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Math Help - Please help with PDE in polar coordinates.

  1. #1
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    Please help with PDE in polar coordinates.

    For Polar coordinates, r^2=x^2+y^2 and x=r\cos \theta, y=r\sin\theta

    r^2=x^2+y^2\Rightarrow\; r\frac{\partial {r}}{\partial {x}}=x+y\frac{\partial {y}}{\partial {x}}
    \Rightarrow\; \frac{\partial {r}}{\partial {x}}=\frac{x}{r}+\frac{y}{r}\frac{\partial {y}}{\partial {x}}

    \frac{\partial {y}}{\partial {x}}=0, then \frac{\partial {r}}{\partial {x}}=\frac{x}{r}


    \frac{\partial {r}}{\partial {x}}=\cos\theta as \frac{\partial {y}}{\partial {x}}=0

    You can now put in 90^o\;>\;\theta\;>45^o, you'll get r>x and \frac{\partial {r}}{\partial {x}}>1


    Let's just use an example where ##\theta =60^o##, so to every unit change of x, rwill change for 2 unit. So \frac{\partial {r}}{\partial {x}}=2

    The same reasoning, r=2x. So using this example, \frac{\partial {r}}{\partial {x}}=\cos 60^o=0.5 which does not agree with the example I gave.

    I am missing something, please help.
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    Re: Please help with PDE in polar coordinates.

    I am puzzled as to what your question is . Your title refers to a "PDE in polar coordinates" but there is no PDE, just relations between Cartesian Coordinates and Polar Coordinates. In any case, from x= r cos(\theta), r= \dfrac{x}{cos(\theta)} so \dfrac{\partial r}{\partial x}= \dfrac{1}{cos(\theta)}, not cos(\theta). Similarly, \dfrac{\partial r}{\partial x}= \dfrac{r}{x}= \dfrac{r}{rcos(\theta)}= \dfrac{1}{cos(\theta)}.
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    Re: Please help with PDE in polar coordinates.

    Quote Originally Posted by HallsofIvy View Post
    I am puzzled as to what your question is . Your title refers to a "PDE in polar coordinates" but there is no PDE, just relations between Cartesian Coordinates and Polar Coordinates. In any case, from x= r cos(\theta), r= \dfrac{x}{cos(\theta)} so \dfrac{\partial r}{\partial x}= \dfrac{1}{cos(\theta)}, not cos(\theta). Similarly, \dfrac{\partial r}{\partial x}= \dfrac{r}{x}= \dfrac{r}{rcos(\theta)}= \dfrac{1}{cos(\theta)}.
    Thanks for the reply, but according to the derivation shown in my first post, \frac{\partial {r}}{\partial {x}}=\frac {x}{r}=\cos\theta
    This is verified by more than one book.
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    Re: Please help with PDE in polar coordinates.

    Quote Originally Posted by Alan0354 View Post
    For Polar coordinates, r^2=x^2+y^2 and x=r\cos \theta, y=r\sin\theta

    r^2=x^2+y^2\Rightarrow\; r\frac{\partial {r}}{\partial {x}}=x+y\frac{\partial {y}}{\partial {x}}
    \Rightarrow\; \frac{\partial {r}}{\partial {x}}=\frac{x}{r}+\frac{y}{r}\frac{\partial {y}}{\partial {x}}

    \frac{\partial {y}}{\partial {x}}=0, then \frac{\partial {r}}{\partial {x}}=\frac{x}{r}


    \frac{\partial {r}}{\partial {x}}=\cos\theta as \frac{\partial {y}}{\partial {x}}=0

    You can now put in 90^o\;>\;\theta\;>45^o, you'll get r>x and \frac{\partial {r}}{\partial {x}}>1


    Let's just use an example where ##\theta =60^o##, so to every unit change of x, rwill change for 2 unit. So \frac{\partial {r}}{\partial {x}}=2

    The same reasoning, r=2x. So using this example, \frac{\partial {r}}{\partial {x}}=\cos 60^o=0.5 which does not agree with the example I gave.

    I am missing something, please help.
    \frac{\partial {r}}{\partial {x}}=\cos\theta if \frac{\partial {y}}{\partial {x}}=0. If r=2x=\sqrt{x^2+y^2}, then y = \pm x\sqrt{3}, so \frac{\partial {y}}{\partial {x}} = \pm \sqrt{3} \neq 0.
    Thanks from topsquark, LimpSpider and Alan0354
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    Re: Please help with PDE in polar coordinates.

    Quote Originally Posted by SlipEternal View Post
    \frac{\partial {r}}{\partial {x}}=\cos\theta if \frac{\partial {y}}{\partial {x}}=0. If r=2x=\sqrt{x^2+y^2}, then y = \pm x\sqrt{3}, so \frac{\partial {y}}{\partial {x}} = \pm \sqrt{3} \neq 0.
    Thanks for the reply. This is just so confusing.


