Resonant p-Laplacian problems with functional boundary conditions

Abstract

By constructing a suitable projection scheme and using the extension of Mawhin’s continuation theorem, the existence of solution for functional p-Laplacian boundary value problems at resonance is studied. The paper is a generalization of some current results to a fully nonlinear case.

1 Introduction

A boundary value problem is said to be at resonance if the corresponding homogeneous boundary value problem has a non-trivial solution. Mawhin’s continuation theorem [1] and its extension by Ge and Ren [2] are effective tools in the study of boundary value problems at resonance (see [3–10] and the references therein). In Refs. [6, 9], the authors studied the existence of solutions for functional boundary value problems with a linear differential operator by using Mawhin’s continuation theorem. In [6], we extended the results of [9] to include new resonance scenarios. Since the p-Laplacian operator occurs in many applications such as non-Newtonian mechanics, nonlinear elasticity and glaciology, combustion theory, we would like to further extend the results of [6] to the third-order functional p-Laplacian boundary value problem at resonance

$$ \textstyle\begin{cases} (\varphi_{p}(u''))'(t)=f(t,u(t),u'(t),u''(t)), & t \in(0,1),\\ u''(0)=0, \qquad B_{1}(u)=B_{2}(u)=0, \end{cases} $$

(1.1)

where $f:[0,1]\times\mathbb{R}^{3}\rightarrow\mathbb{R}$ is continuous, $p>1$, $\varphi_{p}(s)=|s|^{p-2}s$, $B_{1},B_{2}:C^{2}[0,1]\rightarrow\mathbb{R}$ are linear bounded functions with $B_{2}(t)B_{1}(1)=B_{2}(1)B_{1}(t)$. Although the paper by Han and Kang [11] explores positive solutions in the non-resonant setting of a dynamic equation on a measure chain with Sturm–Liouville boundary conditions in place of our functional conditions, it is also relevant since it bears some similarity to the boundary value problem considered herein. Finally, one can easily extend the scheme used in this paper to include fractional analogs of [4] and thus to extend the findings of [10].

2 Preliminaries

We introduce the theoretical foundations of the method; for more details, see [2].

Definition 2.1

Let X and Y be two Banach spaces with norms $\Vert \cdot \Vert _{X}$, $\|\cdot \|_{Y}$, respectively. An operator $M: X\cap\operatorname{dom} M\rightarrow Y$ is said to be quasi-linear if

• (i) $\operatorname{Im} M =M(X \cap\operatorname{dom} M)$ is a closed subset of Y;

• (ii) $\operatorname{Ker} M =\{x\in X \cap\operatorname{dom}M: Mx=0\}$ is linearly homeomorphic to $\mathbb{R}^{n}$.

In this paper, an operator $T:X\rightarrow Y$ is said to be bounded if $T(V)\subset Y$ is bounded for any bounded subset $V \subset X$.

Definition 2.2

A linear operator $P: X\rightarrow X$, where X is a vector space, is a projector if $P^{2}x=Px$.

Let $X_{1}=\operatorname{Ker} M$, $P:X\rightarrow X_{1}$ be a projector and $X_{2}$ be the complement space of $X_{1}$ in X with $X=X_{1}\oplus X_{2}$. Let $\Omega\subset X$ be an open and bounded set with the origin $0 \in\Omega$.

Definition 2.3

Suppose that $N_{\lambda}:\overline{\Omega}\rightarrow Y$, $\lambda\in[0,1]$ is a continuous and bounded operator and $N_{1}$ is denoted by N. Let $\Sigma_{\lambda}=\{x\in\overline{\Omega}:Mx=N_{\lambda}x\}$. The operator $N_{\lambda}$ is said to be M-quasi-compact in Ω̅ if there exists a vector subspace $Y_{1}$ of Y satisfying $\operatorname{dim} Y_{1}=\operatorname{dim} X_{1}$ and the operators Q and R such that the following conditions hold:

• (a) $\operatorname{Ker}Q=\operatorname{Im}M$;

• (b) $QN_{\lambda} x=0$, $\lambda\in(0,1)$ ⇔ $QNx=0$;

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

• (d) $M[P+R(\cdot,\lambda)]=(I-Q)N_{\lambda}$, where $Q:Y\rightarrow Y_{1}$ is continuous, bounded with $Q(I-Q)=0$, $QY=Y_{1}$ and $R:\overline{\Omega}\times[0,1]\rightarrow X_{2}$ is continuous and compact with $Pu+R(u,\lambda)\in\operatorname{dom} M$, $u\in\overline{\Omega}$, $\lambda\in[0,1]$.

We use the result of Ge and Ren [2].

Theorem 2.4

Let X and Y be Banach spaces and $\Omega \subset X$ be an open and bounded nonempty set. Suppose that $M:X\cap\operatorname{dom} M\rightarrow Y$ is a quasi-linear operator and $N_{\lambda}:\overline{\Omega}\rightarrow Y$, $\lambda\in[0,1]$, is M-quasi-compact. In addition, if the following conditions hold:

$(C_{1})$ :

$Mx\neq N_{\lambda}x$, $x\in\partial\Omega\cap \operatorname{dom}M$, $\lambda\in(0,1)$;

$(C_{2})$ :

$\operatorname{deg}(JQN, \Omega\cap\operatorname{Ker}M, 0)\neq0$, where $N=N_{1}$, $J:\operatorname{Im}Q\rightarrow\operatorname{Ker}M$ is a homeomorphism with $J(0)=0$ and deg is the Brouwer degree,

then the abstract equation $Mx=Nx$ has at least one solution in $\operatorname{dom}M\cap\overline{\Omega}$.

