Input-Output Differential Equations for Identifiability

Why Eliminate an Unobserved State?

A dynamical model may describe several biological quantities even when only one is measured. An input-output differential equation expresses the observed output and its derivatives without the hidden states. Its coefficients can expose which parameter combinations the output can distinguish, even when the individual parameters cannot be separated.

This is an algebraic route to structural identifiability: it asks what exact, continuous, noise-free output information could reveal under the stated model assumptions. It does not by itself guarantee that coefficients can be estimated reliably from sparse or noisy measurements. The earlier material on global and local structural identifiability distinguishes these idealized questions from practical estimation.

Definition: An input-output differential equation is a differential relation involving observed outputs, their time derivatives, known inputs if present, and model parameters, with unobserved state variables eliminated. Its coefficients are parameter-dependent quantities. Matching those coefficients to the model parameters can reveal identifiable parameter combinations, provided the output trajectory distinguishes the terms in the equation.

A Nonlinear Two-State Model

Consider the following hypothetical model. The measured quantity is \(y(t)\); the unobserved state is \(x(t)\). The parameters are \(a,b,c,d\), with \(b\ne0\) and \(c\ne0\), as required by the equations. Assume the parameters are constant in time. No input is applied in this example.

\begin{aligned} y' &= b x,\\ x' &= -\frac{y}{b}+d x^2+\frac{ad}{b}x+\frac{d}{b^3c}. \end{aligned}

This is a deliberately compact algebraic example, not a proposed biological mechanism. Its nonlinear term \(d x^2\) makes the elimination more than a linear transfer-function calculation. For the coefficient analysis below, assume an ideal exact output trajectory over an interval. Initial conditions are not being used as extra information to recover parameters; the question is what can be learned from the input-output equation and its coefficients.

Differentiate the measured-state equation \(y'=bx\). Since \(b\) is constant, \(y''=bx'\). Substitute the equation for \(x'\), then replace \(x\) by \(y'/b\), which is valid because \(b\ne0\).

\begin{aligned} y'' &=b\left(-\frac{y}{b}+d x^2+\frac{ad}{b}x+\frac{d}{b^3c}\right)\\ &=-y+bd x^2+ad x+\frac{d}{b^2c}\\ &=-y+\frac{d}{b}(y')^2+\frac{ad}{b}y'+\frac{d}{b^2c}. \end{aligned}

The result contains only the observed output, its derivatives, and parameters. Write its three parameter-dependent coefficients as \(A,B,C\):

y''=-y+A(y')^2+B y'+C,\qquad A=\frac{d}{b},\quad B=\frac{ad}{b},\quad C=\frac{d}{b^2c}.

The coefficient of \(y\) is fixed at \(-1\) in this model; it is not another unknown coefficient to estimate. When the trajectory distinguishes the three features \((y')^2\), \(y'\), and \(1\), the coefficients \(A,B,C\) can be matched to the parameter combinations. If \(d\ne0\), then \(A\ne0\) and \(C\ne0\), so

\frac{B}{A}=a,\qquad \frac{A}{C}=bc,\qquad A=\frac{d}{b}.

Thus the equation can identify \(a\), \(bc\), and \(d/b\), under the stated conditions and provided the terms are distinguishable. It does not generally identify \(b,c,d\) individually. The parameter combination \(d/b\) and product \(bc\) leave multiple possible triples \((b,c,d)\). This is a concrete instance of identifiable combinations rather than individually identifiable parameters.

Worked Examples

Worked Example: Recovering Combinations from Coefficients

Suppose exact input-output analysis of the model gives \(A=0.2500\), \(B=0.5000\), and \(C=0.1250\). Assume \(b\ne0\), \(c\ne0\), \(d\ne0\), and that the trajectory distinguishes the three coefficient terms. Recover the combinations supported by the equation.

From the coefficient definitions,

\frac{d}{b}=A=0.2500,\qquad a=\frac{B}{A}=\frac{0.5000}{0.2500}=2.000,\qquad bc=\frac{A}{C}=\frac{0.2500}{0.1250}=2.000.

