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I am trying to obtain an expression for the concentration $C$ based on this stationary equation :

$\frac{\partial C}{\partial t} = \frac{1}{r} \frac{d}{dr} \left(r \frac{\partial C}{\partial r}\right) - \frac{k}{D} C = 0 $

Being in cylindrical/radial coordinate, I can't simplify the $r$, which would make the calculation more easy.

I tried to resolve this in Mathematica. Analytical solution can be obtained based on Bessel's functions, to which I'm not really used.

Can someone help me to manually get to these analytical solution ?

NB : problem quite similar to this one (I think) : Reaction-diffusion PDE in cylindrical coordinates - Green's function method

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1 Answer 1

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$$ \partial_t C = \frac{1}{r}\partial_r \left(r\partial_r C\right) - \frac{k}{D}C $$ Seperation of variables $C(r,t) = R(r)T(t)$

$$ \frac{\dot{T}}{T} = \frac{1}{R}\frac{1}{r}\frac{d}{dr} \left(r\frac{dR}{dr}\right) - \frac{k}{D} = -\lambda_n^2 $$

this leads to $$ \frac{\dot{T}}{T} = -\lambda_n^2\\ \frac{1}{R}\frac{1}{r}\frac{d}{dr} \left(r\frac{dR}{dr}\right) - \frac{k}{D} = -\lambda_n^2 $$ we can write $$ \frac{1}{r}\frac{d}{dr} \left(r\frac{dR}{dr}\right) - \lambda_1 R = 0\\ R'' + \frac{1}{r}R' - \lambda_1 R \implies r^2R'' +rR' -\lambda_1r^2R = 0 $$ the last equation has solutions of the form $$ R(r) = A_1J_{0}\left(\sqrt{-\lambda_1}r\right) + A_2Y_{0}\left(\sqrt{-\lambda_1}r\right) $$ where $$ -\lambda_1 = \lambda_n^2-\frac{k}{D} $$

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  • $\begingroup$ Thanks for the answer ! Just to be sure I clearly understood : now I have to put the found $R(r)$ value into the expression of $C(r,t)=R(r) T(t)$ in order to derive the value of $\lambda^{2}_{n}$ ? $\endgroup$
    – Valacar
    Dec 12, 2014 at 13:33
  • $\begingroup$ Yes that's correct. From here you use the standard method for computing the solution from separation of variables technique :). $\endgroup$
    – Chinny84
    Dec 12, 2014 at 14:14

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