近年来, 随着自然科学和工程技术中许多非线性问题的不断出现, Sobolev空间表现出了其应用范围的局限性.例如, 对一类有变指数增长条件的非线性问题的研究.具有变指数增长性条件的非线性问题是一个新兴的研究课题.在对这类非线性问题进行研究时, 变指数Lebesgue空间及Sobolev空间发挥着重要的作用^[1-4].

在本文中, 我们主要研究一类$p(t)$-Laplace中立型微分泛函方程

$ \begin{equation}\label{eq:w1} (\phi_{p(t)}(x(t)-cx(t-\sigma))')'+g(t, x(t-\tau(t)))=e(t) \end{equation}$

(1.1)

周期解的存在性, 其中$\phi_{p(t)}:\mathbb{R}\rightarrow \mathbb{R}$, $\phi_{p(t)}(u)=|u|^{p(t)-2}u$, $g\in C(\mathbb{R}^2, \mathbb{R})$, $1 < p^-\leq p(t)\leq p^+ < \infty$且Lipschitz连续, $\tau(t), e(t)$是定义在$\mathbb{R}$上的具有周期$T$的函数, $\sigma, c\in\mathbb{R}$且$|c|\neq 1$.

由于指数$p(t)$为函数, $p(t)$-Laplace算子较之$p$-Laplace具有更复杂的非线性性, 许多对于$p$-Laplace问题成立的方法和结果不再适用于$p(t)$-Laplace问题.本文结合了Mawhin连续定理与不等式技巧的应用, 克服了变指数$p(t)$产生的困难, 得到了新的研究成果.关于常指数增长条件下的周期解存在性的问题, 可参考文献[5-6].

首先回忆一些关于变指数Lebesgue空间$L^{p(t)}(\Omega)$的定理^[7].

定义2.1  设$\mathcal{P}$是所有Lebesgue可测函数集, 且$p:\Omega\rightarrow(1, \infty)$.对于$p(t)\in \mathcal{P}(\Omega)$, 设

$ \| u\|_{p(t)} = \inf \{ {\lambda > 0:\displaystyle\int_{\Omega}|\frac{u}{\lambda}|^{p(t)}\;dt \le 1} \}.$

变指数Lebesgue空间$L^{p(t)}(\Omega)$是由一类满足$\displaystyle\int_{\Omega}|u(t)|^{p(t)}dt<\infty$的函数$u$组成, 且$L^{p(t)}(\Omega)$是由上述范数诱导的Banach空间.定义$p(t)\in \mathcal{P}(\Omega)$的共轭函数$q(t)$为$q(t)=\frac{p(t)}{p(t)-1}.$

定理2.2令$p\in \mathcal{P}(\Omega)$, 那么

$ \displaystyle\int_\Omega |f(t) \cdot g(t)| dt \le 2\| f\|_{p(t)}\| g \|_{q(t)} $

对于每个$f\in L^{p(t)}(\Omega)$和$g\in L^{p'(t)}(\Omega)$都成立.

定理2.3设$u, \, u_k\in L^{p(t)}(\Omega)$, 则

1) $\| u\|_{p(t)} < 1({ = 1; > 1})$当且仅当$\rho(u) <1({ = 1; >1})$.

2) 如果$\|u\|_{p(t)}>1$, 则$\|u\|_{p(t)}^{p^-}\le\rho(u)\le\|u\|_{p(t)}^{p^+} $.

3) 如果$\|u\|_{p(t)}<1$, 则$\|u\|_{p(t)}^{p^+}\le\rho(u)\le\|u\|_{p(t)}^{p^-}$.

4) $\lim\limits_{k\rightarrow \infty}\|u_k\|_{p(t)}=0$当且仅当$\lim\limits_{k\rightarrow\infty}\rho(u_k)=0$.

5) $\|u_k\|_{p(t)}\rightarrow\infty$当且仅当$\rho(u_k)\rightarrow\infty$.

对于Mawhin连续性定理^[8].令$X, Y$是实Banach空间, $L:D(L)\subset X\rightarrow Y$是一个Fredholm算子并且指数为0, $D(L)$是$L$的值域, 这意味着Im$ L$在$Y$上是一个闭域并且有

$ {\hbox{dimKer}}L={\hbox{dim}}(Y/{\hbox{Im}} L)<+\infty.$

考虑补充子空间$X_1, Y_1$, 其中$ X={\hbox{Ker}} L \oplus X_1, ~Y={\hbox{Im}} L\oplus Y_1.$令$ P: X \rightarrow {\hbox{Ker}} L, ~Q:Y\rightarrow Y_1$是自然投影.显然有${\hbox{Ker}} L\cap (D(L)\cap X_1)=\{0\}. $因此限制条件$L_P:=L\mid _ {D(L)\cap X_1}$是可逆的, 由$K$表示$L_P$的逆.

