the density (4)ρt=μα1+αdiv⁡kx∇ρ1+α.If we additionally assume that p may explicitly depend on x and has the form p=μργ(x), then equation for ρ becomes (5)ρt=μdiv⁡kxρ∇ργx,and can be written as (6)ρt=μdiv⁡kxγxργx∇ρ)+ρlog⁡ρkx∇γ.If k(x)=a(x)I, where a(x) is a function and I is the unit matrix, then (4) becomes (7)ρt=μα1+αdiv⁡ax∇ρ1+α=μαdiv⁡axρα∇ρ,and (6) has the form (8)ρt=μdiv⁡axγxργx∇ρ+ρlog⁡ρaxI·∇γ.In this paper, we generalized (8) to the following type: (9)ut=div⁡axumx∇u+∑i=1N∂biumx+1∂xi,x,t∈QT=Ω×0,T,and consider the initial-boundary value problem, where m(x)>0 is a C1(Ω¯) function, bi(s)∈C1(R), Ω⊂RN is a bounded domain with a smooth boundary ∂Ω.Ifa(x)≡1, bi=0, m(x)=m-1 is a constant, (9) is equivalent to the so-called porous medium equation (10)ut=△um.In this case, there exists an abundant literature; one can refer to the survey books [1–6] and the references therein.Ifa(x)≥0, in one way, (9) can be regarded as a special case of reaction-diffusion equation (11)ut=div⁡au,x,t∇u+div⁡bu→,there are also many papers devoted to its well-posedness problem. The most striking part of this equation is that if there is an interior point of the set (12)x∈Ω:a·,x,t=0,then the uniqueness of weak solution can be proved only under the entropy condition; one can refer to [7–15]. Moreover, if a(·,x,t) is degenerate on the boundary, how to impose a suitable boundary value condition to study the well-posedness of weak solutions to (11) has attracted extensive attentions and has been widely studied for a long time. In the other word, though the initial value (13)ux,0=u0x,x∈Ω,is always imposed, the Dirchilet boundary condition (14)ux,t=0,x,t∈ST=∂Ω×0,T,may not be imposed or be imposed in a weaker sense than the traditional trace. One can refer to [7–12] for the details.In another way, the evolutionary equations with variable exponents, especially the so-called electrorheological fluids equations with the form(15)ut=div⁡∇upx-2∇u+fx,t,u,∇u,have been brought to the forefront by many scholars since the beginning of this century; one can refer to [16–23] and the references therein. But we noticed that, compared with (15), the papers devoted to the equations with the type (16)ut=div⁡uγx∇u+fx,t,u,∇u,seem much fewer. The existence, uniqueness, and localization properties of solutions to (16) have been studied by Antontsev-Ahmarev in [24]. The free boundary problem and the numerical study were researched in [25] by Duque et al. Different from these papers [16–20, 24, 25], we enable the diffusion a(x) in (9) to be degenerate on the boundary. In detail, we suppose that (17)ax>0,x∈Ω;ax=0,x∈∂Ω,bi(s) is a C1(R function, and (18)0≤u0∈L∞Ω,ax∇u0m∈L2Ω.Definition 1.
If a nonnegative functionu(x,t) satisfies(19)u∈L∞QT,axumx∇u∈L∞0,T;L2Ω,and for any function φ∈C1(QT), φ|t=T=0, φ|∂Ω=0, there holds (20)∬QT-∂φ∂tu+axumx∇u∇φdxdt+∑i=1N∬QTbiumx+1φxix,tdxdt=∫Ωu0φx,0dx,then we say u(x,t) is weak solution of (9) with the initial value (13) in the sense (21)limt→0∫Ωux,t-u0xdx=0.If u(x,t) satisfies (14) in the sense of the trace in addition, then we say it is a weak solution of the initial-boundary value problem of (9).Theorem 2.
Ifm(x)>0 is a C1(Ω¯) function, bi(s)∈C1(R), bi(0)=0, i=1,2,…,N, u0(x)≥0 satisfies (18), then (9) with initial value (13) has a nonnegative solution.Based on the usual Dirichlet boundary value condition, we have the following.Theorem 3.
Ifm(x)>0 is a C1(Ω¯) function, bi(s)∈C1(R), i=1,2,…,N,(22)∫Ωa-1xdx<∞,then the initial-boundary value problems (9), (13) and (14) have a uniqueness solution.In some cases, we can establish the stability of the weak solutions without any boundary value condition.Theorem 4.
