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On the Cauchy problem for the evolution p -Laplacian equations with gradient term and source and measures as initial data
Nonlinear Analysis 72 (2010) 3396–3411
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On the Cauchy problem for the evolution p-Laplacian equations with gradient term and source and measures as initial dataI Haifeng Shang, Fengquan Li ∗ Department of Applied Mathematics, Dalian University of Technology, Dalian, 116024, China
article
abstract
info
Article history: Received 14 November 2008 Accepted 10 December 2009
In this paper we study the Cauchy problem for the evolution p-Laplacian equations with gradient term and source on the assumption that measures as initial conditions. © 2009 Elsevier Ltd. All rights reserved.
MSC: 35K15 35K55 35K65 Keywords: Cauchy problem p-Laplacian equation Gradient term Source Measures as initial data
1. Introduction and statement of the main results Here we will consider the following Cauchy problem ut − div(|Du|p−2 Du) = λ0 uq + λ|Du|l u(x, 0) = µ
in ST = RN × (0, T ),
on R , N
(1.1) (1.2)
where p ≥ 2, q ≥ 1, 1 ≤ l < p, λ ∈ R, λ0 ≥ 0, T > 0 and µ is a nonnegative Radon measure. In the case of λ0 = 0 and λ = 0, Di Benedetto and Herrero proved in [1] that for every Radon measure µ satisfying some polynomial growth at infinity there is a solution. Also they showed that every nonnegative solution u has a unique initial trace µ. Concerning the case λ = 0 and λ0 > 0, the Cauchy problem and initial trace problem were studied in [2]. For the subcritical case q < p − 1 + p/N, the existence of solution can be obtained for any Radon measure. But, for the supercritical case q ≥ p − 1 + p/N, to obtain the existence result, some higher integrability for the initial data need to be imposed. They assume
Z
µh (y)dy < ∞,
sup x∈RN
|x−y|<1
h>
N p
(q − p + 1).
I Project supported by NSFC (No: 10401009) and NCET of China (No: 060275).
∗
Corresponding author. E-mail addresses: [email protected] (H. Shang), [email protected] (F. Li).
0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.12.023
(1.3)
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3397
For the case p = 2, the equation has been studied by many authors (see [3–9]). This equation was introduced in [5] to investigate the effect of the damping term |Du|l on the existence or nonexistence of a global solution. Also, it has been proposed as a model in population dynamics (see [7]). For the case p > 2 and 1 ≤ l ≤ p − 1, the existence of solutions was studied in [10]. They proved the existence of solutions with initial data measures for the subcritical case. Moreover, in the supercritical case, they need the regularity condition (1.3). From the statement above, we see that for the supercritical case q ≥ p − 1 + p/N, to ensure the existence of solution, some regularity condition need to be imposed on the initial data µ. In contrast, here we are able to overcome this difficulty, essentially by choosing the correct geometry for the domains, and obtained existence for measures as initial data, even for the supercritical case. For the porous medium equation, the Cauchy problem and initial trace problem with strongly nonlinear sources were studied in [11]. The existence of solutions of the Cauchy problem with gradient term has been studied by [12] under optimal assumption on initial data. On the other hand, the Cauchy problem and initial traces of evolution p-Laplacian equation with absorption term was studied in [13]. In this paper, we study the problem (1.1)–(1.2) motivated by the ideas in [12,11,2,10] and give the existence of solutions under optimal assumptions on the initial data µ for Eq. (1.1). Definition 1.1. A nonnegative measurable function u(x, t ) defined in ST is called a weak solution of (1.1)–(1.2), if for every bounded open set Ω , with smooth boundary ∂ Ω , u ∈ Cloc (0, T ; L1 (Ω )) ∩ Lloc (0, T ; W 1,p (Ω )) ∩ L∞ loc (ST ), p
and
Z Ω
u(x, t )ϕ(x, t )dx +
Z tZ Ω
s
[−uϕτ + |Du|p−2 DuDϕ]dxdτ =
Z Ω
u(x, s)ϕ(x, s)dx + λ0
+λ
s
Z tZ s
1, ∞
for all 0 < s < t < T and all ϕ ∈ Wloc
Z lim
t →0
RN
u(x, t )η(x)dx =
Z RN
Z tZ
Ω
Ω
uq ϕ dxdτ
|Du|l ϕ dxdτ ,
(1.4)
(0, T ; L∞ (Ω )) ∩ Lploc (0, T ; W01,p (Ω )). Moreover
ηdµ,
∀η ∈ C01 (RN ).
(1.5)
Weak subsolutions (resp. supersolutions) are defined in the same way except that the = in (1.4) is replaced by ≤ (resp. ≥) and ϕ is taken to be nonnegative. We introduce some notations as in [12]. Let µ be any nonnegative Radon measure in RN , and u ∈ L∞ loc (ST ), u ≥ 0. Suppose that 0 ≤ θ ≤ N is given. Set
Z [µ] = sup sup ρ θ − x∈RN 0<ρ<1
Bρ (x)
dµ,
Z [u]t = sup sup sup ρ θ − 0<τ
Bρ (x)
u(y, τ )dy,
0 < t < T,
(1.6)
where
Z Z 1 − dµ = dµ, |E | E E
|E | is the Lebesgue measure of E .
We use γ (a1 , a2 , . . . , an ) to denote positive constants depending only on specified quantities a1 , a2 , . . . , an . First we state our main existence results as follows. Theorem 1.1 (The Case p − 1 ≤ l < p). Let [µ] be finite, θ (q − 1) < p + θ (p − 2) and θ (l − p + 1) < p − l. Then there exists a solution to (1.1)–(1.2) defined in RN × (0, T0 ), where T0 = T0 ([µ], N , p, q, |λ|, l, θ , λ0 ), such that for ∀ 0 < t < T0 , we have
[u]t ≤ γ [µ], ku(·, t )k∞,RN ≤ γ
(1.7) θ − t p+θ(p−2)
here γ = γ (N , p, q, l, |λ|, λ0 , θ ).
[µ],
(1.8)
3398
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
Theorem 1.2 (The Case 1 ≤ l < p − 1). Let [µ] be finite, θ (q − 1) < p + θ (p − 2). Then there exists a solution to (1.1)–(1.2) defined in RN × (0, T00 ), where T00 = T00 ([µ], N , p, q, |λ|, l, θ , λ0 ), such that for ∀ 0 < t < T00 , we have
[u]t ≤ γ ([µ] + 1), ku(·, t )k∞,RN ≤ γ t
(1.9)
− p+θ(θp−2)
([µ] + 1),
(1.10)
here γ = γ (N , p, q, l, |λ|, λ0 , θ ). Remark 1.1. The dependence of T0 and T00 on the quantities specified in the statement of Theorems 1.1 and 1.2 can be made explicit. We refer to the proofs of Lemmas 2.5 and 2.6. Remark 1.2. In [10], they only considered the case of p > 2, 1 ≤ l ≤ p − 1 and proved the existence of solutions with initial data measures for 1 ≤ q < p − 1 + p/N. In this paper, we extend their results to the case of 1 ≤ q < p − 1 + p/θ for initial data measures. Moreover, we obtain the result in the case of p − 1 < l < p. Next we prove that the critical threshold for θ in the gradient term made explicit in Theorem 1.1 is actually optimal for the existence of solutions in the class considered here. Theorem 1.3. Let p − 1 < l < p, λ > 0, u be a nonnegative solution to (1.1)–(1.2), in RN × (0, T ) such that [u]t < ∞, 0 < t < T , and
µ(Bρ (x0 )) ≥ ρ N −θ ,
0<ρ<
where > 0, x0 ∈ RN are given. Then θ (l − p + 1) ≤ p − l. Remark 1.3. We remark here that the positivity of the right-hand side of (1.1) is exploited in the proof of Theorem 1.3. Remark 1.4. If u is a weak solution of (1.1) and also u ∈ Cloc (0, T ; L2loc (RN )), then by the results in [14] and [15], u is actually locally Hölder continuous in ST . For the case λ < 0, this result was given in the Remark 1.1 in [16]. Remark 1.5. We note that our methods also yield existence of solutions of variable sign, approaching a signed measure µ as t tends to zero, because the estimates below hold separately for the positive and negative parts of solutions. This paper is organized as follows: In Section 2, a priori estimates will be given. In Section 3, we will finish the proof of Theorems 1.1 and 1.2. In Section 4, Theorem 1.3 will be proved. 2. A priori estimates In this section, we will prove some estimates for solutions of (1.1). ∗ ∗ For any u ∈ L∞ loc (ST ), u ≥ 0, 0 < T ≤ T , we define, following [12]
huit = sup sup
sup
0<τ
Z ρ θ − u(y, τ )dy,
1
R(t ) = Γ t p+θ(p−2) ,
(2.1)
Bρ
for all 0 < t < T ∗ , where Γ is a positive constant which can be chosen a priori dependent of N , p, q, l, [µ], λ0 , λ, and which will be specified later. We also assume that T ∗ is chosen so that R(T ∗ ) ≤ 1. Note that the last assumption is meaningful because Γ is independent of T ∗ . The connection between [u]t and huit will be commented upon in Remarks 2.1 and 2.2. The following sup-estimate will play an important role in proving the existence results. Lemma 2.1 (The Case p − 1 ≤ l < p). Let u be a nonnegative continuous weak subsolution of (1.1) in ST ∗ . Assume also that a time 0 < T 0 < T ∗ is given such that θ(p−2)
p−2
l
q −1
−1
p−l Γ −p t p+θ(p−2) ku(·, t )k∞,RN + t ku(·, t )k∞,RN + t ku(·, t )k∞, ≤ 1, RN
∀0 < t < T 0 .
(2.2)
Then
ku(·, t )k∞,RN ≤ γ t
− p+θ(θp−2)
Γ
p
κ (N −θ)
p
huitκ ,
∀0 < t < T 0 ,
where γ = γ (N , p, q, l, |λ|, θ , λ0 ), κ = N (p − 2) + p.
(2.3)
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3399
Proof. This proof will be divided into two steps. Step 1: Let ρ > 0, σ ∈ 0,
ρ
ρn =
σ
+
2n+1
2 Bn = Bρn (x0 ),
1 2
, and x0 ∈ RN be fixed, let k > 0 to be chosen. For n = 0, 1, 2, . . . , set
σ p
k t, kn = k − n+1 , 2n+1 2 Qn = Bn × (tn , t ), 0 < tn < t ≤ T 0 .
ρ,
tn =
t
2
−
Let ζn (x, τ ) be a smooth cut-off function in Qn with 0 ≤ ζn (x, τ ) ≤ 1, such that
ζn ≡ 1 in Qn+1 ,
0≤
∂ζn 2(n+2)p ≤γ , ∂τ σ pt
|Dζn | ≤ γ
2n+2
σρ
.
p
Taking ϕ = (u − kn+1 )+ ζn as a test function in (1.4), we get
Z
1 2
(u − kn+1 )2+ (x, t 0 )ζnp dx + Bn
Z
t0
Z
tn
Z t0 Z
t0
Z
Bn
Z
tn
Z t0 Z
(u − kn+1 )2+ ζnp−1 ζnτ dxdτ + λ0
=p tn
|D(u − kn+1 )+ |p ζnp dxdτ + p
Bn
tn
ζnp−1 (u − kn+1 )+ |Du|p−2 Du · Dζn dxdτ Bn
Z t0 Z
uq (u − kn+1 )+ ζnp dxdτ + λ Bn
tn
|Du|l (u − kn+1 )+ ζnp dxdτ , (2.4) Bn
where tn < t 0 < t. By Young’s inequality, we obtain
Z 0 Z Z 0Z t 1 t p−1 p−2 p ζ (u − kn+1 )+ |Du| DuDζn dxdτ ≤ |D(u − kn+1 )+ |p ζnp dxdτ tn Bn n 4 tn Bn Z t0 Z +γ (u − kn+1 )p+ |Dζn |p dxdτ . tn
(2.5)
Bn
Moreover t0
Z
Z
|Du| (u − l
|λ| tn
) ζ
kn+1 + np dxd
τ ≤
4
Bn
Z
tn
4
τ +γ
Z
|D(u − kn+1 )+ |p ζnp dxdτ + γ
Z
|D(u − kn+1 )+ | ζ p
p n dxd
Bn
t0
Z
1
≤
t0
Z
1
tn
Z
t0
tn t0
p
Z
(u − kn )+p−l dxdτ Bn
Z
p
u p−l
Bn
tn
−2
(u − kn )2+ dxdτ .
(2.6)
Bn
If u > 2kn , then
(u − kn )2+ ≥
u 2
(u − kn+1 )+ .
If kn+1 ≤ u ≤ 2kn
(u − kn )2+ ≥ (u − kn )+ (kn+1 − kn ) ≥ 2−n−3 u(u − kn+1 )+ . Hence
λ0
t0
Z
Z
tn
uq (u − kn+1 )+ ζnp dxdτ ≤ 2n γ
Z
t0 tn
Bn
Z
uq−1 (u − kn )2+ dxdτ .
(2.7)
Bn
Substituting (2.5)–(2.7) into (2.4), we obtain
Z
(u − kn+1 )2+ ζnp (x, τ )dx +
sup
tn <τ
ZZ
Bn
|D((u − kn+1 )+ ζn )|p dxdτ ≤ γ
Qn
2np
σ pt
(1 + M )
ZZ
(u − kn )2+ dxdτ ,
(2.8)
Qn
where
M = sup
0<τ
l −1 −2 q −1 p−l τ ρ −p ku(·, τ )kp∞, + τ k u (·, τ )k + τ k u (·, τ )k ∞,Bρ (x0 ) . Bρ (x0 ) ∞,Bρ (x0 )
By Gagliado–Nirenberg inequality [17, p. 64],
ZZ
ζ (u − kn+1 )+ dxdτ ≤ γ d n
b
ZZ
Qn
where b = p +
Qn 2p N
and d is large enough.
|D((u − kn+1 )+ ζn )| dxdτ p
Z
(u −
sup
tn <τ
Bn
) ζ (x, τ )dx 2
kn+1 + np
Np
,
(2.9)
3400
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
Set An = {(x, t ) ∈ Qn−1 : u(x, t ) ≥ kn }, and observe that
ZZ
(u − kn )2+ dxdτ ≥ γ 2−2n k2 |An+1 |.