令$\Omega$是$X$的一个开的有界子集并且$D(L)\cap \Omega\not=\phi.$如果$QN(\bar \Omega)$是有界的, 算子$ K(I - Q)N:\bar \Omega \rightarrow X $是致密的, 则在$\bar \Omega$中$ N:\bar \Omega \to Y $是$L$ -紧致的.

引理2.4  假设$X, Y$是两个Banach空间, $ L:D(L) \subset X \to Y $是一个Fredholm算子并且指数为0.此外$ \Omega \subset X$是一个开的有界集合, 在$\bar \Omega$上$ N:\bar \Omega \to Y $是$L$ -紧致的.如果有

(1) $ Lx \ne \lambda Nx, \forall x \in \partial \Omega \cap D(L), \lambda \in \left({0, 1} \right) $.

(2) $Nx \notin {\mathop{\rm Im}\nolimits} L, \forall x \in \partial \Omega \cap {\hbox{Ker}} L$.

(3) $\deg \left\{ {JQN, \Omega \cap \hbox{Ker} L, 0} \right\} \ne 0, $其中$ J:{\mathop{\rm Im}\nolimits} Q \to {\hbox{Ker}} L $是同构的.则$ Lx = Nx $在$ \bar \Omega \cap D(L) $上有一个解.

引理2.5  如果$ \left| c \right| \ne 1 $, 且$A$在$ {C_T} $上有连续有界映射, 那么

(1) $ \left\| {{A^{ - 1}}x} \right\| \le \frac{{\left\| x \right\|}}{{\left| {\left| c \right| - 1} \right|}}, \forall x \in {C_T}. $

(2) $ \displaystyle\int_0^T {\left| {({A^{ - 1}}f)(t)} \right|dt \le \frac{1}{{\left| {1 - \left| c \right|} \right|}}\displaystyle\int_0^T {\left| {f(s)} \right|ds, \forall f \in {C_T}.} } $

(3) $ \displaystyle\int_0^T {{{\left| {({A^{ - 1}}f)(t)} \right|}^2}} dt \le \frac{1}{{{{(1 - \left| c \right|)}^2}}}\displaystyle\int_0^T {{{\left| {f(s)} \right|}^2}ds}, \forall f \in {C_T}. $

首先把方程(1.1)写成如下形式

$ \label{eq:w2} \left\{\begin{array}{lll} {(A{x_1})'(t) = {\varphi _{q(t)}}({x_2}(t)) = {{\left| {{x_2}(t)} \right|}^{q(t) - 2}}{x_2}(t), }\\ {{x_2}^\prime (t) = - g(t, {x_1}(t - \tau (t))) + e(t), } \end{array}\right. $

(P)

其中$\frac{1}{{p\left(t \right)}} + \frac{1}{{q\left(t \right)}} = 1 $.显然如果$ x(t) = {({x_1}(t), {x_2}(t))^T} $是(P)的一个周期解, 那么$x_1(t)$也是方程(1.1)的一个周期解.

假设$T$是一个大于0的常数$ {C_T} = \left\{ {\varphi :\varphi \in C\left({R, R} \right), \varphi \left({t + T} \right) \equiv \varphi \left(t \right)} \right\} $且具有范数$ {\left| \varphi \right|_0} = {\max\limits_{t \in \left[{0, T} \right]}}\left| {\varphi \left(t \right)} \right| $, $ X = Y = \left\{ {x = \left({{x_1}(t), {x_2}(t)} \right) \in C\left({R, {R^2}} \right):x\left({t + T} \right) \equiv x\left(t \right)} \right\} $具范数$ \left\| x \right\| = \max \left\{ {{{\left| {{x_1}} \right|}_0}, {{\left| {{x_2}} \right|}_0}} \right\} $, 显然$X$和$Y$是Banach空间.令

$ \begin{eqnarray}&& {A:{C_T} \to {C_T}, \;\;\;\;\;\;(Ax)(t) = x(t) - cx(t - \sigma ), }\\ && {L:D(L) \subset {C_T} \to {C_{T, }}\;\;\;\;Lx = \left(\begin{array}{*{20}{c}} {(A{x_1})'}\\ {{x_2}^\prime } \end{array}\right), } \end{eqnarray} $

其中$ x = \left({\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right).$易得

$ \mbox{Ker}L = {R^2}, {\mathop{\rm Im}\nolimits} L = \left\{ {x:x \in Y, \displaystyle\int_0^T {x\left( s \right)ds} = 0} \right\}, $

所以$L $是一个指标为零的Fredholm算子.