We make use of well-known inequalities [12] in the context of the p-Laplacian $\varphi_{p}(s)$, $p > 1$. For $u, v \geq0$, we have

$$ \varphi_{p}(u+v) \leq \textstyle\begin{cases} \varphi_{p}(u)+\varphi_{p}(v), & \mbox{if } 1 < p \leq2, \\ 2^{p-2}(\varphi_{p}(u)+\varphi_{p}(v)), & \mbox{if } p > 2. \end{cases} $$

(2.1)

3 Main results

We work in the Banach spaces $X=\{u\in C^{2}[0,1]:u''(0)=0\}$ with the norm $\Vert u \Vert _{X}=\max\{ \Vert u \Vert _{0}, \Vert u' \Vert _{0}, \Vert u'' \Vert _{0}\}$ and $Y=C[0,1]$ with the norm $\Vert y \Vert _{Y}= \Vert y \Vert _{0}$, where $\Vert \cdot \Vert _{0}$ is the max-norm and introduce the following assumptions:

$(A_{0})$ :

The linear functionals $B_{i}: X \to\mathbb{R}$, $i=1,2$, satisfy $B_{1}(t)=\beta$, $B_{1}(1)=\alpha$, $B_{2}(t)=k\beta$, $B_{2}(1)=k\alpha$, where $\alpha,\beta,k \in \mathbb{R}$ with $\alpha^{2}+\beta^{2}\neq0$.

$(A_{1})$ :

$\Vert B_{i}u \Vert \leq k_{i} \Vert u \Vert _{X}$, $k_{i}\in\mathbb{R^{+}}$, $u\in X$, $i=1,2$.

$(A_{2})$ :

The functional $F: Y \to\mathbb{R}$ defined by

$$ F(y) = (B_{2}-kB_{1}) \biggl( \int_{0}^{t}(t-s)\varphi_{q} \biggl( \int _{0}^{s}y(r) \, dr \biggr) \, ds \biggr), $$

(3.1)

where $\frac{1}{p}+\frac{1}{q}=1$ is increasing, that is, if $y_{1},y_{2}\in Y$, $y_{1}(t)\leq y_{2}(t)$, $t\in[0,1]$, $y_{1}\not\equiv y_{2}$, then $F(y_{1}) < F(y_{2})$.

Define operators $M:X\cap\operatorname{dom}M\rightarrow Y$ and $N_{\lambda}:X\rightarrow Y$ by

$$Mu(t)=\bigl(\varphi_{p}\bigl(u''\bigr) \bigr)'(t), $$

where $\operatorname{dom}M=\{u\in X:B_{1}(u)=B_{2}(u)=0,(\varphi _{p}(u''))' \in C[0,1]\}$, and

$$N_{\lambda}u(t)=\lambda f\bigl(t,u(t),u'(t),u''(t) \bigr), \quad\lambda\in[0,1]. $$

It is easy to see recalling (3.1) that

$$\operatorname{Ker}M=\bigl\{ c(\alpha t-\beta):c\in\mathbb{R}\bigr\} \quad\mbox{and} \quad\operatorname{Im}M=\bigl\{ y\in Y: F(y)=0\bigr\} . $$

In fact, if $y\in\operatorname{Im}M$, there exists a function $u\in \operatorname{dom}M$ with $Mu=y$. So,

$$u(t)= \int_{0}^{t}(t-s)\varphi_{q} \biggl( \int_{0}^{s}y(r) \, dr \biggr) \, ds+at+b, \quad a,b \in\mathbb{R}. $$

By $B_{i}(u)=0$, we get

$$\begin{aligned} &B_{1}(u)=B_{1} \biggl( \int_{0}^{t}(t-s)\varphi_{q} \biggl( \int_{0}^{s}y(r) \, dr \biggr) \, ds \biggr)+a \beta+b\alpha=0, \\ &B_{2}(u)=B_{2} \biggl( \int_{0}^{t}(t-s)\varphi_{q} \biggl( \int_{0}^{s}y(r) \, dr \biggr) \, ds \biggr)+a k \beta+b k\alpha=0. \end{aligned}$$

Thus, $F(y) =0$.

Conversely, if $y\in Y$ satisfies $F(y) =0$, we let

$$\begin{aligned} u(t) =& \int_{0}^{t}(t-s)\varphi_{q} \biggl( \int_{0}^{s}y(r) \, dr \biggr) \, ds \\ &{}- \frac{1}{\alpha^{2}+\beta^{2}}B_{1} \biggl( \int_{0}^{t}(t-s)\varphi_{q} \biggl( \int_{0}^{s}y(r) \, dr \biggr) \, ds \biggr) (\beta t+\alpha). \end{aligned}$$

Clearly, $u''(0)=0$, $(\varphi_{p}(u''))'=y$ and $B_{1}(u)=B_{2}(u)=0$. Therefore, $u\in\operatorname{dom}M$ and $Mu=y$, that is, $y\in\operatorname{Im}M$.

