Solvability of boundary value problem with p-Laplacian at resonance

Abstract

By generalizing the extension of the continuous theorem of Ge and Ren and constructing suitable Banach spaces and operators, we investigate the existence of solutions for p-Laplacian boundary value problems at resonance. An example is given to illustrate our results.

MSC:34B15.

1 Introduction

In this paper, we will study the boundary value problem

$$\{\begin{array}{c}{({\phi }_p(u^{″}))}^{\prime }(t)=f(t,u(t),u^{\prime }(t),u^{″}(t)),\hfill \\ u(0)=u^{″}(0)=0,u^{\prime }(1)={\int }_0^{1}k(t)u^{\prime }(t)dt,\hfill \end{array}$$

(1.1)

and

$$\{\begin{array}{c}{({\phi }_p(u^{″}))}^{\prime }(t)=f(t,u(t),u^{\prime }(t),u^{″}(t)),\hfill \\ u^{″}(0)=0,u^{\prime }(0)={\int }_0^{1}g(t)u^{\prime }(t)dt,u^{\prime }(1)={\int }_0^{1}h(t)u^{\prime }(t)dt,\hfill \end{array}$$

(1.2)

where ${\phi }_p(s)={|s|}^{p-2}s$, $p>1$, ${\int }_0^{1}k(t)dt=1$, ${\int }_0^{1}g(t)dt=1$, ${\int }_0^{1}h(t)dt=1$.

A boundary value problem is said to be a resonance one if the corresponding homogeneous boundary value problem has a non-trivial solution. Mawhin’s continuous theorem [1] is an effective tool to solve this kind of problems when the differential operator is linear, see [2–10] and references cited therein. But it does not work for nonlinear cases such as boundary value problems with a p-Laplacian, which attracted the attention of mathematicians in recent years [11–15]. Ge and Ren extended Mawhin’s continuous theorem [15] and many authors used their results to solve boundary value problems with a p-Laplacian, see [16, 17]. In this new theorem, two projectors P and Q must be constructed. But it is difficult to give the projector Q in many boundary value problems with a p-Laplacian. In this paper, we generalize the extension of the continuous theorem and show that the p-Laplacian problem is solvable when Q is not a projector. And we will use this new theorem to discuss problems (1.1) and (1.2), respectively.

In this paper, we will always suppose that

(H₁) $k(t),g(t),h(t)\in L^1[0,1]$ are nonnegative and ${\parallel k\parallel }_1={\parallel g\parallel }_1={\parallel h\parallel }_1=1$, where ${\parallel k\parallel }_1:={\int }_0^1|k(t)|dt$.

(H₂) $f(t,u,v,w)$ is continuous in $[0,1]\times {\mathbb{R}}^3$.

2 Preliminaries

Definition 2.1 [15]

Let X and Y be two Banach spaces with norms ${\parallel \cdot \parallel }_X$, ${\parallel \cdot \parallel }_Y$, respectively. A continuous operator $M:X\cap domM\to Y$ is said to be quasi-linear if

• (i) $ImM:=M(X\cap domM)$ is a closed subset of Y,

• (ii) $KerM:=\{x\in X\cap domM:Mx=0\}$ is linearly homeomorphic to ${\mathbb{R}}^n$, $n<\mathrm{\infty }$,

where domM denote the domain of the operator M.

Let $X_1=KerM$ and $X_2$ be the complement space of $X_1$ in X, then $X=X_1\oplus X_2$. Let $P:X\to X_1$ be a projector and $\mathrm{\Omega }\subset X$ an open and bounded set with the origin $\theta \in \mathrm{\Omega }$.

Definition 2.2 Suppose $N_{\lambda }:\overline{\mathrm{\Omega }}\to Y$, $\lambda \in [0,1]$ is a continuous and bounded operator. Denote $N_1$ by N. Let ${\mathrm{\Sigma }}_{\lambda }=\{x\in \overline{\mathrm{\Omega }}\cap domM:Mx=N_{\lambda }x\}$. $N_{\lambda }$ is said to be M-quasi-compact in $\overline{\mathrm{\Omega }}$ if there exists a vector subspace $Y_1$ of Y satisfying $dim{Y}_1=dim{X}_1$ and two operators Q, R with $Q:Y\to Y_1$, $QY=Y_1$, being continuous, bounded, and satisfying $Q(I-Q)=0$, $R:\overline{\mathrm{\Omega }}\times [0,1]\to X_2\cap domM$ continuous and compact such that for $\lambda \in [0,1]$,

• (a) $(I-Q)N_{\lambda }(\overline{\mathrm{\Omega }})\subset ImM\subset (I-Q)Y$,

• (b) $Q{N}_{\lambda }x=\theta$, $\lambda \in (0,1)\iff QNx=\theta$,

• (c) $R(\cdot ,0)$ is the zero operator and $R(\cdot ,\lambda ){|}_{{\mathrm{\Sigma }}_{\lambda }}=(I-P){|}_{{\mathrm{\Sigma }}_{\lambda }}$,

• (d) $M[P+R(\cdot ,\lambda )]=(I-Q)N_{\lambda }$.

