以复合函数为桥梁，将“偏导”变为“导”，进而转化为微分方程

一、题目

已知，$u$ $=$ $u\left(\sqrt{x^{2}+y^{2}}\right)$ 其中，$r=\sqrt{x^{2}+y^{2}}>0$ 有二阶连续的偏导数，且满足：

$$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}-\frac{1}{x} \frac{\partial u}{\partial x}+u=x^{2}+y^{2}$$

则 $u\left(\sqrt{x^{2}+y^{2}}\right)=?$

难度评级：

二、解析

可以将 $u\left(\sqrt{x^{2}+y^{2}}\right)$ 看作是 $u(r)$ 和 $r(x, y)=\sqrt{x^{2}+y^{2}}$ 的复合函数，于是：

$$
\frac{\partial u}{\partial x}=\frac{\mathrm{~ d} u}{\mathrm{~ d} r} \cdot \frac{\partial r}{\partial x}=\frac{\mathrm{~ d} u}{\mathrm{~ d} r} \cdot \frac{2 x}{2}\left(x^{2}+y^{2}\right)^{\frac{-1}{2}} \Rightarrow
$$

$$
\frac{\partial u}{\partial x}=\frac{\mathrm{~ d} u}{\mathrm{~ d} r} \cdot \frac{x}{r} \Rightarrow
$$

$$
\frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right)=\frac{\partial}{\partial x}\left(\frac{\mathrm{~ d} u}{\mathrm{~ d} r} \cdot \frac{x}{r}\right)=
$$

$$
\frac{\mathrm{d}}{\mathrm{~ d} x}\left(\frac{\mathrm{~ d} u}{\mathrm{~ d} r}\right) \cdot \frac{x}{r}+\frac{\mathrm{~ d} u}{\mathrm{~ d} r} \cdot \frac{r-x \cdot \frac{x}{r}}{r^{2}}=
$$

$$
\frac{\mathrm{d}}{\mathrm{~ d} r} \frac{\mathrm{~ d} u}{\mathrm{~ d} r} \cdot \frac{\partial r}{\partial x} \cdot \frac{x}{r}+\frac{\mathrm{~ d} u}{\mathrm{~ d} r}\left(\frac{1}{r}-\frac{x^{2}}{r^{3}}\right)=
$$

$$
\frac{\mathrm{d}^{2} u}{\mathrm{~ d} r^{2}} \cdot \frac{x^{2}}{r^{2}}+\frac{\mathrm{~ d} u}{\mathrm{~ d} r}\left(\frac{1}{r}-\frac{x^{2}}{r^{3}}\right)
$$

又由 $r = \sqrt{x^{2} + y^{2}}$ 中 $x$ 和 $y$ 的对称性可知：

$$
\frac{\partial^{2} u}{\partial y^{2}}=\frac{\mathrm{d}^{2} u}{\mathrm{~ d} r^{2}} \cdot \frac{y^{2}}{r^{2}}+\frac{\mathrm{~ d} u}{\mathrm{~ d} r}\left(\frac{1}{r}-\frac{y^{2}}{r^{3}}\right)
$$

于是：

$$
\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}-\frac{1}{x} \frac{\partial u}{\partial x}+u=x^{2}+y^{2} \Rightarrow
$$

$$
\frac{\mathrm{d}^{2} u}{\mathrm{~ d} r^{2}} \cdot \frac{x^{2}}{r^{2}}+\frac{\mathrm{~ d} u}{\mathrm{~ d} r}\left(\frac{1}{r}-\frac{x^{2}}{r^{3}}\right)+
$$

$$
\frac{\mathrm{d}^{2} u}{\mathrm{~ d} r^{2}} \cdot \frac{y^{2}}{r^{2}}+\frac{\mathrm{~ d} u}{\mathrm{~ d} r}\left(\frac{1}{r}-\frac{y^{2}}{r^{3}}\right)-\frac{1}{x} \cdot \frac{\mathrm{~ d} u}{\mathrm{~ d} r} \cdot \frac{x}{r}+u=r^{2} \Rightarrow
$$

$$
\frac{\mathrm{d}^{2} u}{\mathrm{~ d} r^{2}}\left(\frac{x^{2}+y^{2}}{r^{2}}\right)+\frac{\mathrm{~ d} u}{\mathrm{~ d} r}\left(\frac{2}{r}-\frac{x^{2}+y^{2}}{r^{3}}\right)-\frac{1}{r} \cdot \frac{\mathrm{~ d} u}{\mathrm{~ d} r}+u= r^{2} \Rightarrow
$$

$$
\frac{\mathrm{d}^{2} u}{\mathrm{~ d} r^{2}}+\frac{1}{r} \frac{\mathrm{~ d} u}{\mathrm{~ d} r}-\frac{1}{r} \frac{\mathrm{~ d} u}{\mathrm{~ d} r}+u=r^{2} \Rightarrow
$$

$$
\frac{\mathrm{d}^{2} u}{\mathrm{~ d} r^{2}}+u=r^{2} \Rightarrow
$$

我们可以将 $u$ 和 $r$ 写成 $y$ 与 $x$ 的形式，以符合我们常见的形式，从而更方便的完成计算：

$$
y^{\prime \prime}+y=x^{2} \Rightarrow \lambda^{2}+1=0 \Rightarrow \lambda=0 \pm 1
$$

$$
y^{*}=e^{0 x} \cdot\left(C_{1} \cos x+C_{2} \sin x\right) \Rightarrow
$$

$$
y^{*}=C_{1} \cos r+C_{2} \sin r
$$

$$
Y^{*}=x^{k} Q_{n}(x) e^{\mu x} \Rightarrow \mu=0, \mu^{\mu} \neq \lambda_{1}, \mu \neq \lambda_{2}
$$

$$
\Rightarrow k=0 \Rightarrow Y^{*}=Q_{n}(x)=A x^{2}+B x+C \Rightarrow
$$

$$
Y^{* \prime}=2 A x+B
$$

$$
Y^{* \prime \prime}=2 A \Rightarrow
$$

$$
2 A+A x^{2}+B x+C=x^{2} \Rightarrow
$$

$$
\left\{\begin{array}{l}A=1 \\ B=0 \\ C=-2\end{array} \Rightarrow Y^{*}=x^{2}-2 \Rightarrow Y^{*}=r^{2}-2\right.
$$

$$
Y=u(r) = C_{1} \cos r+C_{2} \sin r+r^{2}-2 \Rightarrow
$$

$$
u \sqrt{x^{2}+y^{2}}=
$$

$$
C_{1} \cos \sqrt{x^{2}+y^{2}}+C_{2} \sin \sqrt{x^{2}+y^{2}}+x^{2}+y^{2}-2
$$

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