The existence and multiplicity of positive solutions of nonlinear sixth-order boundary value problem with three variable coefficients

Abstract

In this article, we discuss the existence and multiplicity of positive solutions for the sixth-order boundary value problem with three variable parameters as follows:

$$\left\{\begin{array}{c}\hfill u^{\left(6\right)}+A\left(t\right)u^{\left(4\right)}+B\left(t\right)u^{\left(2\right)}+C\left(t\right)u+f\left(x,u\right)=0,\hfill \\ \hfill u\left(0\right)=u\left(1\right)=u^{″}\left(0\right)=u^{″}\left(1\right)=u^{\left(4\right)}\left(0\right)=u^{\left(4\right)}\left(1\right)=0,\hfill \end{array}\right.$$

where A(t), B(t), C(t) ∈ C[0,1], f(t, u) : [0,1] × [0, ∞) → [0. ∞) is continuous. The proof of our main result is based upon spectral theory of operators and fixed point theorem in cone.

1 Introduction

In this article, we study the existence and multiplicity of positive solution for the following nonlinear sixth-order boundary value problem (BVP for short) with three variable parameters

$$\left\{\begin{array}{c}\hfill -u^{\left(6\right)}-C\left(t\right)u^{\left(4\right)}+B\left(t\right)u^{″}-A\left(t\right)u=f\left(t,u\right),t\in \left(0,1\right),\hfill \\ \hfill u\left(0\right)=u\left(1\right)=u^{″}\left(0\right)=u^{″}\left(1\right)=u^{\left(4\right)}\left(0\right)=u^{\left(4\right)}\left(1\right)=0,\hfill \end{array}\right.$$

(1.1)

where A(t), B(t), C(t) ∈ C[0,1], f(t, u) : [0,1] × [0, ∞) → [0. ∞) is continuous.

In recent years, BVPs for sixth-order ordinary differential equations have been studied extensively, see [1–7] and the references therein. For example, Tersian and Chaparova [1] have studied the existence of positive solutions for the following systems (1.2):

$$\left\{\begin{array}{c}\hfill u^{\left(6\right)}+A{u}^{\left(4\right)}+B{u}^{″}+Cu-f\left(t,u\right)=0.0<x<L,\hfill \\ \hfill u\left(0\right)=u\left(L\right)=u^{″}\left(0\right)=u^{″}\left(L\right)=u^{\left(4\right)}\left(0\right)=u^{\left(4\right)}\left(L\right)=0,\hfill \end{array}\right.$$

(1.2)

where A, B, and C are some given real constants and f(x, u) is a continuous function on R², is motivated by the study for stationary solutions of the sixth-order parabolic differential equations

$$\begin{array}{c}\hfill \frac{\partial u}{\partial t}=\frac{{\partial }^{6}u}{\partial x^6}+A\frac{{\partial }^{4}u}{\partial x^4}+B\frac{{\partial }^{2}u}{\partial x^2}+f\left(x,u\right).\hfill \end{array}$$

This equation arose in the formation of the spatial periodic patterns in bistable systems and is also a model for describing the behaviour of phase fronts in materials that are undergoing a transition between the liquid and solid state. When f(x, u) = u - u³, it was studied by Gardner and Jones [2] as well as by Caginalp and Fife [3]. In [1], existence of nontrivial solutions for (1.2) is proved using a minimization theorem and a multiplicity result using Clarks theorem when C = 1 and f(x, u) = u³. The authors have studied also the homoclinic solutions for (1.2) when C = -1 and f(x, u) = -a(x)u|u|^σ , where a(x) is a positive periodic function and σ is a positive constant by the mountain-pass theorem of Brezis-Nirenberg and concentration-compactness arguments. In [4], by variational tools, including two Brezis-Nirenbergs linking theorems, Gyulov et al. have studied the existence and multiplicity of nontrivial solutions of BVP (1.2).

Recently, in [5], the existence and multiplicity of positive solutions of sixth-order BVP with three parameters

$$\left\{\begin{array}{c}\hfill -u^{\left(6\right)}-\gamma u^{\left(4\right)}+\beta u^{″}-\alpha u=f\left(t,u\right),t\in \left[0,1\right],\hfill \\ \hfill u^{\left(i\right)}\left(0\right)=u^{\left(i\right)}\left(1\right)=0,i=0,1,2,3,4,5\hfill \end{array}\right.$$

(1.3)

has been studied under the hypothesis of

(A₁) f : [0,1] × [0, ∞) → [0. ∞) is continuous.

(A₂) α, β, γ ∈ R and under the condition of satisfying

$$\begin{array}{c}\frac{\alpha }{{\pi }^6}+\frac{\beta }{{\pi }^4}+\frac{\gamma }{{\pi }^2}<1,\\ 3{\pi }^4-2\gamma {\pi }^2-\beta >0,\gamma <3{\pi }^2,\\ 18\alpha \beta \gamma -{\beta }^2{\gamma }^2+4\alpha {\gamma }^3+27{\alpha }^2-4{\beta }^3\le 0,\end{array}$$

the existence and multiplicity for positive solution of BVP (1.3) are established by using fixed point index theory. In this article, we consider more general BVP (1.1), based upon spectral theory of operators and fixed point theorem in cone, we will establish the existence and multiplicity positive solution of BVP (1.1) and extend the result of [5] under appropriate conditions. Our ideas mainly come from [5, 8–10].

We list the following conditions for convenience:

(H₁) f : [0,1] × [0, +∞) → [0. +∞) is continuous.

(H₂) A(t), B(t), C(t) ∈ C[0,1], α = min_0≤t≤1A(t), β = min_0≤t≤1B(t), γ = min_0≤t≤1C(t), and satisfies

$$\begin{array}{c}\frac{\alpha }{{\pi }^6}+\frac{\beta }{{\pi }^4}+\frac{\gamma }{{\pi }^2}<1,\\ 3{\pi }^4-2\gamma {\pi }^2-\beta >0,\gamma <3{\pi }^2,\\ 18\alpha \beta \gamma -{\beta }^2{\gamma }^2+4\alpha {\gamma }^3+27{\alpha }^2-4{\beta }^3\le 0.\end{array}$$

Let Y = C[0,1], Y₊ = {u ∈ Y : u(t) ≥ 0, t ∈ [0,1]}. It is well known that Y is a Banach space equipped with the norm ||u||₀ = sup_0≤t≤1|u(t)|, u ∈ Y. Set X = { u ∈ C⁴[0,1] : u(0) = u(1) = u''(0) = u''(1) = 0}, then X also is a Banach space equipped with the norm ||u|| _X = max {||u(t)||₀, ||u"(t)||₀, ||u⁽⁴⁾(t)||₀}. If u ∈ C⁴[0,1] ∩ C⁶(0,1) fulfils BVP (1.1), then we call u is a solution of BVP (1.1). If u is a solution of BVP (1.1), and u(t) > 0, t ∈ (0, 1), then we say u is a positive solution of BVP (1.1).

2 Preliminaries

In this section, we will make some preliminaries which are needed to show our main results.

Lemma 2.1. Let u ∈ X, then ||u||₀ ≤ ||u"||₀ ≤ ||u⁽⁴⁾||₀ ≤ ||u|| _X .

Proof. The proof is similar to the Lemma 1 in [8], so we omit it. □

Lemma 2.2. [5] Let λ₁, λ₂, and λ₃ be the roots of the polynomial P (λ) = λ³ + γλ² - βλ + α. Suppose that condition (H₂) holds, then λ₁, λ₂, and λ₃ are real and greater than -π².

Note : Based on Lemma 2.2, it is easy to learn that when the three parameters satisfy the condition of (H₂), they satisfy the condition of non-resonance.

Let G _i (t, s)(i = 1, 2, 3) be the Green's function of the linear BVP

• u"(t) + λ _i u(t) = 0, u(0) = u(1) = 0,

Lemma 2.3. [10]G _i (t, s)(i = 1, 2, 3) has the following properties

(c1) G _i (t, s) > 0, ∀t, s ∈ (0, 1).

(c2) G _i (t, s) <C _i G _i (s, s), ∀t, s ∈ [0,1], in which C _i > 0 is constant.

