该文考虑如下带有$ p $-Laplacian算子的脉冲耦合系统

(1.1) $ \left\{\begin{array}{lll} -(\Phi_{p}(u'(t)))'+g(t)\Phi_{p}(u(t)) = f_{u}(u(t), v(t)),~ &a.e.\; t\in[0, T], \\ -(\Phi_{p}(v'(t)))'+h(t)\Phi_{p}(v(t)) = f_{v}(u(t), v(t)),~ &a.e.\; t\in[0, T], \\ \Delta_{p}u'(t_{j}) = \phi_{p}(u'(t_{j}^{+}))-\phi_{p}(u'(t_{j}^{-})) = I_{j}(u(t_{j})),~ &j = 1, 2, \cdots , n, \\ \Delta_{p}v'(t_{j}) = \phi_{p}(v'(t_{j}^{+}))-\phi_{p}(v'(t_{j}^{-})) = J_{j}(v(t_{j})),~ &j = 1, 2, \cdots , n, \\ u(0) = u(T) = v(0) = v(T) = 0, \end{array}\right. $

其中$ \Phi_{p}(x) = |x|^{p-2}x, \; p > 1 $, $ 0 = t_{0} < t_{1} < t_{2} < \cdot\cdot\cdot < t_{n} < t_{n+1} = T, \; g, h\in L^{\infty}[0, T] $, $ f_{u}, f_{v}:\mathbb{R} ^{2}\rightarrow \mathbb{R} $为连续函数,且函数$ I_{j}, J_{j}:\mathbb{R} \rightarrow \mathbb{R} $, $ j = 1, 2, \cdots, n $是连续的.

随着科学与技术的发展,越来越多的科学家发现由脉冲微分方程所建立的数学模型能更加精确的描述现实世界中事物发展的瞬时突变现象.由于脉冲现象的广泛存在,使得脉冲微分方程理论在生物学、医学、物理学等学科中都具有广泛的应用.对于脉冲微分方程的研究,已经有众多的学者做过深入的探讨,其中有许多经典的方法,如不动点理论,拓扑度理论和比较法(包含上下解法和单调迭代法)等.近20多年来,用变分法研究脉冲微分方程得到广泛的关注,具体可参考文献[1-4, 7-10, 12, 15-17].

对于带有$ p $-Laplacian算子的微分方程的研究可以追溯到上世纪80年代,最早研究的是Esteban-Vazquez^[5]和Herrero-Vazquez^[6],研究结果所归结的模型为

$ \left\{\begin{array}{lll} -(\Phi_{p}(u'))' = q(t)f(t, u, u'), \\ u(0) = a, \; \; u(1) = b \end{array}\right. $

和

$ \left\{\begin{array}{lll} -(\Phi_{p}(u'))' = q(t)f(t, u, u'), \\ u'(0) = a, \; \; u(1) = b. \end{array}\right. $

此后,关于带有$ p $-Laplacian算子的问题的研究备受数学家们的重视并不断发展.最早用变分法研究带有$ p $-Laplacian算子的脉冲微分方程是田雨和葛渭高教授,在文献[13]中,他们研究了一类带有Sturm-Liouville边界的脉冲微分方程,通过变分法证明方程至少存在两个正解.之后, Nieto-O'Regan^[10]用变分法研究了一类二阶脉冲微分方程的线性问题和非线性问题,对这一领域的发展做出了标志性的贡献.此后,变分法作为一种新的方法研究非连续的问题(如脉冲问题)得到迅速发展,并取得了丰富的成果,具体如下:

在文献[2]中, Chen-Tang研究了如下带有$ p $-Laplacian算子的脉冲微分方程

(1.2) $ \left\{\begin{array}{lll} -(|u'(t)|^{p-2}u'(t))'+g(t)|u'(t)|^{p-2}u(t) = f(t, u(t)),~ &a.e.t\in[0, T], \\ \bigtriangleup u'(t_{j}) = |u'(t)|^{p-2}u'(t_{j}^{+})-|u'(t_{j}^{-})|^{p-2}u'(t_{j}^{-}) = I_{j}(u(t_{j})),~ & j = 1, \cdots , n, \\ u(0) = u(T) = 0, \end{array}\right. $

其中$ p\geq2 $, $ 0 = t_{_{0}} < t_{1} < t_{2} < \cdot\cdot\cdot < t_{n} < t_{n+1} = T $, $ f:[0, T]\times \mathbb{R} \rightarrow \mathbb{R} $是连续的, $ g\in L^{\infty}[0, T] $, $ I_{j}:\mathbb{R} \rightarrow \mathbb{R}, \; j = 1, 2, \cdots, n $是连续的.他们通过临界点理论得到问题(1.2)存在一个非平凡解,至少存在一个非平凡解和有无穷多解.

在文献[14]中, Wu-Liu第一次研究了如下脉冲微分方程的耦合系统

(1.3) $ \left\{\begin{array}{lll} -u''(t)+g(t)u(t) = f_{u}(u(t), v(t)),~ &a.e.\; t\in[0, T], \\ -v''(t)+h(t)v(t) = f_{v}(u(t), v(t)),~ &a.e.\; t\in[0, T], \\ u(0) = u(T) = v(0) = v(T) = 0, \\ \Delta u'(t_{j}) = u'(t_{j}^{+})-u'(t_{j}^{-}) = I_{j}(u(t_{j})),~ &j = 1, \cdots , n, \\ \Delta v'(t_{j}) = v'(t_{j}^{+})-v'(t_{j}^{-}) = J_{j}(v(t_{j})),~ &j = 1, \cdots , n, \end{array}\right. $

其中$ 0 = t_{0} < t_{1} < t_{2} < \cdot\cdot\cdot < t_{n} < t_{n+1} = T $, $ g, h\in L^{\infty}[0, T] $, $ f_{u}, f_{v}:\mathbb{R} ^{2}\rightarrow \mathbb{R} $是连续函数,并且函数$ I_{j}, J_{j}:\mathbb{R} \rightarrow \mathbb{R}, \; j = 1, 2, \cdots, n $是连续函数, $ u'(t_{j}^{+}), v'(t_{j}^{+}) $分别表示$ u', v' $在$ t_{j} $处的右极限, $ u'(t_{j}^{-}), v'(t_{j}^{-}) $分别表示$ u', v' $在$ t_{j} $处的左极限, $ j = 1, 2, \cdots, n. $作者通过极小化方法证明问题(1.3)至少存在一个解.

