Positive solutions for elastic beam equations with nonlinear boundary conditions and a parameter

Abstract

This paper is concerned with the existence, nonexistence, and uniqueness of convex monotone positive solutions of elastic beam equations with a parameter λ. The boundary conditions mean that the beam is fixed at one end and attached to a bearing device or freed at the other end. By using fixed point theorem of cone expansion, we show that there exists {\lambda }^{\ast }\ge {\lambda }_{\ast }>0 such that the beam equation has at least two, one, and no positive solutions for , and \lambda >{\lambda }^{\ast }, respectively; furthermore, by using cone theory we establish some uniqueness criteria for positive solutions for the beam and show that such solution x_{\lambda } depends continuously on the parameter λ. In particular, we give an estimate for critical value of parameter λ.

MSC:34B18, 34B15.

1 Introduction and preliminaries

In this paper, we consider the following nonlinear fourth-order two-point boundary value problem (BVP) for elastic beam equation:

(1)

where \lambda \ge 0 is a parameter. Throughout this paper, we assume that f\in C([0,1]\times R_{+},R_{+}), q\in C(R_{+},R_{+}), R_{+}=[0,+\mathrm{\infty }). x\in C[0,1] is called a positive solution of BVP (1) if x is a solution of BVP (1) and x(t)>0, . A convex monotone positive solution means convex nondecreasing positive solution.

Because of characterization of the deformation of the equilibrium state, fourth-order boundary value problems for elastic beam equations are extensively applied to mechanics and engineering; see [1–3]. Some nonlinear elastic beam equations have been studied extensively. For a small sample of such work, we refer the reader to the work of Bai and Wang [4], Bai [5], Bonanno and Bellaa [6], Li [7], Liu and Li [8], Liu [9], Ma and Xu [10], and Ma and Thompson [11] on an elastic beam whose two ends are simply supported, the works of Yang [12] and Zhang [13] on an elastic beam of which one end is embedded and another end is fastened with a sliding clamp, and the work of Graef et al. [14] on multipoint boundary value problems.

BVP (1) with q(x)\equiv 0 is called a cantilever beam equation, it describes the deflection of the elastic beam fixed at the left end and free at the right end. Existence and multiplicity of positive solutions of cantilever beam problems without parameter have been studied by some authors; see Yao [15, 16] and references therein. BVP (1) with q(x)\not\equiv 0 describes the deflection of the elastic beam fixed at the left end and attached to a bearing device given by the function −q at the right end. When the elastic beam equation does not contain parameter λ, the existence of multiple positive solutions and unique positive solution was presented in [17] by variational methods and in [18] by a fixed point theorem, respectively; monotone positive solutions were obtained by using the monotone iteration method in [19]. However, there are few papers concerned with positive solutions for BVP (1) with parameter, especially with the solution’s dependence on parameter λ in the existing literature. The aim of this paper is to show that the existence and number of convex monotone positive solutions of BVP (1) are affected by the parameter λ.

The paper is organized as follows. In Section 2, we present that a nontrivial and nonnegative solution of BVP (1) is convex monotone positive solution. In Section 3, we obtain some results on the existence, multiplicity and nonexistence of positive solutions for BVP (1). These results show that the number of positive solutions for BVP (1) depends on the parameter λ. In Section 4, we establish some uniqueness criteria for positive solutions for BVP (1) and show that such a positive solution x_{\lambda } depends continuously on the parameter λ. In particular, we give an estimate for the critical value of the parameter λ.

In the rest of this section, we introduce some notations and known results. For the reader’s convenience, we suggest that one refer to [20–22], and [23] for details.

Let E be a real Banach space and θ denote the zero element of E. A nonempty closed convex set P\subset E is called a cone of E if it satisfies (i) x\in P, r>0\Rightarrow rx\in P; (ii) x\in P, -x\in P\Rightarrow x=\theta. E is partially ordered by the cone P, i.e., x\le y iff y-x\in P. A cone P is said to be normal if there exists a positive number N, called the normal constant of P, such that \theta \le x\le y implies \parallel x\parallel \le N\parallel y\parallel. For u,v\in E, u\le v, denote [u,v]=\{x\in E\mid u\le x\le v\}.

For all x,y\in E, the notation x\sim y means that there exist {\mu }_1>0 and {\mu }_2>0 such that {\mu }_{1}x\le y\le {\mu }_{2}x. Clearly, ∼ is an equivalence relation. Given e>0 (i.e., e\in P and e\ne \theta), we denote by P_e the set P_e=\{x\in E\mid x\sim e\}. It is easy to see that P_e\subset P.

Let D\subseteq E. An operator T:D\to E is said to be increasing if for x,y\in D, x\le y\Rightarrow Tx\le Ty. An element x^{\ast }\in D is called a fixed point of T if T{x}^{\ast }=x^{\ast }.

Lemma 1.1 (Fixed point theorem of cone expansion) [21, 22]

Assume that {\mathrm{\Omega }}_1 and {\mathrm{\Omega }}_2 are bounded open subsets of E with \theta \in {\mathrm{\Omega }}_1\subset {\overline{\mathrm{\Omega }}}_1\subset {\mathrm{\Omega }}_2. Let T:P\cap ({\overline{\mathrm{\Omega }}}_2-{\mathrm{\Omega }}_1)\to P be a completely continuous operator such that \parallel Tx\parallel \le \parallel x\parallel if x\in P\cap \partial {\mathrm{\Omega }}_1 and \parallel Tx\parallel \ge \parallel x\parallel if x\in P\cap \partial {\mathrm{\Omega }}_2. Then T has a fixed point in P\cap ({\overline{\mathrm{\Omega }}}_2-{\mathrm{\Omega }}_1).

Lemma 1.2 [23]

Let P be a normal cone in E, T:P_e\to P_e be increasing and for all x\in P_e and t\in (0,1), there exists \alpha (t)\in (0,1) such that T(tx)\ge t^{\alpha (t)}Tx. Then T has a unique fixed point x^{\ast } in P_e. Moreover, constructing successively the sequence w_n=T{w}_{n-1} (n=1,2,\dots) for any w_0\in P_e, we have {lim}_{n\to +\mathrm{\infty }}\parallel w_n-x^{\ast }\parallel =0.

2 Solutions

In what follows, set E=C[0,1], the Banach space of all continuous functions on [0,1] with the norm \parallel x\parallel =max\{|x(t)|\mid t\in [0,1]\}. P=\{x\in C[0,1]\mid x(t)\ge 0,t\in [0,1]\}. It is clear that P is a normal cone and its normality constant is 1.

From [18] and [19], it is evident that BVP (1) has an integral formulation given by

x(t)=\lambda {\int }_0^{1}G(t,s)f(s,x(s))ds+q(x(1))(\frac{1}{2}{t}^2-\frac{1}{6}{t}^3),

(2)

where

G(t,s)=\frac{1}{6}\{\begin{array}{ll}{t}^2(3s-t),& 0\le t\le s\le 1,\\ s^2(3t-s),& 0\le s\le t\le 1.\end{array}

(3)

It is easy to see that G(t,s)\ge 0 and

\frac{1}{3}{t}^2{s}^2\le G(t,s)\le \frac{1}{2}{t}^{2}s,t,s\in [0,1].

(4)

Define operators A,B,C_{\lambda }:P\to C[0,1] by

(Ax)(t)={\int }_0^{1}G(t,s)f(s,x(s))ds,(Bx)(t)=q(x(1))(\frac{1}{2}{t}^2-\frac{1}{6}{t}^3),C_{\lambda }=\lambda A+B.

Then A(P)\subset P, B(P)\subset P, and C_{\lambda }(P)\subset P.

It is clear from (2) that solving BVP (1) is equivalent to finding fixed points of the operator C_{\lambda }. In particular, x is a fixed point of B iff x is a solution of the following BVP:

and x is a fixed point of λA iff x is a solution of the following cantilever beam problem:

(5)

Lemma 2.1 If x\in C^4[0,1] satisfies

\{\begin{array}{l}{x}^{(4)}(t)\ge 0,t\in (0,1),\\ x(0)=x^{\prime }(0)=x^{″}(1)=0,x^{‴}(1)\le 0,\end{array}

(6)

then

• (i) x(t) is nondecreasing in t\in [0,1], moreover, 0\le x(t)\le x(1), t\in [0,1];

• (ii) x^{″}(t)\ge 0, t\in [0,1], that is, x(t) is a convex function on [0,1].

