# Extinction and Decay Estimates of Solutions for thep-Laplacian Equations with Nonlinear Absorptions and Nonlocal Sources

Authors: Xiulan Wu; Jun Fu
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

We investigate the extinction and decay estimates of thep-Laplacian equations with nonlinear absorptions and nonlocal sources. By Gagliardo-Nirenberg inequality, we obtain the sufficient conditions of extinction solutions, and we also give the precise decay estimates of the extinction solutions.


---

## Body

## 1. Introduction

In this paper, we consider the following fast diffusivep-Laplacian equation:
(1)ut=div(|∇u|p-2∇u)+λ∫Ω‍uq(x,t)dx-kur,x∈Ω,t>0,(2)u(x,t)=0,x∈∂Ω,t>0,u(x,0)=u0(x),x∈Ω,
where 1<p<2, k,q,λ>0, 0<r<1, Ω⊂RN (N≥2) is a bounded domain with smooth boundary and u0(x)∈L∞(Ω)∩W01,p(Ω) is a nonnegative function. Equation (1) is a class of nonlinear singular parabolic equations and appears to be relevant in the theory of non-Newtonian fluids perturbed by both nonlocal sources and nonlinear absorptions; see [1–4], for instance. Extinction is the phenomenon whereby the evolution of some nontrivial initial data u0(x) produces a nontrivial solution u(x,t) in a time interval 0<t<T and u(x,t)→0 as t→T. As an important property of solutions of developing equations, the extinction recently has been studied intensively by several authors in [5–9]. In paper [10], the authors discussed the extinction behavior of solutions for Problem (1)-(2) when r=1. In this paper, we investigated the extinction of solutions when 0<r<1. Due to the nature of our problem, we would like to use the following lemmas by [11].Lemma 1 (Gagliardo-Nirenberg inequality).
Suppose thatβ≥0, N>p≥1, β+1≤q≤(β+1)Np/(N-p); then for u such that |u|βu∈W1,p(Ω), one has
(3)∥u∥q≤C1/(β+1)∥u∥r1-θ∥∇(|u|βu)∥pθ/(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.


