方企勤 第五章 多元函数微分学 第5.2题
📝 题目
5.2.23 设 $x = f\left( {u,v}\right) ,y = g\left( {u,v}\right)$ 满足方程
$$
\frac{\partial f}{\partial u} = \frac{\partial g}{\partial v},\;\frac{\partial f}{\partial v} = - \frac{\partial g}{\partial u},
$$
又设 $w = w\left( {x,y}\right)$ 满足方程 $\frac{{\partial }^{2}w}{\partial {x}^{2}} + \frac{{\partial }^{2}w}{\partial {y}^{2}} = 0$ . 证明:
(1) 函数 $w = w\left\lbrack {f\left( {u,v}\right) ,g\left( {u,v}\right) }\right\rbrack$ 满足方程: $\frac{{\partial }^{2}w}{\partial {u}^{2}} + \frac{{\partial }^{2}w}{\partial {v}^{2}} = 0$ ;
(2) $\frac{{\partial }^{2}\left( {fg}\right) }{\partial {u}^{2}} + \frac{{\partial }^{2}\left( {fg}\right) }{\partial {v}^{2}} = 0$ .
💡 答案解析
5.2.22
题目：证明函数
$$
u = \frac{1}{2a\sqrt{\pi t}} e^{-\frac{(x-b)^2}{4a^2 t}} \quad (t>0)
$$
满足热传导方程
$$
\frac{\partial u}{\partial t} = a^2 \frac{\partial^2 u}{\partial x^2}.
$$
证明步骤：
1. **先对 $x$ 求一阶偏导**
令 $\displaystyle z = \frac{(x-b)^2}{4a^2 t}$，则 $\displaystyle u = \frac{1}{2a\sqrt{\pi t}} e^{-z}$。
$$
\frac{\partial u}{\partial x} = \frac{1}{2a\sqrt{\pi t}} e^{-z} \cdot \left(-\frac{2(x-b)}{4a^2 t}\right)
= -\frac{x-b}{2a^3 \sqrt{\pi t^{3/2}}} e^{-z}.
$$
2. **再对 $x$ 求二阶偏导**
使用乘积法则：
$$
\frac{\partial^2 u}{\partial x^2}
= -\frac{1}{2a^3 \sqrt{\pi t^{3/2}}} \left[ e^{-z} + (x-b) e^{-z} \cdot \left(-\frac{2(x-b)}{4a^2 t}\right) \right].
$$
化简括号内：
第一项：$e^{-z}$，
第二项：$\displaystyle (x-b)\cdot \left(-\frac{x-b}{2a^2 t}\right) e^{-z} = -\frac{(x-b)^2}{2a^2 t} e^{-z}$。
所以：
$$
\frac{\partial^2 u}{\partial x^2} = -\frac{e^{-z}}{2a^3\sqrt{\pi t^{3/2}}} \left[ 1 - \frac{(x-b)^2}{2a^2 t} \right].
$$
3. **对 $t$ 求偏导**
$u = C t^{-1/2} e^{-(x-b)^2/(4a^2 t)}$，其中 $\displaystyle C = \frac{1}{2a\sqrt{\pi}}$。
对 $t$ 求导：
$$
\frac{\partial u}{\partial t} = C\left[ -\frac12 t^{-3/2} e^{-z} + t^{-1/2} e^{-z} \cdot \frac{(x-b)^2}{4a^2 t^2} \right]
= \frac{C e^{-z}}{t^{3/2}} \left[ -\frac12 + \frac{(x-b)^2}{4a^2 t} \right].
$$
4. 比较两边
右边 $\displaystyle a^2 \frac{\partial^2 u}{\partial x^2}$：
$$
a^2 \cdot \left( -\frac{e^{-z}}{2a^3\sqrt{\pi t^{3/2}}} \left[1 - \frac{(x-b)^2}{2a^2 t}\right] \right)
= -\frac{e^{-z}}{2a\sqrt{\pi} t^{3/2}} \left[1 - \frac{(x-b)^2}{2a^2 t}\right].
$$
而左边 $\displaystyle \frac{\partial u}{\partial t}$ 为：
$$
\frac{1}{2a\sqrt{\pi}} \frac{e^{-z}}{t^{3/2}} \left[ -\frac12 + \frac{(x-b)^2}{4a^2 t} \right]
= \frac{e^{-z}}{2a\sqrt{\pi} t^{3/2}} \left( -\frac12 + \frac{(x-b)^2}{4a^2 t} \right).
$$
注意：
$$
-\frac12 + \frac{(x-b)^2}{4a^2 t} = -\left(1 - \frac{(x-b)^2}{2a^2 t}\right) \cdot \frac12?
