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,
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),
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.
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 .
The valid solution is
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
Integrating it with respect to ,
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.
It shows that the system is asymptotically stable.