Electronic Journal of Qualitative Theory of Differential Equations 2010, No. 15, 1-12; http://www.math.u-szeged.hu/ejqtde/

Extinction and non-extinction of solutions for a nonlocal

reaction-diffusion problem

Wenjun Liu

College of Mathematics and Physics, Nanjing University of Information Science and

Technology, Nanjing 210044, China. E-mail: wjliu@nuist.edu.cn.

Department of Mathematics, Southeast University, Nanjing 210096, China.

Abstract We investigate extinction properties of solutions for the homogeneous Dirichlet boundary value problem of the nonlocal reaction-diffusion equation u_{t}−dΔu+ku^{p} = ∫_{Ω} u^{q}(x, t) dx with p, q ∈ (0, 1) and k, d > 0. We show that q = p is the critical extinction exponent. Moreover, the precise decay estimates of solutions before the occurrence of the extinction are derived.

Keywords: reaction-diffusion equation; extinction; non-extinction.

AMS Subject Classification (2000): 35K20, 35K55.

1 Introduction and main results

This paper is devoted to the extinction properties of solutions for the following diffusion equation with nonlocal reaction

u_{t} − dΔu + ku^{p} = ∫_{Ω} u^{q}(x, t) dx, x ∈ Ω, t > 0,  (1.1)

subject to the initial and boundary value conditions

u(x, t) = 0, x ∈ ∂Ω, t > 0,  (1.2)
u(x, 0) = u_{0}(x), x ∈ Ω,  (1.3)

where p, q ∈ (0, 1), k, d > 0, Ω ⊂ ℝ^{N}(N > 2) is an bounded domain with smooth boundary and u_{0}(x) ∈ L^{∞}(Ω) ∩ W_{0}^{1,2}(Ω) is a nonzero non-negative function.

Many physical phenomena were formulated into nonlocal mathematical models ([2, 3, 6, 7]) and there are a large number of papers dealing with the reaction-diffusion equations with nonlocal reactions or nonlocal boundary conditions (see [18, 20, 21, 23] and the references therein). In particular, M. Wang and Y. Wang [23] studied problem (1.1)–(1.3) for p, q ∈ [1, +∞) and concluded that: the blow-up occurs for large initial data if q > p ≥ 1 while all solutions exist globally if 1 ≤ q < p; in case of p = q, the issue depends on the comparison of |Ω| and k. For further studies of problem (1.1)–(1.3) we refer the read to [1, 13, 14, 19, 26] and the references therein. In all the above works, p, q ∈ [1, +∞) was assumed.

Extinction is the phenomenon whereby the evolution of some nontrivial initial data u_{0}(x) produces a nontrivial solution u(x, t) in a time interval 0 < t < T and then u(x, t) ≡ 0 for all (x, t) ∈ Ω × [T, +∞). It is an important property of solutions for many evolution equations which have been studied extensively by many researchers. Especially, there are some papers concerning the extinction for the following semilinear parabolic equation for special cases

u_{t} − dΔu + ku^{p} = λu^{q}, x ∈ Ω, t > 0,  (1.4)

where p ∈ (0, 1) and q ∈ (0, 1]. In case λ = 0, it is well-known that solutions of problem (1.2)–(1.4) vanishes within a finite time. Evans and Knerr [9] established this for the Cauchy problem by constructing a suitable comparison function. Fukuda [10] studied problem (1.2)– (1.4) with λ > 0 and q = 1 and concluded that: when λ < λ_{1}, the term Δu dominates the term λu so that solutions of problem (1.2)–(1.4) behave the same as those of (1.2)–(1.4) with λ = 0; when λ > λ_{1} and ∫_{Ω} u_{0}ϕ(x)d x > (λ − λ_{1})^{− \frac{1}{1−p}}, solutions of problem (1.2)–(1.4) grow up to infinity as t → ∞. Here, λ_{1} is the first eigenvalue of −Δ with zero Dirichlet boundary condition and ϕ(x) > 0 in Ω with max_{x∈Ω} ϕ(x) = 1 is the eigenfunction corresponding to the eigenvalue λ_{1}. Yan and Mu [24] investigated problem (1.2)–(1.4) with 0 < p < q < 1 and N > 2(q − p)/(1 − p) and obtained that the non-negative weak solution of problem (1.2)–(1.4) vanishes in finite time for any initial data provided that k is appropriately large. For papers concerning the extinction for the porous medium equation or the p-Laplacian equation, we refer the reader to [8, 11, 12, 15, 16, 22, 25] and the references therein. Recently, the present author [17] considered the extinction properties of solutions for the homogeneous Dirichlet boundary value problem of the p-Laplacian equation

