「常微分方程」ODE 期末复习习题集
复习一下作业题 QWQ 若有错欢迎指出.
PS: 为统一, 无论是证明题还是计算题, 均使用 "证明" 开头.
例 1
下列方程的解在什么区域存在唯一:
\[ y' = \sqrt[3]{ y }. \]
证明.
令 \(f(x, y) = \sqrt[3]{ y }\), 显然 \(f(x, y)\) 在 \(\mathbb{R}^{2}\) 连续. 又有
\[ \frac{ \partial f }{ \partial y } = \frac{1}{3}y^{-\frac{2}{3}}. \]
故其在 \(y \neq 0\) 时连续, 满足 Lipschitz 条件, 故解唯一. 而当 \(y = 0\) 时, 至少存在 \(y \equiv 0\) 和 \(y = \frac{2\sqrt{ 6 }}{3}x^{\frac{3}{2}}\) 两种解, 不满足.
综上, 区域为 \(\{ (x, y) : y \neq 0 \}\).
Q.E.D.
例 2
设函数 \(f(x, y)\) 在 \((x_{0}, y_{0})\) 的邻域内是 \(y\) 的非增函数, 试证初始值问题
\[ y' = f(x, y), \quad y(x_{0}) = y_{0} \]
的解在此邻域内 \(x \geq x_{0}\) 一侧最多只有一个.
证明.
若存在两个解, 设其为 \(y_{1}, y_{2}\), 令 \(F(x) = (y_{1}(x) - y_{2}(x))^{2}\), 则有 \(F(x_{0}) = 0\). 有
\[ F'(x) = 2(y_{1}(x) - y_{2}(x))(y_{1}'(x) - y_{2}'(x)) = 2(y_{1}(x) - y_{2}(x))(f(x, y_{1}(x)) - f(x, y_{2}(x))). \]
又有 \(f(x, y)\) 对 \(y\) 非增, 则若对某个 \(x\), 有 \(y_{1}(x) < y_{2}(x)\), 则有 \(f(x, y_{1}(x)) - f(x, y_{2}(x)) \geq 0\). 同理, 有 \(y_{1}(x) - y_{2}(x)\) 和 \(f(x, y_{1}(x)) - f(x, y_{2}(x))\) 总是异号, 则 \(F'(x) \leq 0, \ \forall x \geq x_{0}\).
又有 \(F(x_{0}) = 0, \ F(x) \geq 0\), 故 \(F(x) = 0, \ \forall x \geq x_{0}\), 即 \(y_{1} = y_{2}\). 故最多只有一个解.
Q.E.D.
例 3
求下列方程的通解:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^{4} + y^{3}}{xy^{2}}. \]
证明.
此为 Bernoulli 方程. 变形得
\[ \frac{\mathrm{d}y}{\mathrm{d}x} - \frac{1}{x}y = x^{3}y^{-2}. \]
令 \(z = y^{1 - (-2)} = y^{3}\), 有
\[ \begin{align*} \frac{\mathrm{d}z^{\frac{1}{3}}}{\mathrm{d}x} - \frac{1}{x}z^{\frac{1}{3}} &= x^{3}z^{-\frac{2}{3}} \\ \frac{1}{3}z^{-\frac{2}{3}}\frac{\mathrm{d}z}{\mathrm{d}x} - \frac{1}{x}z^{\frac{1}{3}} &= x^{3}z^{-\frac{2}{3}} \\ \frac{\mathrm{d}z}{\mathrm{d}x} - 3x^{-1}z &= 3x^{3} \\ x^{-3}\frac{\mathrm{d}z}{\mathrm{d}x} - 3x^{-4}z &= 3 \\ \frac{\mathrm{d}}{\mathrm{d}x} zx^{-3} &= 3 \\ zx^{-3} &= 3x + C \\ z &= 3x^{4} + Cx^{3}, & C \in \mathbb{R}. \end{align*} \]
故解为 \(y = \sqrt[3]{ 3x^{4} + Cx^{3} }, \ C \in \mathbb{R}\).
Q.E.D.
例 4
求下列方程的通解:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x + 3y - 5}{x - y - 1}. \]
证明.
