Nondimensionalizing QSP Balance Equations

Why Nondimensionalize a Balance Equation?

A compartmental balance equation states how a quantity changes through inputs, outputs, and transfers. Its dimensional form is essential for interpreting amounts, concentrations, and rates. But the numerical values of those quantities depend on the units chosen: an amount could be reported in milligrams or micrograms, and time in hours or minutes. Nondimensionalization separates these unit choices from the underlying structure of the model.

The procedure replaces each dimensional state and time with a dimensionless variable scaled by a stated reference value. Parameters then enter through dimensionless groups, such as a ratio of two clearances or the ratio of an input rate to a maximum elimination rate. These groups help identify which processes set the model’s characteristic scales and make dimensional errors easier to spot.

Definition: A dimensionless state is a state divided by a positive reference scale with the same units. A dimensionless time is time divided by a positive reference time. A dimensionless parameter group is a combination of model parameters whose units cancel, so that it is a pure number.

Nondimensionalization changes the form in which a model is expressed; by itself, it does not approximate the model, discard a process, or change its predictions. The scales are choices that must be stated, justified, and used consistently.

Transforming a General Balance

Let \(x_i(t)\) be a model state with units \(U_i\), and let its balance be \(dx_i/dt=f_i(x,p,t)\), where \(p\) denotes parameters and the right-hand side has units \(U_i/\text{time}\). Choose a positive state scale \(X_i\) with units \(U_i\) for each state, and a positive time scale \(T\) with units of time. Define \(z_i=x_i/X_i\) and \(\tau=t/T\). Since \(x_i=X_i z_i\) and \(t=T\tau\), the chain rule gives the transformed equation below.

\frac{dx_i}{dt}=\frac{X_i}{T}\frac{dz_i}{d\tau} \qquad\Longrightarrow\qquad \frac{dz_i}{d\tau} =\frac{T}{X_i}f_i(X_1z_1,\ldots,X_nz_n,p,T\tau).

The factor \(T/X_i\) converts the dimensional rate to a rate per unit dimensionless time and state scale. Each term in the resulting equation must be dimensionless. If it is not, either the original balance, the chosen substitution, or the algebra needs correction.

A useful scale often makes a leading coefficient equal to one. For example, the inverse of a first-order rate constant is a natural time scale for a process with rate \(k\), while a concentration such as a half-saturation constant can be a natural state scale in a saturable rate law. Other choices can be valid: a scale is not unique, but a useful one makes the dimensionless equation and its parameter groups easy to interpret.

Worked Examples

Worked Example: Constant Input and First-Order Elimination

Consider a hypothetical one-compartment amount \(A(t)\) with fixed volume \(V\), constant input rate \(R_0\), and linear elimination clearance \(CL\). Assume the input and parameters remain constant over the time interval considered, and that elimination rate is \(CL\) times concentration. The balance is

\frac{dA}{dt}=R_0-\frac{CL}{V}A=R_0-kA, \qquad k=\frac{CL}{V}.

Here \(R_0\) has units mg/h, \(CL\) has units L/h, \(V\) has units L, and \(k\) has units h\(^{-1}\). Choose \(A_c=R_0/k\) as the amount scale and \(T=1/k\) as the time scale. Define \(a=A/A_c\) and \(\tau=t/T\). Substitution gives

\frac{A_c}{T}\frac{da}{d\tau}=R_0-kA_c a \qquad\Longrightarrow\qquad \frac{da}{d\tau}=\frac{TR_0}{A_c}-Tk\,a=1-a.

For hypothetical values \(R_0=12\) mg/h, \(CL=3\) L/h, and \(V=6\) L, \(k=3/6=0.500\) h\(^{-1}\). Therefore \(T=1/0.500=2.00\) h and \(A_c=12/0.500=24.0\) mg. If \(A(0)=14.4\) mg, then \(a(0)=14.4/24.0=0.600\). The transformed equation has solution \(a(\tau)=1-0.400e^{-\tau}\). At \(t=2.00\) h, \(\tau=1\), so \(a=1-0.400e^{-1}=0.853\) and \(A=24.0(0.853)=20.5\) mg. The scales make the balance’s equilibrium amount and its first-order time scale explicit.

