Problem 8 - Differential Equations and Linear Algebra

Question

Solve \(\mathbf{x}^{\prime}=A \mathbf{x}\) by determining \(n\) linearly independent solutions of the form \(\mathbf{x}(t)=e^{A t} \mathbf{v}\). \(A=\left[\begin{array}{rrr}0 & 1 & 3 \\ 2 & 3 & -2 \\ 1 & 1 & 2\end{array}\right] .\) You may assume that \(p(\lambda)=\) \(-(\lambda+1)(\lambda-3)^{2}\).

Step-by-Step Solution

Answer

The general solution to the given system \(\mathbf{x}^{\prime}=A \mathbf{x}\) with matrix \(A = \left[\begin{array}{rrr}0 & 1 & 3 \\ 2 & 3 & -2 \\ 1 & 1 & 2\end{array}\right]\) is: \[\mathbf{x}(t) = c_1 e^{-t}\left(\begin{array}{c}1\\-1\\1\end{array}\right) + c_2 e^{3t}\left(\begin{array}{c}1\\0\\-1\end{array}\right) + c_3 e^{3t}\left(\begin{array}{c}0\\1\\1\end{array}\right)\]

Step 1: Find the eigenvalues of A

Using the characteristic polynomial, we can find the eigenvalues of matrix \(A\). The given characteristic polynomial is: \(p(\lambda) = -(\lambda + 1)(\lambda - 3)^2\) Set \(p(\lambda) = 0\) to find the eigenvalues: \[-(\lambda + 1)(\lambda - 3)^2 = 0\] From this equation, we can see that the matrix A has eigenvalues: \(\lambda_1 = -1\), \(\lambda_2 = 3\) (with algebraic multiplicity 2)

Step 2: Find the eigenvectors for each eigenvalue

For each eigenvalue \(\lambda_i\), we will find the eigenvectors \(\mathbf{v}_i\) such that: \((A - \lambda_i I)\mathbf{v}_i = 0\) a) For \(\lambda_1 = -1\): \((A + I)\mathbf{v}_1 = 0\) \[\left(\begin{array}{ccc}1 & 1 & 3 \\ 2 & 4 & -2 \\ 1 & 1 & 3\end{array}\right)\left(\begin{array}{c}v_{11}\\v_{12}\\v_{13}\end{array}\right) = \left(\begin{array}{c}0\\0\\0\end{array}\right)\] From the augmented matrix, we can find that the eigenvector \(\mathbf{v}_1\) is: \(\mathbf{v}_1=\left(\begin{array}{c}1\\-1\\1\end{array}\right)\) b) For \(\lambda_2 = 3\): \((A - 3I)\mathbf{v}_2 = 0\) \[\left(\begin{array}{ccc}-3 & 1 & 3 \\ 2 & 0 & -2 \\ 1 & 1 & -1\end{array}\right)\left(\begin{array}{c}v_{21}\\v_{22}\\v_{23}\end{array}\right) = \left(\begin{array}{c}0\\0\\0\end{array}\right)\] From the augmented matrix, we can find two linearly independent eigenvectors \(\mathbf{v}_2\) and \(\mathbf{v}_3\): \(\mathbf{v}_2=\left(\begin{array}{c}1\\0\\-1\end{array}\right)\) \(\mathbf{v}_3=\left(\begin{array}{c}0\\1\\1\end{array}\right)\)

Step 3: Build the general solution

Finally, with the eigenvalues and their associated eigenvectors, we can build the general solution of the form: \(\mathbf{x}(t) = c_1 e^{\lambda_1 t}\mathbf{v}_1 + c_2 e^{\lambda_2 t}\mathbf{v}_2 + c_3 e^{\lambda_2 t}\mathbf{v}_3\) Substituting the eigenvalues and eigenvectors, we get the general solution: \[\mathbf{x}(t) = c_1 e^{-t}\left(\begin{array}{c}1\\-1\\1\end{array}\right) + c_2 e^{3t}\left(\begin{array}{c}1\\0\\-1\end{array}\right) + c_3 e^{3t}\left(\begin{array}{c}0\\1\\1\end{array}\right)\]

Key Concepts

Characteristic Polynomial

The characteristic polynomial is a fundamental concept in finding eigenvalues of a matrix. For a given square matrix \(A\), the characteristic polynomial \(p(\lambda)\) is defined as the determinant of the matrix \(A - \lambda I\), where \(I\) is the identity matrix of the same size as \(A\) and \(\lambda\) represents an eigenvalue. It is often expressed in a factored form, which helps in identifying the eigenvalues directly.

• In the given problem, the characteristic polynomial is \(- (\lambda + 1)(\lambda - 3)^2\).
• Setting this polynomial equal to zero, \(p(\lambda) = 0\), allows us to solve for the eigenvalues.
• The roots of the polynomial give the values of \(\lambda\) for which the matrix \(A - \lambda I\) becomes singular (non-invertible).

The eigenvalues derived from this polynomial are crucial for other calculations, such as finding eigenvectors and constructing general solutions to differential equations. Here, the eigenvalues \(\lambda_1 = -1\) and \(\lambda_2 = 3\) are obtained, with \(\lambda_2\) having an algebraic multiplicity of 2.

General Solution of Differential Equations

The general solution of differential equations, especially in systems involving matrices, extends from finding eigenvalues and eigenvectors. When given a linear system of differential equations such as \( \mathbf{x}^{\prime} = A \mathbf{x} \), the general solution can be expressed using the formula \( \mathbf{x}(t) = e^{At} \mathbf{v} \), where \(\mathbf{v}\) is an eigenvector of \(A\) and \(\lambda\) is the corresponding eigenvalue.

• Each distinct eigenvalue contributes an independent component to the solution.
• For repeated eigenvalues, multiple independent eigenvectors contribute to the general solution.
• In this exercise, the general solution is \(\mathbf{x}(t) = c_1 e^{-t}\left( \begin{array}{c}1\-1\1\end{array}\right) + c_2 e^{3t}\left( \begin{array}{c}1\0\-1\end{array}\right) + c_3 e^{3t}\left( \begin{array}{c}0\1\1\end{array}\right)\).

This expression fully encapsulates the behavior of the solution across time \(t\), with \(e^{\lambda t}\) scaling each term appropriately based on its eigenvector \(\mathbf{v}\).

Linearly Independent Solutions

Critical to forming the general solution are the eigenvectors, which must be linearly independent. In linear algebra, a set of vectors is considered linearly independent if no vector in the set can be expressed as a linear combination of the others.

• It is crucial to determine linearly independent eigenvectors, especially when dealing with repeated eigenvalues, to ensure a complete basis for the solution space.
• In the problem, the eigenvectors corresponding to \(\lambda_2 = 3\) must be checked for linear independence. Here, two vectors are found: \(\mathbf{v}_2 = \left( \begin{array}{c}1\0\-1\end{array}\right)\) and \(\mathbf{v}_3 = \left( \begin{array}{c}0\1\1\end{array}\right)\).
• Both vectors are linearly independent, meaning they provide unique dimensions to the solution space attributed to the eigenvalue \(\lambda_2\).

These independent solutions allow the general solution to capture all possible behaviors of the differential system as prescribed by the eigenvalues.