(c3) G _i (t, s) ≥ δ _i G _i (t, t)G _i (s, s), ∀t, s ∈ [0,1], in which δ _i > 0 is constant.

We set

$$M_i=\underset{0\le t\le 1}{\text{max}}{G}_i\left(s,s\right),m_i=\underset{\frac{1}{4}\le t\le \frac{3}{4}}{\text{min}}{G}_i\left(s,s\right),i=1,2,3.$$

(2.1)

$$C_{ij}=\underset{0}{\overset{1}{\int }}{G}_i\left(\delta ,\delta \right)G_j\left(\delta ,\delta \right)d\delta ,c_{ij}=\underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}{G}_i\left(\delta ,\delta \right)G_j\left(\delta ,\delta \right)d\delta ,i,j=1,2,3.$$

(2.2)

$$D_i=\underset{0\le t\le 1}{\text{max}}\underset{0}{\overset{1}{\int }}{G}_i\left(t,s\right)ds,d_i=\underset{\frac{1}{4}\le t\le \frac{3}{4}}{\text{max}}\underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}{G}_i\left(t,s\right)dsi=1,2,3,$$

(2.3)

then starting from Lemma 2.3 we know M _i , m _i , C _ij > 0.

For any h ∈ Y, take into consideration of linear BVP:

$$\left\{\begin{array}{c}\hfill -u^{\left(6\right)}-\gamma u^{\left(4\right)}+\beta u^{″}-\alpha u=h\left(t\right),t\in \left[0,1\right],\hfill \\ \hfill u\left(0\right)=u\left(1\right)=u^{″}\left(0\right)=u^{″}\left(1\right)=u^{\left(4\right)}\left(0\right)=u^{\left(4\right)}\left(1\right)=0,\hfill \end{array}\right.$$

(2.4)

where α, β, γ satisfy assumption (H₂). Since

$$-u^{\left(6\right)}-\gamma u^{\left(4\right)}+\beta u^{″}-\alpha u=\left(-\frac{d^2}{d{t}^2}+{\lambda }_1\right)\left(-\frac{d^2}{d{t}^2}+{\lambda }_2\right)\left(-\frac{d^2}{d{t}^2}+{\lambda }_3\right)u,$$

(2.5)

then for any h ∈ Y, the LBVP(2.4) has a unique solution u, which we denoted by Ah = u. The operator A can be expressed by

$$u\left(t\right)=\left(Ah\right)\left(t\right):=\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{G}_1\left(t,\delta \right)G_2\left(\delta ,\tau \right)G_3\left(\tau ,s\right)h\left(t\right)dsd\tau d\delta .$$

(2.6)

Lemma 2.4. The linear operator A : Y → X is completely continuous and ||A|| ≤ ϖ, where ϖ = |λ₂+λ₃|(C₁C₂C₃M₁M₂M₃|λ₃|+C₁C₂M₁M₂)+| λ₂λ₃|(C₁C₂C₃M₁M₂M₃+C₁M₁).

Proof. It is easy to show that the operator A : Y → X is linear operator. ∀h ∈ Y, u = Ah ∈ X, u(0) = u(1) = u"(0) = u"(1) = u⁽⁴⁾(0) = u⁽⁴⁾(1) = 0. Let $v=\left(-\frac{d^2}{d{t}^2}+{\lambda }_2\right)\left(-\frac{d^2}{d{t}^2}+{\lambda }_3\right)u,$ that is

$$v=\left(-\frac{d^2}{d{t}^2}+{\lambda }_2\right)\left(-\frac{d^2}{d{t}^2}+{\lambda }_3\right)u=u^{\left(4\right)}-\left({\lambda }_2+{\lambda }_3\right)u^{″}+{\lambda }_2{\lambda }_{3}u,$$

(2.7)

by (2.5) and (2.7), we have

$$\left\{\begin{array}{c}\hfill -v^{″}+{\lambda }_{1}v=h\left(t\right),t\in \left(0,1\right),\hfill \\ \hfill v\left(0\right)=v\left(1\right)=0,\hfill \end{array}\right.$$

and $v\left(t\right)={\int }_0^1{G}_1\left(t,s\right)h\left(s\right)ds,t\in \left[0,1\right],$ so

$$u^{\left(4\right)}-\left({\lambda }_2+{\lambda }_3\right)u^{″}+{\lambda }_2{\lambda }_{3}u=\underset{0}{\overset{1}{\int }}{G}_1\left(t,s\right)h\left(s\right)ds,t\in \left[0,1\right].$$

(2.8)

By (2.6), for any t ∈ [0,1], we have

$$\left|u\left(t\right)\right|\le \underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{G}_1\left(t,\delta \right)G_2\left(\delta ,\tau \right)G_3\left(\tau ,s\right)\left|h\left(t\right)\right|dsd\tau d\delta \le C_1{C}_2{C}_3{M}_1{M}_2{M}_3\left|\right|h|{|}_0.$$

(2.9)

Again, let ω = -u" + λ₃u, then ω(0) = ω(1) = ω"(0) = ω"(1) = 0, by (2,5), we have

$$\left\{\begin{array}{c}\hfill {\omega }^{\left(4\right)}-\left({\lambda }_1+{\lambda }_2\right){\omega }^{″}+{\lambda }_1{\lambda }_2\omega =h\left(t\right),t\in \left(0,1\right),\hfill \\ \hfill \omega \left(0\right)=\omega \left(1\right)={\omega }^{″}\left(0\right)={\omega }^{″}\left(1\right)=0.\hfill \end{array}\right.$$

(2.10)

Then $\omega \left(t\right)={\int }_0^1{\int }_0^1{G}_1\left(t,\tau \right)G_2\left(\tau ,s\right)h\left(s\right)dsd\tau ,t\in \left[0,1\right],$ that is

$$-u^{″}+{\lambda }_{3}u=\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{G}_1\left(t,\tau \right)G_2\left(\tau ,s\right)h\left(s\right)dsd\tau ,t\in \left[0,1\right].$$

(2.11)

So

$$\left|u^{″}\right|\le C_1{C}_2{M}_1{M}_2\left(1+\left|{\lambda }_3\right|C_3{M}_3\right)\left|\right|h|{|}_0,t\in \left[0,1\right].$$

(2.12)

Based on (2.8), (2.9), and (2.12), we have

$$\begin{array}{ll}\hfill \left|u^{\left(4\right)}\left(t\right)\right|& \le |{\lambda }_2+{\lambda }_3|\left|u^{″}\left(t\right)\right|+\left|{\lambda }_2{\lambda }_3\right|\left|u\left(t\right)\right|+\underset{0}{\overset{1}{\int }}{G}_1\left(t,s\right)\left|h\left(s\right)\right|ds\\ \le |{\lambda }_2+{\lambda }_3|\left(C_1{C}_2{C}_3{M}_1{M}_2{M}_3\left|{\lambda }_3\right|+C_1{C}_2{M}_1{M}_2\right)\left|\right|h|{|}_0\\ +\left|{\lambda }_2{\lambda }_3\right|\left(C_1{C}_2{C}_3{M}_1{M}_2{M}_3\left|\right|h|{|}_0+C_1{M}_1\right)\left|\right|h|{|}_0\\ \le \varpi \left|\right|h|{|}_0,t\in \left[0,1\right],\end{array}$$

(2.13)

where

$$\begin{array}{ll}\hfill \varpi & =|{\lambda }_2+{\lambda }_3|\left(C_1{C}_2{C}_3{M}_1{M}_2{M}_3\left|{\lambda }_3\right|+C_1{C}_2{M}_1{M}_2\right)\\ +\left|{\lambda }_2{\lambda }_3\right|\left(C_1{C}_2{C}_3{M}_1{M}_2{M}_3+C_1{M}_1\right).\end{array}$$

(2.14)

So, ||u⁽⁴⁾(t)|| ≤ ϖ||h||₀, by Lemma2.1, ||u|| _X ≤ ϖ||h||₀, then

$$\left|\right|Ah|{|}_X\le \varpi |\left|h\right|{|}_0,$$

(2.15)

so A is continuous, and ||A|| ≤ ϖ.

Next, we will show that A is compact with respect to the norm ||·|| _X on X.

