In this article, we give conditions on parameters k, l that the generalized eigenvalue problem x″″ + kx″ + lx = λh(t)x, 0 < t < 1, x(0) = x(1) = x′(0) = x′(1) = 0 possesses an infinite number of simple positive eigenvalues and to each eigenvalue there corresponds an essential unique eigenfunction ψ_k which has exactly k - 1 simple zeros in (0,1) and is positive near 0. It follows that we consider the fourth-order two-point boundary value problem x″″ + kx″ + lx = f(t,x), 0 < t < 1, x(0) = x(1) = x′(0) = x′(1) = 0, where f(t, x) ∈ C([0,1] × ℝ, ℝ) satisfies f(t, x)x > 0 for all x ≠ 0, t ∈ [0,1] and lim_|x|→0 f(t,x)/x = a(t), lim_|x|→+∞ f(t,x)/x = b(t) or lim_x→-∞ f(t,x)/x = 0 and lim_x→+∞f(t,x)/x = c(t) for some a(t), b(t), c(t) ∈ C([0,1], (0,+∞)) and t ∈ [0,1]. Furthermore, we obtain the existence and multiplicity results of nodal solutions for the above problem. The proofs of our main results are based upon disconjugate operator theory and the global bifurcation techniques.

MSC (2000): 34B15.

The deformations of an elastic beam in equilibrium state with fixed both endpoints can be described by the fourth-order boundary value problem

where f: ℝ → ℝ is continuous, λ ∈ ℝ is a parameter and l is a given constant. Since the problem (1.1) cannot transform into a system of second-order equation, the treatment method of second-order system does not apply to the problem (1.1). Thus, existing literature on the problem (1.1) is limited. Recently, when l = 0, the existence and multiplicity of positive solutions of the problem (1.1) has been studied by several authors, see Agarwal and Chow [1], Ma and Wu [2], Yao [3,4] and Korman [5]. Especially, when l ≠ 0, l satisfying (H1) and h(t) satisfying (H2), Xu and Han [6] studied the existence of nodal solutions of the problem (1.1) by applying bifurcation techniques, where

(H1) l ∈ (-π⁴, π⁴/64) is given constant.

(H2) h ∈ C([0,1], [0, ∞)) with h(t) ≢ 0 on any subinterval of [0,1].

Motivated by [6], we consider the existence of nodal solutions of general fourth-order boundary value problem

and under the assumptions:

(A1) One of following conditions holds

(i) k, l satisfying are given constants with

(ii) k, l satisfying are given constants with

(A2) f(t, x) ∈ C([0,1] × ℝ, ℝ) satisfies f(t, x)x > 0 for all x ≠ 0 and t ∈ [0,1].

(A3) There exists a(t) ∈ C([0,1], (0, ∞)) such that

(A4) There exists b(t) ∈ C([0,1], (0, ∞)) such that

(A5) There exists c(t) ∈ C([0,1], (0, ∞)) such that

However, in order to use bifurcation technique to study the nodal solutions of the problem (1.2), we first prove that the generalized eigenvalue problem

(where h satisfies (H2)) has an infinite number of positive eigenvalues

and each eigenvalue corresponding an essential unique eigenfunction ψ_k which has exactly k - 1 simple zeros in (0,1) and is positive near 0. Fortunately, Elias [7] developed a theory on the eigenvalue problem

where

and ρ_i ∈ C^n-i[a, b] with ρ_i > 0 (i = 0,1,..., n) on [a, b]. ℒ₀y,...., ℒ_n-1y are called the quasi-derivatives of y(t). To apply Elias's theory, we have to prove that (1.8) can be rewritten to the form of (1.10), that is, the linear operator

has a factorization of the form

on [0,1], where l_i ∈ C^4-i[0,1] with l_i > 0 (i = 0, 1, 2, 3, 4) on [0, 1], and x(0) = x(1) = x′(0) = x′(1) = 0 if and only if

This can be achieved under (A1) by using the disconjugacy theory in [8].

The rest of the article is arranged as follows: In Section 2, we state some disconjugacy theory which can be used in this article, and then show that (A1) implies the equation

is disconjugate on [0, 1], and establish some preliminary properties on the eigenvalues and eigenfunctions of the generalized eigenvalue problem (1.8). Finally in Section 3, we state and prove our main results (Theorems 3.1 and 3.2 ).

