Positive solutions for the third-order boundary value problems with the second derivatives

Abstract

By using the fixed-point index theory in a cone and defining a linear operator, we obtain the existence of at least one positive solution for the third-order boundary value problem with integral boundary conditions

$$\left\{\begin{array}{c}{u}^{‴}\left(t\right)+f\left(t,u\left(t\right),u^{″}\left(t\right)\right)=0,t\in \left(0,1\right),\\ u\left(0\right)=0,u^{″}\left(0\right)=0,u\left(1\right)={\int }_0^{1}g\left(t\right)u\left(t\right)dt,\end{array}\right.$$

where f : [0, 1] × R⁺ × R^-→ R⁺ is a nonnegative function. The associated Green's function for the above problem is also used, and a new reproducing cone also used.

1 Introduction

By eigenvalue criteria, Webb [1] obtained the existence of multiple positive solutions of a Hammerstein integral equation of the form

$$u\left(t\right)=\underset{0}{\overset{1}{\int }}k\left(t,s\right)g\left(s\right)f\left(s,u\left(s\right)\right)ds,$$

where k can have discontinuities and g ∈ L¹. Then, some articles have studied different BVPs by this way (see [2–5]). Webb [4] introduced an unified method to study existence of at least one nonzero solution for higher order boundary value problems

$$\left\{\begin{array}{c}{u}^{\left(n\right)}\left(t\right)+g\left(t\right)f\left(t,u\left(t\right)\right)=0,0<t<1,\\ u^{\left(n\right)}\left(0\right)=0,0\le k\le n-2,u\left(1\right)={\int }_0^{1}u\left(s\right)dA\left(s\right).\end{array}\right.$$

In 2010, Hao [5] considered the existence of positive solutions of the n th-order BVP

$$\left\{\begin{array}{c}{u}^{\left(n\right)}\left(t\right)+\lambda a\left(t\right)f\left(t,u\left(t\right)\right)=0,t\in \left(0,1\right),\\ u\left(0\right)=u^{\prime }\left(0\right)=\cdots =u^{\left(n-2\right)}\left(0\right)=0,u\left(1\right)={\int }_0^{1}u\left(s\right)dA\left(s\right).\end{array}\right.$$

Guo [6] studied the existence of positive solutions for the there-point boundary problem with the first-order derivative.

$$\left\{\begin{array}{c}{x}^{″}+f\left(t,x,x^{\prime }\right)=0,0<t<1,\\ x\left(0\right)=0,x\left(1\right)=\alpha x\left(\eta \right),\end{array}\right.$$

where f is a nonnegative continuous function. In 2011, Zhao [7] studied third-order differential equations:

$$x^{‴}+f\left(t,x\left(t\right)\right)=\theta ,t\in \left[0,1\right],$$

subject to integral boundary condition of the form

$$x\left(0\right)=\theta ,x^{″}\left(0\right)=\theta ,x\left(1\right)=\underset{0}{\overset{1}{\int }}g\left(t\right)x\left(t\right)dt,$$

where f ∈ C([0, 1] × P, P).

In this article, we study the existence of positive solutions for the following boundary value problem

$$\left\{\begin{array}{c}{u}^{‴}\left(t\right)+f\left(t,u\left(t\right),u^{″}\left(t\right)\right)=0,t\in \left(0,1\right),\\ u\left(0\right)=0,u^{″}\left(0\right)=0,u\left(1\right)={\int }_0^{1}g\left(t\right)u\left(t\right)dt.\end{array}\right.$$

(1.1)

The results are proved by applying the fixed point index theory in a cone and spectral radius of a linear operator. Unlike reference [7], the nonlinear part f involves the second-order derivative and just satisfies Caratheodory conditions.

The following conditions are satisfied throughout this article:

(H₁) f : [0, 1] × R⁺ × R^- → R⁺ satisfies Caratheodory conditions, that is, f(·,u, v) is measurable for each fixed u ∈ R⁺, v ∈ R^-, and f(t, ·,·) is continuous for a.e. t ∈ [0, 1]. For any r, r' > 0, there exists ${\phi }_{r,r^{\prime }}\left(t\right)\in L^{\infty }\left[0,1\right]$, such that $0\le f\left(t,u,v\right)\le {\phi }_{r,r^{\prime }}\left(t\right)$, where (u, v) ∈ [0, r] × [-r', 0], a.e. t ∈ [0, 1];

(H₂) g ∈ L[0, 1] is nonnegative, b ∈ [0, 1), where $b={\int }_0^{1}sg\left(s\right)ds$.