u_{t} − div (|∇u|^{p−2}∇u) + βu^{q} = λu^{r}, x ∈ Ω, t > 0.

But as far as we know, no work is found to deal with the extinction properties of solutions for problem (1.1)–(1.3) which contains a nonlocal reaction term.

The purpose of the present paper is to investigate the extinction properties of solutions for the nonlocal reaction-diffusion problem (1.1)–(1.3). Our results below show that q = p is the critical extinction exponent for the weak solution of problem (1.1)–(1.3): if 0 < p < q < 1, the non-negative weak solution vanishes in finite time provided that |Ω| is appropriately small or k is appropriately large; if 0 < q < p < 1, the weak solution cannot vanish in finite time for any non-negative initial data; if 0 < q = p < 1, the weak solution cannot vanish in finite time for any non-negative initial data when k < ∫_{Ω} ψ^{q}(x)dx/M^{q} (≤ |Ω|), while it vanishes in finite time for any initial data u_{0} when k > |Ω|. Here ψ(x) is the unique positive solution of the linear elliptic problem

−Δψ = 1 in Ω; ψ = 0 on ∂Ω  (1.5)

and M = max_{x∈Ω} ψ(x). This is quite different from that of local reaction case, in which the first eigenvalue of the Dirichlet problem plays a role in the critical case (see [8, 12, 15, 22, 25]).

Moreover, the precise decay estimates of solutions before the occurrence of the extinction will be derived.

We now state our main results.

Theorem 1 Assume that 0 < p < q < 1.

1) If N < 4(q − p)/[(1 − p)(1 − q)], the non-negative weak solution of problem (1.1)–(1.3) vanishes in finite time provided that the initial data u_{0} (or |Ω|) is appropriately small or k is appropriately large.

2) If N = 4(q − p)/[(1 − p)(1 − q)], the non-negative weak solution of problem (1.1)–(1.3) vanishes in finite time for any initial data provided that |Ω| is appropriately small or k is appropriately large.

3) If N > 4(q − p)/[(1 − p)(1 − q)], the non-negative weak solution of problem (1.1)–(1.3) vanishes in finite time for any initial data provided that |Ω| is appropriately small or k is appropriately large.

Moreover, one has

\{ \begin{matrix}∥u(·, t)∥_{2} ≤ ∥u_{0}∥_{2} e^{−α_{1}t}, & t ∈ [0, T_{1}), \\ ∥u(·, t)∥_{2} ≤ [(||u(·, T_{1})||_{2}^{2−θ_{2}} + \frac{k_{2}}{d_{1}λ_{1}} ) e^{−(2−θ_{2})d_{1}λ_{1} (t−T_{1})} − \frac{k_{2}}{d_{1}λ_{1}} ]^{\frac{1}{2−θ_{2}}} , & t ∈ [T_{1}, T_{1}^{∗}), \\ ∥u(·, t)∥_{2} ≡ 0, & t ∈ [T_{1}^{∗}, +∞), \end{matrix}

for N < 4(q − p)/[(1 − p)(1 − q)],

\{ \begin{matrix}∥u(·, t)∥_{2} ≤ [(||u_{0}||_{2}^{1−q} + \frac{k_{1} − |Ω|^{\frac{3−q}{2}}}{d_{1}λ_{1}} ) e^{−(1−q)d_{1}λ_{1} t} − \frac{k_{1} − |Ω|^{\frac{3−q}{2}}}{d_{1}λ_{1}} ]^{\frac{1}{1−q}} , & t ∈ [0, T_{2}^{∗}), \\ ∥u(·, t)∥_{2} ≡ 0, & t ∈ [T_{2}^{∗}, +∞), \end{matrix}

