Let us list some conditions for convenience.

(H

_{1}) There exists a constant

$b>0$ such that

$f(t,x)\ge -b,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in [0,1],x\ge 0.$

(H

_{2}) There exists a constant

$\u03f5>0$ such that

$\underset{|x|\to +\mathrm{\infty}}{lim\hspace{0.17em}sup}\frac{f(t,x)}{x}\le {\lambda}_{1}-\u03f5\phantom{\rule{1em}{0ex}}\text{uniformly on}t\in [0,1],$

where ${\lambda}_{1}$ is defined by (H_{0}).

(H_{3}) $f(t,0)=0$ uniformly on $t\in [0,1]$.

(H

_{4})

$\underset{x\to 0}{lim}\frac{f(t,x)}{x}=\lambda \phantom{\rule{1em}{0ex}}\text{uniformly on}t\in [0,1],$

where ${\lambda}_{n}<\lambda <{\lambda}_{n+1}$, ${\lambda}_{n}$, ${\lambda}_{n+1}$ is defined by (H_{0}).

**Theorem 3.1** *Suppose that* (H_{0}), (H_{1}), (H_{2}), (H_{3}), (H_{4}) *are satisfied and* *n* *is an odd number in* (H_{4}). *Then the boundary value problem* (1.1) *has at least a nontrivial solution*.

*Proof* Choose

$0<\delta <\u03f5$, then

$h={\lambda}_{1}-\u03f5+\delta <{\lambda}_{1}$. By (H

_{2}), there exists a constant

$M>0$ such that

$\frac{f(t,x)}{x}\le h,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in [0,1],|x|\ge M.$

(3.1)

So, by (3.1) and (H

_{1}), we know that

where $C={sup}_{0\le t\le 1,|x|<M}|f(t,x)|$.

By (3.2) and (3.3), we have

where ${C}_{1}=C{\int}_{0}^{1}G(t,s)\phantom{\rule{0.2em}{0ex}}ds$, ${C}^{\ast}=b{\int}_{0}^{1}G(t,s)\phantom{\rule{0.2em}{0ex}}ds$, $t\in [0,1]$, $\overline{B}=hB$, *B* is defined by (2.1). Obviously, ${C}_{1}\in P$, ${C}^{\ast}\in P$, $\overline{B}:P\to P$ is a positive completely continuous operator. By Lemma 2.3, we have $r(B)=\frac{1}{{\lambda}_{1}}$, so $r(\overline{B})=hr(B)<{\lambda}_{1}r(B)=1$.

By (H

_{3}), we have

$A\theta =\theta $, and

$\left({A}_{\theta}^{\prime}u\right)(t)={\int}_{0}^{1}G(t,s){f}_{x}^{\prime}(s,0)u(s)\phantom{\rule{0.2em}{0ex}}ds=\lambda {\int}_{0}^{1}G(t,s)u(s)\phantom{\rule{0.2em}{0ex}}ds,$

(3.6)

*i.e.*, ${A}_{\theta}^{\prime}=\lambda B$. By Lemma 2.3, 1 is not an eigenvalue of the linear operator ${A}_{\theta}^{\prime}$. Since ${\lambda}_{n}<\lambda <{\lambda}_{n+1}$, *n* is an odd number, the sum of the algebraic multiplicities for all eigenvalues of ${A}_{\theta}^{\prime}$ lying in $(1,+\mathrm{\infty})$ is an odd number. By Lemma 2.1, the operator *A* has at least one nonzero fixed point. So, the boundary value problem (1.1) has at least one nontrivial solution. □

**Theorem 3.2** *Suppose* (H

_{0}), (H

_{2}), (H

_{3}), (H

_{4})

*are satisfied and* *n* *is an even number in* (H

_{4}).

*In addition*,

*assume that* $f(t,x)x>0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in [0,1],x\in (-\mathrm{\infty},+\mathrm{\infty}),x\ne 0.$

(3.7)

*Then the boundary value problem* (1.1) *has at least three nontrivial solutions*: *one positive solution*, *one negative solution and one sign*-*changing solution*.

*Proof* By (3.7), we have

$f(t,x)>0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in [0,1],\mathrm{\forall}x>0;\phantom{\rule{2em}{0ex}}f(t,x)<0,\phantom{\rule{1em}{0ex}}\mathrm{\forall}t\in [0,1],\mathrm{\forall}x<0.$

(3.8)

By (3.1) and (3.8), (3.4) and (3.5) hold. From (H_{3}), (3.6) holds, and by Lemma 2.3, 1 is not an eigenvalue of the linear operator ${A}_{\theta}^{\prime}$. Since ${\lambda}_{n}<\lambda <{\lambda}_{n+1}$, *n* is an even number, the sum of the algebraic multiplicities for all eigenvalues of ${A}_{\theta}^{\prime}$ lying in $(1,+\mathrm{\infty})$ is an even number.

