# On a Partial Boundary Value Condition of a Porous Medium Equation with Exponent Variable

Authors: Huashui Zhan
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

The initial-boundary value problem of a porous medium equation with a variable exponent is considered. Both the diffusion coefficientax,t and the variable exponent px,t depend on the time variable t, and this makes the partial boundary value condition not be expressed as the usual Dirichlet boundary value condition. In other words, the partial boundary value condition matching up with the equation is based on a submanifold of ∂Ω×0,T. By this innovation, the stability of weak solutions is proved.


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## Body

## 1. Introduction

The porous medium equation with a constant exponent is widely used to model several real-life problems, and it has been extensively studied, one can refer to the survey books [1–6] and the references therein. The dynamics system of a partially nonlocal and inhomogeneous nonlinear medium has been considered in [7–9]. The case where the exponent of nonlinearity is not constant was proposed by Antontsev and Shmarev in [10], where the existence, uniqueness, and some properties of the solution in a bounded fixed domain were researched. By using the Galerkin finite element method, Duque et al. [11] proved the convergence of a fully discrete solution for this problem in a fixed domain. Based on one of the properties proved in [11] that the solution is with the finite speed of propagation, Duque et al. [12] considered the free boundary problem by using the moving mesh method to the porous medium. However, the moving mesh method was first introduced by Huang and Russell in [13].In this paper, we consider the initial-boundary value problem of a generalized porous medium equation with a variable exponent:(1)ut=divax,tumx,t∇u+∑i=1N∂biumx,t+1∂xi,x,t∈QT=Ω×0,T,where mx,t>0 is a C1QT¯ function, ax,t≥0 is a C1QT¯ function, bis∈C1ℝ, and Ω⊂ℝN is a bounded domain with a smooth boundary ∂Ω.Equation (1) is a special case of the reaction-diffusion equation:(2)ut=divau,x,t∇u+divb⟶u,x,t.Becausea⋅,x,t may degenerate on the boundary, how to impose a suitable boundary value condition to study the well posedness of weak solutions to equation (2) has attracted extensive attention and has been widely studied for a long time. In some details, though the initial value(3)ux,0=u0x,x∈Ω,is always imposed, the Dirichlet boundary condition(4)ux,t=0,x,t∈∂Ω×0,T,may not be imposed or be imposed in a weaker sense than the traditional trace. One can refer to the references [14–19] for the details.Naturally, besides the porous medium equation with variable exponents, the so-called electrorheological fluid equations with the form(5)ut=div∇upx−2∇u+fx,t,u,∇u,have been brought to the forefront by many more scholars. Since the beginning of this century, there are a great deal of papers devoting to the well-posedness problem, the intrinsic Harnack inequalities, the long-time behavior, and the Hölder regularity of weak solutions, one can refer to the literatures [20–31] and the references therein.Ifmx,t=mx, then ax,t=ax satisfies(6)ax>0,x∈Ω,ax=0,x∈∂Ω,(7)0≤u0∈L∞Ω,ax∇u0mx∈L2Ω.The existence and the stability of weak solutions to equation(8)ut=divaxumx∇u+∑i=1N∂biumx+1∂xi,x,t∈Ω×0,T,has been studied in [32]. It is found that the degeneracy of ax on boundary (6) may replace the usual boundary value condition (4). In other words, if ax satisfies (6), the stability of weak solutions may be proved without the usual boundary value condition (4).In this paper, we will use some ideas of [32] to study the well posedness of weak solutions to equation (1). Because both ax,t and px,t are dependent on the time variable t, the problem becomes more difficult and the question of existence of such solutions is still open for equation (8), as well as for evolution p-Laplace equation with the exponent p depending on t [22]. Instead of condition (6), we only assume that ax,t≥0. By this assumption, we find out a partial boundary value condition matching up with the equation. Moreover, because both ax,t and px,t are dependent on the time variable t, the partial boundary value condition cannot be expressed as the usual Dirichlet boundary value condition. In other words, the partial boundary value condition is based on a submanifold of ∂Ω×0,T. By this innovation, the stability of weak solutions is proved.