Suppose {h _n }(n = 1, 2, . . .) an arbitrary bounded sequence in Y, then there exists K₀> 0 such that ||h _n ||₀ ≤ K₀, n = 1, 2, . . . . Let u _n = Ah _n , 1, 2, ...By (2.8), ∀t₁, t₂ ∈ [0, 1], t₁< t₂, we have

$$\begin{array}{c}\left|u_n^{\left(4\right)}\left(t_2\right)-u_n^{\left(4\right)}\left(t_1\right)\right|\\ \le |{\lambda }_2+{\lambda }_3||{u^{″}}_n\left(t_2\right)-{u^{″}}_n\left(t_1\right)|+\left|{\lambda }_2{\lambda }_3\right||u_n\left(t_2\right)-u_n\left(t_1\right)|+\underset{0}{\overset{1}{\int }}|G_1\left(t_2,s\right)-G_1\left(t_1,s\right)|\left|h_n\left(s\right)\right|ds.\\ \le |{\lambda }_2+{\lambda }_3|\left(\left|{\lambda }_3\right||u_n\left(t_2\right)-u_n\left(t_1\right)|+\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}|G_1\left(t_2,\tau \right)-G_1\left(t_1,\tau \right)|\left|G_2\left(\tau ,s\right)\right|\left|h_n\left(s\right)\right|dsd\tau \right)\\ +\left|{\lambda }_2{\lambda }_3\right||u_n\left(t_2\right)-u_n\left(t_1\right)|+\underset{0}{\overset{1}{\int }}|G_1\left(t_2,s\right)-G_1\left(t_1,s\right)|\left|h_n\left(s\right)\right|ds.\\ \le \left({\lambda }_3^2+2\left|{\lambda }_2{\lambda }_3\right|\right)\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}|G_1\left(t_2,\delta \right)-G_1\left(t_1,\delta \right)|\left|G_2\left(\delta ,\tau \right)\right|\left|G_3\left(\tau ,s\right)\right|\left|h_n\left(s\right)\right|dsd\tau d\delta .\\ +|{\lambda }_2+{\lambda }_3|\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}|G_1\left(t_2,\tau \right)-G_1\left(t_1,\tau \right)|\left|G_2\left(\tau ,s\right)\right|\left|h_n\left(s\right)\right|dsd\tau \\ +\underset{0}{\overset{1}{\int }}|G_1\left(t_2,s\right)-G_1\left(t_1,s\right)|\left|h_n\left(s\right)\right|ds.\\ \le \left(\left({\lambda }_3^2+2\left|{\lambda }_2{\lambda }_3\right|\right)\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}|G_1\left(t_2,\delta \right)-G_1\left(t_1,\delta \right)|G_2\left(\delta ,\tau \right)G_3\left(\tau ,s\right)dsd\tau d\delta .\right.\\ \left.+|{\lambda }_2+{\lambda }_3|\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}|G_1\left(t_2,\tau \right)-G_1\left(t_1,\tau \right)G_2\left(\tau ,s\right)dsd\tau +\underset{0}{\overset{1}{\int }}|G_1\left(t_2,s\right)-G_1\left(t_1,s\right)|ds\right)K_0.\end{array}$$

Because G _i (t, s)(i = 1, 2, 3) is uniform continuity on [0,1] × [0,1], based on the above demonstration, it is easy to proof that ${\left\{u_n^{\left(4\right)}\right\}}_{n=1}^{\infty }$ is equicontinuous on [0,1]. From (2.15), we know ||u||₀, ||u"||₀, ||u⁽⁴⁾||₀ ≤ ||u|| _X ≤ ϖ||h _n ||₀ ≤ ϖK₀, so $\left\{u_n\left(t\right)\right\},\left\{u_n^{″}\left(t\right)\right\}$ and $\left\{u_n^{\left(4\right)}\left(t\right)\right\}$ are relatively compact in R. Based on Lemma 1.2.7 in [11], we know ${\left\{u_n\right\}}_{n=1}^{\infty }$ is the relatively compact in X, so A is compact operator. □

The main tools of this article are the following well-known fixed point index theorems.

Let E be a Banach Space and K ⊂ E be a closed convex cone in E. Assume that Ω is a bounded open subset of E with boundary ∂Ω, and K ∩ Ω ≠ ∅. Let $A:K\cap \bar{\mathrm{\Omega }}\to K$ be a completely continuous mapping. If Au ≠u for every u ∈ K ∩ ∂Ω, then the fixed point index i(A, K ∩ Ω, K) is well defined. We have that if i(A, K ∩ Ω, K) ≠0, then A has a fixed point in K ∩ Ω.

Let K _r = {u ∈ K |||u|| <r} and ∂K _r = {u ∈ K |||u|| <r} for every r > 0.

Lemma 2.5. [12] Let A : K → K be a completely continuous mapping. If μAu ≠u for every u ∈ ∂K _r and 0 < μ ≤ 1, then i(A, K _r , K) = 1.

Lemma 2.6. [12] Let A : K → K be a completely continuous mapping. Suppose that the following two conditions are satisfied:

• (i) $\underset{u\in \partial K_r}{\text{inf}}\left|\right|Au\left|\right|>0,$

• (ii) μAu ≠u for every u ∈ ∂K _r and μ ≥ 1,

then i(A, K _r , K) = 0.

Lemma 2.7. [12] Let X be a Banach space, and let K ⊆ X be a cone in X. For p > 0, define $K_p=\left\{u\in K\left|\right|\left|u\right||<p\right\}$. Assume that A : K _p → K is a completely continuous mapping such that Au ≠u for every u ∈ ∂K _p = {u ∈ K|||u|| = p}.

• (i) If ||u|| ≤ ||Au||, for every u ∈ ∂K _p , then i(A, K _p , K) = 0.

• (ii) If ||u|| ≥ ||Au||, for every u ∈ ∂K _p , then i(A, K _p , K) = 1.

3 Main results

We bring in following notations in this section:

$$\begin{array}{c}{\underset{}{f}}_0=\underset{u\to 0_{+}}{\text{lim}}\text{inf}\underset{0\le t\le 1}{\text{min}}\left(f\left(t,u\right)/u\right),{\bar{f}}_{\infty }=\underset{u\to +\infty }{\text{lim}}\text{sup}\underset{0\le t\le 1}{\text{max}}\left(f\left(t,u\right)/u\right),\\ {\bar{f}}_0=\underset{u\to 0_{+}}{\text{lim}}\text{sup}\underset{0\le t\le 1}{\text{max}}\left(f\left(t,u\right)/u\right),{\underset{}{f}}_{\infty }=\underset{u\to +\infty }{\text{lim}}\text{inf}\underset{0\le t\le 1}{\text{min}}\left(f\left(t,u\right)/u\right).\\ a\left(t\right)=A\left(t\right)-\alpha ,b\left(t\right)=B\left(t\right)-\beta ,c\left(t\right)=C\left(t\right)-\gamma ,\\ \Gamma ={\pi }^6-\gamma {\pi }^4-\beta {\pi }^2-\alpha ,K=\underset{0\le t\le 1}{\text{max}}\left[a\left(t\right)+b\left(t\right)+c\left(t\right)\right],\end{array}$$

Suppose that:

(H₃) L = ϖK < 1, where ϖ is defined as in (2.14).

Theorem 3.1. Assume that (H₁)-(H₃) hold, and $b\left(t\right)\ge \left({\lambda }_2+{\lambda }_3\right)c\left(t\right),{\lambda }_{3}b\left(t\right)-a\left(t\right)\le {\lambda }_3^{2}c\left(t\right),$ then in each of the following cases:

• (i) ${\underset{}{f}}_0>\Gamma ,{\bar{f}}_{\infty }<\left(1-L\right)\Gamma ,$ (ii) ${\bar{f}}_0<\left(1-L\right)\Gamma ,{\underset{}{f}}_{\infty }>\Gamma ,$the BVP (1.1) has at least one positive solution.

