Solution to Laplace's Equation: f = r^-n-1 * cos(n+1)θ

[ibysaiyan]

Homework Statement

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

Homework Equations

P.D.E

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 ?

 

[HACR]

What happened to partial differential equation?

 

[ibysaiyan]

What happened to partial differential equation?

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

 

[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!

 

[Bread18]

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

 

[ibysaiyan]

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

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...

 

[Bread18]

Ha yeah, no problem.