for N = 4(q − p)/[(1 − p)(1 − q)],

\{ \begin{matrix}∥u(·, t)∥_{2} ≤ [(||u_{0}||_{2}^{2−θ_{2}} + \frac{k_{3}}{d_{3}λ_{1}} ) e^{−(2−θ_{2})d_{3}λ_{1} t} − \frac{k_{3}}{d_{3}λ_{1}} ]^{\frac{1}{2−θ_{2}}} , & t ∈ [0, T_{3}^{∗}), \\ ∥u(·, t)∥_{2} ≡ 0, & t ∈ [T_{3}^{∗}, +∞), \end{matrix}

for N > 4(q −p)/[(1−p)(1−q)], where d_{1}, d_{3}, T_{1}, T_{i}^{∗} and k_{i} (i = 1, 2, 3) are positive constants to be given in the proof, α_{1} > d_{1}λ_{1} and

θ_{2} = \frac{2N(1 − p) + 4(1 + p)}{N(1 − p) + 4} ∈ (1, 2).

Remark 1 One can see from the proof below that the restriction N > 4(q − p)/[(1 − p)(1 − q)] in the case 3) can be extended to N > 2(q − p)/(1 − p). This has been proved in [24] for the local reaction case.

Theorem 2 Assume that 0 < p = q < 1.

1) If k > |Ω|, the non-negative weak solution of problem (1.1)–(1.3) vanishes in finite time for any initial data u_{0}. Moreover, one has

\{ \begin{matrix}∥u(·, t)∥_{2} ≤ [(||u_{0}||_{2}^{1−q} + \frac{k_{4}}{d_{4}λ_{1}} ) e^{−(1−q)d_{4}λ_{1} t} − \frac{k_{4}}{d_{4}λ_{1}} ]^{\frac{1}{1−q}} , & t ∈ [0, T_{2}^{∗}), \\ ∥u(·, t)∥_{2} ≡ 0, & t ∈ [T_{2}^{∗}, +∞), \end{matrix}

where d_{4} and k_{4} are positive constants to be given in the proof.

2) If k < ∫_{Ω} ψ^{q}(x)dx/M^{q} (≤ |Ω|), then the weak solution of problem (1.1)–(1.3) cannot vanish in finite time for any non-negative initial data.

3) If k = ∫_{Ω} ψ^{q}(x)dx/M^{q}, then the weak solution of problem (1.1)–(1.3) cannot vanish in finite time for any identically positive initial data.

Theorem 3 Assume that 0 < q < p < 1, then the weak solution of (1.1)–(1.3) cannot vanish in finite time for any non-negative initial data.

Remark 2 One can conclude from Theorems 1–3 that q = p is the critical extinction exponent of solutions for problem (1.1)–(1.3).

The rest of the paper is organized as follows. In Section 2, we will give some preliminary lemmas. We will prove Theorems 1–3 in Section 3-5.

2 Preliminary

Let ∥ · ∥_{p} and ∥ · ∥_{1,p} denote L^{p}(Ω) and W^{1,p}(Ω) norms respectively, 1 ≤ p ≤ ∞. Before proving our main results, we will give some preliminary lemmas which are of crucial importance in the proofs. We first give the following comparison principle, which can be proved as in [22, 23, 25].

Lemma 1 Suppose that \underline{u}(x, t), \overline{u}(x, t) are a subsolution and a supersolution of problem (1.1)– (1.3) respectively, then \underline{u}(x, t) ≤ \overline{u}(x, t) a.e. in Ω_{T}.

The following inequality problem is often used to derive extinction of solutions (see [22, 25]).

\frac{dy}{dt} + αy^{k} ≤ 0, t ≥ 0; y(0) ≥ 0,

where α > 0 is a constant and k ∈ (0, 1). Due to the nature of our problem, we would like to use the following lemmas which are of crucial importance in the proofs of decay estimates.