在硕士论文^[18]中,张然在第一部分研究一类带有$ p $-Laplacian算子的脉冲耦合系统的边值问题

(1.4) $ \left\{\begin{array}{lll} -(\Phi_{p}(u'(t)))'+\lambda\Phi_{p}(u(t)) = f_{u}(u(t), v(t)),~ &a.e.\; t\in[0, T], \\ -(\Phi_{p}(v'(t)))'+\lambda\Phi_{p}(v(t)) = f_{v}(u(t), v(t)),~ &a.e.\; t\in[0, T], \\ u(0) = u(T) = v(0) = v(T) = 0, \\ \Delta_{p}u'(t_{j}) = \phi_{p}(u'(t_{j}^{+}))-\phi_{p}(u'(t_{j}^{-})) = I_{j}(u(t_{j})),~ &j = 1, \cdots , n, \\ \Delta_{p}v'(t_{j}) = \phi_{p}(v'(t_{j}^{+}))-\phi_{p}(v'(t_{j}^{-})) = J_{j}(v(t_{j})),~ &j = 1, \cdots , n, \end{array}\right. $

其中$ 0 = t_{0} < t_{1} < t_{2} < \cdot\cdot\cdot < t_{n} < t_{n+1} = T, \; \Phi_{p}(x) = |x|^{p-2}x $, $ p > 1 $, $ f_{u}, f_{v}:\mathbb{R} ^{2}\rightarrow \mathbb{R} $为连续函数,且函数$ I_{j}, J_{j}:\mathbb{R} \rightarrow \mathbb{R} $是连续的.作者通过对称山路引理证明问题(1.4)有无穷多个弱解.

受到以上文献启发,该文研究问题(1.1),考虑脉冲函数满足两类不同的超线性条件时,通过山路引理和对称山路引理,证明问题(1.1)弱解的存在性和多重性.

接下来建立如下条件:

(H1) (ⅰ)对所有的$ x\in \mathbb{R} $, $ I_{j} $ ($ j = 1, 2, \cdots, n $)满足$ I_{j}(x)x\geq 0, $ ($ J_{j}(x)x\geq 0) $;

(ⅱ)对所有的$ x\in\mathbb{R} \backslash\{0\} $,存在常数$ v_{j} > 0 $, $ \delta_{j} > 0 $, $ j = 1, 2, \cdots, n $,使得$ \int_{0}^{x}I_{j}(s){\rm d}s\leq \delta_{j}|x|^{v_{j}} $, $ (\int_{0}^{x}J_{j}(s){\rm d}s\leq \delta_{j}|x|^{v_{j}}) $;

(ⅲ)对所有的$ x\in\mathbb{R} \backslash\{0\} $, $ I_{j}(x)x\leq v_{j}\int_{0}^{x}I_{j}(s){\rm d}s $, ($ J_{j}(x)x\leq v_{j}\int_{0}^{x}J_{j}(s){\rm d}s $);

(H1)' (ⅰ)存在常数$ \mu > p $,使得$ I_{j}(x)x\leq\mu\int_{0}^{x}I_{j}(s){\rm d}s < 0 $, ($ J_{j}(x)x\leq\mu\int_{0}^{x}J_{j}(s){\rm d}s < 0 $);

(ⅱ)存在常数$ \delta_{j} > 0 $,使得$ \int_{0}^{x}I_{j}(s){\rm d}s\geq -\delta_{j}|x|^{\mu} $, ($ \int_{0}^{x}J_{j}(s){\rm d}s\geq -\delta_{j}|x|^{\mu} $),其中$ j = 1, $$ 2, \cdots, n; $

(H2) $ f_{s}(s, t), f_{t}(s, t) = o(|s|^{p-1}+|t|^{p-1}) $,当$ |s|+|t|\rightarrow 0. $

(H3)存在$ c > 0, \; M > 0 $且有$ \beta > \theta > p $,使得

$ \mbox{当$|s|+|t|\geq M $时},~ f(s, t)\geq c(|s|^{\beta}+|t|^{\beta}), $

以及

$ \theta f(s, t)\leq sf_{s}(s, t)+tf_{t}(s, t); $

(H3)'存在$ c > 0, \; M > 0 $且有$ \beta > \mu > p $,使得

$ \mbox{当$|s|+|t|\geq M $时},~ f(s, t)\geq c(|s|^{\beta}+|t|^{\beta}), $

以及

$ \mu f(s, t)\leq sf_{s}(s, t)+tf_{t}(s, t); $

(H4) $ f_{s}(s, t) $和$ f_{t}(s, t) $分别关于$ s, t $为奇函数, $ I_{j}(x) $和$ J_{j}(x) $分别关于$ x $为奇函数, $ j = 1, 2, \cdots, n. $

在该文中,总假设$ \gamma > -\lambda_{1} $,其中$ \gamma = \min\{\mathop{{\rm ess}\inf}\limits_{t\in[0, T]}g(t), \mathop{{\rm ess}\inf}\limits_{t\in[0, T]}h(t)\} $,且$ \lambda_{1} $是如下Dirichlet问题的第一特征值

(1.5) $ \left\{\begin{array}{lll} -(\Phi_{p}(u'(t)))'+\lambda\Phi_{p}(u(t)) = 0,~ t\in[0, T], \\ u(0) = u(T) = 0. \end{array}\right.$

由以上条件,得到如下结论:

定理1.1  假设条件(H1), (H2), (H3)成立,则问题(1.1)至少存在一个弱解.

定理1.2  假设条件(H1), (H2), (H3), (H4)成立,则问题(1.1)有无穷多个弱解.

定理1.3  假设条件(H1)', (H2), (H3)'成立,则问题(1.1)至少存在一个弱解.

定理1.4  假设条件(H1)', (H2), (H3)', (H4)成立,则问题(1.1)有无穷多个弱解.

这部分,先介绍下面需要用到的一些定义和引理.

