examples of growth of perturbations in chemical organizations


We will examine several simple examples of chemical systems where we start with one species of molecules (or a closed subset of species) then intoruduce a small perturbation and evolve the system using mass action dynamics. We want to know whether this perturbation will grow and, if so, at what rate. Ultimately, we would like to link the behavor to some feature of the reaction system, perhaps related to Rosen’s theory of M-R systems.

To get started, consider a trivial case, A+BB+B. The system of equations which describes this system is:

dxdt =-kxy
dydt =kxy

It is easy enough to solve this system. We begin by noting that ddt(x+y)=0, hence x+y=x0+y0. Substituting this back in to the second equation, we conclude that

dydt=k(x0+y0-y)(y).

This equation can readily be solved to yield the implicit solution

kt=1x0+y0logyx0+y0-yx0y0

which can be solved to produce the explicit solution

y=x0+y0-x0+y01+y0x0exp(k(x0+y0)t).

Looking at the solution, we see that it starts out at y=y0 and grows towards y=x0+y0 as t. This is as we expect — as time goes on, whatever A’s there are left react with B’s to turn into B’s until we are left with nothing but B’s.

If we suppose that, at the initial time t=0, there is only a tiny proportion of B’s, i.e. y0x0, then we may make an expansion of the fraction and conclude that y grows exponentially for small values of t:

11+y0x0exp(k(x0+y0)t)1-y0x0exp(k(x0+y0)t)
yy0x0exp(k(x0+y0)t)

We can also come to this conclusionMathworldPlanetmath by bounding y without solving the equation first. For a simple equation like this which is readily solved, this is hardly needed but, for larger, more complicated equations, it becomes important and this simple example can serve as a illustration of the general technique.

Theorem 1.

Let C be a real number such that 0<C<1 and let x0 and y0 be strictly positive real numbers such that 0<y0<C(x0+y0). Set t1=1k(x0+y0)logCx0+y0y0. Then there exists a function f:[0,t1)[0,C(x0+y0)) such that f satisfies the differential equationMathworldPlanetmath

df(t)dt=k(x0+y0-f(t))f(t).

and, for all t[0,t1),

y0exp((1-C)k(x0+y0)t)f(t)y0exp(k(x0+y0)t).
Proof.

By the existence theoremMathworldPlanetmath, there exists a positive real number t0 and a function f:[0,t0) such that f(0)=y0 and f satifies the differential equation. Since f(0)=y0<C(x0+y0), by continuity there exists a positive real number t2 and a function f:[0,t0)[0,C(x0+y0)) which satisfies the same differential equation with the same initial conditionMathworldPlanetmath. Furthermore, we assume that t2 is maximal.

Starting with this condition f(t)<C(x0+y0) and doing some algebra, we conclude that

(1-C)k(x0+y0)1ydydtk(x0+y0)

Now, 1ydydt=ddt(logy) so, by the mean value theorem, we conclude that

(1-C)k(x0+y0)tlogyy0k(x0+y0)t.

Exponentiating,

y0exp((1-C)k(x0+y0)t)yy0exp(k(x0+y0)t).

We can ensure that the bound on y is satisfied if the condition C(x0+y0)y0exp(k(x0+y0)t) is met, which amounts to demanding that 0tt1 where t1=1k(x0+y0)logCx0+y0y0.

Title examples of growth of perturbations in chemical organizations
Canonical name ExamplesOfGrowthOfPerturbationsInChemicalOrganizations
Date of creation 2014-05-31 15:55:20
Last modified on 2014-05-31 15:55:20
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 33
Author rspuzio (6075)
Entry type Example