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, unstable, or semistable (see Problem 7 ). $$ d y / d t=y^{2}\left(4-y^{2}\right), \quad-\infty

Short answer

Question: Given the function \(f(y) = y^2(4 - y^2)\), determine the stability of its equilibrium points. Answer: The equilibrium points of the function are \(y=-2\), \(y=0\), and \(y=2\). Based on the behavior of the solutions around these points, \(y=-2\) and \(y=2\) are asymptotically stable, while \(y=0\) is unstable.

Step-by-step solution

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

To sketch the graph of \(f(y) = y^2(4-y^2)\), first observe that this is a quartic function (fourth degree polynomial) with a negative leading coefficient. Also note that it's symmetric with respect to the \(y\)-axis, meaning it's even. Therefore, we expect the plot to have two extremal points with a single root between them. Now, let's find the points where \(f'(y) = 0\), as these are potential extremal points (maxima or minima). $$ f'(y) = \frac{d}{dy} (y^2(4-y^2)) = 2y(4-2y^2) $$ Setting \(f'(y) = 0\), we get \(2y(4-2y^2) = 0\). So, the points of interest are \(y=0\) and \(y=\pm\sqrt{2}\). We can now sketch the graph of \(f(y)\) versus \(y\).

Determine the critical (equilibrium) points.

The critical points or equilibrium points are the values of \(y\) for which \(f(y)=0\). From the function \(f(y)=y^2(4-y^2)\), we can find the equilibrium points by setting \(f(y)\) equal to zero and solving for \(y\): $$y^2(4-y^2) = 0$$ The solutions of this equation are \(y=0\), \(y=2\), and \(y=-2\).

Classify each critical point as asymptotically stable, unstable, or semistable.

To classify the stability of the equilibrium points, we will analyze the behavior of the solutions \(y(t)\) as \(t\rightarrow\infty\). The equation \(dy/dt = f(y)\) gives us information on the sign of the solutions. Specifically, - if \(dy/dt > 0\), \(y\) is increasing; - if \(dy/dt < 0\), \(y\) is decreasing; - if \(dy/dt = 0\), \(y\) is at a critical point. For \(y < -2\), \(f(y) > 0\), so the derivative is positive and \(y\) is increasing. For \(-2 0\), so the derivative is positive and \(y\) is increasing. For \(2<y\), \(f(y) < 0\), so the derivative is negative and \(y\) is decreasing. Now, we can classify the critical points based on the function behavior around them: - \(y=-2\): As \(y\) is increasing to the left of this point and decreasing to the right, this point is asymptotically stable. - \(y=0\): As \(y\) is decreasing to the left of this point and increasing to the right, this point is unstable. - \(y=2\): As \(y\) is increasing to the left of this point and decreasing to the right, this point is asymptotically stable.

Key concepts

Critical Points

When analyzing differential equations, identifying the critical points is crucial. These are the values at which the rate of change of the system, given by the derivative, is zero. In the context of our equation, this translates to points on the graph where the slope of the tangent line is horizontal. Mathematically, we determine these by setting the derivative of our function to zero (in our case, setting \(dy/dt\) to zero). By calculating these points, we gain insights into the behavior of a dynamic system without solving the entire equation. These points are where the system can be in 'equilibrium,' meaning it doesn't change unless perturbed.

For example, if we look at our quartic function \(f(y) = y^2(4 - y^2)\), the critical points occur where this function intersects the y-axis. They serve as the foundation to understand how a system behaves over time and are indispensable in studying the system's long-term behavior.

Equilibrium Solutions

Equilibrium solutions, often intertwined with the concept of critical points, are where a system rests when no external forces are applied. These points on the function graph represent the states where the dynamic system, once arrived at them, will remain indefinitely in the absence of any disturbances. Equilibrium solutions can be stable, unstable, or semistable depending on their surrounding conditions.

To illustrate, for the given differential equation \(dy/dt = y^2(4 - y^2)\), the equilibrium solutions are found when the equation equals zero, revealing the points \(y = 0\), \(y = 2\), and \(y = -2\). Knowing the equilibrium solutions allows us to predict the points towards which a system might naturally evolve or the points it may try to avoid.

Quartic Function

In mathematics, a quartic function is a fourth-degree polynomial, a type of function with significant implications in various areas, especially in differential equations. They often represent more complex behaviors than quadratic or cubic functions, including multiple extrema and points of inflection.

Take the function \(f(y) = y^2(4 - y^2)\) as an example; this quartic function helps us visualize the dynamic system's potential behavior. By studying its shape and turning points, we can intuit how a variable will change over time concerning its derivative. Its graph provides a clear visual representation of where the function increases or decreases, and thus, where the system's critical and equilibrium points are positioned.

Asymptotically Stable

An asymptotically stable point in the context of differential equations is one where, if the system is disturbed slightly from the equilibrium position, it will return to it over time as \(t \rightarrow \infty\). This property is significant for determining the long-term behavior of dynamic systems, indicating a structure that self-corrects in the face of small disruptions.

For instance, in the differential equation \(dy/dt = y^2(4 - y^2)\), the equilibrium points \(y=-2\) and \(y=2\) are classified as asymptotically stable. As seen from the behavior of the function around these points, any small deviation will result in the system eventually settling back to these points. This characterizes a system that, despite perturbations, inherently tends to return to a stable state.