Worked Example: Saturable Elimination and an Input Group

Consider a hypothetical well-mixed compartment with fixed volume \(V\), constant amount input \(R_{\mathrm{in}}\), and saturable elimination at rate \(V_{\max}C/(K_m+C)\), where \(C\) is concentration. Assume \(V_{\max}\) has units amount/time, \(K_m\) has units amount/volume, and the input and volume are constant. The concentration balance is

V\frac{dC}{dt}=R_{\mathrm{in}}-\frac{V_{\max}C}{K_m+C}.

Use \(C_c=K_m\), define \(c=C/K_m\), and choose \(T=VK_m/V_{\max}\). This time scale has units of time because \(VK_m\) is an amount. With \(\tau=t/T\), division by \(V_{\max}\) yields

\frac{dc}{d\tau}=\rho-\frac{c}{1+c}, \qquad \rho=\frac{R_{\mathrm{in}}}{V_{\max}}.

Both terms in this equation are dimensionless. The group \(\rho\) compares the constant input rate with the maximum elimination rate; it is not a concentration or a rate with units.

For hypothetical values \(V=3.00\) L, \(K_m=2.00\) mg/L, \(V_{\max}=18.0\) mg/h, and \(R_{\mathrm{in}}=6.00\) mg/h, \(T=(3.00\text{ L})(2.00\text{ mg/L})/(18.0\text{ mg/h})=0.333\) h, and \(\rho=6.00/18.0=0.333\). At a positive steady state, \(0=\rho-c/(1+c)\). Thus \(c=\rho(1+c)\), so \(c=\rho/(1-\rho)=0.500\). The corresponding concentration is \(C=K_m c=(2.00\text{ mg/L})(0.500)=1.00\) mg/L. This positive steady state exists here because \(\rho<1\); input is below the maximum elimination rate.

Worked Example: Exchange Between Compartments With Unequal Volumes

Suppose hypothetical amounts \(A_1=V_1C_1\) and \(A_2=V_2C_2\) exchange through a flow \(Q\), with flux \(Q(C_1-C_2)\) from compartment 1 to compartment 2 when \(C_1>C_2\). Compartment 1 also has linear elimination clearance \(CL_1\). Assume the volumes and flow are fixed. The concentration balances are

V_1\frac{dC_1}{dt}=-Q(C_1-C_2)-CL_1C_1, \qquad V_2\frac{dC_2}{dt}=Q(C_1-C_2).

The transfer terms have units \((\text{volume}/\text{time})(\text{amount}/\text{volume})=\text{amount}/\text{time}\), as does \(CL_1C_1\). Choose a shared concentration scale \(C_c\) and \(T=V_1/Q\); set \(c_i=C_i/C_c\) and \(\tau=t/T\). Then

\frac{dc_1}{d\tau}=-(c_1-c_2)-\frac{CL_1}{Q}c_1, \qquad \frac{dc_2}{d\tau}=\frac{V_1}{V_2}(c_1-c_2).

The groups \(CL_1/Q\) and \(V_1/V_2\) are dimensionless. With hypothetical \(V_1=3.00\) L, \(V_2=6.00\) L, \(Q=0.600\) L/h, and \(CL_1=0.300\) L/h, the chosen time scale is \(T=3.00/0.600=5.00\) h, \(CL_1/Q=0.500\), and \(V_1/V_2=0.500\). For \(c_1=1.00\) and \(c_2=0.400\), the dimensionless slopes are

\frac{dc_1}{d\tau}=-(1.00-0.400)-0.500(1.00)=-1.10, \qquad \frac{dc_2}{d\tau}=0.500(1.00-0.400)=0.300.

These signs agree with the specified processes at those concentrations: the first compartment loses material by net transfer and elimination, while the second gains material by transfer. The shared concentration scale cancels from both equations, but the volume ratio remains because equal concentration changes do not correspond to equal amount changes in unequal volumes.

Choosing and Interpreting Scales

A practical scale should reflect the balance being transformed, not merely make the numbers in a parameter table look small. A first-order process with rate constant \(k\) suggests \(1/k\) as a time scale. A transfer process with rate \(Q/V\) suggests \(V/Q\). A state scale may be a characteristic baseline or input-driven level, or a parameter intrinsic to the rate law. State scales need not be equilibrium values; they are reference quantities used to express the states.

