非线性阿贝尔方程

$ \begin{align} \frac{dx}{dt} = a(t)x^{3}+b(t)x^{2}+c(t)x+d(t) \end{align} $

(1.1)

在物理和工程技术等许多领域有着重要应用^[1-2], 方程(1.1)的数学性质已被数学和物理学者^[3-15]进行了深入研究. 文献[14, 15]提出了得到阿贝尔方程的通解的一种方法, 他们都假定$ y = y_{1}(t) $是方程(1.1)的一个特解, 然后通过变量代换方法, 给出了阿贝尔方程的通解; 文献[16]假设$ \gamma = \gamma(t) $是阿贝尔方程的一个周期特解, 然后, 利用变量代换法和不动点定理, 得到阿贝尔方程的其他周期解的存在性.

本文首先考虑下列阿贝尔型方程:

$ \begin{align} \frac{dx}{dt} = a(t)x^{3}+b(t)x^{2}, \end{align} $

(1.2)

文献[15]给出了方程(1.2)可积的充分必要条件, 如下:

命题1.1^[15]   阿贝尔型方程(1.2)可积的充分必要条件是$ a(t), b(t) $满足下列条件:

$ \begin{align} \frac{d}{dt}\Big(\frac{a(t)}{b(t)}\Big) = kb(t), \end{align} $

(1.3)

其中$ k $是常数.

在条件(1.3)成立时, $ b(t)\neq 0 $, (1.3)两边从$ t_{0} $到$ t(t>t_{0}) $积分, 可得:

$ \begin{align} \frac{a(t)}{b(t)} = \frac{a(t_{0})}{b(t_{0})}+\int_{t_{0}}^{t}kb(s)ds, \end{align} $

(1.4)

如果$ k = 0, $则$ \frac{a(t)}{b(t)}\equiv C, $此时, $ a(t) $和$ b(t) $是线性相关的; 如果$ C = 0 $, 则$ a(t) = 0 $, 此时方程(1.2)只有零周期解; 如果$ C\neq 0 $, 则容易验证方程(1.2)有两个常数周期解$ x_{1}(t) = 0 $(二重), $ x_{2}(t) = -\frac{1}{C} $.

如果$ k\neq0, $由(1.4)可知,

$ \begin{align} \frac{a(t)}{b(t)}\rightarrow \infty(t\rightarrow +\infty), \end{align} $

(1.5)

此时, $ a(t), b(t) $不可能都是周期函数. 因此, 当$ a(t) $和$ b(t) $都是周期函数并且线性无关时, 方程(1.2)是不可积的.

本文首先考虑$ a(t) $和$ b(t) $是周期函数时的微分方程(1.2), 此时除了$ a(t) $和$ b(t) $线性相关外, (1.2) 是不可积的. 本文研究在不求出(1.2)的解的情况下, (1.2)的周期解的存在性. 文献[16]利用不动点定理, 得到(1.2)的唯一非零周期解的存在性; 本文受文献[16]的启发, 利用不同于文献[16]的方法, 得到方程(1.2)的唯一非零周期解的存在性; 然后, 讨论了方程(1.1), 在一定条件下, 利用变量代换法, 将方程(1.1)转化为方程(1.2), 从而得到阿贝尔方程(1.1)的两个周期解的存在性.

本文余下部分安排如下: 第二节, 我们给出四个引理以方便以后使用; 第三节, 利用不动点定理得到阿贝尔型方程存在唯一非零周期解的四个定理; 第四节, 当方程的系数函数满足一定条件时, 我们得到了阿贝尔方程的两个周期解的存在性.

$ E^{n} $表示$ n $维实数空间或$ n $维复数空间, $ R $表示实数集合, $ C(R, E^{n}) $表示$ R $到$ E^{n} $的连续向量函数所构成的集合, $ C(R, R) $表示$ R $到$ R $的连续函数所构成的集合.

定义2.1^[17]    设函数$ f(t)\in C(R, E^{n}) $是$ \omega- $周期的, $ a(f, \lambda) = \int_{0}^{\omega}f(t)e^{-i\lambda t}dt $一定存在, $ a(f, \lambda) $称为$ f(t) $的傅里叶系数, 使$ a(f, \lambda)\neq0 $的实数$ \lambda $称为$ f(t) $的傅里叶指数; 存在可数集$ \Lambda_{f} $, 当$ \lambda\in\Lambda_{f} $时, $ a(f, \lambda)\neq0 $, 只要$ \lambda\not\in \Lambda_{f} $, 必有$ a(f, \lambda) = 0 $, $ \Lambda_{f} $称为$ f(t) $的指数集.

定义2.2^[17]   $ \Lambda_{f} $中元素的整系数线性组合所构成的实数集合称为$ f(t) $的模(module)或频率模, 记作$ \mod(f) $, 即

$ \mod(f) = \Big\{\mu|\mu = \sum\limits_{j = 1}^{N}n_{j}\lambda_{j}, n_{j}, N\in Z^{+}, N\geq 1, \lambda_{j}\in \Lambda_{f}\Big\}. $

引理2.1 ^[18]   考虑如下方程:

$ \begin{align} \frac{dx}{dt} = a(t)x+b(t), \end{align} $

(2.1)

这里, $ a(t), b(t) $是$ R $上的$ \omega $-周期连续函数, 如果$ \int_{0}^{\omega}a(t)dt\neq 0, $则方程(2.1)有唯一的$ \omega $-周期连续解$ \eta(t) $, $ \mod(\eta)\subseteq\mod(a(t), b(t)) $, 并且$ \eta(t) $可表示如下:

$ \begin{align}\eta(t) = \left\{ \begin{array}{l} \int_{-\infty}^{t}e^{\int_{s}^{t}a(\tau)d\tau}b(s)ds, \int_{0}^{\omega}a(t)dt< 0\\ - \int_{t}^{+\infty}e^{\int_{s}^{t}a(\tau)d\tau}b(s)ds. \int_{0}^{\omega}a(t)dt> 0. \end{array} \right.\end{align} $

(2.2)

引理2.2 ^[18]   假设$ \omega $-周期连续函数序列$ \{f_{n}(t)\} $在$ R $的任一紧集上收敛, $ f(t) $是一个$ \omega $-周期连续函数, 并且$ \mod(f_{n})\subseteq\mod(f)(n = 1, 2, \cdots) $, 那么$ \{f_{n}(t)\} $在$ R $上一致收敛.