## 2. Main Results and Proofs

Theorem 2.
Assume thatp-1=q with r<1; then the non-negative nontrivial weak solution of Problem (1)-(2) vanishes in finite time for any non-negative initial data provided that |Ω| or λ is sufficiently small.(1)
For the case  2N/(N+2)≤p<2, one has(4)∥u(·,t)∥2≤(∥u0∥22-k1-M1(2-k1)t)1/(2-k1),t∈[0,T1),∥u(·,t)∥2≡0,t∈[T1,+∞),where  k1,M1, and  T1  are given by (11), (16), and (17), respectively.(2)
For the case  1<p<2N/(N+2), one has(5)∥u(·,t)∥1+s≤(∥u0∥1+s1+s-k2-M2(1+s-k2)t)1/(1+s-k2),t∈[0,T2),∥u(·,t)∥2≡0,t∈[T2,+∞),where  s,  k2,  M2, and  T2  are given by (18), (22), (26), and (28), respectively.Proof.
(1) For the case2N/(N+2)≤p<2, multiplying (1) by u and integrating over Ω, we deduce from the Hölder inequality that
(6)12ddt∥u∥22+∥∇u∥pp+k∥u∥r+1r+1≤λ|Ω|∥u∥pp.
inequality
(7)∥u∥p≤B∥∇u∥p,
where B denotes the optimal embedding constant, combining (6) and (7) we have
(8)12ddt∥u∥22+(1-λBp|Ω|)∥∇u∥pp+k∥u∥r+1r+1≤0.
By Lemma 1, we have
(9)∥u∥2≤C1(N,p,r)∥∇u∥pθ∥u∥1+r1-θ,
where θ1=(1/(1+r)-1/2)(1/N-1/p+1/(1+r))-1.
It is easy to check thatθ1∈(0,1]; using Young's inequality with ε, it follows from (9) that
(10)∥u∥2k1≤C1k1(N,p,r)(ε1∥∇u∥pp+C(ε1)∥u∥1+rpk1(1-θ1)/(p-k1θ1)),
where ε1>0 and k1>0 will be determined later. We choose
(11)k1=2[(1+r)p+N(p-1-r)]2p+N(p-1-r).
Then we can conclude that k1∈(1,2) and pk1(1-θ1)/(p-k1θ1)=1+r. Therefore, it follows from (10) that
(12)∥u∥1+r1+r≥(C1-k1(N,p,r)∥u∥2k1-ε1∥∇u∥pp)1C(ε1).
By combining (8) and (12), we have
(13)12ddt∥u∥22+(1-λBp|Ω|-kε1C(ε1))∥∇u∥pp+kC1-k1(N,p,r)C(ε1)∥u∥2k1≤0.
Choosing ε1 small enough such that 1-kε1/C(ε1)>0 and |Ω|≤(1-kε1/C(ε1))/λBp, then we have 1-kε1/C(ε1)-Bpλ|Ω|>0. Therefore, we deduce from k1∈(1,2) that
(14)ddt∥u∥2+M1∥u∥2k1-1≤0,
which implies that
(15)∥u(·,t)∥2≤(∥u0∥22-k1-M1(2-k1)t)1/(2-k1),t∈[0,T1),∥u(·,t)∥2≡0,t∈[T1,+∞),
where
(16)M1=kC1-k1(N,p,r)C(ε1),(17)T1=∥u0∥22-k1M1(2-k1).
(2) For the case 1<p<2N/(N+2), multiplying (1) by us, where
(18)s>l=2N-p(1+N)p>1,
integrating over Ω, we deduce from the Hölder inequality that
(19)11+sddt∥u∥1+s1+s+spp(p+s-1)p∥∇u(p+s-1)/p∥pp+k∥u∥s+rs+r≤λ|Ω|∥u∥p+s-1p+s-1.
By Lemma 1 and s>1, we have
(20)∥u∥s+1≤C2(N,p,r)∥∇u(p+s-1)/p∥ppθ2/(p+s-1)∥u∥s+r1-θ2,
where θ2=N(1-r)(p+s-1)/(s+1)[p(s+r)+N(p-1-r)]. By (18) and r<1, it is easy to check that θ2∈(0,1). By Young's inequality with ε, it follows from (19) that
(21)∥u∥s+1k2≤C2k2(N,p,r,s)(ε2∥∇u(p+s-1)/p∥pp+C(ε2)×∥u∥s+r(1-θ2)k2(p+s-1)/(p+s-1-k2θ2)),
where ε2>0 and k2>0 will be determined later. We choose
(22)k2=(s+1)[(s+r)p+N(p-1-r)](s+1)p+N(p-1-r);
then it follows that k2∈(s,s+1) and (p+s-1)k2(1-θ2)/(p+s-1-k2θ2)=s+r. Therefore, it follows from (21) that
(23)∥u∥s+rs+r≥C2-k2(N,p,r,s)C(ε2)∥u∥s+1k2-ε2C(ε2)∥∇u(p+s-1)/p∥pp.
By combining (19) and (23), we have by poincare inequality
