Chemical Reactions 1

Problem 1.

The functions $x(t)$ and $y(t)$ describe the concentration of two substances, respectively. The first component transforms into the second one with the rate coefficient $k\gt 0$, and the second component decomposes at the rate $\mu\gt 0$. Describe the system.

Show Solution

We have to investigate the system $$\begin{align*} x'(t) &=-k x(t),\\ y'(t) &=k x(t)-\mu y(t);\\ x(0) &=x_0 ,\,y(0)=y_0. \end{align*}$$ From the first equation $$x(t)=x_0e^{-kt}.$$ Substituting it into the second equation we get $$\begin{align*} y'(t)+\mu y(t) &=kx_0e^{-kt},\\ y(0) &=y_0. \end{align*}$$

Using the general formula for solution

we obtain $$y(t)=e^{-\mu t}\left[kx_0\int e^{\mu t-kt}dt+C \right].$$ Case (a). $\boldsymbol{\mu=k}$.
Then $$y(t)=(kx_0 t+C)e^{-kt}.$$ Since $y(0)=y_0$ it follows $$y(t)=(kx_0 t+y_0)e^{-kt}.$$ We note that if $x_0\gt 0$ then $$\frac{x(t)}{y(t)}=\frac{x_0}{kx_0 t+y_0},$$ and $$\lim_{t\to\infty}\frac{x(t)}{y(t)}=0.$$ Case (b). $\boldsymbol{\mu\neq k}$.
Then $$y(t)=\left(-\frac{kx_0}{k-\mu}e^{-t(k-\mu)} +C \right)e^{-\mu t}.$$ Since $y(0)=y_0$ it follows $$y(t)=\left(-\frac{kx_0 e^{-t(k-\mu)}}{k-\mu} +y_0+\frac{x_0 k}{k-\mu} \right)e^{-\mu t}.$$ We note that if $x_0\gt 0$ then $$\begin{align*} \frac{x(t)}{y(t)} &=\frac{x_0 e^{-t(k-\mu)}}{-\frac{kx_0 e^{-t(k-\mu)}}{k-\mu} +y_0+\frac{x_0 k}{k-\mu}}\\ &=\frac{x_0}{-\frac{k x_0}{k-\mu}+e^{t(k-\mu)}\left(y_0+\frac{k x_0}{k-\mu} \right)}. \end{align*}$$ If $k\gt \mu$ then $$\lim_{t\to\infty}\frac{x(t)}{y(t)}=0.$$ If $k\lt\mu$ then $$\lim_{t\to\infty}\frac{x(t)}{y(t)}=\frac{\mu-k}{k}.\quad\blacksquare$$