2 Preliminaries

Lemma 2.1[7]. Let y ∈ L¹[0, 1] and y ≥ 0, the problem

$$\left\{\begin{array}{c}{u}^{‴}\left(t\right)+y\left(t\right)=0,t\in \left(0,1\right),\\ u\left(0\right)=0,u^{″}\left(0\right)=0,u\left(1\right)={\int }_0^{1}g\left(t\right)u\left(t\right)dt\end{array}\right.$$

(2.1)

has a unique solution

$$u\left(t\right)=\underset{0}{\overset{1}{\int }}H\left(t,s\right)y\left(t\right)ds,$$

where $H\left(t,s\right)=G\left(t,s\right)+\frac{t}{1-b}{\int }_0^{1}G\left(\tau ,s\right)y\left(\tau \right)d\tau ,b={\int }_0^{1}sg\left(s\right)ds$,

$$G\left(t,s\right)=\left\{\begin{array}{c}\frac{1}{2}t{\left(1-s\right)}^2-\frac{1}{2}{\left(t-s\right)}^2,0\le s\le t\le 1,\\ \frac{1}{2}t{\left(1-s\right)}^2,0\le t\le s\le 1.\end{array}\right.$$

Lemma 2.2. Let y ∈ L¹[0, 1] and y ≥ 0, the unique solution of the boundary value problem (2.1) satisfies the following conditions: u(t) ≥ 0, u"(t) ≤ 0, for t ∈ [0, 1].

Proof. By Lemma 2.1, u(t) ≥ 0. By differential equations u'"(t) + y(t) = 0, t ∈ (0, 1), we get

$$\begin{array}{c}{u}^{″}\left(t\right)-u^{″}\left(0\right)=-\underset{0}{\overset{1}{\int }}y\left(s\right)ds,\\ u^{″}\left(t\right)=-\underset{0}{\overset{1}{\int }}y\left(s\right)ds\le 0.\end{array}$$

Let X = C²[0, 1] with $∥u∥=\underset{0\le t\le 1}{\max }\left|u\left(t\right)\right|+\underset{0\le t\le 1}{\max }\left|u^{″}\left(t\right)\right|$. Obviously, (X, ||·||) is a Banach space. Define the cone P ⊂ X by

$$P=\left\{u\in X\left|u\left(t\right)\ge 0,u^{″}\left(t\right)\le 0\right|\right\},P_r=\left\{u\in P\left|∥u∥\right.<r,r>0\right\}.$$

Obviously P is a cone in X, and P _r is a bounded open subset in P.

Definition 2.1[1]. Let P be a cone in a Banach space X. If for any x ∈ X and x⁺, x^- ∈ P, writing x = x⁺ + x^- shows that P is a reproducing cone.

Lemma 2.3. P is a reproducing cone in X.

Proof. Suppose u ∈ X, so u" ∈ C[0, 1] and

$$u^{″}=u^{-}-u^{+},$$

(2.2)

where u^- = min{u"(t), 0}, u⁺ = min{-u"(t), 0}. Obviously u⁺,u^- ∈ C[0, 1] and u⁺ ≤ 0,u^- ≤ 0. For (2.2), we get

$$\begin{array}{c}{u}^{\prime }\left(t\right)=\underset{0}{\overset{t}{\int }}{u}^{-}\left(s\right)ds-\underset{0}{\overset{t}{\int }}{u}^{+}\left(s\right)ds+u^{\prime }\left(0\right),\\ u\left(t\right)=\underset{0}{\overset{t}{\int }}ds\underset{0}{\overset{s}{\int }}{u}^{-}\left(\tau \right)d\tau -\underset{0}{\overset{t}{\int }}ds\underset{0}{\overset{s}{\int }}{u}^{+}\left(\tau \right)d\tau +u^{\prime }\left(0\right)t+u\left(0\right).\end{array}$$

