We consider existence of symmetric positive solutions for a fourth-order singular boundary value problem:

$ \left\{\begin{array}{ll}x^{(4)}(t)+f(t, x(t))=0, \;0 < t < 1, \\ x(0)=x(1)=x^{'}(0)=x^{'}(1)=0, \end{array}\right. $

(1)

which describes the deformations of an elastic beam with both endpoints fixed, where $f:(0, 1)\times(0, +\infty)\rightarrow (0, +\infty)$ is conditions and $f(t, x)=f(1-t, x)$ for each $(0, 1)\times(0, +\infty).$ $f(t, x(t))$ may be singular at $t=0$ and/or $t=1.$

Here symmetric positive solutions for a fourth-order singular boundary value problem (1) satisfying $x(t)=x(1-t)$ and $x(t)>0, t\in(0, 1).$

Boundary value problems arise in a variety of different areas of applied mathematics and physics (see [1, 2] and the references therein). Recently many authors studied the existence of positive solutions for four-order singular boundary value problems for example [3-13] and the references therein. Most of these results are obtained via transforming the four-order boundary value problems into a second-order boundary value problems, and then applying the Leray-Schauder continuation method, the topologial degree theory, the fixed point theorems on cones, the critical point theory, or the lower and upper solution method. However results about the existence of symmetric positive solutions to singular boundary value problem (1) are few. Motivated by the results in [9, 11] we try to establish optimal existence of symmetric positive solutions to problem (1) by applying Leggett-Williams fixed point theorem.

We consider problem (1) in a Banach space $C[0, 1]$ equipped with the norm $\|x\|=\max\limits_{0\leq t\leq 1}|x(t)|$. A function $x(t)\in C[0, 1]$ is said to be a concave function if $x(\tau t_{1}+(1-\tau)t_{2})\geq \tau x(t_{1})+(1-\tau)x(t_{2})$ for all $t_{1}, t_{2}, \tau\in [0, 1].$ We denote

$ C^{+}[0, 1]=\{x\in C[0, 1]:x(t)\geq 0, t\in[0, 1]\}. $

Let $K$ be a cone of $C[0, 1]$ and $m, n$ be constants, $0 < m < n$. Define

$ \begin{eqnarray*} &&K_{r}=\{x\in K\mid \|x\| < r\}, \overline{K_{r}}=\{x\in K\mid \|x\|\leq r\}, \\ &&K(u, m, n)=\{x\in K\mid m\leq u(x), \|x\|\leq n\}.\end{eqnarray*} $

Let $G(t, s)$ be the Green's function of the corresponding boundary value problem (1), i.e.,

$ G(t, s)=\left\{\begin{array}{ll} \frac {1}{6}t^{2}(1-s)^{2}[(s-t)+2(1-t)s], \;\;\;\;&0\leq t\leq s\leq 1, \\ \frac {1}{6}s^{2}(1-t)^{2}[(t-s)+2(1-s)t], \;\;\;\;\;\;&0 \leq s \leq t\leq 1, \end{array}\right. $

and $G(\tau(s), s)=\max\limits_{0\leq t\leq 1}G(t, s)$, where

$ \tau(s)=\left\{\begin{array}{ll} \frac {1}{3-2s}, &0\leq s\leq \frac{1}{2}, \\ \frac {2s}{1+2s}, &\frac{1}{2}\leq s \leq 1.\end{array}\right. $

After a simple calculation, we get

(Ⅰ) $\displaystyle\int^{1}_{0}G(\tau(s), s)ds=\max\limits_{0\leq t\leq 1}\int^{1}_{0}G(t, s)ds=\max\limits_{0\leq t\leq 1}\frac{1}{24}t^{2}(1-t)^{2}=\frac{1}{384};$

(Ⅱ) $G(1-t, 1-s)=G(t, s);$

(Ⅲ) $\min\limits_{0\leq c\leq 1}\displaystyle\frac{G(\frac{1}{4}, c)}{G(\frac{1}{2}, c)}=\frac{1}{4};$

(Ⅳ)(see [9]) $q(t)G(\tau(s), s)\leq G(t, s)\leq G(\tau(s), s), q(t)=\min\{t^{2}, (1-t)^{2}\}, t\in [0, 1].$

Lemma 2.1 (see [14])  Let $A:K\rightarrow K$ be a completely continuous operator, $u$ be a nonnegative continuous concave function on $K$, and satisfies $u(x)\leq \|x\|$ for all $x\in \overline{K_{r}}$. In addition, assume that there exist $0 < d < m < n\leq r$ satisfy the following conditions:

(ⅰ) $\{x\in K(u, m, n)\mid u(x)>m\}\neq\varnothing$, and $u(Ax)>m$ for $x\in K(u, m, n)$;

(ⅱ) $\|Ax\|\leq d$ for $x\in \overline{K_{d}}$;

(ⅲ) $u(Ax)>m$ for $x\in K(u, m, r)$ and $\|Ax\|>n$;

then $A$ has at least three fixed points $x_{1}, x_{2}, x_{3}$ on $\overline{K_{r}}$ satisfy $\|x_{1}\| < d, m < u(x_{2})$, and $\|x_{3}\|>d$ for $u(x_{3}) < m.$

