Problem 41 - Advanced Engineering Mathematics

Question

(a) Solving \(r-k x=0\) for \(x\) we find the equilibrium solution \(x=r / k\). When \(x0\) and when \(x>r / k, d x / d t<0 .\) From the phase portrait we see that \(\lim _{t \rightarrow \infty} x(t)=r / k\). (b) From \(d x / d t=r-k x\) and \(x(0)=0\) we obtain \(x=r / k-(r / k) e^{-k t}\) so that \(x \rightarrow r / k\) as \(t \rightarrow \infty .\) If \(x(T)=r / 2 k\) then \(T=(\ln 2) / k\).

Step-by-Step Solution

Answer

The stable equilibrium is \( x = \frac{r}{k} \) with \( T = \frac{\ln 2}{k} \) when \( x(T) = \frac{r}{2k} \).

Step 1: Set up the differential equation

Start with the given differential equation: \( \frac{dx}{dt} = r - kx \). We need to find the equilibrium solution where the rate of change is zero.

Step 2: Find the equilibrium solution

For equilibrium, set \( \frac{dx}{dt} = 0 \), which gives us the equation \( r - kx = 0 \). Solving for \( x \), we get \( x = \frac{r}{k} \).

Step 3: Analyze the stability around the equilibrium

For \( x < \frac{r}{k} \), \( \frac{dx}{dt} > 0 \), meaning \( x \) is increasing. For \( x > \frac{r}{k} \), \( \frac{dx}{dt} < 0 \), meaning \( x \) is decreasing. This indicates that \( x = \frac{r}{k} \) is a stable equilibrium.

Step 4: Solve the differential equation with initial condition

Given \( \frac{dx}{dt} = r - kx \) with the initial condition \( x(0) = 0 \), we solve this first-order linear differential equation. The solution is \( x(t) = \frac{r}{k} - \frac{r}{k} e^{-kt} \).

Step 5: Limit behavior of the solution

As \( t \rightarrow \infty \), the exponential term \( e^{-kt} \rightarrow 0 \), so \( x(t) \rightarrow \frac{r}{k} \). This confirms the stability at \( x = \frac{r}{k} \).

Step 6: Solve for time when specific solution value reached

We want to determine \( T \) such that \( x(T) = \frac{r}{2k} \). Set \( \frac{r}{2k} = \frac{r}{k} - \frac{r}{k} e^{-kT} \) and solve for \( T \).

Step 7: Solve the equation for T

Rearrange to \( \frac{1}{2} = 1 - e^{-kT} \), which simplifies to \( e^{-kT} = \frac{1}{2} \). Taking the natural logarithm of both sides gives \( -kT = \ln \frac{1}{2} \). Therefore, \( T = \frac{\ln 2}{k} \).

Key Concepts

Equilibrium Solution

An equilibrium solution in the context of differential equations is a condition where the system is at rest. For a differential equation, it means the rate of change of the dependent variable is zero. To find the equilibrium solution, we set the differential equation equal to zero and solve for the variable of interest.
In the case of the differential equation \( \frac{dx}{dt} = r - kx \), the equilibrium condition is when \( r - kx = 0 \). Solving this gives the equilibrium solution \( x = \frac{r}{k} \).
This value represents a state where, once reached, the system no longer changes; the dependent variable \( x \) remains constant over time.
Recognizing equilibrium solutions is crucial when analyzing stability and behavior of systems given by differential equations.

Stability Analysis

Stability analysis examines the behavior of solutions to differential equations near equilibrium points. In this context, we look at how solutions behave when slightly perturbed from the equilibrium value.
For the equation \( \frac{dx}{dt}= r - kx \), we have computed the equilibrium solution \( x = \frac{r}{k} \). To analyze stability, we observe the derivative's sign around this point:

• When \( x < \frac{r}{k} \), we have \( \frac{dx}{dt} > 0 \), indicating that \( x \) is increasing towards \( \frac{r}{k} \).
• When \( x > \frac{r}{k} \), \( \frac{dx}{dt} < 0 \), indicating that \( x \) is decreasing towards \( \frac{r}{k} \).

This means that small deviations from the equilibrium will decay over time, bringing the system back to equilibrium. Hence, \( x = \frac{r}{k} \) is a stable equilibrium.

First-order Linear Differential Equation

First-order linear differential equations are a type where the highest derivative is the first. They have a standard form of \( \frac{dx}{dt} + p(t)x = q(t) \). When \( p(t) \) and \( q(t) \) are constants, the equations are easier to solve.
The equation \( \frac{dx}{dt} = r - kx \) is a simple example of this type. Here, \( p(t) = -k \) and \( q(t) = r \). To solve these equations, one often uses an integrating factor or finds a particular solution and the complementary homogeneous solution.
In our case, a solution of the form \( x(t) = \frac{r}{k} - \frac{r}{k} e^{-kt} \) is obtained, combining both transient and steady-state behaviors.

Phase Portrait

A phase portrait is a visual representation of the trajectories of a dynamical system in the phase plane. It allows for an intuitive understanding of how solutions evolve over time.
For the differential equation \( \frac{dx}{dt} = r - kx \), plotting \( x \) against its rate of change across different initial conditions can reveal the system's long-term behavior.
The phase portrait typically shows that all trajectories approach the line \( x = \frac{r}{k} \) as \( t \to \infty \). This trend implies that the solution stabilizes around the equilibrium.

• Below \( x = \frac{r}{k} \), curves rise towards this line.
• Above it, curves descend toward this line.

By analyzing phase portraits, one gains insight into the system's stability and general behavior over time, enhancing the understanding gained from analytical solutions alone.