Zhan and Xu Journal of Inequalities and Applications 2012, 2012:120 http://www.journalofinequalitiesandapplications.com/content/2012/1/120
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The asymptotic behavior of the solution of a doubly degenerate parabolic equation with the convection term Huashui Zhan1* and Bifen Xu2 * Correspondence:
[email protected] 1 School of Sciences, Jimei University, Xiamen 361021, People’s Republic of China Full list of author information is available at the end of the article
Abstract The objective of this article is to study the large time asymptotic behavior of the nonnegative weak solution of the following nonlinear parabolic equation ut = div |Dum |p−2 Dum + div B(um )
with initial condition u(x, 0) = u0(x). By using Moser iteration technique, assuming that the uniqueness of the Barenblatt-type solution Ec of the equation ut = div(|Dum| p-2 Dum) is true, then the solution u may satisfy 1 t μ u(x, t) − Ec (x, t) → 0, as t → ∞,
which is uniformly true on the sets
x∈R : N
|x| < at
1
μN
,
a > 0 . Here B(um) =
(b1(um), b2(um), ..., bN(um)) satisfies some growth order conditions, the exponents m and p satisfy m(p - 1) >1. Mathematics Subject Classification 2000: 35K55; 35K65; 35B40. Keywords: degenerate parabolic equation, convection term, weak solution, asymptotic behavior.
1. Introduction The objective of this article is to study the large time asymptotic behavior of the nonnegative weak solution of the nonlinear parabolic equation with the following type ut = div(|Dum |p−2 Dum ) + div(B(um )), u(x, 0) = u0 (x),
on RN ,
in S = RN × (0, ∞),
(1:1) (1:2)
where m(p - 1) >1, N ≥ 1, u0(x) Î L1(RN), D is the spatial gradient operator, and the N ∂(b (um )) i . convection term div(B(um )) = ∂xi i=1
Equation (1.1) appears in a number of different physical situations [1]. For example, in the study of water infiltration through porous media, Darcy’s linear relation
© 2012 Zhan and Xu; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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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 j 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 a ranges from 1 for laminar flow to 2 for completely turbulent flow (cf. [2-4] and references therein). If it is assumed that infiltration takes place in a horizontal column of the medium, according to the continuity equation ∂θ ∂V + = 0, ∂t ∂x
then (1.4) and (1.5) give ∂θ ∂ D(θ )p |θx |p−1 θx = ∂t ∂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. [7]), where u stands for the density, and the pressure is proportional to um-1 (see also [8]). 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 [9] 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; L1 (RN )) ∩ L∞ (RN × (τ , T)), p
um ∈ L1oc (0, T; W 1.p (RN )),
ut ∈ L1 (RN × (τ , T)),
(1:6) ∀τ > 0;
(1:7)
(ii) S
p−2 [uϕt − Dum Dum · Dϕ − B(um ) · Dϕ] dxdt = 0,
∀ϕ ∈ C10 (S);
(1:8)
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(iii) lim t→0
u(x, t) − u0 (x) dx = 0.
(1:9)
RN
If there exist the positive constants k1, a such that B(s) ≤ k1 |s|1+α , B (s) ≤ k1 |s|α , ∀s ∈ R1 = (−∞, +∞),
(1:10)
Chen-Wang [10] 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 [11-17]. 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 p−1 −1 ⎧⎡ p ⎤ ⎫ ⎨ ⎬ m(p − 1) − 1 −1 −1 p−1 m(p − 1) − 1 ⎦ Ec = t μ ⎣b − (Nμ) p−1 |x| t Nμ ⎩ ⎭ mp +
is the Barenblatt-type solution of the Cauchy problem ut = div |Dum |p−2 Dum , in S = RN × (0, ∞), u(x, 0) = cδ(x),
on
(1:11)
RN ,
(1:12)
where μ = m(p − 1) − 1 + Np , c = RN u0 (x)dx , b is a constant b= Ec (x, t)dx , and δ denotes the Dirac mass centered at the origin.