令投影$ P:X \to {\hbox{Ker}}L $与$ Q:Y \to {\mathop{\rm Im}\nolimits} Q $定义如下

$ Px = \frac{1}{T}\displaystyle\int\limits_0^T {x\left( s \right)ds}; Qy = \frac{1}{T}\displaystyle\int\limits_0^T {y\left( s \right)ds}.$

记$ {L_p}^{ - 1} $为$ L\left| {_{{\hbox{Ker}}P \cap D\left(L \right)}} \right. $的逆.显然$ {\mathop{\rm Im}\nolimits} P = {\hbox{Ker}}\; L, {\hbox{Ker}} L = {\mathop{\rm Im}\nolimits} Q = {R^2} $,

$ \begin{eqnarray}&& \left[ {{L_p}^{ - 1}y} \right](t) = \left( {\begin{array}{*{20}{c}} {({A^{ - 1}}F{y_1})(t)}\\ {(F{y_2})(t)} \end{array}} \right), \nonumber\\ &&\label{eq:w3} \left[ {Fy} \right](t) = - \displaystyle\int\limits_t^T {y(s)ds + \displaystyle\int\limits_0^T {\frac{s}{T}y(s)ds} }, \end{eqnarray} $

其中$ y(t) = \left({\begin{array}{*{20}{c}} {{y_1}(t)}\\ {{y_2}(t)} \end{array}} \right). $

给出以下假设条件

$[{{\hbox{H}}_1}]\; \; x[g(t.x)-e(t)] > 0, \forall t \in R, \left| x \right| > D; $

$ [{{\hbox{H}}_2}]\; \; {\lim\limits_{x \to - \infty }}{\sup\limits_{t \in [0, T]}}\frac{{\left| {g(t, x) - e(t)} \right|}}{{{{\left| x \right|}^{\mu (t) - 1}}}} \le r, $其中$ \mu(t) \in C(R), 1 < {\mu ^ - } \le \mu (t) \le {\mu ^ + } < {p^ - }.$

引理3.1  如果$ p(t) > 2 $且存在正常数$ D, r \ge 0 $满足$ [{\hbox{H}}_1]$-$[{\hbox{H}}_2]$, 则(1.1)至少有一个$T$型周期解.

证  显然当且仅当$ Lx = Nx $成立时, 方程(P)有$T$型周期解, 其中$ N:{C_T} \to {C_T}, $

$ (Nx)(t) = \left( {\begin{array}{*{20}{l}} {{\varphi _{q(t)}}({x_2}(t))}\\ { - g(t, {x_1}(t - \tau (t))) + e(t)} \end{array}} \right).$

由已知条件, 可得$N$在$\overline{\Omega}$上是$L$ -紧致的, 其中$\Omega$是$C_T$上的任意开且有界闭集.记

$ {\Omega _1} = \{ x \in {C_T}:Lx = \lambda Nx, \lambda \in \left( {0, T} \right)\} , $

如果$\forall x = \left({\begin{array}{*{20}{c}} {{x_1}(t)}\\ {{x_2}(t)} \end{array}} \right) \in {\Omega _1}$, 那么$x$一定满足

$ \begin{equation}\label{eq:w4} \left\{ {\begin{array}{*{20}{l}} {{{(A{x_1})}^\prime }(t) = \lambda {\varphi _{q(t)}}({x_2}(t)) = \lambda {{\left| {{x_2}(t)} \right|}^{q(t) - 2}}{x_2}(t), }\\ {{x_2}^\prime (t) = - \lambda g(t, {x_1}(t - \tau (t))) + \lambda e(t). } \end{array}} \right. \end{equation}$

(3.1)

由(3.1)式的第一个等式可得$ {x_2}(t) = {\varphi _{p(t)}}(\frac{1}{\lambda }{(A{x_1})^\prime }(t)) $, 将其代入(3.1)式的第二个等式, 可得