定义2.1^[11] ((PS)条件)  设$ E $是一个实自反Banach空间,对于任意序列$ \{u_{k}\}\subset E $,若$ \varphi(u_{k}) $有界且$ \varphi'(u_{k})\rightarrow 0, (k\rightarrow +\infty) $,则$ \{u_{k}\} $有收敛子列,则称$ \varphi $满足Palais-Smale(PS)条件.

引理2.1^[11] (山路引理)  设$ E $是实Banach空间, $ \varphi\in C^{1}(E, \mathbb{R} ) $且满足Palais-Smale条件, $ \varphi(0) = 0 $,若$ \varphi $满足如下条件：

(1)存在常数$ \rho , \alpha >0 $,使得$ \varphi|_{\partial B_{\rho}}\geq \alpha $;

(2)存在$ e\in E\backslash B_{\rho} $,使得$ \varphi(e)\leq 0 $.

那么$ \varphi $有临界值$ c\geq \alpha $且

$ c = \inf \limits_{g\in \Gamma}\max \limits_{s\in [0, 1]}I(g(s)), $

其中

$ \Gamma = \{g\in C([0, T], E)|g(0) = 0, g(T) = e\}. $

引理2.2^[11] (对称山路引理)  设$ E $是一个无限维实Banach空间, $ \varphi\in C^{1}(E, \mathbb{R}) $是一个满足Palais-Smale的偶泛函,其中$ \varphi(0) = 0 $,若$ E = V\bigoplus X $,其中$ V $是有限维的,且$ \varphi $满足:

(1)存在$ \rho > 0 $和$ \alpha > 0 $,使得对于满足$ \|u\| = \rho $的任意$ u\in X $有$ \varphi(u)\geq \alpha $;

(2)对于$ E $中的任意有限维子空间$ W $,存在常数$ R = R(W) $使得在$ W\backslash B_{R(W)} $上有$ \varphi(u)\leq 0 $.

那么$ \varphi $有一个无界的临界值序列.

在$ C[0, T] $, $ L^{p}[0, T] $空间和Sobolev空间$ W_{0}^{1, p}(0, T) $上,分别定义如下范数：

$ \|u\|_{\infty} = \max\limits_{t\in[0, T]}|u(t)|,~ \|u\|_{p} = \bigg(\int_{0}^{T}|u(t)|^{p}\bigg)^{\frac{1}{p}}, $

$ \|u\|_{W_{0}^{1, p}(0, T)} = \bigg(\int_{0}^{T}|u'(t)|^{p}+g(t)|u(t)|^{p}{\rm d}t\bigg)^{\frac{1}{p}}. $

假设空间$ X = W_{0}^{1, p}(0, T)\times W_{0}^{1, p}(0, T) $,则在$ X $上的范数定义为

$ \|(u, v)\|_{X}^{p} = \|u\|_{W_{0}^{1, p}(0, T)}^{p}+\|v\|_{W_{0}^{1, p}(0, T)}^{p}. $

接下来,介绍Poincaré不等式:对任意$ u\in W_{0}^{1, p}(0, T) $,

$ \int_{0}^{T}|u'(t)|^{p}{\rm d}t\geq \lambda_{1}\int_{0}^{T}|u(t)|^{p}{\rm d}t, $

其中$ \lambda_{1} $是问题(1.5)的第一特征值.

引理2.3^[2]  如果$ g\in L^{\infty}[0, T] $, $ \mathop{{\rm ess}\inf}\limits_{t\in [0, T]}g(t) = d > -\lambda_{1} $.则范数$ \|\cdot\|_{p} $和范数$ \|\cdot\|_{W_{0}^{1, p}(0, T)} $是等价的.

由引理2.3可得如下引理.

引理2.4  假设$ \gamma > -\lambda_{1} $, $ \gamma = \min\{\mathop{{\rm ess}\inf}\limits_{t\in[0, T]}g(t), \mathop{{\rm ess}\inf}\limits_{t\in[0, T]}h(t)\} $,则范数$ \|\cdot\|_{X} $和范数$ \|\cdot\| $在空间$ X $上等价,其中$ \|(u, v)\|^{p} = \|u\|_{p}^{p}+\|v\|_{p}^{p}. $

引理2.5^[18]  对$ \forall(u, v)\in X $,存在常数$ c_{1} > 0 $,使得

$ \|u\|_{\infty}, \|v\|_{\infty}\leq c_{1}\|(u, v)\|_{X}. $

取$ (\varphi, \psi)\in X $,同时乘以

$ -(\Phi_{p}(u'(t)))'+g(t)\Phi_{p}(u(t)) = f_{u}(u(t), v(t)) $

和

$ -(\Phi_{p}(v'(t)))'+h(t)\Phi_{p}(v(t)) = f_{v}(u(t), v(t)) $

左右两边,从$ 0 $到$ T $上作积分,可得

$ \begin{eqnarray*} -\int_{0}^{T}(\Phi_{p}(u'(t)))'\varphi(t){\rm d}t+\int_{0}^{T}g(t)\Phi_{p}(u(t))\varphi(t){\rm d}t = \int_{0}^{T}f_{u}(u(t), v(t))\varphi(t){\rm d}t \end{eqnarray*} $

和

$ \begin{eqnarray*} -\int_{0}^{T}(\Phi_{p}(v'(t)))'\psi(t){\rm d}t+\int_{0}^{T}h(t)\Phi_{p}(v(t))\psi(t){\rm d}t = \int_{0}^{T}f_{v}(u(t), v(t))\psi(t){\rm d}t, \end{eqnarray*} $

其中

$ \begin{eqnarray*} && \int_{0}^{T}(\Phi_{p}(u'(t)))'\varphi(t){\rm d}t = \sum \limits_{j = 0}^{n}\int_{t_{j}}^{t_{j+1}}(\Phi_{p}(u'(t)))'\varphi(t){\rm d}t\\ & = &\sum \limits_{j = 0}^{n}\left(\Phi_{p}(u'(t_{j+1}^{-})\varphi(t_{j+1}^{-})-\Phi_{p}(u'(t_{j}^{+})\varphi(t_{j}^{+})-\int_{t_{j}}^{t_{j+1}}\Phi_{p}(u'(t))\varphi'(t){\rm d}t\right)\\ & = &\Phi_{p}(u'(T))\varphi(T)-\Phi_{p}(u'(0))\varphi(0)-\sum \limits_{j = 1}^{n}\left(\Phi_{p}(u'(t_{j}^{+}))-\Phi_{p}(u'(t_{j}^{-}))\right)\varphi(t_{j})\\ &&-\int_{0}^{T}\Phi_{p}(u'(t))\varphi'(t){\rm d}t\\ & = &-\sum \limits_{j = 1}^{n}I_{j}(u(t_{j}))\varphi(t_{j})-\int_{0}^{T}\Phi_{p}(u'(t))\varphi'(t){\rm d}t. \end{eqnarray*} $