Ifa(x) satisfies (17) and (22)(23)∇a=0,x∈∂Ω,u(x,t) is a solution of (9) with the initial value (13) but without the boundary value condition, u(x,t) satisfies(24)∫Ωax1+mx+1log⁡u2∇m2dx≤c,then u(x,t) is the unique solution.At last, we assume that(25)bis1-bis2≤cis1-s2,i=1,2,…,N.and probe the stability of weak solutions based on a partial boundary value condition.Theorem 5.
Letu,v be two solutions of (9) with the initial values u0(x),v0(x), respectively, and with a partial boundary value condition(26)ux,t=vx,t=0,x,t∈Σ1×0,T.It is supposed that, for every i∈1,2,…,N, either bi′(s)≥0 or bi′(s)≤0, a(x) satisfies (17) and(27)1λ∫Ω∖Ωλax∇a2dx1/2≤c,u and v satisfy(28)∫Ωax1+mx+1log⁡u2∇m2dx<∞,∫Ωax1+mx+1log⁡v2∇m2dx<∞.Then(29)∫Ωux,t-vx,tdx≤c∫Ωu0x-v0xdx.Here, if bi′(s)≥0,1≤i≤N, then(30)Σ1=x∈∂Ω:∑i=1Nciaxi<0.However, if bi′(s)≥0,1≥i≤N, then(31)Σ1=x∈∂Ω:∑i=1Nciaxi>0.To show that the partial boundary value condition (26) with the expression (30) or (31) is reasonable, let us review the equation (32)ut-div⁡ax∇u-∑i=1NbixDiu+cx,tu=fx,t.According to Fichera-Oleinik theory [26–29], the boundary value condition matching up with (32) is (33)ux,t=0,x,t∈Σ×0,T,with that (34)Σ=x∈∂Ω:bixnix<0,where n→=ni is the inner normal vector of Ω. Since (9) is nonlinear, Fichera-Oleinik theory is invalid; whether the partial boundary Σ1 in (26) can be expressed similar to (34) has become an interesting problem. Theorem 5 partially answers this question. One can see that if a(x)=d(x)=dist⁡x,∂Ω is the distance function from the boundary, axi=dxi=ni, the expression (30) or (31) is similar to (34). In fact, instead of (9), if we consider the equation (35)ut-div⁡dxumx∇u-∑i=1NbixDiu=0,by a similar method as the proof of Theorem 5, we can show that the partial boundary value condition matching up with (35) has the same expression as (34). Thus, the partial boundary value condition (26) with the expression (30) or (31) is reasonable.At the end of the Introduction section, we would like to suggest that ifm(x)=m is a constant, then condition (24) in Theorem 4 and condition (28) in Theorem 5 are naturally true. Actually, when m(x)=m is a constant, a(x)=dα(x); (9) has been studied by the author in [29]. But, one can see that, the results (Theorems 4 and 5) are much better and clearer than the results in [29].


## 2. The Proof of Theorem2

Proof of Theorem2.
We suppose thatu0∈C0∞(Ω) and 0≤u0≤M, and consider the following regularized problem:(36)unt=div⁡ax+1nunmx+1n∇u+∑i=1N∂biunmx+1∂xi,x,t,x,t∈QT,unx,t=1n,x,t∈∂Ω×0,T,unx,0=u0nx=u0x+1n,x∈Ω.
According to the standard parabolic equation theory, there is a weak solution(37)un∈L∞QT,ax+1n1/2unmx+1n1/2∇u∈L2QT,and(38)1n⩽unx,t≤u0L∞Ω+1n,x,t∈QT.
Moreover, by comparison theorem, we clearly have(39)un+1x,t⩽unx,t,which yields(40)ux,t=limn→∞unx,t,and(41)ux,t≤M+1.In what follows, we are able to prove that the limit function u is a weak solution of (9) with the initial value (13).
Multiplying both sides of the first equation in (36) by ϕ=unm(x)+1-1/nm(x)+1, and integrating it over QT, we have(42)∬QTuntunmx+1-1nmx+1dxdt=∬QTdiv⁡ax+1nunmx+1n∇uunmx+1-1nmx+1dxdt+∑i=1N∬QT∂biunmx+1∂xiunmx+1-1nmx+1dxdt.