$ \begin{equation}\label{eq:w5} {\left[ {{\varphi _{p(t)}}(\frac{1}{\lambda }{{(A{x_1})}^\prime }(t))} \right]^\prime } + \lambda g(t, {x_1}(t - \tau (t))) = \lambda e(t). \end{equation}$

(3.2)

在区间$[0, T]$上积分(3.2)式的两边, 由牛顿-莱布尼兹定理, 可得

$ \begin{eqnarray}&& \displaystyle\int_0^T {{{\left[ {{\varphi _{p(t)}}(\frac{1}{\lambda }{{(A{x_1})}^\prime }(t))} \right]}^\prime }{\rm{dt = }}0}, \nonumber \\ && \label{eq:w6} \displaystyle\int_0^T {[g(t, {x_1}(t - \tau (t))) - e(t)]dt = 0} . \end{eqnarray} $

(3.3)

由积分中值定理, 存在常数$ \xi \in \left[{0, T} \right] $满足$ g(\xi, {x_1}(\xi - \tau (\xi))) - e(t) = 0.$由假设$[{\hbox{H}}_1]$可得$\left| {{x_1}(\xi - \tau (\xi))} \right| \le D.$令$ \xi - \tau \left(\xi \right) = kT + {t_0} $, 其中$k\in {\cal Z}, t_0\in [0, T)$, 因此

$\left| {{x_1}({t_0})} \right| = \left| {{x_1}\left( {\xi - \tau \left( \xi \right)} \right)} \right| \le d, $

这意味着

$ {\left| {{x_1}} \right|_0} \le D + \displaystyle\int_0^T {\left| {{x_1}(t)} \right|} dt .$

另一方面, 将(3.2)式两端同时乘$ (A{x_1})(t), $可得

$ \begin{equation}\label{eq:w7} [{\varphi _{p(t)}}(\frac{1}{\lambda }{x_1}^\prime (t))]'{(A{x_1})^\prime }(t) + \lambda g(t, {x_1}(t - \tau (t))){(A{x_1})^\prime }(t) = \lambda (A{x_1})(t)e(t). \end{equation}$

(3.4)

在区间$[0, T]$上积分(3.4)式的两边, 由于

$ {\displaystyle\int_0^T {[{\varphi _{p(t)}}(\frac{1}{\lambda }{{(A{x_1})}^\prime }(t))]'(A{x_1})(t)dt = - \displaystyle\int_0^T {\frac{1}{{{\lambda ^{p(t) - 1}}}}\left| {\frac{1}{\lambda }{{(A{x_1})}^\prime }(t)} \right|} } ^{p(t)}}dt .$

因为$\lambda>1$, 有

$ \begin{array} \displaystyle\int_0^T {{{\left| {(A{{x'}_1})(t)} \right|}^{p(t)}}dt} &\le {\displaystyle\int_0^T {\frac{1}{{{\lambda ^{p(t)}}}}\left| {(A{x_1}^\prime )(t)} \right|} ^{p(t)}}dt\\ &\leq \displaystyle\int_0^T {\left| {g(t, {x_1} (t - \tau (t))) - e(t)} \right|\left| {(A{x_1})(t)} \right|} dt. \end{array}$

(3.5)

即

$ \begin{array} \displaystyle\int_0^T {{{\left| {(A{{x'}_1})(t)} \right|}^{p(t)}}dt} \le&\displaystyle\int_0^T {\left| {g(t, {x_1}(t - \tau (t))) - e(t)} \right|\left| {{x_1}(t) - c{x_1}(t - \sigma )} \right|} dt \\ & \le{\left| {{x_1}} \right|_0}\displaystyle\int_0^T {\left| {g(t, {x_1}(t - \tau (t))) - e(t)} \right|} dt \\ & + \left| c \right|{\left| {{x_1}} \right|_0}\displaystyle\int_0^T {\left| {g(t, {x_1}(t - \tau (t))) - e(t)} \right|} dt \\ & \le (1 + \left| c \right|){\left| {{x_1}} \right|_0}\displaystyle\int_0^T {\left| {g(t, {x_1}(t - \tau (t))) - e(t)} \right|} dt. \end{array}$

(3.6)

由假设$[{\hbox{H}}_2]$可知, $ \exists \rho > 0 $满足

$ \begin{equation}\label{eq:w10} \left| {g(t, x) - e(t)} \right| \le (r + \varepsilon ){\left| x \right|^{\mu (t) - 1}}, \;\;\forall t \in R, ~x < - \rho. \end{equation}$

(3.7)

