# The asymptotic behavior of the solution of a doubly degenerate parabolic equation with the convection term

## Abstract

$u t = div D u m p - 2 D u m + div B ( u m )$

with initial condition u(x, 0) = u0(x). By using Moser iteration technique, assuming that the uniqueness of the Barenblatt-type solution E c of the equation u t = div(|Dum|p-2Dum) is true, then the solution u may satisfy

$t 1 μ u ( x , t ) - E c ( x , t ) → 0 , as t → ∞ ,$

which is uniformly true on the sets $x ∈ R N : | x | < a t 1 μ N , a > 0$. Here B(um) = (b1(um), b2(um), ..., b N (um)) satisfies some growth order conditions, the exponents m and p satisfy m(p - 1) > 1.

Mathematics Subject Classification 2000: 35K55; 35K65; 35B40.

## 1. Introduction

$u t = div ( D u m p - 2 D u m ) + div ( B ( u m ) ) , in S= R N × ( 0 , ∞ ) ,$
(1.1)
$u ( x , 0 ) = u 0 ( x ) , on R N ,$
(1.2)

where m(p - 1) > 1, N ≥ 1, u0(x) L1(RN), D is the spatial gradient operator, and the convection term $div ( B ( u m ) ) = ∑ i = 1 N ∂ ( b i ( u m ) ) ∂ x i$.

Equation (1.1) appears in a number of different physical situations .

For example, in the study of water infiltration through porous media, Darcy's linear relation

$V=-K ( θ ) ∇ϕ,$
(1.3)

satisfactorily describes flow conditions provided the velocities are small. Here V represents the seepage velocity of water, θ is the volumetric moisture content, K(θ) is the hydraulic conductivity and ϕ is the total potential, which can be expressed as the sum of a hydrostatic potential ψ(θ) and a gravitational potential z

$ϕ=ψ ( θ ) +z.$
(1.4)

However, (1.3) fails to describe the flow for large velocities. To get a more accurate description of the flow in this case, several nonlinear versions of (1.3) have been proposed. One of these versions is

$V α =-K ( θ ) ∇ϕ,$
(1.5)

where α ranges from 1 for laminar flow to 2 for completely turbulent flow (cf.  and references therein). If it is assumed that infiltration takes place in a horizontal column of the medium, according to the continuity equation

$∂ θ ∂ t + ∂ V ∂ x =0,$

then (1.4) and (1.5) give

$∂ θ ∂ t = ∂ ∂ x D ( θ ) p θ x p - 1 θ x$

with $1 p =α$ and D(θ) = K(θ)ψ'(θ). Choosing D(θ) = D0θm-1(cf. [5, 6]), one obtains (1.1) with B(s) 0, u being the volumetric moisture content.

Another example where Equation (1.1) appears is the one-dimensional turbulent flow of gas in a porous medium (cf. ), where u stands for the density, and the pressure is proportional to um-1(see also ). Typical values of p are again 1 for laminar (non-turbulent) flow and $1 2$ for completely turbulent flow.

The existence of nonnegative solution of (1.1)-(1.2) without the convection term div(B(um)), defined in some weak sense, had been well established (see  etc.). Here we quote the following definition.

Definition 1.1. A nonnegative function u(x, t) is called a weak solution of (1.1)-(1.2) if u satisfies

(i)

$u∈C ( 0 , T ; L 1 ( R N ) ) ∩ L ∞ ( R N × ( τ , T ) ) ,$
(1.6)
$u m ∈ L 1 oc p ( 0 , T ; W 1 . p ( R N ) ) , u t ∈ L 1 ( R N × ( τ , T ) ) , ∀ τ > 0 ;$
(1.7)

(ii)

$∬ S [ u φ t - D u m p - 2 D u m ⋅ D φ - B ( u m ) ⋅ D φ ] d x d t = 0 , ∀ φ ∈ C 0 1 ( S ) ;$
(1.8)

(iii)

$lim t → 0 ∫ R N u ( x , t ) - u 0 ( x ) d x = 0 .$
(1.9)

If there exist the positive constants k1, α such that

$B ( s ) ≤ k 1 s 1 + α , B ′ ( s ) ≤ k 1 s α , ∀ s ∈ R 1 = ( - ∞ , + ∞ ) ,$
(1.10)

Chen-Wang  had proved the existence and the uniqueness of the weak solutions of (1.1) and (1.2) in the sense of Definition 1.1.