If u(0) ≥ 0, u'(0)t ≥ 0, let

$$u_1=-\underset{0}{\overset{t}{\int }}ds\underset{0}{\overset{s}{\int }}{u}^{+}\left(\tau \right)d\tau +u^{\prime }\left(0\right)t+u\left(0\right),u_2=-\underset{0}{\overset{t}{\int }}ds\underset{0}{\overset{s}{\int }}{u}^{-}\left(\tau \right)d\tau .$$

So u₁ ≥ 0, u₂ ≥ 0, then u₁, u₂ ∈ P and u = u₁ - u₂.

If u(0) ≤ 0, u'(0)t ≤ 0, let

$$u_1=-\underset{0}{\overset{t}{\int }}ds\underset{0}{\overset{s}{\int }}{u}^{+}\left(\tau \right)d\tau ,u_2=-\underset{0}{\overset{t}{\int }}ds\underset{0}{\overset{s}{\int }}{u}^{-}\left(\tau \right)d\tau -u^{\prime }\left(0\right)t-u\left(0\right).$$

So u₁ ≥ 0, u₂ ≥ 0, then u₁, u₂ ∈ P and u = u₁ - u₂.

If u(0) ≥ 0, u'(0)t ≤ 0, let

$$u_1=-\underset{0}{\overset{t}{\int }}ds\underset{0}{\overset{s}{\int }}{u}^{+}\left(\tau \right)d\tau +u\left(0\right),u_2=-\underset{0}{\overset{t}{\int }}ds\underset{0}{\overset{s}{\int }}{u}^{-}\left(\tau \right)d\tau -u^{\prime }\left(0\right)t.$$

So u₁ ≥ 0, u₂ ≥ 0, then u₁, u₂ ∈ P and u = u₁ - u₂.

If u(0) ≤ 0, u'(0)t ≥ 0, let

$$u_1=-\underset{0}{\overset{t}{\int }}ds\underset{0}{\overset{s}{\int }}{u}^{+}\left(\tau \right)d\tau +u^{\prime }\left(0\right)t,u_2=-\underset{0}{\overset{t}{\int }}ds\underset{0}{\overset{s}{\int }}{u}^{-}\left(\tau \right)d\tau -u\left(0\right).$$

So u₁ ≥ 0, u₂ ≥ 0, then u₁, u₂ ∈ P and u = u₁ - u₂.

Then P is a reproducing cone in X.

Lemma 2.4 (Krein-Rutman) [8]. Let K be a reproducing cone in a real Banach space X and let L : K → K be a compact linear operator with L(K) ⊂ K. r(L) is the spectral radius of L. If r(L) > 0, then there is φ₁ ∈ K\{0} such that Lφ₁ = r(L)φ₁.

Lemma 2.5[9]. Let X be a Banach space, P be a cone in X and Ω(P) be a bounded open subset in P. Suppose that $A:\overline{\mathrm{\Omega }\text{(}P)}\to P$ is a completely continuous operator. Then the following results hold

• (1) If there exists u ₀ ∈ P\{0} such that u ≠ Au + λu ₀, for any u ∈ ∂ Ω(P), λ ≥ 0, then the fixed-point index i(A, Ω(P), P) = 0.

• (2) If 0 ∈ Ω(P), Au ≠ λu, for any u ∈ ∂ Ω(P), λ ≥ 1, then the fixed-point index i(A, Ω(P), P) = 1.

Define the operator A: X → X, L: X → X, by

$$Au\left(t\right)=\underset{0}{\overset{1}{\int }}H\left(t,s\right)f\left(s,u\left(s\right),u^{″}\left(s\right)\right)ds,$$

$$Lu\left(t\right)=\underset{0}{\overset{1}{\int }}H\left(t,s\right)\left(u\left(s\right)-u^{″}\left(s\right)\right)ds,$$

So A : P → P is completely continuous operator; L : P → P is a compact linear operator.

