2. Preliminaries

In this section, we present some preliminary concepts, a new Sobolev inequality and an important proposition.

For convenience, we first introduce some necessary basic knowledge and signs. Throughout the paper, $\left \langle a,\, b \right \rangle$ denotes the inner product for any $a,\,b \in \mathbb {R}^N$, while ${\left | a \right |}$ denotes the Euclidean norm for $a \in \mathbb {R}^N$. $Q \in O(N)$ and $O(N)$ denotes the orthogonal group on $\mathbb {R}^N$.

Set $C = C^0( {[0,\,T],\,{\mathbb {R}^N}} )$ with the norm $\left \| u \right \|_0 = \mathop {\max }\limits _{0 \le t \le 1} \left | {u(t)} \right |$, $C^m = C^m( {[0,\,T],\,{\mathbb {R}^N}} )$ with the norm ${\left \| u \right \|_m}=\max \{ {\left \| u \right \|_0},\,{\left \| {u'} \right \|_0},\, \cdots,\,{\left \| {{u^{(m)}}} \right \|_0}\}$, $L^p=L^p(0,\,T;{\mathbb {R}^N})$ with the norm ${\left \| u \right \|_{{L^p}}} = {( {\int _0^T {{{\left | {u(t)} \right |}^p}\,{\rm d}t} } )^{{1}/{p}}}$.

Let ${C_Q} = \left \{ {u \in C:u(T)=Qu(0)} \right \}$ , $C_Q^1 = \{ u \in {C^1}:u(T) = Qu(0),\,u'(T) = Qu'(0) \}$,

$X = \left \{ {u \in C_Q^1:{\phi _p}(u') \ {\rm is} \ {\rm absolutely}\ {\rm continuous}} \right \}$ and $Y=L^1(0,\,T;{\mathbb {R}^N})$.

The function $f:[0,\,T] \times \mathbb {R}^N \times \mathbb {R}^N \to \mathbb {R}^N$ is assumed to be Carath$\acute {e}$odory, which satisfies

1. (1) the function $f(t,\, \cdot,\, \cdot )$ is continuous on $\mathbb {R}^N \times \mathbb {R}^N$ for a.e. $t \in [0,\,T]$;

2. (2) the function $f(\cdot,\, x ,\, y )$ is measurable on $[0,\,T]$ for each $(x,\,y) \in \mathbb {R}^N \times \mathbb {R}^N$;

3. (3) for each $r>0$ there exists $a_r \in {L^1}( {(0,\,T);\mathbb {R}} )$ such that, for a.e. $t \in [0,\,T]$ and each $(x,\,y) \in \mathbb {R}^N \times \mathbb {R}^N$ with $\left | x \right | \le r$, $\left | y \right | \le r$, one has

\begin{align*} \left| {f(t,x,y)} \right| \le {a_r}(t). \end{align*}

Let $Q \in O(N)$ and ${\rm I}$ be the identity operator. By the orthogonal decomposition theorem in linear algebra, we have

(2.1)\begin{equation} {\mathbb{R}^N} = \ker {\rm(I - Q)} \oplus {\mathop{\rm Im}\nolimits} {\rm(I - Q)}. \end{equation}

Define the orthogonal projector

(2.2)\begin{equation} {\mathcal{P}}:{\mathbb{R}^N} \to \ker {\rm(I - Q)}. \end{equation}

If $\ker {\rm (I}\hbox{-}{\rm Q)} \ne \{0\}$ and $Q \ne {\rm I}$, define ${L_P} = {\left. { {\rm (I\ }\hbox{-}{\rm \ Q)} } \right |_{\ker \mathcal {P}}}$. Then $L_P$ is a bijection from $\ker \mathcal {P}$ to ${\mathop {\rm Im}\nolimits } {\rm (I\ }\hbox{-}{\rm \ Q)}$.

If $Q = {\rm I}$, then let ${\mathcal {P}} = {\rm I}$ and ${L_P}={\rm I}$.

Let $H_{T,Q}^{1}=\left \{ {u \in {H^{1}}:u(T) = Qu(0)} \right \} \subset {H^{1}}$ with the inner product

\[ {\left\langle {u,v} \right\rangle } = \int_0^T {\left\langle {u(t),v(t)} \right\rangle + \left\langle {\dot u(t),\dot v(t)} \right\rangle \,{\rm d}t} , \]

and corresponding norm ${\left \| u \right \|^2} = {\left \langle {u,\,u} \right \rangle }$, where $Q \in O(N)$. It is easy to show that $H_{T,Q}^{1}$ is a Hilbert space and the embedding $H_{T,Q}^{1}\hookrightarrow C$ is compact. Next we will check Wirtinger inequality and Sobolev inequality still hold on $H_{T,Q}^{1}$.

Theorem 2.1 If $u \in H_{T,Q}^{1}$ and $\int _0^T {u(t)\,{\rm d}t} \in {\mathop {\rm Im}\nolimits } (I - Q),$ then there exist constants ${\lambda _1}>0,$ $c_1>0$ such that

\[ \int_0^T {{{\left| {u(t)} \right|}^2}\,{\rm d}t} \le {\lambda _1}\int_0^T {{{\left| {\dot u(t)} \right|}^2}\,{\rm d}t}, \]

(Wirtinger inequality) and

\[ {\left\| u \right\|_0} \le c_1{\left\| {\dot u} \right\|_{{L^2}}}, \]

(Sobolev inequality).

In order to prove theorem 2.1, we first prove the following lemma.

Lemma 2.3 Define the functional $J: H_{T,Q}^{1} \to \mathbb {R}$ by

\[ J(u) = \int_0^T {{\left| {\dot u(t)} \right|}^2\,{\rm d}t}. \]

Then ${c_2} = \mathop {\min }\limits _{u \in E} J(u) > 0,$ where

\[ E = \left\{ {u \in H_{T,Q}^{1}:\int_0^T {{{\left| u \right|}^2}\,{\rm d}t} = 1,\int_0^T {u(t)\,{\rm d}t} \in {\mathop{\rm Im}\nolimits} (I - Q)} \right\}. \]

Proof. Let ${c _2} = \mathop {\inf }\limits _{u \in E} J(u)$. Obviously, $J$ is coercive on $E$. Then there exists bounded sequence $\{ {u_n}\} \in E$ such that $J(u_n) \to {c _2}$. Because $H_ {T,\, Q} ^ {1}$ is a Hilbert space, there is a subsequence of $\{ {u_n}\}$, which we rename the same, which satisfies ${u_n} \rightharpoonup u (n \to \infty )$. The set $E$ is weakly sequentially closed, as follows easily from the compact embedding of $H_{T,Q}^{1}$ in $C$. Then $u \in E$. Because $J$ is continuous and convex on $E$, then $J$ is weakly lower semi-continuous on $E$. It follows that

\[ {c_2} = \underline{\lim}J(u_n) \ge J(u) \ge 0. \]

Then we have that $J$ gets the minimum value ${c_2}$ at $u$, i.e., $J(u)=c_2$. If $c _2 = 0$, then $u = a \in \mathbb {R}^N$ with $(I-Q) a=0$. However $\int _0^T {u(t)\,{\rm d}t}=T a \in {\mathop {\rm Im}\nolimits } \rm (I - Q)$ from the definition of $E$. Then by (2.1), $a =0$, which contradicts $\int _0^T {{{\left | u \right |}^2}\,{\rm d}t} = 1$. Thus, $c _2 > 0$.

Proof of theorem 2.1 Suppose that $u \in H_{T,Q}^{1}$ with $\int _0^T {u(t)\,{\rm d}t} \in {\mathop {\rm Im}\nolimits } \rm (I - Q)$. If $\int _0^T {{{\left | {u(t)} \right |}^2}\,{\rm d}t} = 0$, then the result is obviously true. Assume $\int _0^T {{{\left | {u(t)} \right |}^2}\,{\rm d}t} \ne 0$. Let

\[ v = \frac{u}{{{{\left( {\int_0^T {{{\left| {u(t)} \right|}^2}\,{\rm d}t} } \right)}^{{1}/{2}}}}}. \]

Then $v \in E$. By lemma 2.3, we have

\[ \int_0^T {{{\left| {\dot v(t)} \right|}^2}\,{\rm d}t} \ge {c_2}, \]

and hence

\[ \int_0^T {{{\left| {\dot u(t)} \right|}^2}\,{\rm d}t} \ge {c_2}\int_0^T {{{\left| {u(t)} \right|}^2}\,{\rm d}t}. \]

Taking ${\lambda _1} = \frac {1}{{{c_2}}}$, Wirtinger inequality holds. Because $H_{T,Q}^{1}\hookrightarrow C$, there exists $c>0$ such that ${\left \| u \right \|_0} \le c{\left \| u \right \|}$. Then we obtain Sobolev inequality ${\left \| u \right \|_0} \le c_1{\left \| {\dot u} \right \|_{{L^2}}}$.

Proof. (i) As $Q \in O(N)$, for $\alpha \in \mathbb {R}^N$, one has $\left | {Q\alpha } \right | = \left | \alpha \right |$, and

\[ {\phi _p}(Q\alpha ) = {\left| {Q\alpha } \right|^{p - 2}}(Q\alpha ) = {\left| \alpha \right|^{p - 2}}(Q\alpha ) = Q{\left| \alpha \right|^{p - 2}}\alpha = Q{\phi _p}(\alpha ). \]

(ii) It can be checked by simple calculation.

Assume that $\ker {\rm (I}\hbox{-}{\rm Q)} \ne \{0\}$. Consider the simple rotating periodic boundary value problem

(2.3)\begin{align} - ({\phi _p}(u'))' & = f(t), \end{align}

(2.4)\begin{align} u(T) & = Qu(0),u'(T) = Qu'(0), \end{align}

where $f(t) \in Y$ satisfying $f(t+T)=Qf(t)$.

Suppose that $u(t) \in X$ is a solution to (2.3) (2.4). By integrating (2.3) over $[0,\,T]$, we have that

\[ {\phi _p}(u'(0)) - {\phi _p}(u'(T)) = \int_0^T {\left( { - ({\phi _p}(u'))'} \right)\,{\rm d}t} = \int_0^T {f(t)\,{\rm d}t}. \]

Using (2.4) and lemma 2.4(i), we get

(2.5)\begin{equation} {\rm(I-Q)}({\phi _p}(u'(0))) = \int_0^T {f(t)\,{\rm d}t}. \end{equation}

So

\[ {\mathcal{P}}({\rm(I-Q)}({\phi _p}(u'(0)))) = {\mathcal{P}}\left(\int_0^T {f(t)\,{\rm d}t}\right), \]

which yields

(2.6)\begin{equation} {\mathcal {P}}\left(\int_0^T {f(t)\,{\rm d}t}\right)=0. \end{equation}

On the other hand, since

\[ {\phi _p}(u'(0)) = {\mathcal{P}}({\phi _p}(u'(0))) + { (\rm{I} - \mathcal{P})}({\phi _p}(u'(0))), \]

and

\[ \int_0^T {f(t)\,{\rm d}t} = {\mathcal{P}}\left(\int_0^T {f(t)\,{\rm d}t}\right) + { (\rm{I} - \mathcal{P})}\left(\int_0^T {f(t)\,{\rm d}t}\right), \]

then

(2.7)\begin{align} & {\rm(I-Q)}({\mathcal{P}}({\phi _p}(u'(0))) + { (\rm{I} - \mathcal{P})}({\phi _p}(u'(0))))\nonumber\\ & \quad = {\mathcal {P}}\left(\int_0^T {f(t)\,{\rm d}t} \right) + { (\rm{I} - \mathcal{P})}\left(\int_0^T {f(t)\,{\rm d}t}\right). \end{align}

