# The Stability of the Solutions for a Porous Medium Equation with a Convection Term

Authors: Huashui Zhan; Miao Ouyang
Journal: Discrete Dynamics in Nature and Society
 (2018)

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## Abstract

This paper studies the initial-boundary value problem of a porous medium equation with a convection term. If the equation is degenerate on the boundary, then only a partial boundary condition is needed generally. The existence of the weak solution is proved by the monotone convergent method. Moreover, according to the different boundary value conditions, the stability of the solutions is studied. In some special cases, the stability can be proved without any boundary value condition.


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## Body

## 1. Introduction

Consider the motion of the ideal barotropic gas through a porous medium. Letρ be the gas density, V the velocity, and p the pressure. 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 p=P(ρ). Here, k(x) is a given matrix. We usually assume that Ps=μsα with μ,α=const. The above laws then lead to a semilinear parabolic equation for the density ρ: (3)ρt=μα1+αdivkx∇ρ1+α.If k(x)=axI, where ax is a function and I is the unit matrix, then (3) becomes (4)ρt=μα1+αdivax∇ρ1+α=μαdivaxρα∇ρ.Also, (4) can be regarded as the generalization of the nonlinear heat equation (5)ut=divhu,x∇u,where the function hu,x has the meaning of nonlinear thermal conductivity dependent on the temperature u=ux,t. If ax≡1 in (4) or hu,x≡hu in (5), that is, (6)ut=Δum,which is called the porous medium equation, there are well-known monographs or textbooks devoting to the well-posedness problem of (6); one can refer to [1–6] and the references therein. If ax≥0 in (4) or hu,x depending on x in (5), the situation may be different from that of (6). For example, if a(x)∣x∈∂Ω=0, we consider the equation (7)ut=∇ax∇u,and suppose that there are two classical solutions u and v of (7) with the initial values u0 and v0, respectively. Then it is easy to show that (8)∫Ωux,t-vx,t2dx≤∫Ωu0x-v0x2dx,which implies that the classical solutions (if there are) of (7) are controlled by the initial value completely. In other words, the stability of the classical solutions of (7) is true without any boundary value condition. Yin and Wang [7] also showed that the non-Newtonian fluid equation with the type (9)ut=divdαx∇up-2∇u,x,t∈Ω×0,Thas similar properties, where Ω is a bounded domain in RN with appropriately smooth boundary, dx=distx,∂Ω, and α>0 is a constant. Since the diffusion coefficient dα(x) vanishes on the boundary, it seems that there is no heat flux across the boundary. However, Yin and Wang [7] showed that the fact might not coincide with what we image. In fact, the exponent α, which characterizes the vanishing ratio of the diffusion coefficient near the boundary, does determine the behavior of the heat transfer near the boundary. They proved that, if 0<α<p-1, the solution of (9), u∈H0γ for some constant γ>1, and the trace of u on the boundary can be defined in the traditional way; then, in physics sense, there is no heat flux across the boundary actually, while, if α≥p-1, the existence and uniqueness of solutions were proved without any boundary conditions, which means that whether there is heat flux across the boundary is uncertain. Later, Yin and Wang [8] had shown that only a partial boundary value condition matches up with the equation (10)∂u∂t-div⁡ax∇up-2∇u-fixDiu+cx,tu=gx,t,x,t∈Ω×0,T.Inspired by Yin and Wang [7, 8], we will study the porous medium equation with a convection term, (11)ut=div⁡dα∇um+∑i=1N∂bium∂xi,x,t∈QT=Ω×0,T,with the initial value (12)ux,0=u0x,x∈Ω,and with the partial boundary condition (13)ux,t=0,x,t∈Σm×0,T,where Σm is defined as follows. When 0<α<1, Σm=∂Ω; when α≥1, Σm=x∈∂Ω:bi′0nix<0 and {ni} is the inner normal vector of Ω. The expression of Σm is derived in [9], we do not repeat the details here.We suppose thatbi(s) is a C1 function, and (14)dα/2∇u0m∈L2Ω,0≤u0∈L∞Ω.Definition 1.
