Inequalities of Green’s functions and positive solutions to nonlocal boundary value problems

Abstract

In this paper, we discuss the positive solutions of beam equations with the nonlinearities including the slope and bending moment under nonlocal boundary conditions involving Stieltjes integrals. We pose some inequality conditions on nonlinearities and the spectral radius conditions on associated linear operators. These conditions mean that the nonlinearities have superlinear or sublinear growth. The existence of positive solutions is obtained by fixed point index on cones in $C^{2}[0,1]$, and some examples are given for beam equations subject to mixed integral and multi-point boundary conditions with sign-changing coefficients.

1 Introduction and preliminaries

In this paper, we discuss the existence of positive solutions to fourth-order boundary value problems (BVPs):

$$ \textstyle\begin{cases} u^{(4)}(t)=f(t,u(t),u'(t),u''(t)),\quad t\in [0,1], \\ u(0)=\beta _{1}[u],\quad\quad u'(1)=\beta _{2}[u], \\ u''(0)+\beta _{3}[u]=0,\quad\quad u''(1)+\beta _{4}[u]=0,\end{cases} $$

(1.1)

and

$$ \textstyle\begin{cases} -u^{(4)}(t)=g(t,u(t),u'(t),u''(t)),\quad t\in [0,1], \\ u(0)=\alpha _{1}[u],\quad\quad u'(0)=\alpha _{2}[u], \quad\quad u''(0)=\alpha _{3}[u],\quad\quad u''(1)=\alpha _{4}[u],\end{cases} $$

(1.2)

where $\beta _{i}[u]$, $\alpha _{i}[u]$ ($i=1,2,3,4$) are given by

$$ \beta _{i}[u]= \int _{0}^{1}u(t)\,dB_{i}(t)\quad \text{and}\quad \alpha _{i}[u]= \int _{0}^{1}u(t)\,dA_{i}(t) $$

involving Stieltjes integrals with $B_{i}$, $A_{i}$ of bounded variation. They model the deflection of beam equations with the nonlinearities including the slope $u'$ and bending moment $u''$. The boundary conditions of Stieltjes integrals imply that the mechanism at the end points depends on the feedback along parts of the beam to control the displacement.

By monotone iteration method, the cantilever beam equation containing the slope term

$$ u^{(4)}(t)=f\bigl(t,u(t),u'(t)\bigr) $$

was considered by Alves et al. [1] and Yao [18] separately under the boundary conditions

$$ u(0)=u'(0)=0,\qquad u'''(1)=g \bigl(u(1)\bigr),\qquad u'(1)=0 \quad \text{or}\quad u''(1)=0, $$

where g is a continuous function, and

$$ u(0)=u'(0)=u''(1)=u'''(1)=0. $$

Respectively based on fixed point index method and global bifurcation technique, Li [10] and Ma [12] were devoted to the beam equations involving the bending moment with the hinged ends

$$ \textstyle\begin{cases} u^{(4)}(t)=f(t,u(t),u''(t)),\quad t\in (0,1), \\ u(0)=u''(0)=u(1)=u''(1)=0.\end{cases} $$

Li [11] was concerned with the cantilever beam equation

$$ \textstyle\begin{cases} u^{(4)}(t)=f(t,u(t),u'(t),u''(t),u'''(t)),\quad t\in [0,1], \\ u(0)=u'(0)=u''(1)=u'''(1)=0,\end{cases} $$

where $f: [0,1]\times \mathbb{R}_{+}^{3}\times \mathbb{R}_{-}\to \mathbb{R}_{+}$ is continuous. The existence of positive solutions is obtained if the superlinear or sublinear growth conditions are satisfied for the nonlinearity. However, the boundary conditions in [1, 10–12, 18] are all local. Webb et al. [17] considered the existence of positive solutions for the beam equation

$$ u^{(4)}(t)=g(t)f\bigl(t,u(t)\bigr),\quad \text{a.e. }t\in (0,1) $$

respectively subject to nonlocal boundary conditions such as

$$\begin{aligned}& u(0)=0, \quad\quad u(1)=\alpha [u],\quad\quad u'(0)=0, \quad\quad u'(1)=0, \\& u(0)=0, \quad\quad u(1)=0, \quad\quad u'(0)=0, \quad\quad u'(1)+\alpha [u]=0, \\& u(0)=0, \quad\quad u''(0)=0, \quad\quad u(1)=\alpha [u], \quad\quad u''(1)=0, \end{aligned}$$

and

$$ u(0)=0, \quad\quad u(1)=0, \quad\quad u''(0)=0, \quad \quad u''(1)+\alpha [u]=0. $$

In these conditions $\alpha [u]=\int _{0}^{1}u(s)\,dA(s)$ is given by Stieltjes integral. Infante and Pietramala [6] studied the existence of positive solutions for cantilever beam equation

$$ \textstyle\begin{cases} u^{(4)}(t)=g(t)f(t,u(t)),\quad t\in (0,1), \\ u(0)=u'(0)=u''(1)=0,\quad\quad u'''(1)+k_{0}+B(\alpha [u])=0,\end{cases} $$

where $k_{0}$ is a nonnegative constant, $\alpha [u]$ is Stieltjes integral, and B is a nonnegative continuous function. The nonlinearity f in [6, 17] is not affected by the slope and bending moment, and the authors used the method of fixed point index on cone. We also refer to some other articles, for instance, [2, 5, 7, 9, 14, 19].

Recently, the authors in [13] investigated the existence of positive solutions to the following problems:

$$ \textstyle\begin{cases} u^{(4)}(t)=f(t,u(t),u'(t),u''(t),u'''(t)),\quad t\in [0,1], \\ u'(0)+\beta _{1}[u]=0,\quad\quad u''(0)+\beta _{2}[u]=0,\quad\quad u(1)=\beta _{3}[u], \quad\quad u'''(1)=0,\end{cases} $$

(1.3)

and

$$ \textstyle\begin{cases} -u^{(4)}(t)=g(t,u(t),u'(t),u''(t),u'''(t)),\quad t\in [0,1], \\ u(0)=\alpha _{1}[u],\quad\quad u'(0)=\alpha _{2}[u],\quad\quad u''(0)=\alpha _{3}[u], \quad\quad u'''(1)=0,\end{cases} $$

(1.4)

where $\beta _{i}[u]$, $\alpha _{i}[u]$ ($i=1,2,3$) are linear functionals involving Stieltjes integrals of signed measures, and the nonlinearities f, g satisfy the Nagumo condition, which restricts f and g on $u'''$ to quadric growth for the superlinear case, as in Li [11].

If the nonlinearities in (1.3) and (1.4) are independent of $u'''$, the restriction of quadric growth can certainly be rid of. However, the boundary conditions in (1.1) and (1.2) are different from those in (1.3) and (1.4). Especially the third derivatives with respect to t of their Green’s functions corresponding to (1.1) and (1.2) may be sign-changing while they are not corresponding to (1.3) and (1.4), which plays an important part when estimating the norms in [13]. When BVPs are converted to integral equations, a general method due to Webb and Infante [16] is applied to use the theory of fixed point index on cones in $C^{2}[0,1]$. Some examples are given for beam equations subject to mixed integral and multi-point boundary conditions with sign-changing coefficients. We also cite [15] in which a different approach is applied to the existence of positive solutions for the problem

$$ \textstyle\begin{cases} u^{(4)}(t)=h(t)f(t,u(t),u'(t),u''(t)),\quad t\in (0,1), \\ u(0)=u(1)=\beta _{1}[u],\quad\quad u''(0)+\beta _{2}[u]=0, \quad\quad u''(1)+\beta _{3}[u]=0,\end{cases} $$

where $f: [0,1]\times \mathbb{R}_{+}\times \mathbb{R}\times \mathbb{R}_{-} \to \mathbb{R}_{+}$ is continuous, $h\in L^{1}(0,1)$, and $\beta _{i}[u]$ is Stieltjes integral ($i=1,2,3$).

If the nonempty subset P in Banach space X satisfies the following conditions: (i) it is a closed convex set, (ii) $\lambda x\in P$ for any $\lambda >0$, $x\in P$, and (iii) $\pm x\in P\Leftrightarrow x=0$ (0 stands for the zero element in X), then P is said to be a cone in X. A cone P is called reproducing if $X=P-P$. It is well known that if P is a solid cone, i.e., the interior point set $\mathring{P}\neq \emptyset$, P is reproducing. Now we state some properties of fixed point index (see [3, 4]).

Lemma 1.1

LetΩbe a bounded open subset of Banach spaceXwith$0\in \varOmega$andPbe a cone in X. If$S:P\cap \overline{\varOmega }\to P$is a completely continuous operator and$\mu Su\neq u$for$u\in P\cap \partial \varOmega$and$\mu \in [0,1]$, then the fixed point index$i(S,P\cap \varOmega ,P)=1$.

Lemma 1.2

LetΩbe a bounded open subset of Banach spaceXandPbe a cone in X. If$S:P\cap \overline{\varOmega }\to P$is a completely continuous operator and there exists$v_{0}\in {P\setminus \{0\}}$such that$u-Su\neq \nu v_{0}$for$u\in P\cap \partial \varOmega$and$\nu \geq 0$, then the fixed point index$i(S,P\cap \varOmega ,P)=0$.

Lemma 1.3

(Krein–Rutman)

LetPbe a reproducing cone in Banach spaceXand$L:X\to X$be a completely continuous linear operator with$L(P)\subset P$. If the spectral radius$r(L)>0$, then there exists$\varphi \in {P\setminus \{0\}}$such that$L\varphi =r(L)\varphi$.

Lemma 1.4

([8])

LetPbe a cone in Banach spaceXand$L:X\to X$be a completely continuous linear operator with$L(P)\subset P$. If there exist$v_{0}\in {P\setminus \{0\}}$and$\lambda _{0}>0$such that$Lv_{0}\ge \lambda _{0}v_{0}$in the sense of partial ordering induced byP, then there exist$u_{0}\in {P\setminus \{0\}}$and$\lambda _{1}\ge \lambda _{0}$such that$Lu_{0}=\lambda _{1}u_{0}$.

Throughout this paper, denote the Banach space that consists of all second-order continuously differentiable functions on $[0,1]$ by $X=C^{2}[0,1]$ and the norm by $\Vert {u} \Vert _{C^{2}}=\max \{ \Vert {u} \Vert _{C}, \Vert {u' \Vert _{C}}, \Vert {u'' \Vert _{C}}\}$.

2 Inequalities of Green’s function and positive solutions for (1.1)

For BVP (1.1) we make the assumption:

$(C_{1})$:

$f: [0,1]\times \mathbb{R}_{+}^{2}\times \mathbb{R}_{-} \to \mathbb{R}_{+}$ is continuous, here $\mathbb{R}_{+}=[0,\infty )$ and $\mathbb{R}_{-}=(-\infty ,0]$.

