Question

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{1} & {-1} \\ {5} & {-3}\end{array}\right) \mathbf{x} $$

Short answer

Question: Find a fundamental matrix for the given system of linear differential equations satisfying the initial condition \(\Phi(0)=\mathbf{1}\): $$ \frac{d \mathbf{X}(t)}{dt} = \left(\begin{array}{ll}{1} & {-1} \\\ {5} & {-3}\end{array}\right) \mathbf{X}(t) $$ Answer: The fundamental matrix satisfying \(\Phi(0)=\mathbf{1}\) is: $$ \mathbf{\Phi}(0) = \begin{pmatrix} 1 & 1 \\ 5 & -1 \end{pmatrix} $$

Step-by-step solution

Calculate the eigenvalues and eigenvectors of the coefficient matrix

First, we need to calculate the eigenvalues and corresponding eigenvectors of the coefficient matrix. The given matrix is: $$ A=\left(\begin{array}{ll}{1} & {-1} \\\ {5} & {-3}\end{array}\right) $$ To find the eigenvalues \(\lambda\) of the matrix, we must solve the characteristic equation for each eigenvalue: $$ \det(A-\lambda I) = 0 $$ where \(I\) is the identity matrix: $$ \left|\begin{array}{cc}{1-\lambda} & {-1} \\\ {5} & {-3-\lambda}\end{array}\right|=0 $$ Calculate the determinant: $$ (1-\lambda)((-3)-\lambda) - (-1)(5) = 0 $$ This result in: $$ \lambda^2 + 2\lambda -8 = 0 $$ Solving the quadratic equation produces the eigenvalues: \(\lambda_1 = -4\) and \(\lambda_2 = 2\). Now, we will find the corresponding eigenvectors for each eigenvalue. For \(\lambda_1 = -4\), \((A-\lambda_1 I)\mathbf{v}=\mathbf{0}\) becomes: $$ \left(\begin{array}{ll}{5} & {-1} \\\ {5} & {1}\end{array}\right)\mathbf{v}=\mathbf{0} $$ From the first row of the equations, we have: $$ 5v_1 - v_2 = 0 $$ Choosing the simplest eigenvector, we can set \(v_1 = 1\) and \(v_2 = 5\). Thus, the eigenvector for \(\lambda_1\) is \(\mathbf{v}_1 = \begin{pmatrix} 1 \\ 5 \end{pmatrix}\). For \(\lambda_2 = 2\), \((A-\lambda_2 I)\mathbf{v}=\mathbf{0}\) becomes: $$ \left(\begin{array}{ll}{-1} & {-1} \\\ {5} & {-5}\end{array}\right)\mathbf{v}=\mathbf{0} $$ From the first row of the equations, we have: $$ -v_1 - v_2 = 0 $$ Choosing the simplest eigenvector, we can set \(v_1 = 1\) and \(v_2 = -1\). Thus, the eigenvector for \(\lambda_2\) is \(\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\).

Form the fundamental matrix

Now, use the eigenvectors to create the fundamental matrix, which consists of linearly independent solutions in form of \(X(t) = \mathbf{v}e^{\lambda t}\). Solution \(1\): \(X_1(t) = \mathbf{v}_1e^{\lambda_1 t} = \begin{pmatrix}1 \\ 5\end{pmatrix}e^{-4t}\) Solution \(2\): \(X_2(t) = \mathbf{v}_2e^{\lambda_2 t} = \begin{pmatrix}1 \\ -1\end{pmatrix}e^{2t}\) We can then create the fundamental matrix (\(\mathbf{\Phi}(t)\)) from the solutions: $$ \mathbf{\Phi}(t) = \begin{pmatrix} e^{-4t} & e^{2t} \\ 5e^{-4t} & -e^{2t} \end{pmatrix} $$

Find the fundamental matrix \(\Phi(t)\) satisfying the initial condition \(\Phi(0)=\mathbf{1}\)