解方程组 \(\begin{cases} x + 3y - 5 = 0 \\ x - y - 1 = 0 \end{cases}\), 得 \((x, y) = (2, 1)\). 故令 \(X = x - 2, \ Y = y - 1\), 有
\[ \begin{align*} \frac{\mathrm{d}Y}{\mathrm{d}X} &= \frac{X + 3Y}{X - Y} \\ \frac{\mathrm{d}Y}{\mathrm{d}X} &= \frac{1 + 3\frac{Y}{X}}{1 - \frac{Y}{X}} \end{align*} \]
令 \(Z = \frac{Y}{X}\), 有
\[ \begin{align*} Z + X\frac{\mathrm{d}Z}{\mathrm{d}X} &= \frac{1 + 3Z}{1 - Z} \\ \frac{1}{X} \, \mathrm{d}X &= \frac{1 - Z}{Z^{2} + 2Z + 1} \, \mathrm{d}Z \\ \log|X| &= \int \frac{1 - Z}{Z^{2} + 2Z + 1} \, \mathrm{d}Z \\ \log|X| &= \int \left( \frac{2}{(Z + 1)^{2}} - \frac{1}{Z + 1} \right) \, \mathrm{d}(Z + 1) \\ \log|X| &= -\frac{2}{Z + 1} - \log|Z + 1| + C \\ \log|X| &= -\frac{2X}{X + Y} - \log\left| \frac{X + Y}{X} \right| + C \\ \log|X + Y| + \frac{2X}{X + Y} &= C, & C \in \mathbb{R}. \end{align*} \]
故解为 \(\log|x + y - 3| + \frac{2x - 4}{x + y - 3} = C, \ C \in \mathbb{R}\).
Q.E.D.
例 5
验证下列方程是否为全微分方程, 并且对全微分方程求解:
\[ xf(x^{2} + y^{2}) \, \mathrm{d}x + yf(x^{2} + y^{2}) \, \mathrm{d}y = 0. \]
其中, \(f(u)\) 是连续可微函数.
证明.
有 \(\frac{\mathrm{d}}{\mathrm{d}y} xf(x^{2} + y^{2}) = \frac{\mathrm{d}}{\mathrm{d}x} yf(x^{2} + y^{2}) = 2xy \frac{\mathrm{d} f(x^{2} + y^{2})}{\mathrm{d} (x^{2} + y^{2})}\), 故该方程为全微分方程.
令 \(u = x^{2} + y^{2}\), 故 \(\mathrm{d}u = 2x \, \mathrm{d}x + 2y \, \mathrm{d}y\), 故
\[ xf(x^{2} + y^{2}) \, \mathrm{d}x + yf(x^{2} + y^{2}) \, \mathrm{d}y = \frac{1}{2}f(u) \, \mathrm{d}u = 0 \implies f(u) \, \mathrm{d}u = 0. \]
令 \(F(u)\) 满足 \(F'(u) = f(u)\), 则有 \(F(u) = C, \ C \in \mathbb{R}\).
代入, 得通解为 \(F(x^{2} + y^{2}) = C, \ C \in \mathbb{R}\), 其中 \(F'(u) = f(u)\).
Q.E.D.
例 6
求下列方程的通解:
\[ x(4y \, \mathrm{d}x + 2x \, \mathrm{d}y) + y^{3}(3y \, \mathrm{d}x + 5x \, \mathrm{d}y) = 0. \]
证明.
整理得 \((4xy + 3y^{4}) \, \mathrm{d}x + (2x^{2} + 5xy^{3}) \, \mathrm{d}y = 0\). 易验证该方程不是全微分方程, 需寻找积分因子.
令积分因子为 \(x^{m}y^{n}\), 则有 \((4x^{m + 1}y^{n + 1} + 3x^{m}y^{n + 4}) \, \mathrm{d}x + (2x^{m + 2}y^{n} + 5x^{m + 1}y^{n + 3}) = 0\).
故有 \(\frac{ \partial }{ \partial y } (4x^{m + 1}y^{n + 1} + 3x^{m}y^{n + 4}) = 4(n + 1)x^{m + 1}y^{n} + 3(n + 4)x^{m}y^{n + 3}\), \(\frac{ \partial }{ \partial x } (2x^{m + 2}y^{n} + 5x^{m + 1}y^{n + 3}) = 2(m + 2)x^{m + 1}y^{n} + 5(m + 1)x^{m}y^{n + 3}\).