Obviously, KerM is linearly homeomorphic to $\mathbb {R}$. Let $y_{n}\in\operatorname{Im}M\subset Y$, $y_{n}\rightarrow y\in Y$. Since

$$\biggl\Vert \int_{0}^{t}(t-s)\varphi_{q} \biggl( \int_{0}^{s}y_{n}(r) \,dr \biggr)\,ds - \int_{0}^{t}(t-s)\varphi_{q} \biggl( \int_{0}^{s}y(r) \,dr \biggr)\,ds \biggr\Vert _{X} \to0, \quad\mbox{as } n \to\infty, $$

then, by ($A_{1}$), $|F(y_{n})-F(y)| \to0$ as $n \to\infty$. This, together with $y_{n}\in\operatorname{Im}M$, shows that $y\in \operatorname{Im}M$. Hence, ImM is a closed subset of Y. Thus, M is quasi-linear. Set $X_{1}=\operatorname{Ker}M$. Define operators $P:X\rightarrow X$ and $Q:Y\rightarrow Y$ by

$$Pu=\frac{\alpha u'(0)-\beta u(0)}{\alpha^{2}+\beta^{2}}(\alpha t-\beta), $$

and $Qy=c$, where c satisfies $F(y-c) =0$.

Clearly, P is a projector and $\operatorname{Ker}Q=\operatorname{Im}M$. Set $Y_{1}=\mathbb{R}$.

Lemma 3.1

The operator $Q:Y\rightarrow Y_{1}$ is continuous, bounded and $Q(I-Q)=0$, $QY=Y_{1}$, $|Qy|\leq \Vert y \Vert _{Y}$.

Proof

For $y\in Y$, by $(A_{1})$ and $(A_{2})$, it follows that the function $F(y -\cdot):\mathbb{R}\rightarrow\mathbb{R}$, defined in terms of (3.1), is continuous and decreasing. Choose $a,b\in\mathbb{R}$ and $y \in Y$ such that $a> \Vert y \Vert _{Y},b<- \Vert y \Vert _{Y}$. By $(A_{2})$, $F(y-a) <0 < F(y-b)$. So, there exists a unique constant c with $|c|\leq \Vert y \Vert _{Y}$ such that $F(y-c)=0$. Thus, Q is well defined and $|Qy|\leq \Vert y \Vert _{Y}$. For $y_{1},y_{2}\in Y$, $Q(y_{1})=c_{1}$, $Q(y_{2})=c_{2}$, if $c_{2}-c_{1}> \Vert y_{2}-y_{1} \Vert _{Y}$, it follows from $(A_{2})$ that

$$0 = F(y_{1}-c_{1}) = F \bigl(y_{2}-c_{2}- \bigl[(y_{2}-y_{1})-(c_{2}-c_{1})\bigr] \bigr) > F(y_{2} -c_{2}) = 0, $$

which is a contradiction. If $c_{2}-c_{1}<- \Vert y_{2}-y_{1} \Vert _{Y}$, then

$$0=F(y_{1}-c_{1}) =F \bigl(y_{2}(r)-c_{2}- \bigl[(y_{2}-y_{1})-(c_{2}-c_{1})\bigr] \bigr) < F(y_{2}-c_{2})=0, $$

which is a contradiction, again. Thus, $|Q(y_{2})-Q(y_{1})|=|c_{2}-c_{1}|\leq \Vert y_{2}-y_{1} \Vert _{Y}$, that is, Q is continuous.

Obviously, $Q(I-Q)=0$ and $QY=Y_{1}$. □

We define $R(u,\lambda):X\times[0,1]\rightarrow X_{2}$ by

$$\begin{aligned} R(u,\lambda) (t) ={}& \int_{0}^{t}(t-s)\varphi_{q} \biggl( \int _{0}^{s}(I-Q)N_{\lambda}u(r)\, dr \biggr)\,ds \\ &{}-\frac{1}{\alpha^{2}+\beta^{2}}B_{1} \biggl( \int_{0}^{t}(t-s)\varphi_{q} \biggl( \int_{0}^{s}(I-Q)N_{\lambda}u(r)\, dr \biggr)\,ds \biggr) (\beta t+\alpha), \end{aligned}$$

where $X_{1}\oplus X_{2}=X$.

Lemma 3.2

The operator $R:\overline{\Omega}\times[0,1]\rightarrow X_{2}$ is continuous and compact with $Pu+R(u,\lambda)\in\operatorname{dom}M$, $u\in\overline{\Omega}$, $\lambda\in[0,1]$, where $\Omega\subset X$ is bounded.

Proof

Since $PR(u,\lambda)=0$, $R(u,\lambda)\in X_{2}$. For $u\in X$, $\lambda\in[0,1]$, it follows from the continuity of $B_{1}$, Q and f that $R(u,\lambda)$ is continuous. Clearly, $(\varphi _{p}((Pu+R(u,\lambda))''))'=(I-Q)N_{\lambda}u\in C[0,1]$, $(Pu+R(u,\lambda))''(0)=0$ and $B_{1}(Pu+R(u,\lambda))=0$. Considering $(I-Q)N_{\lambda}u\in \operatorname{Ker}Q=\operatorname{Im}M$, we get $B_{2}(Pu+R(u,\lambda))=0$. So, $Pu+R(u,\lambda)\in\operatorname{dom}M$. Now, we prove that R is compact.