From (2.6),(2.7) and the definition of $\mathcal {P}$, it follows that

\[ {\rm(I-Q)}{\mathcal{P}}({\phi _p}(u'(0)))=0= {\mathcal {P}}\left(\int_0^T {f(t)\,{\rm d}t}\right), \]

and

\[ {\rm(I-Q)} {(\rm{I} - \mathcal{P})}({\phi _p}(u'(0))) = {(\rm{I} - \mathcal{P})}\left(\int_0^T {f(t)\,{\rm d}t}\right). \]

Taking ${L_P}$ to act on above equation, one has

(2.8)\begin{equation} {(\rm{I} - \mathcal{P})}({\phi _p}(u'(0))) = L_P^{{-}1} {(\rm{I} - \mathcal{P})}\left(\int_0^T {f(t)\,{\rm d}t}\right)=L_P^{{-}1}\left(\int_0^T {f(t)\,{\rm d}t}\right). \end{equation}

Integrating (2.3) and combining (2.8), we have

\[ {\phi _p}(u'(t)) ={-} \int_0^t {f(s)\,{\rm d}s} + {\mathcal {P}}({\phi _p}(u'(0))) + L_P^{ - 1}\left(\int_0^T {f(s)\,{\rm d}s}\right), \]

i.e.,

(2.9)\begin{equation} u'(t) = {\phi _q}\left( { - \int_0^t {f(s)\,{\rm d}s} + {\mathcal {P}}({\phi _p}(u'(0))) + L_P^{ - 1}\left(\int_0^T {f(s)\,{\rm d}s}\right)} \right) \buildrel \Delta \over = a(t) , \end{equation}

where $\frac {1}{p} + \frac {1}{q} = 1(p,\,q > 1)$. Integrating (2.9) over $[0,\,T]$, we obtain

\[ u(0) - u(T) = \int_0^T {\left( { - a(t)} \right)\,{\rm d}t}. \]

Similar to the previous discussion, we have

\[ {\mathcal {P}}\left(\int_0^T {\left( { - a(t)} \right)\,{\rm d}t} \right) = 0, \]

i.e.,

(2.10)\begin{equation} {\mathcal{P}}\left(\int_0^T -{{\phi _q}\left( { - \int_0^t {f(s)\,{\rm d}s} + {\mathcal {P}}{\phi _p}(u'(0)) + L_P^{ - 1}\int_0^T {f(s)\,{\rm d}s} } \right)\,{\rm d}t} \right) = 0. \end{equation}

Following (2.5)–(2.8), and

\[ {\rm(I-Q)}{(\rm{I} - \mathcal{P})}(u(0)) = {(\rm{I} - \mathcal{P})}\left(\int_0^T {\left( { - a(t)} \right)\,{\rm d}t} \right)=\int_0^T {\left( { - a(t)} \right)\,{\rm d}t} , \]

we get

(2.11)\begin{equation} {(\rm{I} - \mathcal{P})}(u(0)) = L_P^{ - 1}\left(\int_0^T {\left( { - a(t)} \right)\,{\rm d}t} \right). \end{equation}

Integrating (2.9), we obtain

(2.12)\begin{equation} u(t) = {\mathcal {P}}(u(0))+L_P^{ - 1}\left( \int_0^T {\left( { - a(s)} \right)\,{\rm d}s}\right)+\int_0^t {a(s)\,{\rm d}s}. \end{equation}

On the basis of (2.10), we define mapping $G_h : {\ker \rm (I-Q)} \to {\ker \rm (I-Q)}$ by

(2.13)\begin{equation} {G_h}(\gamma )={\mathcal {P}}\int_0^T {{\phi _q}\left( { - \int_0^t {h(s)\,{\rm d}s} + \gamma+ L_P^{ - 1}\int_0^T {h(s)\,{\rm d}s} } \right)\,{\rm d}t}, \quad \gamma \in {\ker \rm(I-Q)}, \end{equation}

where $h \in {Y_1} = \left \{ {h \in Y\left | {\int _0^T {h(s)\,{\rm d}s} \in {\mathop {\rm Im}\nolimits } {\rm (I\ }\hbox{-}{\rm \ Q)}} \right.} \right \}$. Next we discuss the properties of $G_h$.

Proof. (i) Because ${\mathcal {P}}:{\mathbb {R}^N} \to \ker {\rm (I\ }\hbox{-}{\rm \ Q)}$ is the orthogonal projector, then for any $\gamma _1,\, \gamma _2 \in \ker {\rm (I}\hbox{-}{\rm Q)}$, we have

\[ \left\langle {{(\rm{I} - \mathcal{P})}\left(\int_0^T {{\phi _q}\left( { - \int_0^t {h(s)\,{\rm d}s} + \gamma_1 + L_P^{ - 1}\int_0^T {h(s)\,{\rm d}s} } \right)\,{\rm d}t}\right) ,\gamma_2 } \right\rangle = 0. \]

Let

\begin{align*} {K_h}(\gamma ) & = \int_0^T {{\phi _q}\left( { - \int_0^t {h(s)\,{\rm d}s} + \gamma + L_P^{ - 1}\int_0^T {h(s)\,{\rm d}s} } \right)\,{\rm d}t} \\ & = {(\rm{I} - \mathcal{P})}\left(\int_0^T {{\phi _q}\left( { - \int_0^t {h(s)\,{\rm d}s} + \gamma + L_P^{ - 1}\int_0^T {h(s)\,{\rm d}s} } \right)\,{\rm d}t}\right) \\ & \quad + {\mathcal{P}}\left(\int_0^T {{\phi _q}\left( { - \int_0^t {h(s)\,{\rm d}s} + \gamma + L_P^{ - 1}\int_0^T {h(s)\,{\rm d}s} } \right)\,{\rm d}t}\right). \end{align*}

From the definition of ${G_h}$ and ${\mathcal {P}}$, it follows that

(2.15)\begin{equation} \left\langle {{G_h}(\gamma_1 ),\gamma_2 } \right\rangle = \left\langle {{K_h}(\gamma_1 ),\gamma_2 } \right\rangle. \end{equation}

According to lemma 2.4 (ii), for $\gamma _1 \ne \gamma _2$, we have

\begin{align*} & \left\langle {{K_h}({\gamma _1}) - {K_h}({\gamma _2}),{\gamma _1} - {\gamma _2}} \right\rangle\\ & \quad = \int_0^T {\left\langle {{\phi _q}\left( {{\gamma _1} + l_h (t)} \right) - {\phi _q}\left( {{\gamma _2} + l_h(t)} \right),{\gamma _1} - {\gamma _2}} \right\rangle \,{\rm d}t} > 0, \end{align*}

where $l_h (t)=- \int _0^t {h(s)\,{\rm d}s} + L_P^{ - 1}\int _0^T {h(s)\,{\rm d}s} \in C$. Combining (2.15), we obtain

(2.16)\begin{equation} \left\langle {{G_h}(\gamma_1 )-{G_h}(\gamma_2 ),\gamma_1- \gamma_2} \right\rangle = \left\langle {{K_h}(\gamma_1)-{K_h}(\gamma_2),\gamma_1- \gamma_2} \right\rangle>0. \end{equation}

And hence, if (2.14) has a solution then it is unique.

To prove existence of solutions, we will show that $\left \langle {{G_h}(\gamma ),\,\gamma } \right \rangle > 0$ for $\left | \gamma \right |$ sufficiently large. Indeed, we have

\begin{align*} \left\langle {{G_h}(\gamma ),\gamma } \right\rangle & = \left\langle {{K_h}(\gamma ),\gamma } \right\rangle \\ & = \int_0^T {\left\langle {{\phi _q}\left( {\gamma + l_h (t)} \right),\gamma } \right\rangle \,{\rm d}t} \\ & = \int_0^T {\left\langle {{\phi _q}\left( {\gamma + l_h (t)} \right),\gamma + l_h(t)} \right\rangle \,{\rm d}t} - \int_0^T {\left\langle {{\phi _q}\left( {\gamma + l_h(t)} \right),l_h (t)} \right\rangle \,{\rm d}t}. \end{align*}

So

(2.17)\begin{equation} \left\langle {{G_h}(\gamma ),\gamma } \right\rangle \ge \int_0^T {\left\langle {{\phi _q}\left( {\gamma + l_h (t)} \right),\gamma + l_h (t)} \right\rangle \,{\rm d}t} - {\left\| l_h \right\|_0}\int_0^T {\left| {{\phi _q}\left( {\gamma + l_h (t)} \right)} \right|\,{\rm d}t} . \end{equation}

Due to the definition of ${\phi _q}$, we have

(2.18)\begin{equation} \left\langle {{G_h}(\gamma ),\gamma } \right\rangle \ge \int_0^T {\left( {\left| {\left( {\gamma + l_h (t)} \right)} \right| - {{\left\| l_h \right\|}_0}} \right){{\left| {\left( {\gamma + l_h (t)} \right)} \right|}^{q - 1}}\,{\rm d}t} . \end{equation}

Since $q>1$ and ${{\left | { {\gamma + l_h (t)} } \right |}} \to \infty$ as $\left | \gamma \right | \to \infty$, there exists $r>0$ such that

(2.19)\begin{equation} \left\langle {{G_h}(\gamma ),\gamma } \right\rangle >0 \ {\rm for} \ {\rm all} \gamma \in \ker {\rm(I-Q)} \ {\rm with}\ \left| \gamma \right| \ge r. \end{equation}

It follows from the properties of topological degree that the equation ${G_h}(\gamma )=0$ has a solution for each $h \in {Y_1}$, which is unique by our previous argument.

(ii) From (i), we can define a functional $\tilde \gamma :{Y_1} \to \ker {\rm (I}\hbox{-}{\rm Q)}$ which satisfies

(2.20)\begin{equation} {G_h}(\tilde \gamma) ={\mathcal{P}}\left(\int_0^T {{\phi _q}\left( { - \int_0^t {h(s)\,{\rm d}s} + \tilde \gamma (h)+ L_P^{ - 1}\int_0^T {h(s)\,{\rm d}s} } \right)\,{\rm d}t}\right)=0, \end{equation}

for any $h \in {Y_1}$. Hence,

\[ 0 = \left\langle {{G_h}(\tilde \gamma ),\tilde \gamma } \right\rangle = \left\langle {{K_h}(\tilde \gamma ),\tilde \gamma } \right\rangle . \]

Then

(2.21)\begin{equation} \int_0^T {\left\langle {{\phi _q}\left( {\tilde \gamma + l_h (t)} \right),\tilde \gamma + l_h(t)} \right\rangle \,{\rm d}t} = \int_0^T {\left\langle {{\phi _q}\left( {\tilde \gamma + l_h (t)} \right),l_h (t)} \right\rangle \,{\rm d}t}. \end{equation}

Let $\Omega \subset Y_1$ be a bounded subset. Then there is $M_1 >0$ such that ${\left \| h \right \|_{{L^1}}} \le {M_1}$ and $\left | {\int _0^t {h(s)\,{\rm d}s} } \right | \le {M_1}$, for any $h \in \Omega$. Due to the definition of $L_P^{ - 1}$ and $l_h (t)$, there exists a constant $M_2>0$ such that $\left | {L_P^{ - 1}\int _0^T {h(s)\,{\rm d}s}} \right | \le {M_2}$, and

\[ \left| {{{l_h (t)}}} \right| \le \left| {\int_0^t {h(s)\,{\rm d}s} } \right| + \left| {L_P^{ - 1}\int_0^T {h(s)\,{\rm d}s} } \right| \le {M_1} + {M_2}, \]

that is, ${\left \| l_h \right \|_0} \le \sqrt N ({M_1} + {M_2})$, for any $h \in \Omega$.

