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Abstract and Applied Analysis
Volume 2011 (2011), Article ID 142128, 13 pages
http://dx.doi.org/10.1155/2011/142128
Research Article

On Asymptotic Behavior for Reaction Diffusion Equation with Small Time Delay

School of Electrical Engineering and Automation, Tianjin University, Tianjin 300072, China

Received 11 October 2011; Revised 5 November 2011; Accepted 5 November 2011

Academic Editor: Elena Litsyn

Copyright © 2011 Xunwu Yin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We investigate the asymptotic behavior of scalar diffusion equation with small time delay 𝑢𝑡Δ𝑢=𝑓(𝑢(𝑡),𝑢(𝑡𝜏)). Roughly speaking, any bounded solution will enter and stay in the neighborhood of one equilibrium when the equilibria are discrete.

1. Introduction

With delay systems appearing frequently in science, engineering, physics, biology, economics, and so forth, many authors have recently devoted their interests to the effect of small delays on the dynamics of some system. This problem is relatively well understood for linear systems, including both finite-dimensional and infinite-dimensional situations, see [15]. However, for nonlinear systems, the problem is much more complicated, but there are some very nice results in [610].

In this paper, we consider the following scalar reaction-diffusion equation with a time delay 𝑢𝑡Δ𝑢=𝑓(𝑢(𝑡),𝑢(𝑡𝜏))𝑥Ω𝑅𝑁.(1.1) It is proved in [1113] that for such diffusion equation without delay, 𝑢𝑡Δ𝑢=𝑓(𝑢),(1.2) subject to homogeneous boundary conditions, all globally defined bounded solutions must approach the set of equilibria as 𝑡 tends to infinity. This depends heavily on the fact that (1.2) is a gradient system with the Lyapunov function 1𝑉(𝑢)=2Ω||||𝑢2Ω𝐹(𝑢),(1.3) where 𝐹 is a primitive of 𝑓. It is well known that solutions of (1.1) will typically oscillate in 𝑡 as 𝑡 if the delay is not sufficiently small. However, we will point out such interesting result that oscillations do not happen for sufficiently small delay. Specifically we obtain the conclusion that for given 𝑅,𝜀>0 there exists a sufficiently small 𝜏>0 such that any solution of (1.1) satisfying limsup𝑡𝑢(𝑥,𝑡)𝐻10(Ω)𝑅 will ultimately enter and stay in the 𝜀 neighborhood of some equilibrium.

As a matter of fact, for the finite-dimensional situation, in [6] Li and Wang considered the general nonlinear gradient system with multiple small time delays𝑥𝑥(𝑡)=𝑓𝑡𝑟1(𝑡),,𝑥𝑡𝑟𝑛(𝑡).(1.4) Making use of the Morse structure of invariant sets of gradient systems, he obtained a similar result. Following this idea, we investigate (1.1) in the infinite-dimensional situation. The difference between the two situations is very great. For example, under the finite-dimensional situation there must exist convergent subsequence for any bounded sequence. This is not correct in the infinite-dimensional situation. We only have weak compactness. In other words, bounded sequences in a reflexive Banach space are weakly precompact. In order to overcome this difficulty, we apply the famous Aubin-Lions lemma [14].

2. Preliminaries

In this paper, we assume Ω to be an open, bounded subset of 𝑅𝑁 and 𝜏 to be a positive parameter (the delay). Consider the following scalar delayed initial boundary value problem:𝑢𝑡],Δ𝑢=𝑓(𝑢(𝑡),𝑢(𝑡𝜏))inΩ×(0,𝑇𝑢=𝑢0[],],onΩ×𝜏,0𝑢=0on𝜕Ω×(0,𝑇(2.1) where the nonlinear 𝑓𝑅2𝑅 is assumed to be continuous and to satisfy ||𝑓||(𝑢,𝑣)𝐶1+|𝑢|𝜌+|𝑣|𝜌,(2.2)𝑓(𝑢,𝑣)𝑢𝐶1𝑢2+𝑢𝑣+𝐶2.(2.3) Here 𝐶,𝐶1,and𝐶2 are all constants, 𝜌=1+2/𝑁. Firstly we will give the definition of weak solution for (2.1).