Remark 1.1. If we let k = 0, then the condition (A1) reduces to (H1) in [6].

Remark 1.2. Since the function f(t, x) is more general than the function h(t)f(x) in [6], then the problem considered in this article is more general than the problem in [6].

Remark 1.3. If we let k = 0 and f(t, x) = λh(t)f(x), then Theorem 3.2 reduces to [[6], Theorem 3.1].

Remark 14. For other results on the existence and multiplicity of positive solutions and nodal solutions for the boundary value problems of fourth-order ordinary differential equations based on bifurcation techniques, see [9-14]s and their references.

Let

be nth-order linear differential equation whose coefficients p_k(⋅) (k = 1,..., n) are continuous on an interval I.

Definition 2.1 [[8], Definition 0.2, p. 2]. Equation (2.1) is said to be disconjugate on an interval I if no nontrivial solution has n zeros on I, multiple zeros being counted according to their multiplicity.

Lemma 2.2 [[8], Theorem 0.7, p. 3]. Equation (2.1) is disconjugate on a compact interval I if and only if there exists a basis of solutions y₀, ...,y_n-1 such that

on I. A disconjugate operator L[y] = y⁽ⁿ⁾ + p₁(t)y^(n-1)+ ⋯ + p_n(t)y can be written as

where ρ_k ∈ C^n-k(I) (k = 0,1,..., n) and

and ρ₀ρ₁ ⋯ ρ_n ≡ 1.

Lemma 2.3 [[8], Theorem 0.13, p. 9]. Green's function G(t,s) of the disconjugate equation (2.3) and the two-point boundary value conditions

satisfies

Now using Lemmas 2.2 and 2.3, we will prove some preliminary results.

Theorem 2.4. Let (A1) hold. Then

(i) L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

where ρ_k ∈ C^4-k[0,1] with ρ_k > 0 (k = 0, 1, 2, 3, 4).

(ii) x(0) = x(1) = x′(0) = x′(1) = 0 if and only if

where

Proof of Theorem 2.4. We divide the proof into nine cases.

Case 1.

In the case, we have corresponding L[x] = x″″ + kx″ + lx = 0 that the equation λ⁴ + kλ² + lx = 0 has 4 roots λ₁ = m₁ + m₂i, λ₂ = m₁- m₂i, λ₃ = -m₁ + m₂i, and λ₄ = -m₁- m₂i, where

Combining with (2.10), we have . Thus, we get that either the following (1) or (2) holds:

(1) , for ;

(2) , for .

Furthermore, it is easy to check that

Take

It is easy to check that x₀(t), x₁(t), x₂(t), and x₃(t) form a basis of solutions of L[x] = 0. By simple computation, we have

This together with (2.11) implies that w_i > 0(i=1, 2, 3, 4) on [0,1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

and accordingly

Using (2.15), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 2. k ∈ (-∞, 0) and .

In the case, applying the similar method used in Case 1, we take

where .

It is easy to check that x₀(t), x₁(t), x₂(t), and x₃(t) form a basis of solutions of L[x] = 0.

By simple computation, we have

Clearly, w_i > 0 (i = 1, 2, 3, 4) on [0,1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

and accordingly

Using (2.19), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 3.k ∈ (-∞, 0) and .

In the case, we take

where ,

It is easy to check that x₀(t), x₁(t), x₂(t), and x₃(t) form a basis of solutions of L[x] = 0.

By simple computation, we have

Clearly, w_i > 0 (i = 1, 2, 3, 4) on [0, 1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0, 1], and L[x] has a factorization

and accordingly

Using (2.23), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 4. k ∈ (-∞,0), l = 0.

In the case, we take

where .

It is easy to check that x₀(t), x₁(t), x₂(t), and x₃(t) form a basis of solutions of L[x] = 0.

By simple computation, we have

Clearly, w_i > 0 (i = 1, 2, 3, 4) on [0, 1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

and accordingly

Using (2.27), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 5. k = 0, l = 0. The case is obvious.

Case 6. k ∈ (0,π²), l = 0.

