Transforming the Laplace operator from Polar to Cartesian coordinates

Question

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.

Answer 1

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).

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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.

Answer 2

Using a bit of differential geometry. The key is the writing $\Delta f = \text{div grad}f$: if we manage to express div and grad in a coordinate-independent manner, we can write $\Delta$ easily in any coordinate system, be it cartesian or polar.

The metric and the inverse metric are $$ g_{ab}=\left(\begin{array}[cc] \ 1&0\\ 0&1 \end{array} \right)=g_{ab}^{-1} $$ in cartesian coordinates and $$ g'_{ab}=\left(\begin{array}[cc] \ 1&0\\ 0&r^2 \end{array} \right),\ \ \ g'^{-1}_{ab}=\left(\begin{array}[cc] \ 1&0\\ 0&r^{-2} \end{array} \right) $$ in polar coordinates. The volume form is $\omega=dx\wedge dy=r\, dr\wedge d\varphi$ and the differential of a function is $$ df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy,\ \ \ dF= \frac{\partial F}{\partial r}dr + \frac{\partial F}{\partial \varphi}d\varphi. $$ Now, div is given by $ d[\omega(E)]=\text{div}(E) \omega, $ for any vector field $E=E^x \partial_x + E^y\partial_y$ or $B=B^r \partial_r + B^\varphi\partial_\varphi$; so $$ d[\omega(E)] = d[E^x dy - E^y dx]=\left(\frac{\partial E^x}{\partial x}+\frac{\partial E^y}{\partial y}\right) dx\wedge dy\implies\text{div}(E)=\frac{\partial E^x}{\partial x}+\frac{\partial E^y}{\partial y}. $$ Or $$ d[\omega(B)] = d[B^r r\,d\varphi - B^\varphi r\,dr] =\left(\frac{1}{r}\frac{\partial (rB^r)}{\partial r}+\frac{\partial B^\varphi}{\partial \varphi}\right)r\, dr\wedge d\varphi\implies\text{div}(B)= \frac{1}{r}\frac{\partial (rB^r)}{\partial r}+\frac{\partial B^\varphi}{\partial \varphi}. $$ And grad$f=g^{-1}df$, so $$ \text{grad}f = \frac{\partial f}{\partial x}\partial_x + \frac{\partial f}{\partial y}\partial_y \ \text{ and } \ \text{grad}F = \frac{\partial F}{\partial r}\partial_r + \frac{1}{r^2}\frac{\partial F}{\partial \varphi}\partial_\varphi. $$ Finally, $\Delta=$div grad: $$ \Delta f = \frac{\partial^2 f}{\partial x^2}+ \frac{\partial^2 f}{\partial y^2} \ \text{ and } \ \Delta F = \frac{1}{r}\left( r \frac{\partial F}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 F}{\partial \varphi^2}. $$