A nonnegative functionu(x,t) is said to be the weak solution of (11) with the initial value (12), if (15)dα/2∇um∈L∞0,T;L2Ω,u∈L∞QT;for any function φ∈C1QT, φ∣t=T=0, φ∣∂Ω=0, there holds (16)∬QT-∂φ∂tu+dα∇um∇φdxdt+∬QTbiumφxix,tdxdt=∫Ωu0φx,0dx,
and the initial condition is satisfied in the sense that(17)limt→0⁡∫Ωux,t-u0xdx=0.If u(x,t) satisfies (13) in the sense of the trace in addition, then we say it is a weak solution of the initial-boundary value problem of (11).First of all, we will study the well-posedness problem of (11).Theorem 2.
Ifm>0, 2>α>0, u0x≥0 satisfy (15), then (11) with initial value (12) has a nonnegative solution. Moreover, if 0<α<1, then Σm=∂Ω; the solution is unique.Then, we will study the stability of the solutions.Theorem 3.
Ifbi(u)≡0, i.e. equation (11) is not with the convection term, u and v are two solutions of equation (11) with the initial value u0(x), v0(x) respectively, α>1, then (18)∫Ωux,t-vx,t≤c∫Ωu0x-v0xdx.Sincebi(u)≡0 in Theorem 3, there are some regrets more or less. For (11) itself, we can not prove the same conclusion for the time being. However, as compensation, we can consider a more complicate equation than (11), (19)ut=divdα∇um+∑i=1N∂bium,x,t∂xi.Theorem 4.
Letu and v be two solutions of (19) with the initial values u0x, v0(x), respectively, if 1<α<2, and (20)bi·,x,t≤ax,axx∈∂Ω=0;a(x) satisfies (21)∫Ωaxd-αdx⩽c;then the stability of the weak solutions is true in the sense of (18).It is more or less strange that the caseα=1 is not included in Theorems 3 and 4.At last, we will probe the stability of the weak solutions based on the partial boundary value condition.Theorem 5.
Letu, v be two solutions of (11) with the initial values u0(x), v0(x), respectively. If 2>α≥1, m>0, (22)∫Ωdα-1∇um<∞,∫Ωdα-1∇vm<∞,and the partial boundary condition (12) is satisfied in the sense of trace, then (23)∫Ωux,t-vx,tdx≤c∫Ωu0x-v0xdx+∫Σm′um-vmdΣ,where Σm′=∂Ω\Σm.Theorem 6.
IfΩ is a C2 domain, α≥2, and m≥1, then (11) with the initial value u0 and the partial boundary condition (13) has a BV solution. Moreover, let u,v be two solutions of (11) with the different initial values u0(x),v0(x), respectively. Then (24)∫Ωux,t-vx,t≤∫Ωu0x-v0xdx+∫Σm′u-vdΣ,where Σm′=∂Ω\Σm, Σm=x∈∂Ω:bi′0nix<0, and {ni} is the inner normal vector of Ω.Ifbi≡0, Theorem 6 has been included in Theorem 3, while, if bi≡0 is not true, then Theorem 6 has its independent sense. Such phenomena that the solution of a degenerate parabolic equation may be free from the limitation of the boundary condition also can be found in [7–14]. We will use some ideas in [9, 14]. The uniqueness of the weak solutions when Σm=∂Ω had been proved in [14]. Since [14] was written in Chinese, for the completeness of the paper, we still give its proof in what follows. In addition, how to obtain the stability (23) without condition (22) is a very interesting problem. Last but not least, roughly speaking, in this paper, we can show that if o<α<1 or α≥2, then the weak solution u can be defined the trace on the boundary in the traditional sense; it is surprising that if 1≤α<2, whether u can be defined the trace on the boundary is unknown for the time being.


