Second order non linear differential equation: Central force question

Question

The problem is as below:

I have derived that the particle satisfies the motion equation $$ \frac{d^2u}{d \theta ^2 } + u = \frac{F(1/u)}{mh^2u^2} $$ by Newton's Law, $u= 1/r$ and $h = r^2 \frac{d \theta }{dt}$ is constant. Hence, by substitution, we obtain from initial conditions that $h=av$, and that from the motion equation, $$ \frac{d^2 u }{ d \theta ^2 } = 6au^2$$ And this as far I could get. I could not really solve this equation - I tried multiplying by $u'$ both sides but nothing useful turns out. Also I could not find a place to use the information on radial velocity. (Probably used for solving the differential equation). Any ideas?

Thank you!

Answer 1

I know this is going to sound dumb, but all I knew how to do was to show that the candidate solution satisfied the initial conditions and the differential equation. initially $r=a$, so $u=\frac1r=\frac1a$ as prescribed. Then $\frac{dr}{dt}=-2v$, $\frac{d\theta}{dt}=\frac va$, so $$\frac{dr}{d\theta}=-2a=-\frac1{u^2}\frac{du}{d\theta}=-r^2\frac{du}{d\theta}=-a^2\frac{du}{d\theta}$$ at $t=0$, and $$\frac{du}{d\theta}=\frac d{d\theta}\frac1{a(1-\theta)^2}=\frac2{a(1-\theta)^3}$$ And this matches the initial condition $\frac{du}{d\theta}=\frac2a$. Then $$\frac{d^2u}{d\theta^2}=\frac6{a(1-\theta)^4}=\frac{6a}{\left[a(1-\theta)^2\right]^2}6au^2$$ So no insights, but the solution just works.

EDIT: Reviewing the problem, I can now see how close you were to solution. You had $$\frac{d^2u}{d\theta^2}\frac{du}{d\theta}=6au^2\frac{du}{d\theta}$$ Then you integrated to get $$\frac12\left(\frac{du}{d\theta}\right)^2=2au^3+c_1$$ And here you were blocked because you didn't see how to apply the initial conditions to get $c_1=\frac12\left(\frac2a\right)^2-2a\left(\frac1a\right)^3=0$. From that point it's easy to wrap up $$\frac{du}{u^{3/2}}=2a^{1/2}\,d\theta$$ $$-2u^{-1/2}=2a^{1/2}\theta+c_2$$ $$-2a^{1/2}=c_2$$ $$-2u^{-1/2}=-2r^{1/2}=2a^{1/2}(\theta-1)$$ $$r=a(1-\theta)^2$$

Answer 2

Taking it from : $$ \frac{d^2 u }{ d \theta ^2 } = 6au^2$$ $u(\theta)$ can be expressed on closed form thanks to the Weiertras's elliptic function, as shown below : $$ 2\frac{d^2 u }{ d \theta ^2 }\frac{d u }{ d \theta } - 12au^2\frac{d u }{ d \theta }=0$$ $$\left(\frac{d u }{ d \theta }\right)^2 - 4au^3+c_1$$ $$\frac{d u }{ d \theta } =\pm \sqrt{4au^3-c_1} \quad\to\quad d\theta=\pm\frac{du}{\sqrt{4au^3+c_1}}$$ $$\theta=\pm\int \frac{du}{\sqrt{4au^3-c_1}}$$ This is an elliptic integral which inverse is a Weierstrass's elliptic function : https://en.wikipedia.org/wiki/Weierstrass%27s_elliptic_functions#Differential_equation $$\pm a^{1/3}\theta= \int \frac{ds}{\sqrt{4s^3-c_1}}+c_2=\Theta+c_2 \qquad\qquad \begin{cases} s=a^{1/3}u \\ \Theta=\pm a^{1/3}\theta-c_2 \end{cases}$$ $$s=\wp\left(\Theta\:;\: 0\:,\: c_1 \right) \quad\to\quad u(\theta)=a^{-1/3}\wp\left(\pm a^{1/3}\theta-c_2\:;\: 0\:,\: c_1 \right)$$ $c_1$ and $c_2$ are constants ; $\wp$ is the Weiertrass's elliptic function.