故有
\[ \begin{align*} 4(n + 1)x^{m + 1}y^{n} + 3(n + 4)x^{m}y^{n + 3} &= 2(m + 2)x^{m + 1}y^{n} + 5(m + 1)x^{m}y^{n + 3} \\ \implies & \begin{cases} 4(n + 1) = 2(m + 2) \\ 3(n + 4) = 5(m + 1) \end{cases} \end{align*} \]
解得 \(m = 2, n = 1\). 故积分因子为 \(x^{2}y\), 原方程为 \((4x^{3}y^{2} + 3x^{2}y^{5}) \, \mathrm{d}x + (2x^{4}y + 5x^{3}y^{4}) \, \mathrm{d}y = 0\). 则有
\[ M(x, y) = \int (4x^{3}y^{2} + 3x^{2}y^{5}) \, \mathrm{d}x + \varphi(y) = x^{4}y^{2} + x^{3}y^{5} + \varphi(y). \]
故
\[ \frac{ \partial M(x, y) }{ \partial y } = 2x^{4}y + 5x^{3}y^{4} + \varphi'(y) = 2x^{4}y + 5x^{3}y^{4} \implies \varphi'(y) = 0 \implies \varphi(y) = C. \]
故有
\[ \mathrm{d}M(x, y) = 0 \iff x^{4}y^{2} + x^{3}y^{5} = C, \ C \in \mathbb{R}. \]
Q.E.D.
例 7
求下列微分方程的通解:
\[ y(1 + xy) \, \mathrm{d}x + x(xy - 1) \, \mathrm{d}y = 0. \]
证明.
该方程满足 \(yf(xy) \, \mathrm{d}x + xg(xy) \, \mathrm{d}y = 0\) 的形式. 考虑变量替换 \(z = xy\), 有 \(y = \frac{z}{x}\), 故
\[ \begin{align*} \frac{z}{x}(1 + z) \, \mathrm{d}x + x(z - 1) \frac{x \, \mathrm{d}z - z \, \mathrm{d}x}{x^{2}} &= 0 \\ 2z \, \mathrm{d}x + (z - 1)x \, \mathrm{d}z &= 0 \\ \frac{2}{x} \, \mathrm{d}x &= \frac{1 - z}{z} \, \mathrm{d}z \\ 2\log|x| &= \log|z| - z + C \\ \log|x^{2}| &= \log|xy| - xy + C \\ \log\left| \frac{x}{y} \right| + xy &= C, & C \in \mathbb{R}. \end{align*} \]
Q.E.D.
例 8
求下列微分方程的通解 (\(p = \frac{\mathrm{d}y}{\mathrm{d}x}\)):
\[ 2y = p^{2} + 4px + 2x^{2}. \]
证明.
两边求导, 得
\[ \begin{align*} 2p &= 2pp' + 4p + 4xp' + 4x \\ (2x + p)p' + 2x + p &= 0 \\ (2x + p)(p' + 1) &= 0. \end{align*} \]
故有 \(2x + p = 0\) 或 \(p' + 1 = 0\).
若 \(2x + p = 0\), 有 \(p = -2x\), 带回, 有 \(y = -x^{2}\). 其为奇解.
若 \(p' + 1 = 0\), 有 \(p = -x + C\), 带入有 \(y = -\frac{x^{2}}{2} + Cx + \frac{C^{2}}{2}, \ C \in \mathbb{R}\).
Q.E.D.
例 9
用参数法求解下列微分方程 (\(p = \frac{\mathrm{d}y}{\mathrm{d}x}\)):
\[ p(x - \log p) = 1. \]
证明.
有 \(x = \frac{1}{p} + \log p\), 故有 \(\mathrm{d}x = \mathrm{d}\left( \frac{1}{p} + \log p \right) = \frac{p - 1}{p^{2}} \, \mathrm{d}p\). 带入 \(\mathrm{d}y = p \, \mathrm{d}x\), 有
\[ \mathrm{d}y = \left( 1 - \frac{1}{p} \right) \, \mathrm{d}p \iff y = p - \log p + C. \]
故通解为 \(\begin{cases} x = \frac{1}{p} + \log p \\ y = p - \log p + C \end{cases}, \ p > 0, C \in \mathbb{R}\).
Q.E.D.
例 10
用逐次迭代法求下列初始值问题的近似解:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = x + y^{2}, \quad y(0) = 0. \]
证明.
有公式
\[ y_{n + 1}(x) = y_{0} + \int_{x_{0}}^{x} f(t, y_{n}(t)) \, \mathrm{d}t. \]
又有 \(x_{0} = y_{0} = 0\), 故有
\[ \begin{align*} y_{1}(x) &= \int_{0}^{x} (t + 0^{2}) \, \mathrm{d}t = \frac{1}{2}x^2; \\ y_{2}(x) &= \int_{0}^{x} \left( t + \left( \frac{1}{2}t^{2} \right)^{2} \right) \, \mathrm{d}t = \frac{1}{2}x^{2} + \frac{1}{20}x^{5}; \\ y_{3}(x) &= \int_{0}^{x} \left( t + \left( \frac{1}{2}t^{2} + \frac{1}{20}t^{5} \right)^{2} \right) \, \mathrm{d}t = \frac{1}{2}x^{2} + \frac{1}{20}x^{5} + \frac{1}{160}x^{8} + \frac{1}{4400}x^{11}. \end{align*} \]
故近似解为 \(y(x) \approx \frac{1}{2}x^{2} + \frac{1}{20}x^{5} + \frac{1}{160}x^{8} + \frac{1}{4400}x^{11}\).