## 2. The Well-Posedness Problem

We consider the following regularized problem:(25)unt=divd+1nα∇unm+∑i=1N∂biunm∂xi,x,t∈QT,unx,t=1n,x,t∈ST=∂Ω×0,T,unx,0=u0nx=u0x+1n,x∈Ω.According to the standard parabolic equation theory, there is a weak solution(26)un∈L∞QT,∇unm∈L2QT,which satisfies (27)1n⩽unx,t≤u0L∞Ω+1n,x,t∈QT,by the maximum principle.Theorem 7.
Ifm>0, 2>α>0, and u0(x)≥0 satisfy (14), then (11) with initial value (12) has a nonnegative solution.Proof.
First we suppose thatu0∈C0∞Ω and 0≤u0≤M, and consider the following normalized problem (28)unt=divanun∇un+∑i=1N∂biunm∂xi,x,t∈QT,unx,t=1n,x,t∈∂Ω×0,T,unx,0=u0nx,x∈Ω.Here, anu⩾cn>0, and (29)anu=md+1nαum-1,ifu∈1n,M+1n.
Thus, the solution of the problemun is also a solution of problem (25). Moreover, by comparison theorem, we clearly have (30)un+1x,t⩽unx,t,which yields (31)ux,t=limn→∞⁡unx,t.Now, we can prove that the limit function u is a weak solution of (6) with the initial value (8).
Multiplying both sides of the first equation in (25) by ϕ=unm-1/nm and integrating it over QT, we have (32)∫QTuntunm-1nmdxdt=∫Ωununm-1nmt=0t=Tdx-∬QTun∂∂tunm-1nmdxdt=∫Ωunx,Tunmx,T-1nmdx-∫Ωu0nxu0nmx-1nmdx-∬QTmunmuntdxdt=∬QTdiv⁡d+1nα∇unmunm-1nmdxdt+∑i=1N∬QT∂biunm∂xiunm-1nmdxdt.By the fact (33)∬QT∂biunm∂xiunm-1nmdxdt=-∬QTbiunm∂∂xiunm-1nmdxdt=-∬QT∂∂xi∫1/nmunmbisdsdxdt=0,then we have (34)m+1∬QTd+1nα∇unm2dxdt=∫Ωu0nmx-1nmu0nxdx-∫Ωunx,Tunmx,T-1nmdx+∬QT∂biunm∂xiunm-1nmdxdt⩽∫Ωu0mx+1nm+1dx+1nM+1n∫Ωdx.Thus, we obtain (35)d+1nα/2∇unmL2QT≤c.By choosing a subsequence, we can assume that (36)d+1nα/2∇unm⇀ζ,weakly in L2. We need to prove that (37)ζ=dα/2∇um.
For any∀ψ∈C0∞Ω, denoting that dn≜d+1/n, we have (38)∬QTdnα/2∇unm·ψdxdt=∬QT∇dnα/2unm·ψdxdt-α2∬QTdnα/2-1∇d·unmψdxdt=-∬QTdnα/2unm·∇ψdxdt-α2∬QTdnα/2-1∇d·unmψdxdt.
Letn→∞. The left hand side is (39)limn→∞⁡∬QTdnα/2∇unm·ψdxdt=∬QTζψdxdt.while on the right hand side, by (40)∇d·unm≤c∇d·unm≤cM+1nm≤c,and by the condition 0<α<2, (41)∫Ωdnα/2-1dx≤c,using the control convergent theorem, (42)α2limn→∞⁡∬QTdnα/2-1∇d·unm·ψdxdt=α2∬QTdα/2-1∇d·umψdxdt,we have (43)limn→∞⁡-∬QTdnα/2unm·∇ψdxdt-α2∬QTdnα/2-1∇d·unmψdxdt=-∬QTdα/2um·∇ψdxdt-α2∬QTdα/2-1∇d·umψdxdt=∬QTdα/2∇um·ψdxdt.Thus we obtain (37).
At the same time, sincebi∈C1, by (31), we have (44)limn→∞⁡biunm=bium.Thus, u is a solution of (11) with the initial value (12).
Ifu0 only satisfies (14), by considering the problem of (25) 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 (15) and (17). Theorem 7 is proved.Lemma 8.