Now we show $\left | {\tilde \gamma (h) } \right |$ is bounded on $\Omega$. Assume on contrary that $\left \{ {\tilde \gamma (h):h \in \Omega } \right \}$ is not bounded. Then for any given $M_3>\sqrt N ({M_1} + {M_2})$, there is $h \in \Omega$ such that

\[ {M_3} \le \left| {\tilde \gamma (l_h) + l_h(t)} \right|,\quad t \in [0,T]. \]

Hence by (2.21), we find that

\begin{align*} {M_3}\int_0^T {{{\left| {\tilde \gamma (l_h) + l_h (t)} \right|}^{q - 1}}\,{\rm d}t} & \le \int_0^T {{{\left| {\tilde \gamma (l_h) + l_h(t)} \right|}^q}\,{\rm d}t} \\ & =\int_0^T {\left\langle {{\phi _q}\left( {\tilde \gamma + l_h (t)} \right),\tilde \gamma + l_h(t)} \right\rangle \,{\rm d}t}\\ & = \int_0^T {\left\langle {{\phi _q}\left( {\tilde \gamma + l_h (t)} \right),l_h (t)} \right\rangle \,{\rm d}t}\\ & \le {\left\| l_h \right\|_0}\int_0^T {{{\left| {\tilde \gamma (l_h) + l_h(t)} \right|}^{q - 1}}\,{\rm d}t}. \end{align*}

Thus ${M_3} \le {\left \| l_h \right \|_0}$, a contradiction. Therefore $\tilde \gamma$ sends bounded set in $Y_1$ into bounded set in $\ker {\rm (I}\hbox{-}{\rm Q)}$.

Finally we show the continuity of $\tilde \gamma$. Let $\{h_n\}$ be a convergent sequence in $Y_1$, i.e., ${h_n} \to h$, as $n \to \infty$. It is easy to show that ${l_{{h_n}}} \to l_h$ in $C[0,\,T]$ as $n \to \infty$. Since $\{ \tilde \gamma ({l_{{h_n}}})\}$ is bounded sequence, there exists a subsequence $\{\tilde \gamma ({l_{{h_j}}})\}$ such that $\tilde \gamma ({l_{{h_j}}}) \to \hat \gamma (j \to \infty )$. Letting $j \to \infty$ in

\[ {\mathcal{P}}\left(\int_0^T {{\phi _q}\left( {\tilde \gamma ({l_{{h_j}}}) + {{l}_{{h_j}}}(t)} \right)\,{\rm d}t} \right)= 0, \]

we find that

\[ {\mathcal{P}}\left(\int_0^T {{\phi _q}\left( {\hat \gamma + l_h (t)} \right)\,{\rm d}t} \right) = 0, \]

and $\tilde \gamma (l_{h_j})=\hat \gamma$ from the definition of $\hat \gamma$, which show the continuity of $\tilde \gamma$.

Define the projectors ${\hat {\mathcal {P}}}:X \to X$ and ${\hat {\mathcal {Q}}}:Y \to Y$ respectively by

\begin{align*} {\hat {\mathcal {P}}}(u) & = {\mathcal {P}}(u(0)),\\ {\hat {\mathcal {Q}}}(f) & ={\mathcal {P}} \left(\frac{1}{T} \int_0^T {f(t)\,{\rm d}t}\right). \end{align*}

For $h \in Y$, let $\gamma : Y \to \ker {\rm (I}\hbox{-}{\rm Q)}$ be defined by

(2.22)\begin{equation} \gamma (h) = \tilde \gamma ({(\rm{I}-\hat{\mathcal{Q}})}h). \end{equation}

Then, it is clear that $\gamma$ is a continuous function and sends bounded set into bounded set. Noting that $\dim \ker {\rm (I}\hbox{-}{\rm Q)} < \infty$, so $\gamma$ is a completely continuous mapping.

3. An equivalent operator equation

In this section, we give an equivalent operator equation with RPBVP (H_Q).

Firstly, set Nemytski operator ${N_f}:X \to Y$ by

(3.1)\begin{equation} {N_f}u = f(t,u(t),u'(t)), \end{equation}

where $f:\mathbb {R} \times \mathbb {R}^N \times \mathbb {R}^N \to \mathbb {R}^N$ is Carath$\acute {e}$odory with $f(t + T,\,x,\,y) = Qf(t,\,{Q^{ - 1}}x, {Q^{ - 1}}y)$.

Next, we define the operator $H$ on $X$ by

(3.2)\begin{equation} (Hu)(t) = {\hat {\mathcal {P}}}(u) +{\hat {\mathcal {Q}}}({N_f}u) - L_P^{ - 1}\left( \int_0^T{c(s)\,{\rm d}s}\right)+\int_0^t {c(s)\,{\rm d}s}, \end{equation}

where

\begin{align*} c(s) & = {\phi _q}\left(\vphantom{\int_0^T} - \int_0^s {({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t}\right.\\ & \left.\quad + L_P^{ - 1}\int_0^T {({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t} +\tilde \gamma \left( ({\rm{I} - \hat{\mathcal{Q}}})({N_f}u) \right) \right), \end{align*}

$\tilde \gamma ( ({\rm {I}\ }\hbox{-}{\rm \hat {\mathcal {Q}}})({N_f}u) )$ is defined in (2.22) and $\frac {1}{p} + \frac {1}{q} = 1(p,\,q > 1)$.

By (2.20) and the definition of $\tilde \gamma$, we have ${\mathcal {P}}( \int _0^T{c(s)\,{\rm d}s})=G_{({\rm {I}\ }\hbox{-}{\rm \hat {\mathcal {Q}}})({N_f}u)} (\tilde \gamma )=0$. So the definition of $H$ is fine.

Proof. Obviously, $H$ is continuous in $C$ from the continuity of ${\hat {\mathcal {P}}}$, ${\hat {\mathcal {Q}}}$, ${N_f}$ and $L_P^{ - 1}$. Writing $H(t) \buildrel \Delta \over = (H(u))(t)$ and $\tilde \gamma = \tilde \gamma ( ({\rm {I}\ }\hbox{-}{\rm \hat {\mathcal {Q}}})({N_f}u) )$, we have that

\[ H'(t)= {\phi _q}\left( { - \int_0^t {({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(s)\,{\rm d}s} + L_p^{ - 1}\int_0^T {({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t} +\tilde \gamma } \right), \]

and

(3.3)\begin{equation} {\phi _p}\left(H'(t)\right)={ - \int_0^t {({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(s)\,{\rm d}s} + L_p^{ - 1}\int_0^T {({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t} +\tilde \gamma }, \end{equation}

which show $H \in C^1$ and ${\phi _p}(H'(t))$ is absolutely continuous, where $\frac {1}{p} + \frac {1}{q} = 1(p,\,q > 1)$.

Next, we will prove $H(T)=QH(0)$ and $H'(T)=QH'(0)$.

By (3.2), we have

(3.4)\begin{equation} H(T) ={\hat {\mathcal {P}}}(u) +{\hat {\mathcal {Q}}}({N_f}u) - L_P^{ - 1}\left( \int_0^T{c(s)\,{\rm d}s}\right)+\int_0^T {c(s)\,{\rm d}s}, \end{equation}

and

\[ H(0) = {\hat {\mathcal {P}}}(u) +{\hat {\mathcal {Q}}}({N_f}u) - L_P^{ - 1}\left( \int_0^T{c(s)\,{\rm d}s}\right). \]

Then

(3.5)\begin{equation} QH(0) = Q\left({\hat {\mathcal {P}}}(u) +{\hat {\mathcal {Q}}}({N_f}u) - L_P^{ - 1}\left( \int_0^T{c(s)\,{\rm d}s}\right)\right). \end{equation}

From the definition of $\tilde \gamma$, we find that

\begin{align*} & {\mathcal{P}}\left(\int_0^T {c(s)\,{\rm d}s}\right)\\& \quad={\mathcal{P}}\left(\int_0^T {{\phi _q}\left( { - \int_0^s {({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t} + L_P^{ - 1}\int_0^T {({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t} +\tilde \gamma } \right)\,{\rm d}s}\right)\\ & \quad= 0. \end{align*}

The definitions of ${\hat {\mathcal {P}}}$ and ${\hat {\mathcal {Q}}}$ yield that

\[ {\rm(I-Q)}{\hat {\mathcal {P}}}(u)=0,\quad {\rm(I-Q)}{\hat {\mathcal {Q}}}({N_f}u)=0, \]

i.e.,

(3.6)\begin{equation} {\hat {\mathcal {P}}}(u)=Q({\hat {\mathcal {P}}}(u)), \quad {\hat {\mathcal {Q}}}({N_f}u)=Q({\hat {\mathcal {Q}}}({N_f}u)). \end{equation}

According to the definition of $L_P^{ - 1}$, we get that

(3.7)\begin{align} QL_P^{ - 1}\left( \int_0^T{c(s)\,{\rm d}s}\right)& ={-}{\rm(I-Q)}L_P^{ - 1}\left( \int_0^T{c(s)\,{\rm d}s}\right)+L_P^{ - 1}\left( \int_0^T{c(s)\,{\rm d}s}\right)\nonumber\\ & ={-} \int_0^T{c(s)\,{\rm d}s}+L_P^{ - 1}\left( \int_0^T{c(s)\,{\rm d}s}\right). \end{align}

Substituting (3.6)–(3.7) into (3.5), we obtain

\[ QH(0)=H(T). \]

On the other hand, we see from (3.3) that

\[ {\phi _p}\left(H'(T)\right)={ - \int_0^T {({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t} + L_P^{ - 1}\int_0^T {({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t} +\tilde \gamma }, \]

and

\[ {\phi _p}\left(H'(0)\right)={ L_P^{ - 1}\int_0^T {({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t} +\tilde \gamma }. \]

Because $\tilde \gamma \in \ker {\rm (I}\hbox{-}{\rm Q)}$, i.e., $Q\tilde \gamma =\tilde \gamma$, then we have that

\begin{align*} Q{\phi _p}\left(H'(0)\right)& = QL_P^{ - 1}\left( \int_0^T{({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t}\right)+ Q\tilde \gamma\\ & ={-}{\rm(I-Q)}L_P^{ - 1}\left( \int_0^T{({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t}\right)\\ & \qquad +L_P^{ - 1}\left( \int_0^T{({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t}\right)+\tilde \gamma\\ & ={-} \int_0^T{({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t}+L_P^{ - 1}\left( \int_0^T{({\rm{I} - \hat{\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t}\right)+\tilde \gamma. \end{align*}

It follows from lemma 2.4 that

\[ {\phi _p}\left(H'(T)\right)=Q{\phi _p}\left(H'(0)\right)={\phi _p}\left(QH'(0)\right), \]

which yields that $H'(T)=QH'(0)$. Hence the $H$ is a continuous operator from $X$ to $X$.