Lemma 2.6[7]. Assume that (H₂) holds, then choose $\delta \in \left(0,\frac{1}{2}\right)$, for all t ∈ [δ, 1 - δ],v, s ∈ [0, 1], we have

$$\begin{array}{c}G\left(t,s\right)\ge \rho G\left(v,s\right),\\ H\left(t,s\right)\ge \rho H\left(v,s\right),\end{array}$$

where ρ = 4δ²(1 - δ).

Note: r(L) is the spectral radius of L. $h=\underset{t\in \left[\delta ,1-\delta \right]}{\min }\underset{\delta }{\overset{1-\delta }{\int }}H\left(t,s\right)ds$, where $\delta \in \left(0,\frac{1}{2}\right)$. By Lemma 2.6, obviously h > 0.

Lemma 2.7. Suppose conditions (H₁), (H₂) hold, then r(L) > 0.

Proof. Take u(t) ≡ 1, then u"(t) = 0, for any t ∈ [δ, 1 - δ] we get

$$\begin{array}{c}Lu\left(t\right)\ge \underset{\delta }{\overset{1-\delta }{\int }}H\left(t,s\right)ds\ge h>\left(0\right).\\ L^{2}u\left(t\right)\ge \underset{\delta }{\overset{1-\delta }{\int }}H\left(t,s\right)Lu\left(s\right)ds\ge h\underset{\delta }{\overset{1-\delta }{\int }}H\left(t,s\right)ds\ge h^2>0.\end{array}$$

Repeating the process gives

$$L^{k}u\left(t\right)\ge h^k.$$

So, we get $∥L^k∥\ge h^k,r\left(L\right)=\underset{k\to \infty }{\lim }{∥L^k∥}^{\frac{1}{k}}\ge h>0$. The proof is completed.

By Lemma 2.4, then there is φ₁ ∈ P\{0} such that Lφ₁ = r(L)φ₁.

3 Main results

In the following, we use the notation:

$$\begin{array}{l}\overline{f}(u,v)=\underset{t\in [0,1]\backslash E}{\sup }f(t,u,v),f(u,v)=\underset{t\in [0,1]\backslash E}{\inf }f(t,u,v),\\ f^{\infty }=\max \left\{\underset{u\to \infty }{\lim }\sup \left\{\underset{v\in R^{-}}{\sup }\frac{\overline{f}(u,v)}{u-v}\right\},\underset{v\to -\infty }{\lim }\sup \left\{\underset{u\in R^{+}}{\sup }\frac{\overline{f}(u,v)}{u-v}\right\}\right\},\\ f_0^d=\max \left\{\underset{u\to 0+}{\lim }\inf \left\{\underset{v\in [-d,0]}{\inf }\frac{f(u,v)}{u-v}\right\},\underset{v\to 0-}{\lim }\inf \left\{\underset{u\in [0,d]}{\inf }\frac{f(u,v)}{u-v}\right\}\right\},\end{array}$$

where E is a fixed subset of [0, 1] of measure zero, d > 0.

Lemma 3.1. Suppose

$$0\le f^{\infty }<\mu ,$$

(3.1)

where μ = 1/r(L), then there exists R₀ > 0 such that i(A, P _r , P) = 1 for each r > R₀.

Proof. Let ε > 0 satisfy f^∞ ≤ μ - ε, then there exist r₁ > 0 such that

$$f\left(t,u,v\right)\le \left(\mu -\epsilon \right)\left(u-v\right),$$

for all u > r₁ or v < -r₁ and a.e. t ∈ [0, 1].

By (H₁), there exists ${\phi }_{r_1}\in L^{\infty }\left[0,1\right]$ such that

$$0\le f\left(t,u,v\right)\le {\phi }_{r_1}\left(t\right),$$

for all (u, v) ∈ [0, r₁] × [-r₁, 0] and a.e. t ∈ [0, 1]. Hence, we have

$$f\left(t,u,v\right)\le \left(\mu -\epsilon \right)\left(u-v\right)+{\phi }_{r_1}\left(t\right),$$

(3.2)

for all u ∈ R⁺, v ∈ R^- and a.e. t ∈ [0, 1].

