VC3.34

From Exampleproblems

$\nabla^2(\frac{1}{r})=\nabla^2{(x^2+y^2+z^2)^{\frac{-1}{2}}}\,$

= $\sum \frac{\partial^2}{\partial x^2}{(x^2+y^2+z^2)^{\frac{-1}{2}}}\,$ --(1)

Now, $\frac{\partial}{\partial x}{(x^2+y^2+z^2)^{\frac{-1}{2}}}=-\frac{1}{2}(x^2+y^2+z^2)^{\frac{-3}{2}}\cdot 2x\,$ --(2)

Therefore, $\frac{\partial^2}{\partial x^2}{(x^2+y^2+z^2)^{\frac{-1}{2}}}=\frac{\partial}{\partial x}[\frac{\partial}{\partial x}{(x^2+y^2+z^2)^{\frac{-1}{2}}}]=\frac{\partial}{\partial}{-x(x^2+y^2+z^2)^{\frac{-3}{2}}}\,$ ,using (2)

= $(-1)(x^2+y^2+z^2)^{\frac{-3}{2}}+(-x)(\frac{-3}{2})(x^2+y^2+z^2)^{\frac{-5}{2}}\cdot (2x)\,$

= $(x^2+y^2+z^2)^{\frac{-5}{2}}[3x^2-(x^2+y^2+z^2)]\,$

= $(2x^2-y^2-z^2)(x^2+y^2+z^2)^{\frac{-5}{2}}\,$ --(3)

Proceeding similarly,we have

$\frac{\partial^2}{\partial y^2}{(x^2+y^2+z^2)^{\frac{-1}{2}}}=(2y^2-x^2-z^2)(x^2+y^2+z^2)^{\frac{-5}{2}}\,$ --(4)

And $\frac{\partial^2}{\partial z^2}{(x^2+y^2+z^2)^{\frac{-1}{2}}}=(2z^2-x^2-y^2)(x^2+y^2+z^2)^{\frac{-5}{2}}\,$ --(5)

Adding (3),(4),(5),we obtain

$\sum \frac{\partial^2}{\partial x^2}{(x^2+y^2+z^2)^{\frac{-1}{2}}}\,$

= $(x^2+y^2+z^2)^{\frac{-5}{2}}[2x^2-y^2-z^2+2y^2-x^2-z^2+2z^2-x^2-y^2]=0\,$

or $\nabla^2(\frac{1}{r})=0\,$ using (1)

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