Extinction and Decay Estimates of Solutions for the 𝑝-Laplacian Equations with Nonlinear Absorptions and Nonlocal Sources

Abstract

We investigate the extinction and decay estimates of the 𝑝-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.

1. Introduction

In this paper, we consider the following fast diffusive 𝑝-Laplacian equation: 𝑢_𝑡=div|∇𝑢|^𝑝−2∇𝑢+𝜆_Ω𝑢^𝑞(𝑥,𝑡)𝑑𝑥−𝑘𝑢^𝑟,𝑥∈Ω,𝑡>0,𝑢()𝑢()𝑥,𝑡=0,𝑥∈𝜕Ω,𝑡>0,𝑥,0=𝑢₀(𝑥),𝑥∈Ω, (1)(2) where 1<𝑝<2, 𝑘,𝑞,𝜆>0, 0<𝑟<1, Ω⊂𝑅^𝑁 (𝑁≥2) is a bounded domain with smooth boundary and 𝑢₀(𝑥)∈𝐿^∞(Ω)∩𝑊₀^1,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 that 𝛽≥0, 𝑁>𝑝≥1, 𝛽+1≤𝑞≤(𝛽+1)𝑁𝑝/(𝑁−𝑝); then for 𝑢 such that |𝑢|^𝛽𝑢∈𝑊^1,𝑝(Ω), one has ‖𝑢‖_𝑞≤𝐶^1/(𝛽+1)‖𝑢‖_𝑟^1−𝜃∇|𝑢|^𝛽𝑢_𝑝^{𝜃/(𝛽+1)} (3) 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 that 𝑝−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‖𝑢(⋅,𝑡)‖₂≤𝑢₀^2−𝑘1₂−𝑀₁2−𝑘₁𝑡^{1/(2−𝑘1)},𝑡∈0,𝑇₁,‖𝑢(⋅,𝑡)‖₂≡0,𝑡∈𝑇₁,+∞, (4) where  𝑘₁,   𝑀₁, and  𝑇₁  are given by (11), (16), and (17), respectively. (2)For the case  1<𝑝<2𝑁/(𝑁+2), one has‖𝑢(⋅,𝑡)‖_1+𝑠≤𝑢₀^{1+𝑠−𝑘2}_1+𝑠−𝑀₂1+𝑠−𝑘₂𝑡^{1/(1+𝑠−𝑘2)},𝑡∈0,𝑇₂,‖𝑢(⋅,𝑡)‖₂≡0,𝑡∈𝑇₂,+∞, (5) where  𝑠,  𝑘₂,  𝑀₂, and  𝑇₂  are given by (18), (22), (26), and (28), respectively.

