Problem 28 - 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. $$ \frac{d y}{d x}=\frac{y e^{y}-9 y}{e^{y}} $$

Step-by-Step Solution

Answer

Critical points: \( y = 0 \) (unstable), \( y = \ln 9 \) (asymptotically stable).

Step 1: Simplify the Differential Equation

The given differential equation is \( \frac{d y}{d x} = \frac{y e^y - 9y}{e^y} \). Simplify by factoring the expression:\[ \frac{d y}{d x} = \frac{y(e^y - 9)}{e^y} = y \cdot \left(1 - \frac{9}{e^y}\right) \]This gives us the simplified form: \( \frac{d y}{d x} = y \cdot \left( 1 - \frac{9}{e^y} \right) \).

Step 2: Determine Critical Points

Critical points occur where \( \frac{d y}{d x} = 0 \). This means:\[ y \cdot \left( 1 - \frac{9}{e^y} \right) = 0 \]From this equation, the critical points are determined by \( y = 0 \) or \( 1 - \frac{9}{e^y} = 0 \).Solving \( 1 - \frac{9}{e^y} = 0 \) gives \( e^y = 9 \) which implies \( y = \ln 9 \).Thus, the critical points are \( y = 0 \) and \( y = \ln 9 \).

Step 3: Analyze Stability of Critical Points

To classify the stability, examine the sign of \( \frac{d y}{d x} \) around each critical point:- **For** \( y = 0 \): If \( y > 0 \), \( \frac{d y}{d x} = y \cdot (1 - \frac{9}{e^y}) < 0 \) (since \( 1 - \frac{9}{e^y} < 0 \)). If \( y < 0 \), \( \frac{d y}{d x} > 0 \). Thus, \( y = 0 \) is unstable.- **For** \( y = \ln 9 \): If \( y > \ln 9 \), \( \frac{d y}{d x} > 0 \) (since \( 1 - \frac{9}{e^y} > 0 \)). If \( y < \ln 9 \), \( \frac{d y}{d x} < 0 \). Thus, \( y = \ln 9 \) is asymptotically stable.

Step 4: Sketch the Phase Portrait

Draw the \(y\) axis and mark the critical points \( y = 0 \) and \( y = \ln 9 \). Use arrows to indicate the direction of \( \frac{d y}{d x} \):- For \( y < 0 \): Arrows point right (increasing).- Between \( y = 0 \) and \( y = \ln 9 \): Arrows point left (decreasing).- For \( y > \ln 9 \): Arrows point right (increasing).This indicates solutions starting from above \( y = \ln 9 \) or below \( y = 0 \) will move towards \( y = \ln 9 \), while near \( y = 0 \) solutions diverge from this point. This confirms the stability classification.

Key Concepts

Critical Points

In the study of differential equations, critical points are pivotal to understanding how a function behaves. They are where the derivative of the function equals zero, implying no change at that point. For the equation \( \frac{d y}{d x} = y \cdot \left( 1 - \frac{9}{e^y} \right) \), we find critical points by setting \( \frac{d y}{d x} = 0 \). This gives us the equation \( y \left( 1 - \frac{9}{e^y} \right) = 0 \). From this, the critical points are determined by \( y = 0 \) and \( e^y = 9 \), and solving for \( y \) gives us \( y = \ln 9 \). These are the points where changes in the function's behavior are most pronounced.

• \( y = 0 \)
• \( y = \ln 9 \)

Understanding critical points allows for the classification of the system's stability, which leads us to the next important topic, phase portraits.

Phase Portrait

A phase portrait provides a visual representation of the solutions to a differential equation. It helps in illustrating how solutions progress over time in relation to their critical points. In our scenario, the critical points are at \( y = 0 \) and \( y = \ln 9 \). When sketching a phase portrait, you want to denote these points on the \( y \) axis and show the direction in which the solutions move.First, for values less than \( y = 0 \), the derivative \( \frac{d y}{d x} \) is positive, indicating that the values increase toward zero.
Between \( y = 0 \) and \( y = \ln 9 \), the solution decreases, as \( \frac{d y}{d x} \) becomes negative.
For \( y > \ln 9 \), the derivative is again positive, so values increase indefinitely.

• Arrows indicating direction are crucial.
• Label the critical points clearly.
• Observe the direction of flow of solutions.

Stability Analysis

Stability analysis tells us how perturbations in the system behave near the critical points. Essentially, it classifies whether these points attract or repel nearby solutions. For this equation, we have two critical points: \( y = 0 \) and \( y = \ln 9 \). Here's how to classify them:- At \( y = 0 \): When \( y > 0 \), since \( 1 - \frac{9}{e^y} < 0 \), \( \frac{d y}{d x} \) is negative, suggesting solutions move away, and similar divergence occurs for \( y < 0 \). Therefore, \( y = 0 \) is considered an unstable critical point.- At \( y = \ln 9 \): When \( y > \ln 9 \), \( 1 - \frac{9}{e^y} > 0 \), indicating \( \frac{d y}{d x} > 0 \); when \( y < \ln 9 \), \( \frac{d y}{d x} < 0 \). In both cases, solutions approach \( y = \ln 9 \), making it an asymptotically stable point.Understanding stability assesses the long-term behavior of solutions, which is key for differential equations.

Autonomous Differential Equations

Autonomous differential equations are those in which the independent variable, often time \( t \), does not appear explicitly in the equation. This characteristic means the equation is expressed solely in terms of the dependent variable and its derivatives.
Our example, \( \frac{d y}{d x} = y \cdot \left( 1 - \frac{9}{e^y} \right) \), is a first-order autonomous equation because \( x \) is not part of the equation itself.

• The time-independence property leads to constant behavior regardless of where and when the process starts.
• The phase space formed is essential for visualizing the trajectory of solutions.
• Studying autonomous equations helps predict how systems naturally evolve over time.

Such equations have widespread applications, especially when processes inherently depend only on the current state and not on the sequence of time evolution.