Proof From (6), we have x^{‴}(t)\le x^{‴}(1)\le 0. Moreover, x^{″}(t)\ge x^{″}(1)=0. So, x^{\prime }(t)\ge x^{\prime }(0)=0. Thus, we complete the proof of the lemma. □

Now, let

K=\{x\in P|x(t)\text{ is nondecreasing},x(t)\ge \frac{2}{3}{t}^{2}x(1),t\in [0,1]\},

then, it is easy to show that K\subset P is also a cone in E, and if x\in K, then \parallel x\parallel =x(1).

Lemma 2.2 C_{\lambda }(P)\subset K, A(P)\subset K, B(P)\subset K.

Proof x\in P implies x(t)\ge 0, so f(t,x(t))\ge 0 and q(x(1))\ge 0. Moreover, for x\in P,

\begin{array}{c}{(C_{\lambda }x)}^{(4)}(t)=\lambda f(t,x(t))\ge 0,t\in (0,1),\hfill \\ (C_{\lambda }x)(0)={(C_{\lambda }x)}^{\prime }(0)={(C_{\lambda }x)}^{″}(1)=0,\hfill \\ {(C_{\lambda }x)}^{‴}(1)=-q(x(1))\le 0.\hfill \end{array}

By Lemma 2.1, (C_{\lambda }x)(t) is convex and nondecreasing in t\in [0,1]. From (3) and (4) we have

(C_{\lambda }x)(t)-\frac{2}{3}{t}^2(C_{\lambda }x)(1)\ge \frac{\lambda }{9}{t}^2{\int }_0^1{s}^{3}f(s,x(s))ds+q(x(1))(\frac{5}{18}{t}^2-\frac{1}{6}{t}^3)\ge 0,

that is, \frac{2}{3}{t}^2(C_{\lambda }x)(1)\le (C_{\lambda }x)(t) for t\in [0,1]. Thus, we obtain C_{\lambda }(P)\subset K. From the above proof, we can show that A(P)\subset K and B(P)\subset K. This ends the proof. □

Lemma 2.3

• (i) A:P\to K is a completely continuous operator;

• (ii) if q(x) is nondecreasing, then B:P\to K is a completely continuous operator.

Proof Similarly to the proof of Theorem 1 in [19], applying the Arzela-Ascoli Theorem, the proof can be completed. □

From the proof of Lemma 2.2 we can show the following result.

Theorem 2.4 If x\in P\mathrm{\setminus }\{\theta \} is a solution for BVP (1), then x is a convex monotone positive solution for BVP (1).

So, in the following sections, we only need to study solutions for BVP (1) in P\mathrm{\setminus }\{\theta \}.

3 Existence and nonexistence results

It is obvious from Lemma 2.2 that if x\in P\mathrm{\setminus }\{\theta \} is a solution for BVP (1) then x\in K\mathrm{\setminus }\{\theta \}. So in this section, we will apply Lemma 1.1 to study the existence, multiplicity and nonexistence of solutions for BVP (1) in K\mathrm{\setminus }\{\theta \}. It is reasonable that the domain of C_{\lambda } is restricted on K. The following conditions will be assumed:

(H1) f(t,x) is nondecreasing in x\in [0,+\mathrm{\infty }) for fixed t\in [0,1];

(H2) q(x) is nondecreasing in x\in [0,+\mathrm{\infty });

(H3) F_0:={\int }_0^1{s}^{2}f(s,0)ds>0;

(H4) ;

(H5) f_{\mathrm{\infty }}:={lim}_{x\to +\mathrm{\infty }}{min}_{t\in [\frac{1}{2},1]}\frac{f(t,x)}{x}=+\mathrm{\infty };

(H6) q_{\mathrm{\infty }}:={lim\hspace{0.17em}inf}_{x\to +\mathrm{\infty }}\frac{q(x)}{x}>3.

Set

\mathrm{\Lambda }=\{\lambda >0\mid \text{there exists }{x}_{\lambda }\in K\mathrm{\setminus }\{\theta \}\text{ such that }{C}_{\lambda }{x}_{\lambda }=x_{\lambda }\}

(7)

and {\lambda }^{\ast }=sup\mathrm{\Lambda }.

Lemma 3.1 Suppose that (H1)-(H3) hold. If {\lambda }^{\prime }\in \mathrm{\Lambda }, then (0,{\lambda }^{\prime }]\subset \mathrm{\Lambda }.

Proof {\lambda }^{\prime }\in \mathrm{\Lambda } means that there exists x_{{\lambda }^{\prime }}\in K\mathrm{\setminus }\{\theta \} such that C_{{\lambda }^{\prime }}{x}_{{\lambda }^{\prime }}=x_{{\lambda }^{\prime }}. Therefore, for any \lambda \in (0,{\lambda }^{\prime }], we have C_{\lambda }{x}_{{\lambda }^{\prime }}\le C_{{\lambda }^{\prime }}{x}_{{\lambda }^{\prime }}=x_{{\lambda }^{\prime }}. Set w_0=x_{{\lambda }^{\prime }}, w_n=C_{\lambda }{w}_{n-1}, n=1,2,\dots . From (H1) and (H2) we obtain w_0(t)\ge w_1(t)\ge \cdots \ge w_n(t)\ge \cdots \ge \frac{F_0\lambda }{3}{t}^2. By Lemma 2.3 and (H3), \{w_n\} converges to a fixed point of C_{\lambda } in K\mathrm{\setminus }\{\theta \}. Thus (0,{\lambda }^{\prime }]\subset \mathrm{\Lambda }. This completes the proof. □

Let {\lambda }_{\ast }=\frac{2-q(1)}{F_1}, F_1={\int }_0^{1}sf(s,1)ds, u_0(t)=\frac{F_0\lambda }{3}{t}^2, v_0(t)=t^2 and

F_{\mathrm{\infty }}=\underset{x\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{t\in [0,1]}{max}\frac{f(t,x)}{x},Q_{\mathrm{\infty }}=\underset{x\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{q(x)}{x}.

Theorem 3.2 Suppose that (H1)-(H3) hold.

• (i) If (H4) holds, then C_{\lambda } has minimal and maximal fixed points in [u_0,v_0] for \lambda \in (0,{\lambda }_{\ast }]. Moreover, there exists {\lambda }^{\ast }\ge {\lambda }_{\ast }>0 such that C_{\lambda } has at least one and has no fixed points in K\mathrm{\setminus }\{\theta \} for and \lambda >{\lambda }^{\ast }, respectively.

• (ii) If , , then when F_{\mathrm{\infty }}>0, there exists {\lambda }^{\ast }\ge \frac{2(2-Q_{\mathrm{\infty }})}{F_{\mathrm{\infty }}}>0 such that C_{\lambda } has at least one and no fixed points in K\mathrm{\setminus }\{\theta \} for and \lambda >{\lambda }^{\ast }, respectively; when F_{\mathrm{\infty }}=0, C_{\lambda } has at least one fixed point in K\mathrm{\setminus }\{\theta \} for \lambda >0.

Proof (i) From (H1), (H3), and (H4) we have {\lambda }_{\ast }>0. For any \lambda \in (0,{\lambda }_{\ast }], we obtain

(C_{\lambda }{u}_0)(t)\ge \frac{\lambda }{3}{t}^2{\int }_0^1{s}^{2}f(s,0)ds=u_0(t),(C_{\lambda }{v}_0)(t)\le \frac{1}{2}{t}^2({\lambda }_{\ast }{F}_1+q(1))\le v_0(t).

Set u_n=C_{\lambda }{u}_{n-1}, v_n=C_{\lambda }{v}_{n-1}, n=1,2,\dots , then from (H1) and (H2) we have

u_0(t)\le u_1(t)\le \cdots \le u_n(t)\le \cdots \le v_n(t)\le \cdots \le v_1(t)\le v_0(t).

(8)

Lemma 2.3 implies that \{u_n\} and \{v_n\} converge to fixed points u_{\lambda } and v_{\lambda } of C_{\lambda }, respectively. From (8) it is evident that u_{\lambda },v_{\lambda }\in K\mathrm{\setminus }\{\theta \} are the minimal fixed point and maximal fixed point of C_{\lambda } in [u_0,v_0], respectively. From the definition of {\lambda }^{\ast } we can complete the rest of the proof.

