Question Number 47189 by 23kpratik last updated on 06/Nov/18
$$\:{show}\:{that}\:\bigtriangledown^{\mathrm{2}} \left({log}\:{r}\right)=\mathrm{1}/{r} \\ $$
Commented by Joel578 last updated on 06/Nov/18
$${what}\:{is}\:\bigtriangledown\:{symbol}\:{mean}? \\ $$
Commented by ajfour last updated on 06/Nov/18
$$\bigtriangledown\:{is}\:{gradient} \\ $$$$\bigtriangledown^{\mathrm{2}} \:{is}\:{laplacian}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Nov/18
$$\overset{\rightarrow} {{r}}={ix}+{jy}+{kz} \\ $$$${r}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\:\: \\ $$$${lnr}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right) \\ $$$$\bigtriangledown^{\mathrm{2}} =\frac{\partial^{\mathrm{2}} }{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} }{\partial{z}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} }{\partial{z}^{\mathrm{2}} } \\ $$$$\frac{\partial^{\mathrm{2}} }{\partial{x}^{\mathrm{2}} }\left\{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\right. \\ $$$$\frac{\partial}{\partial{x}}\left\{\frac{\partial}{\partial{x}}\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\right\} \\ $$$$\frac{\partial}{\partial{x}}\left\{\frac{\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{x}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\right\} \\ $$$$=\frac{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)×\mathrm{1}−{x}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${thus}\:{on}\:{addition}\:{we}\:{get} \\ $$$$\frac{{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{z}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }=\frac{\mathrm{1}}{{r}^{\mathrm{2}} } \\ $$