Abstract
We investigate the asymptotic behavior of scalar diffusion equation with small time delay . 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 [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 It is proved in [11–13] that for such diffusion equation without delay, subject to homogeneous boundary conditions, all globally defined bounded solutions must approach the set of equilibria as tends to infinity. This depends heavily on the fact that (1.2) is a gradient system with the Lyapunov function where is a primitive of . It is well known that solutions of (1.1) will typically oscillate in as 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 there exists a sufficiently small such that any solution of (1.1) satisfying will ultimately enter and stay in the 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 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 to be an open, bounded subset of and to be a positive parameter (the delay). Consider the following scalar delayed initial boundary value problem: where the nonlinear is assumed to be continuous and to satisfy Here are all constants, . Firstly we will give the definition of weak solution for (2.1).
Definition 2.1. A function is called a weak solution of (2.1) if and only if(i), (ii), (iii),
for each . Here and denote the pair of and , the inner product in , respectively. Next we will give two very important lemmas many times used in the proof of two theorems.
Lemma 2.2. If is bounded in , then is bounded in .
Proof. Let , and because , . Before testing the boundedness of , we firstly estimate Here we utilize the Hlder inequality, the fact of continuously and is bounded in . So In view of (2.2), we can easily see Integrating the above inequality with and , we complete the proof.
Remark 2.3. If is bounded in , 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 , , and be three Banach spaces with . Suppose that is compactly embedded in and is continuously embedded in . Suppose also that and are reflexive spaces. For , let . Then the embedding of W into is also compact.
Finally we give the definition of equilibrium solution of (1.2) and omega limit set , where is a bounded solution of (1.1). Selecting as our phase space, we denote by the limit set
As usual, an equilibrium solution of (1.2) is defined as a solution which does not depend on ; the equilibrium states are thus the functions satisfying the elliptic boundary value problem in the weak sense.
Let each equilibrium be isolated and let be the bounded complete solution of (1.2). Then we have for some equilibrium and with , where is the Lyapunov function (1.3). A complete solution of (1.2) means a solution 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 , 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 of the form
where is an orthogonal basis of and is an orthonormal basis of . We get from solving the following ODES:
According to standard existence theory of ODES, we can obtain the local existence of .
Next we will establish some priori estimates for . Multiplying (2.1) by and integrating over , we have
Because of (2.3) and the Cauchy inequality, we can get
Getting rid of the term , from the differential form of Gronwall's inequality, we yield the estimate
Returning once more to inequality (3.4), we integrate from 0 to and employ the inequality above to find
Multiplying (2.1) by and then integrating over , we have
Using the Cauchy inequality and Lemma 2.2, we get
Again from the differential form of Gronwall's inequality, we integrate from 0 to
Since , so
According to estimates (3.6), (3.10), Lemma 2.2, and weak compactness, we see that
Here a subsequence of is still denoted by . Applying Lemma 2.2, we can conclude that strongly in . Hence A.E. in . Since is continuous, it follows that A.E. in . Thanks to (3.11) and Lemma 1.3 in [14], one has
Next fix an integer and choose a function having the form
where are given smooth functions. Choosing and multiplying (3.2) by sum , and then integrating with respect to , we can find
Recalling (3.11) and (3.12) and passing to weak limits, we get
Because functions of the form are dense in , so the above equality holds for all functions .
Lastly we will show . Notice that for each with we get the following from (3.15):
Similarly, from (3.14), we deduce
In view of (3.2), in ; once again employ (3.11) and (3.12) to find
As is arbitrary, so we get the result . Since for in , we can obtain the result. As for being arbitrary, we see the global existence of (2.1).
Theorem 3.2. Assume that each equilibrium of (1.2) is isolated. Let be given arbitrarily. Then there exists a sufficiently small such that any solution of (1.1) with will eventually enter and stay in the neighborhood of some equilibrium.
Proof. Here we select as our phase space. For simplicity, we will verify the correctness of the conclusion for such bounded solutions of (1.1) as for all . That is to say they are in .
Assume there are equilibria of (1.2) , ordered by , where is the Lyapunov function (1.3). We will follow two steps to prove our result.
Step 1. We firstly verify that for any , there exists a sufficiently small such that
for any solution of (1.1) in .
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 and a corresponding solution sequence of (1.1) in satisfying
for all and . According to the definition of , for each we can take a such that for
Let for . It is easy to see is the weak solution of
Next we will show there is a strong convergent subsequence of in for . Still denoting , we can also prove the limit is in fact the weak solution of (1.2). From the elliptic equation regularity theorem, we can multiply (3.22) by and integrate over
Because of the remark in Section 2, we can get
Integrating from to , from the boundedness of and , we conclude that is bounded in . Multiplying (3.22) by and utilizing the same method above, we can also conclude that is bounded in . Applying the Aubin-Lions lemma, we can conclude that there is a strong convergent subsequence of in for . We may set strongly in . Of course strongly in . Hence a.e. in . Since is continuous, it follows that a.e. in . Thanks to the weak convergence of in and lemma 1.3 in [14], one has
So we prove that is the weak solution of (1.2). Considering (3.21), we have the following estimate for :
From the above inequality, we can surely know for all . However, because (1.2) is a gradient system, this contradicts the fact that for some . We obtain the correctness of (3.19).Step 2. We will complete the proof of the theorem that if is sufficiently small, then for any bounded solution of (1.1) there must exist a and sufficiently large such that for
Here we also adopt contradiction method to prove the result. If the desired conclusion was not correct, there would be a decreasing sequence and a corresponding solution sequence of (1.1) in which does not satisfy (3.27).
In view of , it is easy to infer that
Without loss of generality, we can assume that for all
Denote by the smallest satisfying
It is easy to see that there exists a subsequence of such that for some , we have .
We will claim if , then there exists a and such that for
Indeed, if the fact did not hold, there would be a subsequence of (for simplicity still denoted by ) such that
According to the definition of and (3.32), we can choose a sequence satisfying
Now we define
Obviously . Let
From (3.33) and the definition of , it is clear to see
Obviously is the weak solution of
Following the method above, we can also prove there is a strong convergent subsequence of in for . Still denoting and letting , the limit defined on is indeed the weak solution of (1.2).
Next we will show that . In fact, if , then can be well defined at . In view of (3.32), we see . Hence for . That is to say, strongly in . Because and , it follows from Theorem 4 in of [15] that . That is to say
By the definition of continuity, there exists such that
So
Hence
Thus
Obviously
This contradicts the fact strongly in . So it must be .
Let . Then there must be . Otherwise for
where we use (3.37) and the fact strongly in . So it is impossible that for .
Lastly we need to verify . Considering (3.36), for any sufficiently small such that , we have
Because strongly in , we easily get the result. In a word we conclude that
This obviously contradicts (2.9). So we get the correctness of (3.31).
According the definition of , we can conclude that
For convenience we may assume that (3.48) holds for all .
Fix a , and denote by the smallest satisfying
From (3.48) we know for all . Similarly there are a subsequence of and a such that for all . Following the same process above, we can prove that if , then there exists a and such that for
By the choice of , it is easy to see that
Repeating the same argument again and again, we finally get sequences
and such that
In particular, of course we have
This clearly contradicts (3.28). And the proof is completed.
Acknowledgment
This work was supported by NNSF of China (10771159).