such
that
RN
By using some ideas of [9,14], we have the following Theorem 1.2. Suppose m(p - 1) >1, B satisfies (1.10) with a < p - 1 and m(1 + α) ≥ 1 + μ = m(p − 1) +
p . N
If Ec is a unique solution of (1.11) and (1.12), then
the solution u of (1.1) and (1.2) satisfies 1
t μ |u(x, t) − Ec (x, t)| → 0,
uniformly on the sets
as t → ∞,
x ∈ RN :
1 |x| < at μN , a > 0 , where c = RN u0 dx as
before. Remark 1.3. For m = 1, the uniqueness of solutions of (1.11) and (1.12) is known (see [18]). By assuming that the uniqueness of the Barenblatt-type solution of (1.11) is true, Yang and Zhao [14] had established the similar large time behavior of solution of the Cauchy problem of the following equation
Zhan and Xu Journal of Inequalities and Applications 2012, 2012:120 http://www.journalofinequalitiesandapplications.com/content/2012/1/120
ut = div |Dum |p−2 Dum − uq ,
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in S = RN × (0, ∞),
(1:13)
While Zhan [17] had considered the Cauchy problem of the following equation p−2 p ut = div Dum Dum − Dum 1 − uq ,
in S = RN × (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(u m )). The absorption term -uq in (1.13), or −|Dum |p1 − uq in (1.14), is always less than 0. This fact made us be able to draw it away in many estimates in [14] or [17]. 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 uk = kN u(kx, kNμ t),
k > 0.
It is easy to see that they are the solutions of the problems p−2 ut = div Dum Dum + kN(1+μ) div(B(k−Nm um )),
on RN ,
u(x, 0) = u0k (x),
where μ = m(p − 1) +
in S = RN × (0, ∞),
p N
(2:1)
(2:2)
− 1 as before and u0k(x) = kN u0(kx).
Lemma 2.1 For any s Î (ma, m(p - 1)), the nonnegative solution uk satisfies T 0 BR
T
m 2 us−m k Duk dxdt ≤ c(s, R, |u0 |L1 ), s 2 1 + uk
(2:3)
p
um(p−1)+ N −s dxdt ≤ c(s, R, |u0 |L1 ).
(2:4)
0 BR
¯ , Proof. From Definition 1.1, we are able to deduce that (see [19]): for ∀ϕ ∈ C1 (S)
= 0 when |x| is large enough, for any t Î [0, T], 0 < h < t, RN
t p−2 m Du · Dϕ − kN(1+μ) B k−Nm um Dϕ dxdt = uk ϕ(x, h)dx. uk ϕ(x, t)dx− uk ϕt − Dum k k k h RN
RN
(2:5)
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Let ψR ∈ C ∞ 0 (B2R ),
0 ≤ ψR ≤ 1,
usk p 1+usk ψR
By an approximate procedure, we can choose ϕ =
u k (x,t)
zs p dzψR dx + s 1 + zs
0
RN
|DψR | ≤ cR−1 .
ψR = 1 on BR ,
t h RN
us−m k 1+
2 usk
in (2.5), then
m p p Du ψ dxdτ R k
uk (x,t) s usk m p−2 p−1 z p Du ψ Duk · DψR dxdτ + dzψR dx R k 1 + usk 1 + zs RN 0 t s uk p . dxdτ +kN(1+μ) B(k−Nm um )D ψ k s R 1 + uk N
t ≤ −p h RN
(2:6)
(2:7)
h R
Noticing t usk m p−2 p−1 m Du ψ (x)Du · DψR dxdτ R k k s N 1 + uk h R
⎧ ⎫ ⎪ ⎡ ⎡ ⎤ p ⎤p ⎪ ⎪ ⎪ (s−m)(p−1) p−1 ⎪ ⎪ ⎪ ⎪ p − 1 (s−m) s− ⎪ t ⎪ p p ⎨ ⎢ u ⎬ ⎢ ⎥ ⎥ u k k m p−1 p−1 ⎥ ⎢ ⎢ ⎥ |Dψ | ε⎣ ≤ Du + c(ε) ψ R R k ⎣ ⎦ ⎦ ⎪ dxdτ ⎪ p−1 p−1 ⎪ ⎪ s 2 p s 1−2 p ⎪ 1 + u 1 + u ⎪ ⎪ h RN ⎪ k k ⎪ ⎪ ⎩ ⎭ t =ε h RN
m p p us−m k Duk ψR dxdτ + c(ε) s 2 1 + uk
t N(1+μ)
k
B(k−Nm um k )D
h RN
t = kN(1+μ)
B(k−Nm um k) −
h RN
Dus (1 + us )2
t
m(p−1)+s
uk
(2:8)
|DψR |p dxdτ ,
h RN
usk p dxdτ ψ 1 + usk R pus
p
ψR +
1 + us
p−1
(2:9)
ψR DψR dxdτ .