同理可得

$ \int_{0}^{T}(\Phi_{p}(v'(t)))'\psi(t){\rm d}t = -\sum \limits_{j = 1}^{n}J_{j}(v(t_{j}))\psi(t_{j})-\int_{0}^{T}\Phi_{p}(v'(t))\psi'(t){\rm d}t. $

综上可得

(2.1) $ \begin{eqnarray} && \int_{0}^{T}\Phi_{p}(u'(t))\varphi'(t){\rm d}t+\int_{0}^{T}g(t)\Phi_{p}(u(t))\varphi(t){\rm d}t +\sum \limits_{j = 1}^{n}I_{j}(u(t_{j}))\varphi(t_{j})\\ & = &\int_{0}^{T}f_{u}(u(t), v(t))\varphi(t){\rm d}t \end{eqnarray} $

和

(2.2) $ \begin{eqnarray} && \int_{0}^{T}\Phi_{p}(v'(t))\psi'(t){\rm d}t+\int_{0}^{T}h(t)\Phi_{p}(v(t))\psi(t){\rm d}t +\sum \limits_{j = 1}^{n}J_{j}(v(t_{j}))\psi(t_{j})\\ & = &\int_{0}^{T}f_{v}(u(t), v(t))\psi(t){\rm d}t. \end{eqnarray} $

由(2.1)–(2.2)式可得

$ \begin{eqnarray*} &&\int_{0}^{T}\Phi_{p}(u'(t))\varphi'(t){\rm d}t+\int_{0}^{T}\Phi_{p}(v'(t))\psi'(t){\rm d}t+\int_{0}^{T}g(t)\Phi_{p}(u(t))\varphi(t){\rm d}t\\ &&+\int_{0}^{T}h(t)\Phi_{p}(v(t))\psi(t){\rm d}t+\sum \limits_{j = 1}^{n}I_{j}(u(t_{j}))\varphi(t_{j})+\sum \limits_{j = 1}^{n}J_{j}(v(t_{j}))\psi(t_{j})\\ & = &\int_{0}^{T}f_{u}(u(t), v(t))\varphi(t){\rm d}t+\int_{0}^{T}f_{v}(u(t), v(t))\psi(t){\rm d}t. \end{eqnarray*} $

若对于任意的$ (\varphi, \psi)\in X $,上式成立,则称$ (u, v)\in X $是问题(1.1)的弱解.

定义泛函$ \Phi $如下:

(2.3) $ \begin{eqnarray} \Phi(u, v)& = &\frac{1}{p}\int_{0}^{T}|u'(t)|^{p}+g(t)|u(t)|^{p}{\rm d}t+\frac{1}{p}\int_{0}^{T}|v'(t)|^{p}+h(t)|v(t)|^{p}{\rm d}t\\ &&+\sum \limits_{j = 1}^{n}\int_{0}^{u(t_{j})}I_{j}(t){\rm d}t+\sum \limits_{j = 1}^{n}\int_{0}^{v(t_{j})}J_{j}(t){\rm d}t-\int_{0}^{T}f(u(t), v(t)){\rm d}t\\ & = &\frac{1}{p}\|(u, v)\|_{X}^{p}+\sum \limits_{j = 1}^{n}\int_{0}^{u(t_{j})}I_{j}(t){\rm d}t+\sum \limits_{j = 1}^{n}\int_{0}^{v(t_{j})}J_{j}(t){\rm d}t-\int_{0}^{T}f(u(t), v(t)){\rm d}t. \end{eqnarray} $

由$ f_{u}, f_{v}, I_{j} $和$ J_{j}, j = 1, 2, \cdots, n $的连续性可知, $ \Phi\in C^{1}(X, \mathbb{R}) $.对于任意的$ (\varphi, \psi)\in X $,可得

(2.4) $ \begin{eqnarray} \Phi'(u, v)(\varphi, \psi)& = &\int_{0}^{T}\Phi_{p}(u'(t))\varphi'(t){\rm d}t+\int_{0}^{T}\Phi_{p}(v'(t))\psi'(t){\rm d}t+\int_{0}^{T}g(t)\Phi_{p}(u(t))\varphi(t){\rm d}t\\ &&+\int_{0}^{T}h(t)\Phi_{p}(v(t))\psi(t){\rm d}t+\sum \limits_{j = 1}^{n}I_{j}(u(t_{j}))\varphi(t_{j})+\sum \limits_{j = 1}^{n}J_{j}(v(t_{j}))\psi(t_{j})\\ &&-\int_{0}^{T}f_{u}(u(t), v(t))\varphi(t){\rm d}t-\int_{0}^{T}f_{v}(u(t), v(t))\psi(t){\rm d}t. \end{eqnarray} $

所以问题(1.1)的弱解就是泛函$ \Phi $对应的临界点.

引理2.6  假设条件(H1), (H2), (H3)成立,则泛函$\Phi $满足(PS)条件.