## 2. The Partial Boundary Value Condition and the Main Results

For any givent∈0,T and small enough λ>0, we set(9)Ωλt=x∈Ω:ax,t>λ,Ωλi1t=x∈Ω\Ωλt:axi<0,Σ1it=limλ⟶0Ωλi1t,i=1,2,…,N,(10)Ωλi2t=x∈Ω\Ωλt:axi≥0,Σ2it=limλ⟶0Ωλi2t,i=1,2,…,N.The most important and essential improvement is that instead of the usual boundary value condition (4), the stability of weak solutions is proved based on a partial boundary value condition:(11)ux,t=0,x,t∈Σ1∪Σ2,where(12)Σ2=x∈∂Ω:ax,t≠0,and for any given i∈1,2,…,N, if bi′s≥0,(13)Σ1=∪i=1NΣ1it×0,T,and if bi′s≤0, (14)Σ1=∪i=1NΣ2it×0,T.Thus, if for everyi∈1,2,…,N, either bi′s≥0 or bi′s≤0, then one can deduce an expression Σ1 from the above discussion. For example, bis≥0 when 1≤i≤k and bi′s≤0 when k+1≤i≤N; then,(15)Σ1=∪i=1kΣ1it×0,T∪∪i=k+1NΣ2it×0,T.The most characteristic out of the ordinary is thatΣ1 or Σ2 is just a submanifold of ∂Ω×0,T, and it cannot be expressed as a cylinder with the form Γ×0,T and Γ⊂∂Ω.Definition 1.
Ifux,t≥0 and satisfies(16)u∈L∞QT,ax,0umx,0∇u∈L∞0,T;L2Ω,and for ∀φ∈C01QT,(17)∬QT−∂φ∂tu+ax,tumx,t∇u∇φdxdt+∑i=1N∬QTbiumx,t+1φxix,tdxdt=0, and then ux,t is said to be a weak solution of equation (1) with the initial value (3) in the sense(18)limt⟶0∫Ωux,t−u0xdx=0.
Moreover, ifux,t satisfies (4) or (3) in the sense of the trace in addition, then it is said to be a weak solution of the initial-boundary value problem of equation (1).Theorem 1.
Ifmx,t>0 is a C1QT¯ function, bis satisfies(19)bis1−bis2≤cs1−s2,i=1,2,…,N.Ifu0x≥0 satisfies (16), then equation (1) with initial value (3) has a nonnegative solution.Theorem 2.
Letmx,t>0 be a C1QT¯ function, then bis∈C1ℝ satisfy (19):(20)∫Ωa−1x,tdx≤cT.Then, the initial-boundary value problem (1), (3), and (4) has a uniqueness solution.Theorem 3.
Letux,tandvx,t be two solutions of equation (1) with the initial values u0xandv0x, respectively, and with a partial boundary value condition(21)ux,t=vx,t=0,x,t∈Σ1∪Σ2.It is supposed that, for everyi∈1,2,…,N, either bi′s≥0 or bi′s≤0, ax,t satisfies(22)1λ∫Ω\Ωλtax,t∇a2dx1/2≤cT,and ux,t and vx,t satisfy(23)∫Ωax,t1+mx,t+1logu2∇m2dx≤cT,∫Ωax,t1+mx,t+1logv2∇m2dx≤cT.Then,(24)∫Ωux,t−vx,tdx≤cT∫Ωu0x−v0xdx.Hereinafter, the constantcT represents a constant which depends on T.At the end of this section, we would like to suggest that ifmx,t=m is a constant, then condition (22) in Theorem 3 is naturally true.