The supported conclusions are \(d/b=0.2500\), \(a=2.000\), and \(bc=2.000\). For example, \(b=4\), \(c=0.5\), and \(d=1\) produce these combinations. So do \(b=8\), \(c=0.25\), and \(d=2\): both choices have \(d/b=0.25\) and \(bc=2\). The coefficients therefore do not choose between those parameter triples.

No physical units were assigned to this hypothetical model’s variables or parameters, so the numerical combinations are reported without units. In an applied model, units must be carried through the coefficient definitions and reported with each inferred combination.

Worked Example: The Special Case \(d=0\)

Set \(a=3\), \(b=2\), \(c=0.5\), and \(d=0\). The model’s input-output equation has \(A=B=C=0\), because each of these coefficients contains \(d\). Therefore,

$$ y''=-y. $$

The general solution is \(y(t)=Y_0\cos(t)+V_0\sin(t)\), where \(Y_0\) and \(V_0\) are constants determined by the output’s initial value and slope. The output equation contains no \(a\); changing \(a\) cannot be detected from these coefficients. In fact, the term involving \(a\) in the state equation is multiplied by \(d\), so it vanishes from the state dynamics when \(d=0\) as well.

This case shows why it would be incorrect to say that \(a\) is recoverable merely because \(a\ne0\). Here \(a=3\ne0\), but the coefficient ratio \(B/A\) is \(0/0\), which is undefined, and the equation gives no value for \(a\). Recovery of \(a\) by this ratio requires \(d\ne0\), not \(a\ne0\).

Worked Example: A Trajectory That Hides Coefficients

Use \(a=2\), \(b=4\), \(c=0.5\), and \(d=1\), giving \(A=0.25\), \(B=0.5\), and \(C=0.125\). Consider the model at the equilibrium \(x=0\), \(y=0.125\). The measured output is constant, so \(y'=0\) and \(y''=0\).

Substitution into the input-output equation gives

$$ 0=-0.125+0.25(0)^2+0.5(0)+0.125. $$

The equality holds, but this constant trajectory reveals \(C=0.125\) only. The terms multiplied by \(A\) and \(B\) are both zero throughout the observation interval, so this output cannot determine either coefficient. The parameter values used to generate the trajectory do not change that limitation.

The equilibrium is consistent with the state equations: \(y'=b x=0\), and \(x'=-y/b+d/(b^3c)=-0.125/4+1/(64\cdot0.5)=0\). Thus the example is not an arbitrary substitution into the output equation; it is a trajectory of the model.

When Does Coefficient Matching Work?

Even with exact observations, an equation’s coefficients can be recovered from a particular trajectory only if the corresponding terms can be distinguished along that trajectory. For this example, after forming \(y''+y\), the relevant features are \((y')^2\), \(y'\), and \(1\). In sampled exact derivative data, their columns must be linearly independent to determine all three coefficients by a linear coefficient fit. The equilibrium in the last example fails this condition: its first two feature columns are zero.

This is a trajectory-specific requirement. It is separate from whether the parameter-to-coefficient map allows parameter combinations to be recovered. A useful analysis therefore has two stages: first determine whether the output can distinguish the coefficients; then examine whether the coefficients determine the parameters or combinations of interest. A coefficient ratio is not justified until both stages have been checked.

Eliminate the hidden state.
Differentiate the measured-state equation as needed, substitute the hidden-state dynamics, and use the measured-state equation to remove hidden variables. Record restrictions, such as \(b\ne0\), required for each substitution.

Collect like output terms.
Write the result as an input-output equation and label each parameter-dependent coefficient. Keep fixed coefficients distinct from unknown ones.

Check whether the trajectory distinguishes the terms.
Determine whether the output features multiplying unknown coefficients vary independently over the interval or sampled points. A constant or otherwise special trajectory may remove information.

Invert the coefficient relations conditionally.
Use coefficient ratios or other algebra only where their denominators are nonzero. State the resulting combinations and any remaining parameter ambiguity.

