Differential-Algebra Identifiability Workshop

Why Eliminate a State?

In a mechanistic model, some states may be difficult or impossible to measure directly. Structural identifiability asks whether ideal, noise-free observations of the measured output over time can determine model parameters. Differential-algebra methods address this by eliminating unobserved states and deriving a differential equation involving only the output, its derivatives, known inputs, and parameters. Matching that equation’s coefficients to an observed output equation can reveal which parameter combinations are identifiable.

The calculation is exact only under explicit assumptions. The model structure and observation definition must be correct; time is measured on a known scale; parameters are constant; and the output is observed continuously and without error for the structural question being considered. In practice, derivatives estimated from noisy, sparse data can be unreliable. This tutorial focuses on the structural calculation, not on fitting noisy data.

A Nonlinear Two-State Pathway

Consider a pathway in which an unobserved state \(x(t)\) decays by a first-order process, while producing an observed state \(y(t)\) through a quadratic rate:

\begin{aligned} \frac{dx}{dt} &= -a x,\\ \frac{dy}{dt} &= b x^2-cy,\\ Y(t)&=y(t). \end{aligned}

Here \(Y(t)\) denotes the measured output, which equals the state \(y(t)\). The parameters satisfy \(a>0\), \(b>0\), and \(c>0\). If \(x\) and \(y\) are concentrations, then \(a\) and \(c\) have units of inverse time, while \(b\) has units of inverse concentration per unit time. There is no external input in this example. The initial states may be known or unknown; that distinction matters when interpreting what the output can identify.

Definition: An input-output equation is a differential equation involving measured outputs, their time derivatives, known inputs, and model parameters, but not unobserved state variables. Coefficient matching means equating the coefficients in this equation to coefficients determined from an ideal output trajectory, then solving the resulting parameter equations.

Eliminating the Hidden State

The first equation gives a convenient relation for the derivative of the hidden contribution \(b x^2\). Differentiate the output equation and substitute the state equation:

\begin{aligned} y'' &= 2bxx'-cy'\\ &= -2abx^2-cy'. \end{aligned}

The original output equation also gives \(b x^2=y'+cy\). Substituting this expression removes \(x\):

y''=-2a(y'+cy)-cy', \qquad y''+(2a+c)y'+2ac\,y=0.

Since \(Y=y\), the input-output equation is

Y''+A Y'+B Y=0, \qquad A=2a+c,\quad B=2ac.

The equation contains no \(b\). That absence is informative: the differential equation alone cannot recover \(b\). It does not, by itself, prove that no other information in an experiment could constrain \(b\); initial-state knowledge and the output’s initial slope can add information.

What Coefficient Matching Can Establish

Suppose the ideal output trajectory allows the two coefficients \(A\) and \(B\) to be determined uniquely from \(Y''+A Y'+B Y=0\). Then the model parameters must satisfy \(A=2a+c\) and \(B=2ac\). The coefficients identify these combinations, but they do not generally identify \(a\) and \(c\) separately and uniquely.

To see the ambiguity, form a quadratic whose roots are \(c\) and \(2a\):

\lambda^2-A\lambda+B=0 \quad\Longleftrightarrow\quad (\lambda-c)(\lambda-2a)=0.

Thus the output coefficients determine the unordered pair \(\{c,2a\}\). If its two roots are distinct, assigning one root to \(c\) and the other to \(2a\) gives two parameter pairs. Both pairs have the same \(A\) and \(B\), so both satisfy the same input-output equation. This is a finite ambiguity, not a continuous family of solutions.

Coefficient recovery itself also needs adequate output information. For example, if \(Y'\) and \(Y\) are proportional throughout the observed interval, the equation may not uniquely determine both coefficients from that trajectory. A coefficient-matching conclusion is conditional on being able to determine the coefficients from the ideal output; it is not a substitute for checking that condition.

Worked Example: Recovering the Coefficients and Candidate Parameters

Suppose an ideal output trajectory, measured in hours, determines the input-output coefficients \(A=0.90\ \mathrm{h}^{-1}\) and \(B=0.080\ \mathrm{h}^{-2}\). Find the candidate values of \(a\) and \(c\).

The roots of \(\lambda^2-A\lambda+B=0\) are

\lambda =\frac{0.90\pm\sqrt{(0.90)^2-4(0.080)}}{2} =\frac{0.90\pm\sqrt{0.49}}{2} =\frac{0.90\pm0.70}{2}.