Lemma 2 ^{[5]} Let y(t) be a non-negative absolutely continuous function on [0, +∞) satisfying

\frac{dy}{dt} + αy^{k} + β y ≤ 0, t ≥ T_{0}; y(T_{0}) ≥ 0,  (2.1)

where α, β > 0 are constants and k ∈ (0, 1). Then we have decay estimate

\{ \begin{matrix}y(t) ≤ [(y^{1−k}(T_{0}) + \frac{α}{β}) e^{(k−1)β (t−T_{0})} − \frac{α}{β}]^{\frac{1}{1−k}} , & t ∈ [T_{0}, T_{∗}), \\ y(t) ≡ 0, & t ∈ [T_{∗}, +∞), \end{matrix}

where T_{∗} = \frac{1}{(1 − k)β} ln (1 + \frac{β}{α}y^{1−k}(T_{0})).

Lemma 3 ^{[15]} Let 0 < k < m ≤ 1, y(t) ≥ 0 be a solution of the differential inequality

\frac{dy}{dt} + αy^{k} + βy ≤ γ y^{m}, t ≥ 0; y(0) = y_{0} > 0,  (2.2)

where α, β > 0, γ is a positive constant such that γ < αy_{0}^{k−m}. Then there exist η > β, such that

0 ≤ y(t) ≤ y_{0} e^{−ηt}, t ≥ 0.

Consider the following ODE problem

\frac{dy}{dt} + αy^{k} + βy = γ y^{m}, t ≥ 0; y(0) = y_{0} ≥ 0; y(t) > 0, t > 0.  (2.3)

If α = 0, β > 0 and γ > 0, we can easily derive that the non-constant solution of this problem is

y(t) = [(y_{0}^{1−m} − \frac{γ}{β} ) e^{−(1−m)β t} + \frac{γ}{β} ]^{\frac{1}{1−m}} > 0, ∀ t > 0.

If α, β, γ > 0, we have

Lemma 4 ^{[17]} Let α, β, γ > 0 and 0 < m < k < 1. Then there exists at least one non-constant solution of the ODE problem (2.3).

Proof. It is easy to prove that the following algebraic equation

αy^{k} + βy = γ y^{m}

has unique positive solution (denoted by y_{∗}).

We first consider the case y_{0} > 0. By considering the sign of y′(t) via y(t) at [0, y_{∗}), we see that: if 0 < y_{0} < y_{∗}, then y(t) is increasing with respect to t > 0; if y_{0} > y_{∗}, then y(t) is decreasing with respect to t > 0. Therefore, solution with non-negative initial value y_{0} remains positive and of course approaches y_{∗} as t → +∞.

When y_{0} = 0, we choose a sufficiently small constant ε ∈ (0, y_{∗}) and consider the following problem

\frac{dz}{dt} + αz^{k} + βz = γ z^{m}, t ≥ 0; z(0) = ε > 0; z(t) > 0, t > 0.  (2.4)

Then problem (2.4) exists at least one non-constant solution z = z(t) satisfying z′(t) > 0 for all t ∈ ℝ. We continue the proof based on the following claim: there is a time t_{0} ∈ (−∞, 0), such that z(t_{0}) = 0. By setting y(t) = z(t + t_{0}), ∀ t ≥ 0, we get that y(t) is a non-constant solution satisfying (2.3).

We now only need to prove the above mentioned claim. Indeed, if it is not true, then 0 < z(t) < ε for all t ∈ (−∞, 0). Since 0 < m < k < 1 and z′(t) > 0 for all t ∈ ℝ, there is a t_{1} ∈ (−∞, 0) so that αz^{k} + βz ≤ \frac{γ}{2} z^{m} for all t ∈ (−∞, t_{1}], i.e.,

\frac{dz}{dt} ≥ \frac{γ}{2} z^{m} for all t ∈ (−∞, t_{1}].

Integrating the above inequality on (t, t_{1}), we get

z^{1−m}(t_{1}) − z^{1−m}(t) > \frac{γ}{2} (1 − m)(t_{1} − t),

which causes a contradiction as t → −∞.