    \frac{\partial y}{\partial x}=\frac{\partial y(r,\theta)}{\partial r}\frac{\partial r}{\partial x(r,\theta)}+\frac{\partial y(r,\theta)}{\partial \theta}\frac{\partial \theta}{\partial x(r,\theta)}=\cos\theta \frac{\partial y(r,\theta)}{\partial r}-\frac{\sin\theta}{r}\frac{\partial y(r,\theta)}{\partial \theta }= 0


    So what is correct?
    Last edited by Alan0354; October 30th 2013 at 12:45 PM.
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    Re: Please help with PDE in polar coordinates.

    Just a thought...are we to assume that you have a DEq that is being integrated over a region? And that you are trying to convert that region from rectangular coordinates to polar coordinates? In that case x and y are independent (except on the boundaries) and thus \frac{\partial y}{\partial x} = 0.

    It might help to post the entire problem.

    -Dan
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    Re: Please help with PDE in polar coordinates.

    Quote Originally Posted by topsquark View Post
    Just a thought...are we to assume that you have a DEq that is being integrated over a region? And that you are trying to convert that region from rectangular coordinates to polar coordinates? In that case x and y are independent (except on the boundaries) and thus \frac{\partial y}{\partial x} = 0.

    It might help to post the entire problem.

    -Dan
    Thanks for the reply. This is the complete question. I just find this that I cannot explain. It is just simple Polar coordinates and I listed the derivation according to more than one sources in the first post. Then I just use an example to show a different result that I cannot explain.

    As you can see from the derivation, \frac{\partial{r}}{\partial{x}}=\frac{x}{r} if \frac{\partial{y}}{\partial{x}}=0 as I showed my work. But that does not explain the example I gave when 45^o<\theta<90^o.


    And the plot even thickens when SlipEternal showed that \frac{\partial{y}}{\partial{x}}\neq 0
    Thanks
    Last edited by Alan0354; October 30th 2013 at 03:04 PM.
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    Re: Please help with PDE in polar coordinates.

    Quote Originally Posted by Alan0354 View Post
    Thanks for the reply. This is the complete question. I just find this that I cannot explain. It is just simple Polar coordinates and I listed the derivation according to more than one sources in the first post. Then I just use an example to show a different result that I cannot explain.

    As you can see from the derivation, \frac{\partial{r}}{\partial{x}}=\frac{x}{r} if \frac{\partial{y}}{\partial{x}}=0 as I showed my work. But that does not explain the example I gave when 45^o<\theta<90^o.


    And the plot even thickens when SlipEternal showed that \frac{\partial{y}}{\partial{x}}\neq 0
    Thanks
    Well, that's why I'm curious. First about the DEq tag in your thread title. Why is that there if you don't have a DEq you are working with? And your comment that you found that \frac{\partial y}{\partial x} = 0 "verified by more than one book" leads me to believe you are converting a domain in xy to a domain in r \theta. At the very least you don't seem to be converting an equation (differential or otherwise) in this thread.

    This whole question is just confusing!

    -Dan
    Thanks from Alan0354
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    Re: Please help with PDE in polar coordinates.

    Quote Originally Posted by topsquark View Post
    Well, that's why I'm curious. First about the DEq tag in your thread title. Why is that there if you don't have a DEq you are working with? And your comment that you found that \frac{\partial y}{\partial x} = 0 "verified by more than one book" leads me to believe you are converting a domain in xy to a domain in r \theta. At the very least you don't seem to be converting an equation (differential or otherwise) in this thread.

    This whole question is just confusing!

    -Dan
    I might have use the wrong tittle, but my question is still valid. I triple checked my post before I posted it. Maybe, you can change it to a better tittle for me.
    Thanks
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    Re: Please help with PDE in polar coordinates.

    Now I am officially lost!!! I since posted this question in two different math forums, people there both asked and clarified, then no response!!!

    I have one book and at least one article derived the equations like in my first post, but obviously the example I gave does not agree with the first post and I triple checked my example. I don't think I did anything wrong.......apparently I have not get any suggestion otherwise from three forums!!! I am pretty sure I am missing something as the book I used is a text book used in San Jose State and I studied through 7 chapters and yet to find a single mistake until the question here.

    Anyone has anything to say?

    From my example,
    \frac{\partial{r}}{\partial{x}}=\frac{r}{x}=\frac{  1}{\cos\theta}
    And this answer makes a lot more sense.

    Thanks
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    Re: Please help with PDE in polar coordinates.

    Quote Originally Posted by Alan0354 View Post
    Now I am officially lost!!! I since posted this question in two different math forums, people there both asked and clarified, then no response!!!

    I have one book and at least one article derived the equations like in my first post, but obviously the example I gave does not agree with the first post and I triple checked my example. I don't think I did anything wrong.......apparently I have not get any suggestion otherwise from three forums!!! I am pretty sure I am missing something as the book I used is a text book used in San Jose State and I studied through 7 chapters and yet to find a single mistake until the question here.