Let us analyse every term in (42):(43)∬Qtuntunmx+1-1nmx+1dxdt=∫Ω1mx+2unmx+2x,t-unmx+2x,0dx-∫Ω1nmx+1unx,t-unx,0dx.(44)∬QTdiv⁡ax+1nunmx+1n∇ununmx+1-1nmx+1dxdt=-∬QTax+1nunmx+1n∇un·∇unmx+1-1nmx+1dxdt=-∬QTax+1nunmx+1n∇un·log⁡unumx+1∇mx+mx+1unmx∇un-1nmx+1log⁡n∇mxdxdt=-∬QTax+1nunmx+1nmx+1unmx∇un2dxdt-∬QTax+1nunmx+1n∇un∇mx·log⁡unumx+1-1nmx+1log⁡ndxdt≤-∬QTax+1nunmx+1nmx+1unmx∇un2dxdt+∬QTax+1nunmx+1n·12mx+1unmx∇un2+12mx+1unmx-1∇mx2·log⁡ununmx+1-1nmx+1log⁡ndxdt=-12∬QTax+1nunmx+1nmx+1unmx∇un2dxdt+12∬QTax+1nunmx+1nmx+1unmx-1∇mx2·log⁡ununmx+1-1nmx+1log⁡ndxdt≤-12∬QTax+1nunmx+1nmx+1unmx∇un2dxdt+c.Here, we have used (41) and the fact (45)limn→∞ax+1nunmx+1nmx+1unmx-1∇mx2log⁡ununmx+1-1nmx+1log⁡n=axmx+1-1∇mx2umx+1log⁡u<∞.In addition, by the fact(46)∬QT∂biunmx+1∂xiunmx+1dxdt=-∬QTbiunmx+1∂∂xiunmx+1dxdt=-∬QT∂∂xi∫1/nmx+1unmx+1bisds+bin-mx-11nmx+1log⁡nmxixdxdt=-∬QTbin-mx-11nmx+1log⁡nmxixdxdt,using the assumption that bi(0)=0, we have(47)∑i=1N∬QT∂biunmx+1∂xiunmx+1-1nmx+1dxdt=-∑i=1N∬QTbin-mx-11nmx+1log⁡nmxixdxdt-∑i=1N∬QTbiunmx+11nmx+1log⁡nmxixdxdt,accordingly(48)limn→∞∑i=1N∬QT∂biunmx+1∂xiunmx+1-1nmx+1dxdt=0,and so(49)∑i=1N∬QT∂biunmx+1∂xiunmx+1-1nmx+1dxdt≤c.Then by (41), (42), (43), (44), and (49), we have(50)∬QTax+1nunmx+1nunmx∇un2dxdt≤c.
There is aζ→∈L2(QT) and(51)ax+1nunmx+1nunmx1/2∇un⇀ζ→,weakly in L2(QT). We now prove that(52)ζ→=ax1/2umx∇u.
For any∀ψ∈C0∞(Ω), we have(53)∬QTax+1nunmx+1nunmx1/2∇unψdxdt=∬QT∇ax+1nunmx+1nunmx1/2unψdxdt-12∬QTψunax+1nunmx+1nunmx-1/2·∇ax+1nunmx+1nunmxdxdt,(54)limn→∞∬QT∇ax+1nunmx+1nunmx1/2unψdxdt=-limn→∞∬QTax+1nunmx+1nunmx1/2un∇ψdxdt=-∬QTaxumx+1∇ψdxdt.Denoting that An=a(x)+1/n|un|m(x)+1/nunm(x), then(55)∇An=∇ax+1nunmx+1nunmx=∇aunmx+1nunmx+ax+1n∇unmx+1nunmx=∇aunmx+1nunmx+ax+1n2unmx+1numxlog⁡un∇m+umx-1mx∇u=∇aunmx+1nunmx+ax+1n2unmx+1numxlog⁡un∇m+ax+1n2unmx+1numx-1mx∇u=I1+I2+I3.(56)-12∬QTψunax+1nunmx+1nunmx-1/2·∇ax+1nunmx+1nunmxdxdt=-12∬QTψunAn-1/2∇Andxdt=-12∬QTψunAn-1/2I1+I2+I3dxdt(57)-limn→∞12∬QTψunAn-1/2I3dxdt=limn→∞12∬QTψunAn1/2∇umx2unmx+1/nunmx+1/n=-∬QTmxψζ→dxdt.(58)-limn→∞12∬QTψunAn-1/2I2dxdt=-limn→∞12∬QTax+1nunmx+1nunmx-1/2·ψunax+1n2unmx+1numxlog⁡un∇mdxdt=-∬QTψax1/2umxlog⁡u∇mdxdt.(59)-limn→∞12∬QTψunAn-1/2I1dxdt=-limn→∞12∬QTax+1nunmx+1nunmx-1/2ψun∇aunmx+1nunmxdxdt=-12∬QTψax-1/2umx+1∇adxdt.
