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Reaction diffusion problem concentric spheresby scg08
Tags: reaction diffusion 
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#1
Apr3009, 07:54 AM

P: 1

Hello,
I am trying to (numerically) solve the following reactiondiffusion equation for the probability density of the a pair, [tex]\rho (\vec{r}_1,\vec{r}_2)[/tex]: [tex]\dot{\rho} (\vec{r}_1,\vec{r}_2,t) = D_1 \nabla^2_1 \rho (\vec{r}_1,\vec{r}_2,t) + D_2 \nabla^2_2 \rho (\vec{r}_1,\vec{r}_2,t)  k \left( \left\ \vec{r}_1  \vec{r}_2 \right\ \right) [/tex], where the subscripts refer to the first and second particle, respectively. In 2D and polar coordinates, [tex]r_i[/tex] and [tex]\theta_i [/tex]: [tex] \nabla^2_i = \frac{1}{r_i} \frac{\partial}{\partial r_i} r_i \frac{\partial}{\partial r_i} + \frac{1}{r_i^2} \frac{\partial}{\partial \theta_i} [/tex]. The domain is confined by two concentric spheres: [tex] 0 \leq \left\ \vec{r}_1 \right\ \leq R [/tex] and [tex] \left\ \vec{r}_2 \right\ \geq R [/tex]. The initial condition are spherically symmetric, i.e. only depends on the [tex]r_i[/tex]s. The reaction term is a function of the distance of the two particles, i.e. in 2D [tex] k( \left\ \vec{r}_1  \vec{r}_2 \right\ ) = k( \sqrt{r_1^2 + r_2^2  2 r_1 r_2 \cos ( \theta_1\theta_2)} ) [/tex]. I hoped to get rid of at least 1 coordinate by a variable transformation and separation of variables. However, so far I just could not come up with a separable problem. Do I really have to retain all 4 variables? Any suggestions of how to reduce this problem to something manageable are highly welcome. Eventually I will be interested in 3D and 4D as well. Thank you, Daniel 


#2
May1409, 10:24 AM

P: 333

Just throwing idea.
If the forces between particles 1 and 2 are conservative, try working in the centre of mass frame. In mechanics we use this frame to solve the central force motion and scattering problem. 


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