Time-delay differential-difference equation

Question

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

Answer 1

If a $C^2$ solution $q$ exists, the function $r$ defined on $[0,3]$ by $r(t)=\dot q(t)+\dot q(t-1)$ is such that $r(t)+r(t+1)=k$ on $[0,2]$ and $r(t)=c$ on $[2,3]$. Hence $k=r(2)+r(3)=2c$ and $r(t)=c$ on $[0,3]$. Since $r(0)=-2$, $c=-2$. Since $\dot q=-1$ on $[-1,0]$, $\dot q=-1$ on $[-1,3]$. Thus $q(t)=-t$ on $[-1,3]$, in particular $q(3)=-3$ and the condition $q(3)=2$ makes $q$ nonexistent.