## 3. The Proof of Theorem1

First, we suppose thatu0∈C0∞Ω and 0≤u0≤M and consider the following regularized problem:(25)unt=divax,t+1nunmx,t+1n∇u+∑i=1N∂biunmx,t+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 solutionun∈L∞QT satisfying(26)ax,t+1n1/2unmx,t+1n1/2∇u∈L2QT,1n≤unx,t≤u0L∞Ω+1n,x,t∈QT.Moreover, by comparison theorem, we clearly have(27)un+1x,t≤unx,t,which yields(28)ux,t=limn⟶∞unx,t,(29)ux,t≤M+1.In what follows, we are able to prove that the limit functionu is a weak solution of (1) with the initial value (3).Multiplying both sides of the first equation in (25) by ϕ=mx,t+2unmx,t+1−1/nmx,t+1 and integrating it over QT, we have(30)∬QTuntunmx+1−1nmx,t+1mx,t+2dxdt=∬QTdivax,t+1nunmx,t+1n∇uunmx,t+1−1nmx,t+1mx,t+2dxdt+∑i=1N∬QT∂biunmx,t+1∂xiunmx,t+1−1nmx+1mx,t+2dxdt.For the left-hand side of (29),(31)∬Qtuntunmx,t+1−1nmx,t+1mx,t+2dxdt=∫Ωunmx,t+2x,t−unmx+2x,0dx−∫Ω1nmx,t+1unx,t−unx,0mx,t+2dx+∬QTun∂∂t1nmx,t+1mx,t+2dxdt.For the first term of the right-hand side of (29), because(32)limn⟶∞ax,t+1nunmx,t+1nmx,t+1unmx,t−1∇m2⋅logununmx,t+1−1nmx,t+1logn=ax,tmx,t+1−1∇mx,t2umx,t+1logu<∞,∬QTax,t+1nunmx,t+1nunmx,t+1−1nmx,t+1∇m2dxdt<∞,by a complicated calculation and using the Young inequality, we can deduce that(33)∬QTdivax,t+1nunmx,t+1n∇ununmx,t+1−1nmx,t+1mx,t+2dxdt≤−c∬QTax,t+1nunmx,t+1nmx,t+1unmx,t∇un2dxdt+c.For the second term of the right-hand side of (29), because(34)∬QT∂biunmx,t+1∂xiunmx,t+1mx,t+2dxdt=−∬QTbiunmx,t+1∂∂xiunmx+1mx,t+2dxdt=−∬QT∂∂xi∫​1/nmx,t+1unmx,t+1bisds+bin−mx,t−11nmx,t+1lognmxix,tmx,t+2dxdt−∬QTbiunmx,t+1unmx,t+1mxidxdt=−∬QTbin−mx,t−11nmx,t+1lognmxix,tdxdt−∬QTbiunmx,t+1unmx,t+1mxidxdt,we can deduce that(35)∑i=1N∬QT∂biunmx,t+1∂xiunmx,t+1−1nmx,t+1dxdt≤c.From (28)–(34), we extrapolate(36)∬QTax,t+1nunmx,t+1nunmx,t∇un2dxdt≤c.Accordingly, there isζ⟶∈L2QT such that(37)ax,t+1nunmx,t+1nunmx,t1/2∇un⇀ζ⟶,weakly in L2QT. We now can prove(38)ζ⟶=ax,t1/2umx,t∇u,as in a similar way as that in [32].The last but not the least, by thatbi∈C1ℝ, using (27), we have(39)limn⟶∞biunmx+1=biumx+1.Lettingn⟶∞ in (30), by (37), (38), and (39), we know ux,t satisfies (18).Secondly, ifu0 satisfies only (16), we should consider equation (9) with the initial value u0ε which is the mollified function of u0, from the above that there is a weak solution uε satisfying (18). Letting ε⟶0, the limit function ux,t is a solution of (1) satisfying (17) and (18), but generally is not continuous at t=0 as in the case u0∈C0∞Ω.Thirdly, the initial value (4) can be proved in a similar way as that when mx,t=m−1 is a constant, one can refer to [5] for the details. Thus, u is a solution of equation (1) with the initial value (4). Thus, Theorem 1 is proved.