(2.10)
Qn
Combining (2.9)–(2.10) with (2.8), we obtain
ZZ
(u − kn+1 )+ dxdτ ≤ 2
ZZ
Qn+1
2d
(u − kn+1 )2+ ζn b dxdτ
Qn
Z Z
(u −
≤ Qn
≤γ
) ζ b
kn+1 + nd dxd
2b (1+ Np )
1+M
k
σ pt
τ
2b
− 2(bb−2)
· |An+1 | 2
b−2 b
2 2 (b−2+p+ pN b
)
n Z Z
(u − kn )+ dxdτ 2
2p 1+ bN
.
Qn
If k is chosen to satisfy
ZZ u− Q0
k
2
2 +
dxdτ ≤ γ
1+M
−(1+ Np ) k
σ pt
N (b−2) p
.
(2.11)
Then by Lemma 5.6 of [17, p. 95], we get
ZZ
(u − kn )2+ dxdτ → 0,
as n → ∞,
Qn
i.e. kuk∞,Q∞ ≤ k. We select k = γ
N 1 +M 1 + p σ pt
RR
u dxdτ 2
Q0
p N (b−2)
, then (2.11) holds. It follows from this and Young’s
inequality that
kuk∞,Q∞ ≤ γ ≤γ
1+ Np Z Z
1+M
σ pt
1+M
! N (bp−2) u2 dxdτ
Q0
N +p N (b−2)
p ZZ N (b−2) kuk∞,Q0 udxdτ
σ pt
Q0
≤ kuk∞,Q0 + γ ()
1+M
N +κ p Z Z
σ pt
udxdτ
κp
,
Q0
where 0 < < 1. By an iteration process similar to [11, p. 393], we obtain
ku(·, t )k∞,B ρ ≤ γ
2
1+M
N +κ p Z Z
σ pt
udxdτ
κp
.
(2.12)
Q0
Step 2: For all 0 < t < T 0 , Bρ ⊂ RN , ρ = R(t ), (2.12) and (2.2) imply that
ku(·, t )k∞,B ρ ≤ γ t
− Nκ
2
0 N
! κp
Z tZ −
udxdτ
BR(t ) p
p
≤ γ t − κ R(t ) κ (N −θ) huitκ = γt
− p+θ(θp−2)
Γ
p
κ (N −θ)
p
huitκ .
Lemma 2.2 (The Case 1 ≤ l < p − 1). Let u be a nonnegative continuous weak subsolution of (1.1) in ST ∗ . Assume also that a time 0 < T 00 < min{T ∗ , 1} is given such that θ(p−2)
p−2
q −1
Γ −p t p+θ(p−2) ku(·, t )k∞,RN + t ku(·, t )k∞,RN ≤ 1,
∀0 < t < T 00 .
(2.20 )
Then
θ − ku(·, t )k∞,RN ≤ γ t p+θ(p−2) Γ
p
κ (N −θ)
p huitκ + t ,
where γ = γ (N , p, q, l, |λ|, θ , λ0 ), κ = N (p − 2) + p.
∀0 < t < T 00 ,
(2.30 )
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3401
Proof. For the case of 1 ≤ l < p − 1, we take the same test function as in Lemma 2.1 and let R(t ) ≤ ρ < 1, the differences between these two cases are the estimate of term |Du|l and the selection of k. Similarly to (2.6), we have
Z
t0
Z
tn
1
|Du|l (u − kn+1 )+ ζnp dxdτ ≤
Z
4
Bn
t0
Z
tn
|D(u − kn+1 )+ |p ζnp dxdτ + γ
t0
Z
Bn
tn
l 1+ p− l
Z
(u − kn )+
dxdτ .
Bn
By virtue of 1 ≤ l < p − 1, we obtain 1 p−1
≤
l p−l
< p − 1.
l If 1 ≤ p− < p − 1, by (2.20 ) and note that R(t ) ≤ ρ < 1, then for all tn < τ < t, we have l θ(p−2)
p−2
1 ≥ Γ −p τ p+θ(p−2) ku(·, τ )k∞,RN −2 = τ R(τ )−p ku(·, τ )kp∞, RN −2 ≥ τ R(t )−p ku(·, τ )kp∞, RN −2 ≥ τ ρ −p ku(·, τ )kp∞, , RN p
i.e., ku(·, τ )k∞,RN ≤ ρ p−2 τ
γ
Z
t0
Z
tn
1 − p− 2
l 1+ p− l
(u − k n )+
dxdτ ≤ γ ρ p−2
−1 p−l l
p−1 2
If p−l < 1, and k > t l
t0
Z
tn
t
l −1 p−l
Z
t t0
Z
1
1
Bn
1 since ρ < 1, 0 ≤ p− 2
Z
1
p
≤γ
γ
p
1
≤ 4 p−2 ρ p−2 t − p−2 , since τ > tn ≥ 4t . Hence
Z
tn
t0
tn
Z
(u − kn )2+ dxdτ
Bn
(u − kn )2+ dxdτ ,
Bn
− 1 < 1 and 0 < t < 1. in fact, here we take k = γ
l p−l
l 1+ p− l
(u − k n )+
dxdτ ≤ γ
N 1 +M 1 + p σ pt
1− p−l l Z
1 kn+1 − kn
Bn
≤ γ2
n1
Z
t
t0
Z
tn
RR
t0
Q0
Z
tn
u dxdτ 2
p N (b−2)
! + t , then
(u − kn )2+ ζ p dxdτ Bn
(u − kn )2+ dxdτ .
Bn
The above estimates, together with (2.4)–(2.5) and (2.7), still yield (2.8). But here M = sup
0<τ
−2 q −1 τ ρ −p ku(·, τ )kp∞, + τ k u (·, τ )k Bρ (x0 ) ∞,Bρ (x0 ) .
The following proof is similar as in Lemma 2.1 except the selection of k, and we obtain the iteration inequality
kuk∞,Q∞ ≤ kuk∞,Q0 + γ ()
1+M
N +κ p Z Z
σ pt
udxdτ
κp
+ t.
Q0
Then the rest of proof is similar to the proof of (2.12) and Step 2 in Lemma 2.1.
We also need the estimates of |Du|l as follows: Lemma 2.3 (The Case p − 1 ≤ l < p). If the assumptions of Lemma 2.1, p − 1 ≤ σ < p and θ (σ − p + 1) < p − σ hold, then for every Bρ (x0 ) ⊂ RN , 0 < t < T 0 , R(t ) ≤ ρ ≤ 1, if p ≥ 2 and σ 6= 1, we have
Z tZ 0
B ρ (x0 )
|Du|σ dxdτ ≤ γ G(t ) Γ
θ (N −θ)p p−σ − κ t 2σ −p p+θ(p−2)
p
huitκ
2σp−p
,
2
where G(t ) = sup0<τ
(2.13)
3402
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
If p = 2 and σ = 1, we have
Z tZ
1
|Du|dxdτ ≤ γ t 2 −h ρ 2h G(t ),
B ρ (x0 )
0
(2.130 )
2
where 0 < h <
1 , 2
γ = γ (N , p, q, l, |λ|, λ0 , h).
Proof. Assume p > 2, and set Bρ = Bρ (x0 ). The calculations to follow are formal in that u is required to be strictly positive. The calculations can be made rigorous by replacing u with u + and letting → 0. Take ϕ = t β ζ p ur as a test function in (1.4), where ζ is a piecewise smooth cut-off function in Bρ , such that 0≤ζ ≤1
in Bρ ,
ζ = 1 in B ρ ,
|Dζ | ≤
2
2
ρ
and β > 0, r > 0 are to be chosen. Thus we obtain
Z
1 1+r
=
t β ζ p ur +1 (x, t )dx + r
Z tZ 0
Bρ
β
Z tZ
1+r
0
τ β ζ p ur −1 |Du|p dxdτ + p
Z tZ 0
Bρ
τ β−1 ζ p ur +1 dxdτ + λ0
Z tZ
Bρ
0
τ β ζ p−1 ur |Du|p−2 DuDζ dxdτ Bρ
τ β ζ p uq+r dxdτ + λ
Z tZ
Bρ
0
τ β ζ p ur |Du|l dxdτ .
(2.14)
Bρ
Young’s inequality implies that
Z Z Z tZ Z tZ t r β p−1 r p−2 τ ζ u |Du| DuDζ dxdτ ≤ p τ β ζ p ur −1 |Du|p dxdτ + γ τ β ur +p−1 |Dζ |p dxdτ . 0 Bρ 4 0 Bρ 0 Bρ
(2.15)
Moreover
Z tZ
τ β ζ p ur |Du|l dxdτ ≤
|λ| 0
Bρ
r
Z tZ
4
0
τ β ζ p ur −1 |Du|p dxdτ + γ
Z tZ
Bρ
0
l
τ β ur + p−l dxdτ .
(2.16)
Bρ
Substituting (2.15)–(2.16) into (2.14), and by (2.2) and (2.3), we obtain
Z tZ 0
β p r −1
τ ζ u
|Du| dxdτ ≤ γ p
Z tZ
Bρ
0
≤γ
l
τ β−1 ur +1 (1 + τ ρ −p up−2 + τ uq−1 + τ u p−l −1 )dxdτ
Bρ
Z tZ 0
τ β−1 ur +1 dxdτ
Bρ
≤ γ G(t )
t
Z 0
≤ γ G(t )t
τ β−1 ku(·, τ )kr∞,Bρ dτ
β− p+θ(θ pr −2)
Γ
p(N −θ)r
κ
pr
huitκ ,
(2.17)
provided
β>
θr p + θ (p − 2)
.
(2.18)
Next we estimate by Hölder inequality,
Z tZ 0
|Du|σ ζ σ dxdτ =
Z tZ
Bρ
0
≤γ
σ
σ
σ
σ
τ β p ζ σ u(r −1) p |Du|σ τ −β p u−(r −1) p dxdτ Bρ
! σp
Z tZ 0
β p r −1
τ ζ u Bρ
|Du| dxdτ p
Z tZ 0
τ
σ −β p−σ
u
σ −(r −1) p−σ
!1− σp dxdτ
.
(2.19)
Bρ
Assume from now on
σ rσ −1≥ . p−σ p−σ
(2.20)
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3403
By Lemma 2.1, we have
Z tZ 0
σ
Z tZ
σ
τ −β p−σ u−(r −1) p−σ dxdτ =
0
Bρ
σ
σ
rσ
τ −β p−σ u1+ p−σ −1− p−σ dxdτ
Bρ
≤ G(t )
t
Z 0
σ −1− r σ p−σ
σ
p−σ τ −β p−σ ku(·, τ )k∞, Bρ
σ − θ 1−β p−σ p+θ(p−2) t
≤ γ G(t )
σ −1− r σ p−σ
dτ
p−σ
Γ
p(N −θ)
κ
σ −1− r σ p−σ
p−σ
p
κ t
hui
σ −1− r σ p−σ
p−σ
, (2.21)
provided
σ θ 1−β − p−σ p + θ (p − 2)
σ rσ −1− p−σ p−σ
> 0.
(2.22)
Thus by (2.17), (2.21) and (2.19), it is easy to obtain (2.13). In this following text, we will choose β and r such that (2.18), (2.20) and (2.22) hold. This is trivial task if θ = 0. Assume that θ > 0, and 0<β <
p−σ
σ
.
(2.23)
Then (2.18) and (2.20) are implied by r < min
p + θ (p − 2)
θ
β,
2σ − p
σ
, z.
Rewrite (2.22) as r >
p−σ
σ
θ 2σ − p p + θ (p − 2) · −1 +β , z0. p + θ (p − 2) p − σ θ
By virtue of θ (σ − p + 1) < p − σ , we obtain
θ
2σ − p
p + θ (p − 2) p − σ
− 1 < 0.
(2.24)
It is easy to check that z 0 < z. Finally we choose r > 0 such that r ∈ (z 0 , z ). As p = 2, the proof above also holds for the case 1 < σ < 2. If σ = 1, by (2.17), we have
Z tZ 0
τ β ζ 2 ur −1 |Du|2 dxdτ ≤ γ
Bρ
Z tZ 0
τ β−1 ur +1 dxdτ . Bρ
Using the method of the proof of Proposition 7.4 in [11], we can prove
Z tZ 0
1
B ρ (x0 )
|Du|dxdτ ≤ γ t 2 −h ρ 2h G(t ).
2
This finishes the proof.
Lemma 2.4 (The Case 1 ≤ l < p − 1). If the assumptions of Lemma 2.2 hold. Then for every Bρ (x0 ) ⊂ RN , 0 < t < T 00 , R(t ) ≤ ρ ≤ 1, we have
Z tZ 0
Bρ
|Du|p−1 dxdτ ≤ γ G(t ) Γ
1 (N −θ)(p−2) κ t p+θ(p−2)
p−2
huit κ + t
p−1 p
,
(x ) 2 0
where G(t ) = sup0<τ
Γ = C [µ] p+θ(p−2) , where C > 0 will be chosen depending on N , p, q, l, |λ|, λ0 , θ . (Obviously we may assume [µ] > 0 throughout.)
