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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 138623, 9 pages
Heteroclinic Solutions for Nonautonomous EFK Equations
1School of Mathematics and Systems Science, Beijing University of Aeronautics and Astronautics, Beijing 100191, China
2School of Mathematical Sciences, Peking University, Beijing 100875, China
Received 30 August 2012; Accepted 12 February 2013
Academic Editor: Nikolaos Papageorgiou
Copyright © 2013 Y. L. Yeun. 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.
We explore the nonautonomous fourth-order differential equation which has important applications in materials science. By variational approach, we find heteroclinic solutions of the equation. The conditions on the potential function are mild enough to include a broad class of equations. We also consider a separate case where is periodic in .
The goal of this paper is to study the nonautonomous extended Fisher-Kolmogorov (EFK) equation where is a time-dependent potential function and denotes the partial derivative with respect to the second variable .
Fourth-order differential equations, which often appear in nonlinear elasticity, fluid mechanics, and relating physical problems, have received growing attention from researchers. For example, Peletier and Troy [1–4] studied the EFK equation with odd nonlinearity mainly using shooting argument and variational method. They found the existence of periodic, heteroclinic, and homoclinic solutions and proved the existence of chaotic solutions oscillating between 1 and −1 with all critical points being either minima or maxima. They also explored the boundedness, monotonicity, and other quantitative properties of solutions. Part of Peletier and Troy’s work was generalized by Kalies and VanderVorst ; they considered the EFK equation with general nonlinear term where is supposed to have symmetric wells with equal depth and grow superquadratically. Then, they investigated the multitransition structure of heteroclinic and homoclinic solutions. Primary examples are The symmetric property of is rather restrictive, and many physical problems do not satisfy such conditions. Later, Kalies et al. [6, 7] obtained heteroclinic and homoclinic solutions connecting saddle-focus equilibria, and the potential function is assumed to grow superquadratically and has at least two nondegenerate global minima, not necessarily symmetric. In case of (3) with , the presence of saddle-focus equilibria amounts to the requiring of . Periodic and chaotic solutions of (3) are also explored. These results were generalized in recent works [8, 9], where the author obtained heteroclinic solutions for (3) under very mild conditions of ; in particular, they do not assume that is symmetric or that saddle-focus equilibria are present.
We have just mentioned a few works that are closely related to our results. A good summary of typical researches in fourth-order differential equations can be found in .
In this paper we step forward to study heteroclinic solutions of the nonautonomous EFK equation (1). We are inspired by Yeun , Rabinowitz [11, 12], and Izydorek and Janczewska . However, we are emphasizing that the argument of [9, 11] relies on the fact that the equation is autonomous; therefore, the method there cannot be reproduced here to tackle the time-dependent version (1) which is no longer autonomous. The nonautonomous case necessitates careful analysis. Another point should be made is that we are working on the phase plane; this is essentially different from [11, 13]. In our argument, we also benefit from analysis of [3, 4] and comments of .
In spite of its practical significance , there has been few researches on (1) with time-dependent potentials; the paper seems to be the first attempt in the direction and the methods here can be applied to similar equations.
2. Heteroclinic Solutions
By a heteroclinic solution connecting and , we mean a solution verifying
Throughout the paper, let denote the neighborhood of a set defined as below
We will make the following assumptions:, ; let be a set which has at least two points and we assume the points in do not collapse together, one of the following is satisfied:(a) there is , such that for any ;(b) uniformly for . uniformly for in any compact subsets of . For each , where is defined in with .
Theorem 1. Let satisfy . Then for every , there is a pair of heteroclinic solutions of (1), one emanating from and the other terminating at .
Remark 2. To be noted, we do not assume has wells of equal depth. By the assumptions of the theorem, we necessarily have . However, may contain other points which are not global minima of but still verify . In the special case , is automatically satisfied.
Notation. Define . For an element in , we denote it by for brevity. Whence, without further notice, we will use and where it is appropriate to denote element in and its counterpart in .
We will obtain heteroclinic solutions of (1) by minimizing the energy functional on an appropriate subset Denote the energy on a bounded interval by
We will work on the space equipped with the norm It is readily seen that is a Hilbert space and implies .
Let , . Denote by the set of verifying(i); (ii), for all .
That is, consists of functions which join and , and at least keep a distance of to the set . Clearly, is not empty, we may consider the minimization problem
We will find for each a candidate in and then send to zero to obtain the minimizer, which is actually a heteroclinic solution.
