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. $$ \left[\begin{array}{rr} 9 & -1 \\ 9 & 3 \end{array}\right] $$

Short answer

The origin is a defective node (improper node) with eigenvalue 6.

Step-by-step solution

Understand the Problem

The system of differential equations is given as \( \mathbf{x}^{\prime}=\mathbf{A} \mathbf{x} \), where \( \mathbf{A} = \begin{bmatrix} 9 & -1 \ 9 & 3 \end{bmatrix} \). We need to find the type of the equilibrium point at the origin by determining the eigenvalues of \( \mathbf{A} \).

Find the Eigenvalues of A

To find the eigenvalues, solve the characteristic equation: \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). \( \mathbf{I} \) is the identity matrix. This gives:\[\begin{vmatrix}9 - \lambda & -1 \9 & 3 - \lambda\end{vmatrix} = (9-\lambda)(3-\lambda) + 9 = 0\]\[\lambda^2 - 12\lambda + 36 = 0\]Solve the quadratic equation to get the eigenvalues.

Solve the Quadratic Equation

The characteristic equation is \( \lambda^2 - 12\lambda + 36 = 0 \). Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), with \( a = 1, b = -12, c = 36 \), compute:\[\lambda = \frac{12 \pm \sqrt{144 - 144}}{2} = \frac{12}{2} = 6\]Both eigenvalues are 6, indicating a repeated, or double, eigenvalue.

Determine the Type of Equilibrium Point

With a repeated eigenvalue of 6, check the matrix \( \mathbf{A} \) for diagonalizability. If it's not diagonalizable, we classify the point based on the Jordan form. Since 6 is real and the eigenvalue algebraic multiplicity (2) equals geometric multiplicity (1 if \( \mathbf{A} \) is not diagonalizable), it's a defective node.

Use Technology for Plotting

Using pplane6 or similar, input the linear system to plot solution curves. Observing straight line solutions will confirm the type of equilibrium as determined without technological aid. A non-diagonalizable system with these eigenvalues will exhibit behavior typical of a proper or improper node allowing verification using the technology.

Key concepts

Eigenvalues

To better understand a linear dynamical system, begin by focusing on eigenvalues. **Eigenvalues** are special numbers associated with a linear transformation represented by a matrix. They provide invaluable insights into the system's behavior.

In our problem, we are working with a matrix \[\mathbf{A} = \begin{bmatrix} 9 & -1 \ 9 & 3 \end{bmatrix}\]To find eigenvalues, we solve the **characteristic equation**, which involves setting up and solving the determinant equation \(\det(\mathbf{A} - \lambda \mathbf{I}) = 0\). This step helps reveal critical properties of the matrix.

For the given problem, solving this ultimately yields \(\lambda = 6\), a **repeated** or **double eigenvalue**. Knowing eigenvalues allows us to move on to deducing the nature of the equilibrium point and foresee the general behavior of the system.

Equilibrium Point

Once eigenvalues are found, our attention shifts to understanding the **equilibrium point**. An equilibrium point for a system \(\mathbf{x}^{\prime} = \mathbf{A}\mathbf{x}\) occurs where \(\mathbf{x}^{\prime} = 0\), leading us to the source, sink, or saddle characteristics of the system.

In this exercise, because the eigenvalue is real and repeated, the equilibrium point at the origin is determined to be a defective node. Equilibrium point analysis without technology helps us predict system stability and possible types of solution behavior.

Typically, **equilibrium points** can be categorized as:

• **Stable Node**: All trajectories converge to the point.
• **Unstable Node**: Trajectories diverge from the point.
• **Center**: All trajectories are periodic.
• **Saddle Point**: Contains both converging and diverging directions.

For our specific case, monitoring these characteristics lends critical insight when plotting solution curves.

Characteristic Equation

The path to finding eigenvalues starts with constructing and solving the **characteristic equation**. This is essentially a polynomial equation derived from the determinant of \(\mathbf{A} - \lambda \mathbf{I}\), where \(\mathbf{I}\) is the identity matrix.

The characteristic equation for our matrix is:\[(9 - \lambda)(3 - \lambda) + 9 = 0\]Simplifying, we arrive at \[\lambda^2 - 12\lambda + 36 = 0\]To solve, use the quadratic formula \(\lambda = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\). This formula efficiently finds eigenvalues for any second-degree polynomial.

In our example, the solution yields a repeated eigenvalue of 6. Recognizing that the characteristic equation provides these insights is crucial for unlocking deeper understanding, influencing our realization of system behavior and equilibrium.

Solution Curves

After determining all theoretical aspects, visualizing system behavior with **solution curves** completes the picture. Solution curves illustrate how trajectories evolve over time within the phase plane of the system.

Using the previously determined eigenvalues and equilibrium point, solution curves can be drawn using tools like pplane6, which plots these trajectories. For our defective node, the curves show paths' convergence and reveal any possible complicated patterns.

Plotting solution curves gives:

• A visual exhibit of system stability.
• A confirmation of theoretical results.
• Diagnosis of critical trajectories such as straight line solutions.

In our problem, the graphs corroborate the equilibrium type and provide an intuitive understanding that complements the numeric and algebraic results, bridging the gap between calculation and conceptual comprehension.