Theorem 2.1 Let X and Y be two Banach spaces with the norms ${\parallel \cdot \parallel }_X$, ${\parallel \cdot \parallel }_Y$, respectively, and let $\mathrm{\Omega }\subset X$ be an open and bounded nonempty set. Suppose

$$M:X\cap domM\to Y$$

is a quasi-linear operator and that $N_{\lambda }:\overline{\mathrm{\Omega }}\to Y$, $\lambda \in [0,1]$ is M-quasi-compact. In addition, if the following conditions hold:

(C₁) $Mx\ne N_{\lambda }x$, $\mathrm{\forall }x\in \partial \mathrm{\Omega }\cap domM$, $\lambda \in (0,1)$,

(C₂) $deg\{JQN,\mathrm{\Omega }\cap KerM,0\}\ne 0$,

then the abstract equation $Mx=Nx$ has at least one solution in $domM\cap \overline{\mathrm{\Omega }}$, where $N=N_1$, $J:ImQ\to KerM$ is a homeomorphism with $J(\theta )=\theta$.

Proof The proof is similar to the one of Lemma 2.1 and Theorem 2.1 in [15]. □

We can easily get the following inequalities.

Lemma 2.1 For any $u,v\ge 0$, we have

• (1) ${\phi }_p(u+v)\le {\phi }_p(u)+{\phi }_p(v)$, $1<p\le 2$.

• (2) ${\phi }_p(u+v)\le 2^{p-2}({\phi }_p(u)+{\phi }_p(v))$, $p\ge 2$.

In the following, we will always suppose that q satisfies $1/p+1/q=1$.

3 The existence of a solution for problem (1.1)

Let $X=C^2[0,1]$ with norm $\parallel u\parallel =max\{{\parallel u\parallel }_{\mathrm{\infty }},{\parallel u^{\prime }\parallel }_{\mathrm{\infty }},{\parallel u^{″}\parallel }_{\mathrm{\infty }}\}$, $Y=C[0,1]\times C[0,1]$ with norm $\parallel (y_1,y_2)\parallel =max\{{\parallel y_1\parallel }_{\mathrm{\infty }},{\parallel y_2\parallel }_{\mathrm{\infty }}\}$, where ${\parallel y\parallel }_{\mathrm{\infty }}={max}_{t\in [0,1]}|y(t)|$. We know that $(X,\parallel \cdot \parallel )$ and $(Y,\parallel \cdot \parallel )$ are Banach spaces.

Define operators $M:X\cap domM\to Y$, $N_{\lambda }:X\to Y$ as follows:

$$Mu=\left[\begin{array}{c}{({\phi }_p(u^{″}))}^{\prime }(t)\\ T{({\phi }_p(u^{″}))}^{\prime }(t)\end{array}\right],N_{\lambda }u=\left[\begin{array}{c}\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\\ 0\end{array}\right],$$

where $Ty=c$, $y\in C[0,1]$, c satisfying

$$\begin{array}{r}{\int }_0^{1}k(t){\int }_t^1{\phi }_q({\int }_0^{s}y(r)-cdr)dsdt=0,\\ domM=\{u\in X\mid {\phi }_p\left(u^{″}\right)\in C^1[0,1],u(0)=u^{″}(0)=0\}.\end{array}$$

(3.1)

Lemma 3.1 For $y\in C[0,1]$, there is only one constant $c\in \mathbb{R}$ such that $Ty=c$ with $|c|\le {\parallel y\parallel }_{\mathrm{\infty }}$ and that $T:C[0,1]\to \mathbb{R}$ is continuous.

Proof For $y\in C[0,1]$, let

$$F(c)={\int }_0^{1}k(t){\int }_t^1{\phi }_q\left({\int }_0^s(y(r)-c)dr\right)dsdt.$$

Obviously, $F(c)$ is continuous and strictly decreasing in ℝ. Take $a={min}_{t\in [0,1]}y(t)$, $b={max}_{t\in [0,1]}y(t)$. It is easy to see that $F(a)\ge 0$, $F(b)\le 0$. Thus, there exists a unique constant $c\in [a,b]$ such that $F(c)=0$, i.e. there is only one constant $c\in \mathbb{R}$ such that $Ty=c$ with $|c|\le {\parallel y\parallel }_{\mathrm{\infty }}$.

For $y_1,y_2\in C[0,1]$, assume $T{y}_1=c_1$, $T{y}_2=c_2$. By $k(t)\ge 0$, ${\int }_0^{1}k(t)dt=1$ and ${\phi }_q$ being strictly increasing, we obtain, if $c_2-c_1>{max}_{t\in [0,1]}(y_2(t)-y_1(t))$, then

$$\begin{array}{rcl}0& =& {\int }_0^{1}k(t){\int }_t^1{\phi }_q\left({\int }_0^s(y_2(r)-c_2)dr\right)dsdt\\ =& {\int }_0^{1}k(t){\int }_t^1{\phi }_q\left({\int }_0^s[(y_1(r)-c_1)+(y_2(r)-y_1(r)-(c_2-c_1))dr]\right)dsdt\\ <& {\int }_0^{1}k(t){\int }_t^1{\phi }_q\left({\int }_0^s(y_1(r)-c_1)dr\right)dsdt=0.\end{array}$$

This is a contradiction. On the other hand, if $c_2-c_1<{min}_{t\in [0,1]}(y_2(t)-y_1(t))$, then

$$\begin{array}{rcl}0& =& {\int }_0^{1}k(t){\int }_t^1{\phi }_q\left({\int }_0^s(y_2(r)-c_2)dr\right)dsdt\\ =& {\int }_0^{1}k(t){\int }_t^1{\phi }_q\left({\int }_0^s[(y_1(r)-c_1)+(y_2(r)-y_1(r)-(c_2-c_1))dr]\right)dsdt\\ >& {\int }_0^{1}k(t){\int }_t^1{\phi }_q\left({\int }_0^s(y_1(r)-c_1)dr\right)dsdt=0.\end{array}$$

This is a contradiction, too. So, we have ${min}_{t\in [0,1]}(y_2(t)-y_1(t))\le c_2-c_1\le {max}_{t\in [0,1]}(y_2(t)-y_1(t))$, i.e. $|c_2-c_1|\le {\parallel y_2-y_1\parallel }_{\mathrm{\infty }}$. So, $T:C[0,1]\to \mathbb{R}$ is continuous. The proof is completed. □

It is clear that $u\in domM$ is a solution if and only if it satisfies $Mu=Nu$, where $N=N_1$. For convenience, let ${(a,b)}^L:=\left[\begin{array}{c}a\\ b\end{array}\right]$.

