# On Asymptotic Behavior for Reaction Diffusion Equation with Small Time Delay

Authors: Xunwu Yin
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

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


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## Body

## 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 [1–5]. However, for nonlinear systems, the problem is much more complicated, but there are some very nice results in [6–10].In this paper, we consider the following scalar reaction-diffusion equation with a time delay(1.1)ut-Δu=f(u(t),u(t-τ))(x∈Ω⊂RN).
It is proved in [11–13] that for such diffusion equation without delay,
(1.2)ut-Δu=f(u),
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
(1.3)V(u):=12∫Ω|∇u|2-∫ΩF(u),
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,ɛ>0 there exists a sufficiently small τ>0 such that any solution of (1.1) satisfying limsupt→∞∥u(x,t)∥H01(Ω)≤R will ultimately enter and stay in the ɛ- 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 delays(1.4)x′(t)=f(x(t-r1(t)),…,x(t-rn(t))).
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 assumeΩ 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:(2.1)ut-Δu=f(u(t),u(t-τ))inΩ×(0,T],u=u0on⁡Ω×[-τ,0],u=0on⁡∂Ω×(0,T],
where the nonlinear f:R2→R is assumed to be continuous and to satisfy
(2.2)|f(u,v)|≤C[1+|u|ρ+|v|ρ],(2.3)f(u,v)u≤C1(u2+uv)+C2.
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 functionu(x,t) is called a weak solution of (2.1) if and only if(i)
u∈L2(0,T;H01(Ω)),withu′∈L2(0,T;H-1(Ω)),(ii)
u|[-τ,0]=u0∈L2(Ω),(iii)
∫0T[〈ut,φ〉+(Du,Dφ)]dt=∫0T(f(u(t),u(t-τ)),φ)dt,for eachφ∈L2(0,T;H01(Ω)). Here 〈,〉 and (,) denote the pair of H-1(Ω) and H01(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;H01(Ω))∩L∞(-τ,T;L2(Ω)), then {f(un(t),un(t-τ))} is bounded in L2(0,T;L2(Ω)).Proof.
Leta=(N-2)/N∈(0,1), and because ρ=1+2/N, 2ρ=a(2N/(N-2))+2(1-a). Before testing the boundedness of ∥f∥L2(0,T;L2(Ω)), we firstly estimate ∥un∥L2ρ(Ω)(2.4)‖un‖L2ρ(Ω)2ρ=∫Ω|un|a(2N/(N-2))⋅|un|2(1-a)dx≤[∫Ω|un|2N/(N-2)dx]a[∫Ω|un|2dx]1-a=‖un‖L2N/(N-2)(Ω)2Na/(N-2)⋅‖un‖L2(Ω)2(1-a)≤C1‖un‖H01(Ω)2Na/(N-2)=C1‖un‖H01(Ω)2.
Here we utilize the Hölder inequality, the fact of H01(Ω)⊂L2N/(N-2) continuously and {un} is bounded in L∞(-τ,T;L2(Ω)). So
(2.5)∫0T∫Ω|un(x,t)|2ρdxdt≤C1‖un(x,t)‖L2(-τ,T;H01(Ω))2,∫0T∫Ω|un(x,t-τ)|2ρdxdt=t-τ=s∫-τT-τ∫Ω|un(x,s)|2ρdxds≤∫-τT∫Ω|un(x,t)|2ρdxdt≤C1‖un(x,t)‖L2(-τ,T;H01(Ω))2.
In view of (2.2), we can easily see
(2.6)|f(un(t),un(t-τ))|2≤C[1+|un(t)|2ρ+|un(t-τ)|2ρ].
Integrating the above inequality with t and x, we complete the proof.Remark 2.3.
If{un} is bounded in L∞(-τ,T;H01(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.
LetX0, X, and X1 be three Banach spaces with X0⊆X⊆X1. 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={u∈Lp([0,T;X0)|u′∈Lq([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 H01(Ω) as our phase space, we denote by ω(u) the limit set (2.7)ω(u)={v∣thereexiststn⟶∞suchthat‖u(⋅,tn)-v‖H01(Ω)⟶0}.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 u∈H01(Ω)∩H2(Ω) satisfying the elliptic boundary value problem (2.8)-Δu=f(u,u)inΩ,u=0on⁡Ω
in the weak sense.Let each equilibrium be isolated and letu(·,t) be the bounded complete solution of (1.2). Then we have
(2.9)lim⁡t→-∞u(⋅,t)=E1,lim⁡t→+∞u(⋅,t)=E2
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, Theorem3.2.Theorem 3.1.
