How to obtain approximation solution for this differential equations?

Question

I am looking for approximation solution for this differential equations \begin{align*} & \ddot{x}=-2\,\Omega\,\cos \left( \phi \right) {\dot z}+2\,\Omega\,\sin \left( \phi \right) {\dot y}\\ &\ddot{y}=-2\,\Omega\,\sin \left( \phi \right) {\dot x}\\ &\ddot{z}=2\,\Omega\,\cos \left( \phi \right) {\dot x}+g \end{align*}

I found this solution:

first transfer the equation with $~\xi=x+i\,y~$

to \begin{align*} & \ddot{\xi}=-2\,i\,\Omega\,\sin\phi\,\dot{\xi}-2\Omega\,\cos(\phi)\dot{z}\\ &\ddot{z}=2\Omega\cos(\phi)\,\dot{x}+g \end{align*}

then

"This coupled system can be easily expanded into a series up to the third order in time t with initial values $~x[0] = y[0] = z[0] = 0~$ and $~\dot{\xi}(0)=D\xi_0,~\dot{z}(0)=Dz_0~$"

the solution is \begin{align*} &\xi(t)=\sum_{i}^{3}\,a_i\,t^i\\ &z(t)=\sum_{i}^{3}\,b_i\,t^i \end{align*} where $~a_i~,b_i~$ are constant.

Question

how we find this solution? and what is the solution for $~x(t)~,y(t)~$?

Answer 1

Some questions: is $g$ constant? Is $\phi$ constant?

Hint: if so, I'd do a derivative of the first equation, and replace $\ddot{y}$ and $\ddot{z}$ from the second and third equation. You get a linear second-order equation in $\dot{x}$ that you can easily solve. Then, go back to the other equations to find $y$, $z$. Eventually, prescribe the initial conditions to the general expressions you have.

Answer 2

Just identify the coefficients of the power series: $$ \dot \xi =\sum (n+1)\xi_{n+1}t^n\\ \ddot \xi =\sum (n+2)(n+1)\xi_{n+2}t^n $$ so you get: $$ (n+2)(n+1)\xi_{n+2} = -2i\Omega\sin\phi(n+1)\xi_{n+1} - 2\Omega\cos\phi(n+1)z_{n+1}\\ (n+2)(n+1)z_{n+2} = 2\Omega\cos\phi(n+1)x_{n+1}-g\delta_{n}\\ $$ This enables you to solve for the coefficients by induction and you can get them by hand for the first terms.

Note that the indices start at $1$ which is why with no loss of generality you assume zero initial positions.

Btw a standard approximation with the Coriolis force is to neglect vertical displacement like for the Foucault pendulum.

Hope this helps.