Lemma 3.2 M is a quasi-linear operator.

Proof It is easy to see that $KerM=\{bt\mid b\in \mathbb{R}\}:=X_1$.

For $u\in X\cap domM$, if $Mu={(y,c)}^L$, then c satisfies (3.1). On the other hand, if $y\in C[0,1]$, $Ty=c$, take

$$u(t)={\int }_0^t(t-s){\phi }_q({\int }_0^{s}y(r)dr)ds.$$

By a simple calculation, we get $u\in X\cap domM$ and $Mu={(y,c)}^L$. Thus

$$ImM=\{{(y,c)}^L\mid y\in C[0,1],c\text{ satisfies (3.1)}\}.$$

By the continuity of T, we find that $ImM\subset Y$ is closed. So, M is quasi-linear. The proof is completed. □

Lemma 3.3 $T(c)=c$, $T(y+c)=T(y)+c$, $T(cy)=cT(y)$, $c\in \mathbb{R}$, $y\in C[0,1]$.

Proof The proof is simple. Therefore, we omit it. □

Take a projector $P:X\to X_1$ and an operator $Q:Y\to Y_1$ as follows:

$$(Pu)(t)=u^{\prime }(0)t,Q{(y,y_1)}^L={(0,T{y}_1-Ty)}^L,$$

where $Y_1=\{{(0,c)}^L\mid c\in \mathbb{R}\}$. Obviously, $QY=Y_1$, and $dim{Y}_1=dim{X}_1$.

By the continuity and boundedness of T, we can easily see that Q is continuous and bounded in Y. It follows from Lemma 3.3 that $Q(I-Q){(y,y_1)}^L={(0,0)}^L$, $y,y_1\in C[0,1]$.

Define an operator $R:X\times [0,1]\to X_2$ as

$$R(u,\lambda )(t)={\int }_0^t(t-s){\phi }_q\left({\int }_0^s\lambda f(r,u(r),u^{\prime }(r),u^{″}(r))dr\right)ds,$$

where $KerM\oplus X_2=X$. By (H₂) and the Arzela-Asscoli theorem, we can easily see that $R:\overline{\mathrm{\Omega }}\times [0,1]\to X_2\cap domM$ is continuous and compact, where $\mathrm{\Omega }\subset X$ is an open bounded set.

Lemma 3.4 Assume that $\mathrm{\Omega }\subset X$ is an open bounded set. Then $N_{\lambda }$ is M-quasi-compact in $\overline{\mathrm{\Omega }}$.

Proof It is clear that $ImP=KerM$, $Q{N}_{\lambda }x=\theta$, $\lambda \in (0,1)\iff QNx=\theta$ and $R(\cdot ,0)=0$. For $u\in \overline{\mathrm{\Omega }}$,

$$\begin{array}{rcl}(I-Q)N_{\lambda }u& =& \left[\begin{array}{c}\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ -T[\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))]\end{array}\right]\\ =& \left[\begin{array}{c}\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\\ T[\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))]\end{array}\right]\in ImM.\end{array}$$

Since $ImM\subset KerQ$ and $y=Qy+(I-Q)y$, we obtain $ImM\subset (I-Q)Y$. Thus, $(I-Q)N_{\lambda }(\overline{\mathrm{\Omega }})\subset ImM\subset (I-Q)Y$.

For $u\in {\mathrm{\Sigma }}_{\lambda }=\{u\in \overline{\mathrm{\Omega }}\cap domM:Mu=N_{\lambda }u\}$, we get

$$\begin{array}{rcl}R(u,\lambda )& =& {\int }_0^t(t-s){\phi }_q\left({\int }_0^s\lambda f(r,u(r),u^{\prime }(r),u^{″}(r))dr\right)ds\\ =& {\int }_0^t(t-s){\phi }_q\left({\int }_0^s{\left({\phi }_p\left(u^{″}\right)\right)}^{\prime }\right)ds\\ =& u(t)-u^{\prime }(0)t=(I-P)u,\end{array}$$

i.e. Definition 2.2(c) holds. For $u\in \overline{\mathrm{\Omega }}$, we have

$$M[Pu+R(u,\lambda )]=\left[\begin{array}{c}\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\\ T[\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))]\end{array}\right]=(I-Q)N_{\lambda }u.$$

So, Definition 2.2(d) holds. Therefore, $N_{\lambda }$ is M-quasi-compact in $\overline{\mathrm{\Omega }}$. The proof is completed. □

Theorem 3.1 Assume that the following conditions hold.

(H₃) There exists a nonnegative constant K such that one of (1) and (2) holds:

• (1) $Bf(t,A,B,C)>0$, $t\in [0,1]$, $|B|>K$, $A,C\in \mathbb{R}$,

• (2) $Bf(t,A,B,C)<0$, $t\in [0,1]$, $|B|>K$, $A,C\in \mathbb{R}$.

