Abstract
This article is concerned with the existence and multiplicity of nontrival solutions for a fourth-order elliptic equation
by using the mountain pass theorem.
Keywords:
fourth-order elliptic equation; nontrivial solutions; mountain pass theorem1 Introduction
In this article we study the existence of nontrivial solutions for the fourth-order boundary value problem
where Ω ⊂ RN is a bounded smooth domain, f : Ω × R → R and M : R → R are continuous functions. The existence and multiplicity results for Equation (1) are considered in [1-3] by using variational methods and fixed point theorems in cones of ordered Banach space with space dimension is one.
On the other hand, The four-order semilinear elliptic problem
arises in the study of traveling waves in a suspension bridge, or the study of the static deflection of an elastic plate in a fluid, and has been studied by many authors, see [4-10] and the references therein.
Inspired by the above references, the object of this article is to study existence and multiplicity of nontrivial solution of a fourth-order elliptic equation under some conditions on the function M(t) and the nonlinearity. The proof is based on the mountain pass theorem, namely,
Lemma 1.1. Let E be a real Banach space, and I ∈ C1(E, R) satisfy (PS)-condition. Suppose
(1) There exist ρ > 0, α > 0 such that
where Bp = {u ∈ E|∥u∥ ≤ ρ}.
(2) There is an e ∈ E and ∥e∥ > ρ such that
Then I(u) has a critical value c which can be characterized as
where Γ = {γ ∈ C([0, 1],E)|γ(0) = 0,γ(1) = e}.
The article is organized as follows: Section 2 is devoted to giving the main result and proving the existence of nontrivial solution of Equation (1). In Section 3, we deal with the multiplicity results of Equation (1) whose nonlinear term is asymptotically linear at both zero and infinity
2 Main result I
Theorem 2.1. Assume the function M(t) and the nonlinearity f(x, t) satisfying the following conditions:
(H1) M(t) is continuous and satisfies
for some m0 > 0. In addition, that there exist m' > m0 and t0 > 0, such that
(H2) f(x, t) ∈ C(Ω × R); f(x, t) ≡ 0, ∀x ∈ Ω, t ≤ 0, f(x, t) ≥ 0, ∀x ∈ Ω, t > 0;
(H3) |f(x, t)| ≤ a(x) + b|t|p, ∀t ∈ R and a.e. x in Ω, where a(x) ∈ Lq (Ω), b ∈ R
and
(H4) f(x, t) = o(|t|) as t → 0 uniformly for x ∈ Ω ;
(H5) There exists a constant Θ > 2 and R > 0, such that
Then Equation (1) has at least one nonnegative solution.
Let Ω ⊂ RN be a bounded smooth open domain,
and the deduced norm
Let λ1 be the positive first eigenvalue of the following second eigenvalue problem
Then from [4], it is clear to see that Λ1 = λ1(λ1 - c) is the positive first eigenvalue of the following fourth-order eigenvalue problem
where c < λ1. By Poincare inequality, for all u ∈ H, we have
A function u ∈ H is called a weak solution of Equation (1) if
holds for any v ∈ H. In addition, we see that weak solutions of Equation (1) are critical points of the functional I : H → R defined by
where
Lemma 2.2. Assume that (H1)-(H5) hold, then I(u) satisfies the (PS)-condition.
Proof. Let {un} ⊂ H be a (PS)-sequence. In particular, {un} satisfies
Since f(x, t) is sub-critical by (H3), from the compactness of Sobolev embedding and, following the standard processes we know that to show that I verifies (PS)-condition it is enough to prove that {un} is bounded in H. By contradiction, assume that ∥un∥ → +∞.
Case I. If
(H1), (H3), and (H5), we have
On the other hand, it is easy to obtain that
Then, from above, we can have
which contradicts ∥un∥ → +∞. Therefore {un} is bounded in H.
Case II. if
This lemma is completely proved.
Lemma 2.3. Suppose that (H1)-(H5) hold, then we have
(1) there exist constants ρ > 0, α > 0 such that
(2) I(tφ1) → -∞ as t → +∞.
