Simple existence/uniqueness proof for Newton's 2nd?

[Fredrik]

Is there a simple proof of the existence and uniqueness of a solution of x''(t)=f(x'(t),x(t),t), where f is an appropriately "nice" function? (Given an initial condition x(t_0)=x_0,\ x'(t_0)=v_0, or a boundary condition x(t_0)=x_0,\ x(t_1)=x_1).

 

[Office_Shredder]

It's called Picard's theorem. It relates to first order differential equations, but through using vectors can solve a system of them. Any second order differential equation can be turned into a system of first order differential equations by the following (using your differential equation as the starting point)

z_1(t)=x(t) and z_2(t)=x'(t) we get

z_1'(t)=z_2(t)

and

z_2'(t)=f(z_1(t), z_2(t),t)

Here's a wikipedia link to the theorem
http://en.wikipedia.org/wiki/Picard–Lindelöf_theorem

 

[Fredrik]

Thanks. I've seen references to Picard's theorem and I own a book that proves it, but I thought it was more general than that. I was hoping that there would exist a theorem with a particularly simple proof that covers the cases that are interesting for physics, but I guess not.

What theorem do we use when we've been given boundary conditions instead of an initial condition? There's still a unique solution, right?

 

[Office_Shredder]

When you're given boundary conditions it's a lot tougher. I forgot to mention a general technique for solving odes:
http://en.wikipedia.org/wiki/Method_of_variation_of_parameters

But you can have problems. Let's take the easy equation y''+y=0. So y=Acos(x)+Bsin(x). Let's furthermore take the boundary conditions y(0)=0 and y(\pi)=0. Then all we get is that the solution is something of the form y=Asin(x)

If instead we had y(0)=0 and y(\pi)=1 there would be no solution.