The roots are \(0.80\ \mathrm{h}^{-1}\) and \(0.10\ \mathrm{h}^{-1}\). One assignment is \(c=0.80\ \mathrm{h}^{-1}\), \(2a=0.10\ \mathrm{h}^{-1}\), giving \(a=0.050\ \mathrm{h}^{-1}\). The swapped assignment is \(c=0.10\ \mathrm{h}^{-1}\), \(2a=0.80\ \mathrm{h}^{-1}\), giving \(a=0.40\ \mathrm{h}^{-1}\).

Both pairs reproduce the matched coefficients: in either case \(2a+c=0.90\ \mathrm{h}^{-1}\) and \(2ac=0.080\ \mathrm{h}^{-2}\). Coefficient matching alone therefore does not select a unique pair.

Initial Conditions Can Add Information

The initial output slope satisfies \(y'(0)=b x(0)^2-cy(0)\). If \(x(0)\) and \(y(0)\) are known and the output trajectory determines \(y'(0)\), then each candidate value of \(c\) implies a corresponding value of \(b\):

b=\frac{y'(0)+c\,y(0)}{x(0)^2}.

This equation may rule out a candidate if it implies a value inconsistent with the model assumptions, such as \(b\leq0\). But if both candidate values are allowed, the known initial states and initial slope need not remove the ambiguity. If \(x(0)\) is unknown, this slope relation involves the product \(b x(0)^2\), so it does not separate those quantities.

Worked Example: Known Initial States Do Not Always Remove the Ambiguity

Use the two parameter pairs from the preceding example. Suppose the known initial states are \(x(0)=2\ \mu\mathrm{M}\) and \(y(0)=1\ \mu\mathrm{M}\), and the observed initial slope is \(y'(0)=0.20\ \mu\mathrm{M}/\mathrm{h}\). Calculate the corresponding \(b\) for each candidate \(c\).

For \(c=0.80\ \mathrm{h}^{-1}\),

b=\frac{0.20\ \mu\mathrm{M}/\mathrm{h} +(0.80\ \mathrm{h}^{-1})(1\ \mu\mathrm{M})} {(2\ \mu\mathrm{M})^2} =\frac{1.00\ \mu\mathrm{M}/\mathrm{h}}{4\ \mu\mathrm{M}^2} =0.25\ \mu\mathrm{M}^{-1}\mathrm{h}^{-1}.

For \(c=0.10\ \mathrm{h}^{-1}\),

b=\frac{0.20+0.10}{4} =0.075\ \mu\mathrm{M}^{-1}\mathrm{h}^{-1},

where the numerator is \(0.30\ \mu\mathrm{M}/\mathrm{h}\). Both values of \(b\) are positive. Each candidate model has the same \(A\), \(B\), \(y(0)\), and \(y'(0)\), and therefore the same solution of the second-order input-output equation. The known initial states and slope in this example do not distinguish the two parameter sets.

Worked Example: Unknown Initial State Confounds the Production Rate

Suppose \(y(0)=1\ \mu\mathrm{M}\), \(y'(0)=0.20\ \mu\mathrm{M}/\mathrm{h}\), and \(c=0.80\ \mathrm{h}^{-1}\), but \(x(0)\) is not known. Determine what the initial slope constrains, and give two compatible choices of \(b\) and \(x(0)\).

Rearranging the initial-slope equation gives

b x(0)^2=y'(0)+cy(0) =0.20+0.80 =1.00\ \mu\mathrm{M}/\mathrm{h}.

For \(x(0)=1\ \mu\mathrm{M}\), this requires \(b=1.00\ \mu\mathrm{M}^{-1}\mathrm{h}^{-1}\). For \(x(0)=2\ \mu\mathrm{M}\), it requires \(b=0.25\ \mu\mathrm{M}^{-1}\mathrm{h}^{-1}\). Both satisfy the same slope constraint. More generally, when \(a\) and \(c\) are fixed, the output forcing term is \(b x(t)^2=b x(0)^2 e^{-2at}\). Thus changing \(b\) and \(x(0)\) while preserving their product \(b x(0)^2\) preserves that forcing term and the output trajectory for the same \(y(0)\).

A Short Identifiability Justification

The coefficient map is

(a,c)\longmapsto(A,B)=(2a+c,\,2ac).

