We will study a simple chemical reaction described by

where two molecules of are combined reversibly to form and, are the reaction rates.

If is the concentration of , of , then according to the Law of Mass Action,

or equivalently,

We seek first the equilibrium points that represent the steady state of the system. They are the constant solutions where and , simultaneously.

From and , it is apparent that

is an equilibrium point.

To find the value of , we solve for from (0-1),

Substitute it in (0-2),

,

i.e.,

This is a *2nd order* nonlinear differential equaion. Since it has no direct dependence on , we can reduce its order by appropriate substitution of its first order derivative.

Let

,

we have

so that (0-6) is reduced to

a *1st order* differential eqution. It follows that either or .

The second case gives

Integrate it with respect to ,

Hence, the equilibrium points of (0-1) and (0-2) can be obtained by solving a quadratic equation

Notice in order to have as a solution, must be non-negative .

Fig. 1

The valid solution is

Fig. 2

By (0-4),

and so, the equilibrium point is

Next, we turn our attentions to the phase-plan trajectories that describe the paths traced out by the pairs over the course of time, depending on the initial values.

For . Dividing (0-2) by (0-1) yields

i.e.,

Integrating it with respect to ,

By (0-3),

Therefore,

Moreover, by (0-5)

As a result,

Substitute in (1-1), we have

This is the trajectory of the system. Clearly, all trajectories are monotonically decreaseing lines.

At last, let us examine how the system behaves in the long run.

If then (see Fig. 2) and will increase. As a result, will decrease. Similarly, if ensures that will decrease. Consequently, will increase.

Fig. 3 Trajectories and Equilibriums

It is evident that as time advances, on the trajectory approaches the equilibrium point

A phase portrait of the system is illustrated in Fig. 4.

Fig. 4

It shows that the system is asymptotically stable.

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