Similar to Webb and Infante [16], BVP (1.1) can be converted to integral equation in $C^{2}[0,1]$

$$ u(t)=(Su) (t)=: \int _{0}^{1}k_{S}(t,s)f \bigl(s,u(s),u'(s),u''(s)\bigr)\,ds, $$

(2.1)

where

$$ k_{S}(t,s)=\bigl\langle \bigl(I-[B] \bigr)^{-1}\mathcal{K}(s),\gamma (t)\bigr\rangle +k_{0}(t,s) = \sum_{i=1}^{4}\kappa _{i}(s)\gamma _{i}(t)+k_{0}(t,s),$$

(2.2)

$\langle (I-[B])^{-1}\mathcal{K}(s),\gamma (t)\rangle$ is the inner product in $\mathbb{R}^{4}$,

$$\begin{aligned}& \mathcal{K}_{i}(s)= \int _{0}^{1}k_{0}(t,s) \,dB_{i}(t)\quad (i=1,2,3,4), \\& \gamma _{1}(t)=1, \quad\quad \gamma _{2}(t)=t,\quad\quad \gamma _{3}(t)=\frac{1}{6}t\bigl(t^{2}-3t+3\bigr), \quad\quad \gamma _{4}(t)=\frac{1}{6}t\bigl(3-t^{2} \bigr), \end{aligned}$$

$\kappa _{i}(s)$ is the ith component of $(I-[B])^{-1}\mathcal{K}(s)$,

$$ k_{0}(t,s)= \textstyle\begin{cases} \frac{1}{6}t(1-s)(3s-t^{2}), & 0\le t\le s\le 1, \\ \frac{1}{6}s(t^{3}-3t^{2}+3t-s^{2}), & 0\le s\le t\le 1. \end{cases} $$

We put forward the following hypotheses:

$(C_{2})$:

$B_{i}$ is of bounded variation and $\mathcal{K}_{i}(s)\ge 0$, $\forall s\in [0,1]$ ($i=1,2,3,4$);

$(C_{3})$:

The $4\times 4$ matrix $[B]$ is positive and its $(i,j)$th entry is $\beta _{i}[\gamma _{j}]$, i.e., it has nonnegative entries. In addition, its spectrum radius $r([B])<1$.

Remark

The integral operator S in (2.1) and the corresponding Green’s function $k_{S}(t,s)$ in (2.2) are obtained completely following the method in Webb and Infante [16, pp. 241–243] though f is independent of the derivatives of u there.

Lemma 2.1

If$(C_{2})$and$(C_{3})$hold, then$\kappa _{i}(s)\ge 0$ ($i=1,2,3,4$) and, for$t, s\in [0,1]$,

$$ c_{0}(t)\varPhi _{0}(s)\le k_{S}(t,s)\leq \varPhi _{0}(s),$$

(2.3)

where

$$ \varPhi _{0}(s)=\sum_{i=1}^{4} \kappa _{i}(s)+\frac{1}{6}s(1-s) (1+s),\qquad c_{0}(t)=\frac{1}{6}t\bigl(t^{2}-3t+3\bigr), $$

and

$$ c_{1}(t)\varPhi _{1}(s)\le \frac{\partial k_{S}(t,s)}{\partial t} \leq \varPhi _{1}(s),\qquad c_{2}(t)\varPhi _{2}(s) \le - \frac{\partial ^{2}k_{S}(t,s)}{\partial t^{2}}\leq \varPhi _{2}(s),$$

(2.4)

where

$$\begin{aligned}& \varPhi _{1}(s)=\sum_{i=2}^{4} \kappa _{i}(s)+\frac{1}{2}s(1-s),\qquad c_{1}(t)= \frac{1}{2}\bigl(1-t^{2}\bigr), \\& \varPhi _{2}(s)=\sum_{i=3}^{4} \kappa _{i}(s)+s(1-s),\qquad c_{2}(t)=\min \{t,1-t\}. \end{aligned}$$

Proof

For $s\in [0,1]$, $\kappa _{i}(s)\ge 0$ ($i=1,2,3,4$) are due to [16, proof of Theorem 2.4] since both $(I-[B])^{-1}$ and $\mathcal{K}(s)$ are nonnegative.

According to the following two inequalities:

$$\begin{aligned}& \frac{1}{6}t\bigl(t^{2}-3t+3\bigr)\sum _{i=1}^{4}\kappa _{i}(s)\le \sum _{i=1}^{4} \kappa _{i}(s)\gamma _{i}(t)\le \sum_{i=1}^{4}\kappa _{i}(s), \\& \begin{gathered} \frac{1}{6}t\bigl(t^{2}-3t+3\bigr) \frac{1}{6}s(1-s) (1+s) \\ \quad \le \frac{1}{2}t\bigl(3-t^{2}\bigr)\frac{1}{6}s(1-s) (1+s) \\ \quad \le k_{0}(t,s)\le \frac{1}{6}s(1-s) (1+s),\end{gathered} \end{aligned}$$

we have, for $t,s\in [0,1]$,

$$ c_{0}(t)\varPhi _{0}(s)\le k_{S}(t,s)=\sum _{i=1}^{4}\kappa _{i}(s) \gamma _{i}(t)+k_{0}(t,s)\leq \varPhi _{0}(s). $$

Moreover, the next two inequalities

$$\begin{aligned}& \frac{1}{2}(1-t)^{2}\sum_{i=2}^{4} \kappa _{i}(s)\le \sum_{i=1}^{4} \kappa _{i}(s)\gamma '_{i}(t)\le \sum _{i=2}^{4}\kappa _{i}(s), \\& \frac{1}{2}(1-t)^{2}\frac{1}{2}s(2-s)\le (1-t)^{2}\frac{1}{2}s(2-s) \le \frac{\partial k_{0}(t,s)}{\partial t}\le \frac{1}{2}s(1-s) \end{aligned}$$

imply, for $t,s\in [0,1]$,

$$ c_{1}(t)\varPhi _{1}(s)\le \frac{\partial k_{S}(t,s)}{\partial t}=\sum _{i=1}^{4} \kappa _{i}(s)\gamma '_{i}(t)+\frac{\partial k_{0}(t,s)}{\partial t} \leq \varPhi _{1}(s). $$

Finally from the two inequalities

$$\begin{aligned}& \begin{aligned} \min \{t,1-t\}\sum_{i=3}^{4} \kappa _{i}(s)&\le -\sum_{i=1}^{4} \kappa _{i}(s)\gamma ''_{i}(t) \\ & = (1-t)\kappa _{3}(s)+t\kappa _{4}(s)\le \sum _{i=3}^{4} \kappa _{i}(s),\end{aligned} \\& \min \{t,1-t\}s(1-s)\le - \frac{\partial ^{2}k_{0}(t,s)}{\partial t^{2}}\le s(1-s), \end{aligned}$$

it follows that

$$ c_{2}(t)\varPhi _{2}(s)\le - \frac{\partial ^{2}k_{S}(t,s)}{\partial t^{2}}=-\sum _{i=1}^{4} \kappa _{i}(s)\gamma ''_{i}(t)- \frac{\partial ^{2}k_{0}(t,s)}{\partial t^{2}}\leq \varPhi _{2}(s) $$

for $t,s\in [0,1]$. □

Define the subsets in $C^{2}[0,1]$ as follows:

$$\begin{aligned}& P= \bigl\{ u\in C^{2}[0,1]: u(t)\geq 0, u'(t)\ge 0, u''(t)\le 0, \forall t\in [0,1] \bigr\} , \end{aligned}$$

(2.5)

$$\begin{aligned}& \begin{aligned}[b] K= {}& \bigl\{ u\in P: u(t)\ge c_{0}(t) \Vert u \Vert _{C}, u'(t)\ge c_{1}(t) \bigl\Vert u' \bigr\Vert _{C}, \\ &{} -u''(t)\ge c_{2}(t) \bigl\Vert u'' \bigr\Vert _{C}, \forall t\in [0,1]; \beta _{i}[u] \geq 0\ (i=1,2,3,4) \bigr\} .\end{aligned} \end{aligned}$$

(2.6)

Clearly both P and K are cones, and it is easy to check that P is a solid cone. Denote the cone ordering induced by P, $u\preceq v$ for $u,v\in X$ if and only if $v-u\in P$ and equivalently $v\succeq u$.

Now we define linear operators in $C^{2}[0,1]$:

$$\begin{aligned}& (L_{i}u) (t)= \int _{0}^{1}k_{S}(t,s) \bigl(a_{i}u(s)+b_{i}u'(s)-c_{i}u''(s) \bigr)\,ds \quad (i=1,2), \end{aligned}$$

(2.7)

$$\begin{aligned}& (L_{3}u) (t)=a_{1} \int _{0}^{1}k_{S}(t,s)u(s) \,ds, \end{aligned}$$

(2.8)

where $a_{i}$, $b_{i}$, $c_{i}$ ($i=1,2$) are nonnegative constants.

Similar to [16], we have the following Lemma 2.2 by Lemma 2.1.

Lemma 2.2

If$(C_{1})$–$(C_{3})$hold, then$S:P\to K$and$L_{i}: C^{2}[0,1]\to C^{2}[0,1]$are completely continuous operators with$L_{i}(P)\subset K$ ($i=1,2,3$).

Theorem 2.1

Under hypotheses $(C_{1})$ – $(C_{3})$ suppose that

$(F_{1})$ :

there exist constants$a_{2}, b_{2}, c_{2}\ge 0$and$r>0$such that

$$ f(t,x_{1},x_{2},x_{3})\leq a_{2}x_{1}+b_{2}x_{2}-c_{2}x_{3}$$

(2.9)

for all$(t,x_{1},x_{2},x_{3})\in [0,1]\times [0,r]^{2}\times [-r,0]$; moreover, the spectral radius$r(L_{2})<1$, where$L_{2}$is defined by (2.7),

$(F_{2})$ :

there exist positive constants$a_{1}$, $b_{1}$, $c_{1}$, $C_{0}$satisfying

$$ \begin{aligned}[b] &\min \biggl\{ \frac{a_{1}}{6} \int _{0}^{1}c_{0}(s)\varPhi _{0}(s)\,ds, \frac{b_{1}}{2} \int _{0}^{1}c_{1}(s)\varPhi _{1}(s)\,ds, \\ &\quad \frac{c_{1}}{2} \int _{0}^{1}\bigl(\kappa _{3}(s)+\kappa _{4}(s)\bigr)c_{2}(s)\,ds \biggr\} >1\end{aligned} $$

(2.10)

such that

$$ f(t,x_{1},x_{2},x_{3})\geq a_{1}x_{1}+b_{1}x_{2}-c_{1}x_{3}-C_{0}$$

(2.11)

for all$(t,x_{1},x_{2},x_{3})\in [0,1]\times \mathbb{R}_{+}^{2}\times \mathbb{R}_{-}$.

Then BVP (1.1) has one positive solution in K.

Proof

(i) First we prove that $\mu Su\ne u$ for $u\in K\cap \partial \varOmega _{r}$ and $\mu \in [0,1]$, where $\varOmega _{r}=\{u\in C^{2}[0,1]: \Vert {u} \Vert _{C^{2}}< r\}$.

In fact, if there exist $u_{1}\in K\cap \partial \varOmega _{r}$ and $\mu _{0}\in [0,1]$ such that $u_{1}=\mu _{0}Su_{1}$, then we deduce from

$$ 0\leq u_{1}(t), u_{1}'(t)\leq r, \quad\quad 0 \le -u_{1}''(t)\leq r, \quad \forall t\in [0,1] $$

and (2.9) that, for $t\in [0,1]$,

$$ u_{1}(t)\le (L_{2}u_{1}) (t), \quad\quad u_{1}'(t)\le (L_{2}u_{1})'(t), \quad\quad u_{1}''(t)\ge (L_{2}u_{1})''(t), $$

thus $(I-L_{2})u_{1}\preceq 0$. Because of the spectral radius $r(L_{2})<1$, we know that $I-L_{2}$ has a bounded inverse operator $(I-L_{2})^{-1}: P\to P$ and $u_{1}\preceq (I-L_{2})^{-1}0=0$, which contradicts $u_{1}\in K\cap \partial \varOmega _{r}$.

Therefore, $i(S,K\cap \varOmega _{r},K)=1$ follows from Lemma 1.1.

(ii) In this step we construct a homotopy and find a subset $\varOmega _{R}$ in order to compute the fixed point index later.

