Question

Involve equations of the form $dy/dt=f(y).$ 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(y^{2}-1\right), \quad-\infty

Short answer

Short Answer: The critical points for the given differential equation, $y^2(y^2-1)$, are 0 and $\pm 1$. The equilibrium point 0 is unstable, while the equilibrium points 1 and -1 are asymptotically stable.

Step-by-step solution

Sketch the graph of $f(y)$ versus $y$ for the given differential equation

Create a graph for the given differential equation, $f(y)=\frac{dy}{dt}=y^2(y^2-1)$. This function is a polynomial of degree 4, specifically a quartic function, which means it will have up to 4 roots. You can plot it using any graphing calculator or software.

Determine the critical (equilibrium) points

Critical or equilibrium points are the points where $\frac{dy}{dt}=0$. In this case, 0 = $y^2(y^2-1)$. We can find these points by setting each factor equal to zero: (1) $y^2=0$, which implies $y=0$. (2) $y^2-1=0$, which implies $y^2=1$. Hence, $y=\pm 1$. So, the critical points are 0 and $\pm 1$.

Classify each critical point as asymptotically stable, unstable, or semi-stable

To determine the stability of each critical point, we need to analyze the behavior of $f(y)$ for all critical points in $y$. (a) For $y=0$, 1. If $y<0$, then $f(y)>0$ (right side of the differential equation is positive), so $\frac{dy}{dt}>0$, implying that $y$ is increasing over time. 2. If $0<y<1$, then $f(y)>0$ (right side of the differential equation is positive), so $\frac{dy}{dt}>0$, implying that $y$ is increasing over time. Since in the region of $y=0$, $\frac{dy}{dt}>0$ always holds, $y=0$ is an unstable equilibrium point. (b) For $y=1$, 1. If $0<y<1$, then $f(y)>0$ (right side of the differential equation is positive), so $\frac{dy}{dt}>0$, implying that $y$ is increasing over time. 2. If $y>1$, then $f(y)<0$ (right side of the differential equation is negative), so $\frac{dy}{dt}<0$, implying that $y$ is decreasing over time. As the behavior in the surrounding regions has opposite trends, $y=1$ is an asymptotically stable equilibrium point. (c) For $y=-1$, 1. If $y<-1$, then $f(y)<0$ (right side of the differential equation is negative), so $\frac{dy}{dt}<0$, implying that $y$ is decreasing over time. 2. If $-1<y<0$, then $f(y)<0$ (right side of the differential equation is negative), so $\frac{dy}{dt}<0$, implying that $y$ is decreasing over time. As before, the behavior in the surrounding regions is opposite, so $y=-1$ is an asymptotically stable equilibrium point. In summary, the critical points 1 and -1 are asymptotically stable, while the critical point 0 is unstable.

Key concepts

Equilibrium Points

Equilibrium points are vital when analyzing differential equations, specifically when they describe how a quantity changes over time. In mathematical terms, an equilibrium point is a condition where the derivative of the function, such as $\frac{dy}{dt}$, equals zero. This means that at equilibrium, the system does not change; instead, it remains constant over time. To find these points, we set the function equal to zero and solve for $y$.
In the given exercise, the differential equation is of the form $\frac{dy}{dt}=y^2(y^2-1)$. To identify the equilibrium points, we solve $y^2(y^2-1)=0$, providing us with the points: $y=0,\pm 1$. Each of these corresponds to a condition where the rate of change is zero, making them potential candidates for equilibrium.
The importance of these points goes beyond finding them; understanding their behavior helps determine system stability. At these points, scientists and engineers decide if solutions will remain steady, deviate, or oscillate, forming a foundation for analysis of systems from ecosystems to economics.

Stability Analysis

After locating equilibrium points, we move to stability analysis to determine the nature of each point. Stability tells us how a system behaves following small disturbances. Simply put, a stable system returns to equilibrium after a slight shift, whereas an unstable one moves away. We categorize each equilibrium as asymptotically stable, unstable, or semi-stable.
In our step-by-step solution:

• Unstable: If $y=0$, the surrounding values of $y$ indicate that the system moves away from zero—implications that $y$ increases no matter what slight change occurs. Hence, the equilibrium is unstable.
• Asymptotically Stable: For $y=1$ and $y=-1$, the behavior is opposite. For $y=1$, $f(y)<0$ beyond $y=1$ and $f(y)>0$ within $0<y<1$, guiding the system back to equilibrium. The same logic holds for $y=-1$, ensuring that the system eventually stabilizes in both directions.

Stability sheds light on whether an equilibrium state will hold in the long run or its tendency to change over time. This knowledge assists in fields ranging from flight dynamics to drug dosage models.

Quartic Functions

Quartic functions are polynomial functions of degree four. They come into play often in differential equations, especially when examining higher-order behaviors and multiple equilibrium points.
The function in the exercise, $f(y)=y^2(y^2-1)$, is a quartic because it involves $y^4$ as the highest power of $y$. Such a function can have up to four roots, informing us of its complexity and the potential number of equilibrium points.
These roots, derived in solving $y^2(y^2-1)=0$, give insight into where a system naturally stabilizes. This polynomial structure also affects how the graph of $f(y)$ shapes—sometimes strikingly different from simpler quadratic or cubic functions, presenting intriguing symmetry and inflection points.
Explorations of quartic functions help deepen the understanding of dynamic behavior, allowing for better graphical representation and a clearer visualization of how complex systems evolve.