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

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 $x(t)>0$, $0<t<1$. 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.

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

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. □

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

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. □

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) $q(1)<2$;

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

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

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. □

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. □

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.

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 ${lim}_{n\to +\mathrm{\infty }}\parallel x_{{\lambda }_n}\parallel =+\mathrm{\infty }$.

which is a contradiction.

which is a contradiction.

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 $0<\lambda \le {\lambda }_{\ast }$, ${\lambda }_{\ast }<\lambda \le {\lambda }^{\ast }$ 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 $\parallel x_{\lambda ,1}\parallel \le 1$.

When (H6) holds, from the proof of Lemma 3.3 we can set $K_2^{\prime }=\{x\in K\mid \parallel x\parallel <N_2\}$. Then ${\overline{K}}_1\subset K_2^{\prime }$. If $x\in \partial K_2^{\prime }$, we have $\parallel C_{\lambda }x\parallel =C_{\lambda }x(1)\ge \frac{1}{3}q(x(1))\ge \frac{1}{3}(q_{\mathrm{\infty }}-ϵ)x(1)>\parallel x\parallel$.

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 $0<\parallel x_{\lambda ,1}\parallel <1<\parallel x_{\lambda ,2}\parallel$. 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 $0\le \lambda <\frac{6-2b}{a_1^{\ast }}$.

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 $0<F_{\mathrm{\infty }}<+\mathrm{\infty }$, 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 $0<\lambda <{\lambda }^{\ast }$ 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 $0<\lambda \le \frac{2}{F_1}$, $\frac{2}{F_1}<\lambda \le {\lambda }^{\ast }$ 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 $0\le \lambda <\frac{6}{a_1^{\ast }}$.

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 $0<\lambda \le 7.361$, has at least one positive solution for $7.361<\lambda <{\lambda }^{\ast }$ and has no positive solutions for $\lambda >{\lambda }^{\ast }$, where $u_0(t)=\frac{\lambda }{18}{t}^2$ and $v_0(t)=t^2$.

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 $0\le \lambda <\frac{80}{11}$.

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 $f(t,rx)\ge r^{\beta (r)}f(t,x)$.

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 $x\in P_e$.

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. □

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