Obviously, from (3.8) and (2.2), we easily get

$F(P\mathrm{\setminus}\{\theta \})\subset P\mathrm{\setminus}\{\theta \},\phantom{\rule{2em}{0ex}}F((-P)\mathrm{\setminus}\{\theta \})\subset (-P)\mathrm{\setminus}\{\theta \}.$

(3.9)

From (2.1), we easily know that $B(P\mathrm{\setminus}\{\theta \})\subset P\mathrm{\setminus}\{\theta \}$, $B((-P)\mathrm{\setminus}\{\theta \})\subset (-P)\mathrm{\setminus}\{\theta \}$.

So, by (3.9), we have

$A(P\mathrm{\setminus}\{\theta \})\subset intP,\phantom{\rule{2em}{0ex}}A((-P)\mathrm{\setminus}\{\theta \})\subset int(-P).$

By Lemma 2.2, the boundary value problem (1.1) has at least three nontrivial solutions containing a positive solution, a negative solution and a sign-changing solution. □

**Remark** By Theorem 3.1 and Theorem 3.2, we can see that the methods used in this paper are different from [11–21], and the results are different from [11–21].

**Example 3.1** We consider the following integral boundary value problem:

$\{\begin{array}{c}-{x}^{\u2033}(t)=f(t,x(t)),\phantom{\rule{1em}{0ex}}0\le t\le 1,\hfill \\ x(0)=0,\phantom{\rule{2em}{0ex}}x(1)={\int}_{0}^{1}sx(s)\phantom{\rule{0.2em}{0ex}}ds,\hfill \end{array}$

(3.10)

where

$f(t,x)=\{\begin{array}{cc}5x+t\sqrt{x}-40,\hfill & t\in [0,1],x\in (4,+\mathrm{\infty}),\hfill \\ \frac{t-31}{3}(x-1)+11+t,\hfill & t\in [0,1],x\in (1,4],\hfill \\ 10x+(1+t){x}^{\frac{5}{3}},\hfill & t\in [0,1],x\in [-1,1],\hfill \\ \frac{t+29}{7}(x+1)-11-t,\hfill & t\in [0,1],x\in (-8,-1],\hfill \\ 5x+t\sqrt[3]{x},\hfill & t\in [0,1],x\in (-\mathrm{\infty},-8].\hfill \end{array}$

By simple calculations, we get that ${\lambda}_{1}\approx 7.53$, ${\lambda}_{2}\approx 37.41$, ${\lambda}_{3}\approx 86.80$, $\lambda =10$. So, ${\lambda}_{1}<\lambda <{\lambda}_{2}$. It is easy to know that the nonlinear term *f* satisfies (H_{1}), (H_{2}), (H_{3}), (H_{4}). Thus, the boundary value problem (3.10) has at least a nontrivial solution by Theorem 3.1.

**Example 3.2** We consider the following integral boundary value problem:

$\{\begin{array}{c}-{x}^{\u2033}(t)=f(t,x(t)),\phantom{\rule{1em}{0ex}}0\le t\le 1,\hfill \\ x(0)=0,\phantom{\rule{2em}{0ex}}x(1)={\int}_{0}^{1}sx(s)\phantom{\rule{0.2em}{0ex}}ds,\hfill \end{array}$

(3.11)

where

$f(t,x)=\{\begin{array}{cc}6x+t\sqrt{x},\hfill & t\in [0,1],x\in (4,+\mathrm{\infty}),\hfill \\ \frac{t-18}{3}(x-1)+41+t,\hfill & t\in [0,1],x\in (1,4],\hfill \\ 40x+(1+t){x}^{\frac{5}{3}},\hfill & t\in [0,1],x\in [-1,1],\hfill \\ \frac{t+7}{7}(x+1)-41-t,\hfill & t\in [0,1],x\in (-8,-1],\hfill \\ 6x+t\sqrt[3]{x},\hfill & t\in [0,1],x\in (-\mathrm{\infty},-8].\hfill \end{array}$

By simple calculations, we get that ${\lambda}_{1}\approx 7.53$, ${\lambda}_{2}\approx 37.41$, ${\lambda}_{3}\approx 86.80$, $\lambda =40$. So ${\lambda}_{2}<\lambda <{\lambda}_{3}$. It is easy to know that the nonlinear term *f* satisfies (H_{2}), (H_{3}), (H_{4}) and $f(t,x)x>0$, $\mathrm{\forall}t\in [0,1]$, $x\ne 0$. The boundary value problem (3.11) has at least three nontrivial solutions containing a positive solution, a negative solution and a sign-changing solution by Theorem 3.2.