= Exact polynomial steps for a delay equation
{title2=$y'(t)=a y(t-1),\quad \phi=1$}
Use the <Malthusian delay differential equation> with time measured in units of its constant delay:
$$
y'(t)=a y(t-1),\qquad y(t)=1\quad(-1\leq t\leq0).
$$
Here $a$ is a real constant. On the first interval $y(t-1)=1$, hence $y(t)=1+at$. On the next interval the delayed input is $1+a(t-1)$, already known, so the <method of steps for a delay differential equation> gives
$$
y(t)=\begin{cases}
1+at,&0\leq t\leq1,\\
1+at+\dfrac{a^2}{2}(t-1)^2,&1\leq t\leq2,\\
1+at+\dfrac{a^2}{2}(t-1)^2+\dfrac{a^3}{6}(t-2)^3,&2\leq t\leq3.
\end{cases}
$$
Each <polynomial> integrates the preceding interval's delayed <polynomial>, and its integration constant ensures <continuity>. At $a=1$, the step endpoints are $y(1)=2$, $y(2)=7/2$, and $y(3)=37/6$. At $t=3/2$, the value is $21/8$, but its <derivative> is the earlier value $y(1/2)=3/2$. This is the geometric distinction from the <ordinary differential equation> $y'=y$, whose <derivative> would equal its current value.
\Image[/delay-equation-history-and-steps.png]
{title=A prescribed past becomes successive exact polynomial steps}
{description=The <history function of a delay differential equation> is $1$ and $a=1$. Each colored interval supplies the delayed input for the next one; the lower panel gives that next interval's <derivative>. The marked <tangent line> at $t=3/2$ has <slope> $y(1/2)=3/2$. All curves are evaluated from exact <polynomials>.}
More generally, for any integer $n\geq0$,
$$
y(t)=1+\sum_{k=1}^{n+1}\frac{a^k}{k!}(t-k+1)^k,\qquad n\leq t\leq n+1.
$$
At an integer join the newly introduced term vanishes. Inside a step, <differentiation> lowers each power by one and gives exactly $a$ times the preceding step evaluated at $t-1$. Together with $y(0)=1$, this verifies the formula by <mathematical induction>; it is an exact construction, not a truncated <power series>.
The history and the new equation need not prescribe matching <derivatives> at the starting time; their mismatch can reappear later in higher <derivatives>.<Delay-differential equations by Skip Thompson>[Skip Thompson, Delay-differential equations, Derivative Discontinuities: initial incompatibility and its propagation]{ref} In this example, $y'(0^-)=0$ and $y'(0^+)=a$. For $a\ne0$, $y'$ is continuous at $t=1$ but $y''$ jumps there by $a^2$; at $t=2$, the first jump is in $y'''$, of size $a^3$. The <solution of a differential equation> is therefore continuously differentiable for positive time, although it need not have continuous <derivatives> of every order at step boundaries.
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