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Topic: Radially symmetric point source diffusion with decay
Replies: 7   Last Post: Apr 27, 2012 6:08 PM

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 Gib Bogle Posts: 42 Registered: 3/28/11
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 Newton-Raphson 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.

Date Subject Author
4/25/12 Gib Bogle
4/26/12 Nicolas Neuss
4/26/12 Gib Bogle
4/27/12 Nicolas Neuss
4/27/12 Peter Spellucci
4/27/12 Gib Bogle