Since $\frac{1}{\mu }$ is the spectrum radius of L. It follows from ${\left(\frac{1}{\mu -\epsilon }I-L\right)}^{-1}=\sum _{n=0}^{\infty }{\left(\mu -\epsilon \right)}^{n+1}{L}^n$, (I/(μ - ε) - L)^-l exists, let

$$C=∥\underset{0}{\overset{1}{\int }}H\left(t,s\right){\phi }_{r_1}\left(s\right)ds∥,R_0=∥{\left(\frac{1}{\mu -\epsilon }I-L\right)}^{-1}\frac{C}{\mu -\epsilon }\left(\frac{3}{2}-\frac{1}{2}{t}^2\right)∥.$$

For r > R₀, by Lemma 2.5 we will prove

$$Au\ne \lambda u,$$

for each u ∈ ∂P _r and λ ≥ 1.

In fact, if not, there exist u₀ ∈ ∂P _r and λ₀ ≥ 1 such that Au₀ = λ₀u₀.

Together with (3.2) implies

$$\begin{array}{l}{u}_0(t)\le A{u}_0(t)\le {\displaystyle \underset{0}{\overset{1}{\int }}H(t,s)[(\mu -\epsilon )u_0(s)+{\phi }_{r_1}(t)]ds}\hfill \\ \le {\displaystyle \underset{0}{\overset{1}{\int }}H(t,s)(\mu -\epsilon )[u_0(s)-v_0(s))+{\phi }_{r_1}(s)]ds}.\hfill \end{array}$$

So

$$\begin{array}{c}{u}_0\left(t\right)\le \left(\mu -\epsilon \right)L{u}_0\left(t\right)+C,\\ u_0^{{}^{″}}\left(t\right)\ge {\lambda }_0{u}_0^{{}^{″}}\left(t\right)={\left(A{u}_0\left(t\right)\right)}^{″}\ge \left(\mu -\epsilon \right){\left(L{u}_0\left(t\right)\right)}^{″}-C.\end{array}$$

Then

$$\left(\frac{1}{\mu -\epsilon }I-L\right)u_0\left(t\right)\le \frac{C}{\mu -\epsilon }\left(\frac{3}{2}-\frac{1}{2}{t}^2\right),{\left(\left(\frac{1}{\mu -\epsilon }I-L\right)u_0\left(t\right)\right)}^{{}^{″}}\ge {\left(\frac{C}{\mu -\epsilon }\left(\frac{3}{2}-\frac{1}{2}{t}^2\right)\right)}^{{}^{″}}.$$

So

$$\frac{C}{\mu -\epsilon }\left(\frac{3}{2}-\frac{1}{2}{t}^2\right)-\left(\frac{1}{\mu -\epsilon }I-L\right)u_0\left(t\right)\in P.$$

Then

$$\begin{array}{l}{u}_0(t)\le {\left(\frac{I}{\mu -\epsilon }-L\right)}^{-1}\frac{C}{\mu -\epsilon }\left(\frac{3}{2}-\frac{1}{2}{t}^2\right),u_0^{\text{'}\text{'}}(t)\ge {\left[{\left(\frac{I}{\mu -\epsilon }-L\right)}^{-1}\frac{C}{\mu -\epsilon }\left(\frac{3}{2}-\frac{1}{2}{t}^2\right)\right]}^{\text{'}\text{'}},\\ \Vert u_0(t)\Vert \le R_0<r.\end{array}$$

This is a contradiction. By Lemma 2.5 (2), we get that i(A, P _r , P) = 1 for each r > R₀. The proof is completed.

Lemma 3.2. Suppose there exists d > 0 such that

$$\mu <f_0^d\le \infty .$$

(3.3)

Then there exists ρ₀ > 0 and d ≥ ρ₀ such that for each ρ∈ (0, ρ₀], if u ≠ Au for u ∈ ∂Pρ, then i(A, P _ρ , P) = 0.

Proof. Let ε > 0 satisfy $f_0^d\ge \mu +\epsilon$, there exist d ≥ ρ₀ > 0 such that

$$f\left(t,u,v\right)\ge \left(\mu +\epsilon \right)\left(u-v\right),$$

(3.4)

for u ∈ [0, ρ₀],v ∈ [-ρ₀,0] and a.e. t ∈ [0, 1].