引理2.3 ^[19]   假设$ V $是度量空间, $ C $是$ V $的闭凸集, 其边界是$ \partial C $, 如果$ T:V\rightarrow V $是连续的紧映射, 且使得$ T(\partial C)\subseteq C $, 则$ T $在$ C $上至少有一个不动点.

考虑一维周期微分方程:

$ \begin{align} \frac{dx}{dt} = f(t, x), \end{align} $

(2.3)

这里, $ f:R\times I\rightarrow R $是一个连续函数, $ f(t+\omega, x) = f(t, x), \omega>0.I\subseteq R. $

引理2.4^[20]   如果$ f(t, x) $关于$ x $有三阶连续偏导数, 并且$ f_{xxx}^{'''}(t, x)\neq 0, $则(2.3)最多有三个连续周期解.

为了方便起见, 假设$ f(t) $是$ R $上的$ \omega $-周期连续函数, 我们用下列记号表示:

$ \begin{align} f_{M} = \sup\limits_{t\in [0, \omega]}f(t), \, \, f_{L} = \inf\limits_{t\in [0, \omega]}f(t). \end{align} $

(2.4)

这一节, 我们考虑阿贝尔型方程, 给出了阿贝尔型方程唯一非零周期解的存在性的四个结论.

定理3.1   考虑方程(1.2), $ a(t), b(t) $是$ R $上的$ \omega $-周期连续函数, 如果以下条件成立:

$ (H_{1})\, \, a(t)<0, \quad (H_{2})\, \, b(t)<0, $

则方程(1.2)存在唯一负$ \omega $-周期连续解$ \gamma(t) $, 且有$ -\Big(\frac{b}{a}\Big)_{M}\leq \gamma(t)\leq-\Big(\frac{b}{a}\Big)_{L}. $

证   (1) 证明方程(1.2)的负$ \omega $-周期连续解的存在性.

假设

$ \begin{align} S = \Bigg\{\varphi(t)\in C(R, R)\Big|\varphi(t+\omega) = \varphi(t)\Bigg\}, \end{align} $

(3.1)

任取$ \varphi(t), \psi(t)\in S, $定义度量如下:

$ \begin{align} \rho(\varphi, \psi) = \sup\limits_{t\in [0, \omega]}|\varphi(t)-\psi(t)|, \end{align} $

(3.2)

则$ (S, \rho) $是一个完备度量空间, 取$ S $的一个闭凸集如下:

$ \begin{align} B = \Bigg\{\varphi(t)\in S\Big|-\Big(\frac{b}{a}\Big)_{M}\leq \varphi(t)\leq-\Big(\frac{b}{a}\Big)_{L}, \mod(\varphi)\subseteq \mod(a, b)\Bigg\}. \end{align} $

(3.3)

任意给定$ \varphi(t)\in B, $考虑如下微分方程:

$ \begin{align} \frac{dx}{dt} = a(t)\varphi^{2}(t)x+b(t)\varphi^{2}(t), \end{align} $

(3.4)

因为$ a(t), b(t) $和$ \varphi(t) $都是$ \omega $-周期连续函数, 故$ a(t)\varphi^{2}(t), b(t)\varphi^{2}(t) $也都是$ \omega $-周期连续函数.

由(3.3), $ (H_{1}) $和$ (H_{2}) $, 可得

$ \begin{align} a_{L}\Big(-\Big(\frac{b}{a}\Big)_{M}\Big)^{2}\leq a(t)\varphi^{2}(t)\leq a_{M}\Big(-\Big(\frac{b}{a}\Big)_{L}\Big)^{2}<0, \end{align} $

(3.5)

即

$ \begin{align} a_{L}\Big(\big(\frac{b}{a}\big)_{M}\Big)^{2}\leq a(t)\varphi^{2}(t)\leq a_{M}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}<0, \end{align} $

(3.6)

所以有

$ \begin{align} \frac{1}{\omega}\int_{0}^{\omega}a(t)\varphi^{2}(t)dt<0, \end{align} $

(3.7)

根据引理2.1, 方程(3.4)存在唯一的如下的$ \omega $-周期连续解:

$ \begin{align} \eta(t) = \int_{-\infty}^{t}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}b(s)\varphi^{2}(s)ds, \end{align} $

(3.8)

并且

$ \begin{align} \mod(\eta)\subseteq \mod(a(t)\varphi^{2}(t), b(t)\varphi^{2}(t)), \end{align} $

(3.9)

由(3.3), 可得

$ \begin{align} \mod(a(t)\varphi^{2}(t))\subseteq \mod(a, b), \end{align} $

(3.10)

$ \begin{align} \mod(b(t)\varphi^{2}(t))\subseteq \mod(a, b), \end{align} $

(3.11)

因此有

$ \begin{align} \mod(\eta)\subseteq \mod(a, b). \end{align} $

(3.12)

