拉普拉斯方程：
验证函数

\begin{align*} z=ln \sqrt{x^2+y^2} \end{align*}

满足方程

\begin{align*} {\partial^2{z} \over \partial{x^2}}+{\partial^2{z} \over \partial{y^2}}=0 \end{align*}

证:因为

\begin{align*} z=ln \sqrt{x^2+y^2} =1/2 ln(x^2+y^2) \end{align*}

所以

\begin{align*} {\partial{z} \over \partial{x}}={x \over(x^2+y^2)} \end{align*} \begin{align*} {\partial^2{z} \over \partial{x^2}}={(x^2+y^2)-2x \cdot x \over(x^2+y^2)^2}={y^2-x^2 \over(x^2+y^2)^2} \end{align*} \begin{align*} {\partial{z} \over \partial{y}}={y \over(x^2+y^2)} \end{align*} \begin{align*} {\partial^2{z} \over \partial{y^2}}={(x^2+y^2)-2y \cdot y \over(x^2+y^2)^2}={x^2-y^2 \over(x^2+y^2)^2} \end{align*}

Q.E.D.

验证函数

\begin{align*} u={1 \over r} \end{align*}

满足方程

\begin{align*} {\partial^2{u} \over \partial{x^2}}+{\partial^2{u} \over \partial{y^2}}+{\partial^2{u} \over \partial{z^2}}=0 \end{align*}

其中

\begin{align*} r =\sqrt{x^2+y^2+z^2} \end{align*}

证：

\begin{align*} {\partial{u} \over \partial{x}}={-1 \over r^2} \cdot {1 \over {2r}} \cdot 2x=-{x \over r^3} \end{align*} \begin{align*} {\partial^2{u} \over \partial{x^2}}=-{1 \over r^3}+ {3 \over r^4}x \cdot {1 \over {2r}} \cdot 2x=-{1 \over r^3}+{3x^2 \over r^5} \end{align*}

同理

\begin{align*} {\partial^2{u} \over \partial{y^2}}=-{1 \over r^3}+ {3 \over r^4}y \cdot {1 \over {2r}} \cdot 2y=-{1 \over r^3}+{3y^2 \over r^5} \end{align*} \begin{align*} {\partial^2{u} \over \partial{z^2}}=-{1 \over r^3}+ {3 \over r^4}z \cdot {1 \over {2r}} \cdot 2z=-{1 \over r^3}+{3z^2 \over r^5} \end{align*}

于是

\begin{align*} {\partial^2{u} \over \partial{x^2}}+{\partial^2{u} \over \partial{y^2}}+{\partial^2{u} \over \partial{z^2}}={-3 \over r^3}+{(3x^2+3y^2+3z^2) \over r^5}=0 \end{align*}