Question

What can be done with this?

$\frac{dx}{dt} = \alpha(x+y)(M-x)$

$\frac{dy}{dt} = \alpha(N-y)x$

$x(0) = c$, $y(0) = d$

$\alpha, M, N, c, d$ are positive reals. The system dynamics exists only until $x$ reaches $M$ or $y$ reaches $N$.

Can we get $x$ and $y$ as functions of $t$?

Answer 1

I have no idea where this system comes from or what you have tried or are interested in trying.

My initial thought is to look at this numerically using a simple finite-difference approximation for the time derivative in order to express the functions $x$ and $y$ in terms of $t$.

A central-difference approximation for the time derivatives looks like this: $$ \frac{dx}{dt} \approx \frac{x_{i+1} - x_{i}}{\Delta t}, \text{ and similarly for }\frac{dy}{dt}$$

where $\Delta t = t_{i+1} - t_{i}$ is a time step we can choose later, and $x_{i+1}$ is a symbol for $x(t_{i+1})$.

So, the system can be expressed as:

$$\begin{align} x_{i+1} &= \Delta t \left[ \alpha(x_{i} + y_{i})(M + x_{i})\right] + x_{i} \\ & \\ y_{i+1} &= \Delta t \left[ \alpha(N - y_{i})x_{i}\right] + y_{i} \end{align}$$

This RHS of this formulation is expressed in terms of known quantities (coming from the initial conditions). We also essentially have expressions for $x(t)$ and $y(t)$, but are not in general reliable in the long term.

If this looks interesting to you, take this question over to math.stackexchange.com and see what they (we) come up with. We can run this through matlab or whatever if you like. I am not so sure it is appropriate here.

Answer 2

You have two (nonnegative) fixed points (0,0) and (M,N). The first one is a saddle and the second one is a stable node. The (nonnegative part of the) unstable manifold of the fisrt one converges to the second one. A trivial dynamical analysis of the phase plane can be done easily. I recommend the plotting of the solutions in the phase plane using Mathematica's NDSolve.