以下是一道偏微分考研题：

设 $u=u(x,y)$，$x=r\cos heta$，$y=r\sin heta$，试证明：

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

解答：

首先，根据链式法则有：

$$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial heta}\frac{\partial heta}{\partial x}=-\sin heta\frac{\partial u}{\partial r}+\frac{\cos heta}{r}\frac{\partial u}{\partial heta}$$

$$\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial u}{\partial heta}\frac{\partial heta}{\partial y}=\cos heta\frac{\partial u}{\partial r}+\frac{\sin heta}{r}\frac{\partial u}{\partial heta}$$

然后，对 $\frac{\partial u}{\partial x}$ 和 $\frac{\partial u}{\partial y}$ 分别求偏导数，有：

$$\frac{\partial^2u}{\partial x^2}=-\sin heta\frac{\partial}{\partial r}\left(\frac{\partial u}{\partial x}ight)+\frac{\cos heta}{r}\frac{\partial}{\partial heta}\left(\frac{\partial u}{\partial x}ight)=-\sin heta\frac{\partial}{\partial r}\left(-\sin heta\frac{\partial u}{\partial r}+\frac{\cos heta}{r}\frac{\partial u}{\partial heta}ight)+\frac{\cos heta}{r}\frac{\partial}{\partial heta}\left(-\sin heta\frac{\partial u}{\partial r}+\frac{\cos heta}{r}\frac{\partial u}{\partial heta}ight)$$

$$\frac{\partial^2u}{\partial y^2}=\cos heta\frac{\partial}{\partial r}\left(\frac{\partial u}{\partial y}ight)+\frac{\sin heta}{r}\frac{\partial}{\partial heta}\left(\frac{\partial u}{\partial y}ight)=\cos heta\frac{\partial}{\partial r}\left(\cos heta\frac{\partial u}{\partial r}+\frac{\sin heta}{r}\frac{\partial u}{\partial heta}ight)+\frac{\sin heta}{r}\frac{\partial}{\partial heta}\left(\cos heta\frac{\partial u}{\partial r}+\frac{\sin heta}{r}\frac{\partial u}{\partial heta}ight)$$

将上式中的 $\frac{\partial u}{\partial r}$ 和 $\frac{\partial u}{\partial heta}$ 用 $\frac{\partial u}{\partial x}$ 和 $\frac{\partial u}{\partial y}$ 表示，有：

$$\frac{\partial^2u}{\partial x^2}=-\sin^2 heta\frac{\partial^2u}{\partial r^2}-\sin heta\cos heta\frac{1}{r}\frac{\partial u}{\partial r}+\frac{\cos^2 heta}{r^2}\frac{\partial^2u}{\partial heta^2}$$

$$\frac{\partial^2u}{\partial y^2}=\cos^2 heta\frac{\partial^2u}{\partial r^2}+\sin heta\cos heta\frac{1}{r}\frac{\partial u}{\partial r}+\frac{\sin^2 heta}{r^2}\frac{\partial^2u}{\partial heta^2}$$

将上式相加，即可得到所要证明的式子：

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