How to use method of multiple scale in coupled differential equation?

Question

Lets consider a coupled differential equation-

$\ddot{z} + \omega_{o}^{2}z + 2\beta\dot{z} + \frac{\epsilon}{2m}\theta = 0$

and,

$\ddot{\theta} + \omega_{o}^{2}\theta + 2\gamma\dot{\theta} + \frac{\epsilon}{2I}z = 0$

where, $\omega_{o}^{2}$, $I$, $\beta$, $\gamma$ and, $m$ are constants.

My approach:

Very similar to the undamped version, substituting one of the coordinates into the other, but equations become more messy, like (even when assuming $\gamma=0$) -

$\ddddot{\theta}+2\beta\dddot{\theta}+2\omega_{o}^{2}\ddot{\theta}+2\beta\omega_{o}^{2}\dot{\theta}+(\omega_{o}^{4}-\frac{\epsilon^{2}}{4mI})\theta=0$

Assuming solution, $\theta(t)=Ae^{\alpha t}e^{i\omega t}$, because $\theta$ should decay as there is damping in the system, I get following -

$g(\omega,\alpha)=(\alpha+i\omega)^{4}+2\beta\cdot(\alpha+i\omega)^{3}+2\omega_{o}^{2}\cdot(\alpha+i\omega)^{2}+2\beta\omega_{o}^{2}\cdot(\alpha+i\omega)+(\omega_{o}^{4}-\frac{\epsilon^{2}}{4mI})=0$

this would be satisfied only when,

$Re(g(\omega,\alpha))=0$ & $Im(g(\omega,\alpha))=0$,

I get -

$Re(g(\omega,\alpha))=0\Rightarrow\omega^{4}-(6\alpha^{2}+3\alpha \cdot 2\beta-2\omega_{o}^{2})\cdot\omega^{2}+(\alpha^{4}+2\beta\alpha^{3}-2\omega_{o}^{2}\alpha^{2}+2\beta\omega_{o}^{2}\alpha+\omega_{o}^{4}-\frac{\epsilon^{2}}{4mI})=0$

and,

$Im(g(\omega,\alpha))=0\Rightarrow4\alpha^{3}\omega+6\alpha^{2}\beta\omega-\alpha\cdot(4\omega^{3}+4\omega\omega_{o}^{2})+2\beta\omega_{o}^{2}\omega-2\beta\omega^{3}=0$

Equations are too complicated to solve, although $Re(g(\omega,\alpha))= 0$ can be solved with quadratic formula and I get,

$\omega_{1,2}^{2}=\frac{1}{2}[6\alpha^{2}+6\beta\alpha+2\omega_{o}^{2}\pm\sqrt{32\alpha^{4}+64\alpha^{3}\beta+16\alpha^{3}\omega_{o}^{2}+36\alpha^{2}\beta^{2}+16\alpha \beta\omega_{o}^{2}+\frac{\epsilon^{2}}{4mI}}]$

Now I am stuck, how should I proceed with $Im(g(\omega,\alpha))=0$ to get $\alpha$ in terms of other constants?

Also, how can I apply the method of multiple scales (or Poincare-Lindstedt method) in these equations?

Can $(2\beta\dot{z} + \frac{\epsilon}{2m}\theta)$ and $(2\gamma\dot{\theta} + \frac{\epsilon}{2I}z)$ be considered as perturbations if $\beta$, $\gamma$, $\epsilon$ are small?

Answer 1

$$\begin{cases}\ddot{z} + \omega^{2}z + \frac{\epsilon}{2m}\theta = 0\\ \ddot{\theta} + \omega^{2}\theta + \frac{\epsilon}{2I}z = 0\end{cases}$$ Without approximation (whatever $\epsilon$ is), the system can be solved classically :

$\epsilon\theta =-2m(\ddot{z} + \omega^{2}z)$ that we put into the second equation :

$$-2m(\ddddot{z} + \omega^{2}\ddot z)-2m\omega^2(\ddot{z} + \omega^{2}z) +\frac{\epsilon^2}{2I}z=0$$

$$\ddddot{z} +2\omega^{2}\ddot z +\omega^4z =\frac{\epsilon^2}{4mI}z$$

This is a fourth order linear ODE which can be classically solved.

Nevertheless, if $\epsilon$ is small, you can consider $\frac{\epsilon^2}{4mI}z$ as a perturbation term.

Since it is a single ODE, I suppose that you known how to proceed.

You will obtain the result on the form $\quad z(t)\simeq z_0(t)+\epsilon z_1(t)$.

Then solve $\quad\ddot{\theta} + \omega^{2}\theta + \frac{\epsilon}{2I}z_0 = 0\quad$ on the same manner, in order to obtain $\quad \theta(t)\simeq \theta_0(t)+\epsilon \theta_1(t)$

Answer 2

To put it in more general terms, you have a system $$ p(D)x = ϵay\\ q(D)y = ϵbx $$ where $p,q$ are polynomials with constant real coefficients. Eliminating $y$ you get $$ q(D)p(D)x=ϵ^2abx $$ If $r$ is a simple root of $p(r)=0$ but not a root of $q$, $q(r)\ne 0$, then there is a root $r+ϵ^2h+O(ϵ^4)$ of the combined system close-by where $h=\frac{ab}{p'(r)q(r)}$. The basis solution to that root is $$ x=Ae^{(r+ϵ^2h)t}+O(ϵ^4),\\ y=Be^{(r+ϵ^2h)t}+O(ϵ^4) $$ where $B=ϵA\frac{p'(r)h}{a}=ϵC\frac{b}{q(r)}$. Similar for simple roots of $q$ that are not roots of $p$.

This should cover your situation as long as the friction constants are different. If they are equal, then a simple root $r$ of $p$ is also a simple root of $q$ and the combined polynomial has a root $r+ϵh+O(ϵ^2)$ where $$ p'(r)q'(r)h^2=ab\implies h=\pm\sqrt{\frac{ab}{p'(r)q'(r)}} $$ and the basis solution $$ x=Ae^{(r+ϵh)t}+O(ϵ^2),\\ y=Be^{(r+ϵh)t}+O(ϵ^2) $$ has its coefficients connected via $p'(r)h\,A=a\,B$, $B=A\frac{p'(r)h}a$.