• (ii) For any , there exists N_0>0 such that f(t,x)\le (F_{\mathrm{\infty }}+ϵ)x and q(x)\le (Q_{\mathrm{\infty }}+ϵ)x for x>N_0, t\in [0,1]. Let w_0(t)=2N_0{t}^2 and {\lambda }_0=\frac{2(2-Q_{\mathrm{\infty }}-ϵ)}{F_{\mathrm{\infty }}+ϵ}, then {\lambda }_0>0 and

  (C_{{\lambda }_0}{w}_0)(t)\le \frac{1}{2}{w}_0(t)(\frac{{\lambda }_0}{2}(F_{\mathrm{\infty }}+ϵ)+Q_{\mathrm{\infty }}+ϵ)\le w_0(t).

Similarly to the proof of Lemma 3.1, we can show {\lambda }_0\in \mathrm{\Lambda }. The conclusion (ii) follows from Lemma 3.1 and the definition of {\lambda }^{\ast }. This completes the proof of Theorem 3.2. □

Lemma 3.3 Suppose that (H1)-(H3) hold and that one of (H5) and (H6) holds. If Λ is nonempty, then

• (i) Λ is bounded from above, that is, ;

• (ii) {\lambda }^{\ast }\in \mathrm{\Lambda }.

Proof (i) Suppose to the contrary that there exists an increasing sequence {\{{\lambda }_n\}}_1^{+\mathrm{\infty }}\subset \mathrm{\Lambda } such that {lim}_{n\to +\mathrm{\infty }}{\lambda }_n=+\mathrm{\infty }. Set x_{{\lambda }_n}\in K\mathrm{\setminus }\{\theta \} is a fixed point of C_{{\lambda }_n}, that is, C_{{\lambda }_n}{x}_{{\lambda }_n}=x_{{\lambda }_n}. There are two cases to be considered.

Case 1. {\{x_{{\lambda }_n}\}}_1^{+\mathrm{\infty }} is bounded, that is, there exists a constant M>0 such that \parallel x_{{\lambda }_n}\parallel \le M for n=1,2,\dots . Hence, from (H1), (H3), and (4) we have

M\ge \parallel x_{{\lambda }_n}\parallel =(C_{{\lambda }_n}{x}_{{\lambda }_n})(1)\ge \frac{1}{3}{\lambda }_n{\int }_0^1{s}^{2}f(s,x_{{\lambda }_n}(s))ds\ge \frac{F_0}{3}{\lambda }_n\to +\mathrm{\infty },

which is a contradiction.

Case 2. {\{x_{{\lambda }_n}\}}_1^{+\mathrm{\infty }} is unbounded, that is, there exists a subsequence of {\{x_{{\lambda }_n}\}}_1^{+\mathrm{\infty }}, still denoted by {\{x_{{\lambda }_n}\}}_1^{+\mathrm{\infty }}, such that {lim}_{n\to +\mathrm{\infty }}\parallel x_{{\lambda }_n}\parallel =+\mathrm{\infty }.

When (H5) holds, take L>\frac{72}{{\lambda }_1}, there exists N_1>0 such that f(t,x)\ge Lx for x\ge N_1, t\in [\frac{1}{2},1]. Choose n_1 such that \parallel x_{{\lambda }_{n_1}}\parallel >6N_1. Thus, f(t,\frac{1}{6}\parallel x_{{\lambda }_{n_1}}\parallel )\ge \frac{1}{6}L\parallel x_{{\lambda }_{n_1}}\parallel, t\in [\frac{1}{2},1]. Moreover, from (H1) and the definition of K, we have

\parallel x_{{\lambda }_{n_1}}\parallel =(C_{{\lambda }_{n_1}}{x}_{{\lambda }_{n_1}})(1)\ge \frac{1}{3}{\lambda }_1{\int }_{\frac{1}{2}}^1{s}^{2}f(s,\frac{1}{6}\parallel x_{{\lambda }_{n_1}}\parallel )ds>\frac{1}{72}{\lambda }_{1}L\parallel x_{{\lambda }_{n_1}}\parallel >\parallel x_{{\lambda }_{n_1}}\parallel ,

which is a contradiction.

When (H6) holds, choose ϵ>0 such that \frac{1}{3}(q_{\mathrm{\infty }}-ϵ)>1. There exists N_2>1 such that q(x)\ge (q_{\mathrm{\infty }}-ϵ)x for x\ge N_2. Choose n_2 such that \parallel x_{{\lambda }_{n_2}}\parallel >N_2, so

q(x_{{\lambda }_{n_2}}(1))=q(\parallel x_{{\lambda }_{n_2}}\parallel )\ge (q_{\mathrm{\infty }}-ϵ)\parallel x_{{\lambda }_{n_2}}\parallel .

Moreover,

\parallel x_{{\lambda }_{n_2}}\parallel =(C_{{\lambda }_{n_2}}{x}_{{\lambda }_{n_2}})(1)\ge \frac{1}{3}q(x_{{\lambda }_{n_2}}(1))\ge \frac{1}{3}(q_{\mathrm{\infty }}-ϵ)\parallel x_{{\lambda }_{n_2}}\parallel >\parallel x_{{\lambda }_{n_2}}\parallel ,

which is a contradiction.

Consequently, we find that Λ is bounded from above.

• (ii) By the definition of {\lambda }^{\ast }, there exists a nondecreasing sequence {\{{\lambda }_n\}}_1^{+\mathrm{\infty }} such that {lim}_{n\to +\mathrm{\infty }}{\lambda }_n={\lambda }^{\ast }. Let x_{{\lambda }_n}\in K\mathrm{\setminus }\{\theta \} be a fixed point of C_{{\lambda }_n}. Arguing similarly as above in case 2, we can show that {\{x_{{\lambda }_n}\}}_1^{+\mathrm{\infty }} is a bounded subset in K, that is, there exists a constant M>0 such that \parallel x_{{\lambda }_n}\parallel \le M, n=1,2,\dots ; on the other hand, note that

  |x_{{\lambda }_n}(t_1)-x_{{\lambda }_n}(t_2)|\le {\lambda }^{\ast }{\int }_0^1|G(t_1,s)-G(t_2,s)|f(s,M)ds+\frac{4}{3}q(M)|t_1-t_2|,

we see that {\{x_{{\lambda }_n}\}}_1^{+\mathrm{\infty }} is an equicontinuous subset in K. Consequently, by an application of the Arzela-Ascoli Theorem we conclude that {\{x_{{\lambda }_n}\}}_1^{+\mathrm{\infty }} is a relatively compact set in K. So, there exists a subsequence \{x_{{\lambda }_{n_i}}\}\subset \{x_{{\lambda }_n}\} converging to x^{\ast }\in K. Note that

x_{{\lambda }_{n_i}}(t)={\lambda }_{n_i}{\int }_0^{1}G(t,s)f(s,x_{{\lambda }_{n_i}}(s))ds+q(x_{{\lambda }_{n_i}}(1))(\frac{1}{2}{t}^2-\frac{1}{6}{t}^3).

By taking the limit we have x^{\ast }(t)=(C_{{\lambda }^{\ast }}{x}^{\ast })(t)\ge \frac{{\lambda }_1}{3}{F}_0{t}^2, that is, {\lambda }^{\ast }\in \mathrm{\Lambda }. The proof is complete. □

Theorem 3.4 Suppose that (H1)-(H4) hold and that one of (H5) and (H6) holds. Then, there exists a {\lambda }^{\ast }\ge {\lambda }_{\ast }>0 such that BVP (1) has at least two, one, and no positive solutions for , and \lambda >{\lambda }^{\ast }, respectively.

Proof Theorem 3.2 implies (0,{\lambda }_{\ast }]\subset \mathrm{\Lambda }, so {\lambda }^{\ast }\ge {\lambda }_{\ast }>0. From Lemmas 3.1 and 3.3, we have (0,{\lambda }^{\ast }]=\mathrm{\Lambda }. Therefore, from the definition of {\lambda }^{\ast } we only to prove that C_{\lambda } has at least two fixed points in K\mathrm{\setminus }\{\theta \} for \lambda \in (0,{\lambda }_{\ast }].

Now, given \lambda \in (0,{\lambda }_{\ast }]. Theorem 3.2 means that C_{\lambda } has at least one fixed point x_{\lambda ,1}\in K\mathrm{\setminus }\{\theta \} which satisfies \parallel x_{\lambda ,1}\parallel \le 1.