In the case, we take

where , σ is a positive constant. Clearly, m ∈ (0,π) and then

It is easy to check that x₀(t), x₁(t), x₂(t), and x₃(t) form a basis of solutions of L[x] = 0.

By simple computation, we have

Clearly, w_i > 0 (i = 1, 2, 3, 4) on [0, 1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

and accordingly

Using (2.32), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 7. π² (k - π²) < l < 0.

In the case, we take

where , σ is a positive constant. Clearly, m₂ ∈ (0,π) and then

It is easy to check that x₀(t), x₁(t), x₂(t), and x₃(t) form a basis of solutions of L[x] = 0. By simple computation, we have

Clearly, w_i > 0 (i = 1, 2, 3, 4) on [0, 1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

and accordingly

Using (2.37), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 8. and

In the case, we take

where , σ is a positive constant. Clearly, and then

It is easy to check that x₀(t), x₁(t), x₂(t), and x₃(t) form a basis of solutions of L[x] = 0. By simple computation, we have

Clearly, w_i > 0 (i = 1, 2, 3, 4) on [0, 1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

and accordingly

Using (2.42), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

Case 9.

In the case, we take

where Clearly, , , m₁ < m₂ and then

It is easy to check that x₀(t), x₁(t), x₂(t), and x₃(t) form a basis of solutions of L[x] = 0.

By simple computation, we have

Clearly, w_i > 0 (i = 1, 2, 3, 4) on [0, 1].

By Lemma 2.2, L[x] = 0 is disconjugate on [0,1], and L[x] has a factorization

and accordingly

Using (2.47), we conclude that x(0) = x(1) = x′(0) = x′(1) = 0 is equivalent to (2.8).

This completes the proof of Theorem 2.4.

Remark 2.5. If condition (A1) does not hold, the results of Theorem 2.4 cannot be obtained. For example, in the case of L[x] = 0 with , we have . Applying the similar method to prove case 1 in Theorem 2.4, we conclude that

form a basis of solutions of L[x] = 0. By simple computation, we have

From , we easily get that tan . Furthermore, w₃ < 0. Thus, Theorem 2.4 does not hold in this case.

Remark 2.6. In the following, consider L[x] = 0, for k, l are given constants, by the similar method in Remark 2.5, we may gain the location of (k,l) in the (k, l)-plane and the results of w₃ or w₁ corresponding and and and and and and and and ω₃ < 0. Furthermore, it follows that the conclusion of Theorem 2.4 cannot be yielded in the cases.

Theorem 2.7. Let (A1) hold and h satisfy (H2). Then

(i) The problem (1.8) has an infinite number of positive eigenvalue

(ii) λ_k(h) → ∞ as k → ∞.

(iii) To each eigenvalue λ_k(h) there corresponds an essential unique eigenfunction ψ_k which has exactly k - 1 simple zeros in (0,1) and is positive near 0.

(iv) Given an arbitrary subinterval of [0,1], then an eigenfunction which belongs to a sufficiently large eigenvalue change its sign in that subinterval.

(v) For each k ∈ ℕ, the geometric multiplicity of λ_k(h) is 1.

Proof of Theorem 2.7. (i)-(iv) are immediate consequences of Elias [[7], Theorem 1-5] and Theorem 2.4, we only prove (υ).

Let

with

To show (υ), it is enough to prove

Clearly

Suppose on the contrary that the geometric multiplicity of λ_k(h) is greater than 1. Then there exists and subsequently

for some γ ≠ 0. Multiplying both sides of (2.50) by ψ_k(t) and integrating from 0 to 1, we deduce that

which is a contradiction !

Theorem 2.8 (Maximum principle). Let (A1) hold. Let e ∈ C[0,1] with e ≥ 0 on [0,1] and e ≢ 0 on any compact subinterval in [0,1]. If x ∈ C⁴[0,1] satisfies

Then x > 0 on (0,1).

Proof. When (A1) holds, the homogeneous problem

has only trivial solution. So the boundary value problem (2.52) has a unique solution which may be represented in the form

where G(t, s) is Green's function. By Theorem 2.4 and Lemma 2.3 (take n = 4, k = 2), we have

that is, G(t, s) > 0, for all (t, s) ∈ (0,1) × (0,1).