For any , define then by our assumptions. For , we take which is defined in .
From now on, we always assume that
Lemma 3. Let . Then for any , such that and for any , In particular, if , then for any such that for any ,
Proof. Denote and . Then
Since , by assumptions, we have Proof for the second part is similar.
Remark 4. The proof of Lemma 3 indicates that any function with cannot oscillate too often between different points in when the time approaches infinity; precisely, we will show exist and have limits in .
Lemma 5. Let . implies .
Proof. Suppose to the contrary that . Let satisfy (15).
Case 1 (). If is an infinite set, then we can find infinitely many nonoverlapping time intervals such that for and for , by Lemma 3, for any , a contradiction.
If is a finite set, let , then . Since , there is a sequence of nonoverlapping time intervals, also denoted by , such that for and for . Again Lemma 3 shows for any , also a contradiction.
Case 2 ( uniformly in ). There are and such that for and . Let ( denotes the integral part of ). Since , there is a sequence of nonoverlapping time intervals with , and on this is possible via the continuity of . We claim that . Indeed, for , so which implies Hence, for any , a contradiction.
Therefore, in either case, we must have .
The preceding lemma has an important corollary which will be used later; we prove it here.
Lemma 6. For and , define Then is a bounded set in with the usual uniform norm.
Proof. Without loss of generality, assume . Let satisfy (15). Suppose to the contrary that, for any , there is an orbit such that , and . Since is continuous, there is for which and for ; this can be accomplished by taking as the first hitting time of on the level .
Case 1 (). If is an infinite set, then for each , , there are nonoverlapping time intervals such that for , for ; moreover, the number of intervals as . Then by Lemma 3, for any , a contradiction.
If is a finite set, let , then . For each , , there is an with and for . Then Lemma 3 indicates for any , also a contradiction.
Case 2 ( uniformly in ). There are and such that for and . For each , there are nonoverlapping time intervals such that , and on . The same argument that we used to obtain (24) shows Hence for any , , a contradiction.
Therefore, a constant depending only on and exists, for which for all .
Lemma 7. Let . If , then there are such that and .
Proof. By Lemma 5, . Since , then for some . Therefore, the set of accumulation points of at is not empty. Moreover, . Indeed, suppose there are and such that for all . Then, by definition of in (14) with and , which shows , a contradiction. Hence, is not empty and there exists an . Finally we claim . Suppose otherwise, there must be some such that intersects and infinitely many times, then Lemma 3 implies that the energy along the orbit would tend to infinity, contradicting the hypothesis. Thus, we have shown tends to as . The other limit can be proved similarly.
Theorem 8. For any , . There exists such that minimizes the functional on the set , that is, .
Proof. Let be a minimizing sequence for (13),
For notational convenience, we will suppress subscripts in what follows and simply write . Now since is a minimizing sequence, we have for some . Moreover, Lemma 6 shows that is a bounded sequence; therefore, is bounded in . Going if necessary to a subsequence, we may suppose that converges weakly in to an element . Sobolev imbedding theorem shows in . To show is the minimizer we are looking for, we need firstly to prove
Indeed, for any ,
by the weakly lower semicontinuity of ,
Clearly, the constant is independent of and . Since , the above inequality implies , and
Hence, (34) holds.
Next we show Inequality (34) and Lemma 7 imply that there are verifying But for each , for all and in , so for all . Therefore, . For each , there is such that
We claim . Suppose to the contrary that Then, there is such that Hence, whenever and . In particular
Let Since , are well defined for each . The same argument that we used to obtain (24) shows for some constant . Whence, for , by the definition of and , , and the mean value theorem, where . Inequality (44) also implies . Hence, by , as , which leads to a contradiction that as . Similar argument shows that and the proof is completed.
Theorem 9. The minimizer obtained in Theorem 8 is a classical solution on .
Proof. For any , since is an open set, there is a maximal open interval containing . Let , then for sufficiently small . Since minimize on , then, for all , Fix any , this equality also holds for any . Hence, is a weak solution of the following system: the system has a unique solution, denoted by . Then Combining (49) and (51) yields for all . Since belongs to , it follows that is constant. Noting the boundary conditions that coincides with at and , we have . Since are arbitrary, it follows that , which completes the proof.