As we have said before, we are mainly interested in the behavior of solution of (1.1) and (1.2) as t → ∞. According to the different properties of the initial function u0(x), the corresponding nonnegative solutions may have different large time asymptotic behaviors, one can refer to the references . In our article, we are going to study the large time asymptotic behavior for the solution of (1.1) and (1.2) by comparing it to the Barenblatt-type solution, let us give some details.

It is not difficult to verify that

$E c = t - 1 μ b - m ( p - 1 ) - 1 m p N μ - 1 p - 1 x t - 1 N μ p p - 1 + p - 1 m ( p - 1 ) - 1$

is the Barenblatt-type solution of the Cauchy problem

(1.11)
$u ( x , 0 ) = c δ ( x ) , on R N ,$
(1.12)

where $μ=m ( p - 1 ) -1+ p N ,c= ∫ R N u 0 ( x ) dx$, b is a constant such that $b = ∫ R N E c ( x , t ) d x$, and δ denotes the Dirac mass centered at the origin.

By using some ideas of [9, 14], we have the following

Theorem 1.2. Suppose m(p - 1) > 1, B satisfies (1.10) with α < p - 1 and $m ( 1 + α ) ≥1+μ=m ( p - 1 ) + p N$. If E c is a unique solution of (1.11) and (1.12), then the solution u of (1.1) and (1.2) satisfies

uniformly on the sets $x ∈ R N : x < a t 1 μ N , a > 0$, where $c= ∫ R N u 0 dx$ as before.

Remark 1.3. For m = 1, the uniqueness of solutions of (1.11) and (1.12) is known (see ).

By assuming that the uniqueness of the Barenblatt-type solution of (1.11) is true, Yang and Zhao  had established the similar large time behavior of solution of the Cauchy problem of the following equation

$u t = div D u m p - 2 D u m - u q , in S= R N × ( 0 , ∞ ) ,$
(1.13)

While Zhan  had considered the Cauchy problem of the following equation

$u t = div D u m p - 2 D u m - D u m p 1 - u q , in S = R N × ( 0 , ∞ ) ,$
(1.14)

and also had got the similar result as Theorem 1.2. Comparing (1.1) with (1.13) or (1.14), the most difficulty comes from that the convection term div(B(um)). The absorption term -uq in (1.13), or $- D u m p 1 - u q$ in (1.14), is always less than 0. This fact made us be able to draw it away in many estimates in  or . But the convection term div(B(um)) plays important role in this article, and it can not be drawn away randomly in the estimates we needed, we have to deal with it by some special techniques.

At the end of this introduction section, we would like to point that the condition m(p - 1) > 1 in Theorem 1.2, which means that the Equation (1.1) or (1.11) is a doubly degenerate parabolic equation, plays an important role in the proof of the theorem. In other words, if it is not true, (1.1) is in singular case, then the large time behavior of the solution in this case is still an open problem.

## 2. Some important lemmas

Let u be a nonnegative solution of (1.1) and (1.2). We define the family of functions

$u k = k N u ( k x , k N μ t ) , k > 0 .$

It is easy to see that they are the solutions of the problems

(2.1)
$u ( x , 0 ) = u 0 k ( x ) , on R N ,$
(2.2)

where $μ=m ( p - 1 ) + p N -1$ as before and u0k(x) = kN u0(kx).

Lemma 2.1 For any s (, m(p - 1)), the nonnegative solution u k satisfies

$∫ 0 T ∫ B R u k s - m 1 + u k s 2 D u k m 2 d x d t ≤ c ( s , R , u 0 L 1 ) ,$
(2.3)
$∫ 0 T ∫ B R u m ( p - 1 ) + p N - s d x d t ≤ c ( s , R , u 0 L 1 ) .$
(2.4)

Proof. From Definition 1.1, we are able to deduce that (see ): for $∀φ∈ C 1 ( S ̄ )$, φ = 0 when |x| is large enough, for any t [0, T], 0 < h < t,

$∫ R N u k φ ( x , t ) d x - ∫ h t ∫ R N u k φ t - D u k m p - 2 D u k m ⋅ D φ - k N ( 1 + μ ) B k - N m u k m D φ d x d t = ∫ R N u k φ ( x , h ) d x .$
(2.5)