(2.25)
3404
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
Remark 2.1. For the case p − 1 ≤ l < p, it follows from (1.6) and (2.1) that huit ≤ [u]t . Using Lemma 2.1 and (2.25), we obtain p
N
[u]t ≤ huit + γ (C )[µ] κ (p−2) huitκ ,
∀0 < t < T 0 ,
(2.26)
where T 0 is as in Lemma 2.1. Remark 2.2. For the case 1 ≤ l < p − 1, we may choose
Γ = C [µ] +
p−1−l
p−2 p+θ(p−2)
p−1
.
Similar to Remark 2.1, we also have
N (p−2) p p−1−l κ [u]t ≤ huit + γ (C ) [µ] + huitκ , p−1
∀0 < t < T 00 ,
(2.260 )
where T 00 is as in Lemma 2.2. Let us conclude this section with the following lemmas, giving a priori bounds of a solution to (1.1)–(1.2) in terms of the initial data. Lemma 2.5 (The Case p − 1 ≤ l < p). Let u ≥ 0 be a bounded and uniformly continuous solution to (1.1)–(1.2) in ST ∗ . If θ(q − 1) < p + θ (p − 2) and θ (l − p + 1) < p − l. Then there exists T0 = T0 ([µ], N , p, q, |λ|, l, θ , λ0 ) < T ∗ such that
[u]t ≤ γ [µ],
∀0 < t < T0 ,
(2.27)
and (2.2)–(2.3) hold for all 0 < t < T0 , where γ = γ ([µ], N , p, q, |λ|, l, θ , λ0 ). Proof. As p > 2, define
t0 = sup 0 < T 0 < T ∗ |(2.2) holds, where Γ is given by (2.25) . Choose 0 < t < t0 and let Bρ ⊂ RN be any ball with radius R(t ) ≤ ρ ≤ 1, center at x0 ∈ RN . Take ζ as test function in (1.4), where ζ is a standard cut-off function in Bρ , with 0≤ζ ≤1
in Bρ ,
ζ = 1 in B ρ ,
2
|Dζ | ≤
2
ρ
.
Direct calculation shows that
Z Bρ
u(x, t )dx ≤
Z
dµ + Bρ
2
2
ρ
Z tZ 0
|Du|p−1 dxdτ + |λ|
Z tZ
Bρ
0
|Du|l dxdτ + λ0 Bρ
Z tZ 0
uq dxdτ .
(2.28)
Bρ
Multiplying ρ θ |Bρ |−1 on both sides of the above inequality and using Lemmas 2.1 and 2.3, we get
( p−2 Z θ(q−1) p+θ(p−2) huit κ 1− N ρ − u(x, t )dx ≤ 2 [µ] + γ huit C − κ + t p+θ(p−2) Γ [µ] Bρ 2 ) θ
+t
p−l θ(2l−p) p − p(p+θ(p−2))
Γ
(N −θ)(2l−p) κ
p(N −θ)(q−1)
κ
p(q−1)
huit
κ
2l−p
huit
κ
for all 0 < t < t0 , R(t ) ≤ ρ ≤ 1. By virtue of x0 ∈ RN is arbitrarily, it is immediately seen that
( huit ≤ 2 [µ] + γ huit C N
+t
p−l θ(2l−p) p − p(p+θ(p−2))
− p+θ(κp−2)
Γ
(N −θ)(2l−p) κ
huit [µ]
p−κ 2
2l−p
huit
+t
θ(q−1) 1− p+θ(p−2)
Γ
p(N −θ)(q−1)
κ
p(q−1)
huit
κ
)
κ
.
(2.29)
Set t1 = sup{0 < t < T ∗ |huit ≤ 2N +2 [µ]}
θ(q−1) 1− t2 = sup 0 < t < T ∗ |t p+θ(p−2) Γ
p(N −θ)(q−1)
κ
p(q−1)
huit
κ
+t
p−l θ(2l−p) p − p(p+θ(p−2))
Γ
(N −θ)(2l−p) κ
2l−p
huit
κ
<δ ,
(2.30)
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3405
where δ > 0 (small) is to be chosen. Note that t1 , t2 are well defined because the stipulated assumptions make sure that huit is continuous in [0, T ∗ ], and the exponent of t in (2.30) is positive. Let t3 = min{t0 , t1 , t2 }. Then for 0 < t < t3 , we have
γC
− p+θ(κp−2)
huit [µ]
p−κ 2
≤ γ C−
p+θ(p−2)
κ
(2N +2 )
p−2
κ
≤
1 4
,
provided C is suitably chosen. Then if we choose δ < 41γ , it follows from (2.29),
∀0 < t < t3 .
huit ≤ 2N +1 [µ],
(2.31)
By (2.26) and (2.31), we get
[u]t ≤ γ [µ],
∀0 < t < t3 .
Therefore all the claims made in the statement will follow, using the sup-estimates proven in this section, provided we show that we can indeed find a quantitative estimates below t3 ≥ T0 , with T0 as above. We may assume t3 < T ∗ , since the estimate is otherwise trivial. First we note that (2.31) implies t3 < t1 . Next, let us rule out the case t3 = t0 . To this end, we show that, for a suitable choice of C and δ , the left-hand side of (2.2) is bounded away from 1 up to t = t3 . As a matter of fact, Lemma 2.1, together with (2.31), implies for t ≤ t3 , x ∈ RN , θ(p−2)
l
Γ −2 t p+θ(p−2) u(x, t )p−2 + tu(x, t )q−1 + tu(x, t ) p−l
−1
p+θ(p−2)
1−
θ(q−1)
p(N −θ)(q−1)
p(q−1)
κ + t p+θ(p−2) Γ huit ≤ γ C− κ p−p l 2l − p θ(2l−p) p−l (N −θ)(2l−p) κ p − p(p+θ(p−2)) κ + t huit Γ p+θ(p−2) p + δ + δ p−l ≤ γ C− κ
≤
1 2
κ
.
(2.32)
If C is chosen large enough, and δ in the definition of t2 is chosen small enough, we can guarantee the last inequality in (2.32). Finally, we are left with the task of estimating below t3 = t2 . This can be accomplished at once, by replacing huit with 2γ [µ] in the definition of t2 in (2.30), owing to (2.31). As p = 2, similar to the above proof. Lemma 2.6 (The Case 1 ≤ l < p − 1). Let u ≥ 0 be a bounded and uniformly continuous solution to (1.1)–(1.2) in ST ∗ . If θ(q − 1) < p + θ (p − 2), then there exists T00 = T00 ([µ], N , p, q, |λ|, l, θ , λ0 ) < T ∗ such that
[u]t ≤ γ ([µ] + 1),
∀0 < t < T00 ,
and (2.20 ), (2.30 ) hold for all 0 < t < T00 , where γ = γ ([µ], N , p, q, |λ|, l, θ , λ0 ). Proof. Define t00 = sup{0 < T 00 < min{T ∗ , 1}|(2.20 ) holds, where Γ is given in Remark 2.2}. Choose 0 < t < t00 and let Bρ ⊂ RN be any ball with radius R(t ) ≤ ρ ≤ 1, center at x0 ∈ RN . Take ζ as test function in (1.4), where ζ is as in Lemma 2.5. We also obtain (2.28), then Young’s inequality implies that
Z Bρ
u(x, t )dx ≤
Z Bρ
2
γ dµ + ρ
Z tZ |Du| 0
p−1
dxdτ + λ0
Bρ
Z tZ 0
uq dxdτ +
p−1−l
Bρ
p−1
|Bρ |,
since ρ < 1. Multiplying ρ θ |Bρ |−1 on both sides of the above inequality, using Lemmas 2.2 and 2.4 and by virtue of x0 ∈ RN is arbitrarily, we get
huit ≤ 2N [µ] + γ huit
C−
+ 2N
p−1−l p−1
for all 0 < t < t00 , R(t ) ≤ ρ ≤ 1.
p+θ(p−2)
κ
huit [µ] +
p−1−l p−1
! p−κ 2 + Γ −1 t
p−1 1 p − p+θ(p−2)
+t
θ(q−1) 1− p+θ(p−2)
Γ
p(N −θ)(q−1)
κ
p(q−1)
huit
κ
(2.33)
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H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
Set
t10 = sup 0 < t < min{T ∗ , 1}|huit ≤ 2N +2
θ(q−1) 1− t p+θ(p−2)
t20 = sup 0 < t < min{T ∗ , 1}|
Γ
[µ] +
p−1−l
p(N −θ)(q−1)
κ
p−1 p(q−1)
huit
κ
+ Γ −1 t
p−1 1 p − p+θ(p−2)
<δ ,
(2.34)
where δ > 0 (small) is to be chosen. Note that t10 , t20 are well defined because the stipulated assumptions make sure that huit is continuous in [0, T ∗ ], and the exponent of t in (2.34) is positive. Let t30 = min{t00 , t10 , t20 }. Then for 0 < t < t30 , we have
γ C−
! p−κ 2
huit
p+θ(p−2)
κ
[µ] +
≤ γ C−
p−1−l p−1
p+θ(p−2)
κ
(2N +2 )
p−2
κ
≤
1 4
,
provided C is suitably chosen. Then if we choose δ < 41γ , it follows from (2.33),
p−1−l huit ≤ 2N +1 [µ] + , p−1
∀0 < t < t30 .
(2.35)
By (2.260 ) and (2.35), we get
[u]t ≤ γ ([µ] + 1),
∀0 < t < t30 .
Therefore all the claims made in the statement will follow, using the sup-estimates proven in this section, provided we show that we can indeed find a quantitative estimates below t30 ≥ T00 , with T00 as above. We may assume t30 < T ∗ , since the estimate is otherwise trivial. First we note that (2.35) implies t30 < t10 . Then the rest of the proof is similar as in Lemma 2.5, we omit the details. 3. Proof of Theorems 1.1 and 1.2 In this section, we will give the proof of Theorems 1.1 and 1.2 by an approximation method. Proof of Theorem 1.1. Consider the approximating problems:
(P)
unt − div(|Dun |p−2 |Dun |) = λ0 min{uqn , n} + λ min{|Dun |l , n} u (x, t ) = 0 un (x, 0) = u (x) n 0n
in Bn × (0, T ), in ∂ Bn × (0, T ), on Bn ,
where Bn = {x ∈ RN | |x| < n} and u0n ∈ C0∞ (RN ) is nonnegative and has compact support in Bn , and satisfies
Z lim
n→∞
RN
u0n η(x)dx =
Z RN
η(x)dµ,
∀η ∈ C0∞ (RN ),
and
[u0n ] ≤ γ (N )[µ]. By Theorem 1.1 of Chapter 3 in [18], problem (P) has a nonnegative solution un ∈ C (0, T ; L2 (Bn )) ∩ Lp (0, T ; W 1,p (Bn )). By Theorem 3.2 of Chapter 5 in [15], we have un ∈ L∞ (Bn × (0, T )), here the bounds depend upon n. By the results of [14, 15] and Remark 1.4, one obtains that un is Hölder continuous in B¯ n × [0, T ). We will regard un as defined in the whole ST (= RN × (0, T )) by extending them to be zero outside Bn . From the above arguments we obtain that every un satisfies the conditions in Lemmas 2.1, 2.3 and 2.5. Therefore by Lemma 2.1, for any compact set K ⊂ ST0 , we have
kun k∞,K ≤ γ (K , [µ]),
(3.1)
where T0 is defined as in Lemma 2.5. Note that T0 and γ in (3.1) are independent of n. For each compact set K contained in ST0 , there exists a natural number n0 such that K ⊂ Bn0 ×(0, T0 ). Combining (3.1) with the results in [14,15] and Remark 1.4, we get uniformly Hölder estimates for the sequence {un }n>n0 in each K , we may assume that un → u
uniformly on K .
(3.2)
For every bounded open set Ω and all 0 < s < t < T0 , then there exists a compact set K ⊂ ST0 such that Ω × (s, t ) ⊂⊂ K . Take ϕ = un ζ 2 as test function in (P), where ζ ∈ C0∞ (K ) with 0≤ζ ≤1
in K ,
ζ = 1 in Ω × (s, t ),
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3407
we can deduce that
ZZ
|Dun |p ζ 2 dxdτ ≤ 2
ZZ
|Dun |p−1 ζ |Dζ |dxdτ +
ZZ
u2n ζ |ζτ |dxdτ
K
K
K
ZZ
|Dun |l un ζ 2 dxdτ + λ0
+ |λ|
ZZ
unq+1 ζ 2 dxdt .
(3.3)
K
K
By Lemma 2.3, (3.1) and (3.3), we obtain
kDukp,Ω ×(s,t ) ≤ γ (K , [µ]).
(3.4)
Assume that un , uk are two solutions of (P) with initial value u0n and u0k respectively, we have
(un − uk )t − div(|Dun |p−2 Dun − |Duk |p−2 Duk ) = λ0 (min{uqn , n} − min{uqk , k}) + λ(min{|Dun |l , n} − min{|Duk |l , k}).
(3.5)
Multiply (3.5) by (un − uk )ζ 2 , we get
ZZ
ZZ |un − uk |2 ζ |ζτ |dxdτ + γ (|Dun |p−1 + |Duk |p−1 )ζ |Dζ | |un − uk |dxdτ K K ZZ + λ0 | min{uqn , n} − min{uqk , k}| |un − uk |ζ 2 dxdτ Z ZK + |λ| | min{|Dun |l , n} − min{|Duk |l , k}| |un − uk |ζ 2 dxdτ K ZZ ≤ γ {|un − uk | |ζτ | + (|Dun |p−1 + |Duk |p−1 )|Dζ | + uqn + uqk + |Dun |l
|Dun − Duk |p ζ 2 dxdτ ≤ γ
K
ZZ
K
+ |Duk |l }|un − uk |ζ dxdτ → 0 as n, k → ∞.