## 4. The Proof of Theorem2

One can see that, in Definition1, there is not any definition on the general derivative ut. At the beginning of this section, we first answer this question.For anyt∈0,T, the Banach space VtΩ is defined by(40)VtΩ=ux,t:ux,t∈L2Ω∩W01,1Ω,∇ux,t2∈L1Ω,uVtΩ=u2,Ω+∇u2,Ω,and Vt′Ω is denoted as its dual space. The Banach space WQT is defined by(41)WQT=u:0,T⟶VtΩu∈L2QT,∇u2∈L1QT,u=0onΓ=∂Ω,uWQT=∇u2,QT+u2,QT,and W′QT is denoted as its dual space. From [21], we have (42)w∈W′QT⇔w=w0+∑i=1NDiwi,w0∈L2QT,wi∈L2QT,∀ϕ∈WQT,≪w,ϕ≫=∬QTw0ϕ+∑iNwiDiϕdxdt.It is easy to prove the following lemmas, so we omit the details here.Lemma 1.
Ifux,t is a weak solution of equation (1) with the initial value (3), then ut∈W′QT.Lemma 2.
Suppose thatu∈WQT and ut∈W′QT. For any continuous function hs, let Hs=∫0shsds. For a.e.t1,t2∈0,T, there holds(43)∫t1t2∫Ωhuutdxdt=∫ΩHux,t2−Hux,t1dx.Lemma 3.
If∫Ωax,t−1dx≤cT, then ∫Ω∇umx,t+1dx≤cT. So, the weak solution of equation (1) ux,t can be defined as the homogeneous boundary value condition u|∂Ω=0 in the sense of the trace.Theorem 4.
Letux,t and vx,t be two solutions of equation (1) with the initial values u0xandv0x respectively, and with(44)ux,t=vx,t=0,x,t∈∂Ω×0,T.If∫Ωax,t−1xdx≤cT, then(45)∫Ωux,t−vx,t≤c∫Ωu0x−v0xdx.Proof.
For any given positive integern, let gns=∫0shnτdτ and hns=2n1−ns+. Then, hns∈Cℝ, and we have(46)limn⟶∞gns=sgns,limn⟶∞sgn′s=0.
Becauseux,t=vx,t=0 on the boundary ∂Ω×0,T, we choose gnumx,t+1−vmx,t+1 as the test function and integrate over Qt=Ω×0,t. Then,(47)∫0t∫Ωgnumx,s+1−vmx,s+1∂u−v∂tdxds+∬Qtax,smx,s+1∇umx,s+1−∇vmx,s+12hnumx,s+1−vmx,s+1dxds−∬Qtax,smx,s+1umx,s+1logu−vmx,s+1logu⋅hnumx,s+1−vmx,s+1∇m∇umx,s+1−vmx,s+1dxds+∑i=1N∬Qtbiumx,s+1−bivmx,s+1umx,s+1−vmx,s+1xi⋅hnumx,s+1−vmx,s+1dxds=0.