    Anyone has anything to say?

    From my example,
    \frac{\partial{r}}{\partial{x}}=\frac{r}{x}=\frac{  1}{\cos\theta}
    And this answer makes a lot more sense.

    Thanks
    The reason no one has a suggestion for you is because you are not asking any valid questions. If \frac{\partial{y}}{\partial{x}}=0, then \frac{\partial{r}}{\partial{x}} = \cos \theta. But, if \frac{\partial{y}}{\partial{x}}\neq 0, then \frac{\partial{r}}{\partial{x}} \neq \cos \theta.

    Also, as HallsofIvy showed, \frac{\partial{r}}{\partial{x}} = \dfrac{1}{\cos\theta}, so setting those two equal, we get \cos^2\theta = 1, so \frac{\partial{y}}{\partial{x}}=0 only if \theta = n\pi where n is an integer. Hence, \frac{\partial{y}}{\partial{x}}=0 only if \frac{\partial{r}}{\partial{x}} = \pm 1.
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    Re: Please help with PDE in polar coordinates.

    Quote Originally Posted by SlipEternal View Post
    The reason no one has a suggestion for you is because you are not asking any valid questions. If \frac{\partial{y}}{\partial{x}}=0, then \frac{\partial{r}}{\partial{x}} = \cos \theta. But, if \frac{\partial{y}}{\partial{x}}\neq 0, then \frac{\partial{r}}{\partial{x}} \neq \cos \theta.

    Also, as HallsofIvy showed, \frac{\partial{r}}{\partial{x}} = \dfrac{1}{\cos\theta}, so setting those two equal, we get \cos^2\theta = 1, so \frac{\partial{y}}{\partial{x}}=0 only if \theta = n\pi where n is an integer. Hence, \frac{\partial{y}}{\partial{x}}=0 only if \frac{\partial{r}}{\partial{x}} = \pm 1.
    But I posted in post #5 proof \frac{\partial{y}}{\partial{x}}= 0 for polar coordinates.

    On top of that, you showed \frac{\partial{y}}{\partial{x}}\neq 0, which I agree with you. But the proof in post #5 say otherwise, so what am I missing?
    Last edited by Alan0354; October 31st 2013 at 12:54 PM.
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    Re: Please help with PDE in polar coordinates.

    Quote Originally Posted by Alan0354 View Post
    But I posted in post #5 proof \frac{\partial{y}}{\partial{x}}= 0 for polar coordinates.

    On top of that, you showed \frac{\partial{y}}{\partial{x}}\neq 0, which I agree with you. But the proof in post #5 say otherwise, so what am I missing?
    In post 5, you assumed \frac{\partial{x(r,\theta)}}{\partial{r}} = \cos\theta, which means you were already assuming \frac{\partial{y}}{\partial{x}} = 0. You never proved it.
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    Re: Please help with PDE in polar coordinates.

    Quote Originally Posted by SlipEternal View Post
    In post 5, you assumed \frac{\partial{x(r,\theta)}}{\partial{r}} = \cos\theta, which means you were already assuming \frac{\partial{y}}{\partial{x}} = 0. You never proved it.
    Oops, I wrote that upside down. I meant you assumed \frac{\partial{r}}{\partial{x(r,\theta)}} = \cos\theta
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    Re: Please help with PDE in polar coordinates.

    Quote Originally Posted by SlipEternal View Post
    Oops, I wrote that upside down. I meant you assumed \frac{\partial{r}}{\partial{x(r,\theta)}} = \cos\theta
    I think you find my problem. Let me work this out:

    r^2=x^2+y^2\;\Rightarrow\;\frac{\partial{r}}{\part  ial{x}}=\frac {x}{r}+\frac{y}{r}\frac{\partial{y}}{\partial{x}}

    x=r\cos\theta\;\Rightarrow \partial{x}=\partial{r}\cos\theta\;\Rightarrow\;\f  rac{\partial{r}}{\partial{x}}=\frac{1}{\cos\theta}  =\frac{r}{x}

    \Rightarrow\;\frac{r}{x}=\frac {x}{r}+\frac{y}{r}\frac{\partial{y}}{\partial{x}}

    \Rightarrow\;\frac{\partial{y}}{\partial{x}}=\left  (\frac{r}{x}-\frac{x}{r}\right)\frac{r}{y}

    y=r\sin\theta and x=r\cos\theta

    \Rightarrow\;\frac{\partial{y}}{\partial{x}}=\left  (\frac{r}{x}-\frac{x}{r}\right)\frac{r}{y} = \frac{r^2}{r^2 \cos\theta \sin\theta}-\frac{\cos\theta}{\sin\theta}=\tan\theta


    I really appreciate your help, I know it must be my mistake, I don't believe for a moment the book is wrong. Just small little mistake cause me over a day!!! Now everything comes together.

    Thanks
    Last edited by Alan0354; October 31st 2013 at 03:07 PM.
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