Modeling — Question 9

Question 9 – Solution

Strategy. Each forward reaction consumes one A and creates two B; the resulting weighted conservation law reduces the nonlinear system.

Step 1: Respect the reaction stoichiometry. A net forward reaction rate $a-b^2$ gives $$\boxed{a'=-a+b^2,\qquad b'=2a-2b^2}$$ in the chosen numerical units. Hence $(2a+b)'=0$ and $\boxed{2a+b=2}$ mol/L. The sum $a+b$ has derivative $a-b^2$ and is not conserved: the reaction changes the number of molecules, while preserving the weighted building-block total.

Step 2: Reduce within the physical domain. With $a=1-b/2$, the scalar equation is $\boxed{b'=2-b-2b^2}$, with $0\le b\le 2$ required by $a,b\ge 0$. At $b=0$ the derivative is $2>0$, and at $b=2$ it is $-8<0$. Uniqueness and inward boundary directions make the interval invariant. The concentrations cannot be extended to negative values merely because an algebraic equation has another root.

Step 3: Identify the attracting physical equilibrium. Let $d=\sqrt{17}$ and $r_\pm=(-1\pm d)/4$. Then $b'=-2(b-r_+)(b-r_-)$, with $r_-<0<r_+<2$. The only physical equilibrium is $$\boxed{b_*=r_+,\qquad a_*=1-r_+/2.}$$ For $0\le b<r_+$, $b'>0$. Uniqueness prevents crossing the equilibrium, so the solution from zero increases and stays below $r_+$. Its limit must be a zero of the scalar right side; otherwise its positive derivative would remain bounded away from zero. Thus it tends to $r_+$.

Step 4: Separate and impose the initial concentration. Since $2(r_+-r_-)=d$, integration gives $$\log\left|\frac{b-r_+}{b-r_-}\right|=-dt+ \log\left|\frac{r_+}{r_-}\right|.$$ The ratio is negative on the physical branch starting at zero. Solving gives $$\boxed{b(t)=\frac{r_+(1-e^{-dt})}{1-(r_+/r_-)e^{-dt}},\qquad a(t)=1-\frac{b(t)}2.}$$ The denominator exceeds $1$ at finite times because $r_+/r_-<0$. The formulas give $(a,b)=(1,0)$ initially and the stated equilibrium as $t\to\infty$. For every finite $t$, $b<r_+$, so arrival is asymptotic.