Definition 2.1. A function 𝑢(𝑥,𝑡) is called a weak solution of (2.1) if and only if(i)𝑢𝐿2(0,𝑇;𝐻10(Ω)),with𝑢𝐿2(0,𝑇;𝐻1(Ω)), (ii)𝑢|[𝜏,0]=𝑢0𝐿2(Ω), (iii)𝑇0[𝑢𝑡,𝜑+(𝐷𝑢,𝐷𝜑)]𝑑𝑡=𝑇0(𝑓(𝑢(𝑡),𝑢(𝑡𝜏)),𝜑)𝑑𝑡,

for each 𝜑𝐿2(0,𝑇;𝐻10(Ω)). Here , and (,) denote the pair of 𝐻1(Ω) and 𝐻10(Ω), the inner product in 𝐿2(Ω), respectively. Next we will give two very important lemmas many times used in the proof of two theorems.

Lemma 2.2. If {𝑢𝑛} is bounded in 𝐿2(𝜏,𝑇;𝐻10(Ω))𝐿(𝜏,𝑇;𝐿2(Ω)), then {𝑓(𝑢𝑛(𝑡),𝑢𝑛(𝑡𝜏))} is bounded in 𝐿2(0,𝑇;𝐿2(Ω)).

Proof. Let 𝑎=(𝑁2)/𝑁(0,1), and because 𝜌=1+2/𝑁, 2𝜌=𝑎(2𝑁/(𝑁2))+2(1𝑎). Before testing the boundedness of 𝑓𝐿2(0,𝑇;𝐿2(Ω)), we firstly estimate 𝑢𝑛𝐿2𝜌(Ω)𝑢𝑛𝐿2𝜌2𝜌(Ω)=Ω||𝑢𝑛||𝑎(2𝑁/(𝑁2))||𝑢𝑛||2(1𝑎)𝑑𝑥Ω||𝑢𝑛||2𝑁/(𝑁2)𝑑𝑥𝑎Ω||𝑢𝑛||2𝑑𝑥1𝑎=𝑢𝑛𝐿2𝑁𝑎/(𝑁2)2𝑁/(𝑁2)(Ω)𝑢𝑛𝐿2(1𝑎)2(Ω)𝐶1𝑢𝑛𝐻2𝑁𝑎/(𝑁2)10(Ω)=𝐶1𝑢𝑛2𝐻10(Ω).(2.4) Here we utilize the Ḧolder inequality, the fact of 𝐻10(Ω)𝐿2𝑁/(𝑁2) continuously and {𝑢𝑛} is bounded in 𝐿(𝜏,𝑇;𝐿2(Ω)). So 𝑇0Ω||𝑢𝑛||(𝑥,𝑡)2𝜌𝑑𝑥𝑑𝑡𝐶1𝑢𝑛(𝑥,𝑡)2𝐿2𝜏,𝑇;𝐻10(Ω),𝑇0Ω||𝑢𝑛||(𝑥,𝑡𝜏)2𝜌𝑑𝑥𝑑𝑡𝑡𝜏=𝑠=𝑇𝜏𝜏Ω||𝑢𝑛||(𝑥,𝑠)2𝜌𝑑𝑥𝑑𝑠𝑇𝜏Ω||𝑢𝑛||(𝑥,𝑡)2𝜌𝑑𝑥𝑑𝑡𝐶1𝑢𝑛(𝑥,𝑡)2𝐿2𝜏,𝑇;𝐻10(Ω).(2.5) In view of (2.2), we can easily see ||𝑓𝑢𝑛(𝑡),𝑢𝑛||(𝑡𝜏)2||𝑢𝐶1+𝑛||(𝑡)2𝜌+||𝑢𝑛||(𝑡𝜏)2𝜌.(2.6) Integrating the above inequality with 𝑡 and 𝑥, we complete the proof.

Remark 2.3. If {𝑢𝑛} is bounded in 𝐿(𝜏,𝑇;𝐻10(Ω)), we can also get the same conclusion. The underlying lemma is the famous Aubin-Lions lemma. We only give the statement of the lemma.

Lemma 2.4. Let 𝑋0, 𝑋, and 𝑋1 be three Banach spaces with 𝑋0𝑋𝑋1. Suppose that 𝑋0 is compactly embedded in 𝑋 and 𝑋 is continuously embedded in 𝑋1. Suppose also that 𝑋0 and 𝑋1 are reflexive spaces. For 1<𝑝,𝑞<, let 𝑊={𝑢𝐿𝑝([0,𝑇;𝑋0)|𝑢𝐿𝑞([0,𝑇];𝑋1)}. Then the embedding of W into 𝐿𝑝[0,𝑇;𝑋) is also compact.