(H₄) There exist nonnegative functions $a(t),b(t),c(t),e(t)\in L^1[0,1]$ such that

$$|f(t,x,y,z)|\le a(t){\phi }_p(|x|)+b(t){\phi }_p(|y|)+c(t){\phi }_p(|z|)+e(t),t\in [0,1],x,y,z\in \mathbb{R},$$

where ${\phi }_q({\parallel a\parallel }_1+{\parallel b\parallel }_1+{\parallel c\parallel }_1)<2^{2-q}$, if $1<p\le 2$; ${\phi }_q(2^{p-2}{\parallel a\parallel }_1+2^{p-2}{\parallel b\parallel }_1+{\parallel c\parallel }_1)<1$, if $p\ge 2$.

Then boundary value problem (1.1) has at least one solution.

In order to prove Theorem 3.1, we show two lemmas.

Lemma 3.5 Suppose (H₃) and (H₄) hold. Then the set

$${\mathrm{\Omega }}_1=\{u\in domM\mid Mu=N_{\lambda }u,\lambda \in (0,1)\}$$

is bounded in X.

Proof For $u\in {\mathrm{\Omega }}_1$, we have $Q{N}_{\lambda }u=0$, i.e. $Tf(t,u(t),u^{\prime }(t),u^{″}(t))=0$. By (H₃), there exists a constant $t_0\in [0,1]$ such that $|u^{\prime }(t_0)|\le K$. Since $u(t)={\int }_0^t{u}^{\prime }(s)ds$, $u^{\prime }(t)=u^{\prime }(t_0)+{\int }_{t_0}^t{u}^{″}(s)ds$, we have

$$|u(t)|\le {\parallel u^{\prime }\parallel }_{\mathrm{\infty }},|u^{\prime }(t)|\le K+{\parallel u^{″}\parallel }_{\mathrm{\infty }},t\in [0,1].$$

(3.2)

It follows from $Mu=N_{\lambda }u$, (H₄), and (3.2) that

$$\begin{array}{rcl}|u^{″}(t)|& =& \left|{\phi }_q\left({\int }_0^t\lambda f(s,u(s),u^{\prime }(s),u^{″}(s))ds\right)\right|\\ \le & {\phi }_q({\int }_0^{1}a(t){\phi }_p(|u|)+b(t){\phi }_p\left(\left|u^{\prime }\right|\right)+c(t){\phi }_p\left(\left|u^{″}\right|\right)+e(t)dt)\\ \le & {\phi }_q[({\parallel a\parallel }_1+{\parallel b\parallel }_1){\phi }_p(K+{\parallel u^{″}\parallel }_{\mathrm{\infty }})+{\parallel c\parallel }_1{\phi }_p\left({\parallel u^{″}\parallel }_{\mathrm{\infty }}\right)+{\parallel e\parallel }_1].\end{array}$$

If $1<p\le 2$, by Lemma 2.1, we get

$$|u^{″}(t)|\le {\phi }_q(B_1+A_1{\phi }_p\left({\parallel u^{″}\parallel }_{\mathrm{\infty }}\right))\le 2^{q-2}[{\phi }_q(B_1)+{\phi }_q(A_1){\parallel u^{″}\parallel }_{\mathrm{\infty }}],$$

thus

$${\parallel u^{″}\parallel }_{\mathrm{\infty }}\le \frac{2^{q-2}{\phi }_q(B_1)}{1-2^{q-2}{\phi }_q(A_1)},$$

where $B_1=({\parallel a\parallel }_1+{\parallel b\parallel }_1){\phi }_p(K)+{\parallel e\parallel }_1$, $A_1={\parallel a\parallel }_1+{\parallel b\parallel }_1+{\parallel c\parallel }_1$.

If $p>2$, by Lemma 2.1, we get

$$|u^{″}(t)|\le {\phi }_q(B_2+A_2{\phi }_p\left({\parallel u^{″}\parallel }_{\mathrm{\infty }}\right))\le [{\phi }_q(B_2)+{\phi }_q(A_2){\parallel u^{″}\parallel }_{\mathrm{\infty }}],$$

thus

$${\parallel u^{″}\parallel }_{\mathrm{\infty }}\le \frac{{\phi }_q(B_2)}{1-{\phi }_q(A_2)},$$

where $B_2=2^{p-2}({\parallel a\parallel }_1+{\parallel b\parallel }_1){\phi }_p(K)+{\parallel e\parallel }_1$, $A_2=2^{p-2}({\parallel a\parallel }_1+{\parallel b\parallel }_1)+{\parallel c\parallel }_1$.

These, together with (3.2), mean that ${\mathrm{\Omega }}_1$ is bounded in X. □

Lemma 3.6 Assume (H₃) holds. Then

$${\mathrm{\Omega }}_2=\{u\in KerM\mid QNu=0\}$$

is bounded in X, where $N=N_1$.

Proof For $u\in {\mathrm{\Omega }}_2$, we have $u=bt$ and $Tf(t,bt,b,0)=0$. By (H₃), we get $|b|\le K$. So, ${\mathrm{\Omega }}_2$ is bounded. The proof is completed. □

Proof of Theorem 3.1 Let $\mathrm{\Omega }=\{u\in X\mid \parallel u\parallel <r\}$, where r is large enough such that $K<r<+\mathrm{\infty }$ and $\mathrm{\Omega }\supset \overline{{\mathrm{\Omega }}_1}$.

