Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
Wikipedia says the third term in your expression for the polar Laplace operator should have an additional factor of $\frac{1}{r^2}$. –  Qiaochu Yuan Mar 15 '12 at 7:27
    
@Qiaochu: Indeed-- twas quite late when I posted this. I have corrected it now. –  tentaclenorm Mar 15 '12 at 17:46

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
    
Beni, thanks for this. I feel like I'm missing something here, though. Is there a reason why the mixed partial derivatives do not appear in the second equation? [which I believe you meant to write as $\frac{\partial}{\partial x}\left(\frac{\partial}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial}{\partial \theta}\frac{\partial \theta}{\partial x} \right) $] –  tentaclenorm Mar 18 '12 at 6:12
    
@AmeliaYzaguirre: Yes, I forgot the mixed derivatives. Sorry about that. –  Beni Bogosel Mar 18 '12 at 11:22
    
thanks for your answer. Thought I accepted this earlier. cheers –  tentaclenorm Apr 1 '12 at 3:30
    
@BeniBogosel Isn't the mixed derivative supposed to have a factor 2? –  Michiel Aug 20 at 6:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.