令

$ \begin{eqnarray}&& {E_1} = \left\{ {t \in \left[ {0, T} \right]:{x_1}(t - \tau (t)) < - \rho } \right\}, \\ &&{E_2} = \left\{ {t \in \left[ {0, T} \right]:\left| {{x_1}(t - \tau (t))} \right| \le \rho } \right\}, \\ && {E_3} = \left\{ {t \in \left[ {0, T} \right]:{x_1}(t - \tau (t)) > \rho } \right\}, \end{eqnarray}$

由(3.3)式可得

$ \begin{equation}\label{eq:w11} \left( {\displaystyle\int_{{E_1}} {} + \displaystyle\int_{{E_2}} {} + \displaystyle\int_{{E_3}} {} } \right)[g(t, {x_1}(t - \tau (t)) - e(t))]dt = 0, \end{equation}$

(3.8)

因此

$ \begin{array} \displaystyle\int\limits_{{E_3}} {\left| {g(t, {x_1}(t - \tau (t)) - e(t))} \right|} dt =& \displaystyle\int\limits_{{E_3}} {\left[ {g(t, {x_1}(t - \tau (t)) - e(t))} \right]dt} %\\=& - \left( {\displaystyle\int\limits_{{E_1}} {}} + \displaystyle\int\limits_{{E_2}} {}\right) \\ &\le\left( {\displaystyle\int\limits_{{E_1}} {} + \displaystyle\int\limits_{{E_2}} {} } \right)\left| {g(t, {x_1}(t - \tau (t)) - e(t))} \right|dt. \end{array}$

(3.9)

由(3.8)式和(3.9)式可得

$ \begin{array} \displaystyle\int_0^T {\left| {g(t, {x_1}(t - \tau (t)) - e(t))} \right|} dt =\left( {\displaystyle\int_{{E_1}} {} + \displaystyle\int_{{E_2}} {} + \displaystyle\int_{{E_3}} {} } \right)[g(t, {x_1}(t - \tau (t)) - p(t))]dt\\ \le 2\left( {\displaystyle\int\limits_{{E_1}} {} + \displaystyle\int\limits_{{E_2}} {} } \right)\left| {g(t, {x_1}(t - \tau (t)) - e(t))} \right|dt\\ \le 2\displaystyle\int\limits_{{E_1}} {(r + \varepsilon ){{\left| {{x_1}(t - \tau (t))} \right|}^{\mu (t) - 1}}dt} + 2{\tilde g_\rho }T, \end{array}$

(3.10)

其中$ {\tilde g_\rho } = {\max\limits_{t \in {E^2}}}\left| {g(t, {x_1}(t - \tau (t))) - e(t)} \right|.$由(3.6)式和(3.10)式, 可得

$ \begin{array} &\displaystyle\int_0^T {{{\left| {(A{{x'}_1})(t)} \right|}^{p(t)}}dt} \le (1 + \left| c \right|){\left| {{x_1}} \right|_0}\displaystyle\int_0^T {\left| {g(t, {x_1}(t - \tau (t))) - e(t)} \right|} dt \\ = 2(1 + \left| c \right|){\left| {{x_1}} \right|_0}{\tilde g_\rho }T+ 2(1 + \left| c \right|)(r + \varepsilon ){\left| {{x_1}} \right|_0}\displaystyle\int\limits_{{E_1}} {{{\left| {{x_1}(t - \tau (t))} \right|}^{\mu (t) - 1}}dt}. \end{array}$

(3.11)

显然, 可以得到

$ \exists {\tilde M_1} > 0 ~\mbox{满足} ~ {\left| {{x_1}} \right|_0} < {\tilde M_1} . $

下面讨论$ {\left| {{x_1}} \right|_0}. $因为

$ {\left| {{x_1}} \right|_0} \le D + \displaystyle\int_0^T {\left| {{x_1}(t)} \right|} dt.$

如果$ \exists K \ge 0 $, 满足$ \displaystyle\int_0^T {\left| {{x_1}(t)} \right|} dt < K $, 则$ {\left| {{x_1}} \right|_0} \le D + \displaystyle\int_0^T {\left| {{x_1}(t)} \right|} dt = D + K = {M_1}.$当$ {\left| {{x_1}} \right|_0} \le 1 ~\mbox{时}, M_1=1; $当$ {\left| {{x_1}} \right|_0} > 1 ~\mbox{时}, {\left| {{x_1}(t - \tau (t))} \right|^{\mu (t) - 1}} \le {\left| x \right|_0}^{{\mu ^ + } - 1}. $由(3.11)式, 可得