Proof. ∀h ∈ Y, consider the LBVP

$$\left\{\begin{array}{c}\hfill -u^{\left(6\right)}-C\left(t\right)u^{\left(4\right)}+B\left(t\right)u^{″}-A\left(t\right)u=h\left(t\right),\hfill \\ \hfill u\left(0\right)=u\left(1\right)=u^{″}\left(0\right)=u^{″}\left(1\right)=u^{\left(4\right)}\left(0\right)=u^{\left(4\right)}\left(1\right)=0,\hfill \end{array}\right.0<t<1$$

(3.1)

It is easy to prove (3.1) is equivalent to the following BVP

$$\left\{\begin{array}{c}\hfill -u^{\left(6\right)}-\gamma u^{\left(4\right)}+\beta u^{″}-\alpha =Gu+h\left(t\right),\hfill \\ \hfill u\left(0\right)=u\left(1\right)=u^{″}\left(0\right)=u^{″}\left(1\right)=u^{\left(4\right)}\left(0\right)=u^{\left(4\right)}\left(1\right)=0,\hfill \end{array}\right.0<t<1$$

(3.2)

where Gv := (C(t) - γ)v⁽⁴⁾ - (B(t)- β)v" + (A(t) - α)v, ∀v ∈ X. Obviously, the operator G : X → Y is linear, and ∀v ∈ X, t ∈ [0,1], we have |Gv(t)| ≤ K ||v|| _X . Hence ||Gv||₀ ≤ K ||v|| _X , and so ||G|| ≤ K. On the other hand, u ∈ C⁴[0,1]⋂C⁶(0,1), t ∈ [0,1] is a solution of (3.2) iff u ∈ X satisfies u = A(Gu + h), i.e.,

$$u\in X,\left(I-AG\right)u=Ah.$$

(3.3)

Owing to G : X → Y and A : Y → X, the operator I - AG maps X into Y. From A ≤ ϖ (by Lemma 2.4) together with ||G|| ≤ K and condition (H₃), applying operator spectral theorem, we have that the operator (I - AG)^-1 exists and is bounded. Let H = (I - AG)^-1A, then (3.3) is equivalent to u = Hh. By the Neumann expansion formula, H can be expressed by

$$H=\left(I+AG+\cdots +{\left(AG\right)}^n+\cdots \right)A=A+\left(AG\right)A+\cdots +{\left(AG\right)}^{n}A+\cdots .$$

(3.4)

The complete continuity of A with the continuity of (I - AG)^-1 yields that the operator H : Y → X is completely continuous. If we restrict H : Y₊ → Y, ∀h ∈ Y₊ and mark u = Ah, then u ∈ X ∩ Y₊. Based on equation (2.8), (2.11) and Lemma 2.4, we have

$$\begin{array}{c}{u}^{″}={\lambda }_{3}u-\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{G}_1\left(t,\tau \right)G_2\left(\tau ,s\right)h\left(s\right)dsd\tau \le {\lambda }_{3}u,t\in \left[0,1\right],\\ u^{\left(4\right)}=\left({\lambda }_2+{\lambda }_3\right)u^{″}-{\lambda }_2{\lambda }_{3}u+\underset{0}{\overset{1}{\int }}{G}_1\left(t,s\right)h\left(s\right)ds\ge \left({\lambda }_2+{\lambda }_3\right)u^{″}-{\lambda }_2{\lambda }_{3}u,t\in \left[0,1\right],\end{array}$$

by b(t) ≥ (λ₂ + λ₃)c(t) and${\lambda }_{3}b\left(t\right)-a\left(t\right)\le {\lambda }_3^{2}c\left(t\right)$, we have

$$\begin{array}{ll}(Gu)(t)\hfill & =c(t)u^{(4)}-b(t)u\text{'}\text{'}+a(t)u\hfill \\ \ge \left[({\lambda }_2+{\lambda }_3)c(t)-b(t)\right]u\text{'}\text{'}-\left[{\lambda }_2{\lambda }_{3}c(t)-a(t)\right]u\hfill \\ \ge {\lambda }_3\left[({\lambda }_2+{\lambda }_3)c(t)-b(t)\right]u-\left[{\lambda }_2{\lambda }_{3}c(t)-a(t)\right]u\hfill \\ \ge \left[{\lambda }_3^{2}c(t)-{\lambda }_{3}b(t)+a(t)\right]u\ge 0,t\in [0,1].\hfill \end{array}$$

Hence

$$\forall h\in Y_{+},\left(GAh\right)\left(t\right)\ge 0,\forall t\in \left[0,\text{ 1}\right],$$

(3.5)

and so (AG)(Ah)(t) = A(GAh)(t) ≥ 0, ∀t ∈ [0,1]. Suppose that ∀h ∈ Y₊, (AG) ^k (Ah)(t) ≥ 0, ∀t ∈ [0,1]. For any h ∈ Y₊, let h₁ = GAh, by (3.5) we have h₁ ∈ Y₊, and so

$$\begin{array}{c}\hfill {\left(AG\right)}^{k+1}\left(Ah\right)\left(t\right)={\left(AG\right)}^k\left(AGAh\right)\left(t\right)={\left(AG\right)}^k\left(A{h}_1\right)\left(t\right)\ge 0,\forall t\in \left[0,1\right].\hfill \end{array}$$

Thus by induction it follows that ∀n ≥ 1, ∀h ∈ Y₊, (AG) ⁿ (Ah)(t) ≥ 0, ∀t ∈ [0,1]. By (3.4), we have

$$\begin{array}{ll}\hfill \forall h\in Y_{+},\left(Hh\right)\left(t\right)& =\left(Ah\right)\left(t\right)+\left(AG\right)\left(Ah\right)\left(t\right)+\cdots +{\left(AG\right)}^n\left(Ah\right)\left(t\right)+\cdots \\ \ge \left(Ah\right)\left(t\right),\forall t\in \left[0,1\right].\end{array}$$

(3.6)

So H : Y₊ → Y₊ ∩ X.

On the other hand, we have

$$\begin{array}{ll}\hfill \forall h\in Y_{+},\left(Hh\right)\left(t\right)& \le \left(Ah\right)\left(t\right)+\left|\right|\left(AG\right)\left|\right|\left(Ah\right)\left(t\right)+\cdots +\left|\right|{\left(AG\right)}^n\left|\right|\left(Ah\right)\left(t\right)+\cdots \\ \le \left(1+L+\cdots +L^n+\cdots \right)\left(Ah\right)\left(t\right)\\ \le \frac{1}{1-L}\left(Ah\right)\left(t\right),\forall t\in \left[0,1\right].\end{array}$$

(3.7)

So the following inequalities hold

$$\begin{array}{cc}\hfill \left|\right|Hh|{|}_0\hfill & \hfill \le \frac{1}{1-L}\left|\right|Ah|{|}_0,\forall t\in \left[0,1\right].\hfill \end{array}$$

(3.8)

For any u ∈ Y₊, define Fu = f(t, u). Based on condition (H₁), it is easy to show F : Y₊ → Y₊ is continuous. By (3.1)-(3.3), It is easy to see that u ∈ C⁴[0,1] ∩ C⁶(0, 1) is a positive solution of BVP (1.1) iff u ∈ Y₊ is a nonzero solution of an operator equation as follows

$$\begin{array}{c}\hfill u=HFu.\hfill \end{array}$$

(3.9)

Let Q = HF. Obviously, Q : Y₊ → Y₊ is completely continuous. We next show that the operator Q has at least one nonzero fixed point in Y₊.

Let

$$\begin{array}{c}\hfill P=\left\{u\in Y_{+}|u\left(t\right)\ge \sigma |\left|u\right|{|}_0,\forall t\in \left[0,1\right]\right\}.\hfill \end{array}$$

In which

$$\sigma =\frac{{\delta }_1{\delta }_2{\delta }_3{C}_{12}{C}_{23}}{C_1{C}_2{C}_3{M}_1{M}_2}\left(1-L\right)G_1\left(t,t\right).$$

(3.10)

Here M₁ and M₂ can be defined as that in (2.1), C₁₂ and C₂₃ can be defined as that in (2.2), C _i ,δ _i (i = 1, 2, 3) can be defined as that in Lemma 2.3. It is easy to prove that P is a cone in Y. We will prove QP ⊂ P next.

