# Find bounded solutions of second order ODE system

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.

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.

• I can't integrate the $E$ equation to know $\dot{p}$ because of the norm, so I should do a change of coordinates? Sep 14 '20 at 18:17
• For a given set of initial conditions, $E$ is constant all along the trajectory. For instance, for $p_0 = (x_0, y_0)$ and $\dot p_0 = (v_{x_0}, v_{y_0})$ we have $E_0 = \frac 12(v_{x_0}^2+ v_{y_0}^2)-\frac{k}{\sqrt{x_0^2+y_0^2}}$ Sep 14 '20 at 19:36
• so, going with the energetic argument, for certain values of the energy the trajectory will be bounded and for other it will not. Sep 14 '20 at 19:56
• Please. Have a look at ocw.mit.edu/courses/aeronautics-and-astronautics/… Sep 14 '20 at 20:05