Large time behavior of free boundary for a degenerate parabolic equation with absorption

Nonlinear Analysis 74 (2011) 5069–5080

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Large time behavior of free boundary for a degenerate parabolic equation with absorption Jiaqing Pan Institute of Mathematics, Jimei University, Xiamen, 361021, PR China

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abstract

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Article history: Received 6 January 2011 Accepted 28 April 2011 Communicated by Enzo Mitidieri

This work studies the large time behavior of free boundary and continuous dependence on nonlinearity for the Cauchy problem of a degenerate parabolic partial differential equation with absorption. Our objective is to give an explicit expression of speed of propagation of the solution and to show that the solution depends on the nonlinearity of the equation continuously. © 2011 Elsevier Ltd. All rights reserved.

MSC: 35K10 35K15 35K55 35K65 Keywords: Free boundary Dependence on nonlinearity of equation Degenerate parabolic Absorption

1. Introduction Consider the Cauchy problem of the nonlinear degenerate parabolic equation



ut = ∆(um ) − θ up u(x, 0) = u0 (x)

in Q , on RN ,

(1.1)

where θ is a given positive constant, Q = RN × R+ , N ≥ 1 and 1 < p < m,

0 ≤ u0 (x) ≤ L,

0<

∫ RN

u0 (x)dx < ∞.

(1.2)

We know that (1.1) has received extensive attentions during the past several decades, and many interesting results, such as the existence, uniqueness, blow-up property and large time behavior of the solutions, are obtained ([1–6] and therein). But the aim of the present work is to consider two other questions. The first one we are interested in is the continuous dependence on the nonlinearity of equation in (1.1). To be exact, if v(x, t ) is a solution to the Cauchy problem of linear heat equation with the same initial value

(P )



vt = ∆v v(x, 0) = u0 (x)

in Q , on RN ,

we want to discuss the difference between u and v in Q and give an explicit error-estimate. The earliest contribution to the subject was made, maybe, by Benilan and Crandall in 1981. If φn : R −→ R is continuous and nondecreasing for all E-mail addresses: [email protected], [email protected]. 0362-546X/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2011.04.072

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J. Pan / Nonlinear Analysis 74 (2011) 5069–5080

n = 1, 2, 3, . . ., φn (0) = 0, they proved that (p. 162 in [5])

‖un (·, t ) − u∞ (·, t )‖L1 (RN ) −→ 0,

as φn −→ φ∞ ,

where un are the solutions of the Cauchy problem



ut = ∆(φn (u)) u(x, 0) = u0 (x)

in Q , on RN .

However, as pointed in [1], the result of [5] is not written in terms of explicit estimates. To study the problem more precisely, Cockburn and Gripenberg [1] in 1999, discussed the Cauchy problem of the degenerate parabolic equation



vt = ∆(φ(v)) + ∇ · (Φ (v)) v(0) = h

in Q , on RN ,

(1.3)

where,

Φj ∈ C 1 (R, RN ),

Φj (0) = 0,

φj (0) = 0,

φj′ (t ) > 0 when t > 0

for j = 1, 2. The explicit estimate in [1] is

‖v1 (·, t ) − v2 (·, t )‖L1 (RN ) ≤ ‖h1 − h2 ‖L1 (RN ) + ‖h1 ‖TV (RN )     √   × t · sup ‖Φ1′ (s) − Φ2′ (s)‖L∞ (RN ) + 4 tN sup  φ1′ (s) − φ2′ (s) s∈I (h ) s∈I (h ) 1

(1.4)

1

for any t > 0, where I (h) = (−‖h− ‖∞ , ‖h+ ‖∞ ). Clearly, (1.4) tells us that the error-estimate ‖v1 (·, t ) − v2 (·, t )‖L1 (RN ) is not necessarily bounded as t −→ ∞. However, if we set φ(v) = v m (m > 1) and Φ = 0 in (1.3), the conservation of total mass [2] claims

‖vi (·, t )‖L1 (RN ) = ‖hi ‖L1 (RN ) i = 1, 2 for all t > 0. Therefore, the error-estimate (1.4) in this case should be bounded with respect to t > 0 uniformly and in fact,

‖v1 (·, t ) − v2 (·, t )‖L1 (RN ) ≤ ‖h1 ‖L1 (RN ) + ‖h2 ‖L1 (RN ) t > 0.

(1.5)

Comparing (1.4) and (1.5), we suspect that the power of t of the last term on the right hand side in (1.4) might be lower. Thus the present work will discuss this question more precisely and give an explicit error-difference formula. If u(x, t ) is a solution to (1.1), then the first result of the present work is

‖v − u‖L2 (RN ×(0,T )) ≤ [(m − 1) + T θ ]A,   N (m−2)+2 where A = O(T β ) with β = max 0, 2[N (m−1)+2] and v(x, t ) is the solution to problem (P ), v=



1

N ∫

√ 2 πt

RN

u0 (ξ )e−

(x−ξ )2 4t

dξ .

