1. Introduction
Consider the motion of the ideal barotropic gas through a porous medium. Let ρ be the gas density, V the velocity, and p the pressure. The motion is governed by the mass conservation law (1)ρt+divρV=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=μα divaxρα∇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-divax∇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=divdα∇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 that bi(s) is a C1 function, and (14)dα/2∇u0m∈L2Ω, 0≤u0∈L∞Ω.

Definition 1.
A nonnegative function u(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∇φdx dt+∬QTbiumφxix,tdx dt=∫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.
If m>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.
If bi(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.

Since bi(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.
Let u 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)∫Ω axd-αdx⩽c;then the stability of the weak solutions is true in the sense of (18).

It is more or less strange that the case α=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.
Let u, 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.
If Ω 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.

If bi≡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.
If m>0, 2>α>0, and u0(x)≥0 satisfy (14), then (11) with initial value (12) has a nonnegative solution.

Proof.
First we suppose that u0∈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, if u∈1n,M+1n.

Thus, the solution of the problem un 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-1nmdx dt=∫Ωununm-1nmt=0t=Tdx-∬QTun∂∂tunm-1nmdx dt=∫Ωunx,Tunmx,T-1nmdx-∫Ωu0nxu0nmx-1nmdx-∬QTmunmuntdx dt=∬QTdivd+1nα∇unmunm-1nmdx dt+∑i=1N∬QT∂biunm∂xiunm-1nmdx dt.By the fact (33)∬QT∂biunm∂xiunm-1nmdx dt=-∬QTbiunm∂∂xiunm-1nmdx dt=-∬QT∂∂xi∫1/nmunmbisds dx dt=0,then we have (34)m+1∬QTd+1nα∇unm2dx dt=∫Ωu0nmx-1nmu0nxdx-∫Ωunx,Tunmx,T-1nmdx+∬QT∂biunm∂xiunm-1nmdx dt⩽∫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·ψ dx dt=∬QT∇dnα/2unm·ψ dx dt-α2∬QTdnα/2-1∇d·unmψ dx dt=-∬QTdnα/2unm·∇ψ dx dt-α2∬QTdnα/2-1∇d·unmψ dx dt.

Let n→∞. The left hand side is (39)limn→∞∬QTdnα/2∇unm·ψ dx dt=∬QTζψ dx dt.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·ψ dx dt=α2∬QTdα/2-1∇d·umψ dx dt,we have (43)limn→∞-∬QTdnα/2unm·∇ψ dx dt-α2∬QTdnα/2-1∇d·unmψ dx dt=-∬QTdα/2um·∇ψ dx dt-α2∬QTdα/2-1∇d·umψ dx dt=∬QTdα/2∇um·ψ dx dt.Thus we obtain (37).

At the same time, since bi∈C1, by (31), we have (44)limn→∞biunm=bium.Thus, u is a solution of (11) with the initial value (12).

If u0 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.
Let u0 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∇umγdx dt≤c.

Proof.
Since α<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∇umγdx dt=∬x,t∈QT;dβ∇um⩽1∇umγdx dt+∬x,t∈QT;dβ∇um>1∇umγdx dt⩽∬QTd-βγdx dt+∬QTdα∇umα/β+γdx dt⩽∬QTd-βγdx dt+∬QTdα1+∇um2dx dt⩽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.
If m>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 Theorem 7 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 all 0≤φ∈C01(QT), (48)∬QTφ∂u-v∂tdx dt=-∬QTdα∇um-∇vm·∇φ dx dt-∑i=1N∬QTbium-bivmφxidx dt.

For any given positive integer n, 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=sgns,limn→∞sgn′s=0.

Since 0<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-vmdx dt=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=∫Ωsgnum-vm∂u-v∂tdx=∫Ωsgnu-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 Theorems 7 and 9, we clearly have the following.

Corollary 10.
Theorem 2 is true.

3. The Stability without the Boundary Value Condition
Consider a simpler equation than (11). (58)ut=divdα∇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,if x∈Ω2λ,1λdx-λ,x∈Ωλ\Ω2λ0,if x∈Ω\Ωλ.

Proof of Theorem <xref ref-type="statement" rid="thm1.3">3</xref>.
Suppose u0, 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∂tdx dt=-∬QTdα∇um-∇vm·∇φ dx dt.

By a process of limit, we can choose ϕgn(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→∞lim λ→0∫Ωϕxgnum-vm∂u-v∂tdx=limn→∞∫Ωgnum-vm∂u-v∂tdx=∫Ωsgnum-vm∂u-v∂tdx=∫Ωsgnu-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=4 limλ→0∫Ωλ\Ω2λλα∇d2λ2dx≤climλ→0λα-1=0.

Now, after letting λ→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=divdα∇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 Theorem <xref ref-type="statement" rid="thm1.4">4</xref>.
Suppose u0, 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∂tdx dt=-∬QTdα∇um-∇vm·∇φ dx dt-∑i=1N∬QTbium,x,t-bivm,x,tφxidx dt.

By a process of limit, we can choose gn(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-vmdx dt=0.

Let us analyze every term in the left hand side of (70). For the first term, we clearly have (71)limn→∞ lim λ→0∫Ωgnϕum-vm∂u-v∂tdx=limn→∞∫Ωgnum-vm∂u-v∂tdx=∫Ωsgnum-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 function bi 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→0cλ∫Ωλ\Ω2λaxum-vmgn′ϕum-vm dx ≤c∫∂ΩaxdΣ=0.

Moreover, we can prove that (77)limn→∞lim λ→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 letting λ→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 Theorem 5; the proof is similar as that of Theorem 4

Proof of Theorem <xref ref-type="statement" rid="thm1.5">5</xref>.
Suppose u0, 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 all 0≤φ∈C01(QT), (84)∬QTφ∂u-v∂tdx dt=-∬QTdα∇um-∇vm·∇φ dx dt-∑i=1N∬QTbium-bivmφxidx dt.

By a process of limit, we can choose gnϕum-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-vmdx dt=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→0cλ∫Ωλ\Ω2λbium-bivmgn′ϕum-vmum-vm dx ≤c∫∂Ωbium-bivmdΣ≤c∫Σm′um-vmdΣ.

Moreover, we can prove that (87)limn→∞ lim λ→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≤c limλ→0∫Ωλ\Ω2λdα-1∇um-∇vmum-vmgn′ϕum-vmdx≤c limλ→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→∞lim λ→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,tdx dt.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+divbu, 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 function u∈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 exists gi∈L2(QT), i=1,2,…,N, such that, for any φ(x,t)∈C01(QT), (102)∬QTφx,tgix,tdx dt=∬QTφx,tau,x,t^∂u∂xidx dt,where a(u,x,t)^ is composite mean value of a(u,x,t).

(2) For any φ∈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-kgi2φdx dt+∬QT∫kuaxis,x,tSηs-kdsφxidx dt≥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, in QT,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 [<xref ref-type="bibr" rid="B20">9</xref>]).
Let uε 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)graduεL1Ω≤c,where gradu2=∑i=1N∂u/∂xi2+∂u/∂t2 and c is independent of ε.

By the theorem, we can prove the existence of the entropy solution u∈BV(QT) of equation (98) in the sense of Definition 11.

Theorem 13 (see [<xref ref-type="bibr" rid="B20">9</xref>]).
Suppose that A(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.

If bi(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.
If Ω 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.
If α≥2, m≥1, by Theorem 12, u∈BV(QT), we can define the trace of u on the boundary ∂Ω.