Lemma 5 ^{[4]} (Gagliardo-Nirenberg) Let β ≥ 0, N > p ≥ 1, β + 1 ≤ q, and 1 ≤ r ≤ q ≤ (β + 1)Np/(N − p), then for u such that |u|^{β}u ∈ W^{1,p}(Ω), we have

∥u∥_{q} ≤ C∥u∥_{r}^{1−θ} ∥∇ (|u|^{β}u)∥_{p}^{θ/(β+1)}

with θ = (β + 1)(r^{−1} − q^{−1})/{N^{−1} − p^{−1} + (β + 1)r^{−1}}, where C is a constant depending only on N, p and r.

3 The case 0 < p < q < 1: proof of Theorem 1

Multiplying (1.1) by u and integrating over Ω, we have

\frac{1}{2} \frac{d}{dt}∥u∥_{2}^{2} + d∥∇u∥_{2}^{2} = ∫_{Ω} u dx ∫_{Ω} u^{q}(y, t) dy − k∥u∥_{p+1}^{p+1}.  (3.1)

By Hölder inequality, we have

∫_{Ω} u dx ∫_{Ω} u^{q}(y, t) dy ≤ |Ω|^{\frac{2s−1−q}{s}} ∥u∥_{s}^{q+1},  (3.2)

where s ≥ 1 to be determined later. we substitute (3.2) into (3.1) to get

\frac{1}{2} \frac{d}{dt}∥u∥_{2}^{2} + d∥∇u∥_{2}^{2} = |Ω|^{\frac{2s−1−q}{s}} ∥u∥_{s}^{q+1} − k∥u∥_{p+1}^{p+1}.  (3.3)

1) For the case N < 4(q − p)/[(1 − p)(1 − q)], we set s = 2 in (3.3). By lemma 5, one can get

∥u∥_{2} ≤ C_{1}(N, p)∥u∥_{p+1}^{1−θ_{1}}∥∇u∥_{2}^{θ_{1}},  (3.4)

where

θ_{1} = (\frac{1}{p + 2} − \frac{1}{2}) (\frac{1}{N} − \frac{1}{2} + \frac{1}{p + 1})^{−1} = \frac{N(1 − p)}{2(p + 1) + N(1 − p)}.

0 < p < 1 implies that 0 < θ_{1} < 1. It follows from (3.4) and Young’s inequality that

∥u∥_{2}^{θ_{2}} ≤C_{1}(N, p)^{θ_{2}}∥u∥_{p+1}^{(1−θ_{1})θ_{2}}∥∇u∥_{2}^{θ_{1}θ_{2}}
≤C_{1}(N, p)^{θ_{2}} (ε_{1}∥∇u∥_{2}^{2} + C(ε_{1})∥u∥_{p+1}^{2(1−θ_{1})θ_{2}/(2−θ_{1}θ_{2})}) ,  (3.5)

for ε_{1} > 0 and θ_{2} > 1 to be determined. We choose θ_{2} = \frac{2N(1−p)+4(1+p)}{N(1−p)+4}, then 1 < θ_{2} < 2 and 2(1 − θ_{1})θ_{2}/(2 − θ_{1}θ_{2}) = p + 1. Thus, (3.5) becomes

\frac{C_{1}(N, p)^{−θ_{2}}}{C(ε_{1})} ∥u∥_{2}^{θ_{2}} − \frac{ε_{1}}{C(ε_{1})}∥∇u∥_{2}^{2} ≤ ∥u∥_{p+1}^{p+1}.  (3.6)

We substitute (3.6) into (3.3) to get

\frac{1}{2} \frac{d}{dt}∥u∥_{2}^{2} + (d − \frac{kε_{1}}{C(ε_{1})}) ∥∇u∥_{2}^{2} + \frac{kC_{1}(N, p)^{−θ_{2}}}{C(ε_{1})} ∥u∥_{2}^{θ_{2}} ≤ |Ω|^{\frac{3−q}{2}} ∥u∥_{2}^{q+1}.