Letn→∞ in (53). We obtain that(60)∬QTmx+1ζψdxdt=-∬QTψax1/2umxlog⁡u∇mψdxdt-12∬QTψax-1/2umx+1∇aψdxdt=-∬QTu∇ax1/2umxψdxdt+∬QTax1/2mxumx∇uψdxdt,which implies(61)∬QTζψdxdt=∬QTax1/2umx∇uψdxdt,
Sincebi∈C1(R), by (40), we have(62)limn→∞biunmx+1=biumx+1.Letting n→∞ in (42), by (51), (52), and (62), we know u(x,t) satisfies (20). At the same time, the initial value (13) can be proved in a similar way as that when m(x)=m-1 is a constant; one can refer to [5] for the details. Thus, u is a solution of (9) with the initial value (13). If u0 only satisfies (18), by considering the problem of (36) with the initial value u0ε which is the mollified function of u0, then we can get the conclusion by a process of limitation. Certainly, the solution u(x,t) generally is not continuous at t=0, but satisfies (19) and (20). Theorem 2 is proved.


## 3. The Stability Based on the Dirichlet Boundary Value Condition

Lemma 6.
If∫Ωa(x)-1dx⩽c, then ∫Ω|∇um(x)+1|dx⩽c.Proof.
Sinceaxumx∇uL2(QT)≤c,(63)∬QTaxu2mx∇u2dxdt≤c,which yields(64)∬QTax∇umx+12dxdt≤2∬QTaxumx+1log⁡u∇m2dxdt+2∬QTaxmx+1umx∇u2dxdt≤c.
Then(65)∫Ω∇umx+1dx=∫x∈Ω:a1/2∇umx+1⩽1∇umx+1dx+∫x∈Ω:a1/2∇umx+1>1∇umx+1dx⩽∫Ωax-1/2dx+∫Ωax∇umx+12dx⩽c.Thus um(x)+1 can be defined by the trace on the boundary in the traditional way. By the definition of the trace, we also know that u can be defined by the trace on the boundary in the traditional way. The lemma is proved.For every fixedt∈[0,T), we define the Banach space Vt(Ω) by(66)VtΩ=ux,t:ux,t∈L2Ω⋂W01,1Ω,∇ux,t2∈L1Ω,uVtΩ=u2,Ω+∇u2,Ω,and denote by Vt′(Ω) its dual space. In addition, we denote the Banach space W(QT) by(67)WQT=u:0,T→VtΩu∈L2QT,∇u2∈L1QT,u=0onΓ=∂Ω,uWQT=∇u2,QT+u2,QT,and denote by W′(QT) its dual space. According to [18], we know that(68)w∈W′QT⇔w=w0+∑i=1NDiwi,w0∈L2QT,wi∈L2QT,∀ϕ∈WQT,≪w,ϕ≫=∬QTw0ϕ+∑iNwiDiϕdxdt.The norm inW′(QT) is defined by (69)vW′QT=sup⁡≪v,ϕ≫∣ϕ∈WQT,ϕWQT≤1.Lemma 7.
Ifu(x,t) is a weak solution of (9) with the initial value (13), then ut∈W′(QT).Proof.
For anyv∈W(QT) and vW(QT)=1, there holds (70)ut,v=-∬QTaxumx∇u∇vdxdt-∑i=1N∬QTvxibiumx+1dxdt.By Young’s inequality, it follows from (70) that (71)uεt,v≤c∬QTaxu2mx∇u2dxdt+∬QTv2+∇v2dxdt+1≤c,we have (72)utW′QT≤c.Lemma 8.
Suppose thatu∈W(QT) and ut∈W′(QT). For any continuous function h(s), let H(s)=∫0sh(s)ds. For a.e. t1,t2∈(0,T), there holds(73)∫t1t2∫Ωhuutdxdt=∫ΩHux,t2-Hux,t1dx.This lemma can be found in [18].Theorem 9.
Let∫Ωa-1(x)dx<∞ and let u and v be two solutions of (9) with the initial values u0(x),v0(x), respectively, and with the same homogeneous boundary value conditions(74)ux,t=vx,t=0,x,t∈∂Ω×0,T.Then(75)∫Ωux,t-vx,t≤c∫Ωu0x-v0xdx.Proof.
For any given positive integern, let gn(s)=∫0shn(τ)dτ, hn(s)=2n1-n|s|+. Then hn(s)∈C(R), and we have(76)hns≥0,shns≤1,gns≤1,and(77)limn→∞gns=sgn⁡s,limn→∞sgn′s=0.