Let ρ ∈ (0,ρ₀], by Lemma 2.5 (1), we prove that: u ≠ Au + λφ₁ for all u ∈ ∂Pρ, λ > 0, where φ₁ ∈ P\{0} is the eigenfunction of L corresponding to the eigenvalue $\frac{1}{\mu }$. In fact, if not, there exist u₀ ∈ ∂P _ρ , λ₀ > 0 such that u₀ = Au₀ + λ₀φ₁. This implies

$$u_0\ge {\lambda }_0{\phi }_{1}and{u}_0^{{}^{″}}\le {\lambda }_0{\phi }_1^{{}^{″}}.$$

Let: ${\lambda }^{*}=\sup \left\{\lambda |u_0\ge \lambda {\phi }_{\mathsf{\text{1}}},u_0^{"}\le \lambda {\phi }_1^{"}\right\}$.

So 0 < λ₀ < λ* < ∞ and $u_0\ge {\lambda }^{*}{\phi }_{\mathsf{\text{1}}},u_0^{{}^{″}}\le {\lambda }^{*}{\phi }_1^{{}^{″}}$. Then, u₀- λ*φ₁ ∈ P.

For L(P) ⊂ P, we get

$$\mu L{u}_0\ge {\lambda }^{*}\mu L{\phi }_{\mathsf{\text{1}}}={\lambda }^{*}{\phi }_{\mathsf{\text{1}}},\mu {\left(L{u}_0\right)}^{{}^{″}}\le {\lambda }^{*}\mu {\left(L{\phi }_{\mathsf{\text{1}}}\right)}^{{}^{″}}={\lambda }^{*}{\phi }_1^{{}^{″}}.$$

By (3.4), we get

$$\begin{array}{l}A{u}_0={\displaystyle \underset{0}{\overset{1}{\int }}H(t,s)f(s,u_0(s),u_0^{\text{'}\text{'}}(s))ds\ge (\mu +\epsilon )L{u}_0}.\hfill \\ (A{u}_0)″\le (\mu +\epsilon )(L{u}_0)″.\hfill \end{array}$$

So, we know

$$\begin{array}{c}{u}_0=A{u}_0+{\lambda }_0{\phi }_{\mathsf{\text{1}}}\ge \left(\mu +\epsilon \right)L{u}_0+{\lambda }_0{\phi }_{\mathsf{\text{1}}}\ge \left({\lambda }^{*}+{\lambda }_0\right){\phi }_{\mathsf{\text{1}}}.\\ {\left(u_0\right)}^{{}^{″}}={\left(A{u}_0\right)}^{{}^{″}}+{\lambda }_0{\phi }_1^{{}^{″}}\le \left(\mu +\epsilon \right){\left(L{u}_0\right)}^{{}^{″}}+{\lambda }_0{\phi }_1^{{}^{″}}\le \left({\lambda }^{*}+{\lambda }_0\right){\phi }_1^{{}^{″}}.\end{array}$$

which contradicts the definition of λ*.

Lemma 3.3. Suppose there is ρ₁ > 0 such that

$$f\left(t,u,v\right)\le d_1{\rho }_{\mathsf{\text{1}}},$$

(3.5)

for u ∈ [0, ρ₁] and v ∈ [-ρ₁, 0] a.e. t ∈ [0, 1], where $d_1=\frac{1}{∥{\int }_0^{1}H\left(t,s\right)ds∥}$, if Au ≠ u for $u\in \partial P_{{\rho }_1}$, then $i\left(A,P_{{\rho }_1},P\right)=1$.