−Nm(1+α) m 1+α , Since s >am, B(k−Nm um uk k ) ≤ k1 k p s−m+(1+α)mp ψ u R k ≤1 s 2 1+u k
is always true, we have t t −Nm um )us−m B(k−Nm um ) B(k s p p k k k s m − ψR Duk dxdτ ψR Duk dxdτ = 2 s 2 s m 1 + uk 1 + uk N N h R t
h R
≤ε h RN
p m p sψR us−m k Duk dxdτ s 2 1 + uk
t
≤ε h RN
t + c(ε)s h RN
p −Nm m p sψR us−m k uk ) dxdτ , (2:10) B(k s 2 1 + uk
m p sψR us−m k Du dxdτ + c(ε, R)k−Nm(1+α)p , p = p , k 2 s p−1 1 + uk p
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and
u k (x,h)
0
RN
zs p dzψR dx ≤ 1 + zs
u(x, kNμ h)dx.
(2:11)
RN
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
u k (x,t)
sup 0
zs dzdx + 1 + zs
0
RN
T h RN
T
≤c
u(x, kNμ h)dx + c
RN
m p p us−m k Du ψ dxdτ R k 2 1 + usk m(p−1)+s
uk
(2:12)
|DψR |p dxdτ + c.
h RN
Since uk Î L∞ (RN × (h, T)) ∩ L1(ST), T
m(p−1)+s
lim
uk
R→∞ h
|DψR |p dxdτ = 0.
(2:13)
RN
Let h ® 0 in (2.12). Then
u k (x,t)
sup 0
zs dzdx + 1 + zs
sT
0
m p Du dxdτ ≤ c k 2 1 + usk us−m k
u0 dx.
(2:14)
RN
From this inequality, it is clear of that sup
T uk (x, t)dx +
0
0 B2R
m p us−m k Du dxdτ ≤ c(R). k 2 1 + usk
(2:15)
So (2.3) is true. Let m(p − 1) − s p w = u1 .
u1 = max uk (x, t), 1 ,
By Sobolev’s imbedding inequality (see [19]), for ξ ∈ C10 (B2R ) , ξ ≥ 0, we have ⎛ ⎝
⎞1
r
⎡
ξ w dx⎠ ≤ c⎣ p
B2R
r
⎤s⎡ p
D(ξ w)p dx⎦ ⎣
B2R
⎤ (1−θ )[m(p−1)−s]
p
⎦ m(p−1)−s dx p
w
,
B2R
where θ=
m(p − 1) − s 1 − p r
1 1 m(p − 1) − s − + N p p
−1 ,
r=
p[m(p − 1) + Np − s] . m(p − 1) − s
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It follows that ⎛ p D(ξ w) dxdt sup ⎝
ξ p wr dxdt ≤ c ST
t∈(0,T)
ST
⎞ (r−p)[m(p−1)−s] p
dx⎠ wm(p−1)−s p
,
(2:16)
B2R
where we denote ST = RN × (0, T). Since |Dw|p ≤ c
us−m k 1+
2 usk
m p Du a.e. on {uk ≥ 1} and |Dw| = 0 on {uk ≤ 1}, k
we have
|D(ξ w)|p dxdt ≤ c ST
⎡
≤ c⎣
(ξ p |Dw|p + wp |Dξ |p )dxdt ST
m(p−1)−s |Dξ |p u1 dxdt
T + 0 B2R
ST
us−m k (1 + usk )2
⎤
(2:17)
p ⎦ |Dum k | dxdt .
Hence, by (2.16), (2.(17) and (2.15), we get
p m(p−1)+ N −s ξ p u1 dxdt
⎛ ≤ c(s, R, |u0 |L1 ) ⎝1 +
ST
⎞
dxdt ⎠ .