By the definition of weak solutions, we have(78)∬QTutφx,tdxdt+∬QTaxmx+1∇umx+1∇φdxdt-∬QTaxmx+1umx+1log⁡u∇m∇φdxdt+∑i=1N∬QTbiumx+1φxidxdt=0.
Since∫Ωa(x)-1dx⩽c, u=v=0 on the boundary, we can choose gn(um(x)+1-vm(x)+1) as the test function. Then(79)∫QTgnumx+1-vmx+1∂u-v∂tdxdt+∬QTaxmx+1∇umx+1-∇vmx+12gn′umx+1-vmx+1dxdt-∬QTaxmx+1umx+1log⁡u-vmx+1log⁡u·gn′umx+1-vmx+1∇m∇umx+1-vmx+1dxdt+∑i=1N∬QTbiumx+1-bivmx+1umx+1-vmx+1xign′umx+1-vmx+1dxdt=0.(80)∬QTaxmx+1∇umx+1-∇vmx+12gn′umx+1-vmx+1dxdt≥0.and(81)-∬QTaxmx+1umx+1log⁡u-vmx+1log⁡u·gn′umx+1-vmx+1∇m∇umx+1-vmx+1dxdt≥-12∬QTaxmx+1∇umx+1-∇vmx+12gn′umx+1-vmx+1dxdt-12∬QTaxmx+1umx+1log⁡u-vmx+1log⁡ugn′umx+1-vmx+12dxdt.For simplism, in what follows, we denote that(82)Dn=Ω:umx+1-vmx+1<1n,D0=x∈Ω:u-v=0and, clearly,(83)limn→∞Dn=D0.We have(84)∫Ωbiumx+1-bivmx+1umx+1-vmx+1xign′umx+1-vmx+1dx=∫Dnbiumx+1-bivmx+1·gn′umx+1-vmx+1umx+1-vmx+1xidx⩽c∫Dnbiumx+1-bivmx+1umx+1-vmx+1umx+1-vmx+1xidx=c∫Dna-1/2biumx+1-bivmx+1umx+1-vmx+1axumx+1-vmx+1xidx⩽c∫Dna-1/2biumx+1-bivmx+1umx+1-vmx+12dx1/2·∫Dnax∇umx+1-vmx+12dx1/2.Since ∫Ωa-1(x)dx<∞,(85)∫Dna-1/2biumx+1-bivmx+1umx+1-vmx+12dx⩽c∫Dnax-1dx⩽c.
IfD0=x∈Ω:umx+1-vmx+1=0 is with 0 measure, we have(86)limn→∞∫Dnax-1dx=∫D0ax-1dx=0.If the set D0=x∈Ω:umx+1-vmx+1=0 has a positive measure, then,(87)limn→∞∫Dnax∇umx+1-vmx+12dx=∫D0ax∇umx+1-vmx+12dx=0.Therefore, in both cases,(88)limn→∞∫Ωbiumx+1-bivmx+1gn′umx+1-vmx+1umx+1-vmx+1xidx=0.
At last,(89)limn→∞∫Ωgnumx+1-vmx+1∂u-v∂tdx=∫Ωsgn⁡umx+1-vmx+1∂u-v∂tdx=∫Ωsgn⁡u-v∂u-v∂t=ddtu-v1.Let n→∞ in (79). By (80), (81), (88), and (89), we have(90)∫Ωux,t-vx,tdx⩽∫Ωu0x-v0xdx=0.Corollary 10.
Theorem3 is true.


## 4. The Stability without the Boundary Value Condition

In this section, we will prove Theorem4.Theorem 11.
Letu,v be two nonnegative solutions of (9) with the initial values u0(x),v0(x), respectively. If a(x) satisfies (17) and(91)∫Ωax-1∇a2dx≤c,∇a=0,x∈∂Ω,u and v satisfy(92)∫Ωax1+mx+1log⁡u2∇m2dx≤c,∫Ωax1+mx+1log⁡v2∇m2dx≤c,then(93)∫Ωux,t-vx,tdx≤c∫Ωu0x-v0xdx.Proof.
For all0≤φ∈C01(QT), by the definition of weak solutions, for all 0≤φ∈C01(QT), we have(94)∬QTutφx,tdxdt+∬QTaxmx+1∇umx+1∇φdxdt-∬QTaxmx+1umx+1log⁡u∇m∇φdxdt+∑i=1N∬QTbiumx+1φxidxdt=0.
