Abstract
In this paper we study the existence of even positive homoclinic solutions for pLaplacian ordinary differential equations (ODEs) of the type , where , and the functions a and b are strictly positive and even. First, we prove a result on symmetry of positive solutions of pLaplacian ODEs. Then, using the mountainpass theorem, we prove the existence of symmetric positive homoclinic solutions of the considered equations. Some examples and additional comments are given.
MSC: 34B18, 34B40, 49J40.
Keywords:
pLaplacian ODEs; homoclinic solution; weak solution; PalaisSmale condition; mountainpass theorem1 Introduction and main results
In this paper we prove the existence of positive homoclinic solutions for pLaplacian ODEs of the type
(H) the functions are are continuously differentiable, strictly positive, and . Let, moreover, and be even functions on ℝ, and for .
By a solution of (1), we mean a function such that , and Eq. (1) holds for every . We are looking for positive solutions of (1) which are homoclinic, i.e., and as .
In the case , and , similar problems are considered in [13] using variational methods. Note that in [2] and [3] the following secondorder differential equations are considered:
and
where a, b and c are periodic, bounded functions and a and c are positive. These equations come from a biomathematics model suggested by Austin [4] and Cronin [5]. Further results and the phase plane analysis of these equations with constant coefficients are given in [6]. Note that the periodic and homoclinic solutions of pLaplacian ODEs are considered in [7,8].
The present work is an extension of these studies to pLaplacian ODEs. Let be the Sobolev space of pintegrable absolutely continuous functions such that
We use a variational treatment of the problem considering the functional
Using the wellknown mountainpass theorem, we conclude that the functional has a nontrivial critical point , which is a solution of the restricted problem
Further, we obtain uniform estimates for the solutions , extended by 0 outside . Then, a positive homoclinic solution of (1) is found as a limit of , as in . The function is also an even function.
To obtain the property, we extend the symmetry lemma of Korman and Ouyang [9] to the pLaplacian equations. The result is formulated and proved in Section 2.
Our main result is:
Theorem 1Suppose that, and assumptions (H) hold. Then Eq. (1) has a positive solutionsuch thatandas. Moreover, the solutionis an even function, asandfor.
Theorem 1 is proved in Section 3. From its proof we have
from which it follows that as . Observe that if , the problem
has a unique solution . Indeed, multiplying the equation by u and integrating by parts over ℝ, we obtain
A simplified method can be applied to the equations
under assumptions (H) and , . Note that in this case, the even homoclinic solution of Eq. (3) satisfies
and again as . If a and b are constants, Eq. (3) is a conservative system and one can plot the phase curves in the phase plane . An example is given at the end of Section 3.
2 Preliminary results
Let , and . It is clear that is a differentiable function and . Moreover, exists and for .
Let , be the space of Lebesgue measurable functions such that the norm .
The dual space of is , where . Let be the duality pairing between and . By the Hölder inequality, for any and . We will use the following lemmata in further considerations.
Lemma 2For any, the following inequality holds:
Proof of Lemma 2. Note that for , . From the Hölder inequality, we have
□
The statement of Lemma 3 follows simply from the identity
The onedimensional pLaplacian operator for a differentiable function u on the interval I is introduced as . Let us consider the problem
A function is said to be a solution of the problem (4) if with is such that is absolutely continuous and holds a.e. in .
We formulate an extension of Lemma 1 of [9] for pLaplacian nonlinear equations. The result of Korman and Ouyang is onedimensional analogue of the result of Gidas, Ni and Nirenberg [10] for symmetry of positive solutions of semilinear Laplace equations. In the case of pLaplacian equations, the symmetry of solutions in higher dimensions is discussed by Reihel and Walter [11].
Theorem 4Assume thatsatisfies (5). Then any positive solutionuof (4) is an even function such thatandfor.
Remark 1 Let us note that if the function f satisfies (5), but u is not a positive solution of (4), then u is not necessarily an even function. A simple counter example in the case is the problem
The term satisfies (5) in the interval , but the solution of the problem is negative in and not an even function. Its graph is presented in Figure 1. It would be more interesting to show an example for the case and f satisfying the additional assumption .
Figure 1. Graph of the functions.
Sketch of Proof of Theorem 4 Suppose that the function u has only one global maximum on .
Assume that the function has a finite number of local minima in the interval , and let be the largest local minimum. Let be the local maximum and be such that . Denote and , and let and be the inverse functions of the function in the intervals and , respectively. Multiplying the equation in (4) by and integrating in , we obtain by Lemma 3 and (5):
which leads to contradiction. One can prove the last fact using other arguments; see, for instance, Theorem 2.1 of [12]. Suppose now that u has infinitely many local minima in . Further, we can follow the steps of the proof of Lemma 1 of [9] with corresponding modifications based on Lemma 3. □
3 Proof of the main result
Let be the Sobolev space of pintegrable absolutely continuous functions such that
and . Note that if is strictly positive and bounded, i.e., there exist a and A such that , then is an equivalent norm in .
We need an extension to the pcase of the following proposition by Rabinowitz [13].
Proof of Proposition 5 Let . It follows
Integrating with respect to and using the Hölder and Jensen inequalities, we obtain
(ii) Take . Since , there exists such that by (i)
□
We are looking for positive solutions of (1), which are homoclinic, i.e., and as . Firstly, we look for positive solutions of the problem
A function is said to be a solution of the problem () if with is such that is absolutely continuous and holds a.e. in .