由$ (H_{1}) $, $ (H_{2}) $, (3.6) 和(3.8), 我们有

$ \begin{align}\eta(t) = &\int_{-\infty}^{t}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}\frac{b(s)}{a(s)}a(s)\varphi^{2}(s)ds \geq(\frac{b}{a})_{M}\int_{-\infty}^{t}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}a(s)\varphi^{2}(s)ds\\ = &-(\frac{b}{a})_{M}\int_{-\infty}^{t}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}d\Big(\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau\Big) = -(\frac{b}{a})_{M}\Big[e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}\Big]_{-\infty}^{t}\\ = &-(\frac{b}{a})_{M}\Big[1-e^{\int_{-\infty}^{t}a(\tau)\varphi^{2}(\tau)d\tau}\Big], (-\infty<t<+\infty)\\ = &-(\frac{b}{a})_{M}, \end{align} $

(3.13)

且

$ \begin{align}\eta(t)& = \int_{-\infty}^{t}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}\frac{b(s)}{a(s)}a(s)\varphi^{2}(s)ds \leq(\frac{b}{a})_{L}\int_{-\infty}^{t}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}a(s)\varphi^{2}(s)ds\\ & = -(\frac{b}{a})_{L}\int_{-\infty}^{t}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}d\Big(\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau\Big) = -(\frac{b}{a})_{L}\Big[e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}\Big]_{-\infty}^{t}\\ & = -(\frac{b}{a})_{L}\Big[1-e^{\int_{-\infty}^{t}a(\tau)\varphi^{2}(\tau)d\tau}\Big], (-\infty<t<+\infty)\\ & = -(\frac{b}{a})_{L}, \end{align} $

(3.14)

故

$ \begin{align} -(\frac{b}{a})_{M}\leq\eta(t)\leq-(\frac{b}{a})_{L}, \end{align} $

(3.15)

因此$ \eta(t)\in B $.

定义映射:

$ \begin{align} (T\varphi)(t) = \eta(t) = \int_{-\infty}^{t}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}b(s)\varphi^{2}(s)ds, \end{align} $

(3.16)

所以任意给定$ \varphi(t)\in B, $有$ (T\varphi)(t)\in B $, 因此$ T:B\rightarrow B $.

现在, 我们证明映射$ T $是紧映射. 在$ B $中任取一序列$ \{\varphi_{n}(t)\}\subseteq B(n = 1, 2, \cdots) $, 则有

$ \begin{align} -\Big(\frac{b}{a}\Big)_{M}\leq\varphi_{n}(t)\leq -\Big(\frac{b}{a}\Big)_{L}, \mod(\varphi_{n})\subseteq \mod(a, b), (n = 1, 2, \cdots) \end{align} $

(3.17)

另外, $ (T\varphi_{n})(t) = x_{\varphi_{n}}(t) $满足

$ \begin{align} \frac{dx_{\varphi_{n}}(t)}{dt} = a(t)\varphi_{n}^{2}(t)x_{\varphi_{n}}(t)+b(t)\varphi_{n}^{2}(t), \end{align} $

(3.18)

因此我们有

$ \begin{align} \Big|\frac{dx_{\varphi_{n}}(t)}{dt}\Big|\leq a_{L}\Big(-\big(\frac{b}{a}\big)_{M}\Big)^{3}-b_{L}\Big(\big(\frac{b}{a}\big)_{M}\Big)^{2}, \mod(x_{\varphi_{n}}(t))\subseteq\mod(a, b), \end{align} $

(3.19)

故$ \Big\{\frac{dx_{\varphi_{_{n}}}(t)}{dt}\Big\} $是一致有界的, 所以, $ \Big\{x_{\varphi_{n}}(t)\Big\} $在$ R $上是一致有界、等度连续的, 根据Ascoli-arzela定理, 对于$ B $中的任一序列$ \Big\{x_{\varphi_{n}}(t)\Big\}\subseteq B $, 总存在一个子列(仍用$ \Big\{x_{\varphi_{n}}(t)\Big\} $表示)使得$ \Big\{x_{\varphi_{n}}(t)\Big\} $在$ R $的任一紧集上一致收敛, 再由(3.19), 根据引理2.2, $ \Big\{x_{\varphi_{n}}(t)\Big\} $在$ R $上一致收敛, 也就是说, $ T $是$ B $上的紧映射.

接下来, 我们证明$ T $是连续映射. 假设$ \{\varphi_{n}(t)\}\subseteq B, \varphi(t)\in B $, 且有

$ \begin{align} \varphi_{n}(t)\rightarrow \varphi(t), (n\rightarrow \infty) \end{align} $

(3.20)

由(3.16), 我们有

$ \begin{eqnarray*} &&|(T\varphi_{n})(t)-(T\varphi)(t)|\\ & = &\Bigg|\int_{-\infty}^{t}e^{\int_{s}^{t}a(\tau)\varphi_{n}^{2}(\tau)d\tau}b(s)\varphi_{n}^{2}(s)ds-\int_{-\infty}^{t}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}b(s)\varphi^{2}(s)ds\Bigg|\\ & = &\Bigg|\int_{-\infty}^{t}e^{\int_{s}^{t}a(\tau)\varphi_{n}^{2}(\tau)d\tau}b(s)\Big(\varphi_{n}^{2}(s)-\varphi^{2}(s)\Big)ds\\ &&+\int_{-\infty}^{t}\Big(e^{\int_{s}^{t}a(\tau)\varphi_{n}^{2}(\tau)d\tau}-e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}\Big)b(s)\varphi^{2}(s)ds\Bigg|\\ & = &\Bigg|\int_{-\infty}^{t}e^{\int_{s}^{t}a(\tau)\varphi_{n}^{2}(\tau)d\tau}b(s)\Big(\varphi_{n}^{2}(s)-\varphi^{2}(s)\Big)ds\\ &&+\int_{-\infty}^{t}e^{\xi}\Big(\int_{s}^{t}a(\tau)\big(\varphi_{n}^{2}(\tau)-\varphi^{2}(\tau)\big)d\tau\Big) b(s)\varphi^{2}(s)ds\Bigg|, \\ \end{eqnarray*} $

