同济高数 第9章 第9-2-10题
📝 题目
10．验证：
（1）$y=\mathrm{e}^{-k n^{2} t} \sin n x$ 满足 $\displaystyle \frac{\partial y}{\partial t}=k \frac{\partial^{2} y}{\partial x^{2}}$ ；
（2）$r=\sqrt{x^{2}+y^{2}+z^{2}}$ 满足 $\displaystyle \frac{\partial^{2} r}{\partial x^{2}}+\frac{\partial^{2} r}{\partial y^{2}}+\frac{\partial^{2} r}{\partial z^{2}}=\frac{2}{r}$ ．
💡 答案解析
[AI解答]
[AI解答]
（1）验证
设
$$
y = e^{-k n^{2} t} \sin(n x)
$$
先求对 $t$ 的偏导数：
$$
\frac{\partial y}{\partial t} = \frac{\partial}{\partial t}\left( e^{-k n^{2} t} \sin(n x) \right)
= \sin(n x) \cdot \left( -k n^{2} e^{-k n^{2} t} \right)
= -k n^{2} e^{-k n^{2} t} \sin(n x)
$$
再求对 $x$ 的二阶偏导数：
一阶偏导：
$$
\frac{\partial y}{\partial x} = e^{-k n^{2} t} \cdot n \cos(n x)
$$
二阶偏导：
$$
\frac{\partial^{2} y}{\partial x^{2}} = e^{-k n^{2} t} \cdot \left( -n^{2} \sin(n x) \right)
= -n^{2} e^{-k n^{2} t} \sin(n x)
$$
于是
$$
k \frac{\partial^{2} y}{\partial x^{2}} = k \cdot \left( -n^{2} e^{-k n^{2} t} \sin(n x) \right)
= -k n^{2} e^{-k n^{2} t} \sin(n x)
$$
比较得
$$
\frac{\partial y}{\partial t} = k \frac{\partial^{2} y}{\partial x^{2}}
$$
成立。
---
（2）验证
设
$$
r = \sqrt{x^{2} + y^{2} + z^{2}}
$$
先求对 $x$ 的一阶偏导：
$$
\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}} = \frac{x}{r}
$$
再求二阶偏导：
$$
\frac{\partial^{2} r}{\partial x^{2}} = \frac{\partial}{\partial x}\left( \frac{x}{r} \right)
= \frac{1 \cdot r - x \cdot \frac{\partial r}{\partial x}}{r^{2}}
= \frac{r - x \cdot \frac{x}{r}}{r^{2}}
= \frac{r - \frac{x^{2}}{r}}{r^{2}}
= \frac{r^{2} - x^{2}}{r^{3}}
$$
同理可得：
$$
\frac{\partial^{2} r}{\partial y^{2}} = \frac{r^{2} - y^{2}}{r^{3}}, \quad
\frac{\partial^{2} r}{\partial z^{2}} = \frac{r^{2} - z^{2}}{r^{3}}
$$
求和：
$$
\frac{\partial^{2} r}{\partial x^{2}} + \frac{\partial^{2} r}{\partial y^{2}} + \frac{\partial^{2} r}{\partial z^{2}}
= \frac{(r^{2} - x^{2}) + (r^{2} - y^{2}) + (r^{2} - z^{2})}{r^{3}}
= \frac{3r^{2} - (x^{2}+y^{2}+z^{2})}{r^{3}}
$$
而 $x^{2}+y^{2}+z^{2} = r^{2}$，代入得：
$$
\frac{3r^{2} - r^{2}}{r^{3}} = \frac{2r^{2}}{r^{3}} = \frac{2}{r}
$$
因此原式成立。
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难度：★★☆☆☆