证  令$ \theta>\max\{v_{1}, v_{2}, \cdots , v_{n}\}$,由(2.3)–(2.4)式可知

$ \begin{eqnarray*} &&\theta\Phi(u_{k}, v_{k})-\Phi'(u_{k}, v_{k})(u_{k}, v_{k})\\ & = &(\frac{\theta}{p}-1)\|(u_{k}, v_{k})\|_{X}^{p}+(\theta\sum \limits_{j = 1}^{n}\int_{0}^{u_{k}(t_{j})}I_{j}(t){\rm d}t-\sum \limits_{j = 1}^{n}I_{j}(u_{k}(t_{j}))u_{k}(t_{j}))\\ &&+(\theta\sum \limits_{j = 1}^{n}\int_{0}^{v_{k}(t_{j})}J_{j}(t){\rm d}t-\sum \limits_{j = 1}^{n}J_{j}(v_{k}(t_{j}))v_{k}(t_{j}))\\ &&+\int_{0}^{T}(f_{u}(u_{k}, v_{k})u_{k}+f_{v}(u_{k}, v_{k})v_{k}-\theta f(u_{k}, v_{k})){\rm d}t. \end{eqnarray*} $

由条件(H1), (H3)可知

$ \theta\sum \limits_{j = 1}^{n}\int_{0}^{u_{k}(t_{j})}I_{j}(t){\rm d}t-\sum \limits_{j = 1}^{n}I_{j}(u_{k}(t_{j}))u_{k}(t_{j})\geq 0, $

$ \theta\sum \limits_{j = 1}^{n}\int_{0}^{v_{k}(t_{j})}J_{j}(t){\rm d}t-\sum \limits_{j = 1}^{n}J_{j}(v_{k}(t_{j}))v_{k}(t_{j})\geq 0, $

$ \int_{0}^{T}(f_{u}(u_{k}, v_{k})u_{k}+f_{v}(u_{k}, v_{k})v_{k}-\theta f(u_{k}, v_{k})){\rm d}t\geq 0. $

所以

$ \theta\Phi(u_{k}, v_{k})-\Phi(u_{k}, v_{k})(u_{k}, v_{k})\geq(\frac{\theta}{p}-1)\|(u_{k}, v_{k})\|_{X}^{p}. $

因为$ \theta > p $ ,所以序列$\{(u_{k}, v_{k})\} $是有界的.所以存在$ \{(u_{k}, v_{k})\}$的子序列(仍记为$ \{(u_{k}, v_{k})\} $ )使得

$ \begin{eqnarray*} &&(u_{k}, v_{k})\rightharpoonup (u, v), \; \; \mbox{在$X$中, }\\ &&(u_{k}, v_{k})\rightarrow (u, v), \; \; \mbox{在$C[0, T]$中,当$k\rightarrow +\infty$时.} \end{eqnarray*} $

因此

(2.5) $ \begin{equation} (\Phi'(u_{k}, v_{k})-\Phi'(u, v))(u_{k}-u, v_{k}-v)\rightarrow 0, \end{equation} $

(2.6) $ \begin{equation} \sum \limits_{j = 1}^{n}(I_{j}(u_{k}(t_{j}))-I_{j}(u(t_{j})))(u_{k}(t_{j})-u(t_{j}))\rightarrow 0, \end{equation} $

(2.7) $ \begin{equation} \sum \limits_{j = 1}^{n}(J_{j}(v_{k}(t_{j}))-J_{j}(v(t_{j})))(v_{k}(t_{j})-v(t_{j}))\rightarrow 0, \end{equation} $

(2.8) $ \begin{equation} \int_{0}^{T}(f_{u}(u, v)-f_{u}(u_{k}, v_{k}))(u_{k}-u){\rm d}t\rightarrow 0, \end{equation} $

(2.9) $ \begin{equation} \mbox{当$k\rightarrow +\infty$时}, \ \int_{0}^{T}(f_{v}(u, v)-f_{v}(u_{k}, v_{k}))(v_{k}-v){\rm d}t\rightarrow 0. \end{equation} $

所以

(2.10) $ \begin{eqnarray} &&(\Phi'(u_{k}, v_{k})-\Phi'(u, v))(u_{k}-u, v_{k}-v)\\ & = &\int_{0}^{T}(\Phi_{p}(u_{k}')-\Phi_{p}(u'))(u_{k}'-u'){\rm d}t+\int_{0}^{T}g(t)(\Phi_{p}(u_{k})-\Phi_{p}(u))(u_{k}-u){\rm d}t\\ &&+\int_{0}^{T}(\Phi_{p}(v_{k}')-\Phi_{p}(v'))(v_{k}'-v'){\rm d}t+\int_{0}^{T}h(t)(\Phi_{p}(v_{k})-\Phi_{p}(v))(v_{k}-v){\rm d}t\\ &&+\sum\limits_{j = 1}^{n}(I_{j}(u_{k}(t_{j}))-I_{j}(u(t_{j})))(u_{k}(t_{j})-u(t_{j}))\\ &&+\sum\limits_{j = 1}^{n}(J_{j}(v_{k}(t_{j}))-J_{j}(v(t_{j})))(v_{k}(t_{j})-v(t_{j}))\\ &&+\int_{0}^{T}(f_{u}(u, v)-f_{u}(u_{k}, v_{k}))(u_{k}-u){\rm d}t+\int_{0}^{T}(f_{v}(u, v)-f_{v}(u_{k}, v_{k}))(v_{k}-v){\rm d}t. \end{eqnarray} $

可知,当$ p\geq2 $时,对任意的$ x,y\in \mathbb{R} $,存在$ c_{p}>0 $使得

$ (|x|^{p-2}x-|y|^{p-2}y)(x-y)\geq c_{p}|x-y|^{p}. $

由(2.10)式可得

(2.11) $ \begin{eqnarray} c_{p}\|u_{k}-u,v_{k}-v\|^{p} &\leq &\left\|\Phi'(u_{k},v_{k})-\Phi'(u,v)\right\|\|u_{k}-u,v_{k}-v\|\\ &&-\sum\limits_{j = 1}^{n}(I_{j}(u_{k}(t_{j}))-I_{j}(u(t_{j})))(u_{k}(t_{j})-u(t_{j}))\\ &&-\sum\limits_{j = 1}^{n}(J_{j}(v_{k}(t_{j}))-J_{j}(v(t_{j})))(v_{k}(t_{j})-v(t_{j}))\\ &&-\int_{0}^{T}(f_{u}(u,v)-f_{u}(u_{k},v_{k}))(u_{k}-u){\rm d}t\\ &&-\int_{0}^{T}(f_{v}(u,v)-f_{v}(u_{k},v_{k}))(v_{k}-v){\rm d}t. \end{eqnarray} $

由(2.5)–(2.11)式可知:当$ p\geq 2,\; k\rightarrow+\infty $时, $ (u_{k},v_{k})\rightarrow (u,v) $,即在$ X $中, $ (u_{k},v_{k}) $强收敛到$ (u,v) $.

