Question

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

Short answer

Short Answer: The fundamental matrix for the given system of differential equations is: $$ \mathbf{\Phi}(t) = \frac{1}{4} e^{-3t} \begin{pmatrix} 4 & 2 \\ 2 & 1 \end{pmatrix} +\frac{3}{4} e^{t} \begin{pmatrix} 4 & -2 \\ -2 & 1 \end{pmatrix}. $$

Step-by-step solution

Find eigenvalues and eigenvectors of the matrix

To find the eigenvalues, we need to solve the following equation for \(\lambda\): \( \text{det}(A - \lambda I) = 0 \) where \(A = \begin{pmatrix} -1 & -4\\ 1& -1 \end{pmatrix}\) and \(I\) is the identity matrix. So, $$ \text{det}\begin{pmatrix} (-1-\lambda) & -4 \\ 1 & (-1-\lambda) \end{pmatrix} = ( -1 -\lambda)( -1 -\lambda) - (-4)(1)= \lambda^2 +2\lambda -3 = (\lambda+3)(\lambda-1) = 0 $$ Here, we find two eigenvalues: \(\lambda_1 = -3\) and \(\lambda_2 = 1\). Now, let's find the eigenvectors corresponding to each eigenvalue. For \(\lambda_1 = -3\), we have \((A-\lambda_1 I) \mathbf{v}_1 = 0\): $$ \begin{pmatrix} 2 & -4 \\ 1 & 2 \end{pmatrix} \mathbf{v}_1 = 0 $$ As row-2 of the matrix is a multiple of row-1, we can consider only the first row. From the first row, \(2v_{1,1}-4v_{1,2}=0 \Rightarrow v_{1,2}= \frac{1}{2}v_{1,1}\). Taking \(v_{1,1}=2,\) we have \(v_{1,2} = 1\), so \(\mathbf{v}_1 =\begin{pmatrix} 2 \\1 \end{pmatrix}\) For \(\lambda_2 = 1\), we have \((A-\lambda_2 I) \mathbf{v}_2 = 0\): $$ \begin{pmatrix} -2 & -4 \\ 1 & -2 \end{pmatrix} \mathbf{v}_2 = 0 $$ Again, the row-2 of the matrix is a multiple of row-1, so we can consider only the first row. From the first row, \(-2v_{2,1}-4v_{2,2}=0 \Rightarrow v_{2,1}=-2v_{2,2}\). Taking \(v_{2,2}=1,\) we have \(v_{2,1} = -2\), so \(\mathbf{v}_2 =\begin{pmatrix} -2 \\1 \end{pmatrix}\)

Form the matrix exponential

Now, we will form the matrix exponential using the eigenvalues and eigenvectors we found. We can express the matrix exponential as: $$ \mathbf{\Phi}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 \mathbf{v}_1^T + c_2 e^{\lambda_2 t} \mathbf{v}_2 \mathbf{v}_2^T, $$ where \(c_1\) and \(c_2\) are constants. Now, we have to find the values of \(c_1\) and \(c_2\) such that \(\Phi(0)=\mathbf{1}\), where \(\mathbf{1}\) is the identity matrix.

Find \(c_1\) and \(c_2\)

By substituting \(t=0\) in the matrix exponential, we get: $$ \mathbf{1} = c_1 e^{0} \mathbf{v}_1 \mathbf{v}_1^T + c_2 e^{0} \mathbf{v}_2 \mathbf{v}_2^T =c_1 \begin{pmatrix} 2 \\1 \end{pmatrix}\begin{pmatrix} 2&1 \end{pmatrix} + c_2 \begin{pmatrix} -2 \\1 \end{pmatrix}\begin{pmatrix} -2&1 \end{pmatrix} $$ This gives us: $$ \mathbf{1} = c_1 \begin{pmatrix} 4 & 2 \\ 2 & 1 \end{pmatrix} + c_2 \begin{pmatrix} 4 & -2 \\ -2 & 1 \end{pmatrix}. $$ By equating the matrices, we obtain the following system of linear equations: $$ \begin{cases} 4c_1+4c_2=1 \\ 2c_1-2c_2=0 \\ 2c_1-2c_2=0 \\ c_1+c_2=1 \end{cases} $$ Solving this system, we find \(c_1=\frac{1}{4}\) and \(c_2=\frac{3}{4}\).