Let M be

$$ \begin{aligned}[b] &\max \biggl\{ \frac{6C_{0}\int _{0}^{1}\varPhi _{0}(s)\,ds}{a_{1}\int _{0}^{1}c_{0}(s)\varPhi _{0}(s)\,ds-6}, \frac{2C_{0}\int _{0}^{1}\varPhi _{1}(s)\,ds}{b_{1}\int _{0}^{1}c_{1}(s)\varPhi _{1}(s)\,ds-2}, \\ &\quad \frac{C_{0}\int _{0}^{1}(\kappa _{3}(s)+\kappa _{4}(s))\,ds}{c_{1}\int _{0}^{1}(\kappa _{3}(s)+\kappa _{4}(s))c_{2}(s)\,ds-2} \biggr\} .\end{aligned} $$

(2.12)

Obviously, $M>0$ if we notice that

$$ C_{0} \int _{0}^{1}\bigl(\kappa _{3}(s)+\kappa _{4}(s)\bigr)\,ds>0 $$

since $\frac{c_{1}}{2}\int _{0}^{1}(\kappa _{3}(s)+\kappa _{4}(s))c_{2}(s)\,ds>1$ by (2.10).

For $u\in K$, define the homotopy $H(\lambda ,u)=Su+\lambda v$, where

$$ v(t)=C_{0} \int _{0}^{1}k_{S}(t,s)\,ds, $$

then $v\in K$ and $H: [0,1]\times K\to K$ is completely continuous.

Let $R>\max \{r, M\}$ and we will show that

$$ H(\lambda ,u)\ne u, \quad \forall u\in K\cap \partial \varOmega _{R}, \lambda \in [0,1],$$

(2.13)

where $\varOmega _{R}=\{u\in C^{2}[0,1]: \Vert {u} \Vert _{C^{2}}< R\}$.

If it does not hold, there exist $u_{2}\in K\cap \partial \varOmega _{R}$ and $\lambda _{0}\in [0,1]$ such that

$$ H(\lambda _{0},u_{2})=u_{2},$$

(2.14)

thus (2.11) and Lemma 2.1 yield that

$$\begin{aligned}& \begin{aligned}[b] \Vert u_{2} \Vert _{C}&=u_{2}(1) \\ & = \int _{0}^{1}k_{S}(1,s)f \bigl(s,u_{2}(s),u'_{2}(s),u''_{2}(s) \bigr)\,ds+ \lambda _{0}C_{0} \int _{0}^{1}k_{S}(1,s)\,ds \\ & \ge \int _{0}^{1}k_{S}(1,s) \bigl[a_{1}u_{2}(s)+b_{1}u'_{2}(s)-c_{1}u''_{2}(s)-C_{0}+ \lambda _{0}C_{0}\bigr]\,ds \\ & \ge \int _{0}^{1}k_{S}(1,s) \bigl[a_{1}u_{2}(s)-C_{0}\bigr]\,ds \\ & \ge \frac{a_{1}}{6} \int _{0}^{1}\varPhi _{0}(s)u_{2}(s) \,ds-C_{0} \int _{0}^{1}\varPhi _{0}(s)\,ds \\ & \ge \frac{a_{1}}{6} \Vert u_{2} \Vert _{C} \int _{0}^{1}c_{0}(s) \varPhi _{0}(s)\,ds-C_{0} \int _{0}^{1}\varPhi _{0}(s) \,ds,\end{aligned} \end{aligned}$$

(2.15)

$$\begin{aligned}& \begin{aligned}[b] \bigl\Vert u'_{2} \bigr\Vert _{C}&=u'_{2}(0) \\ & = \int _{0}^{1}\frac{\partial k_{S}(0,s)}{\partial t}f \bigl(s,u_{2}(s),u'_{2}(s),u''_{2}(s) \bigr)\,ds+ \lambda _{0}C_{0} \int _{0}^{1}\frac{\partial k_{S}(0,s)}{\partial t}\,ds \\ & \ge \int _{0}^{1} \frac{\partial k_{S}(0,s)}{\partial t} \bigl[a_{1}u_{2}(s)+b_{1}u'_{2}(s)-c_{1}u''_{2}(s)-C_{0}+ \lambda _{0}C_{0}\bigr]\,ds \\ &\ge \int _{0}^{1} \frac{\partial k_{S}(0,s)}{\partial t} \bigl[b_{1}u'_{2}(s)-C_{0}\bigr]\,ds \\ & \ge \frac{b_{1}}{2} \int _{0}^{1}\varPhi _{1}(s)u'_{2}(s) \,ds-C_{0} \int _{0}^{1}\varPhi _{1}(s)\,ds \\ & \ge \frac{b_{1}}{2} \bigl\Vert u'_{2} \bigr\Vert _{C} \int _{0}^{1}c_{1}(s) \varPhi _{1}(s)\,ds-C_{0} \int _{0}^{1}\varPhi _{1}(s) \,ds.\end{aligned} \end{aligned}$$

(2.16)

Since $u_{2}^{(4)}(t)=f(t,u_{2}(t),u_{2}'(t),u''_{2}(t))+\lambda _{0}C_{0} \ge 0$ for $t\in [0,1]$, we know that if $0\le u'''(0)\le u'''(1)$, then

$$ \bigl\Vert u'' \bigr\Vert _{C}=-u''_{2}(0) \ge -\frac{1}{2}\bigl(u''_{2}(0)+u''_{2}(1) \bigr); $$

if $u'''(0)\le u'''(1)\le 0$, then

$$ \bigl\Vert u'' \bigr\Vert _{C}=-u''_{2}(1) \ge -\frac{1}{2}\bigl(u''_{2}(0)+u''_{2}(1) \bigr); $$

if $u'''(0)\le 0\le u'''(1)$, then there exists $\xi \in [0,1]$ such that

$$ \bigl\Vert u''_{2} \bigr\Vert _{C}=-u''_{2}(\xi )\ge - \frac{1}{2}\bigl(u''_{2}(0)+u''_{2}(1) \bigr). $$

Therefore the proof of Lemma 2.1 leads to

$$\begin{aligned} \bigl\Vert u''_{2} \bigr\Vert _{C}&\ge -\frac{1}{2}\bigl(u''_{2}(0)+u''_{2}(1) \bigr) \\ & = -\frac{1}{2} \int _{0}^{1} \biggl( \frac{\partial ^{2}k_{S}(0,s)}{\partial t^{2}}+ \frac{\partial ^{2}k_{S}(1,s)}{\partial t^{2}} \biggr)\bigl[f\bigl(s,u_{2}(s),u'_{2}(s),u''_{2}(s) \bigr)+ \lambda _{0}C_{0}\bigr]\,ds \\ & = \frac{1}{2} \int _{0}^{1}\bigl(\kappa _{3}(s)+\kappa _{4}(s)\bigr)\bigl[f\bigl(s,u_{2}(s),u'_{2}(s),u''_{2}(s) \bigr)+ \lambda _{0}C_{0}\bigr]\,ds \\ & \ge \frac{1}{2} \int _{0}^{1}\bigl(\kappa _{3}(s)+\kappa _{4}(s)\bigr)\bigl[a_{1}u_{2}(s)+b_{1}u'_{2}(s)-c_{1}u''_{2}(s)-C_{0}+ \lambda _{0}C_{0}\bigr]\,ds \\ & \ge \frac{1}{2} \int _{0}^{1}\bigl(\kappa _{3}(s)+\kappa _{4}(s)\bigr)\bigl[-c_{1}u''_{2}(s)-C_{0} \bigr]\,ds \\ & \ge \frac{c_{1}}{2} \bigl\Vert u''_{2} \bigr\Vert _{C} \int _{0}^{1}\bigl( \kappa _{3}(s)+\kappa _{4}(s)\bigr)c_{2}(s)\,ds -\frac{C_{0}}{2} \int _{0}^{1}\bigl( \kappa _{3}(s)+\kappa _{4}(s)\bigr)\,ds, \end{aligned}$$

which implies by (2.12), (2.15), and (2.16) that

$$ \Vert u_{2} \Vert _{C}\le M, \quad\quad \bigl\Vert u'_{2} \bigr\Vert _{C}\le M, \quad\quad \bigl\Vert u''_{2} \bigr\Vert _{C} \le M, $$

a contradiction to $\Vert u_{2} \Vert _{C^{2}}=R>M$.

From (2.13) it follows that

$$ i(S,K\cap \varOmega _{R},K)=i\bigl(H(0,\cdot ),K\cap \varOmega _{R},K\bigr)=i\bigl(H(1, \cdot ),K\cap \varOmega _{R},K\bigr)$$

(2.17)

by the homotopy invariance property of fixed point index.

(iii) Now we search for an appropriate element in K for the sake of the next step. For the function $c_{0}(t)=\frac{1}{6}t(t^{2}-3t+3)$, we have from (2.8) and Lemma 2.1 that

$$ (L_{3}c_{0}) (t)=a_{1} \int _{0}^{1}k_{S}(t,s)c_{0}(s) \,ds\ge \biggl(a_{1} \int _{0}^{1}c_{0}(s)\varPhi _{0}(s)\,ds \biggr)c_{0}(t). $$

From (2.10) it follows that $a_{1}\int _{0}^{1}c_{0}(s)\varPhi _{0}(s)\,ds>6$. Since $L_{3}$ is a completely continuous linear operator in $C[0,1]$, we consider the nonnegative cone $C^{+}[0,1]=\{u\in C[0,1]: u(t)\ge 0, \forall t\in [0,1]\}$ in Lemma 1.4. There exist $\lambda _{1}>6$ and $\varphi _{0}\in C^{+}[0,1]\backslash \{0\}$ such that $\varphi _{0}=\lambda _{1}^{-1}L_{3}\varphi _{0}$. Obviously $\varphi _{0}\in P$ can be directly verified, and thus $\varphi _{0}\in K$ by Lemma 2.2.

(iv) Now we prove that $u-H(1,u)\neq \nu \varphi _{0}$ for $u\in K\cap \partial \varOmega _{R}$ and $\nu \ge 0$, where $\varphi _{0}$ is as in step (iii), and hence

$$ i\bigl(H(1,\cdot ),K\cap \varOmega _{R},K \bigr)=0$$

(2.18)

holds by Lemma 1.2.

If there exist $u_{0}\in K\cap \partial \varOmega _{R}$ and $\nu _{0}\geq 0$ such that $u_{0}-H(1,u_{0})=\nu _{0}\varphi _{0}$. Clearly $\nu _{0}>0$ by (2.13), and thus

$$ u_{0}(t)=\bigl(H(1,u_{0})\bigr) (t)+\nu _{0} \varphi _{0}(t)\ge \nu _{0}\varphi _{0}(t) $$

for $t\in [0,1]$. Set

$$ \nu ^{*}=\sup \bigl\{ \nu >0: u_{0}(t)\geq \nu \varphi _{0}(t), \forall t \in [0,1]\bigr\} , $$

then $\nu _{0}\le \nu ^{*}<+\infty$ and $u_{0}(t)\geq \nu ^{*}\varphi _{0}(t)$ for $t\in [0,1]$. From (2.11) we have that, for $t\in [0,1]$,

$$ \begin{aligned} u_{0}(t)&=\bigl(H(1,u_{0}) \bigr) (t)+\nu _{0}\varphi _{0}(t)\ge (L_{3}u_{0}) (t)+ \nu _{0}\varphi _{0}(t) \\ &\ge \nu ^{*}(L_{3}\varphi _{0}) (t)+\nu _{0}\varphi _{0}(t) = \lambda _{1}\nu ^{*}\varphi _{0}(t)+\nu _{0}\varphi _{0}(t). \end{aligned} $$

Since $\lambda _{1}>6$, we have $\lambda _{1}\nu ^{*}+\nu _{0}>\nu ^{*}$, which contradicts the definition of $\nu ^{*}$.