Let . Note that t(3-t)\le 2 for t\in [0,1], so for x\in K with \parallel x\parallel =1, i.e., x\in \partial K_1, we have

(9)

When (H5) holds, take L^{\prime }>\frac{72}{\lambda }, there exists N_1^{\prime }>1 such that f(t,x)\ge L^{\prime }x for x\ge N_1^{\prime }, t\in [\frac{1}{2},1]. Set . Then {\overline{K}}_1\subset K_2. If x\in \partial K_2, we have

\parallel C_{\lambda }x\parallel =(C_{\lambda }x)(1)\ge \frac{\lambda }{3}{\int }_{\frac{1}{2}}^1{s}^{2}f(s,\frac{1}{6}\parallel x\parallel )ds>\frac{\lambda L^{\prime }}{72}\parallel x\parallel >\parallel x\parallel .

When (H6) holds, from the proof of Lemma 3.3 we can set . Then {\overline{K}}_1\subset K_2^{\prime }. If x\in \partial K_2^{\prime }, we have \parallel C_{\lambda }x\parallel =C_{\lambda }x(1)\ge \frac{1}{3}q(x(1))\ge \frac{1}{3}(q_{\mathrm{\infty }}-ϵ)x(1)>\parallel x\parallel.

Consequently, in virtue of Lemma 1.1 we find that C_{\lambda } has another fixed point x_{\lambda ,2} with

x_{\lambda ,2}\in \{\begin{array}{ll}{\overline{K}}_2-K_1,& \text{as (H5) holds},\\ {\overline{K}}_2^{\prime }-K_1,& \text{as (H6) holds}.\end{array}

Equation (9) implies that C_{\lambda } has no fixed points in \partial K_1. In conclusion, for \lambda \in (0,{\lambda }_{\ast }], C_{\lambda } has at least two fixed points x_{\lambda ,1} and x_{\lambda ,2} in K with . The proof is complete. □

Remark 3.1 In the above results, we can replace (H5) with the following condition: there exists {ϵ}_0\in (0,1) such that {lim}_{x\to +\mathrm{\infty }}{min}_{t\in [{ϵ}_0,1]}\frac{f(t,x)}{x}=+\mathrm{\infty }.

In the following, we give some sufficient conditions that BVP (1) has no positive solutions.

Theorem 3.5 Suppose that there exists a nonnegative integrable function a(t) such that f(t,x)\ge a(t)x, t\in [0,1], x\in [0,+\mathrm{\infty }) and a^{\ast }:={\int }_0^1{s}^4(3-s)a(s)ds>0. Then BVP (1) has no positive solutions for \lambda >\frac{9}{a^{\ast }}.

Proof Assume to the contrary that x_{\lambda }\in K\mathrm{\setminus }\{\theta \} is a solution of BVP (1), then \parallel x_{\lambda }\parallel =(C_{\lambda }{x}_{\lambda })(1)\ge \frac{1}{9}\parallel x_{\lambda }\parallel \lambda {\int }_0^1{s}^4(3-s)a(s)ds>\parallel x_{\lambda }\parallel, which is a contradiction. The proof is complete. □

Similarly to the proof of Theorem 3.5, we can easily obtain the following results.

Theorem 3.6 Suppose that there exist an integrable function a_1(t)\ge 0 and a number b\in [0,3) such that f(t,x)\le a_1(t)x, q(x)\le bx, t\in [0,1], x\in [0,+\mathrm{\infty }) and a_1^{\ast }:={\int }_0^1{s}^2(3-s)a_1(s)ds>0. Then BVP (1) has no positive solutions for .

Theorem 3.7 Suppose that q(x)>3x, x\in [0,+\mathrm{\infty }). Then BVP (1) has no positive solutions for \lambda \ge 0.

Remark 3.2 When q(x)\equiv 0, BVP (1) becomes a cantilever beam problem (5). In this case, we can delete the conditions on q in Theorems 3.2, 3.4-3.6 and obtain the following corresponding results for BVP (5).

Suppose that (H1) and (H3) hold. Then BVP (5) has minimal and maximal solutions in [u_0,v_0] for \lambda \in (0,\frac{2}{F_1}]. Further, if , then there exists {\lambda }^{\ast }\ge max\{\frac{2}{F_1},\frac{4}{F_{\mathrm{\infty }}}\} such that BVP (5) has at least one and has no positive solutions for and \lambda >{\lambda }^{\ast }, respectively; if F_{\mathrm{\infty }}=0 then BVP (5) has at least one positive solution for \lambda >0.

Suppose that (H1), (H3), and (H5) hold. Then {\lambda }^{\ast }\ge \frac{2}{F_1} and BVP (5) has at least two, one and has no positive solutions for , and \lambda >{\lambda }^{\ast }, respectively.

Under the conditions in Theorem 3.5, BVP (5) has no positive solutions for \lambda >\frac{9}{a^{\ast }}.

Suppose that a_1(t) and a_1^{\ast } satisfy the conditions in Theorem 3.6, then BVP (5) has no positive solutions for .

Remark 3.3 (i) We give an example to illustrate Theorem 3.2. Let f(t,x)=t^3+\frac{1}{5}ln(1+x), and

q(x)=\{\begin{array}{ll}sinx,& 0\le x\le \frac{\pi }{2},\\ \frac{2}{\pi }x,& \frac{\pi }{2}\le x\le \pi ,\frac{3\pi }{2}\le x\le 2\pi ,\\ 2+|sinx|,& \pi \le x\le \frac{3\pi }{2},\\ 4,& x\ge 2\pi .\end{array}

By straightforward calculations we see that F_0=\frac{1}{6}, F_1=\frac{1}{5}(1+\frac{ln2}{2}), q(1)=sin1, {\lambda }_{\ast }=\frac{2-q(1)}{F_1}\doteq 7.361, F_{\mathrm{\infty }}=\frac{1}{5}, and Q_{\mathrm{\infty }}=0. So the conditions in Theorem 3.2 are satisfied. Therefore, by Theorem 3.2 we find that there exists {\lambda }^{\ast }\ge \frac{4}{F_{\mathrm{\infty }}}=20 such that BVP (1) has minimal and maximal solutions in [u_0,v_0] for , has at least one positive solution for and has no positive solutions for \lambda >{\lambda }^{\ast }, where u_0(t)=\frac{\lambda }{18}{t}^2 and v_0(t)=t^2.

We give another example to illustrate Theorem 3.4. Let f(t,x)=\frac{t}{2}(1+x^2)+\frac{1}{8}{e}^t\sqrt{x}, and

q(x)=\{\begin{array}{ll}\frac{1}{8}{x}^{\frac{4}{3}},& 0\le x\le 8,\\ 2,& x\ge 8.\end{array}

A straightforward calculation can show that F_0=\frac{1}{8}, f_{\mathrm{\infty }}=+\mathrm{\infty }, q(1)=\frac{1}{8}, F_1=\frac{11}{24}, and {\lambda }_{\ast }=\frac{45}{11} Therefore, the conditions (H1)-(H5) hold. Thus, by Theorem 3.4 we see that there exists {\lambda }^{\ast }\ge \frac{45}{11} such that BVP (1) has at least two, one, and no positive solutions for , , and \lambda >{\lambda }^{\ast }, respectively.

• (ii) In Theorems 3.5-3.7, we do not require f and q to be monotone in x. For example, let f(t,x)=\frac{tx}{1+|cosx|} and

  q(x)=\{\begin{array}{ll}x|sinx|,& 0\le x\le 2\pi ,\\ 0,& x\ge 2\pi .\end{array}

Take a_1(t)=t, b=1, then the conditions in Theorem 3.6 are satisfied and a_1^{\ast }=\frac{11}{20}. So by Theorem 3.6 we find that BVP (1) has no positive solutions for .

4 Uniqueness and dependence on parameter

In this section, we will apply cone theory to further study the uniqueness of solution for BVP (1) in P\mathrm{\setminus }\{\theta \} and the dependence of such a positive solution on the parameter λ. The following hypotheses are needed:

(H7) q(1)\ne 0 and for all x\in [0,+\mathrm{\infty }) and r\in (0,1), there exists \alpha (r)\in (0,1) such that q(rx)\ge r^{\alpha (r)}q(x);

(H8) f(t,1)\not\equiv 0 and f(t,rx)\ge rf(t,x) for r\in (0,1), t\in [0,1], x\in [0,+\mathrm{\infty });

(H9) for all t\in [0,1], x\in [0,+\mathrm{\infty }) and r\in (0,1), there exists \beta (r)\in (0,1) such that f(t,rx)\ge r^{\beta (r)}f(t,x).