这里, $ \xi $介于$ \int_{s}^{t}a(\tau)\varphi^{2}_{n}(\tau)d\tau $和$ \int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau $之间, 因此, $ \xi $在$ a_{L}\Big(\big(\frac{b}{a}\big)_{M}\Big)^{2}(t-s) $和$ a_{M}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s) $之间, 所以有

$ \begin{eqnarray*} &&|(T\varphi_{n})(t)-(T\varphi)(t)|\\ &\leq&\int_{-\infty}^{t}e^{a_{M}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s)}|b(s)|\Big|\varphi_{n}(s)+\varphi(s)\Big|\Big|\varphi_{n}(s)-\varphi(s)\Big|ds\\ &&+\int_{-\infty}^{t}e^{a_{M}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s)}\Big(\int_{s}^{t}|a(\tau)|\big|\varphi_{n}(\tau)+\varphi(\tau)\big|\big|\varphi_{n}(\tau)-\varphi(\tau)\big|d\tau\Big) |b(s)|\varphi^{2}(s)ds\\ &\leq&\Bigg(\int_{-\infty}^{t}e^{a_{M}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s)}|b(s)|\Big|\varphi_{n}(s)+\varphi(s)\Big|ds\\ &&+\int_{-\infty}^{t}e^{a_{M}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s)}\Big(\int_{s}^{t}|a(\tau)|\big|\varphi_{n}(\tau)+\varphi(\tau)\big|d\tau\Big) |b(s)|\varphi^{2}(s)ds\Bigg)\rho(\varphi_{n}, \varphi)\\ &\leq&\Bigg(-2b_{L}\big(\frac{b}{a}\big)_{M}\int_{-\infty}^{t}e^{a_{M}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s)}ds\\ &&+2a_{L}b_{L}\Big(\big(\frac{b}{a}\big)_{M}\Big)^{3}\int_{-\infty}^{t}e^{a_{M}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s)}\big(t-s\big) ds\Bigg)\rho(\varphi_{n}, \varphi)\\ & = &\Bigg(\frac{2b_{L}\big(\frac{b}{a}\big)_{M}}{a_{M}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}}+\frac{2a_{L}b_{L}\Big(\big(\frac{b}{a}\big)_{M}\Big)^{3}}{\Big(a_{M}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}\Big)^{2}}\Bigg)\rho(\varphi_{n}, \varphi), \\ \end{eqnarray*} $

由(3.20)及上式, 可得

$ \begin{align} (T\varphi_{n})(t)\rightarrow (T\varphi)(t), (n\rightarrow \infty) \end{align} $

(3.21)

因此, $ T $是连续映射, 由(3.16), 易知, $ T(\partial B)\subseteq B, $根据引理2.3, $ T $在$ B $上至少有一个不动点, 这个不动点即为方程(1.2)的负$ \omega $-周期连续解$ \gamma(t), $且有

$ \begin{align} -\Big(\frac{b}{a}\Big)_{M}\leq\gamma(t)\leq-\Big(\frac{b}{a}\Big)_{L}. \end{align} $

(3.22)

(2) 我们证明方程(1.2)恰有唯一非零周期解$ \gamma(t) $.

令

$ \begin{align} f(t, x) = a(t)x^{3}+b(t)x^{2}, \end{align} $

(3.23)

则有

$ \begin{align} f_{xxx}^{'''}(t, x) = 6a(t)<0, \end{align} $

(3.24)

由(3.24), 根据引理2.4, 方程(1.2)至多有三个连续周期解, 我们已经知道, 方程(1.2)有三个连续周期解: $ \gamma(t) $和二重零解$ \gamma_{1}(t) = \gamma_{2}(t) = 0 $, 所以方程(1.2)只有唯一的非零$ \omega $-周期连续解$ \gamma(t) $, 且有

$ -\Big(\frac{b}{a}\Big)_{M}\leq \gamma(t)\leq-\Big(\frac{b}{a}\Big)_{L}. $

定理3.1证毕.

定理3.2   考虑方程(1.2), $ a(t), b(t) $是$ R $上的$ \omega $-周期连续函数, 如果以下条件成立:

$ (H_{1})\, \, a(t)<0, \quad (H_{2})\, \, b(t)>0, $

则方程(1.2)存在唯一正$ \omega $-周期连续解$ \gamma(t) $, 且有$ -\Big(\frac{b}{a}\Big)_{M}\leq \gamma(t)\leq-\Big(\frac{b}{a}\Big)_{L}. $

证   令

$ \begin{align} x = -u, \end{align} $

(3.25)

则方程(1.2)化为

$ \begin{align} \frac{du}{dt} = a(t)u^{3}-b(t)u^{2}, \end{align} $

(3.26)

由$ (H_{1}) $和$ (H_{2}), $方程(3.26)满足定理3.1的所有条件, 根据定理3.1, 方程(3.26)有唯一的负$ \omega $-周期连续解$ u_{1}(t), $并且

$ \begin{align} -\Big(-\frac{b}{a}\Big)_{M}\leq u_{1}(t)\leq-\Big(-\frac{b}{a}\Big)_{L}, \end{align} $

(3.27)

即

$ \begin{align} \Big(\frac{b}{a}\Big)_{L}\leq u_{1}(t)\leq\Big(\frac{b}{a}\Big)_{M}. \end{align} $

(3.28)

由(3.25), 可得方程(1.2)存在唯一的正$ \omega $-周期连续解$ \gamma(t), $并且

$ -\Big(\frac{b}{a}\Big)_{M}\leq \gamma(t)\leq-\Big(\frac{b}{a}\Big)_{L}. $

定理3.2证毕.