(vi) From (2.17) and (2.18) it follows that $i(S,K\cap \varOmega _{R},K)=0$ and

$$ i\bigl(S,K\cap (\varOmega _{R}\setminus \overline{\varOmega }_{r}),K\bigr)=i(S,K\cap \varOmega _{R},K)-i(S,K\cap \varOmega _{r},K)=-1. $$

Hence S has one fixed solution, i.e., BVP (1.1) has one positive solution in K. □

Theorem 2.2

Under hypotheses $(C_{1})$ – $(C_{3})$ suppose that

$(F_{3})$ :

there exist constants$a_{1}, b_{1}, c_{1}, C_{0}\ge 0$such that

$$ f(t,x_{1},x_{2},x_{3})\leq a_{1}x_{1}+b_{1}x_{2}-c_{1}x_{3}+C_{0}$$

(2.19)

for all$(t,x_{1},x_{2},x_{3})\in [0,1]\times \mathbb{R}_{+}^{2}\times \mathbb{R}_{-}$, moreover the spectral radius$r(L_{1})<1$;

$(F_{4})$ :

there exist constants$a_{2}, b_{2}, c_{2}\ge 0$and$r>0$such that

$$ f(t,x_{1},x_{2},x_{3})\geq a_{2}x_{1}+b_{2}x_{2}-c_{2}x_{3}$$

(2.20)

for all$(t,x_{1},x_{2},x_{3})\in [0,1]\times [0,r]^{2}\times [-r,0]$, moreover the spectral radius$r(L_{2})\ge 1$; where$L_{i}: C^{2}[0,1]\to C^{2}[0,1]$ ($i=1,2$) are defined by (2.7).

Then BVP (1.1) has one positive solution in K.

Proof

Let $W=\{u\in K: {\text{there exists a }\mu \in [0,1]\text{ with }} u=\mu Su \}$ where S and K are respectively defined in (2.1) and (2.6).

We first assert that W is a bounded set. In fact, if $u\in W$, then $u=\mu Su$ for some $\mu \in [0,1]$. From (2.7) and (2.19) we have that

$$\begin{aligned} u(t)&=\mu (Su) (t)= \mu \int _{0}^{1}k_{S}(t,s)f \bigl(s,u(s),u'(s),u''(s)\bigr)\,ds \\ &\leq \int _{0}^{1}k_{S}(t,s) \bigl[a_{1}u(s)+b_{1}u'(s)-c_{1}u''(s)+C_{0} \bigr]\,ds \\ &= (L_{1}u) (t)+C_{0} \int _{0}^{1}k_{S}(t,s)\,ds \end{aligned}$$

and

$$ \bigl((I-L_{1})u\bigr) (t)\leq C_{0} \int _{0}^{1}k_{S}(t,s)\,ds=:v(t),\quad t \in [0,1]. $$

Obviously $v\in P$ and it is easy to see from (2.19) that, for $t\in [0,1]$,

$$ u'(t)\le (L_{1}u)'(t)+v'(t), \qquad u''(t)\ge (L_{1}u)''(t)+v''(t), $$

thus $(I-L_{1})u\preceq v$. Because of the spectral radius $r(L_{1})<1$, we know that $I-L_{1}$ has a bounded inverse operator $(I-L_{1})^{-1}$, which can be written as

$$ (I-L_{1})^{-1}=I+L_{1}+L_{1}^{2}+ \cdots +L_{1}^{n}+\cdots . $$

Since $L_{1}(P)\subset K\subset P$ by Lemma 2.2, we have $(I-L_{1})^{-1}(P)\subset P$, which implies the inequality $u\preceq (I-L_{1})^{-1}v$. Therefore, for $t\in [0,1]$,

$$\begin{aligned}& 0\le u(t)\leq \bigl((I-L_{1})^{-1}v\bigr) (t),\qquad 0\le u'(t)\le \bigl((I-L_{1})^{-1}v \bigr)'(t), \\& 0\ge u''(t)\ge \bigl((I-L_{1})^{-1}v \bigr)''(t), \end{aligned}$$

and hence $\Vert u \Vert _{C^{2}}\le \Vert (I-L_{1})^{-1}v \Vert _{C^{2}}$, i.e., W is bounded.

Now select $R>\max \{r,\sup W\}$, then $\mu Su\neq u$ for $u\in K\cap \partial \varOmega _{R}$ and $\mu \in [0,1]$, and $i(S,K\cap \varOmega _{R}, K)=1$ follows from Lemma 1.1.

Since $L_{2}: P\to K\subset P$ and $r(L_{2})\ge 1$, it follows from Lemma 1.3 that there exists $\varphi _{0}\in P\setminus \{0\}$ such that $L_{2}\varphi _{0}=r(L_{2})\varphi _{0}$. Furthermore, $\varphi _{0}=(r(L_{2}))^{-1}L_{2}\varphi _{0}\in K$.

We may suppose that S has no fixed points in $K\cap \partial \varOmega _{r}$ and will show that $u-Su\neq \nu \varphi _{0}$ for $u\in K\cap \partial \varOmega _{r}$ and $\nu \ge 0$.

Otherwise, there exist $u_{0}\in K\cap \partial \varOmega _{r}$ and $\nu _{0}\ge 0$ such that $u_{0}-Su_{0}=\nu _{0}\varphi _{0}$, and it is clear that $\nu _{0}>0$. Since $u_{0}\in K\cap \partial \varOmega _{r}$, we have

$$ 0\leq u_{0}(t), u_{0}'(t)\leq r,\qquad -r\le u_{0}''(t)\le 0,\quad \forall t \in [0,1]. $$

It follows from (2.2), (2.7), and (2.20) that $\forall t\in [0,1]$,

$$ (Su_{0}) (t)\geq (L_{2}u_{0}) (t),\qquad (Su_{0})'(t)\ge (L_{2}u_{0})'(t), \qquad (Su_{0})''(t) \le (L_{2}u_{0})''(t), $$

which imply that

$$ u_{0}=\nu _{0}\varphi _{0}+Su_{0} \succeq \nu _{0}\varphi _{0}+ L_{2}u_{0} \succeq \nu _{0}\varphi _{0}.$$

(2.21)

Set $\nu ^{*}=\sup \{\nu >0: u_{0}\succeq \nu \varphi _{0}\}$, then $\nu _{0}\le \nu ^{*}<+\infty$ and $u_{0}\succeq \nu ^{*}\varphi _{0}$. Thus it follows from (2.21) that

$$ u_{0}\succeq \nu _{0}\varphi _{0}+ L_{2}u_{0}\succeq \nu _{0}\varphi _{0}+ \nu ^{*}L_{2}\varphi _{0}=\nu _{0}\varphi _{0}+\nu ^{*}r(L_{2}) \varphi _{0}. $$

But $r(L_{2})\ge 1$, so $u_{0}\succeq (\nu _{0}+\nu ^{*})\varphi _{0}$, which is a contradiction to the definition of $\nu ^{*}$. Therefore $u-Su\neq \nu \varphi _{0}$ for $u\in K\cap \partial \varOmega _{r}$ and $\nu \ge 0$.

From Lemma 1.2 it follows that $i(S,K\cap \varOmega _{r}, K)=0$.

Making use of the properties of fixed point index, we have that

$$ i\bigl(S,K\cap (\varOmega _{R}\setminus \overline{\varOmega }_{r}),K\bigr)=i(S,K\cap \varOmega _{R}, K)-i(S,K\cap \varOmega _{r}, K)=1, $$

and hence S has one fixed point in K. Therefore, BVP (1.1) has one positive solution in K. □

As an example, we consider the fourth-order boundary problem under mixed multi-point and integral boundary conditions with sign-changing coefficients:

$$ \textstyle\begin{cases} u^{(4)}(t)=f(t,u(t),u'(t),u''(t)),\quad t\in [0,1], \\ u(0)=\frac{1}{4}u(\frac{1}{4})-\frac{1}{12}u(\frac{3}{4}),\quad \quad u'(1)= \int _{0}^{1}u(t)(t-\frac{1}{8})\,dt, \\ u''(0)+\frac{1}{2}u(\frac{1}{2})-\frac{1}{4}u(\frac{3}{4})=0,\quad \quad u''(1)+ \frac{1}{2}u(\frac{1}{4})-\frac{1}{4}u(\frac{1}{2})=0,\end{cases} $$

(2.22)

thus $\beta _{1}[u]=\frac{1}{4}u(\frac{1}{4})-\frac{1}{12}u(\frac{3}{4})$, $\beta _{2}[u]=\int _{0}^{1}u(t)(t-\frac{1}{8})\,dt$, $\beta _{3}[u]= \frac{1}{2}u(\frac{1}{2})-\frac{1}{4}u(\frac{3}{4})$, and $\beta _{4}[u]=\frac{1}{2}u(\frac{1}{4})-\frac{1}{4}u(\frac{1}{2})$. We estimate some coefficients, and Matlab is used in some places.

$$\begin{aligned} \mathcal{K}_{1}(s)&= \int _{0}^{1}k_{0}(t,s) \,dB_{1}(t) = \frac{1}{4}k_{0} \biggl( \frac{1}{4},s \biggr)-\frac{1}{12}k_{0} \biggl( \frac{3}{4},s \biggr) \\ &= \textstyle\begin{cases} - \frac{1}{36}s^{3}+\frac{1}{96}s, & 0 \leq s\le \frac{1}{4}, \\ \frac{1}{72}s^{3}-\frac{1}{32}s^{2}+\frac{7}{384}s- \frac{1}{1536}, & \frac{1}{4}< s\le \frac{3}{4}, \\ -\frac{1}{192}s+\frac{1}{192}, & \frac{3}{4}< s\le 1,\end{cases}\displaystyle \end{aligned}$$

and hence $0\le \mathcal{K}_{1}(s)<0.0026$;

$$ \mathcal{K}_{2}(s)= \int _{0}^{1}k_{0}(t,s) \biggl(t- \frac{1}{8} \biggr)\,dt= \frac{1}{120}s^{5}- \frac{1}{192}s^{4}-\frac{1}{16}s^{3}+ \frac{57}{960}s\quad (0\le s\le 1), $$

and hence $0\le \mathcal{K}_{2}(s)<0.0223$;

$$\begin{aligned} \mathcal{K}_{3}(s)&= \int _{0}^{1}k_{0}(t,s) \,dB_{3}(t) = \frac{1}{2}k_{0} \biggl( \frac{1}{2},s \biggr)-\frac{1}{4}k_{0} \biggl( \frac{3}{4},s \biggr) \\ &= \textstyle\begin{cases} - \frac{1}{24}s^{3}+\frac{49}{1536}s, & 0 \leq s\le \frac{1}{2}, \\ \frac{1}{24}s^{3}-\frac{1}{8}s^{2}+\frac{145}{1536}s- \frac{1}{96}, & \frac{1}{2}< s\le \frac{3}{4}, \\ -\frac{1}{32}s^{2}+\frac{37}{1536}s+\frac{11}{1536}, & \frac{3}{4}< s\le 1,\end{cases}\displaystyle \end{aligned}$$

and hence $0\le \mathcal{K}_{3}(s)<0.0108$;

$$\begin{aligned} \mathcal{K}_{4}(s)&= \int _{0}^{1}k_{0}(t,s) \,dB_{4}(t) = \frac{1}{2}k_{0} \biggl( \frac{1}{4},s \biggr)-\frac{1}{4}k_{0} \biggl( \frac{1}{2},s \biggr) \\ &= \textstyle\begin{cases} - \frac{1}{24}s^{3}+\frac{3}{256}s, & 0 \leq s\le \frac{1}{4}, \\ \frac{1}{24}s^{3}-\frac{1}{16}s^{2}+\frac{7}{256}s- \frac{1}{768}, & \frac{1}{4}< s\le \frac{1}{2}, \\ -\frac{1}{256}s+\frac{1}{256}, & \frac{1}{2}< s\le 1, \end{cases}\displaystyle \end{aligned}$$

and hence $0\le \mathcal{K}_{4}(s)<0.0025$.

