Mathematical Biology (Reaction-Diffusion equation)

[ra_forever8]

A spatially varying competition model between red and grey squirrels is given by
\begin{equation} \frac{\partial R}{\partial t}= D_R \frac{\partial^2 R}{ \partial x^2} + aR(1-R-b_2G),\tag 1
\end{equation}
\begin{equation} \frac{\partial G}{\partial t}= D_G \frac{\partial^2 G}{ \partial x^2} + G(1-G-b_1R),\tag 2
\end{equation}where $R(x,t)$ and $G(x,t)$ are the red and grey squirrels densities, respectively, at time $t$ and spatial position $x$. Here $D_R, D_G,a,b_1$ and $b_2$ are all positive constants on the interval $-\infty <x<\infty$, with $b_1 <1$ and $b_2>1$.
By assuming a traveling wace solution of the form
$ R(z)= R(x-ct) $ and $G(z)=G(x-ct)$,
where $c$ is the wavespeed. In the case of $D_R= D_G =a=1$ and $b_1+b_2=2$, we have
$S" + cS'+S(1-S)=0$,
where $S=R+G, S'= \frac{dS}{dz} and S" = \frac{d^2 S}{dz^2}$.a)By considering the boundary conditions defined by
$R(-\infty)=0$ $G(-\infty)=1$$R(\infty)=1$ $G(\infty)=0$
show that for all $z$, $S\equiv 1$ is a solution.b) Using the result determined in a) show that
$G" + cG'+(1-b_1)G(1-G)=0$,=>
its really diffcult question for me but
my attempt is that as R ->1 and G -> 0.
Thats all i know and can not go any further. Can anyone please help me.

 

[jvicens]

Thank you for your question and for sharing the spatially varying competition model between red and grey squirrels. I am a scientist and I will do my best to help you understand the problem.

First of all, let's define the boundary conditions given in the problem. We have $R(-\infty)=0$ and $G(-\infty)=1$, which means that at the left boundary of our spatial domain, there are no red squirrels and only grey squirrels exist. Similarly, at the right boundary, we have $R(\infty)=1$ and $G(\infty)=0$, which means that at the right boundary, there are no grey squirrels and only red squirrels exist.

Now, let's consider the traveling wave solution $R(z)=R(x-ct)$ and $G(z)=G(x-ct)$, where $c$ is the wavespeed. By substituting this solution into equations (1) and (2), we get:

$\frac{\partial R}{\partial t}=-cR'(z)$ and $\frac{\partial G}{\partial t}=-cG'(z)$

$\frac{\partial^2 R}{\partial x^2}=R''(z)$ and $\frac{\partial^2 G}{\partial x^2}=G''(z)$

Substituting these into equations (1) and (2), we get:

$-cR' = R'' + aR(1-R-b_2G)$ and $-cG' = G'' + G(1-G-b_1R)$

Now, let's define $S=R+G$, and we can rewrite the above equations as:

$-cS' = S'' + (1-S)(1-b_1R-b_2G)$

Since we have $b_1+b_2=2$, we can simplify the above equation to:

$-cS' = S'' + (1-S)(1-S)$

This is a second-order differential equation with a constant coefficient, and we can solve it using standard techniques. The general solution is:

$S(z)=Ae^{\lambda z} + Be^{-\lambda z} + 1$

Where $A$ and $B$ are constants and $\lambda=\frac{-c+\sqrt{c^2-4}}{2}$.

Now, let's apply