定理3.3    考虑方程(1.2), $ a(t), b(t) $是$ R $上的$ \omega $-周期连续函数, 如果以下条件成立:

$ (H_{1})\, \, a(t)>0, \quad (H_{2})\, \, b(t)>0, $

则方程(1.2)存在唯一负$ \omega $-周期连续解$ \gamma(t) $, 且有$ -\Big(\frac{b}{a}\Big)_{M}\leq \gamma(t)\leq-\Big(\frac{b}{a}\Big)_{L}. $

证   (1) 我们证明方程(1.2)的负$ \omega $-周期连续解的存在性.

假设

$ \begin{align} S = \Bigg\{\varphi(t)\in C(R, R)\Big|\varphi(t+\omega) = \varphi(t)\Bigg\}, \end{align} $

(3.29)

任取$ \varphi(t), \psi(t)\in S, $定义度量如下:

$ \begin{align} \rho(\varphi, \psi) = \sup\limits_{t\in [0, \omega]}|\varphi(t)-\psi(t)|, \end{align} $

(3.30)

则$ (S, \rho) $是一个完备度量空间, 取$ S $的一个闭凸集如下:

$ \begin{align} B = \Bigg\{\varphi(t)\in S\Big| -\Big(\frac{b}{a}\Big)_{M}\leq \varphi(t)\leq-\Big(\frac{b}{a}\Big)_{L}, \mod(\varphi)\subseteq \mod(a, b)\Bigg\}. \end{align} $

(3.31)

任意给定$ \varphi(t)\in B, $考虑如下微分方程:

$ \begin{align} \frac{dx}{dt} = a(t)\varphi^{2}(t)x+b(t)\varphi^{2}(t), \end{align} $

(3.32)

因为$ a(t), b(t) $和$ \varphi(t) $都是$ \omega $-周期连续函数, 故$ a(t)\varphi^{2}(t), b(t)\varphi^{2}(t) $也都是$ \omega $-周期连续函数.

由(3.31), $ (H_{1}) $和$ (H_{2}) $, 可得

$ \begin{align} 0<a_{L}\Big(-\Big(\frac{b}{a}\Big)_{L}\Big)^{2}\leq a(t)\varphi^{2}(t)\leq a_{M}\Big(-\Big(\frac{b}{a}\Big)_{M}\Big)^{2}, \end{align} $

(3.33)

即

$ \begin{align} 0<a_{L}\Big(\frac{b}{a}\Big)_{L}^{2}\leq a(t)\varphi^{2}(t)\leq a_{M}\Big(\frac{b}{a}\Big)_{M}^{2}, \end{align} $

(3.34)

所以有

$ \begin{align} \frac{1}{\omega}\int_{0}^{\omega}a(t)\varphi^{2}(t)dt>0, \end{align} $

(3.35)

根据引理2.1, 方程(3.32) 存在唯一如下的$ \omega $-周期连续解:

$ \begin{align} \eta(t) = -\int_{t}^{+\infty}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}b(s)\varphi^{2}(s)ds, \end{align} $

(3.36)

且

$ \begin{align} \mod(\eta)\subseteq \mod(a(t)\varphi^{2}(t), b(t)\varphi^{2}(t)), \end{align} $

(3.37)

由(3.31), 可得

$ \begin{align} \mod(a(t)\varphi^{2}(t))\subseteq \mod(a, b), \end{align} $

(3.38)

$ \begin{align} \mod(b(t)\varphi^{2}(t))\subseteq \mod(a, b), \end{align} $

(3.39)

因此我们有

$ \begin{align} \mod(\eta)\subseteq \mod(a, b). \end{align} $

(3.40)

由$ (H_{1}) $, $ (H_{2}) $, (3.36)和(3.38), 我们有

$ \begin{align}\eta(t)& = -\int_{t}^{+\infty}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}\frac{b(s)}{a(s)}a(s)\varphi^{2}(s)ds \geq-(\frac{b}{a})_{M}\int_{t}^{+\infty}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}a(s)\varphi^{2}(s)ds\\ & = (\frac{b}{a})_{M}\int_{t}^{+\infty}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}d\Big(\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau\Big) = (\frac{b}{a})_{M}\Big[e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}\Big]_{t}^{+\infty}\\ & = (\frac{b}{a})_{M}\Big[e^{\int_{+\infty}^{t}a(\tau)\varphi^{2}(\tau)d\tau}-1\Big], (-\infty<t<+\infty)\\ & = -(\frac{b}{a})_{M}, \end{align} $

(3.41)

且

$ \begin{align}\eta(t)& = -\int_{t}^{+\infty}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}\frac{b(s)}{a(s)}a(s)\varphi^{2}(s)ds \leq-(\frac{b}{a})_{L}\int_{t}^{+\infty}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}a(s)\varphi^{2}(s)ds\\ & = (\frac{b}{a})_{L}\int_{t}^{+\infty}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}d\Big(\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau\Big) = (\frac{b}{a})_{L}\Big[e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}\Big]_{t}^{+\infty}\\ & = (\frac{b}{a})_{L}\Big[e^{\int_{+\infty}^{t}a(\tau)\varphi^{2}(\tau)d\tau}-1\Big], (-\infty<t<+\infty)\\ & = -(\frac{b}{a})_{L}, \end{align} $

(3.42)

故有

$ \begin{align} -(\frac{a}{b})_{M}\leq\eta(t)\leq-(\frac{a}{b})_{L}, \end{align} $

(3.43)

因此$ \eta(t)\in B $.

定义映射:

$ \begin{align} (T\varphi)(t) = \eta(t) = -\int_{t}^{+\infty}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}b(s)\varphi^{2}(s)ds, \end{align} $

(3.44)

所以任意给定$ \varphi(t)\in B, $有$ (T\varphi)(t)\in B $, 因此 $ T:B\rightarrow B $.

