- #1

Carla1985

- 94

- 0

\[

\begin{align}

\frac{dr}{dt}= &\ - \left(\alpha+\frac{\epsilon}{2}\right)r + \left(1-\frac{\epsilon}{2}\right)\alpha p - \alpha^2\beta r p + \frac{\epsilon}{2} \\

\frac{dp}{dt}= &\ \left(1-\frac{\epsilon}{2\alpha}\right)r - \left(1+\frac{\epsilon}{2}\right)p - \alpha\beta r p + \frac{\epsilon}{2\alpha}

\end{align}

\]

with initial conditions $r(0)=1, p(0)=0$ and $\epsilon$ is a small parameter. After substituting in the asymptotic approximations $r\approx r_0+\epsilon r_1+...$, $p\approx p_0+\epsilon p_1+...$ I get a leading order problem as

\[

\begin{align}

\frac{dr_0}{dt}= &\ - \alpha r_0 + \alpha p_0 -\alpha^2\beta r_0 p_0 \\

\frac{dp_0}{dt}= &\ r_0 - p_0 -\alpha\beta r_0 p_0

\end{align}

\]

Clearly, this is still of no help as it's still nonlinear. So my question is, is there any way of getting past this?

The only other piece of information I have is that at equilibrium the solution for both will be $O(\epsilon^{1/2})$. I did attempt to rescale for this but I then get a singular perturbation problem in which the limit of the inner solution is infinity so I could not do the matching (I will explain in more detail what I did for this if needed). If anyone could suggest a way in which I can solve this problem I would be very grateful.

Thank you!

Carla