Remark 4.1 The inequalities in (H7), (H8), and (H9) are equivalent to the following inequalities, respectively:

\begin{array}{c}q(sx)\le s^{\alpha (\frac{1}{s})}q(x),s>1,x\in [0,+\mathrm{\infty }),\hfill \\ f(t,sx)\le sf(t,x),s>1,t\in [0,1],x\in [0,+\mathrm{\infty }),\hfill \\ f(t,sx)\le s^{\beta (\frac{1}{s})}f(t,x),s>1,t\in [0,1],x\in [0,+\mathrm{\infty }).\hfill \end{array}

Let e(t)=t^2 and define P_e as in Section 1. It is obvious that P_e\subset P and if x\in P_e then x(0)=0 and x(t)>0, t\in (0,1].

Remark 4.2 (H2) and (H7) imply q(x)>0 for x>0. Moreover, q(x(1))>0 for x\in P_e.

Remark 4.3 Let x_{\lambda } be a solution for BVP (1) in P\mathrm{\setminus }\{\theta \}. If (H2) and (H7) hold, then x_{\lambda }\in P_e. Indeed, from Theorem 2.4 we have x_{\lambda }(1)=\parallel x_{\lambda }\parallel. So Remark 4.2 implies q(x_{\lambda }(1))>0. Note that

\frac{q(x_{\lambda }(1))}{3}{t}^2\le x_{\lambda }(t)=(C_{\lambda }{x}_{\lambda })(t)\le \frac{1}{2}(\lambda {\int }_0^{1}sf(s,x_{\lambda }(s))ds+q(x_{\lambda }(1)))t^2,

we conclude x_{\lambda }\in P_e.

So, in this section, we only need to consider the unique solution for BVP (1) in P_e.

Lemma 4.1 Assume that (H2) and (H7) hold. Then B has a unique fixed point x_0 in P_e, moreover, constructing successively the sequence w_n=B{w}_{n-1} (n=1,2,\dots) for any initial value w_0\in P_e, we have {lim}_{n\to +\mathrm{\infty }}\parallel w_n-x_0\parallel =0.

Proof For any x\in P_e, we have \frac{1}{3}q(x(1))t^2\le Bx(t)\le \frac{1}{2}q(x(1))t^2, which means B(P_e)\subset P_e. For all x\in P_e, r\in (0,1), from (H7) we have B(rx)(t)\ge r^{\alpha (r)}Bx(t). Consequently, the conclusion follows from Lemma 1.2. This completes the proof. □

Lemma 4.2 Assume that (H1), (H2), (H7), and (H8) hold. Then

• (i) C_{\lambda }:P_e\to P_e is an increasing operator;

• (ii) for any \lambda \ge 0 and x\in P_e, there exists \phi (\lambda ,x)\in (0,1) such that Bx\ge \phi (\lambda ,x)C_{\lambda }x;

• (iii) for [u,v]\subset P_e and r\in (0,1), there exists \eta (r,u,v)>0 such that

  C_{\lambda }(rx)\ge r(1+\eta (r,u,v))C_{\lambda }x,\mathrm{\forall }x\in [u,v].

Proof The conclusion (i) follows from (H1), (H2), (H7), and (4).

The proof of (ii). For given \lambda \ge 0, x\in P_e, from (H1) and (4) we have

(C_{\lambda }x)(t)\le \frac{1}{2}{t}^2(\lambda {\int }_0^{1}sf(s,\parallel x\parallel )ds+q(x(1)))\le \frac{3(\lambda {\int }_0^{1}sf(s,\parallel x\parallel )ds+q(x(1))}{2q(x(1))}Bx(t).

Let

\phi (\lambda ,x)=\frac{2q(x(1))}{3(\lambda {\int }_0^{1}sf(s,\parallel x\parallel )ds+q(x(1)))},

(10)

then and

Bx\ge \phi (\lambda ,x)C_{\lambda }x.

(11)

The proof of (iii). For any x\in [u,v], u,v\in P_e, from (10) and (11) we have

Bx\ge \frac{2q(x(1))}{3(\lambda {\int }_0^{1}sf(s,\parallel x\parallel )ds+q(x(1))}{C}_{\lambda }x\ge \frac{2q(u(1))}{3(\lambda {\int }_0^{1}sf(s,\parallel v\parallel )ds+q(v(1)))}{C}_{\lambda }x.

Moreover, from (H7) and (H8) we have

C_{\lambda }(rx)\ge \lambda rAx+r^{\alpha (r)}Bx\ge r(1+\eta (r,u,v))C_{\lambda }x,\mathrm{\forall }r\in (0,1),x\in [u,v],

where \eta (r,u,v)=\frac{2(r^{\alpha (r)}-r)q(u(1))}{3r(\lambda {\int }_0^{1}sf(s,\parallel v\parallel )ds+q(v(1)))}>0. This completes the proof. □

Lemma 4.3 Assume that (H1), (H2), (H7), and (H8) hold. Then C_{\lambda } has a unique fixed point x_{\lambda } in P_e iff there exists y_{\lambda }\in P_e such that C_{\lambda }{y}_{\lambda }\le y_{\lambda }. Moreover, constructing successively the sequence w_n=C_{\lambda }{w}_{n-1} (n=1,2,\dots) for any initial value w_0\in P_e, we have

\underset{n\to +\mathrm{\infty }}{lim}\parallel w_n-x_{\lambda }\parallel =0.

(12)

Proof ‘⇒’ Let x_{\lambda } be a fixed point of C_{\lambda } in P_e, i.e., C_{\lambda }{x}_{\lambda }=x_{\lambda }. Taking y_{\lambda }=x_{\lambda }, we obtain C_{\lambda }{y}_{\lambda }\le y_{\lambda }.

‘⇐’ By virtue of Lemma 4.1, B has a unique fixed point x_0 in P_e. Moreover,

x_0\le C_{\lambda }{x}_0,\lambda \ge 0.

(13)

Now, we are going to prove

x_0\le y_{\lambda },\lambda \ge 0.

(14)

Let {\tau }_0=inf\{\tau >0\mid x_0\le \tau y_{\lambda }\}, then {\tau }_0\le 1. Otherwise, {\tau }_0>1, from Lemma 4.2 we have

x_0\le C_{\lambda }{x}_0\le C_{\lambda }({\tau }_0{y}_{\lambda })\le \frac{{\tau }_0}{1+\eta (\frac{1}{{\tau }_0},x_0,{\tau }_0{y}_{\lambda })}{C}_{\lambda }{y}_{\lambda }\le \frac{{\tau }_0}{1+\eta (\frac{1}{{\tau }_0},x_0,{\tau }_0{y}_{\lambda })}{y}_{\lambda }.

By the definition of {\tau }_0, we get a contradiction {\tau }_0\le \frac{{\tau }_0}{1+\eta (\frac{1}{{\tau }_0},x_0,{\tau }_0{y}_{\lambda })}. Thus, (14) holds.

Set x_n=C_{\lambda }{x}_{n-1}, y_n=C_{\lambda }{y}_{n-1}, y_0=y_{\lambda }, n=1,2,\dots . From (13) and (14) we have

x_0\le x_1\le \cdots \le x_n\le \cdots \le y_n\le \cdots \le y_1\le y_0=y_{\lambda }.

(15)

Lemma 2.3 implies that \{x_n\} and \{y_n\} converge to fixed points x^{\ast } and y^{\ast } of C_{\lambda }, respectively. From (15), we have

x_0\le x_1\le \cdots \le x_n\le \cdots \le x^{\ast }\le y^{\ast }\le \cdots \le y_n\le \cdots \le y_1\le y_{\lambda }.

