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How can I use Fourier Transform to solve Fisher-Kolmogorov Equation in 1D?

\begin{equation} u_t(x,t) = u_{xx}(t) + u(1-u) \end{equation}

\begin{equation} u(0,x) = \phi(x) \end{equation}

with Dirichlet \begin{equation} u(0,t)=0 \\ u(1,t)=0 \end{equation}

and Neumann boundary conditions \begin{equation} u_x(0,t)=0\\ u_x(1,t)=0 \end{equation}

Can I just do the following?

\begin{equation} F\{u(x,t)\}=\hat{u}(k,t) = \int_{-\infty}^{+\infty}u(x,t)e^{-ikx}dx \end{equation}

\begin{equation} \hat{u}_t(k,t) = (ik)^2\hat{u}(k) + F\{u(x,y)(1-u(x,t))\} \\ u_t(x,t)=F^{-1} \{ (ik)^2\hat{u}(k) \} + u(x,y)(1-u(x,t)) \end{equation}

This is a simple Matlab implementation in 1D:

After 250 iterations using forward Euler with $\Delta t = 0.01$ , the solution looks something like Fisher-Kolmogorov

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  • $\begingroup$ I don't understand what the plot shows. The function you are seeking, $u(x,t)$ depends also on time, but your plot only shows the $x$-dependence. $\endgroup$ Jul 11 '15 at 19:21
  • $\begingroup$ It's one frame after 250 iterations using forward Euler and $\Delta t = 0.01$ $\endgroup$
    – ilciavo
    Jul 11 '15 at 19:24
  • $\begingroup$ OK, but so is the shown solution correct or wrong? $\endgroup$ Jul 13 '15 at 16:29
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The problem can be partially solved following this tutorial.

Given Fisher-Kolmogorov equation

\begin{equation} u_t=u_{xx}+u(1-u) \end{equation}

It can also be written as

\begin{equation} u_t = u_{xx} + u\,v\\ v_t = v_{xx} - u\,v \end{equation}

where $v = (1-u)$

Solving the equation \begin{equation} u_t = u_{xx} \\ \end{equation}

from $t$ to $\Delta t$ gives

\begin{equation} \hat{u}_t = (ik)^2 \hat{u} \\ \tilde{u}(x,t+\Delta t) = F^{-1} \left\{e^{-k^2 \Delta t} \hat{u} \right\} \end{equation}

Now solving the equation

\begin{equation} u_t = u \, v \\ \end{equation}

Using the splitting operator method we can write

\begin{equation} u(x,t+\Delta t) = e^{-v\Delta t} \tilde{u}(x,t+\Delta t) \end{equation}

Doing that likewise for $v$ gives the solution

Solution

Diffusion_Reaction.m

Boundary conditions still must be imposed.

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