Subsection \[reduced\_epm\], we have that $$E(x_0,(u_{10}, \ldots,u_{m0}),\left(\overline{t}_1,i_1\right), \ldots, \left(\overline{t}_N,i_N\right), t_f) = x_f.$$ Or, considering that the final time $t_f$, the initial point $x_0$, the components $(i_1, \ldots,i_N)$ and the initial values $(u_{10}, \ldots,u_{m0})$ are fixed, $$E(\overline{t}_1, \ldots, \overline{t}_N) = x_f.$$ Let $\delta x_f$ be some perturbation of the final point $x_f$. We look for a vector $\delta \mathcal{T} = (\delta t_1, \ldots, \delta t_N)$ so that the system reaches the target point $x_f + \delta x_f$: $$E(\overline{t}_1 + \delta t_1, \ldots, \overline{t}_N + \delta t_N) = x_f + \delta x_f.$$ As we have shown in Proposition \[prop\_differentiability\_epm\] the differentiability of the reduced end-point mapping, we can write $$E(\overline{t}_1 + \delta t_1, \ldots, \overline{t}_N + \delta t_N) = E(\overline{t}_1, \ldots,\overline{t}_N) + dE(\overline{t}_1, \ldots,\overline{t}_N) \cdot \delta \mathcal{T} + o(\| \delta \mathcal{T}\|).$$ At order one, the solution is given by the solution of the linear equation $$dE(\overline{t}_1, \ldots,\overline{t}_N) \cdot \delta \mathcal{T} = \delta x_f.$$ It is natural to target the final point $x_f +\delta x_f$ while shifting the switching times as little as possible. That is, we look for the solution of minimal euclidian norm of the previous equation, which is given by $\delta \mathcal{T} = dE(\overline{t}_1, \ldots, \overline{t}_N)^{\dagger} \cdot \delta x_f$.

Therefore, 