Diagnostics, Alternatives, and Limitations

In exact structural analysis, a warning sign is a trajectory on which one or more coefficient features vanish or become redundant. In numerical work, the same issue appears as an ill-conditioned coefficient fit: small errors in derivative estimates can then cause large changes in the fitted coefficients. These are distinct problems. Exact feature dependence prevents coefficient recovery even without noise; near-dependence can make recovery unstable even when it is theoretically possible.

Differentiating measured data is especially sensitive to noise, and estimating \(y''\) from sparse observations can be unreliable. Smoothing or numerical differentiation can reduce some noise but introduces choices and potential bias. An alternative is to fit the original ODE directly to the observed output, integrating the states numerically rather than estimating derivatives point by point. Direct fitting avoids explicit second-derivative estimates, but it does not remove structural non-identifiability: parameter sets that imply the same ideal output remain indistinguishable from that output alone.

The earlier material on compartmental identifiability through transfer functions treats a related input-output strategy for linear compartmental systems. Transfer functions are suited to linear systems and their input-output relations; the elimination shown here accommodates a nonlinear state term. Neither representation should be interpreted as automatically identifying every model parameter. The input-output coefficients may identify only combinations, and a particular input or trajectory may fail to expose even those coefficients.

An input-output equation is also not a substitute for checking the model assumptions. The derivation above requires constant parameters and \(b\ne0\), and the state equation itself requires \(c\ne0\). If parameters vary over time, differentiating \(y'=bx\) introduces an additional term involving \(b'\). If the measured output is scaled, delayed, or only indirectly related to the state, the output equation must be re-derived for that observation model rather than reused unchanged.

Common Pitfall

Common mistake: “The coefficient ratio \(B/A\) gives \(a\), so \(a\) is identifiable whenever \(a\ne0\).” This reverses the needed condition. Since \(A=d/b\) and \(B=ad/b\), the ratio is valid when \(d\ne0\) (and \(b\ne0\)); then \(B/A=a\), including when \(a=0\). If \(d=0\), both coefficients vanish and the ratio is undefined, regardless of whether \(a\) itself is nonzero. Also check that the trajectory distinguishes the coefficient terms before claiming the coefficients have been recovered.

Why This Matters in QSP

Mechanistic models often contain latent states for quantities such as unobserved target occupancy or intracellular drug amount. Eliminating those states can clarify what a measured response could, even in principle, reveal about model parameters. The resulting combinations can guide reparameterization and prevent an estimation exercise from assigning separate biological interpretations to quantities the output cannot distinguish.

The method has limits: elimination can create complicated equations, obscure state or initial-condition constraints, and produce coefficients that are hard to estimate from real measurements. Its value is a precise algebraic bridge between a state-space model and observable behavior. Treat the inferred combinations as conclusions under the specified model and observation assumptions, not as proof of biological causation.

Check Your Understanding

Answer each question, then compare with the solutions.

For the model in this tutorial, what must be nonzero for the coefficient ratio \(B/A\) to recover \(a\)? Does that condition require \(a\ne0\)?
If \(A=0.40\), \(B=0\), and \(C=0.10\), and the trajectory distinguishes the terms, what values of \(a\), \(d/b\), and \(bc\) follow?
Why can the equilibrium trajectory \(y=0.125\), \(y'=0\) fail to identify \(A\) and \(B\) even when the model parameters used to generate it have \(d\ne0\)?
What is one practical limitation of estimating the input-output equation by differentiating noisy observations?

Solutions. (1) Since \(A=d/b\), require \(d\ne0\) and \(b\ne0\); \(a\) may equal zero, in which case \(B/A=0\). (2) \(a=B/A=0/0.40=0\), \(d/b=A=0.40\), and \(bc=A/C=0.40/0.10=4\). (3) With \(y'=0\), both \((y')^2\) and \(y'\) are zero throughout, so their coefficients make no contribution to the observed equation; the constant output constrains \(C\), not \(A\) or \(B\). (4) Noise is amplified by numerical differentiation, particularly when estimating the second derivative, making fitted coefficients unstable or biased.