Question

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rrr}{1} & {-1} & {4} \\ {3} & {2} & {-1} \\ {2} & {1} & {-1}\end{array}\right) \mathbf{x} $$

Short answer

Answer: The fundamental matrix \(\mathbf{\Phi}(t)\) is given by: \(\mathbf{\Phi}(t) = \begin{pmatrix}0 & e^{2t} & 2e^{-t} \\ 0 & (2-t)e^{2t} & 2(-t-1)e^{-t} \\ 0 & (-1-t)e^{2t} & (e^{-t}-t-1)\end{pmatrix}\).

Step-by-step solution

Finding Eigenvalues and Eigenvectors

Given the matrix: $$ A = \left(\begin{array}{rrr}{1} & {-1} & {4} \\\ {3} & {2} & {-1} \\\ {2} & {1} & {-1}\end{array}\right) $$ We can find the eigenvalues by solving the following equation: $$ \det(A - \lambda I) = 0 $$ This results in the characteristic equation: $$ -(\lambda - 1)\left[(2-\lambda)(-1-\lambda) + 1\right] + (-1)\left(3(-1-\lambda)+2\right) = 0 $$ Solving this equation, we get the eigenvalues: $$ \lambda_1 = 1, \lambda_2 = 2, \lambda_3 = -1. $$ Next, we find the eigenvectors associated with each eigenvalue by solving the equation: $$ (A - \lambda_k I)\mathbf{v}_k = 0 $$ where \(k=1,2,3\). The eigenvectors are: $$ \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \mathbf{v}_3 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}. $$

Constructing the General Solution

To construct the general solution to the system, we will first form the matrix \(\mathbf{V}(t)\) whose columns are the solutions of the form \(\mathbf{v}_k e^{\lambda_k t}\): $$ \mathbf{V}(t) = \begin{pmatrix} \ e^{t} & \ \ e^{2t} & \ e^{-t} \\ \ e^{t} & \ \ 0 & \ -2e^{-t} \\ \ 0 & \ -e^{2t} & \ \ e^{-t} \\ \end{pmatrix} $$ Next, we will compute the matrix exponential of \(A\), denoted as \(e^{At}\): $$ e^{At} = \mathbf{V}(t) \begin{pmatrix} e^{\lambda_1t} & 0 & 0 \\ 0 & e^{\lambda_2t} & 0 \\ 0 & 0 & e^{\lambda_3t} \end{pmatrix} \mathbf{V}(t)^{-1} $$ The matrix \(\mathbf{V}(t)^{-1}\) can be computed using Gaussian elimination or Cramer's rule: $$ \mathbf{V}(t)^{-1} = \dfrac{1}{t-1}\begin{pmatrix}2 & e^{t} & e^{-t} \\ -1 & 0 & -e^{t}+e^{-t} \\ 0 & -e^{t} & -1-e^{t}+e^{-t}\end{pmatrix} $$ The general solution is then given by: $$ \mathbf{x}(t) = \mathbf{V}(t) \begin{pmatrix} e^{\lambda_1t} \\ e^{\lambda_2t} \\ e^{\lambda_3t} \end{pmatrix} = \begin{pmatrix}xe^{t}+ye^{2t}+2ze^{-t}\\xe^{t}+y(2-t)e^{2t}+2z(-t-1)e^{-t}\\y(-1-t)e^{2t}+z(e^{-t}-t-1)\end{pmatrix} $$ where \(x,y,z\) are constants.

Finding the Fundamental Matrix Satisfying the Initial Conditions

To find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\), we need to determine the constants \(x,y,z\) such that: $$ \mathbf{1} = \mathbf{x}(0) $$ This gives us the system of equations: $$ \begin{cases} x+y+\phantom{t}2z = \phantom{-}1 \\ x-2y+(1-2t)z = -1 \\ (-1-t)y+(1-t-z)z = \phantom{-}1 \\ \end{cases} $$ At \(t = 0\), we have: $$ \begin{cases} x+y+2z = 1 \\ x-2y+z = -1 \\ -y+z = 1 \\ \end{cases} $$ Solving this system of equations, we obtain \(x=0, y=-1, z=0\). Thus, the fundamental matrix satisfying the initial condition is given by: $$ \mathbf{\Phi}(t) = \begin{pmatrix}0 & e^{2t} & 2e^{-t} \\ 0 & (2-t)e^{2t} & 2(-t-1)e^{-t} \\ 0 & (-1-t)e^{2t} & (e^{-t}-t-1)\end{pmatrix}. $$

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial in the study of linear algebra and its application to systems of differential equations. When working with a system of linear equations, eigenvalues represent how the matrix of the system can scale a vector, while eigenvectors are the specific vectors that are scaled by the matrix.

Eigenvalues are found by solving the characteristic equation, which leads to values that express certain fundamental properties of the linear transformation described by the matrix. For a given eigenvalue, the corresponding eigenvector is computed by solving the equation \(A - \(lambda_k\) I)\mathbf{v}_k = 0\), where \((\lambda_k\)) is an eigenvalue, \(I\) is the identity matrix, and \((\mathbf{v}_k\)) is the eigenvector linked to \((\lambda_k\)). In the context of the given exercise, computing these values allows us to determine the behavior of the system over time.

Eigenvectors can also be particularly helpful when forming the general solution to a system of differential equations, as they can be combined with the eigenvalues in an exponential function to express how the system evolves.

Characteristic Equation

The characteristic equation is a fundamental concept in determining the eigenvalues of a matrix. It is obtained by setting the determinant of \( (A - \(lambda\) I) \) equal to zero, where \(A\) is the matrix in question, \(lambda\) represents a scalar (eigenvalue), and \(I\) is the identity matrix of corresponding size. This equation results in a polynomial whose roots are the eigenvalues of the matrix.

In our exercise, the characteristic equation has the form \(\det(A - \lambda I) = 0\), which leads to a cubic polynomial. Solving this polynomial entails finding the values of \((\lambda\)) that satisfy the equation. These values are the keys to unlocking the system's dynamics as they provide the necessary constants to construct the fundamental matrix.

Once the eigenvalues are computed, eigenvectors are subsequently found, which in turn assist in determining the solution to the system of linear differential equations by directly influencing the structure of the solution.

Matrix Exponential

The concept of the matrix exponential is a powerful tool that arises in the analysis of linear systems of differential equations. The matrix exponential, denoted as \( e^{At} \) for a matrix \(A\), is used to find solutions of such systems over time. It represents the continuous-time version of raising a matrix to a power when dealing with dynamical systems.

In the exercise provided, \( e^{At} \) is found by multiplying the matrix of eigenvectors, \( \mathbf{V}(t) \), by a diagonal matrix with the exponential eigenvalues along the diagonal, and then multiplying this by the inverse of \( \mathbf{V}(t) \).

The calculation of the matrix exponential provides us with the fundamental matrix of the system, which encapsulates the full behavior of the system over time when coupled with initial conditions. By analyzing the matrix exponential, we gain insight into the long-term behavior of the system and how perturbations in the initial state will evolve as time progresses.