History Functions and Consistent Initial Conditions

Why a Delayed Model Needs a Past

An ordinary differential equation determines the next change in a state from its value at the current time. A delay differential equation (DDE) can also depend on earlier values. Consequently, specifying only the state at the start of simulation is not generally enough: the model needs the relevant part of the trajectory before that time. In a QSP model, this past may represent receptor occupancy before the observation window, a cell population entering a delayed process, or a concentration that drives a downstream response after a lag.

The earlier material on specifying a delay differential equation establishes how delayed terms enter the model. Here the key issue is what values those terms require at the initial time, and when the chosen past is compatible with the state and smoothness of the model.

Definition: Consider a retarded DDE with fixed positive delays \(\tau_1,\ldots,\tau_m\):

\frac{d y(t)}{dt} = f\!\left(t,y(t),y(t-\tau_1),\ldots,y(t-\tau_m),\theta\right), \qquad t\geq t_0.

Here \(y(t)\) is a vector of model states, \(f\) is the rate function, and \(\theta\) denotes parameters. Let \(\tau_{\max}=\max_j \tau_j\). A history function is a specified function \(\phi(t)\) giving the state over the interval \([t_0-\tau_{\max},t_0]\). It must provide values for every delayed state argument the model uses. The initial state is the endpoint value \(y(t_0)=\phi(t_0)\), not a replacement for the entire function.

For a model with several states, the history is vector-valued and supplies the past of each state that appears in a delayed term. The history need not supply more time than the largest delay for the fixed-delay formulation above. A delayed external input also needs past values if the model evaluates that input before \(t_0\).

Continuity and Compatibility at the Initial Time

For the usual retarded DDE with a continuous rate function, a standard starting point is a continuous history \(\phi\) on the required interval. Its endpoint must agree with the initial state:

y(t_0)=\phi(t_0).

This endpoint match is the basic continuity condition. The DDE then specifies the derivative immediately after the start, using the current state and the delayed values from the history. If the history is differentiable from the left at \(t_0\), a stronger condition is needed for the combined trajectory to have a continuous first derivative there.

Definition: First-derivative compatibility at \(t_0\) holds when the derivative of the history approaching from the left equals the derivative required by the DDE immediately to the right:

\phi'(t_0^-) = f\!\left(t_0,\phi(t_0),\phi(t_0-\tau_1),\ldots, \phi(t_0-\tau_m),\theta\right).

This condition is not required merely to define a continuous-history solution. If it fails, the state can still be continuous, and the DDE still determines its right-hand evolution, but the derivative has a jump at \(t_0\). Whether that loss of smoothness is a problem depends on the intended model and analysis. If a continuously differentiable trajectory is required, the condition must hold. Higher smoothness requires corresponding higher-order compatibility conditions, provided the history and rate function are sufficiently smooth.

Why the Derivative Condition Has This Form

The compatibility equation follows directly from matching the left and right derivatives. On the history interval, the trajectory is defined by \(y(t)=\phi(t)\), so its left derivative at the initial time is \(\phi'(t_0^-)\). On the simulated interval, the DDE gives the right derivative. At \(t_0\), the current value is \(\phi(t_0)\), and every delayed value is \(\phi(t_0-\tau_j)\), because those times lie in the history interval. Therefore the right derivative equals the right-hand side of the compatibility equation. Equality of these two one-sided derivatives is exactly what it means for the derivative to be continuous at \(t_0\).

This reasoning also shows why a model can have a correct endpoint state but an incompatible derivative: the endpoint value does not determine the values of the history at \(t_0-\tau_j\), and those values can change the rate at the start.

Worked Example: Checking a Constant History

Suppose a concentration \(y(t)\), measured in nM, follows \(y'(t)=-a y(t)+b y(t-\tau)\), with \(a=0.30\ \mathrm{h}^{-1}\), \(b=0.20\ \mathrm{h}^{-1}\), and \(\tau=2\ \mathrm{h}\). The hypothetical pre-simulation history is constant at \(10\ \mathrm{nM}\) over \([t_0-2\ \mathrm{h},t_0]\).

