Deriving First-Order Transit-Chain Equations

From a Cascade Diagram to Differential Equations

A cascade represents a signal passing through successive stages before it appears at an output. To use that structure in a PK/PD model, each arrow must become a term in an equation: what enters a stage, what leaves it, and how quickly does the amount or signal in that stage change? Writing those terms explicitly makes assumptions about input placement and delay visible rather than hiding them in a diagram.

Here, a stage is represented by a state variable, and transfer out of each stage is first order: at any time, the outgoing rate is proportional to the current state. The equations below describe a linear, unbranched chain. They do not specify how a drug generates the external input; that input is supplied by the surrounding model.

Definition: A first-order transit chain with \(n\) stages consists of states \(x_1(t),\ldots,x_n(t)\), positive rate constants \(k_1,\ldots,k_n\), and an external input flux \(q_0(t)\) entering stage 1. Each \(x_i\) is expressed in a common signal or amount unit, each \(k_i\) has units of inverse time, and \(q_0\) has units of state per time. For \(i=1,\ldots,n\), stage \(i\) loses material at rate \(k_i x_i(t)\); for \(i>1\), it receives the outflow from stage \(i-1\).

The input is a flux, not necessarily a signal level. For example, if an upstream input signal \(u(t)\), measured in the same units as the states, feeds stage 1 through a first-order transfer with rate \(k_0\), the corresponding flux is \(q_0(t)=k_0u(t)\). Its units are signal per time. Keeping this distinction clear prevents adding quantities with incompatible units.

Deriving the Coupled Equations

For each stage, use the balance principle: rate of change equals total inflow minus total outflow. Stage 1 receives the external flux \(q_0(t)\) and loses \(k_1x_1(t)\). Every later stage receives the preceding stage’s outflow and loses its own outflow. Thus the equations are

\begin{aligned} \frac{dx_1}{dt} &= q_0(t)-k_1x_1(t),\\ \frac{dx_i}{dt} &= k_{i-1}x_{i-1}(t)-k_ix_i(t), \qquad i=2,\ldots,n. \end{aligned}

The coupling term in stage \(i\) is \(k_{i-1}x_{i-1}\), not \(k_ix_{i-1}\): it is the rate at which the preceding stage loses signal. The rate constant \(k_i\) instead determines how quickly signal leaves stage \(i\). This distinction matters when the rates differ.

An initial-value model also specifies every state at a reference time, conventionally \(t=0\):

x_i(0)=x_{i,0},\qquad i=1,\ldots,n.

The initial values need not be zero. Setting all \(x_{i,0}=0\) describes an initially empty chain; nonzero values describe signal already present at the start of the modeled interval. Initial conditions are part of the model, not a consequence of the differential equations.

These equations assume constant, positive rate constants; linear outflow proportional to the state; transfer only to the next stage; and no feedback or additional production or loss within the stages. The external input may vary with time. With common state units, every term in each equation has units of state per time. The equations also require that the modeled states and input be defined over the time interval of interest.

Worked Examples

Worked Example: Writing a Three-Stage Chain

Suppose three signal states \(x_1,x_2,x_3\) are measured in arbitrary signal units. Their outflow rate constants are \(k_1=0.50\ \text{min}^{-1}\), \(k_2=0.25\ \text{min}^{-1}\), and \(k_3=0.20\ \text{min}^{-1}\). A constant external flux \(q_0=0.60\) signal units/min enters stage 1, and the initial states are \(x_1(0)=0.40\), \(x_2(0)=0.20\), and \(x_3(0)=0.10\) signal units. Derive the equations and calculate the initial derivatives.