Letχτ,s(t) be the characteristic function of [τ,s]⊂(0,T). By a process of limit, we can choose(95)χτ,stgnarumx+1-vmx+1as the test function; then(96)∫τs∫Ωgnarumx+1-vmx+1∂u-v∂tdxdt+∫τs∫Ωaxr+1mx+1∇umx+1-∇vmx+1·∇umx+1-vmx+1gn′arumx+1-vmx+1dxdt+r∫τs∫Ωaxrmx+1∇umx+1-∇v(mx+1·∇aumx+1-vmx+1gn′arumx+1-vmx+1dxdt-∫τs∫Ωaxr+1mx+1umx+1log⁡u-vmx+1log⁡v∇m·∇umx+1-vmx+1·gn′aumx+1-vmx+1dxdt-r∫τs∫Ωaxrmx+1umx+1log⁡u-vmx+1log⁡v∇m·∇aumx+1-vmx+1·gn′arumx+1-vmx+1dxdt+∑i=1N∫τs∫Ωbiumx+1-bivmx+1rar-1axixumx+1-vmx+1·gn′arumx+1-vmx+1+arumx+1-vmx+1xign′arumx+1-vmx+1dxdt=0.Let us analyse every term in (96). Firstly, we have(97)∫Ωaxr+1mx+1∇umx+1-∇vmx+1·∇umx+1-vmx+1gn′arumx+1-vmx+1dx≥0.Secondly, since ∫Ωa(x)-1|∇a|2dx≤c,(98)∫Ωaxrmx+1∇umx+1-∇v(mx+1·∇aumx+1-vmx+1gn′arumx+1-vmx+1dx≤∫Ωaxrmx+1∇umx+1+∇v(mx+1)∇aumx+1-vmx+1·gn′arumx+1-vmx+1dx≤c∫Ωax∇umx+12+∇vmx+12dx1/2·∫Ωax∇aaarumx+1-vmx+1gn′arumx+1-vmx+12dx1/2→0,as λ→0.
By that(99)limn→0⁡∫Dnaxumx+1log⁡umx+1-vmx+1log⁡vmx+12gn′umx+1-vmx+12∇m2dx≤limn→0⁡∫Dnaxumx+1log⁡umx+1-vmx+1log⁡vmx+12umx+1-vmx+12∇m2dx≤∫D0ax1+mx+1log⁡ζ2∇m2dx<∞.where ζ∈(u,v) in the mean value, we have(100)∫Dnaxumx+1log⁡umx+1-vmx+1log⁡vmx+12gn′umx+1-vmx+12∇m2dx≤∫Dnaxumx+1log⁡umx+1-vmx+1log⁡vmx+12umx+1-vmx+12∇m2·umx+1-vmx+12gn′umx+1-vmx+12dx<∞.Using the Lebesgue dominated convergence theorem, we have(101)limn→∞∫Dnaxumx+1log⁡umx+1-vmx+1log⁡vmx+12gn′umx+1-vmx+12∇m2dx=0,and so(102)∫Ωaxmx+1umx+1log⁡u-vmx+1log⁡v∇m·ar∇umx+1-vmx+1gn′arumx+1-vmx+1dxdt≤∫Dnax∇umx+1-∇vmx+12dx1/2·c∫Dnaxumx+1log⁡umx+1-vmx+1log⁡vmx+12gn′umx+1-vmx+12∇m2dx1/2→0,as λ→0.
Thirdly, since|∇a|x∈∂Ω=0,(103)limλ→0limn→∞∫Ωbiumx+1-bivmx+1ϕλxixgnumx+1-vmx+1dx≤limλ→0∫Ωbiumx+1-bivmx+1ϕλxixdx≤climλ→01λ∫Ω∖Ωλ∇adx=∫∂Ω∇adΣ=0.Moreover, as in the proof of Theorem 3, we can show that(104)limλ→0limn→∞∫τs∫Ωbiumx+1-bivmx+1ϕλumx+1-vmx+1xign′umx+1-vmx+1dxdt=0,and, clearly,(105)limn→∞∫τs∫Ωgnarumx+1-vmx+1ϕλ∂u-v∂tdx=∫Ωux,s-vx,sdx-∫Ωux,τ-vx,τdx.At last, let n→∞ in (96). Then(106)∫Ωux,s-vx,sdx≤c∫Ωux,τ-vx,τdx.By the arbitraries of τ, we have(107)∫Ωux,t-vx,tdx⩽∫Ωu0x-v0xdx.


