ODE on the plane

wofsy

Can someone tell me how to solve the ODE,

d2x/dt2 = -cosx d2y/dt2 = -cosy in the plane?

dirk_mec1

You can use seperation of variables and then integrate twice (with an integration constant!)

wofsy

thanks

HallsofIvy

Those two equations are completely independent- so it is not necessary to "separate variables". You are just solving two separate second order differential equations. And, in fact you just need to integrate each twice.

Mute

Hrm? They don't look independent to me. The derivatives are with respect to t, not x or y. Integrating the x equation would give, for instance,

[tex]\dot{x(t)} - \dot{x(t_0)} = \int_{t_0}^{t}d\tau~\cos y(\tau)[/tex]

which isn't so useful if you can't solve for what y(t) is.

HallsofIvy

If d²x/dt²= cos(x) is a single equation in the dependent variable x as a function of t. There is absolutely no reason to introduce y. Since the independent variable "t" does not appear in the equation, I would use "quadrature":

Let v= dx/dt so that d²x/dt²= dv/dt= (dv/dx)(dx/dt)= v dv/dx. Now you have vdv/dx= cos(x) or vdv= cos(x)dx. Integrating, (1/2)v²= sin(x)+ C. dx/dt= v= [itex]\sqrt{2(sin(x)+ C)}[/itex] or
[tex]\frac{dx}{\sqrt{2(sin(x)+ C)}}= dt[/tex]
That left side is an "elliptical integral".

Of course, y will be exactly the same, though possibly with different constants of integration.

Mute

Ah, I see, I didn't parse the problem the way it was intended to be read. I read it as

[tex]\frac{d^2x}{dt^2} = -\cos x \frac{d^2y}{dt^2} = -\cos y [/tex]

i.e.,

[tex]\frac{d^2x}{dt^2} = -\cos y[/tex]
and
[tex]\cos x \frac{d^2y}{dt^2} = \cos y [/tex]


This is why I always use some sort of punctuation in between separate equations written on the same line. =P

HallsofIvy

And how do you know that was how it was "intended to be read"?

wofsy

I was able to get to the elliptic integral. But I have no idea what it looks like. Further, if x and y are both the same elliptic integral then the orbits in the plane should be fairly simple. But what do they look like?