Proof. We only need to prove that $H$ is a compact operator. Let $S \subset X$ be an open bounded subset such that ${\left \| u \right \|_1} \le {M_1}$ for any $u \in S$. It is easy to see that ${N_f}$ is continuous and sends bounded set into equi-integrable set. Then there exists $l(t) \in {L^1}( {(0,\,T);\mathbb {R}} )$ such that $\left | {N_f}{(u)} \right | \le l(t)$ for any $u \in S$. Taking $\left \{ {{u_n}} \right \} \subset S$, we have

(3.8)\begin{equation} (H{u_n})(t) = {\hat {\mathcal {P}}}({u_n}) +{\hat {\mathcal {Q}}}({N_f}{u_n}) - L_P^{ - 1}\left( \int_0^T{{c_n}(s)\,{\rm d}s}\right)+\int_0^t {{c_n}(s)\,{\rm d}s}, \end{equation}

where

\[ {c_n}(s) ={\phi _q}\left( { - \int_0^s {({\rm{I} - \hat {\mathcal {Q}}})({N_f}{u_n})(t)\,{\rm d}t} + L_p^{ - 1}\int_0^T {({\rm{I} - \hat {\mathcal {Q}}})({N_f}{u_n})(t)\,{\rm d}t} +\tilde \gamma } \right) , \]

and $\tilde \gamma = \tilde \gamma ( ({\rm {I}\ }\hbox{-}{\rm \hat {\mathcal {Q}}})({N_f}u_n) )$ is defined in (2.22). Obviously, $\left | {\hat {\mathcal {P}}}({u_n}) \right | \le {M_1}$ and

\[ \left| {\hat {\mathcal {Q}}}({N_f}{u_n}) \right| \le \left| {\frac{1}{T}\int_0^T {({N_f}{u_n})(\tau )\,{\rm d}\tau } } \right| \le {\frac{1}{T}\int_0^T {l(\tau )\,{\rm d}\tau } }= \frac{1}{T} M_2. \]

Then

\begin{align*} & \left| {\int_0^s {({\rm{I} - \hat {\mathcal {Q}}})({N_f}{u_n})(t)\,{\rm d}t} } \right|\\ & \quad \le \int_0^T {\left| {({N_f}{u_n})(t)} \right|\,{\rm d}t} + \int_0^T {\left| {\hat {\mathcal{Q}}({N_f}{u_n})(t)} \right|\,{\rm d}t} \le 2\int_0^T {l(\tau )\,{\rm d}\tau } = 2{M_2}, \end{align*}

uniformly in $s \in [0,\,T]$. By the continuity of the $L_P^{ - 1}$ and $\tilde \gamma$, there exists $M_3 > 0$ such that

\[ \left| L_P^{ - 1}\int_0^T {({\rm{I} - \hat {\mathcal {Q}}})({N_f}{u_n})(t)\,{\rm d}t} \right| \le {M_3}, \]

and

\[ \left| \tilde \gamma \right| \le {M_3}. \]

Hence we have that

\begin{align*} \left| {{c_n}(s)} \right| & = {\left| { - \int_0^s {({\rm{I} - \hat {\mathcal {Q}}})({N_f}{u_n})(t)\,{\rm d}t} + L_P^{ - 1}\int_0^T {({\rm{I} - \hat {\mathcal {Q}}})({N_f}{u_n})(t)\,{\rm d}t} + \tilde \gamma } \right|^{q - 1}} \\ & \le {\left| {\left| {\int_0^s {({\rm{I} - \hat {\mathcal {Q}}})({N_f}{u_n})(t)\,{\rm d}t} } \right| + \left| {L_P^{ - 1}\int_0^T {({\rm{I} - \hat {\mathcal {Q}}})({N_f}{u_n})(t)\,{\rm d}t} } \right| + \left| {\tilde \gamma } \right|} \right|^{q - 1}}\\ & \le {\left( 2(M_2+M_3) \right)^{q - 1}}, \end{align*}

uniformly in $s \in [0,\,T]$. Therefore, we obtain that there is a constant $M_4 >0$ such that

\[ \left| {{{(H{u_n})(t)}}} \right| \le {M_4},\quad \forall\ u_n \in S, \]

uniformly in $t \in [0,\,T]$, which shows that $\left \{ {H({u_n})} \right \}$ is uniformly bounded in $C$. Since $(H{u_n})'(t)= {c_n}(t)$, then $\left \{ {H({u_n})'(t)} \right \}$ is uniformly bounded in $C$. Hence, $\left \{ {H({u_n})} \right \}$ is equi-continuous. According to the Arzel$\grave {a}$-Ascoli theorem, $\left \{ {H({u_n})} \right \}$ is sequentially compact.

For any $u \in S$ and $s_1,\,s_2 \in [0,\,T]$, we have

\begin{align*} \left| w(s_2)-w(s_1) \right| & \le \left| {\int_{{s_1}}^{{s_2}} {({N_f}u)(t)\,{\rm d}t} } \right|+ \left| {{{\hat {\mathcal{Q}}}}({N_f}u)} \right|\left| {{s_2} - {s_1}} \right| \\ & \le {\int_{{s_1}}^{{s_2}} {l(\tau )\,{\rm d}\tau } }+\left| {{s_2} - {s_1}} \right|{\int_0^T {l(\tau )\,{\rm d}\tau } }, \end{align*}

where $w(s)=\int _0^s {({\rm {I\ }\hbox{-}{\rm \ }}\widehat {\mathcal {Q}})({N_f}u)(t)\,{\rm d}t}$.

Taking sequence $\{u_n\} \subset S$, then $\left \{ { - \int _0^s {({\rm {I}\ }\hbox{-}{\rm \hat {\mathcal {Q}}})({N_f}{u_n})(t)\,{\rm d}t} } \right \}$ is uniformly bounded and equi-continuous. By Arzel$\grave {a}$-Ascoli theorem there is a subsequence of $\left \{ { - \int _0^s {({\rm {I}\ }\hbox{-}{\rm \hat {\mathcal {Q}}})({N_f}{u_n})(t)\,{\rm d}t} } \right \}$, which we rename the same, which is convergent in $C$. Then, passing to a subsequence if necessary, we obtain that the sequence

\[ \left\{ { - \int_0^s {({\rm{I} - \hat {\mathcal {Q}}})({N_f}{u_n})(t)\,{\rm d}t} + L_p^{ - 1}\int_0^T {({\rm{I} - \hat {\mathcal {Q}}})({N_f}{u_n})(t)\,{\rm d}t} + \tilde \gamma } \right\}, \]

is convergent in $C$. Using that ${\phi _q}:C \to C$ is continuous it follows that $\left \{ {{c_{n}}} \right \}$ is convergent in $C$. Hence the mapping $H$ is a completely continuous operator.

Proof. Assume that $u \in X$ is a fixed point of $H$: $H(u)=u$, i.e.,

(3.9)\begin{equation} u(t) = {\hat {\mathcal {P}}}(u) +{\hat {\mathcal {Q}}}({N_f}u) - L_P^{ - 1}\left( \int_0^T{{c}(s)\,{\rm d}s}\right)+\int_0^t {{c}(s)\,{\rm d}s}), \end{equation}

where

\begin{align*} {c}(s) & ={\phi _q}\left(\vphantom{\int_0^T} - \int_0^s {({\rm{I} - \hat {\mathcal {Q}}})({N_f}u)(t)\,{\rm d}t}\right.\\& \left.\quad + L_P^{ - 1}\int_0^T {({\rm{I} - \hat {\mathcal {Q}}})({N_f}u)(t)\,{\rm d}t} +\tilde \gamma\left(({\rm{I} - \hat {\mathcal {Q}}})({N_f}u)\right) \right) , \end{align*}

and $\tilde \gamma ( ({\rm {I}\ }\hbox{-}{\rm \hat {\mathcal {Q}}})({N_f}u))$ as (2.22). Hence

\[ u(0) = {\hat {\mathcal {P}}}(u) +{\hat {\mathcal {Q}}}({N_f}u) - L_P^{ - 1}\left( \int_0^T{{c}(s)\,{\rm d}s}\right). \]

From the definitions of ${\hat {\mathcal {P}}}$, ${\hat {\mathcal {Q}}}$ and $L_P^{ - 1}$, it follows that

\begin{align*} {\mathcal {P}}(u(0))& ={\mathcal {P}}\left({\mathcal {P}}u(0) +\frac{1}{T}{\mathcal {P}}\int_0^T {({N_f}u)(\tau )\,{\rm d}\tau } - L_P^{ - 1}\left( \int_0^T{{c}(s)\,{\rm d}s}\right)\right)\\ & ={\mathcal {P}}(u(0)) +\frac{1}{T}{\mathcal {P}}\left(\int_0^T {({N_f}u)(\tau )\,{\rm d}\tau }\right), \end{align*}

which yields

(3.10)\begin{equation} \frac{1}{T}{\mathcal {P}}\left(\int_0^T {({N_f}u)(\tau )\,{\rm d}\tau }\right)=0. \end{equation}

By (3.9), (3.10) and the definition of $\hat {\mathcal {Q}}$, we have

\[ {\left( {{\phi _p}({{u'(t)}})} \right)^\prime } ={-} ({\rm{I} - \hat {\mathcal {Q}}})({N_f}u)(t) ={-} {f (t,u(t),u'(t))} + { \hat {\mathcal {Q}}}({N_f}u)={-} {f (t,u(t),u'(t))}. \]

Noting the definition of $X$, we know that $u$ is a solution of (H_Q).

On the other hand, assume that $u \in X$ is a solution of RPBVP (H_Q), i.e.,

\[ \left\{ \begin{array}{@{}l} -({\phi _p}(u'))' = f(t,u,u'),\\ u(T) = Qu(0),u'(T) = Qu'(0). \end{array} \right. \]

Similar to the previous discussion, we obtain that

(3.11)\begin{align} & {\mathcal {P}}\int_0^T {({N_f(u))(t)}\,{\rm d}t} =0, \end{align}

(3.12)\begin{align} & {\mathcal {P}}\left(\int_0^T {{\phi _q}\left( { - \int_0^t {(N_f(u))(s)\,{\rm d}s} + {\mathcal {P}}{\phi _p}(u'(0)) + L_P^{ - 1}\int_0^T {(N_f(u))(s)\,{\rm d}s} } \right)\,{\rm d}t}\right) = 0, \end{align}

and

(3.13)\begin{equation} u(t) = {\hat {\mathcal {P}}}(u) - L_P^{ - 1}\left( \int_0^T{a(s)\,{\rm d}s}\right)+\int_0^t {a(s)\,{\rm d}s}, \end{equation}

where

\[ a(t)={\phi _q}\left( { - \int_0^t {(N_f(u))(s)\,{\rm d}s} + {\mathcal{P}}{\phi _p}(u'(0)) + L_P^{ - 1}\int_0^T {(N_f(u))(s)\,{\rm d}s} } \right). \]

Due to (3.11), we have

\[ ({\rm{I} - \hat {\mathcal {Q}}})({N_f}u)(t)=({N_f}u)(t). \]

According to the definition of $\tilde \gamma$ and (3.12), we get that

\[ {\mathcal {P}}{\phi _p}(u'(0))=\tilde \gamma \left( {({\rm{I} - \hat {\mathcal {Q}}})({N_f}u)(t)} \right). \]

From (3.13), it follows that

\[ u(t) = {\hat {\mathcal {P}}}(u) +{\hat {\mathcal {Q}}}({N_f}u) - L_P^{ - 1}\left( \int_0^T{{c}(s)\,{\rm d}s}\right)+\int_0^t {{c}(s)\,{\rm d}s}, \]

where

\begin{align*} {c}(s) & ={\phi _q}\left( \vphantom{\int_0^T} - \int_0^s {({\rm{I} - \hat {\mathcal {Q}}})({N_f}u)(t)\,{\rm d}t}\right.\\ & \left.\quad + L_P^{ - 1}\int_0^T {({\rm{I} - \hat {\mathcal {Q}}})({N_f}u)(t)\,{\rm d}t} +\tilde \gamma \left( {({\rm{I} - \hat {\mathcal {Q}}})({N_f}u)(t)} \right)\right) . \end{align*}

Hence, we obtain

\[ u=H(u), \]

i.e., $u$ is a fixed point of operator $H$.

4. A new continuation theorem

In this section, we build a new continuation theorem for studying the existence of solutions of RPBVP (H_Q).

Theorem 4.1 Suppose that $\Omega$ is an open bounded set in $X$ such that the following conditions hold.

