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Math Help - [SOLVED] Laplace's equation

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Laplace's equation

    Transform Laplace's equation to polar coordinates.

    \frac{\partial^2{u}}{\partial{x^2}} + \frac{\partial^2{u}}{\partial{y^2}} = 0

    u(x, y) = u(rcos\varphi + rsin\varphi)

    \left\{\begin{array}{cc}\frac{\partial{u}}{\partia  l{r}} = cos\varphi \frac{\partial{u}}{\partial{x}} + sin\varphi \frac{\partial{u}}{\partial{y}}\\ \frac{\partial{u}}{\partial{\varphi}} = rcos\varphi \frac{\partial{u}}{\partial{y}} - rsin\varphi \frac{\partial{u}}{\partial{x}}\end{array}\right.

    Now what?
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  2. #2
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    Quote Originally Posted by Spec View Post
    Transform Laplace's equation to polar coordinates.

    \frac{\partial^2{u}}{\partial{x^2}} + \frac{\partial^2{u}}{\partial{y^2}} = 0

    u(x, y) = u(rcos\varphi + rsin\varphi)

    \left\{\begin{array}{cc}\frac{\partial{u}}{\partia  l{r}} = cos\varphi \frac{\partial{u}}{\partial{x}} + sin\varphi \frac{\partial{u}}{\partial{y}}\\ \frac{\partial{u}}{\partial{\varphi}} = rcos\varphi \frac{\partial{u}}{\partial{y}} - rsin\varphi \frac{\partial{u}}{\partial{x}}\end{array}\right.

    Now what?
    First read this: http://www.math.unc.edu/Faculty/pbe/...ations/HW8.pdf


    I'll derive the first result (using your notation):


    \frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \cdot \frac{\partial r}{\partial x} + \frac{\partial u}{\partial \varphi} \cdot \frac{\partial \varphi}{\partial x}


    To get \frac{\partial r}{\partial x}, differentiate implicitly with respect to x (use the chain rule on each side):

    x^2 + y^2 = r^2 \Rightarrow 2x = 2r \frac{\partial r}{\partial x} \Rightarrow \frac{\partial r}{\partial x} = \frac{x}{r} = \cos \varphi.


    To get \frac{\partial \varphi}{\partial x}, differentiate implicitly with respect to x (use the quotient rule on the right hand side and the chain rule on the left hand side):

    \tan \varphi = \frac{y}{x} \Rightarrow \sec^2 \varphi {\partial \varphi}{\partial x} = -\frac{y}{x^2} \Rightarrow {\partial \varphi}{\partial x} = \cos^2 \varphi \, \frac{r \sin \varphi}{r^2 \cos^2 \varphi} = - \frac{\sin \varphi}{r}.



    Therefore \frac{\partial u}{\partial x} = \cos \varphi \, \frac{\partial u}{\partial r} - \frac{\sin \varphi}{r} \, \frac{\partial u}{\partial \varphi} .


    Now there's a bit of work left for you to do.
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