Computer Algebra Recipes

150 CHAPTER 4. NONLINEAR ODE MODELS
4.1 First-Order Models
4.1.1 An Irreversible Reaction
I shall use the phrase \time's arrow" to express this one-way property
of time which has no analogue in space.
Arthur Eddington, British astrophysicist (1882{1944)
As our ¯rst example, we consider the following problem, which the reader was
previously asked to solve numerically (see Problem 2-29).
Consider the irreversible chemical reaction
2K 2Cr 2O 7 +2H 2O+3S ! 4KOH+2Cr 2O 3 +3SO 2
with initially N1 molecules of potassium dichromate (K 2Cr 2O 7), N2 molecules of
water (H2O), N3 atoms of sulphur (S), and 0 molecules of potassium hydroxide
(KOH). The number x of KOH molecules at time t seconds is given by the
nonlinear rate equation
_x = k (2 N1 ¡ x) 2 (2 N2 ¡ x) 2 (4 N3=3 ¡ x) 3
with k =1:64 £ 10 ¡20 s ¡1 . Taking N1 = 2000, N2 = 2000, and N3 = 3000,
analytically determine the number of KOH molecules at arbitrary time t>0
and plot the solution for the ¯rst second after the chemical reaction begins.
How many KOH molecules are present at 0:2 seconds? How many are present
in the limit t !1?
To observe Maple's method of solution, the infolevel[dsolve] command
is set equal to 2, and the rate equation entered.
> restart: infolevel[dsolve]:=2:
> de:=diff(x(t),t)=k*(2*N1-x(t))^2*(2*N2-x(t))^2
*(4*N3/3-x(t))^3;
de := d
dt x (t) =k (2 N1 ¡ x(t))2 (2 N2 ¡ x(t)) 2
μ
4 N3
3
3
¡ x(t)
Theratecoe±cientk and the initial molecule numbers are speci¯ed.
> k:=1.64*10^(-20): N1:=2000: N2:=2000: N3:=3000:
The di®erential equation de is analytically solved for x(t), subject to x(0) = 0.
> sol:=dsolve(fde,x(0)=0g,x(t));
Methods for ¯rst order ODEs:
| Trying classi¯cation methods |
trying a quadrature
trying 1st order linear
trying Bernoulli
trying separable