The $4\times 4$ matrix

$$[B]=\left(\begin{array}{cccc}{\beta }_1[{\gamma }_1]& {\beta }_1[{\gamma }_2]& {\beta }_1[{\gamma }_3]& {\beta }_1[{\gamma }_4]\\ {\beta }_2[{\gamma }_1]& {\beta }_2[{\gamma }_2]& {\beta }_2[{\gamma }_3]& {\beta }_2[{\gamma }_4]\\ {\beta }_3[{\gamma }_1]& {\beta }_3[{\gamma }_2]& {\beta }_3[{\gamma }_3]& {\beta }_3[{\gamma }_4]\\ {\beta }_4[{\gamma }_1]& {\beta }_4[{\gamma }_2]& {\beta }_4[{\gamma }_3]& {\beta }_4[{\gamma }_4]\end{array}\right)=\left(\begin{array}{cccc}\frac{1}{6}& 0& \frac{1}{96}& \frac{1}{192}\\ \frac{3}{8}& \frac{13}{48}& \frac{19}{320}& \frac{103}{960}\\ \frac{1}{4}& \frac{1}{16}& \frac{49}{1536}& \frac{59}{1536}\\ \frac{1}{4}& 0& \frac{3}{256}& \frac{1}{256}\end{array}\right)$$

and its spectrum radius $r([B])\thickapprox 0.2976<1$. Those mean that $(C_{2})$ and $(C_{3})$ are satisfied.

Now we take stock of some constants in Theorem 2.1 and Theorem 2.2.

$$ \bigl(I-[B]\bigr)^{-1}< \begin{pmatrix} 1.2066 & 0.0012 & 0.0132 & 0.0070 \\ 0.6958 & 1.3796 & 0.0941 & 0.1559 \\ 0.3688 & 0.0895 & 1.0431 & 0.0519 \\ 0.3073 & 0.0014 & 0.0157 & 1.0064 \end{pmatrix} $$

and

$$ \bigl(I-[B]\bigr)^{-1}\mathcal{K}(s)< \begin{pmatrix} 0.0033 \\ 0.0340 \\ 0.0143 \\ 0.0035 \end{pmatrix}, $$

thus $k_{S}(t,s)<0.0033+0.0340t+0.0143\times \frac{1}{6}t(t^{2}-3t+3)+0.0035 {\times} {\frac{1}{6}t(3-t^{2})}+k_{0}(t,s)<0.1051$. So, for $u\in C^{2}[0,1]$ and $t\in [0,1]$,

$$ \begin{aligned} \bigl\vert (L_{i}u) (t) \bigr\vert \le {}& 0.1051 \int _{0}^{1}\bigl(a_{i} \bigl\vert u(s) \bigr\vert +b_{i} \bigl\vert u'(s) \bigr\vert +c_{i} \bigl\vert u''(s) \bigr\vert \bigr)\,ds \\ \le {}& 0.1051(a_{i}+b_{i}+c_{i}) \Vert u \Vert _{C^{2}}\quad (i=1,2),\end{aligned} $$

here $L_{i}$ ($i=1,2$) are defined in (2.7). Since all the terms are nonnegative in the first derivative of $k_{S}(t,s)$ with respect to t and they are non-positive in the second derivative of $k_{S}(t,s)$, we also have that, for $u\in C^{2}[0,1]$ and $t\in [0,1]$,

$$\begin{aligned}& \begin{aligned} \bigl\vert (L_{i}u)'(t) \bigr\vert \le {}& 0.1680 \int _{0}^{1}\bigl(a_{i} \bigl\vert u(s) \bigr\vert +b_{i} \bigl\vert u'(s) \bigr\vert +c_{i} \bigl\vert u''(s) \bigr\vert \bigr)\,ds \\ \le {}& 0.1680(a_{i}+b_{i}+c_{i}) \Vert u \Vert _{C^{2}}\quad (i=1,2),\end{aligned} \\& \begin{aligned} \bigl\vert (L_{i}u)''(t) \bigr\vert \le {}& 0.2590 \int _{0}^{1}\bigl(a_{i} \bigl\vert u(s) \bigr\vert +b_{i} \bigl\vert u'(s) \bigr\vert +c_{i} \bigl\vert u''(s) \bigr\vert \bigr)\,ds \\ \le{} & 0.2590(a_{i}+b_{i}+c_{i}) \Vert u \Vert _{C^{2}}\quad (i=1,2).\end{aligned} \end{aligned}$$

Therefore the radius $r(L_{i})\le \Vert L_{i} \Vert \le 0.2590(a_{i}+b_{i}+c_{i})<1$ if

$$ a_{i}+b_{i}+c_{i}< 0.2590^{-1} \quad (i=1,2).$$

(2.23)

On the other hand, we have from Lemma 2.1 and Lemma 2.2 that, for $u\in K\setminus \{0\}$ and $t\in [0,1]$,

$$ \begin{aligned} (L_{2}u) (t)\ge {}& \int _{0}^{1}k_{S}(t,s)a_{2}u(s) \,ds \geq a_{2}c_{0}(t) \int _{0}^{1}\varPhi _{0}(s)u(s)\,ds \\ \geq {}& a_{2}c_{0}(t) \int _{0}^{1}\varPhi _{0}(s)c_{0}(s) \Vert {u} \Vert _{C}\,ds=a_{2}c_{0}(t) \Vert {u} \Vert _{C} \int _{0}^{1}c_{0}(s)\varPhi _{0}(s)\,ds\end{aligned} $$

and

$$ \bigl\Vert (L_{2}u) \bigr\Vert _{C}=(L_{2}u) (1)\ge \frac{1}{6}a_{2} \Vert {u} \Vert _{C} \int _{0}^{1}c_{0}(s) \varPhi _{0}(s)\,ds, $$

hence

$$\begin{aligned} \bigl(L_{2}^{2}u\bigr) (t)&\ge a_{2} \int _{0}^{1}k_{S}(t,s) (L_{2}u) (s)\,ds \\ &\ge a_{2}c_{0}(t) \int _{0}^{1}\varPhi _{0}(s) (L_{2}u) (s)\,ds \ge a_{2}c_{0}(t) \int _{0}^{1}\varPhi _{0}(s)c_{0}(s) \bigl\Vert (L_{2}u) \bigr\Vert _{C}\,ds \\ &\ge \frac{1}{6}a_{2}^{2}c_{0}(t) \Vert {u} \Vert _{C} \biggl( \int _{0}^{1}c_{0}(s) \varPhi _{0}(s)\,ds \biggr)^{2} \end{aligned}$$

and

$$ \bigl\Vert \bigl(L_{2}^{2}u\bigr) \bigr\Vert _{C}=\bigl(L_{2}^{2}u\bigr) (1)\ge \frac{1}{36}a_{2}^{2} \Vert {u} \Vert _{C} \biggl( \int _{0}^{1}c_{0}(s)\varPhi _{0}(s)\,ds \biggr)^{2}. $$

By induction,

$$ \bigl\Vert \bigl(L_{2}^{n}u\bigr) \bigr\Vert _{C}=\bigl(L_{2}^{n}u\bigr) (1)\ge \biggl( \frac{a_{2}}{6} \biggr)^{n} \Vert {u} \Vert _{C} \biggl( \int _{0}^{1}c_{0}(s)\varPhi _{0}(s)\,ds \biggr)^{n}. $$

As a result, it follows that, for $u\in K\setminus \{0\}$,

$$ \bigl\Vert {L_{2}^{n}} \bigr\Vert \Vert {u} \Vert _{C^{2}}\geq \bigl\Vert {L_{2}^{n}}u \bigr\Vert _{C^{2}}\geq \bigl\Vert {L_{2}^{n}}u \bigr\Vert _{C} \ge \biggl(\frac{a_{2}}{6} \biggr)^{n} \Vert {u} \Vert _{C} \biggl( \int _{0}^{1}c_{0}(s) \varPhi _{0}(s)\,ds \biggr)^{n}, $$

and according to Gelfand’s formula, the spectral radius

$$ \begin{aligned} r(L_{2})&= \lim_{n\to \infty } \bigl\Vert {L_{2}^{n}} \bigr\Vert ^{1/n} \\ &\ge \frac{a_{2}}{6} \biggl( \int _{0}^{1}c_{0}(s)\varPhi _{0}(s)\,ds \biggr)\lim_{n\to \infty } \biggl( \frac{ \Vert {u} \Vert _{C}}{ \Vert {u} \Vert _{C^{2}}} \biggr)^{1/n} =\frac{a_{2}}{6} \biggl( \int _{0}^{1}c_{0}(s)\varPhi _{0}(s)\,ds \biggr), \end{aligned} $$

which implies that $r(L_{2})\ge 1$ when

$$ \begin{aligned}[b] a_{2}&\ge \frac{30240}{29} = \frac{6}{\int _{0}^{1}\frac{1}{6}s(1-s)(1+s)\times \frac{1}{6}s(s^{2}-3s+3)\,ds} \\ &\ge \frac{6}{\int _{0}^{1}c_{0}(s)\varPhi _{0}(s)\,ds}.\end{aligned} $$

(2.24)

Example 2.1

If

$$ f(t,x_{1},x_{2},x_{3})= \frac{x_{1}^{5}+x_{2}^{5}-x_{3}^{5}}{1+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}, $$

then BVP (2.22) has a positive solution.

Proof

Take $a_{2}=b_{2}=c_{2}=1$, $r<1$, it is easy to check that (2.9) and (2.23) for $i=2$ are satisfied. Now take $a_{1}=1043$, $b_{1}=69$, $c_{1}=903$, it is clear that

$$\begin{aligned}& \begin{aligned} \frac{a_{1}}{6} \int _{0}^{1}c_{0}(s)\varPhi _{0}(s)\,ds&= \frac{a_{1}}{36} \int _{0}^{1}s\bigl(s^{2}-3s+3\bigr) \varPhi _{0}(s)\,ds \\ &> \frac{a_{1}}{36} \int _{0}^{1}s\bigl(s^{2}-3s+3\bigr) \frac{1}{6}s(1-s) (1+s)\,ds \\ &> \frac{30240}{29}\times \frac{1}{216} \int _{0}^{1}s^{2}\bigl(s^{2}-3s+3 \bigr) (1-s) (1+s)\,ds=1,\end{aligned} \\& \begin{aligned} \frac{b_{1}}{2} \int _{0}^{1}c_{1}(s)\varPhi _{1}(s)\,ds&= \frac{b_{1}}{4} \int _{0}^{1}\bigl(1-s^{2}\bigr)\varPhi _{1}(s)\,ds \\ &> \frac{b_{1}}{4} \int _{0}^{1}\bigl(1-s^{2}\bigr) \frac{1}{2}s(1-s)\,ds \\ &> \frac{480}{7}\times \frac{1}{8} \int _{0}^{1}\bigl(1-s^{2}\bigr)s(1-s) \,ds=1.\end{aligned} \end{aligned}$$

We also have

$$ \bigl(I-[B]\bigr)^{-1}> \begin{pmatrix} 1.2064 & 0.0010 & 0.0130 & 0.0068 \\ 0.6956 & 1.3794 & 0.0939 & 0.1557 \\ 0.3686 & 0.0893 & 1.0429 & 0.0517 \\ 0.3071 & 0.0012 & 0.0155 & 1.0062 \end{pmatrix}$$

(2.25)

and

$$ \int _{0}^{1}\mathcal{K}(s)s(1-s)\,ds> \begin{pmatrix} 31.1\times 10^{-5} \\ 288.9\times 10^{-5} \\ 137.5\times 10^{-5} \\ 27.4\times 10^{-5}\end{pmatrix}.$$

(2.26)

It follows from Lemma 2.1, (2.25), and (2.26) that

$$ \frac{c_{1}}{2} \int _{0}^{1}\bigl(\kappa _{3}(s)+\kappa _{4}(s)\bigr)c_{2}(s)\,ds \ge \frac{c_{1}}{2} \int _{0}^{1}\bigl(\kappa _{3}(s)+\kappa _{4}(s)\bigr)s(1-s)\,ds>1 $$

since $\kappa _{3}(s)$ and $\kappa _{4}(s)$ are the third and the fourth components in $(I-[B])^{-1}\mathcal{K}(s)$ respectively, so (2.10) is valid. It can be seen that (2.11) is satisfied for $C_{0}$ large enough. Then BVP (2.22) has a positive solution by Theorem 2.1. □

Example 2.2

If $f(t,x_{1},x_{2},x_{3})=\sqrt[3]{x_{1}}-\sqrt[3]{x_{3}}$, then BVP (2.22) has a positive solution.

