# Laplace's equation w/ polar coordinates

1. May 6, 2012

### Jamin2112

1. The problem statement, all variables and given/known data

The lecture notes say that ∇ = urr + (1/r)ur + (1/r2)uθθ. I'm not sure how this comes about. The notes never explain it.

2. Relevant equations

(???)

3. The attempt at a solution

No attempts on the actual homework problem until this ∇ thing is cleared up.

2. May 6, 2012

### HallsofIvy

Staff Emeritus
That is the "Laplacian" operator, $\partial^2/\partial x^2+ \partial^2/\partial x^2$ in two dimensional xy- coordinates, in polar coordinates.

For $\phi$ any differentiable function of x and y, by the chain rule,
$$\frac{\partial \phi}{\partial x}= \frac{\partial \phi}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial \phi}{\partial \theta}\frac{\partial \theta}{\partial x}$$

Since $r= (x^2+ y^2)^{1/2}$, $\frac{\partial r}{\partial x}= (1/2)(x^2+ y^2)^{-1/2}(2x)= (1/2)2rcos(\theta)/r= cos(\theta)$. Since $\theta= artctan(y/x)$,
$$\frac{\partial \theta}{\partial x}= (1./(1+ y^2/x^2))(-(y/x^2)= -y/(x^2+ y^2)= -rsin(\theta)/r^2= -sin(\theta)/r$$.

So
$$\frac{\partial \phi}{\partial x}= cos(\theta)\frac{\partial \phi}{\partial r}- \frac{sin(\theta)}{r} \frac{\partial \phi}{\partial \theta}$$

Then
$$\frac{\partial^2\phi}{\partial x^2}= cos(\theta)\frac{\partial}{\partial r}\left(cos(\theta)\frac{\partial \phi}{\partial r}- \frac{sin(\theta)}{r} \frac{\partial \phi}{\partial \theta}\right)$$$$+ \frac{sin(\theta)}{r}\left(cos(\theta)\frac{\partial \phi}{\partial r}- \frac{sin(\theta)}{r} \frac{\partial \phi}{\partial \theta}\right)$$

Complete that then do the same for $\partial^2\phi/\partial \theta^2$.

Last edited: May 6, 2012
3. May 6, 2012

### Jamin2112

Bingo. Got it.

4. May 6, 2012

### Jamin2112

HallsofIvy, I have another question if you have the time. I'm trying to solve

2u=0 ​

with the boundary conditions

u(x,0)=0, u(0,y)=0, u(x,1)=x, u(1,y)=y.​

I've found that u(x,0)=0 and u(0,y)=0 imply

u(x,y)= D * sinh(√kx) * sin(√ky)​

where k was my separation constant.

I then did

u(x,1)=x → uxx(x,1)=0=D * sinh(√kx) * sin(√ky)​

to conclude that √k = n∏ and that our solution is of the form

Ʃn≥1 Dn * sinh(n∏x) * sin(n∏y).​

Finally, I solved for Dn by

u(1,y)=y=Ʃn≥1 Dn * sinh(n∏) * sin(n∏y)
→ Dn * sinh(n∏) = (2/∏)∫[0,∏] y * sin(n∏y) dy.​

Since that gives me a very messy solution for u, I'm assuming it's wrong. Thoughts?