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Thread: Additive PDE Help

  1. #1
    Mar 2012

    Question Additive PDE Help

    Hey Guys

    Literally do not know where to start on this one. I've only learnt Multiplicative separation but I think I might be over thinking it.

    I've uploaded a capture below

    Attached Thumbnails Attached Thumbnails Additive PDE Help-capture.png  
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  2. #2
    MHF Contributor
    Prove It's Avatar
    Aug 2008

    Re: Additive PDE Help

    Assuming that $\displaystyle \displaystyle \begin{align*} E \end{align*}$ is a constant, you have $\displaystyle \displaystyle \begin{align*} \left(\frac{\partial S}{\partial r}\right)^2 + \frac{1}{r^2} \left(\frac{\partial S}{\partial \theta}\right)^2 + \frac{a\cos{\theta}}{r^2} = E \end{align*}$ and if we let $\displaystyle \displaystyle \begin{align*} S(r, \theta) = R(r) + \Theta(\theta) \end{align*}$, since the first term is only a function of $\displaystyle \displaystyle \begin{align*} r \end{align*}$ and the second is only a function of $\displaystyle \displaystyle \begin{align*} \theta \end{align*}$, then evaluating derivatives gives $\displaystyle \displaystyle \begin{align*} \frac{\partial S}{\partial r} = \frac{dR}{dr} \end{align*}$ and $\displaystyle \displaystyle \begin{align*} \frac{\partial S}{\partial \theta} = \frac{d\Theta}{d\theta} \end{align*}$, and substituting gives

    $\displaystyle \displaystyle \begin{align*} \left(\frac{dR}{dr}\right)^2 + \frac{1}{r^2}\left(\frac{d\Theta}{d\theta}\right)^ 2 + \frac{a\cos{\theta}}{r^2} &= E \\ \frac{1}{r^2}\left(\frac{d\Theta}{d\theta}\right)^ 2 + \frac{a\cos{\theta}}{r^2} &= E - \left(\frac{dR}{dr}\right)^2 \\ \left(\frac{d\Theta}{d\theta}\right)^2 + a\cos{\theta} &= r^2E - r^2\left(\frac{dR}{dr}\right)^2 \end{align*}$

    Since the left hand side is only a function of $\displaystyle \displaystyle \begin{align*} r \end{align*}$ and the right hand side is only a function of $\displaystyle \displaystyle \begin{align*} \theta \end{align*}$, the only way they can be equal is if they are both equal to some constant value $\displaystyle \displaystyle \begin{align*} \lambda \end{align*}$. This gives

    $\displaystyle \displaystyle \begin{align*} r^2E - r^2\left(\frac{dR}{dr}\right)^2 &= \lambda \\ \left(\frac{d\Theta}{d\theta}\right)^2 + a\cos{\theta} &= \lambda \end{align*}$

    which are two ordinary differential equations
    Last edited by Prove It; Mar 27th 2012 at 06:19 AM.
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