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Thread: Differential equation for two-dimensional diffusion/heat transfer

  1. #1
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    Question Differential equation for two-dimensional diffusion/heat transfer

    Assume there is a hollow cylinder with no concentration gradient over r, and then, the concentration gradient is along the z axis and the cylinder angle as

    $$ \frac{\partial c}{\partial t} = D \Biggl(\frac{1}{r^2}\frac{\partial^2 c}{\partial \phi^2}\ + \frac{\partial^2 c} {\partial z^2}\Biggl)$$

    How should the Laplace transform be applied to solve this differential equation?
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  2. #2
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    Re: Differential equation for two-dimensional diffusion/heat transfer

    Personally, I have never liked to use "Laplace transform". I would separate variables: Let $\displaystyle c= T(t)P(\phi)Z(z)$ so that the equation becomes $\displaystyle PZ\frac{dT}{dt}= D\left(TZ\frac{1}{r^2}\frac{d^2P}{d\phi^2}+ PT\frac{d^2Z}{dz^2}\right)$ and then, dividing by PTZ, $\displaystyle \frac{1}{T}\frac{dT}{dt}= D\left(\frac{1}{r^2P}\frac{d^2P}{d\phi^2}+ \frac{1}{Z}\frac{d^2Z}{dz^2}\right)$.

    So we must have $\displaystyle \frac{1}{T}\frac{dT}{dt}= \lambda$ and $\displaystyle D\left(\frac{1}{r^2P}\frac{d^2P}{d\phi^2}+ \frac{1}{Z}\frac{d^2Z}{dz^2}\right)= \lambda$ for some constant $\displaystyle \lambda$. Further, we must have $\displaystyle D\frac{1}{r^2P}\frac{d^2P}{d\phi^2}= \mu$ and $\displaystyle D\frac{1}{Z}\frac{d^2Z}{dz^2}= \nu$ for constants $\displaystyle \mu$ and $\displaystyle \nu$ such that $\displaystyle \mu+ \nu= \lambda$. Use boundary initial conditions to determine possible values for those constants.
    Last edited by HallsofIvy; Aug 13th 2017 at 05:06 AM.
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