I'm trying to find the error in my logic here.
Let's say we are given the Laplace operator in polar coordinates:
$$ \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial}{\partial \theta^2} \tag{1} $$
and we're interested in transforming back to Cartesian coordinates. We could first make use of the usual coordinate transformation, namely
$$ u = r \sin(\theta), ~~v = r \cos(\theta), \tag{2}$$
to write the partial derivatives in polar coordinates as follows,
$$ \frac{\partial}{\partial r} = \sin(\theta) \frac{\partial}{\partial u} + \cos(\theta)\frac{\partial}{\partial v}, \tag{3}$$
$$ \frac{\partial}{\partial \theta} = r \cos(\theta) \frac{\partial}{\partial u} – r \sin(\theta)\frac{\partial}{\partial v}. \tag{4}$$
I think we should then rewrite all $\theta$ and $r$ in terms of $u$ and $v$, making use of $r^2 = u^2 + v^2$, so that in particular
$$ \sin(\theta) = \frac{u}{\sqrt{u^2 + v^2}}, \tag{5}$$
$$ \cos(\theta) = \frac{v}{\sqrt{u^2 + v^2}} \tag{6}.$$
With this, (3) and (4) then become
$$ \frac{\partial}{\partial r} = \frac{u}{\sqrt{u^2 + v^2}} \frac{\partial}{\partial u} + \frac{v}{\sqrt{u^2 + v^2}} \frac{\partial}{\partial v}, \tag{3*}$$
$$ \frac{\partial}{\partial \theta} = v \frac{\partial}{\partial u} – u \frac{\partial}{\partial v}. \tag{4*}$$
Our next task is to then compute $\frac{\partial^2}{\partial r^2}$ and $\frac{\partial^2}{\partial \theta^2}$, which we can carefully do by multiplying the right-hand sides of (3*) and (4*), keeping in mind that the operators will act on the coefficients.
However, executing this and adding the terms together yields nothing that looks remotely near the known form, namely $$\frac{\partial^2}{\partial u^2} + \frac{\partial^2}{\partial v^2}.$$ So I'm wondering, what am I missing here?
In addition to this particular example, I'm interested in any general information you have about transforming partial derivatives from polar to Cartesian coordinates.
Best Answer
Here is my proof: $$ \frac{\partial}{\partial x}=\frac{\partial}{\partial r} \frac{\partial r}{\partial x} +\frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x}$$ so by applying the product rule
$$ \frac{\partial^2}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial}{\partial r} \frac{\partial r}{\partial x}+\frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x}\right)= \frac{\partial^2}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2+ \frac{\partial}{\partial r}\frac{\partial^2 r}{\partial x^2} + \frac{\partial^2}{\partial \theta^2}\left(\frac{\partial \theta}{\partial x}\right)^2+ \frac{\partial}{\partial \theta}\frac{\partial^2 \theta}{\partial x^2}$$ $$ +\frac{\partial^2 }{\partial r \partial \theta}\frac{\partial r}{\partial x} \frac{\partial \theta}{\partial x}$$
You get the same relation with $y$. Now you just have to calculate the derivatives of $r,\theta$ with respect to $x,y$. You have $$ r=\sqrt{x^2+y^2},\ \theta=\arctan \frac{y}{x}$$.
What follows is a simple calculus exercise on derivatives. You just need to prove that $$\left(\frac{\partial r}{\partial x}\right)^2+\left(\frac{\partial r}{\partial y}\right)^2=1, $$ etc.(the relations you need so that when you sum $\frac{\partial^2}{\partial x^2}$ and $\frac{\partial^2}{\partial y^2}$ you'll get the polar form of the laplacian).
Going on the lines you started, you shouldn't change $\sin \theta,\cos \theta$ in terms of $x,y$. Just calculate the next derivative with respect to $r,\theta$, using appropriately the formula of the partial derivative of composition of functions.