Question

The following two systems both possess zero-valued Jacobians. Construct a phase diagram for each, and deduce the locations of all the equilibrium's that exist: (a) \(x^{\prime}=x+y\) \(y^{\prime}=-x-y\) (b) \(x^{\prime}=0\) \(y^{\prime}=0\)

Short answer

System (a) has a line of equilibrium \(x+y=0\), while system (b) has equilibrium everywhere.

Step-by-step solution

Determine the Jacobian

For the first system, \( x' = x + y \) and \( y' = -x - y \), we first find the Jacobian matrix by calculating the partial derivatives with respect to \(x\) and \(y\). The Jacobian \(J\) is \[J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 1 & 1 \ -1 & -1 \end{bmatrix}.\] The determinant of this Jacobian is zero, indicating potential equilibrium points. Repeat the same for the second system, which trivially gives a zero Jacobian as all derivatives are zero.

Find Equilibrium Points for System (a)

An equilibrium point occurs where \(x' = 0\) and \(y' = 0\). For the system \(x' = x + y = 0\) and \(y' = -x - y = 0\), we can write these equations as:1. \(x + y = 0\)2. \(-x - y = 0\)Solving these two equations simultaneously, both equations are essentially identical, leading to no unique solution, thus indicating a line of equilibrium along \( x + y = 0 \). This confirms a line of equilibrium along which any point can satisfy both equations.

Find Equilibrium Points for System (b)

For system (b) \(x' = 0\) and \(y' = 0\), the system does not contain \(x\) and \(y\) terms, indicating that every point in the phase space is an equilibrium. This forms an entire plane of equilibrium.

Construct Phase Diagram for System (a)

Using the line of equilibrium \(x + y = 0\), plot the line in the phase diagram. Vector field analysis around this line shows behavior increasing or decreasing perpendicular to the line. Stability analysis indicates solutions neither converge nor diverge rapidly from this line.

Construct Phase Diagram for System (b)

System (b) simply indicates that the entire plane is in equilibrium. The phase diagram is filled with arrows representing no movement at any point, thus it is in equilibrium stationary state everywhere.

Key concepts

Phase Diagram

A phase diagram is an essential tool for visualizing the behavior of dynamic systems. It uses arrows, known as vectors, to represent the state of a system at various points in the phase space, which in our context are the values of variables like \(x\) and \(y\).

These diagrams allow you to see how the system evolves from one state to another over time. For instance, in system (a) from the exercise, the line \(x + y = 0\) acts as an equilibrium. When we draw it on a phase diagram, any combination of \(x\) and \(y\) that satisfies this line doesn't change over time, as both derivatives are zero.

For system (b), as every point in the field where \(x' = 0\) and \(y' = 0\), the whole diagram is filled with zero vectors. This shows no dynamics, meaning the entire plane is an equilibrium where nothing changes.

Jacobian Matrix

The Jacobian matrix is a fundamental concept when analyzing systems of differential equations. It is a matrix of first-order partial derivatives of the system's functions. In simpler terms, it provides information about how the functions change with respect to each of their variables.

For example, in system (a) from the exercise, the Jacobian matrix is calculated as \[ J = \begin{bmatrix} \frac{\partial x'}{\partial x} & \frac{\partial x'}{\partial y} \ \frac{\partial y'}{\partial x} & \frac{\partial y'}{\partial y} \end{bmatrix} = \begin{bmatrix} 1 & 1 \ -1 & -1 \end{bmatrix}. \]

The determinant of this matrix is zero. This often indicates potential equilibrium points, as the linear approximation provided by the Jacobian cannot be used to predict behavior beyond the proximity of equilibrium, necessitating further phase analysis.

System of Differential Equations

Systems of differential equations, like the ones in the exercise, involve finding solutions that describe how variables change over time. This pair of equations relates changes in \(x\) and \(y\) over time to their current values.

For instance, in system (a), the equations \(x' = x + y\) and \(y' = -x - y\) depict how changes in one variable influence the other. When analyzing such systems, you seek equilibrium points where these changes stabilize, meaning \(x' = 0\) and \(y' = 0\).

In system (b), \(x' = 0\) and \(y' = 0\) indicate that changes in \(x\) and \(y\) are constant, suggesting an entirely static system where every point is an equilibrium.

Stability Analysis

Stability analysis helps determine how a system behaves near its equilibrium points. It provides insights into whether small disturbances or changes will dissipate over time or cause the system to diverge away from equilibrium.

In the exercise, system (a) presents a line of equilibrium along \(x + y = 0\). Any perturbation along this line does not diverge, suggesting a neutral kind of stability. Meanwhile, system (b) suggests static stability across its entire phase space since all points are equilibrium.

• If small deviations return to equilibrium, the system is stable.
• If deviations amplify away from equilibrium, the system is unstable.
• In case of no net force towards or away, the stability is neutral as with system (a).