Abstract
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 fourthorder twopoint 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.
Keywords:
disconjugacy theory; bifurcation; nodal solutions; eigenvalue1 Introduction
The deformations of an elastic beam in equilibrium state with fixed both endpoints can be described by the fourthorder 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 secondorder equation, the treatment method of secondorder 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 ∈ (π^{4}, π^{4}/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 fourthorder 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^{ni}[a, b] with ρ_{i }> 0 (i = 0,1,..., n) on [a, b]. ℒ_{0}y,...., ℒ_{n1}y are called the quasiderivatives 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^{4i}[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 fourthorder ordinary differential equations based on bifurcation techniques, see [914]s and their references.
2 Preliminary results
Let
be nthorder 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 y_{0}, ...,y_{n1 }such that
on I. A disconjugate operator L[y] = y^{(n) }+ p_{1}(t)y^{(n1)}+ ⋯ + p_{n}(t)y can be written as
where ρ_{k }∈ C^{nk}(I) (k = 0,1,..., n) and
and ρ_{0}ρ_{1} ⋯ 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 twopoint 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^{4k}[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.
In the case, we have corresponding L[x] = x″″ + kx″ + lx = 0 that the equation λ^{4 }+ kλ^{2 }+ lx = 0 has 4 roots λ_{1 }= m_{1 }+ m_{2}i, λ_{2 }= m_{1} m_{2}i, λ_{3 }= m_{1 }+ m_{2}i, and λ_{4 }= m_{1} m_{2}i, where
Combining with (2.10), we have . Thus, we get that either the following (1) or (2) holds:
Furthermore, it is easy to check that
Take
It is easy to check that x_{0}(t), x_{1}(t), x_{2}(t), and x_{3}(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).
In the case, applying the similar method used in Case 1, we take
It is easy to check that x_{0}(t), x_{1}(t), x_{2}(t), and x_{3}(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).
In the case, we take
It is easy to check that x_{0}(t), x_{1}(t), x_{2}(t), and x_{3}(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
It is easy to check that x_{0}(t), x_{1}(t), x_{2}(t), and x_{3}(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,π^{2}), 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 x_{0}(t), x_{1}(t), x_{2}(t), and x_{3}(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. π^{2 }(k  π^{2}) < l < 0.
In the case, we take
where , σ is a positive constant. Clearly, m_{2 }∈ (0,π) and then
It is easy to check that x_{0}(t), x_{1}(t), x_{2}(t), and x_{3}(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).
In the case, we take
where , σ is a positive constant. Clearly, and then
It is easy to check that x_{0}(t), x_{1}(t), x_{2}(t), and x_{3}(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).
In the case, we take
where Clearly, , , m_{1 }< m_{2 }and then
It is easy to check that x_{0}(t), x_{1}(t), x_{2}(t), and x_{3}(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, w_{3 }< 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 w_{3 }or w_{1 }corresponding and and and and and and and and ω_{3 }< 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 15] 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 ∈ C^{4}[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).
3 Main results
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+i1 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 ω_{1},...,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 z_{2},...,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 ∈ C^{2}[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(⋅,⋅), ξ_{1}(⋅,⋅), ξ_{2}(⋅,⋅) ∈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., nondegenerate) 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 C^{2}[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,ρ_{0}) such that
which, together with the fact y_{n }→ y, implies that exists n_{0 }∈ 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 ρ_{2}, such that
Using (3.22), (3.24), and the fact that , we conclude that there exists a closed interval I_{1 }⊂ (τ(0, ∞), τ(1, ∞)) such that
uniformly for t ∈ I_{1}.
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 I_{1}. However, this contradicts the fact that for all m sufficiently large we have I_{1 }⊂ (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 I_{2 }⊂ (τ(1, ∞), τ(2, ∞)) such that
uniformly for t ∈ I_{2}.
This implies the solution of the equation
must change sign on I_{2}. 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,...,k1},
Taking a subsequence and relabeling it as {(µ_{n}, x_{n})}, if necessary, it follows that for each l ∈ {0,..., k1},
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).
and
then
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.
Competing interests
The author declares that they have no competing interests.
Acknowledgements
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. 111404).
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