(3.6)
The limit relation in (3.6) is due to Lemma 2.3, (3.1) and (3.2). Therefore by extracting a subsequence if necessary, we have Dun → Du,
a.e. in Ω × (s, t ) and strongly in Lp (Ω × (s, t )).
(3.7)
Thus by (3.1)–(3.2), (2.3) and (3.7), (2.27), let n → ∞, we can prove (1.4), (1.7) and (1.8). The proof of (1.5) as follows: For ∀ η ∈ C01 (RN ), assume that there exists some natural number n0 such that suppη ⊂ Bn0 , taking η as a test function in (3.5) (here n, k ≥ n0 ), we obtain
Z
(un (x, t ) − uk (x, t ))ηdx = −
Z tZ
Bn0
0
+ λ0
(|Dun |p−2 |Dun | − |Duk |p−2 |Duk |)Dηdxdτ Bn0
Z tZ 0
(min{uqn , n} − min{uqk , k})ηdxdτ + λ Bn0
Z tZ 0
− min{|Duk |l , k})ηdxdτ +
Z
(min{|Dun |l , n} Bn0
(u0n − u0k )ηdx. Bn0
Using Young’s inequality, Lemmas 2.1, 2.3 and 2.5 (with ρ = 1), furthermore, we can find finite balls B1 (x1 ), B1 (x2 ), . . . , B1 (xk0 ) covering B¯ n0 , then we have
Z Z Z Z tZ t (un (x, t ) − uk (x, t ))ηdx ≤ (|Dun |p−1 + |Duk |p−1 )|Dη|dxdτ + λ0 (uqn + uqk )|η|dxdτ Bn 0 B 0 B n n 0 0 0 Z Z tZ + |λ| (|Dun |l + |Duk |l )|η|dxdτ + (u − u0k )ηdx Bn 0n 0 Bn0 0 Z Z tZ Z tZ q l l q ≤γ (|Dun | + |Duk | )dxdτ + γ (un + uk )dxdτ + γ t + (u0n − u0k )ηdx Bn 0 Bn0 0 Bn0 0 ! ! Z Z Z Z θ(q−1) p−l 1− ≤ γ t p sup un dx + uk dx + γ t p+θ(p−2) sup un dx + uk dx 0<τ
Bn0
Bn0
0<τ
Bn0
Bn0
3408
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
Z +γt + (u0n − u0k )ηdx Bn 0 Z ≤ C (t ) + (u0n − u0k )ηdx Bn 0
where C (t ) is a positive constant independent of n and k, and limt →0 C (t ) = 0. Hence as n, k ≥ n0 , we have
Z
RN
u(x, t )ηdx −
Z RN
Z Z ηdµ ≤ (u(x, t ) − un (x, t ))ηdx + (un (x, t ) − uk (x, t ))ηdx Bn Bn Z0 Z 0 Z + (uk (x, t ) − u0k )ηdx + u0k ηdx − ηdµ Bn Bn Bn0 0 Z Z 0 ≤ (u(x, t ) − un (x, t ))ηdx + (u0n − u0k )ηdx + C (t ) Bn Bn Z 0 Z0 Z u0k ηdx − ηdµ . (uk (x, t ) − u0k )ηdx + + Bn Bn Bn 0
0
0
Let n → ∞, t → 0, k → ∞ in the indicated order in the above inequality, we get (1.5). Thus the proof of Theorem 1.1 is completed. Proof of Theorem 1.2. With minor changes, Theorem 1.2 can be proved using the method above too, we omit the details. 4. Proof of Theorem 1.3 To prove Theorem 1.3, we need to prove the following lemma. Lemma 4.1. Let u ∈ C (ST ) be a nonnegative solution of (1.1)–(1.2), and let u(·, t ) converge to µ in the sense of measures, with
µ(Bρ (x0 )) ≥ ρ N −θ ,
0 < ρ < σ,
(4.1)
for some given x0 ∈ RN , , ρ, σ > 0, 0 ≤ θ ≤ N. Then p
u(x, t ) ≥ γ0 p+θ(p−2) t
− p+θ(θp−2)
,
(4.2)
p−2
1
where |x − x0 | ≤ ρ < σ , ρ = γ0 p+θ(p−2) t p+θ(p−2) , γ0 = γ0 (N , p, θ ), 0 < t < T . Proof. Let v(x, t ) be the weak solution of the following Cauchy problem,
vt − div(|Dv|p−2 Dv) = 0 v(x, 0) = µ
in ST = RN × (0, T ), on RN ,
(4.3)
where µ is a nonnegative Radon measure. By the result of [1], we have
Z − Bρ (x0 )
dµ ≤ γ
(
ρp
p−1 2
+
t
t
Np
ρp
inf
|x−x0 |<ρ
κp ) v(x, t ) ,
where κ = N (p − 2) + p. Combining (4.1), (4.4) and the definition of ρ , we obtain p
v(x, t ) ≥ γ0 p+θ(p−2) t
− p+θ(θp−2)
,
where choose γ0 suitably small. Note that u is the supersolution of (4.3), thus p
u(x, t ) ≥ v(x, t ) ≥ γ0 p+θ(p−2) t which yields (4.2).
− p+θ(θp−2)
,
(4.4)
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3409
Proof of Theorem 1.3. If θ = 0, the conclusion is trivial. In the following we may assume θ > 0 and x0 = 0. From now on, all the balls are understood to be center at 0. Let ω, λ be the first eigenfunction and eigenvalue of the eigenvalue problem,
−1ω = λω ω=0
in Bρ , on ∂ Bρ .
The first eigenfunctions have a fixed sign and we may select a positive one normalized so that
Z
ωdx = 1. Bρ
It is known by classical results that λ = γ (N )ρ −2 , and
−N ω(x) ≤ γ (N )ρ , −1 ω(x) ≥ γ (N )ρ −N , |Dω(x)| ≤ γ (N )ρ −N −1 ,
∀x ∈ Bρ , ∀x ∈ B ρ , 2 ∀x ∈ Bρ .
(4.5)
Taking ϕ = ωζ s as a test function in (1.4), where s > l−pl+1 , ζ (x) is a cut-off function in B ρ and satisfy 2
0 ≤ ζ (x) ≤ 1
in B ρ , 2
ζ (x) ≡ 1 in B ρ ,
|Dζ | ≤
4
ζ is radially nonincreasing.
4
ρ
(4.6)
Then we have
Z
Z
uωζ s dx −
B ρ (t )
B ρ (t 0 )
2
uωζ s dx = −
Z tZ t0
|Du|p−2 Du(ζ s Dω + sωζ s−1 Dζ )dxdτ
Bρ
2
2
+λ
Z tZ t0
|Du| ωζ dxdτ + λ0 l
Bρ
s
Z tZ t0
2
uq ωζ s dxdτ ,
Bρ
2
where 0 < t < t < T . Applying Young’s inequality and using (4.5), noting s > l−pl+1 , we obtain 0
Z B ρ (t )
uωζ s dx −
Z
2
B ρ (t 0 )
uωζ s dx ≥ −γ
Z tZ t0
2
l l l l ω1− l−p+1 |Dω| l−p+1 + ωζ s− l−p+1 |Dζ | l−p+1 dxdτ
Bρ
2
+
λ
Z tZ
2
t0
|Du| ωζ dxdτ + λ0 l
Bρ
s
Z tZ t0
2
≥ −γ ρ
− l−pl+1
Bρ
uq ωζ s dxdτ
2
(t − t 0 ) +
λ
Z tZ
2
t0
Bρ
|Du|l ωζ s dxdτ +
2
Z tZ t0
Bρ
uq ωζ s dxdτ .
(4.7)
2
Due to the assumption [u]t < ∞, then
Z
Z udx ≤ B ρ (t )
Bρ (t )
2
udx ≤ γ ρ N −θ ,
0 < t < t∗ , 0 < ρ < 1,
(4.8)
for a suitable t∗ > 0 (t∗ < T ) to be chosen. Define A(t ) = {x ∈ B ρ |u(x, t ) > C0 ρ −θ }, 2
E (t ) = B ρ \ A(t ), 2
where C0 > 0 is to be chosen. Hence for 0 < t < t∗ , 0 < ρ < 1, (4.8) yields
|A(t )| ≤ γ C0−1 |B ρ |. 2
If C0 is chosen large enough, then we have
|E (t ) ∩ B ρ | ≥ 4
1 2
|B ρ |. 4
(4.9)
Let us use the following result in [12, p. 3921]: Assume that f ∈ W 1,r (B ρ ), f ≥ 0, r ≥ 1, then for all measurable sets 2 E0 ⊂ B ρ , f ≡ 0 in E0 , 4
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H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
Z Bρ
ρ −1 r Z r N +1 −1 f G(|x|)dx ≤ γ (N , r ) ρ |E0 | G(0)G 4
2
Bρ
|Df |r G(|x|)dx,
(4.10)
2
ρ
where G(·) : [0, 2 ) → (0, ∞) is nonincreasing. Taking f = (u − C0 ρ −θ )+ , r = l, E0 = E (t ) ∩ B ρ , G(|x|) = ωζ s in (4.10), by (4.5)–(4.6) and integrating (4.10) over (t 0 , t ), 4 we obtain
ρ
Z tZ
−l
t0
Bρ
(u −
C0 ρ −θ )l+ ωζ s dxdτ
≤γ
Z tZ t0
2
|Du|l ωζ s dxdτ ,
Bρ
(4.11)
2
for 0 < t 0 < t < t∗ , 0 < ρ < 1. Moreover, we note that (4.2) implies t
u(x, τ ) ≥ 2C0 ρ −θ ,
2
< τ < t,
1
|x| ≤ γ0 ρ,
for ρ = ρ(t ) = C1 t p+θ(p−2) ,
(4.12)
provided C1 > 0 is fixed large enough (in (4.12) γ0 depends on C1 ). From now on, we choose ρ as in (4.12). We also choose t∗ to ensure ρ < 1 for t < t∗ . Thus direct calculations lead us to
Z
γ
B ρ (τ )
(u − C0 ρ −θ )l+ ωζ s dx ≥
Z
2
B ρ (τ )
ul ωζ s dx,
t 2
< τ < t < t∗ .
(4.13)
2
By (4.7), (4.11), (4.13) and using (4.5), noting that λ0 ≥ 0, with the help of Hölder inequality, obtaining
Z
uωζ dx − s
B ρ (z ) 2
Z
uωζ dx ≥ −γ ρ s
B ρ ( 2t )
t 2
z−
t
2
+ γρ
−l
Z zZ t 2
2
≥ −γ ρ
for
− l−pl+1
< z < t. Thus
R
B ρ (z )
− l−pl+1
z−
t 2
+ γ ρ −l
Bρ
2
z
Z t 2
ul ωζ s dxdτ
l
Z Bρ
uωζ s dx dτ ,
2
uωζ s dx majorizes the solution y(z ) to
2
− l 0 −l l y = γ ρ y − γ ρ l−p+1 Z t y( ) = uωζ s dx. 2
in
t 2
,t , (4.14)
Bρ ( 2t )
Therefore, the solution y of the problem (4.14) is bounded over ( 2t , t ). It follows from Lemma 4.1 in [19] that
Z
n
B ρ ( 2t )
l
uωζ s dx ≤ γ max ρ l−1 t
− l−11
o p−l , ρ − l−p+1 .
(4.15)
2
By (4.5) and (4.12), we have
Z B ρ ( 2t )
uωζ s dx ≥ γ ρ −θ ,
(4.16)
2
for t small and ρ = ρ(t ) defined above. Let t → 0, it is easily seen that (4.15) and (4.16) imply θ (l − p + 1) ≤ p − l. This completes the proof of Theorem 1.3.
Acknowledgement The authors would like to thank the referee for some helpful suggestions which correct some mistakes and improve the presentation of the original manuscript. References [1] E. Di Benedetto, M.A. Herrero, On the Cauchy problem and initial traces for a degenerate parabolic equation, Trans. Amer. Math. Soc. 134 (1989) 187–224. [2] Junning Zhao, On the Cauchy problem and initial traces for the evolution p-Laplacian equations with strongly nonlinear sources, J. Differential Equations 121 (1995) 329–383.
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
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[3] L. Amour, M. Ben-Artzi, Global existence and decay for viscous Hamilton–Jacobi equations, Nonlinear Anal. 31 (1998) 621–628. [4] P. Baras, M. Pierre, Critère d’existence de solutions positives pour des équations semi-linéaires non nonotones, Ann. Inst. H. Poincaré Anal. Non Linéaire 2 (1985) 185–212. [5] M. Chipot, F.B. Weissler, Some blow up results for a nonlinear parabolic problem with a gradient term, SIAM J. Math. Anal. 20 (4) (1989) 886–907. [6] M. Hesaaraki, A. Moameni, Blow-up of positive solutions for a family of nonlinear parabolic equations in general domain in RN , Michigan Math. 52 (2004) 375–389. [7] P. Souplet, Finite time blow-up for a nonlinear parabolic equation with a gradient term and applications, Math. Methods Appl. Sci. 19 (1996) 1317–1333. [8] P. Souplet, Recent results and open problem on parabolic equations with gradient nonlinearities, Electron. J. Differential Equations 20 (2001) 1–19. [9] P. Souplet, F.B. Weissler, Poincaré’s inequality and global solutions of a nonlinear parabolic equation, Ann. Inst. H. Poincaré Anal. Non Linéaire 16 (1999) 337–373. [10] Songzhe Lian, Hongjun Yuan, Chunling Cao, Wenjie Gao, Xiaojing Xu, On the Cauchy problem for the evolution p-Laplacian equations with gradient term and source, J. Differential Equations 235 (2007) 544–585. [11] D. Andreucci, E. Di Benedetto, On the Cauchy problem and initial traces for a class of evolution equations with strongly nonlinear sources, Ann. Sc. Norm. Super. Pisa Cl. Sci. (4) 18 (3) (1991) 393–441. [12] D. Andreucci, Degenerate parabolic equations with initial data measures, Trans. Amer. Math. Soc. 349 (10) (1997) 3911–3923. [13] M.F. Bidaut-Véron, E. Chasseigne, L. Véron, Initial trace of solutions of some quasilinear parabolic equations with absorption, J. Funct. Anal. 193 (2002) 140–205. [14] E. Di Benedetto, On the local behaviour of solutions of degenerate parabolic equations with measurable coefficients, Ann. Sc. Norm. Super. Pisa Cl. Sci. (4) 13 (1986) 487–535. [15] E. Di Benedetto, Degenerate Parabolic Equations, Springer-Verlag, New York, 1993. [16] D. Andreucci, A.F. Tedeev, M. Ughi, The Cauchy problem for degenerate parabolic equations with source and damping, Ukr. Math. Bull. 1 (1) (2004) 1–23. [17] O.A. Ladyzhenskaja, V.A. Solonikov, N.N. Ural’cema, Linear and Quasilinear Equations of Parabolic Type, Amer. Math. Soc., Providence, RI, 1968. [18] J.L. Lions, Quelques Méthodes de Résolution des Problémes aux Limites Nonlinéaires, Dunod, Paris, 1969. [19] D. Andreucci, New results on the Cauchy problem for parabolic systems and equations with strongly nonlinear sources, Manuscripta Math. 77 (1992) 127–159.