(16)

To prove that C_{\lambda } has only one fixed point in [x_0,y_{\lambda }], let

{\mu }_n=sup\{\tau >0\mid x_n\ge \tau y_n\},n=0,1,2,\dots ,

(17)

then

(18)

From (16)-(18) we infer that , which means that {lim}_{n\to +\mathrm{\infty }}{\mu }_n=\mu \le 1. We assert that \mu =1. Otherwise, for n\ge 1, then by Lemma 4.2 we deduce that

x_{n+1}\ge C_{\lambda }({\mu }_n{y}_n)\ge C_{\lambda }\left(\frac{{\mu }_n}{\mu }\mu y_n\right)\ge \frac{{\mu }_n}{\mu }{C}_{\lambda }(\mu y_n)\ge {\mu }_n(1+\eta (\mu ,x_0,y_{\lambda }))y_{n+1}.

By (17), we have {\mu }_{n+1}\ge {\mu }_n(1+\eta (\mu ,x_0,y_{\lambda })), moreover, \mu \ge \mu (1+\eta (\mu ,x_0,y_{\lambda })), which is a contradiction. So \mu =1. Thus, by (16) and (18) we have

\parallel y^{\ast }-x^{\ast }\parallel \le \parallel y_n-x_n\parallel \le (1-{\mu }_n)\parallel y_0\parallel \to 0\text{as }n\to +\mathrm{\infty },

which means that x^{\ast }=y^{\ast }:=x_{\lambda } is the unique fixed point of C_{\lambda } in [x_0,y_{\lambda }].

Now, we prove that x^{\ast } is the unique fixed point of C_{\lambda } in P_e. By the above proof, we only need to show that C_{\lambda } does not have any fixed point in P_e\setminus [x_0,y_{\lambda }]. If \overline{x} is a fixed point of C_{\lambda } in P_e\setminus [x_0,y_{\lambda }]. Let

\overline{\mu }=sup\{\tau >0|\tau x^{\ast }\le \overline{x}\le \frac{1}{\tau }{x}^{\ast }\}.

(19)

It is evident that . If , then x_0\le x^{\ast }\le \frac{1}{\overline{\mu }}{x}^{\ast }\le \frac{1}{\overline{\mu }}{y}_{\lambda }. By Lemma 4.2 we have

\begin{array}{rl}\overline{\mu }(1+\eta (\overline{\mu },x_0,\frac{1}{\overline{\mu }}{y}_{\lambda }))x^{\ast }& \le C_{\lambda }\left(\overline{\mu }{x}^{\ast }\right)\le \overline{x}=C_{\lambda }\overline{x}\le C_{\lambda }\left(\frac{1}{\overline{\mu }}{x}^{\ast }\right)\\ \le \frac{1}{\overline{\mu }(1+\eta (\overline{\mu },x_0,\frac{1}{\overline{\mu }}{y}_{\lambda }))}{x}^{\ast }.\end{array}

Thus, from (19) we have \overline{\mu }\ge \overline{\mu }(1+\eta (\overline{\mu },x_0,\frac{1}{\overline{\mu }}{y}_{\lambda })), which is a contradiction. So \overline{\mu }=1. Moreover, \overline{x}=x^{\ast }, which implies the contradiction: \overline{x}=x^{\ast }\in [x_0,y_{\lambda }] and \overline{x}\in P_e\setminus [x_0,y_{\lambda }].

Finally, the iterative scheme and (12) can be proved in a similar way to the proof of Theorem 3.4 of [21], here it is omitted. The proof is complete. □

Theorem 4.4 Assume that (H1), (H2), (H7), and (H8) hold. Then there exists a {\lambda }^{\ast }>0 such that BVP (1) has a unique solution x_{\lambda } in P_e for \lambda \in [0,{\lambda }^{\ast }) and does not have any solution in P_e for \lambda \ge {\lambda }^{\ast }. Moreover, set w_n=\lambda A{w}_{n-1}+B{w}_{n-1} (n=1,2,\dots) for any w_0\in P_e, then (12) holds.

Proof By Lemma 4.1, B has the unique fixed point x_0 in P_e. So x_0(t)=q(x_0(1))(\frac{1}{2}{t}^2-\frac{1}{6}{t}^3), moreover, \parallel x_0\parallel =x_0(1)=\frac{1}{3}q(x_0(1))>0. Let {\rho }_0=\frac{1}{2}{\int }_0^1\frac{sf(s,x_0(1))}{x_0(1)}ds, we have

(A{x}_0)(t)\le \frac{t^2}{2}{\int }_0^{1}sf(s,x_0(1))ds\le q(x_0(1))(\frac{1}{2}{t}^2-\frac{1}{6}{t}^3){\rho }_0={\rho }_0{x}_0(t).

(20)

Set \mathrm{\Delta }=\{\lambda \ge 0\mid \text{ there exists }{x}_{\lambda }\in P_e\text{ such that }{C}_{\lambda }{x}_{\lambda }=x_{\lambda }\}. Lemma 4.3 implies that

\mathrm{\Delta }=\{\lambda \ge 0\mid \text{ there exists }{y}_{\lambda }\in P_e\text{ such that }{C}_{\lambda }{y}_{\lambda }\le y_{\lambda }\}.

(21)

Similarly to the proof of Lemma 3.1, we can show that \lambda \in \mathrm{\Delta } implies [0,\lambda ]\subset \mathrm{\Delta }.

Now, take s_0>1 and let {\lambda }_0=\frac{1}{{\rho }_0}(1-s_0^{\alpha (\frac{1}{s_0})-1}) and y_{{\lambda }_0}=s_0{x}_0, then {\lambda }_0>0 and y_{{\lambda }_0}\in P_e. By (H7), (H8), and (20), we have C_{{\lambda }_0}{y}_{{\lambda }_0}\le {\lambda }_0{s}_0{\rho }_0{x}_0+s_0^{\alpha (\frac{1}{s_0})}{x}_0\le y_{{\lambda }_0}, that is, {\lambda }_0\in \mathrm{\Delta }. Moreover, [0,{\lambda }_0]\subset \mathrm{\Delta }.

Let {\lambda }^{\ast }=sup\mathrm{\Delta }, then {\lambda }^{\ast }\ge {\lambda }_0>0. We assert that {\lambda }^{\ast }\notin \mathrm{\Lambda }. Indeed, if {\lambda }^{\ast }=+\mathrm{\infty }, from the definition of {\lambda }^{\ast } it is obvious that {\lambda }^{\ast }\notin \mathrm{\Lambda }. Suppose that and {\lambda }^{\ast }\in \mathrm{\Lambda }. Then by (14) and (21) there exists x_0\le y_{{\lambda }^{\ast }}\in P_e such that C_{{\lambda }^{\ast }}{y}_{{\lambda }^{\ast }}\le y_{{\lambda }^{\ast }}. Similarly to the proof of (20), we have

(A{y}_{{\lambda }^{\ast }})(t)\le \left(\frac{1}{2}{\int }_0^1\frac{sf(s,\parallel y_{{\lambda }^{\ast }}\parallel )}{x_0(1)}ds\right)x_0(t)\le \left(\frac{1}{2}{\int }_0^1\frac{sf(s,\parallel y_{{\lambda }^{\ast }}\parallel )}{x_0(1)}ds\right)(B{y}_{{\lambda }^{\ast }})(t).

Denote {\rho }_1=\frac{1}{2}{\int }_0^1\frac{sf(s,\parallel y_{{\lambda }^{\ast }}\parallel )}{x_0(1)}ds, then

(22)

Set v=s_1{y}_{{\lambda }^{\ast }} for given s_1>1, then v\in P_e. Since , we can choose \delta >0 such that . Therefore, from (22) we have

\begin{array}{rcl}{C}_{{\lambda }^{\ast }+\delta }(v)& \le & ({\lambda }^{\ast }+\delta )s_{1}A{y}_{{\lambda }^{\ast }}+s_1^{\alpha (\frac{1}{s_1})}B{y}_{{\lambda }^{\ast }}\\ \le & s_1({\lambda }^{\ast }A{y}_{{\lambda }^{\ast }}+\delta {\rho }_{1}B{y}_{{\lambda }^{\ast }}+B{y}_{{\lambda }^{\ast }}-\delta {\rho }_{1}B{y}_{{\lambda }^{\ast }})\le v.\end{array}

This means that {\lambda }^{\ast }+\delta \in \mathrm{\Delta }, which is a contradiction to the definition of {\lambda }^{\ast }. So, \mathrm{\Delta }=[0,{\lambda }^{\ast }). Consequently, an application of Lemma 4.3 completes the proof. □

In what follows, we assume that x_0 is the unique fixed point of B in P_e, x_{\lambda } is the unique fixed point of C_{\lambda } in P_e and {\lambda }^{\ast }=sup\mathrm{\Delta }.