There exists a constant $C>0$ such that $\Vert N_{\lambda}u \Vert _{Y}\leq C$ in Ω̅ for all $\lambda\in[0,1]$. Note that

$$\bigl\vert \bigl(R(u,\lambda)\bigr)''(t) \bigr\vert = \biggl\vert \varphi_{q} \biggl( \int_{0}^{t}(I-Q)N_{\lambda }u(s)\,ds \biggr) \biggr\vert \leq(2C)^{q-1}, $$

$(R(u,\lambda))''$ is uniformly bounded in Ω̅ together with $R(u,\lambda)$ and $(R(u,\lambda))'$. Also, since $\varphi_{q}(\cdot)$ is uniformly continuous in $[-2C,2C ]$ and, for $0 \leq t_{1} < t_{2} \leq1$,

$$\biggl\vert \int_{0}^{t_{2}}(I-Q)N_{\lambda}u(s)\,ds- \int _{0}^{t_{1}}(I-Q)N_{\lambda}u(s)\,ds \biggr\vert \leq2C(t_{2}-t_{1}), $$

it follows that $\{R(u,\lambda))'': u \in\overline{\Omega}, \lambda\in[0,1]\}$ is equicontinuous. By the mean value theorem, $\{(R(u,\lambda))': u \in\overline{\Omega},\lambda\in[0,1]\}$ and $\{R(u,\lambda): u \in\overline{\Omega}, \lambda\in [0,1]\}$ are also equicontinuous. The compactness of the operator R follows from the Arzela–Ascoli theorem. □

Now, we will show that $N_{\lambda}$ is M-quasi-compact in Ω̅, where $\Omega\subset X$ is an open and bounded set with $0\in\Omega$.

Obviously, $N_{\lambda}$ is continuous, bounded and $\operatorname{dim} X_{1}=\operatorname{dim} Y_{1}$.

Lemma 3.3

The operator $N_{\lambda}$ is M-quasi-compact in Ω̅.

Proof

Obviously, $\operatorname{Ker}Q=\operatorname{Im}M$, $QN_{\lambda}u=0,~\lambda\in (0,1)\Leftrightarrow QNu=0$, $R(\cdot,0)$ is the zero operator and $M(Pu+R(u,\lambda))=(I-Q)N_{\lambda}u$. Considering Lemmas 3.1 and 3.2, we need only to prove that $R(\cdot,\lambda)|_{\sum _{\lambda}}=(I-P)|_{\sum_{\lambda}}$.

To this end, $u\in{\sum_{\lambda}}$ implies $N_{\lambda}u=Mu$, $u''(0)=0$, $B_{i}(u)=0$, $i=1,2$. Thus, $QN_{\lambda}u=0$ and

$$\begin{aligned} R(u,\lambda) =& \int_{0}^{t}(t-s)\varphi_{q} \biggl( \int_{0}^{s}N_{\lambda }u(r)\, dr \biggr)\,ds\\ &{}- \frac{ B_{1} (\int_{0}^{t}(t-s)\varphi_{q}(\int _{0}^{s}N_{\lambda}u(r)\, dr)\,ds )}{\alpha^{2}+\beta^{2}}(\beta t+\alpha ) \\ =& \int_{0}^{t}(t-s)\varphi_{q} \biggl( \int_{0}^{s}\bigl(\varphi _{p} \bigl(u''\bigr)\bigr)'(r)\, dr \biggr)\,ds\\ &{}- \frac{ B_{1} (\int_{0}^{t}(t-s)\varphi _{q}(\int_{0}^{s}(\varphi_{p}(u''))'(r)\, dr)\,ds )}{\alpha^{2}+\beta ^{2}}(\beta t+\alpha) \\ =& \int_{0}^{t}(t-s)u''(s)\,ds- \frac{ B_{1} (\int _{0}^{t}(t-s)u''(s)\,ds )}{\alpha^{2}+\beta^{2}}(\beta t+\alpha) \\ =& u(t)-u'(0)t-u(0)-\frac{-u'(0)\beta-u(0)\alpha}{\alpha^{2}+\beta ^{2}}(\beta t+\alpha) \\ =& u(t)-\frac{\alpha u'(0)-\beta u(0)}{\alpha^{2}+\beta^{2}}(\alpha t-\beta ) \\ =& (I-P)u. \end{aligned}$$

The proof is completed. □

In order to obtain our main results, we need the following hypotheses:

$(H_{1})$ :

There exists a constant $M_{0}>0$ such that if $|u(t)|+|u'(t)|>M_{0}$, then $F(Nu) \neq0$.

$(H_{2})$ :

There exist functions $a,b,c,d \in C[0,1]$ with $\Vert b \Vert _{1}+ \Vert c \Vert _{1}+ \Vert d \Vert _{1}<1$, if $1 < p \leq2$ and $2^{p-2}( \Vert b \Vert _{1}+ \Vert c \Vert _{1})+ \Vert d \Vert _{1}<1$, if $p > 2$, such that

$$\bigl\vert f(t,u,v,w) \bigr\vert \leq a(t)+b(t)\varphi_{p}\bigl( \vert u \vert \bigr)+c(t)\varphi _{p}\bigl( \vert v \vert \bigr)+d(t)\varphi_{p}\bigl( \vert w \vert \bigr), \quad t\in[0,1], u,v,w\in\mathbb{R}, $$

where $\Vert y \Vert _{1}=\int_{0}^{1} \vert y(t) \vert \,dt$.