Contents lists available at ScienceDirect
Nonlinear Analysis journal homepage: www.elsevier.com/locate/na
On the Cauchy problem for the evolution p-Laplacian equations with gradient term and source and measures as initial dataI Haifeng Shang, Fengquan Li ∗ Department of Applied Mathematics, Dalian University of Technology, Dalian, 116024, China
article
abstract
info
Article history: Received 14 November 2008 Accepted 10 December 2009
In this paper we study the Cauchy problem for the evolution p-Laplacian equations with gradient term and source on the assumption that measures as initial conditions. © 2009 Elsevier Ltd. All rights reserved.
MSC: 35K15 35K55 35K65 Keywords: Cauchy problem p-Laplacian equation Gradient term Source Measures as initial data
1. Introduction and statement of the main results Here we will consider the following Cauchy problem ut − div(|Du|p−2 Du) = λ0 uq + λ|Du|l u(x, 0) = µ
in ST = RN × (0, T ),
on R , N
(1.1) (1.2)
where p ≥ 2, q ≥ 1, 1 ≤ l < p, λ ∈ R, λ0 ≥ 0, T > 0 and µ is a nonnegative Radon measure. In the case of λ0 = 0 and λ = 0, Di Benedetto and Herrero proved in [1] that for every Radon measure µ satisfying some polynomial growth at infinity there is a solution. Also they showed that every nonnegative solution u has a unique initial trace µ. Concerning the case λ = 0 and λ0 > 0, the Cauchy problem and initial trace problem were studied in [2]. For the subcritical case q < p − 1 + p/N, the existence of solution can be obtained for any Radon measure. But, for the supercritical case q ≥ p − 1 + p/N, to obtain the existence result, some higher integrability for the initial data need to be imposed. They assume
Z
µh (y)dy < ∞,
sup x∈RN
|x−y|<1
h>
N p
(q − p + 1).
I Project supported by NSFC (No: 10401009) and NCET of China (No: 060275).
∗
Corresponding author. E-mail addresses: [email protected] (H. Shang), [email protected] (F. Li).
0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.12.023
(1.3)
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3397
For the case p = 2, the equation has been studied by many authors (see [3–9]). This equation was introduced in [5] to investigate the effect of the damping term |Du|l on the existence or nonexistence of a global solution. Also, it has been proposed as a model in population dynamics (see [7]). For the case p > 2 and 1 ≤ l ≤ p − 1, the existence of solutions was studied in [10]. They proved the existence of solutions with initial data measures for the subcritical case. Moreover, in the supercritical case, they need the regularity condition (1.3). From the statement above, we see that for the supercritical case q ≥ p − 1 + p/N, to ensure the existence of solution, some regularity condition need to be imposed on the initial data µ. In contrast, here we are able to overcome this difficulty, essentially by choosing the correct geometry for the domains, and obtained existence for measures as initial data, even for the supercritical case. For the porous medium equation, the Cauchy problem and initial trace problem with strongly nonlinear sources were studied in [11]. The existence of solutions of the Cauchy problem with gradient term has been studied by [12] under optimal assumption on initial data. On the other hand, the Cauchy problem and initial traces of evolution p-Laplacian equation with absorption term was studied in [13]. In this paper, we study the problem (1.1)–(1.2) motivated by the ideas in [12,11,2,10] and give the existence of solutions under optimal assumptions on the initial data µ for Eq. (1.1). Definition 1.1. A nonnegative measurable function u(x, t ) defined in ST is called a weak solution of (1.1)–(1.2), if for every bounded open set Ω , with smooth boundary ∂ Ω , u ∈ Cloc (0, T ; L1 (Ω )) ∩ Lloc (0, T ; W 1,p (Ω )) ∩ L∞ loc (ST ), p
and
Z Ω
u(x, t )ϕ(x, t )dx +
Z tZ Ω
s
[−uϕτ + |Du|p−2 DuDϕ]dxdτ =
Z Ω
u(x, s)ϕ(x, s)dx + λ0
+λ
s
Z tZ s
1, ∞
for all 0 < s < t < T and all ϕ ∈ Wloc
Z lim
t →0
RN
u(x, t )η(x)dx =
Z RN
Z tZ
Ω
Ω
uq ϕ dxdτ
|Du|l ϕ dxdτ ,
(1.4)
(0, T ; L∞ (Ω )) ∩ Lploc (0, T ; W01,p (Ω )). Moreover
ηdµ,
∀η ∈ C01 (RN ).
(1.5)
Weak subsolutions (resp. supersolutions) are defined in the same way except that the = in (1.4) is replaced by ≤ (resp. ≥) and ϕ is taken to be nonnegative. We introduce some notations as in [12]. Let µ be any nonnegative Radon measure in RN , and u ∈ L∞ loc (ST ), u ≥ 0. Suppose that 0 ≤ θ ≤ N is given. Set
Z [µ] = sup sup ρ θ − x∈RN 0<ρ<1
Bρ (x)
dµ,
Z [u]t = sup sup sup ρ θ − 0<τ
Bρ (x)
u(y, τ )dy,
0 < t < T,
(1.6)
where
Z Z 1 − dµ = dµ, |E | E E
|E | is the Lebesgue measure of E .
We use γ (a1 , a2 , . . . , an ) to denote positive constants depending only on specified quantities a1 , a2 , . . . , an . First we state our main existence results as follows. Theorem 1.1 (The Case p − 1 ≤ l < p). Let [µ] be finite, θ (q − 1) < p + θ (p − 2) and θ (l − p + 1) < p − l. Then there exists a solution to (1.1)–(1.2) defined in RN × (0, T0 ), where T0 = T0 ([µ], N , p, q, |λ|, l, θ , λ0 ), such that for ∀ 0 < t < T0 , we have
[u]t ≤ γ [µ], ku(·, t )k∞,RN ≤ γ
(1.7) θ − t p+θ(p−2)
here γ = γ (N , p, q, l, |λ|, λ0 , θ ).
[µ],
(1.8)
3398
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
Theorem 1.2 (The Case 1 ≤ l < p − 1). Let [µ] be finite, θ (q − 1) < p + θ (p − 2). Then there exists a solution to (1.1)–(1.2) defined in RN × (0, T00 ), where T00 = T00 ([µ], N , p, q, |λ|, l, θ , λ0 ), such that for ∀ 0 < t < T00 , we have
[u]t ≤ γ ([µ] + 1), ku(·, t )k∞,RN ≤ γ t
(1.9)
− p+θ(θp−2)
([µ] + 1),
(1.10)
here γ = γ (N , p, q, l, |λ|, λ0 , θ ). Remark 1.1. The dependence of T0 and T00 on the quantities specified in the statement of Theorems 1.1 and 1.2 can be made explicit. We refer to the proofs of Lemmas 2.5 and 2.6. Remark 1.2. In [10], they only considered the case of p > 2, 1 ≤ l ≤ p − 1 and proved the existence of solutions with initial data measures for 1 ≤ q < p − 1 + p/N. In this paper, we extend their results to the case of 1 ≤ q < p − 1 + p/θ for initial data measures. Moreover, we obtain the result in the case of p − 1 < l < p. Next we prove that the critical threshold for θ in the gradient term made explicit in Theorem 1.1 is actually optimal for the existence of solutions in the class considered here. Theorem 1.3. Let p − 1 < l < p, λ > 0, u be a nonnegative solution to (1.1)–(1.2), in RN × (0, T ) such that [u]t < ∞, 0 < t < T , and
µ(Bρ (x0 )) ≥ ρ N −θ ,
0<ρ<
where > 0, x0 ∈ RN are given. Then θ (l − p + 1) ≤ p − l. Remark 1.3. We remark here that the positivity of the right-hand side of (1.1) is exploited in the proof of Theorem 1.3. Remark 1.4. If u is a weak solution of (1.1) and also u ∈ Cloc (0, T ; L2loc (RN )), then by the results in [14] and [15], u is actually locally Hölder continuous in ST . For the case λ < 0, this result was given in the Remark 1.1 in [16]. Remark 1.5. We note that our methods also yield existence of solutions of variable sign, approaching a signed measure µ as t tends to zero, because the estimates below hold separately for the positive and negative parts of solutions. This paper is organized as follows: In Section 2, a priori estimates will be given. In Section 3, we will finish the proof of Theorems 1.1 and 1.2. In Section 4, Theorem 1.3 will be proved. 2. A priori estimates In this section, we will prove some estimates for solutions of (1.1). ∗ ∗ For any u ∈ L∞ loc (ST ), u ≥ 0, 0 < T ≤ T , we define, following [12]
huit = sup sup
sup
0<τ
Z ρ θ − u(y, τ )dy,
1
R(t ) = Γ t p+θ(p−2) ,
(2.1)
Bρ
for all 0 < t < T ∗ , where Γ is a positive constant which can be chosen a priori dependent of N , p, q, l, [µ], λ0 , λ, and which will be specified later. We also assume that T ∗ is chosen so that R(T ∗ ) ≤ 1. Note that the last assumption is meaningful because Γ is independent of T ∗ . The connection between [u]t and huit will be commented upon in Remarks 2.1 and 2.2. The following sup-estimate will play an important role in proving the existence results. Lemma 2.1 (The Case p − 1 ≤ l < p). Let u be a nonnegative continuous weak subsolution of (1.1) in ST ∗ . Assume also that a time 0 < T 0 < T ∗ is given such that θ(p−2)
p−2
l
q −1
−1
p−l Γ −p t p+θ(p−2) ku(·, t )k∞,RN + t ku(·, t )k∞,RN + t ku(·, t )k∞, ≤ 1, RN
∀0 < t < T 0 .
(2.2)
Then
ku(·, t )k∞,RN ≤ γ t
− p+θ(θp−2)
Γ
p
κ (N −θ)
p
huitκ ,
∀0 < t < T 0 ,
where γ = γ (N , p, q, l, |λ|, θ , λ0 ), κ = N (p − 2) + p.
(2.3)
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3399
Proof. This proof will be divided into two steps. Step 1: Let ρ > 0, σ ∈ 0,
ρ
ρn =
σ
+
2n+1
2 Bn = Bρn (x0 ),
1 2
, and x0 ∈ RN be fixed, let k > 0 to be chosen. For n = 0, 1, 2, . . . , set
σ p
k t, kn = k − n+1 , 2n+1 2 Qn = Bn × (tn , t ), 0 < tn < t ≤ T 0 .
ρ,
tn =
t
2
−
Let ζn (x, τ ) be a smooth cut-off function in Qn with 0 ≤ ζn (x, τ ) ≤ 1, such that
ζn ≡ 1 in Qn+1 ,
0≤
∂ζn 2(n+2)p ≤γ , ∂τ σ pt
|Dζn | ≤ γ
2n+2
σρ
.
p
Taking ϕ = (u − kn+1 )+ ζn as a test function in (1.4), we get
Z
1 2
(u − kn+1 )2+ (x, t 0 )ζnp dx + Bn
Z
t0
Z
tn
Z t0 Z
t0
Z
Bn
Z
tn
Z t0 Z
(u − kn+1 )2+ ζnp−1 ζnτ dxdτ + λ0
=p tn
|D(u − kn+1 )+ |p ζnp dxdτ + p
Bn
tn
ζnp−1 (u − kn+1 )+ |Du|p−2 Du · Dζn dxdτ Bn
Z t0 Z
uq (u − kn+1 )+ ζnp dxdτ + λ Bn
tn
|Du|l (u − kn+1 )+ ζnp dxdτ , (2.4) Bn
where tn < t 0 < t. By Young’s inequality, we obtain
Z 0 Z Z 0Z t 1 t p−1 p−2 p ζ (u − kn+1 )+ |Du| DuDζn dxdτ ≤ |D(u − kn+1 )+ |p ζnp dxdτ tn Bn n 4 tn Bn Z t0 Z +γ (u − kn+1 )p+ |Dζn |p dxdτ . tn
(2.5)
Bn
Moreover t0
Z
Z
|Du| (u − l
|λ| tn
) ζ
kn+1 + np dxd
τ ≤
4
Bn
Z
tn
4
τ +γ
Z
|D(u − kn+1 )+ |p ζnp dxdτ + γ
Z
|D(u − kn+1 )+ | ζ p
p n dxd
Bn
t0
Z
1
≤
t0
Z
1
tn
Z
t0
tn t0
p
Z
(u − kn )+p−l dxdτ Bn
Z
p
u p−l
Bn
tn
−2
(u − kn )2+ dxdτ .
