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I need to solve the following system:

$$\left\{ \begin{array} { l l } { (u')^2 + (v')^2 = 1 } \\ {u'v'' - u''v' = uu' + vv' } \end{array} \right.$$

and it's a task that's proven to be quite hard by analythical methods. I had asked for help plotting the solutions to a similar system before, but I haven't been able to adapt the method to this case. My ultimate goal is to plot the curves given by:

$\alpha(s) = (\cos(u(s)), \sin(u(s)), v(s))$

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Does this help?

Clear[u, v];
{u, v} = {u[t], v[t]} /. NDSolve[{
  u'[t]^2 + v'[t]^2 == 1, 
  u'[t] v''[t] - u''[t] v'[t] == u[t] u'[t] + v[t] v'[t], 
  u'[0] == 1, v'[0] == -1}, {u[t], v[t]}, {t, 0, 64}][[1]];
ParametricPlot3D[{Cos[u], Sin[u], v}, {t, 0, 64}]

enter image description here

That does display some warnings which should be carefully considered.

If this is not what you are looking for then perhaps you could explain more about why you are not able to use the solution and what you need a solution to do.

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  • $\begingroup$ That helps a lot, actually! Thank you. I'm a beginner at this, so some trivial stuff like this still evades me. $\endgroup$ – Matheus Andrade Aug 22 '18 at 0:48
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    $\begingroup$ Wonderful. There is a lot of detail inside that for someone new to understand. Look at the initial conditions you need and any other changes needed. If you can't get those changes to work then let us know and someone might be able to help. You can also position the mouse cursor, press the left mouse button and drag to reorient the graphic inside MMA. Looking at this "from the end" does appear that the solution is on the surface of a cylinder. Asking whether those warnings are serious might be a good thing to do $\endgroup$ – Bill Aug 22 '18 at 1:08
  • $\begingroup$ (+!) @Bill, your initial conditions violate the first equation which is the cause for one of the error messages. Something like the following should work better: \[Phi] = Pi/3; {u, v} = {u[t], v[t]} /. NDSolve[{ u'[t]^2 + v'[t]^2 == 1, u'[t] v''[t] - u''[t] v'[t] == u[t] u'[t] + v[t] v'[t], u'[0] == Cos[\[Phi]], v'[0] == Sin[\[Phi]]}, {u[t], v[t]}, {t, 0, 64}][[1]]; $\endgroup$ – Henrik Schumacher Aug 22 '18 at 7:41

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