📝 题目
5.2.22 证明: 函数 $u = \frac{1}{{2a}\sqrt{\pi t}}{\mathrm{e}}^{-\frac{{\left( x - b\right) }^{2}}{4{a}^{2}t}}$ ( $a,b$ 为实数) 当 $t > 0$ 时满足方程
$$ \frac{\partial u}{\partial t} = {a}^{2}\frac{{\partial }^{2}u}{\partial {x}^{2}}. $$
💡 答案与解析
### 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 $z = \frac{(x-b)^2}{4a^2 t}$，则 $u = \frac{1}{2a\sqrt{\pi t}} e^{-z}$。 $$ \frac{\partial u}{\partial x} = \frac{1}{2a\sqrt{\pi t}} e^{-z} \cdot \left(-\frac{2(x-b)}{4a^2 t}\right) = -\frac{x-b}{2a^3 \sqrt{\pi t^{3/2}}} e^{-z}. $$
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：$(x-b)\cdot \left(-\frac{x-b}{2a^2 t}\right) e^{-z} = -\frac{(x-b)^2}{2a^2 t} e^{-z}$。 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 $C = \frac{1}{2a\sqrt{\pi}}$。 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 $a^2 \frac{\partial^2 u}{\partial x^2}$： $$ a^2 \cdot \left( -\frac{e^{-z}}{2a^3\sqrt{\pi t^{3/2}}} \left[1 - \frac{(x-b)^2}{2a^2 t}\right] \right) = -\frac{e^{-z}}{2a\sqrt{\pi} t^{3/2}} \left[1 - \frac{(x-b)^2}{2a^2 t}\right]. $$ 而左边 $\frac{\partial u}{\partial t}$ 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。
---
### 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 $\frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial y^2}=0$。
**(1)** 证明 $w(f(u,v), g(u,v))$ 满足 $\frac{\partial^2 w}{\partial u^2} + \frac{\partial^2 w}{\partial v^2}=0$。
**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 $\frac{\partial^2 (fg)}{\partial u^2} + \frac{\partial^2 (fg)}{\partial v^2} = 0$。
**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。
---
### 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^