$$
实际上：
$$
-\frac12 + \frac{(x-b)^2}{4a^2 t} = -\frac12 \left(1 - \frac{(x-b)^2}{2a^2 t}\right).
$$
因此左右两边相等。证毕。
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5.2.23
已知 $x=f(u,v), y=g(u,v)$ 满足柯西-黎曼方程：
$$
\frac{\partial f}{\partial u} = \frac{\partial g}{\partial v}, \quad \frac{\partial f}{\partial v} = -\frac{\partial g}{\partial u}.
$$
且 $w(x,y)$ 满足 Laplace 方程 $\displaystyle \frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial y^2}=0$。
(1) 证明 $w(f(u,v), g(u,v))$ 满足 $\displaystyle \frac{\partial^2 w}{\partial u^2} + \frac{\partial^2 w}{\partial v^2}=0$。
证明：
由链式法则：
$$
\frac{\partial w}{\partial u} = w_x f_u + w_y g_u,
$$
$$
\frac{\partial w}{\partial v} = w_x f_v + w_y g_v.
$$
再求二阶：
$$
\frac{\partial^2 w}{\partial u^2} = (w_{xx} f_u + w_{xy} g_u) f_u + w_x f_{uu} + (w_{yx} f_u + w_{yy} g_u) g_u + w_y g_{uu}.
$$
由于 $w_{xy}=w_{yx}$，整理得：
$$
\frac{\partial^2 w}{\partial u^2} = w_{xx} f_u^2 + 2 w_{xy} f_u g_u + w_{yy} g_u^2 + w_x f_{uu} + w_y g_{uu}.
$$
同理：
$$
\frac{\partial^2 w}{\partial v^2} = w_{xx} f_v^2 + 2 w_{xy} f_v g_v + w_{yy} g_v^2 + w_x f_{vv} + w_y g_{vv}.
$$
相加：
$$
\frac{\partial^2 w}{\partial u^2} + \frac{\partial^2 w}{\partial v^2} = w_{xx}(f_u^2+f_v^2) + 2 w_{xy}(f_u g_u + f_v g_v) + w_{yy}(g_u^2+g_v^2) + w_x(f_{uu}+f_{vv}) + w_y(g_{uu}+g_{vv}).
$$
由柯西-黎曼条件可得：
- $f_u^2+f_v^2 = g_u^2+g_v^2$，且 $f_u g_u + f_v g_v = 0$（因为 $f_u g_u + f_v g_v = f_u g_u + (-g_u)(-f_u) = 0$? 检查：$f_v = -g_u$，$g_v = f_u$，所以 $f_u g_u + f_v g_v = f_u g_u + (-g_u)(f_u)=0$）。
- 另外，由柯西-黎曼条件可推出 $f_{uu}+f_{vv}=0$，$g_{uu}+g_{vv}=0$（因为 $f$ 和 $g$ 是调和函数）。
因此上式化简为：
$$
\frac{\partial^2 w}{\partial u^2} + \frac{\partial^2 w}{\partial v^2} = (w_{xx}+w_{yy})(f_u^2+f_v^2) = 0.
$$
得证。
(2) 证明 $\displaystyle \frac{\partial^2 (fg)}{\partial u^2} + \frac{\partial^2 (fg)}{\partial v^2} = 0$。
证明：
令 $w(x,y)=xy$，显然 $w_{xx}+w_{yy}=0$。由(1)结论，$w(f,g)=fg$ 满足 Laplace 方程在 $(u,v)$ 下形式，即得证。
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5.2.24
方程：
$$
\frac{\partial^2 u}{\partial t^2} = \frac{\partial^2 u}{\partial x^2}.
$$
作变量替换 $\xi = x+t$，$\eta = x-t$。
步骤：
1. 计算一阶偏导：
$$
\frac{\partial u}{\partial t} = u_\xi \cdot 1 + u_\eta \cdot (-1) = u_\xi - u_\eta,
$$
$$
\frac{\partial u}{\partial x} = u_\xi \cdot 1 + u_\eta \cdot 1 = u_\xi + u_\eta.
$$
2. 二阶偏导：
$$
\frac{\partial^2 u}{\partial t^2} = \frac{\partial}{\partial t}(u_\xi - u_\eta) = (u_{\xi\xi} - u_{\xi\eta}) - (u_{\eta\xi} - u_{\eta\eta}) = u_{\xi\xi} - 2u_{\xi\eta} + u_{\eta\eta}.
$$
$$
\frac{\partial^2 u}{\partial x^