How to solve two-body system equation that find $L$ and $u$ simultaneously make $(Lu^{\prime})^{\prime} + Lu = \frac{1}{L}$ satisified?

Question

The dynamical system of two-body system is written as (the mass and other constants is assumed to be 1) :

$\ddot{r} - r{\dot{\theta}}^{2} = \frac{1}{r^2} \tag{1}$

Consider the two-body system is under the influence of torque $\Gamma$:

$ \frac{dL }{dt} = \frac{d r^2 \dot{\theta}}{dt} = \Gamma \tag{2}$

where the angular momentum is $L = r^2 \dot{\theta}$.

With the substitution $u = \frac{1}{r}$ and following calculation

$\dot{r} = -\frac{\dot{u}}{u^2} = -L\frac{d u}{d\theta}$

$\ddot{r} = -\frac{L^2}{u^2} \frac{d^2u}{d\theta^2} - L\frac{dL}{d\theta}\frac{du}{d\theta}$

$\dot{\theta}^2r = L^2u^3 $

Then Equation.1 turns to be:

$(Lu^\prime)^\prime + Lu = \frac{1}{L} , ^\prime = \frac{d}{d\theta} \tag{3}$

which describes the evolution of two-body system. More information can be found in

https://iopscience.iop.org/article/10.1088/2399-6528/ab9c30,

https://en.wikipedia.org/wiki/Sturm–Liouville_theory

https://en.wikipedia.org/wiki/Two-body_problem.

Thank you for sharing ideas and inspiration.

Answer 1

Assume angular momentum L is a constant that no torque is at present. Then equation.3 turns to be

$Lu^{\prime\prime} + Lu = \frac{1}{L}$

$u^{\prime\prime} + u = \frac{1}{L^2}$

Conic sections solutions are:

$ u = \frac{1}{L^2} + \epsilon cos(\theta+\phi)$

where $\epsilon$ is eccentricity.

Assume angular momentum is not a constant (some torque is present). Further assume $u$ is of the form $u = c_1 e^{\alpha_1 \theta}$ and angular momentum $L$ is of the form $L=c_2 e^{\alpha_2 \theta}$ :

For equation.3, we have

$(c_2 e^{\alpha_2 \theta} (c_1 e^{\alpha_1 \theta})^\prime)^\prime + c_2 e^{\alpha_2 \theta} c_1 e^{\alpha_1 \theta} = \frac{1}{c_2 e^{\alpha_2 \theta}}$

$(\alpha_1 c_1 c_2 e^{(\alpha_1 + \alpha_2) \theta} )^\prime + c_1 c_2 e^{(\alpha_1 +\alpha_2) \theta} = \frac{1}{c_2} e^{-\alpha_2 \theta}$

$\alpha_1 (\alpha_1 + \alpha_2) c_1 c_2 e^{(\alpha_1 + \alpha_2) \theta} + c_1 c_2 e^{(\alpha_1 +\alpha_2) \theta} = \frac{1}{c_2} e^{-\alpha_2 \theta}$

$(\alpha_1 (\alpha_1 + \alpha_2) +1) c_1 c_2 e^{(\alpha_1 + \alpha_2) \theta} = \frac{1}{c_2} e^{-\alpha_2 \theta}$

Then the equation requests

$\alpha_1 = -2 \alpha_2$ and $(1+\frac{\alpha_1^2}{2})c_1 c_2^2 = 1$.