How to get this variational equation and this linearization?

Question

Consider the following system: $$ u'=v,~~~~~v'=-cv-f(u)+w,~~~~~w'=-(\epsilon / c)(u-\gamma w). $$

Here, $f(u)=u(u-a)(1-u),~a< 1/2$ and $\varepsilon,\gamma$ are positive.

Here are two results I do not understand how to get them:

(1)

Let $S_{\epsilon}=(u_{\epsilon}, v_{\epsilon}, w_{\epsilon})$ be a solution. Consider the variational equations: $$ \delta u'=\delta v',~~~~~\delta v'=-c\delta v - f'(u_{\epsilon})\delta u+\delta w,~~~~~\delta w'=-(\epsilon /c) (\delta u - \gamma\delta w).~~~ (*) $$

(2)

$(*)$ is well approximated by the system linearized at $U_1=(u_1,v_1,w_1)$ (this is a special point of the phase space) with $\epsilon =0$: $$ \delta u' = \delta v,~~~~~\delta v'=-c\delta v-f'(u_1)\delta u+\delta w,~~~~~\delta w'=0.~~~(**) $$

1. I would like to know how to get $(*)$ and $(**)$.

2. Is variational equation and linearization the same?

Answer 1

So you have a system

$$u'=v,~~~~~v'=-cv-f(u)+w,~~~~~w'=-(\epsilon / c)(u-\gamma w)$$

and some trajectory $(u_{\rm ref} (t), v_{\rm ref} (t), w_{\rm ref} (t) )$ which of course satisfies system of ODEs. You want to study how nearby solutions behave w.r.t. to $(u_{\rm ref} (t), v_{\rm ref} (t), w_{\rm ref} (t) )$. For that you introduce quantitites $\delta u = u_{\rm ref} - u$, $\delta v = v_{\rm ref} - v$ and $\delta w = w_{\rm ref} - w$ -- these are your variations. You can differentiate all these (taking into account that you also have system of ODEs):

$$ \dot{\delta u} = \dot{u_{\rm ref}} - \dot{u} = v_{\rm ref} - v = \delta v, $$

$$ \dot{\delta w} = \dot{w_{\rm ref}} - \dot{w} = - \frac{\varepsilon}{c}\bigl ((u_{\rm ref} - u) - \gamma \cdot (w_{\rm ref} - w) \bigr) = - \frac{\varepsilon}{c}(\delta u - \gamma \cdot \delta w). $$

So far so good, except that the last expression will cause troubles:

$$ \dot{\delta v} = \dot{v_{\rm ref}} - \dot{v} = (-cv_{\rm ref} + w_{\rm ref} - f(u_{\rm ref})) - (-cv + w - f(u)) = -c \cdot {\delta v} + \delta w - \left ( f(u_{\rm ref}) - f(u) \right ). $$

Everything is right and great, but we can't express it in terms of variations only. But we can linearize it and see what happens. We just write $f(u_{\rm ref}) - f(u) = f'(u_{\rm ref}) (u_{\rm ref} - u) + o \left ( \| u_{\rm ref} - u \| \right ) \approx f'(u_{\rm ref}) \cdot {\delta u}$. And this gives us a system of variation equations linearized along $(u_{\rm ref}, v_{\rm ref}, w_{\rm ref})$ . Depending on what solution $(u_{\rm ref} (t), v_{\rm ref} (t), w_{\rm ref} (t) )$ you plug in, you might get a system with constant coefficients (when you plug in equilibrium solution), periodic coefficients (when you plug in periodic solution) or just non-autonomous system of equations.