Proof. Suppose $u\in \partial P_{{\rho }_1}$, by Lemma 2.2, we get

$$\begin{array}{ll}\hfill ∥Au∥& =\underset{0\le t\le 1}{\max }Au\left(t\right)-\underset{0\le t\le 1}{\min }{\left(Au\left(t\right)\right)}^{{}^{″}}\\ =\underset{0\le t\le 1}{\max }\underset{0}{\overset{1}{\int }}H\left(t,s\right)f\left(t,u\left(t\right),u^{{}^{″}}\left(t\right)\right)ds+\underset{0\le t\le 1}{\max }{\left(\underset{0}{\overset{1}{\int }}H\left(t,s\right)f\left(t,u\left(t\right),u^{{}^{″}}\left(t\right)\right)ds\right)}^{{}^{″}}\\ \le d_1{\rho }_{\mathsf{\text{1}}}\left[\underset{0\le t\le 1}{\max }\underset{0}{\overset{1}{\int }}H\left(t,s\right)ds+\underset{0\le t\le 1}{\max }{\left(\underset{0}{\overset{1}{\int }}H\left(t,s\right)ds\right)}^{{}^{″}}\right]\le {\rho }_{\mathsf{\text{1}}}.\end{array}$$

That is Au ≠ λu for each $u\in \partial P_{{\rho }_1}$, λ > 1. If Au ≠ u for $u\in \partial P_{{\rho }_1}$, by Lemma 2.5, then $i\left(A,P_{{\rho }_1},P\right)=1$.

Lemma 3.4. Suppose there is ρ₂ > 0 such that

$$f\left(t,u,v\right)\ge d_{\mathsf{\text{2}}}{\rho }_{\mathsf{\text{2}}},$$

(3.6)

for u ∈ [0, _{ρ 2}] and v ∈ [-_{ρ 2}, 0] a.e. t ∈ [0, 1], where $d_2=\frac{1}{\underset{t\in \left[\delta ,1-\delta \right]}{\min }{\int }_0^{1}H\left(t,s\right)ds-\underset{t\in \left[\delta ,1-\delta \right]}{\max }{\left({\int }_0^{1}H\left(t,s\right)ds\right)}^{{}^{″}}}.$. If Au ≠ u for $u\in \partial P_{{\rho }_2}$, then $i\left(A,P_{{\rho }_2},P\right)=0$.

Proof. For $u\in \partial P_{{\rho }_2}$, t ∈ [δ, 1 - δ], by Lemma 2.2, we get

$$\begin{array}{ll}\hfill Au+{\left(Au\right)}^{{}^{″}}& =\underset{0}{\overset{1}{\int }}H\left(t,s\right)f\left(t,u\left(t\right),u^{{}^{″}}\left(t\right)\right)ds+{\left(\underset{0}{\overset{1}{\int }}H\left(t,s\right)f\left(t,u\left(t\right),u^{{}^{″}}\left(t\right)\right)ds\right)}^{{}^{″}}\\ \ge d_2{\rho }_2\left[\underset{0}{\overset{1}{\int }}H\left(t,s\right)ds+{\left(\underset{0}{\overset{1}{\int }}H\left(t,s\right)ds\right)}^{{}^{″}}\right]\ge {\rho }_2.\end{array}$$

This implies that u ≠ Au + λφ for each $u\in \partial P_{{\rho }_2}$, λ > 0, where φ ∈ P\{0} is the eigenfunction of L corresponding to r(L). Suppose u ≠ Au for $u\in \partial P_{{\rho }_2}$, by Lemma 2.5, then $i\left(A,P_{{\rho }_2},P\right)=0$.

Theorem 3.1. The boundary value problem (1.1) has at least one positive solution if one of the following conditions holds.

(C1) There exists d > 0 such that (3.3) and (3.1) hold.

(C2) There exists d > 0, ρ₁> 0 such that (3.3) and (3.5) hold.

(C3) There exists ρ₂ > 0 such that (3.6) and (3.1) hold.

(C4) There exists ρ₁, ρ₂ > 0 with 0 < ρ₂ < ρ₁d₁/d₂ such that (3.5) and (3.6) hold.

Proof. When condition (C1) holds, by Lemma 3.1 and 0 ≤ f^∞ < μ, we get that there exists r > 0 such that i(A, P _r , P) = 1. It follows from Lemma 3.2 and $\mu <f_0^b\le \infty$, then there exists 0 < ρ < min{r, d} such that either there exists u ∈ ∂P _ρ that i(A, P _ρ , P) = 0 or u = Au. So BVP (1.1) has at least one positive solution u ∈ P with ρ ≤ ||u|| < r.

When one of other conditions holds, the results can be proved similarly.