I have a system of ODEs:
$$\dot{x} = \frac{m_1 x (1-x-y)}{a_1 + 1-x-y} - x$$ $$\dot{y} = \frac{m_2 y (1-x-y)}{a_2 + 1-x-y} - y$$
I'm trying to show that all the flows converge to the point $(1-\lambda_1,0)$ but I'm having some difficulties. (Here $\lambda_{1,2} = \frac{a_{1,2}}{m_{1,2}-1}$, $0 < \lambda < 1$ and $m>1$)
I can see that on the line $x+y = 1$ the flow is pointing "inwards" and along the axis $x=0$ $y=0$ the flow is along those axes and so the particle will stay in the upper right quadrant region bounded by those lines.
I can also see there is a critical point at (0,0) which I think gives an unstable node.
I've also found critical point at $(0,1-\lambda_2)$ and $(1-\lambda_1,0)$ but I can't figure out how to analyse these.
I'm struggling to put this together to get the convergence claim.
Thanks for any help
To check fixpoint stability we linearize the equations around the fixpoint $(x,y) = (\overline{x} ,\overline{y})$. Put $x= \overline{x} + \delta x$, $y= \overline{y} + \delta y$ and expand the equations to first order in the perturbations. For a general system
$$\dot{x} = F(x,y)$$ $$\dot{y} = G(x,y)$$
the fixpoints are determined by $F(\overline{x},\overline{y})=G(\overline{x},\overline{y})=0$. Linearizing around such a point gives us the perturbations equations
$$\dot{\delta x} = \left.\frac{\partial F(x,y)}{\partial x}\right|_{(x,y)=(\overline{x} ,\overline{y})}\delta x + \left.\frac{\partial F(x,y)}{\partial y}\right|_{(x,y)=(\overline{x} ,\overline{y})}\delta y$$
$$\dot{\delta y} = \left.\frac{\partial G(x,y)}{\partial x}\right|_{(x,y)=(\overline{x} ,\overline{y})}\delta x + \left.\frac{\partial G(x,y)}{\partial y}\right|_{(x,y)=(\overline{x} ,\overline{y})}\delta y$$
Now try to solve this system and see if it has exponentially growing or decaying solutions. Exponentially decaying solutions indicates stability while exponentially increasing solutions indicates that the fixpoint is unstable.
As an example: For your problem (for simplicity taking $m_1=m_2=m$ and $a_1=a_2=a$) and the fixpoint $(x,y) = (0,1-\lambda)$ we find
$$\dot{\delta x} = -\frac{1}{m+1}\delta x$$
$$\dot{\delta y} = \frac{(a-m)m^2}{a(1+m)^2}\delta x - \frac{m^3 + a(1+m-m^2)}{a(1+m)^2}\delta y$$
The first equation has the solution $\delta x(t) = \delta x(0) e^{-\frac{t}{m+1}}$ and is exponentially decaying. The second equation has the solution
$$\delta y = \delta x(0) e^{-\frac{t}{(1+m)}} + (\delta y(0) - \delta x(0))e^{-\frac{m^3 + a(1+m-m^2)}{a(1+m)^2}t}$$
which is also exponentially decaying since $m^3 + a(1+m-m^2) = m^2(m-a) + a(1+m) > 0$ since $a+1<m$. This shows that the fixpoint $(0,1-\lambda)$ is a stable fixpoint.