Determine the fundamental matrix \(\mathbf{\Phi}(t)\)

Finally, substitute the values of \(c_1\) and \(c_2\) into the matrix exponential to obtain the fundamental matrix \(\mathbf{\Phi}(t)\): $$ \mathbf{\Phi}(t) = \frac{1}{4} e^{-3t} \begin{pmatrix} 2 \\1 \end{pmatrix}\begin{pmatrix} 2&1 \end{pmatrix} + \frac{3}{4} e^{t} \begin{pmatrix} -2 \\1 \end{pmatrix}\begin{pmatrix} -2&1 \end{pmatrix} $$ Therefore, the fundamental matrix \(\mathbf{\Phi}(t)\) is: $$ \mathbf{\Phi}(t) = \frac{1}{4} e^{-3t} \begin{pmatrix} 4 & 2 \\ 2 & 1 \end{pmatrix} +\frac{3}{4} e^{t} \begin{pmatrix} 4 & -2 \\ -2 & 1 \end{pmatrix}. $$

Key concepts

Eigenvalues and Eigenvectors

The concepts of eigenvalues and eigenvectors are fundamental in understanding how matrices operate, especially within systems of differential equations.

An **eigenvalue** is a scalar that indicates how much the direction of an eigenvector is scaled during a matrix transformation. To find eigenvalues, solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \), where \( A \) is your matrix and \( I \) is the identity matrix.

For example, given \( A = \begin{pmatrix} -1 & -4 \ 1 & -1 \end{pmatrix} \), solving the determinant equation helps locate two eigenvalues: \( \lambda_1 = -3 \) and \( \lambda_2 = 1 \).

An **eigenvector** is a non-zero vector that changes only in scale, not direction, when the matrix is applied. To find an eigenvector, insert each eigenvalue back into \( (A - \lambda I) \mathbf{v} = 0 \). For \( \lambda_1 = -3 \), an eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 2 \ 1 \end{pmatrix} \), and for \( \lambda_2 = 1 \), it's \( \mathbf{v}_2 = \begin{pmatrix} -2 \ 1 \end{pmatrix} \).

These vectors are key in forming the fundamental matrix that solves differential systems.

System of Differential Equations

A system of differential equations involves multiple equations that relate functions and their derivatives. These are widely used to model real-world problems like population growth or electrical circuits.

Consider the given system \( \mathbf{x}' = A\mathbf{x} \), where \( A = \begin{pmatrix} -1 & -4 \ 1 & -1 \end{pmatrix} \). Our goal is to find the solution matrix \( \mathbf{x}(t) \).

The solution begins with finding the eigenvalues and eigenvectors of matrix \( A \). These help construct the fundamental matrix, which in turn makes it possible to express the general solution of the system.

The fundamental matrix \( \mathbf{\Phi}(t) \) offers a way to convert a complex system into manageable computations. It describes the effect of the matrix on the state vector as it evolves over time.

• First, calculate eigenvalues and eigenvectors of the matrix \( A \).
• Use these to solve the homogeneous system and find \( \mathbf{\Phi}(t) \).

Solving such systems involves combining linear algebra concepts within the framework of calculus.

Matrix Exponential

The matrix exponential provides a powerful method to solve systems of linear differential equations. It is analogous to the exponential function for scalars.

To construct the matrix exponential \( \mathbf{\Phi}(t) \) for a matrix \( A \), employ the formula using eigenvalues and eigenvectors: \[ \mathbf{\Phi}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 \mathbf{v}_1^T + c_2 e^{\lambda_2 t} \mathbf{v}_2 \mathbf{v}_2^T \] where \( c_1 \) and \( c_2 \) are constants determined by initial conditions.

The calculation starts with transforming the system by finding \( \lambda_1 \), \( \lambda_2 \), \( \mathbf{v}_1 \), and \( \mathbf{v}_2 \). The goal is to make \( \mathbf{\Phi}(t) \) satisfy \( \Phi(0) = \mathbf{1} \).

• Identify eigenvalues and corresponding eigenvectors.
• Insert these into the exponential format.
• Apply the initial conditions to determine constants.

This results in the solution for the system's dynamic behavior over time, offering insights into its stability and responses.