Retarded delay differential equation ($y^{\prime }(t)=f(t,y(t),y(t-\tau ))$)

For a single constant delay $\tau >0$, a retarded delay differential equation has the displayed form: delayed values of the unknown function enter, but delayed first derivatives do not. The unknown can also be a vector.^[ref]

Here $t$ denotes time, and the rule applies for $t>0$, with a right derivative at $0$. On $0\le t\le \tau$, the argument $t-\tau$ lies in the prescribed past. Once this interval is known, it supplies the delayed values for $\tau \le t\le 2\tau$. This is the method of steps for a delay differential equation: solve successive ordinary differential equations with a known delayed input, keeping each new endpoint as the next initial condition. For a continuous function $f$ locally Lipschitz continuous in its two state arguments, locally uniformly in $t$, this gives local existence and uniqueness for each continuous history function of a delay differential equation, while the solution of a differential equation remains in the domain of $f$. A nonlinear rule can still produce finite-time blowup.

A history function of a delay differential equation supplies the past interval needed to start a retarded delay differential equation. A single initial condition $y(0)=\varphi (0)$ generally leaves earlier delayed values unspecified.^[ref]

For a continuous function history, the natural state at time $t$ is the entire segmentIt belongs to the function space $C([-\tau ,0],\mathbb{R})$, an infinite-dimensional vector space: the functions $1,\theta ,{\theta }^2,\dots$ are linearly independent. Shifting $y_t$ forward retains the part of the past that the equation can still read. Thus even one scalar unknown can require a state in an infinite-dimensional vector space.

The insufficiency of one value is visible without any approximation. Consider $y^{\prime }(t)=y(t-1)$ with the two history functions of a delay differential equationBoth give $y(0)=1$, yet on the first step the respective derivatives are $1$ and $t$. Direct integration givesIn particular $y_A(1)=2$ while $y_B(1)=3/2$. On the next step,At $t=2$ they are $7/2$ and $8/3$, a gap of $5/6$. These different futures reflect different prescribed histories; each fully specified problem has a unique solution of a differential equation. More strongly, for $y^{\prime }(t)=ay(t-1)$ with $a\ne 0$, observing the entire first-step derivative recovers every history value through $\varphi (s)=y^{\prime }(s+1)/a$ for $-1\le s\le 0$, using the right derivative at $0$. The future can depend on the whole function, rather than a finite list of its samples.

Use the Malthusian delay differential equation with time measured in units of its constant delay: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 givesEach 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 $y^{\prime }=y$, whose derivative would equal its current value.

More generally, for any integer $n\ge 0$,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.^[ref] In this example, $y^{\prime }(0^{-})=0$ and $y^{\prime }(0^{+})=a$. For $a\ne 0$, $y^{\prime }$ is continuous at $t=1$ but $y^{\prime \prime }$ jumps there by $a^2$; at $t=2$, the first jump is in $y^{\prime \prime \prime }$, 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.