Denote the positivity set of u by Hu (t ) = {x ∈ RN : u(x, t ) > 0} t > 0.

(1.6)

The second aim of this paper is to discuss the large time behavior of Hu (t ) with respect to t. It is well known that the study on this topic has a long history. Clearly, if we consider the linear heat equation ut = ∆u, we see that u(x, t ) > 0 everywhere in Q if only its initial value u0 satisfies (1.2). That is to say, the speed of propagation of u(x, t ) is infinite in this case. However, this fact is not true for degenerate parabolic equations. For example, Gilding and Peletier [7,8] discussed the free boundary problems of degenerate parabolic equations

∂u ∂ 2 φ(u) = ∂t ∂ x2 and

∂u ∂ 2 um ∂ un = + 2 ∂t ∂x ∂x respectively. They proved that the speeds of propagation of the solutions are finite. But they got no explicit formulas. As to the case of the dimension N ≥ 1, Barenblatt function [9] B(x, t , C ) = t

−λ

[

C −κ

|x|2

] m−1 1

t 2µ +

(1.7)

J. Pan / Nonlinear Analysis 74 (2011) 5069–5080

5071

gives a source type solution to the Cauchy problem



Bt = ∆(Bm ) B(x, 0) = M δ(x)

in Q , on RN ,

where m > 1, [h]+ = max{h, 0} and

λ=

N N (m − 1) + 2

,

µ=

λ N

and C is such a positive constant that

 HB ( t ) =



x ∈ RN : |x| <



C µ t

,

κ=

RN

λ(m − 1)

(1.8)

2mN

Bdx = M. By (1.7), we see that



κ

and we get the exact expanding behavior of HB (t ):

 |x| =

C µ t

κ

x ∈ ∂ HB (t ).

(1.9)

This fact tells us that the solution B(x, t ) may retain its positivity at any given point when time increases. To extend the result of [9], Vazquez [2], employing the Barenblat function and comparison theorem, proved that the solution to the Cauchy problem



ut = ∆ um u(x, 0) = u0 (x)

in Q , on RN

also has bounded positivity set Hu (t ): c1 t µ ≤ |x| ≤ c2 t µ

x ∈ ∂ Hu (t ),

(1.10)

where u0 (x) satisfies (1.2) and, u0 (x) is supported in a bounded set of R . Here we see that the expanding behavior of Hu (t ) and HB (t ) is very similar. Moreover, if the initial value u0 satisfies some conditions [10], thereby the solution u(x, t ) is continuous, then (1.10) shows that every point in RN is eventually reached by the diffusing substance. However, for general ∑N ∑N ∂ ∂u ∂u parabolic equation ut = i,j=1 ∂ x (aij ∂ x ) + i=1 bi ∂ x + cu, whether the property will be remained or not, there are no N

j

i

i

other explicit results up until now to the knowledge of the author. It seems that the positivity set Hu (t ) will be smaller as t increased (or even, Hu (t0 ) will be empty for some t0 > 0) owing to the absorption −θ up of (1.1). Thus, the present work will investigate this problem and give an explicit formula: sup |x| ≥ (χ (t , θ ))µ

x∈Hu (t )

t > 0,

(1.11)

where χ (t , θ ) is a positive function with respect to t ∈ (0, ∞) and µ is defined by (1.8). In particular, χ (t , θ ) −→ C ∗ t as θ −→ 0, so (1.11) gives (1.9) in this case. In addition, (1.11) tells us that the positivity set of u(x, t ) will never be extinguished even if the equation has absorption −θ up . Besides the two conclusions mentioned above, the present work intends to prove some useful estimates. It is well known [11,2,12] that if u(x, t ) is a solution to the Cauchy problem of the equation ut = ∆(um ), then

∂u u ≥− . ∂t (m − 1)t

(1.12)

This inequality is very important for the study of the equation ut = ∆(um ), many interesting results can be obtained from it. However, few such estimates can be found for the general parabolic equations. We intend to consider this problem in this paper and will establish a similar one: ut ≥

θ (m − p)Lp−1 (m − 1)(e−θ(m−p)Lp−1 t − 1)

u.

Clearly, letting θ −→ 0 in this inequality gives ut ≥ (m−−11)t u, which is just (1.12). We say that a nonnegative function u(x, t ) ∈ C ([0, ∞) : L1 (RN )) is a solution to the Cauchy problem (1.1) with the initial value (1.2) in Q if (i) ut , um , ∆um ∈ L1loc ((0, ∞) : L1 (RN )); (ii) ut = ∆um − θ up in the sense of distributions in Q ; (iii) u(x, t ) −→ u0 (x) in L1 (RN ) as t −→ 0.