$(H_{3})$ :

There exists a constant $M_{1}>0$ such that for $|c| >M_{1}$ one of the following inequalities holds:

$$\begin{aligned} &cQN\bigl(c(\alpha t-\beta)\bigr)>0, \end{aligned}$$

(3.2)

$$\begin{aligned} &cQN\bigl(c(\alpha t-\beta)\bigr)< 0. \end{aligned}$$

(3.3)

Lemma 3.4

Assume that $(H_{1})$ and $(H_{2})$ hold. Then the set

$$\Omega_{1}=\bigl\{ u\in\operatorname{dom} M:Mu=N_{\lambda}u, \lambda\in(0,1)\bigr\} $$

is bounded.

Proof

Since $u\in\Omega_{1}$, $QN_{\lambda}u=0$. By $(H_{1})$, there exists $t_{0} \in[0,1]$ such that $|u(t_{0})|\leq M_{0}$, $|u'(t_{0})|\leq M_{0}$. It follows from

$$u'(t)= \int_{t_{0}}^{t}u''(s) \, ds+u'(t_{0})\quad \mbox{and} \quad u(t)= \int _{t_{0}}^{t}u'(s) \, ds+u(t_{0}) $$

that

$$ \bigl\vert u'(t) \bigr\vert \leq M_{0}+ \bigl\Vert u'' \bigr\Vert _{0}, \qquad \bigl\vert u(t) \bigr\vert \leq2M_{0}+ \bigl\Vert u'' \bigr\Vert _{0}. $$

(3.4)

Based on $Mu=N_{\lambda}u$ and $(H_{2})$, we get

$$\begin{aligned} \bigl\vert \varphi_{p}\bigl(u''\bigr) \bigr\vert =& \biggl\vert \lambda \int_{0}^{t}Nu(s) \, ds \biggr\vert \\ \leq &\Vert a \Vert _{1}+ \Vert b \Vert _{1}\varphi_{p} \bigl(2M_{0}+ \bigl\Vert u'' \bigr\Vert _{0}\bigr)+ \Vert c \Vert _{1}\varphi_{p} \bigl(M_{0}+ \bigl\Vert u'' \bigr\Vert _{0}\bigr)+ \Vert d \Vert _{1}\varphi_{p} \bigl( \bigl\Vert u'' \bigr\Vert _{0} \bigr). \end{aligned}$$

If $1< p\leq2$, by (2.1), we have

$$\bigl\vert \varphi_{p}\bigl(u''\bigr) \bigr\vert \leq \Vert a \Vert _{1}+\bigl(2 \Vert b \Vert _{1}+ \Vert c \Vert _{1}\bigr)M_{0}^{p-1}+ \bigl( \Vert b \Vert _{1}+ \Vert c \Vert _{1}+ \Vert d \Vert _{1}\bigr)\varphi_{p}\bigl( \bigl\Vert u'' \bigr\Vert _{0}\bigr). $$

Thus,

$$\bigl\Vert u'' \bigr\Vert _{0}\leq \varphi_{q} \biggl( \frac{ \Vert a \Vert _{1}+(2 \Vert b \Vert _{1}+ \Vert c \Vert _{1})M_{0}^{p-1}}{1-( \Vert b \Vert _{1}+ \Vert c \Vert _{1}+ \Vert d \Vert _{1})} \biggr). $$

Similarly, if $p>2$, then

$$\bigl\Vert u'' \bigr\Vert _{0}\leq \varphi_{q} \biggl( \frac{ \Vert a \Vert _{1}+(2^{p-1} \Vert b \Vert _{1}+ \Vert c \Vert _{1})2^{p-2}M_{0}^{p-1}}{1-2^{p-2}( \Vert b \Vert _{1}+ \Vert c \Vert _{1})- \Vert d \Vert _{1})} \biggr). $$

The above inequalities, together with (3.4), imply that $\Omega _{1}$ is bounded. □

Lemma 3.5

Assume that $(H_{3})$ holds. Then the set

$$\Omega_{2}=\{u\in\operatorname{Ker}M:QNu=0\} $$

is bounded.

Proof

If $u\in\Omega_{2}$, then $u_{c}(t)=c(\alpha t-\beta)$ and $F(Nu_{c}) = 0$. By $(H_{3})$, we get $|c|\leq M_{1}$. This means that $\Omega _{2}$ is bounded. □

Theorem 3.6

Assume that $(A_{0})$–$(A_{2})$ and $(H_{1})$–$(H_{3})$ hold. Then the functional boundary value problem (1.1) has at least one solution.

Proof

Choose $R_{0}$ large enough such that $\Omega=\{u\in X: \Vert u \Vert < R_{0}\}\supset\overline{\Omega}_{1}\cup\overline{\Omega}_{2}$ and $R_{0}>M_{1}(|\alpha|+|\beta|)$. By Lemma 3.4, $Mu \neq N_{\lambda}u$ for $u\in\partial\Omega \cap\operatorname{Ker}M$, $\lambda\in(0,1)$. So, $(C_{1})$ of Theorem 2.4 holds.