Let us analyse every term in (47). In the first place,(48)∬Qtax,smx,s+1∇umx,s+1−∇vmx,s+12hnumx,s+1−vmx,s+1dxdt≥0,(49)−∬Qtax,smx,s+1umx,s+1logu−vmx,s+1logu⋅hnumx,s+1−vmx,s+1∇m∇umx,s+1−vmx,s+1dxds≥−12∬Qtax,smx,s+1∇umx,s+1−∇vmx,s+12hnumx,s+1−vmx,s+1dxds−12∬Qtax,smx,s+1umx,s+1logu−vmx,s+1loguhnumx,s+1−vmx,s+12dxds.
In the second place, we deal with the fourth term on the left-hand side of (47). For any given t∈0,T, we set(50)Dnt=Ω:umx,t+1−vmx,t+1<1n,D0t=x∈Ω:u−v=0.
Clearly,(51)limn⟶∞Dnt=D0t.
Based on these denotations, we have(52)∫0t∫Ωbiumx,s+1−bivmx,s+1umx,s+1−vmx,s+1xihnumx,s+1−vmx,s+1dxds=∫0t∫Dnsbiumx,s+1−bivmx,s+1⋅hnumx,s+1−vmx,s+1umx,s+1−vmx,s+1xidxds≤c∫0t∫Dnsbiumx,s+1−bivmx,s+1umx,s+1−vmx,s+1umx,s+1−vmx,s+1xidxds=c∫0t∫Dnsax,s−1/2biumx,s+1−bivmx,s+1umx,s+1−vmx,s+1ax,sumx,s+1−vmx,s+1xidxds≤c∫0s∫Dnsax,s−1/2biumx,s+1−bivmx,s+1umx,s+1−vmx,s+12dx1/2⋅∫0t∫Dnsax,t∇umx,s+1−vmx,s+12dxds1/2.
Because∫Ωax,s−1xdx≤cT,(53)∫0t∫Dnsax,s−1/2biumx,s+1−bivmx,s+1umx,s+1−vmx,s+12dxds≤c∫Dnsax,s−1dxds≤cT.
IfD0s=x∈Ω:umx,s+1−vmx,s+1=0 has 0 measure, then(54)limn⟶∞∫0t∫Dnsax,s−1dx=∫0t∫D0sax,s−1dxds=0.
IfD0s=x∈Ω:umx,s+1−vmx,s+1=0 has a positive measure, then(55)limn⟶∞∫0t∫Dnsax,s∇umx,s+1−vmx,s+12dxds=∫0t∫D0sax,s∇umx,s+1−vmx,s+12dxds=0.
Thus, in both cases, we always have(56)limn⟶∞∫0t∫Ωbiumx,s+1−bivmx,s+1hnumx,s+1−vmx,s+1⋅umx,s+1−vmx,s+1xidxds=0.
In the third place, for the first term on the left-hand side of (47), by Lemma 2,(57)limn⟶∞∫0t∫Ωgnumx,s+1−vmx,s+1∂u−v∂tdxds=∫0t∫Ωsgnumx,s+1−vmx,s+1∂u−v∂tdxds=∫0t∫Ωsgnu−v∂u−v∂sds=∫0tddtu−v1ds.
Letn⟶∞ in (47). Formulas (48)–(57) yield(58)∫Ωux,t−vx,tdx≤∫Ωu0x−v0xdx.Corollary 1.
Theorem2 is true.