Letu0 satisfy (14). If 0<α<1 and u is a solution of (11) with the initial value (12), then there exists a constant γ>1 such that (45)∬QT∇umγdxdt≤c.Proof.
Sinceα<1, there exists constant β∈(α,1), β<α+1/2 such that 2-α/β>1. Therefore, there exists γ∈(1,2-α/β) such that βγ<1. Therefore, (46)∬QT∇umγdxdt=∬x,t∈QT;dβ∇um⩽1∇umγdxdt+∬x,t∈QT;dβ∇um>1∇umγdxdt⩽∬QTd-βγdxdt+∬QTdα∇umα/β+γdxdt⩽∬QTd-βγdxdt+∬QTdα1+∇um2dxdt⩽c.Thus um can be defined the trace on the boundary in the traditional way. By the definition of the trace, we also know that u can be defined as the trace on the boundary in the traditional way. The lemma is proved.Theorem 9.
Ifm>0,1>α>0 and u0(x)≥0 satisfies (14), then Σm=∂Ω, and the solution of the initial-boundary value problem (11)–(13) is unique.Proof.
First of all, by Theorem7 and Lemma 8, there is a nonnegative solution of the initial-boundary value problem (11)–(13). Then, we prove its uniqueness. Let u,v be two solutions of equation (11) with (47)ux,0=vx,0,ux,t=vx,t=0,x,t∈∂Ω×0,T.
For all0≤φ∈C01(QT), (48)∬QTφ∂u-v∂tdxdt=-∬QTdα∇um-∇vm·∇φdxdt-∑i=1N∬QTbium-bivmφxidxdt.
For any given positive integern, let gn(s)=∫0shn(τ)dτ, hn(s)=2n1-n|s|+. Then hn(s)∈C(R), and we have (49)hns≥0,shns≤1,gns≤1,limn→∞gns=sgn⁡s,limn→∞sgn′s=0.
Since0<α<1, by Lemma 8, we can define the traces of u,v on the boundary. By a process of limit, we can choose gn(um-vm) as the test function; then (50)∫Ωgnum-vm∂u-v∂tdx+∫Ωdα∇um-∇vm·∇um-vmgn′um-vmdx+∑i=1N∬QTbium-bivmum-vmxign′um-vmdxdt=0.Moreover, we can prove that (51)limn→∞⁡∫Ωbium-bivmgn′um-vmum-vmxidx=0.In detail, the limitation (51) is established by the following calculations.(52)∫Ωbium-bivmgn′um-vmum-vmxidx=∫x∈Ω:um-vm<1/nbium-bivmgn′um-vmum-vmxidx⩽c∫x∈Ω:um-vm<1/nbium-bivmum-vmum-vmxidx=c∫x∈Ω:um-vm<1/nd-α/2bium-bivmum-vmdα/2um-vmxidx⩽c∫x∈Ω:u-v<1/nd-α/2bium-bivmum-vm2dx1/2·∫x∈Ω:um-vm<1/ndα∇um-vm2dx1/2.Since 0<α<1, (53)∫x∈Ω:um-vm<1/nd-α/2bium-bivmum-vm2dx⩽c∫x∈Ω:um-vm<1/nd-αxdx⩽c∫Ωd-αdx⩽c.
In (52), let n→∞. If {x∈Ω:|um-vm|=0} is a set with 0 measure, by 0<α<1, we have (54)limn→∞⁡∫x∈Ω:um-vm<1/nd-αxdx=∫x∈Ω:um-vm=0d-αxdx=0.If the set {x∈Ω:|um-vm|=0} has a positive measure, then, (55)limn→∞∫x∈Ω:um-vm<1/ndα∇um-vm2dx=∫x∈Ω:um-vm=0dα∇um-vm2dx=0.Therefore, in both cases, the right hand side of inequality (52) goes to 0 as n→∞.
Clearly,(56)limn→∞⁡∫Ωgnum-vm∂u-v∂tdx=∫Ωsgn⁡um-vm∂u-v∂tdx=∫Ωsgn⁡u-v∂u-v∂t=ddtu-v1.Now, let n→∞ in (50). Then (57)∫Ωux,t-vx,tdx⩽∫Ωu0x-v0xdx=0.We have the conclusion.By Theorems7 and 9, we clearly have the following.Corollary 10.
Theorem2 is true.


