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=-k(y-1)^{2}, \quad k>0, \quad-\infty

Short answer

Short Answer: The function \(f(y) = -k(y-1)^{2}\) represents an inverted parabola with vertex (1, 0). The equilibrium point is \(y = 1\). Since the function \(f(y)\) is negative for both \(y < 1\) and \(y > 1\), the slope of the tangent is negative, and the solutions will move towards the critical point. Therefore, the equilibrium point \(y=1\) is asymptotically stable.

Step-by-step solution

Graph the function \(f(y)\)

To graph the function, we plot the expression \(f(y) = -k(y-1)^{2}\), where k is a positive constant. We know that the graph will be an inverted parabola with vertex at the point (1, 0). Since k is a positive constant, the parabola will open downwards. Due to the value of k, the steepness of the graph might change; however, it doesn't affect the equilibrium point and its stability.

Find the critical points

To find the critical (equilibrium) points, we need to find the values of y for which \(f(y) = 0\). Set \(f(y)\) to zero and solve for y: $$ -k(y - 1)^2 = 0 $$ Divide both sides by \(-k\) (since \(k > 0\)): $$ (y-1)^2 = 0 $$ Therefore, the equilibrium point is \(y=1\).

Analyze the stability of the critical points

To classify the critical point as asymptotically stable, unstable, or semistable, we need to understand how the function behaves around the equilibrium point. In other words, we will evaluate the sign of \(f(y)\) for \(y < 1\) and \(y > 1\). For \(y < 1\), \((y-1)^2 > 0\): $$ f(y) = -k(y-1)^2 < 0 \implies \frac{dy}{dt} < 0 $$ For \(y > 1\), \((y-1)^2 > 0\): $$ f(y) = -k(y-1)^2 < 0 \implies \frac{dy}{dt} < 0 $$ As the function \(f(y)\) is negative for both \(y < 1\) and \(y > 1\), the slope of the tangent is negative, and the solutions of the differential equation will move towards the critical point. This means that the equilibrium point \(y=1\) is asymptotically stable.

Key concepts

Stability Analysis

Stability analysis in differential equations is a fundamental concept for understanding how solutions behave over time. When we examine a dynamical system given by a differential equation, we often want to know if, and how, solutions will settle down to a steady state as time progresses. This steady state is often referred to as an equilibrium point or a critical point.

In the context of the given exercise, stability analysis involves examining the behavior of solutions near the equilibrium point, which is found at the graph's vertex. Here, we look at the sign of the derivative \( dy/dt \) to determine the direction in which the solutions are moving.

By observing that the function \( f(y) = -k(y-1)^2 \) is always negative unless \( y = 1 \) (where it equals zero), we deduce that the solutions approach the equilibrium point from both directions, indicating that it is stable. If the solutions eventually reach and stay at this point, we call it asymptotically stable, meaning that over time, the solutions will get arbitrarily close to the equilibrium point.

Critical Points

A critical point in a differential equation is a value of the independent variable where the derivative of the function is zero or undefined, indicating a potential equilibrium state. For the equation \( dy/dt = -k(y-1)^2 \) with \( k>0 \), we find a critical point by setting the derivative to zero and solving for \( y \).

Specifically, we find that \( (y-1)^2 = 0 \) which simplifies to \( y = 1 \), revealing that this point is where the rate of change of \( y \) is zero. This is a key step to conduct a stability analysis since knowing the location of critical points lets us concentrate on their behavior without solving the entire differential equation universally.

Critical points are essential to determine the qualitative behavior of the system because they can be attractors, repellers, or neutral, influencing how nearby solutions behave.

Phase Plane Analysis

Phase plane analysis is a graphical method used to study the behavior of two-dimensional dynamical systems, usually systems of first-order differential equations. Even though our given exercise involves a single differential equation, the underlying ideas can be similarly applied.

One way to visualize the stability of an equilibrium point on a phase plane is to plot the direction field, which illustrates the slope of the tangent to the solution curves at various points in the plane. Here, because we have a simple one-dimensional case, we look at the sign of the derivative \( dy/dt \) to determine the direction of motion and infer the behavior of solutions as they move towards or away from the critical point.

From the information given by observing the sign of \( dy/dt \), we can draw conclusions about the system's behavior without explicitly solving the equation, which is analogous to visualizing trajectories on a phase plane.

Asymptotic Stability

Asymptotic stability is a stronger form of stability for an equilibrium point in a dynamical system. An equilibrium point is asymptotically stable if solutions that start close enough not only remain close to the equilibrium but also eventually converge to it as time approaches infinity.

In our exercise, this concept is illustrated by the behavior of the solutions of the differential equation \( dy/dt = -k(y-1)^2 \). As the derivative is negative on both sides of the equilibrium point \( y = 1 \), all trajectories lead toward this point, which means that solutions starting near \( y = 1 \) will inevitably be drawn to this value.

This convergence towards the equilibrium point characterizes the asymptotic stability of the point, which is a desirable property in many physical and engineering systems because it implies that the system will return to equilibrium even after small disturbances.