(24)11+sddt∥u∥1+s1+s+(spp(p+s-1)p-kε2C(ε2)-λ|Ω|Bp)×∥∇u(p+s-1)/p∥pp+kC2k2(N,p,r,s)C(ε2)∥u∥s+1k2≤0.
Choosing ε2>0 small enough such that spp/(p+s-1)p-kε2/C(ε2)>0 and |Ω|≤(spp/(p+s-1)p-kε2/C(ε2))/λ|Ω|Bp, then we have spp/(p+s-1)p-kε2/C(ε2)-λ|Ω|Bp>0. Therefore, we deduce from k2∈(s,s+1) that
(25)ddt∥u∥1+s+M2∥u∥1+sk2-s≤0,
where
(26)M2=kC2-k2(N,p,r,s)C(ε2),
which implies that
(27)∥u(·,t)∥1+s≤[∥u0∥1+s1+s-k2-M2(1+s-k2)t]1/(1+s-k2),t∈[0,T2),∥u(·,t)∥1+s≡0,t∈[T2,+∞),
where
(28)T2=∥u0∥1+s1+s-k2M2(1+s-k2).
The proof of Theorem 2 is complete.Theorem 3.
Assume thatr<1.(1)
If  2N/(N+2)≤p<2  with  q>k1-1=(2rp+N(p-1-r))/(2p+N(p-1-r)), then the non-negative nontrivial weak solution of Problem (1)-(2) vanishes in finite time provided that  u0  (or  |Ω|  or  λ) is sufficiently small and(29)∥u(·,t)∥2≤(∥u0∥22-k1-(2-k1)M3t)1/(2-k1),t∈[0,T3),∥u(·,t)∥2≡0,t∈[T3,+∞),where  k1,  M3, and  T3  are given by (11), (35), and (33), respectively.(2)
If  1<p<2N/(N+2)  with  q>k2-s=((s+1)rp+N(p-1-r))/((s+1)p+N(p-1-r)), then the non-negative nontrivial weak solution of Problem (1)-(2) vanishes if finite time provided that  u0  (or  |Ω|  or  λ) is sufficiently small and(30)∥u(·,t)∥s+1≤(∥u0∥s+1s+1-k2-(s+1-k2M4)t)1/(s+1-k2),t∈[0,T4),∥u(·,t)∥s+1≡0,t∈[T4,+∞),where  s,  k2,  M4, and  T4  are given by (18), (22), (39), and (41), respectively.Proof.
(1)  If2N/(N+2)≤p<2, multiplying (1) by u and integrating over Ω, we deduce from (12) and the Hölder inequality that
(31)12ddt∥u∥22+(1-kε1C(ε1))∥∇u∥pp+kC1-k1(N,p,r)C(ε1)×∥u∥2k1-λ|Ω|(3-q)/2∥u∥2q+1≤0.
By choosing ε1>0 small enough such that 1-kε1/C(ε1)≥0, we obtain that
(32)ddt∥u∥2+M3∥u∥2k1-1≤0,
provided that ∥u0∥2≤(kC1-k1(N,p,r)/C(ε1)λ|Ω|(3-q)/2)1/(q-k1+1) and q>k1-1=(2rp+N(p-1-r))/(2p+N(p-1-r)), where
(33)M3=kC1-k1(N,p,r)C(ε1)-λ|Ω|(3-q)/2∥u0∥2q-k1+1>0.
From (32) and k1∈(1,2), we can derive that
(34)∥u(·,t)∥2≤(∥u0∥22-k1-(2-k1)M3t)1/(2-k1),t∈[0,T3),∥u(·,t)∥2≡0,t∈[T3,+∞),
where
(35)T3=∥u0∥22-k1(2-k1)M3.
(2)  If 1<p<2N/(N+2), multiplying (1) by us, where s is given by (18) and integrating over Ω, we deduce from the Hölder inequality and (23) that
(36)11+sddt∥u∥1+s1+s+(spp(p+s-1)p-kε2C(ε2))∥∇u(p+s-1)/p∥pp+kC2-k2(N,p,r,s)C(ε2)∥u∥s+1k2≤λ∥u∥1+sq+s|Ω|(2+s-q)/(1+s).
Choosing ε2>0 small enough such that spp/(p+s-1)p-kε2/C(ε2)>0, we have
(37)ddt∥u∥1+s+∥u∥1+sk2-s(kC2k2(N,p,r,s)C(ε2)-λ|Ω|(2+s-q)/(1+s)∥u∥1+sq+s-k2kC2k2(N,p,r,s)C(ε2))≤0.
Therefore, we have(38)ddt∥u∥1+s+M4∥u∥1+sk2-s≤0,
provided that ∥u0∥1+s≤(kC2-k2(N,p,r,s)/C(ε2)λ|Ω|(2+s-q)/(1+s))1/(q+s-k2) and q>k2-s=((s+1)rp+N(p-1-r))/((s+1)p+N(p-1-r)), where
(39)M4=kC2-k2(N,p,r,s)C(ε2)-λ|Ω|(2+s-q)/(1+s)∥u0∥1+sq+s-k2>0.
It follows from (38) and k2∈(s,s+1) that
(40)∥u(·,t)∥1+s≤(∥u0∥s+1s+1-k2-M4(s+1-k2)t)1/(s+1-k2),t∈[0,T4),∥u(·,t)∥s+1≡0,t∈[T4,+∞),
where
(41)T4=∥u0∥1+s1+s-k2M4(s+1-k2).
The proof of Theorem 3 is complete.



---

*Source: 928080-2013-08-18.xml*