Differential equation resembling to cycloid

[tom-73]

What is the function corresponding to this ODE:
http://home.arcor.de/luag/math/dgl.jpg

In complex notation it obviously shows up like this:

a * z''(t) + b * |z'(t)| * z'(t) + c = 0;

The numerical solution shows a graph resembling to a cycloid.

Thanks for any help!
Tom

 

[haruspex]

What is the function corresponding to this ODE:
http://home.arcor.de/luag/math/dgl.jpg

In complex notation it obviously shows up like this:

a * z''(t) + b * |z'(t)| * z'(t) + c = 0;

It does? I don't see how.
If you divide through by the surd and subtract the 1st eqn from the second, I believe you get something integrable.

 

[tom-73]

Thank you for your comment. I tried to divide and subtract. The problem is the term in the middle: (y' - eps*x') vs. (x' + eps*y')
It makes the situation even worse - I did not succeed in finding a simplified pattern.

The complex notation in my first post has been derived by simply multiplying the 2nd equation by i and adding the result to the first equation.

What I investigated in the meanwhile:

The cycloid ODE in complex notation should be

a * z''(t) + b * z'(t) + c = 0;

The only difference is the multiplication with |z'| in the middle which in fact produces a value near 1 for curtate cycloids with r1 << r0 (the point tracing out the curve is inside the circle, which rolls on a line AND it is close to the center).

The ODEs in my first posts describe a phugoid, a more general form of the cycloid I suppose.

It seems that the phugoid has no analytic solution. Any suggestions?

Tom

 

[haruspex]

Sorry, I overlooked what happens to the RHS. My original suggestion was nonsense.

The complex notation in my first post has been derived by simply multiplying the 2nd equation by i and adding the result to the first equation.

Ah yes, I see it now. Sorry for the noise.