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J. Pan / Nonlinear Analysis 74 (2011) 5069–5080

In this paper, we will use the symbols: Q = RN × (0, ∞);

QT = RN × (0, T );

Bk = {x ∈ RN : |x| < k};

Qk = Bk × (0, ∞) for k > 0.

As is well known, |Bk | is the volume of Bk and



N 2

|Bk | = π Γ

1+

N

−1

2

kN .

(1.13)

We state our main conclusions below. Theorem 1. There exists a unique global solution u(x, t ) to the Cauchy problem (1.1) with (1.2), which has the properties: ut ≥

θ (m − p)Lp−1 (m − 1)(e−θ(m−p)Lp−1 t − 1)

u

(1.14)

in the sense of distributions. Theorem 2. If u(x, t ) is the solution to (1.1) with (1.2) on QT , then

‖v − u‖L2 (QT ) ≤ [(m − 1) + T θ]A

(1.15)

when T is large sufficiently, where v(x, t ) is the solution to the Cauchy problem (P) and A = O(T β ), N (m − 2) + 2



β = max 0,



2[N (m − 1) + 2]

.

Theorem 3. Let u0 satisfy (1.2) and supp u0 = Bε with ε > 0, u(x, t ) be the solution to the Cauchy problem (1.1) with (1.2). Then there is a positive function χ (t , θ ) such that sup |x| ≥ (χ (t , θ ))µ ,

(1.16)

x∈Hu (t )

where Hu (t ) and µ are defined by (1.6) and (1.8) respectively. In particular, χ (t , θ ) −→ C ∗ t as θ −→ 0, where

[ ∫ C = (m − 1) ∗

RN

u0 dx · π

− N2

Γ

 1+

N

]m−1

2

.

We organize the paper as follows. We first prove Theorem 1 in Section 2, and then, prove the continuous dependence on the nonlinearity of the equation in Section 3. Finally, we discuss the large time behavior of free boundary in Section 4. 2. Proof of Theorem 1 Although the proof of the existence of the solution to the problem (1.1) has been established by some works [2,13] with a standard procedure, we have to show the main steps in Lemma 2.1 because we need them to prove our main conclusions. Lemma 2.1 (The Existence and Uniqueness). For every given T > 0, the Cauchy problem (1.1) with (1.2) has a nonnegative solution u(x, t ) in QT . Proof. For every k > 2 and T > 0, set Qk,T = Bk × (0, T ),

Sk,T = ∂ Bk × (0, T ),

and

∫ u0,η =

RN

u0 (y)Jη (x, y)dy,

where,

 J (x) =

1

e |x|2 −1 0

|x| < 1, |x| ≥ 1,

u0η,k = u0η ζk ,

J. Pan / Nonlinear Analysis 74 (2011) 5069–5080

and Jη (x) = σ η1N J ( ηx ) with σ = properties: ζk (x) ∈ C0∞ (RN ),

 ζk (x) = 1 0 < ζk (x) < 1 ζk (x) = 0

 |x|<1

1 |x|2 −1

e

5073

dx for η > 0 and, {ζk }k>2 is a smooth cutoff sequence with the following

|x| ≤ k − 1, k − 1 < |x| < k, |x| ≥ k.

Clearly, u0η,k (x) −→ u0 (x) in L1 (RN )as η −→ 0 and k −→ ∞. Moreover, it is not difficult to see that the derivatives of the functions ζk up to second order are bounded with respect to x ∈ RN uniformly. Specially, there is a positive constant γ such that

|∇ζk | ≤

γ

γ

and |∆ζk | ≤

k

.

k2

Consider the Dirichlet problem

 ut = ∆(um ) − θ|u|p−1 u u(x, t ) = η u(x, 0) = u (x) + η 0η,k

in Qk,T , on Sk,T , in Bk .

(2.1)

The similar procedure (Th. 4 in Ch. II of [2]) yields that the Dirichlet problem (2.1) has a unique solution uη,k ∈ C ∞ (Qk,T ) ∩ C (Q k,T ). Letting k −→ ∞, η −→ 0 and employing the similar procedure of Ch. III in [2], we see that there exists a nonnegative function u(x, t ), which is the solution of the Cauchy problem (1.1) with (1.2) in Q and 0 ≤ u(x, t ) ≤ L

in QT .

To prove the uniqueness, we take two functions p(s) and j(r ) as [2]. If u and  u are two solutions to (1.1), then the similar procedure [2] gives

∫ RN

[u(x, t ) −  u(x, t )]+ ζk dx ≤

∫ RN

+

[u(x, t0 ) −  u(x, t0 )]+ ζk dx

γ k2

∫ t∫ t0

RN

|w|dxdt − θ

∫ t∫ t0

RN

[ up −  up ]+ ζk dxdτ t > t0 > 0.

