Question

When certain kinds of chemicals are combined, the rate at which the new compound is formed is modeled by the autonomous differential equation dX/dt = k(a-X)(B-X) where k > 0 is a constant of proportionality and B > a > 0. Here X(t) denotes the number of grams of the new compound formed in time t. (a) Use a phase portrait of the differential equation to predict the behavior of X(t) as t -> infinity. (b) Consider the case when a = B. Use a phase portrait of the differential equation to predict the behavior of X(t) as t -> infinity when X(0) < a. When X(0) > a. (c) Verify that an explicit solution of the DE in the case when k=1 and a=B is X(t)=a-1/(t+c). Find a solution that satisfies X(0) = a/2. Then find a solution that satisfies X(0)=2a. Graph these two solutions. Does the behavior of the solutions as t->infinity agree with your answers to part (b)?

Answer 1

$\dfrac{\mathrm dX}{\mathrm dt}=k(a-X)(B-X)$

(a) We're given that $0<a<B$. We have four cases to consider: (1) $X<0$; (2) $0<X<a$; (3) $a<X<B$; and (4) $B<X$.

In case (1), both $a-X>0$ and $B-X>0$, so $k(a-X)(B-X)>0$, which means $X(t)$ is increasing and converging to $X=0$ as $t\to\infty$.

In case (2), $a-X>0$ and $B-X>0$ again, so $k(a-X)(B-X)>0$ as well, so $X(t)$ is increasing again and converging to $X=a$ from below.

In case (3), $a-X<0$ and $B-X>0$, so $k(a-X)(B-X)<0$, which means $X(t)$ is decreasing and converging to $X=a$ from above as $t\to\infty$.

Finally, in case (4), both $a-X<0$ and $B-X<0$, so tha t$k(a-X)(B-X)>0$, and so $X(t)$ is increasing and diverging as $t\to\infty$.

So a simplified phase portrait for the solution $X(t)$ might look something like

$\begin{matrix}&\uparrow\\X=B&-\\&\downarrow\\X=a&-\\&\uparrow\\X=0&-\\&\uparrow\end{matrix}$

where the vertical axis represents $X(t)$.

(b) If $a=B$, then $\dfrac{\mathrm dX}{\mathrm dt}=k(a-X)^2$, which has the phase portrait

$\begin{matrix}&\uparrow\\X=a&-\\&\uparrow\\X=0&-\\&\uparrow\end{matrix}$

so that $X(t)\to\infty$ when $X>a$; $X(t)\to a$ from below when $0<X<a$; and $X\to0$ from below as $t\to\infty$ in either case of $X(0)>a$ or $X(0)<a$.

(c) With $k=1$ and $a=B$, we have the separable ODE

$\dfrac{\mathrm dX}{\mathrm dt}=(a-X)^2$
$\dfrac{\mathrm dX}{(a-X)^2}=\mathrm dt$
$\displaystyle\int\frac{\mathrm dX}{(a-X)^2}=\int\mathrm dt$
$\dfrac1{a-X}=t+C$
$a-X=\dfrac1{t+C}$
$X=a+\dfrac1{t+C}$

With the initial value $X(0)=\dfrac a2$, we have

$\dfrac a2=a+\dfrac1C\implies C=-\dfrac2a$

With $X(0)=2a$, we would get

$2a=a+\dfrac1C\implies C=\dfrac1a$

In both cases, we're assuming that $X(0)>a$, since we are given that $0<a<B$ from the start. In part (b), we found that $X(t)\to\infty$ as $t\to\infty$ (regardless of whether $X(0)$ is smaller or larger than $a$, in fact), so we only need to verify that the solution we found also conforms to this trend. We have

$\displaystyle\lim_{t\to\infty}X(t)=\lim_{t\to\infty}\left(a+\dfrac1{t+C}\right)=a$

which confirms the conclusion of the phase portrait.