$\begin{eqnarray} \displaystyle\int_0^T {{{\left| {(A{{x'}_1})(t)} \right|}^{p(t)}}dt} &\le 2(1 + \left| c \right|){\left| {{x_1}} \right|_0}{\tilde g_\rho }T + 2(1 + \left| c \right|)(r + \varepsilon ){\left| {{x_1}} \right|_0}\displaystyle\int\limits_{{E_1}} {{{\left| {{x_1}(t - \tau (t))} \right|}^{\mu (t) - 1}}dt} \\ &\le 2(1 + \left| c \right|){\left| {{x_1}} \right|_0}{\tilde g_\rho }T + 2T(1 + \left| c \right|)(r + \varepsilon ){\left| {{x_1}} \right|_0}^{{\mu ^ + }}. \end{eqnarray}$

因为

$ \begin{eqnarray}{\left| {{x_1}} \right|_0} &\le& D + \displaystyle\int_0^T {\left| {{x_1}(t)} \right|} dt, \nonumber\\ \displaystyle\int_0^T {{{\left| {(A{{x'}_1})(t)} \right|}^{p(t)}}dt} &\leq&(1 + \left| c \right|)(2{\tilde g_\rho }TD + 2{\tilde g_\rho }T\displaystyle\int_0^T {\left| {{x_1}^\prime (t)} \right|dt} \nonumber\\&& \label{eq:w15}+ 2(r + \varepsilon )T{(D + \displaystyle\int_0^T {\left| {{x_1}^\prime (t)} \right|} dt)^{{\mu ^ + }}}). \end{eqnarray} $

(3.12)

如果$\displaystyle\int_0^T {\left| {{x_1}^\prime (t)} \right|} dt = 0, $则$ {\left| {{x_1}} \right|_0} \le D~ \mbox{取}~M_1=D.$如果$ \displaystyle\int_0^T {\left| {{x_1}^\prime (t)} \right|} dt > 0, $则由定理2.2, 有

$ \begin{equation}\label{eq:w16} {\displaystyle\int_0^T {\left| {A{x_1}^\prime (t)} \right|} ^2}dt \le {\left\| 1 \right\|_{\frac{{p(t)}}{{p(t) - 2}}}}.{\left\| {{{(A{x_1}^\prime (t))}^2}} \right\|_{\frac{{p(t)}}{2}}}. \end{equation}$

(3.13)

由引理2.5, 可得

$ \begin{array} \displaystyle\int_0^T {{\left| {{x_1}^\prime (t)} \right|}^2}dt &= &\displaystyle\int_0^T {{{\left| {({A^{ - 1}}A{x_1}^\prime )(t)} \right|}^2}dt \le \frac{{\displaystyle\int_0^T {{{\left| {(A{x_1})(t)} \right|}^2}dt} }}{{{{(1 - \left| c \right|)}^2}}}}\nonumber\\ &\le& \frac{1}{{{{(1 - |c|)}^2}}}{\left\| 1 \right\|_{\frac{{p(t)}}{{p(t) - 2}}}}\cdot{\left\| {{{(A{x_1}^\prime (t))}^2}} \right\|_{\frac{{p(t)}}{2}}}. \end{array} $

(3.14)

当$ \displaystyle\int_0^T {{{\left| {(A{{x'}_1})(t)} \right|}^{p(t)}}dt} \le 1 $时, 即$ {\left\| {{{(A{x_1}^\prime (t))}^2}} \right\|_{\frac{{p(t)}}{2}}} \le 1 $时, 可得

$ \displaystyle\int_0^T {{{\left| {{x_1}^\prime (t)} \right|}^2}dt \le \frac{1}{{{{(1 - |c|)}^2}}}{{\left\| 1 \right\|}_{\frac{{p(t)}}{{p(t) - 2}}}}} .$

因此取$ {M_1} = \frac{1}{{{{(1 - |c|)}^2}}}{\left\| 1 \right\|_{\frac{{p(t)}}{{p(t) - 2}}}}, $当$ \displaystyle\int_0^T {{{\left| {(A{{x'}_1})(t)} \right|}^{p(t)}}dt} > 1 $时, 即$ {\left\| {{{(A{x_1}^\prime (t))}^2}} \right\|_{\frac{{p(t)}}{2}}} > 1. $由(3.14)式, 可得