## 3. The Stability without the Boundary Value Condition

Consider a simpler equation than (11). (58)ut=div⁡dα∇um,x,t∈QT=Ω×0,T,with the initial value (12), but without any boundary value condition. For a small positive constant λ>0, let (59)Ωλ=x∈Ω:dx=distx,∂Ω>λ,and let (60)ϕx=1,ifx∈Ω2λ,1λdx-λ,x∈Ωλ\Ω2λ0,ifx∈Ω\Ωλ.Proof of Theorem3.
Supposeu0, v0 only satisfy (7), α>1. Let u, v be two solutions of (58) with the initial-boundary values u0, v0, respectively. For all 0≤φ∈C01(QT), (61)∬QTφ∂u-v∂tdxdt=-∬QTdα∇um-∇vm·∇φdxdt.
By a process of limit, we can chooseϕgn(um-vm) as the test function; then (62)∫Ωϕxgnum-vm∂u-v∂tdx+∫Ωdα∇um-∇vm·ϕ∇um-vmgn′um-vmdx+∫Ωdα∇um-∇vm·∇ϕgnum-vmdx=0.
Clearly, we have(63)∫Ωdα∇um-∇vm·ϕ∇um-vmgn′um-vmdx≥0,limn→∞⁡limλ→0⁡∫Ωϕxgnum-vm∂u-v∂tdx=limn→∞⁡∫Ωgnum-vm∂u-v∂tdx=∫Ωsgn⁡um-vm∂u-v∂tdx=∫Ωsgn⁡u-v∂u-v∂t=ddtu-v1.As for the term (64)∫Ωdα∇um-∇vm·∇ϕgnum-vmdx,we have (65)limλ→0⁡∫Ωdα∇um-∇vm·∇ϕgnum-vmdx≤∫Ωdα∇um2+∇vm2dx1/2∫Ωλ\Ω2λdα∇ϕ2dx1/2=0.
The last equality of (65) is due to that since α>1, we have (66)limλ→0⁡∫Ωλ\Ω2λdα∇ϕ2dx=4limλ→0⁡∫Ωλ\Ω2λλα∇d2λ2dx≤climλ→0⁡λα-1=0.
Now, after lettingλ→0, let n→∞ in (62). Then (67)∫Ωux,t-vx,tdx⩽∫Ωu0x-v0xdx.Theorem 3 is proved.Consider a more complicated equation than (11).(68)ut=div⁡dα∇um+∑i=1N∂bium,x,t∂xi,x,t∈QT=Ω×0,T,with the initial value (12), but without any boundary value condition.Proof of Theorem4.
Supposeu0, v0 only satisfy (7), 1<α<2. Let u,v be two solutions of equation (11) with the initial-boundary values u0, v0, respectively. For all 0≤φ∈C01(QT),(69)∬QTφ∂u-v∂tdxdt=-∬QTdα∇um-∇vm·∇φdxdt-∑i=1N∬QTbium,x,t-bivm,x,tφxidxdt.
By a process of limit, we can choosegn(ϕ(um-vm)) as the test function; then (70)∫Ωgnϕum-vm∂u-v∂tdx+∫Ωdα∇um-∇vm·ϕ∇um-vmgn′ϕum-vmdx+∫Ωdα∇um-∇vm·∇ϕum-vmgn′um-vmdx+∑i=1N∬QTbium,x,t-bivm,x,tϕxium-vm+ϕum-vmxign′ϕum-vmdxdt=0.
