Question

In Exercises \(21-30\), perform steps i) and ii) for the linear system $$ \mathbf{x}^{\prime}=\mathbf{A x}, $$ where \(\mathbf{A}\) is the given matrix. i) Find the type of the equilibrium point at the origin. Do this without the aid of any technology or pplane6. You may, of course, check your answer using pplane. ii) Use pplane6 to plot several solution curves - enough to fully illustrate the behavior of the system. You will find it convenient to use the "linear system" choice from the Gallery menu. If the eigenvalues are real, use the "Keyboard input" option to include straight line solutions starting at \(\pm 10\) times the eigenvectors. If the equilibrium point is a saddle point, compute and plot the separatrices. $$ \left[\begin{array}{rr} -1 & 0 \\ 3 & -3 \end{array}\right] $$

Short answer

The equilibrium point at the origin is a stable node.

Step-by-step solution

Identify matrix \( \mathbf{A} \)

The matrix \( \mathbf{A} \) given in the problem is \( \begin{bmatrix} -1 & 0 \ 3 & -3 \end{bmatrix} \). Our goal is to analyze this system at the equilibrium point \((0,0)\).

Find Eigenvalues

To find the eigenvalues, we solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). The identity matrix \( \mathbf{I} \) for our system is \( \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \). Thus, \( \mathbf{A} - \lambda \mathbf{I} = \begin{bmatrix} -1 - \lambda & 0 \ 3 & -3 - \lambda \end{bmatrix} \). Calculating the determinant gives us \((-1 - \lambda)(-3 - \lambda) = 0\). Solve for \( \lambda \) to find: \(\lambda_1 = -1\), \(\lambda_2 = -3\).

Determine Equilibrium Point Type

The eigenvalues \(\lambda_1 = -1\) and \(\lambda_2 = -3\) are both real and negative. This implies that the equilibrium point at the origin is a **stable node**. This is because negative eigenvalues indicate solutions will decay towards zero.

Plot Phase Plane (with pplane6)

In pplane6, the phase portrait can be plotted by entering the system matrix and using the linear system choice from the Gallery menu. Since the equilibrium point is a stable node, the trajectories will spiral in towards the origin, following the directions determined by the eigenvectors.

Key concepts

Equilibrium Point

In the context of linear systems, the **equilibrium point** is where the system rests naturally, meaning no changes occur over time. For a 2-dimensional linear system like the one in this exercise, the equilibrium point is at the origin \((0,0)\). At this point, the system of differential equations sets to zero, indicating a stationary state. To determine the type of equilibrium point, we examine the eigenvalues of the system's matrix. Specifically, the signs of the eigenvalues tell us about the behavior of system trajectories near this point:

• If all eigenvalues are negative, as in our example, the equilibrium point is a **stable node**. All trajectories converge to this point, implying decay towards it over time.
• Positive eigenvalues indicate an **unstable node**, where trajectories diverge away from the point.
• If eigenvalues have mixed signs, the equilibrium point could be a **saddle point**.

In our exercise, both eigenvalues are negative, confirming a stable node. Understanding these types helps predict system behavior without plotting every trajectory.

Eigenvalues

**Eigenvalues** are critical in determining the stability and nature of a linear system's equilibrium point. They are derived from the system matrix via the characteristic equation. In this exercise, the matrix is \[\begin{bmatrix} -1 & 0 \ 3 & -3 \end{bmatrix}.\]To calculate the eigenvalues, we subtract \( \lambda \mathbf{I} \) from \(\mathbf{A} \), where \(\lambda \) is a scalar and \(\mathbf{I} \) is the identity matrix. Solving the determinant of this modified matrix gives us the eigenvalues.In this case, solving the determinant of \(\mathbf{A} - \lambda \mathbf{I} = 0\) results in \(\lambda_1 = -1\)and \(\lambda_2 = -3\).The negative signs of both eigenvalues are important. They indicate that over time, the system will stabilize at the equilibrium point, making the origin a stable node. The concept of eigenvalues extends to various fields, providing insight into system dynamics quickly and succinctly.

Phase Plane Analysis

**Phase Plane Analysis** provides a visual representation of a system's dynamics. It maps out the trajectories of differential equations on a coordinate plane, offering an intuitive grasp of the system's behavior.For our exercise, plotting this for the matrix \(\mathbf{A} = \begin{bmatrix} -1 & 0 \ 3 & -3 \end{bmatrix}\) reveals a stable node at the origin. The direction and curvature of trajectories illustrate how solutions approach this point over time.In practice, we use tools like pplane6 to automatically generate phase portraits. This involves feeding the system matrix into the software and instructing it to display the trajectories. The trajectories will indicate paths solutions take as time progresses:

• Solutions spiral or curve depending on eigenvectors and their directions.
• Trajectories heading towards the origin confirm stability.
• Systems with complex eigenvalues show spiraling behavior.

Phase plane analysis is invaluable for predicting long-term behavior without solving the differential equations manually. It simplifies understanding how different initial conditions can evolve within the system.