We choose ε_{1} small enough such that d_{1} := d − \frac{kε_{1}}{C(ε_{1})} > 0. Once ε_{1} is fixed, we set k_{1} = \frac{kC_{1}(N,p)^{−θ_{2}}}{C(ε_{1})} . Then, by Poincare’s inequality, we get

\frac{d}{dt}∥u∥_{2} + k_{1}∥u∥_{2}^{θ_{2}−1} + d_{1}λ_{1}∥u∥_{2} ≤ |Ω|^{\frac{3−q}{2}} ∥u∥_{2}^{q}.  (3.7)

Since N < 4(q − p)/[(1 − p)(1 − q)], we further have 0 < θ_{2} − 1 < q. By Lemma 3, there exists α_{1} > d_{1}λ_{1}, such that

0 ≤ ∥u∥_{2} ≤ ∥u_{0}∥_{2} e^{−α_{1}t}, t ≥ 0,

provided that

∥u_{0}∥_{2} < (\frac{k_{1}}{|Ω|^{\frac{3−q}{2}}} )^{\frac{1}{q−θ_{2}+1}} = (\frac{kC_{1}(N, p)^{−θ_{2}}}{C(ε_{1})|Ω|^{\frac{3−q}{2}}} )^{\frac{1}{q−θ_{2}+1}} .  (3.8)

Furthermore, there exists T_{1}, such that

k_{1} − |Ω|^{\frac{3−q}{2}} ∥u∥_{2}^{q−θ_{2}+1}
≥k_{1} − |Ω|^{\frac{3−q}{2}} (∥u_{0}∥_{2} e^{−α_{1}T_{1}})^{q−θ_{2}+1} := k_{2} > 0,  (3.9)

holds for t ∈ [T_{1}, +∞). Therefore, when t ∈ [T_{1}, +∞), (3.7) turns to

\frac{d}{dt}∥u∥_{2} + k_{2}∥u∥_{2}^{θ_{2}−1} + d_{1}λ_{1}∥u∥_{2} ≤ 0.  (3.10)

By Lemma 2, we can obtain the desired decay estimate for

T_{1}^{∗} = \frac{1}{(2 − θ_{2})d_{1}λ_{1}} ln (1 + \frac{d_{1}λ_{1}}{k_{2}} ||u(·, T_{1})||_{2}^{2−θ_{2}}) .  (3.11)

2) When N = 4(q − p)/[(1 − p)(1 − q)], we still choose s = 2 in (3.3), and then θ_{2} − 1 = q. Thus, (3.7) becomes

\frac{d}{dt}∥u∥_{2} + (k_{1} − |Ω|^{\frac{3−q}{2}} ) ∥u∥_{2}^{q} + d_{1}λ_{1}∥u∥_{2} ≤ 0.  (3.12)

By Lemma 2, we can obtain the desired decay estimate for

T_{2}^{∗} = \frac{1}{(1 − q)d_{1}λ_{1}} ln (1 + \frac{d_{1}λ_{1}}{k_{1} − |Ω|^{\frac{3−q}{2}}} ||u_{0}||_{2}^{1−q}) ,  (3.13)

provided that |Ω| < k1^{\frac{2}{3−q}} = (\frac{kC_{1}(N,p)^{−θ_{2}}}{C(ε_{1})} )^{\frac{2}{3−q}}.

3) For the case N > 4(q − p)/[(1 − p)(1 − q)], we back to (3.3). By lemma 5, one can get

∥u∥_{s} ≤ C_{2}(N, p)∥u∥_{p+1}^{1−θ_{3}}∥∇u∥_{2}^{θ_{3}},  (3.14)

where

θ_{3} = (\frac{1}{p + 1} − \frac{1}{s}) (\frac{1}{N} − \frac{1}{2} + \frac{1}{p + 1})^{−1} = \frac{2N(s − p − 1)}{s[2(p + 1) + N(1 − p)]}.

If N > 2, one further needs p + 1 < s < 2N/(N − 2). The choice of s implies that 0 < θ_{3} < 1. It follows from (3.14) and Young’s inequality that

∥u∥_{s}^{q+1} ≤C_{2}(N, p)^{q+1}∥u∥_{p+1}^{(1−θ_{3})(q+1)}∥∇u∥_{2}^{θ_{3}(q+1)}
≤C_{2}(N, p)^{q+1} (ε_{2}∥∇u∥_{2}^{2} + C(ε_{2})∥u∥_{p+1}^{2(1−θ_{3})(q+1)/[2−θ_{3}(q+1)]}) ,  (3.15)

for ε_{2} > 0 to be determined later. We choose s = \frac{N(q+1)(1−p)}{N(1−p)−2(q−p)}, then θ_{3} = \frac{2(q−p)}{(q+1)(1−p)} and 2(1 − θ_{3})(q + 1)/[2 − θ_{3}(q + 1)] = p + 1. We substitute (3.15) into (3.3) to get