For any u ∈ P, let h = Fu, then h ∈ Y₊. By (3.6) and Lemma 2.3, we have

$$\left(Qu\right)\left(t\right)=\left(HFu\right)\left(t\right)\ge \left(AFu\right)\left(t\right),\forall t\in \left[0,1\right]\}.$$

By Lemma 2.3, for all u ∈ P, we have

$$\begin{array}{c}\left(AFu\right)\left(t\right)=\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{G}_1\left(t,\delta \right)G_2\left(\delta ,\tau \right)G_3\left(\tau ,s\right)\left(Fu\right)\left(s\right)dsd\tau d\delta \\ \le C_1{C}_2{C}_3{M}_1{M}_2\underset{0}{\overset{1}{\int }}{G}_3\left(s,s\right)\left(Fu\right)\left(s\right)ds.\end{array}$$

And accordingly we have$\left|\right|AFu|{|}_0\le C_1{C}_2{C}_3{M}_1{M}_2{\int }_0^1{G}_3\left(s,s\right)\left(Fu\right)\left(s\right)ds$, that is

$${\int }_0^1{G}_3\left(s,s\right)\left(Fu\right)\left(s\right)ds\ge \frac{\left|\right|AFu|{|}_0}{C_1{C}_2{C}_3{M}_1{M}_2}.$$

(3.11)

By using (c3) in Lemma 2.3, (3.8) and (3.11), we have

$$\begin{array}{ll}\hfill \left(AFu\right)\left(t\right)& \ge {\delta }_1{\delta }_2{\delta }_3{C}_{12}{C}_{23}{G}_1\left(t,t\right)\underset{0}{\overset{1}{\int }}{G}_3\left(s,s\right)\left(Fu\right)\left(s\right)ds\\ \ge \frac{{\delta }_1{\delta }_2{\delta }_3{C}_{12}{C}_{23}{G}_1\left(t,t\right)}{C_1{C}_2{C}_3{M}_1{M}_2}\left|\right|AFu|{|}_0\\ \ge \frac{{\delta }_1{\delta }_2{\delta }_3{C}_{12}{C}_{23}{G}_1\left(t,t\right)}{C_1{C}_2{C}_3{M}_1{M}_2}\left(1-L\right)\left|\right|HFu|{|}_0\\ \ge \frac{{\delta }_1{\delta }_2{\delta }_3{C}_{12}{C}_{23}{G}_1\left(t,t\right)}{C_1{C}_2{C}_3{M}_1{M}_2}\left(1-L\right)\left|\right|Qu|{|}_0.\end{array}$$

So$\left(Qu\right)\left(t\right)\ge \frac{{\delta }_1{\delta }_2{\delta }_3{C}_{12}{C}_{23}{G}_1\left(t,t\right)}{C_1{C}_2{C}_3{M}_1{M}_2}\left(1-L\right)\left|\right|Qu|{|}_0=\sigma |\left|Qu\right|{|}_0$. Thus QP ⊂ P.

Let

$$\rho =\frac{{\delta }_1{\delta }_2{\delta }_3{C}_{12}{C}_{23}{m}_1\left(1-L\right)}{C_1{C}_2{C}_3{M}_1{M}_2},$$

(3.12)

in which m₁ can be defined as that in (2.1). It's easy to prove

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \forall u\in P\Rightarrow u\left(t\right)\ge \rho \left|\right|u|{|}_0,\forall t\in \left[\frac{1}{4},\frac{3}{4}\right].\hfill \end{array}\hfill \end{array}$$

(3.13)

Case (i), since ${\underset{}{f}}_0>\Gamma$, there exist ε > 0 and r₀> 0 such that f(t, x) ≥ (Γ + ε)x, 0 ≤ t ≤ 1, 0 < × ≤ r₀. Let r ∈ (0, r₀) and Ω _r = {u ∈ P | ||u||₀ ≤ r}, then for every u ∈ ∂Ω _r , we have ||u||₀ = r, 0 < u(t) ≤ r, t ∈ (0, 1), and so f(t, u(t)) ≥ (Γ + ε)u(t), t ∈ (0,1). By (3.13), it follows that

$$\begin{array}{c}\hfill f\left(t,u\left(t\right)\right)>\left(\Gamma +\epsilon \right)u\left(t\right)\ge \left(\Gamma +\epsilon \right)\rho r,\forall t\in \left[\frac{1}{4},\frac{3}{4}\right].\hfill \end{array}$$

(3.14)

From (3.6) and (3.14), we have

$$\begin{array}{ll}\hfill \left|\right|Qu|{|}_0& \ge Qu\left(\frac{1}{2}\right)=\left(HFu\right)\left(\frac{1}{2}\right)\ge \left(AFu\right)\left(\frac{1}{2}\right)\\ =\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{G}_1\left(\frac{1}{2},\delta \right)G_2\left(\delta ,\tau \right)G_3\left(\tau ,s\right)f\left(s,u\left(s\right)\right)dsd\tau d\delta \\ \ge \underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}\underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}\underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}{G}_1\left(\frac{1}{2},\delta \right)G_2\left(\delta ,\tau \right)G_3\left(\tau ,s\right)\left(\Gamma +\epsilon \right)\rho rdsd\tau d\delta \\ \ge {\delta }_1{\delta }_2{\delta }_3{m}_1{C}_{12}{C}_{23}\left(\Gamma +\epsilon \right)\rho r\underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}{G}_3\left(s,s\right)ds.\\ \ge \frac{1}{2}{\delta }_1{\delta }_2{\delta }_3{m}_1{m}_3{C}_{12}{C}_{23}\left(\Gamma +\epsilon \right)\rho r>0.\end{array}$$

Therefore, $\underset{u\in \partial {\mathrm{\Omega }}_r}{\text{inf}}{∥Qu∥}_0>0$. Now we shall prove ∀u ∈ ∂Ω _r , μ ≥ 1, μQu ≠ u. In fact, suppose the contrary, then there exist $u_0\in \partial {\mathrm{\Omega }}_r$, and μ₀ ≥ 1 such that μ₀Qu₀ = u₀. By (3.6), we have $u_0\left(t\right)\ge \frac{1}{{\mu }_0}{u}_0\left(t\right)=Q{u}_0\left(t\right)\ge \left(AF{u}_0\right)\left(t\right)$. Let ω₀ = AFu₀, then u₀ ≥ ω₀ and ω₀(t) satisfies BVP (2.4) with h = Fu₀. Hence

$$\left\{\begin{array}{c}\hfill -{\omega }_0^{\left(6\right)}-\gamma {\omega }_0^{\left(4\right)}+\beta {{\omega }^{″}}_0-\alpha {\omega }_0=f\left(t,u_0\right),t\in \left(0,1\right),\hfill \\ \hfill {\omega }_0\left(0\right)={\omega }_0\left(1\right)={{\omega }^{″}}_0\left(0\right)={{\omega }^{″}}_0\left(1\right)={\omega }_0^{\left(4\right)}\left(0\right)={\omega }_0^{\left(4\right)}\left(1\right)=0,\hfill \end{array}\right.$$

(3.15)

After multiplying the two sides of the first equation in (3.15) by sin &#960t and integrating on [0,1], we have

$$\Gamma \underset{0}{\overset{1}{\int }}{\omega }_0\left(t\right)\text{sin}\pi tdt=\underset{0}{\overset{1}{\int }}f\left(t,u_0\left(t\right)\right)\text{sin}\pi tdt,$$

then

$$\begin{array}{ll}\hfill \left(\Gamma +\epsilon \right)\underset{0}{\overset{1}{\int }}{u}_0\left(t\right)\text{sin}\pi tdt& \le \underset{0}{\overset{1}{\int }}f\left(t,u_0\left(t\right)\right)\text{sin}\pi tdt\\ =\Gamma \underset{0}{\overset{1}{\int }}{\omega }_0\left(t\right)\text{sin}\pi tdt\le \Gamma \underset{0}{\overset{1}{\int }}{u}_0\left(t\right)\text{sin}\pi tdt.\end{array}$$

(3.16)

Since $u_0\left(t\right)\ge \rho {∥u_0∥}_0=\rho r,\forall t\in \left[\frac{1}{4},\frac{3}{4}\right]$, so ${\int }_0^1{u}_0\left(t\right)\text{sin}\pi tdt>0$ and we see that Γ + ε < Γ, which is a contradiction. Then based on Lemma 2.6, we come to

$$i\left(Q,{\mathrm{\Omega }}_r,P\right)=0.$$

(3.17)