1. (a) For $\forall \ \lambda \in (0,\,1),$ the problem

   (4.1)\begin{equation} \left\{ \begin{array}{@{}l} -({\phi _p}(u'))' = \lambda f(t,u,u'),\\ u(T) = Qu(0),u'(T) = Qu'(0), \end{array} \right. \end{equation}
   has no solution on $\partial \Omega$.

2. (b) Assume that $\ker {\rm {(I\ }\hbox{-}{\rm \ Q)}} \ne \{0\},$ the equation

(4.2)\begin{equation} F(a): = \frac{1}{T} {\mathcal {P}}\left( \int_0^T {f(t,a,0)\,{\rm d}t} \right ) = 0, \end{equation}

has no solution on $\partial \Omega \cap \ker {\rm {(I\ }\hbox{-}{\rm \ Q)}},$ and the Brouwer degree

(4.3)\begin{equation} {\deg _B}(F,\Omega \cap \ker {\rm{(I - Q)}},0) \ne 0, \end{equation}

where the orthogonal projector ${\mathcal {P}}:{\mathbb {R}^N} \to \ker {\rm (I\ }\hbox{-}{\rm \ Q)}$.

Then RPBVP (H_Q) has at least one solution in $\bar \Omega$.

Proof. Consider the following homotopy boundary value problem with (H_Q)

(4.4)\begin{equation} \left\{ \begin{array}{@{}l} -({\phi _p}(u'))' = \lambda({N_f}u)(t) + (1-\lambda)\dfrac{1}{T}{\mathcal {P}}\int_0^T {({N_f}u)(\tau )\,{\rm d}\tau },\\ u(T) = Qu(0),u'(T) = Qu'(0), \end{array} \right. \end{equation}

where $({N_f}u)(t)=f(t,\,u,\,u'),\, \lambda \in [0,\,1]$. For $\lambda \in (0,\,1]$, if $u$ is a solution to problem (4.4), then by integrating both sides of (4.4) over $[0,\,T]$, we have

\begin{align*} {\phi _p}(u'(0)) - {\phi _p}(u'(T)) & = {\rm(I-Q)}({\phi _p}(u'(0))) \\ & = \lambda \int_0^T {({N_f}u)(\tau )\,{\rm d}\tau } +(1-\lambda){\mathcal {P}} \left( \int_0^T {({N_f}u)(\tau )\,{\rm d}\tau }\right ). \end{align*}

Taking ${ \mathcal {P}}$ to act on above equation, one has

(4.5)\begin{equation} {\mathcal {P}} \left( \int_0^T {({N_f}u)(\tau )\,{\rm d}\tau }\right )=0. \end{equation}

Similarly, if $u$ is a solution to problem (4.1), then

\[ {\mathcal {P}} \left( \int_0^T {({N_f}u)(\tau )\,{\rm d}\tau }\right )=0. \]

Hence, for $\lambda \in (0,\,1]$, problems (4.1) and (4.4) have the same solutions. Define homotopy operator ${N}:X \times [0,\,1] \to Y$ by

\[ N(u,\lambda )=\lambda({N_f}u)(t) + (1-\lambda)\frac{1}{T}{\mathcal {P}}\int_0^T {({N_f}u)(\tau )\,{\rm d}\tau }=\lambda {N_f}u + (1-\lambda)\hat {\mathcal {Q}}({N_f}u)(t) . \]

From lemma 3.3, problem (4.4) can be written by the equivalent operator equation

(4.6)\begin{equation} u={H_\lambda}u, \end{equation}

where

\[ {H_\lambda}u = {\hat {\mathcal {P}}}(u) +{\hat {\mathcal {Q}}}({N_f}u) - L_P^{ - 1}\left( \int_0^T{{c_\lambda}(s)\,{\rm d}s}\right)+\int_0^t {{c_\lambda}(s)\,{\rm d}s}), \]

and

\[ {c_\lambda}(s) ={\phi _q}\left( { - \int_0^s {\lambda({\rm{I} - \hat {\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t} + L_P^{ - 1}\int_0^T {\lambda({\rm{I} - \hat {\mathcal{Q}}})({N_f}u)(t)\,{\rm d}t} +\tilde \gamma } \right), \]

$\tilde \gamma = \tilde \gamma (\lambda ({\rm {I}\ }\hbox{-}{\rm \hat {\mathcal {Q}}})({N_f}u))$ is defined in (2.22).

Assume that for ${\lambda }=1$, the problem (4.6) has no solution on $\partial \Omega$ otherwise the proof is complete. Due to hypothesis (i) we know that (4.6) has no solutions for $(u,\,\lambda ) \in \partial \Omega \times (0,\,1]$. For $\lambda =0$, (4.4) is the form

(4.7)\begin{equation} \left\{ \begin{array}{@{}l} -({\phi _p}(u'))' = \dfrac{1}{T}{\mathcal {P}}\left(\int_0^T {({N_f}u)(\tau )\,{\rm d}\tau }\right),\\ u(T) = Qu(0),u'(T) = Qu'(0). \end{array} \right. \end{equation}

Now we claim the problem (4.7) has no solution on $\partial \Omega \times 0$. If $u$ is a solution of problem (4.7), then $u$ satisfies (4.5), which shows $u'(t) = {\phi _q}(\alpha )$, where $\alpha \in {\mathbb {R}^N}$. By lemma 2.5, we have $\alpha =0$ and $u(t)=\beta \, (\beta \in \ker {\rm {(I\ }\hbox{-}{\rm \ Q)}})$. It follows from (4.5) that

\[ {\mathcal {P}}\left(\int_0^T {f(\tau ,\beta,0)\,{\rm d}\tau } \right) = 0, \]

which, together with hypothesis (ii), implies that $u=\beta \notin \partial \Omega$. Thus we have proved that (4.6) has no solution $(u,\,\lambda ) \in \partial \Omega \times [0,\,1]$.

By lemma 3.2, ${H_\lambda }$ is a completely continuous operator. Then we have that for each $\lambda \in [0,\,1]$, the Leray-Schauder degree ${\deg _{LS}}(I - {H_\lambda },\,\Omega,\,0)$ is well defined, and

(4.8)\begin{equation} {\deg _{LS}}(I - {H_1},\Omega ,0) = {\deg _{LS}}(I - {H_0},\Omega ,0). \end{equation}

It is clear that the operator equation

(4.9)\begin{equation} u = {H_1}(u) \end{equation}

is equivalent to the problem (H_Q). Now, we only prove that ${\deg _{LS}}(I - {H_0}, \Omega,\,0) \ne 0$.

Because

\[ (I - {H_0})u=u- {\hat {\mathcal {P}}}(u) -\frac{1}{T}{\mathcal{P}}\left(\int_0^T {({N_f}u)(\tau )\,{\rm d}\tau } \right ), \]

$(I - {H_0})u=0$ deduces to $u= {\hat {\mathcal {P}}}(u) -\frac {1}{T}{\mathcal {P}}(\int _0^T {({N_f}u)(\tau )\,{\rm d}\tau })$, which from the definitions of ${\mathcal {P}}$ and ${\hat {\mathcal {P}}}$ yields $u=c$ in $\Omega$. On the basis of lemma 2.5, we have $c \in \ker \rm {(I - Q)}$. Hence by the properties of the Leray-Schauder degree and (4.3), we get that

\begin{align*} {\deg _{LS}}(I - {H_0},\Omega ,0) & ={\deg _{LS}}(I - {H_0},\Omega \cap \ker {\rm{(I - Q)}}),0) \\ & ={\deg _B}({-}F,\Omega \cap \ker {\rm{(I - Q)}}) ,0) \ne 0, \end{align*}

where the function $F$ is defined in (4.2). Then ${\deg _{LS}}(I - {H_1},\,\Omega,\,0) \ne 0,$ that is, RPBVP (H_Q) has at least one solution in $\bar \Omega$.

Theorem 4.3 Suppose that $\ker {\rm {(I\ }\hbox{-}{\rm \ Q)}} = \{0\},$ and $\Omega$ is an open bounded set in $X$ such that $0 \in \Omega$ and the problem

(4.10)\begin{equation} \left\{ \begin{array}{@{}l} -({\phi _p}(u'))' = \lambda f(t,u,u'),\\ u(T) = Qu(0),u'(T) = Qu'(0), \end{array} \right. \end{equation}

has no solution on $\partial \Omega,$ for $\forall \ \lambda \in (0,\,1)$.

Then RPBVP (H_Q) has at least one solution in $\bar \Omega$.

Proof. Since $\ker {\rm {(I\ }\hbox{-}{\rm \ Q)}} = \{0\}$, there exists ${\rm {(\ {I\ }\hbox{-}{\rm \ Q}\ )}}^{-1}$. Define $\tilde H : X \to X$ by

(4.11)\begin{align} \tilde Hu & = \int_0^t {{\phi _q}\left( { - \int_0^s {({N_f}u)(\tau )\,{\rm d}\tau } + {{\left( {I - Q} \right)}^{ - 1}}\int_0^T {({N_f}u)(\tau )\,{\rm d}\tau } } \right)\,{\rm d}s}\nonumber\\ & \quad - {\left( {I - Q} \right)^{ - 1}}\int_0^T {{\phi _q}\left( { - \int_0^s {({N_f}u)(\tau )\,{\rm d}\tau } + {{\left( {I - Q} \right)}^{ - 1}}\int_0^T {({N_f}u)(\tau )\,{\rm d}\tau } } \right)\,{\rm d}s}. \end{align}

Then the RPBVP (H_Q) is equivalent to the operator equation

\[ u=\tilde Hu. \]

Similar to the lemmas 3.1–3.2, we can prove that $\tilde H$ is a completely continuous operator from $X$ to $X$. Furthermore, define ${\tilde H_\lambda }$ by

(4.12)\begin{align} \tilde H_\lambda u & = \int_0^t {{\phi _q}\left( { - \int_0^s {\lambda ({N_f}u)(\tau )\,{\rm d}\tau } + {{\left( {I - Q} \right)}^{ - 1}}\int_0^T {\lambda ({N_f}u)(\tau )\,{\rm d}\tau } } \right)\,{\rm d}s}\nonumber\\ & \quad - {\left( {I \,{-}\, Q} \right)^{ - 1}}\int_0^T {{\phi _q}\left(\!{ - \int_0^s {\lambda ({N_f}u)(\tau )\,{\rm d}\tau } \,{+}\, {{\left( {I \,{-}\, Q} \right)}^{ - 1}}\int_0^T {\lambda ({N_f}u)(\tau )\,{\rm d}\tau } } \right){\rm d}s}, \end{align}

for $\lambda \in [0,\,1]$. We assume that for ${\lambda }=1$, (4.10) has no solution on $\partial \Omega$ otherwise the proof is complete. For ${\lambda }=0$, the Eqn (4.10) only has zero solution via lemma 2.5 and $\ker {\rm {(I\ }\hbox{-}{\rm \ Q)}} = \{0\}$. By hypothesis, for each $\lambda \in [0,\,1]$, the Leray-Schauder degree ${\deg _{LS}}(I - {\tilde H_\lambda },\,\Omega,\,0)$ is well defined, and

\[ {\deg _{LS}}(I - {\tilde H_1},\Omega ,0) = {\deg _{LS}}(I - {\tilde H_0},\Omega ,0)={\deg _{LS}}(I,\Omega ,0)=1. \]

Hence, the RPBVP (H_Q) has at least one solution in $\bar \Omega$.

5. Applications

In this section, we take useful of theorem 4.1 to further discuss the sufficient conditions of existence of solutions for the two kinds of RPBVP (H_Q).

5.1 Existence of solutions for a kind of the RPBVP (H_Q)

Theorem 5.1 Assume that $\ker \rm (I-Q) \ne \{0\}$ and the following conditions hold.