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

Theorem 5.
Letux,t and vx,t be two solutions of equation (1) with the initial values u0x and v0x respectively, and with a partial boundary value condition(59)ux,t=vx,t=0,x,t∈Σ1∪Σ2,where(60)Σ2=x∈∂Ω:ax,t≠0,and for any given i∈1,2,…,N, if bi′s≥0,(61)Σ1=∪i=1NΣ1it×0,T,and if bi′s≤0,(62)Σ1=∪i=1NΣ2it×0,T.It is supposed that(63)1λ∫Ω\Ωλtax,t∇a2dx1/2≤cT,and ux,t and vx,t satisfy(64)∫Ωax,t1+mx,t+1logu2∇m2dx≤cT,∫Ωax,t1+mx,t+1logv2∇m2dx≤cT.Then,(65)∫Ωux,t−vx,tdx≤cT∫Ωu0x−v0xdx.Proof.
According to the definition of weak solutions, for all0≤φ∈C01QT, we have(66)∬QTutφx,tdxdt+∬QTax,tmx,t+1∇umx,t+1∇φdxdt−−∬QTax,tmx,t+1umx,t+1logu∇m∇φdxdt+∑i=1N∬QTbiumx,t+1φxidxdt=0.
For anyt∈0,T and a small positive constant λ>0, based on(67)Ωλt=x∈Ω:ax,t>λ,we define that(68)ϕλtx=1,ifx∈Ωλt,ax,tλ,x∈Ω\Ωλt.
Letχτ,st be the characteristic function of τ,s⊂0,T. Because(69)ux,t=0=vx,t,x∈x∈Ω:ax,t≠0,we can choose(70)χτ,stϕλtxgnumx,t+1−vmx,t+1,as the test function. Then,(71)∫τs∫Ωϕλtxgnumx,t+1−vmx,t+1∂u−v∂tdxdt+∫τs∫Ωax,tmx,t+1∇umx,t+1−∇vmx,t+1·ϕλtx∇umx,t+1−vmx,t+1·hnumx,t+1−vmx,t+1dxdt+∫τs∫Ωax,tmx,t+1∇umx,t+1−∇vmx,t+1·∇ϕλx,tgnumx,t+1−vmx,t+1dxdt−∫τs∫Ωax,tmx,t+1umx,t+1logu−vmx,t+1logv∇m·ϕλtx∇umx,t+1−vmx,t+1·hnumx,t+1−vmx,t+1dxdt−∫τs∫Ωax,tmx,t+1umx,t+1logu−vmx,t+1logv∇m·∇ϕλtxgnumx,t+1−vmx,t+1dxdt+∑i=1N∫τs∫Ωbiumx,t+1−bivmx,t+1ϕλxixgnumx,t+1−vmx,t+1+ϕλtumx,t+1−vmx,t+1xihnumx,t+1−vmx,t+1dxdt=0.
Firstly, we still have(72)∫τs∫Ωax,tmx,t+1∇umx,t+1−∇vmx,t+1⋅ϕλt∇umx,t+1−vmx,t+1hnumx,t+1−vmx,t+1dxdt≥0.
Secondly, for the third term on the left-hand side of (71) by that 1/λ∫Ω\Ωλtax,t∇a2dx1/2≤cT, we have(73)∫τs∫Ωax,tmx,t+1∇umx,t+1−∇vmx,t+1·∇ϕλtgnumx,t+1−vmx,t+1dxdt≤∫τs∫Ω\Ωλtax,tmx,t+1∇umx,t+1−∇vmx,t+1·∇ϕλtgnumx,t+1−vmx,t+1dxdt≤∫τs∫Ω\Ωλtax,tmx,t+1∇umx,t+1+∇vmx,t+1∇ϕλtdxdt≤∫τs1λ∫Ω\Ωλtax,tmx,t+1∇umx,t+1+∇vmx,t+1∇adxdt≤c∫τs∫Ω\Ωλtax,t∇umx,t+12+∇vmx,t+12dx1/2·1λ∫Ω\Ωλtax,t∇a2dx1/2dt⟶0,as λ⟶0. Thirdly, for the fourth term on the left-hand side of (71), we can show(74)∫τs∫Ωax,tmx+1umx,t+1logu−vmx,t+1logv∇m·ϕλtx∇umx,t+1−vmx,t+1hnumx,t+1−vmx,t+1dxdt≤∫τs∫Dntax,t∇umx,t+1−∇vmx,t+12dx1/2·c∫Dntax,tumx,t+1logumx,t+1−vmx,t+1logvmx,t+12hnumx,t+1−vmx,t+12∇m2dx1/2dt⟶0,as λ⟶0. The corresponding details are given below.
IfD0t=x∈Ω:u−v=0 is with 0 measure, then(75)limn⟶0∫τs∫Dntax,t∇umx,t+1−∇vmx,t+12dxdt≤∫τs∫D0tax,t∇umx,t+12+∇vmx,t+12dxdt=0,and by (64),(76)∫Ωax,t1+mx,t+1logu2∇m2dx≤cT,we obtain(77)limn⟶0∫τs∫Dntax,tumx,t+1logumx,t+1−vmx,t+1logvmx,t+12·hnumx,t+1−vmx,t+12∇m2dxdt≤limn⟶0∫τs∫Dntax,tumx,t+1logumx,t+1−vmx,t+1logvmx,t+12umx,t+1−vmx,t+12∇m2dxdt≤∫τs∫D0tax,t1+mx,t+1logv2∇m2dx≤cT.