Define Since is nonincreasing as a function of , exists and we may choose a decreasing sequence such that By Lemma 6, for each , the set of candidates for the infimum is finite; hence, it is achieved by an element , namely, Theorem 8 shows, for each , there is a such that Again by Lemma 6, must be a finite set and thus contains a constant subsequence ; in other words, there is a for which , for all . Without loss of generality, we also denote this constant sequence by (correspondingly ), then where .
Theorem 10. For sufficiently large, is a heteroclinic solution connecting and .
Proof. To finish the proof, we show that never touches for large enough. Assume not, then there are and such that and for ; precisely, is the first hitting time of on . An application of Lemma 6 shows that is bounded, so it must contain a constant subsequence which we still denote by , that is, for . Now, two situations may occur, for infinitely many , (i) reaches before or (ii) touches before .
Since , is a decreasing sequence of and is nonincreasing in ; hence, which by Lemma 6 implies that must be uniformly bounded on , say for all .
Case 1 ( for all ). Before stepping forward, we need to show is bounded both from above and below. Suppose is unbounded from above. Let For large enough, , so by definition of . Hence, is unbounded from above. Without loss of generality; we may suppose . Similar to (46) it can be shown that for some positive constant . Hence, where . Now assumption and the uniform boundedness of indicate that is bounded, which is a contradiction. Thus, must have an upper bound. Similarly it can be proved that is also bounded from below by considering and . Hence, is bounded.
Define where is a polynomial defined as It is easy to verify the boundary conditions In addition, ,,, and is bounded.
Comparing the energy of and , we have By assumption , Moreover, since is uniformly bounded on and is bounded, it readily follows that Meanwhile, by construction, So Therefore, for large enough, which is impossible.
Case 2 ( for some ). A contradiction can also be reached by similar arguments.
Either case leads to a contradiction; therefore, for sufficiently large, the orbit of must keep some distance away from , and hence it is a classical solution, and the proof is complete.
Proof of Theorem 1. Theorem 10 shows for every , there is heteroclinic solution of (1) emanating from . If we consider the set of function for which , and , for all ; then the proof for works equally well for , and so, for every , there is heteroclinic solution of (1) terminating at .
3. Heteroclinic Solution in Periodic Case
In the last section, we consider the case where is periodic in with period . In this case, the assumptions and proof can be simplified. For completeness, we devote this section to the periodic case. Since most results in the last section can be carried out verbatim, we just present those that are different. We make the following assumptions: , ; the set is discrete and has at least two points, and We also assume without no loss of generality . There is a constant such that uniformly for . For each ,
Note that is no longer needed in the periodic case. is indeed a special case of , we are using instead of only for illustrative purpose. As before, we work on the space with the norm
We will use the shift operator defined for each as
Clearly, if is a heteroclinic solution, so is for all . Recall , is the set of verifying(i);(ii), for all ; where .
is not empty, we may consider the minimization problem
Theorem 11. For any , . There exists such that minimizes the functional on the set , that is, .
Proof. Let be a minimizing sequence for (77). As in the preceding section, it is bounded in and we may choose a subsequence, also denoted by , converging weakly in to some and in . Moreover,
It remains to show Inequality (78) and Lemma 7 imply that there are verifying Since is periodic in , if is the minimizing sequence, so is for any sequence and . Whence, with being appropriately chosen and replaced with , we are safe to suppose that for and for some . Noting for each , for all and in , we have for all and so . Since for , , so .
It remains to show . Note . Suppose that . Choose , for the time being we only assume that . Since , there is a such that for all . Since in , so there is , which depends on , such that for larger than . But for some , so intersects and . By Lemma 3, we have Now construct a sequence as follows: where and is a polynomial with coefficients It is easy to check that satisfies Moreover, tends to zero if and . By construction . So far, we only used the fact , we can fix smaller still such that, for all , Then for . This implies which is impossible. The proof is complete.
Theorem 12. There is a pair of orbits which connect and , one emanating from and the other terminating at .
Proof. The proof is similar to Theorems 8 and 10 in .
Lastly, we observe the following.
Theorem 13. For the solution or obtained above, ,.
Proof. Consider . By , there are positive constants such that
whenever . Since , for larger than some .
From (89) and the Euler equation, we have An application of the interpolation inequality yields where is a positive constant.
Therefore, we have proved that The limit for negative infinity can be derived similarly. The proof is complete.
This work was supported by NSFC under Grant no. 11126035 and partly NSFC Grant no. 11201016.
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