Let be the Sobolev space of p-integrable 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 p-case of the following proposition by Rabinowitz [13].
Proposition 5 Let . Then:
-
(i)
If , for ,
(6)
-
(ii)
For every ,
(7)
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 mountain-pass theorem of Ambrosetti and Rabinowitz [16].
Theorem 6 (Mountain-pass theorem)
Let X be a Banach space with norm , , and I satisfy the (PS) condition. Suppose that there exist , and such that
-
(i)
if ,
-
(ii)
. Let , where
Then c is a critical value of I, i.e., there exists such that and .
Next, denote by several positive constants.
Lemma 7 Let , and assumptions (H) hold. Then for every , the problem () has a positive solution . Moreover, there is a constant , independent of T, such that
Proof Step 1. satisfies the (PS) condition.
Let be a sequence, and suppose there exist and such that for
(9)
and
(10)
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
if . Then .
Let be such that if and also . Consider the function
Then
for μ large enough.
By the mountain-pass theorem, there exists a solution such that
(11)
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
(12)
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 functions is uniformly bounded and equicontinuous.
By (8) and the embedding of in , there is such that . Then by the equation of (), it follows that
(13)
By the mean value theorem for every natural n and every , there exists such that
Then, as a consequence of (13), we obtain
(14)
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 functions is 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
(15)
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 Arzelà-Ascoli 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
(16)
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
Note, that this implies as .
From (16) and Proposition 5, it follows
so .
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
considering the functional
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.