Problem 23 - Advanced Engineering Mathematics

Question

Find the critical points and phase portrait of the given autonomous first- order differential equation. Classify each critical point as asymptotically stable, unstable, or semi-stable. By hand, sketch typical solution curves in the regions in the \(x y\) -plane determined by the graphs of the equilibrium solutions. $$ \frac{d y}{d x}=(y-2)^{4} $$

Step-by-Step Solution

Answer

The only critical point is at \( y = 2 \), and it is semi-stable.

Step 1: Identify Equilibrium Solutions

Equilibrium solutions occur where the derivative is zero. For the equation \( \frac{dy}{dx} = (y-2)^4 \), set \( (y-2)^4 = 0 \). This gives the equilibrium solution \( y = 2 \).

Step 2: Determine Stability of Critical Point

A critical point is asymptotically stable if \( f'(y) < 0 \) immediately positive on both sides of the equilibrium, unstable if increasing away from the equilibrium, and semi-stable if \( f'(y) = 0 \). Calculate the derivative \( \frac{d}{dy}(y-2)^4 = 4(y-2)^3 \) at \( y = 2 \), which is \( 0 \), confirming semi-stability.

Step 3: Analyze Behavior Around Critical Point

Since \( \frac{dy}{dx} = (y-2)^4 \) is always non-negative and specifically zero only at \( y = 2 \), solution curves will move towards \( y = 2 \) from either side, specifically approaching but not crossing 2.

Step 4: Phase Portrait Illustration

In the phase portrait, plot \( y = 2 \) as a horizontal line, since it is the equilibrium solution. Draw solution curves that approach \( y = 2 \) from both above and below the line, indicating they never cross \( y = 2 \) due to the behavior of \((y-2)^4\) as positive or zero.

Step 5: Classify Critical Point

Since the critical point does not converge towards \( y = 2 \) nor do solutions diverge away, it is classified as a semi-stable equilibrium.

Key Concepts

Critical Points

In the context of differential equations, finding critical points is an essential step in understanding the system's behavior. Critical points occur where the derivative of the function equals zero. For our equation, \[ \frac{dy}{dx} = (y-2)^4, \] we find a critical point by setting \[ (y-2)^4 = 0. \]This leads us to the critical point at \( y = 2 \). This point is also referred to as an equilibrium solution because the system is "at rest" here — meaning that there is no change occurring at this point. Recognizing these points helps in determining how the system behaves in their vicinity, which is crucial for the overall system analysis.

Phase Portraits

Phase portraits offer a visual illustration of the trajectory of solutions in the plane, showing how they evolve over time. For the equation \[ \frac{dy}{dx} = (y-2)^4, \]the phase portrait is a helpful tool. In this example, the phase portrait portrays the equilibrium solution \( y = 2 \) as a horizontal line. Since \( (y-2)^4 \) is non-negative — and only equals zero exactly at \( y = 2 \) — solution curves will approach this line from either side but will never cross it. These curves in the phase portrait reveal how solutions behave as they tend towards the equilibrium without crossing it, aiding our understanding of the long-term behavior of the system. Remember that in this scenario with a semi-stable point, solution trajectories appear to "stick" to the \( y = 2 \) line without diverging once they touch it.

Equilibrium Solutions

Equilibrium solutions, also known as steady-state solutions, are critical in the study of differential equations. These solutions occur at points where the system does not change, meaning the rate of change of the variable is zero. This can be seen in the equation \[ \frac{dy}{dx} = (y-2)^4. \]The value \( y = 2 \) serves as an equilibrium solution because substituting \( y = 2 \) into the equation results in zero, indicating no change. Solutions near an equilibrium will generally display behaviors dependent on the stability of that point. Identifying these solutions allows us to categorize aspects of the system's dynamics, like stability and visualizing the expected long-term behavior of solutions.

Stability Analysis

Stability analysis reveals the tendency of a system to return to equilibrium after a small perturbation. For our equation \[ \frac{dy}{dx} = (y-2)^4, \] we assess stability by examining the derivative of the function around the equilibrium point.Compute the derivative: \[ \frac{d}{dy}(y-2)^4 = 4(y-2)^3. \]Evaluating this at \( y = 2 \) gives us zero, leading to the conclusion that the point is semi-stable. In this scenario, the system's solutions approach the equilibrium but neither fully converge nor diverge. Semi-stability means that solutions can linger at this equilibrium but are sensitive to initial conditions. The system's response to disturbances will be crucial for practical applications and understanding how stable these critical points are.