ST
Let ξ = ψRb , ψR. be the function satisfying (2.6) and b =
m(p−1)−s
|Dξ |p u1
p pb m(p−1)+ N −s ψR u1 dxdt
⎛ ≤ c(s, R, |u0 |L1 )⎝1 +
sT
p N[m(p−1)+ N −s] . p
⎞
Then
m(p − 1) − s − 1) − s + Np
p m(p pb m(p−1)+ N −s ψR u1 dxdt ⎠
,
sT
by Moser iteration technique, the above inequality implies (2.4) is true. Let Qr = Br (x0) × (t0 - rp, t0) with t0 >(2r)p and uk1 = max{uk, 1}. Also by Moser iteration technique, we have Lemma 2.2 The nonnegative solution uk satisfies ⎛ ⎞1/s1 ⎜ ⎟ m(p−1)−1+s1 sup uk ≤ c(ρ, s1 )⎝ uk1 dxdt ⎠ , Qρ
(2:18)
Q2ρ
where c(r, s 1 ) depends on r and s 1 , and s 1 can be any number satisfying 0 < s1 < 1 +
p . N
¯ , = 0 when |x| is large enough, we have Proof. For ∀ϕ ∈ C1 (S)
T uk (x, t)ϕdx −
RN
p−2 m Du · Dϕ − kN(1+μ) B(k−Nm um )Dϕ]dxdt [uk ϕt − Dum k k k
0 RN
(2:19)
=
u0k (x)ϕ(x, 0)dx. RN
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Let ξ be the cut function on Qr, i.e. 0 ≤ ξ ≤ 1, ξ |Qρ = 1, ξ |RN \Q2ρ = 0. −1 We choose the testing function in (2.19) as ϕ = ξ p u2γ , where γ > k
1 2
is a constant.
Then
1 2γ
2γ
ξ p uk (x, t)dx +
2γ − 1 m
ξ
=p
2γ −1−m
p Dum k dxds
ξ p uk 0 B2ρ
B2ρ
t
t
p−1
p−1 2γ −1 dxds |Dξ | uk Dum k
t
p + 2γ
2γ
ξ p−1 |ξt | uk dxds
0 B2ρ
0 B2ρ
t +pkN(1+μ)
2γ −1
ξ p−1 B(k−Nm um k )uk
(2:20)
Dξ dxdt
0 B2ρ N(1+μ) 2γ
+k
−1 m
t
2γ −m−1
ξ p uk
m B(k−Nm um k )Duk dxds.
0 B2ρ
Using Young inequality, by (1.10), p−1 p−1 2γ −1 2γ −1−m p−1 |Dξ | um Dum Dum ξ p−1 |Dξ | uk = uk ξ k k k 2γ −1−m p + c(ε)ump |Dξ |p ), ≤ uk (εξ p Dum k k p 2γ −m−1 −Nm m m −Nm(1+α) p mα+2γ −1 B(k uk )Duk ≤ k Dum ξ uk ξ uk k 2γ −1−m mα+m m ≤ k−Nm(1+α) ξ p uk uk Duk p 2γ −1−m m(1+α)p ≤ k−Nm(1+α) ξ p uk + ε Dum (c(ε)uk k ), from (2.20), we have 1 2γ
2γ ξ p uk (x,
t p 2γ − 1 2γ − 1 2γ −1−m ∇um t)dx+ ξ p uk −ε 1+ k dxds m m
B2ρ
0 B2ρ
t ≤c
2γ −1+m(p−1)
uk
|Dξ |p dxds +
0 B2ρ
+c(ε)k−Nm(1+α)+N(1+μ)
p 2γ
t
2γ
ξ p−1 |ξt | uk dxds
(2:21)
0 B2ρ
t
|uk |2γ −1−m+m(1+α)p dxds.
0 B2ρ
By the fact of that ⎛ p ⎞ 2γ − 1 + m(p − 1) p 2γ −1+m(p−1) 2γ − 1 − m 2γ − 1 + m(p − 1) ⎜ ⎟ p m p p Dξ + Duk ξ uk D ⎝ξ u ⎠ = uk mp p 2γ −1+m(p−1) p m p 2γ −1−m ≤ c|Dξ | uk + cξ Duk uk ,
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from (2.21), we have sup
t0 −2ρ p
t
⎛ ⎞ 2γ − 1 + m(p − 1) p ⎜ ⎟ p 2γ D ⎜ξ u ⎟ dxds ξ p uk dxds+ ⎝ k ⎠ Q2ρ
2γ −1+m(p−1)
≤c
uk
t |Dξ |p dxds + c
0 B2ρ
(2:22)
2γ
ξ p−1 |ξt | uk dxds 0 B2ρ
+ c(ε)k−Nm(1+α)+N(1+μ)
2γ − 1 m
t
2γ −1−m+m(1+α)p
|uk |
dxds.
