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Re: Radially symmetric point source diffusion with decay
Posted:
Apr 25, 2012 5:16 PM


On 26/04/2012 1:16 a.m., Peter Spellucci wrote: > Gib Bogle<g.bogle@auckland.ac.nz> writes: >> On 25/04/2012 4:23 p.m., Gib Bogle wrote: >>> I'm interested in solving for the concentration field around an object >>> (a cell) that is secreting a substance (chemokine) that diffuses and >>> decays. I can solve the equation in 3D on a finite difference grid, but >>> it occurs to me that radial symmetry reduces this to a 1D equation. >>> >>> According to me, the second order 1D DE is >>> >>> (1/r)d/dr[r.dC/dr] = K.C >> >> I see that this is wrong (it is the 2D case). In 3D it should be >> >> (1/r^2)d/dr[r^2.dC/dr] = K.C >> >> I need to correct my Matlab code. My question stands, however. > > as I understand, you try to solve (originally in 3D) a diffusion > problem with a finite ?(non zero measure) source, the concentration given > on the surface of the source, i.e. you have initial values only for C. > I miss the time derivative: is this a steady state? or is your > real question: given C(at r=1 in your case) find r_end > such that there C=given value ? > (Since you try to vary the initial slope until your requests are > sufficiently satisfied) > Then this is a free boundary value problem which however could be transformed > into a standard one (and solved as such). > Warning: boundary value problems, solved as initial value problems, can exhibit > enormous trouble since the general solution of the ode might have strongly > growing components. > > hth > peter >
Sorry Peter, I meant to say that it's the steady state solution that I want. I know that for the decay rates of interest the concentration is very close to zero at r_end = 30, and for my purposes I don't care about the tiny difference between the solutions for r_end = 20 and r_end = 30. As it turns out the shooting method with NewtonRaphson works very well, and is quite robust with a wide range of starting slope. From a bit of reading I can see that there probably will not be a tractable analytical solution  even Laplace's eqtn for the 3D case has a rather complicated series solution involving positive and negative powers of r.