Let $H(u,\delta)=\rho\delta u+(1-\delta)JQNu$, $u\in\operatorname{Ker}M\cap\overline{\Omega}$, $\delta\in[0,1]$, where $J:\operatorname{Im}Q\rightarrow\operatorname{Ker}M$ is a homeomorphism with $J(c)=c(\alpha t-\beta)$, and $\rho=1$ or $\rho =-1$, if (3.2) or (3.3) hold, respectively.

For $u\in\operatorname{Ker}M\cap\partial\Omega$, $u=c(\alpha t-\beta )\neq0$, $H(u,1)=\rho c(\alpha t-\beta)\neq0$. By Lemma 3.5, we know that $H(u,0)=QN(c(\alpha t-\beta))(\alpha t-\beta)\neq0$. For $\delta\in (0,1)$, $u=c(\alpha t-\beta)\in\operatorname{Ker}M\cap\partial\Omega$, $\Vert u \Vert =R_{0}\leq|c|(|\alpha|+|\beta|)$, we have $|c|>M_{1}$. If $H(c(\alpha t- \beta),\delta)=\rho\delta c(\alpha t-\beta)+(1-\delta)QN(c(\alpha t-\beta))(\alpha t-\beta)=0$, by $(H_{3})$, we obtain

$$c^{2}=-\frac{1-\delta}{\delta}\rho c\cdot QN\bigl(c(\alpha t-\beta) \bigr)< 0, $$

which is a contradiction. Thus, $H(u,\delta)\neq0$, $u\in\operatorname{Ker}M\cap\partial\Omega$, $\delta\in[0,1]$.

By invariance of degree under a homotopy,

$$\begin{aligned} \operatorname{deg}(JQN,\Omega\cap\operatorname{Ker}M,0) &=\operatorname{deg}\bigl(H( \cdot,0),\Omega\cap\operatorname{Ker}M,0\bigr) \\ &=\operatorname{deg}\bigl(H(\cdot,1),\Omega\cap\operatorname{Ker}M,0\bigr) \\ &=\operatorname{deg}(\rho I,\Omega\cap\operatorname{Ker}M,0)=\pm1\neq0. \end{aligned}$$

By Theorem 2.4, the problem (1.1) has at least one solution in Ω̅. □

In the next results the inequality $|u(t)|+u'(t)| >M$ of $(H_{1})$ is replaced with either $|u(t)| >M$ or $|u'(t)| >M$, which will lead to slight modifications of the proof of Lemma 3.4. We recall that $\alpha^{2}+\beta^{2} \neq0$.

Lemma 3.7

Assume that $\alpha\neq0$ and the following conditions hold:

$(H_{4})$ :

There exists a constant $M_{2}>0$ such that if $|u'(t)|>M_{2}$, then $F(Nu)\neq0$.

$(H_{5})$ :

There exist functions $a,b,c,d \in C[0,1]$ such that

$$\bigl\vert f(t,u,v,w) \bigr\vert \leq a(t)+b(t)\varphi_{p}\bigl( \vert u \vert \bigr)+c(t)\varphi _{p}\bigl( \vert v \vert \bigr)+d(t)\varphi_{p}\bigl( \vert w \vert \bigr), \quad t\in[0,1], u,v,w\in\mathbb{R}, $$

and

$$\begin{aligned} &\biggl(2+\frac{ \vert \beta \vert }{ \vert \alpha \vert } \biggr) \biggl(1+\frac{k_{1}( \vert \alpha \vert + \vert \beta \vert )}{\alpha^{2}+\beta^{2}} \biggr) \bigl( \Vert b \Vert _{0}+ \Vert c \Vert _{0}+ \Vert d \Vert _{0}\bigr)^{q-1}\\ &\quad < \textstyle\begin{cases} 2^{3-2q}, & \textit{if } 1 < p \leq2, \\ 2^{1-q}, & \textit{if } p > 2. \end{cases}\displaystyle \end{aligned}$$

Then the set

$$\Omega_{1}=\bigl\{ u\in\operatorname{dom} M: Mu=N_{\lambda}u, \lambda\in(0,1)\bigr\} $$

is bounded.

Proof

For $u\in\Omega_{1}$, $QNu=0$. Following the proof of Lemma 3.3 and applying $(H_{4})$, we obtain $R(u,\lambda)=(I-P)u$ and a constant $t_{2}\in[0,1]$ such that $|u'(t_{2})|\leq M_{2}$.

Since $u(t)=Pu(t)+(I-P)u(t)=Pu(t)+R(u,\lambda)$, $|(Pu)'(t_{2})|\leq M_{2}+ \Vert R(u,\lambda) \Vert _{X}$. By the definition of P, we have

$$\biggl\vert \frac{\alpha u'(0)-\beta u(0)}{\alpha^{2}+\beta^{2}} \biggr\vert \leq\frac {1}{ \vert \alpha \vert } \bigl(M_{2}+ \bigl\Vert R(u,\lambda) \bigr\Vert _{X} \bigr). $$