Let us analyze every term in the left hand side of (70). For the first term, we clearly have (71)limn→∞⁡limλ→0⁡∫Ωgnϕum-vm∂u-v∂tdx=limn→∞⁡∫Ωgnum-vm∂u-v∂tdx=∫Ωsgn⁡um-vm∂u-v∂tdx=∫Ωsgn⁡u-v∂u-v∂t=ddtu-v1.For the second term, we have (72)∫Ωdα∇um-∇vm·ϕ∇um-vmgn′ϕum-vmdx≥0.For the third term, since (73)limλ→0⁡∫Ωλ\Ω2λdα/2-1um-vmgn′ϕum-vm2dx≤limλ→0⁡∫Ωdα-2um-vmgn′ϕum-vm2dx≤∫Ωdα-2dx≤c,by 0>α-2>-1, we have (74)limλ→0⁡∫Ωdα∇um-∇vm·∇ϕum-vmgn′ϕum-vmdx=limλ→0⁡∫Ωλ∖Ω2λcλdα∇um-∇vmum-vmgn′ϕum-vmdx≤climλ→0⁡∫Ωλ∖Ω2λdα-1∇um-∇vmum-vmgn′ϕum-vmdx≤climλ→0⁡∫Ωλ∖Ω2λdα/2∇um-∇vm2dx1/2∫Ωλ∖Ω2λdα/2-1um-vmgn′ϕum-vm2dx1/2=0,by limn→∞gn′(s)s=0.
Now, we deal with the terms related to the convection functionbi in (70). In the first place, by (20), (75)bium,x,t-bivm,x,t≤2ax;according to the definition of the trace, we have (76)limλ→0∫Ωbium,x,t-bivm,x,tgn′ϕum-vmum-vmϕxixdx≤limλ→0⁡cλ∫Ωλ\Ω2λaxum-vmgn′ϕum-vmdx≤c∫∂ΩaxdΣ=0.
Moreover, we can prove that(77)limn→∞⁡limλ→0⁡∫Ωbium,x,t-bivm,x,tgn′ϕum-vmum-vmxiϕxdx=0.In detail, the limitation (77) is established by the following calculations. (78)limλ→0⁡∫Ωbium,x,t-bivm,x,tgn′ϕum-vmum-vmxiϕxdx=∫Ωbium,x,t-bivm,x,tgn′um-vmum-vmxidx=∫x∈Ω:um-vm<1/nbium,x,t-bivm,x,tgn′um-vmum-vmxidx⩽c∫x∈Ω:um-vm<1/nbium,x,t-bivm,x,tum-vmum-vmxidx=c∫x∈Ω:um-vm<1/nd-α/2bium,x,t-bivm,x,tum-vmdα/2um-vmxidx⩽c∫x∈Ω:u-v<1/nd-α/2bium,x,t-bivm,x,tum-vm2dx1/2·∫x∈Ω:um-vm<1/ndα/2∇um-vm2dx1/2.By (21), (79)∫x∈Ω:um-vm<1/nd-α/2bium,x,t-bivm,x,tum-vm2dx⩽c∫Ωaxd-αdx⩽c.Let n→∞ in (78). If {x∈Ω:|um-vm|=0} is a set with 0 measure, then (80)limn→∞∫x∈Ω:um-vm<1/naxd-αdx=∫x∈Ω:um-vm=0axd-αdx=0.
If the set{x∈Ω:|um-vm|=0} has a positive measure, then, (81)limn→∞⁡∫x∈Ω:um-vm<1/ndα∇um-vm2dx=∫x∈Ω:um-vm=0dα∇um-vm2dx=0.Therefore, in both cases, the right hand side of inequality (74) goes to 0 as n→∞.
At last, Now, after lettingλ→0, let n→∞ in (71). By (72), (73), (74), (76), (77), (78), (79), (80), and (81), then (82)∫Ωux,t-vx,tdx⩽∫Ωu0x-v0xdx.Theorem 4 is proved.


