Problem 16 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

How to solve $$ P_{0} \quad \text { known } \quad P_{t+1}-P_{t}=r P_{t}+D e^{k t} $$ a. Find an 'equilibrium' exponential sequence \(E_{t}=C e^{k t}\) such that $$ \begin{aligned} E_{t+1}-E_{t} &=r E_{t}+D e^{k t} \\ C e^{k(t+1)}-C e^{k t} &=r C e^{k t}+D e^{k t} \\ \left(C e^{k}-C-r C\right) e^{k t} &=D e^{k t} \end{aligned} $$ Choose \(C=D /\left(e^{k}-1-r\right)\). b. Subtract the two equations: $$ \begin{aligned} P_{t+1}-P_{t} &=r P_{t}+D e^{k t} \\ E_{t+1}-E_{t} &=r E_{t}+D e^{k t} \end{aligned} $$ to get $$ P_{t+1}-E_{t+1}=(1+r)\left(P_{t}-E_{t}\right) $$ Argue that \(P_{t}-E_{t}=\left(P_{0}-E_{0}\right)(1+r)^{t}\) c. Conclude that $$ P_{t}=\frac{D}{e^{k}-1-r} e^{k t}+\left(P_{0}-\frac{D}{e^{k}-1-r}\right)(1+r)^{t} $$

Step-by-Step Solution

Answer

Find the sequence \(C = \frac{D}{e^k - 1 - r}\), derive \(P_{t+1} - E_{t+1} = (1 + r)(P_t - E_t)\), and combine results to express \(P_t\).

Step 1: Identify the Equilibrium Sequence

Given the equilibrium exponential sequence \(E_t = C e^{kt}\), we need to find \(C\) such that it satisfies the equation: \(E_{t+1} - E_t = r E_t + D e^{kt}\). This leads to the equation: \((Ce^k - C - rC)e^{kt} = De^{kt}\). By dividing through by \(e^{kt}\), we obtain \(C(e^k - 1 - r) = D\). Solving for \(C\) gives \(C = \frac{D}{e^k - 1 - r}\).

Step 2: Subtract Equilibrium from Original Equation

Given the equations: \(P_{t+1} - P_t = rP_t + De^{kt}\) and \(E_{t+1} - E_t = rE_t + De^{kt}\), we subtract the second from the first to obtain \(P_{t+1} - E_{t+1} = r(P_t - E_t) + De^{kt} - De^{kt}\). Simplifying this, we arrive at: \(P_{t+1} - E_{t+1} = (1 + r)(P_t - E_t)\).

Step 3: Solve the Homogeneous Equation for the Difference

From the equation \(P_{t+1} - E_{t+1} = (1 + r)(P_t - E_t)\), we recognize a homogeneous linear difference equation of the form \(x_{t+1} = (1 + r)x_t\), where \(x_t = P_t - E_t\). The solution to this is \(x_t = x_0(1 + r)^t\), where \(x_0 = P_0 - E_0\). Thus, \(P_t - E_t = (P_0 - E_0)(1 + r)^t\).

Step 4: Combine Results to Solve for \(P_t\)

Since we have \(P_t = E_t + (P_t - E_t)\) and we know \(E_t = \frac{D}{e^k - 1 - r}e^{kt}\) and \(P_t - E_t = (P_0 - E_0)(1 + r)^t\), we substitute to get \(P_t = \frac{D}{e^k - 1 - r}e^{kt} + (P_0 - \frac{D}{e^k - 1 - r})(1 + r)^t\).

Key Concepts

Equilibrium Exponential Sequence

An equilibrium exponential sequence is a concept often used in solving difference equations like the one presented in the exercise. It involves finding a particular solution that maintains a consistent exponential change over time. This sequence is typically expressed in the form \(E_t = C e^{kt}\), where \(C\) is a constant coefficient and \(k\) is the exponential growth rate.

To find this sequence, you need to satisfy the balance equation \(E_{t+1} - E_t = rE_t + De^{kt}\). By doing so, you determine the value of \(C\). To solve for \(C\), set the equation into a form where both sides involving terms with \(e^{kt}\) can be equalized. In the exercise, we arrived at \((Ce^k - C - rC)e^{kt} = De^{kt}\), and upon simplifying, the solution for \(C\) was determined as \(C = \frac{D}{e^k - 1 - r}\).

This solution reflects the nature of equilibrium in the system, where \(E_t\) evolves regularly according to the exponential factor included.

Homogeneous Equations

In the solving process, we encounter a homogeneous equation, which is a type of difference equation where each term is a function of the dependent variable, but there are no constant or other independent terms in the equation. These types of equations take the form \(x_{t+1} = ax_t\), where \(a\) is a constant coefficient. In this problem, it was related to \(P_{t+1} - E_{t+1} = (1+r)(P_t - E_t)\).

Solving homogeneous equations helps determine how the difference between the actual sequence \(P_t\) and the equilibrium \(E_t\) evolves. Here, finding the solution involves recognizing that the structure \(P_{t+1} - E_{t+1} = (1+r)(P_t - E_t)\) is a classic linear homogeneous equation. Consequently, the solution pattern \(x_t = x_0 (1+r)^t\) can be applied. By setting \(x_t = P_t - E_t\), we find that \(x_0\), the initial difference at \(t=0\), evolves according to the factor \((1+r)^t\).

This relationship illustrates how an initial difference in values can either grow or decay depending on the value of \(1+r\). If \(1+r > 1\), the difference will increase over time, whereas if \(1+r < 1\), it will decrease, indicating stability or convergence.

Difference Equation Solutions

The concept of solving difference equations is pivotal in understanding the dynamic behavior of sequences. Difference equations describe how quantities evolve over discrete time intervals, differing from continuous time equations that use derivative functions.

In this exercise, the purpose is to find an expression for \(P_t\) by combining both the equilibrium sequence and the homogeneous equation solution. The step-by-step solution achieved this by stating that \(P_t = E_t + (P_t - E_t)\). Knowing both parts individually allows us to construct a complete solution.

Specifically, using \(E_t = \frac{D}{e^k - 1 - r}e^{kt}\) and \(P_t - E_t = (P_0 - E_0)(1+r)^t\), we substitute these into the expression for \(P_t\):

• \(E_t\): Represents the steadily growing equilibrium component.
• \(P_t - E_t\): Captures the deviation component that grows or shrinks over time.

Combining them gives:\[ P_t = \frac{D}{e^k - 1 - r}e^{kt} + \left(P_0 - \frac{D}{e^k - 1 - r}\right)(1+r)^t \]

This solution reflects both the persistent exponential growth modeled by the equilibrium sequence and any initial disparities that affect future outcomes depending on \((1+r)^t\). This dual approach is essential for predicting future states accurately.