Question 4 of 7: Discrete Growth Model Solved with the Shift (E) Operator

Question 4: Discrete Growth Model Solved with the Shift (E) Operator (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The nonlinear recurrence $y_{{k+2}}\,y_{{k+1}}^{{-3/2}} = y_k^{{-1/2}}$ in the coded concentration $y_k$, the substitution $x_k = \log_a y_k$, and the shift operator $E$. A target value $y_k = 85$.

Find. The observation instant $k$ (or the two consecutive instants that bracket it) at which $y_k$ first reaches 85.

Approach. Take logarithms to linearize the recurrence, solve the resulting linear difference equation with the $E$-operator characteristic equation, then use the closed form to locate the instant where $y_k = 85$.

1. Linearize by logarithms. Taking $\log_a$ of $y_{{k+2}} = y_{{k+1}}^{{3/2}}y_k^{{-1/2}}$ gives a linear constant-coefficient difference equation in $x_k=\log_a y_k$: $$x_{{k+2}} - \tfrac{3}{2}x_{{k+1}} + \tfrac{1}{2}x_k = 0.$$
2. Characteristic equation (E-operator). Writing the equation as $\big(E^2 - \tfrac{3}{2}E + \tfrac{1}{2}\big)x_k = 0$ and clearing the fraction, $$2E^2 - 3E + 1 = (2E-1)(E-1) = 0 \;\Rightarrow\; E = 1,\ \tfrac{1}{2}.$$
3. General solution. With roots $1$ and $\tfrac12$, $$x_k = A + B\left(\tfrac{1}{2}\right)^k \;\Rightarrow\; y_k = a^{{x_k}} = y_\infty\left(\frac{y_0}{y_\infty}\right)^{{(1/2)^k}},\quad y_\infty = \frac{y_1^2}{y_0},$$ where the result is independent of the base $a$ (it cancels), and $y_\infty = a^A$ is the steady state.
4. Qualitative behaviour. One root equals 1 (a constant, steady mode) and the other is $\tfrac12$ (a decaying transient). Because neither root exceeds 1 in magnitude, the coded concentration does not grow without bound — it approaches the finite steady value $y_\infty = y_1^2/y_0$ monotonically. Whether 85 is actually reached depends on the initial pair $(y_0,y_1)$: it is attained only if 85 lies between $y_0$ and $y_\infty$.
5. Instant where $y_k = 85$. Setting $y_k = 85$ and solving the closed form for $k$, $$\left(\tfrac{1}{2}\right)^k = \frac{\ln(85/y_\infty)}{\ln(y_0/y_\infty)} \;\Rightarrow\; k = -\log_2\!\left[\frac{\ln(85/y_\infty)}{\ln(y_0/y_\infty)}\right].$$ The concentration reaches 85 between instants $\lfloor k\rfloor$ and $\lceil k\rceil$.
6. Worked demonstration. The printed problem gives no initial observations, so — per the exam's instruction to "state any assumption made" — take a representative growing pair $y_0 = 10$, $y_1 = 30$. Then $y_\infty = 30^2/10 = 90$, and $$k = -\log_2\!\left[\frac{\ln(85/90)}{\ln(10/90)}\right] = -\log_2(0.0260) = 5.26.$$ Since $y_5 = 84.0$ and $y_6 = 87.0$, the coded concentration reaches 85 $\;\boxed{\text{between observation instants }k=5\text{ and }k=6}$.