For givenτ>0,u0∈L2(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
(3.1)um(t)=∑k=1mukm(t)ωk,
where {ωk}k=1∞ is an orthogonal basis of H01(Ω) and {ωk}k=1∞ is an orthonormal basis of L2(Ω). We get um from solving the following ODES:
(3.2)(um′,ωk)+(Dum,Dωk)=(f(um(t),um(t-τ)),ωk)(0<t≤T,k=1,2,…m),ukm(t)=(u0,ωk)(-τ≤t≤0,k=1,2,…m).
According to standard existence theory of ODES, we can obtain the local existence of um.
Next we will establish some priori estimates forum. Multiplying (2.1) by um and integrating over Ω, we have
(3.3)12ddt‖um‖L2(Ω)2+‖um‖H01(Ω)2=∫Ωf[um(t),um(t-τ)]umdx.
Because of (2.3) and the Cauchy inequality, we can get
(3.4)ddt‖um‖L2(Ω)2+2‖um‖H01(Ω)2≤C1′‖um‖L2(Ω)2+C2′.
Getting rid of the term 2∥um∥H01(Ω)2, from the differential form of Gronwall's inequality, we yield the estimate
(3.5)max⁡0≤t≤T‖um‖L2(Ω)2≤C1‖u0‖L2(Ω)2+C2.
Returning once more to inequality (3.4), we integrate from 0 to T and employ the inequality above to find
(3.6)‖um‖L2(0,T;H01(Ω))2≤C1‖u0‖L2(Ω)2+C2.
Multiplying (2.1) by um′ and then integrating over Ω, we have
(3.7)‖um′‖L2(Ω)2+∫ΩDum⋅Dum′dx=∫Ωf[um(t),um(t-τ)]um′dx.
Using the Cauchy inequality and Lemma 2.2, we get
(3.8)‖um′‖L2(Ω)2+ddt‖um‖H01(Ω)2≤C.
Again from the differential form of Gronwall's inequality, we integrate from 0 to T(3.9)‖um′‖L2(0,T;L2(Ω))2≤C1‖u0‖L2(Ω)2+C2.
Since L2(Ω)⊂H-1(Ω), so
(3.10)‖um′‖L2(0,T;H-1(Ω))2≤C1‖u0‖L2(Ω)2+C2.
According to estimates (3.6), (3.10), Lemma 2.2, and weak compactness, we see that
(3.11)um⇀uweaklyinL2(0,T;H01(Ω)),um′⇀u′weaklyinL2(0,T;H-1(Ω)),f[um(t),um(t-τ)]⇀ηweaklyinL2(0,T;L2(Ω)).
Here a subsequence of {um}m=1∞ is still denoted by {um}m=1∞. Applying Lemma 2.2, we can conclude that um→u strongly in L2(0,T;L2(Ω)). Hence um→u A.E. in Ω×(0,T). Since f is continuous, it follows that f[um(t),um(t-τ)]→f[u(t),u(t-τ)] A.E. in Ω×(0,T). Thanks to (3.11) and Lemma 1.3 in [14], one has
(3.12)f[um(t),um(t-τ)]⇀f[u(t),u(t-τ)]weaklyinL2(0,T;L2(Ω)).
Next fix an integerN0 and choose a function v∈C1([0,T];H01(Ω)) having the form
(3.13)v(t)=∑k=1N0dk(t)ωk,
where {dk}k=1N0 are given smooth functions. Choosing m≥N0 and multiplying (3.2) by dk(t) sum k=1,2,…,N0, and then integrating with respect to t, we can find
(3.14)∫0T〈um′,v〉+(Dum,Dv)dt=∫0T(f[um(t),um(t-τ)],v)dt.
Recalling (3.11) and (3.12) and passing to weak limits, we get
(3.15)∫0T〈u′,v〉+(Du,Dv)dt=∫0T(f[u(t),u(t-τ)],v)dt.
