# Laplacian(1/r)

For the given function $\phi(r)=1/r=1/\sqrt{x^2+y^2+z^2}$ demonstrate that $\nabla^2\phi(r)=0$ on $\Re\setminus{0}$ and that the function itself tends to zero when each $x,y,z->\infty$

$\phi=\frac{1}{\sqrt{x^2+y^2+z^2}}$

$\frac{d\phi}{dx}=-\frac{2x}{2(x^2+y^2+z^2)^{3/2}}=-\frac{x}{(x^2+y^2+z^2)^{3/2}}$   ,$\frac{d\phi}{dy}=-\frac{y}{(x^2+y^2+z^2)^{3/2}}$  and $\frac{d\phi}{dz}=-\frac{z}{(x^2+y^2+z^2)^{3/2}}$

$\frac{d^2\phi}{dx^2}=-\frac{1}{(x^2+y^2+z^2)^{3/2}} +(3/2)*\frac{2x^2}{(x^2+y^2+z^2)^{5/2}}=-\frac{1}{(x^2+y^2+z^2)^{3/2}}+\frac{3x^2}{(x^2+y^2+z^2)^{5/2}}$

$\frac{d^2\phi}{dy^2}=-\frac{1}{(x^2+y^2+z^2)^{3/2}}+\frac{3y^2}{(x^2+y^2+z^2)^{5/2}}$

$\frac{d^2\phi}{dy^2}=-\frac{1}{(x^2+y^2+z^2)^{3/2}}+\frac{3z^2}{(x^2+y^2+z^2)^{5/2}}$

$\nabla^2\phi=-\frac{3}{(x^2+y^2+z^2)^{3/2}}+\frac{3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{5/2}}=0$

If each of the variables goes to $\infty$ then

$\lim_{x\to\infty} \phi=\frac{1}{\sqrt{\infty^2+y^2+z^2}}=1/\infty=0$ and $\lim_{x\to-\infty} \phi=\frac{1}{\sqrt{(-\infty)^2+y^2+z^2}}=1/\infty=0$