$(f_1)$ There exist $h \in {L^1}([0,\,T],\,{\mathbb {R}_ + })$ and $n \in {C^1}({\mathbb {R}^N},\,{\mathbb {R}^N})$ satisfying $n'(x)$ is negative semi-definite and $n(Qx)=Qn(x)$ for each $x \in {\mathbb {R}^N},$ such that

(5.1)\begin{equation} \left| {f(t,x,y)} \right| \le \left\langle {f(t,x,y),n(x)} \right\rangle + h(t), \end{equation}

for any $x,\,y \in {\mathbb {R}^N},$ and a.e. $t \in [0,\,T]$.

$(f_2)$ $f$ satisfies a generalized Villari-type condition, i.e. there exists a constant $M>0$ such that for all $u \in X$ with $\mathop {\min }\limits _{t \in [0,T]} \left | {{u}(t)} \right | > M$,

(5.2)\begin{equation} {\mathcal{P}}\left(\int_0^T {f(t,u,u')\,{\rm d}t}\right) \ne 0, \end{equation}

where ${\mathcal {P}}:{\mathbb {R}^N} \to \ker {\rm (I\ }\hbox{-}{\rm \ Q)}$.

Then the problem (H_Q) has at least one solution.

Proof. First we take a priori estimate for solutions of (4.1). Let $(u,\,\lambda ) \in X \times (0,\,1)$ be a solution to problem (4.1). Then we have

(5.3)\begin{equation} {\phi _p}(u'(t)) ={-} \int_0^t {\lambda f(s,u,u')\,{\rm d}s} + {\mathcal {P}}({\phi _p}(u'(0))) + L_P^{ - 1}\int_0^T {\lambda f(s,u,u')\,{\rm d}s}, \end{equation}

and

\begin{align*} {\rm(I-Q)}u(0)& =\int_0^T {\phi _q}\left( -\int_0^t { \lambda f(s,u,u')\,{\rm d}s} + {\mathcal{P}}({\phi _p}(u'(0)) )\right.\\& \quad \left.+ L_P^{ - 1}\int_0^T {\lambda f(s,u,u')\,{\rm d}s} \right)\,{\rm d}t\in {\rm Im} (I-Q), \end{align*}

where $\frac {1}{p} + \frac {1}{q} = 1(p,\,q > 1)$. Hence,

(5.4)\begin{equation} {\mathcal{P}}\left(\int_0^T {{\phi _q}\left( -{\int_0^t { \lambda f(s,u,u')\,{\rm d}s} + {\mathcal{P}}({\phi _p}(u'(0)) )+ L_P^{ - 1}\int_0^T {\lambda f(s,u,u')\,{\rm d}s} } \right){\rm d}t} \right)\! = 0. \end{equation}

Because $n'(u)$ is negative semi-definite, we obtain that

(5.5)\begin{align} 0 & \ge \int_0^T {\left\langle {{\phi _p}(u'(t)),n'(u(t))u'(t)} \right\rangle \,{\rm d}t} \nonumber\\ & = \left\langle {{\phi _p}(u'(t)),n(u(t))} \right\rangle \left| {_0^T} \right. - \int_0^T {\left\langle {{{\left( {{\phi _p}(u'(t))} \right)}^\prime },n(u(t))} \right\rangle \,{\rm d}t} \nonumber\\ & = \left\langle {{\phi _p}(u'(T)),n(u(T))} \right\rangle - \left\langle {{\phi _p}(u'(0)),n(u(0))} \right\rangle - \int_0^T {\left\langle {{{\left( {{\phi _p}(u'(t))} \right)}^\prime },n(u(t))} \right\rangle \,{\rm d}t}\nonumber\\ & = \left\langle {Q{\phi _p}(u'(0)),Qn(u(0))} \right\rangle - \left\langle {{\phi _p}(u'(0)),n(u(0))} \right\rangle + \int_0^T {\left\langle {\lambda f(t,u,u'),n(u(t))} \right\rangle \,{\rm d}t} \nonumber\\ & = \int_0^T {\left\langle {\lambda f(t,u,u'),n(u(t))} \right\rangle \,{\rm d}t}. \end{align}

Furthermore, we have that

(5.6)\begin{equation} {\mathcal{P}}\left(\int_0^T \lambda{f(t,u,u')\,{\rm d}t}\right) = 0. \end{equation}

By (5.1) and (5.5), we have that

(5.7)\begin{equation} \lambda \int_0^T {\left| {f(t,u,u')} \right|\,{\rm d}t} \le \int_0^T {\left\langle {f(t,u,u'),n(u)} \right\rangle \,{\rm d}t} + \int_0^T {h(t)\,{\rm d}t} \le \int_0^T {h(t)\,{\rm d}t} \buildrel \Delta \over = {M_1}. \end{equation}

According to the definition of $L_P$ , there exists $M_2>0$ such that

(5.8)\begin{equation} \left| {L_P^{ - 1}\int_0^T {\lambda f(t,u,u')\,{\rm d}t} } \right| \le {M_2}. \end{equation}

From (5.6) and (5.7), it follows that $\lambda f(t,\,u,\,u') \in Y_1$ is $L^1$-bounded for any solution of (4.1). According to the definition of $G_h (\gamma )$ , proposition 2.6 , (5.4),(5.7) and (5.8), we have that $\tilde \gamma = {\mathcal {P}}({\phi _p}(u'(0)))$ is bounded, i.e., there exists $M_3>0$ such that

\[ \left| {{\mathcal{P}}({\phi _p}(u'(0))}) \right| \le {M_3}. \]

Hence for any $t \in [0,\,T]$, we have

\begin{align*} \left| {{\phi _p}(u'(t))} \right| & \le \left| {\int_0^t {\lambda f(s,u,u')\,{\rm d}s} } \right| + \left| {{\mathcal{P}}({\phi _p}(u'(0)))} \right| + \left| {L_P^{ - 1}\int_0^T {\lambda f(s,u,u')\,{\rm d}s} } \right|\\ & \le {M_1} + {M_2} + {M_3}. \end{align*}

In the light of the definition of ${\phi _q}$ , there exists $M_4>0$ such that

\[ {\left\| {u'} \right\|_0} \le {M_4}. \]

Thanks to (5.6) and hypothesis $(f_2)$, there exists $t_j \in [0,\,T]$ such that $\left | {u({t_j})} \right | < M$, and

\[ \left| {u(t)} \right| = \left| {u({t_j})} \right| + \left| {\int_{{t_j}}^t {u'(s)\,{\rm d}s} } \right| \le M + T{M_4}={M_5}. \]

It follows that

\[ {\left\| {u} \right\|_1} \le M_4+M_5 \buildrel \Delta \over = {r}. \]

Let $\Omega _0 =\left \{ {u \in X\left | {{{\left \| u \right \|}_1} < r + 1} \right.} \right \}$. Then condition (i) of theorem 4.1 is satisfied.

Take constant $\alpha \in X$, then $\alpha \in \ker {\rm (I}\hbox{-}{\rm Q)}$. By hypothesis $(f_2)$, one of the following conditions holds:

(5.9)\begin{equation} \left\langle {{\mathcal{P}}\left(\int_0^T {f(t,\alpha ,0)\,{\rm d}t}\right),\alpha } \right\rangle > 0, \quad \left| \alpha \right| > M, \end{equation}

or

(5.10)\begin{equation} \left\langle {{\mathcal{P}}\left(\int_0^T {f(t,\alpha ,0)\,{\rm d}t}\right),\alpha } \right\rangle <0, \quad \left| \alpha \right| > M. \end{equation}

In the case (5.9), define the following homotopy mapping:

\[ {H_\mu }(\alpha )=\mu \alpha + (1 - \mu ){\mathcal{P}}\left(\int_0^T {f(t,\alpha ,0)\,{\rm d}t }\right); \]

in the case (5.10), define the following homotopy mapping:

\[ {H_\mu }(\alpha )={-}\mu \alpha + (1 - \mu ){\mathcal{P}}\left(\int_0^T {f(t,\alpha ,0)\,{\rm d}t }\right), \]

where $\mu \in [0,\,1]$. It is easy to check that the solution of ${H_\mu }(\alpha )=0$ must be in $\Omega _1 \cap \ker \rm (I - Q)$, where $\Omega _1 = \left \{ {u \in X\left | {{{\left \| u \right \|}_1} < M + 1} \right.} \right \}$. Then we have that

\begin{align*} {\deg _B}({H_\mu }(\alpha ) ,\Omega_1 \cap \ker \rm (I - Q),0)& ={\deg _B}({\mathcal{P}}\left(\int_0^T {f(t,\alpha ,0)\,{\rm d}t}\right),\Omega_1 \cap \ker \rm (I - Q),0)\\ & = {\deg _B}({\pm} {\rm I},\Omega_1 \cap \ker \rm (I - Q),0) \ne 0. \end{align*}

Thus the condition (ii) of theorem 4.1 is satisfied with $\Omega _1$.

Finally, take

\[ \Omega = \left\{ {u \in X\left| {{{\left\| u \right\|}_1} < \max \{ r+1,M + 1\} } \right.} \right\}. \]

Then conditions (i) and (ii) of theorem 4.1 are satisfied on $\Omega$, which leads to the problem (H_Q) has at least one solution.

Remark 5.2 The Villari condition was first introduced for the scalar case by Villari in [Reference Villari13], i.e., there exists a $k>0$ such that for all $u \in C^1([0,\,T],\, \mathbb {R})$ with $\mathop {\min }\limits _{t \in [0,T]} \left | {u(t)} \right | \ge k$,

\[ {\mathop{\rm sgn}} (u)\int_0^T {f(t,u,u')\,{\rm d}t} \ge 0. \]

Obviously, the above condition requires $u$ and $\int _0^T {f(t,\,u,\,u')\,{\rm d}t}$ to be the same sign. But our conditions do not require that.

In [Reference Manásevich and Mawhin8], Man$\acute {a}$sevich and Mawhin gave the generalized Villari condition for periodic problem, i.e., there exists a $k>0$ such that for all $u \in C_T^1$, $u = ({u_1},\, \cdots,\,{u_N})$, with $\mathop {\min }\limits _{t \in [0,T]} \left | {u_j(t)} \right | \ge k$, for some $j \in \{ 1,\, \cdots,\,N\}$,

\[ \int_0^T {{f_i}(t,u,u')\,{\rm d}t} \ne 0, \]

for some $i \in \{ 1,\, \cdots,\,N\}$. However, this does not lead to the condition (ii) of theorem 4.1.

Corollary 5.3 Assume that $\ker \rm (I-Q) \ne \{0\}$ and the following conditions hold.

1. (1) The condition $(f_1)$ of theorem 5.1 holds.

2. (2) There exist $h_1 \in {L^1}([0,\,T],\,{\mathbb {R}_ + })$ and $\alpha :[0,\, + \infty ) \to [0,\, + \infty )$ such that $\alpha (s) \to +\infty$ as $s \to +\infty$ and

   (5.11)\begin{equation} \alpha (\left| x \right|) - {h_1}(t) \le \left| {{\mathcal{P}} f(t,x,y)} \right| \end{equation}
   for almost all $t \in [0,\,T]$ and all $x ,\,y \in \mathbb {R}^N$.

3. (3) Condition (4.3) holds.

Then the problem (H_Q) has at least one solution.