0 B2ρ
Let 2γ − 1 + m(p − 1) β = max 1, , γ
and 2γ −1+m(p−1) p
w = ξ β uk
.
By the embedding theorem, from (2.22), we have wh dxdt ≤ c
⎧ ⎪ ⎨
sup ⎪ ⎩t0 −2ρ p
B2ρ
Q2ρ
⎫ 2γ p ⎪ ⎬ 2γ − 1 + m(p − 1) w dx ⎪ ⎭
2γ − 1 + m(p − 1) (1−δ)h 2γ p
. t0 −(2ρ p )
⎛ ⎜ ⎝
⎞ δh p ⎟ |Dw|p dx⎠ dx,
(2:23)
B2ρ
where δ=
1 2γ − 1 + m(p − 1) 1 1 2γ − 1 + m(p − 1) −1 . . − − + 2γ p h N p 2γ p
In particular, we choose h = p[1 +
2γ p ], N(2γ − 1 + m(p − 1))
then from (2.23), we have
2γ p 2γ −1+m(p−1)+ N
ξ βh uk
Q2ρ
⎛ ⎜ ≤ c1 ⎝
⎞p
sup
t0 −2ρ p
ξ
2γ pβ 2γ −1+m(p−1) u2γ dx⎟ ⎠ k
N
dxdt
⎛ ⎞ 2γ − 1 + m(p − 1) p ⎜ β ⎟ p D ⎜ξ u ⎟ dxdt . k ⎝ ⎠ Q2ρ
⎛ ⎫1+ Np ⎞ 2γ − 1 + m(p − 1) p ⎪ ⎪ ⎬ 2γ pβ ⎜ β ⎟ p 2γ 2γ −1+m(p−1) ⎜ ⎟ sup D ⎝ξ uk ≤ c2 ξ uk dx + dxdt . ⎠ ⎪ ⎪ p ⎪ ⎪ ⎩t0 −2ρ
(2:24)
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Now, for τ ∈ [ 12 , 1] , we denote that 1−τ , l = 1, 2, . . . , ρ l = 2ρ τ + 2l
and choose the cut functions ξl(x, t) of Qrl, such that on Qr(l+1), ξl = 1. Denote p , 2γ = K l . N
K =1+
and let u1k = max{1, uk }.
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
m(p−1)−1+K l+1 uk dxdt
Qρ(l+1)
≤
m(p−1)−1+K l+1
≤
u1k
m(p−1)−1+K l+1
uk ⎧ ⎪ ⎨
dxdt
Qρ(l+1)
≤
dxdt + mesQρ(l+1)
Qρ(l+1)
ccl1 p ⎪ ⎩ [(1 − τ )ρ]
m(p−1)−1+K l
u1k
⎫K ⎪ ⎬ dxdt
Qρl
⎪ ⎭
.
Using Moser iteration technique, we have ' sup u1k ≤ 2τρ
(
1 (1 − τ )ρ
)N+p
Q2ρ
m(p−1)−1+K
u1k
dxdt
*1 K
.
Then, we have
sup u1k ≤ 2τρ
K−r (sup u1k ) K . Q2ρ
' (
1 (1 − τ )ρ
)N+p
Q2ρ
m(p−1)−1+r
u1k
dxdt
*1 K
.
By Schwarz inequality,
sup u1k 2τρ
' *1 m(p−1)−1+r r 1 1 ≤ sup u1k + c(r) ( u1k dxdt . )N+p 2 2ρ (1 − τ )ρ Q2ρ
By the Lemma 3.1 in [19], for any τ ∈ [ 12 , 1) , we have + sup u1k ≤ c(r, p) 2τρ
Q2ρ
m(p−1)−1+r
u1k
dxdt
,1 r
,
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and from this inequality, we get the conclusion of the lemma. Lemma 2.3 The nonnegative solution uk satisfies T
m p Du dxdt ≤ c(τ , R). k
(2:25)
|ukt |p dxdt ≤ c(τ , R).