(2.2)

Clearly, [up −  up ]+ ≤ pLp−1 [u −  u]+ thanks to 0 ≤ u, u ≤ L and p > 1. Using this inequality in (2.2) and letting k −→ ∞ and t0 −→ 0 yields

∫ RN

[ up −  up ]+ dx ≤ pLp−1

∫ RN

[u0 −  u0 ]+ dx − θ pLp−1

∫ t∫ RN

0

[up −  up ]+ dxdτ .

Finally, the Gronwall inequality gives

∫ RN





∫

[ u0 −  u0 ]+ dx e−pθ L

p−1 t

.

(2.3)

  ∫ p−1 [ up −  up ]− dx ≤ pLp−1 [ u0 −  u0 ]− dx e−pθ L t .

(2.4)

p

p

[u −  u ]+ dx ≤ pL

p−1

RN

Similarly,

∫ RN

RN

Combining (2.3) and (2.4) gives a L1 -contraction principle:

∫

p

RN

p



|u −  u |dx ≤ pL

p−1

Now (2.5) shows the uniqueness.



∫ RN

| u0 −  u0 |dx · e−pθ L

p−1 t

t > 0.



Lemma 2.2. If u(x, t ) is the nonnegative solution of problem (1.1) with (1.2), then ut ≥

θ (m − p)Lp−1 (m − 1)(e−θ(m−p)Lp−1 t − 1)

in the sense of distributions in Q .

u

(2.5)

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J. Pan / Nonlinear Analysis 74 (2011) 5069–5080

Proof. For every given T > 0, suppose that uη,k is the solution of the Dirichlet problem (2.1) in Qk,T . We first prove

∂ θ (m − p)(L + η)p−1 uη,k ≥ uη,k in Qk,T . ∂t (m − 1)(e−θ(m−p)(L+η)p−1 t − 1)

(2.6)

To do this, we set V = (uη,k )m

q=

and

Vt V

.

Clearly, q(x, t ) = 0

on Sk,T .

(2.7)

For every given t > 0, set

Ωk+ = {x ∈ Ω : q(x, t ) > 0}, Ωk− = {x ∈ Ω : q(x, t ) < 0}, Ωk0 = {x ∈ Ω : q(x, t ) = 0}. Thereby, Qk,T = (Ωk+ ∪ Ωk0 ) × (0, T ) ∪ Ωk− × (0, T ). Owing to m > p > 1, the right hand side of (2.6) is negative, so (2.6) is true for (x, t ) ∈ (Ωk+ ∪ Ωk0 ) × (0, T ). Therefore, we next prove (2.6) for (x, t ) ∈ Ωk− × (0, T ) only. p

It follows from Vt = V ′ (∆V − θ uη,k ) that q = qt =

=

V′ V

[∆V − θ (upη,k )]. Thus,

  V ′′ · (u ) · V − V ′ · V  η,k t t p p ∆Vt − θ (uη,k )t + ∆ V − θ ( u ) η, k 2

V′  V V V

V

[∆V − θ (upη,k )]2

′

∆Vt +

V2

[VV ′′ − (V ′ )2 ] − θ

V′ V

(upη,k )t .

Since ∆Vt = ∆(qV ) = V ∆q + 2∇ V · ∇ q + q∆V , we obtain qt = V ′ ∆q + 2

V′ V V′

∇V · ∇q +

V′ V

q∆V +

[∆V − θ (upη,k )]2 V2

[VV ′′ − (V ′ )2 ] − θ p

V′ V

p−1

uη,k (uη,k )t

[ ] V′ q2 ∇ V · ∇ q + q q + θ (uη,k )p + ′ 2 [VV ′′ − (V ′ )2 ] − θ pqupη,−k1 V V (V ) [ ′ ] ′ ′′ V VV V = V ′ ∆q + 2 ∇ V · ∇ q + q2 ′ 2 + θ q (uη,k )p − pupη,−k1 V (V ) V = V ′ ∆q + 2

= V ′ ∆q + 2

V′ V

∇V · ∇q +

m−1 m

p−1

q2 + θ (m − p)uη,k q.

(2.8) p−1

Recalling q < 0 in this case and uη,k ≤ L + η and m > p > 1, we have (m − p)uη,k q ≥ (m − p)(L + η)p−1 q. Thus, qt ≥ V ′ ∆q + 2

V′ V

∇V · ∇q +

m−1 m

q2 + θ (m − p)(L + η)p−1 q.

(2.9)

q + θ (m − p)(L + η)p−1 q.

(2.10)

Moreover, q=0

on ∂ Ωk− × (0, T ).