Finally we give the definition of equilibrium solution of (1.2) and omega limit set 𝜔(𝑢), where 𝑢(𝑥,𝑡) is a bounded solution of (1.1). Selecting 𝐻10(Ω) as our phase space, we denote by 𝜔(𝑢) the limit set 𝜔(𝑢)=𝑣thereexists𝑡𝑛𝑢suchthat,𝑡𝑛𝑣𝐻10(Ω)0.(2.7)

As usual, an equilibrium solution of (1.2) is defined as a solution which does not depend on 𝑡; the equilibrium states are thus the functions 𝑢𝐻10(Ω)𝐻2(Ω) satisfying the elliptic boundary value problem Δ𝑢=𝑓(𝑢,𝑢)inΩ,𝑢=0onΩ(2.8) in the weak sense.

Let each equilibrium be isolated and let 𝑢(,𝑡) be the bounded complete solution of (1.2). Then we have lim𝑡𝑢(,𝑡)=𝐸1,lim𝑡+𝑢(,𝑡)=𝐸2(2.9) for some equilibrium 𝐸1 and 𝐸2 with 𝑉(𝐸1)>𝑉(𝐸2), where 𝑉 is the Lyapunov function (1.3). A complete solution of (1.2) means a solution 𝑢(,𝑡) defined on (,+). Now we will introduce our main results.

3. Main Results

In this section, we will prove two theorems. One is the existence of global solution. The other is our core, Theorem 3.2.

Theorem 3.1. For given 𝜏>0,𝑢0𝐿2(Ω), problem (2.1) has a global weak solution.