(2.6)
Bn
If u > 2kn , then
(u − kn )2+ ≥
u 2
(u − kn+1 )+ .
If kn+1 ≤ u ≤ 2kn
(u − kn )2+ ≥ (u − kn )+ (kn+1 − kn ) ≥ 2−n−3 u(u − kn+1 )+ . Hence
λ0
t0
Z
Z
tn
uq (u − kn+1 )+ ζnp dxdτ ≤ 2n γ
Z
t0 tn
Bn
Z
uq−1 (u − kn )2+ dxdτ .
(2.7)
Bn
Substituting (2.5)–(2.7) into (2.4), we obtain
Z
(u − kn+1 )2+ ζnp (x, τ )dx +
sup
tn <τ
ZZ
Bn
|D((u − kn+1 )+ ζn )|p dxdτ ≤ γ
Qn
2np
σ pt
(1 + M )
ZZ
(u − kn )2+ dxdτ ,
(2.8)
Qn
where
M = sup
0<τ
l −1 −2 q −1 p−l τ ρ −p ku(·, τ )kp∞, + τ k u (·, τ )k + τ k u (·, τ )k ∞,Bρ (x0 ) . Bρ (x0 ) ∞,Bρ (x0 )
By Gagliado–Nirenberg inequality [17, p. 64],
ZZ
ζ (u − kn+1 )+ dxdτ ≤ γ d n
b
ZZ
Qn
where b = p +
Qn 2p N
and d is large enough.
|D((u − kn+1 )+ ζn )| dxdτ p
Z
(u −
sup
tn <τ
Bn
) ζ (x, τ )dx 2
kn+1 + np
Np
,
(2.9)
3400
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
Set An = {(x, t ) ∈ Qn−1 : u(x, t ) ≥ kn }, and observe that
ZZ
(u − kn )2+ dxdτ ≥ γ 2−2n k2 |An+1 |.
(2.10)
Qn
Combining (2.9)–(2.10) with (2.8), we obtain
ZZ
(u − kn+1 )+ dxdτ ≤ 2
ZZ
Qn+1
2d
(u − kn+1 )2+ ζn b dxdτ
Qn
Z Z
(u −
≤ Qn
≤γ
) ζ b
kn+1 + nd dxd
2b (1+ Np )
1+M
k
σ pt
τ
2b
− 2(bb−2)
· |An+1 | 2
b−2 b
2 2 (b−2+p+ pN b
)
n Z Z
(u − kn )+ dxdτ 2
2p 1+ bN
.
Qn
If k is chosen to satisfy
ZZ u− Q0
k
2
2 +
dxdτ ≤ γ
1+M
−(1+ Np ) k
σ pt
N (b−2) p
.
(2.11)
Then by Lemma 5.6 of [17, p. 95], we get
ZZ
(u − kn )2+ dxdτ → 0,
as n → ∞,
Qn
i.e. kuk∞,Q∞ ≤ k. We select k = γ
N 1 +M 1 + p σ pt
RR
u dxdτ 2
Q0
p N (b−2)
, then (2.11) holds. It follows from this and Young’s
inequality that
kuk∞,Q∞ ≤ γ ≤γ
1+ Np Z Z
1+M
σ pt
1+M
! N (bp−2) u2 dxdτ
Q0
N +p N (b−2)
p ZZ N (b−2) kuk∞,Q0 udxdτ
σ pt
Q0
≤ kuk∞,Q0 + γ ()
1+M
N +κ p Z Z
σ pt
udxdτ
κp
,
Q0
where 0 < < 1. By an iteration process similar to [11, p. 393], we obtain
ku(·, t )k∞,B ρ ≤ γ
2
1+M
N +κ p Z Z
σ pt
udxdτ
κp
.
(2.12)
Q0
Step 2: For all 0 < t < T 0 , Bρ ⊂ RN , ρ = R(t ), (2.12) and (2.2) imply that
ku(·, t )k∞,B ρ ≤ γ t
− Nκ
2
0 N
! κp
Z tZ −
udxdτ
BR(t ) p
p
≤ γ t − κ R(t ) κ (N −θ) huitκ = γt
− p+θ(θp−2)
Γ
p
κ (N −θ)
p
huitκ .
Lemma 2.2 (The Case 1 ≤ l < p − 1). Let u be a nonnegative continuous weak subsolution of (1.1) in ST ∗ . Assume also that a time 0 < T 00 < min{T ∗ , 1} is given such that θ(p−2)
p−2
q −1
Γ −p t p+θ(p−2) ku(·, t )k∞,RN + t ku(·, t )k∞,RN ≤ 1,
∀0 < t < T 00 .
(2.20 )
Then
θ − ku(·, t )k∞,RN ≤ γ t p+θ(p−2) Γ
p
κ (N −θ)
p huitκ + t ,
where γ = γ (N , p, q, l, |λ|, θ , λ0 ), κ = N (p − 2) + p.
∀0 < t < T 00 ,
(2.30 )
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3401
Proof. For the case of 1 ≤ l < p − 1, we take the same test function as in Lemma 2.1 and let R(t ) ≤ ρ < 1, the differences between these two cases are the estimate of term |Du|l and the selection of k. Similarly to (2.6), we have
Z
t0
Z
tn
1
|Du|l (u − kn+1 )+ ζnp dxdτ ≤
Z
4
Bn
t0
Z
tn
|D(u − kn+1 )+ |p ζnp dxdτ + γ
t0
Z
Bn
tn
l 1+ p− l
Z
(u − kn )+
dxdτ .
Bn
By virtue of 1 ≤ l < p − 1, we obtain 1 p−1
≤
l p−l
< p − 1.
l If 1 ≤ p− < p − 1, by (2.20 ) and note that R(t ) ≤ ρ < 1, then for all tn < τ < t, we have l θ(p−2)
p−2
1 ≥ Γ −p τ p+θ(p−2) ku(·, τ )k∞,RN −2 = τ R(τ )−p ku(·, τ )kp∞, RN −2 ≥ τ R(t )−p ku(·, τ )kp∞, RN −2 ≥ τ ρ −p ku(·, τ )kp∞, , RN p
i.e., ku(·, τ )k∞,RN ≤ ρ p−2 τ
γ
Z
t0
Z
tn
1 − p− 2
l 1+ p− l
(u − k n )+
dxdτ ≤ γ ρ p−2
−1 p−l l
p−1 2
If p−l < 1, and k > t l
t0
Z
tn
t
l −1 p−l
Z
t t0
Z
1
1
Bn
1 since ρ < 1, 0 ≤ p− 2
Z
1
p
≤γ
γ
p
1
≤ 4 p−2 ρ p−2 t − p−2 , since τ > tn ≥ 4t . Hence
Z
tn
t0
tn
Z
(u − kn )2+ dxdτ
Bn
(u − kn )2+ dxdτ ,
Bn
− 1 < 1 and 0 < t < 1. in fact, here we take k = γ
l p−l
l 1+ p− l
(u − k n )+
dxdτ ≤ γ
N 1 +M 1 + p σ pt
1− p−l l Z
1 kn+1 − kn
Bn
≤ γ2
n1
Z
t
t0
Z
tn
RR
t0
Q0
Z
tn
u dxdτ 2
p N (b−2)
! + t , then
(u − kn )2+ ζ p dxdτ Bn
(u − kn )2+ dxdτ .
Bn
The above estimates, together with (2.4)–(2.5) and (2.7), still yield (2.8). But here M = sup
0<τ
−2 q −1 τ ρ −p ku(·, τ )kp∞, + τ k u (·, τ )k Bρ (x0 ) ∞,Bρ (x0 ) .
The following proof is similar as in Lemma 2.1 except the selection of k, and we obtain the iteration inequality
kuk∞,Q∞ ≤ kuk∞,Q0 + γ ()
1+M
N +κ p Z Z
σ pt
udxdτ
κp
+ t.
Q0
Then the rest of proof is similar to the proof of (2.12) and Step 2 in Lemma 2.1.
We also need the estimates of |Du|l as follows: Lemma 2.3 (The Case p − 1 ≤ l < p). If the assumptions of Lemma 2.1, p − 1 ≤ σ < p and θ (σ − p + 1) < p − σ hold, then for every Bρ (x0 ) ⊂ RN , 0 < t < T 0 , R(t ) ≤ ρ ≤ 1, if p ≥ 2 and σ 6= 1, we have
Z tZ 0
B ρ (x0 )
|Du|σ dxdτ ≤ γ G(t ) Γ
θ (N −θ)p p−σ − κ t 2σ −p p+θ(p−2)
p
huitκ
2σp−p
,
2
where G(t ) = sup0<τ
(2.13)
3402
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
If p = 2 and σ = 1, we have
Z tZ
1
|Du|dxdτ ≤ γ t 2 −h ρ 2h G(t ),
B ρ (x0 )
0
(2.130 )
2
where 0 < h <
1 , 2
γ = γ (N , p, q, l, |λ|, λ0 , h).
Proof. Assume p > 2, and set Bρ = Bρ (x0 ). The calculations to follow are formal in that u is required to be strictly positive. The calculations can be made rigorous by replacing u with u + and letting → 0. Take ϕ = t β ζ p ur as a test function in (1.4), where ζ is a piecewise smooth cut-off function in Bρ , such that 0≤ζ ≤1
in Bρ ,
ζ = 1 in B ρ ,
|Dζ | ≤
2
2
ρ
and β > 0, r > 0 are to be chosen. Thus we obtain
Z
1 1+r
=
t β ζ p ur +1 (x, t )dx + r
Z tZ 0
Bρ
β
Z tZ
1+r
0
τ β ζ p ur −1 |Du|p dxdτ + p
Z tZ 0
Bρ
τ β−1 ζ p ur +1 dxdτ + λ0
Z tZ
Bρ
0
τ β ζ p−1 ur |Du|p−2 DuDζ dxdτ Bρ
τ β ζ p uq+r dxdτ + λ
Z tZ
Bρ
0
τ β ζ p ur |Du|l dxdτ .
(2.14)
Bρ
Young’s inequality implies that
Z Z Z tZ Z tZ t r β p−1 r p−2 τ ζ u |Du| DuDζ dxdτ ≤ p τ β ζ p ur −1 |Du|p dxdτ + γ τ β ur +p−1 |Dζ |p dxdτ . 0 Bρ 4 0 Bρ 0 Bρ
(2.15)
Moreover
Z tZ
τ β ζ p ur |Du|l dxdτ ≤
|λ| 0
Bρ
r
Z tZ
4
0
τ β ζ p ur −1 |Du|p dxdτ + γ
Z tZ
Bρ
0
l
τ β ur + p−l dxdτ .
(2.16)
Bρ
Substituting (2.15)–(2.16) into (2.14), and by (2.2) and (2.3), we obtain
Z tZ 0
β p r −1
τ ζ u
|Du| dxdτ ≤ γ p
Z tZ
Bρ
0
≤γ
l
τ β−1 ur +1 (1 + τ ρ −p up−2 + τ uq−1 + τ u p−l −1 )dxdτ
Bρ
Z tZ 0
τ β−1 ur +1 dxdτ
Bρ
≤ γ G(t )
t
Z 0
≤ γ G(t )t
τ β−1 ku(·, τ )kr∞,Bρ dτ
β− p+θ(θ pr −2)
Γ
p(N −θ)r
κ
pr
huitκ ,
(2.17)
provided
β>
θr p + θ (p − 2)
.
(2.18)
Next we estimate by Hölder inequality,
Z tZ 0
|Du|σ ζ σ dxdτ =
Z tZ
Bρ
0
≤γ
σ
σ
σ
σ
τ β p ζ σ u(r −1) p |Du|σ τ −β p u−(r −1) p dxdτ Bρ
! σp
Z tZ 0
β p r −1
τ ζ u Bρ
|Du| dxdτ p
Z tZ 0
τ
σ −β p−σ
u
σ −(r −1) p−σ
!1− σp dxdτ
.
(2.19)
Bρ
Assume from now on
σ rσ −1≥ . p−σ p−σ
(2.20)
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3403
By Lemma 2.1, we have
Z tZ 0
σ
Z tZ
σ
τ −β p−σ u−(r −1) p−σ dxdτ =
0
Bρ
σ
σ
rσ
τ −β p−σ u1+ p−σ −1− p−σ dxdτ
Bρ
≤ G(t )
t
Z 0
σ −1− r σ p−σ
σ
p−σ τ −β p−σ ku(·, τ )k∞, Bρ
σ − θ 1−β p−σ p+θ(p−2) t
≤ γ G(t )
σ −1− r σ p−σ
dτ
p−σ
Γ
p(N −θ)
κ
σ −1− r σ p−σ
p−σ
p
κ t
hui
σ −1− r σ p−σ
p−σ
, (2.21)
provided
σ θ 1−β − p−σ p + θ (p − 2)
σ rσ −1− p−σ p−σ
> 0.
(2.22)
Thus by (2.17), (2.21) and (2.19), it is easy to obtain (2.13). In this following text, we will choose β and r such that (2.18), (2.20) and (2.22) hold. This is trivial task if θ = 0. Assume that θ > 0, and 0<β <
p−σ
σ
.
(2.23)
Then (2.18) and (2.20) are implied by r < min
p + θ (p − 2)
θ
β,
2σ − p
σ
, z.
Rewrite (2.22) as r >
p−σ
σ
θ 2σ − p p + θ (p − 2) · −1 +β , z0. p + θ (p − 2) p − σ θ
By virtue of θ (σ − p + 1) < p − σ , we obtain
θ
2σ − p
p + θ (p − 2) p − σ
− 1 < 0.