## 5. The Stability Based on the Partial Boundary Value Condition

In this section, we will prove Theorem5. We assume that(108)bis1-bis2≤cis1-s2,i=1,2,…,N.we denote that Ωλ=x∈Ω:ax>λ as before and denote that(109)Ωλi1=x∈Ω∖Ωλ:axi<0,Σ1i=limλ→0Ωλi1,(110)Ωλi2=x∈Ω∖Ωλ:axi≥0,Σ2i=limλ→0Ωλi2.Theorem 12.
Letu,v be two solutions of (9) with the initial values u0(x),v0(x), respectively, and with a partial boundary value condition(111)ux,t=vx,t=0,x,t∈Σ1×0,T.Here, for any given i∈1,2,…,N, if bi′(s)≥0,(112)Σ1=x∈∂Ω:x∈Σ1i;if bi′(s)≥0,(113)Σ1=x∈∂Ω:x∈Σ2i.It is supposed that a(x) satisfies (17) and(114)1λ∫Ω∖Ωλax∇a2dx1/2≤c,u and v satisfy(115)∫Ωax1+mx+1log⁡u2∇m2dx<∞,∫Ωax1+mx+1log⁡v2∇m2dx<∞.Then(116)∫Ωux,t-vx,tdx≤c∫Ωu0x-v0xdx,a.e.∈0,T.Proof.
For all0≤φ∈C01(QT), by the definition of weak solutions, for all 0≤φ∈C01(QT), we have(117)∬QTutφx,tdxdt+∬QTaxmx+1∇umx+1∇φdxdt-∬QTaxmx+1umx+1log⁡u∇m∇φdxdt+∑i=1N∬QTbiumx+1φxidxdt=0.
For a small positive constantλ>0, let(118)Ωλ=x∈Ω:ax>λ,and(119)ϕλx=1,ifx∈Ωλ,axλ,x∈Ω∖Ωλ.Let χτ,s(t) be the characteristic function of [τ,s]⊂(0,T). By a process of limit, we can choose (120)χτ,stϕλxgnumx+1-vmx+1as the test function; then(121)∫τs∫Ωϕλxgnumx+1-vmx+1∂u-v∂tdxdt+∫τs∫Ωaxmx+1∇umx+1-∇vmx+1·ϕλx∇umx+1-vmx+1·gn′umx+1-vmx+1dxdt+∫τs∫Ωaxmx+1∇umx+1-∇v(mx+1·∇ϕλxgnumx+1-vmx+1dxdt-∫τs∫Ωaxmx+1umx+1log⁡u-vmx+1log⁡v∇m·ϕλx∇umx+1-vmx+1·gn′umx+1-vmx+1dxdt-∫τs∫Ωaxmx+1umx+1log⁡u-vmx+1log⁡v∇m·∇ϕλxgnumx+1-vmx+1dxdt+∑i=1N∫τs∫Ωbiumx+1-bivmx+1ϕλxixgnumx+1-vmx+1+ϕλumx+1-vmx+1xign′umx+1-vmx+1dxdt=0.Let us analyse every term in (121):(122)∫Ωaxmx+1∇umx+1-∇vmx+1·ϕλ∇umx+1-vmx+1gn′umx+1-vmx+1dx≥0.By (114), 1/λ∫Ω∖Ωλa(x)|∇a|2dx1/2≤c,(123)∫Ωaxmx+1∇umx+1-∇v(mx+1·∇ϕλgnumx+1-vmx+1dx≤∫Ω∖Ωλaxmx+1∇umx+1-∇v(mx+1·∇ϕλgn(umx+1-vmx+1dx≤∫Ω∖Ωλaxmx+1∇umx+1+∇v(mx+1)∇ϕλdx≤1λ∫Ω∖Ωλaxmx+1∇umx+1+∇v(mx+1)∇adx≤c∫Ω∖Ωλax∇umx+12+∇vmx+12dx1/2·1λ∫Ω∖Ωλax∇a2dx1/2→0,as λ→0.(124)∫Ωaxmx+1umx+1log⁡u-vmx+1log⁡v∇m·ϕλx∇umx+1-vmx+1gn′umx+1-vmx+1dxdt≤∫Dnax∇umx+1-∇vmx+12dx1/2·c∫Dnaxumx+1log⁡umx+1-vmx+1log⁡vmx+12gn′umx+1-vmx+12∇m2dx1/2→0,as λ→0.