Let

$ψ R ∈ C 0 ∞ ( B 2 R ) ,0≤ ψ R ≤1, ψ R =1 on B R , D ψ R ≤c R - 1 .$
(2.6)

By an approximate procedure, we can choose $φ = u k s 1 + u k s ψ R p$ in (2.5), then

$∫ R N ∫ 0 u k ( x , t ) z s 1 + z s d z ψ R p d x + s ∫ h t ∫ R N u k s - m 1 + u k s 2 D u k m p ψ R p d x d τ ≤ - p ∫ h t ∫ R N u k s 1 + u k s D u k m p - 2 ψ R p - 1 D u k ⋅ D ψ R d x d τ + ∫ R N ∫ 0 u k ( x , t ) z s 1 + z s d z ψ R p d x + k N ( 1 + μ ) ∫ h t ∫ R N B ( k - N m u k m ) D u k s 1 + u k s ψ R p d x d τ .$
(2.7)

Noticing

$∫ h t ∫ R N u k s 1 + u k s D u k m p - 2 ψ R p - 1 ( x ) D u k m ⋅ D ψ R d x d τ ≤ ∫ h t ∫ R N ε u k ( s - m ) p - 1 p 1 + u k s 2 p - 1 p D u k m p - 1 ψ R p - 1 p p - 1 + c ( ε ) u k s - ( s - m ) ( p - 1 ) p 1 + u k s 1 - 2 p - 1 p D ψ R p d x d τ = ε ∫ h t ∫ R N u k s - m 1 + u k s 2 D u k m p ψ R p d x d τ + c ( ε ) ∫ h t ∫ R N u k m ( p - 1 ) + s D ψ R p d x d τ ,$
(2.8)
$k N ( 1 + μ ) ∫ h t ∫ R N B ( k - N m u k m ) D u k s 1 + u k s ψ R p d x d τ = k N ( 1 + μ ) ∫ h t ∫ R N B ( k - N m u k m ) - D u s 1 + u s 2 ψ R p + p u s 1 + u s ψ R p - 1 D ψ R d x d τ .$
(2.9)

Since s > αm, $B ( k - N m u k m ) ≤ k 1 k - N m ( 1 + α ) u k m 1 + α$,

$ψ R p u k s - m + ( 1 + α ) m p ′ 1 + u k s 2 ≤1$

is always true, we have

$- ∫ h t ∫ R N B ( k - N m u k m ) 1 + u k s 2 ψ R p D u k s d x d τ = s m ∫ h t ∫ R N B ( k - N m u k m ) u k s - m 1 + u k s 2 ψ R p D u k m d x d τ ≤ ε ∫ h t ∫ R N s ψ R p u k s - m 1 + u k s 2 D u k m p d x d τ + c ( ε ) s ∫ h t ∫ R N s ψ R p u k s - m 1 + u k s 2 B ( k - N m u k m ) p ′ d x d τ , ≤ ε ∫ h t ∫ R N s ψ R p u k s - m 1 + u k s 2 D u k m p d x d τ + c ( ε , R ) k - N m ( 1 + α ) p ′ , p ′ = p p - 1 ,$
(2.10)

and

$∫ R N ∫ 0 u k ( x , h ) z s 1 + z s d z ψ R p d x ≤ ∫ R N u ( x , k N μ h ) d x .$
(2.11)

Noticing that the condition m(p - 1) > 1 and

$m ( 1 + α ) ≥1+μ=m ( p - 1 ) + p N ,$

then by (2.7)-(2.11), we obtain

$sup 0 < t < T ∫ R N ∫ 0 u k ( x , t ) z s 1 + z s d z d x + ∫ h T ∫ R N u k s - m 1 + u k s 2 D u k m p ψ R p d x d τ ≤ c ∫ R N u ( x , k N μ h ) d x + c ∫ h T ∫ R N u k m ( p - 1 ) + s D ψ R p d x d τ + c .$
(2.12)

Since u k L (RN × (h, T)) ∩ L1(S T ),

$lim R → ∞ ∫ h T ∫ R N u k m ( p - 1 ) + s D ψ R p d x d τ = 0 .$
(2.13)

Let h → 0 in (2.12). Then

$sup 0 < t < T ∫ B 2 R ∫ 0 u k ( x , t ) z s 1 + z s d z d x + ∬ s T u k s - m 1 + u k s 2 D u k m p d x d τ ≤ c ∫ R N u 0 d x .$
(2.14)

From this inequality, it is clear of that

$sup 0 < t < T ∫ B 2 R u k ( x , t ) d x + ∫ 0 T ∫ B 2 R u k s - m 1 + u k s 2 D u k m p d x d τ ≤ c ( R ) .$
(2.15)

So (2.3) is true.

