Problem 24 - Differential Equations and Linear Algebra

Question

Determine the general solution to the linear system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\). $$\left[\begin{array}{rrrr} 7 & 0 & 0 & -1 \\ 0 & 6 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 2 & 0 & 0 & 5 \end{array}\right]$$

Step-by-Step Solution

Answer

The general solution to the linear system \(\mathbf{x}^{\prime}=A \mathbf{x}\) with the matrix \(A = \left[\begin{array}{rrrr} 7 & 0 & 0 & -1 \\ 0 & 6 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 2 & 0 & 0 & 5 \end{array}\right]\) is: \[ \mathbf{x}(t) = \begin{bmatrix} C_1e^{7t} \\ C_2e^{6t} \\ C_3e^{-t} \\ C_4e^{5t} \end{bmatrix} \]

Step 1: Identify the eigenvalues

Since the given matrix is in diagonal form, the eigenvalues are the diagonal elements: \[ \lambda_1 = 7, \lambda_2 = 6, \lambda_3 = -1, \lambda_4 = 5 \]

Step 2: Determine the eigenvectors

For this particular problem, finding the eigenvectors is straightforward. Each eigenvalue corresponds to a unique eigenvector, which is a vector with all zeroes except for a one in the position of the corresponding eigenvalue in the diagonal matrix. So, the eigenvectors are: \[ \mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \mathbf{v}_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \mathbf{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \mathbf{v}_4 = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \]

Step 3: Compute the matrix exponential

Using the matrix exponential formula, the general solution to the linear system is given by: \[ \mathbf{x}(t) = e^{At} \mathbf{x}(0) =\begin{bmatrix} e^{7t} & 0 & 0 & 0 \\ 0 & e^{6t} & 0 & 0 \\ 0 & 0 & e^{-t}& 0 \\ 0 & 0 & 0 & e^{5t} \end{bmatrix} \mathbf{x}(0) \] Now, let \(\mathbf{x}(0) = \begin{bmatrix} C_1 \\ C_2 \\ C_3 \\ C_4 \end{bmatrix}\), where \(C_1, C_2, C_3,\) and \(C_4\) are constants.

Step 4: Find the general solution

Multiply the matrix exponential by the initial conditions vector to get the general solution: \[ \mathbf{x}(t) = \begin{bmatrix} e^{7t} & 0 & 0 & 0 \\ 0 & e^{6t} & 0 & 0 \\ 0 & 0 & e^{-t}& 0 \\ 0 & 0 & 0 & e^{5t} \end{bmatrix} \begin{bmatrix} C_1 \\ C_2 \\ C_3 \\ C_4 \end{bmatrix} = \begin{bmatrix} C_1e^{7t} \\ C_2e^{6t} \\ C_3e^{-t} \\ C_4e^{5t} \end{bmatrix} \] Thus, the general solution to the given linear system is: \[ \mathbf{x}(t) = \begin{bmatrix} C_1e^{7t} \\ C_2e^{6t} \\ C_3e^{-t} \\ C_4e^{5t} \end{bmatrix} \]

Key Concepts

Eigenvalues

Understanding eigenvalues is essential in solving linear systems. An eigenvalue is a special scalar associated with a linear system of equations; it is a value for which there exists a non-zero vector (eigenvector) that, when multiplied by the matrix, results in a vector that is simply a multiple of the original vector. This concept is fundamental in various fields, including stability analysis, vibrations analysis, and quantum mechanics.

In the given exercise, identifying eigenvalues for matrix A is simplified as the matrix is diagonal, meaning the eigenvalues are the entries along the diagonal. Therefore, the eigenvalues are λ₁ = 7, λ₂ = 6, λ₃ = -1, and λ₄ = 5. These values play a crucial role in determining the behavior of the system, as they influence the rate and direction of growth or decay of the system's solutions over time.

Eigenvalues are indicative of the underlying system's dynamics. For instance, positive eigenvalues suggest growth in the corresponding direction, while negative eigenvalues imply decay.

Eigenvectors

When solving for the dynamics of a system, along with eigenvalues, we must also find the eigenvectors. An eigenvector is a non-zero vector that does not change direction when a linear transformation is applied to it; only its magnitude is scaled by the corresponding eigenvalue. In essence, an eigenvector points in a direction in which it's merely stretched or compressed by the matrix transformation, not redirected.

In our linear system, since the matrix is diagonal, the eigenvectors are particularly easy to determine. Each eigenvector corresponds to one of the diagonal elements and is a unit vector with a '1' in the position where the eigenvalue appears in the matrix, and zeroes elsewhere. Thus, the exercise yields the eigenvectors as 𝐯₁,𝐯₂,𝐯₃, and 𝐯₄, coinciding with the standard basis of ℝ⁴.

These eigenvectors are pivotal in constructing the general solution of the system, as they form a basis upon which any initial state vector can be expressed and evolved over time.

Matrix Exponential

The matrix exponential is a concept used to solve systems of linear differential equations. It extends the idea of exponentiating a single number to exponentiating an entire matrix. For a given matrix A, the matrix exponential, denoted as e^At, defines the solution of the system dx/dt = Ax when x(0) - the initial condition - is known.

For diagonal matrices, the computation simplifies, and the matrix exponential of A becomes a diagonal matrix with the exponentials of the original diagonal entries as its new entries. This is seen in the step-by-step solution where matrix exponential of A is calculated and directly provides the time-evolved states of the system.

Initial Condition

The initial condition of a system tells us the state of the system at the beginning of our observation - time t = 0. This is usually given as a vector x(0), which contains the starting values for each variable in the system. In the general solution of our exercise, the initial condition is represented as a vector of constants: C₁, C₂, C₃, and C₄.

By multiplying the matrix exponential by the initial condition vector, we can find how this state evolves over time. Each constant C_i scales the solution in the direction of the corresponding eigenvector. This is a crucial step in the process because the initial condition affects the trajectory of the solution, even though the system's dynamics are governed by the eigenvectors and eigenvalues of matrix A.