\frac{1}{2} \frac{d}{dt}∥u∥_{2}^{2}+(d − ε_{2}C_{2}(N, p)^{q+1}|Ω|^{\frac{2s−1−q}{s}} ) ∥∇u∥_{2}^{2}+(k − C(ε_{2})C_{2}(N, p)^{q+1}|Ω|^{\frac{2s−1−q}{s}} ) ∥u∥_{p+1}^{p+1} ≤ 0.

We choose ε_{2} small enough such that d_{2} := d − ε_{2}C_{2}(N, p)^{q+1}|Ω|^{\frac{2s−1−q}{s}} > 0. Once ε_{2} is fixed, we set k_{0} = C(ε_{2})C_{2}(N, p)^{q+1}|Ω|^{\frac{2s−1−q}{s}}. When k > k_{0} = C(ε_{2})C_{2}(N, p)^{q+1}|Ω|^{\frac{2s−1−q}{s}}, we get

\frac{1}{2} \frac{d}{dt}∥u∥_{2}^{2} + d_{2}∥∇u∥_{2}^{2} + (k − k_{0}) ∥u∥_{p+1}^{p+1} ≤ 0.  (3.16)

We note (3.6) holds provided that 0 < q < 1 and is independent of the relation of N and 4(q − p)/[(1 − p)(1 − q)]. So, we substitute (3.6) into (3.16) to get

\frac{1}{2} \frac{d}{dt}∥u∥_{2}^{2} + (d_{2} − \frac{(k − k_{0})ε_{1}}{C(ε_{1})} ) ∥∇u∥_{2}^{2} + \frac{(k − k_{0})C_{1}(N, p)^{−θ_{2}}}{C(ε_{1})} ∥u∥_{2}^{θ_{2}} ≤ 0.

We recall that θ_{2} = \frac{2N(1−p)+4(1+p)}{N(1−p)+4} ∈ (1, 2). We choose ε_{1} small enough such that d_{3} := d_{2} − \frac{(k−k_{0})ε_{1}}{C(ε_{1})} > 0. Once ε_{1} is fixed, we set k_{3} = \frac{(k−k_{0})C_{1}(N,p)^{−θ_{2}}}{C(ε_{1})} . Thus, we get

\frac{d}{dt}∥u∥_{2} + k_{3}∥u∥_{2}^{θ_{2}−1} + d_{3}λ_{1}∥u∥_{2} ≤ 0.

By Lemma 2, we can obtain the desired decay estimate for

T_{3}^{∗} = \frac{1}{(2 − θ_{2})d_{1}λ_{1}} ln (1 + \frac{d_{3}λ_{1}}{k_{3}} ∥u_{0}∥_{2}^{2−θ_{2}}) .  (3.17)

4 The case 0 < p = q < 1: proof of Theorem 2

In this section, we consider the case 0 < p = q < 1.

1) If k > |Ω|, we choose s = p + 1 in (3.3) to get

\frac{1}{2} \frac{d}{dt}∥u∥_{2}^{2} + d∥∇u∥_{2}^{2} + (k − |Ω|)∥u∥_{p+1}^{p+1} ≤ 0.  (4.1)

We substitute (3.6) into (4.1) to obtain

\frac{1}{2} \frac{d}{dt}∥u∥_{2}^{2} + (d − \frac{(k − |Ω|)ε_{1}}{C(ε_{1})} ) ∥∇u∥_{2}^{2} + \frac{(k − |Ω|)C_{1}(N, p)^{−θ_{2}}}{C(ε_{1})} ∥u∥_{2}^{θ_{2}} ≤ 0.