Thus,

$$ \Vert u \Vert _{X}\leq \Vert Pu \Vert _{X}+ \bigl\Vert R(u,\lambda) \bigr\Vert _{X}\leq \biggl(1+\frac{ \vert \beta \vert }{ \vert \alpha \vert } \biggr)M_{2}+ \biggl(2+\frac{|\beta|}{|\alpha|} \biggr) \bigl\Vert R(u,\lambda) \bigr\Vert _{X}. $$

(3.5)

Since

$$\begin{aligned} \bigl\Vert R(u,\lambda) \bigr\Vert _{X} &\leq \biggl(1+ \frac{k_{1}(|\alpha|+|\beta|)}{\alpha ^{2}+\beta^{2}} \biggr) \biggl\Vert \int_{0}^{t}(t-s)\varphi_{q} \biggl( \int _{0}^{s}(I-Q)N_{\lambda}u(r)\, dr \biggr)\,ds \biggr\Vert _{X} \\ &\leq \biggl(1+\frac{k_{1}(|\alpha|+|\beta|)}{\alpha^{2}+\beta^{2}} \biggr)2^{q-1}\varphi_{q} \bigl( \Vert N_{\lambda}u \Vert _{Y}\bigr), \end{aligned}$$

by $(H_{5})$, we have

$$\begin{aligned} \Vert u \Vert _{X} \leq& \biggl(1+\frac{|\beta|}{|\alpha|} \biggr)M_{2} \\ &{}+2^{q-1} \biggl(2+\frac{|\beta|}{|\alpha|} \biggr) \biggl(1+ \frac {k_{1}(|\alpha|+|\beta|)}{\alpha^{2}+\beta^{2}} \biggr)\\ &{}\times \varphi_{q} \bigl( \Vert a \Vert _{0}+\bigl( \Vert b \Vert _{0}+ \Vert c \Vert _{0}+ \Vert d \Vert _{0}\bigr) \varphi_{p} \bigl( \Vert u \Vert _{X}\bigr) \bigr). \end{aligned}$$

By $(H_{5})$, $\Omega_{1}$ is bounded, if $p > 2$. With a different constant, the same inequality shows that $\Omega_{1}$ is bounded, if $1 < p \leq2$. □

Example

Consider

$$\bigl(\phi_{p}\bigl(u''(t)\bigr) \bigr)' = f\bigl(t,u(t),u'(t),u''(t) \bigr), \quad t \in(0,1), $$

where $p=3/2$ and

$$f\bigl(t,u(t),u'(t),u''(t)\bigr)= t+A \sin \bigl(\sqrt{ \bigl\vert u(t) \bigr\vert } \bigr)+A\frac{u'(t)+1}{|u'(t)|+1} \sqrt{ \bigl\vert u'(t) \bigr\vert } +A \sin \bigl(\sqrt{ \bigl\vert u''(t) \bigr\vert } \bigr), $$

where $A=0.043$.

We impose the functional conditions

$$u''(0)=0, \qquad B_{1}(u)= u'(0)+ 2 \int_{0}^{1}u(s)\, ds =0, \qquad B_{2}(u)=u(1)=0. $$

Then the functional problem is at resonance with $B_{1}(1)=B_{1}(t)=2$, $B_{2}(1)=B_{2}(t)=1$, $k=1/2$, $k_{1}=3$, $Ker M =\{c(t-1): c \in\mathbb {R}\}$. In this case, $\alpha=\beta=2$ and $\Vert b \Vert _{0}= \Vert c \Vert _{0}= \Vert d \Vert _{0}=A$ and $q=3$. Moreover,

$$2^{2q-3} \biggl(2+\frac{|\beta|}{|\alpha|} \biggr) \biggl(1+\frac{k_{1}(|\alpha |+|\beta|)}{\alpha^{2}+\beta^{2}} \biggr) \bigl( \Vert b \Vert _{0}+ \Vert c \Vert _{0}+ \Vert d \Vert _{0}\bigr)^{q-1} = 540 A^{2} < 1. $$

Clearly,

$$\begin{aligned} \bigl\vert f(t,u,v,w) \bigr\vert &\leq t+A \sqrt{ \vert u \vert }+A\sqrt{ \vert v \vert } +A \sqrt{ \vert w \vert }\\ &= t+ A \phi _{p} \bigl( \vert u \vert \bigr)+A\phi_{p}\bigl( \vert v \vert \bigr)+A \phi_{p}\bigl( \vert w \vert \bigr), \quad t \in(0,1). \end{aligned}$$

For convenience, introduce

$$Y(s) = \phi_{q} \biggl( \int_{0}^{s} f\bigl(r,u(r),u'(r),u''(r) \bigr) \, dr \biggr). $$

Hence

$$\begin{aligned} F(Nu) & = (B_{2}-kB_{1}) \biggl( \int_{0}^{t}(t-s) Y(s) \,ds \biggr) \\ &= \int_{0}^{1} (1-s)Y(s) \,ds - \int_{0}^{1} \biggl( \int_{0}^{s} (s-r)Y(r) \, dr \biggr) \,ds \\ & = \int_{0}^{1} (1-s)Y(s) \,ds - \frac{1}{2} \int_{0}^{1} (1-s)^{2} Y(s) \,ds \\ & = \frac{1}{2} \int_{0}^{1} \bigl(1-s^{2}\bigr) Y(s) \,ds. \end{aligned}$$

