Question

Involve equations of the form \(d y / d t=f(y)\). In each problem sketch the graph of \(f(y)\) versus \(y,\) determine the critical (equilibrium) points, and classify each one as asymptotically stable or unstable. $$ d y / d t=y(y-1)(y-2), \quad y_{0} \geq 0 $$

Short answer

Based on the solution, we have determined the critical points and their stability for the given first-order differential equation. The critical points are: 1. \(y=0\) is an unstable equilibrium point. 2. \(y=1\) is an asymptotically stable equilibrium point. 3. \(y=2\) is an unstable equilibrium point. The graph of \(f(y)\) versus \(y\) shows a positive slope when \(y < 0\), a negative slope when \(0 < y < 1\), a positive slope when \(1 < y < 2\), and a negative slope when \(y > 2\). The critical points \((0,0)\), \((1,0)\), and \((2,0)\) in the graph represent their respective equilibria and stability conditions described above.

Step-by-step solution

Find the critical points

The critical points occur when \(\frac{dy}{dt} = f(y) = 0\). Our equation is: $$ \frac{dy}{dt} = y(y-1)(y-2) $$ To find the critical points, we set this expression equal to zero and solve for \(y\): $$ y(y-1)(y-2) = 0 $$ This equation has three solutions: \(y = 0\), \(y = 1\), and \(y = 2\). These are the critical points.

Evaluate the stability of the critical points

To determine the stability of each critical point, we need to evaluate the derivative of \(f(y)\) at the critical points: $$ f'(y) = \frac{d}{dy}(y(y-1)(y-2)) $$ First, let's find \(f'(y)\). To do this, we can use the product rule for differentiation: $$ f'(y) = (y(y-1)(y-2))'+ y'(y-1)(y-2)\\ $$ $$ = (y'(y-1)+y(y-1)')(y-2) \\ $$ $$ = (1(1)(y-1)+y(1)(y-2) + 0(1)(y-1)(y-2)\\ $$ $$ = (y-1)(2y-2) $$ Now, let's evaluate \(f'(y)\) at each critical point: 1. \(y=0\): \(f'(0) = (-1)(-2) > 0\), so \(y=0\) is an unstable critical (equilibrium) point. 2. \(y=1\): \(f'(1) = (0)(0) = 0\), the second derivative test is inconclusive for \(y = 1\). 3. \(y=2\): \(f'(2) = (1)(2) > 0\), so \(y=2\) is an unstable critical (equilibrium) point. For \(y=1\), since the second derivative test is inconclusive, we analyze the behavior of \(f(y)\) near \(y=1\) and observe that \(f(y) > 0\) for \(y < 1\) and \(f(y) < 0\) for \(y > 1\). Therefore, \(y=1\) is an asymptotically stable equilibrium point.

Sketch the graph of \(f(y)\) versus \(y\)

To sketch the graph, we should consider the sign of \(\frac{dy}{dt}\) in different intervals: - For \(y < 0\), \(y(y-1)(y-2) > 0\) and \(\frac{dy}{dt} > 0\). - For \(0 < y < 1\), \(y(y-1)(y-2) < 0\) and \(\frac{dy}{dt} < 0\). - For \(1 < y < 2\), \(y(y-1)(y-2) > 0\) and \(\frac{dy}{dt} > 0\). - For \(y > 2\), \(y(y-1)(y-2) < 0\) and \(\frac{dy}{dt} < 0\). Now, we can sketch the graph of \(f(y)\) versus \(y\). The graph should show that \(f(y)\) is positive when \(y < 0\), negative when \(0 < y < 1\), positive when \(1 < y < 2\), and negative when \(y > 2\). We should also mark the critical points \((0,0)\), \((1,0)\), and \((2,0)\) and their stability: - Unstable critical point at \(y=0\). - Asymptotically stable critical point at \(y=1\). - Unstable critical point at \(y=2\). The graph will resemble a cubic function with a local minimum at \(y=1\) and local maximum at \(y=0\) and \(y=2\).

Key concepts

Equilibrium Points

In the context of differential equations, equilibrium points, also known as critical points, are values where the rate of change of a system is zero. This means that the system is at a standstill concerning that particular variable. In mathematical terms, we find these points by setting the derivative equal to zero and solving for the variable.

Consider the function derived from the differential equation: \[\frac{dy}{dt} = y(y-1)(y-2)\]
To find the equilibrium points, we solve \(y(y-1)(y-2) = 0\). The solutions to this equation are the points where the function is zero, leading us to the equilibrium points \(y = 0\), \(y = 1\), and \(y = 2\).

These points are crucial because they help us understand various states where the system does not change, providing insights into the behavior of the system.

Stability Analysis

Determining whether an equilibrium point is stable or unstable is vital in understanding the behavior of a dynamic system over time. The stability indicates if the system will return to the equilibrium point after a small disturbance or if it will move away from it.

For our differential equation, we take the derivative of the function at each equilibrium point to assess stability: \[f'(y) = \frac{d}{dy}(y(y-1)(y-2))\]Using differentiation rules, we find \(f'(y)\) and then evaluate it at each equilibrium point:

• \(y=0\): \(f'(0) > 0\), unstable because small disturbances cause movements away.
• \(y=1\): \(f'(1) = 0\), further analysis shows it's asymptotically stable as disturbances lead back to it.
• \(y=2\): \(f'(2) > 0\), unstable.

An asymptotically stable point, like \(y=1\), acts like an attractor where solutions tend to settle. In contrast, unstable points repel solutions, causing them to diverge.

Graphical Solutions

Graphical solutions help visualize how a system evolves over time by examining the sign and values of the derivative function. For the given system,\[y(y-1)(y-2)\]we need to evaluate \(\frac{dy}{dt}\) across various intervals of \(y\) to sketch its behavior.

The key insight from the graphical analysis is understanding where \(\frac{dy}{dt}\) is positive or negative, which informs us whether the system is moving upwards or downwards at a particular \(y\):

• \(y < 0\): \(\frac{dy}{dt} > 0\), so the system increases.
• \(0 < y < 1\): \(\frac{dy}{dt} < 0\), indicating a decrease.
• \(1 < y < 2\): \(\frac{dy}{dt} > 0\), showing an increase again.
• \(y > 2\): \(\frac{dy}{dt} < 0\), leading to a decrease.

Drawing this out, the graph resembles a cubic curve with turning points at the equilibrium points, enabling us to visually deduce the stability and behavior of the system around these points. This graphical tool is instrumental in highlighting the dynamic transitions across different regions of \(y\).