Problem 10 - Advanced Engineering Mathematics

Question

Find the general solution of the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{lll} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right) \mathbf{X} $$

Step-by-Step Solution

Answer

The general solution is \( \mathbf{X}(t) = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 e^{2t} \) for corresponding eigenvectors.

Step 1: Define the Matrix Equation

The system of differential equations is represented by \( \mathbf{X}' = A\mathbf{X} \), where \( A \) is the coefficient matrix given by \( \begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 1 \end{pmatrix} \). We need to find the general solution, which involves solving for \( \mathbf{X} \).

Step 2: Find Eigenvalues

To find the eigenvalues, solve the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. Calculate \( A - \lambda I = \begin{pmatrix} 1-\lambda & 0 & 1 \ 0 & 1-\lambda & 0 \ 1 & 0 & 1-\lambda \end{pmatrix} \). Then, compute the determinant and find the eigenvalues \( \lambda \).

Step 3: Calculate the Determinant

Compute the determinant of \( A - \lambda I \): \[\det(A - \lambda I) = \begin{vmatrix} 1-\lambda & 0 & 1 \ 0 & 1-\lambda & 0 \ 1 & 0 & 1-\lambda \end{vmatrix} = (1-\lambda)((1-\lambda)^2 - 0) - 0 - (1)(1(1-\lambda)) \]Simplifying gives \((1-\lambda)^2 - 1 = 0\). Solve to find \( \lambda = 0, 2\).

Step 4: Find the Eigenvectors

For each eigenvalue, solve \((A - \lambda I)\mathbf{v} = \mathbf{0} \) to find the corresponding eigenvectors. Starting with \( \lambda = 0 \), solve: \[\begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0}\]Upon solving, find eigenvectors. Repeat the process for \( \lambda = 2 \).

Step 5: Construct the General Solution

Using the eigenvalues and eigenvectors, the general solution is of the form \( \mathbf{X}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} \). Substitute \( \lambda_1 = 0 \) with eigenvector \( \mathbf{v}_1 \), and \( \lambda_2 = 2 \) with eigenvector \( \mathbf{v}_2 \).

Step 6: Verify the Solution

Verify that the solution satisfies the original matrix differential equation \( \mathbf{X}' = A\mathbf{X} \). Differentiate \( \mathbf{X}(t) \) and plug into the equation to check consistency.

Key Concepts

Eigenvalues and Eigenvectors

In linear algebra, eigenvalues and eigenvectors are fundamental concepts in understanding the dynamics of matrix transformations. An eigenvector of a square matrix \( A \) is a non-zero vector \( \mathbf{v} \) such that when \( A \) is multiplied by \( \mathbf{v} \), the result is a scalar multiple of \( \mathbf{v} \). This scalar is called the eigenvalue \( \lambda \). Mathematically, this is expressed as:

• \( A\mathbf{v} = \lambda\mathbf{v} \)

To find eigenvalues, you look for the \( \lambda \) values that satisfy the characteristic equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. After finding the eigenvalues, you solve \( (A - \lambda I)\mathbf{v} = \mathbf{0} \) to find the corresponding eigenvectors. These act as the building blocks for solutions to differential equations involving matrices.

Matrix Differential Equations

Matrix differential equations are crucial in understanding systems that can be modeled by multiple interrelated rates of change. They are represented in the form:

• \( \mathbf{X}' = A\mathbf{X} \)

where \( \mathbf{X} \) is the vector of unknown functions, and \( A \) is a coefficient matrix. Solving these types of equations often involves eigenvalues and eigenvectors. By identifying the eigenvalues and eigenvectors of the matrix \( A \), these equations can be tackled more systematically. The solutions tend to be expressed as combinations of terms involving exponentials, with each term corresponding to a different eigenvalue and eigenvector pair.

Characteristic Equation

The characteristic equation plays a central role in solving matrix differential equations. It is derived from the matrix \( A \) by the equation:

• \( \det(A - \lambda I) = 0 \)

This equation provides the eigenvalues of the matrix. For the given matrix in our problem, substituting \( A = \begin{pmatrix} 1 & 0 & 1 \ 0 & 1 & 0 \ 1 & 0 & 1 \end{pmatrix} \), we find that the characteristic equation is \( (1-\lambda)^2 - 1 = 0 \). Solving it reveals the eigenvalues, which are crucial for solving the system. These eigenvalues, when associated with their corresponding eigenvectors, help in constructing the solution to the differential equation.

General Solution of Differential Equations

The general solution to a system of matrix differential equations incorporates eigenvalues and eigenvectors. After identifying these, the solution can be expressed in the form:

• \( \mathbf{X}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} + \ldots \)

where \( c_1, c_2 \) are constants determined by initial conditions, \( \mathbf{v}_1, \mathbf{v}_2 \) are the eigenvectors, and \( \lambda_1, \lambda_2 \) are the eigenvalues. For the specific exercise, the solution involved eigenvalues \( \lambda = 0 \) and \( \lambda = 2 \), paired with their respective eigenvectors. This form not only solves the system of equations but also provides insight into the behavior of the system over time, like stability and oscillations.