Applying inflow minus outflow to each stage gives

\begin{aligned} \frac{dx_1}{dt} &= 0.60-0.50x_1,\\ \frac{dx_2}{dt} &= 0.50x_1-0.25x_2,\\ \frac{dx_3}{dt} &= 0.25x_2-0.20x_3. \end{aligned}

At \(t=0\), substitute the stated initial values:

\begin{aligned} \frac{dx_1}{dt}(0)&=0.60-(0.50)(0.40)=0.40,\\ \frac{dx_2}{dt}(0)&=(0.50)(0.40)-(0.25)(0.20)=0.15,\\ \frac{dx_3}{dt}(0)&=(0.25)(0.20)-(0.20)(0.10)=0.030. \end{aligned}

The initial rates of change are therefore \(0.40\), \(0.15\), and \(0.030\) signal units/min, respectively. Each is positive because the specified inflow exceeds the outflow at that instant. All terms have the required units: for example, \(k_2x_2\) has units \(\text{min}^{-1}\) times signal units, or signal units/min.

Worked Example: Checking a Constant-Input Equilibrium

Consider a three-stage chain with a constant input flux \(q_0=0.80\) signal units/min and rates \(k_1=0.40\ \text{min}^{-1}\), \(k_2=0.20\ \text{min}^{-1}\), and \(k_3=0.10\ \text{min}^{-1}\). Find the state values required for equilibrium, where all derivatives are zero, and check the result in the equations.

At equilibrium, stage 1’s inflow must equal its outflow:

0=q_0-k_1x_1 \quad\Longrightarrow\quad x_1=\frac{q_0}{k_1} =\frac{0.80\ \text{signal units/min}}{0.40\ \text{min}^{-1}} =2.00\ \text{signal units}.

For stage 2, set its derivative to zero and use \(x_1=2.00\):

0=k_1x_1-k_2x_2 \quad\Longrightarrow\quad x_2=\frac{(0.40)(2.00)}{0.20} =4.00\ \text{signal units}.

For stage 3:

0=k_2x_2-k_3x_3 \quad\Longrightarrow\quad x_3=\frac{(0.20)(4.00)}{0.10} =8.00\ \text{signal units}.

Substitution verifies the balance: \(0.80-(0.40)(2.00)=0\), \((0.40)(2.00)-(0.20)(4.00)=0\), and \((0.20)(4.00)-(0.10)(8.00)=0\), all in signal units/min. At equilibrium the same flux passes through every stage, while the state values differ because the rates differ. This calculation checks the equations; it does not claim that equilibrium has been reached from any particular initial condition.

Worked Example: Placing the External Input at Stage 2

Suppose a flux \(q_2(t)\) enters stage 2 rather than stage 1 in a three-stage chain. Let \(k_1=0.50\ \text{min}^{-1}\), \(k_2=0.25\ \text{min}^{-1}\), and \(k_3=0.20\ \text{min}^{-1}\). Initially all states are zero, and the input flux at \(t=0\) is \(0.30\) signal units/min. Write the equations and find the initial derivatives.

No external input enters stage 1, so it has only its outflow. Stage 2 receives both the outflow from stage 1 and the external flux. Stage 3 receives the outflow from stage 2:

\begin{aligned} \frac{dx_1}{dt} &= -0.50x_1,\\ \frac{dx_2}{dt} &= 0.50x_1+q_2(t)-0.25x_2,\\ \frac{dx_3}{dt} &= 0.25x_2-0.20x_3. \end{aligned}

Since \(x_1(0)=x_2(0)=x_3(0)=0\) and \(q_2(0)=0.30\), direct substitution gives

\begin{aligned} \frac{dx_1}{dt}(0)&=-(0.50)(0)=0,\\ \frac{dx_2}{dt}(0)&=(0.50)(0)+0.30-(0.25)(0)=0.30,\\ \frac{dx_3}{dt}(0)&=(0.25)(0)-(0.20)(0)=0. \end{aligned}

Only stage 2 has a nonzero initial derivative because it alone receives the input at that instant. The input’s location is encoded in the equation, not inferred from the rate constants. If the model instead specifies an input level \(u_2(t)\) that transfers into stage 2 at rate \(k_{\mathrm{in}}\), the flux term would be \(k_{\mathrm{in}}u_2(t)\); the units and meaning of that term must be stated.