(2.24)
It is easy to check that z 0 < z. Finally we choose r > 0 such that r ∈ (z 0 , z ). As p = 2, the proof above also holds for the case 1 < σ < 2. If σ = 1, by (2.17), we have
Z tZ 0
τ β ζ 2 ur −1 |Du|2 dxdτ ≤ γ
Bρ
Z tZ 0
τ β−1 ur +1 dxdτ . Bρ
Using the method of the proof of Proposition 7.4 in [11], we can prove
Z tZ 0
1
B ρ (x0 )
|Du|dxdτ ≤ γ t 2 −h ρ 2h G(t ).
2
This finishes the proof.
Lemma 2.4 (The Case 1 ≤ l < p − 1). If the assumptions of Lemma 2.2 hold. Then for every Bρ (x0 ) ⊂ RN , 0 < t < T 00 , R(t ) ≤ ρ ≤ 1, we have
Z tZ 0
Bρ
|Du|p−1 dxdτ ≤ γ G(t ) Γ
1 (N −θ)(p−2) κ t p+θ(p−2)
p−2
huit κ + t
p−1 p
,
(x ) 2 0
where G(t ) = sup0<τ
Γ = C [µ] p+θ(p−2) , where C > 0 will be chosen depending on N , p, q, l, |λ|, λ0 , θ . (Obviously we may assume [µ] > 0 throughout.)
(2.25)
3404
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
Remark 2.1. For the case p − 1 ≤ l < p, it follows from (1.6) and (2.1) that huit ≤ [u]t . Using Lemma 2.1 and (2.25), we obtain p
N
[u]t ≤ huit + γ (C )[µ] κ (p−2) huitκ ,
∀0 < t < T 0 ,
(2.26)
where T 0 is as in Lemma 2.1. Remark 2.2. For the case 1 ≤ l < p − 1, we may choose
Γ = C [µ] +
p−1−l
p−2 p+θ(p−2)
p−1
.
Similar to Remark 2.1, we also have
N (p−2) p p−1−l κ [u]t ≤ huit + γ (C ) [µ] + huitκ , p−1
∀0 < t < T 00 ,
(2.260 )
where T 00 is as in Lemma 2.2. Let us conclude this section with the following lemmas, giving a priori bounds of a solution to (1.1)–(1.2) in terms of the initial data. Lemma 2.5 (The Case p − 1 ≤ l < p). Let u ≥ 0 be a bounded and uniformly continuous solution to (1.1)–(1.2) in ST ∗ . If θ(q − 1) < p + θ (p − 2) and θ (l − p + 1) < p − l. Then there exists T0 = T0 ([µ], N , p, q, |λ|, l, θ , λ0 ) < T ∗ such that
[u]t ≤ γ [µ],
∀0 < t < T0 ,
(2.27)
and (2.2)–(2.3) hold for all 0 < t < T0 , where γ = γ ([µ], N , p, q, |λ|, l, θ , λ0 ). Proof. As p > 2, define
t0 = sup 0 < T 0 < T ∗ |(2.2) holds, where Γ is given by (2.25) . Choose 0 < t < t0 and let Bρ ⊂ RN be any ball with radius R(t ) ≤ ρ ≤ 1, center at x0 ∈ RN . Take ζ as test function in (1.4), where ζ is a standard cut-off function in Bρ , with 0≤ζ ≤1
in Bρ ,
ζ = 1 in B ρ ,
2
|Dζ | ≤
2
ρ
.
Direct calculation shows that
Z Bρ
u(x, t )dx ≤
Z
dµ + Bρ
2
2
ρ
Z tZ 0
|Du|p−1 dxdτ + |λ|
Z tZ
Bρ
0
|Du|l dxdτ + λ0 Bρ
Z tZ 0
uq dxdτ .
(2.28)
Bρ
Multiplying ρ θ |Bρ |−1 on both sides of the above inequality and using Lemmas 2.1 and 2.3, we get
( p−2 Z θ(q−1) p+θ(p−2) huit κ 1− N ρ − u(x, t )dx ≤ 2 [µ] + γ huit C − κ + t p+θ(p−2) Γ [µ] Bρ 2 ) θ
+t
p−l θ(2l−p) p − p(p+θ(p−2))
Γ
(N −θ)(2l−p) κ
p(N −θ)(q−1)
κ
p(q−1)
huit
κ
2l−p
huit
κ
for all 0 < t < t0 , R(t ) ≤ ρ ≤ 1. By virtue of x0 ∈ RN is arbitrarily, it is immediately seen that
( huit ≤ 2 [µ] + γ huit C N
+t
p−l θ(2l−p) p − p(p+θ(p−2))
− p+θ(κp−2)
Γ
(N −θ)(2l−p) κ
huit [µ]
p−κ 2
2l−p
huit
+t
θ(q−1) 1− p+θ(p−2)
Γ
p(N −θ)(q−1)
κ
p(q−1)
huit
κ
)
κ
.
(2.29)
Set t1 = sup{0 < t < T ∗ |huit ≤ 2N +2 [µ]}
θ(q−1) 1− t2 = sup 0 < t < T ∗ |t p+θ(p−2) Γ
p(N −θ)(q−1)
κ
p(q−1)
huit
κ
+t
p−l θ(2l−p) p − p(p+θ(p−2))
Γ
(N −θ)(2l−p) κ
2l−p
huit
κ
<δ ,
(2.30)
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3405
where δ > 0 (small) is to be chosen. Note that t1 , t2 are well defined because the stipulated assumptions make sure that huit is continuous in [0, T ∗ ], and the exponent of t in (2.30) is positive. Let t3 = min{t0 , t1 , t2 }. Then for 0 < t < t3 , we have
γC
− p+θ(κp−2)
huit [µ]
p−κ 2
≤ γ C−
p+θ(p−2)
κ
(2N +2 )
p−2
κ
≤
1 4
,
provided C is suitably chosen. Then if we choose δ < 41γ , it follows from (2.29),
∀0 < t < t3 .
huit ≤ 2N +1 [µ],
(2.31)
By (2.26) and (2.31), we get
[u]t ≤ γ [µ],
∀0 < t < t3 .
Therefore all the claims made in the statement will follow, using the sup-estimates proven in this section, provided we show that we can indeed find a quantitative estimates below t3 ≥ T0 , with T0 as above. We may assume t3 < T ∗ , since the estimate is otherwise trivial. First we note that (2.31) implies t3 < t1 . Next, let us rule out the case t3 = t0 . To this end, we show that, for a suitable choice of C and δ , the left-hand side of (2.2) is bounded away from 1 up to t = t3 . As a matter of fact, Lemma 2.1, together with (2.31), implies for t ≤ t3 , x ∈ RN , θ(p−2)
l
Γ −2 t p+θ(p−2) u(x, t )p−2 + tu(x, t )q−1 + tu(x, t ) p−l
−1
p+θ(p−2)
1−
θ(q−1)
p(N −θ)(q−1)
p(q−1)
κ + t p+θ(p−2) Γ huit ≤ γ C− κ p−p l 2l − p θ(2l−p) p−l (N −θ)(2l−p) κ p − p(p+θ(p−2)) κ + t huit Γ p+θ(p−2) p + δ + δ p−l ≤ γ C− κ
≤
1 2
κ
.
(2.32)
If C is chosen large enough, and δ in the definition of t2 is chosen small enough, we can guarantee the last inequality in (2.32). Finally, we are left with the task of estimating below t3 = t2 . This can be accomplished at once, by replacing huit with 2γ [µ] in the definition of t2 in (2.30), owing to (2.31). As p = 2, similar to the above proof. Lemma 2.6 (The Case 1 ≤ l < p − 1). Let u ≥ 0 be a bounded and uniformly continuous solution to (1.1)–(1.2) in ST ∗ . If θ(q − 1) < p + θ (p − 2), then there exists T00 = T00 ([µ], N , p, q, |λ|, l, θ , λ0 ) < T ∗ such that
[u]t ≤ γ ([µ] + 1),
∀0 < t < T00 ,
and (2.20 ), (2.30 ) hold for all 0 < t < T00 , where γ = γ ([µ], N , p, q, |λ|, l, θ , λ0 ). Proof. Define t00 = sup{0 < T 00 < min{T ∗ , 1}|(2.20 ) holds, where Γ is given in Remark 2.2}. Choose 0 < t < t00 and let Bρ ⊂ RN be any ball with radius R(t ) ≤ ρ ≤ 1, center at x0 ∈ RN . Take ζ as test function in (1.4), where ζ is as in Lemma 2.5. We also obtain (2.28), then Young’s inequality implies that
Z Bρ
u(x, t )dx ≤
Z Bρ
2
γ dµ + ρ
Z tZ |Du| 0
p−1
dxdτ + λ0
Bρ
Z tZ 0
uq dxdτ +
p−1−l
Bρ
p−1
|Bρ |,
since ρ < 1. Multiplying ρ θ |Bρ |−1 on both sides of the above inequality, using Lemmas 2.2 and 2.4 and by virtue of x0 ∈ RN is arbitrarily, we get
huit ≤ 2N [µ] + γ huit
C−
+ 2N
p−1−l p−1
for all 0 < t < t00 , R(t ) ≤ ρ ≤ 1.
p+θ(p−2)
κ
huit [µ] +
p−1−l p−1
! p−κ 2 + Γ −1 t
p−1 1 p − p+θ(p−2)
+t
θ(q−1) 1− p+θ(p−2)
Γ
p(N −θ)(q−1)
κ
p(q−1)
huit
κ
(2.33)
3406
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
Set
t10 = sup 0 < t < min{T ∗ , 1}|huit ≤ 2N +2
θ(q−1) 1− t p+θ(p−2)
t20 = sup 0 < t < min{T ∗ , 1}|
Γ
[µ] +
p−1−l
p(N −θ)(q−1)
κ
p−1 p(q−1)
huit
κ
+ Γ −1 t
p−1 1 p − p+θ(p−2)
<δ ,
(2.34)
where δ > 0 (small) is to be chosen. Note that t10 , t20 are well defined because the stipulated assumptions make sure that huit is continuous in [0, T ∗ ], and the exponent of t in (2.34) is positive. Let t30 = min{t00 , t10 , t20 }. Then for 0 < t < t30 , we have
γ C−
! p−κ 2
huit
p+θ(p−2)
κ
[µ] +
≤ γ C−
p−1−l p−1
p+θ(p−2)
κ
(2N +2 )
p−2
κ
≤
1 4
,
provided C is suitably chosen. Then if we choose δ < 41γ , it follows from (2.33),
p−1−l huit ≤ 2N +1 [µ] + , p−1
∀0 < t < t30 .
(2.35)
By (2.260 ) and (2.35), we get
[u]t ≤ γ ([µ] + 1),
∀0 < t < t30 .
Therefore all the claims made in the statement will follow, using the sup-estimates proven in this section, provided we show that we can indeed find a quantitative estimates below t30 ≥ T00 , with T00 as above. We may assume t30 < T ∗ , since the estimate is otherwise trivial. First we note that (2.35) implies t30 < t10 . Then the rest of the proof is similar as in Lemma 2.5, we omit the details. 3. Proof of Theorems 1.1 and 1.2 In this section, we will give the proof of Theorems 1.1 and 1.2 by an approximation method. Proof of Theorem 1.1. Consider the approximating problems:
(P)
unt − div(|Dun |p−2 |Dun |) = λ0 min{uqn , n} + λ min{|Dun |l , n} u (x, t ) = 0 un (x, 0) = u (x) n 0n
in Bn × (0, T ), in ∂ Bn × (0, T ), on Bn ,
where Bn = {x ∈ RN | |x| < n} and u0n ∈ C0∞ (RN ) is nonnegative and has compact support in Bn , and satisfies
Z lim
n→∞
RN
u0n η(x)dx =
Z RN
η(x)dµ,
∀η ∈ C0∞ (RN ),
and
[u0n ] ≤ γ (N )[µ]. By Theorem 1.1 of Chapter 3 in [18], problem (P) has a nonnegative solution un ∈ C (0, T ; L2 (Bn )) ∩ Lp (0, T ; W 1,p (Bn )). By Theorem 3.2 of Chapter 5 in [15], we have un ∈ L∞ (Bn × (0, T )), here the bounds depend upon n. By the results of [14, 15] and Remark 1.4, one obtains that un is Hölder continuous in B¯ n × [0, T ). We will regard un as defined in the whole ST (= RN × (0, T )) by extending them to be zero outside Bn . From the above arguments we obtain that every un satisfies the conditions in Lemmas 2.1, 2.3 and 2.5. Therefore by Lemma 2.1, for any compact set K ⊂ ST0 , we have
kun k∞,K ≤ γ (K , [µ]),
(3.1)
where T0 is defined as in Lemma 2.5. Note that T0 and γ in (3.1) are independent of n. For each compact set K contained in ST0 , there exists a natural number n0 such that K ⊂ Bn0 ×(0, T0 ). Combining (3.1) with the results in [14,15] and Remark 1.4, we get uniformly Hölder estimates for the sequence {un }n>n0 in each K , we may assume that un → u
uniformly on K .
(3.2)
For every bounded open set Ω and all 0 < s < t < T0 , then there exists a compact set K ⊂ ST0 such that Ω × (s, t ) ⊂⊂ K . Take ϕ = un ζ 2 as test function in (P), where ζ ∈ C0∞ (K ) with 0≤ζ ≤1
in K ,
ζ = 1 in Ω × (s, t ),
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3407
we can deduce that
ZZ
|Dun |p ζ 2 dxdτ ≤ 2
ZZ
|Dun |p−1 ζ |Dζ |dxdτ +
ZZ
u2n ζ |ζτ |dxdτ
K
K
K
ZZ
|Dun |l un ζ 2 dxdτ + λ0
+ |λ|
ZZ
unq+1 ζ 2 dxdt .
