We begin with the fourth-order m-point boundary value problem
(3.1)
(3.2)
where 0 < ξ1 < ξ2 < ⋯ < ξm-2< 1, β
i
> 0, i = 1, 2, ..., m - 2.
The following assumption will stand throughout this section:
Lemma 3.1 Denote ξ0 = 0, ξm -1= 1, β0 = βm -1= 0, and y(t) ∈ C[0, 1]. Problem (3.1), (3.2) has the unique solution
where
for ξi-1≤ s ≤ ξ
i
, i = 1, 2, ..., m -1.
Proof Let G(t, s) be the Green's function of problem x(4)(t) = 0 with boundary condition (3.2). We can suppose
Considering the definition and properties of Green's function together with the boundary condition (3.2), we have
A straightforward calculation shows that
These give the explicit expression of the Green's function and the proof of Lemma 3.1 is completed.
Lemma 3.2 One can see that G(t, s) ≥ 0, t, s ∈ [0, 1].
Proof For ξi-1≤ s ≤ ξ
i
, i = 1, 2, ..., m - 1,
Then , 0 ≤ t, s ≤ 1. Thus G(t, s) is increasing on t. By a simple computation, we see
These ensures that G(t, s) ≥ 0, t, s ∈ [0, 1].
Lemma 3.3 Suppose x(t) ∈ C3[0, 1] and
Furthermore x(4)(t) ≥ 0 and there exist t0 such that x(4)(t0) > 0. Then x(t) has the following properties:
-
(1)
-
(2)
-
(3)
where are positive constants.
Proof Since x(4)(t) ≥ 0, t ∈ [0, 1], then x'''(t) is increasing on [0, 1]. Considering x'''(1) = 0, we have x'''(t) ≤ 0, t ∈ [0, 1]. Thus x''(t) is decreasing on [0, 1]. Considering this together with the boundary condition x''(1) = 0, we conclude that x''(t) ≥ 0. Then x(t) is convex on [0, 1]. Taking into account that x'(0) = 0, we get that
-
(1)
From the concavity of x(t), we have
Multiplying both sides with β
i
and considering the boundary condition, we have
(3.3)
Thus
-
(2)
Considering the mean-value theorem together with the concavity of x(t), we have
(3.4)
Multiplying both sides with β
i
and considering the boundary condition, we have
(3.5)
which yields that .
-
(3)
For and x'(0) = 0, we get
For and x"(1) = 0, we get
Consequently
These give the proof of Lemma 3.3.
Remark Lemma 3.3 ensures that
Let Banach space E = C3[0, 1] be endowed with the norm
Define the cone P ⊂ E by
Let the nonnegative continuous concave functional α, the nonnegative continuous convex functionals γ, θ and the nonnegative continuous functional ψ be defined on the cone by
By Lemma 3.3, the functionals defined above satisfy
(3.6)
Denote
Assume that there exist constants 0 < a, b, d with a < b < λd such that
Theorem 3.1 Under assumptions (A1)-(A3), problem (1.1), (1.2) has at least three positive solutions x1, x2, x3 satisfying
Proof Problem (1.1, 1.2) has a solution x = x(t) if and only if x solves the operator equation
Then
For , considering Lemma 3.3 and assumption (A1), we have f(t, x(t), x'(t), x''(t), x'''(t)) ≤ d. Thus
Hence . An application of the Arzela-Ascoli theorem yields that T is a completely continuous operator. The fact that the constant function x(t) = b/δ ∈ P(γ, θ, α, b, c, d) and α(b/δ) > b implies that
For x ∈ P(γ, θ, α, b, c, d), we have b ≤ x(t) ≤ b/δ and |x'''(t)| < d. From assumption (A2), we see
Hence, by definition of α and the cone P, we can get
which means α(Tx) > b, ∀x ∈ P(γ, θ, α, b, b/δ, d). This ensures that condition (S1) of Lemma 2.1 is fulfilled.
Second, with (3.4) and b < λd, we have
for all x ∈ P(γ, α, b, d) with .
Thus, condition (S2) of Lemma 2.1 holds. Finally we show that (S3) also holds. We see ψ(0) = 0 < a and 0 ∉ R(γ, ψ, a, d). Suppose that x ∈ R(γ, ψ, a, d) with ψ(x) = a, then by the assumption of (A3),
which ensures that condition (S3) of Lemma 2.1 is fulfilled. Thus, an application of Lemma 2.1 implies that the fourth-order m-point boundary value problem (1.1, 1.2) has at least three positive convex increasing solutions x1, x2, x3 with the properties that