Proof. We will use classical Galerkin's method to build a weak solution of (2.1). Consider the approximate solution 𝑢𝑚(𝑡) of the form 𝑢𝑚(𝑡)=𝑚𝑘=1𝑢𝑚𝑘(𝑡)𝜔𝑘,(3.1) where {𝜔𝑘}𝑘=1 is an orthogonal basis of 𝐻10(Ω) and {𝜔𝑘}𝑘=1 is an orthonormal basis of 𝐿2(Ω). We get 𝑢𝑚 from solving the following ODES: 𝑢𝑚,𝜔𝑘+𝐷𝑢𝑚,𝐷𝜔𝑘=𝑓𝑢𝑚(𝑡),𝑢𝑚(𝑡𝜏),𝜔𝑘𝑢(0<𝑡𝑇,𝑘=1,2,𝑚),𝑚𝑘(𝑢𝑡)=0,𝜔𝑘(𝜏𝑡0,𝑘=1,2,𝑚).(3.2) According to standard existence theory of ODES, we can obtain the local existence of 𝑢𝑚.
Next we will establish some priori estimates for 𝑢𝑚. Multiplying (2.1) by 𝑢𝑚 and integrating over Ω, we have 12𝑑𝑢𝑑𝑡𝑚2𝐿2(Ω)+𝑢𝑚2𝐻10(Ω)=Ω𝑓𝑢𝑚(𝑡),𝑢𝑚(𝑢𝑡𝜏)𝑚𝑑𝑥.(3.3) Because of (2.3) and the Cauchy inequality, we can get 𝑑𝑢𝑑𝑡𝑚2𝐿2(Ω)𝑢+2𝑚2𝐻10(Ω)𝐶1𝑢𝑚2𝐿2(Ω)+𝐶2.(3.4) Getting rid of the term 2𝑢𝑚2𝐻10(Ω), from the differential form of Gronwall's inequality, we yield the estimate max0𝑡𝑇𝑢𝑚2𝐿2(Ω)𝐶1𝑢02𝐿2(Ω)+𝐶2.(3.5) Returning once more to inequality (3.4), we integrate from 0 to 𝑇 and employ the inequality above to find 𝑢𝑚2𝐿20,𝑇;𝐻10(Ω)𝐶1𝑢02𝐿2(Ω)+𝐶2.(3.6) Multiplying (2.1) by 𝑢𝑚 and then integrating over Ω, we have 𝑢𝑚2𝐿2(Ω)+Ω𝐷𝑢𝑚𝐷𝑢𝑚𝑑𝑥=Ω𝑓𝑢𝑚(𝑡),𝑢𝑚(𝑢𝑡𝜏)𝑚𝑑𝑥.(3.7) Using the Cauchy inequality and Lemma 2.2, we get 𝑢𝑚2𝐿2(Ω)+𝑑𝑢𝑑𝑡𝑚2𝐻10(Ω)𝐶.(3.8) Again from the differential form of Gronwall's inequality, we integrate from 0 to 𝑇𝑢𝑚2𝐿20,𝑇;𝐿2(Ω)𝐶1𝑢02𝐿2(Ω)+𝐶2.(3.9) Since 𝐿2(Ω)𝐻1(Ω), so 𝑢𝑚2𝐿20,𝑇;𝐻1(Ω)𝐶1𝑢02𝐿2(Ω)+𝐶2.(3.10) According to estimates (3.6), (3.10), Lemma 2.2, and weak compactness, we see that 𝑢𝑚𝑢weaklyin𝐿20,𝑇;𝐻10,𝑢(Ω)𝑚𝑢weaklyin𝐿20,𝑇;𝐻1,𝑓𝑢(Ω)𝑚(𝑡),𝑢𝑚(𝑡𝜏)𝜂weaklyin𝐿20,𝑇;𝐿2.(Ω)(3.11) Here a subsequence of {𝑢𝑚}𝑚=1 is still denoted by {𝑢𝑚}𝑚=1. Applying Lemma 2.2, we can conclude that 𝑢𝑚𝑢 strongly in 𝐿2(0,𝑇;𝐿2(Ω)). Hence 𝑢𝑚𝑢 A.E. in Ω×(0,𝑇). Since 𝑓 is continuous, it follows that 𝑓[𝑢𝑚(𝑡),𝑢𝑚(𝑡𝜏)]𝑓[𝑢(𝑡),𝑢(𝑡𝜏)] A.E. in Ω×(0,𝑇). Thanks to (3.11) and Lemma 1.3 in [14], one has 𝑓𝑢𝑚(𝑡),𝑢𝑚[](𝑡𝜏)𝑓𝑢(𝑡),𝑢(𝑡𝜏)weaklyin𝐿20,𝑇;𝐿2.(Ω)(3.12)
Next fix an integer 𝑁0 and choose a function 𝑣𝐶1([0,𝑇];𝐻10(Ω)) having the form 𝑣(𝑡)=𝑁0𝑘=1𝑑𝑘(𝑡)𝜔𝑘,(3.13) where {𝑑𝑘}𝑁0𝑘=1 are given smooth functions. Choosing 𝑚𝑁0 and multiplying (3.2) by 𝑑𝑘(𝑡) sum 𝑘=1,2,,𝑁0, and then integrating with respect to 𝑡, we can find 𝑇0𝑢𝑚+,𝑣𝐷𝑢𝑚,𝐷𝑣𝑑𝑡=𝑇0𝑓𝑢𝑚(𝑡),𝑢𝑚(𝑡𝜏),𝑣𝑑𝑡.(3.14) Recalling (3.11) and (3.12) and passing to weak limits, we get 𝑇0𝑢,𝑣+(𝐷𝑢,𝐷𝑣)𝑑𝑡=𝑇0[](𝑓𝑢(𝑡),𝑢(𝑡𝜏),𝑣)𝑑𝑡.(3.15) Because functions of the form 𝑣(𝑡) are dense in 𝐿2(0,𝑇;𝐻10(Ω)), so the above equality holds for all functions 𝑣𝐿2(0,𝑇;𝐻10(Ω)).
Lastly we will show 𝑢|[𝜏,0]=𝑢0𝐿2(Ω). Notice that for each 𝑣𝐶1([0,𝑇];𝐻10(Ω)) with 𝑣(𝑇)=0 we get the following from (3.15): 𝑇0𝑣,𝑢+(𝐷𝑢,𝐷𝑣)𝑑𝑡=𝑇0[](𝑓𝑢(𝑡),𝑢(𝑡𝜏),𝑣)𝑑𝑡+(𝑢(0),𝑣(0)).(3.16) Similarly, from (3.14), we deduce 𝑇0𝑣,𝑢𝑚+𝐷𝑢𝑚,𝐷𝑣𝑑𝑡=𝑇0𝑓𝑢𝑚(𝑡),𝑢𝑚𝑢(𝑡𝜏),𝑣𝑑𝑡+𝑚.(0),𝑣(0)(3.17) In view of (3.2), 𝑢𝑚(0)𝑢0 in 𝐿2(Ω); once again employ (3.11) and (3.12) to find 𝑇0𝑣,𝑢+(𝐷𝑢,𝐷𝑣)𝑑𝑡=𝑇0[]𝑢(𝑓𝑢(𝑡),𝑢(𝑡𝜏),𝑣)𝑑𝑡+0.,𝑣(0)(3.18) As 𝑣(0) is arbitrary, so we get the result 𝑢(0)=𝑢0. Since for 𝑡[𝜏,0],𝑢𝑚(𝑡)𝑢0 in 𝐿2(Ω), we can obtain the result. As for 𝑇 being arbitrary, we see the global existence of (2.1).