现在, 我们证明映射$ T $是紧映射. 在$ B $中任取一序列$ \{\varphi_{n}(t)\}\subseteq B(n = 1, 2, \cdots) $, 则有

$ \begin{align} -\Big(\frac{b}{a}\Big)_{M}\leq\varphi_{n}(t)\leq -\Big(\frac{b}{a}\Big)_{L}, \mod(\varphi_{n})\subseteq \mod(a, b), \quad(n = 1, 2, \cdots), \end{align} $

(3.45)

另外, $ (T\varphi_{n})(t) = x_{\varphi_{n}}(t) $满足

$ \begin{align} \frac{dx_{\varphi_{n}}(t)}{dt} = a(t)\varphi_{n}^{2}(t)x_{\varphi_{n}}(t)+b(t)\varphi_{n}^{2}(t), \end{align} $

(3.46)

因此我们有

$ \begin{align} \Big|\frac{dx_{\varphi_{n}}(t)}{dt}\Big|\leq a_{M}\Big(\big(\frac{b}{a}\big)_{M}\Big)^{3}+b_{M}\Big(\big(\frac{b}{a}\big)_{M}\Big)^{2}, \mod(x_{\varphi_{n}}(t))\subseteq\mod(a, b), \end{align} $

(3.47)

故$ \Big\{\frac{dx_{\varphi_{_{n}}}(t)}{dt}\Big\} $是一致有界的, 所以, $ \Big\{x_{\varphi_{n}}(t)\Big\} $在$ R $上是一致有界、等度连续的, 根据Ascoli-arzela定理, 对于$ B $中的任一序列$ \Big\{x_{\varphi_{n}}(t)\Big\}\subseteq B $, 总存在一个子列(仍用$ \Big\{x_{\varphi_{n}}(t)\Big\} $表示)使得$ \Big\{x_{\varphi_{n}}(t)\Big\} $在$ R $的任一紧集上一致收敛, 由(3.47), 根据根据引理2.2, $ \Big\{x_{\varphi_{n}}(t)\Big\} $在$ R $上一致收敛, 也就是说, $ T $是$ B $上的紧映射.

接下来, 我们证明$ T $是连续映射. 假设$ \{\varphi_{n}(t)\}\subseteq B, \varphi(t)\in B $, 且有

$ \begin{align} \varphi_{n}(t)\rightarrow \varphi(t), (n\rightarrow \infty) \end{align} $

(3.48)

由(3.44), 我们有

$ \begin{eqnarray*} &&|(T\varphi_{n})(t)-(T\varphi)(t)|\\ & = &\Bigg|\int_{t}^{+\infty}e^{\int_{s}^{t}a(\tau)\varphi_{n}^{2}(\tau)d\tau}b(s)\varphi_{n}^{2}(s)ds-\int_{t}^{+\infty}e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}b(s)\varphi^{2}(s)ds\Bigg|\\ & = &\Bigg|\int_{t}^{+\infty}e^{\int_{s}^{t}a(\tau)\varphi_{n}^{2}(\tau)d\tau}b(s)\Big(\varphi_{n}^{2}(s)-\varphi^{2}(s)\Big)ds\\ &&+\int_{t}^{+\infty}\Big(e^{\int_{s}^{t}a(\tau)\varphi_{n}^{2}(\tau)d\tau}-e^{\int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau}\Big)b(s)\varphi^{2}(s)ds\Bigg|\\ & = &\Bigg|\int_{t}^{+\infty}e^{\int_{s}^{t}a(\tau)\varphi_{n}^{2}(\tau)d\tau}b(s)\Big(\varphi_{n}^{2}(s)-\varphi^{2}(s)\Big)ds\\ &&+\int_{t}^{+\infty}e^{\xi}\Big(\int_{s}^{t}a(\tau)\big(\varphi_{n}^{2}(\tau)-\varphi^{2}(\tau)\big)d\tau\Big) b(s)\varphi^{2}(s)ds\Bigg|.\\ \end{eqnarray*} $

这里, $ \xi $介于$ \int_{s}^{t}a(\tau)\varphi^{2}(\tau)d\tau $和$ \int_{s}^{t}a(\tau)\varphi^{2}_{n}(\tau)d\tau $之间, 因此, $ \xi $在$ a_{M}\Big(\big(\frac{b}{a}\big)_{M}\Big)^{2}(t-s) $和$ a_{L}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s) $之间, 故我们有

$ \begin{eqnarray*} &&|(T\varphi_{n})(t)-(T\varphi)(t)|\\ &\leq&\int_{t}^{+\infty}e^{a_{L}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s)}|b(s)|\Big|\varphi_{n}(s)+\varphi(s)\Big|\Big|\varphi_{n}(s)-\varphi(s)\Big|ds\\ &&+\int_{t}^{+\infty}e^{a_{L}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s)}\Big(\int_{s}^{t}|a(\tau)|\big|\varphi_{n}(\tau)+\varphi(\tau)\big|\big|\varphi_{n}(\tau)-\varphi(\tau)\big|d\tau\Big) |b(s)|\varphi^{2}(s)ds\\ &\leq&\Bigg(\int_{t}^{+\infty}e^{a_{L}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s)}|b(s)|\Big|\varphi_{n}(s)+\varphi(s)\Big|ds\\ &&+\int_{t}^{t+\infty}e^{a_{L}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s)}\Big(\int_{s}^{t}|a(\tau)|\big|\varphi_{n}(\tau)+\varphi(\tau)\big|d\tau\Big) |b(s)|\varphi^{2}(s)ds\Bigg)\rho(\varphi_{n}, \varphi)\\ &\leq&\Bigg(2b_{M}\big(\frac{b}{a}\big)_{M}\int_{t}^{+\infty}e^{a_{L}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s)}ds\\ &&+2a_{M}b_{M}\Big(\big(\frac{b}{a}\big)_{M}\Big)^{3}\int_{t}^{+\infty}e^{a_{L}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}(t-s)}\big(s-t\big) ds\Bigg)\rho(\varphi_{n}, \varphi)\\ & = &\Bigg(\frac{2b_{M}\big(\frac{b}{a}\big)_{M}}{a_{L}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}}+\frac{2a_{M}b_{M}\Big(\big(\frac{b}{a}\big)_{M}\Big)^{3}}{\Big(a_{L}\Big(\big(\frac{b}{a}\big)_{L}\Big)^{2}\Big)^{2}}\Bigg)\rho(\varphi_{n}, \varphi), \\ \end{eqnarray*} $