Let

$u 1 = max u k ( x , t ) , 1 , w = u 1 m ( p - 1 ) - s p .$

By Sobolev's imbedding inequality (see ), for $ξ∈ C 0 1 ( B 2 R )$, ξ ≥ 0, we have

$∫ B 2 R ξ p w r d x 1 r ≤ c ∫ B 2 R D ( ξ w ) p d x s p ∫ B 2 R p w m ( p - 1 ) - s d x ( 1 - θ ) [ m ( p - 1 ) - s ] p ,$

where

$θ = m ( p - 1 ) - s p - 1 r 1 N - 1 p + m ( p - 1 ) - s p - 1 , r = p [ m ( p - 1 ) + p N - s ] m ( p - 1 ) - s .$

It follows that

$∬ S T ξ p w r d x d t ≤ c ∬ S T D ( ξ w ) p d x d t sup t ∈ ( 0 , T ) ∫ B 2 R p w m ( p - 1 ) - s d x ( r - p ) [ m ( p - 1 ) - s ] p ,$
(2.16)

where we denote S T = RN × (0, T). Since

we have

$∬ S T ∫ | D ( ξ w ) | p d x d t ≤ c ∬ S T ( ξ p | D w | p + w p | D ξ | p ) d x d t ≤ c ∬ S T | D ξ | p u 1 m ( p - 1 ) - s d x d t + ∫ 0 T ∫ B 2 R u k s - m ( 1 + u k s ) 2 | D u k m | p d x d t .$
(2.17)

Hence, by (2.16), (2.(17) and (2.15), we get

$∬ S T ξ p u 1 m ( p - 1 ) + p N - s d x d t ≤ c ( s , R , u 0 L 1 ) 1 + ∬ S T D ξ p u 1 m ( p - 1 ) - s d x d t .$

Let $ξ= ψ R b$, ψ R . be the function satisfying (2.6) and $b= N [ m ( p - 1 ) + p N - s ] p$. Then

$∬ s T ψ R p b u 1 m ( p - 1 ) + p N - s d x d t ≤ c ( s , R , u 0 L 1 ) 1 + ∬ s T ψ R p b u 1 m ( p - 1 ) + p N - s d x d t m ( p - 1 ) - s m ( p - 1 ) - s + p N ,$

by Moser iteration technique, the above inequality implies (2.4) is true.

Let Q ρ = B ρ (x0) × (t0 - ρp, t0) with t0 > (2ρ)p and uk 1= max{u k , 1}. Also by Moser iteration technique, we have

Lemma 2.2 The nonnegative solution u k satisfies

(2.18)

where c(ρ, s1) depends on ρ and s1, and s1 can be any number satisfying $0< s 1 <1+ p N$.

Proof. For $∀φ∈ C 1 ( S ̄ )$, φ = 0 when |x| is large enough, we have

$∫ R N u k ( x , t ) φ d x - ∫ 0 T ∫ R N [ u k φ t - D u k m p - 2 D u k m ⋅ D φ - k N ( 1 + μ ) B ( k - N m u k m ) D φ ] d x d t = ∫ R N u 0 k ( x ) φ ( x , 0 ) d x .$
(2.19)

Let ξ be the cut function on Q ρ , i.e.

$0≤ξ≤1,ξ | Q ρ =1,ξ | R N \ Q 2 ρ =0.$

We choose the testing function in (2.19) as $φ= ξ p u k 2 γ - 1$, where $γ > 1 2$ is a constant. Then

$1 2 γ ∫ B 2 ρ ξ p u k 2 γ ( x , t ) d x + 2 γ - 1 m ∫ 0 t ∫ B 2 ρ ξ p u k 2 γ - 1 - m D u k m p d x d s = p ∫ 0 t ∫ B 2 ρ ξ p - 1 D ξ u k 2 γ - 1 D u k m p - 1 d x d s + p 2 γ ∫ 0 t ∫ B 2 ρ ξ p - 1 ξ t u k 2 γ d x d s + p k N ( 1 + μ ) ∫ 0 t ∫ B 2 ρ ξ p - 1 B ( k - N m u k m ) u k 2 γ - 1 D ξ d x d t + k N ( 1 + μ ) 2 γ - 1 m ∫ 0 t ∫ B 2 ρ ξ p u k 2 γ - m - 1 B ( k - N m u k m ) D u k m d x d s .$
(2.20)