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

In this section, we will prove Theorem5; the proof is similar as that of Theorem 4Proof of Theorem5.
Supposeu0, v0 only satisfy (7), 1≤α<2. Let u, v be two solutions of (11) with the initial-boundary values u0, v0, respectively, and with the same homogeneous partial boundary value condition (83)ux,t=vx,t=0,x,t∈Σm×0,T,
For all0≤φ∈C01(QT), (84)∬QTφ∂u-v∂tdxdt=-∬QTdα∇um-∇vm·∇φdxdt-∑i=1N∬QTbium-bivmφxidxdt.
By a process of limit, we can choosegnϕum-vm as the test function as in Theorem 4; then(85)∫Ωgnϕum-vm∂u-v∂tdx+∫Ωdα∇um-∇vm·ϕ∇um-vmgn′ϕum-vmdx+∫Ωdα∇um-∇vm·∇ϕum-vmgn′um-vmdx+∑i=1N∬QTbium-bivmϕxium-vm+ϕum-vmxign′ϕum-vmdxdt=0.Let us analyze every term in (85). By bium-bivm≤cum-vm and then according to the definition of the trace, by (83), we have (86)limλ→0⁡∫Ωbium-bivmgn′ϕum-vmum-vmϕxixdx≤limλ→0⁡cλ∫Ωλ\Ω2λbium-bivmgn′ϕum-vmum-vmdx≤c∫∂Ωbium-bivmdΣ≤c∫Σm′um-vmdΣ.
Moreover, we can prove that(87)limn→∞⁡limλ→0⁡∫Ωbium-bivmgn′ϕum-vmum-vmxiϕxdx=0.In detail, the limitation (87) is established by the following calculations. (88)limλ→0∫Ωbium-bivmgn′ϕum-vmum-vmxiϕxdx=∫Ωbium-bivmgn′um-vmum-vmxidx=∫x∈Ω:um-vm<1/nbium-bivmgn′um-vmum-vmxidx⩽c∫x∈Ω:um-vm<1/nbium-bivmum-vmum-vmxidx=c∫x∈Ω:um-vm<1/nd-α-1/2bium-bivmum-vmdα-1/2um-vmxidx⩽c∫x∈Ω:u-v<1/nd-α-1/2bium-bivmum-vm2dx1/2·∫x∈Ω:um-vm<1/ndα-1∇um-vm2dx1/2.Since 1≤α<2,(89)∫x∈Ω:um-vm<1/nd-α-1/2bium-bivmum-vm2dx⩽c∫Ωd1-αdx⩽c.In (88), let n→∞. If {x∈Ω:|um-vm|=0} is a set with 0 measures, then (90)limn→∞∫x∈Ω:um-vm<1/nd1-αxdx=∫x∈Ω:um-vm=0d1-αxdx=0.If the set {x∈Ω:|um-vm|=0} has a positive measure, then, by (22), (91)limn→∞∫x∈Ω:um-vm<1/ndα-1∇um-vm2dx≤∫x∈Ω:um-vm=0dα-1∇um2+∇vm2dx=0.Therefore, in both cases, the right hand side of inequality (88) goes to 0 as n→∞.
At the same time, then,(92)limλ→0∫Ωdα∇um-∇vm·∇ϕum-vmgn′ϕum-vmdx=limλ→0∫Ωλ\Ω2λcλdα∇um-∇vmum-vmgn′ϕum-vmdx≤climλ→0∫Ωλ\Ω2λdα-1∇um-∇vmum-vmgn′ϕum-vmdx≤climλ→0∫Ωλ\Ω2λdα-1/2∇um-∇vm2dx1/2∫Ωλ\Ω2λdα-1/2um-vmgn′ϕum-vm2dx1/2.Since (93)limλ→0∫Ωλ\Ω2λdα-1/2um-vmgn′ϕum-vm2dx≤limλ→0∫Ωdα-1/2um-vmgn′ϕum-vm2dx≤∫Ωdα-1dx≤c,
and by (22) (94)limλ→0∫Ωλ\Ω2λdα-1/2∇um-∇vm2dx=0,then we have (95)limλ→0⁡∫Ωdα∇um-∇vm·∇ϕum-vmgn′ϕum-vmdx=0.
Clearly,(96)limn→∞limλ→0∫Ωgnϕum-vm∂u-v∂tdx=ddtu-v1.Now, after letting λ→0, let n→∞ in (85). Then (97)∫Ωux,t-vx,tdx⩽∫Ωu0x-v0xdx+c∫σm′u-vdΣ+c∫0t∫Ωux,t-vx,tdxdt.By Gronwall Lemma, the stability (23) is true. Theorem 5 is proved.


