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Math Help - Transform Polar

  1. #16
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    \displaystyle u_{xx}=\left(\cos(\theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)\left(\cos(  \theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)

    Why can't this be computed like u_{yy}\text{?}

    If I take

    \displaystyle u_{yy}=\left(\frac{y}{r}u_r+\frac{x}{r^2}u_{\theta  }\right)^2

    and solve, the solution is correct but it doesn't work with u_{xx}. How come?

    Also, when I approach

    \displaystyle u_{xx}=\left(\cos(\theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)\left(\cos(  \theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)

    \displaystyle\Rightarrow \cos(\theta)u_r\cos(\theta)u_r-2u_{r\theta}\frac{\sin(\theta)\cos(\theta)}{r^2}+\  frac{\sin(\theta)}{r}u_{\theta}\frac{\sin(\theta)}  {r}u_{\theta}

    and

    \displaystyle u_{yy}=\sin(\theta)u_r\sin(\theta)u_r+2u_{r\theta}  \frac{\sin(\theta)\cos(\theta)}{r^2}+\frac{\cos(\t  heta)}{r}u_{\theta}\frac{\cos(\theta)}{r}u_{\theta  }

    Now, I am wary of saying I have cosine and sine squared since I need

    \displaystyle\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right).

    How can I put those two equations together in order to yield the intended results?
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  2. #17
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    Quote Originally Posted by dwsmith View Post
    \displaystyle u_{xx}=\left(\cos(\theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)\left(\cos(  \theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)

    Why can't this be computed like u_{yy}\text{?}

    If I take

    \displaystyle u_{yy}=\left(\frac{y}{r}u_r+\frac{x}{r^2}u_{\theta  }\right)^2

    and solve, the solution is correct but it doesn't work with u_{xx}. How come?

    Also, when I approach

    \displaystyle u_{xx}=\left(\cos(\theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)\left(\cos(  \theta)u_r-\frac{\sin(\theta)}{r}u_{\theta}\right)

    \displaystyle\Rightarrow \cos(\theta)u_r\cos(\theta)u_r-2u_{r\theta}\frac{\sin(\theta)\cos(\theta)}{r^2}+\  frac{\sin(\theta)}{r}u_{\theta}\frac{\sin(\theta)}  {r}u_{\theta}

    and

    \displaystyle u_{yy}=\sin(\theta)u_r\sin(\theta)u_r+2u_{r\theta}  \frac{\sin(\theta)\cos(\theta)}{r^2}+\frac{\cos(\t  heta)}{r}u_{\theta}\frac{\cos(\theta)}{r}u_{\theta  }

    Now, I am wary of saying I have cosine and sine squared since I need

    \displaystyle\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right).

    How can I put those two equations together in order to yield the intended results?
    I have it now.

    We have

    \displaystyle u_{xx}=\left(\frac{x}{r}u_r-\frac{y}{r^2}u_{\theta}\right)^2=\frac{x}{r}u_r\le  ft(\frac{x}{r}u_r\right)-\frac{2}{r}u_{r}\left(u_{\theta}\frac{xy}{r^2}\rig  ht)+\frac{y^2}{r^4}u_{\theta\theta}=\frac{x^2}{r}u  _r\left(\frac{1}{r}u_r\right)-\frac{2}{r}u_{r}\left(u_{\theta}\frac{xy}{r^2}\rig  ht)+\frac{y^2}{r^4}u_{\theta\theta}

    \displaystyle u_{yy}=\left(\frac{y}{r}u_r+\frac{x}{r^2}u_{\theta  }\right)^2=\frac{y}{r}u_r\left(\frac{y}{r}u_r\righ  t)+\frac{2}{r}u_{r}\left(u_{\theta}\frac{xy}{r^2}\  right)+\frac{x^2}{r^4}u_{\theta\theta}=\frac{y^2}{  r}u_r\left(\frac{1}{r}u_r\right)+\frac{2}{r}u_{r}\  left(u_{\theta}\frac{xy}{r^2}\right)+\frac{x^2}{r^  4}u_{\theta\theta}

    \displaystyle u_r\left(\frac{1}{r}u_r\right)\left(\frac{x^2+y^2}  {r}\right)+\left(\frac{x^2+y^2}{r^4}\right)u_{\the  ta\theta}=ru_r\left(\frac{1}{r}u_r\right)+\frac{1}  {r^2}u_{\theta\theta}

    \displaystyle\Rightarrow u_t=k\left(ru_r\left(\frac{1}{r}u_r\right)+\frac{1  }{r^2}u_{\theta\theta}\right)

    My r and 1/r are swapped, hmmm....
    Last edited by dwsmith; February 4th 2011 at 10:22 PM.
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