When multiple processes compete, one scale cannot make every coefficient equal to one. The remaining dimensionless groups quantify their relative strengths. For example, \(CL_1/Q\) compares elimination clearance with exchange flow. A small or large group can indicate a substantial difference in process scales, but it does not by itself justify deleting a term or replacing the model with an approximation.

Derivation check: If a dimensional term in \(dx_i/dt\) has units \(U_i/\text{time}\), multiplying it by \(T/X_i\) gives \((\text{time}/U_i)(U_i/\text{time})=1\). Thus each term in \(dz_i/d\tau\) must be dimensionless. A remaining unit indicates an inconsistency, not a parameter group.

Diagnostics, Alternatives, and Limitations

A dimensional audit and nondimensionalization complement one another. A dimensional audit checks that each term can be added to the others in the original balance. Nondimensionalization goes further by exposing the combinations of parameters that govern the scaled equations. Keeping the original dimensional equations is often clearer for reporting amounts, concentrations, and rates; a dimensionless form is often clearer for comparing process strengths or identifying characteristic scales. Neither form replaces the other.

Check the transformation by confirming that every chosen scale has the stated units, applying the chain rule explicitly, and verifying that each transformed term is unitless. Also check signs and directions against the original balance. If \(V_1/V_2\) in an unequal-volume exchange model is accidentally replaced by one, that is not a harmless scaling choice: it changes the balance.

A poor scale can leave all coefficients extremely large or small, obscuring the model’s characteristic behavior. That may signal a scale that is unrelated to the processes of interest, or genuinely different time scales in the system. Conversely, a neat equation does not establish that its biological assumptions are sound. Nondimensionalization cannot correct a missing flux, an invalid rate law, or an inappropriate fixed-volume assumption; those must be evaluated in the dimensional model.

A common mistake is to divide a balance by a parameter and call every resulting coefficient dimensionless. For instance, dividing a rate by \(V_{\max}\) alone does not guarantee a dimensionless equation if the left side has not also been scaled. The correction is to define state and time variables first, apply the chain rule, and then check the units of every resulting term and group.

Write the dimensional balance.
State the units of each state, parameter, and rate before transforming the equation.

Choose explicit scales.
Give each state scale the same units as its state, and give the time scale units of time.

Substitute and apply the chain rule.
Express states and time using their dimensionless variables; do not omit the scale factors in the derivative.

Audit the transformed equation.
Confirm that all terms and parameter groups are dimensionless and that their signs and dependence match the original balance.

Check Your Understanding

Answer each question, then compare your work with the solutions below.

A balance is \(dX/dt=u-kX\), where \(X\) has units mg, \(u\) has units mg/h, and \(k\) has units h\(^{-1}\). With \(X_c=u/k\) and \(T=1/k\), what is the dimensionless equation for \(x=X/X_c\) and \(\tau=t/T\)?
For the saturable balance \(V\,dC/dt=R_{\mathrm{in}}-V_{\max}C/(K_m+C)\), state the units of \(R_{\mathrm{in}}/V_{\max}\) and interpret the group.
In an exchange model, \(V_1=4\) L and \(Q=0.8\) L/h. What time scale follows from \(T=V_1/Q\)?
Why does the dimensionless exchange equation for compartment 2 contain \(V_1/V_2\) when the concentration scale is the same in both compartments?

Solutions. (1) Substitution gives \((X_c/T)\,dx/d\tau=u-kX_cx\). Since \(X_c=u/k\) and \(T=1/k\), both coefficients reduce to \(u\), so \(dx/d\tau=1-x\). (2) The ratio is dimensionless because both rates have units amount/time. It compares the input rate with the maximum saturable elimination rate. (3) \(T=(4\text{ L})/(0.8\text{ L/h})=5\) h. (4) The amount balance for compartment 2 is \(V_2\,dC_2/dt=Q(C_1-C_2)\). Scaling time by \(V_1/Q\) leaves the coefficient \(V_1/V_2\); the unequal volumes convert the shared concentration scale into different amount scales.