Modeling — Question 10

Question 10 – Solution

Strategy. Initial states equal to coordinate vectors reveal matrix columns; conservation and nonnegative transfer rates then provide physical checks.

Step 1: Identify the directed rates. Since $X'(0)=Ae_j$, the $j$th experimental vector is column $j$. Thus $$\boxed{A=\begin{pmatrix}-2&1&0\\2&-4&2\\0&3&-2\end{pmatrix}.}$$ The rates are $k_{1\to 2}=2$, $k_{2\to 1}=1$, $k_{2\to 3}=3$, $k_{3\to 2}=2$ h$^{-1}$. Each diagonal entry is minus the total outgoing rate from its compartment. Transposing the measurements would incorrectly replace these column balances with row balances.

Step 2: Check the physical state space. Every column sums to zero, so $(x_1+x_2+x_3)'=0$. At a boundary $x_j=0$ with the other fractions nonnegative, the corresponding derivative is a sum of nonnegative incoming fluxes. The simplex $\boxed{x_j\ge 0,\ x_1+x_2+x_3=1}$ is therefore invariant. Conservation bounds every fraction by $1$.

Step 3: Balance the equilibrium fluxes. The end-compartment equations require $x_2=2x_1$ and $2x_3=3x_2$. Normalization then gives $\boxed{X_*=(1,2,3)^T/6}$. Indeed the opposing fluxes on the first edge are $2(1/6)=1(2/6)$, and on the second are $3(2/6)=2(3/6)$. These two relations and the total also prove uniqueness of the normalized equilibrium.

Step 4: Resolve the first experiment and its interpretation. The eigenpairs are $0,(1,2,3)^T$, $-2,(1,0,-1)^T$ and $-6,(1,-4,3)^T$. Resolving $(1,0,0)^T$ gives $$X=X_*+\tfrac 34e^{-2t}(1,0,-1)^T+\tfrac 1{12}e^{-6t}(1,-4,3)^T.$$ Equivalently, $$\boxed{x_1=\tfrac 16+\tfrac 34e^{-2t}+\tfrac 1{12}e^{-6t},\quad x_2=\tfrac 13(1-e^{-6t}),\quad x_3=\tfrac 12-\tfrac 34e^{-2t}+\tfrac 14e^{-6t}.}$$ The coefficients reproduce the initial state, and each mode satisfies the matrix equation. Since $x_2<1/3$ at finite times, equilibrium is approached but never reached then. The experiments uniquely identify this constant linear transfer matrix; they do not rule out nonlinear or time-dependent models sharing the same three measured initial slopes.