当$ 1<p<2 $时,根据文献[1]的结果可知:存在$ d_{p}>0 $使得

$ \begin{eqnarray*} &&\int_{0}^{T}(\Phi_{p}(u_{k}')-\Phi_{p}(u'))(u_{k}'-u'){\rm d}t+\int_{0}^{T}g(t)(\Phi_{p}(u_{k})-\Phi_{p}(u))(u_{k}-u){\rm d}t\\ &&+\int_{0}^{T}(\Phi_{p}(v_{k}')-\Phi_{p}(v'))(v_{k}'-v'){\rm d}t+\int_{0}^{T}h(t)(\Phi_{p}(v_{k})-\Phi_{p}(v))(v_{k}-v){\rm d}t\\ &\geq& d_{p}\left(\frac{2^{p-2}\|u_{k}-u\|^{2}}{(\|u_{k}\|+\|u\|)^{2-p}}+\frac{2^{p-2}\|v_{k}-v\|^{2}}{(\|v_{k}\|+\|v\|)^{2-p}}\right). \end{eqnarray*} $

类似可得,在$ X $中, $ (u_{k},v_{k})\rightarrow (u,v) $,所以泛函$ \Phi $满足(PS)条件.

首先先证明定理1.1.

证  由条件(H2)可知

$ f(u,v) = o(|u|^{p}+|v|^{p}),\; \mbox{当}\ |u|+|v|\rightarrow 0. $

故$ \forall\varepsilon>0 $, $ \exists\delta>0 $,对于空间$ X $中的$ (u,v) $,当$ |u|,|v|<\delta $时,有

(3.1) $ \begin{eqnarray} |f(u,v)|\leq \varepsilon(|u|^{p}+|v|^{p}). \end{eqnarray} $

由Sobolev嵌入定理可知: $ \forall(u,v)\in X $,存在$ c_{2}>0 $,使得

(3.2) $ \begin{eqnarray} \|(u,v)\|^{p}\leq c_{2}\|(u,v)\|_{X}^{p}. \end{eqnarray} $

令$ \varepsilon = \frac{1}{2pc_{2}} $, $ \forall (u,v)\in X $,其中$ \|(u,v)\|_{X}\leq \frac{\delta}{c_{1}} $, $ \|u\|_{\infty}\leq\delta $以及$ \|v\|_{\infty}\leq\delta $,由条件(H1)和(2.3), (3.1), (3.2)式可知

$ \begin{eqnarray*} \Phi(u,v)& = &\frac{1}{p}\|(u,v)\|_{X}^{p}+\sum \limits_{j = 1}^{n}\int_{0}^{u(t_{j})}I_{j}(t){\rm d}t+\sum \limits_{j = 1}^{n}\int_{0}^{v(t_{j})}J_{j}(t){\rm d}t-\int_{0}^{T}f(u,v){\rm d}t\\ &\geq & \frac{1}{p}\|(u,v)\|_{X}^{p}-\int_{0}^{T}f(u,v){\rm d}t\\ &\geq & \frac{1}{p}\|(u,v)\|_{X}^{p}-\varepsilon\int_{0}^{T}(|u|^{p}+|v|^{p}){\rm d}t\\ & = &\frac{1}{p}\|(u,v)\|_{X}^{p}-\varepsilon(\|u\|_{p}^{p}+\|v\|_{p}^{p})\\ &\geq & \frac{1}{p}\|(u,v)\|_{X}^{p}-\frac{1}{2p}\|(u,v)\|_{X}^{p}\\ & = &\frac{1}{2p}\|(u,v)\|_{X}^{p}. \end{eqnarray*} $

取$ \alpha = \frac{\delta^{p}}{2p(c_{1})^{p}} $, $ \rho = \frac{\delta}{c_{1}} $时,则$ \Phi(u,v)\geq \alpha, $$ \forall u\in X\bigcap \partial B_{\rho}. $满足山路引理条件(1),接下来,证明山路引理条件(2),由条件(H2), (H3)可知,对任意的$ (u,v)\in \mathbb{R} ^{2} $,存在$ c_{3}>0 $,使得

(3.3) $ \begin{eqnarray} f(u,v)\geq c(|u|^{\beta}+|v|^{\beta})-c_{3}. \end{eqnarray} $

因此

$ \begin{eqnarray*} \Phi(u,v)& = &\frac{1}{p}\|(u,v)\|_{X}^{p}+\sum \limits_{j = 1}^{n}\int_{0}^{u(t_{j})}I_{j}(t){\rm d}t+\sum \limits_{j = 1}^{n}\int_{0}^{v(t_{j})}J_{j}(t){\rm d}t-\int_{0}^{T}f(u,v)\\ &\leq& \frac{1}{p}\|(u,v)\|_{X}^{p}+\delta_{j}\sum \limits_{j = 1}^{n}|u|^{v_{j}}+\delta_{j}\sum \limits_{j = 1}^{n}|v|^{v_{j}}-\int_{0}^{T}c(|u|^{\beta}+|v|^{\beta})-c_{3}{\rm d}t\\ &\leq& \frac{1}{p}\|(u,v)\|_{X}^{p}+\delta_{j}\sum \limits_{j = 1}^{n}\|u\|_{\infty}^{v_{j}}+\delta_{j}\sum \limits_{j = 1}^{n}\|v\|_{\infty}^{v_{j}}-c\int_{0}^{T}(|u|^{\beta}+|v|^{\beta}){\rm d}t+c_{3}T\\ &\leq& \frac{1}{p}\|(u,v)\|_{X}^{p}+2c_{1}^{v_{j}}\delta_{j}\sum \limits_{j = 1}^{n}\|(u,v)\|_{X}^{v_{j}}-c\|u\|_{\beta}^{\beta}-c\|v\|_{\beta}^{\beta}+c_{3}T. \end{eqnarray*} $

令$ \|u\|_{W_{0}^{1,p}(0,T)} = \|v\|_{W_{0}^{1,p}(0,T)} = 1 $, $ \delta = \max\{\delta_{1},\delta_{2},\cdots ,\delta_{n}\} $,则