To find the fundamental matrix satisfying \(\Phi(0) = \mathbf{1}\), replace the time variable t with 0: $$ \mathbf{\Phi}(0) = \begin{pmatrix} e^{-4(0)} & e^{2(0)} \\ 5e^{-4(0)} & -e^{2(0)} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 5 & -1 \end{pmatrix} $$ The fundamental matrix satisfying \(\Phi(0)=\mathbf{1}\) is: $$ \mathbf{\Phi}(0) = \begin{pmatrix} 1 & 1 \\ 5 & -1 \end{pmatrix} $$

Key concepts

Eigenvalues

Eigenvalues are a fascinating concept in linear algebra, playing a critical role in understanding the behavior of linear transformations. They are numbers associated with a matrix that reveal significant properties about the matrix, especially in the context of differential equations.
To find eigenvalues, you solve the characteristic equation, which is obtained from the matrix equation \( \det(A - \lambda I) = 0 \), where \(A\) is your matrix, \(\lambda\) represents the eigenvalues, and \(I\) is the identity matrix of the same size as \(A\).
This equation results in a polynomial usually known as the characteristic polynomial, and its roots are the eigenvalues.- **Purpose**: They help us understand the stability and dynamics of a system of differential equations.
- **Calculation**: In our problem, the matrix \(A\) has eigenvalues \(\lambda_1 = -4\) and \(\lambda_2 = 2\). These tell us how the solutions of the differential system behave over time.

Eigenvectors

Once we determine the eigenvalues, the next step is finding the associated eigenvectors. These are vector solutions corresponding to each eigenvalue. They provide direction in which the transformation applied by the matrix \(A\) stretches the vector by a factor equal to the eigenvalue.
For an eigenvalue \(\lambda\), its eigenvector \(\mathbf{v}\) is found by solving \((A - \lambda I) \mathbf{v} = \mathbf{0}\).
In our worked example:

• For \(\lambda_1 = -4\), the corresponding eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \ 5 \end{pmatrix}\).
• For \(\lambda_2 = 2\), the eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1 \ -1 \end{pmatrix}\).

Eigenvectors are vital since they span the solution space and define the direction of each part of the system's motion.

Fundamental Matrix

The fundamental matrix is an essential concept in solving linear systems of differential equations. It is a matrix formed by linearly independent solutions to the differential equation.
To form a fundamental matrix \(\mathbf{\Phi}(t)\), you take the general solutions of the differential equations, expressed as \(X(t) = \mathbf{v} e^{\lambda t}\) for each eigenvector and eigenvalue pair.
From our system:

• \(X_1(t) = \begin{pmatrix} 1 \ 5 \end{pmatrix} e^{-4t}\)
• \(X_2(t) = \begin{pmatrix} 1 \ -1 \end{pmatrix} e^{2t}\)

These solutions populate the columns of the fundamental matrix:
\[\mathbf{\Phi}(t) = \begin{pmatrix} e^{-4t} & e^{2t} \ 5e^{-4t} & -e^{2t} \end{pmatrix}\]
Additionally, it must satisfy the condition \(\Phi(0) = \mathbf{1}\), the identity matrix. Solving this results in the identity, indicating that the fundamental matrix is accurately representing the system's solutions.

System of Equations

A system of linear equations forms the backbone of many mathematical and engineering problems. When dealing with multiple linear equations simultaneously, as in the given differential system, we essentially manage a system of equations.
In our context, the system is described by the matrix equation \(\mathbf{x}' = A\mathbf{x}\), where \(A\) is a matrix, and \(\mathbf{x}\) represents a vector of unknown functions.
Working with a system of equations involves:

• Finding eigenvalues and eigenvectors to understand the geometric nature of the solution space.
• Constructing the fundamental matrix to represent all solutions to the system.
• Using initial conditions, if any, to ensure the solution fits specific scenario requirements.

This matrix-based approach is crucial because it allows for the systematic solution of systems that are too complicated to manage by direct integration.