由(3.48)和上式, 可得

$ \begin{align} (T\varphi_{n})(t)\rightarrow (T\varphi)(t), (n\rightarrow \infty). \end{align} $

(3.49)

因此, $ T $是连续映射, 由(3.44), 易得, $ T(\partial B)\subseteq B, $根据引理2.3, $ T $在$ B $上至少有一个不动点, 这个不动点即为方程(1.2)的$ \omega $-周期连续解, 且有

$ \begin{align} -\Big(\frac{b}{a}\Big)_{M}\leq\gamma(t)\leq-\Big(\frac{b}{a}\Big)_{L}. \end{align} $

(3.50)

(2) 我们证明方程(1.2)恰有唯一非零周期解$ \gamma(t) $.

令

$ \begin{align} f(t, x) = a(t)x^{3}+b(t)x^{2}, \end{align} $

(3.51)

则有

$ \begin{align} f_{xxx}^{'''}(t, x) = 6a(t)>0, \end{align} $

(3.52)

由(3.52), 根据引理2.4, 方程(1.2)至多有三个连续周期解, 我们已经知道, 方程(1.2)有三个连续周期解: $ \gamma(t) $和二重零解$ \gamma_{1}(t) = \gamma_{2}(t) = 0 $, 所以方程(1.2)有唯一的非零$ \omega $-周期连续解$ \gamma(t) $, 且有

$ -\Big(\frac{b}{a}\Big)_{M}\leq \gamma(t)\leq-\Big(\frac{b}{a}\Big)_{L}. $

定理3.3证毕.

定理3.4    考虑方程(1.2), $ a(t), b(t) $是$ R $上的$ \omega $-周期连续函数, 如果以下条件成立:

$ (H_{1})\, \, a(t)>0, \quad (H_{2})\, \, b(t)<0, $

则方程(1.2)有唯一正$ \omega $-周期连续解$ \gamma(t) $, 且有$ -\Big(\frac{b}{a}\Big)_{M}\leq \gamma(t)\leq-\Big(\frac{b}{a}\Big)_{L}. $

证   令

$ \begin{align} x = -u, \end{align} $

(3.53)

则方程(1.2) 化为

$ \begin{align} \frac{du}{dt} = a(t)u^{3}-b(t)u^{2}, \end{align} $

(3.54)

由$ (H_{1}) $和$ (H_{2}), $方程(3.54) 满足定理3.3的所有条件, 根据定理3.3, 方程(3.54) 有唯一负$ \omega $-周期连续解$ u_{1}(t), $并且

$ \begin{align} -\Big(-\frac{b}{a}\Big)_{M}\leq u_{1}(t)\leq-\Big(-\frac{b}{a}\Big)_{L}, \end{align} $

(3.55)

由(3.53), 可知方程(1.2)有唯一的正$ \omega $-周期连续解$ \gamma(t), $且有

$ \begin{align} -\Big(\frac{b}{a}\Big)_{M}\leq \gamma(t)\leq-\Big(\frac{b}{a}\Big)_{L}. \end{align} $

(3.56)

定理3.4证毕.

这一节, 我们讨论阿贝尔方程(1.1), 当方程的系数函数满足一定的条件, 利用变量代换, 方程(1.1)可以化为方程(1.2), 利用第三节的四个结论, 从而得到方程(1.1)的周期解的存在性.

定理4.1    考虑方程(1.1), $ a(t) $是$ R $上的$ \omega $-周期连续函数, 且$ a(t) $在$ R $上具有连续导数, 如果$ (H_{1})\, \, b(t) = b_{1}(t)+b_{2}(t) $条件成立, $ b_{1}(t), b_{2}(t) $是$ R $上的$ \omega $-周期连续函数, 并且$ b_{1}(t) $在$ R $上具有连续导数,

$ (H_{2})\, \, a(t)<0, b_{2}(t)<0, \qquad (H_{3})\, \, c(t) = \frac{b_{1}^{2}(t)+2b_{1}(t)b_{2}(t)}{3a(t)}, $

$ (H_{4})\, \, d(t) = \frac{b_{1}^{3}(t)}{27a^{3}(t)}+\frac{b_{1}^{2}(t)b_{2}(t)}{9a^{2}(t)}-\frac{d}{dt}\Big(\frac{b_{1}(t)}{3a(t)}\Big), $

则方程(1.1)恰有两个$ \omega $-周期连续解.

证   由$ (H_{1}), (H_{2}), (H_{3}), (H_{4}) $, 方程(1.1)化为

$ \begin{align}\begin{array}{l} \frac{dx}{dt} = a(t)x^{3}+b(t)x^{2}+c(t)x+d(t)\\ \, \, \, \, \, \, \, \, = a(t)x^{3}+[b_{1}(t)+b_{2}(t)]x^{2}+\frac{b_{1}^{2}(t)+2b_{1}(t)b_{2}(t)}{3a(t)}x+\frac{b_{1}^{3}(t)}{27a^{3}(t)}+\frac{b_{1}^{2}(t)b_{2}(t)}{9a^{2}(t)}-\frac{d}{dt}\Big(\frac{b_{1}(t)}{3a(t)}\Big), \\ \end{array} \end{align} $

(4.1)

即

$ \begin{align}\begin{array}{l} \frac{d(x+\frac{b_{1}(t)}{3a(t)})}{dt} = a(t)x^{3}+b_{1}(t)x^{2}+\frac{b_{1}^{2}(t)}{3a(t)}x+\frac{b_{1}^{3}(t)}{27a^{3}(t)}+b_{2}(t)x^{2}+\frac{2b_{1}(t)b_{2}(t)}{3a(t)}x+\frac{b_{1}^{2}(t)b_{2}(t)}{9a^{2}(t)}\\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, = a(t)\bigg(x+\frac{b_{1}(t)}{3a(t)}\bigg)^{3}+b_{2}(t)\bigg(x+\frac{b_{1}(t)}{3a(t)}\bigg)^{2}. \end{array} \end{align} $

(4.2)

因此, $ \gamma_{1}(t) = -\frac{b_{1}(t)}{3a(t)} $是方程(1.1)的一个$ \omega $-周期连续解.

