Problem 12 - Differential Equations and Linear Algebra

Question

Determine the general solution to the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\) $$\left[\begin{array}{rrrr} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 2 & 1 \\ 0 & 1 & 0 & 2 \end{array}\right]$$

Step-by-Step Solution

Answer

The general solution to the system \(\mathbf{x'}=A\mathbf{x}\) with the given matrix \(A\) is: \[\mathbf{x}(t) = c_1e^{3t}\begin{bmatrix}1\\2\\1\end{bmatrix} + c_2e^{(-1+\sqrt{2})t}\begin{bmatrix}1\\-1+\sqrt{2}\\1\end{bmatrix} + c_3e^{(-1-\sqrt{2})t}\begin{bmatrix}1\\-1-\sqrt{2}\\1\end{bmatrix}\] where \(c_1\), \(c_2\), and \(c_3\) are constants determined by the initial conditions.

Step 1: Finding the eigenvalues

To find the eigenvalues, we need to solve the characteristic equation which is given by: \[\det(A - \lambda I) = 0\] Where \(A\) is the given matrix, \(\lambda\) represents the eigenvalues, and \(I\) is the identity matrix. Subtracting \(\lambda I\) from matrix A, we get: \[\left[\begin{array}{ccc} 3-\lambda & 0 & -1 \\ 0 & -3-\lambda & -1 \\ 0 & 2 & -1-\lambda \end{array}\right]\] Now, calculate the determinant and set it to zero: \[\det(A-\lambda I) = (3-\lambda)((-3-\lambda)(-1-\lambda)-(-1)(2)) = 0\] This simplifies to: \[(\lambda - 3)(\lambda^2 + 2\lambda - 1) = 0\] The eigenvalues are \(\lambda = 3, \lambda_{1,2} = -1 \pm \sqrt{2}\).

Step 2: Finding the eigenvectors

Now we will find the eigenvectors for each eigenvalue by solving the equation \((A - \lambda I)\mathbf{v} = 0\), where \(\mathbf{v}\) represents the eigenvector. For \(\lambda = 3\), we have: \[\left[\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -6 & -1 \\ 0 & 2 & -4 \end{array}\right]\mathbf{v} = \mathbf{0}\] Let's use the third-row equation to find a relation between the components of the eigenvector \(\mathbf{v} = \begin{bmatrix} x\\y\\z \end{bmatrix}\): \[2y - 4z = 0\implies y=2z\] Now we can choose \(\mathbf{v}=\begin{bmatrix} 1\\2\\1 \end{bmatrix}\) as the eigenvector corresponding to \(\lambda = 3\). For \(\lambda_{1,2}=-1\pm \sqrt{2}\), we do the same process: \[\left[\begin{array}{ccc} 4\mp\sqrt{2} & 0 & -1 \\ 0 & -2\mp\sqrt{2} & -1 \\ 0 & 2 & 2\pm\sqrt{2} \end{array}\right]\mathbf{v}_{1,2} = \mathbf{0}\] It is easy to see that the eigenvectors for these eigenvalues are \(\mathbf{v}_1 = \begin{bmatrix} 1 \\ -1+\sqrt{2} \\ 1 \end{bmatrix}\) and \(\mathbf{v}_2 = \begin{bmatrix}1 \\ -1-\sqrt{2}\\ 1 \end{bmatrix}\).

Step 3: Constructing the general solution

The general solution to the system \(\mathbf{x'}=A\mathbf{x}\) can be written as: \[\mathbf{x}(t) = c_1e^{\lambda_1 t}\mathbf{v_1} + c_2e^{\lambda_2 t}\mathbf{v_2} + c_3e^{\lambda_3 t}\mathbf{v_3}\] Using our previous results, we get the general solution: \[\mathbf{x}(t) = c_1e^{(3)t}\begin{bmatrix}1\\2\\1\end{bmatrix} + c_2e^{(-1+\sqrt{2})t}\begin{bmatrix}1\\-1+\sqrt{2}\\1\end{bmatrix} + c_3e^{(-1-\sqrt{2})t}\begin{bmatrix}1\\-1-\sqrt{2}\\1\end{bmatrix}\]

Key Concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra. They are essential when analyzing matrices, especially in high-dimensional data. An eigenvalue refers to a scalar that indicates how a linear transformation stretches or compresses vectors. Consequently, if there is a non-zero vector so that the transformation only scales it, this vector is called an eigenvector.

• For example, in the equation \(A\mathbf{v} = \lambda\mathbf{v}\), \(\mathbf{v}\) is the eigenvector and \(\lambda\) is the eigenvalue.
• Finding the eigenvalues involves solving a characteristic equation, resulting in values that describe how vectors are transformed.

In the given exercise, we have identified specific eigenvalues \( \lambda = 3, -1 \pm \sqrt{2} \). With each eigenvalue, there is a corresponding eigenvector, which we solve for to fully understand the system's behavior.

Differential Equations

Differential equations involve functions and their derivatives, and they are used to describe various physical phenomena. In practice, when we encounter a situation described by rates of change, differential equations help us model and predict these behaviors.

• In linear algebra, systems of first-order differential equations can often be expressed as matrix equations. This provides a structured way to tackle complex systems.
• The matrix form can transform the analysis into a problem of finding eigenvalues and eigenvectors, which simplifies the problem significantly.

In this exercise, the system \(\mathbf{x}' = A\mathbf{x}\) is a differential equation governed by the matrix \(A\). By finding eigenvalues and eigenvectors, we were able to determine the general solution which involves exponential functions describing how the system behaves over time.

System of Linear Equations

A system of linear equations consists of multiple linear equations involving the same set of variables. Solving these systems is crucial for engineering, physics, and any fields where linear relationships occur.

• Each equation describes a plane, and the solution to the system corresponds to the intersection of all these planes.
• Matrices are a convenient way to represent these systems, allowing for efficient computation and analysis.

When solving for eigenvectors in the exercise, we effectively deal with such systems. Each eigenvalue equation translates into a system of linear equations. Solving these provides insights into the eigenvectors or the directions along which matrices stretch or compress vectors.

Characteristic Equation

The characteristic equation is central to finding eigenvalues. This equation is derived by setting the determinant of \(A - \lambda I\) to zero. Here, \(A\) is our matrix, \(\lambda\) is the eigenvalue, and \(I\) is the identity matrix of the same dimension.

• Solving the characteristic equation yields the eigenvalues, giving us insights into the behavior of a linear transformation.
• A critical step in eigenvalue problems, mastering this process is essential for understanding complex systems in linear algebra.

In our exercise, the characteristic equation \((\lambda - 3)(\lambda^2 + 2\lambda - 1) = 0\) was solved to find the eigenvalues. This foundational step allowed us to further calculate the eigenvectors, completing the understanding needed to describe the system's dynamics.