Consider the equation qt = V ′ ∆q + 2

V′ V

∇V · ∇q +

m−1 m

It is easy to see that the function q∗ =

mθ (m − p)(L + η)p−1

(m − 1)(e−θ(m−p)(L+η)p−1 t − 1)

2

J. Pan / Nonlinear Analysis 74 (2011) 5069–5080

5075

satisfies Eq. (2.10) in Ωk− × (0, T ) and q∗ (x, 0) = −∞, q∗ (x, t ) < 0

on ∂ Ωk− × (0, T ).

Although the domain Ωk− × (0, T ) might not be a cylinder in the space RN × R+ , the comparison theorem (Th. 16 in Ch. 2 of [14]) is also applicable for this situation. For these reasons, the comparison theorem claims q ≥ q∗ , this means

∂ θ (m − p)(L + η)p−1 uη,k ≥ uη,k in Ωk− × (0, T ). ∂t (m − 1)(e−θ(m−p)(L+η)p−1 t − 1)

(2.11)

Thus (2.6) holds in Ωk− × (0, T ). Finally, letting η −→ 0 and k −→ ∞ in (2.6) gives the result of our lemma. Proof of Theorem 1. Combining Lemma 2.1–Lemma 2.2 gives Theorem 1.





In order to prove our Theorems 2 and 3, we first give some Ls -estimates about u(x, t ) for 0 < s ≤ 1 and s = 2 and s = ∞. We first take the cutoff functions ζk defined in Lemma 2.1, and integrate by parts as follows.

∫ RN

∫ t∫

(u − u0 )ζk dx =

RN

0

∫ t∫ = RN

0

[∆(um ) − θ up ]ζk dxdτ [um ∆ζk − θ up ζk ]dxdτ .

(2.12)

Recalling 0 ≤ u ≤ L and p > 1, we get

∫ RN

∫ t∫

(u − u0 )ζk dx ≥

RN

0

[um ∆ζk − θ Lp−1 uζk ]dxdτ .

Since the definition of the solution tells us um ∈ L1 (RN ), we have inequality yields

∫



RN



udx ≥

RN

u0 dx − L

∫

p−1 udx ≥ e−L θ t

θ

RN

0



RN

um ∆ζk dx −→ 0 as k −→ ∞. Therefore, the above

udxdt. Thereby, the Gronwall inequality implies

t > 0.

u0 dx

RN

p−1

t 

RN

(2.13)

As to the case of 0 < s < 1, we can use (2.12) and get

∫

s

u dx ≥ L

s−1

∫ udx

RN

RN s−1 −Lp−1 θ t

≥L

∫

e

u0 dx RN

t > 0.

(2.14)

Recalling (2.13), we know that (2.14) is true for all s ∈ (0, 1]. To obtain L2 -estimate, we multiply the equation in (1.1) by u(x, t ) and integrate by parts in RN , we get d

∫

dt

u2 dx = − RN

∫

2 m

RN

u1−m |∇ um |2 dx − 2θ

∫ RN

up+1 dx.

Recalling 0 ≤ u ≤ L and m > 1, we get

∫

2

u dx + RN

2L1−m

∫

m

|∇ u | dxdt + 2θ m 2

QT

∫ u QT

p+1

∫ dxdt ≤ RN

u20 dx.

(2.15)

Finally, if θ = 0 in Eq. (1.1), we have a well-known L∞ -estimate (Th. 9 in Ch. III in [2]) u(x, t ) ≤ ct

− N (m−N1)+2

,

(2.16)

where the positive constant c is bounded with respect to m > 1. Since θ > 0 in (1.1), the maximum principle claims that the solution to (1.1) is less than that one of θ = 0. Therefore, (2.16) is also true for our solution with θ > 0.

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J. Pan / Nonlinear Analysis 74 (2011) 5069–5080

3. Proof of Theorem 2 We prove Theorem 2 directly. Proof of Theorem 2. For every given T > 0, let G(x, t ) = v − um , t

∫

ψ(x, t ) =

G(x, τ )dτ

0 < t < T.

T

Since (v − u)t = ∆G + θ up in the sense of distributions in QT , we multiply the equation by ψζk (where {ζk }k>2 is defined in our Lemma 2.1) and integrate by parts in QT , we obtain

∫

(ζk ∇ G · ∇ψ + ψ∇ G · ∇ζk − θ u ψζk )dxdt = p

∫

(v − u)Gζk dxdt .

(3.1)

QT

QT

Set T

∫

I (T ) =

∫

ζk ∇ G · ∇ψ dxdt .

RN

0

It is easy to see that I (T ) =

T

∫ 0

=−

1

∫ ∫R

2

N

RN

ζk

∂ (∇ψ) · ∇ψ dxdt ∂t

ζk |∇ψ(x, 0)|2 dx.