## 5. The BV Solution of Equation

Recently, Zhan considered the initial-boundary value problem of the following equation in [9] (98)∂u∂t=∂∂xiau,x,t∂u∂xi+div⁡bu,x,t∈QT,with (99)ux,0=u0x,x∈Ω,(100)ux,t=0,x,t∈Σp×0,T,where (101)Σp=x∈Ω:bi′0nix<0,Au,x,t=∫0uas,x,tds,as,x,t≥0,a0,x,t=0.Definition 11.
A functionu∈BV(QT)∩L∞(QT) is said to be the entropy solution of equation (98) with the initial-boundary values (99)-(100), if we have the following.
(1) There existsgi∈L2(QT), i=1,2,…,N, such that, for any φ(x,t)∈C01(QT), (102)∬QTφx,tgix,tdxdt=∬QTφx,tau,x,t^∂u∂xidxdt,where a(u,x,t)^ is composite mean value of a(u,x,t).
(2) For anyφ∈C02(QT), φ≥0, for any k∈R, and for any small η>0, u satisfies (103)∬QTIηu-kφt-Bηiu,kφxi+Aηu,x,t,kΔφ-∑i=1NSη′u-kgi2φdxdt+∬QT∫kuaxis,x,tSηs-kdsφxidxdt≥0.
(3) The boundary condition (100) is true in the sense of trace. The initial value is true in the sense of (104)limt→0∫Ωux,t-u0xdx=0.
The existence of the BV solution of equation (98) is by considering the following regularized problem: (105)∂u∂t=∂∂xiau,x,t∂u∂xi+εΔu+∑i=1N∂biu∂xi,inQT,with the initial-boundary conditions (106)ux,0=u0x,x∈Ω,(107)ux,t=0,x,t∈∂Ω×0,T.If there is a constant δ>0 such that (108)a·,x,t-δ∑s=1N+1axs2≥0,where xN+1=t, then we have the following important estimate.Theorem 12 (see [9]).
Letuε be the solution of (105) with (106) and (107). If axi=∂au,x,t/∂xi and at are all bounded, assumption (108) is true; then (109)grad⁡uεL1Ω≤c,where grad⁡u2=∑i=1N∂u/∂xi2+∂u/∂t2 and c is independent of ε.By the theorem, we can prove the existence of the entropy solutionu∈BV(QT) of equation (98) in the sense of Definition 11.Theorem 13 (see [9]).
Suppose thatA(s,x,t) is C3, bi(s) is C2, u0(x)∈L∞(Ω)∩C1(Ω), and there is a constant δ>0 such that (108) is true. Then (98) with the initial condition (99) has an entropy solution in the sense of Definition 11. Moreover, let u, v be two solutions of (6) with the initial value u0(x), v0(x) satisfying (7). Then (110)∫Ωux,t-vx,t≤∫Ωu0x-v0xdx+∫Σm′u-vdΣ,where Σm′=∂Ω\Σm, Σm={x∈∂Ω:bi′(0)ni(x)<0}, and {ni} is the inner normal vector of Ω. In particular, if bi(s)≡0, then Σp=∅; we have (111)∫Ωux,t-vx,t≤∫Ωu0x-v0xdx.Ifbi(s)≡0, Theorem 13 implies that the solution of (98) is controlled by the initial condition. In other words, no boundary value condition is needed.Now, let(112)au,x,t=mdαum-1.Then, for any 0≤u≤M, (113)at=0,axi=mαdα-1dxium-1,by the fact that |∇d|2=1 a.e. in Ω, (114)au,x,t-δ∑s=1N+1axs2=mdαum-1-δm2α2d2α-2u2m-1=mdαum-11-δmαdα-2um-1.If α≥2, m≥1, there exists δ such that inequality (108) is true. But, in general, the distance function d only is a continuous function and is differential for almost everywhere in Ω; then (115)as,x,t=mdαsm-1does not belong to C3, so we can not have Theorems 12 and 13 directly. However, if we check the proof of Theorems 12 and 13, only if we assume that Ω is a appropriately smooth such that dxi(x) is integral on ∂Ω, then similar to the proof of Theorems 12 and 13, we can prove Theorem 4; we omit the details here.Remark 14.
IfΩ is a C2 domain, then d(x) is differential near the boundary ∂Ω, so dxi(x) is a continuous function on ∂Ω and is integral on ∂Ω.Remark 15.
Ifα≥2, m≥1, by Theorem 12, u∈BV(QT), we can define the trace of u on the boundary ∂Ω.



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*Source: 5364746-2018-01-10.xml*