(2:26)
τ BR
T τ BR
Proof. By Lemmas 2.1 and 2.2, {uk} are uniformly bounded on every compact set K ⊂ ST . Let ψR be a function satisfying (2.6) and ξ ∈ C10 (0, T) with 0 ≤ ξ ≤ 1, ξ = 1 if t Î (τ, T). We choose η = ψRp ξ um in (2.5) to obtain k m p p 1 p Du ψ ξ dxdt um+1 (x, T)ψ dx + R R k k m+1 ST RN m p−2 m 1 p Du · DψR ψ p−1 ξ dxdt um+1 ξ ψ dxdt − p um = R R k k Duk k m+1 ST ST p p−1 m B(k−Nm um DψR um +kN(1+μ) k )ξ (pψ k + ψR Duk )dxdt.
(2:27)
ST
Noticing ≤ε
m p−1 p−1 |DψR | ψR ξ dxdt um k Duk
ST
m p p Du ψ ξ dxdt + c(ε) k
ST
pm
uk |DψR |p ξ dxdt,
R
ST
T c −Nm(1+α) m(1+α) −Nm m p−1 m ≤ B(k u )ξ ψ Dψ u dxdt ξ uk dxdτ , k R k k R 0 Br ST m p p p −mN m m ≤ c(ε) B(k−Nm um )p ξ ψ p dxdτ + ε Du ψ ξ dxdt B(k u )ξ ψ Du )dxdt R R R k k k k ST ST ST m p p mp (1+α) p Du ψ ξ dxdt, uk ξ ψR dxdτ + ε ≤ c(ε)k−Nm(1+α) R k ST
ST
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) = ukr (x, t) = ruk (x, r m(p−1)−1 t),
r ∈ (0, 1).
(2:28)
Then vt (x, t) = div(|Dvm |p−2 Dvm ) + r m(p−1) kN(1+μ) div(B(k−Nm r −m vm )),
(2:29)
v(x, 0) = ruk (x, 0).
(2:30)
Zhan and Xu Journal of Inequalities and Applications 2012, 2012:120 http://www.journalofinequalitiesandapplications.com/content/2012/1/120
Page 12 of 16
By (2.1) and (2.29), for any ϕ ∈ C10 (ST ) , we have
m p−2 m m p−2 m ∂ Du Du − Dv Dv Dϕdxdt (uk − v)dxdt + k k ∂t ST ST m(p−1) [B(k−Nm um B(k−Nm r −m vm )]Dϕdxdt = 0. +kN(1+μ) k)−r ϕ
(2:31)
ST
Let gn(s) = 1 when s > 1n ; gn(s) = ns when 0 ≤ s ≤ 1n ; gn(s) = 0 when s <0, and let m in (2.31) be substituted by ϕgn (um k − v ) . Then
m ϕgn (um k −v ) ST
∂ (uk − v)dxdt + ∂t
[B(k−Nm um k)
N(1+μ)
+k
−r
m(p−1)
p−2 m m p−2 m Du − Dv Dv ][g D(um − vm )ϕ + gn Dϕ]dxdt [Dum n k k k
ST m B(k−Nm r −m vm )][gn D(um k − v )ϕ + gn Dϕ]dxdt = 0.