On the other hand, since ${\bar{f}}_{\infty }<\left(1-L\right)\Gamma$, there exist ε ∈ (0, (1 - L)Γ) and R₀> 0 such that f(t, x) ≤ [(1-L)Γ - ε] x, 0 ≤ t ≤ 1, x >R₀. Let $M_{R_0}=\underset{0\le t\le 1,0\le x\le R_0}{\text{sup}}f\left(t,x\right)$. Then

$$f\left(t,x\right)<\left[\left(1-L\right)\Gamma -\epsilon \right]x+M_{R_0},0\le t\le 1,x\ge 0.$$

We choose $R>\text{max}\left\{R_0,r,\frac{\sqrt{2}{M}_{R_0}}{\rho \epsilon }\right\}$ and let ${\mathrm{\Omega }}_R=\left\{u\in P\left|\right|\left|u\right|{|}_0<R\right\}.$ Next we prove ∀u ∈ ∂Ω _R , μ ≥ 1, μu ≠ Qu. Assume on the contrary that ∃μ₀ ≥ 1, u₀ ∈ ∂Ω _R , such that μ₀u₀ = Qu₀. Let ω₁ = AFu₀, by (3.6), we have $u_0\le {\mu }_0{u}_0=Q{u}_0\le \frac{1}{1-L}AF{u}_0\le \frac{1}{1-L}{\omega }_1$ and ω₁(t) satisfies BVP (2.4) with h = Fu₀. Similarly to (3.16), we can prove

$$\begin{array}{ll}\hfill \left(1-L\right)\Gamma \underset{0}{\overset{1}{\int }}{u}_0\left(t\right)\text{sin}\pi tdt& \le \Gamma \underset{0}{\overset{1}{\int }}{\omega }_1\left(t\right)\text{sin}\pi tdt=\underset{0}{\overset{1}{\int }}f\left(t,u_0\left(t\right)\right)\text{sin}\pi tdt\\ \le \left[\left(1-L\right)\Gamma -\epsilon \right]\underset{0}{\overset{1}{\int }}{u}_0\left(t\right)\text{sin}\pi tdt+M_{R_0}\underset{0}{\overset{1}{\int }}\text{sin}\pi tdt,\end{array}$$

(3.18)

and so

$$\begin{array}{ll}\hfill M_{R_0}\underset{0}{\overset{1}{\int }}\text{sin}\pi tdt& \ge \epsilon \underset{0}{\overset{1}{\int }}{u}_0\left(t\right)\text{sin}\pi tdt\ge \epsilon \underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}{u}_0\left(t\right)\text{sin}\pi tdt\\ \ge \rho \epsilon \left|\right|u_0|{|}_0\underset{0}{\overset{1}{\int }}\text{sin}\pi tdt,\end{array}$$

(3.19)

Thus, by (3.19), we have $R=\left|\right|u_0|{|}_0\le \frac{\sqrt{2}{M}_{R_0}}{\rho \epsilon }$ which is contradictory with $R>\frac{\sqrt{2}{M}_{R_0}}{\rho \epsilon }.$

Then by Lemma 2.5 we know

$$i\left(Q,{\mathrm{\Omega }}_R,P\right)=1.$$

(3.20)

Now, by the additivity of fixed point index, combine (3.17) and (3.20) to conclude that

$$i\left(Q,{\mathrm{\Omega }}_R\backslash {\bar{\mathrm{\Omega }}}_r,P\right)=i\left(Q,{\mathrm{\Omega }}_R,P\right)-i\left(Q,{\mathrm{\Omega }}_r,P\right)=1.$$

Therefore Q has a fixed point in ${\mathrm{\Omega }}_R\backslash {\bar{\mathrm{\Omega }}}_r,$ which is the positive solution of BVP (1.1).

Case (ii), since ${\bar{f}}_0<\left(1-L\right)\Gamma ,$ based on the definition of ${\bar{f}}_0,$ we may choose ε > 0 and ω > 0, so that

$$f\left(t,u\right)\le \left[\left(1-L\right)\Gamma -\epsilon \right]u,0\le t\le 1,0\le u\le \omega .$$

(3.21)

Let r ∈ (0, ω), we now prove that μQu ≠ u for every u ∈ ∂Ω _r , and 0 < μ ≤ 1. In fact, suppose the contrary, then there exist u₀ ∈ ∂Ω _r , and 0 < μ₀ ≤ 1 such that μ₀Qu₀ = u₀. Let ω₂ = AFu₀, by (3.6), we have $u_0={\mu }_{0}Q{u}_0\le \frac{1}{1-L}AF{u}_0\le \frac{1}{1-L}{\omega }_2$ and ω₂(t) satisfies BVP (2.4) with h = Fu₀. Similarly to (3.18), we have

$$\begin{array}{ll}\hfill \left(1-L\right)\Gamma \underset{0}{\overset{1}{\int }}{u}_0\left(t\right)\text{sin}\pi tdt& \le \Gamma \underset{0}{\overset{1}{\int }}{\omega }_2\left(t\right)\text{sin}\pi tdt=\underset{0}{\overset{1}{\int }}f\left(t,u_0\left(t\right)\right)\text{sin}\pi tdt\\ \le \left[\left(1-L\right)\Gamma -\epsilon \right]\underset{0}{\overset{1}{\int }}{u}_0\left(t\right)\text{sin}\pi tdt.\end{array}$$

(3.22)

Since ${\int }_0^1{u}_0\left(t\right)\text{sin}\pi tdt>0,$ We see that (1 - L)Γ ≤ (1 - L)Γ - ε, which is a contradiction. By Lemma 2.5, we have

$$i\left(Q,{\mathrm{\Omega }}_r,P\right)=1.$$

(3.23)

On the other hand, because ${\underset{}{f}}_{\infty }>\Gamma ,$ there exist ε ∈ (0, Γ) and H > 0 such that

$$f\left(t,x\right)\ge \left(\Gamma +\epsilon \right)x,t\in \left[0,1\right],x>H.$$

(3.24)

Let C = max_{0≤t≤1,0≤x≤H}|f(t,x) - (Γ + ε)x| + 1, then it is clear that

$$f\left(t,x\right)\ge \left(\Gamma +\epsilon \right)x-C,t\in \left[0,1\right],x\ge 0.$$

(3.25)

Choose R > R₀ = max {H/ρ, ω}, ∀u ∈ ∂Ω _r . By (3.13) and (3.25), we have

$$u\left(s\right)\ge \rho \left|\right|u|{|}_0>H,\forall s\in \left[\frac{1}{4},\frac{3}{4}\right].$$

And so

$$f\left(s,u\left(s\right)\right)\ge \left(\Gamma +\epsilon \right)u\left(s\right)\ge \left(\Gamma +\epsilon \right)\rho \left|\right|u|{|}_0,\forall s\in \left[\frac{1}{4},\frac{3}{4}\right].$$

(3.26)

From (3.6) and (3.26), we get

$$\begin{array}{ll}\hfill \left|\right|Qu|{|}_0\ge Qu\left(\frac{1}{2}\right)& =\left(HFu\right)\left(\frac{1}{2}\right)\ge \left(AFu\right)\left(\frac{1}{2}\right)\\ =\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{G}_1\left(\frac{1}{2},\delta \right)G_2\left(\delta ,\tau \right)G_3\left(\tau ,s\right)f\left(s,u\left(s\right)\right)dsd\tau d\delta \\ \ge \underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}\underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}\underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}{G}_1\left(\frac{1}{2},\delta \right)G_2\left(\delta ,\tau \right)G_3\left(\tau ,s\right)\left(\Gamma +\epsilon \right)\rho \left|\right|u|{|}_{0}dsd\tau d\delta \\ \ge {\delta }_1{\delta }_2{\delta }_3{m}_1{C}_{12}{C}_{23}\left(\Gamma +\epsilon \right)\rho \left|\right|u|{|}_0\underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}{G}_3\left(s,s\right)ds.\\ \ge \frac{1}{2}{\delta }_1{\delta }_2{\delta }_3{m}_1{m}_3{C}_{12}{C}_{23}\left(\Gamma +\epsilon \right)\rho \left|\right|u|{|}_0>0,\end{array}$$