Proof. By (H1)-(H4), we see that for any ε > 0, there exist constants C 1 > 0, C2 such that for all (x, s) ∈ Ω × R, one have
Choosing ε > 0 small enough, we have
by (3), (5), (7) and the Sobolev inequality. So, part 1 is proved if we choose ∥u∥ ρ > 0 small enough.
On the other hand, we have
using (4) and (H5). Hence,
as t → +∞ and part 2 is proved.
Proof of Theorem 2.1. From Lemmas 2.2 and 2.3, it is clear to see that I(u) satisfies the hypotheses of Lemma 1.1. Therefore I(u) has a critical point.
3 Existence result II
Theorem 3.1. Assume that (H1) holds. In addition, assume the following conditions are hold:
(H6) f(x, t)t ≥ 0 for x ∈ Ω, t ∈ R;
(H7)
Then Equation (1) has at least two nontrivial solutions, one of which is positive and the other is negative.
Let u+ = max{u, 0}, u- = min{u, 0}. Consider the following problem
where
Define the corresponding functional I+ : H → R as follows:
where
Similarly, we also can define
and
where
Lemma 3.2. Assume that (H1), (H6), and (H7) hold, then I± satisfies the (PS) condition.
Proof. We just prove the case of I+. The arguments for the case of I- are similar. Since Ω is bounded and (H7) holds, then if {un} is bounded in H, by using the Sobolve embedding and the standard procedures, we can get a convergent subsequence. So we need only to show that {un} is bounded in H.
Let {un} ⊂ H be a sequence such that
By (H7), it is easy to see that
Now, (9) implies that, for all ϕ ∈ H, we have
Set ϕ = un, we have
Next, we will show that
thus {ωn} is bounded in H. Passing to a subsequence, we may assume that there exists ω ∈ H with
On the other hand,
Then ω ∈ H is a weak solution of the equation
The weak maximum principle implies that ω = ω+ ≥ 0. Choosing ϕ (x) = φ1(x) > 0, which is the corresponding eigenfunctions of λ1. From (10), we get
On the other hand, we can easily see that Λ = λ1(λ1 + m') is the eigenvalue of the problem
and the corresponding eigenfunction is still φ1(x). If ω(x) > 0, we also have
which follows that ω ≡ 0 by Λ < β But this conclusion contradicts
Hence {un} is bounded in H.
Now we prove that the functionals I± has a mountain pass geometry.
Lemma 3.3. Assume that (H1), (H7) hold, then we have
(1) there exists ρ, R > 0 such that I±(u) > R, if ∥u∥ = ρ;
(2) I±(u) are unbounded from below.
Proof. By (H7), for any ε > 0, there exists C 1 > 0, C2 > 0 such that ∀(x, s) ∈ Ω × R, we have
and
where
We just prove the case of I+. The arguments for the case of I- are similar. Let ϕ = tφ1. When t is sufficiently large, by (16) and (H1), it is easy to see that
On the other hand, by (17), (H1), the Poincare inequality and the Sobolve embedding, we have
where C4 is a constant. Choosing ∥u∥ = ρ small enough, we can obtain I+(u) ≥ R > 0 if ∥u∥ = ρ.
Proof of Theorem 3.1. From Lemma 3.3, it is easy to see that there exists e ∈ H with ∥e∥ > ρ such that I±(e) < 0.
Define
and
From Lemma 3.3, we have
Moreover, by Lemma 3.2, the functions I± satisfies the (PS)-condition. By Lemma 1.1, we know that c+ is a critical value of I+ and there is at least one nontrivial critical point in H corresponding to this value. This critical in nonnegative, then the strong maximum principle implies that is a positive solution of Equation (1). By an analogous way we know there exists at least one negative solution, which is a nontrivial critical point of I- Hence, Equation (1) admits at least a positive solution and a negative solution.
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
Acknowledgements
The authors' would like to thank the referees for valuable comments and suggestions for improving this article.
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