By Lemmas 3.5 and 3.6, we know $Mu\ne N_{\lambda }u$, $u\in domM\cap \partial \mathrm{\Omega }$ and $QNu\ne 0$, $u\in KerM\cap \partial \mathrm{\Omega }$.

Let $H(u,\delta )=\rho \delta u+(1-\delta )JQNu$, $\delta \in [0,1]$, $u\in KerM\cap \overline{\mathrm{\Omega }}$, where $J:ImQ\to KerM$ is a homeomorphism with $J{(0,b)}^L=bt$, $\rho =\{\begin{array}{ll}-1,& {\text{if (H}}_3\text{)(1) holds},\\ 1,& {\text{if (H}}_3\text{)(2) holds}.\end{array}$

Define a function $Sgn(x)=\{\begin{array}{ll}1,& \text{if }x>0,\\ -1,& \text{if }x<0.\end{array}$

For $u\in KerM\cap \partial \mathrm{\Omega }$, we have $u=bt\ne 0$. Thus

$$H(u,\delta )=\rho \delta bt+(1-\delta )(-Tf(t,bt,b,0))t.$$

If $\delta =1$, $H(u,1)=\rho bt\ne 0$. If $\delta =0$, by $QNu\ne 0$, we get $H(u,0)=JQN(bt)\ne 0$. For $0<\delta <1$, we now prove that $H(u,\delta )\ne 0$. Otherwise, if $H(u,\delta )=0$, then

$$Tf(t,bt,b,0)=\frac{\rho \delta }{1-\delta }b.$$

(3.3)

Since $\parallel u\parallel =r>K$, we have $|b|>K$. Thus, $T[bf(t,bt,b,0)]=bTf(t,bt,b,0)=\frac{\rho \delta }{1-\delta }{b}^2$. So, we have $Sgn(bf(t,bt,b,0))=Sgn\{T[bf(t,bt,b,0)]\}=Sgn(\frac{\rho \delta }{1-\delta }{b}^2)=Sgn(\rho )$. A contradiction with the definition of ρ. So, $H(u,\delta )\ne 0$, $u\in KerM\cap \partial \mathrm{\Omega }$, $\delta \in [0,1]$.

By the homotopy of degree, we get

$$\begin{array}{rcl}deg(JQN,\mathrm{\Omega }\cap KerM,0)& =& deg(H(\cdot ,0),\mathrm{\Omega }\cap KerM,0)\\ =& deg(H(\cdot ,1),\mathrm{\Omega }\cap KerM,0)=deg(\rho I,\mathrm{\Omega }\cap KerM,0)\ne 0.\end{array}$$

By Theorem 2.1, we can see that $Mu=Nu$ has at least one solution in $\overline{\mathrm{\Omega }}$. The proof is completed. □

Example Let us consider the following boundary value problem at resonance:

$$\{\begin{array}{c}{({\phi }_p(u^{″}))}^{\prime }(t)=\frac{1}{8}tsin{x}^3+\frac{1}{16}{y}^3+t^{3}sin{z}^3+cost,\hfill \\ u(0)=u^{″}(0)=0,u^{\prime }(1)=2{\int }_0^{1}t{u}^{\prime }(t)dt,\hfill \end{array}$$

(3.4)

where $p=4$.

Corresponding to problem (1.1), we have $q=\frac{4}{3}$, $a(t)=\frac{1}{8}t$, $b(t)=\frac{1}{16}$, $c(t)=t^3$, $e(t)=cost$, $k(t)=2t$.

Take $K=4$. By a simple calculation, we find that the conditions (H₁)-(H₄) hold. By Theorem 3.1, we obtain the result that problem (3.4) has at least one solution.

4 The existence of a solution for problem (1.2)

Let $X=C^2[0,1]$ with norm $\parallel u\parallel =max\{{\parallel u\parallel }_{\mathrm{\infty }},{\parallel u^{\prime }\parallel }_{\mathrm{\infty }},{\parallel u^{″}\parallel }_{\mathrm{\infty }}\}$, $Y=C[0,1]\times C[0,1]\times C[0,1]$ with norm $\parallel (y_1,y_2,y_3)\parallel =max\{{\parallel y_1\parallel }_{\mathrm{\infty }},{\parallel y_2\parallel }_{\mathrm{\infty }},{\parallel y_3\parallel }_{\mathrm{\infty }}\}$, where ${\parallel y\parallel }_{\mathrm{\infty }}={max}_{t\in [0,1]}|y(t)|$. We know that $(X,\parallel \cdot \parallel )$ and $(Y,\parallel \cdot \parallel )$ are Banach spaces.

Define operators $M:X\cap domM\to Y$, $N_{\lambda }:X\to Y$ as follows:

$$Mu=\left[\begin{array}{c}{({\phi }_p(u^{″}))}^{\prime }(t)\\ T_1{({\phi }_p(u^{″}))}^{\prime }(t)\\ T_2{({\phi }_p(u^{″}))}^{\prime }(t)\end{array}\right],N_{\lambda }u=\left[\begin{array}{c}\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\\ 0\\ 0\end{array}\right],$$

where $T_{1}y=c_1$, $T_{2}y=c_2$, $y\in C[0,1]$, $c_1$, $c_2$ satisfy

$$\begin{array}{r}{\int }_0^{1}g(t){\int }_0^t{\phi }_q({\int }_0^{s}y(r)-c_{1}dr)dsdt=0,\\ {\int }_0^{1}h(t){\int }_t^1{\phi }_q({\int }_0^{s}y(r)-c_{2}dr)dsdt=0,\\ domM=\{u\in X\mid {\phi }_p\left(u^{″}\right)\in C^1[0,1],u^{″}(0)=0\}.\end{array}$$

(4.1)

Lemma 4.1 For $y\in C[0,1]$, there is only one constant $c_i\in \mathbb{R}$ such that $T_{i}y=c_i$ with $|c_i|\le {\parallel y\parallel }_{\mathrm{\infty }}$. And $T_i:C[0,1]\to \mathbb{R}$ are continuous, $i=1,2$.