Q.E.D.
例 11
给定方程
\[ y'' + 8y' + 7y = q(x), \]
其中, \(q(x)\) 在 \(0 \leq x \leq +\infty\) 上连续, 试利用常数变易公式证明:
- 若 \(q(x)\) 在 \(0 \leq x \leq + \infty\) 上有界, 则此方程的每一个解在 \(0 \leq x < +\infty\) 上有界;
- 若 \(\lim\limits_{ x \to +\infty } q(x) = 0\), 则此方程的每一个解 \(y(x)\) 都有 \(\lim\limits_{ x \to +\infty } y(x) = 0\).
证明.
对于 \(y'' + 8y' + 7y = 0\), 设 \(\lambda^{2} + 8\lambda + 7 = 0\), 解得 \(\lambda_{1} = -7, \ \lambda_{2} = -1\). 故有通解 \(y = C_{1}e^{-7x} + C_{2}e^{-x}\).
对于原方程, 设特解为 \(y^{*} = c_{1}(x)e^{-7x} + c_{2}(x)e^{-x}\), 则有
\[ \begin{cases} c_{1}'(x)e^{-7x} + c_{2}'(x)e^{-x} = 0 \\ c_{1}'(x)(e^{-7x})' + c_{2}'(x)(e^{-x})' = q(x) \end{cases} \implies \begin{cases} c_{1}'(x)e^{-7x} + c_{2}'(x)e^{-x} = 0 \\ -7c_{1}'(x)e^{-7x} - c_{2}'(x)e^{-x} = q(x) \end{cases} \]
解得 \(c_{1}'(x) = -\frac{e^{7x}}{6}q(x), \ c_{2}'(x) = \frac{e^{x}}{6}q(x)\), 则有特解
\[ y^{*} = \frac{1}{6}\left( -e^{-7x}\int e^{7x}q(x) \, \mathrm{d}x + e^{-x}\int e^{x}q(x) \, \mathrm{d}x \right). \]
故该方程有通解
\[ y = C_{1}e^{-7x} + C_{2}e^{-x} + \frac{1}{6}\left( -e^{-7x}\int e^{7x}q(x) \, \mathrm{d}x + e^{-x}\int e^{x}q(x) \, \mathrm{d}x \right), \quad C_{1}, C_{2} \in \mathbb{R}. \]
接下来解答问题:
若 \(q(x) = O(1)\), 则有
\[ \begin{align*} y &= C_{1}e^{-7x} + C_{2}e^{-x} + \frac{1}{6}\left( -e^{-7x}\int e^{7x}q(x) \, \mathrm{d}x + e^{-x}\int e^{x}q(x) \, \mathrm{d}x \right) \\ &= C_{1}e^{-7x} + C_{2}e^{-x} + \frac{1}{6}\left( -e^{-7x}\int_{0}^{x} e^{7t}q(t) \, \mathrm{d}t + e^{-x}\int_{0}^{x} e^{t}q(t) \, \mathrm{d}t + O(1) \right) \\ &= C_{1}e^{-7x} + C_{2}e^{-x} + \frac{1}{6}\left( \int_{0}^{x} q(t)(-e^{-7x}e^{7t} + e^{-x}e^{t}) \, \mathrm{d}t + O(1) \right) \\ &= C_{1}e^{-7x} + C_{2}e^{-x} + \frac{1}{6}\left( O(1) \int_{0}^{x} (-e^{-7x}e^{7t} + e^{-x}e^{t}) \, \mathrm{d}t + O(1) \right) \\ &= C_{1}e^{-7x} + C_{2}e^{-x} + \frac{1}{6}\left( O(1) \left( -e^{-7x}\frac{e^{7x} - 1}{7} + e^{-x}(e^{x} - 1) \right) + O(1) \right) \\ &= C_{1}e^{-7x} + C_{2}e^{-x} + \frac{1}{6}\left( O(1) \left( \frac{e^{-7x}}{7} - e^{-x} + \frac{6}{7} \right) + O(1) \right). \end{align*} \]
故此时 \(y\) 在 \(0 \leq x < +\infty\) 有界;
此时有 \(C_{1}e^{-7x} + C_{2}e^{-x} \to 0 \ (x \to +\infty)\), 故只需考虑 \(e^{-7x}\int e^{7x}q(x) \, \mathrm{d}x\) 和 \(e^{-x}\int e^{x}q(x) \, \mathrm{d}x\).
