2D solution curves for each of the 2 variables,the second graph can be considered as a pair of implicit functions! A 3 variable system.I've been give the behaviors on the 2 projected planes. How do I arrive at the final 3d curve?

My question is a very simple one.I have a system of 3 variables:- X,Y,Z.I have been provided the solution curves for a couple of the 3 variables with respect to the other(fig.X vs Y,X vs Z). I know that for the X vs Z parabola graph,the system gradually changes as the time changes(like a limit cycle except that it is a open curve) thereby making it possible for the existence of 2 implicit functions at different times.

Is it possible to go from just these 2D solution plots

option (A) to a pair of 2 d.o.f system of autonomous/NON-autonomous differential equations? I'm thinking this would be on the same lines as the Kolmogorov theorem and hence easier to solve,since the theorem makes it clear about the existence of stability points and limit cycles.

$\frac{\mathrm{dX} }{\mathrm{d} t}=\\ \frac{\mathrm{dY} }{\mathrm{d} t}=\\$

$\frac{\mathrm{dX} }{\mathrm{d} t}=\\ \frac{\mathrm{dZ} }{\mathrm{d} t}=$

OR option(B) to a 3 system differential equation(like the Lorenz system)?

$\frac{\mathrm{dX} }{\mathrm{d} t}=\\ \frac{\mathrm{dY} }{\mathrm{d} t}=\\ \frac{\mathrm{dZ} }{\mathrm{d} t}=$

Can I think of these as the projected curves of the 3d phase plot and proceed from there on to maybe arrive at a 3dof system of d.e's(like the Lorenz system)?

  • $\begingroup$ I believe this problem may not have a unique solution. If you consider the curve to be given as X(t), Y(t), Z(t) where t \in (0,1) and a continuous monotonously increasing function f so that f(0) = 0, and f(1) = 1 then X(f(t)), Y(f(t)), Z(f(t)) will as well be a solution to your problem. $\endgroup$ – peterm Mar 6 '13 at 9:43
  • $\begingroup$ I thought so too when the 'time' variable was included. But,I have edited my post.I am afraid to add time t as the 4th variable of the system as it will make things more complicated.So,I keep it simple with just the 3 variables X,Y,Z. $\endgroup$ – Sunny Marella Mar 6 '13 at 10:10

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