$ {\displaystyle\int_0^T {{{\left| {{x_1}^\prime (t)} \right|}^2}dt \le \frac{1}{{{{(1 - |c|)}^2}}}{{\left\| 1 \right\|}_{\frac{{p(t)}}{{p(t) - 2}}}}\cdot\left( {\displaystyle\int_0^T {{{\left| {(A{{x'}_1})(t)} \right|}^{p(t)}}dt} } \right)} ^{\frac{2}{{{p^ - }}}}}.$

因为$ \frac{2}{{{p^ - }}} < 1, $所以

$ \displaystyle\int_0^T {{{\left| {{x_1}^\prime (t)} \right|}^2}dt \le \frac{1}{{{{(1 - |c|)}^2}}}{{\left\| 1 \right\|}_{\frac{{p(t)}}{{p(t) - 2}}}}}\cdot \displaystyle\int_0^T {{{\left| {(A{{x'}_1})(t)} \right|}^{p(t)}}dt}.$

由(3.12)得

$ \begin{array} \displaystyle\int_0^T {{\left| {{x_1}^\prime (t)} \right|}^2}dt \le& \frac{1}{{{{(1 - |c|)}^2}}}{{\left\| 1 \right\|}_{\frac{{p(t)}}{{p(t) - 2}}}}\cdot [ 2(1 + c)(r + \varepsilon )T(D + \displaystyle\int_0^T|x_1(t)| dt)^{\mu ^ + } \\ &+ 2(1 + c){{\tilde g}_\rho }T(D + \displaystyle\int_0^T {|{x_1}^\prime (t)|} dt) ]\\ &= {C_1}{\left( {D + \displaystyle\int_0^T {|{x_1}^\prime (t)|dt} } \right)^{{\mu ^ + }}} + {C_2}(D + \displaystyle\int_0^T {|{x_1}^\prime (t)|dt} ) , \end{array}$

(3.15)

其中

$ \begin{eqnarray}&& {C_1} = \frac{1}{{{{(1 - |c|)}^2}}}{\left\| 1 \right\|_{\frac{{p(t)}}{{p(t) - 2}}}} \cdot 2(1 + c)(r + \varepsilon )T , \\ &&{C_2} = \frac{1}{{{{(1 - |c|)}^2}}}{\left\| 1 \right\|_{\frac{{p(t)}}{{p(t) - 2}}}} \cdot 2(1 + c){\tilde g_\rho }T .\end{eqnarray}$

又由

$ \displaystyle\int_0^T {\left| {{x_1}^\prime (t)} \right|dt \le {T^{\frac{1}{2}}}{{(\displaystyle\int_0^T {{{\left| {{x_1}^\prime (t)} \right|}^2}dt} )}^{\frac{1}{2}}}} , $

因此

$ \begin{eqnarray}{(D + \displaystyle\int_0^T {\left| {{x_1}(t)} \right|} dt)^{{\mu ^ + }}} &\le& {2^{{\mu ^ + }}}{D^{{\mu ^ + }}} + {2^{{\mu ^ + }}}{(\displaystyle\int_0^T {\left| {{x_1}^\prime (t)} \right|dt} )^{{\mu ^ + }}}\\ & \le& {2^{{\mu ^ + }}}{D^{{\mu ^ + }}} + {2^{{\mu ^ + }}}{T^{\frac{{{\mu ^ + }}}{2}}}{(\displaystyle\int_0^T {{{\left| {{x_1}^\prime (t)} \right|}^2}dt} )^{\frac{{{\mu ^ + }}}{2}}}.\end{eqnarray}$

由(3.15)式, 得

$ \displaystyle\int_0^T {{{\left| {{x_1}^\prime (t)} \right|}^2}dt \le {C_1}\left[ {{2^{{\mu ^ + }}}{D^{{\mu ^ + }}} + {2^{{\mu ^ + }}}{T^{\frac{{{\mu ^ + }}}{2}}}{{(\displaystyle\int_0^T {{{\left| {{x_1}^\prime (t)} \right|}^2}} dt)}^{\frac{{{\mu ^ + }}}{2}}}} \right]} + {C_2}(D + {T^{\frac{1}{2}}}{(\displaystyle\int_0^T {{{\left| {{x_1}^\prime (t)} \right|}^2}dt} )^{\frac{1}{2}}}). $

由于$ {\mu ^ + } < 2$可得, 存在$C>0$, 使得$ \displaystyle\int_0^T {{{\left| {{x_1}^\prime (t)} \right|}^2}dt \le C}.$因此$ {\left| {{x_1}} \right|_0} \le D + C + T = {M_1}. $综上所述, $ \exists {M_1} > 0, $使得$ |{x_1}{|_0} < {M_1} $.