Theorem 3.1  Suppose the following conditions hold:

(H1) $f\in C((0, 1)\times[0, +\infty), [0, +\infty)), f(t, x)\leq g(t)h(x), g\in C((0, 1), [0, +\infty)),\\ h\in C( [0, +\infty), [0, +\infty));$

(H2) $0 < \displaystyle\int^{1}_{0}G(\tau(s), s)g(s)ds < +\infty;$

(H3) There exist $ 0 < d < m < \frac{r}{2}$ such that

1) $h(x)\leq d[\displaystyle\int^{1}_{0}G(\tau(s), s)g(s)ds]^{-1}$ for $0\leq x\leq d;$

2) $f(t, x)>16m[\displaystyle\int^{1}_{0}G(\tau(s), s)ds]^{-1}=6144m$ for $m\leq x\leq 2m;$

3) $h(x) < r[\displaystyle\int^{1}_{0}G(\tau(s), s)g(s)ds]^{-1}$ for $0\leq x\leq r;$

then problem(1) has triple symmetric positive solutions $x_{1}, x_{2}, x_{3}$ satisfy $\|x_{1}\| < d, m < u(x_{2})$, and $\|x_{3}\|>d$ for $u(x_{3}) < m.$

Proof Denote $K=\{x\in C^{+}[0, 1]:x(t)$is convex function and $x(t)=x(1-t), t\in [0, 1]\}$, then $K$ is a cone of $C^{+}[0, 1].$

Let $u(x)=\min\limits_{\frac{1}{4}\leq t\leq \frac{3}{4}}x(t)$ for $x\in K.$ then $u(x)=x(\frac{1}{4})\leq x(\frac{1}{2})=\|x\|$. It is well known that $x(t)$ is a positive solution of problem (1) if only if $x(t)$ is a positive solution of the equation $x(t)=\displaystyle\int^{1}_{0}G(t, s)f(s, x(s))ds$.

Define operator $A:K\rightarrow K$ by $Ax(t)=\displaystyle\int^{1}_{0}G(t, s)f(s, x(s))ds$. Obviously $Ax(t)\geq 0, (Ax)^{''}(t) < 0$ for $0 < t < 1, $ and for $x\in K$,

$ \begin{eqnarray*} &&Ax(1-t) =\int^{1}_{0}G(1-t, s)f(s, x(s))ds\\ &=&\int^{1-t}_{0}G(1-t, s)f(s, x(s))ds+\int^{1}_{1-t}G(1-t, s)f(s, x(s))ds\\ &=&-\int^{t}_{1}G(1-t, 1-r)f(1-r, x(1-r))dr-\int^{0}_{t}G(1-t, 1-r)f(1-r, x(1-r))dr\\ &=&\int^{1}_{t}G(t, r)f(r, x(r))dr+\int^{t}_{0}G(t, r)f(r, x(r))dr =Ax(t)\end{eqnarray*} $

consequently $Ax\in K$, that is $A:K\rightarrow K.$ By Arzela-Ascoli theorem, we can prove $A:K\rightarrow K$ is completely continuous.

From (H1) and 3) in (H3), for any $x\in \overline{K_{r}}, $ we know that

$ \begin{eqnarray*} \|Ax\|&\leq& \int^{1}_{0}G(\tau(s), s)g(s)h(x(s))ds\\ &\leq& \int^{1}_{0}G(\tau(s), s)g(s)ds\cdot r[\int^{1}_{0}G(\tau(s), s)g(s)ds]^{-1}=r.\end{eqnarray*} $

So $A(\overline{K_{r}})\subset \overline{K_{r}}.$ Choose $x(t)=2m, 0 < t < 1, $ then $x(t)\in K(u, m, 2m), $ and $u(x)=u(2m)>m.$ Then $\{x\in K(u, m, 2m)\mid u(x)>m\}\neq\varnothing$. And for $x\in K(u, m, 2m), u(x)=x(\frac{1}{4})\geq m$. Hence $m\leq x(s)\leq 2m, \frac{1}{4}\leq s \leq \frac{3}{4}.$ Thus for any $x\in K(u, m, 2m), $ from 2) in (H3), we obtain

$ \begin{eqnarray*} u(Ax)&=&\min\limits_{\frac{1}{4}\leq t\leq \frac{3}{4}}Ax(t)=\min\limits_{\frac{1}{4}\leq t\leq \frac{3}{4}}\int^{1}_{0}G(t, s)f(s, x(s))ds\\ &\geq& \min\limits_{\frac{1}{4}\leq t\leq \frac{3}{4}}\int^{1}_{0}q(t)G(\tau(s), s)f(s, x(s))ds\\ &\geq& \frac{1}{16}\int^{1}_{0}G(\tau(s), s)f(s, x(s))ds\\ &\geq& \frac{1}{16}\cdot 6144m \int^{1}_{0}G(\tau(s), s)ds=m.\end{eqnarray*} $

Thus condition (ⅰ) of Lemma 2.1 holds.

