Infinitely Many Homoclinic Solutions for Nonperiodic Fourth Order Differential Equations with General Potentials

and Applied Analysis 3 It is obvious that K(u) is linear. Now we show that K(u) is bounded. In fact, for any u ∈ X, by the Hölder inequality and (F1), we can obtain that

The assumption ( 0 ) is too strict to be satisfied by many general functions (); for example, () = 1.In addition, although there is perturbation, the right of (2) is superlinear or asymptotically linear as || → +∞.In the present paper we study the infinitely many homoclinic solutions for (1) under more general assumption than  0 and sublinear condition on .
Before stating our results we introduce some notations.Throughout this paper, we denote by ‖‖  the   -norm, 2 ≤  ≤ ∞.  ∞ (R) is the Banach space of essentially bounded functions equipped with the norm If we take a subsequence of a sequence {  } we will denote it again by {  }.Now we state our main result.
In addition, if  is odd symmetry in , that is, then one gets the existence of infinitely many nontrivial homoclinic solutions.
The paper is organized as follows.In Section 2, we present some preliminaries.In Section 3, we give the proof of our main results.

Preliminaries
In order to prove our main results, we first give some properties of space  on which the variational setting for problem (1) is defined.Lemma 4 (see [5]).Assume that () hold.Then there exists a constant where then  is a Hilbert space with the inner product and the corresponding norm ‖‖ 2 = (, ).Note that for all  ∈ [2, +∞], with the embedding being continuous.Hence, for any  ∈ [2, +∞], there is   > 0 such that Now we begin describing the variational formulation of problem (1).Consider the functional  :  → R defined by Lemma 5.Under the conditions of (), ( 1 )-( 2 ),  ∈  1 (, R) and its derivative is given by the following; for all , V ∈ .In addition, any critical point of  on  is a classical solution of problem (1).
Proof.We firstly show that  :  → R. From ( 1 ), one has By the Hölder inequality and (14), we have where  2 is constant in (11).Hence,  defined by ( 12) is well defined on .
Next we prove that  ∈  1 (, R).To this end, we rewrite  as follows: It is easy to check that  ∈  1 (, R), and we have On the other hand, we will show that  ∈  1 (, R) and for any given , V ∈ .For any given  ∈ , let us define It is obvious that () is linear.Now we show that () is bounded.In fact, for any  ∈ , by the Hölder inequality and ( 1 ), we can obtain that Moreover, for , V ∈ , using the Mean Value Theorem, we get for some 0 < () < 1.On the other hand, () ∈  2/(2−) (R, R + ), for any  > 0, there exists  > 0 such that Therefore, on account of the Sobolev compact theorem (| [−,] is compactly embedded in  ∞ ([−, ], R)) and Hölder inequality, we have which, together with (19), implies that (17) holds.It remains to show that   is continuous.Suppose that  →  0 in , then we have sup uniformly with respect to V, which implies that   is continuous.Therefore, we obtain  ∈  1 (, R) and its derivative is given by for all , V ∈ .In addition, from [8], we can know that any critical point of  on  is a classical solution of problem (1).

Lemma 7. Let 𝐸 be a real Banach space and let
To obtain the existence of infinitely many homoclinic solutions for problem (1) under the assumptions of Theorem 2, we will employ the "genus" properties in critical point theory; see [11].
Let  be a Banach space,  ∈  1 (, R), and  ∈ R. We set is closed in  and symmetric with respect to 0} , Definition 8.For  ∈ Σ, we say the genus of  is  (denoted by () = ) if there is an odd map  ∈ ((, R)  \ {0}) and  is the smallest integer with this property.

Lemma 9.
Let  be an even  1 functional on  and satisfy the (PS) condition.For any  ∈ N, set (ii) if there exists  ∈ N such that and  ̸ = (0), then (  ) ≥  + 1.
Remark 10.From Remark 7.3 in [11], we know that if   ∈ Σ and (  ) > 1, then  has infinitely many distinct critical points in .

Proof of the Main Results
To prove our main results, we first give the following Lemma.
Proof.By ( 12) and ( 14) and Hölder inequality, one has Since 1 <  < 2, (28) implies that () → +∞ as ‖‖ → +∞.Consequently,  is bounded from below.Now, we show that  satisfies the (PS) condition.Assume that {  } ∈N ⊂  is a sequence such that {(  )} ∈N is bounded and   (  ) → 0 as  → +∞.Then by (28), there exists a constant  > 0 such that So passing to a subsequence if necessary, it can be assumed that   ⇀  0 in .Since () ∈  2/(2−) (R, R + ), for any  > 0, there exists  > 0 such that Since the embedding of  → In view of the definition of weak convergence, we have Therefore, we can obtain that   →  0 in .Hence,  satisfies (PS) condition.Now we are in the position to complete the proof of Theorems 1 and 2.
Proof of Theorem 1.It is obvious that (0) = 0, and by Lemmas 5 and 11 we know that  is a  1 functional on  satisfying the (PS) condition.In view of (28), we have  is bounded below on .Hence, by Lemma 7,  = inf  () is a critical value of ; that is, there exists a critical point  * ∈  such that ( * ) = .