Its symmetry makes the global ambiguity explicit. Define a second parameter pair by \(a^\ast=c/2\) and \(c^\ast=2a\). Then

2a^\ast+c^\ast=c+2a=A, \qquad 2a^\ast c^\ast=c(2a)=B.

Consequently, the coefficient map cannot generally distinguish a parameter pair from its swapped counterpart. The map is locally one-to-one away from the boundary where the two assigned roots coincide: its Jacobian determinant is

\det\begin{pmatrix} 2&1\\ 2c&2a \end{pmatrix} =4a-2c=2(2a-c).

A nonzero determinant means small changes in the two parameters produce locally distinguishable changes in the two coefficients. It does not remove the global swap: local distinguishability near one candidate is different from unique recovery over the full parameter space. The parameter \(b\) is absent from this coefficient map and cannot be recovered by matching \(A\) and \(B\) alone.

Diagnostics, Alternatives, and Limitations

A practical calculation should keep separate the stages of the argument: derive the state-free equation, determine whether its coefficients are recoverable from the output, and solve the parameter-to-coefficient equations. A failure at any stage invalidates a claim of unique identification. Common warning signs include:

Using an unverified output equation. The derivation depends on the stated model and on differentiability. A misspecified pathway or an incorrect observation map can produce a misleading equation.
Assuming coefficients are uniquely determined. If the observed trajectory does not provide independent information about \(Y'\) and \(Y\), matching may not yield unique \(A\) and \(B\).
Calling the parameters identifiable because the equations can be solved. Solving for roots gives an unordered pair; both assignments must be considered.
Ignoring initial-state assumptions. A known value of \(x(0)\) changes what can be learned from the initial slope. An unknown value leaves a \(b x(0)^2\) confounding.
Applying exact differentiation to noisy data. Structural analysis assumes ideal output information. Numerical second derivatives can amplify measurement noise, so this derivation is not itself a reliable estimation procedure for sparse observations.

State elimination and coefficient matching are a direct way to expose combinations such as \(2a+c\) and \(2ac\). Augmented-state observability analysis is an alternative formulation that treats parameters as constant states and asks whether they can be distinguished from output behavior. Symmetry analysis, familiar from the earlier material on proving nonidentifiability with symmetry groups, can establish ambiguity by displaying parameter transformations that preserve the output. These approaches can support one another, but each must use the same model, observation assumptions, and parameter domain.

The main limitation is that an input-output equation may contain less parameter information than the full set of model and initial-condition constraints. Conversely, finding a parameter combination in the equation does not automatically guarantee that the combination is recoverable from a particular output trajectory. Structural conclusions also do not measure how precisely parameters can be estimated from finite, noisy data.

Key takeaway: For this pathway, the state-free equation exposes the combinations \(2a+c\) and \(2ac\). Their coefficient matching identifies an unordered pair \(\{c,2a\}\), not generally a unique assignment, and it does not involve \(b\). Initial-state information can add constraints, but its effect depends on which initial states are known.

Check Your Understanding

Answer each question, then compare with the solutions below.

Starting from the model equations, what substitution removes \(x^2\) from the differentiated output equation?
If \(A=1.0\ \mathrm{h}^{-1}\) and \(B=0.16\ \mathrm{h}^{-2}\), find the two candidate pairs \((a,c)\).
Why does matching \(A\) and \(B\) not identify \(b\)?
If \(x(0)\) is unknown but \(y(0)\), \(y'(0)\), and \(c\) are known, what combination involving \(b\) and \(x(0)\) is constrained by the initial slope?

Solutions. (1) Use \(b x^2=y'+cy\) in \(y''=-2abx^2-cy'\), giving \(y''+(2a+c)y'+2ac\,y=0\). (2) The roots of \(\lambda^2-\lambda+0.16=0\) are \((1\pm0.60)/2\), or \(0.80\) and \(0.20\ \mathrm{h}^{-1}\). Assigning them to \((c,2a)\) gives \((a,c)=(0.10,0.80)\) or \((0.40,0.20)\), with units \(\mathrm{h}^{-1}\). (3) The parameter \(b\) does not appear in \(A=2a+c\) or \(B=2ac\), so matching those coefficients supplies no equation for \(b\). (4) The slope relation gives \(b x(0)^2=y'(0)+cy(0)\); without \(x(0)\), it constrains the product \(b x(0)^2\), not \(b\) alone.