Consider a control system of the form $\dot x(t) = f(x(t),u(t))$ where $x$ takes his values in $\mathbb{R}^n$ and $u$ in $\mathbb{R}^m$. Here is what I read:
Let $x^*$ be a reference curve associated to the control $u^*$. Take a $L^1$-perturbation of $u^*$ defined by $u_\varepsilon(t) := v \in U$ on $[s,s+\varepsilon[$, where $s \in [0,T[$ is a Lebesgue time of the function $t \mapsto f(x^*(t),u^*(t))$, and $u_\varepsilon(t) := u^*(t) $ elsewhere. The corresponding variation vector satisfies the linear equation $$ \dot w(t) = \frac{\partial f}{\partial x}(x^*(t),u^*(t)) \ w(t), \quad \forall a.e. t \in [s,T] , \qquad \qquad (1) $$ and the initial condition $$w(s)=f(x^*(s),v)-f(x^*(s),u^*(s)).$$
how do they get the equation $(1)$ ? If I understand $w=x_\varepsilon - x$, hence using a Taylor expansion for each $t$ at $(x^*,u^*)$, we have
$$ \dot w = \frac{\partial f}{\partial x}^\intercal(x^*,u^*) \ w + \frac{\partial f}{\partial u}^\intercal(x^*,u^*)\ (u_\varepsilon-u^*) $$ why they do not have the term in $u$ ?
Define $$w(t) = \lim_{\varepsilon \to 0}1/\varepsilon\ (x(t,u_\varepsilon) - x(t,u^*))= \frac{\partial x}{\partial u}(v-u^*)+... \qquad (1)$$. where $u_\varepsilon=u^* + \varepsilon (v-u^*)$ We have $$ x(t,u_\varepsilon) = x_0 + \int_0^t f(x(s,u_\varepsilon(s)),u_\varepsilon)ds $$ and $$ f(x(s,u_\varepsilon(s)),u_\varepsilon) = f(x(s,u^*),u^*) + \varepsilon \frac{\partial f}{\partial x}(x(s,u^*),u^*)\ \frac{\partial x}{\partial u}(v-u^*) + \varepsilon \frac{\partial f}{\partial u}(x(s,u^*),u^*)(v-u^*) + o(\varepsilon^2) $$ hence $$ x(t,u_\varepsilon) = x(t,u^*) + \varepsilon\int_0^t (\frac{\partial f}{\partial x}(x^*,u^*)\ \frac{\partial x}{\partial u}(v-u^*) + \frac{\partial f}{\partial u}(x^*,u^*)(v-u^*))ds +... $$
and finally, using (1), $$ \dot w = \frac{\partial f}{\partial x}(x^*,u^*)\ w + \varepsilon\frac{\partial f}{\partial u}(x^*,u^*)(v-u^*). $$
With the same spirit we can get (1) using the notion of Lebesgue time.