A function is said to be a weak solution of the problem () if
Standard arguments show that a weak solution of the problem () is a solution of () (see [14] and [15]). Consider the modified problem
where . It is easy to see that solutions of the problem () are positive solutions of the problem (). Indeed, if is a solution of () and has negative minimum at , since for , , by the equation , we reach a contradiction
Then and u is a solution of (). We use a variational treatment of the problem (), considering the functional
Critical points of are weak solutions of (), i.e.,
and, by a standard way, they are solutions of (). We show that satisfies the assumptions of the mountainpass theorem of Ambrosetti and Rabinowitz [16].
Theorem 6 (Mountainpass theorem)
LetXbe a Banach space with norm, , andIsatisfy the (PS) condition. Suppose that there exist, andsuch that
Thencis a critical value ofI, i.e., there existssuch thatand.
Next, denote by several positive constants.
Lemma 7Let, and assumptions (H) hold. Then for every, the problem () has a positive solution. Moreover, there is a constant, independent ofT, such that
ProofStep 1.satisfies the (PS) condition.
Let be a sequence, and suppose there exist and such that for
and
Let us denote . From (9) and (10), it follows that
and
Then
and
We have
which implies that the sequence is bounded in . By the compact embedding , there exist and the subsequence of , still denoted by , such that weakly in and strongly in . We will show that strongly in using Lemma 2. By uniform convergence of to u in , it follows that
and
Then
and by Lemma 2,
which implies that . Then and by the uniform convexity of the space , it follows that , as .
Step 2. Geometric conditions.
Obviously, . By assumption (H) it follows
Let be such that if and also . Consider the function
Then
for μ large enough.
By the mountainpass theorem, there exists a solution such that
where
Moreover, using the variational characterization (11), we have
Therefore, is a nontrivial and positive solution of (). By Theorem 4, and for .
Step 3. Uniform estimates.
Let . By continuation with zero of a function to , we have and . Using the variational characterization (11), we infer that and then
Multiplying the equation of () by and integrating by parts, we have
Then by (12),
We get (8) with , which completes the proof. □
Proof of Theorem 1 Take and let be the solution of the problem () given by Lemma 2. Consider the extension of to ℝ with zero outside and denote it by the same symbol.
Claim 1. The sequence of functionsis uniformly bounded and equicontinuous.
By (8) and the embedding of in , there is such that . Then by the equation of (), it follows that
By the mean value theorem for every natural n and every , there exists such that
Then, as a consequence of (13), we obtain
from which it follows and the sequence of functions is equicontinuous. Further, we claim that the sequence is also equicontinuous.
Claim 2. The sequence of functionsis equicontinuous.
To prove this statement, we follow the method given by Tang and Xiao [7]. For completeness, we present it in details.
Suppose that is not an equicontinuous sequence in . Then there exist an and sequences and such that and
By (14), there are numbers and and the subsequence such that and as . By (15), . On the other hand, by (13) we have
Then passing to a limit as , we obtain . Hence, which contradicts . Thus, the sequence is equicontinuous.
Let . By Claim 1 and Claim 2 and the ArzelàAscoli theorem, there is a subsequence of , still denoted by , and functions and of such that and . Trivially, it follows that , and . Repeating this procedure as in [7], we obtain that there is a subsequence of , still denoted by , and such that in . The function satisfies Eq. (1). Indeed, let be an interval of ℝ and such that . By the above considerations, taking a limit as in the equation
equivalent to
we obtain
and hence
Since and are arbitrary, is a solution of (1). Moreover, we have
It remains to show that is nonzero and and .
By Theorem 4, is an even function and attains its maximum at 0. Then by Eq. (1),
By assumption (H)
independently of n. Hence, passing to a limit as , we obtain
From (16) and Proposition 5, it follows
Now, we will show that . The arguments for are similar.
If , there exist and a monotone increasing sequence such that . Then for ,
which contradicts (16).
Moreover, u is an even function that attains its only maximum at 0, since the same holds for the functions . Arguing as in the proof of Theorem 4, we easily obtain that if . □
Remark 2 A simplified method can be applied to the equations
under assumptions (H) and , . Namely, first one looks for the even positive solutions of the problem
where is the Sobolev space of square integrable functions such that
Since is a Hilbert space, compactly embedded in the proof of the (PS)condition is easier. Similar considerations are made in [1] and [3]. Then, the even homoclinic solution is obtained as a limit of the sequence . Note that in this case, the even homoclinic solution of Eq. (3) satisfies
and again as . If a and b are constants, Eq. (3) is a conservative system and one can plot the phase curves in the phase plane . Consider the equation . The phase portrait in a plane, for in the rectangle , is plotted on Figure 2.
Figure 2. Phase portrait of, in.
Competing interests
The author declares that he has no competing interests.
Acknowledgements
Dedicated to Professor Jean Mawhin on the occasion of his 70th anniversary.
The author thanks Prof. Alberto Cabada and Prof. Luis Sanchez for helpful remarks concerning Theorem 4. The author would like to thank the Department of Mathematics and Theoretical Informatics at the Technical University of Kosice, Slovakia, where the paper was prepared during his visit on the SAIA Fellowship programme. The author is thankful to the editor and anonymous referee for their comments and suggestions on the article.
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