Hence, it follows from (3.1) that

∫

(v − u)Gζk dxdt ≤

∫

  |ψ∇ G · ∇ζk | + θ |up ψζk | dxdt QT ∫ γ |up ψζk |dxdt . ≤ ‖∇ G‖L2 (QT ) · ‖ψ‖L2 (QT ) + θ

QT

k

QT

Recalling (2.15), we have a ν > 0 such that

∫

ν

(v − u)Gζk dxdt ≤

+θ

k

QT

∫

|up ψζk |dxdt , QT

where ν does not depend on k. On the other hand,

∫

(v − u)Gζk dxdt =

∫

QT

∫

(v − u)2 ζk dxdt + QT

(v − u)(u − um )ζk dxdt . QT

Letting k −→ ∞, we conclude that

∫

(v − u) dxdt ≤ 2

QT

∫

∫

|(v − u)(u − u )|dxdt + θ m

u

QT

≤

1



|v − u |dτ dxdt m

t

[(v − u) + (u − u ) ]dxdt + θ 2

T

∫

QT

∫

2

p

m 2

QT

∫ u

p

T

∫

QT



|v − u |dτ dxdt . m

0

Thus,

∫

(v − u)2 dxdt ≤ QT

∫

(u − um )2 dxdt + 2θ

∫

QT

up

T

∫

QT

 |v − um |dτ dxdt .

(3.2)

0

To estimate the second term of right hand side of (3.2), we see that 2θ

∫

up QT

T

∫

T

 ∫ |v − um |dτ dxdt = 2θ

0

0

≤ 2θ

0

T

∫ RN

0 T

∫

∫

up (x, t )[v(x, τ ) − um (x, τ )]dxdτ dt

] 21

[∫

u2p dx RN

T

∫

[∫

|v − um |2 dx

dt · 0

RN

] 21

dτ

J. Pan / Nonlinear Analysis 74 (2011) 5069–5080

5077

 ∫ ∫ ∫ ∫  21 2  21 2 T T 1 2p m 2 u dx |v − u | dx dτ ≤ θ dt +ε ε 0 RN 0 RN  ∫ ∫  1 2 ∫ ≤

≤ for every ε > 0. Letting ε = 2θ

∫ u

p

1 4T

θ

ε

 ∫ θ

ε

T

|v − um |2 dxdt

QT

∫

2

 21

2p

+ 2T ε

dt

u dx RN

0



|v − u |dτ dxdt ≤ 4T θ m

2

∫

∫

(|v − u|2 + |u − um |2 )dxdt QT

T

∫

2

 12

2p

u dx

+

dt

RN

0

0

QT

+ Tε

dt

RN

0

1

2

u2p dx

in this inequality, we have

T

∫

T

1

1 2

∫

(|v − u|2 + |u − um |2 )dxdt . QT

Using this inequality in (3.2) yields

∫

∫

(v − u) dxdt ≤ 3 2

(u − u ) dxdt + 8T θ m 2

2

∫

∫

= 3(m − 1)2

∫

2p

u dx

(uδ ln u)2 dxdt + 8T θ 2

∫

QT

≤ 3(m − 1)2

∫

2

 21 dt

RN

0

QT

QT

T

T

∫

QT

u2p dx

dt

RN

0

(uδ ln u)2 dxdt + 8T 2 θ 2

2

 21

∫

u2p dxdt

(3.3)

QT

with 1 < δ < m. To estimate the first term of the right hand side of (3.3), we see that there is a C1 > 0 such that u2δ−2 | ln u|2 ≤ C1 thanks to δ > 1 and 0 ≤ u ≤ L. Hence,

∫

(uδ ln u)2 dxdt ≤ C1

∫

QT

u2 dxdt QT 1

∫

∫

u2 dxdt +

= C1 RN

0

Letting k −→ ∞ in (2.12) we have 1∫

∫

RN

0



RN

T

∫

∫

u2 dxdt RN

1



.

(3.4)

udx ≤ ‖u0 ‖L1 (RN ) , thus,

u2 dxdt ≤ L‖u0 ‖L1 (RN ) .

