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Math Help - Polar Coordinates

  1. #1
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    Polar Coordinates

    Solving a PDE with Polar coordinates

    yu_x-xu_y=0

    x=r\cos{\theta} \ \mbox{and} \ y=r\sin{\theta}

    u(r,\theta)

    Does u_x\Rightarrow u_r \ \mbox{or} \ u_{\theta} \ \mbox{and why?}

    Thanks.
    Last edited by dwsmith; December 24th 2010 at 06:44 PM. Reason: added request for explanation
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  2. #2
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    You need to use the chain rule

    u_x = u_r r_x + u_{\theta} \theta_x

    u_y = u_r r_y + u_{\theta} \theta_y

    or if you use Jacobians (the way I'd do it)

    u_x =\dfrac{\partial(u,y) }{\partial(r,\theta)}/\dfrac{\partial(x,y)}{\partial(u,\theta)}

    u_y =\dfrac{\partial(x,u) }{\partial(r,\theta)}/\dfrac{\partial(x,y)}{\partial(u,\theta)}.
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  3. #3
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    Danny,

    I haven't done anything with the Jacobian in a long time. Could you show me how it is done?

    Thanks.

    \displaystyle u_x=\cos{\theta}u_r-\frac{\sin{\theta}}{r}u_{\theta}

    \displaystyle u_y=\sin{\theta}u_r+\frac{\cos{\theta}}{r}u_{\thet  a}

    \displaystyle yu_x-xu_y=r\sin{\theta}\left[\cos{\theta}u_r-\frac{\sin{\theta}}{r}u_{\theta}\right]-r\cos{\theta}\left[\sin{\theta}u_r+\frac{\cos{\theta}}{r}u_{\theta}\r  ight]

    \displaystyle r\sin{\theta}\cos{\theta}u_r-\sin^2{\theta}u_{\theta}-r\cos{\theta}\sin{\theta}-\cos^2{\theta}u_{\theta}=0

    \displaystyle -u_{\theta}(\sin^2{\theta}+\cos^2{\theta})=0\Righta  rrow -u_{\theta}=0\Rightarrow u_{\theta}=0

    \displaystyle\int u_{\theta}d\theta=\int 0d\theta\Rightarrow u(r,\theta)=f(r)

    The solution I am supposed to conclude is u(x,y)=f(x^2+y^2), f arbitrary.

    How do I finish off from u(r,\theta)=f(r)\mbox{?}

    Thanks.
    Last edited by dwsmith; December 25th 2010 at 02:07 PM. Reason: Partial Solution added
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  4. #4
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    From
    <br />
u_{\theta}=0 \; \Rightarrow \; u(r,\theta)=f(r).<br />
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  5. #5
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    Quote Originally Posted by zzzoak View Post
    From
    <br />
u_{\theta}=0 \; \Rightarrow \; u(r,\theta)=f(r).<br />
    How does that transform to u(x,y)=f(x^2+y^2)\mbox{?}
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  6. #6
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    <br />
f(r)=f( \; \sqrt{x^2+y^2} \; )=g( \; x^2+y^2 \; ).<br />
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  7. #7
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    And making the change from u(r, theta) to u(x,y) is fine?
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  8. #8
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    We convert
    <br />
u(r,\theta)=U(x,y)<br />
    and
    <br />
f(r)=F(x^2+y^2)<br />
    and have finally
    <br />
U(x,y)=F(x^2+y^2).<br />
    We may check inserting it to the initial equation.
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