对于 \(e^{-7x}\int e^{7x}q(x) \, \mathrm{d}x\), 有
\[ \lim\limits_{ x \to +\infty } \frac{\int e^{7x}q(x) \, \mathrm{d}x}{e^{7x}} = \lim\limits_{ x \to +\infty } \frac{\left( \int e^{7x}q(x) \, \mathrm{d}x \right)'}{(e^{7x})'} = \lim\limits_{ x \to +\infty } \frac{e^{7x}q(x)}{7e^{7x}} = 0. \]
对于 \(e^{-x}\int e^{x}q(x) \, \mathrm{d}x\), 有
\[ \lim\limits_{ x \to +\infty } \frac{\int e^{x}q(x) \, \mathrm{d}x}{e^{x}} = \lim\limits_{ x \to +\infty } \frac{\left( \int e^{x}q(x) \, \mathrm{d}x \right)'}{(e^{x})'} = \lim\limits_{ x \to +\infty } \frac{e^{x}q(x)}{e^{x}} = 0. \]
故有 \(y \to 0 \ (x \to +\infty)\).
Q.E.D.
例 12
利用消元法求解下列方程组的解:
\[ \begin{cases} \frac{\mathrm{d}x}{\mathrm{d}t} = y \\ \frac{\mathrm{d}y}{\mathrm{d}t} = x \end{cases} \]
证明.
有
\[ \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}} = \frac{\mathrm{d}y}{\mathrm{d}t} = x \implies x'' - x = 0 \]
对 \(\lambda^{2} - 1 = 0\), 解得 \(\lambda_{1} = -1, \lambda_{2} = 1\), 故有通解 \(x = C_{1}e^{-t} + C_{2}e^{t}\). 带入原式, 有通解
\[ \begin{cases} x = C_{1}e^{-t} + C_{2}e^{t} \\ y = -C_{1}e^{-t} + C_{2}e^{t} \end{cases}, \quad C_{1}, C_{2} \in \mathbb{R}. \]
Q.E.D.
例 13
利用首次积分法求下列方程组的通解:
\[ \begin{cases} x' = \frac{y}{x - y} \\ y' = \frac{x}{x - y} \end{cases} \]
证明.
有
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{x}{x - y}}{\frac{y}{x - y}} = \frac{x}{y} \implies y \, \mathrm{d}y = x \, \mathrm{d}x \implies y^{2} - x^{2} = C_{1}. \]
另外, 有
\[ \frac{\mathrm{d}x}{\mathrm{d}t} - \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{y - x}{x - y} = -1 \implies (\mathrm{d}x - \mathrm{d}y) = -\mathrm{d}t \implies x - y + t = C_{2}. \]
两式联立, 得通解
\[ \begin{cases} x = \frac{1}{2}\left( \frac{C_{1}}{t - C_{2}} - t + C_{2} \right) \\ y = \frac{1}{2}\left( \frac{C_{1}}{t - C_{2}} + t - C_{2} \right) \end{cases}, \quad C_{1}, C_{2} \in \mathbb{R}. \]
Q.E.D.
例 14
利用微分算子求方程
\[ \frac{\mathrm{d}\boldsymbol{x}}{\mathrm{d}t} = \begin{bmatrix} -2 & 1 \\ -3 & 2 \end{bmatrix}\boldsymbol{x} + \begin{bmatrix} -e^{2t} \\ 6e^{2t} \end{bmatrix}. \]
的通解.
证明.
该方程即为
\[ \begin{cases} \frac{\mathrm{d}x}{\mathrm{d}t} = -2x + y - e^{2t} \\ \frac{\mathrm{d}y}{\mathrm{d}t} = -3x + 2y + 6e^{2t} \end{cases} \]
令 \(D = \frac{\mathrm{d}}{\mathrm{d}t}\), 有
\[ \begin{cases} Dx = -2x + y - e^{2t} \\ Dy = -3x + 2y + 6e^{2t} \end{cases} \implies \begin{cases} (D + 2)x - y = e^{2t} \\ 3x + (D - 2)y = 6e^{2t} \end{cases} \implies (D^{2} - 1)x = 6e^{2t}. \]
即 \(x'' - x = 6e^{2t}\). 有 \(\lambda^{2} - 1 = 0\) 的解为 \(\lambda_{1} = -1, \ \lambda_{2} = 1\), 故 \(x'' - x = 0\) 的通解为 \(x = C_{1}e^{-t} + C_{2}e^{t}\).
