How to prove that $\Delta f = 0$ for $f(x)=\frac{1}{||x||}$?

Question

I am given a function $f:\mathbb{R}^3\rightarrow\mathbb{R}:f(x)=\frac{1}{||x||}$, and I am not really sure that the norm must be Euclidian (anyway it wasn't mentioned in the task), and I have to prove that $\Delta f = 0$ (Laplace operator: $\Delta f(x_1,..,x_n) = \sum_{i=1}^n \frac{\partial^2f}{\partial x_i ^2}$). But even if I assume that the norm is Euclidian, I still don't get the right result:

$$\frac{\partial ^2f}{\partial x_i^2} = -\frac{x_i^2}{(x_1^2+x_2^2+x_3^2)^{\frac{3}{2}}} + \frac{1}{(x_1^2+x_2^2+x_3^2)^{\frac{1}{2}}}$$

So:

$$\Delta f = \frac{2}{(x_1^2+x_2^2+x_3^2)^{\frac{1}{2}}}$$

So something is wrong in my calculations. Could you help me?

Thanks in advance!

score 1 · Accepted Answer · edited Jun 05 '13 at 23:11

1

Do you get the same results? If no, try to repeat your calculations once more.

$$\frac{\partial ^2}{\partial x_i^2}\frac{1}{(x_1^2+x_2^2+x_3^2)^{1/2}} = \frac{2x_i^2-x_j^2-x_k^2}{(x_1^2+x_2^2+x_3^2)^{5/2}} $$

After getting all three expressions like this, you will be able to see that sum of numerators is equal to zero.

edited Jun 05 '13 at 23:11

Pedro

122,002

answered Jun 05 '13 at 23:06

tkroman

450

Is it worth mentioning the case of what happens when $|| x || = 0$? – Andrew D Jun 05 '13 at 23:41
@AndrewD: There is a singularity at $0$, so the function is not harmonic on the unit ball. – robjohn Jun 06 '13 at 03:52

How to prove that $\Delta f = 0$ for $f(x)=\frac{1}{||x||}$?

1 Answers1