Method of Steps for Solving DDEs

Why a Delay Can Be Solved One Interval at a Time

A delay differential equation (DDE) uses past state values to determine the rate of change now. For a fixed delay $\tau>0$, the value $x(t-\tau)$ is one delay-length behind $x(t)$. This structure gives a useful way to solve the equation: on the first interval, the delayed value comes from the specified history; on the next interval, it comes from the solution just obtained; and the process continues.

This procedure is called the method of steps. Its central idea is not to approximate the past by the current state. Rather, on each interval of length $\tau$, the delayed state is already known from the history or an earlier solution interval. The remaining equation is an ordinary differential equation (ODE) for the current state.

Definition: Consider a scalar retarded DDE with one constant delay, $$ x'(t)=f\bigl(t,x(t),x(t-\tau)\bigr),\qquad t\geq 0, $$ and a prescribed history $x(t)=\phi(t)$ for $-\tau\leq t\leq 0$. The method of steps solves the DDE successively on $[0,\tau]$, $[\tau,2\tau]$, and so on, using the history or previously computed solution to supply the delayed argument on each interval.

For a well-defined stepwise solution, assume that $\tau$ is fixed and positive, the history is specified over $[-\tau,0]$, and the current-state ODE has a unique solution on each interval under consideration. A standard sufficient condition is that $f$ is continuous in its arguments and locally Lipschitz in the current state $x(t)$. The delayed state may appear nonlinearly in $f$; on a given step it is a known function of time. These assumptions support a retarded equation: the derivative of a delayed state is not itself required to define $x'(t)$.

The state must match the history at the initial point: $x(0)=\phi(0)$. Each later interval starts from the state at the previous interval's endpoint. A history need not have a derivative matching the DDE's right-hand derivative at zero, so the derivative may change there. At positive step boundaries, continuity of the history and solution ensures continuity of the delayed and current states.

Deriving the Stepwise ODEs

On the first interval $0\leq t\leq\tau$, the delayed argument satisfies $t-\tau\in[-\tau,0]$. Therefore $x(t-\tau)=\phi(t-\tau)$, and the first problem is an ODE:

x'(t)=f\bigl(t,x(t),\phi(t-\tau)\bigr),\qquad x(0)=\phi(0),\qquad 0\leq t\leq\tau.

For an interval beginning at $k\tau$, where $k\geq1$, write $t=k\tau+s$ with $0\leq s\leq\tau$. Its delayed argument is $t-\tau=(k-1)\tau+s$, which lies in the preceding interval. Once that preceding solution is known, denote it by $x_{\mathrm{known}}((k-1)\tau+s)$. The new ODE is

\frac{d}{ds}y_k(s) =f\bigl(k\tau+s,y_k(s),x_{\mathrm{known}}((k-1)\tau+s)\bigr), \qquad y_k(0)=x(k\tau).

Here $y_k(s)=x(k\tau+s)$, and its initial value is the endpoint value from the solution already constructed. This follows directly by substituting $t=k\tau+s$ into the DDE. Since $s\in[0,\tau]$, the delayed argument never requires a value from the interval currently being solved. Solving this ODE therefore extends the DDE solution through the next delay-length interval.

If the history is continuous and each interval ODE has a unique solution, this construction produces a continuous solution as far as the interval solutions exist. At each positive step boundary $k\tau$, the current and delayed states are continuous; since $f$ is continuous, the right-hand side and hence the derivative are continuous there. The formula used to represent the delayed input may change, but the formulas agree at the boundary.

Worked Example: A Delayed State as the ODE Input

Use dimensionless time $u$, measured in units of the delay, and solve $dx/du=x(u-1)$ with history $x(u)=1$ for $u\leq0$. Find the solution through $u=2.5$.

First interval, $0\leq u\leq1$. The delayed value is the history $x(u-1)=1$. Thus $x'=1$, $x(0)=1$, and

x(u)=1+u,\qquad 0\leq u\leq1.

In particular, $x(1)=2$.

Second interval, $u=1+s$, $0\leq s\leq1$. Now $x(u-1)=x(s)=1+s$, known from the first interval. Hence $dx(1+s)/ds=1+s$, with $x(1)=2$, so

x(1+s)=2+\int_0^s(1+v)\,dv =2+s+\frac{s^2}{2}.

This gives $x(2)=2+1+\frac12=3.5$.

Third interval, $u=2+s$, $0\leq s\leq0.5$. The delayed value is the second-interval solution $x(1+s)=2+s+\frac{s^2}{2}$. Integrating from $x(2)=3.5$ gives

x(2+s)=3.5+\int_0^s\left(2+v+\frac{v^2}{2}\right)\,dv =3.5+2s+\frac{s^2}{2}+\frac{s^3}{6}.

At $u=2.5$, $s=0.5$, so $x(2.5)=3.5+1+0.125+0.0208333=4.6458333$, or approximately $4.646$ in dimensionless state units. The value used in the third interval is the solution from the second interval, not the third-interval state itself.

Steps with an Instantaneous State Term

The delayed input need not be the only input to the ODE. For example, the current state can also appear linearly. The method remains the same: substitute the known delayed solution, then solve an ODE for the current state.

Worked Example: Solving a Linear DDE Across Two Intervals

In dimensionless time, solve $x'(t)=-x(t)+2x(t-1)$ with constant history $x(t)=1$ for $t\leq0$. Calculate $x(2)$.

First interval, $0\leq t\leq1$. The history gives $x(t-1)=1$, so $x'+x=2$, $x(0)=1$. The solution is

x(t)=2-e^{-t}.

Thus $x(1)=2-e^{-1}$.

