Chemical Reactions 2

Problem 2.

The functions $u(t)$ and $v(t)$ describe the concentration of two substances, respectively. The components transform into each other with the rate coefficient $k\gt 0$, and the following system describes the process $$\begin{align*} u'(t)&=ku(t)v(t),\\ v'(t)&=-ku(t)v(t). \end{align*}$$ For the sake of simplicity let $k:=1$, and $u(0),v(0)=1$. Approximate the solutions by 3rd order polynomials.

Show Solution

From the system, it follows $$u'(0)=1,\quad v'(0)=-1.$$ Since $$\begin{align*} u''(t)&=ku'(t)v(t)+ku(t)v'(t),\\ v''(t)&=-ku'(t)v(t)-ku(t)v'(t) \end{align*}$$ we obtain $$u''(0)=0,\quad v''(0)=0.$$ Similarly, $$\begin{align*} u'''(t)&=ku''(t)v(t)+2ku'(t)v'(t)+ku(t)v''(t),\\ v'''(t)&=-ku''(t)v(t)-2ku'(t)v'(t)-ku(t)v''(t), \end{align*}$$ and $$u'''(0)=-2,\quad v'''(0)=2.$$ Hence $$\begin{align*} u(t)&=1+t-\frac{1}{3}t^3+\ldots, \\ v(t)&=1-t+\frac{1}{3}t^3+\ldots. \quad\blacksquare \end{align*}$$