Theorem 4.5 Assume that (H1), (H2), (H7), and (H8) hold. Then x_{\lambda } depends upon the parameter λ as follows:

• (i) x_{\lambda } is nondecreasing with respect to λ for \lambda \in [0,{\lambda }^{\ast });

• (ii) x_{\lambda } is continuous with respect to λ for \lambda \in [0,{\lambda }^{\ast });

• (iii) {lim}_{\lambda \to 0+}\parallel x_{\lambda }-x_0\parallel =0 and {lim}_{\lambda \to {\lambda }^{\ast }-0}\parallel x_{\lambda }\parallel =+\mathrm{\infty }.

Proof (i) Let {\lambda }_1,{\lambda }_2\in [0,{\lambda }^{\ast }) with {\lambda }_1\le {\lambda }_2. Since C_{{\lambda }_1}{x}_{{\lambda }_2}\le C_{{\lambda }_2}{x}_{{\lambda }_2}=x_{{\lambda }_2}, from the proof of Lemma 4.3, we find that the unique fixed x_{{\lambda }_1} of C_{{\lambda }_1} belongs to [x_0,x_{{\lambda }_2}], which means that x_{{\lambda }_1}\le x_{{\lambda }_2}.

(ii)Let {\lambda }_0\in (0,{\lambda }^{\ast }). In order to prove {lim}_{\lambda \to {\lambda }_0^{-}}\parallel x_{{\lambda }_0}-x_{\lambda }\parallel =0, let sequence \{{\lambda }_n\} satisfy

By virtue of the above conclusion (i) we have

x_{{\lambda }_1}\le x_{{\lambda }_2}\le \cdots \le x_{{\lambda }_n}\le \cdots \le x_{{\lambda }_0},

(23)

which implies that \{x_{{\lambda }_n}\} is a bounded subset in P. Further, similarly to the proof of the conclusion (ii) in Lemma 3.3 we see that \{x_{{\lambda }_n}\} converges to x^{\ast }\in P. From (23) we have x^{\ast }\in [x_{{\lambda }_1},x_{{\lambda }_0}], which leads to x^{\ast }\in P_e. Note that

x_{{\lambda }_n}={\lambda }_{n}A{x}_{{\lambda }_n}+B{x}_{{\lambda }_n}.

By taking the limit we have x^{\ast }={\lambda }_{0}A{x}^{\ast }+B{x}^{\ast }=C_{{\lambda }_0}{x}^{\ast }. Since C_{{\lambda }_0} has only one fixed point in P_e, then x^{\ast }=x_{{\lambda }_0}. This means that \parallel x_{{\lambda }_0}-x_{\lambda }\parallel \to 0 as \lambda \to {\lambda }_0^{-}.

A similar argument can show that for any {\lambda }_0\in [0,{\lambda }^{\ast }), \parallel x_{\lambda }-x_{{\lambda }_0}\parallel \to 0 as \lambda \to {\lambda }_0^{+}. Thus, the proof of (ii) is complete.

• (iii) It is obvious from the above conclusion (ii) that {lim}_{\lambda \to 0+}\parallel x_{\lambda }-x_0\parallel =0.

In order to finish the proof of {lim}_{\lambda \to {\lambda }^{\ast }-0}\parallel x_{\lambda }\parallel =+\mathrm{\infty }, we consider two cases.

Case 1. {\lambda }^{\ast }=+\mathrm{\infty }.

Since x_{\lambda }=\lambda A{x}_{\lambda }+B{x}_{\lambda }\ge \lambda A{x}_0, then \parallel x_{\lambda }\parallel \ge \lambda \parallel A{x}_0\parallel, which means {lim}_{\lambda \to {\lambda }^{\ast }-0}\parallel x_{\lambda }\parallel =+\mathrm{\infty }.

Case 2. .

By the above conclusion (i) we have {lim}_{\lambda \to {\lambda }^{\ast }-0}\parallel x_{\lambda }\parallel \le +\mathrm{\infty }. Suppose to the contrary that . Similarly to the case 2 in the proof of Lemma 3.3, we conclude that C_{{\lambda }^{\ast }} has a fixed point x^{\ast }\in P\mathrm{\setminus }\{\theta \}. From Remark 4.3 we have x^{\ast }\in P_e. So {\lambda }^{\ast }\in [0,{\lambda }^{\ast }), which is a contradiction. This ends the proof. □

Now, we give an estimate for critical value {\lambda }^{\ast } in Theorem 4.4. If (H1) and (H8) hold, then

\frac{f(t,x)}{x}\le f(t,1)\le \underset{t\in [0,1]}{max}f(t,1),x>1,t\in [0,1].

Moreover, F_{\mathrm{\infty }}={lim\hspace{0.17em}sup}_{x\to +\mathrm{\infty }}{max}_{t\in [0,1]}\frac{f(t,x)}{x}\in [0,+\mathrm{\infty }).

Theorem 4.6 Assume that (H1), (H2), (H7), and (H8) hold. Then

(24)

Proof For any ϵ>0, there exists r_0\in (0,1) such that

q(1)\le \frac{1}{r_0}\text{and}f(t,\frac{1}{r})\le \frac{1}{r}(F_{\mathrm{\infty }}+ϵ),r\le r_0,t\in [0,1],

(25)

Note that \frac{r_0^{\alpha (r_0)}}{r_0}>1, we can choose a sufficiently large positive integer number k such that {(\frac{r_0^{\alpha (r_0)}}{r_0})}^k\ge \frac{1}{r_0}, that is,

{\left(\frac{1}{r_0^{\alpha (r_0)}}\right)}^k\le \frac{1}{r_0^{k-1}}.

(26)

Let w(t)={(\frac{1}{r_0})}^{k}e(t), then, from (4), (25), and (26) we have

\begin{array}{c}(Aw)(t)\le \frac{t^2}{2}{\int }_0^{1}sf(s,\frac{1}{r_0^k})ds\le \frac{t^2}{2r_0^{k-1}}{\int }_0^{1}sf(s,\frac{1}{r_0})ds\le \frac{1}{4}(F_{\mathrm{\infty }}+ϵ)w(t),\hfill \\ (Bw)(t)\le \frac{t^2}{2}q\left(\frac{1}{r_0^k}\right)\le \frac{t^2}{2}\left(\frac{1}{r_0^{\alpha (r_0)}}\right)q\left(\frac{1}{r_0^{k-1}}\right)\le \frac{t^2}{2}{\left(\frac{1}{r_0^{\alpha (r_0)}}\right)}^{k}q(1)\le \frac{t^2}{2}{\left(\frac{1}{r_0}\right)}^k=\frac{1}{2}w(t).\hfill \end{array}

Moreover, taking {\lambda }_{ϵ}=\frac{2}{F_{\mathrm{\infty }}+ϵ}, we have

(C_{{\lambda }_{ϵ}}w)(t)={\lambda }_{ϵ}(Aw)(t)+(Bw)(t)\le \frac{1}{4}{\lambda }_{ϵ}(F_{\mathrm{\infty }}+ϵ)w(t)+\frac{1}{2}w(t)\le w(t).

Consequently, from (21) we obtain {\lambda }_{ϵ}=\frac{2}{F_{\mathrm{\infty }}+ϵ}\in [0,{\lambda }^{\ast }), that is, {\lambda }^{\ast }>\frac{2}{F_{\mathrm{\infty }}+ϵ}, which implies that (24) holds. This completes the proof. □

Remark 4.4 Different from Theorems 3.2 and 3.4, the estimate of {\lambda }^{\ast } in Theorem 4.6 does not take into account effect of q(x). This is valuable, because the conditions (H2) and (H7) cannot ensure as {lim\hspace{0.17em}sup}_{r\to 0}\alpha (r)=1. Certainly, if , then {\lambda }^{\ast }\ge \frac{2(2-Q_{\mathrm{\infty }})}{F_{\mathrm{\infty }}}. In particular, if Q_{\mathrm{\infty }}=0, then {\lambda }^{\ast }\ge \frac{4}{F_{\mathrm{\infty }}}.