This is due to the fact that ifD0=x∈Ω:u-v=0 is with 0 measure,(125)limn→0∫Dnax∇umx+1-∇vmx+12dx≤∫D0ax∇umx+12+∇vmx+12dx=0,and by (115)(126)∫Ωax1+mx+1log⁡u2∇m2dx<∞,we have(127)limn→0⁡∫Dnaxumx+1log⁡umx+1-vmx+1log⁡vmx+12gn′umx+1-vmx+12∇m2dx≤limn→0⁡∫Dnaxumx+1log⁡umx+1-vmx+1log⁡vmx+12umx+1-vmx+12∇m2dx≤∫D0ax1+mx+1log⁡ζ2∇m2dx<∞.
whereζ∈(u,v) in the mean value. If D0=x∈Ω:u-v=0 has a positive measure,(128)limn→0∫Dnax∇umx+1-∇vmx+12dx=∫Ωax∇umx+12+∇vmx+12dx<∞,and by (115)(129)limn→0∫Ω:umx+1-vmx+1<1/naxumx+1log⁡umx+1-vmx+1log⁡vmx+12gn′umx+1-vmx+12∇m2dx≤limn→0∫Ω:u-v=0axumx+1log⁡umx+1-vmx+1log⁡vmx+12umx+1-vmx+12∇m2dx=0.Moreover, recall that(130)Ωλi1=x∈Ω∖Ωλ:axi<0,Σ1i=limλ→0Ωλi1,Ωλi2=x∈Ω∖Ωλ:axi≥0,Σ2i=limλ→0Ωλi2.
Ifbi′(s)≥0, by the partial boundary value condition (111)(131)-limλ→0limn→∞∫Ωbiumx+1-bivmx+1ϕλxixgnumx+1-vmx+1dx=-limλ→0∫Ωbiumx+1-bivmx+1ϕλxixsign⁡umx+1-vmx+1dx=-limλ→0∫Ωbiumx+1-bivmx+1ϕλxixsign⁡biumx+1-bivmx+1dx=-limλ→0∫Ω∖Ωλbiumx+1-bivmx+1axiλdx=-limλ→0∫Ω∖Ωλumx+1-vmx+1bi′ξaxiλdx≤limλ→01λ∫Ωλi1-ciaxiumx+1-vmx+1dx=∫Σ1i-ciaxiumx+1-vmx+1dΣ=0.If bi′(s)<0, by the partial boundary value condition (111)(132)-limλ→0limn→∞∫Ωbiumx+1-bivmx+1ϕλxixgnumx+1-vmx+1dx=-limλ→0∫Ωbiumx+1-bivmx+1ϕλxixsign⁡umx+1-vmx+1dx=-limλ→0∫Ωbiumx+1-bivmx+1ϕλxixsign⁡biumx+1-bivmx+1dx=-limλ→0∫Ω∖Ωλbiumx+1-bivmx+1axiλdx=-limλ→0∫Ω∖Ωλumx+1-vmx+1bi′ξaxiλdx≤limλ→01λ∫Ωλi2-ciaxiumx+1-vmx+1dx=∫Σ2i-ciaxiumx+1-vmx+1dΣ=0.
As in the proof of Theorem11 we can show that(133)limλ→0limn→∞∫τs∫Ωbiumx+1-bivmx+1ϕλumx+1-vmx+1xign′umx+1-vmx+1dxdt=0.
Clearly,(134)limn→∞limλ→0∫τs∫Ωgnumx+1-vmx+1ϕλ∂u-v∂tdx=∫Ωux,s-vx,sdx-∫Ωux,τ-vx,τdx.Now, after letting λ→0, let n→∞ in (121). Then(135)∫Ωux,s-vx,sdx≤c∫Ωux,τ-vx,τdx.By the arbitraries of τ, we have(136)∫Ωux,t-vx,tdx⩽∫Ωu0x-v0xdx.Proof of Theorem5.
Since we suppose that, for everyi∈1,2,…,N, either bi′(s)≥0 or bi′(s)≤0, by checking the process of the proof of (131) or (132), we can easily obtain the conclusion of Theorem 5.


## 6. Conclusion

The evolutionary equations with variable exponents, especially the so-called electrorheological fluids equations with the form (15), have been brought to the forefront by many scholars since the beginning of this century. There are more or less beyond one’s imagination; there are only a few references devoted to the porous medium with variable exponents as (16). So, this paper fills the gaps in the related fields. Moreover, the equation considered in this paper is more general than (16). The most important characteristic lies in that there is a degenerate diffusion coefficient a(x) in the equation. This characteristic may make the usual Dirichlet boundary value condition overdetermined and so a partial boundary value condition is expected. The conclusions in this paper answer the problem partially. In addition, since the equation is with variable exponents, there are many technique difficulties to be overcome. This makes our paper contain many cumbersome calculations, but it is necessary.



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