1. Polar coordinates

This is a post by nomad609.

Hi all, my question is essentially the title.

I would like to find the polar coordinates for the Laplace of a particular scalar function.

My question is attached here, if me attaching the image did not work then I will try and drag and drop the image.

Here is the link of the question: Laplace from a scalar function - Album on Imgur

Now my working out for this is here, although I am not entirely sure it is correct.

Again here is the link and the image?

Link: My working out - Album on Imgur

Image:

I would really appreciate if anyone can help me out here.

2. Re: Polar coordinates

Although I do not feel it helps in the context of my question.

Does anyone have any other ideas to help me out please?

3. Re: Polar coordinates

Anyone have a clue :P

4. Re: Polar coordinates

Use the formulas connecting Cartesian and polar coordinates- $x= r cos(\theta)$, $y= r sin(\theta)$, $r= \sqrt{x^2+ y^2}= (x^2+ y^2)^{1/2]$, and $y= arctan\left(\frac{y}{x}\right)$- and the chain rule. It is straight forward but tedious.

$\frac{\partial f}{\partial x}= \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}$

$\frac{\partial r}{\partial x}= \frac{\partial (x^2+ y^2)^{1/2}}{\partial x}= (1/2)(x^2+ y^2)^{-1/2}(2x)= \frac{x}{(x^2+ y^2)^{1/2}}= \frac{r cos(\theta)}{r}= cos(\theta)$

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

So $\frac{\partial f}{\partial x}= cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}$

Now do it again:
$\frac{\partial^2 f}{\partial x^2}= \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)= \frac{\partial}{\partial x}\left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)$
$= cos(\theta)\frac{\partial\left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)}{\partial r}+ \frac{1}{r}sin(\theta)\frac{\partial \left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)}{\partial \theta}$
$= cos(\theta)\left(cos(\theta)\frac{\partial^2 f}{\partial r^2}+ \frac{1}{r^2}sin(\theta)\frac{\partial f}{\partial \theta}- \frac{1}{r}sin(\theta)\frac{\partial^2f}{\partial r\partial \theta}\right)+$ $\frac{1}{r}sin(\theta)\left(-sin(\theta)\frac{\partial f}{\partial r}+ cos(\theta)\frac{\partial^2 f}{\partial r\partial \theta}- \frac{1}{r}cos(\theta)\frac{\partial f}{\partial \theta}- \frac{1}{r}sin(\theta)\frac{\partial^2 f}{\partial \theta^2}\right)$

Well, I said it was tedious! I will leave the rest for you to try.

Notice that this contains a $cos^2(\theta)\frac{\partial^2 f}{\partial r^2}$. The second derivative with respect to y will contain $sin^2(\theta)\frac{\partial^2 f}{\partial r^2}$ so they add to $\frac{\partial^2 f}{\partial r^2}$. Similar things happen with the other derivatives.

5. Re: Polar coordinates

Originally Posted by HallsofIvy
Use the formulas connecting Cartesian and polar coordinates- $x= r cos(\theta)$, $y= r sin(\theta)$, $r= \sqrt{x^2+ y^2}= (x^2+ y^2)^{1/2]$, and $y= arctan\left(\frac{y}{x}\right)$- and the chain rule. It is straight forward but tedious.

$\frac{\partial f}{\partial x}= \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}$

$\frac{\partial r}{\partial x}= \frac{\partial (x^2+ y^2)^{1/2}}{\partial x}= (1/2)(x^2+ y^2)^{-1/2}(2x)= \frac{x}{(x^2+ y^2)^{1/2}}= \frac{r cos(\theta)}{r}= cos(\theta)$

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

So $\frac{\partial f}{\partial x}= cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}$