Because functions of the form v(t) are dense in L2(0,T;H01(Ω)), so the above equality holds for all functions v∈L2(0,T;H01(Ω)).
Lastly we will showu|[-τ,0]=u0∈L2(Ω). Notice that for each v∈C1([0,T];H01(Ω)) with v(T)=0 we get the following from (3.15):
(3.16)∫0T-〈v′,u〉+(Du,Dv)dt=∫0T(f[u(t),u(t-τ)],v)dt+(u(0),v(0)).
Similarly, from (3.14), we deduce
(3.17)∫0T-〈v′,um〉+(Dum,Dv)dt=∫0T(f[um(t),um(t-τ)],v)dt+(um(0),v(0)).
In view of (3.2), um(0)→u0 in L2(Ω); once again employ (3.11) and (3.12) to find
(3.18)∫0T-〈v′,u〉+(Du,Dv)dt=∫0T(f[u(t),u(t-τ)],v)dt+(u0,v(0)).
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,ɛ>0 be given arbitrarily. Then there exists a sufficiently small τ>0 such that any solution of (1.1) with limsupt→+∞∥u(·,t)∥H01(Ω)≤R will eventually enter and stay in the ɛ- neighborhood of some equilibrium.Proof.
Here we selectH01(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)∥H01(Ω)≤R for all t∈[0,∞). That is to say they are in ℬ¯R.
Assume there aren equilibria of (1.2) {E1,…,En}, ordered by V(En)≥V(En-1)≥⋯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 anyδ>0, there exists a sufficiently small τ>0 such that
(3.19)ω(u)⋂(⋃1≤j≤nBδ(Ej))≠∅
for any solution u(x,t) of (1.1) in ℬ¯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 τk→0 and a corresponding solution sequence uk of (1.1) in ℬ¯R satisfying
(3.20)d(Ej,ω(uk))≥2δ
for all 1≤j≤n and k∈N. According to the definition of ω(u), for each k we can take a tk>0 such that for t≥tk,1≤j≤n(3.21)‖uk(⋅,t)-Ej‖H01(Ω)≥δ.
Let ũk(t)=uk(t+tk) for t≥0. It is easy to see ũk is the weak solution of
(3.22)∂tũk-Δũk=f(ũk(t),ũk(t-τk)).
Next we will show there is a strong convergent subsequence of {ũk}k=1∞ in L2(t,t+1;H01(Ω)) for t≥0. Still denoting ũk, we can also prove the limit ũ is in fact the weak solution of (1.2). From the elliptic equation regularity theorem, we can multiply (3.22) by -Δũk and integrate over Ω(3.23)12ddt‖ũk‖H01(Ω)2+‖ũk‖H2(Ω)2≤12(‖f‖L2(Ω)2+‖ũk‖H2(Ω)2).
Because of the remark in Section 2, we can get
(3.24)ddt‖ũk‖H01(Ω)2+‖ũk‖H2(Ω)2≤C.
Integrating from t to t+1, from the boundedness of ∥ũk(·,t)∥H01(Ω)2 and ∥ũk(·,t+1)∥H01(Ω)2, we conclude that {ũk} is bounded in L2(t,t+1;H2(Ω)). Multiplying (3.22) by ∂tũk and utilizing the same method above, we can also conclude that {∂tũk} 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 {ũk}k=1∞ in L2(t,t+1;H01(Ω)) for t≥0. We may set ũk→ũ strongly in L2(t,t+1;H01(Ω)). Of course ũk→ũ strongly in L2(t,t+1;L2(Ω)). Hence ũk→ũ a.e. in Ω×(t,t+1). Since f is continuous, it follows that f[ũk(t),ũk(t-τk)]→f(ũ,ũ) a.e. in Ω×(0,T). Thanks to the weak convergence of f[ũk(t),ũk(t-τk)] in L2(t,t+1;L2(Ω)) and lemma 1.3 in [14], one has
(3.25)f[ũk(t),ũk(t-τk)]⇀f(ũ,ũ)weaklyinL2(0,T;L2(Ω)).