对于特解, 有 \(x^{*} = Ae^{2t}\), 带入得 \(A = 2\). 故原方程的通解为 \(x = C_{1}e^{-t} + C_{2}e^{t} + 2e^{2t}\). 带入原方程, 有通解
\[ \begin{cases} x = C_{1}e^{-t} + C_{2}e^{t} + 2e^{2t} \\ y = C_{1}e^{-t} + 3C_{2}e^{t} + 9e^{2t} \end{cases} \implies \boldsymbol{x} = C_{1}e^{-t}\begin{bmatrix} 1 \\ 1 \end{bmatrix} + C_{2}e^{t}\begin{bmatrix} 1 \\ 3 \end{bmatrix} + e^{2t}\begin{bmatrix} 2 \\ 9 \end{bmatrix}, \quad C_{1}, C_{2} \in \mathbb{R}. \]
Q.E.D.
例 15
设方程组为 \(\frac{\mathrm{d}\boldsymbol{x}}{\mathrm{d}t} = A(t)\boldsymbol{x}\), 其中, \(A(t) = (a_{ij}(t))_{n \times n}\), \(a_{ij}(t)\) 均在 \((-\infty, +\infty)\) 连续. 若 \(\int_{t_{0}}^{+\infty} \sum \limits_{i=1}^{n} a_{ii}(t) \, \mathrm{d}t = +\infty\), 则该方程组中至少有一个解在 \([t_{0}, +\infty)\) 无界.
证明.
取基解矩阵 \(\Phi(t)\), 有 \(|\Phi(t)| = W(t)\). 由 Liouville 公式知
\[ W(t) = W(t_{0})\exp\left( \int_{t_{0}}^{t} \sum \limits_{i=1}^{n} a_{ii}(\tau) \, \mathrm{d}\tau \right). \]
故有 \[ \lim\limits_{ t \to +\infty } |W(t)| = |W(t_{0})| \lim_{ t \to +\infty } \exp\left( \int_{t_{0}}^{t} \sum \limits_{i=1}^{n} a_{ii}(\tau) \, \mathrm{d}\tau \right) = |W(t_{0})| \exp\left( \int_{t_{0}}^{+\infty} \sum \limits_{i=1}^{n} a_{ii}(\tau) \, \mathrm{d}\tau \right) = +\infty. \]
反设所有解有界, 则 \(\Phi(t)\) 的每个元素有界, 故
\[ W(t) = \sum\limits_{(i_{1} i_{2} \cdots i_{n})} (-1)^{\tau(i_{1} i_{2} \cdots i_{n})} \prod_{k=1}^{n} O(1) = n!(O(1))^{n} = O(1). \]
即 \(W(t)\) 有界, 矛盾. 故至少有一解无界.
Q.E.D.
例 16
用常数变易法求下列非齐次线性方程组的通解:
\[ \boldsymbol{x}' = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}\boldsymbol{x} + \begin{bmatrix} 3 \\ -2t \end{bmatrix}. \]
证明.
对于齐次方程 \(\boldsymbol{x}' = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}\boldsymbol{x}\), 令 \(A = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}\), 有 \(|A - \lambda I| = \lambda^{2} + 4 = 0\), 故 \(\lambda = \pm 2i\).
取 \(\lambda = 2i\), 有 \(A - \lambda I = \begin{bmatrix} -2i & -2 \\ 2 & -2i \end{bmatrix} \to \begin{bmatrix} 1 & -i \\ 0 & 0 \end{bmatrix}\), 故有特征向量 \(\boldsymbol{\alpha} = \begin{bmatrix} 1 \\ -i \end{bmatrix}\). 故有解
\[ \boldsymbol{\alpha}e^{\lambda t} = \begin{bmatrix} 1 \\ -i \end{bmatrix}(\cos 2t + i\sin 2t) = \begin{bmatrix} \cos 2t \\ \sin 2t \end{bmatrix} + i\begin{bmatrix} \sin 2t \\ -\cos 2t \end{bmatrix}. \]
故有基解矩阵 \(\Phi(t) = \begin{bmatrix} \cos 2t & \sin 2t \\ \sin 2t & -\cos 2t \end{bmatrix}\).
