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

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.

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

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.

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

where

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

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

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:

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

then

1. (i)

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

2. (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

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,

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

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

1. (i)

   A:P\to K is a completely continuous operator;

2. (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 \}.

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

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

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

1. (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.

2. (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

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

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.

1. (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

1. (i)

   Λ is bounded from above, that is, ;

2. (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

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

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

Moreover,

which is a contradiction.

Consequently, we find that Λ is bounded from above.

1. (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

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

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

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

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

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

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.

1. (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 .

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:

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

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

1. (i)

   C_{\lambda }:P_e\to P_e is an increasing operator;

2. (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;

3. (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

Let

then and

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

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

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

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,

Now, we are going to prove

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

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

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

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

then

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

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

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

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

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 }].