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
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
(ii) k, l satisfying
(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.
Case 1.
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
(1)
(2)
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).
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 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).
Case 3.k ∈ (∞, 0) and
In the case, we take
where
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
where
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
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
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).
Case 8.
In the case, we take
where
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).
Case 9.
In the case, we take
where
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
form a basis of solutions of L[x] = 0. By simple computation, we have
From
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
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
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
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
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
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
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
Let
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
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
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
In this case ║x_{n}║_{E }→ ∞. We divide the equation
by ║x_{n}║_{E }and set
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
which, together with the fact y_{n }→ y, implies that exists n_{0 }∈ ℕ such that
However, this contradicts the fact that
Now, by Theorem 2.7, we obtain µ = λ_{k}(b).
Thus
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
uniformly for t ∈ I_{1}.
However, since
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,
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
Clearly
We claim that
Suppose, to the contrary, that τ(1,∞) < τ(2,∞). Then, from (3.23), (3.24), and the fact that
uniformly for t ∈ I_{2}.
This implies the solution
must change sign on I_{2}. However, this contradicts the fact that for all j sufficiently large we have
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).
In this case, if
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
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
We have that
Furthermore, 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
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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