How to decouple and solve these coupled differential equations

Question

While solving Euler's equation of motion for an asymmetric top, I cam across the three following equations :

$$\dot{\omega_1}+\omega_2 \omega_3=0$$ $$\dot{\omega_2}-\omega_1 \omega_3=0$$ $$\dot{\omega_3}+\frac{\omega_2 \omega_1}{\mu^2}=0$$

The only information that I have been provided is that at time $t=0$, $\omega=N\mu \hat{i}+N\hat{k}$. So I know $\omega_1, \omega_2 $ and $\omega_3$ at time $0$. How can I decouple and solve the above set of equations ?

I've already been given the solution as follows :

$$\omega_1 =\frac{N\mu}{\cosh(Nt)}$$ $$\omega_2 ={N\mu}\tanh(Nt)$$ $$\omega_3 =\frac{N}{\cosh(Nt)}$$

How do I solve the equations myself ? I've solved linear coupled differential equations using matrices, however this one seems to be nonlinear because of the $\omega_i\omega_j$ terms.

Answer 1

$$\dot{\omega_1}+\omega_2 \omega_3=0$$ $${\omega_1}\dot{\omega_1}+{\omega_1}\omega_2 \omega_3=0 \tag{1} $$ $$\dot{\omega_2}-\omega_1 \omega_3=0$$ $${\omega_2}\dot{\omega_2}-{\omega_2}\omega_1 \omega_3=0 \tag {2}$$ Add both DE and integrate. $${\omega_1}^2+{\omega_2}^2=C_1$$ Third DE gives:

$$\dot{\omega_3}+\frac{\omega_2 \omega_1}{\mu^2}=0$$ $${\omega_3}\dot{\omega_3}+\frac{{\omega_3}\omega_2 \omega_1}{\mu^2}=0$$ $${\omega_3}\dot{\omega_3}-\frac{{\omega_1}\dot{\omega_1}}{\mu^2}=0$$ $${\omega_3}^2-\frac{{\omega_1}^2}{\mu^2}=C_2$$

Answer 2

You can combine the first two equations into one for a complex variable $z=ω_1+iω_2$, $z(0)=μN$ $$ \dot z=iω_3z\implies z(t)=z(0)\exp(i\int_0^tω_3(s)\,ds)=z(0)\bigl(\cos(u(t))+i\sin(u(t))\bigr) $$ Then use, as per comment, $ ω_1\dot ω_1 = μ^2ω_3\dotω_3$, so that with initial conditions $ω_1^2=μ^2ω_3^2$, even more $ω_1=μω_3$. Then use $\dot u=ω_3$ and solve the remaining first-order equation $$ \dot u(t)=N\cos(u(t)),~~u(0)=1. $$

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Of course knowing the initial conditions allows to introduce alternative parametrizations of the circle that are valid around the initial point, such as using $$ (1-v^2)^2+(2v)^2=(1+v^2)^2 $$ to set $$ ω_1=μN\frac{1-v^2}{1+v^2},~~~ω_2=μN\frac{2v}{1+v^2},~~~ω_3=N\frac{1-v^2}{1+v^2} $$ so that $$ -\frac{4v\dot v}{(1+v^2)^2}=-N^2\frac{(1-v^2)(2v)}{(1+v^2)^2} $$