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Question Number 47189 by 23kpratik last updated on 06/Nov/18

 show that ▽^2 (log r)=1/r

$$\:{show}\:{that}\:\bigtriangledown^{\mathrm{2}} \left({log}\:{r}\right)=\mathrm{1}/{r} \\ $$

Commented by Joel578 last updated on 06/Nov/18

what is ▽ symbol mean?

$${what}\:{is}\:\bigtriangledown\:{symbol}\:{mean}? \\ $$

Commented by ajfour last updated on 06/Nov/18

▽ is gradient  ▽^2  is laplacian.

$$\bigtriangledown\:{is}\:{gradient} \\ $$$$\bigtriangledown^{\mathrm{2}} \:{is}\:{laplacian}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Nov/18

r^→ =ix+jy+kz  r=(√(x^2 +y^2 +z^2 ))    lnr=(1/2)ln(x^2 +y^2 +z^2 )  ▽^2 =(∂^2 /∂x^2 )+(∂^2 /∂z^2 )+(∂^2 /∂z^2 )  (∂^2 /∂x^2 ){(1/2)ln(x^2 +y^2 +z^2 )  (∂/∂x){(∂/∂x) (1/2)ln(x^2 +y^2 +z^2 )}  (∂/∂x){(((1/2)×2x)/(x^2 +y^2 +z^2 ))}  =(((x^2 +y^2 +z^2 )×1−x(2x))/((x^2 +y^2 +z^2 )^2 ))  =((y^2 +z^2 −x^2 )/((x^2 +y^2 +z^2 )^2 ))  thus on addition we get  ((y^2 +z^2 −x^2 +x^2 −y^2 +z^2 +x^2 +y^2 −z^2 )/((x^2 +y^2 +z^2 )^2 ))  =((x^2 +y^2 +z^2 )/((x^2 +y^2 +z^2 )^2 ))=(1/(x^2 +y^2 +z^2 ))=(1/r^2 )

$$\overset{\rightarrow} {{r}}={ix}+{jy}+{kz} \\ $$$${r}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\:\: \\ $$$${lnr}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right) \\ $$$$\bigtriangledown^{\mathrm{2}} =\frac{\partial^{\mathrm{2}} }{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} }{\partial{z}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} }{\partial{z}^{\mathrm{2}} } \\ $$$$\frac{\partial^{\mathrm{2}} }{\partial{x}^{\mathrm{2}} }\left\{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\right. \\ $$$$\frac{\partial}{\partial{x}}\left\{\frac{\partial}{\partial{x}}\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\right\} \\ $$$$\frac{\partial}{\partial{x}}\left\{\frac{\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{x}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\right\} \\ $$$$=\frac{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)×\mathrm{1}−{x}\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${thus}\:{on}\:{addition}\:{we}\:{get} \\ $$$$\frac{{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{z}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }=\frac{\mathrm{1}}{{r}^{\mathrm{2}} } \\ $$

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