# Solving a simple delayed differential equation

I have a delayed differential equation of the form $$x'(t)=\beta-f\left(x\left(\alpha(t)\right)\right)$$ where $\alpha(t)=t_{j}$ for $t\in[t_{j},t_{j+1})$ for $j=0,1,...$, with $t_{0}=0$ and $t_{j+1}>t_{j}$ and $f$ is continuous. So one can write $x'(t)$ as \begin{eqnarray*} x'(t) & = & \beta-f\left(x\left(0\right)\right),\quad t\in[0,t_{1}),\\ x'(t) & = & \beta-f\left(x\left(t_1\right)\right),\quad t\in[t_{1},t_{2}),\\ x'(t) & = & \beta-f\left(x\left(t_2\right)\right),\quad t\in[t_{2},t_{3}),\\ & \vdots \end{eqnarray*} Then, given the continuity of $f$, starting from the first interval one can sequentially solve for $x$. That is given the initial condition $x(0)$ one can solve the ODE on the first interval, use it together with continuity to obtain $x(t_1)$ as the initial condition of the second equation and so on.

Question: Is this argument correct? If yes, is it also true that the existence and uniqueness of the solution follows from those of the solutions for all intervals?

• At each interval, x(t) is actually a straight line with different slopes. So it is piecewises lines. So solution does exist and it is unique because of the initial value. – user115350 Jun 2 '16 at 2:19
• Thanks, but will they also be continuous or something strange could happen at the boundaries? Intuitively, this shouldn't happen and the piecewise lines should basically be attached together but I'm looking for a rigours argument. – Submartingale Jun 2 '16 at 13:49