(2:32)
ST
Let ϕ(x, t) = θ ( xk )ηj (t) . Where θ ∈ C10 (RN ) , 0 ≤ θ ≤ 1, θ(x) = 1 when x Î B1, and ηj (t) ∈ C10 (0, T) , 0 ≤ hj ≤ 1, which satisfies that hj ® h when j ® ∞, and h is the
characteristic function of (s1, s2), s1 < s2. m p N Since uk, v Î L∞ (RN × (τ, T)), Dum k , Dv Î L (R × (τ, T)), we have
p−2 m m p−2 m Du − Dv Dv ]g θ (x)ηj (t)dxdt ≥ 0, [Dum n k k
ST
x m p−2 m m p−2 m ηj (t)dxdt [ Duk Duk − Dv Dv ]gn Dθ k ST p−1 m p−1 1 ≤ [Dum + Dv ]Dy θ (y) x ηj (t)dxdt → 0, k k y= k ST
as k ® ∞. If we notice that, for any i Î{1,2, ..., N}, m(p−1) −Nm −m m kN(1+μ) lim bi (k−Nm um ) − r b (k r v ) i k r→1 −Nm m = kN(1+μ) bi (k−Nm um v ) k ) − bi (k −Nm m k uk N(1+μ) =k bi (s)ds k−Nm vm m k−Nm uk
sα ds = k1 k−Nm(α+1)+N(1+μ)
≤k
N(1+μ) k−Nm vm
k1 m(α+1) − vm(α+1) . uk α+1
then it is easy to show that m(p−1) m kN(1+μ) lim [B(k−Nm um B(k−Nm r −m vm )]gn ϕ(Dum k)−r k − Dv )dxdt = 0 r→1
ST
Zhan and Xu Journal of Inequalities and Applications 2012, 2012:120 http://www.journalofinequalitiesandapplications.com/content/2012/1/120
Page 13 of 16
At the same time,
N(1+μ)
lim k
k→∞
−Nm m m(p−1) −Nm −m m [B(k u ) − r B(k r v )]g Dϕdxdt n k
= kN(1+μ) lim k→∞ ST ≤ lim kN(1+μ)−1 k→∞
ST
x m(p−1) [B(k−Nm um B(k−Nm r −m vm )]gn Dθ ( )ηj (t)dxdt k)−r k −Nm m uk ) − r m(p−1) B(k−Nm r −m vm ) Dy θ (y)y= x η(t)dxdt B(k
ST
≤ lim kN(1+μ)−1−Nm(1+α) k→∞
k
m(1+α) m(p−2−α) m(1+α) +r v uk Dy θ (y)y= x η(t)dxdt = 0. k
ST
Then, if we let k ® ∞, n ® ∞ and let r ® 1 in (2.32), since μ < a , we have ηj (t)sgn+ (uk − v)
lim r→1
ST
∂(uk − v) dxdt ≤ 0, ∂t
in other words, lim r→1
ηj (t)(uk − v)+ dxdt ≥ 0.
ST
Let j ® ∞. Then lim r→1
(uk (x, s2 ) − v(x, s2 ))+ dx ≤ lim r→1
RN
(uk (x, s1 ) − v(x, s1 ))+ dx.
RN
Similarly, we have lim
r→1 RN
(v(x, s2 ) − uk (x, s2 ))+ dx ≤ lim
r→1 RN
(v(x, s1 ) − uk (x, s1 ))+ dx.
(2:33)
Let s1 ® 0. Then uk ≥ lim ukr . r→1
It follows that uk (x, r m(p−1)−1 t) − uk (x, t) r−1 ≥ lim uk (x, r m(p−1)−1 t), r→1 r→1 (1 − r m(p−1)−1 )t (r m(p−1)−1 − 1)t
lim
which implies that ukt ≥ −
uk . [m(p − 1) − 1]t
(2:34)
Zhan and Xu Journal of Inequalities and Applications 2012, 2012:120 http://www.journalofinequalitiesandapplications.com/content/2012/1/120
Denote w = tguk(x, t), γ = T t
−γ
1 m(p−1)−1 .
By (2.34), wt ≥ 0. By (2.1),
T wt ψR dxdt = −
τ B2R
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m p−2 m Du Du · DψR dxdt k k
τ B2R
T −k
N(1+μ)
B(k−Nm um k )DψR dxdt
T +γ
τ
t−1 uk (x)ψR dxdt
τ B2R
p−1 1 ⎛ T ⎛ T ⎞ ⎞ T p p m p β p Du dxdt ⎠ ⎝ ⎠ |Dψ | ≤ uk dxdt + ⎝ dxdt R k τ τ B2R
τ B2R
(2:35)
τ B2R
+k−Nm(α+1)+N(1+μ)
T
m(α+1)
uk
|DψR |dxdt.
τ B2R
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 {ukj } of {uk} and a function v such that on every compact set K ⊂ S p
m ukj → v in C(K), Dum k Dv in L1oc (ST ), |ukt |L11oc (ST ) ≤ c.
Similar to what was done in the proof of Theorem 2 in [9], we can prove v satisfies (1.11) in the sense of distribution. We now prove v(x, 0) = cδ(x). Let χ ∈ C10 (BR ) . Then we have uk (x, t)χ dx− ϕk χ dx RN
t
=−
RN
m p−2 m Du Du · Dχ dxds − kN(1+μ) k k
0 RN
(3:1) B(k−Nm um k )Dχ dxds.