Second interval, $t=1+s$, $0\leq s\leq1$. The delayed value is $x(s)=2-e^{-s}$, so $dx(1+s)/ds=-x(1+s)+4-2e^{-s}$, with $x(1)=2-e^{-1}$. Multiplying by the integrating factor $e^s$ and integrating from $0$ to $s$ gives

e^s x(1+s)=2-e^{-1}+\int_0^s(4e^v-2)\,dv =2-e^{-1}+4(e^s-1)-2s.

Therefore

x(1+s)=4+e^{-s}\bigl(-2-e^{-1}-2s\bigr).

At $s=1$, the endpoint value is $x(2)=4+e^{-1}(-4-e^{-1})=4-4e^{-1}-e^{-2}\approx2.393$. This is the value at the end of the second step. It agrees with the initial value for any subsequent step, which must be carried forward without resetting the state.

Nonlinear Equations and the Stepwise Logic

A closed-form solution is not required for the method to apply. On each interval, the delayed state is known, but the resulting ODE in the current state may be nonlinear. It can be solved analytically when possible or numerically as an ODE, with the endpoint then passed to the next step.

Worked Example: A Nonlinear DDE with an Exact Step Solution

In dimensionless time, solve $x'(t)=-x(t)x(t-1)$ with $x(t)=1$ for $t\leq0$, through $t=2$.

First interval. Since $x(t-1)=1$, the equation is $x'=-x$, $x(0)=1$. Thus $x(t)=e^{-t}$ on $[0,1]$, and $x(1)=e^{-1}$.

Second interval. Set $t=1+s$, $0\leq s\leq1$. The delayed state is the known value $x(s)=e^{-s}$, so the current state solves $dx(1+s)/ds=-e^{-s}x(1+s)$. Separating variables and using $x(1)=e^{-1}$ yields

\ln x(1+s)-\ln(e^{-1}) =-\int_0^s e^{-v}\,dv =-(1-e^{-s}), \qquad x(1+s)=e^{-2+e^{-s}}.

Consequently $x(2)=e^{-2+e^{-1}}\approx0.1956$. The delayed factor was prescribed by the first-step solution; the current state was still solved as an ODE on the second interval.

Practical Checks, Alternatives, and Limitations

When applying the method by hand or implementing it, check each interval in order. The following sequence makes the dependency structure explicit.

Fix the delay and history.
Record the constant $\tau$, the history on $[-\tau,0]$, and the matching value $x(0)=\phi(0)$.

Form the delayed input.
On each interval, express $x(t-\tau)$ using only the history or the solution from the preceding interval.

Solve the current-state ODE.
Use the already known endpoint as the initial value, and keep the time variable and shifted argument consistent.

Check the endpoint and continue.
Evaluate the solution at the interval boundary, verify it matches the next interval's initial value, and use it to supply the next delayed input.

A common mistake is to replace $x(t-\tau)$ by $x(t)$, or to use the unknown current-interval solution as though it were already known. That changes the equation or introduces an unjustified dependence. The correction is to identify the range of $t-\tau$ on the interval being solved: it lies entirely in the history on the first step, and entirely in the preceding step thereafter.

Other useful checks include confirming that each step begins at the previous endpoint, substituting the proposed solution back into that interval's ODE, and evaluating endpoint arithmetic from the interval formula rather than from a remembered or rounded intermediate value. A mismatch in $x(k\tau)$, an off-by-one delay shift, or an incorrect exponential integrating factor can propagate through every later interval.

The method of steps is a construction and a way to organize the problem, not a claim that every interval has a convenient formula. A direct DDE solver instead handles the history and delayed evaluations numerically, using interpolation or stored solution values as needed. That can be more practical for complicated models, but the stepwise derivation remains useful for understanding which past values drive each interval and for checking a numerical implementation. Solving each interval with an ODE solver is also possible; it still requires correct delayed-state values and careful transfer of endpoint states.

The basic construction assumes a single fixed delay. Time-varying delays, multiple delays, delays that depend on the state, and equations involving delayed derivatives require additional treatment and are not covered by the recurrence derived here. Even with a fixed delay, a stepwise solution can lose validity if an interval ODE ceases to have a finite solution or the assumed regularity conditions fail. The method also does not, by itself, determine long-term stability; that is a separate question from constructing the solution.

In applications, the main value is structural: a fixed delay partitions the time axis into intervals on which the delayed input is already determined. This makes exact solutions, numerical implementations, and model checks easier to reason about. In a QSP model, it also makes clear which earlier state value represents a delayed biological process, rather than treating the delay as an unexplained change to the current rate.

Check Your Understanding

Answer each question, then compare with the solutions below.

For $x'(t)=f(t,x(t),x(t-2))$, what information supplies the delayed input on $0\leq t\leq2$, and on $2\leq t\leq4$?
In dimensionless time, solve $x'(t)=x(t-1)$ on the first interval for history $x(t)=2$ and find $x(1)$.
Under the stated continuity assumptions, what can be concluded about the derivative at a positive step boundary?
For $x'(t)=-x(t)+x(t-1)$ on an interval $[1,2]$, which function should be substituted for the delayed state?
Name one assumption of the basic method of steps that would need reconsideration for a more general DDE.

Solutions. (1) On $[0,2]$, use the prescribed history $x(t-2)=\phi(t-2)$. On $[2,4]$, use the solution already obtained on $[0,2]$. (2) The delayed value is $2$, so $x'=2$, $x(0)=2$, and $x(t)=2+2t$ on $[0,1]$; hence $x(1)=4$. (3) At a positive step boundary, the current and delayed states are continuous; since $f$ is continuous, the derivative is continuous there as well. (4) Substitute the known first-interval solution $x(t-1)$, not the unknown current-interval state. (5) Examples include a single constant delay, a retarded equation without delayed derivatives, or the existence of a unique solution to each interval ODE.