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Is it possible that the system \begin{equation} \begin{cases} 2\dot{q}(t) + \dot{q}(t-1) + \dot{q}(t+1) = c & \text{if} \hspace{5mm} 0 \le t \le 2 , \dot{q}(t) + \dot{q}(t-1) = k & \text{if} \hspace{5mm} 2 \le t \le 3 \end{cases} \end{equation} has, for suitable constants $c$ and $k$, any $\mathcal{C}^2$ solutions $q:[-1,3] \to \mathbb{R}$ satisfying the conditions $q(t) = - t$ for $t \in [-1, 0]$ and $q(3)=2$ ?

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Cross-posted at math.stackexchange.com/questions/342677/… –  Joel Reyes Noche Mar 27 '13 at 22:56
@filippo, posting the same question simultaneously on two or more sites is frowned upon because it leads to duplication of effort on the part of the answerers. At the very least, you should provide links to the other sites you posted the question on. –  Joel Reyes Noche Mar 27 '13 at 22:57

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