Corollary 4.7 Assume that (H1), (H2), (H7), and (H9) hold. Then

• (i) BVP (1) has a unique positive solution x_{\lambda } in P_e for \lambda \in [0,+\mathrm{\infty }). Moreover, for any w_0\in P_e, set w_n=\lambda A{w}_{n-1}+B{w}_{n-1} (n=1,2,\dots), then {lim}_{n\to +\mathrm{\infty }}\parallel w_n-x_{\lambda }\parallel =0;

• (ii) x_{\lambda } is nondecreasing with respect to λ for \lambda \in [0,+\mathrm{\infty });

• (iii) x_{\lambda } is continuous with respect to λ for \lambda \in [0,+\mathrm{\infty });

• (iv) {lim}_{\lambda \to 0+}\parallel x_{\lambda }-x_0\parallel =0 and {lim}_{\lambda \to +\mathrm{\infty }}\parallel x_{\lambda }\parallel =+\mathrm{\infty }.

Proof From (H1), (H2), (H7), and (4), we see that C_{\lambda }:P_e\to P_e is increasing for any given \lambda \ge 0. Further, for any given \lambda \ge 0 we have

C_{\lambda }(rx)=\lambda A(rx)+B(rx)\ge r^{\delta (r)}{C}_{\lambda }x,x\in P_e,r\in (0,1),

where \delta (r)=max\{\alpha (r),\beta (r)\}. Thus, the conclusion (i) follows from Lemma 1.2.

From (H9), we have f(t,rx)\ge rf(t,x) for r\in (0,1), t\in [0,1] and x\in [0,+\mathrm{\infty }). Therefore, in the same way as in the proof of Theorem 4.5, we can complete the rest of the proof. □

When q(x)\equiv c>0 is a constant function, Q_{\mathrm{\infty }}=0 and Bx(t)=c(\frac{1}{2}{t}^2-\frac{1}{6}{t}^3):=x_0(t). It is evident that B satisfies (H2) and (H7). So we can obtain the following two results.

Corollary 4.8 Assume that (H1) and (H8) hold. If F_{\mathrm{\infty }}>0, then

• (i) there exists {\lambda }^{\ast }\ge \frac{4}{F_{\mathrm{\infty }}}>0 such that BVP (1) with q(x)\equiv c has a unique positive solution x_{\lambda } in P_e for \lambda \in [0,{\lambda }^{\ast }) and does not have any solution in P_e for \lambda \ge {\lambda }^{\ast }. Moreover, for any w_0\in P_e, set w_n=x_0+\lambda A{w}_{n-1} (n=1,2,\dots), then {lim}_{n\to +\mathrm{\infty }}\parallel w_n-x_{\lambda }\parallel =0;

• (ii) x_{\lambda } is nondecreasing with respect to λ for \lambda \in [0,{\lambda }^{\ast });

• (iii) x_{\lambda } is continuous with respect to λ for \lambda \in [0,{\lambda }^{\ast });

• (iv) {lim}_{\lambda \to 0+}\parallel x_{\lambda }-x_0\parallel =0 and {lim}_{\lambda \to {\lambda }^{\ast }-}\parallel x_{\lambda }\parallel =+\mathrm{\infty }.

Corollary 4.9 Assume that (H1) and (H8) hold. If F_{\mathrm{\infty }}=0, then

• (i) for any \lambda \in [0,+\mathrm{\infty }), BVP (1) with q(x)\equiv c has a unique positive solution x_{\lambda } in P_e, moreover, for any w_0\in P_e, set w_n=x_0+\lambda A{w}_{n-1} (n=1,2,\dots), then {lim}_{n\to +\mathrm{\infty }}\parallel w_n-x_{\lambda }\parallel =0;

• (ii) x_{\lambda } is nondecreasing in λ for \lambda \in [0,+\mathrm{\infty });

• (iii) x_{\lambda } is continuous with respect to λ for \lambda \in [0,+\mathrm{\infty });

• (iv) {lim}_{\lambda \to 0+}\parallel x_{\lambda }-x_0\parallel =0 and {lim}_{\lambda \to +\mathrm{\infty }}\parallel x_{\lambda }\parallel =+\mathrm{\infty }.

Corollary 4.10 Assume that (H1) and (H9) hold. Then the conclusions (i), (ii), (iii), and (iv) in Corollary  4.9 hold.

Finally, we give two concrete examples to illustrate those results in the section.

Example 1 In BVP (1), let

f(t,x)=\{\begin{array}{ll}\frac{t}{3}x,& 0\le x\le 1,\\ \frac{t}{6}(x+\sqrt{x}),& x>1\end{array}\text{and}q(x)=\{\begin{array}{ll}{x}^{\frac{3}{5}},& 0\le x\le 32,\\ 8,& x>32,\end{array}

it is obvious that the conditions (H1) and (H2) are satisfied. For any r\in (0,1),

as 0\le x\le 1, we have f(t,rx)=\frac{tr}{3}x=rf(t,x);

as x>1 and , we have f(t,rx)=\frac{tr}{3}x\ge \frac{rt}{3}(\frac{1}{2}x+\frac{1}{2}\sqrt{x})\ge \frac{tr}{6}(x+\sqrt{x})=rf(t,x);

as x>1 and rx>1, we have f(t,rx)=\frac{t}{6}(rx+\sqrt{rx})\ge \frac{tr}{6}(x+\frac{1}{\sqrt{r}}\sqrt{x})\ge rf(t,x),

that is, f(t,rx)\ge rf(t,x) for x\in [0,+\mathrm{\infty }) and t\in [0,1]. Similarly, we can obtain q(rx)\ge r^{\frac{3}{5}}q(x) for x\in [0,+\mathrm{\infty }). Therefore, the conditions (H7) and (H8) are satisfied. Note that

F_{\mathrm{\infty }}=\underset{x\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{t\in [0,1]}{max}\frac{f(t,x)}{x}=\frac{1}{6},Q_{\mathrm{\infty }}=\underset{x\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\frac{q(x)}{x}=0.

By Theorems 4.4, 4.5 and Remarks 4.3, 4.4 we see that there exists {\lambda }^{\ast }\ge 24 such that BVP (1) has a unique positive solution x_{\lambda } for \lambda \in [0,{\lambda }^{\ast }) and does not have any positive solution for \lambda \ge {\lambda }^{\ast }. Moreover, for any w_0\in P_e, set w_n(t)=\lambda A{w}_{n-1}(t)+B{w}_{n-1}(t) (n=1,2,\dots), then {lim}_{n\to \mathrm{\infty }}\parallel w_n-x_{\lambda }\parallel =0, and such solution x_{\lambda }(t) satisfies the properties (i), (ii), and (iii) in Theorem 4.5.

Example 2 In BVP (1), let f(t,x)=\frac{t(1-t)x}{1+x}, q(x)=1, t\in [0,1], x\in [0,+\mathrm{\infty }), it is easy to see that (H1) holds. For any r\in (0,1), we have

f(rx)=\frac{t(1-t)rx}{1+rx}\ge r\cdot \frac{t(1-t)x}{1+x}\ge rf(t,x),x\in [0,+\mathrm{\infty }).

So, (H8) holds. Note that

F_{\mathrm{\infty }}=\underset{x\to +\mathrm{\infty }}{lim\hspace{0.17em}sup}\underset{t\in [0,1]}{max}\frac{f(t,x)}{x}=0,

by Corollary 4.9 we find that BVP (1) has a unique positive solution x_{\lambda } for \lambda \ge 0. Moreover, for any w_0\in P_e, set w_n(t)=\frac{1}{2}{t}^2-\frac{1}{6}{t}^3+\lambda A{w}_{n-1}(t) (n=1,2,\dots), then {lim}_{n\to \mathrm{\infty }}\parallel w_n-x_{\lambda }\parallel =0, and such a solution x_{\lambda }(t) satisfies the properties (ii), (iii), and (iv) in Corollary 4.9 with x_0(t)=\frac{1}{2}{t}^2-\frac{1}{6}{t}^3.

In this example, by using Wolfram Mathematica 9.0, we can plot the graphs of solutions x_{\lambda }(t) for BVP (1) with \lambda =0,3,10,15,30,50, as the Figure 1 shows.

Figure 1

Solutions for BVP ( 1 ) with \mathit{f}\mathbf{(}\mathit{t}\mathbf{,}\mathit{x}\mathbf{)}\mathbf{=}\frac{\mathit{t}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{t}\mathbf{)}\mathit{x}}{\mathbf{1}\mathbf{+}\mathit{x}} and \mathit{q}\mathbf{(}\mathit{x}\mathbf{)}\mathbf{=}\mathbf{1}