Use the chain rule to convert the Laplace equation in (x,y) coordinates into an equivilent differental equation in (r,theta) coordinates.

Question

use the equations $r=\sqrt{x^2 +y^2}$ and $\theta=\arctan(\frac{y}{x})$. I was able to get the partial derivative of of $r$ with respect to $x$ and $y$ and the partial derivative of $\theta$ with respect to $x$ and $y$. I have spent hours trying to figure this problem out by looking at the polar form of the Laplace equation as a reference but everything I try does not get me to that point. If anyone is up to the challenge please help me out! I apologize for not putting in Latex format, I have not quite gotten the hang of it yet.

Answer 1

A complete explanation can be found here link, which I summarize below.

To convert the Laplace's equation $ \frac{\partial^2 f}{{\partial x}^2} + \frac{\partial^2 f}{{\partial y}^2} = 0 $ to polar coordinates first we define $x$ and $y$ in terms of $r$ and $\theta$.

$$ x = r\cos{\theta} \ \ \ \ y = r\sin{\theta}$$

Thus $ x $ and $ y $ are functions of $ r $ and $ \theta $ and so we may calculate their partial derivatives.

$$ \frac{\partial x}{\partial r } = \cos{\theta} \ \ \ \ \frac{\partial x}{\partial \theta } = -r\sin{\theta} \ \ \ \ \frac{\partial y}{\partial r} = \sin{\theta} \ \ \ \ \frac{\partial y}{\partial \theta} = r\cos{\theta}$$

Since $ f $ is a function of both $ x $ and $ y $ which are in turn functions of $ u $ and $ \theta $, we use the chain rule to calculate $ \frac{\partial f}{\partial r} $.

$$ \frac{\partial f}{\partial r} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial r}$$

Which after substituting for $ \frac{\partial x}{\partial r} $ and $ \frac{\partial y}{\partial r} $ becomes,

$$ \frac{\partial f}{\partial r} = \cos{\theta}\frac{\partial f}{\partial x}+ \sin{\theta}\frac{\partial f}{\partial y}$$

Now we must calculate $ \frac{\partial^2 f}{{\partial r}^2} $.

$$ \frac{\partial^2 f}{{\partial r}^2} = \frac{\partial}{\partial r} \left[ \cos{\theta}\frac{\partial f}{\partial x}+ \sin{\theta}\frac{\partial f}{\partial y} \right] = \cos{\theta}\frac{\partial}{\partial r}\frac{\partial f}{\partial x} + \sin{\theta}\frac{\partial}{\partial r}\frac{\partial f}{\partial y}$$

Since $ \frac{\partial f}{\partial x} $ and $ \frac{\partial f}{\partial y} $ are both functions of $ x $ and $ y $ we apply the same chain rule as stated above,

$$ = \cos{\theta} \left[ \frac{\partial f}{\partial x}\frac{\partial}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial f}{\partial x}\frac{\partial}{\partial y}\frac{\partial y}{\partial r} \right] + \sin{\theta} \left[ \frac{\partial f}{\partial y}\frac{\partial}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\frac{\partial}{\partial y}\frac{\partial y}{\partial r} \right]$$

Substituting for the partial derivatives of $ x $ and $ y $,

$$ = \cos{\theta} \left[ \cos{\theta}\frac{\partial f}{\partial x}\frac{\partial}{\partial x} + \sin{\theta}\frac{\partial f}{\partial x}\frac{\partial}{\partial y}\right] + \sin{\theta} \left[ \cos{\theta}\frac{\partial f}{\partial y}\frac{\partial}{\partial x} + \sin{\theta}\frac{\partial f}{\partial y}\frac{\partial}{\partial y}\right]$$

Which by Clairaut's Theorem simplifies to,

$$ = \cos^2{\theta}\frac{\partial^2 f}{{\partial x}^2} + 2\sin{\theta}\cos{\theta}\frac{\partial^2 f}{\partial x \partial y} + \sin^2{\theta}\frac{\partial^2 f}{{\partial y}^2}$$

Next in a similar manner we calculate $ \frac{\partial f}{\partial \theta} $ and $ \frac{\partial^2 f}{{\partial\theta}}^2 $. These are more challenging since $ \theta $ does not drop out in the partial derivatives as does $ r $ and so the product rule must be applied in taking the partial derivatives. Thus you should get,

$$ \frac{\partial f}{\partial \theta} = -r\sin{\theta}\frac{\partial f}{\partial x} + r\cos{\theta}\frac{\partial f}{\partial y}$$

and

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

Thus once we divide $ \frac{\partial^2 f}{{\partial \theta}^2}$ by $ \frac{1}{r^2}$, add to $ \frac{\partial^2 f}{{\partial r}^2} $, simplify using $ \sin^2 + \cos^2 = 1 $ and combine like terms we get,

$$ \frac{\partial^2 f}{{\partial x}^2} + \frac{\partial^2 f}{{\partial y}^2} = \frac{\partial^2 f}{{\partial r}^2} + \frac{1}{r}\frac{\partial f}{\partial r} + \frac{1}{r^2}\frac{\partial^2 f}{{\partial\theta}^2} = 0$$

This is my first answer to a question and I skipped a lot of the algebra. If between this and the link there are still difficulties I'll be happy to write out the steps I omitted. Hope this helps.