Help with a partial derivatives problem involving two identical functions with different variables

Question

Let $ v(r, \theta) = u(x, y) $, with $ x = r \cos \theta $ and $ y = r \sin \theta $. Show that

$$ \begin{equation} \dfrac{\partial^2u}{\partial x^2} + \dfrac{\partial^2u}{\partial y^2} = \dfrac{\partial^2 v}{\partial r^2} + \dfrac{1}{r} \dfrac{\partial v}{\partial r} + \dfrac{1}{r^2} \dfrac{\partial^2 v}{\partial \theta^2} \end{equation} $$

I am familiar with partial derivatives, but I have no idea where to start.

Answer 1

We have $r=\sqrt{x^2+y^2}$ and $\theta=\arctan(\frac{y}{x})$. Then by the chain rule

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

and similarly

$$\frac{\partial^2 }{\partial y^2}=\sin^2(\theta)\frac{\partial^2}{\partial r^2}-\frac{\cos(\theta)\sin(\theta)}{r^2}\frac{\partial}{\partial \theta}+\frac{\cos(\theta)\sin(\theta)}{r}\frac{\partial^2}{\partial r\partial\theta}+\frac{\cos^2(\theta)}{r}\frac{\partial}{\partial r}+\frac{\cos(\theta)\sin(\theta)}{r}\frac{\partial^2}{\partial r\partial \theta}-\frac{\sin(\theta)\cos(\theta)}{r^2}\frac{\partial}{\partial \theta}+\frac{\cos^2(\theta)}{r^2}\frac{\partial^2}{\partial \theta^2}$$
Thus combining and using the identity $\cos^2(\theta)+\sin^2(\theta)=1$ gives Laplace's equation in polar coordinates $$\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}$$

See here for another method.

Answer 2

What you're trying to do is called deriving the Laplacian in polar coordinates.

You can start by setting $u(r\cos\theta,r\sin\theta)$ and computing the derivatives for $\dfrac{\partial v}{\partial r}$ ,$\dfrac{\partial^2 v}{\partial r^2}$ etc.

For example $\dfrac{\partial v}{\partial r} =\dfrac{\partial u}{\partial x}\dfrac{\partial x}{\partial r}+\dfrac{\partial u}{\partial y} \dfrac{\partial y}{\partial r}$, by the chain rule. Using that $x=r\cos\theta$ you can compute $\dfrac{\partial x}{\partial r} = \cos\theta$ and so on.

If you want spoilers, a detailed explanation can be found here