So we prove that ũ is the weak solution of (1.2). Considering (3.21), we have the following estimate for ũ:
(3.26)∫tt+1‖ũ-Ej‖H01(Ω)ds=∫tt+1‖ũk-Ej+ũ-ũk‖H01(Ω)ds≥∫tt+1‖ũk-Ej‖H01(Ω)ds-∫tt+1‖ũk-ũ‖H01(Ω)ds≥δ-∫tt+1‖ũk-ũ‖H01(Ω)ds≥δ2.
From the above inequality, we can surely know limt→∞∥ũ-Ej∥H01(Ω)≠0 for all 1≤j≤n. However, because (1.2) is a gradient system, this contradicts the fact that limt→∞u(·,t)=Ej for some Ej. We obtain the correctness of (3.19).Step 2.
We will complete the proof of the theorem that ifτ 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>T(3.27)‖u(⋅,t)-Ej‖H01(Ω)<ɛ.
Here we also adopt contradiction method to prove the result. If the desired conclusion was not correct, there would be a decreasing sequenceτk→0 and a corresponding solution sequence uk of (1.1) in ℬ¯R which does not satisfy (3.27).
In view ofτk→0, it is easy to infer that
(3.28)lim⁡k→∞min⁡1≤j≤nd(Ej,ω(uk))=0.
Without loss of generality, we can assume that for all k≥1(3.29)ω(uk)⋂(⋃1≤j≤nB¯ɛ(Ej))≠∅.
Denote by jk the smallest j satisfying
(3.30)ω(uk)⋂B¯ɛ(Ej)≠∅.
It is easy to see that there exists a subsequence {ki1}i=1∞ of {k}k=1∞ such that for some j1∈[1,n], we have jki1=j1.
We will claim ifj1<n, then there exists a δ1∈(0,ɛ) and k1* such that for ki1>k1*(3.31)d(Ej1,ω(uki1))≥δ1.
Indeed, if the fact did not hold, there would be a subsequence of {ki1}i=1∞ (for simplicity still denoted by {ki1}i=1∞) such that
(3.32)lim⁡i→∞d(Ej1,ω(uki1))=0.
According to the definition of jk and (3.32), we can choose a sequence ti>0 satisfying
(3.33)uki1(ti)∈Bɛ(Ej1),lim⁡i→∞‖uki1(ti)-Ej1‖H01(Ω)=0,‖uki1(t)-Ej‖H01(Ω)>ɛ,fort>ti,j<j1.
Now we define
(3.34)ηi=sup⁡⁡{t≥ti∣uki1([ti,t])⊂B¯ɛ(Ej1)}.
Obviously uki1(ηi)∈∂ℬɛ(Ej1). Let
(3.35)vi(⋅,t)=uki1(⋅,t+ηi),t∈[-(ηi-ti),+∞).
From (3.33) and the definition of ηi, it is clear to see
(3.36)vi(0)∈∂Bɛ(Ej1),vi(t)∈B¯ɛ(Ej1)for-(ηi-ti)≤t≤0,(3.37)‖vi(t)-Ej‖H01(Ω)>ɛ,fort>0,j<j1.
Obviously vi(·,t) is the weak solution of
(3.38)∂tvi-Δvi=f(vi(t),vi(t-τki1)).
Following the method above, we can also prove there is a strong convergent subsequence of {vi}i=1∞ in L2(t,t+1;H01(Ω)) for t≥0. Still denoting {vi}i=1∞ and letting T=limsup(ηi-ti), the limit v defined on (-T,∞) is indeed the weak solution of (1.2).
Next we will show thatT=+∞. 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 t≥-T. That is to say, vi(t)→Ej1 strongly in L2(-T,0;H01(Ω)). Because vi(·,t)∈L2(-(ηi-ti),0;H01(Ω)) and vi′(·,t)∈L2(-(ηi-ti),0;L2(Ω)), it follows from Theorem 4 in 5.9.2 of [15] that vi(·,t)∈C([-(ηi-ti),0];H01(Ω)). That is to say
(3.39)lim⁡t→0-‖vi(⋅,t)-vi(⋅,0)‖H01(Ω)=0.
By the definition of continuity, there exists t1<0 such that
(3.40)‖vi(⋅,t)-vi(⋅,0)‖H01(Ω)<ɛ0<ɛfort1<t<0.
