# 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

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

## 二、解析

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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