# Laplacian of a Smooth Function $f(\vec x) = 1/\|\vec x \|$ for $\| \vec x \| \geq 1$

This is a multivariable calculus problem from a past prelim exam. I have an answer for this written up (posted below), but it seemed rather time-intensive. If there is a slicker way to approach this problem, I'd appreciate seeing it. Thanks!

Recall that for a smooth function $f: \mathbb{R}^3 \to \mathbb{R}$, the Laplacian of $f$ is defined by $$\Delta f = \nabla \cdot ( \nabla f).$$

Suppose that $f: \mathbb{R}^3 \to \mathbb{R}$ is a smooth function satisfying $f(\vec{x}) = 1/\|\vec{x}\|$ for $\|\vec{x}\| \geq 1$.

1. Verify that $\Delta f(\vec{x}) = 0$ for $\|\vec{x}\| \geq 1$.

2. Compute $\int_{\mathbb{R}^3} \Delta f \, dV$.

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For the first part:

$f(\vec{x})=1/r$, with $r=\sqrt{x^2+y^2+z^2}$ (notice that I'm not using polar coordinates here). Now,

$$\frac{\partial r}{\partial x} = \frac{x}{r}$$

So

$$\frac{\partial f}{\partial x} = - \frac{x}{r^3}$$

and

$$\frac{\partial^2 f}{\partial x^2} = - \frac{r^2 - 3 x^2}{r^5}$$

But $$\Delta f = \frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial y^2}+\frac{\partial^2 f}{\partial z^2} = -\frac{1}{r^5} \left[3 r^2 - 3 (x^2+y^2+z^2)\right] = 0$$

For the second part:

Because the laplacian is zero outside the unit sphere, the integral can be restricted to the sphere; and, the laplacian is the divergence of the gradient, and because the function is smooth, we can apply the divergence theorem; hence the integral must be equal to the flux of the gradient over the surface.

But the gradient is given by

$$\nabla f = ( - \frac{x}{r^3}, - \frac{y}{r^3},- \frac{z}{r^3}) = - \frac{\vec{x}}{|\vec{x}|^3}$$

So, it's a vector colinear with $\vec{x}$, pointing towards the origin. And over the surface, it's length is 1, and it's normal to the surface. Hence the flux per surface element is just (minus) the element area; and the total surface integral is just the surface area of the sphere, with negative sign: $- 4 \pi$

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1. In rectangular coordinates, for $\|\vec{x}\| \geq 1$, \begin{align*} f(\vec{x}) &= \frac{1}{\sqrt{x^2+y^2+z^2}} \\ \nabla f &= \left\langle \frac{-x}{(x^2+y^2+z^2)^{3/2}}, \frac{-y}{(x^2+y^2+z^2)^{3/2}}, \frac{-z}{(x^2+y^2+z^2)^{3/2}} \right\rangle \\ &= \frac{-1}{(x^2+y^2+z^2)^{3/2}} \left\langle x,y,z \right\rangle \\ \nabla \cdot \nabla f &= \frac{ (-1) (x^2+y^2+z^2)^{3/2} - (-x)\frac{3}{2}(x^2+y^2+z^2)^{1/2}(2x) }{(x^2+y^2+z^2)^3} \\ &+ \frac{ (-1) (x^2+y^2+z^2)^{3/2} - (-y)\frac{3}{2}(x^2+y^2+z^2)^{1/2}(2y) }{(x^2+y^2+z^2)^3} \\ &+ \frac{ (-1) (x^2+y^2+z^2)^{3/2} - (-z)\frac{3}{2}(x^2+y^2+z^2)^{1/2}(2z) }{(x^2+y^2+z^2)^3} \\ &= \frac{-3(x^2+y^2+z^2) + 3x^2+3y^2+3z^2}{(x^2+y^2+z^2)^{5/2}} \\ &= 0 \end{align*}

2. Since $\Delta f = 0$ for $\|\vec x\| \geq 1$, then $\int_{\mathbb{R}^3} \Delta f \, dV = \int_{\| \vec x \| \leq 1} \Delta f \, dV$. Then by the divergence theorem,

\begin{align*} \int_{\| \vec x \| \leq 1} \nabla \cdot ( \nabla f) dV &= \iint_{\|\vec{x}\|=1} (\nabla f) \cdot \vec n \, dS \\ &= \iint_{\|\vec{x}\|=1} \frac{-1}{(x^2+y^2+z^2)^{3/2}} \left\langle x,y,z \right\rangle \cdot \frac{ \left\langle x,y,z \right\rangle }{(x^2+y^2+z^2)^{1/2}}dS \\ &= \iint_{\|\vec{x}\|=1} \frac{-(x^2+y^2+z^2)}{(x^2+y^2+z^2)^2} dS \\ &= \iint_{\|\vec{x}\|=1}-1 dS \\ &= -4 \pi \end{align*}

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I think you made an error in the last one (where did that -3 come from?). Except that, it's basically the same as my answer, but mine is much more compact and simple to compute; and this always saves times and minimize probability of errors. – leonbloy Jul 2 '11 at 11:11
@leonbloy: Yeah, I caught it after I looked at your answer – I did not compute the dot product correctly. – Michael Chen Jul 2 '11 at 13:09