Proof

Take $a_{1}=1/2$, $b_{1}=0$, $c_{1}=1/3$, $C_{0}=2$ and $a_{2}=1043$, $b_{2}=0$, $c_{2}=1$, ${r=1/33685}$. Obviously, (2.23) for $i=1$ and (2.24) are satisfied, meanwhile conditions (2.19) and (2.20) are fulfilled. Then BVP (2.22) has a positive solution by Theorem 2.2. □

3 Inequalities of Green’s function and positive solutions for (1.2)

For BVP (1.2) we make the assumption:

$(C'_{1})$:

$g: [0,1]\times \mathbb{R}^{3}_{+}\to \mathbb{R}_{+}$ is continuous.

Similar to Webb and Infante [16], BVP (1.2) can be converted to the integral equation in $C^{2}[0,1]$:

$$ u(t)=(\widetilde{S}u) (t)=: \int _{0}^{1}\widetilde{k}_{S}(t,s)g \bigl(s,u(s),u'(s),u''(s)\bigr) \,ds,$$

(3.1)

where

$$ \widetilde{k}_{S}(t,s)=\bigl\langle \bigl(I-[A]\bigr)^{-1} \widetilde{\mathcal{K}}(s), \delta (t)\bigr\rangle +\widetilde{k}_{0}(t,s) =\sum_{i=1}^{4} \widetilde{\kappa }_{i}(s)\delta _{i}(t)+\widetilde{k}_{0}(t,s), $$

$\langle (I-[A])^{-1}\widetilde{\mathcal{K}}(s),\delta (t)\rangle$ is the inner product in $\mathbb{R}^{4}$,

$$\begin{aligned}& \widetilde{\mathcal{K}}_{i}(s):= \int _{0}^{1}\widetilde{k}_{0}(t,s) \,dA_{i}(t) \quad (i=1,2,3,4), \\& \delta _{1}(t)=1, \quad\quad \delta _{2}(t)=t, \quad\quad \delta _{3}(t)=\frac{1}{6}t^{2}(3-t), \quad\quad \delta _{4}(t)=\frac{1}{6}t^{3}, \end{aligned}$$

$\widetilde{\kappa }_{i}(s)$ is the ith component of $(I-[A])^{-1}\widetilde{\mathcal{K}}(s)$,

$$ \widetilde{k}_{0}(t,s)= \textstyle\begin{cases} \frac{1}{6}t^{3}(1-s), & 0\le t\le s\le 1, \\ \frac{1}{6}s(3t^{2}-3ts+s^{2}-t^{3}), & 0\le s\le t\le 1.\end{cases} $$

We put forward the following hypotheses:

$(C'_{2})$:

$A_{i}$ is of bounded variation and $\widetilde{\mathcal{K}}_{i}(s)\ge 0$, $\forall s\in [0,1]$ ($i=1,2,3,4$);

$(C'_{3})$:

The $4\times 4$ matrix $[A]$ is positive whose $(i,j)$th entry is $\alpha _{i}[\delta _{j}]$ and whose spectrum radius $r([A])<1$.

Lemma 3.1

If$(C'_{2})$and$(C'_{3})$hold, then$\widetilde{\kappa }_{i}(s)\ge 0$ ($i=1,2,3,4$) and, for$t, s\in [0,1]$,

$$ \widetilde{c}_{0}(t)\widetilde{\varPhi }_{0}(s)\le \widetilde{k}_{S}(t,s) \leq \widetilde{\varPhi }_{0}(s),$$

(3.2)

where

$$ \widetilde{\varPhi }_{0}(s)=\sum_{i=1}^{4} \widetilde{\kappa }_{i}(s)+ \frac{1}{6}s(1-s) (2-s),\qquad \widetilde{c}_{0}(t)=\frac{1}{6}t^{3}, $$

and

$$ \widetilde{c}_{1}(t)\widetilde{\varPhi }_{1}(s)\le \frac{\partial \widetilde{k}_{S}(t,s)}{\partial t}\leq \widetilde{\varPhi }_{1}(s),\qquad \widetilde{c}_{2}(t)\widetilde{\varPhi }_{2}(s)\le \frac{\partial ^{2}\widetilde{k}_{S}(t,s)}{\partial t^{2}}\leq \widetilde{\varPhi }_{2}(s),$$

(3.3)

where

$$\begin{aligned}& \widetilde{\varPhi }_{1}(s)=\sum_{i=2}^{4} \widetilde{\kappa }_{i}(s)+ \frac{1}{2}s(1-s),\qquad \widetilde{c}_{1}(t)=\frac{1}{2}t^{2}, \\ & \widetilde{\varPhi }_{2}(s)=\sum_{i=3}^{4} \widetilde{\kappa }_{i}(s)+s(1-s), \qquad \widetilde{c}_{2}(t)= \min \{t,1-t\}. \end{aligned}$$

Proof

For $s\in [0,1]$, $\widetilde{\kappa }_{i}(s)\ge 0$ ($i=1,2,3,4$) are due to [16, proof of Theorem 2.4] since both $(I-[A])^{-1}$ and $\widetilde{\mathcal{K}}(s)$ are nonnegative.

According to the following two inequalities:

$$\begin{aligned}& \frac{1}{6}t^{3}\sum_{i=1}^{4} \widetilde{\kappa }_{i}(s)\le \sum_{i=1}^{4} \widetilde{\kappa }_{i}(s)\delta _{i}(t) \le \sum _{i=1}^{4} \widetilde{\kappa }_{i}(s), \\ & \frac{1}{6}t^{3}\frac{1}{6}s(1-s) (2-s)\le t^{3}\frac{1}{6}s(1-s) (2-s) \le \widetilde{k}_{0}(t,s) \le \frac{1}{6}s(1-s) (2-s), \end{aligned}$$

we have, for $t,s\in [0,1]$,

$$ \widetilde{c}_{0}(t)\widetilde{\varPhi }_{0}(s)\le \widetilde{k}_{S}(t,s)= \sum_{i=1}^{4} \widetilde{\kappa }_{i}(s)\delta _{i}(t)+ \widetilde{k}_{0}(t,s) \leq \widetilde{\varPhi }_{0}(s). $$

Moreover, the next two inequalities

$$\begin{aligned}& \frac{1}{2}t^{2}\sum_{i=2}^{4} \widetilde{\kappa }_{i}(s)\le \sum_{i=1}^{4} \widetilde{\kappa }_{i}(s)\delta '_{i}(t)\le \sum_{i=2}^{4} \widetilde{\kappa }_{i}(s), \\ & \frac{1}{2}t^{2}\frac{1}{2}s(1-s)\le t^{2} \frac{1}{2}s(1-s)\le \frac{\partial \widetilde{k}_{0}(t,s)}{\partial t}\le \frac{1}{2}s(1-s) \end{aligned}$$

imply, for $t,s\in [0,1]$,

$$ \widetilde{c}_{1}(t)\widetilde{\varPhi }_{1}(s)\le \frac{\partial \widetilde{k}_{S}(t,s)}{\partial t}=\sum_{i=1}^{4} \widetilde{ \kappa }_{i}(s)\delta '_{i}(t)+ \frac{\partial \widetilde{k}_{0}(t,s)}{\partial t}\leq \widetilde{\varPhi }_{1}(s). $$

Finally, from the two inequalities

$$\begin{aligned}& \min \{t,1-t\}\sum_{i=3}^{4}\kappa _{i}(s)\le \sum_{i=1}^{4} \widetilde{\kappa }_{i}(s)\delta ''_{i}(t)=(1-t) \kappa _{3}(s)+t \kappa _{4}(s) \le \sum _{i=3}^{4}\widetilde{\kappa }_{i}(s), \\ & \min \{t,1-t\}s(1-s)\le \frac{\partial ^{2}\widetilde{k}_{0}(t,s)}{\partial t^{2}}\le s(1-s), \end{aligned}$$

it follows that

$$ \widetilde{c}_{2}(t)\widetilde{\varPhi }_{2}(s)\le \frac{\partial ^{2}\widetilde{k}_{S}(t,s)}{\partial t^{2}}=\sum_{i=1}^{4} \widetilde{ \kappa }_{i}(s)\delta ''_{i}(t)+ \frac{\partial ^{2}\widetilde{k}_{0}(t,s)}{\partial t^{2}}\leq \widetilde{\varPhi }_{2}(s) $$

for $t,s\in [0,1]$. □

Define the subsets in $C^{2}[0,1]$ as follows:

$$\begin{aligned}& \widetilde{P}= \bigl\{ u\in C^{2}[0,1]: u(t)\geq 0, u'(t)\ge 0, u''(t) \ge 0, \forall t\in [0,1] \bigr\} , \end{aligned}$$

(3.4)

$$\begin{aligned}& \begin{aligned}[b] \widetilde{K}= {}& \bigl\{ u\in \widetilde{P}: u(t)\ge c_{0}(t) \Vert u \Vert _{C}, u'(t)\ge c_{1}(t) \bigl\Vert u' \bigr\Vert _{C}, \\ &{} u''(t)\ge c_{2}(t) \bigl\Vert u'' \bigr\Vert _{C}, \forall t\in [0,1]; \alpha _{i}[u] \geq 0\ (i=1,2,3,4) \bigr\} .\end{aligned} \end{aligned}$$

(3.5)

Clearly both P̃ and K̃ are cones, and it is easy to check that P̃ is a solid cone.

Now we define linear operators in $C^{2}[0,1]$:

$$\begin{aligned}& (\widetilde{L}_{i}u) (t) = \int _{0}^{1}\widetilde{k}_{S}(t,s) \bigl( \widetilde{a}_{i}u(s)+\widetilde{b}_{i}u'(s)+ \widetilde{c}_{i}u''(s)\bigr)\,ds \quad (i=1,2), \\& (\widetilde{L}_{3}u) (t)=\widetilde{a}_{1} \int _{0}^{1}\widetilde{k}_{S}(t,s)u(s) \,ds, \end{aligned}$$

(3.6)

where $\widetilde{a}_{i}$, $\widetilde{b}_{i}$, $\widetilde{c}_{i}$ ($i=1,2$) are nonnegative constants.

Similar to [16], we have the following Lemma 3.2 by Lemma 3.1.

Lemma 3.2

If$(C'_{1})$–$(C'_{3})$hold, then$\widetilde{S}:\widetilde{P}\to \widetilde{K}$and$\widetilde{L}_{i}: C^{2}[0,1]\to C^{2}[0,1]$are completely continuous operators with$\widetilde{L}_{i}(\widetilde{P})\subset \widetilde{K}$ ($i=1,2,3$), whereS̃, P̃, K̃are defined separately in (3.1), (3.4), and (3.5).

