examples of growth of perturbations in chemical organizations

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

To get started, consider a trivial case, $A+B\to B+B$. The system of equations whichdescribes this system is:

${\displaystyle \frac{dx}{dt}}$$=-kxy$

${\displaystyle \frac{dy}{dt}}$$=kxy$

It is easy enough to solve this system. We begin by noting that$\frac{d}{dt}(x+y)=0$, hence $x+y=x_0+y_0$.Substituting this back in to the second equation, we concludethat

$$\frac{dy}{dt}=k(x_0+y_0-y)(y).$$

This equation can readily be solved to yield the implicit solution

$$kt=\frac{1}{x_0+y_0}\log \frac{y}{x_0+y_0-y}\cdot \frac{x_0}{y_0}$$

which can be solved to produce the explicit solution

$$y=x_0+y_0-\frac{x_0+y_0}{1+\frac{y_0}{x_0}\exp (k(x_0+y_0)t)}.$$

Looking at the solution, we see that it starts out at $y=y_0$ and growstowards $y=x_0+y_0$ as $t\to \mathrm{\infty }$. This is as we expect — astime goes on, whatever A’s there are left react with B’s to turn into B’suntil we are left with nothing but B’s.

If we suppose that, at the initial time $t=0$, there is only a tinyproportion of B’s, i.e. $y_0\ll x_0$, then we may make an expansionof the fraction and conclude that $y$ grows exponentially for smallvalues of $t$:

$$\frac{1}{1+\frac{y_0}{x_0}\exp (k(x_0+y_0)t)}\approx 1-\frac{y_0}{x_0}\exp (k(x_0+y_0)t)$$

$$y\approx \frac{y_0}{x_0}\exp (k(x_0+y_0)t)$$

We can also come to this conclusion by bounding $y$ without solving theequation first. For a simple equation like this which is readily solved,this is hardly needed but, for larger, more complicated equations, itbecomes important and this simple example can serve as a illustration ofthe general technique.

Theorem 1.

Let $C$ be a real number such that and let $x_{\mathrm{0}}$ and $y_{\mathrm{0}}$ bestrictly positive real numbers such that .Set $t_{\mathrm{1}}\mathrm{=}\frac{\mathrm{1}}{k\mathit{}\mathrm{(}{x}_{\mathrm{0}}\mathrm{+}{y}_{\mathrm{0}}\mathrm{)}}\mathit{}\log \mathit{}C\mathit{}\frac{x_{\mathrm{0}}\mathrm{+}{y}_{\mathrm{0}}}{y_{\mathrm{0}}}$. Thenthere exists a function $f\mathrm{:}\mathrm{[}\mathrm{0}\mathrm{,}{t}_{\mathrm{1}}\mathrm{)}\mathrm{\to }\mathrm{[}\mathrm{0}\mathrm{,}C\mathit{}\mathrm{(}{x}_{\mathrm{0}}\mathrm{+}{y}_{\mathrm{0}}\mathrm{)}\mathrm{)}$ such that$f$ satisfies the differential equation

$$\frac{df(t)}{dt}=k(x_0+y_0-f(t))f(t).$$

and, for all $t\mathrm{\in }\mathrm{[}\mathrm{0}\mathrm{,}{t}_{\mathrm{1}}\mathrm{)}$,

$$y_0\exp ((1-C)k(x_0+y_0)t)\le f(t)\le y_0\exp (k(x_0+y_0)t).$$

Proof.

By the existence theorem, there exists a positive real number$t_0$ and a function $f:[0,t_0)\to ℝ$such that $f(0)=y_0$ and $f$ satifies the differential equation.Since , by continuity there existsa positive real number $t_2$ and a function $f:[0,t_0)\to [0,C(x_0+y_0))$which satisfies the same differential equation with the sameinitial condition. Furthermore, we assume that $t_2$ is maximal.

Starting with this condition anddoing some algebra, we conclude that

$$(1-C)k(x_0+y_0)\le \frac{1}{y}\frac{dy}{dt}\le k(x_0+y_0)$$

Now, $\frac{1}{y}\frac{dy}{dt}=\frac{d}{dt}(\log y)$ so, by the meanvalue theorem, we conclude that

$$(1-C)k(x_0+y_0)t\le \log \frac{y}{y_0}\le k(x_0+y_0)t.$$

Exponentiating,

$$y_0\exp ((1-C)k(x_0+y_0)t)\le y\le y_0\exp (k(x_0+y_0)t).$$

We can ensure that the bound on $y$ is satisfied if the condition$C(x_0+y_0)\ge y_0\exp (k(x_0+y_0)t)$ is met, which amounts todemanding that $0\le t\le t_1$ where $t_1=\frac{1}{k(x_0+y_0)}\log C\frac{x_0+y_0}{y_0}$.

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