WhenD0t=x∈Ω:u−v=0 has a positive measure,(78)limn⟶0∫τs∫Dntax,t∇umx,t+1−∇vmx,t+12dxdt≤∫τs∫Ωax,t∇umx,t+12+∇vmx,t+12dxdt≤cT.
Also by (64),(79)limn⟶0∫τs∫Dntax,tumx,t+1logumx,t+1−vmx,t+1logvmx,t+12hnumx,t+1−vmx,t+12∇m2dxdt≤limn⟶0∫τs∫D0tax,tumx,t+1logumx,t+1−vmx,t+1logvmx,t+12umx,t+1−vmx,t+12∇m2dxdt=0.
Fourthly, whenbi′s≥0, using the partial boundary value condition (59),(80)−limλ⟶0∫τslimn⟶∞∫Ωbiumx,t+1−bivmx,t+1ϕλtxixgnumx,t+1−vmx,t+1dxdt=−limλ⟶0∫τs∫Ωbiumx,t+1−bivmx,t+1ϕλtxixsignumx,t+1−vmx,t+1dxdt=−limλ⟶0∫τs∫Ωbiumx,t+1−bivmx,t+1ϕλtxixsignbiumx,t+1−bivmx,t+1dxdt=−limλ⟶0∫τs∫Ω\Ωλtbiumx,t+1−bivmx,t+1axiλdxdt=−limλ⟶0∫τs∫Ω\Ωλtumx,t+1−vmx,t+1bi′ξaxiλdxdt≤limλ⟶0∫τs1λ∫Ωλi1t−ciaxiumx,t+1−vmx,t+1dxdt=∫τs∫Σi1t−ciaxiumx,t+1−vmx,t+1dΣdt=0.
Whenbi′s≤0, using the partial boundary value condition (59), we have(81)−limλ⟶0limn⟶∞∫τs∫Ωbiumx,t+1−bivmx,t+1ϕλtxixgnumx,t+1−vmx,t+1dxdt=−limλ⟶0∫τs∫Ωbiumx,t+1−bivmx,t+1ϕλtxixsignumx,t+1−vmx,t+1dxdt=−limλ⟶0∫τs∫Ωbiumx,t+1−bivmx,t+1ϕλtxixsignbiumx,t+1−bivmx,t+1dxdt=−limλ⟶0∫τs∫Ω\Ωλtbiumx,t+1−bivmx,t+1axiλdxdt=−limλ⟶0∫τs∫Ω\Ωλtumx,t+1−vmx,t+1bi'ξaxiλdxdt≤limλ⟶0∫τs1λ∫Ωλi2t−ciaxiumx,t+1−vmx,t+1dxdt=∫τs∫Σi2t−ciaxiumx,t+1−vmx,t+1dΣdt=0.
The last but not the least, as in the proof of Theorem2, we can show that(82)limλ⟶0limn⟶∞∫τs∫Ωbiumx,t+1−bivmx,t+1ϕλtumx,t+1−vmx,t+1xi⋅hnumx,t+1−vmx,t+1dxdt=0.
Then,(83)limn⟶∞limλ⟶0∫τs∫Ωgnumx,t+1−vmx,t+1ϕλt∂u−v∂tdxdt=∫Ωux,s−vx,sdx−∫Ωux,τ−vx,τdx.
Lettingλ⟶0 and n⟶∞ in (73), by (74) and (80)–(85), we obtain(84)∫Ωux,s−vx,sdx≤c∫Ωux,τ−vx,τdx.
By the arbitraries of τ, we have(85)∫Ωux,t−vx,tdx≤∫Ωu0x−v0xdx.Proof of Theorem 2.
Because for everyi∈1,2,…,N, either bi′s≥0 or bi′s≤0, by checking the process of the proof of (80) or (81), we easily obtain Theorem 3.


## 6. Conclusion

In this paper, we consider the initial-boundary value problem of a generalized porous medium equation with a variable exponent. Different from the previous related works, both the diffusion coefficientax,t and variable exponent px,t are dependent on the time variable t, we find out a partial boundary value condition matching up with the equation. The most important innovation is that the partial boundary value condition matching up with the equation is based on a submanifold of ∂Ω×0,T. However, because there is an additional condition (23) imposed, Theorem 3 has not answered the problem globally. In other words, how to obtain the same conclusion as that in Theorem 3 without condition (23) is remained to be studied in the future.



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