Theorem 3.1

Under hypotheses $(C'_{1})$ – $(C'_{3})$ suppose that

$(\widetilde{F}_{1})$ :

there exist constants$\widetilde{a}_{2}, \widetilde{b}_{2}, \widetilde{c}_{2}\ge 0$, and$\widetilde{r}>0$such that

$$ g(t,x_{1},x_{2},x_{3})\leq \widetilde{a}_{2}x_{1}+\widetilde{b}_{2}x_{2}+ \widetilde{c}_{2}x_{3}$$

(3.7)

for all$(t,x_{1},x_{2},x_{3})\in [0,1]\times [0,\widetilde{r}]^{3}$, moreover the spectral radius$r(\widetilde{L}_{2})<1$, where$\widetilde{L}_{2}$is defined by (3.6),

[$(\widetilde{F}_{2})$:

there exist positive constants$\widetilde{a}_{1}$, $\widetilde{b}_{1}$, $\widetilde{c}_{1}$, $\widetilde{C}_{0}$satisfying

$$ \begin{aligned}[b] &\min \biggl\{ \frac{\widetilde{a}_{1}}{6} \int _{0}^{1} \widetilde{c}_{0}(s) \widetilde{\varPhi }_{0}(s)\,ds, \frac{\widetilde{b}_{1}}{2} \int _{0}^{1}\widetilde{c}_{1}(s) \widetilde{\varPhi }_{1}(s)\,ds, \\ &\quad \frac{\widetilde{c}_{1}}{2} \int _{0}^{1}\bigl( \widetilde{\kappa }_{3}(s)+\widetilde{\kappa }_{4}(s)\bigr)c_{2}(s) \,ds \biggr\} >1\end{aligned} $$

(3.8)

such that

$$ g(t,x_{1},x_{2},x_{3})\geq \widetilde{a}_{1}x_{1}+\widetilde{b}_{1}x_{2}+ \widetilde{c}_{1}x_{3}-\widetilde{C}_{0}$$

(3.9)

for all$(t,x_{1},x_{2},x_{3})\in [0,1]\times \mathbb{R}_{+}^{3}$.

Then BVP (1.2) has one positive solution inK̃.

Proof

Let

$$\begin{aligned} M={}& \max \biggl\{ \frac{6\widetilde{C}_{0}\int _{0}^{1}\widetilde{\varPhi }_{0}(s)\,ds}{\widetilde{a}_{1}\int _{0}^{1}\widetilde{c}_{0}(s)\widetilde{\varPhi }_{0}(s)\,ds-6}, \frac{2\widetilde{C}_{0}\int _{0}^{1}\widetilde{\varPhi }_{1}(s)\,ds}{\widetilde{b}_{1}\int _{0}^{1}\widetilde{c}_{1}(s)\widetilde{\varPhi }_{1}(s)\,ds-2}, \\ &{} \frac{\widetilde{C}_{0}\int _{0}^{1}(\widetilde{\kappa }_{3}(s)+\widetilde{\kappa }_{4}(s))\,ds}{c_{1}\int _{0}^{1}(\widetilde{\kappa }_{3}(s)+\widetilde{\kappa }_{4}(s))\widetilde{c}_{2}(s)\,ds-2} \biggr\} \end{aligned}$$

and the rest is similar to the proof of Theorem 2.1 in which $\widetilde{c}_{0}(t)=\frac{1}{2}t^{3}$ for step (iv). □

Theorem 3.2

Under hypotheses$(C'_{1})$–$(C'_{3})$, suppose that

$(\widetilde{F}_{3})$:

there exist constants$\widetilde{a}_{1}, \widetilde{b}_{1}, \widetilde{c}_{1}, \widetilde{C}_{0}\ge 0$such that

$$ g(t,x_{1},x_{2},x_{3})\leq \widetilde{a}_{1}x_{1}+\widetilde{b}_{1}x_{2}+ \widetilde{c}_{1}x_{3}+\widetilde{C}_{0}$$

(3.10)

for all$(t,x_{1},x_{2},x_{3})\in [0,1]\times \mathbb{R}_{+}^{3}$, moreover the spectral radius$r(\widetilde{L}_{1})<1$;

$(\widetilde{F}_{4})$:

there exist constants$\widetilde{a}_{2}, \widetilde{b}_{2}, \widetilde{c}_{2}\ge 0$, and$\widetilde{r}>0$such that

$$ g(t,x_{1},x_{2},x_{3})\geq \widetilde{a}_{2}x_{1}+\widetilde{b}_{2}x_{2}+ \widetilde{c}_{2}x_{3}$$

(3.11)

for all$(t,x_{1},x_{2},x_{3})\in [0,1]\times [0,\widetilde{r}]^{3}$, moreover the spectral radius$r(\widetilde{L}_{2})\ge 1$; where$\widetilde{L}_{i}: C^{2}[0,1]\to C^{2}[0,1]$ ($i=1,2$) are defined by (3.6).

Then BVP (1.2) has one positive solution inK̃.

Proof

Denote the cone ordering induced by P̃, $u\preceq v$ for $u,v\in X$ if and only if $v-u\in \widetilde{P}$ and equivalently $v\succeq u$. The rest is similar to the proof of Theorem 2.2. □

As an example, we consider fourth-order boundary problem under mixed multi-point and integral boundary conditions with sign-changing coefficients:

$$ \textstyle\begin{cases} -u^{(4)}(t)=f(t,u(t),u'(t),u''(t)),\quad t\in [0,1], \\ u(0)=\frac{1}{2}u(\frac{1}{4})-\frac{1}{60}u(\frac{3}{4}),\quad \quad u'(0)= \int _{0}^{1}u(t)(t-\frac{1}{8})\,dt, \\ u''(0)=\frac{1}{2}u(\frac{1}{2})-\frac{1}{16}u(\frac{3}{4}),\quad \quad u''(1)= \frac{1}{2}u(\frac{1}{4})-\frac{1}{40}u(\frac{1}{2}),\end{cases} $$

(3.12)

thus $\alpha _{1}[u]=\frac{1}{2}u(\frac{1}{4})-\frac{1}{60}u(\frac{3}{4})$, $\alpha _{2}[u]=\int _{0}^{1}u(t)(t-\frac{1}{8})\,dt$, $\alpha _{3}[u]= \frac{1}{2}u(\frac{1}{2})-\frac{1}{16}u(\frac{3}{4})$, and $\alpha _{4}[u]=\frac{1}{2}u(\frac{1}{4})-\frac{1}{40}u(\frac{1}{2})$. We estimate some coefficients, and Matlab is also used.

$$\begin{aligned} \widetilde{\mathcal{K}}_{1}(s)&= \int _{0}^{1} \widetilde{k}_{0}(t,s) \,dA_{1}(t) =\frac{1}{2}\widetilde{k}_{0} \biggl( \frac{1}{4},s \biggr)-\frac{1}{60}\widetilde{k}_{0} \biggl(\frac{3}{4},s \biggr) \\ &= \textstyle\begin{cases} \frac{29}{360}s^{3}-\frac{9}{160}s^{2}+\frac{83}{7680}s, & 0\leq s\le \frac{1}{4}, \\ -\frac{1}{360}s^{3}+\frac{1}{160}s^{2}-\frac{37}{7680}s+ \frac{1}{768}, & \frac{1}{4}< s\le \frac{3}{4}, \\ -\frac{1}{7680}s+\frac{1}{7680}, & \frac{3}{4}< s\le 1,\end{cases}\displaystyle \end{aligned}$$

and hence $0\le \widetilde{\mathcal{K}}_{1}(s)<0.0007$;

$$ \begin{aligned} \widetilde{\mathcal{K}}_{2}(s)={}& \int _{0}^{1} \widetilde{k}_{0}(t,s) \biggl(t-\frac{1}{8} \biggr)\,dt \\ = {}& -\frac{1}{120}s^{5}+\frac{1}{192}s^{4}+ \frac{1}{16}s^{3}- \frac{13}{96}s^{2}+ \frac{73}{960}s\quad (0\le s\le 1),\end{aligned} $$

and hence $0\le \widetilde{\mathcal{K}}_{2}(s)<0.0129$;

$$\begin{aligned} \widetilde{\mathcal{K}}_{3}(s)&= \int _{0}^{1} \widetilde{k}_{0}(t,s) \,dA_{3}(t) =\frac{1}{2}\widetilde{k}_{0} \biggl( \frac{1}{2},s \biggr)-\frac{1}{16}\widetilde{k}_{0} \biggl(\frac{3}{4},s \biggr) \\ &= \textstyle\begin{cases} \frac{7}{96}s^{3}-\frac{13}{128}s^{2}+\frac{239}{6144}s, & 0\leq s\le \frac{1}{2}, \\ -\frac{1}{96}s^{3}+\frac{3}{128}s^{2}- {\frac{145}{6144}}s+\frac{1}{96}, & \frac{1}{2}< s\le \frac{3}{4}, \\ -{\frac{37}{6144}s+\frac{37}{6144}}, & \frac{3}{4}< s\le 1,\end{cases}\displaystyle \end{aligned}$$

and hence $0\le \widetilde{\mathcal{K}}_{3}(s)<0.0046$;

$$\begin{aligned} \widetilde{\mathcal{K}}_{4}(s)&= \int _{0}^{1} \widetilde{k}_{0}(t,s) \,dB_{4}(t) =\frac{1}{2}\widetilde{k}_{0} \biggl( \frac{1}{4},s \biggr)-\frac{1}{40}\widetilde{k}_{0} \biggl(\frac{1}{2},s \biggr) \\ &= \textstyle\begin{cases} \frac{19}{240}s^{3}-\frac{9}{160}s+\frac{3}{256}s, & 0\leq s\le \frac{1}{4}, \\ -\frac{1}{240}s^{3}+\frac{1}{160}s^{2}-\frac{1}{256}s+ \frac{1}{768}, & \frac{1}{4}< s\le \frac{1}{2}, \\ -\frac{1}{1280}s+\frac{1}{1280}, & \frac{1}{2}< s\le 1,\end{cases}\displaystyle \end{aligned}$$

and hence $0\le \widetilde{\mathcal{K}}_{4}(s)<0.0008$.

The $4\times 4$ matrix

$$[A]=\left(\begin{array}{cccc}{\alpha }_1[{\delta }_1]& {\alpha }_1[{\delta }_2]& {\alpha }_1[{\delta }_3]& {\alpha }_1[{\delta }_4]\\ {\alpha }_2[{\delta }_1]& {\alpha }_2[{\delta }_2]& {\alpha }_2[{\delta }_3]& {\alpha }_2[{\delta }_4]\\ {\alpha }_3[{\delta }_1]& {\alpha }_3[{\delta }_2]& {\alpha }_3[{\delta }_3]& {\alpha }_3[{\delta }_4]\\ {\alpha }_4[{\delta }_1]& {\alpha }_4[{\delta }_2]& {\alpha }_4[{\delta }_3]& {\alpha }_4[{\delta }_4]\end{array}\right)=\left(\begin{array}{cccc}\frac{29}{60}& \frac{9}{80}& \frac{83}{7680}& \frac{1}{7680}\\ \frac{3}{8}& \frac{13}{48}& \frac{73}{960}& \frac{9}{320}\\ \frac{7}{16}& \frac{13}{64}& \frac{239}{6144}& \frac{37}{6144}\\ \frac{19}{40}& \frac{9}{80}& \frac{3}{256}& \frac{1}{1280}\end{array}\right)$$

and its spectrum radius $r([A])\thickapprox 0.6444<1$. Those mean that $(C'_{2})$ and $(C'_{3})$ are satisfied.