Theorem 3.2. Assume that each equilibrium of (1.2) is isolated. Let 𝑅,𝜀>0 be given arbitrarily. Then there exists a sufficiently small 𝜏>0 such that any solution of (1.1) with limsup𝑡+u(,𝑡)𝐻10(Ω)𝑅 will eventually enter and stay in the 𝜀 neighborhood of some equilibrium.

Proof. Here we select 𝐻10(Ω) as our phase space. For simplicity, we will verify the correctness of the conclusion for such bounded solutions 𝑢(𝑥,𝑡) of (1.1) as 𝑢(,𝑡)𝐻10(Ω)𝑅 for all 𝑡[0,). That is to say they are in 𝑅.
Assume there are 𝑛 equilibria of (1.2) {𝐸1,,𝐸𝑛}, ordered by 𝑉(𝐸𝑛)𝑉(𝐸𝑛1)𝑉(𝐸1), where 𝑉 is the Lyapunov function (1.3). We will follow two steps to prove our result.
Step 1. We firstly verify that for any 𝛿>0, there exists a sufficiently small 𝜏>0 such that 𝜔(𝑢)1𝑗𝑛𝛿𝐸𝑗(3.19) for any solution 𝑢(𝑥,𝑡) of (1.1) in 𝑅.
In order to prove (3.19), we proceed by contradiction, which is used repeatedly in the following proof. Assume that there was a decreasing sequence 𝜏𝑘0 and a corresponding solution sequence 𝑢𝑘 of (1.1) in 𝑅 satisfying 𝑑𝐸𝑗𝑢,𝜔𝑘2𝛿(3.20) for all 1𝑗𝑛 and 𝑘𝑁. According to the definition of 𝜔(𝑢), for each 𝑘 we can take a 𝑡𝑘>0 such that for 𝑡𝑡𝑘,1𝑗𝑛𝑢𝑘(,𝑡)𝐸𝑗𝐻10(Ω)𝛿.(3.21) Let ̃𝑢𝑘(𝑡)=𝑢𝑘(𝑡+𝑡𝑘) for 𝑡0. It is easy to see ̃𝑢𝑘 is the weak solution of 𝜕𝑡̃𝑢𝑘Δ̃𝑢𝑘=𝑓̃𝑢𝑘(𝑡),̃𝑢𝑘𝑡𝜏𝑘.(3.22) Next we will show there is a strong convergent subsequence of {̃𝑢𝑘}𝑘=1 in 𝐿2(𝑡,𝑡+1;𝐻10(Ω)) for 𝑡0. Still denoting ̃𝑢𝑘, we can also prove the limit ̃𝑢 is in fact the weak solution of (1.2). From the elliptic equation regularity theorem, we can multiply (3.22) by Δ̃𝑢𝑘 and integrate over Ω12𝑑𝑑𝑡̃𝑢𝑘2𝐻10(Ω)+̃𝑢𝑘2𝐻2(Ω)12𝑓2𝐿2(Ω)+̃𝑢𝑘2𝐻2(Ω).(3.23) Because of the remark in Section 2, we can get 𝑑𝑑𝑡̃𝑢𝑘2𝐻10(Ω)+̃𝑢𝑘2𝐻2(Ω)𝐶.