令

$ \begin{align} x+\frac{b_{1}(t)}{3a(t)} = y, \end{align} $

(4.3)

则方程(4.2)化为

$ \begin{align} \frac{dy}{dt} = a(t)y^{3}+b_{2}(t)y^{2}, \end{align} $

(4.4)

由$ (H_{2}) $, 方程(4.4)满足定理3.1的所有条件, 根据定理3.1, 方程(4.4)有唯一非零$ \omega $-周期连续解$ y_{1}(t) $, 由(4.3), 可得方程(1.1)的另一个$ \omega $-周期连续解$ \gamma_{2}(t) = y_{1}(t)-\frac{b_{1}(t)}{3a(t)}. $定理4.1证毕.

注4.1    定理4.1中, $ (H_{2})\, \, a(t)<0, b_{2}(t)<0 $能被$ (H_{2})\, \, a(t)<0, b_{2}(t)>0 $或$ (H_{2})\, \, a(t)>0, b_{2}(t)<0 $或$ (H_{2})\, \, a(t)>0, b_{2}(t)>0 $代替, 可得类似的结论, 在此, 我们省略了证明过程.

定理4.2    考虑方程(1.1), $ a(t), b(t), c(t) $是$ R $上的$ \omega $-周期连续函数, $ a(t), b(t) $和$ c(t) $在$ R $上具有连续导数, 如果以下条件成立:

$ (H_{1})\, \, a(t)\neq0, \qquad (H_{2})\, \, b^{2}(t)-3a(t)c(t)>0, $

$ \begin{array}{l} (H_{3})\, \, d(t) = -\frac{\Big(b(t)+\sqrt{b^{2}(t)-3a(t)c(t)}\Big)^{3}}{27a^{2}(t)}-\frac{\sqrt{b^{2}(t)-3a(t)c(t)}\Big(b(t)+\sqrt{b^{2}(t)-3a(t)c(t)}\Big)^{2}}{9a^{2}(t)}\\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, +\frac{d}{dt}\Big(\frac{b(t)+\sqrt{b^{2}(t)-3a(t)c(t)}}{3a(t)}\Big), \end{array} $

则方程(1.1)恰有两个$ \omega $-周期连续解.

证   由$ (H_{1}), (H_{2}), (H_{3}) $, 方程(1.1) 化为

$ \begin{align} &\frac{d\Big[x+\frac{b(t)+\sqrt{b^{2}(t)-3a(t)c(t)}}{3a(t)}\Big]}{dt}\\ = &a(t)\Big[x+\frac{b(t)+\sqrt{b^{2}(t)-3a(t)c(t)}}{3a(t)}\Big]^{3} +\sqrt{b_{2}(t)-3a(t)c(t)}\Big[x+\frac{b(t)+\sqrt{b^{2}(t)-3a(t)c(t)}}{3a(t)}\Big]^{2}. \end{align} $

(4.5)

故$ \gamma_{1}(t) = -\frac{b(t)+\sqrt{b^{2}(t)-3a(t)c(t)}}{3a(t)} $是方程(1.1)一个$ \omega $-周期连续解.

令

$ \begin{align} x+\frac{b(t)+\sqrt{b^{2}(t)-3a(t)c(t)}}{3a(t)} = y, \end{align} $

(4.6)

则方程(4.5)化为

$ \begin{align} \frac{dy}{dt} = a(t)y^{3}+\sqrt{b_{2}(t)-3a(t)c(t)}y^{2}, \end{align} $

(4.7)

由$ (H_{1}), (H_{2}) $, 可知方程(4.7)满足定理3.2或定理3.3的所有条件, 根据定理3.2或定理3.3, 方程(4.7) 有唯一非零$ \omega $-周期连续解$ y_{1}(t) $, 由(4.6), 可得方程(1.1)有另一个$ \omega $-周期连续解

$ \gamma_{2}(t) = y_{1}(t)-\frac{b(t)+\sqrt{b^{2}(t)-3a(t)c(t)}}{3a(t)}. $

定理4.2证毕.

定理4.3    考虑方程(1.1), $ a(t), b(t), c(t) $是$ R $上的$ \omega $-周期连续函数, 并且$ a(t), b(t), c(t) $在$ R $上具有连续导数, 如果以下条件成立:

$ (H_{1})\, \, a(t)\neq0, \qquad (H_{2})\, \, b^{2}(t)-3a(t)c(t)>0, $

$ \begin{array}{l} (H_{3})\, \, d(t) = \frac{\Big(-b(t)+\sqrt{b^{2}(t)-3a(t)c(t)}\Big)^{3}}{27a^{2}(t)}-\frac{\sqrt{b^{2}(t)-3a(t)c(t)}\Big(-b(t)+\sqrt{b^{2}(t)-3a(t)c(t)}\Big)^{2}}{9a^{2}(t)}\\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, +\frac{d}{dt}\Big(\frac{b(t)-\sqrt{b^{2}(t)-3a(t)c(t)}}{3a(t)}\Big), \end{array} $

则方程(1.1)恰有两个$ \omega $-周期连续解.

定理4.3的证明和定理4.2的证明相似, 在此省略.