# Time-delay differential-difference equation

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 $q(3)=2$ ?

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Cross-posted at mathoverflow.net/questions/125710/… –  Joel Reyes Noche Mar 27 '13 at 22:55
Shouldn't there be constraints for $q(t), t \in [-2,0]$ sice the first equation depends on $t-1$ as well as $t+1$? And why not rewrite it by shifting the time variable one unit to make it causal, i.e. not involving future $t$? –  Oberdada Mar 28 '13 at 22:17
My first comment was slightly misguided (and the second a mistake). Try this instead: $\dot q(t)=-1, t\in[-1,0]$ implies $\dot q(t-1)=-1; t\in[0,1]$. Next, rewrite the first equation as $2\dot q(t) - 1 + \dot q(t+1)=k, t\in[0,1]$. Go on with similar substitutions and see where it takes you. –  Oberdada Mar 31 '13 at 0:22
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.