有 \(|\Phi(t)| = -1\), 故 \(\Phi^{-1}(t) = \frac{1}{-1}\begin{bmatrix} -\cos 2t & -\sin 2t \\ -\sin 2t & \cos 2t \end{bmatrix} = \begin{bmatrix} \cos 2t & \sin 2t \\ \sin 2t & -\cos 2t \end{bmatrix}\).
令特解为 \(\boldsymbol{x}^{*} = \Phi(t)\boldsymbol{u}(t)\), 故由常数变易公式得
\[ \boldsymbol{u}'(t) = \Phi^{-1}(t)\boldsymbol{f}(t) = \begin{bmatrix} 3\cos 2t - 2t\sin 2t \\ 3\sin 2t + 2t\cos 2t \end{bmatrix}. \]
故
\[ \boldsymbol{u}(t) = \int \begin{bmatrix} 3\cos 2t - 2t\sin 2t \\ 3\sin 2t + 2t\cos 2t \end{bmatrix} \, \mathrm{d}t = \begin{bmatrix} \sin 2t + t\cos 2t \\ t\sin 2t - \cos 2t \end{bmatrix}. \]
故特解 \(\boldsymbol{x}^{*} = \Phi(t)\boldsymbol{u}(t) = \begin{bmatrix} t \\ 1 \end{bmatrix}\). 故通解为
\[ \boldsymbol{x} = C_{1}\begin{bmatrix} \cos 2t \\ \sin 2t \end{bmatrix} + C_{2}\begin{bmatrix} \sin 2t \\ -\cos 2t \end{bmatrix} + \begin{bmatrix} t \\ 1 \end{bmatrix}, \quad C_{1}, C_{2} \in \mathbb{R}. \]
Q.E.D.
例 17
给定极坐标系下的微分方程
\[ \frac{\mathrm{d}\theta}{\mathrm{d}t} = 1, \quad \frac{\mathrm{d}r}{\mathrm{d}t} = \begin{cases} r^{2}\sin \frac{1}{r} & r > 0 \\ 0 & r = 0 \end{cases} \]
证明平衡点 \((0, 0)\) 是稳定的, 但不是渐近稳定的.
证明.
对 \(\forall \varepsilon > 0\), 取足够大的 \(N\), s.t. \(r_{N} = \frac{1}{N\pi} < \varepsilon\), 取 \(\delta = r_{N} = \frac{1}{N\pi}\).
若初始值 \(r(0) < \delta\), 则初始点位于 \(r = r_{N}\) 一圆的内部, 该圆为该系统的一条轨迹, 故从该圆内部出发的点无法超出该圆, 即达到 \(r = r_{N}\) 外部. 故对 \(\forall t \geq 0\), 有 \(r < r_{N} = \varepsilon\), 故稳定.
而 \(N\) 可足够大, 故有无穷个足够小的圆, 即对每个轨迹, 总夹在两圆之间, 即 \(0 < r_{M} < r < r_{N}, \ \forall t \geq 0\). 故有 \(\lim\limits_{ t \to +\infty } r(t) \geq r_{M} > 0\), 故不渐近稳定.
Q.E.D.
例 18
试给出一阶微分方程
\[ \frac{\mathrm{d}x}{\mathrm{d}t} = a(t)x \]
的零解稳定或渐近稳定的充要条件.
证明.
有
\[ \frac{\mathrm{d}x}{x} = a(t) \, \mathrm{d}t \implies \log|x| - \log|x_{0}| = \int_{t_{0}}^{t} a(\tau) \, \mathrm{d}\tau \implies x = x_{0}\exp\left( \int_{t_{0}}^{t} a(\tau) \, \mathrm{d}\tau \right). \]
对于零解稳定, 即对 \(\forall \varepsilon > 0\), 均 \(\exists \delta > 0\), s.t. \(|x_{0}| < \delta\), 则有
\[ |x| = \left| x_{0}\exp\left( \int_{t_{0}}^{t} a(\tau) \, \mathrm{d}\tau \right) \right| < \varepsilon. \]
故当且仅当 \(\int_{t_{0}}^{t} a(\tau) \, \mathrm{d}\tau\) 有上界时, 有 \(|x| = |x_{0}\exp(O(1))| = |x_{0}|O(1) < \delta O(1)\). 此时取 \(\delta = \varepsilon\) 即可.
而当其无界时, \(\delta\) 与 \(t\) 无关, 故无法满足. 综上, 零解稳定的充要条件是 \(\int_{t_{0}}^{t} a(\tau) \, \mathrm{d}\tau\) 有上界.