0 RN
t
To estimate
t
0
m p−2 m Du Du · Dχ dxds k
k
, without loss of the generality, one can
RN
assume that uk >0. By Hölder inequality and Lemma 2.1, t m p−2 m Du Du · Dχ dxdt k k N 0 R
⎡ ≤ c⎣
t 0 B2R
⎤
m p us−m k Du dxdτ ⎦ k 2 1 + usk
1 p−1 ⎡ T ⎤ p p 2(p−1) (p−1)(m−s) .⎣ 1 + usk uk dxdτ ⎦ 0 BR
⎤1 ⎡ t p (p−1)(m−s) (p−1)(s+m) + uk1 dxdτ ⎦ ≤ c⎣ uk1 0 BR
⎡ ≤ c⎣
t 0 BR
⎤ 1 −d p p m(p−1)+ N −s uk1 dxdτ ⎦ td ,
(3:2)
Zhan and Xu Journal of Inequalities and Applications 2012, 2012:120 http://www.journalofinequalitiesandapplications.com/content/2012/1/120
where s ∈ (0,
1 N ),
d=
1−Ns m(p−1)N+p−sN
Page 15 of 16
< P1 , uk1 = max(uk, 1).
Hence from (3.1), we get t m(α+1) uk χ dx − u0k χ dx ≤ ctd + k−Nm(α+1)+N(1+μ) uk |Dχ |dτ ds. N N N R
(3:3)
0 R
R
Letting k ® ∞, t ® 0 in turn, we obtain lim vχ dx = χ (0) ϕdx. t→0R
RN
RN
Thus
v(x, 0) = cδ(x), c =
ϕdx,
RN
v(x, t) is a solution of (1.11) and (1.12). By the assumption on uniqueness of solution, we have v(x, t) = Ec(x, t) and the entire sequence {uk} converges to Ec as k ® ∞. Set t = 1. Then uk (x, 1) = kN u(kx, kNμ ) → Ec (x, 1)
uniformly on every compact subset of RN. Thus by writing kx = k’, kNμ = t’ , and dropping the prime again, we see that 1
1
1
t μ u(x, t) → Ec (xt Nμ , 1) = t μ Ec (x, t)
uniformly on the sets
x∈
RN
: |x| ≤
1 at Nμ
, a > 0. Thus Theorem 1.2 is true.
Acknowledgements The article is supported by NSF (no. 2009J01009) of Fujian Province, Pan Jinlong’s SF of Jimei University in China. Author details 1 School of Sciences, Jimei University, Xiamen 361021, People’s Republic of China 2College of Teacher Education, Jimei University, Xiamen 361021, People’s Republic of China Authors’ contributions All authors read and approved the final manuscript. Competing interests The authors declare that they have no competing interests. Received: 13 January 2012 Accepted: 29 May 2012 Published: 29 May 2012 References 1. Esteban, JR, Vázquez, JL: Homogeneous diffusion in R with power-like nonlinear diffusivity. Arch Ration Mech Anal. 103, 39–88 (1988). doi:10.1007/BF00292920 2. Ahmed, N, Sunada, DK: Nonlinear flows in porous media. J Hydraul Div Proc Soc Civil Eng. 95, 1847–1857 (1969) 3. Kristlanovitch, SA: Motion of ground water which does not con form to Darcy’s Law. Prils Mat Mech (Russian). 4, 33–52 (1940) 4. Volker, RE: Nonlinear flow in porous media by finite elements. J Hydraul Div Proc Soc Civil Eng. 95, 2093–2114 (1969) 5. Muskat, M: The Flow of Homogeneous Fluids Through Porous Media. McGraw Hill, New York (1937) 6. Gilding, BG, Peletier, LA: The Cauchy problem for an equation in the theory of infiltration. Arch Ration Mech Anal. 61, 127–140 (1976) 7. Leibenson, LS: General problem of the movement of a compressible fluid in a porous media. Izv Akad Nauk SSSR Geogr Geophys (Russian). 9, 7–10 (1945)
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doi:10.1186/1029-242X-2012-120 Cite this article as: Zhan and Xu: The asymptotic behavior of the solution of a doubly degenerate parabolic equation with the convection term. Journal of Inequalities and Applications 2012 2012:120.
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