Parameterized math calculations?

[Gribouillis]

1) How did you come up with these definitions?

Well I just felt that due to the fact that the equation has only linear and quadratic terms, it could be possible to eliminate the constants a and b from the equations by scaling the time and the solution with an ad hoc factor. This idea is classical in the field of ODE and PDE, it is related to the techniques of nondimensionalization.

2) Why does this "method" stands mathematically?

If t ⟶ (y(t), e(t)) is a trajectory of the second system (without a and b), then t ⟶ (Y(t), E(t)) := ((b/a) y(bt), (b/a) e(bt)) is a trajectory of the first system and conversely. So there is a one to one correspondance between the trajectories of the two systems. The only problem is that we don't know (a, b), but it is true that the vector (Y, E) at time t is a multiple of the vector (y, e) at time b t. The factor between the two is (b/a)

3) Are y' = y'(t) and e' = e'(t)?

The relation y' = - y e holds at any moment, so it can be y'(t) = - y(t) e(t) or y'(b t) = - y(b t) e(b t) etc. This is not important

4) How am I going to calculate y(bt), e(bt) if I need it in the Runge-Kutta method? You mentioned something about multiple of Y(t), E(t)?

Suppose that you know the initial position (Y0, E0) at time t=0 and the final position (Y1, E1) at time t1. Your problem is to find (a, b) such that the trajectory starting from (Y0, E0) reaches (Y1, E1). In terms of (y, e) we know that initially (y0, e0) is a multiple of (Y0, E0), that is to say (y0, e0) = u * (Y0, E0) but we don't know u which should be equal to a/b. We apply Runge Kutta to the second system, starting from u * (Y0, E0) and we run the method until (y, e) is a multiple of (Y1, E1), say (y1, e1) = v * (Y1, E1). This defines a mapping P: u ⟶ v from (0, inf) to (0, inf). Our problem is to find a value u such that P(u) = u. Indeed in this case we have u = a/b and t = b t1 where t is the time computed by the algorithm to reach this position.