(3.24) Integrating from 𝑡 to 𝑡+1, from the boundedness of ̃𝑢𝑘(,𝑡)2𝐻10(Ω) and ̃𝑢𝑘(,𝑡+1)2𝐻10(Ω), we conclude that {̃𝑢𝑘} is bounded in 𝐿2(𝑡,𝑡+1;𝐻2(Ω)). Multiplying (3.22) by 𝜕𝑡̃𝑢𝑘 and utilizing the same method above, we can also conclude that {𝜕𝑡̃𝑢𝑘} is bounded in 𝐿2(𝑡,𝑡+1;𝐿2(Ω)). Applying the Aubin-Lions lemma, we can conclude that there is a strong convergent subsequence of {̃𝑢𝑘}𝑘=1 in 𝐿2(𝑡,𝑡+1;𝐻10(Ω)) for 𝑡0. We may set ̃𝑢𝑘̃𝑢 strongly in 𝐿2(𝑡,𝑡+1;𝐻10(Ω)). Of course ̃𝑢𝑘̃𝑢 strongly in 𝐿2(𝑡,𝑡+1;𝐿2(Ω)). Hence ̃𝑢𝑘̃𝑢 a.e. in Ω×(𝑡,𝑡+1). Since 𝑓 is continuous, it follows that 𝑓[̃𝑢𝑘(𝑡),̃𝑢𝑘(𝑡𝜏𝑘)]𝑓(̃𝑢,̃𝑢) a.e. in Ω×(0,𝑇). Thanks to the weak convergence of 𝑓[̃𝑢𝑘(𝑡),̃𝑢𝑘(𝑡𝜏𝑘)] in 𝐿2(𝑡,𝑡+1;𝐿2(Ω)) and lemma 1.3 in [14], one has 𝑓̃𝑢𝑘(𝑡),̃𝑢𝑘𝑡𝜏𝑘𝑓(̃𝑢,̃𝑢)weaklyin𝐿20,𝑇;𝐿2.(Ω)(3.25) So we prove that ̃𝑢 is the weak solution of (1.2). Considering (3.21), we have the following estimate for ̃𝑢: 𝑡𝑡+1̃𝑢𝐸𝑗𝐻10(Ω)𝑑𝑠=𝑡𝑡+1̃𝑢𝑘𝐸𝑗+̃𝑢̃𝑢𝑘𝐻10(Ω)𝑑𝑠𝑡𝑡+1̃𝑢𝑘𝐸𝑗𝐻10(Ω)𝑑𝑠𝑡𝑡+1̃𝑢𝑘̃𝑢𝐻10(Ω)𝑑𝑠𝛿𝑡𝑡+1̃𝑢𝑘̃𝑢𝐻10(Ω)𝛿𝑑𝑠2.(3.26) From the above inequality, we can surely know lim𝑡̃𝑢𝐸𝑗𝐻10(Ω)0 for all 1𝑗𝑛. However, because (1.2) is a gradient system, this contradicts the fact that lim𝑡𝑢(,𝑡)=𝐸𝑗 for some 𝐸𝑗. We obtain the correctness of (3.19).
Step 2. We will complete the proof of the theorem that if 𝜏 is sufficiently small, then for any bounded solution 𝑢(,𝑡) of (1.1) there must exist a 𝐸𝑗 and sufficiently large 𝑇 such that for 𝑡>𝑇𝑢(,𝑡)𝐸𝑗𝐻10(Ω)<𝜀.(3.27)
Here we also adopt contradiction method to prove the result. If the desired conclusion was not correct, there would be a decreasing sequence 𝜏𝑘0 and a corresponding solution sequence 𝑢𝑘 of (1.1) in 𝑅 which does not satisfy (3.27).
In view of 𝜏𝑘0, it is easy to infer that lim𝑘min1𝑗𝑛𝑑𝐸𝑗𝑢,𝜔𝑘=0.(3.28) Without loss of generality, we can assume that for all 𝑘1𝜔𝑢𝑘1𝑗𝑛𝜀𝐸𝑗.(3.29) Denote by 𝑗𝑘 the smallest 𝑗 satisfying 𝜔𝑢𝑘𝜀𝐸𝑗.(3.30) It is easy to see that there exists a subsequence {𝑘1𝑖}𝑖=1 of {𝑘}𝑘=1 such that for some 𝑗1[1,𝑛], we have 𝑗𝑘1𝑖=𝑗1.