Proof. (1) For the case 2𝑁/(𝑁+2)≤𝑝<2, multiplying (1) by 𝑢 and integrating over Ω, we deduce from the Hölder inequality that 12𝑑‖𝑢‖𝑑𝑡²²+‖∇𝑢‖^𝑝_𝑝+𝑘‖𝑢‖^𝑟+1_𝑟+1≤𝜆|Ω|‖𝑢‖^𝑝_𝑝. (6) inequality ‖𝑢‖_𝑝≤𝐵‖∇𝑢‖_𝑝, (7) where 𝐵 denotes the optimal embedding constant, combining (6) and (7) we have 12𝑑‖𝑢‖𝑑𝑡²²+1−𝜆𝐵^𝑝|Ω|‖∇𝑢‖^𝑝_𝑝+𝑘‖𝑢‖^𝑟+1_𝑟+1≤0. (8) By Lemma 1, we have ‖𝑢‖₂≤𝐶₁𝑁,𝑝,𝑟‖∇𝑢‖^𝜃_𝑝‖𝑢‖^1−𝜃_1+𝑟, (9) where 𝜃₁=(1/(1+𝑟)−1/2)(1/𝑁−1/𝑝+1/(1+𝑟))⁻¹.
It is easy to check that 𝜃₁∈(0,1]; using Young's inequality with 𝜀, it follows from (9) that ‖𝑢‖^𝑘1₂≤𝐶^𝑘1₁𝑁,𝑝,𝑟𝜀₁‖∇𝑢‖^𝑝_𝑝+𝐶𝜀₁‖𝑢‖^{𝑝𝑘1(1−𝜃1)/(𝑝−𝑘1𝜃1)}_1+𝑟, (10) where 𝜀₁>0 and 𝑘₁>0 will be determined later. We choose 𝑘₁=2(1+𝑟)𝑝+𝑁𝑝−1−𝑟.2𝑝+𝑁𝑝−1−𝑟 (11) Then we can conclude that 𝑘₁∈(1,2) and 𝑝𝑘₁(1−𝜃₁)/(𝑝−𝑘₁𝜃₁)=1+𝑟. Therefore, it follows from (10) that ‖𝑢‖^1+𝑟_1+𝑟≥𝐶^−𝑘1₁𝑁,𝑝,𝑟‖𝑢‖^𝑘1₂−𝜀₁‖∇𝑢‖^𝑝_𝑝1𝐶𝜀₁. (12)
By combining (8) and (12), we have 12𝑑‖𝑢‖𝑑𝑡²²+1−𝜆𝐵^𝑝|Ω|−𝑘𝜀₁𝐶𝜀₁‖∇𝑢‖^𝑝_𝑝+𝑘𝐶^−𝑘1₁𝑁,𝑝,𝑟𝐶𝜀₁‖𝑢‖^𝑘1₂≤0. (13) Choosing 𝜀₁ small enough such that 1−𝑘𝜀₁/𝐶(𝜀₁)>0 and |Ω|≤(1−𝑘𝜀₁/𝐶(𝜀₁))/𝜆𝐵^𝑝, then we have 1−𝑘𝜀₁/𝐶(𝜀₁)−𝐵^𝑝𝜆|Ω|>0. Therefore, we deduce from 𝑘₁∈(1,2) that 𝑑‖𝑢‖𝑑𝑡²+𝑀₁‖𝑢‖^𝑘1₂⁻¹≤0, (14) which implies that ‖𝑢(⋅,𝑡)‖₂≤𝑢₀^2−𝑘1₂−𝑀₁2−𝑘₁𝑡^{1/(2−𝑘1)},𝑡∈0,𝑇₁,‖𝑢(⋅,𝑡)‖₂≡0,𝑡∈𝑇₁,+∞, (15) where 𝑀₁=𝑘𝐶^−𝑘1₁𝑁,𝑝,𝑟𝐶𝜀₁,𝑇₁=𝑢₀^2−𝑘1₂𝑀₁2−𝑘₁. (16)(17) (2) For the case 