Answer to Question #272839 in Calculus for Happpy

Question

Verify that the function u(x,y,z)=1/(√x^2+y^2+z^2) is a solution of the three-dimensional Laplace equation Uxx + Uyy +Uzz = 0.

Answer 1

$u(x,y,z)=\frac{1}{\sqrt{x^2+y^2+z^2}}$

$u_x=-\frac{x}{(x^2+y^2+z^2)^{3/2}}$

$u_{xx}=-\frac{1}{(x^2+y^2+z^2)^{3/2}}+\frac{3x^2}{(x^2+y^2+z^2)^{5/2}}=\frac{3x^2-(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{5/2}}=\frac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}}$

$u_{yy}= \frac{2y^2-x^2-z^2}{(x^2+y^2+z^2)^{5/2}}$

$u_{zz}= \frac{2z^2-x^2-y^2}{(x^2+y^2+z^2)^{5/2}}$

$u_{xx}+u_{yy}+u_{zz}= \frac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}}+ \frac{2y^2-x^2-z^2}{(x^2+y^2+z^2)^{5/2}}+ \frac{2z^2-x^2-y^2}{(x^2+y^2+z^2)^{5/2}}=0$

So, $u(x,y,z)=\frac{1}{\sqrt{x^2+y^2+z^2}}$ is a solution of the three-dimensional Laplace equation $u_{xx}+u_{yy}+u_{zz}=0$