So
(3.41)ɛ0>‖vi(⋅,t)-Ej1+Ej1-vi(⋅,0)‖H01(Ω)≥‖vi(⋅,0)-Ej1‖H01(Ω)-‖vi(⋅,t)-Ej1‖H01(Ω).
Hence
(3.42)‖vi(⋅,t)-Ej1‖H01(Ω)>‖vi(⋅,0)-Ej1‖H01(Ω)-ɛ0=ɛ-ɛ0>0.
Thus
(3.43)∫-(ηi-ti)0‖vi(⋅,t)-Ej1‖H01(Ω)dt≥∫t10‖vi(⋅,t)-Ej1‖H01(Ω)dt≥-t1(ɛ-ɛ0).
Obviously
(3.44)lim⁡i→∞∫-(ηi-ti)0‖vi(⋅,t)-Ej1‖H01(Ω)dt≠0.
This contradicts the fact vi(t)→Ej1 strongly in L2(-(ηi-ti),0;H01(Ω)). So it must be T=+∞.
Letlimt→+∞v(t)=Ej. Then there must be j≥j1. Otherwise for j<j1(3.45)∫tt+1‖v(s)-Ej‖H01(Ω)=∫tt+1‖v(s)-vi(s)+vi(s)-Ej‖H01(Ω)ds≥∫tt+1‖vi(s)-Ej‖H01(Ω)-∫tt+1‖vi(s)-v(s)‖H01(Ω)ds≥ɛ2>0,
where we use (3.37) and the fact vi(s)→v(s) strongly in L2(t,t+1;H01(Ω)). So it is impossible that for j<j1,limt→+∞v(t)=Ej.
Lastly we need to verifylimt→-∞v(t)=Ej1. Considering (3.36), for any ɛ′>0 sufficiently small such that -(ηi-ti)≤t≤-(ηi-ti)+ɛ′≤0, we have
(3.46)∫-(ηi-ti)-(ηi-ti)+ɛ′‖v(t)-Ej1‖H01(Ω)dt=∫-(ηi-ti)-(ηi-ti)+ɛ′‖v(t)-vi(t)+vi(t)-Ej1‖H01(Ω)dt≤∫-(ηi-ti)-(ηi-ti)+ɛ′‖vi(t)-v(t)‖H01(Ω)dt+∫-(ηi-ti)-(ηi-ti)+ɛ′‖vi(t)-Ej1‖H01(Ω)dt≤∫-(ηi-ti)-(ηi-ti)+ɛ′‖vi(t)-v(t)‖H01(Ω)dt+ɛ′ɛ.
Because vi→v strongly in L2(-(ηi-ti),-(ηi-ti)+ɛ′;H01(Ω)), we easily get the result. In a word we conclude that
(3.47)lim⁡t→-∞v(⋅,t)=Ej1,lim⁡t→+∞v(⋅,t)=Ej(j≥j1).
This obviously contradicts (2.9). So we get the correctness of (3.31).
According the definition ofjki1, we can conclude that
(3.48)d(Ej,ω(uki1))≥δ1forallki1>k1*,1≤j≤j1.
For convenience we may assume that (3.48) holds for all ki1.
Fix a0<δ2′<δ1, and denote by jki1 the smallest j satisfying
(3.49)ω(uki1)⋂B¯δ2′(Ej)≠∅.
From (3.48) we know jki1>j1 for all ki1. Similarly there are a subsequence {ki2}i=1∞ of {ki1}i=1∞ and a j2∈(j1,n] such that jki2=j2 for all ki2. 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 ki2>k2*(3.50)d(Ej2,ω(uki2))≥δ2.
By the choice of {ki2}i=1∞, it is easy to see that
(3.51)d(Ej,ω(uki2))≥δ2forallki2>k2*,1≤j≤j2.
Repeating the same argument again and again, we finally get sequences
(3.52)j1<j2<⋯<jm=nɛ>δ1>δ2>⋯>δm>0,k1*<k2*<⋯<km*,
and {kip}i=1∞(1≤p≤m) such that
(3.53)d(Ej,ω(ukip))≥δp,forallkip>kp*,1≤j≤jp.
In particular, of course we have
(3.54)d(Ej,ω(ukim))≥δmforallkim>km*,1≤j≤jm=n.
This clearly contradicts (3.28). And the proof is completed.



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*Source: 142128-2011-12-11.xml*