1. ## [SOLVED] Laplace's equation

Transform Laplace's equation to polar coordinates.

$\frac{\partial^2{u}}{\partial{x^2}} + \frac{\partial^2{u}}{\partial{y^2}} = 0$

$u(x, y) = u(rcos\varphi + rsin\varphi)$

$\left\{\begin{array}{cc}\frac{\partial{u}}{\partia l{r}} = cos\varphi \frac{\partial{u}}{\partial{x}} + sin\varphi \frac{\partial{u}}{\partial{y}}\\ \frac{\partial{u}}{\partial{\varphi}} = rcos\varphi \frac{\partial{u}}{\partial{y}} - rsin\varphi \frac{\partial{u}}{\partial{x}}\end{array}\right.$

Now what?

2. Originally Posted by Spec
Transform Laplace's equation to polar coordinates.

$\frac{\partial^2{u}}{\partial{x^2}} + \frac{\partial^2{u}}{\partial{y^2}} = 0$

$u(x, y) = u(rcos\varphi + rsin\varphi)$

$\left\{\begin{array}{cc}\frac{\partial{u}}{\partia l{r}} = cos\varphi \frac{\partial{u}}{\partial{x}} + sin\varphi \frac{\partial{u}}{\partial{y}}\\ \frac{\partial{u}}{\partial{\varphi}} = rcos\varphi \frac{\partial{u}}{\partial{y}} - rsin\varphi \frac{\partial{u}}{\partial{x}}\end{array}\right.$

Now what?

I'll derive the first result (using your notation):

$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \cdot \frac{\partial r}{\partial x} + \frac{\partial u}{\partial \varphi} \cdot \frac{\partial \varphi}{\partial x}$

To get $\frac{\partial r}{\partial x}$, differentiate implicitly with respect to x (use the chain rule on each side):

$x^2 + y^2 = r^2 \Rightarrow 2x = 2r \frac{\partial r}{\partial x} \Rightarrow \frac{\partial r}{\partial x} = \frac{x}{r} = \cos \varphi$.

To get $\frac{\partial \varphi}{\partial x}$, differentiate implicitly with respect to x (use the quotient rule on the right hand side and the chain rule on the left hand side):

$\tan \varphi = \frac{y}{x} \Rightarrow \sec^2 \varphi {\partial \varphi}{\partial x} = -\frac{y}{x^2} \Rightarrow {\partial \varphi}{\partial x} = \cos^2 \varphi \, \frac{r \sin \varphi}{r^2 \cos^2 \varphi} = - \frac{\sin \varphi}{r}$.

Therefore $\frac{\partial u}{\partial x} = \cos \varphi \, \frac{\partial u}{\partial r} - \frac{\sin \varphi}{r} \, \frac{\partial u}{\partial \varphi}$.

Now there's a bit of work left for you to do.