We choose ε_{1} small enough such that d_{4} := d − \frac{(k−|Ω|)ε_{1}}{C(ε_{1})} > 0. Once ε_{1} is fixed, we set k_{4} = \frac{(k−|Ω|)C_{1}(N,p)^{−θ_{2}}}{C(ε_{1})} . Then, by Poincare inequality, we get

\frac{d}{dt}∥u∥_{2} + k_{4}∥u∥_{2}^{θ_{2}−1} + d_{4}λ_{1}∥u∥_{2} ≤ 0.  (4.2)

By Lemma 2, we can obtain the desired decay estimate for

T_{1}^{∗} = \frac{1}{(2 − θ_{2})d_{4}λ_{1}} ln (1 + \frac{d_{4}λ_{1}}{k_{4}} ||u_{0}||_{2}^{2−θ_{2}}) .  (4.3)

2) If k < ∫_{Ω} ψ^{q}(x)dx/M^{q}, we define

g(t) = [\frac{∫_{Ω} ψ^{q}(x)dx − kM^{q}}{d} (1 − e^{−(1−q) \frac{d}{M} t})]^{\frac{1}{1−q}} ,

which satisfies the ODE problem

g′(t) + \frac{d}{M} g(t) = \frac{∫_{Ω} ψ^{q}(x)dx − kM^{q}}{M} g^{q}(t), t ≥ 0; g(0) = 0.

Let v(x, t) = g(t)ψ(x). Then, we have

v_{t} − dΔv − ∫_{Ω} v^{q}(x, t) dx + kv^{p}
=g′(t)ψ(x) + dg(t) − g^{q}(t) ∫_{Ω} ψ^{q} dx + kg^{q}(t)ψ^{q}
≤g′(t)M + dg(t) − g^{q}(t) ∫_{Ω} ψ^{q} dx + kg^{q}(t)M^{q}
=0.

Moreover, v(x, 0) = g(0)ψ(x) = 0 ≤ u_{0}(x) in Ω, and υ|_{(∂Ω)_{t}} = 0. Therefore, we have u(x, t) ≥ v(x, t) > 0 in Ω × (0, +∞); i.e., v(x, t) is a non-extinction subsolution of problem (1.1)–(1.3).

3) For k = ∫_{Ω} ψ^{q}(x)dx/M^{q}, let w(x, t) = h(t)ψ(x), where h(t) satisfies the ODE problem

\frac{dh}{dt} + \frac{d}{M} h = 0, t ≥ 0; h(0) = h_{0} > 0.

Then, for any identically positive initial data, we can choose h_{0} sufficiently small such that h_{0}ψ(x) ≤ u_{0}(x). According to Lemma 1, we get that w(x, t) is a non-extinction subsolution of problem (1.1)–(1.3).

5 The case 0 < q < p < 1: proof of Theorem 3

Let z(x, t) = j(t)ψ(x), where j(t) satisfies the ODE problem

\frac{dj}{dt} + kM^{p−1}j^{p}(t) + \frac{d}{M} j(t) = \frac{∫_{Ω} ψ^{q}(x)dx}{M}j^{q}(t), t ≥ 0; j(0) = 0; j(t) > 0, t > 0.

Then, we have

z_{t} − dΔz − ∫_{Ω} z^{q}(x, t) dx + kz^{p}
=j′(t)ψ(x) + dj(t) − j^{q}(t) ∫_{Ω} ψ^{q} dx + kj^{p}(t)ψ^{p}
≤j′(t)M + dj(t) − j^{q}(t) ∫_{Ω} ψ^{q} dx + kj^{p}(t)M^{p}
=0.

Moreover, z(x, 0) = j(0)ψ(x) = 0 ≤ u_{0}(x) in Ω, and υ|_{(∂Ω)_{t}} = 0. Therefore, we have u(x, t) ≥ z(x, t) > 0 in Ω × (0, +∞) according to Lemma 1, i.e., z(x, t) is a non-extinction subsolution of problem (1.1)–(1.3).

Acknowledgments

The work was supported by the Science Research Foundation of Nanjing University of Information Science and Technology and the Natural Science Foundation of the Jiangsu Higher Education Institutions (Grant No. 09KJB110005). The author would like to express his sincere gratitude to Professor Mingxin Wang for his enthusiastic guidance and constant encouragement.

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(Received June 18, 2009)