下面讨论$ {\left| {{x_2}} \right|_0} $的边界.由(3.1)式的第一个等式, 可知$ \displaystyle\int_0^T {{{\left| {{x_2}(t)} \right|}^{q(t) - 2}}{x_2}(t)dt = 0}.$这意味$ \exists {t_1} \in [0, T] $着满足$ {x_2}({t_1}) = 0, $因此$ {\left| {{x_2}} \right|_0} \le \displaystyle\int_0^T {\left| {{{x'}_2}(t)} \right|dt}.$由(3.2)式的第二个等式可得

$ \displaystyle\int_0^T {\left| {{x_2}^\prime (t)} \right|} dt \le \lambda \displaystyle\int\limits_0^T {\left| {g(t, {x_1}(t - \tau (t)))} \right|} dt + \lambda \displaystyle\int\limits_0^T {\left| {e(t)} \right|} \le \lambda T{g_{{M_2}}} + \lambda {\left\| e \right\|_1}, $

其中${g_{{M_2}}} = {\max\limits_{\left| x \right| \le {M_2}, t \in [0, T]}}\left| {g(t, x)} \right| $, 因此有$ {\left| {{x_2}} \right|_0} \le \lambda T{g_{{M_2}}} + \lambda {\left\| e \right\|_1} = {M_2}. $

令

$ \begin{eqnarray}&& M = \sqrt {M_2^2 + M_3^2} + 1, ~~ \Omega = \{ x = {({x_1}, {x_2})^T}:{\left| {{x_1}} \right|_0} < M, {\left| {{x_2}} \right|_0} < M\} , \\ && {\Omega _2} = \{ x \in \partial \Omega ;x \in \ker L\} .\end{eqnarray}$

那么

$ QN(x) = \displaystyle\int_0^T {\left( {\begin{array}{*{20}{c}} {{\varphi _{q(t)}}({x_2})}\\ { - g(t, {x_1}(t - \tau (t))) + e(t)} \end{array}} \right)} dt = \left( {\begin{array}{*{20}{c}} {{{\left| {{x_2}} \right|}^{q(t) - 2}}{x_2}}\\ { - g(t, {x_1}) + e(t)} \end{array}} \right).$

若$ QNx = 0 $, 那么$ {x_2} = 0, ~x_1=M$, 或$=M$.但是当$x_1=M$时, 显然$ - g(t, {x_1}) + e(t) < 0 $矛盾.当$x_1=-M$时, 显然$ QNx \ne 0, ~\forall x \in \Omega, x \notin {\mathop{\rm Im}\nolimits} L $, 因此引理2.4中(1)和(2)均得到满足.

下面证明引理2.4 (3)同样成立.定义同构: $ J:{\mathop{\rm Im}\nolimits} Q \to \ker L :$ $ J{({x_1}, {x_2})^T} = {({x_2}, {x_1})^T}. $

令$ H(\mu, x) = \mu x + \frac{{1 - \mu }}{T}JQNx, ~ (\mu, x) \in \Omega \times [0, 1], $那么

$ H(\mu , x) = \left( {\begin{array}{*{20}{c}} {\mu {x_1} + \frac{{1 - \mu }}{T}\left( {\frac{1}{T}\displaystyle\int_0^T {\left[ { - g(t, x) + e(t)} \right]} dt} \right)}\\ {(\mu + \frac{{1 - \mu }}{T}{{\left| {{x_2}} \right|}^{q(t) - 2}}){x_2}} \end{array}} \right), \forall (x, \mu ) \in (\partial \Omega \cap \ker L) \times [0, 1]. $

如果$ H(\mu, x) = 0 $, 那么$x_2=M \mbox{或} -M$.与上面的证明类似, 也可证明$ H(\mu, x) \ne 0 $.因此

$ \begin{eqnarray} &&\deg \{ JQN, \Omega \cap \ker L, 0\} = \deg \{ H(0, x), \Omega \cap \ker L, 0\} \\ &=& \deg \{ H(1, x), \Omega \cap \ker L, 0\} = \deg \{ I, \Omega \cap \ker L, 0\}\neq 0. \end{eqnarray}$

因此引理2.4的以三个条件同样成立.应用引理2.4, 可得$Lx=Nx$在$\overline{\Omega}\bigcap D(L)$上有一个解几乎处处成立, 方程(1.1)存在一个周期解.