System of second-order autonomous ODE

[dr29]

Hello,
I have a system of two second-order autonomous ODEs arising from a population genetics model:
(a-b y)x(1-x) + x''=0
(a-b x)y(1-y) + y''=0
where a and b are constants, and x, y are fonctions of t.
Is there any hope to solve this system?
Thank you for your help.

 

[Eynstone]

One can linearise this if you provide additional details about how fast x,y grow & whether xy is a small quantity.

 

[dr29]

Thank you for your response!
Sorry I should have been more specific; the parameters satisfy 0<a<b<1; attached is the form of the solution I'm looking for (with x/y in blue/red). Here the variable is set at 0 when x=y.
Even if there's no simple solution, the value of x=y at 0, and the slopes at 0 would already be quite interesting...

 

[coelho]

Well... there is a way to de-couple this set of equations, so one gets one ODE for x and one ODE for y, but i think it does more harm than good, as the equations one gets are of fourth-order, enormous, and probably impossible to solve analitically.

Anyway, it seems possible to get some useful informations about this system from the graph you sent. Looking at it, we see that the slope of y (which is y') starts (at t=-100) at almost zero, then increases near t=0, then decreases again to almost zero as t increases past 10. If y' starts at almost zero, increases then returns to almost zero, it reaches a maximum at somewhere between. At the maximum of y', its derivative (y'') is zero.

If we put y''=0 at the second equation, assuming that 0<y<1, we find that this maximum is reached when x=a/b.

The same applies to the curve of x, the only difference is that x' reaches a minimum instead a maximum. Anyway, x'' still is zero, and so we find that at this point y=a/b.

It is very tempting to assume that, as we found equal values for x and y, it means that this value is the one where both curves intercept (at t=0), but there is no reason why it should be. This values only represent the points where x'' and y'' are zero, but nothing ensures that both x'' and y'' are zero at the same instant of time.

Anyway, if you know the values of a and b, you may verify if x (and y) are actually a/b at t=0. If it were, it may be just a coincidence, or it may be a "right" result, even if by pure luck, or guessing.

 

[HallsofIvy]

You can always reduce a system of higher order d.e. s to first order.

Let u= x' and v= y' then x''= u'= (a- by)x(1- x) and y''= v'= (a- bx)y(1- y). Those, together with the equationx x'= u and y'= v give a system of four non-linear equations. It is easy to see that the equilibrium solutions are given by (a- by)x(1- x)= 0, (a- bx)y(1- y)= 0, u= 0, v= 0. The first equation has solutions x= 0, x= 1 and y= a/b. If x=0, the second equation is ay(1- y) which hs solutions y= 0 and y= 1. If x= 1, the second equation is (a- b)y(1- y)= 0 which has the same solutions. if y= a/b then the equation is (a- bx)(a/b)(1- a/b)= 0 which has x= a/b as solution. That is, the equilbrium solutions are
(x, y, u, v)= (0, 0, 0, 0)
(x, y, u, v)= (0, 1, 0, 0)
(x, y, u, v)= (1, 0, 0, 0)
(x, y, u, v)= (1, 1, 0, 0)
(x, y, u, v)= (a/b, a/b, 0, 0).

The Jacobian for the system is
\begin{bmatrix}(a- by)(1- 2x) &amp; -bx(1- x) &amp; 0 &amp; 0 \\ -by(1- y) &amp; (a- bx)(1- 2y) \\ 0 &amp; 0 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 1\end{bmatrix}

Finding the eigenvalues and eigenvectors for that system at each equilibrium point should tell you about the behaviour of the system around that point.

 

[dr29]

OK thanks for your help!
Unfortunately x and y don't reach a/b at the same time, as can be seen in this new figure with a=0.05, b=0.2 (so a/b=0.25).
So with this solution, I have the equilibria (x,y,u,v)=(1,0,0,0), (x,y,u,v)=(0,1,0,0), but not (a/b, a/b, 0,0)...
What would be great would be to have the values of x and x' when y=a/b (and hence x''=0), but I'm not sure if it's feasible...

 

[HallsofIvy]

That's not true. x= a/b, y= a/b, is, as I said before, an equi8librium point- if x= a/b, y= a/b, initially, then they have those values for all t. At the point (a/b, a/b), -a(1- a/b) is an eigenvalue. Whether solutions that start near that will move toward or away from it depends upon whether -a(1- a/b) is positive or negative. In your example, you chose initial values well away from a/b.

 

[dr29]

Yes sorry, it is an equilibrium, but it is never reached in the case I'm interested in. Let me be more explicit: I'm looking for a steady-state solution for the pair of PDEs:
dx/dt=(a-b y)x(1-x)+d^2 x/dz^2
dy/dt=(a-b x)y(1-y)+d^2 y/dz^2
The initial condition (at time t=0) is x=y=0 everywhere (for all values of z, which is a spatial variable), except x=epsilon, y=0 at z=-z_init < 0, x=0, y=epsilon at z=+z_init > 0. Two traveling waves develop for x and y starting from -z_init and z_init, and finally collide at z=0, to take the form shown on the previous figures. I'm trying to describe x(z) and y(z) at this equilibrium.
This new figure shows the system at different times (the x-axis is the spatial z-variable... that was called t in previous posts -- sorry for the confusion)