Using Young inequality, by (1.10),

$ξ p - 1 D ξ u k 2 γ - 1 D u k m p - 1 = u k 2 γ - 1 - m ξ p - 1 D u k m p - 1 D ξ u k m ≤ u k 2 γ - 1 - m ( ε ξ p D u k m p + c ( ε ) u k m p D ξ p ) , ξ p u k 2 γ - m - 1 B ( k - N m u k m ) D u k m ≤ k - N m ( 1 + α ) ξ p u k m α + 2 γ - 1 D u k m ≤ k - N m ( 1 + α ) ξ p u k 2 γ - 1 - m u k m α + m D u k m ≤ k - N m ( 1 + α ) ξ p u k 2 γ - 1 - m ( c ( ε ) u k m ( 1 + α ) p ′ + ε D u k m p ) ,$

from (2.20), we have

$1 2 γ ∫ B 2 ρ ξ p u k 2 γ ( x , t ) d x + 2 γ - 1 m - ε 1 + 2 γ - 1 m ∫ 0 t ∫ B 2 ρ ξ p u k 2 γ - 1 - m ∇ u k m p d x d s ≤ c ∫ 0 t ∫ B 2 ρ u k 2 γ - 1 + m ( p - 1 ) D ξ p d x d s + p 2 γ ∫ 0 t ∫ B 2 ρ ξ p - 1 ξ t u k 2 γ d x d s + c ( ε ) k - N m ( 1 + α ) + N ( 1 + μ ) ∫ 0 t ∫ B 2 ρ u k 2 γ - 1 - m + m ( 1 + α ) p ′ d x d s .$
(2.21)

By the fact of that

$D ξ u 2 γ - 1 + m ( p - 1 ) p p = u k 2 γ - 1 + m ( p - 1 ) p D ξ + 2 γ - 1 + m ( p - 1 ) m p ξ u k 2 γ - 1 - m p D u k m p ≤ c D ξ p u k 2 γ - 1 + m ( p - 1 ) + c ξ p D u k m p u k 2 γ - 1 - m ,$

from (2.21), we have

$sup t 0 - 2 ρ p < t < t 0 ∫ B 2 ρ ξ p u k 2 γ d x d s + ∫ ∫ Q 2 ρ D ξ u k 2 γ - 1 + m ( p - 1 ) p p d x d s ≤ c ∫ 0 t ∫ B 2 ρ u k 2 γ - 1 + m ( p - 1 ) D ξ p d x d s + c ∫ 0 t ∫ B 2 ρ ξ p - 1 ξ t u k 2 γ d x d s + c ( ε ) k - N m ( 1 + α ) + N ( 1 + μ ) 2 γ - 1 m ∫ 0 t ∫ B 2 ρ | u k | 2 γ - 1 - m + m ( 1 + α ) p ′ d x d s .$
(2.22)

Let

$β= max 1 , 2 γ - 1 + m ( p - 1 ) γ ,$

and

$w = ξ β u k 2 γ - 1 + m ( p - 1 ) p .$

By the embedding theorem, from (2.22), we have

$∫ ∫ Q 2 ρ w h d x d t ≤ c sup t 0 - 2 ρ p < t < t 0 ∫ B 2 ρ w 2 γ p 2 γ - 1 + m ( p - 1 ) d x 2 γ - 1 + m ( p - 1 ) 2 γ p ( 1 - δ ) h . ∫ t 0 - ( 2 ρ p ) ∫ B 2 ρ D w p d x δ h p d x ,$
(2.23)

where

$δ= 2 γ - 1 + m ( p - 1 ) 2 γ p - 1 h . 1 N - 1 p + 2 γ - 1 + m ( p - 1 ) 2 γ p - 1 .$

