How to get this variational equation and this linearization? 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 $\epsilon =0$:
    $$
\delta u' = \delta v,~~~~~\delta v'=-c\delta v-f'(u_1)\delta u+\delta w,~~~~~\delta w'=0.~~~(**)
$$




*

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

*Is variational equation and linearization the same?
 A: 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.
