91Ó°ÊÓ

The given equations are as follows.

$\begin{array}{l}{r}^2=x^2+y^2+z^2\\ {∇}^2=\frac{{∂}^2}{∂x^2}+\frac{{∂}^2}{∂y^2}+\frac{{∂}^2}{∂z^2}\end{array}$

The total of the second-order partial derivatives of R , the unknown function, with respect to the Cartesian coordinates, according to Laplace's equation, equals zero.

Put given values in equation${∇}^2\left(\frac{1}{r}\right)=0$.

${∇}^2\left(\frac{1}{r}\right)=(\frac{{∂}^2}{∂x^2}+\frac{{∂}^2}{∂y^2}+\frac{{∂}^2}{∂z^2})\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)$ …â¶Ä¦.(1)

Find the derivative to prove the equation zero.

$\begin{array}{c}\frac{{∂}^2}{∂x^2}\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)=\frac{∂}{∂x}[−\frac{1}{2}\frac{2x}{{(x^2+y^2+z^2)}^{3/2}}]\\ =−\frac{∂}{∂x}\left[\frac{x}{{(x^2+y^2+z^2)}^{3/2}}\right]\end{array}$

Similarly, the other two derivatives are as follows.

$\begin{array}{c}\frac{{∂}^2}{∂x^2}\left(\frac{1}{\sqrt{x^2+y^2+z^2}}\right)=−\frac{1}{{(x^2+y^2+z^2)}^{3/2}}+\frac{3}{2}\frac{x\left(2x\right)}{{(x^2+y^2+z^2)}^{5/2}}\\ =−\frac{1}{{(x^2+y^2+z^2)}^{3/2}}+3\frac{x^2}{{(x^2+y^2+z^2)}^{5/2}}\end{array}$

Put the derivative value in equation(1).

$\begin{array}{c}{∇}^2\left(\frac{1}{r}\right)=−3\frac{1}{{(x^2+y^2+z^2)}^{3/2}}+3\frac{x^2+y^2+z^2}{{(x^2+y^2+z^2)}^{5/2}}\\ =−3\frac{1}{{(x^2+y^2+z^2)}^{3/2}}+3\frac{1}{{(x^2+y^2+z^2)}^{3/2}}\end{array}$

${∇}^2\left(\frac{1}{r}\right)=0$

Hence, the gravitational potential $V=−\frac{Gm}{r}$satisfies the Laplace equation is proved.