&= \overline{x}(t) + d\tilde{E}(t,\overline{\mathcal{T}})\cdot \delta \mathcal{T} + o(\| \delta \mathcal{T}\|).\end{aligned}$$ So, at order one, the problem writes $$d\tilde{E}(t,\overline{\mathcal{T}})\cdot \delta \mathcal{T} = \delta x(t).
\label{eq_perturbations_2}$$ It is well known (see [@allaire2002algebre] for instance), that the solution of minimal norm of this equation is $\delta \mathcal{T} = d\tilde{E}(t,\overline{\mathcal{T}})^{\dagger}\cdot \delta x(t)$. Besides, let $\sigma_{max}(t) > \cdots > \sigma_{min}(t) > 0$ denote the positive singular values of $d\tilde{E}(t,\overline{\mathcal{T}})$. We have that $\left\| d\tilde{E}(t,\overline{\mathcal{T}})^{\dagger} \right\|_2 = {1}/{\sigma_{min}(t)}$ ($\left\| \cdot \right\|_2$ for a matrix denotes the induced norm corresponding to the euclidean norm), so that $$\begin{aligned}
\left\| \delta \mathcal{T} \right\|_2&= \left\| d\tilde{E}(t,\overline{\mathcal{T}})^{\dagger}\cdot \delta x (t)\right\|_2  \\
                    & {\leqslant}\left\| d\tilde{E}(t,\overline{\mathcal{T}})^{\dagger} \right\|_2 \cdot \left\| \delta x (t)\right\|_2 \\
                    & {\leqslant}\frac{\left\| \delta x (t)\right\|_2}{\sigma_{min}},\end{aligned}$$ which concludes the proof.

We have the relation that, for all vector of switching times $\mathcal{T}$ $$E(\tilde{E}(t,\mathcal{T}),\mathcal{T},t_f-t) = x_f.$$ Differentiating this equality with respect to $\mathcal{T}$, we have that, for all $\delta \mathcal{T}$ $$\frac{\partial E}{\partial x_0}(\tilde{E}(t,\mathcal{T}),\mathcal{T},t_f-t)\cdot d\tilde{E}(t,\overline{T})\cdot \delta \mathcal{T} + \frac{\partial E}{\partial \mathcal{T}}(\tilde{E}(t,\mathcal{T}),\mathcal{T},t_f-t)\cdot \delta \mathcal{T} = 0.$$ Replacing the second term by its value in , it follows that $$\frac{\partial E}{\partial x_0}(\tilde{E}(t,\mathcal{T}),\mathcal{T},t_f-t)\cdot d\tilde{E}(t,\mathcal{T})\cdot \delta \mathcal{T} = \frac{\partial E}{\partial x_0}(\tilde{E}(t,\mathcal{T}),\mathcal{T},t_f-t)\cdot \delta x(t).$$

It is easy to show that $\partial E/\partial x_0$ can be expressed as the 