75048 Mathematical modeling 1998

Enzyme kinetics ala Lin & Segel

Harald Hanche-Olsen

                       hanche@math.ntnu.no

Reaction and basic model

The reaction looks like:

s^{∗} + e^{∗} ⇄_{k_{−1}}^{k_{1}} c^{∗} → p^{∗} + e^{∗}

which leads to the model

\frac{ds^{∗}}{dt^{∗}} = −k_{1}s^{∗}e^{∗} + k_{−1}c^{∗}
\frac{dc^{∗}}{dt^{∗}} = k_{1}s^{∗}e^{∗} − k_{−1}c^{∗} − k_{2}c^{∗}
\frac{de^{∗}}{dt^{∗}} = −k_{1}s^{∗}e^{∗} + k_{−1}c^{∗} + k_{2}c^{∗}
\frac{dp^{∗}}{dt^{∗}} = k_{2}c^{∗}

We shall use the initial conditions

s^{∗}(0) = s̄, c^{∗}(0) = 0, e^{∗}(0) = ē, p^{∗}(0) = 0.

Immediate consequences. We get c^{∗} + e^{∗} = ē and s^{∗} + c^{∗} + p^{∗} = s̄. Hence we need only solve for S^{∗} and c^{∗}. Substitute e^{∗} = ē− c^{∗} into the first two equations and get our final, non-scaled model:

\frac{ds^{∗}}{dt^{∗}} = −k_{1}ēs^{∗} + (k_{1}s^{∗} + k_{−1})c^{∗}
\frac{dc^{∗}}{dt^{∗}} = k_{1}ēs^{∗} − (k_{1}s^{∗} + k_{−1} + k_{2})c^{∗}
s^{∗}(0) = s̄, c^{∗}(0) = 0

Scaling

Put

s^{∗} = s̄s, c^{∗} = ēc, t^{∗} = \frac{1}{k_{1}ē}

and, with the nondimensional parameters

κ = \frac{k_{−1} + k_{2}}{k_{1}ē} , λ = \frac{k_{2}}{k_{1}ē}, ε = \frac{ē}{s̄}

we have the problem on its non-dimensional form:

ṡ = −s + (s + κ − λ)c
εċ = s − (s + κ)c
s(0) = 1, c(0) = 0

For later reference, it will be useful to remember that 0 < λ < κ, and we shall assume 0 < ε ≪ 1.

Solution by perturbation

Outer solution. First, put ε = 0 in the differential equations. The second equation becomes the algebraic equation s − (s + κ)c = 0. We solve this for c and substitute in the first equation, which becomes ṡ = −λs/(s + κ). We call the solution (s_{0}, c_{0}):

s_{0} + κ ln s_{0} = Q − λt, c_{0} = \frac{s_{0}}{s_{0} + κ}  (1)

The integration constant Q might be determined by using s_{0}(0) = 1, so Q = 1. But then the other initial condition is not satisified, and so we are not quite sure whether even the first one is satisfied. So presumably, Q must be determined by matching.

Inner solution. Initially we expect s ≈ 1, in which case εċ ≈ 1 − (1 + κ)c. Thus c seems to tend towards 1 with a time constant ε/(1 + κ). Introduce this time constant as a new time scale, and define the inner dimensionless time τ by^{1}

t = \frac{ε}{1 + κ}τ

1Here we depart from Lin & Segel in a small way. It simplifies a few formulae, but changes nothing important.

With s(t) = S(τ) and c(t) = C(τ) the equations become

(1 + κ)S′ = ε( − S + (S + κ − λ)C)
(1 + κ)C′ = S − (S + κ)C

and, of course, we can use the intial conditions as well, so S(0) = 1 and C(0) = 0.

With ε = 0 we get S′ = 0, so the initial conditions imply S = 1. We plug that into the second equation and get C′ = \frac{1}{1+κ} − C, which is easy to solve with the initial condition C = 0. Renaming this solution (S_{0}, C_{0}) we thus have

S_{0}(τ) = 1, C_{0}(τ) = \frac{1 − e^{−τ}}{1 + κ}

Matching. To this order, we can match inner and outer solutions by simply requiring that lim_{τ→∞} S(τ) = lim_{t→0} s(t). Thus Q = 1, as we guessed before. The similar equation lim_{τ→∞} C(τ) = lim_{t→0} c(t) turns out to be automatically satisfied.

To the next order. We try to substitute power series S = S_{0} + εS_{1} + ⋯, C = C_{0} + εC_{1} + ⋯, s = s_{0} + εs_{1} + ⋯, and c = c_{0} + εc_{1} + ⋯ into the inner and outer equations. For S_{0}, C_{0}, s_{0} and c_{0} we find the equations and solutions we have already discovered. Next, we find

S_{1} = −\frac{1}{(1 + κ)^{2}} (λτ + (1 + κ − λ)(1 − e^{−τ}))
C_{1} = −\frac{1}{(1 + κ)^{4}} (λκτ + κ(1 + κ − 2λ) + (1 + κ − λ)e^{−2τ}

+ [\frac{1}{2}λτ^{2} + (1 + κ − λ)(1 − κ)τ − ((1 + κ)^{2} − λ − 2κλ)]e^{−τ} )

while s_{1} and c_{1} are given by

ṡ_{1} = (c_{0} − 1)s_{1} + (κ − λ + s_{0})c_{1}, ċ_{0} = s_{1}(1 − c_{0}) − (κ + s_{0})c_{1}

which requires an initial condition for s_{1}.^{2} So we introduce the intermediate time scale Ψ, put t = Ψτ_{i}, and note that

2At this stage, we do not need an initial condition for c_{1}, as this is system is actually just one differential equation and one algebraic equation, c_{0} being a known function.

τ = \frac{1 + κ}{ε}t = \frac{(1 + κ)Ψ}{ε}τ_{i}

We get, therefore

S(τ) = S_{0}(\frac{(1 + κ)Ψ}{ε}τ_{i}) + εS_{1}(\frac{(1 + κ)Ψ}{ε}τ_{i}) + ⋯

= 1 − \frac{λτ_{i}Ψ}{1 + κ} − \frac{(1 + κ − λ)ε}{(1 + κ)^{2}} + TST + ⋯

which needs to be matched to s(t) = s_{0}(t) + εs_{1}(t) + ⋯ with t = Ψτ_{i}.

Thus we need a power series representation for s_{0} in terms of t: Write s_{0}(t) = 1+γt+⋯ and insert into the first part of (1): 1+γt+κγt+⋯ = 1 − λt, so γ = −λ/(1 + κ), and we can write

s(t) = s_{0}(τ_{i}Ψ) + εs_{1}(τ_{i}Ψ) + ⋯ = 1 − \frac{λτ_{i}Ψ}{1 + κ} + s_{1}(0)ε + ⋯

and we see that the required matching implies

s_{1}(0) = −\frac{(1 + κ − λ)ε}{(1 + κ)^{2}}.

Note: the next term in the approximation for s(t) would contain a factor Ψ^{2}. In order for our argument to remain valid, we need Ψ^{2} ≪ ε as ε → 0; thus we must have

lim_{ε→0} \frac{Ψ}{ε} = ∞, lim_{ε→0} \frac{Ψ}{\sqrt{ε}} = 0.

References

C.C Lin & L.A. Segel: Mathematics applied to deterministic problems in the natural sciences. ISBN 0898712297. Chapter 10 (pp. 302– 320). Lee A. Segel & Marshall Slemrod: The quasi-steady state assumption: A case study in perturbation. SIAM Review 31 #3 (1989), pp. 446-477 (available at the mathematics library).