(3.5)

Moreover, (2.16) yields T

∫ 1

∫

u2 dxdt ≤ c

∫

RN

T

t

− N (m−N1)+2

∫ dt

udx RN

1 T

∫ ≤ c ‖u0 ‖L1 (RN ) ≤ c ‖u0 ‖L1 (RN )

t

− N (m−N1)+2

dt

1

 N (m − 1) + 2  N (m−2)+2 T N (m−1)+2 − 1 . N (m − 2) + 2

(3.6)

Using (3.5) and (3.6) in (3.4), we get

∫

[ ] N (m − 1) + 2  N (m−2)+2 (uδ ln u)2 dxdt ≤ C1 L + c T N (m−1)+2 − 1 ‖u0 ‖L1 (RN ) . N (m − 2) + 2 QT

(3.7)

Employing (3.5) and (3.6) again, we obtain

∫

2p

u dxdt = L QT

2p−2

1

∫

∫

2

∫

T

∫

2

u dxdt + 0

RN



u dxdt 1

RN

[ ] N (m − 1) + 2  N (m−2)+2 ≤ L2p−2 L + c T N (m−1)+2 − 1 ‖u0 ‖L1 (RN ) . N (m − 2) + 2

(3.8)

5078

J. Pan / Nonlinear Analysis 74 (2011) 5069–5080

Using (3.7) and (3.8) in (3.3), we get

∫

  (v − u)2 dxdt ≤ (m − 1)2 + T 2 θ 2 A2

(3.9)

QT

with

[

A2 = max(3C1 , 8L2p−2 )‖u0 ‖L1 (RN ) L + c (3.9) gives (1.15).

] N (m − 1) + 2  N (m−2)+2 T N (m−1)+2 − 1 . N (m − 2) + 2



4. Proof of Theorem 3 If u(x, t ) is the solution to (1.1), we can first get a rough description on the expanding behavior of Hu (t ). In fact, for every t2 > t1 > 0, our Lemma 2.2 implies

 ln

u(x, t2 )





u(x, t1 )

≥ ln

eθ(m−p)L

1

−1

eθ(m−p)L

2

−1

p−1 t

p−1 t

 m−1 1 .

This means u(x, t2 ) · (eθ(m−p)L

1

1

− 1) m−1 ≥ u(x, t1 ) · (eθ(m−p)L t1 − 1) m−1 .   m−1 1 p−1 It follows from m > p > 1 that eθ(m−p)L t1 − 1 > 0. Thus, (4.1) claims the following fact: 

p−1 t

2

p−1

if u(x0 , t0 ) > 0, then u(x0 , t ) > 0 if u(x0 , t0 ) = 0, then u(x0 , t ) = 0

for all t > t0 , x0 ∈ RN , for all t < t0 , x0 ∈ RN .

(4.1)

(4.2)

Although (4.2) has told us that the positivity set Hu (t ) is expanding with respect to t > 0, thereby, the solution u(x, t ) will never vanish even if Eq. (1.1) has an absorption −θ up , we are interested in getting an explicit formula of expanding behavior of Hu (t ). To do this, we need to establish a more precise Poincare´ type inequality first. It is well known that there exists a positive constant α such that α‖φ‖2L2 (Ω ) ≤ ‖∇φ‖2L2 (Ω ) for φ ∈ H01 (Ω ). Recently, Wu (p. 13 in [15]) proved that

α = ρ −2

(4.3)

if Ω = {(x1 , x2 , . . . , xN ) ∈ R : ai < xi < ai + ρ}. In order to prove our Theorem 3 we will show that the choice (4.3) is also right if Ω is a sphere in RN . N

Lemma 4.1. Assume Bρ = {x ∈ RN : |x| < ρ}. If u ∈ H01 (Bρ ), then

‖u‖L2 (Bρ ) ≤ ρ‖∇ u‖L2 (Bρ ) .

(4.4)

Proof. By the similar procedure of [15], we can get (4.4) easily. So we omit the details.



Proof of Theorem 3. For every given t > 0, if Hu (t ) is unbounded, then the proof is finished. Thereby, we next suppose Hu (t ) to be bounded. Denote

|xt | = sup |x| t > 0. x∈Hu (t )

For every λ > 0, set ρ = λ + |xt |. Clearly, u = 0 on ∂ Bρ . By Lemma 2.2,

∫

um ∆(um ) − θ up dx ≥



θ (m − p)Lp−1



Bρ

(m − 1)(e−θ(m−p)Lp−1 t − 1)

∫

u1+m dx. Bρ

It follows from (4.4) that

∫ u

2m

dx + ρ θ 2

∫

Bρ

u

m+p

dx ≤ ρ

Bρ

2

θ (m − p)Lp−1

∫

(m − 1)(1 − e−θ (m−p)Lp−1 t )

u1+m dx

(4.5)

Bρ

for all t > 0. By the Hölder inverse inequality (p. 24 in [16]), we have

∫ u Bρ

m+p

 1m++mp

∫

m+1

dx ≥

u Bρ

dx

1−p

· |Bρ | 1+m

(4.6)

J. Pan / Nonlinear Analysis 74 (2011) 5069–5080

5079

and

∫ u

2m

 12m +m

∫ dx ≥

u

Bρ

m+1

1−m

· |Bρ | 1+m .

dx

(4.7)

Bρ

Using (4.6) and (4.7) in (4.5) gives

 12m +m

∫ u

m+1

1−m 1+m

· |Bρ |

dx

+ρ θ 2

 1m++mp

∫ u

Bρ

m+1

· | Bρ |

dx

1−p 1+m

≤ρ

 m1+−m1

∫ u

m+1

1−m 1+m

· |Bρ |

dx

∫

(m − 1)(1 − e−θ (m−p)Lp−1 t )

Bρ

Owing to u0 > 0 on Bε , (4.2) claims

θ (m − p)Lp−1

2

um+1 dx.