Reading, Checking, and Using the Model

For a chain with input at stage 1, the equations can be written compactly as a linear system. This form makes the direction of coupling explicit:

\frac{d}{dt} \begin{bmatrix} x_1\\x_2\\ \vdots\\x_n \end{bmatrix} = \begin{bmatrix} -k_1&0&\cdots&0\\ k_1&-k_2&\cdots&0\\ 0&k_2&\ddots&0\\ \vdots&\vdots&\ddots&-k_n \end{bmatrix} \begin{bmatrix} x_1\\x_2\\ \vdots\\x_n \end{bmatrix} + \begin{bmatrix} q_0(t)\\0\\ \vdots\\0 \end{bmatrix}.

The matrix is lower triangular because stage \(i\) depends on its own state and the preceding stage, but not on a later stage. If the reported output is the terminal state, define \(y(t)=x_n(t)\). If the reported quantity is the flux leaving the chain, define \(y(t)=k_nx_n(t)\). These are different outputs, with different units in the second case; the choice should match the modeled or measured quantity.

Several checks help detect errors before fitting or simulating a model. First, inspect units: every term in a state equation must be state per time. Second, check the signs: a stage’s own outflow must reduce its state, while an upstream outflow contributes positively to the next stage. Third, evaluate the derivatives at specified initial values, as in the examples. Finally, for a constant input, verify any proposed equilibrium by setting every derivative to zero and checking each balance separately.

With nonnegative input and nonnegative initial states, this model cannot generate negative states. At a boundary where \(x_1=0\), its derivative is \(q_0(t)\), which is nonnegative. At a boundary where \(x_i=0\) for \(i>1\), its derivative is \(k_{i-1}x_{i-1}\), also nonnegative. Thus a computed negative state under these conditions suggests an implementation, numerical, or model-specification problem. Numerical integration can also obscure the expected direction of change if time steps are too large.

The chain is a particular alternative to a single first-order stage and to a model that imposes a fixed time delay. A single stage has only one state balance; a chain represents multiple sequential states. A fixed-delay model represents a different assumption: an input is shifted in time rather than passed through modeled intermediate states. These structures are not interchangeable without considering what the data and scientific model support.

Common Pitfall and Limitations

Wrong reasoning: “The rate constant for stage 2 belongs on the transfer from stage 1 to stage 2, so that term is \(k_2x_1\).” Correction: the transfer is stage 1’s outflow, \(k_1x_1\); stage 2’s own outflow is \(k_2x_2\). Every rate constant labels the outflow from its own state. Checking the units alone will not catch this error, because both expressions can have units of state per time; the flow-balance definition does.

A linear transit chain is limited by its assumptions. It does not represent saturation, state-dependent rates, feedback, branching, or a direct effect of the input on a later stage unless those features are explicitly added. Its states are mathematical representations of sequential signal, and the equations alone do not establish that each state corresponds to a distinct biological mechanism. A good fit cannot by itself validate that interpretation.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

For a two-stage chain with input flux \(q_0(t)\), write the equations and state the units of each rate constant.
For stage 3, why is the inflow term \(k_2x_2\), rather than \(k_3x_2\)?
A constant input flux is \(0.60\) signal units/min and \(k_1=0.30\ \text{min}^{-1}\). What value of \(x_1\) makes the first-stage derivative zero?
If a flux enters stage 2 and every state is initially zero, which state can have a nonzero derivative at the initial instant? Explain.
What is the difference in meaning and units between choosing \(y=x_n\) and \(y=k_nx_n\) as the output?

Solutions. (1) The equations are \(\frac{dx_1}{dt}=q_0(t)-k_1x_1\) and \(\frac{dx_2}{dt}=k_1x_1-k_2x_2\). Both \(k_1\) and \(k_2\) have units of inverse time. (2) Stage 3 receives stage 2’s outflow, which is defined by stage 2’s rate constant \(k_2\). (3) Set \(0=0.60-(0.30)x_1\); hence \(x_1=0.60/0.30=2.00\) signal units. (4) The input-receiving state, stage 2, can have a nonzero derivative; with zero states, its derivative equals the incoming flux. The other stages have zero initial inflow and outflow. (5) \(x_n\) is a terminal state in signal units, whereas \(k_nx_n\) is the terminal outflow flux in signal units per time.