Using (2.54), when e ≥ 0 on [0,1] and e ≢ 0 on any compact subinterval in [0,1], then x > 0 on (0,1).

Theorem 3.1. Let (A1), (A2), (A3) and (A4) hold. Assume that either (i) or (ii) holds for some k ∈ ℕ and j ∈ {0} ∪ ℕ:

Then the problem (1.2) has 2(j + 1) solutions has exactly k+i-1 zeros in (0,1) and is positive near t = 0, and has exactly k + i - 1 zeros in (0,1) and is negative near t = 0.

Theorem 3.2 Let (A1), (A2), (A3) and (A5) hold. Assume that for some k ∈ ℕ,

Then there are at least 2k - 1 nontrivial solutions of the problem (1.2). In fact, there exist solutions ω₁,...,ω_k, such that for 1 ≤ j ≤ k, ω_j has exactly j - 1 simple zeros on the open interval (0,1) and and there exist solutions z₂,...,z_k, such that for 2 ≤ j ≤ k, z_j has exactly j - 1 simple zeros on the open interval (0,1) and .

Let Y = C[0,1] with the norm

Let E = {x ∈ C²[0, 1]|x(0) = x(1) = x′(0) = x′(1) = 0} with the norm

Then is completely continuous. Here is given as in (2.48).

Let ζ(⋅,⋅), ξ₁(⋅,⋅), ξ₂(⋅,⋅) ∈C([0,1] ×ℝ,ℝ) be such that

Here x⁺ = max{x,0}.

Clearly,

uniformly for t ∈ [0,1].

Let

then and are nondecreasing and

Let us consider

as a bifurcation problem from the trivial solution x ≡ 0.

Equation (3.9) can be converted to the equivalent equation

Clearly, the compactness of together with (3.6) imply that

Let denotes the set of functions in E which have exactly k - 1 interior nodal (i.e., non-degenerate) zeros in (0,1) and are positive near t = 0, set , and . They are disjoint and open sets in E. Finally, let and Φ_k = ℝ × S_k.

The results of Rabinowitz [15] for (3.9) can be stated as follows: For each integer k ≥ 1 and each ν = {+, -}, there exists a continuum of solution of (3.9), joining (λ_k(a), 0) to infinity in . Moreover, .

Notice that we have used the fact that if x is a nontrivial solution of (3.9), then all zeros of x on (0, 1) are simple under (A1), (A2), (A3), and (A4).

In fact, (3.9) can be rewritten to

where

Clearly â(t) satisfies (H2). So Theorem 2.7 (iii) yields that all zeros of x on (0,1) are simple.

Proof of Theorem 3.1. We first prove the theorem when j = 0.

It is clear that any solution of (3.9) of the form (1, x) yields solutions x of (1.2). We will show that crosses the hyperplane {1} × E in ℝ × E. To do this, it is enough to show that joins (λ_k(a),0) to (λ_k(b),∞). Let satisfy

We note that µ_n > 0 for all n ∈ ℕ since (0, 0) is the only solution of (3.9) for λ = 0 and .

Case 1. λ_k(b) < 1 < λ_k(a).

In this case, we show that

We divide the proof into two steps.

Step 1. We show that if there exists a constant number M > 0 such that

then joins (λ_k(a),0) to (λ_k(b),∞).

In this case ║x_n║_E → ∞. We divide the equation

by ║x_n║_E and set . Since y_n is bounded in C²[0,1], choosing a subsequence and relabeling if necessary, we have that y_n → y for some y ∈ E with ║y║_E = 1. Moreover, from (3.8) and the fact that is nondecreasing, we have that

since

Thus

where µ: = lim_n→∞μ_n, again choosing a subsequence and relabeling if necessary. Thus

We claim that

Suppose, to the contrary, that . Since y ≠ 0 is a solution of (3.18), all zeros of y in [0,1] are simple. It follows that for some h ∈ ℝ and l ∈ {+, -}. By the openness of we know that there exists a neighborhood U(y,ρ₀) such that

which, together with the fact y_n → y, implies that exists n₀ ∈ ℕ such that

However, this contradicts the fact that Therefore,

Now, by Theorem 2.7, we obtain µ = λ_k(b).