$ \begin{eqnarray*} \Phi(Nu,Nv)&\leq& \frac{1}{p}\|(Nu,Nv)\|_{X}^{p}+2nc_{1}^{v_{j}}\delta\|(Nu,Nv)\|_{X}^{v_{j}}-c\|Nu\|_{\beta}^{\beta}-c\|Nv\|_{\beta}^{\beta}+c_{3}T\\ &\leq&\frac{2}{p}N^{p}+4nc_{1}^{v_{j}}\delta N^{v_{j}}-cN^{\beta}\|u\|_{\beta}^{\beta}-cN^{\beta}\|v\|_{\beta}^{\beta}+c_{3}T\\ &\leq&\frac{2}{p}N^{p}+4nc_{1}^{\theta}\delta N^{\theta}-cN^{\beta}\|u\|_{\beta}^{\beta}-cN^{\beta}\|v\|_{\beta}^{\beta}+c_{3}T. \end{eqnarray*} $

因为$ \beta>\theta>p $, $ \|u\|_{\beta}^{\beta}>0,\; \|v\|_{\beta}^{\beta}>0 $,当$ N\rightarrow +\infty $时, $ \Phi(Nu,Nv)\rightarrow-\infty $.所以存在$ N_{0}>\rho $,使得$ \Phi(N_ {0}u,N_ {0}v)\leq 0 $,满足山路引理条件(2),即泛函$ \Phi $至少存在一个临界点,所以问题(1.1)至少存在一个弱解.

接下来证明定理1.2.

证  显然, $ \Phi(0,0) = 0 $.由条件(H4)可知,由于$ f_{s}(s,t) $和$ f_{t}(s,t) $分别关于$ s,t $为奇函数, $ I_{j}(x) $和$ J_{j}(x) $分别关于$ x $为奇函数,所以泛函$ \Phi $是偶的.对于对称山路引理条件(1)的验证,由于其证明过程与定理1.1的证明类似,这里省略.接下来证明对称山路引理的条件(2).由特征值问题(1.5)可知,该问题所有特征值均为正的,且记为$ \lambda_{n}(p)\ (n = 1,2,3,\cdots ) $ (参见文献[18,引理2.2.3]).设$ E_{n} $分别表示$ \lambda_{n}(p) $相对应的特征空间,则Sobolev空间$ W_{0}^{1,p}(0,T) = \overline{\bigoplus\limits_{i\in \mathbb{N}}E_{i}} $,由此可知, $ X = W_{0}^{1,p}(0,T)\times W_{0}^{1,p}(0,T) = \overline{\bigoplus\limits_{i\in \mathbb{N}}E_{i}\times E_{i}} $.设$ W = \bigoplus\limits_{i = 1}^{2}E_{i}\times E_{i} $,并且$ V = \overline{\bigoplus\limits_{i = 3}^{+\infty}E_{i}\times E_{i}} $,则$ X = V+W $,其中$ W $是有限维的.因为有限维空间的范数等价,所以对$ \forall(u,v)\in W $,存在$ c_{4}>0 $,使得

(3.4) $ \begin{eqnarray} c_{4}\|(u,v)\|_{X}^{\beta}\leq \|(u,v)\|^{\beta} = \|u\|_{\beta}^{\beta}+\|v\|_{\beta}^{\beta}. \end{eqnarray} $

由(2.3), (3.3)和(3.4)式可知

$ \begin{eqnarray*} \Phi(u,v)& = &\frac{1}{p}\|(u,v)\|_{X}^{p}+\sum \limits_{j = 1}^{n}\int_{0}^{u(t_{j})}I_{j}(t){\rm d}t+\sum \limits_{j = 1}^{n}\int_{0}^{v(t_{j})}J_{j}(t){\rm d}t-\int_{0}^{T}f(u,v)\\ &\leq& \frac{1}{p}\|(u,v)\|_{X}^{p}+\delta_{j}\sum \limits_{j = 1}^{n}|u|^{v_{j}}+\delta_{j}\sum \limits_{j = 1}^{n}|v|^{v_{j}}-\int_{0}^{T}[c(|u|^{\beta}+|v|^{\beta})-c_{3}]{\rm d}t\\ &\leq &\frac{1}{p}\|(u,v)\|_{X}^{p}+\delta_{j}\sum \limits_{j = 1}^{n}\|u\|_{\infty}^{v_{j}}+\delta_{j}\sum \limits_{j = 1}^{n}\|v\|_{\infty}^{v_{j}}-c\int_{0}^{T}(|u|^{\beta}+|v|^{\beta}){\rm d}t+c_{3}T\\ & = & \frac{1}{p}\|(u,v)\|_{X}^{p}+\delta_{j}\sum \limits_{j = 1}^{n}\|u\|_{\infty}^{v_{j}}+\delta_{j}\sum \limits_{j = 1}^{n}\|v\|_{\infty}^{v_{j}}-c(\|u\|_{\beta}^{\beta}+\|v\|_{\beta}^{\beta})+c_{3}T\\ &\leq& \frac{1}{p}\|(u,v)\|_{X}^{p}+2nc_{1}^{v_{j}}\delta\|(u,v)\|_{X}^{v_{j}}-cc_{4}\|(u,v)\|_{X}^{\beta}+c_{3}T. \end{eqnarray*} $

因为$ \beta>\theta>p $,当$ \|(u,v)\|_{X}\rightarrow +\infty $时, $ \Phi(u,v)\leq0 $,所以对于$ X $中的任意有限维子空间$ W $,存在常数$ R = R(W) $,使得在$ W\backslash B_{R(W)} $上有$ \varphi(u)\leq 0 $.那么, $ \varphi $有一列无界临界点,所以问题(1.1)有无穷多个弱解.

接下来,证明定理1.3和1.4.显然,根据类似的方法,当满足条件(H1)', (H2), (H3)'时,可证得$ \Phi $满足(PS)条件.先证明定理1.3.