Now do it again:
$\frac{\partial^2 f}{\partial x^2}= \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)= \frac{\partial}{\partial x}\left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)$
$= cos(\theta)\frac{\partial\left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)}{\partial r}+ \frac{1}{r}sin(\theta)\frac{\partial \left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)}{\partial \theta}$
$= cos(\theta)\left(cos(\theta)\frac{\partial^2 f}{\partial r^2}+ \frac{1}{r^2}sin(\theta)\frac{\partial f}{\partial \theta}- \frac{1}{r}sin(\theta)\frac{\partial^2f}{\partial r\partial \theta}\right)+$ $\frac{1}{r}sin(\theta)\left(-sin(\theta)\frac{\partial f}{\partial r}+ cos(\theta)\frac{\partial^2 f}{\partial r\partial \theta}- \frac{1}{r}cos(\theta)\frac{\partial f}{\partial \theta}- \frac{1}{r}sin(\theta)\frac{\partial^2 f}{\partial \theta^2}\right)$

Well, I said it was tedious! I will leave the rest for you to try.

Notice that this contains a $cos^2(\theta)\frac{\partial^2 f}{\partial r^2}$. The second derivative with respect to y will contain $sin^2(\theta)\frac{\partial^2 f}{\partial r^2}$ so they add to $\frac{\partial^2 f}{\partial r^2}$. Similar things happen with the other derivatives.
Wow thank you somuch for this, I will try and finish this off. And if it okay, show you my working out.

Really appreciate this, thank you.

6. Re: Polar coordinates

Originally Posted by HallsofIvy
Use the formulas connecting Cartesian and polar coordinates- $x= r cos(\theta)$, $y= r sin(\theta)$, $r= \sqrt{x^2+ y^2}= (x^2+ y^2)^{1/2]$, and $y= arctan\left(\frac{y}{x}\right)$- and the chain rule. It is straight forward but tedious.

$\frac{\partial f}{\partial x}= \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}$

$\frac{\partial r}{\partial x}= \frac{\partial (x^2+ y^2)^{1/2}}{\partial x}= (1/2)(x^2+ y^2)^{-1/2}(2x)= \frac{x}{(x^2+ y^2)^{1/2}}= \frac{r cos(\theta)}{r}= cos(\theta)$

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

So $\frac{\partial f}{\partial x}= cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}$

Now do it again:
$\frac{\partial^2 f}{\partial x^2}= \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)= \frac{\partial}{\partial x}\left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)$
$= cos(\theta)\frac{\partial\left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)}{\partial r}+ \frac{1}{r}sin(\theta)\frac{\partial \left(cos(\theta) \frac{\partial f}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial f}{\partial \theta}\right)}{\partial \theta}$
$= cos(\theta)\left(cos(\theta)\frac{\partial^2 f}{\partial r^2}+ \frac{1}{r^2}sin(\theta)\frac{\partial f}{\partial \theta}- \frac{1}{r}sin(\theta)\frac{\partial^2f}{\partial r\partial \theta}\right)+$ $\frac{1}{r}sin(\theta)\left(-sin(\theta)\frac{\partial f}{\partial r}+ cos(\theta)\frac{\partial^2 f}{\partial r\partial \theta}- \frac{1}{r}cos(\theta)\frac{\partial f}{\partial \theta}- \frac{1}{r}sin(\theta)\frac{\partial^2 f}{\partial \theta^2}\right)$

Well, I said it was tedious! I will leave the rest for you to try.

Notice that this contains a $cos^2(\theta)\frac{\partial^2 f}{\partial r^2}$. The second derivative with respect to y will contain $sin^2(\theta)\frac{\partial^2 f}{\partial r^2}$ so they add to $\frac{\partial^2 f}{\partial r^2}$. Similar things happen with the other derivatives.

I think I have solved it

Although please, if you don't mind explaining how do you get the bolded part? This is the y=arctan(y/x) part this. Should the "y" on the LHS not be theta? by trig. I think this is a typo maybe?

If you could just clarify this, that would be awesome.

But thanks again for the help.

7. Re: Polar coordinates

Yes, that was a typo. $\theta= arctan(y/x)$.

8. Re: Polar coordinates

Originally Posted by HallsofIvy
Yes, that was a typo. $\theta= arctan(y/x)$.
Thank you bro

I realised after you typed it later on

Yes, that was a typo. $\theta= arctan(y/x)$.