We will claim if 𝑗1<𝑛, then there exists a 𝛿1(0,𝜀) and 𝑘1 such that for 𝑘1𝑖>𝑘1𝑑𝐸𝑗1𝑢,𝜔𝑘1𝑖𝛿1.(3.31) Indeed, if the fact did not hold, there would be a subsequence of {𝑘1𝑖}𝑖=1 (for simplicity still denoted by {𝑘1𝑖}𝑖=1) such that lim𝑖𝑑𝐸𝑗1𝑢,𝜔𝑘1𝑖=0.(3.32) According to the definition of 𝑗𝑘 and (3.32), we can choose a sequence 𝑡𝑖>0 satisfying 𝑢𝑘1𝑖𝑡𝑖𝜀𝐸𝑗1,lim𝑖𝑢𝑘1𝑖𝑡𝑖𝐸𝑗1𝐻10(Ω)𝑢=0,𝑘1𝑖(𝑡)𝐸𝑗𝐻10(Ω)>𝜀,for𝑡>𝑡𝑖,𝑗<𝑗1.(3.33) Now we define 𝜂𝑖=sup𝑡𝑡𝑖𝑢𝑘1𝑖𝑡𝑖,𝑡𝜀𝐸𝑗1.(3.34) Obviously 𝑢𝑘1𝑖(𝜂𝑖)𝜕𝜀(𝐸𝑗1). Let 𝑣𝑖(,𝑡)=𝑢𝑘1𝑖,𝑡+𝜂𝑖𝜂,𝑡𝑖𝑡𝑖.,+(3.35) From (3.33) and the definition of 𝜂𝑖, it is clear to see 𝑣𝑖(0)𝜕𝜀𝐸𝑗1,𝑣𝑖(𝑡)𝜀𝐸𝑗1𝜂for𝑖𝑡𝑖𝑣𝑡0,(3.36)𝑖(𝑡)𝐸𝑗𝐻10(Ω)>𝜀,for𝑡>0,𝑗<𝑗1.(3.37) Obviously 𝑣𝑖(,𝑡) is the weak solution of 𝜕𝑡𝑣𝑖Δ𝑣𝑖𝑣=𝑓𝑖(𝑡),𝑣𝑖𝑡𝜏𝑘1𝑖.(3.38) Following the method above, we can also prove there is a strong convergent subsequence of {𝑣𝑖}𝑖=1 in 𝐿2(𝑡,𝑡+1;𝐻10(Ω)) for 𝑡0. Still denoting {𝑣𝑖}𝑖=1 and letting 𝑇=limsup(𝜂𝑖𝑡𝑖), the limit 𝑣 defined on (𝑇,) is indeed the weak solution of (1.2).
Next we will show that 𝑇=+. In fact, if 𝑇<+, then 𝑣(𝑡) can be well defined at 𝑡=𝑇. In view of (3.32), we see 𝑣(𝑇)=𝐸𝑗1. Hence 𝑣(𝑡)𝐸𝑗1 for 𝑡𝑇. That is to say, 𝑣𝑖(𝑡)𝐸𝑗1 strongly in 𝐿2(𝑇,0;𝐻10(Ω)). Because 𝑣𝑖(,𝑡)𝐿2((𝜂𝑖𝑡𝑖),0;𝐻10(Ω)) and 𝑣𝑖(,𝑡)𝐿2((𝜂𝑖𝑡𝑖),0;𝐿2(Ω)), it follows from Theorem 4 in 5.9.2 of [15] that 𝑣𝑖(,𝑡)𝐶([(𝜂𝑖𝑡𝑖),0];𝐻10(Ω)). That is to say lim𝑡0𝑣𝑖(,𝑡)𝑣𝑖(,0)𝐻10(Ω)=0.(3.39) By the definition of continuity, there exists 𝑡1<0 such that 𝑣𝑖(,𝑡)𝑣𝑖(,0)𝐻10(Ω)<𝜀0<𝜀for𝑡1<𝑡<0.(3.40) So 𝜀0>𝑣𝑖(,𝑡)𝐸𝑗1+𝐸𝑗1𝑣𝑖(,0)𝐻10(Ω)𝑣𝑖(,0)𝐸𝑗1𝐻10(Ω)𝑣𝑖(,𝑡)𝐸𝑗1𝐻10(Ω).(3.41) Hence 𝑣𝑖(,𝑡)𝐸𝑗1𝐻10(Ω)>𝑣𝑖(,0)𝐸𝑗1𝐻10(Ω)𝜀0=𝜀𝜀0>0.(3.42) Thus 0𝜂𝑖𝑡𝑖𝑣𝑖(,𝑡)𝐸𝑗1𝐻10(Ω)𝑑𝑡0𝑡1𝑣𝑖(,𝑡)𝐸𝑗1𝐻10(Ω)𝑑𝑡𝑡1𝜀𝜀0.(3.43) Obviously lim𝑖0𝜂𝑖𝑡𝑖𝑣𝑖(,𝑡)𝐸𝑗1𝐻10(Ω)𝑑𝑡0.(3.44) This contradicts the fact 𝑣𝑖(𝑡)𝐸𝑗1 strongly in 𝐿2((𝜂𝑖𝑡𝑖),0;𝐻10(Ω)). So it must be 𝑇=+.