Proof. Let $(u,\, \lambda )$, $\lambda \in (0,\,1)$ be a solution for problem (4.1). As in the proof of theorem 5.1, it follows from 1) that there is $M_1 >0$ such that ${\left \| {u'} \right \|_0} \le M_1$. We claim that 1) and (5.11) imply that there exists $M_2 >0$ such that ${\left \| {u} \right \|_0} \le M_2$. In fact, by (5.1) and (5.5), we have $\int _0^T {\left | {f(t,\,u,\,u')} \right |\,{\rm d}t} \le {\left \| h \right \|_{{L^1}}}$. From (5.11) and $\left | {{\mathcal {P}}x} \right | \le \left | x \right |$, it follows that

\begin{align*} \int_0^T {\alpha (\left| u \right|)\,{\rm d}t} & \le \int_0^T {\left| {{\mathcal{P}}f(t,u,u')} \right|\,{\rm d}t} + {\left\| {{h_1}} \right\|_{{L^1}}} \le \int_0^T {\left| {f(t,u,u')} \right|\,{\rm d}t} + {\left\| {{h_1}} \right\|_{{L^1}}}\\ & \le {\left\| h \right\|_{{L^1}}} + {\left\| {{h_1}} \right\|_{{L^1}}}. \end{align*}

Since $\alpha (s) \to +\infty$ as $s \to +\infty$, we find the required bound for ${\left \| {u} \right \|_0}$.

Now let $a \in \ker \rm (I-Q)$ such that ${\mathcal {P}}\int _0^T {f(t,\,a,\,0)\,{\rm d}t} = 0$. By (5.11), we get that $\alpha (\left | a \right |) \le {M_3}$, and hence $\left | a \right | \le {M_4}$. Here $M_3$ and $M_4$ are positive constants. Thus there is $r>0$ such that all solution to (4.2) belongs to $\Omega = \left \{ {a \in \ker (I - Q):\left | a \right | < r} \right \}$. The rest of the proof follows from the theorem 5.1.

Example 5.4 Now, we give a simple example for corollary 5.3. Consider the following p-Laplacian differential systems

(5.12)\begin{equation} \left\{ \begin{array}{@{}l} - {\left( {{{\left| {u'} \right|}^{p - 2}}{u'_1}} \right)^\prime } + {u_1}\left( {1 + u_1^2} \right) + {u_1}\left( {1 + u_2^2} \right)\\ \quad = {e_1}(t),\ - {\left( {{{\left| {u'} \right|}^{p - 2}}{u'_2}} \right)^\prime } + {u_2}\left( {1 + u_2^2} \right) = {e_2}(t), \end{array} \right. \end{equation}

with rotating periodic boundary conditions

\[ u(T) = Qu(0),\quad u'(T) = Qu'(0). \]

where $u(t) = \left( \begin{array}{@{}l@{}} {u_1}(t)\\ {u_2}(t) \end{array} \right)$, $e(t) = \left( \begin{array}{@{}l@{}} {e_1}(t)\\ {e_2}(t) \end{array} \right) \in {L^1}(0,\,T; \mathbb {R}^2)$ and $Q = \left(\begin{array}{@{}cc@{}} 1 & 0\\ 0 & { - 1} \end{array}\right)$.

Then we have $\rm {(I - Q)} = \left( {\begin{array}{*{20}{c}} 0 & 0\\ 0 & 2 \end{array}}\right )$, $\ker {\rm (I\ }\hbox{-}{\rm \ Q)} = a\left( \begin{array}{@{}l@{}} 1\\ 0 \end{array} \right)$ and ${{\rm Im}} {\rm (I\ }\hbox{-}{\rm \ Q)} = a\left( \begin{array}{@{}l@{}} 0\\ 1 \end{array} \right)$, $a \in \mathbb {R}$.

Let ${\mathcal {P}}\left( \begin{array}{@{}l@{}} {x_1}\\ {x_2} \end{array} \right) = \left( \begin{array}{@{}l@{}} {x_1}\\ 0 \end{array} \right)$, $n(x) = \left( \begin{array}{@{}l@{}} - 2{x_1} - 2{x_2} \end{array} \right)$ and $\alpha ( {\left | x \right |} ) = \left | x \right |$, for $x \in \mathbb {R}^2$.

Set

\[ f(t,x) = \left( \begin{array}{@{}l@{}} - {x_1}(1 + x_1^2) - {x_1}(1 + x_2^2) + {e_1}(t) - {x_2}(1 + x_2^2) + {e_2}(t) \end{array} \right) . \]

Obviously, for a.e. $t \in [0,\,T]$ and $x \in \mathbb {R}^2$, we have

\[ - \left| {e(t)} \right| + \left| x \right| \le \left| {{\mathcal{P}}f(t,x)} \right| \le \left| {f(t,x)} \right| \le 2{\left| x \right|^3} + \left| x \right| + \left| {e(t)} \right|, \]

as $\left | {{x_1}} \right | \ge 1$ and $\left | {{x_2}} \right | \ge 1$. Hence for some $l(t) \in {L^1}([0,\,T],\,{\mathbb {R}_ + })$, we have

\[ \left| {f(t,x)} \right| \le 2{\left| x \right|^3} + \left| x \right| + \left| {l(t)} \right|, \]

for a.e. $t \in [0,\,T]$ and all $x \in \mathbb {R}^2$. On the other hand,

\[ \left\langle {f(t,x),n(x)} \right\rangle \ge {\left| x \right|^4} + 2{\left| x \right|^2} - 2\left| x \right|\left| e \right|, \]

for a.e. $t \in [0,\,T]$ and all $x \in \mathbb {R}^2$. Thus for a.e. $t \in [0,\,T]$ and all $x \in \mathbb {R}^2$, we can choose $h(t) \in {L^1}([0,\,T],\,{\mathbb {R}_ + })$ such that

\[ \left| {f(t,x)} \right| \le \left\langle {f(t,x),n(x)} \right\rangle + h(t). \]

Next, for $b \in \ker {\rm (I\ }\hbox{-}{\rm \ Q)}$, we have

\[ F(b) = \left( \begin{array}{@{}l@{}} - {b_1}(1 + b_1^2) - {b_1} + \dfrac{1}{T}\int_0^T {{e_1}(t)\,{\rm d}t} \\ 0 \end{array} \right). \]

By the properties of the Brouwer degree, we have for sufficiently large $r>0$

\[ {\deg_B}\left( {F(b),\Omega (r),0} \right) = 1, \]

where $\Omega (r) = \left \{ {b \in \ker (I - Q):\left | b \right | < r} \right \}$. Hence, the RPBVP (5.12) has at least one solution.

5.2 Existence of solutions of RPBVP (H_Q) for the p-Laplacian Liénard-type system

Consider the following p-Laplacian Li$\acute {e}$nard-type system with the rotating periodic boundary conditions:

(5.13)\begin{equation} \left\{ \begin{array}{@{}l@{}} {\left( {{\phi _p}(u'(t))} \right)^\prime } + {\left( {\nabla F(u)} \right)^\prime } + Au(t) = e(t),\\ u(T) = Qu(0),u'(T) = Qu'(0), \end{array} \right. \end{equation}

where $p \ge 2$, $Q$ is an $N \times N$ orthogonal matrix with $\ker {\rm (I}\hbox{-}{\rm Q)} \ne \{0\}$, $A$ is an $N \times N$ matrix with $AQ=QA$, $F \in {C^2}({\mathbb {R}^N},\,\mathbb {R})$ with $F(u) = F(\left | u \right |)$, $e \in L^2$ with $e(t+T)=Qe(t)$.

In [Reference Mawhin10], Mawhin studied the T-periodic solutions of the following p-Laplacian Li$\acute {e}$nard system:

\[ \left\{ \begin{array}{@{}l@{}} {\left( {{\phi _p}(u')} \right)^\prime } + {\left( {\nabla F(u)} \right)^\prime } + Au = e(t),\\ u(0) = u(T),u'(0) = u'(T). \end{array} \right. \]

And the author obtained some existence theorems for the above problem.

Next, we extend periodic boundary value conditions to rotating periodic boundary conditions and give some existence results for (5.13).

Theorem 5.5 Assume that $A$ is a negative definite matrix and satisfies ${\mathcal {P}}A (\alpha )=A{\mathcal {P}} (\alpha )$ for any $\alpha \in \mathbb {R}^N,$ where ${\mathcal {P}}:{\mathbb {R}^N} \to \ker {\rm (I\ }\hbox{-}{\rm \ Q)}$. Then for each $e \in L^2,$ problem (5.13) has at least one solution.

Proof. To apply theorem 4.1, we consider the auxiliary RPBVP:

(5.14)\begin{equation} \left\{ \begin{array}{@{}l@{}} {\left( {{\phi _p}(u'(t))} \right)^\prime } + \lambda {\left( {\nabla F(u)} \right)^\prime } + \lambda Au(t) = \lambda e(t),\\ u(T) = Qu(0),u'(T) = Qu'(0), \end{array} \right. \end{equation}

where $\lambda \in (0,\,1]$.

First, we make a prior estimate. Let $(u,\,\lambda ) \in X \times (0,\,1]$ be solution of (5.14). Integrating (5.14) over $[0,\,T]$, we get that

\[ {\rm(I - Q)}{\phi _p}(u'(0)) + {\rm(I - Q)}\lambda \frac{{d F}}{{d u}}(\left| {u(0)} \right|)\frac{{u(0)}}{{\left| {u(0)} \right|}} + \lambda \int_0^T e (t)\,{\rm d}t = \lambda A\int_0^T u (t)\,{\rm d}t. \]

Taking ${\mathcal {P}}$ to act on the above equation, we have

\[ {\mathcal{P}}\left(\int_0^T e (t)\,{\rm d}t\right) = A {\mathcal{P}}\left( \int_0^T u (t)\,{\rm d}t\right). \]

Let $\bar e ={\mathcal {P}}(\frac {1}{T}\int _0^T {e(t)\,{\rm d}t})$ and $\bar u ={\mathcal {P}}(\frac {1}{T}\int _0^T {u(t)\,{\rm d}t})$, then

(5.15)\begin{equation} \left| {\bar u} \right| = \left| {{A^{ - 1}}\bar e} \right| \le \left| {{A^{ - 1}}} \right|\left| {\bar e} \right|. \end{equation}

Now taking the inner product for the both side of (5.14) by $u$ and integrating over $[0,\,T]$, we obtain

\begin{align*} \int_0^T {\left\langle {{{\left( {{\phi _p}(u'(t))} \right)}^\prime },u(t)} \right\rangle \,{\rm d}t} & = \left\langle {\left( {{\phi _p}(u'(t))} \right),u(t)} \right\rangle \left| {_0^T} \right. - \int_0^T {\left\langle {\left( {{\phi _p}(u'(t))} \right),u'(t)} \right\rangle \,{\rm d}t} \\ & = \left\langle {Q{\phi _p}(u'(0)),Qu(0)} \right\rangle - \left\langle {{\phi _p}(u'(0)),u(0)} \right\rangle\\ & \quad - \int_0^T {{{\left| {u'(t)} \right|}^p}\,{\rm d}t}\\ & ={-} \int_0^T {{{\left| {u'(t)} \right|}^p}\,{\rm d}t},\\ \int_0^T {\left\langle {{{\left( {\nabla F(u)} \right)}^\prime },u(t)} \right\rangle \,{\rm d}t} & = \left\langle {\nabla F(u),u(t)} \right\rangle \left| {_0^T} \right. - \int_0^T {\nabla F(u)du} \\ & = \left\langle {Q\frac{{d F}}{{d u}}(\left| {u(0)} \right|)\frac{{u(0)}}{{\left| {u(0)} \right|}},Qu(0)} \right\rangle\\ & \quad - \left\langle {\frac{{d F}}{{d u}}(\left| {u(0)} \right|)\frac{{u(0)}}{{\left| {u(0)} \right|}},u(0)} \right\rangle \\ & \quad - F(u(T)) + F(u(0))\\ & =F(u(0))-F(\left| {Qu(0)} \right|)=0, \end{align*}

and

\[ \int_0^T {\left\langle {e(t),u(t)} \right\rangle \,{\rm d}t} = T\left\langle {\bar e,\bar u} \right\rangle + \int_0^T {\left\langle {\tilde e(t),\tilde u(t)} \right\rangle \,{\rm d}t}. \]

