# How to decouple and solve these coupled differential equations

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.

• Not sure if it helps, but we have $\omega_1 \omega_2 \omega_3 = -\omega_1 \dot \omega_1 = \omega_2 \dot \omega_2 = -\mu^2 \omega_3 \dot \omega_3$ Dec 9, 2022 at 13:59

$$\dot{\omega_1}+\omega_2 \omega_3=0$$ $${\omega_1}\dot{\omega_1}+{\omega_1}\omega_2 \omega_3=0 \tag{1}$$ $$\dot{\omega_2}-\omega_1 \omega_3=0$$ $${\omega_2}\dot{\omega_2}-{\omega_2}\omega_1 \omega_3=0 \tag {2}$$ Add both DE and integrate. $${\omega_1}^2+{\omega_2}^2=C_1$$ Third DE gives:
$$\dot{\omega_3}+\frac{\omega_2 \omega_1}{\mu^2}=0$$ $${\omega_3}\dot{\omega_3}+\frac{{\omega_3}\omega_2 \omega_1}{\mu^2}=0$$ $${\omega_3}\dot{\omega_3}-\frac{{\omega_1}\dot{\omega_1}}{\mu^2}=0$$ $${\omega_3}^2-\frac{{\omega_1}^2}{\mu^2}=C_2$$
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 $$\dot u(t)=N\cos(u(t)),~~u(0)=1.$$
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 $$-\frac{4v\dot v}{(1+v^2)^2}=-N^2\frac{(1-v^2)(2v)}{(1+v^2)^2}$$