Now we take stock of some constants in Theorem 3.1 and Theorem 3.2.

$$ \bigl(I-[A]\bigr)^{-1}< \begin{pmatrix} 2.2565 & 0.3651 & 0.0545 & 0.0110 \\ 1.3460 & 1.6266 & 0.1445 & 0.0469 \\ 1.3195 & 0.5123 & 1.0962 & 0.0213 \\ 1.2398 & 0.3627 & 0.0551 & 1.0116 \end{pmatrix} $$

and

$$ \bigl(I-[A]\bigr)^{-1}\widetilde{\mathcal{K}}(s)< \begin{pmatrix} 0.0065 \\ 0.0226 \\ 0.0126 \\ 0.0066 \end{pmatrix}, $$

thus $\widetilde{k}_{S}(t,s)<0.0065+0.0226t+0.0126\times \frac{1}{6}t^{2}(3-t)+0.0066 \times \frac{1}{6}t^{3}+\widetilde{k}_{0}(t,s)<0.0987$. So, for $u\in C^{2}[0,1]$ and $t\in [0,1]$,

$$ \begin{aligned} \bigl\vert (\widetilde{L}_{i}u) (t) \bigr\vert \le {}& 0.0987 \int _{0}^{1}\bigl(a_{i} \bigl\vert u(s) \bigr\vert +b_{i} \bigl\vert u'(s) \bigr\vert +c_{i} \bigl\vert u''(s) \bigr\vert \bigr)\,ds \\ \le{} & 0.0987(a_{i}+b_{i}+c_{i}) \Vert u \Vert _{C^{2}}\quad (i=1,2),\end{aligned} $$

here $\widetilde{L}_{i}$ ($i=1,2$) are defined in (3.6). Since all the terms are nonnegative in the first and second derivatives of $k_{S}(t,s)$ with respect to t, we also have that, for $u\in C^{2}[0,1]$ and $t\in [0,1]$,

$$\begin{aligned}& \begin{aligned} \bigl\vert (\widetilde{L}_{i}u)'(t) \bigr\vert \le {}& 0.1573 \int _{0}^{1}\bigl(a_{i} \bigl\vert u(s) \bigr\vert +b_{i} \bigl\vert u'(s) \bigr\vert +c_{i} \bigl\vert u''(s) \bigr\vert \bigr)\,ds \\ \le{} & 0.1573(a_{i}+b_{i}+c_{i}) \Vert u \Vert _{C^{2}}\quad (i=1,2), \end{aligned} \\& \begin{aligned} \bigl\vert (\widetilde{L}_{i}u)''(t) \bigr\vert \le {}& 0.2597 \int _{0}^{1}\bigl(a_{i} \bigl\vert u(s) \bigr\vert +b_{i} \bigl\vert u'(s) \bigr\vert +c_{i} \bigl\vert u''(s) \bigr\vert \bigr)\,ds \\ \le {}& 0.2597(a_{i}+b_{i}+c_{i}) \Vert u \Vert _{C^{2}}\quad (i=1,2). \end{aligned} \end{aligned}$$

Therefore the radius $r(\widetilde{L}_{i})\le \Vert L_{i} \Vert \le 0.2597(a_{i}+b_{i}+c_{i})<1$ if

$$ a_{i}+b_{i}+c_{i}< 0.2597^{-1} \quad (i=1,2).$$

(3.13)

On the other hand, we have from Lemma 3.1 and Lemma 3.2 that, for $u\in \widetilde{K}\setminus \{0\}$ and $t\in [0,1]$,

$$ \begin{aligned} (\widetilde{L}_{2}u) (t)\ge {}& \int _{0}^{1}\widetilde{k}_{S}(t,s) \widetilde{a}_{2}u(s)\,ds\geq \widetilde{a}_{2} \widetilde{c}_{0}(t) \int _{0}^{1}\widetilde{\varPhi }_{0}(s)u(s) \,ds \\ \geq {}& \widetilde{a}_{2}\widetilde{c}_{0}(t) \int _{0}^{1} \widetilde{\varPhi }_{0}(s) \widetilde{c}_{0}(s) \Vert {u} \Vert _{C}\,ds = \widetilde{a}_{2}\widetilde{c}_{0}(t) \Vert {u} \Vert _{C} \int _{0}^{1} \widetilde{c}_{0}(s) \widetilde{\varPhi }_{0}(s)\,ds\end{aligned} $$

and

$$ \bigl\Vert (\widetilde{L}_{2}u) \bigr\Vert _{C}=( \widetilde{L}_{2}u) (1)\ge \frac{1}{6} \widetilde{a}_{2} \Vert {u} \Vert _{C} \int _{0}^{1}\widetilde{c}_{0}(s) \widetilde{\varPhi }_{0}(s)\,ds, $$

hence

$$ \begin{aligned} \bigl(\widetilde{L}_{2}^{2}u \bigr) (t)\ge{}& \widetilde{a}_{2} \int _{0}^{1} \widetilde{k}_{S}(t,s) ( \widetilde{L}_{2}u) (s)\,ds\geq \widetilde{a}_{2} \widetilde{c}_{0}(t) \int _{0}^{1}\widetilde{\varPhi }_{0}(s) ( \widetilde{L}_{2}u) (s)\,ds \\ \ge {}& \widetilde{a}_{2}\widetilde{c}_{0}(t) \int _{0}^{1} \widetilde{\varPhi }_{0}(s) \widetilde{c}_{0}(s) \bigl\Vert (\widetilde{L}_{2}u) \bigr\Vert _{C}\,ds \ge \frac{1}{6}\widetilde{a}_{2}^{2} \widetilde{c}_{0}(t) \Vert {u} \Vert _{C} \biggl( \int _{0}^{1}\widetilde{c}_{0}(s) \widetilde{\varPhi }_{0}(s)\,ds \biggr)^{2} \end{aligned} $$

and

$$ \bigl\Vert \bigl(\widetilde{L}_{2}^{2}u\bigr) \bigr\Vert _{C}=\bigl(\widetilde{L}_{2}^{2}u\bigr) (1)\ge \frac{1}{36}\widetilde{a}_{2}^{2} \Vert {u} \Vert _{C} \biggl( \int _{0}^{1} \widetilde{c}_{0}(s) \widetilde{\varPhi }_{0}(s)\,ds \biggr)^{2}. $$

By induction,

$$ \bigl\Vert \bigl(\widetilde{L}_{2}^{n}u\bigr) \bigr\Vert _{C}=\bigl(\widetilde{L}_{2}^{n}u\bigr) (1) \ge \biggl(\frac{\widetilde{a}_{2}}{6} \biggr)^{n} \Vert {u} \Vert _{C} \biggl( \int _{0}^{1} \widetilde{c}_{0}(s) \widetilde{\varPhi }_{0}(s)\,ds \biggr)^{n}. $$

As a result, it follows that, for $u\in \widetilde{K}\setminus \{0\}$,

$$ \bigl\Vert {\widetilde{L}_{2}^{n}} \bigr\Vert \Vert {u} \Vert _{C^{2}}\geq \bigl\Vert {\widetilde{L}_{2}^{n}}u \bigr\Vert _{C^{2}}\geq \bigl\Vert {\widetilde{L}_{2}^{n}}u \bigr\Vert _{C} \ge \biggl( \frac{\widetilde{a}_{2}}{6} \biggr)^{n} \Vert {u} \Vert _{C} \biggl( \int _{0}^{1} \widetilde{c}_{0}(s) \widetilde{\varPhi }_{0}(s)\,ds \biggr)^{n}, $$

and according to Gelfand’s formula, the spectral radius

$$ \begin{aligned} r(\widetilde{L}_{2})={}&\lim _{n\to \infty } \bigl\Vert {\widetilde{L}_{2}^{n}} \bigr\Vert ^{1/n} \\ \ge {}& \frac{\widetilde{a}_{2}}{6} \biggl( \int _{0}^{1} \widetilde{c}_{0}(s) \widetilde{\varPhi }_{0}(s)\,ds \biggr) \lim_{n\to \infty } \biggl(\frac{ \Vert {u} \Vert _{C}}{ \Vert {u} \Vert _{C^{3}}} \biggr)^{1/n} = \frac{\widetilde{a}_{2}}{6} \biggl( \int _{0}^{1}\widetilde{c}_{0}(s) \widetilde{\varPhi }_{0}(s)\,ds \biggr),\end{aligned} $$

which implies that $r(\widetilde{L}_{2})\ge 1$ when

$$ \widetilde{a}_{2}\ge 5040 = \frac{6}{\int _{0}^{1}\frac{1}{6}s^{3}\times \frac{1}{6}s(1-s)(2-s)\,ds} \ge \frac{6}{\int _{0}^{1}\widetilde{c}_{0}(s)\widetilde{\varPhi }_{0}(s)\,ds}.$$

(3.14)

Example 3.1

If

$$ g(t,x_{1},x_{2},x_{3})= \frac{\frac{3}{2}x_{1}^{5}+x_{2}^{5}+x_{3}^{5}}{1+x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}, $$

then BVP (3.12) has a positive solution.

Proof

Take $\widetilde{a}_{2}=3/2$, $\widetilde{b}_{2}=\widetilde{c}_{2}=1$, $r<1$, it is easy to check that (3.10) and (3.13) for $i=2$ are satisfied. Now take $\widetilde{a}_{1}=5040$, $\widetilde{b}_{1}=160$, $\widetilde{c}_{1}=990$, it is clear that

$$\begin{aligned}& \begin{aligned} \frac{\widetilde{a}_{1}}{6} \int _{0}^{1}\widetilde{c}_{0}(s) \widetilde{\varPhi }_{0}(s)\,ds= {}&\frac{\widetilde{a}_{1}}{36} \int _{0}^{1}s^{3} \widetilde{\varPhi }_{0}(s)\,ds \\ > {}& \frac{\widetilde{a}_{1}}{36} \int _{0}^{1} \frac{1}{6}s^{4}(1-s) (2-s)\,ds \\ ={}& 5040\times \frac{1}{36} \int _{0}^{1} \frac{1}{6}s^{4}(1-s) (2-s)\,ds=1,\end{aligned} \\& \begin{aligned} \frac{\widetilde{b}_{1}}{2} \int _{0}^{1}\widetilde{c}_{1}(s) \widetilde{\varPhi }_{1}(s)\,ds= {}&\frac{\widetilde{b}_{1}}{4} \int _{0}^{1}s^{2} \widetilde{\varPhi }_{1}(s)\,ds \\ > {}& \frac{\widetilde{b}_{1}}{4} \int _{0}^{1}\frac{1}{2}s^{3}(1-s) \,ds= 160\times \frac{1}{4} \int _{0}^{1}\frac{1}{2}s^{3}(1-s) \,ds=1.\end{aligned} \end{aligned}$$

We also have

$$ \bigl(I-[A]\bigr)^{-1}> \begin{pmatrix} 2.2563 & 0.3649 & 0.0543 & 0.0108 \\ 1.3458 & 1.6264 & 0.1443 & 0.0467 \\ 1.3193 & 0.5121 & 1.0960 & 0.0211 \\ 1.2396 & 0.3625 & 0.0549 & 1.0114 \end{pmatrix}$$

(3.15)

and

$$ \int _{0}^{1}\widetilde{\mathcal{K}}(s)s(1-s)\,ds> \begin{pmatrix} 3.2\times 10^{-5} \\ 150\times 10^{-5} \\ 48.6\times 10^{-5} \\ 6.7\times 10^{-5}\end{pmatrix}.$$

(3.16)

It follows from Lemma 3.1, (3.15), and (3.16) that

$$ \frac{\widetilde{c}_{1}}{2} \int _{0}^{1}\bigl(\widetilde{\kappa }_{3}(s)+ \widetilde{\kappa }_{4}(s)\bigr)c_{2}(s) \,ds \ge \frac{\widetilde{c}_{1}}{2} \int _{0}^{1}\bigl(\widetilde{\kappa }_{3}(s)+\widetilde{\kappa }_{4}(s)\bigr)s(1-s)\,ds>1 $$

since $\widetilde{\kappa }_{3}(s)$ and $\widetilde{\kappa }_{4}(s)$ are the third and the fourth components in $(I-[A])^{-1}\widetilde{\mathcal{K}}(s)$ respectively, so (3.8) is valid. It can be seen that (3.9) is satisfied for $\widetilde{C}_{0}$ large enough. Then BVP (3.12) has a positive solution by Theorem 3.1. □

Example 3.2

If $g(t,x_{1},x_{2},x_{3})=\sqrt{x_{1}}+\sqrt{x_{4}}$, then BVP (3.12) has a positive solution.

Proof

Take $\widetilde{a}_{1}=\widetilde{c}_{1}=1$, $\widetilde{b}_{1}=0$, $\widetilde{C}_{0}=1/2$ and ${\widetilde{a}_{2}=5040}$, $\widetilde{b}_{2}=0$, $\widetilde{c}_{2}=1$, ${\widetilde{r}=3.9\times 10^{-8}}$. Obviously, (3.13) for $i=1$ and (3.14) are satisfied, meanwhile conditions (3.10) and (3.11) are fulfilled. Then BVP (3.12) has a positive solution by Theorem 3.2. □