The proof is similar to Lemma 3.1.

It is clear that $u\in domM$ is a solution if and only if it satisfies $Mu=Nu$, where $N=N_1$. For convenience, let ${(a,b,c)}^T:=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$.

Lemma 4.2 M is a quasi-linear operator.

Proof It is easy to get $KerM=\{a+bt\mid a,b\in \mathbb{R}\}:=X_1$.

For $u\in X\cap domM$, if $Mu={(y,c_1,c_2)}^T$, then $c_1$, $c_2$ satisfy (4.1). On the other hand, if $y\in C[0,1]$, $T_{1}y=c_1$, $T_{2}y=c_2$, take

$$u(t)={\int }_0^t(t-s){\phi }_q({\int }_0^{s}y(r)dr)ds.$$

By simple calculation, we get $u\in X\cap domM$ and $Mu={(y,c_1,c_2)}^T$. Thus

$$ImM=\{{(y,c_1,c_2)}^T\mid y\in C[0,1],c_1,c_2\text{ satisfy (4.1)}\}.$$

By the continuity of $T_i$, $i=1,2$, we see that $ImM\subset Y$ is closed. So, M is quasi-linear. The proof is completed. □

Take a projector $P:X\to X_1$ and an operator $Q:Y\to Y_1$ as follows:

$$(Pu)(t)=u(0)+u^{\prime }(0)t,Q{(y,y_1,y_2)}^T={(0,T_1{y}_1-T_{1}y,T_2{y}_2-T_{2}y)}^T,$$

where $Y_1=\{{(0,c_1,c_2)}^T\mid c_i\in \mathbb{R},i=1,2\}$. Obviously, $QY=Y_1$, and $dim{Y}_1=dim{X}_1$.

By the continuity and boundedness of $T_i$, $i=1,2$, we can easily see that Q is continuous and bounded in Y. It follows from Lemma 3.3 that $Q(I-Q){(y,y_1,y_2)}^T={(0,0,0)}^T$, $y,y_1,y_2\in C[0,1]$.

Define an operator $R:X\times [0,1]\to X_2$ as

$$R(u,\lambda )(t)={\int }_0^t(t-s){\phi }_q\left({\int }_0^s\lambda f(r,u(r),u^{\prime }(r),u^{″}(r))dr\right)ds,$$

where $KerM\oplus X_2=X$. By (H₂) and the Arzela-Asscoli theorem, we can easily see that $R:\overline{\mathrm{\Omega }}\times [0,1]\to X_2\cap domM$ is continuous and compact, where $\mathrm{\Omega }\subset X$ is an open bounded set.

Lemma 4.3 Assume that $\mathrm{\Omega }\subset X$ is an open bounded set. Then $N_{\lambda }$ is M-quasi-compact in $\overline{\mathrm{\Omega }}$.

Proof It is clear that $ImP=KerM$, $Q{N}_{\lambda }x=\theta$, $\lambda \in (0,1)\iff QNx=\theta$ and $R(\cdot ,0)=0$. For $u\in \overline{\mathrm{\Omega }}$,

$$\begin{array}{rcl}(I-Q)N_{\lambda }u& =& \left[\begin{array}{c}\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\\ 0\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ -T_1\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\\ -T_2\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\end{array}\right]\\ =& \left[\begin{array}{c}\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\\ T_1\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\\ T_2\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\end{array}\right]\in ImM.\end{array}$$

Since $ImM\subset KerQ$ and $y=Qy+(I-Q)y$, we obtain $ImM\subset (I-Q)Y$. Thus, $(I-Q)N_{\lambda }(\overline{\mathrm{\Omega }})\subset ImM\subset (I-Q)Y$.

For $u\in {\mathrm{\Sigma }}_{\lambda }=\{u\in \overline{\mathrm{\Omega }}\cap domM:Mu=N_{\lambda }u\}$, we get

$$\begin{array}{rcl}R(u,\lambda )& =& {\int }_0^t(t-s){\phi }_q\left({\int }_0^s\lambda f(r,u(r),u^{\prime }(r),u^{″}(r))dr\right)ds\\ =& {\int }_0^t(t-s){\phi }_q\left({\int }_0^s{\left({\phi }_p\left(u^{″}\right)\right)}^{\prime }\right)ds\\ =& u(t)-u(0)-u^{\prime }(0)t=(I-P)u,\end{array}$$

i.e. Definition 2.2(c) holds. For $u\in \overline{\mathrm{\Omega }}$, we have

$$M[Pu+R(u,\lambda )]=\left[\begin{array}{c}\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\\ T_1\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\\ T_2\lambda f(t,u(t),u^{\prime }(t),u^{″}(t))\end{array}\right]=(I-Q)N_{\lambda }u.$$

Thus, Definition 2.2(d) holds. Therefore, $N_{\lambda }$ is M-quasi-compact in $\overline{\mathrm{\Omega }}$. The proof is completed. □