1<𝑝<2𝑁/(𝑁+2), multiplying (1) by 𝑢^𝑠, where ()𝑠>𝑙=2𝑁−𝑝1+𝑁𝑝>1, (18) integrating over Ω, we deduce from the Hölder inequality that 1𝑑1+𝑠‖𝑢‖𝑑𝑡^1+𝑠1+𝑠+𝑠𝑝^𝑝𝑝+𝑠−1^𝑝∇𝑢^{(𝑝+𝑠−1)/𝑝𝑝}_𝑝+𝑘‖𝑢‖^𝑠+𝑟_𝑠+𝑟≤𝜆|Ω|‖𝑢‖^{𝑝+𝑠−1}_{𝑝+𝑠−1}. (19) By Lemma 1 and 𝑠>1, we have ‖𝑢‖_𝑠+1≤𝐶₂𝑁,𝑝,𝑟∇𝑢^{(𝑝+𝑠−1)/𝑝𝑝𝜃2}_𝑝^{/(𝑝+𝑠−1)}‖𝑢‖^1−𝜃2_𝑠+𝑟, (20) 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 ‖𝑢‖^𝑘2_𝑠+1≤𝐶^𝑘2₂𝑁,𝑝,𝑟,𝑠𝜀₂∇𝑢^{(𝑝+𝑠−1)/𝑝𝑝}_𝑝+𝐶𝜀₂×‖𝑢‖^{(1−𝜃2)𝑘2(𝑝+𝑠−1)/(𝑝+𝑠−1−𝑘2𝜃2)}_𝑠+𝑟, (21) where 𝜀₂>0 and 𝑘₂>0 will be determined later. We choose 𝑘₂=(𝑠+1)(𝑠+𝑟)𝑝+𝑁𝑝−1−𝑟();𝑠+1𝑝+𝑁𝑝−1−𝑟 (22) 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 ‖𝑢‖^𝑠+𝑟_𝑠+𝑟≥𝐶^−𝑘2₂𝑁,𝑝,𝑟,𝑠𝐶𝜀₂‖𝑢‖^𝑘2_𝑠+1−𝜀₂𝐶𝜀₂∇𝑢^{(𝑝+𝑠−1)/𝑝𝑝}_𝑝. (23)
By combining (19) and (23), we have by poincare inequality 1𝑑1+𝑠‖𝑢‖𝑑𝑡^1+𝑠1+𝑠+𝑠𝑝^𝑝𝑝+𝑠−1^𝑝−𝑘𝜀₂𝐶𝜀₂−𝜆|Ω|𝐵^𝑝×∇𝑢^{(𝑝+𝑠−1)/𝑝𝑝}_𝑝+𝑘𝐶^𝑘2₂𝑁,𝑝,𝑟,𝑠𝐶𝜀₂‖𝑢‖^𝑘2_𝑠+1≤0. (24) Choosing 𝜀₂>0 small enough such that 𝑠𝑝^𝑝/(𝑝+𝑠−1)^𝑝−𝑘𝜀₂/𝐶(𝜀₂)>0 and |Ω|≤(𝑠𝑝^𝑝/(𝑝+𝑠−1)^𝑝−𝑘𝜀₂/𝐶(𝜀₂))/𝜆|Ω|𝐵^𝑝, then we have 𝑠𝑝^𝑝/(𝑝+𝑠−1)^𝑝−𝑘𝜀₂/𝐶(𝜀₂)−𝜆|Ω|𝐵^𝑝>0. Therefore, we deduce from 𝑘₂∈(𝑠,𝑠+1) that 𝑑‖𝑢‖𝑑𝑡^1+𝑠+𝑀₂‖𝑢‖^𝑘2−𝑠_1+𝑠≤0, (25) where 𝑀₂=𝑘𝐶^−𝑘2₂𝑁,𝑝,𝑟,𝑠𝐶𝜀₂, (26) which implies that ‖𝑢(⋅,𝑡)‖_1+𝑠≤𝑢₀^{1+𝑠−𝑘2}_1+𝑠−𝑀₂1+𝑠−𝑘₂𝑡^{1/(1+𝑠−𝑘2)},𝑡∈0,𝑇₂,‖𝑢(⋅,𝑡)‖_1+𝑠≡0,𝑡∈𝑇₂,+∞, (27) where 𝑇₂=𝑢₀^{1+𝑠−𝑘2}_1+𝑠𝑀₂1+𝑠−𝑘₂. (28) The proof of Theorem 2 is complete.

