show that ▽^2 (log r)=1/r


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

$s h o w t h a t ▽^{2} (l o g r) = 1 / r$
Commented by Joel578 last updated on 06/Nov/18

$w h a t i s ▽ s y m b o l m e a n ?$
Commented by ajfour last updated on 06/Nov/18

$▽ i s g r a d i e n t$ $▽^{2} i s l a p l a c i a n .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 06/Nov/18
	$r^→ =ix+jy+kz r=(√(x^2 +y^2 +z^2 )) lnr=(1/2)ln(x^2 +y^2 +z^2 ) ▽^2 =(∂^2 /∂x^2 )+(∂^2 /∂z^2 )+(∂^2 /∂z^2 ) (∂^2 /∂x^2 ){(1/2)ln(x^2 +y^2 +z^2 ) (∂/∂x){(∂/∂x) (1/2)ln(x^2 +y^2 +z^2 )} (∂/∂x){(((1/2)×2x)/(x^2 +y^2 +z^2 ))} =(((x^2 +y^2 +z^2 )×1−x(2x))/((x^2 +y^2 +z^2 )^2 )) =((y^2 +z^2 −x^2 )/((x^2 +y^2 +z^2 )^2 )) thus on addition we get ((y^2 +z^2 −x^2 +x^2 −y^2 +z^2 +x^2 +y^2 −z^2 )/((x^2 +y^2 +z^2 )^2 )) =((x^2 +y^2 +z^2 )/((x^2 +y^2 +z^2 )^2 ))=(1/(x^2 +y^2 +z^2 ))=(1/r^2 )$
	$\vec{r} = i x + j y + k z$ $r = \sqrt{x^{2} + y^{2} + z^{2}}$ $l n r = \frac{1}{2} l n (x^{2} + y^{2} + z^{2})$ $▽^{2} = \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial z^{2}} + \frac{\partial^{2}}{\partial z^{2}}$ $\frac{\partial^{2}}{\partial x^{2}} {\frac{1}{2} l n (x^{2} + y^{2} + z^{2})$ $\frac{\partial}{\partial x} {\frac{\partial}{\partial x} \frac{1}{2} l n (x^{2} + y^{2} + z^{2})}$ $\frac{\partial}{\partial x} {\frac{\frac{1}{2} \times 2 x}{x^{2} + y^{2} + z^{2}}}$ $= \frac{(x^{2} + y^{2} + z^{2}) \times 1 - x (2 x)}{{(x^{2} + y^{2} + z^{2})}^{2}}$ $= \frac{y^{2} + z^{2} - x^{2}}{{(x^{2} + y^{2} + z^{2})}^{2}}$ $t h u s o n a d d i t i o n w e g e t$ $\frac{y^{2} + z^{2} - x^{2} + x^{2} - y^{2} + z^{2} + x^{2} + y^{2} - z^{2}}{{(x^{2} + y^{2} + z^{2})}^{2}}$ $= \frac{x^{2} + y^{2} + z^{2}}{{(x^{2} + y^{2} + z^{2})}^{2}} = \frac{1}{x^{2} + y^{2} + z^{2}} = \frac{1}{r^{2}}$
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