Theorem 4.1 Assume that the following conditions hold:

(H₅) There exists a nonnegative constant L such that if $|u(t)|>L$, $t\in [0,1]$ then either

$$T_{1}f(t,u(t),u^{\prime }(t),u^{″}(t))\ne 0$$

or

$$T_{2}f(t,u(t),u^{\prime }(t),u^{″}(t))\ne 0.$$

(H₆) There exist nonnegative constants $K_1$, $K_2$ such that one of (1) and (2) holds:

• (1)
  $$Bf(t,A,B,C)>0,t\in [0,1],|B|>K_1,A,C\in \mathbb{R},$$

and

$$Af(t,A,B,C)>0,t\in [0,1],|B|\le K_1,|A|>K_2,C\in \mathbb{R}.$$

• (2)
  $$Bf(t,A,B,C)<0,t\in [0,1],|B|>K_1,A,C\in \mathbb{R},$$

and

$$Af(t,A,B,C)<0,t\in [0,1],|A|>K_2,|B|\le K_1,C\in \mathbb{R}.$$

(H₇) There exist nonnegative functions $a(t),b(t),c(t),e(t)\in L^1[0,1]$ such that

$$|f(t,x,y,z)|\le a(t){\phi }_p(|x|)+b(t){\phi }_p(|y|)+c(t){\phi }_p(|z|)+e(t),t\in [0,1],x,y,z\in \mathbb{R},$$

where ${\phi }_q({\parallel a\parallel }_1+{\parallel b\parallel }_1+{\parallel c\parallel }_1)<2^{2-q}$, if $1<p\le 2$; ${\phi }_q(2^{p-2}{\parallel a\parallel }_1+2^{p-2}{\parallel b\parallel }_1+{\parallel c\parallel }_1)<1$, if $p\ge 2$.

Then boundary value problem (1.2) has at least one solution.

In order to prove Theorem 4.1, we show two lemmas.

Lemma 4.4 Suppose (H₅)-(H₇) hold. Then the set

$${\mathrm{\Omega }}_1=\{u\in domM\mid Mu=N_{\lambda }u,\lambda \in (0,1)\}$$

is bounded in X.

Proof For $u\in {\mathrm{\Omega }}_1$, we have $Q{N}_{\lambda }u=0$, i.e. $T_{i}f(t,u(t),u^{\prime }(t),u^{″}(t))=0$, $i=1,2$. By (H₅) and (H₆), there exist constants $t_0,t_1\in [0,1]$ such that $|u(t_0)|\le L$, $|u^{\prime }(t_1)|\le K_1$. Since $u(t)=u(t_0)+{\int }_{t_0}^t{u}^{\prime }(s)ds$, $u^{\prime }(t)=u^{\prime }(t_1)+{\int }_{t_1}^t{u}^{″}(s)ds$, then

$$|u(t)|\le L+{\parallel u^{\prime }\parallel }_{\mathrm{\infty }},|u^{\prime }(t)|\le K_1+{\parallel u^{″}\parallel }_{\mathrm{\infty }},t\in [0,1].$$

(4.2)

It follows from $Mu=N_{\lambda }u$, (H₇), and (4.2) that

$$\begin{array}{rcl}|u^{″}(t)|& =& \left|{\phi }_q\left({\int }_0^t\lambda f(s,u(s),u^{\prime }{(s)}^{\prime }{u}^{″}(s))ds\right)\right|\\ \le & {\phi }_q({\int }_0^{1}a(t){\phi }_p(|u|)+b(t){\phi }_p\left(\left|u^{\prime }\right|\right)+c(t){\phi }_p\left(\left|u^{″}\right|\right)+e(t)dt)\\ \le & {\phi }_q({\parallel a\parallel }_1{\phi }_p(K_1+L+{\parallel u^{″}\parallel }_{\mathrm{\infty }})+{\parallel b\parallel }_1{\phi }_p(K_1+{\parallel u^{″}\parallel }_{\mathrm{\infty }})\\ +{\parallel c\parallel }_1{\phi }_p\left({\parallel u^{″}\parallel }_{\mathrm{\infty }}\right)+{\parallel e\parallel }_1).\end{array}$$

If $1<p\le 2$, by Lemma 2.1, we get

$$|u^{″}(t)|\le {\phi }_q(B_1+A_1{\phi }_p\left({\parallel u^{″}\parallel }_{\mathrm{\infty }}\right))\le 2^{q-2}[{\phi }_q(B_1)+{\phi }_q(A_1){\parallel u^{″}\parallel }_{\mathrm{\infty }}],$$

thus

$${\parallel u^{″}\parallel }_{\mathrm{\infty }}\le \frac{2^{q-2}{\phi }_q(B_1)}{1-2^{q-2}{\phi }_q(A_1)},$$

where $B_1={\parallel a\parallel }_1{\phi }_p(K_1+L)+{\parallel b\parallel }_1{\phi }_p(K_1)+{\parallel e\parallel }_1$, $A_1={\parallel a\parallel }_1+{\parallel b\parallel }_1+{\parallel c\parallel }_1$.