Theorem 3. Assume that 𝑟<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‖𝑢(⋅,𝑡)‖₂≤𝑢₀^2−𝑘1₂−2−𝑘₁𝑀₃𝑡^{1/(2−𝑘1)},𝑡∈0,𝑇₃,‖𝑢(⋅,𝑡)‖₂≡0,𝑡∈𝑇₃,+∞, (29) 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‖𝑢(⋅,𝑡)‖_𝑠+1≤𝑢₀^{𝑠+1−𝑘2}_𝑠+1−𝑠+1−𝑘₂𝑀₄𝑡^{1/(𝑠+1−𝑘2)},𝑡∈0,𝑇₄,‖𝑢(⋅,𝑡)‖_𝑠+1≡0,𝑡∈𝑇₄,+∞, (30) where  𝑠,  𝑘₂,  𝑀₄, and  𝑇₄  are given by (18), (22), (39), and (41), respectively.

Proof. (1)  If 2𝑁/(𝑁+2)≤𝑝<2, multiplying (1) by 𝑢 and integrating over Ω, we deduce from (12) and the Hölder inequality that 12𝑑‖𝑢‖𝑑𝑡²²+1−𝑘𝜀₁𝐶𝜀₁‖∇𝑢‖^𝑝_𝑝+𝑘𝐶^−𝑘1₁𝑁,𝑝,𝑟𝐶𝜀₁×‖𝑢‖^𝑘1₂−𝜆|Ω|^(3−𝑞)/2‖𝑢‖₂^𝑞+1≤0. (31) By choosing 𝜀₁>0 small enough such that 1−𝑘𝜀₁/𝐶(𝜀₁)≥0, we obtain that 𝑑‖𝑢‖𝑑𝑡²+𝑀₃‖𝑢‖^𝑘1₂⁻¹≤0, (32) provided that ‖𝑢₀‖₂≤(𝑘𝐶^−𝑘1₁(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 𝑀₃=𝑘𝐶^−𝑘1₁𝑁,𝑝,𝑟𝐶𝜀₁−𝜆|Ω|^(3−𝑞)/2𝑢₀^𝑞−𝑘1₂⁺¹>0. (33)
From (32) and 𝑘₁∈(1,2), we can derive that ‖𝑢(⋅,𝑡)‖₂≤𝑢₀^2−𝑘1₂−2−𝑘₁𝑀₃𝑡^{1/(2−𝑘1)},𝑡∈0,𝑇₃,‖𝑢(⋅,𝑡)‖₂≡0,𝑡∈𝑇₃,+∞, (34) where 𝑇₃=𝑢₀^2−𝑘1₂2−𝑘₁𝑀₃. (35) (2)  If 1<𝑝<2𝑁/(𝑁+2), multiplying (1) by 𝑢^𝑠, where 𝑠 is given by (18) and integrating over Ω, we deduce from the Hölder inequality and (23) that 1𝑑1+𝑠‖𝑢‖𝑑𝑡^1+𝑠1+𝑠+𝑠𝑝^𝑝𝑝+𝑠−1^𝑝−𝑘𝜀₂𝐶𝜀₂∇𝑢^{(𝑝+𝑠−1)/𝑝𝑝}_𝑝+𝑘𝐶^−𝑘2₂𝑁,𝑝,𝑟,𝑠𝐶𝜀₂‖𝑢‖^𝑘2_𝑠+1≤𝜆‖𝑢‖^𝑞+𝑠_1+𝑠|Ω|^{(2+𝑠−𝑞)/(1+𝑠)}. (36) Choosing 𝜀₂>0 small enough such that 𝑠𝑝^𝑝/(𝑝+𝑠−1)^𝑝−𝑘𝜀₂/𝐶(𝜀₂)>0, we have 𝑑‖𝑢‖𝑑𝑡^1+𝑠+‖𝑢‖^𝑘2−𝑠_1+𝑠𝑘𝐶^𝑘2₂𝑁,𝑝,𝑟,𝑠𝐶𝜀₂−𝜆|Ω|^{(2+𝑠−𝑞)/(1+𝑠)}‖𝑢‖^{𝑞+𝑠−𝑘2}_1+𝑠≤0. (37)
Therefore, we have 𝑑‖𝑢‖𝑑𝑡^1+𝑠+𝑀₄‖𝑢‖^𝑘2−𝑠_1+𝑠≤0, (38) provided that ‖𝑢₀‖_1+𝑠≤(𝑘𝐶^−𝑘2₂(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 𝑀₄=𝑘𝐶^−𝑘2₂𝑁,𝑝,𝑟,𝑠𝐶𝜀₂−𝜆|Ω|^{(2+𝑠−𝑞)/(1+𝑠)}𝑢₀^{𝑞+𝑠−𝑘2}_1+𝑠>0. (39) It follows from (38) and 𝑘₂∈(𝑠,𝑠+1) that ‖𝑢(⋅,𝑡)‖_1+𝑠≤𝑢₀^{𝑠+1−𝑘2}_𝑠+1−𝑀₄𝑠+1−𝑘₂𝑡^{1/(𝑠+1−𝑘2)},𝑡∈0,𝑇₄,‖𝑢(⋅,𝑡)‖_𝑠+1≡0,𝑡∈𝑇₄,+∞, (40) where 𝑇₄=𝑢₀^{1+𝑠−𝑘2}_1+𝑠𝑀₄𝑠+1−𝑘₂. (41) The proof of Theorem 3 is complete.

Acknowledgments

This work is supported by the Department of Education of Jilin Province (2013445) and by Science and Technology Bureau of Siping, Jilin Province (2012040), and is partially supported by the NSF of China under Grant 11171060.