The endpoint matches the initial state: \(\phi(t_0)=10\ \mathrm{nM}\). The left derivative of a constant history is zero. The DDE's right derivative is

y'(t_0^+) =-(0.30\ \mathrm{h}^{-1})(10\ \mathrm{nM}) +(0.20\ \mathrm{h}^{-1})(10\ \mathrm{nM}) =-3.0+2.0 =-1.0\ \mathrm{nM\,h}^{-1}.

Thus, \(\phi'(t_0^-)=0\) but \(y'(t_0^+)=-1.0\ \mathrm{nM\,h}^{-1}\). The state is continuous at \(t_0\), but the history is not first-derivative compatible. The simulated concentration begins decreasing at \(1.0\ \mathrm{nM\,h}^{-1}\), despite the flat pre-simulation history.

Worked Example: Same Initial State, Different Initial Rates

Let a hypothetical state satisfy \(y'(t)=k y(t-\tau)\), where \(k=0.40\ \mathrm{h}^{-1}\), \(\tau=2\ \mathrm{h}\), and both candidate histories end at \(y(t_0)=5\ \mathrm{nM}\). Compare \(\phi_1(t)=5\ \mathrm{nM}\) with \(\phi_2(t)=5\ \mathrm{nM}+(2\ \mathrm{nM\,h}^{-1})(t-t_0)\). Both histories are continuous and have the same endpoint.

For the constant history, \(y(t_0-\tau)=5\ \mathrm{nM}\), so

y_1'(t_0^+) =(0.40\ \mathrm{h}^{-1})(5\ \mathrm{nM}) =2.0\ \mathrm{nM\,h}^{-1}.

For the second history, \(\phi_2(t_0-2\ \mathrm{h})=5-4=1\ \mathrm{nM}\), so

y_2'(t_0^+) =(0.40\ \mathrm{h}^{-1})(1\ \mathrm{nM}) =0.40\ \mathrm{nM\,h}^{-1}.

The same current state therefore produces different initial rates because the delayed value differs. The second history has left derivative \(2\ \mathrm{nM\,h}^{-1}\), which also differs from its DDE-implied right derivative \(0.40\ \mathrm{nM\,h}^{-1}\). It is continuous but not first-derivative compatible.

Equilibrium Histories and Other Choices

A constant history is convenient, but it should not be chosen solely because it is easy to enter. For an autonomous DDE, a constant history at \(y^*\) is an equilibrium history if the model's rate is zero when all current and delayed states equal \(y^*\). In that case, the state remains constant under the DDE, and both the history derivative and the model-implied derivative are zero. A state that is an equilibrium of the corresponding no-delay model is not necessarily an equilibrium of the DDE, or of the history actually specified.

Worked Example: Testing an Equilibrium History

Consider the hypothetical model \(y'(t)=q-k y(t-\tau)\), where \(y\) is in nM, \(q=2.0\ \mathrm{nM\,h}^{-1}\), \(k=0.25\ \mathrm{h}^{-1}\), and \(\tau=3\ \mathrm{h}\). Set \(y(t_0)=8\ \mathrm{nM}\).

First use the constant history \(\phi(t)=8\ \mathrm{nM}\). The delayed value is \(8\ \mathrm{nM}\), so

y'(t_0^+)=2.0-(0.25)(8)=0\ \mathrm{nM\,h}^{-1}.

The history's left derivative is also zero. Thus the constant history is first-derivative compatible, and the rate remains zero if the same constant history continues to supply every delayed value. This is an equilibrium of this DDE.

Now keep the endpoint at \(8\ \mathrm{nM}\), but use the linear history \(\phi(t)=8\ \mathrm{nM}+(4/3\ \mathrm{nM\,h}^{-1})(t-t_0)\). At \(t_0-3\ \mathrm{h}\), its value is \(4\ \mathrm{nM}\). The initial rate is therefore

y'(t_0^+)=2.0-(0.25)(4)=1.0\ \mathrm{nM\,h}^{-1}.

The left derivative is \(4/3\ \mathrm{nM\,h}^{-1}\), not \(1.0\ \mathrm{nM\,h}^{-1}\). This history has the same initial state as the equilibrium history, but it is neither an equilibrium history nor first-derivative compatible.