If $u'(t) > M_{0} > (2+\frac{1}{A} )^{2}$, then

$$\frac{u'(t)+1}{|u'(t)|+1}\sqrt{ \bigl\vert u'(t) \bigr\vert } > \sqrt{M_{0}} $$

and

$$f\bigl(t,u(t),u'(t),u''(t)\bigr) > -2A+A \sqrt{M_{0}} >0. $$

If $u'(t) < -M_{0}$, then

$$\frac{u'(t)+1}{|u'(t)|+1}\sqrt{ \bigl\vert u'(t) \bigr\vert } < - \sqrt{M_{0}} $$

and

$$f\bigl(t,u(t),u'(t),u''(t)\bigr) < 1+2A-A \sqrt{M_{0}} < 0. $$

Hence, $|u'(t)| > M_{0}$ guarantees $|Y(s)| >0$, which, in turn, implies that $F(Nu)\neq0$. Similarly, one can choose $M_{1} > 0$ such that, for $u_{c}(t)=c(t-1)$,

$$F(Nu_{c}) = (B_{2}-kB_{1}) \biggl( \int_{0}^{t}(t-s) \phi_{q} \biggl( \int_{0}^{s} f\bigl(r,c(r-1),c,0\bigr) \, dr \biggr) \,ds \biggr) \neq0 $$

provided $|c| > M_{1}$.

The above computations show that there is a solution whose existence is governed by Lemma 3.7.

Lemma 3.8

Assume that $\alpha=0$ and the following conditions hold:

$(H_{6})$ :

There exists a constant $M_{3}>0$ such that if $|u(t)|>M_{3}$, then $F(Nu)\neq0$.

$(H_{7})$ :

There exist functions $a,b,c,d\in C[0,1]$ such that

$$\bigl\vert f(t,u,v,w) \bigr\vert \leq a(t)+b(t)\varphi_{p}\bigl( \vert u \vert \bigr)+c(t)\varphi _{p}\bigl( \vert v \vert \bigr)+d(t)\varphi_{p}\bigl( \vert w \vert \bigr), \quad t\in[0,1], u,v,w\in\mathbb{R}, $$

and

$$\biggl(1+\frac{k_{1}}{|\beta|} \biggr) \bigl( \Vert b \Vert _{0}+ \Vert c \Vert _{0}+ \Vert d \Vert _{0} \bigr)^{q-1}< \textstyle\begin{cases} 4^{1-q}, & \textit{if } 1 < p \leq2, \\ 2^{-q}, & \textit{if } p > 2. \end{cases} $$

Then the set

$$\Omega_{1}=\bigl\{ u\in\operatorname{dom} M:Mu=N_{\lambda}u,\lambda \in(0,1)\bigr\} $$

is bounded.

Proof

As in the proof of Lemma 3.3, by $(H_{6})$, we have $R(u,\lambda)=(I-P)u$ and a constant $t_{3}\in[0,1]$ such that $|u(t_{3})|\leq M_{3}$. Since $u(t)=Pu(t)+(I-P)u(t)=Pu(t)+R(u,\lambda)$, $|(Pu)(t_{3})|\leq M_{3}+\| R(u,\lambda)\|_{X}$ and

$$\Vert u \Vert _{X}\leq \Vert Pu \Vert _{X}+ \bigl\Vert (I-P)u \bigr\Vert _{X}\leq M_{3}+2 \bigl\Vert R(u,\lambda) \bigr\Vert _{X}. $$

Since

$$\begin{aligned} \bigl\Vert R(u,\lambda) \bigr\Vert _{X} &\leq \biggl(1+ \frac{k_{1}}{|\beta|} \biggr) \biggl\Vert \int _{0}^{t}(t-s)\varphi_{q} \biggl( \int_{0}^{s}(I-Q)N_{\lambda}u(r)\, dr \biggr)\,ds \biggr\Vert _{X} \\ &\leq \biggl(1+\frac{k_{1}}{|\beta|} \biggr)2^{q-1}\varphi_{q} \bigl( \Vert N_{\lambda }u \Vert _{Y}\bigr), \end{aligned}$$

by $(H_{7})$, we have

$$\Vert u \Vert \leq M_{3}+2^{q} \biggl(1+ \frac{k_{1}}{|\beta|} \biggr)\varphi_{q} \bigl( \Vert a \Vert _{0}+\bigl( \Vert b \Vert _{0}+ \Vert c \Vert _{0}+ \Vert d \Vert _{0}\bigr) \varphi_{p} \bigl( \Vert u \Vert _{X}\bigr) \bigr). $$

This, together with $(H_{7})$, means that $\Omega_{1}$ is bounded in the case $p>2$ and, similarly, for $1 < p \leq2$. □

The proofs of the following theorems are similar to that of Theorem 3.6.

Theorem 3.9

Assume that $\alpha\neq0$, $(A_{0})$–$(A_{2})$ and $(H_{3})$–$(H_{5})$ hold. Then the functional boundary value problem (1.1) has at least one solution.

Theorem 3.10

Assume that $\alpha= 0$, $(A_{0})$–$(A_{2})$ and $(H_{3})$, $(H_{6})$, $(H_{7})$ hold. Then the functional boundary value problem (1.1) has at least one solution.

4 Conclusion

We obtain the existence of solution for a third-order functional p-Laplacian boundary value problem at resonance. This result extends many existent results and generalizes many related problems in the literature.