# Laplace's equation

1. Jan 14, 2012

### ibysaiyan

1. The problem statement, all variables and given/known data
f = r^-n-1 * cos(n+1) θ satisfies laplace's equation.
r^2 $\partial^2 f / \partial r^2$ + r$\partial f / \partial r$ + $\partial^2 f / \partial θ^2$ = 0

2. Relevant equations
P.D.E

3. The attempt at a solution
$\partial f / \partial r$ = nr^n-1 * sin (nθ)
$\partial f/ \partial θ$ = ncos(nθ)*r^n
Are these derivatives right ?

Last edited: Jan 14, 2012
2. Jan 14, 2012

### HACR

What happened to partial differential equation?

3. Jan 14, 2012

### ibysaiyan

Sorry , I have fixed my latex commands.Refresh the webpage.

4. Jan 14, 2012

### ibysaiyan

The function is :

f = $r^{-n-1}$ * cos (n+1)$θ$

Now I have been trying to figure out what bits are exactly related to variable 'r' and 'theta'...
$\frac {\partial f}{\partial θ}$ = {sin (n+1)θ + cos (n+1) } * r^-n-1 or
sin (n+1) * r^-n-1.
Can someone clarify this for me . Thanks!

5. Jan 14, 2012

If $f = cos[(n+1)\theta] r^{-(n+1)}$

then $$\frac{\partial f}{\partial r} = -(n+1)cos[(n+1)\theta] r^{-(n+2)}$$
and $$\frac{\partial f}{\partial \theta} = -(n+1)sin[(n+1)\theta] r^{-(n+1)}$$

Last edited: Jan 14, 2012
6. Jan 14, 2012

### ibysaiyan

Thanks for the help. This makes sense.. I was being sloppy, I didn't realize the fact that differentiating sin nθ is no different than sin (n+1)θ ah....

7. Jan 14, 2012