Next from (H1) and 1) in (H3), for any $x\in \overline{K_{d}}, $ we have

$ \begin{eqnarray*} \|Ax\|&\leq& \max\limits_{0\leq t\leq 1}\int^{1}_{0}G(t, s)g(s)h(x(s))ds\\ &\leq& \int^{1}_{0}G(\tau(s), s)g(s)ds\cdot d[\int^{1}_{0}G(\tau(s), s)g(s)ds]^{-1}=d.\end{eqnarray*} $

So $A:\overline{K_{d}}\rightarrow \overline{K_{d}}.$ Thus condition (ⅱ) of Lemma 2.1 follows.

Finally we prove $u(Ax)>m$ for $x\in K(u, m, r)$ and $\|Ax\|>4m.$

From 2) in (H3), for $x\in K(u, m, r)$ and $\|Ax\|>4m$, we know that

$ \begin{eqnarray*} u(Ax)&=&Ax(\frac{1}{4})=\int^{1}_{0}G(\frac{1}{4}, s)f(s, x(s))ds\\ &=&\int^{1}_{0}\frac{G(\frac{1}{4}, s)}{G(\frac{1}{2}, s)}G(\frac{1}{2}, s)f(s, x(s))ds\\ &\geq& \min\limits_{0\leq c\leq 1}\frac{G(\frac{1}{4}, c)}{G(\frac{1}{2}, c)}\int^{1}_{0}G(\frac{1}{2}, s)f(s, x(s))ds\\ &=&\frac{1}{4}Ax(\frac{1}{2})=\frac{1}{4}\|Ax\|>m.\end{eqnarray*} $

Therefore condition (ⅲ) of Lemma 2.1 holds too. The proof is completed.

Remark Theorem 3.1 also holds when nonlinearity $f(t, x(t))$ is nonsingular at $t=0$ and $t=1$.

Example 4.1 The following boundary value problem:

$ \left\{\begin{array}{ll} x^{(4)}(t)+\frac{h(x)}{t^{2}(1-t)^{2}}=0, \;0 < t < 1, \\ x(0)=x(1)=x^{'}(0)=x^{'}(1)=0\end{array}\right. $

(2)

has triple symmetric positive solutions, where

$ h(x)=\left\{\begin{array}{ll} 4x^{2}, &0\leq x < 2, \\ \frac {5760}{2x-1}, &x \geq 2.\end{array}\right. $

Proof  Let $f(t, x)=h(x)g(t), g(t)=\frac{1}{t^{2}(1-t)^{2}}.$ Obviously $g(t)$ is signular at $t=0$ and $t=1$. $h(x)\in C[0, +\infty)$. So (H1) holds.

Since

$ \begin{eqnarray*} G(\tau(s), s)&=&\left\{\begin{array}{ll}\frac{2s^{2}(1-s)^{3}}{3(3-2s)^{2}}, &0\leq s\leq \frac{1}{2}, \\ \frac{2s^{3}(1-s)^{2}}{3(1+2s)^{2}}, &\frac{1}{2}\leq s \leq 1.\end{array}\right.\\ \int^{1}_{0}G(\tau(s), s)g(s)ds&=&\frac{2}{3}\int^{\frac{1}{2}}_{0}\frac{1-s}{(3-2s)^{2}}ds+\frac{2}{3}\int^{1}_{\frac{1}{2}}\frac{s}{(1+2s)^{2}}ds \\ &=&\frac{1}{3}(\ln3-\ln2)-\frac{1}{12}\approx0.05 >0, \end{eqnarray*} $

then (H2) holds.

1) In (H3) followings from $[\displaystyle\int^{1}_{0}G(\tau(s), s)g(s)ds]^{-1}\approx 20$, we may take $d=\frac{1}{4}$ then

$ h(x)= 4x^{2}\leq 4d^{2}=\frac{1}{4} < d[\displaystyle\int^{1}_{0}G(\tau(s), s)g(s)ds]^{-1}\approx\frac{20}{4}\ \ {\rm for}\ \ 0\leq x \leq d=\frac{1}{4}. $

2) In (H3) is immediate, since we may take $m=2$ then

$ \begin{eqnarray*} f(t, x)&=&h(x)g(t)=\frac{1}{t^{2}(1-t)^{2}}\frac {5760}{2x-1}\\ &\geq& 16 \frac {5760}{2x-1}>16h(4)> 2\times 6144=6144m, \;\;2\leq x\leq 4, 0 < t < 1.\end{eqnarray*} $

3) In (H3) is immediate, since we may take $r=100>2m=4$ then

$ \begin{align} & \underset{0\le x\le 100}{\mathop{\max }}\, h(x)\le h(2)=\frac{5760}{3} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ < 100{{[\int_{0}^{1}{G}(\tau (s), s)g(s)ds]}^{-1}}\approx 2000, \ \ 0\le x\le r=100. \\ \end{align} $

Thus from Theorem 3.1, we know that problen (2) has triple symmetric positive solutions $x_{1}, x_{2}, x_{3}$ satisfy $\|x_{1}\ |< \frac{1}{4}, \;2 < u(x_{2})$, and $\|x_{3}\|>\frac{1}{4}$ for $u(x_{3}) < 2.$