The appropriate history depends on what the simulation represents. Common choices include:

Measured or reconstructed prehistory: use observations or a scientifically justified reconstruction when the pre-treatment trajectory is important.
Constant baseline history: use a constant vector of baseline values when the system is assumed to have remained at that baseline over the full delay window. Check that this assumption is plausible and, for an equilibrium, that the DDE rate is zero there.
Equilibrium history: use a constant equilibrium only when the delayed model's equilibrium condition is satisfied, not merely because the initial state appears stable in an ODE version.
Specified transient history: use a smooth or piecewise-smooth trajectory when the biology implies changing pre-simulation states. Ensure it covers the full required interval and matches the intended state at \(t_0\).

Diagnostics, Alternatives, and Limitations

A history should be treated as part of the model specification, not as an implementation detail. Before simulation, check that the history is defined on the entire required interval, has the right units and state ordering, and gives the intended endpoint. For each delayed term, evaluate the history at \(t_0-\tau_j\) and calculate the DDE-implied initial derivative. If smoothness matters, compare it with the left derivative of the history.

A solver warning near \(t_0\), an unexpected initial slope, or a sharp kink in a simulated trajectory can indicate a history mismatch. A kink is not automatically a numerical failure: it may be the correct consequence of a continuous history whose derivative is incompatible. Conversely, changing solver tolerances cannot correct a scientifically inappropriate history. Tests should distinguish a deliberate derivative jump from an accidental endpoint mismatch or a wrong delayed-state index.

A single point initial condition is simpler, but it cannot determine the past values needed by a DDE. A constant history is a stronger assumption than a point value: it asserts that each state was unchanged throughout the delay window. A measured or reconstructed history can be more informative, but may be uncertain or sparsely observed. These choices affect early predictions, and the data available may not identify the pre-simulation trajectory uniquely. When the history is uncertain, assess whether plausible alternatives materially change predictions rather than presenting one arbitrary history as known.

The conditions described here apply to fixed-delay retarded models with continuous histories. Models with algebraic constraints, state-dependent delays, impulsive events, or prescribed jumps can require additional conditions and specialized numerical treatment. In particular, do not assume that a discontinuous history has the same regularity properties as a continuous one. The model formulation and solver must support the history class being used.

Common pitfall: “The initial state is correct, so the initial condition is complete.” This is wrong for a DDE because the rate at \(t_0\) can depend on values such as \(y(t_0-\tau)\). Specify the full history required by the delays, check its endpoint against the initial state, and check derivative compatibility when a smooth join is required.

Find the required interval.
List the fixed delays and provide history values back to \(t_0-\tau_{\max}\).

Check the endpoint.
Verify that \(\phi(t_0)\) equals the specified initial state for every relevant model state.

Evaluate delayed values and rates.
Substitute \(\phi(t_0-\tau_j)\) into the DDE to obtain the right derivative at \(t_0\).

Check smoothness and meaning.
Compare one-sided derivatives if needed, then confirm that the history represents a defensible biological prehistory.

Check Your Understanding

For each question, first identify the relevant history value or compatibility condition. Solutions are included below.

A DDE has delays of \(1\) h and \(4\) h and begins at \(t_0\). How far back must the history be defined?
For \(y'(t)=0.5y(t-2)\), a constant history is \(6\ \mathrm{nM}\). What is the right derivative at \(t_0\), and is the history first-derivative compatible?
Explain why specifying \(y(t_0)=3\ \mathrm{nM}\) alone cannot determine the initial derivative for \(y'(t)=k y(t-\tau)\).
For \(y'(t)=q-ky(t-\tau)\), state the condition for a constant history \(y(t)=y^*\) to be an equilibrium history.

Solutions. (1) It must cover \([t_0-4\ \mathrm{h},t_0]\), since \(4\) h is the maximum delay. (2) The DDE gives \(y'(t_0^+)=(0.5\ \mathrm{h}^{-1})(6\ \mathrm{nM})=3\ \mathrm{nM\,h}^{-1}\). The constant history has left derivative zero, so it is not first-derivative compatible. (3) The derivative is \(k y(t_0-\tau)\); the point value at \(t_0\) does not specify \(y(t_0-\tau)\). (4) The constant history is an equilibrium when \(q-ky^*=0\), or \(q=ky^*\). Its left derivative and DDE-implied right derivative are then both zero.