Polar form of Laplacian operator.

Question

Prove $$ \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)+\frac{1}{r^2}\left(\frac{\partial^2 u}{\partial \theta^2}\right) $$ for $u=f(x,y)$

Where should I start from? $u$ is a general function of $x$ and $y$?

Answer 1

I would start with $$\frac{\partial u}{\partial x}=\frac{\partial }{\partial r}\frac{\partial r}{\partial x} + \frac{\partial }{\partial \theta}\frac{\partial \theta}{\partial x}$$

$$\frac{\partial u}{\partial y}=\frac{\partial }{\partial r}\frac{\partial r}{\partial y} + \frac{\partial }{\partial \theta}\frac{\partial \theta}{\partial y}$$ Then use that $$r^2=x^2+y^2 \text{ and } x=r\cos\theta \text{ and } y=r\sin\theta$$ to conclude the following:

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

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

Just to make things clear, the above is done so by the following method:

$x=r\cos \theta $ and so $\frac{\partial x}{\partial r}=\cos\theta$.

$y=r\sin\theta$ and so $\frac{\partial y}{\partial r}=\sin\theta.$

$\theta=\arctan(\frac{y}{x})$ and so $\frac{\partial \theta}{\partial x}=\frac{\frac{-y}{x^2}}{1+\frac{y^2}{x^2}}=\frac{-y}{x^2+y^2}=\frac{-\sin\theta}{r} $

$\theta=\arctan(\frac{y}{x})$ and so $\frac{\partial \theta}{\partial y}=\frac{\frac{x}{x^2}}{1+\frac{y^2}{x^2}}=\frac{x}{x^2+y^2}=\frac{\cos\theta}{r} $

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

$$= \cos^2\theta \frac{\partial ^2}{\partial r^2}+\frac{2\sin\theta \cos\theta}{r^2}\frac{\partial}{\partial \theta}-\frac{2 \sin\theta \cos\theta}{r}\frac{\partial ^2}{\partial\theta \partial r}+\frac{\sin^2\theta}{r}\frac{\partial}{\partial r}+\frac{\sin^2\theta}{r^2}\frac{\partial ^2}{\partial\theta^2}$$

Then a very similar calculation gives

$$\frac{\partial ^2}{\partial y^2}= \sin^2\theta \frac{\partial ^2}{\partial r^2}-\frac{2\sin\theta \cos\theta}{r^2}\frac{\partial}{\partial \theta}+\frac{2 \sin\theta \cos\theta}{r}\frac{\partial ^2}{\partial\theta \partial r}+\frac{\cos^2\theta}{r}\frac{\partial}{\partial r}+\frac{\cos^2\theta}{r^2}\frac{\partial ^2}{\partial\theta^2}$$

Adding these two we then get,

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

Answer 2

To make things clear.. $u$ is a function of $x$ and $y$, not of $r$ and $\theta$. When we look at $u$ as a function of $r$ and $\theta$, we are abusing notation and looking at another function $\widetilde{u}$ defined by $$\widetilde{u}(r,\theta) =u(r\cos \theta,r\sin\theta).$$ The left-hand side of the equality you want to prove is the laplacian of $u$ (in $x$ and $y$), and the right-hand side is the laplacian of $\widetilde{u}$ (in $r$ and $\theta $). Once you understand this, the computation boils down to not panicking and just using the chain rule repeatedly.