Resulting from a Classical Mechanics problem, I have the following system of second order ODEs
$$
\begin{align}
\ddot{x} &= \frac{-kx}{(x^2+y^2)^{3/2}} \\
\ddot{y} &= \frac{-ky}{(x^2+y^2)^{3/2}}
\end{align}
$$
With $k>0$ a real constant. I want to find the initial conditions for which the solutions of the system are bounded, and the initial conditions for which the system passes through the origin.
I know the origin is an unstable equilibrium point of the system, but I have no idea how to continue from there. I guess converting into polar coordinates would help with it, but I have only seen examples of converting systems of first order equations, not second order.
Best Answer
Hint.
Calling $p = (x(t),y(t))$ and also
$\cases{ T = \frac 12 ||\dot p||^2\\ U = -\frac{k}{||p||}} $
we have the lagrangian
$$ L = T - U = \frac 12 ||\dot p||^2+\frac{k}{||p||} $$
with the movement equations
$$ \cases{ \ddot x(t) = -\frac{k x(t)}{\left(x(t)^2+y(t)^2\right)^{3/2}}\\ \ddot y(t) = -\frac{k y(t)}{\left(x(t)^2+y(t)^2\right)^{3/2}}} $$
The total energy is given by
$$ E = T + U = \frac 12 ||\dot p||^2-\frac{k}{||p||} $$
NOTE
Making a change of coordinates
$$ \cases{ x(t) = r(t)\cos(\theta(t))\\ y(t) = r(t)\sin(\theta(t)) } $$
on the former lagrangian we arrive at
$$ L = \frac{1}{2} \left(r'(t)^2+r(t)^2 \theta '(t)^2-\frac{2 k}{r(t)}\right) $$
and now deriving the movement equations we arrive at
$$ \left\{ \begin{array}{c} r''(t) = r(t) \theta '(t)^2-\frac{k}{r(t)^2} \\ 2 r(t)r'(t) \theta '(t)+r(t)^2 \theta ''(t)= (r(t)^2\theta'(t))' = 0 \\ \end{array} \right. $$
so we have
$$ r(t)^2\theta'(t) = h_0 $$
and substituting into the first movement equation we get
$$ r''(t) = \frac{h_0^2}{r(t)^3}-\frac{k}{r(t)^2} $$
etc.