In particular, we choose

$h=p [ 1 + 2 γ p N ( 2 γ - 1 + m ( p - 1 ) ) ] ,$

then from (2.23), we have

$∫ ∫ Q 2 ρ ξ β h u k 2 γ - 1 + m ( p - 1 ) + 2 γ p N d x d t ≤ c 1 sup t 0 - 2 ρ p < t < t 0 ∫ B 2 ρ ξ 2 γ p β 2 γ - 1 + m ( p - 1 ) u k 2 γ d x p N . ∫ ∫ Q 2 ρ D ξ β u k 2 γ - 1 + m ( p - 1 ) p p d x d t ≤ c 2 sup t 0 - 2 ρ p < t < t 0 ∫ B 2 ρ ξ 2 γ p β 2 γ - 1 + m ( p - 1 ) u k 2 γ d x + ∫ ∫ Q 2 ρ D ξ β u k 2 γ - 1 + m ( p - 1 ) p p d x d t 1 + p N .$
(2.24)

Now, for $τ ∈ [ 1 2 , 1 ]$, we denote that

$ρ l =2ρ τ + 1 - τ 2 l ,l=1,2,…,$

and choose the cut functions ξ l (x, t) of Q ρl , such that on Qρ(l+1), ξ l = 1.

Denote

$K=1+ p N ,2γ= K l .$

and let

$u 1 k = max { 1 , u k } .$

Then, by (2.23) and (2.24) and the assumption of that a < p - 1, which implies

$2γ-1+m ( 1 + α ) p ′ -m≤2γ-1+m ( p - 1 ) ,$

we have

$∫ ∫ Q ρ ( l + 1 ) u k m ( p - 1 ) - 1 + K l + 1 d x d t ≤ ∫ ∫ Q ρ ( l + 1 ) u 1 k m ( p - 1 ) - 1 + K l + 1 d x d t ≤ ∫ ∫ Q ρ ( l + 1 ) u k m ( p - 1 ) - 1 + K l + 1 d x d t + mes Q ρ ( l + 1 ) ≤ c c 1 l [ ( 1 - τ ) ρ ] p ∫ ∫ Q ρ l u 1 k m ( p - 1 ) - 1 + K l d x d t K .$

Using Moser iteration technique, we have

Then, we have

By Schwarz inequality,

By the Lemma 3.1 in , for any $τ ∈ [ 1 2 , 1 )$, we have

and from this inequality, we get the conclusion of the lemma.

Lemma 2.3 The nonnegative solution u k satisfies

$∫ τ T ∫ B R D u k m p d x d t ≤ c ( τ , R ) .$
(2.25)
$∫ τ T ∫ B R u k t p d x d t ≤ c ( τ , R ) .$
(2.26)

Proof. By Lemmas 2.1 and 2.2, {u k } are uniformly bounded on every compact set K S T . Let ψ R be a function satisfying (2.6) and $ξ∈ C 0 1 ( 0 , T )$ with 0 ≤ ξ ≤ 1, ξ = 1 if t (τ, T). We choose $η= ψ R p ξ u k m$in (2.5) to obtain

$1 m + 1 ∫ R N u k m + 1 ( x , T ) ψ R p d x + ∬ S T D u k m p ψ R p ξ d x d t = 1 m + 1 ∬ S T u k m + 1 ξ ′ ψ R p d x d t - p ∬ S T u k m D u k m p - 2 D u k m ⋅ D ψ R ψ R p - 1 ξ d x d t + k N ( 1 + μ ) ∬ S T B ( k - N m u k m ) ξ ( p ψ p - 1 D ψ R u k m + ψ R p D u k m ) d x d t .$
(2.27)

Noticing

$∬ S T u k m D u k m p - 1 D ψ R ψ R p - 1 ξ d x d t ≤ ε ∬ S T D u k m p ψ R p ξ d x d t + c ( ε ) ∬ S T u k p m D ψ R p ξ d x d t , ∬ S T B ( k - N m u k m ) ξ ψ p - 1 D ψ R u k m d x d t ≤ c R k - N m ( 1 + α ) ∫ 0 T ∫ B r ξ u k m ( 1 + α ) d x d τ , ∬ S T B ( k - m N u k m ) ξ ψ R p D u k m ) d x d t ≤ c ( ε ) ∬ S T B ( k - N m u k m ) p ′ ξ ψ R p d x d τ + ε ∬ S T D u k m p ψ R p ξ d x d t ≤ c ( ε ) k - N m ( 1 + α ) ∬ S T u k m p ′ ( 1 + α ) ξ ψ R p d x d τ + ε ∬ S T D u k m p ψ R p ξ d x d t ,$

from these inequalities, by (2.18) and (2.27), one knows that (2.25) is true. (2.26) is to be proved in what follows.