(4.8)

Bρ

um+1 dx > 0. Consequently, (4.8) yields

 Bρ

+ρ θ 2

 1p+−m1

∫ u

Bρ

m+1

θ (m − p)Lp−1

1−p

· |Bρ | 1+m ≤ ρ 2

dx

Bρ

(m − 1)(1 − e−θ (m−p)Lp−1 t )

.

(4.9)

On the other hand, using the Hölder inverse inequality again yields

∫ u

m+1

 1+s m

∫

s

dx ≥

· |Bρ |

u dx

Bρ

s−1−m s

for all s ∈ (0, 1 + m).

Bρ

Thus, using this estimate on the left of (4.9) gives

 ms−1

∫

1

us1 dx

|Bρ |

(s1 −1−m)(m−1) 1−m + 1+m s1 (1+m)

+ ρ2θ

Bρ

|Bρ |

(s2 −1−m)(p−1) 1−p + 1+m s2 (1+m)

θ (m − p)L

p−1

(m − 1)(1 − e−θ(m−p)Lp−1 t ) N (p−1) N (m−1)+2

for 0 < s1 , s2 ≤ 1. Letting s1 = 1, s2 =

RN

2

us2 dx

Bρ

≤ ρ2

[∫

 ps−1

∫

udx · π

− N2

≤ ρ 2+N (m−1)

Γ

 1+

N

]m−1

+θ

2

θ (m − p)L

and recalling (1.13), we have

[∫ u

N (p−1) N (m−1)+2

RN

p−1

(m − 1)(1 − e−θ(m−p)Lp−1 t )

dxπ

− N2

Γ

 1+

N

 ] N (m−N1)+2

2

.

Using (2.13) and (2.14), we have

[ e

−Lp−1 θ t

∫ RN

u0 dx · π

≤ ρ 2+N (m−1)

− N2

Γ



N

1+

θ (m − p)L

]m−1

[

+θ e

2

p−1

(m − 1)(1 − e−θ(m−p)Lp−1 t )

−Lp−1 θ t

∫ u0 dx · L

N (p−m)−2 N (m−1)+2

RN

π

− N2

Γ

 1+

N

] N (m−N1)+2

2

.

This inequality implies

ρ ≥ χ (t , θ )µ ,

(4.10)

where µ is defined in (1.8) and

(m − 1)(1 − e−θ(m−p)L χ (t , θ ) = θ (m − p)Lp−1

p−1 t

)

[ × e

−θ(m−p)Lp−1 t

+

(m − 1)(1 − e (m − p)Lp−1

−Lp−1 θ t

∫ RN

)

[

p−1 θ t

e− L

u0 dx · π

− N2

Γ



N (p−m)−2

∫ RN

1+ N

N

]m−1

2

u0 dx · L N (m−1)+2 π − 2 Γ

 1+

N 2

] N (m−N1)+2

.

Letting λ −→ 0 in (4.10) gives

|xt | ≥ χ (t , θ )µ ,

(4.11)

5080

J. Pan / Nonlinear Analysis 74 (2011) 5069–5080

which is just (1.16). Moreover, it follows from

p−1 t

(m−1)(1−e−θ(m−p)L θ(m−p)Lp−1

χ (t , θ ) −→ C ∗ t as θ −→ 0  m−1  N with C ∗ = (m − 1) RN u0 dx · π − 2 Γ (1 + N2 ) .

)

−→ (m − 1)t as θ −→ 0 that



To end this paper, we list our conclusions as follows. The Cauchy problem (1.1) with (1.2) has a unique nonnegative solution u(x, t ) which have the following properties:

θ (m − p)Lp−1 (1) ut ≥ u in the sense of distributions; (m − 1)(e−θ(m−p)Lp−1 t − 1) ∫ ∫ p−1 (2) us dx ≥ Ls−1 e−L θ t u0 dx for 0 < s ≤ 1; N R RN  ∫ ∫  ∫ 1 m 2 θ up+1 + u2 dx + 2 |∇ u | dxdt ≤ u20 dx; m−1 RN

QT

mL

RN

(3) ‖v − u‖L2 (QT ) ≤ [(m − 1) + T θ ]A; (4) sup |x| ≥ (χ (t , θ ))µ . x∈Hu (t )

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