from which we see that $\underset{u\in \partial {\mathrm{\Omega }}_r}{\text{inf}}\left|\right|Qu|{|}_0>0,$ namely the hypotheses (i) of Lemma 2.6 holds. Next, we show that if R is large enough, then μQu ≠u for any u ∈ ∂Ω _R and μ ≥ 1. In fact, suppose the contrary, then there exist u₀ ∈ ∂Ω _R and μ₀ ≥ 1 such that μ₀Qu₀ = u₀, then by (3.6), $AF{u}_0\le Q{u}_0\le u_0={\mu }_{0}Q{u}_0\le \frac{{\mu }_0}{1-L}AF{u}_0.$Let ω₀ = AFu₀, then ${\omega }_0\le u_0\le \frac{{\mu }_0}{1-L}{\omega }_0,$ and ω₀ satisfies BVP (2.4), in which h = Fu₀, consequently,

$$\left\{\begin{array}{c}\hfill -{\omega }_0^{\left(6\right)}-\gamma {\omega }_0^{\left(4\right)}+\beta {{\omega }^{″}}_0-\alpha {\omega }_0=f\left(t,u_0\right),t\in \left[0,1\right],\hfill \\ \hfill {\omega }_0\left(0\right)={\omega }_0\left(1\right)={{\omega }^{″}}_0\left(0\right)={{\omega }^{″}}_0\left(1\right)={\omega }_0^{\left(4\right)}\left(0\right)={\omega }_0^{\left(4\right)}\left(1\right)=0,\hfill \end{array}\right.$$

(3.27)

After multiplying the two sides of the first equation in (3.27) by sin πt and integrating on [0,1], we have

$$\begin{array}{ll}\hfill \Gamma \underset{0}{\overset{1}{\int }}{\omega }_0\left(t\right)\text{sin}\pi tdt& =\underset{0}{\overset{1}{\int }}f\left(t,u_0\left(t\right)\right)\text{sin}\pi tdt\ge \left(\Gamma +\epsilon \right)\underset{0}{\overset{1}{\int }}{u}_0\left(t\right)\text{sin}\pi tdt-\frac{2C}{\pi }\\ \ge \left(\Gamma +\epsilon \right)\underset{0}{\overset{1}{\int }}{\omega }_0\left(t\right)\text{sin}\pi tdt-\frac{2C}{\pi }.\end{array}$$

Consequently, we obtain that

$$\begin{array}{c}\hfill \underset{0}{\overset{1}{\int }}{\omega }_0\left(t\right)\text{sin}\pi tdt\le \frac{2C}{\pi \epsilon }.\hfill \end{array}$$

(3.28)

It's easy to prove that ω₀(t), the solution of LBVF (3.27) satisfies

$${\omega }_0\left(t\right)\ge \frac{{\delta }_1{\delta }_2{\delta }_3{C}_{12}{C}_{23}}{C_1{C}_2{C}_3{M}_1{M}_2}{G}_1\left(t,t\right)\left|\right|{\omega }_0|{|}_0,$$

and accordingly,

$$\underset{0}{\overset{1}{\int }}{\omega }_0\left(t\right)\text{sin}\pi tdt\ge \frac{{\delta }_1{\delta }_2{\delta }_3{C}_{12}{C}_{23}\left|\right|{\omega }_0|{|}_0}{C_1{C}_2{C}_3{M}_1{M}_2}\underset{0}{\overset{1}{\int }}{G}_1\left(t,t\right)\text{sin}\pi tdt,$$

(3.29)

by (3.28), we get

$$\left|\right|{\omega }_0|{|}_0\le \frac{2C{C}_1{C}_2{C}_3{M}_1{M}_2}{{\delta }_1{\delta }_2{\delta }_3{C}_{12}{C}_{23}\pi \epsilon }{\left(\underset{0}{\overset{1}{\int }}{G}_1\left(t,t\right)\text{sin}\pi tdt\right)}^{-1}:=\bar{R},$$

(3.30)

Consequently, $\left|\right|u_0|{|}_0\le \frac{{\mu }_0}{1-L}|\left|{\omega }_0\right|{|}_0\le \frac{{\mu }_0}{1-L}\bar{R}.$

We choose $R>\text{max}\left\{\frac{{\mu }_0}{1-L}\bar{R},R_0\right\},$then to any u ∈ ∂ Ω _r , μ ≥ 1, there is always μQu ≠u. Hence, hypothesis (ii) of Lemma 2.6 also holds. By Lemma 2.6, we have

$$i\left(Q,{\mathrm{\Omega }}_R,P\right)=0.$$

(3.31)

Now, by the additivity of fixed point index, combine (3.23) and (3.31) to conclude that

$$i\left(Q,{\mathrm{\Omega }}_R\backslash {\bar{\mathrm{\Omega }}}_r,P\right)=i\left(Q,{\mathrm{\Omega }}_R,P\right)-i\left(Q,{\mathrm{\Omega }}_r,P\right)=-1.$$

Therefore, Q has a fixed poind in ${\mathrm{\Omega }}_R\backslash {\bar{\mathrm{\Omega }}}_r,$ which is the positive solution of BVP (1.1). The proof is completed. □

From Theorem 3.1, we immediately obtain the following.

Corollary 3.1. Assume (H₁)-(H₃) hold, then in each of the following cases:

• (i) ${\underset{}{f}}_0=\infty ,{\bar{f}}_{\infty }=0,$ (ii) ${\bar{f}}_0=0,{\underset{}{f}}_{\infty }=\infty ,$

the BVP (1.1) has at least one positive solution.

4 Multiple solutions

Next, we study the multiplicity of positive solutions of BVP (1.1) and assume in this section that

(H₄) there is a p > 0 such that 0 ≤ u ≤ p and 0 ≤ t ≤ 1 imply f(t, u) < ηp, where $\eta ={\left(\frac{C_1{C}_2{C}_3{M}_1{M}_2}{1-L}{\int }_0^1{G}_1\left(s,s\right)ds\right)}^{-1}.$

(H₅) there is a p > 0 such that σp ≤ u ≤ p and 0 ≤ t ≤ 1 imply f (t, u) ≥ λp, where${\lambda }^{-1}={\delta }_1{\delta }_2{\delta }_3{m}_1{C}_{12}{C}_{23}{\int }_{\frac{1}{4}}^{\frac{3}{4}}{G}_3\left(s,s\right)ds$. Here, σ can be defined as (3.10).

Theorem 4.1. Assume (H₁)-(H₄) hold. If ${\underset{}{f}}_0>\Gamma$ and${\underset{}{f}}_{\infty }>\Gamma$, then BVP (1.1) has at least two positive solution u₁ and u₂ such that 0 ≤ ||u₁||₀ ≤ p ≤ ||u₂||₀.

Proof. According to the proof of Theorem 3.1, there exists 0 < r₀< p < R₁<+∞, such that 0 < r < r₀ implies i(Q, Ω _r , P) = 0 and R ≥ R₁ implies i(Q, Ω _r , P) = 0.

