The original system in (x,c,q):

The transformed system in (u,v,r):

The Jacobian of the transformed system (u,v,r):

Clearly J13=0 and J33=0.

The eigenvalues of (u,v,r):

The Phase Diagrams of the 2-D (u,v) system:

The phase diagram at this scale reveals little.

Zooming on the critical point: u=0, v=0 and placing the nullclines:

Zooming still further:

The 2-D approximation above assumes that r^3*u is small.

My understanding of this diagram is: the critical point (u=0, v=0) may be approached for u>0 and v>0, but not for u<0 and v<0. In terms of the original system, it means that both x and c must be rising and cannot be falling. Is this the correct interpretation?

The 3-D transformed system (u,v,r) exhibits 1 negative eigenvalue for 2 positive eigenvalues at the critical point (u=0,v=0,r=b), which points to a one-dimensional stable manifold onto which the critical point may be approached, for a given r(0)=0 and free u(0) and v(0). The simulation of the 3-D system does indeed yield a converging path, for a very small error tolerance, confirming the insights garnered from the 2-D approximation.

I simulated u(t), v(t), r(t), and then transformed back to x(t), c(t), q(t):

The following is the plot of x(t):

The following is the plot of c(t):

The following is the plot of q(t):

It is clear from the system that the only way, if any, to approach the critical point x=xs, c=cs, q=b is for q<b and that as the system is approaching the critical point, it is most likely that q will rise. The above shows q falling a little initially.

It seems that the critical/stationary point may be approached on a one-dimensional manifold, but only for r>rs and u>0 and v>0 or -- in terms of the original system, only for x>xs and both x and c rising during the transition. For the physical interpretation of my system, that's a little odd, so that's why I'm posting here to seek reassurance about my interpretation.

The system is tricky since the stationary/critical point is not actually defined as v->0 (i.e. xdot->0).

Any pointers will be greatly appreciated. Thanks.

I just noticed that (in Maple 12) you can simply right-click on the animation and "Export As" a gif. That's a lot quicker than exporting as html and retrieving the gif from the folder (a method about which I read somewhere in mapleprimes, possibly about older versions of Maple). The gifs have the same size, so they are surely the same, whether exported directly as gif or via the html conversion. I should have noticed that earlier.

Thanks a lot to both of you, great update.

I did File --> Export As --> HTML and retrieved the gif from Paulina's example:

> animate(plot3d, [sin(x+t)*cos(y-t), x=0..2*Pi, y=0..2*Pi], t=0..Pi, frames=50);

The animation from within Maple is great. I can right-click, select "continuous" to loop the animation and select "slower" or "faster" to adjust the speed of the animation.

But the gif produced with the Maple --> HTML converter produces a gif that cycles through the frames at the highest speed by default. That's the problem I had.

I ought to have been clearer: the problem I had was not with "the number of frames contained within the gif" but with "the speed at which the gif displays the frames, however many there may be". In other words, I was looking for something like:

> animate(plot3d, [sin(x+t)*cos(y-t), x=0..2*Pi, y=0..2*Pi], t=0..Pi, frames=50, speed=1);

where "speed=1" indicates that each frame is to be displayed for 1 second. Just an example.

As remarked in one of the past messages in mapleprimes, if there were a free "MapleViewer.exe" that could fit on a USB flashdisk, there may be less of a need for creating gifs. The gifs are good to show in presentations or post on websites.