设 $u=f(r), r=\sqrt{x^2+y^2+z^2}$, 其中函数 $f$ 二阶可微, 且 $\lim _{x \rightarrow 1} \frac{f(x)-1}{x-1}=1$, 若函数 $u=f(r)$ 满足 $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}=0$, 试求 $f(r)$ 的表达式.
【答案】 根据题意计算得 $\frac{\partial u}{\partial x}=f^{\prime}(r) \frac{x}{r}, \frac{\partial^2 u}{\partial x^2}=f^{\prime \prime}(r) \frac{x^2}{r^2}+f^{\prime}(r)\left(\frac{1}{r}-\frac{x^2}{r^3}\right)$.
同理 $\frac{\partial^2 u}{\partial y^2}=f^{\prime \prime}(r) \frac{y^2}{r^2}+f^{\prime}(r)\left(\frac{1}{r}-\frac{y^2}{r^3}\right) \cdot \frac{\partial^2 u}{\partial z^2}=f^{\prime \prime}(r) \frac{z^2}{r^2}+f^{\prime}(r)\left(\frac{1}{r}-\frac{z^2}{r^3}\right)$.
得 $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}=f^{\prime \prime}(r)+\frac{2}{r} f^{\prime}(r)=0$. 由 $\lim _{x \rightarrow 1} \frac{f(x)-1}{x-1}=1$ 得 $f(1)=1, f^{\prime}(1)=1$.
已知方程 $f^{\prime \prime}(r)+\frac{2}{r} f^{\prime}(r)=0$, 令 $f^{\prime}(r)=p$, 得 $p^{\prime}+\frac{2}{r} p=0$, 解得 $p=\frac{C_1}{r^2}$, 即 $f^{\prime}(r)=\frac{C_1}{r^2}$.
由 $f^{\prime}(1)=1$ 得 $C_1=1$, 故 $f^{\prime}(r)=\frac{1}{r^2}$, 推得 $f(r)=-\frac{1}{r}+C_2$. 再由 $f(1)=1$ 得 $C_2=2$. 所
以 $f(r)=2-\frac{1}{r}$.


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