Thus joins (λ_k(a),0) to (λ_k(b),∞).

Step 2. We show that there exists a constant number M > 0 such that µ_n ∈ (0, M], for all n.

Suppose there is no such M. Choosing a subsequence and relabeling if necessary, it follows that

Let

denotes the zeros of x_n. Then there exists a subsequence {τ(1, n_m)} ⊆ {τ(1, n)} such that

Clearly

We claim that

Suppose, to the contrary, that

Define a function p: [0,1] × [0, ∞) → ℝ by

Then, by (A2), (A3), and (A4), there exist two positive numbers ρ_l and ρ₂, such that

Using (3.22), (3.24), and the fact that , we conclude that there exists a closed interval I₁ ⊂ (τ(0, ∞), τ(1, ∞)) such that

uniformly for t ∈ I₁.

However, since satisfies

the proof of Lemma 4 in [7] (see also the remarks in the final paragraph in [[7], p. 43]), shows that for all n sufficiently large, must change sign on I₁. However, this contradicts the fact that for all m sufficiently large we have I₁ ⊂ (τ(0, n_m),τ(1,n_m)) and

Thus, (3.21) holds.

Next, we work with (τ(1, n_m), τ(2, n_m)). It is easy to see that there is a subsequence such that

Clearly

We claim that

Suppose, to the contrary, that τ(1,∞) < τ(2,∞). Then, from (3.23), (3.24), and the fact that , there exists a closed interval I₂ ⊂ (τ(1, ∞), τ(2, ∞)) such that

uniformly for t ∈ I₂.

This implies the solution of the equation

must change sign on I₂. However, this contradicts the fact that for all j sufficiently large we have and

Therefore, (3.26) holds.

By a similar argument to obtain (3.21) and (3.26), we can show that for each l ∈ {2,...,k-1},

Taking a subsequence and relabeling it as {(µ_n, x_n)}, if necessary, it follows that for each l ∈ {0,..., k-1},

But this is impossible since

for all n. Therefore,

for some constant number M > 0, independent of n ∈ ℕ.

Case 2. λ_k(a) < 1< λ_k(b).

In this case, if is such that

and

then

and, moreover, .

Assume that there exists M > 0 such that for all n ∈ ℕ,

Applying a similar argument to that used in step 1 of Case 1, after taking a subsequence and relabeling if necessary, it follows that

Again joins (λ_k(a), 0) to (λ_k(b), ∞) and the result follows.

Finally, let j ∈ ℕ. By repeating the arguments used in the proof of the case j = 0, we see that for each ν ∈ {+, -} and each i ∈ {k, k + 1,..., k + j},

The result follows.

Proof of Theorem 3.2.

We only need to show that

Suppose on the contrary that

where

Since joins (λ_i,(a), 0) to infinity in and (λ, x) = (0, 0) is the unique solution of (3.9)_{λ = 0} in E, there exists a sequence such that µ_n ∈ (0,1) and ║x_n║_E → ∞ as n → ∞. We may assume that µ_n→µ ∈ [0, 1] as n → ∞. Let y_n = x_n/║x_n║_E, n ≥ 1. From the fact

We have that

Furthermore, since is completely continuous, we may assume that there exists y ∈ E with ║y║_E = 1 such that ║y_n - y║_E → 0 as n →∞. Since

uniformly for t ∈ [0,1], we have from (3.39) and (3.8) that

that is,

By (A1), (A5), and (3.42) and the fact that ║y║_E = 1, we conclude that µc(t)y⁺ ≢ 0 on any compact subinterval in [0,1], and consequently

By Theorem 2.8, we know that y(t) > 0 in (0,1). This means µ is the first eigenvalue of and y is the corresponding eigenfunction. Hence and therefore, since is open and ║y_n - y║_E → 0, we have that for n large. But this contradicts the assumption that and (i,l) ∈ Γ, so (3.36) is wrong, which completes the proof.

The author declares that they have no competing interests.

Thanks are given to Professor R.Y. Ma for his valuable suggestion. The author is also grateful to the anonymous referee for his/her valuable suggestions. This study was supported by: the NSFC (No. 11031003); the Scientific Research Foundation of the Education department of Gansu Province (No. 1114-04).

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