Question

Give a direct definition of an unstable equilibrium of \(y^{\prime \prime}+p(y)=0\).

Short answer

In the context of the given differential equation, \(y^{\prime \prime} + p(y) = 0\), an unstable equilibrium is an equilibrium point \(y_e\) where the function \(p(y)\) satisfies the conditions: \(y > y_e\) implies \(y^{\prime \prime} < 0\) (concave down), and \(y < y_e\) implies \(y^{\prime \prime} > 0\) (concave up). These conditions lead to nearby trajectories moving away from the equilibrium point over time, causing the equilibrium to be unstable.

Step-by-step solution

Understanding Equilibrium Points and Stability

An equilibrium point for a differential equation is a constant solution that makes it true. For the given second-order differential equation, let's find the equilibrium points by setting \(y^{\prime \prime}\) equal to zero: $$y^{\prime \prime} + p(y) = 0 \implies y^{\prime \prime} = -p(y).$$ In this case, at the equilibrium points, \(y^{\prime \prime} = 0\), so \(-p(y) = 0\). Hence, $$p(y) = 0.$$ An equilibrium point is said to be stable if nearby trajectories return toward it over time, unstable if nearby trajectories move away from it over time, or semi-stable if trajectories either converge to or diverge from the equilibrium point. To determine the stability, we'll need to analyze the behavior of solutions around the equilibrium point.

Finding the Equilibrium Points for the Given Equation

To find the equilibrium points for the given equation, we set $$p(y) = 0.$$ Since \(p(y)\) is an arbitrary function, we cannot solve this further. However, let's denote the equilibrium points as \(y_e\).

Checking the Stability of the Equilibrium Points

To check the stability of the equilibrium points, we will analyze the behavior of the second derivative, i.e., \(y^{\prime \prime} = -p(y)\). At an equilibrium point \(y_e\), the second derivative is zero (\(y^{\prime \prime} = 0\)). For the equilibrium points to be unstable, nearby solutions should have the second derivative with opposite signs as the function moves away from the equilibrium point. If the second derivative is positive, the function is concave up, and if it is negative, the function is concave down. So, to put it in simple words, an equilibrium point \(y_e\) is unstable if whenever \(y > y_e\), we have \(y^{\prime \prime} < 0\) (concave down), and whenever \(y < y_e\), we have \(y^{\prime \prime} > 0\) (concave up). This means that if a solution starts near the equilibrium point, it will move further away from it. Hence, the equilibrium point is unstable.

Providing the Direct Definition of an Unstable Equilibrium

So, a direct definition of an unstable equilibrium of the given equation \(y^{\prime \prime} + p(y) = 0\) is as follows: An equilibrium point \(y_e\) for which the function \(p(y)\) is such that $$y > y_e \implies y^{\prime \prime} < 0$$ and $$y < y_e \implies y^{\prime \prime} > 0,$$ which leads to the nearby trajectories moving away from the equilibrium point over time, making it unstable.

Key concepts

Equilibrium Points

Equilibrium points in the context of differential equations are specific solutions where the system remains constant, or in other words, still over time. For the differential equation \(y^{\prime \prime} + p(y) = 0\), these points are crucial in understanding dynamics.

We identify equilibrium points by setting the acceleration (or the second derivative) to zero, which means that the effect of the function \(p(y)\) is nullified:

• Set \(y^{\prime \prime} = 0\), which implies that \(-p(y) = 0\).
• This leads to \(p(y) = 0\).

Equilibrium points provide essential insights into the behavior of the system defined by the differential equation as they define where the system can potentially stabilize.

Stability Analysis

Stability analysis is a method used to determine if an equilibrium point will attract or repel nearby trajectories over time. It's vital for predicting long-term system behavior.

Understanding whether a system will return to stability or diverge further from it relies on stability analysis:

• Stable equilibrium points attract nearby trajectories, meaning perturbations fade over time.
• Unstable equilibrium points cause nearby trajectories to drift away.
• Semi-stable points might see some trajectories converge while others diverge.

By analyzing the surrounding environment of equilibrium points through differential calculus, specifically using the second derivative, one can conclude the type of stability displayed by the system.

Second-Order Differential Equations

Second-order differential equations, like \(y^{\prime \prime} + p(y) = 0\), involve second derivatives and are essential for modeling acceleration-based systems. These equations often define how a system evolves over time across various fields, such as physics and engineering.

The structure of a second-order differential equation includes:

• \(y^{\prime \prime}\): The acceleration or the second derivative of the function.
• \(p(y)\): A function of \(y\) that influences the rate of change.

While solving these equations, finding equilibrium points and analyzing stability is a common approach, as they help elucidate possible system behaviors and responses to perturbations.

Concavity

In the context of differential equations, concavity of a function provides information on how the graph of the function curves, which is integral for determining stability of equilibrium points. Concavity is directly linked to the sign of the second derivative \(y^{\prime \prime}\):

• A positive second derivative \(y^{\prime \prime} > 0\) implies the function is concave up, resembling a U-shape.
• A negative second derivative \(y^{\prime \prime} < 0\) indicates the function is concave down, like an inverted U.

Concavity helps predict how a solution behaves around equilibrium points. For stability analysis, these signs determine whether a solution at equilibrium will revert to the point or diverge, thus classifying the point as stable or unstable.