对于渐近稳定, 即 \(\exists \delta > 0\), s.t. \(|x_{0}| < \delta\), 有
\[ \lim\limits_{ t \to +\infty } |x| = \lim\limits_{ t \to +\infty } \left| x_{0}\exp\left( \int_{t_{0}}^{t} a(\tau) \, \mathrm{d}\tau \right) \right| = 0. \]
故充要条件为 \(\lim\limits_{ t \to +\infty }\exp(\int_{t_{0}}^{t} a(\tau) \, \mathrm{d}\tau) = 0\), 即 \(\int_{t_{0}}^{+\infty} a(\tau) \, \mathrm{d}\tau = -\infty\).
Q.E.D.
例 19
确定下列方程组的周期解, 极限环及其稳定性:
\[ \begin{cases} \frac{\mathrm{d}x}{\mathrm{d}t} = y - \frac{x}{\sqrt{ x^{2} + y^{2} }}(x^{2} + y^{2} - 1) & x^{2} + y^{2} \neq 0 \\ \frac{\mathrm{d}y}{\mathrm{d}t} = -x - \frac{y}{\sqrt{ x^{2} + y^{2} }}(x^{2} + y^{2} - 1) \\ \frac{\mathrm{d}x}{\mathrm{d}t} = 0, \ \frac{\mathrm{d}y}{\mathrm{d}t} = 0 & x^{2} + y^{2} = 0 \end{cases} \]
证明.
令 \(x = r\cos\theta, y = r\sin\theta\), 有 \(x^{2} + y^{2} = r^{2}\), 则有
\[ \begin{align*} r\frac{\mathrm{d}r}{\mathrm{d}t} &= x\frac{\mathrm{d}x}{\mathrm{d}t} + y\frac{\mathrm{d}y}{\mathrm{d}t} \\ &= x\left( y - \frac{x}{r}(r^{2} - 1) \right) + y\left( -x - \frac{y}{r}(r^{2} - 1) \right) \\ &= -r(r^{2} - 1). \end{align*} \]
则有 \(\frac{\mathrm{d}r}{\mathrm{d}t} = 1 - r^{2}\). 对于 \(\theta\), 有
\[ \begin{align*} r^{2}\frac{\mathrm{d}\theta}{\mathrm{d}t} &= x\frac{\mathrm{d}y}{\mathrm{d}t} - y\frac{\mathrm{d}x}{\mathrm{d}t} \\ &= x\left( -x - \frac{y}{r}(r^{2} - 1) \right) - y\left( y - \frac{x}{r}(r^{2} - 1) \right) \\ &= -r^{2}. \end{align*} \]
故有 \(\frac{\mathrm{d}\theta}{\mathrm{d}t} = -1\).
而对于周期解, 需 \(\frac{\mathrm{d}r}{\mathrm{d}t} = 0\), 故有 \(r = 1\), 即有极限环 \(x^{2} + y^{2} = 1\).
对于稳定性, 考察 \(\frac{\mathrm{d}r}{\mathrm{d}t}\) 在 \(r = 1\) 附近的符号:
- 当 \(0 < r < 1\) 时, \(1 - r^{2} > 0\), 故 \(\frac{\mathrm{d}r}{\mathrm{d}t} > 0\), \(r\) 增大趋近于 \(1\);
- 当 \(r > 1\) 时, \(1 - r^{2} < 0\), 故 \(\frac{\mathrm{d}r}{\mathrm{d}t} < 0\), \(r\) 减小趋近于 \(1\);
故无论 \(r\) 的取值, 均趋向于 \(r = 1\), 故为稳定的极限环.
Q.E.D.
例 20
证明下列系统不存在非零周期解:
\[ \begin{cases} \frac{\mathrm{d}x}{\mathrm{d}t} = x + y + x^{3} - y^{3} \\ \frac{\mathrm{d}y}{\mathrm{d}t} = -x + 2y + x^{2}y + \frac{1}{3}y^{3} \end{cases} \]
证明.
取 \(P(x, y) = x + y + x^{3} - y^{3}, \ Q(x, y) = -x + 2y + x^{2}y + \frac{1}{3}y^{3}\). 则有
\[ \frac{ \partial P }{ \partial x } + \frac{ \partial Q }{ \partial y } = 1 + 3x^{2} + 2 + x^{2} + y^{2} = 4x^{2} + y^{2} + 3. \]
故散度正定. 由 Bendixson 判据得不存在非零周期解.
Q.E.D.