Next we prove i(Q, Ω _r , P) = 1 if (H₄) is satisfied. In fact, for every u ∈ ∂Ω _r , based on the preceding definition of Q we come to

$$\begin{array}{ll}\hfill \left(Qu\right)\left(t\right)& =\left(HFu\right)\left(t\right)\le \frac{1}{1-L}\left|\right|AFu|{|}_0\\ =\frac{1}{1-L}\underset{0\le t\le 1}{\text{max}}\left|\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{G}_1\left(t,\delta \right)G_2\left(\delta ,\tau \right)G_3\left(\tau ,s\right)\left(Fu\right)\left(s\right)dsd\tau d\delta \right|\\ \le \frac{C_1{C}_2{C}_3{M}_1{M}_2}{1-L}\left|\underset{0}{\overset{1}{\int }}{G}_3\left(s,s\right)f(s,u\left(s\right)ds\right|.\end{array}$$

Consequently,

$$\begin{array}{ll}\hfill \left|\right|Qu|{|}_0& \le \frac{C_1{C}_2{C}_3{M}_1{M}_2}{1-L}\left|\underset{0}{\overset{1}{\int }}{G}_3\left(s,s\right)f\left(s,u\left(s\right)\right)ds\right|\\ \le \frac{C_1{C}_2{C}_3{M}_1{M}_2}{1-L}\underset{0}{\overset{1}{\int }}{G}_3\left(s,s\right)\eta pds=p=\left|\right|u|{|}_0.\end{array}$$

Therefore, by (ii) of Lemma 2.7 we have

$$i\left(Q,{\mathrm{\Omega }}_p,P\right)=1.$$

(4.1)

Combined with (3.17), (3.31), and (4.1), we have

$$\begin{array}{c}i\left(Q,{\mathrm{\Omega }}_R\backslash {\bar{\mathrm{\Omega }}}_p,P\right)=i\left(Q,{\mathrm{\Omega }}_R,P\right)-i\left(Q,{\mathrm{\Omega }}_p,P\right)=-1.\\ i\left(Q,{\mathrm{\Omega }}_p\backslash {\bar{\mathrm{\Omega }}}_r,P\right)=i\left(Q,{\mathrm{\Omega }}_p,P\right)-i\left(Q,{\mathrm{\Omega }}_r,P\right)=1.\end{array}$$

Therefore, Q has fixed points u₁ and u₂ in ${\mathrm{\Omega }}_p\backslash {\bar{\mathrm{\Omega }}}_r$ and ${\mathrm{\Omega }}_R\backslash {\bar{\mathrm{\Omega }}}_p,$ respectively, which

means that u₁(t) and u₂(t) are positive solutions of BVP (1.1) and 0 ≤ ||u₁||₀ ≤ p ≤ ||u₂||₀. The proof is completed. □

Theorem 4.2. Assume (H₁)-(H₃) and (H₅) can be established, and ${\bar{f}}_0<\left(1-L\right)\Gamma$ and ${\bar{f}}_{\infty }<\left(1-L\right)\Gamma$, then BVP (1.1) has at least two positive solution u₁ and u₂ such that 0 ≤ ||u₁||₀ ≤ p ≤ ||u₂||₀.

Proof. According to the proof of Theorem 3.1, there exists 0 < ω < p < R₂<+ ∞, such that 0 < r < ω implies i(Q, Ω _r , P) = 1 and R ≥ R₂ implies i(Q, Ω _r , P) = 1.

We now prove that i(Q, Ω _r , P) = 0 if (H₅) is satisfied. In fact, for every u ∈ ∂ Ω _r , by (3.13) we come to ρp ≤ ρ||u||₀ ≤ u(t) ≤ ||u||₀ = p, t ∈ [1/4, 3/4], accordingly, by (H₅), we have

$$f\left(t,u\right)\ge \lambda p,t\in \left[\frac{1}{4},\frac{3}{4}\right],\forall u\in \partial {\mathrm{\Omega }}_p.$$

from the proof of (ii) of Theorem 3.1, we have

$$\begin{array}{ll}\hfill \left|\right|Qu|{|}_0& \ge Qu\left(\frac{1}{2}\right)=\left(HFu\right)\left(\frac{1}{2}\right)\ge \left(AFu\right)\left(\frac{1}{2}\right)\\ =\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{G}_1\left(\frac{1}{2},\delta \right)G_2\left(\delta ,\tau \right)G_3\left(\tau ,s\right)f\left(s,u\left(s\right)\right)dsd\tau d\delta \\ \ge \underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}\underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}\underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}{G}_1\left(\frac{1}{2},\delta \right)G_2\left(\delta ,\tau \right)G_3\left(\tau ,s\right)\lambda pdsd\tau d\delta \\ \ge {\delta }_1{\delta }_2{\delta }_3{m}_1{C}_{12}{C}_{23}\underset{\frac{1}{4}}{\overset{\frac{3}{4}}{\int }}{G}_3\left(s,s\right)\lambda pds=p=\left|\right|u|{|}_0.\end{array}$$

Therefore, $\left|\right|Qu|{|}_0\ge Qu\left(\frac{1}{2}\right)\ge |\left|u\right|{|}_0,$according to (i) of Lemma 2.7, we come to

$$i\left(Q,{\mathrm{\Omega }}_p,P\right)=0.$$

(4.2)

Combined with (3.20), (3.23), and (4.2), there exist

$$\begin{array}{c}i\left(Q,{\mathrm{\Omega }}_R\backslash {\bar{\mathrm{\Omega }}}_p,P\right)=i\left(Q,{\mathrm{\Omega }}_R,P\right)-i\left(Q,{\mathrm{\Omega }}_p,P\right)=1.\\ i\left(Q,{\mathrm{\Omega }}_p\backslash {\bar{\mathrm{\Omega }}}_r,P\right)=i\left(Q,{\mathrm{\Omega }}_p,P\right)-i\left(Q,{\mathrm{\Omega }}_r,P\right)=-1.\end{array}$$

Therefore, Q has fixed points u₁ and u₂ in ${\mathrm{\Omega }}_p\backslash {\bar{\mathrm{\Omega }}}_r$ and ${\mathrm{\Omega }}_R\backslash {\bar{\mathrm{\Omega }}}_p,$ respectively, which means that u₁(t) and u₂(t) are positive solutions of BVP (1.1) and 0 ≤ ||u₁||₀ ≤ p ≤ ||u₂||₀. The proof is completed. □

Theorem 4.3. Assume that (H₁)-(H₃) hold. If ${\underset{}{f}}_0>\Gamma$ and ${\bar{f}}_{\infty }<\left(1-L\right)\Gamma$, and there exists p₂> p₁> 0 that satisfies

• (i) f(t, u) < ηp ₁ if 0 ≤ t ≤ 1 and 0 ≤ u ≤ p ₁,

• (ii) f(t, u) ≥ λp ₂ if 0 ≤ t ≤ 1 and σp ₂ ≤ u ≤ p ₂,

where η, σ, λ are just as the above, then BVP (1.1) has at least three positive solutions u₁, u₂, and u₃ such that 0 ≤ ||u₁||₀ ≤ p₁ ≤ ||u₂||₀ ≤ p₂ ≤ ||u₃||₀.

Proof. According to the proof of Theorem 3.1, there exists 0 < r₀< p₁< p₂< R₃<+∞, such that 0 < r < r₀ implies i(Q, Ω _r , P) = 0 and R ≥ R₃ implies i(Q, Ω _r , P) = 1.

From the proof of Theorems 4.1 and 4.2, we have $i\left(Q,{\mathrm{\Omega }}_{p_1},P\right)=1$,$i\left(Q,{\mathrm{\Omega }}_{p_2},P\right)=0$. Combining the four afore-mentioned equations, we have

$$\begin{array}{c}i\left(Q,{\mathrm{\Omega }}_R\backslash {\bar{\mathrm{\Omega }}}_{p_2},P\right)=i\left(Q,{\mathrm{\Omega }}_R,P\right)-i\left(Q,{\mathrm{\Omega }}_{p_2},P\right)=1.\\ i\left(Q,{\mathrm{\Omega }}_{p_2}\backslash {\bar{\mathrm{\Omega }}}_{p_1},P\right)=i\left(Q,{\mathrm{\Omega }}_{p_2},P\right)-i\left(Q,{\mathrm{\Omega }}_{p_1},P\right)=-1.\\ i\left(Q,{\mathrm{\Omega }}_{p_1}\backslash {\bar{\mathrm{\Omega }}}_r,P\right)=i\left(Q,{\mathrm{\Omega }}_{p_1},P\right)-i\left(Q,{\mathrm{\Omega }}_r,P\right)=1.\end{array}$$

Therefore, Q has fixed points u₁, u₂ and u₃ in ${\mathrm{\Omega }}_R\backslash {\bar{\mathrm{\Omega }}}_{p_2},{\mathrm{\Omega }}_{p_2}\backslash {\bar{\mathrm{\Omega }}}_{p_1}$and ${\mathrm{\Omega }}_{p_1}\backslash {\bar{\mathrm{\Omega }}}_r$, respectively, which means that u₁(t), u₂(t) and u₃(t) are positive solutions of BVP (1.1) and 0 ≤ ||u₁||₀ ≤ p₁ ≤ ||u₂||₀ ≤ p₂ ≤ ||u₃||₀. The proof is completed. □