Nonlinear System of ODES

Question

I have been trying to solve this nonlinear system of ODEs analytically with no luck:

$$\frac{d\psi}{dt}=-cot \theta cos \psi$$

$$\frac{d\theta}{dt}=-sin\psi$$

$$\frac{d\phi}{dt}=\frac{cos\psi}{sin\theta}$$

Where of course $\psi$, $\theta$ and $\phi$ are functions of $t$.

Any hints would be appreciated!

Answer 1

$$\frac{d\psi}{dt}=-\cot (\theta) cos (\psi)$$

$$\frac{d\theta}{dt}=-\sin(\psi)$$

$$\frac{d\phi}{dt}=\frac{cos(\psi)}{sin(\theta)}$$

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Hint: notice how there are the same arguments in the first and third equation.

$$\frac{d\psi}{dt}=-\dfrac{cos(\theta)}{\sin(\theta)} \cos (\psi)$$

Let's solve the 3rd equation to $\sin(\theta)$

$$\sin(\theta)=\dfrac{dt}{d\phi}\cos(\psi)$$

And plug it in the 1st one:

$$\frac{d\psi}{dt}=-\dfrac{cos(\theta)}{\dfrac{dt}{d\phi}\cos(\psi)} \cos (\psi)$$

Then $\cos(\psi)$ falls out as long as it doesn't equal zero.

Edit: I made a mistake as i accidentally wrote $d\psi/dt$ instead of $d\phi$ which is fixed now. Further simplification is required, later i'll try to make it simpler.

$$\frac{d\theta}{dt}=-\sin(\psi)$$

$$\frac{d\psi}{dt}=-\dfrac{cos(\theta)}{\dfrac{dt}{d\phi}}$$

Don't forget that $y'\cdot y'\neq y''$ but $(y')^2$

Answer 2

Assuming that $\psi$ can be written in some interval as a function of $\theta$, one can write $$ \frac{d\psi}{d\theta}=\frac{d\psi}{dt}\Big/\frac{d\theta}{dt}=\cot \theta \cot \psi. $$ This gives $$ \frac{d}{d\theta}\log\cos\psi=-\frac1{\cot\psi}\frac{d\psi}{d\theta}=-\cot\theta $$ and so $$ \log\cos\psi=-\log\sin\theta+C\quad\Leftrightarrow\quad \cos\psi=\frac1{\sin\theta}e^C. $$ Now you can potentially integrate explicitly all the equations. For example, the first equation becomes $$ \frac{d\psi}{dt}=-\cot \theta \cos \psi=-e^{-C}\sqrt{1-e^C/\cos^2\psi}\cos^2\psi. $$