Then we have

(5.16)\begin{equation} \int_0^T {{{\left| {u'(t)} \right|}^p}\,{\rm d}t}-\lambda \int_0^T {\left\langle {Au(t),u(t)} \right\rangle \,{\rm d}t}={-} \lambda T\left\langle {\bar e,\bar u} \right\rangle - \lambda \int_0^T {\left\langle {\tilde e(t),\tilde u(t)} \right\rangle \,{\rm d}t}, \end{equation}

where $\tilde e(t) = e(t) - \bar e = e(t) - {\mathcal {P}}(\frac {1}{T}\int _0^T {e(t)\,{\rm d}t} )$ and $\tilde u(t) = u(t) - \bar u = u(t) - {\mathcal {P}}(\frac {1}{T}\int _0^T {u(t)\,{\rm d}t} )$, which yield $\int _0^T {\tilde e(t)\,{\rm d}t},\, \int _0^T {\tilde u(t)\,{\rm d}t} \in {\mathop {\rm Im}\nolimits } {\rm (I\ }\hbox{-}{\rm \ Q)}$. According to the assumption of $A$ and (5.15), we get that

(5.17)\begin{equation} \int_0^T {{{\left| {u'(t)} \right|}^p}\,{\rm d}t} \le T\left| {{A^{ - 1}}} \right|\left| {\bar e} \right|^2 + N{\left\| {\tilde e} \right\|_{{L^1}}}{\left\| {\tilde u} \right\|_0}. \end{equation}

For $\tilde u$, it follows from Sobolev inequality that

(5.18)\begin{equation} {\left\| {\tilde u} \right\|_0} \le M{\left\| {\tilde u'} \right\|_{{L^2}}}= M{\left\| { u'} \right\|_{{L^2}}}. \end{equation}

Next we claim there exists $M_2>0$ such that

(5.19)\begin{equation} {\left\| {u'} \right\|_{{L^2}}} \le {M_2}. \end{equation}

If this is false, there are ${\lambda _n} \in (0,\,1]\,(n = 1,\,2,\, \cdots )$ such that corresponding solutions $u_n$ satisfy ${\left \| {u'_n} \right \|_{{L^2}}} \to \infty$ $(n \to \infty )$. By (5.17), (5.18) and $p \ge 2$, we have

\begin{align*} \left\| {{u'_n}} \right\|_{{L^2}}^2 & \le {\left( T \right)^{{{(p - 2)}}/{p}}}{\left( {\int_0^T {{{\left| {{u'_n}(t)} \right|}^p}\,{\rm d}t} } \right)^{{2}/{p}}}\\ & \le {\left( T \right)^{{{(p - 2)}}/{p}}}{\left( {T\left| {{A^{ - 1}}} \right|{{\left| {\bar e} \right|}^2} + NM{{\left\| {\tilde e} \right\|}_{{L^1}}}{{\left\| {{u'_n}} \right\|}_{{L^2}}}} \right)^{{2}/{p}}}, \end{align*}

which is a contradiction as $n \to \infty$. From (5.15), (5.18) and (5.19), together with $u(t) = \tilde u(t) + \bar u$, it follows that there exists $M_3>0$ such that

(5.20)\begin{equation} {\left\| {u} \right\|_{{0}}} \le {M_3}. \end{equation}

(5.14) implies that

\[ \left| {{{\left( {{\phi _p}(u'(t))} \right)}^\prime }} \right| \le \left| {\frac{{{d ^2}F(u(t))}}{{d {u_i} d {u_j}}}u'(t)} \right| + \left| A \right|\left| {u(t)} \right| + \sum\limits_{i = 1}^N {\left| {{e_i}(t)} \right|}, \]

for a.e. $t \in [0,\,T]$. And owing to (5.20) and the quality of $F(u)$, we obtain

\[ \left| {{{\left( {{\phi _p}(u'(t))} \right)}^\prime }} \right| \le {M_4}\sum\limits_{i = 1}^N {\left| {{u_i}^\prime (t)} \right|} + \left| A \right|{M_3} + \sum\limits_{i = 1}^N {\left| {{e_i}(t)} \right|}, \]

where $\left | {\frac {{{d ^2}F(u(t))}}{{d {u_i} d {u_j}}}} \right | \le {M_4}$. By H$\ddot o$lder inequality, we have

\[ {\left| {{{\left( {{\phi _p}(u'(t))} \right)}^\prime }} \right|^2} \le 3N{({M_4})^2}{\sum\limits_{i = 1}^N {\left| {{u_i}^\prime (t)} \right|} ^2} + 3{\left| A \right|^2}{({M_3})^2} + 3N\sum\limits_{i = 1}^N {{{\left| {{e_i}(t)} \right|}^2}} . \]

Furthermore,

(5.21)\begin{align} \int_0^T {\left| {{{\left( {{\phi _p}(u'(t))} \right)}^\prime }} \right|^2\,{\rm d}t} & \le 3{({M_4})^2}{N^2} {\left\| u' \right\|_{{L^2}}^2} + 3T{\left| A \right|^2}{({M_3})^2} + 3{N^2}{\left\| e \right\|_{{L^2}}^2}\nonumber\\ & \le 3{({M_4})^2}{N^2}{({M_2})^2} + 3T{\left| A \right|^2}{({M_3})^2} + 3{N^2}{\left\| e \right\|_{{L^2}}^2}\buildrel \Delta \over = {M_5}. \end{align}

Write $v(t)={{\phi _p}(u'(t))}$ and decompose it as $v(t)=\tilde v(t) + \bar v$. We have $\int _0^T {\tilde v(t)\,{\rm d}t} \in {\mathop {\rm Im}\nolimits } {\rm (I\ }\hbox{-}{\rm \ Q)}$ and

\[ u'(t) = {\phi _q}\left( {\tilde v(t) + \bar v} \right), \]

where $\frac {1}{p} + \frac {1}{q} = 1(p,\,q > 1)$. Hence,

\[ {\mathcal{P}}\left(\int_0^T {{\phi _q}\left( {\tilde v(t) + \bar v} \right)\,{\rm d}t} \right) = 0. \]

We deduce, from (5.21) and Sobolev inequality, that

\[ \left\| {\tilde v} \right\|_0^2 \le {M^2}\int_0^T {\left| \tilde v'(t) \right|^2\,{\rm d}t}={M^2}\int_0^T {\left| {{{\left( {{\phi _p}(u'(t))} \right)}^\prime }} \right|^2\,{\rm d}t} \le {M^2}{M_5} \buildrel \Delta \over ={M_6}. \]

From proposition 2.6, it follows that $\left | {\bar v} \right | = \left | {\tilde \gamma (\tilde v(t))} \right |$ is bounded. Therefore

\[ {\left\| {{\phi _p}(u')} \right\|_0} = {\left\| v \right\|_0} \le {\left\| {\tilde v} \right\|_0} + \left| {\bar v} \right| \le {M_7}. \]

Then

\[ {\left\| {u'} \right\|_0} \le {M_8}. \]

So there exists $M_0 >0$ independent of $\lambda$ such that

\[ {\left\| u \right\|_1} = \max \{ {\left\| u \right\|_0},{\left\| {u'} \right\|_0}\} \le {M_0}. \]

Secondly, to check the condition (ii) of theorem 4.1, we see that

\[ F(\alpha ): = {\mathcal{P}}\left(\frac{1}{T}\int_0^T {\left( {e(t) - A\alpha } \right)\,{\rm d}t}\right) = \bar e - A\alpha , \]

where $\alpha \in \ker \rm (I-Q)$. Then $F(\alpha )=0$ has the unique solution $\alpha ={A^{ - 1}} \bar e$ which trivially yields that ${\deg _B}(F,\,B(r),\,0)$ is well defined and equal to $\pm 1$ for all sufficiently large $r>0$, so that condition (ii) of theorem 4.1 is satisfied.

Corollary 5.6 If $A$ is a negative definite matrix and satisfies ${\mathcal {P}}A (\alpha )=A{\mathcal {P}} (\alpha )$ for any $\alpha \in \mathbb {R}^N,$ then for each $e \in L^2,$ the RPBVP

(5.22)\begin{equation} \left\{ \begin{array}{@{}l@{}} {\left( {{\phi _p}(u'(t))} \right)^\prime } + Au(t) = e(t),\\ u(T) = Qu(0),u'(T) = Qu'(0), \end{array} \right. \end{equation}

has an unique solution.

Proof. Only the uniqueness has to proved. Let $u$ and $v$ be solutions of (5.22). Then we have

\begin{align*} & {\left( {{\phi _p}(u'(t))} \right)^\prime } - {\left( {{\phi _p}(v'(t))} \right)^\prime } + A(u - v) = 0,\\ & {{u(T) = Qu(0),u'(T) = Qu'(0)}},\quad {{v(T) = Qv(0),v'(T) = Qv'(0)}}. \end{align*}

And hence, after multiplication by $u-v$, and integration by parts over $[0,\,T]$, we get

\begin{align*} & \int_0^T \left\langle {\phi _p}(u'(t)) - {\phi _p}(v'(t)),u'(t) - v'(t) \right\rangle \,{\rm d}t\\ & \quad - \int_0^T {\left\langle {A(u(t) - v(t)),(u(t) - v(t))} \right\rangle \,{\rm d}t} = 0. \end{align*}

The above formula and lemma 2.4 yield that $u=v$.

Corollary 5.7 If $A$ is a negative semi-definite matrix with ${\mathcal {P}}A (\alpha )=A{\mathcal {P}} (\alpha )$ for any $\alpha \in \mathbb {R}^N$, then for each $e \in L^2$ with $\bar e ={\mathcal {P}}(\frac {1}{T}\int _0^T {e(t)\,{\rm d}t})= 0,$ the RPBVP (5.13) has at least one solution $u$ such that $\bar u ={\mathcal {P}}(\frac {1}{T}\int _0^T {u(t)\,{\rm d}t})= 0$.

Proof. Consider the auxiliary RPBVP:

(5.23)\begin{equation} \left\{ \begin{array}{@{}l@{}} {\left( {{\phi _p}(u'(t))} \right)^\prime } +{\left( {\nabla F(u)} \right)^\prime } +Au(t) -\dfrac{1}{n}u(t) = e(t),\\ u(T) = Qu(0),u'(T) = Qu'(0), \end{array} \right. \end{equation}

where $n >0$. By integrating the equation over $[0,\,T]$ and using ${\mathcal {P}}$ to act, then each solution $u$ of (5.23) satisfies

\[ (A-\frac{1}{n} I){\mathcal{P}}\left(\frac{1}{T}\int_0^T {u(t)\,{\rm d}t} \right) = {\mathcal{P}}\left(\frac{1}{T}\int_0^T {e(t)\,{\rm d}t}\right) = \bar e = 0. \]

Notice that $(A-\frac {1}{n}I)$ is negative definite for each $n$. So $\bar u=0$. It follows from theorem 5.5 and its proof that, RPBVP (5.23) has at least one solution $u_n (t)$ for each $n$. Further there is $r_0>0$ independent of $n$ such that ${\left \| u_n \right \|_1} \le {r_0}$. From lemma 3.3, it follows that those $u_n$ are fixed points of the equivalent completely continuous operator. So there exists a subsequence converging to a solution of (5.13) with $\bar u = 0$.

Corollary 5.8 If $A \buildrel \Delta \over = a <0$ is a constant, then for each $e \in L^2,$ the problem (5.13) has at least one solution $u$.

Remark 5.9 If $Q=I$, then we immediately deduce Theorem 6.1 in [Reference Mawhin10] from theorem 5.5.