Let

$v ( x , t ) = u k r ( x , t ) =r u k ( x , r m ( p - 1 ) - 1 t ) ,r∈ ( 0 , 1 ) .$
(2.28)

Then

$v t ( x , t ) = div ( D v m p - 2 D v m ) + r m ( p - 1 ) k N ( 1 + μ ) div ( B ( k - N m r - m v m ) ) ,$
(2.29)
$v ( x , 0 ) = r u k ( x , 0 ) .$
(2.30)

By (2.1) and (2.29), for any $φ∈ C 0 1 ( S T )$, we have

$∬ S T φ ∂ ∂ t ( u k - v ) d x d t + ∬ S T D u k m p - 2 D u k m - D v m p - 2 D v m D φ d x d t + k N ( 1 + μ ) ∬ S T [ B ( k - N m u k m ) - r m ( p - 1 ) B ( k - N m r - m v m ) ] D φ d x d t = 0 .$
(2.31)

Let g n (s) = 1 when $s> 1 n$; g n (s) = ns when $0≤s≤ 1 n$; g n (s) = 0 when s < 0, and let φ in (2.31) be substituted by $φ g n ( u k m - v m )$. Then

$∬ S T φ g n ( u k m - v m ) ∂ ∂ t ( u k - v ) d x d t + ∬ S T [ D u k m p - 2 D u k m - D v m p - 2 D v m ] [ g n ′ D ( u k m - v m ) φ + g n D φ ] d x d t + k N ( 1 + μ ) ∬ S T [ B ( k - N m u k m ) - r m ( p - 1 ) B ( k - N m r - m v m ) ] [ g n ′ D ( u k m - v m ) φ + g n D φ ] d x d t = 0 .$
(2.32)

Let $φ ( x , t ) = θ ( x k ) η j ( t )$. Where $θ∈ C 0 1 ( R N )$, 0 ≤ θ ≤ 1, θ(x) = 1 when x B1, and $η j ( t ) ∈ C 0 1 ( 0 , T )$, 0 ≤ η j ≤ 1, which satisfies that η j η when j → ∞, and η is the characteristic function of (s1, s2), s1 < s2.

Since u k , v L (RN × (τ, T)), $D u k m$, Dvm Lp (RN × (τ, T)), we have

$∬ S T [ D u k m p - 2 D u k m - D v m p - 2 D v m ] g n ′ θ ( x ) η j ( t ) d x d t ≥ 0 , ∬ S T [ D u k m p - 2 D u k m - D v m p - 2 D v m ] g n D θ x k η j ( t ) d x d t ≤ 1 k ∬ S T [ D u k m p - 1 + D v m p - 1 ] D y θ ( y ) y = x k η j ( t ) d x d t → 0 ,$

as k → ∞.

If we notice that, for any i {1,2, ..., N},

$k N ( 1 + μ ) lim r → 1 b i ( k - N m u k m ) - r m ( p - 1 ) b i ( k - N m r - m v m ) = k N ( 1 + μ ) b i ( k - N m u k m ) - b i ( k - N m v m ) = k N ( 1 + μ ) ∫ k - N m v m k - N m u k m b i ′ ( s ) d s ≤ k N ( 1 + μ ) ∫ k - N m v m k - N m u k m s α d s = k 1 k - N m ( α + 1 ) + N ( 1 + μ ) k 1 α + 1 u k m ( α + 1 ) - v m ( α + 1 ) .$

then it is easy to show that

$k N ( 1 + μ ) lim r → 1 ∬ S T [ B ( k - N m u k m ) - r m ( p - 1 ) B ( k - N m r - m v m ) ] g n ′ φ ( D u k m - D v m ) d x d t = 0$

At the same time,

$lim k → ∞ k N ( 1 + μ ) ∬ S T [ B ( k - N m u k m ) - r m ( p - 1 ) B ( k - N m r - m v m ) ] g n D φ d x d t = k N ( 1 + μ ) lim k → ∞ ∬ S T [ B ( k - N m u k m ) - r m ( p - 1 ) B ( k - N m r - m v m ) ] g n D θ ( x k ) η j ( t ) d x d t ≤ lim k → ∞ k N ( 1 + μ ) - 1 ∬ S T B ( k - N m u k m ) - r m ( p - 1 ) B ( k - N m r - m v m ) D y θ ( y ) y = x k η ( t ) d x d t ≤ lim k → ∞ k N ( 1 + μ ) - 1 - N m ( 1 + α ) ∬ S T u k m ( 1 + α ) + r m ( p - 2 - α ) v m ( 1 + α ) D y θ ( y ) y = x k η ( t ) d x d t = 0 .$