(3.3)
K
K
By Lemma 2.3, (3.1) and (3.3), we obtain
kDukp,Ω ×(s,t ) ≤ γ (K , [µ]).
(3.4)
Assume that un , uk are two solutions of (P) with initial value u0n and u0k respectively, we have
(un − uk )t − div(|Dun |p−2 Dun − |Duk |p−2 Duk ) = λ0 (min{uqn , n} − min{uqk , k}) + λ(min{|Dun |l , n} − min{|Duk |l , k}).
(3.5)
Multiply (3.5) by (un − uk )ζ 2 , we get
ZZ
ZZ |un − uk |2 ζ |ζτ |dxdτ + γ (|Dun |p−1 + |Duk |p−1 )ζ |Dζ | |un − uk |dxdτ K K ZZ + λ0 | min{uqn , n} − min{uqk , k}| |un − uk |ζ 2 dxdτ Z ZK + |λ| | min{|Dun |l , n} − min{|Duk |l , k}| |un − uk |ζ 2 dxdτ K ZZ ≤ γ {|un − uk | |ζτ | + (|Dun |p−1 + |Duk |p−1 )|Dζ | + uqn + uqk + |Dun |l
|Dun − Duk |p ζ 2 dxdτ ≤ γ
K
ZZ
K
+ |Duk |l }|un − uk |ζ dxdτ → 0 as n, k → ∞.
(3.6)
The limit relation in (3.6) is due to Lemma 2.3, (3.1) and (3.2). Therefore by extracting a subsequence if necessary, we have Dun → Du,
a.e. in Ω × (s, t ) and strongly in Lp (Ω × (s, t )).
(3.7)
Thus by (3.1)–(3.2), (2.3) and (3.7), (2.27), let n → ∞, we can prove (1.4), (1.7) and (1.8). The proof of (1.5) as follows: For ∀ η ∈ C01 (RN ), assume that there exists some natural number n0 such that suppη ⊂ Bn0 , taking η as a test function in (3.5) (here n, k ≥ n0 ), we obtain
Z
(un (x, t ) − uk (x, t ))ηdx = −
Z tZ
Bn0
0
+ λ0
(|Dun |p−2 |Dun | − |Duk |p−2 |Duk |)Dηdxdτ Bn0
Z tZ 0
(min{uqn , n} − min{uqk , k})ηdxdτ + λ Bn0
Z tZ 0
− min{|Duk |l , k})ηdxdτ +
Z
(min{|Dun |l , n} Bn0
(u0n − u0k )ηdx. Bn0
Using Young’s inequality, Lemmas 2.1, 2.3 and 2.5 (with ρ = 1), furthermore, we can find finite balls B1 (x1 ), B1 (x2 ), . . . , B1 (xk0 ) covering B¯ n0 , then we have
Z Z Z Z tZ t (un (x, t ) − uk (x, t ))ηdx ≤ (|Dun |p−1 + |Duk |p−1 )|Dη|dxdτ + λ0 (uqn + uqk )|η|dxdτ Bn 0 B 0 B n n 0 0 0 Z Z tZ + |λ| (|Dun |l + |Duk |l )|η|dxdτ + (u − u0k )ηdx Bn 0n 0 Bn0 0 Z Z tZ Z tZ q l l q ≤γ (|Dun | + |Duk | )dxdτ + γ (un + uk )dxdτ + γ t + (u0n − u0k )ηdx Bn 0 Bn0 0 Bn0 0 ! ! Z Z Z Z θ(q−1) p−l 1− ≤ γ t p sup un dx + uk dx + γ t p+θ(p−2) sup un dx + uk dx 0<τ
Bn0
Bn0
0<τ
Bn0
Bn0
3408
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
Z +γt + (u0n − u0k )ηdx Bn 0 Z ≤ C (t ) + (u0n − u0k )ηdx Bn 0
where C (t ) is a positive constant independent of n and k, and limt →0 C (t ) = 0. Hence as n, k ≥ n0 , we have
Z
RN
u(x, t )ηdx −
Z RN
Z Z ηdµ ≤ (u(x, t ) − un (x, t ))ηdx + (un (x, t ) − uk (x, t ))ηdx Bn Bn Z0 Z 0 Z + (uk (x, t ) − u0k )ηdx + u0k ηdx − ηdµ Bn Bn Bn0 0 Z Z 0 ≤ (u(x, t ) − un (x, t ))ηdx + (u0n − u0k )ηdx + C (t ) Bn Bn Z 0 Z0 Z u0k ηdx − ηdµ . (uk (x, t ) − u0k )ηdx + + Bn Bn Bn 0
0
0
Let n → ∞, t → 0, k → ∞ in the indicated order in the above inequality, we get (1.5). Thus the proof of Theorem 1.1 is completed. Proof of Theorem 1.2. With minor changes, Theorem 1.2 can be proved using the method above too, we omit the details. 4. Proof of Theorem 1.3 To prove Theorem 1.3, we need to prove the following lemma. Lemma 4.1. Let u ∈ C (ST ) be a nonnegative solution of (1.1)–(1.2), and let u(·, t ) converge to µ in the sense of measures, with
µ(Bρ (x0 )) ≥ ρ N −θ ,
0 < ρ < σ,
(4.1)
for some given x0 ∈ RN , , ρ, σ > 0, 0 ≤ θ ≤ N. Then p
u(x, t ) ≥ γ0 p+θ(p−2) t
− p+θ(θp−2)
,
(4.2)
p−2
1
where |x − x0 | ≤ ρ < σ , ρ = γ0 p+θ(p−2) t p+θ(p−2) , γ0 = γ0 (N , p, θ ), 0 < t < T . Proof. Let v(x, t ) be the weak solution of the following Cauchy problem,
vt − div(|Dv|p−2 Dv) = 0 v(x, 0) = µ
in ST = RN × (0, T ), on RN ,
(4.3)
where µ is a nonnegative Radon measure. By the result of [1], we have
Z − Bρ (x0 )
dµ ≤ γ
(
ρp
p−1 2
+
t
t
Np
ρp
inf
|x−x0 |<ρ
κp ) v(x, t ) ,
where κ = N (p − 2) + p. Combining (4.1), (4.4) and the definition of ρ , we obtain p
v(x, t ) ≥ γ0 p+θ(p−2) t
− p+θ(θp−2)
,
where choose γ0 suitably small. Note that u is the supersolution of (4.3), thus p
u(x, t ) ≥ v(x, t ) ≥ γ0 p+θ(p−2) t which yields (4.2).
− p+θ(θp−2)
,
(4.4)
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
3409
Proof of Theorem 1.3. If θ = 0, the conclusion is trivial. In the following we may assume θ > 0 and x0 = 0. From now on, all the balls are understood to be center at 0. Let ω, λ be the first eigenfunction and eigenvalue of the eigenvalue problem,
−1ω = λω ω=0
in Bρ , on ∂ Bρ .
The first eigenfunctions have a fixed sign and we may select a positive one normalized so that
Z
ωdx = 1. Bρ
It is known by classical results that λ = γ (N )ρ −2 , and
−N ω(x) ≤ γ (N )ρ , −1 ω(x) ≥ γ (N )ρ −N , |Dω(x)| ≤ γ (N )ρ −N −1 ,
∀x ∈ Bρ , ∀x ∈ B ρ , 2 ∀x ∈ Bρ .
(4.5)
Taking ϕ = ωζ s as a test function in (1.4), where s > l−pl+1 , ζ (x) is a cut-off function in B ρ and satisfy 2
0 ≤ ζ (x) ≤ 1
in B ρ , 2
ζ (x) ≡ 1 in B ρ ,
|Dζ | ≤
4
ζ is radially nonincreasing.
4
ρ
(4.6)
Then we have
Z
Z
uωζ s dx −
B ρ (t )
B ρ (t 0 )
2
uωζ s dx = −
Z tZ t0
|Du|p−2 Du(ζ s Dω + sωζ s−1 Dζ )dxdτ
Bρ
2
2
+λ
Z tZ t0
|Du| ωζ dxdτ + λ0 l
Bρ
s
Z tZ t0
2
uq ωζ s dxdτ ,
Bρ
2
where 0 < t < t < T . Applying Young’s inequality and using (4.5), noting s > l−pl+1 , we obtain 0
Z B ρ (t )
uωζ s dx −
Z
2
B ρ (t 0 )
uωζ s dx ≥ −γ
Z tZ t0
2
l l l l ω1− l−p+1 |Dω| l−p+1 + ωζ s− l−p+1 |Dζ | l−p+1 dxdτ
Bρ
2
+
λ
Z tZ
2
t0
|Du| ωζ dxdτ + λ0 l
Bρ
s
Z tZ t0
2
≥ −γ ρ
− l−pl+1
Bρ
uq ωζ s dxdτ
2
(t − t 0 ) +
λ
Z tZ
2
t0
Bρ
|Du|l ωζ s dxdτ +
2
Z tZ t0
Bρ
uq ωζ s dxdτ .
(4.7)
2
Due to the assumption [u]t < ∞, then
Z
Z udx ≤ B ρ (t )
Bρ (t )
2
udx ≤ γ ρ N −θ ,
0 < t < t∗ , 0 < ρ < 1,
(4.8)
for a suitable t∗ > 0 (t∗ < T ) to be chosen. Define A(t ) = {x ∈ B ρ |u(x, t ) > C0 ρ −θ }, 2
E (t ) = B ρ \ A(t ), 2
where C0 > 0 is to be chosen. Hence for 0 < t < t∗ , 0 < ρ < 1, (4.8) yields
|A(t )| ≤ γ C0−1 |B ρ |. 2
If C0 is chosen large enough, then we have
|E (t ) ∩ B ρ | ≥ 4
1 2
|B ρ |. 4
(4.9)
Let us use the following result in [12, p. 3921]: Assume that f ∈ W 1,r (B ρ ), f ≥ 0, r ≥ 1, then for all measurable sets 2 E0 ⊂ B ρ , f ≡ 0 in E0 , 4
3410
H. Shang, F. Li / Nonlinear Analysis 72 (2010) 3396–3411
Z Bρ
ρ −1 r Z r N +1 −1 f G(|x|)dx ≤ γ (N , r ) ρ |E0 | G(0)G 4
2
Bρ
|Df |r G(|x|)dx,
(4.10)
2
ρ
where G(·) : [0, 2 ) → (0, ∞) is nonincreasing. Taking f = (u − C0 ρ −θ )+ , r = l, E0 = E (t ) ∩ B ρ , G(|x|) = ωζ s in (4.10), by (4.5)–(4.6) and integrating (4.10) over (t 0 , t ), 4 we obtain
ρ
Z tZ
−l
t0
Bρ
(u −
C0 ρ −θ )l+ ωζ s dxdτ
≤γ
Z tZ t0
2
|Du|l ωζ s dxdτ ,
Bρ
(4.11)
2
for 0 < t 0 < t < t∗ , 0 < ρ < 1. Moreover, we note that (4.2) implies t
u(x, τ ) ≥ 2C0 ρ −θ ,
2
< τ < t,
1
|x| ≤ γ0 ρ,
for ρ = ρ(t ) = C1 t p+θ(p−2) ,
(4.12)
provided C1 > 0 is fixed large enough (in (4.12) γ0 depends on C1 ). From now on, we choose ρ as in (4.12). We also choose t∗ to ensure ρ < 1 for t < t∗ . Thus direct calculations lead us to
Z
γ
B ρ (τ )
(u − C0 ρ −θ )l+ ωζ s dx ≥
Z
2
B ρ (τ )
ul ωζ s dx,
t 2
< τ < t < t∗ .
(4.13)
2
By (4.7), (4.11), (4.13) and using (4.5), noting that λ0 ≥ 0, with the help of Hölder inequality, obtaining
Z
uωζ dx − s
B ρ (z ) 2
Z
uωζ dx ≥ −γ ρ s
B ρ ( 2t )
t 2
z−
t
2
+ γρ
−l
Z zZ t 2
2
≥ −γ ρ
for
− l−pl+1
< z < t. Thus
R
B ρ (z )
− l−pl+1
z−
t 2
+ γ ρ −l
Bρ
2
z
Z t 2
ul ωζ s dxdτ
l
Z Bρ
uωζ s dx dτ ,
2
uωζ s dx majorizes the solution y(z ) to
2
− l 0 −l l y = γ ρ y − γ ρ l−p+1 Z t y( ) = uωζ s dx. 2
in
t 2
,t , (4.14)
Bρ ( 2t )
Therefore, the solution y of the problem (4.14) is bounded over ( 2t , t ). It follows from Lemma 4.1 in [19] that
Z
n
B ρ ( 2t )
l
uωζ s dx ≤ γ max ρ l−1 t
− l−11
o p−l , ρ − l−p+1 .
(4.15)
2
By (4.5) and (4.12), we have
Z B ρ ( 2t )
uωζ s dx ≥ γ ρ −θ ,
(4.16)
2
for t small and ρ = ρ(t ) defined above. Let t → 0, it is easily seen that (4.15) and (4.16) imply θ (l − p + 1) ≤ p − l. This completes the proof of Theorem 1.3.
Acknowledgement The authors would like to thank the referee for some helpful suggestions which correct some mistakes and improve the presentation of the original manuscript. References [1] E. Di Benedetto, M.A. Herrero, On the Cauchy problem and initial traces for a degenerate parabolic equation, Trans. Amer. Math. Soc. 134 (1989) 187–224. [2] Junning Zhao, On the Cauchy problem and initial traces for the evolution p-Laplacian equations with strongly nonlinear sources, J. Differential Equations 121 (1995) 329–383.
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