Let lim𝑡+𝑣(𝑡)=𝐸𝑗. Then there must be 𝑗𝑗1. Otherwise for 𝑗<𝑗1𝑡𝑡+1𝑣(𝑠)𝐸𝑗𝐻10(Ω)=𝑡𝑡+1𝑣(𝑠)𝑣𝑖(𝑠)+𝑣𝑖(𝑠)𝐸𝑗𝐻10(Ω)𝑑𝑠𝑡𝑡+1𝑣𝑖(𝑠)𝐸𝑗𝐻10(Ω)𝑡𝑡+1𝑣𝑖(𝑠)𝑣(𝑠)𝐻10(Ω)𝜀𝑑𝑠2>0,(3.45) where we use (3.37) and the fact 𝑣𝑖(𝑠)𝑣(𝑠) strongly in 𝐿2(𝑡,𝑡+1;𝐻10(Ω)). So it is impossible that for 𝑗<𝑗1,lim𝑡+𝑣(𝑡)=𝐸𝑗.
Lastly we need to verify lim𝑡𝑣(𝑡)=𝐸𝑗1. Considering (3.36), for any 𝜀>0 sufficiently small such that (𝜂𝑖𝑡𝑖)𝑡(𝜂𝑖𝑡𝑖)+𝜀0, we have (𝜂𝑖𝑡𝑖𝜂)+𝜀𝑖𝑡𝑖𝑣(𝑡)𝐸𝑗1𝐻10(Ω)=𝑑𝑡(𝜂𝑖𝑡𝑖𝜂)+𝜀𝑖𝑡𝑖𝑣(𝑡)𝑣𝑖(𝑡)+𝑣𝑖(𝑡)𝐸𝑗1𝐻10(Ω)𝑑𝑡(𝜂𝑖𝑡𝑖𝜂)+𝜀𝑖𝑡𝑖𝑣𝑖(𝑡)𝑣(𝑡)𝐻10(Ω)𝑑𝑡+(𝜂𝑖𝑡𝑖𝜂)+𝜀𝑖𝑡𝑖𝑣𝑖(𝑡)𝐸𝑗1𝐻10(Ω)𝑑𝑡(𝜂𝑖𝑡𝑖𝜂)+𝜀𝑖𝑡𝑖𝑣𝑖(𝑡)𝑣(𝑡)𝐻10(Ω)𝑑𝑡+𝜀𝜀.(3.46) Because 𝑣𝑖𝑣 strongly in 𝐿2((𝜂𝑖𝑡𝑖),(𝜂𝑖𝑡𝑖)+𝜀;𝐻10(Ω)), we easily get the result. In a word we conclude that lim𝑡𝑣(,𝑡)=𝐸𝑗1,lim𝑡+𝑣(,𝑡)=𝐸𝑗𝑗𝑗1.(3.47) This obviously contradicts (2.9). So we get the correctness of (3.31).
According the definition of 𝑗𝑘1𝑖, we can conclude that 𝑑𝐸𝑗𝑢,𝜔𝑘1𝑖𝛿1forall𝑘1𝑖>𝑘1,1𝑗𝑗1.(3.48) For convenience we may assume that (3.48) holds for all 𝑘1𝑖.
Fix a 0<𝛿2<𝛿1, and denote by 𝑗𝑘1𝑖 the smallest 𝑗 satisfying 𝜔𝑢𝑘1𝑖𝛿2𝐸𝑗.(3.49) From (3.48) we know 𝑗𝑘1𝑖>𝑗1 for all 𝑘1𝑖. Similarly there are a subsequence {𝑘2𝑖}𝑖=1 of {𝑘1𝑖}𝑖=1 and a 𝑗2(𝑗1,𝑛] such that 𝑗𝑘2𝑖=𝑗2 for all 𝑘2𝑖. Following the same process above, we can prove that if 𝑗2<𝑛, then there exists a 𝛿2(0,𝛿2) and 𝑘2>𝑘1 such that for 𝑘2𝑖>𝑘2𝑑𝐸𝑗2𝑢,𝜔𝑘2𝑖𝛿2.(3.50) By the choice of {𝑘2𝑖}𝑖=1, it is easy to see that 𝑑𝐸𝑗𝑢,𝜔𝑘2𝑖𝛿2forall𝑘2𝑖>𝑘2,1𝑗𝑗2.(3.51) Repeating the same argument again and again, we finally get sequences 𝑗1<𝑗2<<𝑗𝑚=𝑛𝜀>𝛿1>𝛿2>>𝛿𝑚>0,𝑘1<𝑘2<<𝑘𝑚,(3.52) and {𝑘𝑝𝑖}𝑖=1(1𝑝𝑚) such that 𝑑𝐸𝑗𝑢,𝜔𝑘𝑝𝑖𝛿𝑝,forall𝑘𝑝𝑖>𝑘𝑝,1𝑗𝑗𝑝.(3.53) In particular, of course we have 𝑑𝐸𝑗𝑢,𝜔𝑘𝑚𝑖𝛿𝑚forall𝑘𝑚𝑖>𝑘𝑚,1𝑗𝑗𝑚=𝑛.(3.54) This clearly contradicts (3.28). And the proof is completed.

Acknowledgment

This work was supported by NNSF of China (10771159).

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