If $p>2$, by Lemma 2.1, we get

$$|u^{″}(t)|\le {\phi }_q(B_2+A_2{\phi }_p\left({\parallel u^{″}\parallel }_{\mathrm{\infty }}\right))\le [{\phi }_q(B_2)+{\phi }_q(A_2){\parallel u^{″}\parallel }_{\mathrm{\infty }}],$$

thus

$${\parallel u^{″}\parallel }_{\mathrm{\infty }}\le \frac{{\phi }_q(B_2)}{1-{\phi }_q(A_2)},$$

where $B_2=2^{p-2}{\parallel a\parallel }_1{\phi }_p(K_1+L)+2^{p-2}{\parallel b\parallel }_1{\phi }_p(K_1)+{\parallel e\parallel }_1$, $A_2=2^{p-2}{\parallel a\parallel }_1+2^{p-2}{\parallel b\parallel }_1+{\parallel c\parallel }_1$.

These, together with (4.2), mean that ${\mathrm{\Omega }}_1$ is bounded in X. □

Lemma 4.5 Assume (H₆) holds. Then

$${\mathrm{\Omega }}_2=\{u\in KerM\mid QNu=0\}$$

is bounded in X, where $N=N_1$.

Proof For $u\in {\mathrm{\Omega }}_2$, we have $u=a+bt$ and $Q(Nu)=0$. By (H₆), we see that there exists a constant $t_0\in [0,1]$ such that $|u(t_0)|=|a+b{t}_0|\le K_2$, $|u^{\prime }(t)|=|b|\le K_1$. So, ${\mathrm{\Omega }}_2$ is bounded. The proof is completed. □

Proof of Theorem 4.1 Let $\mathrm{\Omega }=\{u\in X\mid \parallel u\parallel <r\}$, where r is large enough such that $K_1+K_2<r<+\mathrm{\infty }$ and $\mathrm{\Omega }\supset \overline{{\mathrm{\Omega }}_1}\cup \overline{{\mathrm{\Omega }}_2}$.

By Lemmas 4.4 and 4.5, we know $Mu\ne N_{\lambda }u$, $u\in domM\cap \partial \mathrm{\Omega }$ and $QNu\ne 0$, $u\in KerM\cap \partial \mathrm{\Omega }$.

Let $H(u,\delta )=\rho \delta u+(1-\delta )JQNu$, $\delta \in [0,1]$, $u\in KerM\cap \overline{\mathrm{\Omega }}$, where $J:ImQ\to KerM$ is a homeomorphism with $J{(0,a,b)}^T=a+bt$, $\rho =\{\begin{array}{ll}-1,& {\text{if (H}}_6\text{)(1) holds},\\ 1,& {\text{if (H}}_6\text{)(2) holds}.\end{array}$

Take the function $Sgn(x)$ is the same as the one in Proof of Theorem 3.1.

For $u\in KerM\cap \partial \mathrm{\Omega }$, we have $u=a+bt\ne 0$. Thus

$$H(u,\delta )=\rho \delta (a+bt)+(1-\delta )(-T_{1}f(t,a+bt,b,0)-T_{2}f(t,a+bt,b,0)t).$$

If $\delta =1$, $H(u,1)=\rho (a+bt)\ne 0$. If $\delta =0$, by $QNu\ne 0$, we get $H(u,0)=JQN(a+bt)\ne 0$. For $0<\delta <1$, we now prove that $H(u,\delta )\ne 0$. Otherwise, if $H(u,\delta )=0$, then

$$T_{1}f(t,a+bt,b,0)=\frac{\rho \delta }{1-\delta }a,T_{2}f(t,a+bt,b,0)=\frac{\rho \delta }{1-\delta }b.$$

(4.3)

Since $\parallel u\parallel =max\{{\parallel a+bt\parallel }_{\mathrm{\infty }},|b|\}=r>K_1+K_2$, we have either $|b|>K_1$ or ${\parallel a+bt\parallel }_{\mathrm{\infty }}>K_1+K_2$. If $|b|>K_1$, then $T_{2}bf(t,a+bt,b,0)=b{T}_{2}f(t,a+bt,b,0)=\frac{\rho \delta }{1-\delta }{b}^2$. So, we have $Sgn(bf(t,a+bt,b,0))=Sgn(T_{2}bf(t,a+bt,b,0))=Sgn(\frac{\rho \delta }{1-\delta }{b}^2)=Sgn(\rho )$. This is a contradiction with the definition of ρ. If $|b|\le K_1$, then ${\parallel a+bt\parallel }_{\mathrm{\infty }}>K_1+K_2$. Thus ${min}_{t\in [0,1]}|a+bt|>K_2$ and $Sgn(a)=Sgn(a+bt)$. By $T_{1}af(t,a+bt,b,0)=a{T}_{1}f(t,a+bt,b,0)=\frac{\rho \delta }{1-\delta }{a}^2$, we get $Sgn(T_1(a+bt)f(t,a+bt,b,0))=Sgn(T_{1}af(t,a+bt,b,0))=Sgn(\frac{\rho \delta }{1-\delta }{a}^2)=Sgn(\rho )$. This is a contradiction with the definition of ρ, too. So, $H(u,\delta )\ne 0$, $u\in KerM\cap \partial \mathrm{\Omega }$, $\delta \in [0,1]$.

By the homotopy of degree, we get

$$\begin{array}{rcl}deg(JQN,\mathrm{\Omega }\cap KerM,0)& =& deg(H(\cdot ,0),\mathrm{\Omega }\cap KerM,0)\\ =& deg(H(\cdot ,1),\mathrm{\Omega }\cap KerM,0)=deg(\rho I,\mathrm{\Omega }\cap KerM,0)\ne 0.\end{array}$$

By Theorem 2.1, we find that (1.2) has at least one solution in $\overline{\mathrm{\Omega }}$. The proof is completed. □