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

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

Thank you for sharing ideas and inspiration.

• I did a detailled post on the two-body system here math.stackexchange.com/a/2141482/399263 It is easier to shift $\frac 1r=u+\frac{\mu}{C^2}$ rather than just $\frac 1r=u$ to match better the standard parameters of trajectory.
– zwim
Jan 6, 2021 at 0:54
• Very appreciate your inspiration. Could you please also answer and solve the question under this post. Thank you for the information! Jan 6, 2021 at 0:58

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