0
$\begingroup$

I am new to mathematical-biology and I have to solve the following (diffusion-like) equation \begin{eqnarray} \frac{\partial a(x,t)}{\partial t}= D \frac{\partial^2 a(x,t)}{\partial x^2}\\ \frac{\partial b(x,t)}{\partial t}=-v \frac{\partial b(x,t)}{\partial x}-\mu a(x,t) b(x,t) \\ \frac{\partial c(x,t)}{\partial t}= \gamma \frac{\partial^2 c(x,t)}{\partial x^2}+k b(x,t) \end{eqnarray}

for the functions a(x,t), b(x,t) and c(x,t), subject to boundary conditions $a(x,t\rightarrow \infty)=0$, $a(x\rightarrow \infty,t)=0$, $\frac{\partial c(x,t)}{\partial(x)}|_{x=0}=\frac{\partial a(x,t)}{\partial x}|_{x=0}$, $b(L,t)=b_L$, $c(0,t)=0$, and $\frac{\partial c(x,t)}{\partial x}|_{x=L}=0$, where $D, v, \mu,\gamma, k$ and $b_l$ are constants.

I tried to attack this problem using, first, the separation of variables method. Although I do obtain a "simple" answer for $a(x,t)$ written in terms of exponentials, I am stuck at the second equation. More precisely, this term with $-\mu a(x,t) b(x,t)$ became a nightmare to me. I would NDSolveValue but, so far, only errors pop out from my Mathematica notebook.

New contributor
Gui Volp is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
2
  • 3
    $\begingroup$ Post the actual code that you used (Raw InputForm). You should not use D as a variable name since it already has a defined meaning in Mathematica. $\endgroup$ – Bob Hanlon 2 days ago
  • $\begingroup$ Hi Bob, indeed, in my code I replaced "D" by another constant. pde = {D[aa[x, t],t]==k1*D[aa[x,t],x,x], D[bb[x, t], t]==-k2*D[bb[x,t],x]-k3*aa[x, t]*bb[x, t], D[cc[x, t], t]==k4*D[cc[x,t],x,x]+k5*bb[x, t]}; where k1,k2...k5 are just positive constants (used instead of the greek letters for sake of simplicity). I am then strugling to make the boundary conditions work. I can set a x=xout whete xout would be my "inifity" for the numerical solution $\endgroup$ – Gui Volp 14 hours ago

Your Answer

Gui Volp is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.