证  显然, $ \Phi(0,0) = 0 $.令$ \delta = \max\{\delta_{1},\delta_{2},\cdots ,\delta_{n}\} $, $ \varepsilon = \frac{1}{2pc_{2}} $,由(2.3), (3.1), (3.2)式和条件(H1)', (H2), (H3)'可知

$ \begin{eqnarray*} \Phi(u,v)& = &\frac{1}{p}\|(u,v)\|_{X}^{p}+\sum \limits_{j = 1}^{n}\int_{0}^{u(t_{j})}I_{j}(t){\rm d}t+\sum \limits_{j = 1}^{n}\int_{0}^{v(t_{j})}J_{j}(t){\rm d}t-\int_{0}^{T}f(u,v){\rm d}t\\ &\geq & \frac{1}{p}\|(u,v)\|_{X}^{p}-\sum \limits_{j = 1}^{n}\delta_{j}|u|^{\mu}-\sum \limits_{j = 1}^{n}\delta_{j}|v|^{\mu}-\int_{0}^{T}f(u,v){\rm d}t\\ &\geq & \frac{1}{p}\|(u,v)\|_{X}^{p}-n\delta\|u\|_{\infty}^{^\mu}-n\delta\|v\|_{\infty}^{^\mu}-\varepsilon \int_{0}^{T}(|u|^{p}+|v|^{p}){\rm d}t\\ &\geq & \frac{1}{p}\|(u,v)\|_{X}^{p}-2nc_{1}^{\mu}\delta\|(u,v)\|_{X}^{\mu}-\frac{1}{2pc_{2}}(\|u\|_{p}^{p}+\|v\|_{p}^{p})\\ &\geq & \frac{1}{p}\|(u,v)\|_{X}^{p}-2nc_{1}^{\mu}\delta\|(u,v)\|_{X}^{\mu}-\frac{1}{2p}\|(u,v)\|_{X}^{p}\\ & = &\frac{1}{2p}\|(u,v)\|_{X}^{p}-2nc_{1}^{\mu}\delta\|(u,v)\|_{X}^{\mu}. \end{eqnarray*} $

因为$ \mu>p $,所以当$ \|(u,v)\|_{X} $取足够小时, $ \Phi(u,v)>0 $,满足山路引理条件(1).接下来证明山路引理条件(2),由(2.3)式和条件(H1)'可知

$ \begin{eqnarray*} \Phi(u,v)& = &\frac{1}{p}\|(u,v)\|_{X}^{p}+\sum \limits_{j = 1}^{n}\int_{0}^{u(t_{j})}I_{j}(t){\rm d}t+\sum \limits_{j = 1}^{n}\int_{0}^{v(t_{j})}J_{j}(t){\rm d}t-\int_{0}^{T}f(u,v)\\ &\leq& \frac{1}{p}\|(u,v)\|_{X}^{p}-\int_{0}^{T}[c(|u|^{\beta}+|v|^{\beta})-c_{3}]{\rm d}t\\ &\leq& \frac{1}{p}\|(u,v)\|_{X}^{p}-c\int_{0}^{T}(|u|^{\beta}+|v|^{\beta}){\rm d}t+c_{3}T\\ &\leq& \frac{1}{p}\|(u,v)\|_{X}^{p}-c\|u\|_{\beta}^{\beta}-c\|u\|_{\beta}^{\beta}+c_{3}T. \end{eqnarray*} $

令$ \|u\|_{W_{0}^{1,p}(0,T)} = \|v\|_{W_{0}^{1,p}(0,T)} = 1 $,所以

$ \begin{eqnarray*} \Phi(Nu,Nv)&\leq& \frac{1}{p}\|(Nu,Nv)\|_{X}^{p}-c\|Nu\|_{\beta}^{\beta}-c\|Nv\|_{\beta}^{\beta}+c_{3}T\\ &\leq&\frac{2}{p}N^{p}-cN^{\beta}\|u\|_{\beta}^{\beta}-cN^{\beta}\|v\|_{\beta}^{\beta}+c_{3}T\\ &\leq&\frac{2}{p}N^{p}-cN^{\beta}\|u\|_{\beta}^{\beta}-cN^{\beta}\|v\|_{\beta}^{\beta}+c_{3}T. \end{eqnarray*} $

因为$ \beta>p $,所以当$ N\rightarrow+\infty $时, $ \Phi(Nu,Nv)\rightarrow-\infty $.类似可得,存在$ N_{0}>\rho $,使得$ \Phi(N_ {0}u,N_ {0}v)\leq 0 $,满足山路引理条件(2),即泛函$ \Phi $至少存在一个临界点,所以问题(1.1)至少存在一个弱解.

最后证明定理1.4.

证  显然, $ \Phi(0,0) = 0 $,泛函$ \Phi $满足条件(H4)时, $ \Phi $是偶的.运用对称山路引理可以得到解的多重性.对于对称山路引理条件(1)的验证和定理1.3的证明过程类似,这里省略.接下来证明对称山路引理的条件(2),由条件(H1)', (2.3)和(3.4)式可知

$ \begin{eqnarray*} \Phi(u,v)& = &\frac{1}{p}\|(u,v)\|_{X}^{p}+\sum \limits_{j = 1}^{n}\int_{0}^{u(t_{j})}I_{j}(t){\rm d}t+\sum \limits_{j = 1}^{n}\int_{0}^{v(t_{j})}J_{j}(t){\rm d}t-\int_{0}^{T}f(u,v){\rm d}t\\ &\leq& \frac{1}{p}\|(u,v)\|_{X}^{p}-\int_{0}^{T}[c(|u|^{\beta}+|v|^{\beta})-c_{3}]{\rm d}t\\ &\leq& \frac{1}{p}\|(u,v)\|_{X}^{p}-c\int_{0}^{T}(|u|^{\beta}+|v|^{\beta}){\rm d}t+c_{3}T\\ &\leq& \frac{1}{p}\|(u,v)\|_{X}^{p}-cc_{4}\|(u,v)\|_{X}^{\beta}+c_{3}T. \end{eqnarray*} $

因为$ \beta>p $,当$ \|(u,v)\|_{X}\rightarrow +\infty $时, $ \Phi(u,v)\rightarrow-\infty $,所以对于$ X $中的任意有限维子空间$ W $,存在常数$ R = R(W) $,使得在$ W\backslash B_{R(W)} $上有$ \Phi\leq 0 $.所以泛函$ \Phi(u,v) $存在无穷多个临界点,即问题(1.1)有无穷多个弱解.