Then, if we let k → ∞, n → ∞ and let r → 1 in (2.32), since μ < α , we have

$lim r → 1 ∬ S T η j ( t ) sg n + ( u k - v ) ∂ ( u k - v ) ∂ t d x d t ≤ 0 ,$

in other words,

$lim r → 1 ∬ S T η j ′ ( t ) ( u k - v ) + d x d t ≥ 0 .$

Let j → ∞. Then

$lim r → 1 ∫ R N ( u k ( x , s 2 ) - v ( x , s 2 ) ) + dx≤ lim r → 1 ∫ R N ( u k ( x , s 1 ) - v ( x , s 1 ) ) + dx.$

Similarly, we have

$lim r → 1 ∫ R N ( v ( x , s 2 ) - u k ( x , s 2 ) ) + d x ≤ lim r → 1 ∫ R N ( v ( x , s 1 ) - u k ( x , s 1 ) ) + d x .$
(2.33)

Let s1 → 0. Then

$u k ≥ lim r → 1 u k r .$

It follows that

$lim r → 1 u k ( x , r m ( p - 1 ) - 1 t ) - u k ( x , t ) ( r m ( p - 1 ) - 1 - 1 ) t ≥ lim r → 1 r - 1 ( 1 - r m ( p - 1 ) - 1 ) t u k ( x , r m ( p - 1 ) - 1 t ) ,$

which implies that

$u k t ≥ - u k [ m ( p - 1 ) - 1 ] t .$
(2.34)

Denote w = tγu k (x, t), $γ = 1 m ( p - 1 ) - 1$. By (2.34), w t ≥ 0. By (2.1),

$∫ τ T ∫ B 2 R t - γ w t ψ R d x d t = - ∫ τ T ∫ B 2 R D u k m p - 2 D u k m ⋅ D ψ R d x d t - k N ( 1 + μ ) ∫ τ T B ( k - N m u k m ) D ψ R d x d t + γ ∫ τ T ∫ B 2 R t - 1 u k ( x ) ψ R d x d t ≤ β τ ∫ τ T ∫ B 2 R u k d x d t + ∫ τ T ∫ B 2 R D u k m p d x d t p - 1 p ∫ τ T ∫ B 2 R D ψ R p d x d t 1 p + k - N m ( α + 1 ) + N ( 1 + μ ) ∫ τ T ∫ B 2 R u k m ( α + 1 ) D ψ R d x d t .$
(2.35)

From (2.15), (2.18) and (2.35), we obtain (2.26).

## 3. Proof of Theorem 1.2

Proof of Theorem 1.2. By Lemmas 2.1-2.3, there exists a subsequence ${ u k j }$ of {u k } and a function v such that on every compact set K S

$u k j →v in C ( K ) ,D u k m ⇀D v m in L 1 oc p ( S T ) , u k t L 1 oc 1 ( S T ) ≤c.$

Similar to what was done in the proof of Theorem 2 in , we can prove v satisfies (1.11) in the sense of distribution.

We now prove v(x, 0) = (x). Let $χ ∈ C 0 1 ( B R )$. Then we have

$∫ R N u k ( x , t ) χ d x - ∫ R N φ k χ d x = - ∫ 0 t ∫ R N D u k m p - 2 D u k m ⋅ D χ d x d s - k N ( 1 + μ ) ∫ 0 t ∫ R N B ( k - N m u k m ) D χ d x d s .$
(3.1)

To estimate $∫ 0 t ∫ R N D u k m p - 2 D u k m ⋅ D χ d x d s$, without loss of the generality, one can assume that u k > 0. By Hölder inequality and Lemma 2.1,

$∫ 0 t ∫ R N D u k m p - 2 D u k m ⋅ D χ d x d t ≤ c ∫ 0 t ∫ B 2 R u k s - m 1 + u k s 2 D u k m p d x d τ p - 1 p . ∫ 0 T ∫ B$