Problem 17 - Differential Equations and Linear Algebra

Question

Solve the initial-value problem \(\mathbf{x}^{\prime}=\) \(A \mathbf{x}, \mathbf{x}(0)=\mathbf{x}_{0}\). $$A=\left[\begin{array}{rr} -1 & 4 \\ 2 & -3 \end{array}\right], \quad \mathbf{x}_{0}=\left[\begin{array}{l} 3 \\ 0 \end{array}\right]$$

Step-by-Step Solution

Answer

The short answer is: \[\boldsymbol{x}(t) = \begin{bmatrix} 9e^t - 3e^{-5t} \\ -6e^t \end{bmatrix}\]

Step 1: Compute eigenvalues and eigenvectors of matrix A

To compute the matrix exponential of A, we first need to find the eigenvalues and the eigenvectors of A. Given matrix A: \[A = \begin{bmatrix} -1 & 4 \\ 2 & -3 \end{bmatrix}\] To find the eigenvalues, set the determinant of \(A - \lambda I = 0\), where \(I\) is the identity matrix and \(\lambda\) is an eigenvalue: \[\det(A-\lambda I) = \det\begin{bmatrix} -1-\lambda & 4 \\ 2 & -3-\lambda \end{bmatrix} = (\lambda+1)(\lambda+3) - (4)(2) = \lambda^2 + 4\lambda - 5\] To find the eigenvalues, solve \(\lambda^2 + 4\lambda - 5 = 0\). We get \(\lambda_1 = 1\) and \(\lambda_2 = -5\). Now, find the eigenvectors corresponding to the eigenvalues. For \(\lambda_1 = 1\): \((A-\lambda_1 I)\boldsymbol{v_1} = 0\) \[\begin{bmatrix} -2 & 4 \\ 2 & -4 \end{bmatrix}\boldsymbol{v_1} = 0\] From this, we can find the eigenvector \(\boldsymbol{v_1} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}\). For \(\lambda_2 = -5\): \((A-\lambda_2 I)\boldsymbol{v_2} = 0\) \[\begin{bmatrix} 4 & 4 \\ 2 & 2 \end{bmatrix}\boldsymbol{v_2} = 0\] In this case, we get the eigenvector \(\boldsymbol{v_2} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\).

Step 2: Calculate the matrix exponential of A

We have the eigenvalues and eigenvectors of A, so we can compute the matrix exponential using the formula: \[e^{At} = \boldsymbol{P}e^{\boldsymbol{Dt}}\boldsymbol{P}^{-1}\] where \(\boldsymbol{P}\) is the matrix formed by the eigenvectors, \(\boldsymbol{D}\) is a diagonal matrix with the eigenvalues, and \(e^{Dt}\) is another diagonal matrix with the exponentials of the eigenvalues times t: \[\boldsymbol{P} = \begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix}, \boldsymbol{D} = \begin{bmatrix} 1 & 0 \\ 0 & -5 \end{bmatrix}, e^{\boldsymbol{Dt}} = \begin{bmatrix} e^t & 0 \\ 0 & e^{-5t} \end{bmatrix}\] Now, calculate the inverse of matrix \(\boldsymbol{P}^{-1}\): \[\boldsymbol{P}^{-1} = \frac{1}{(2)(-1) - (1)(1)}\begin{bmatrix} -1 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -2 \end{bmatrix}\] Finally, compute the matrix exponential: \[e^{At} = \boldsymbol{P}e^{\boldsymbol{Dt}}\boldsymbol{P}^{-1} = \begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} e^t & 0 \\ 0 & e^{-5t} \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 3e^t - e^{-5t} & 2e^t + 2e^{-5t} \\ -2e^t - e^{-5t} & -e^t + 3e^{-5t} \end{bmatrix}\]

Step 3: Use the initial condition to find the solution

Calculate the solution vector \(\boldsymbol{x}(t)\) using the matrix exponential \(e^{At}\) and the initial condition \(\boldsymbol{x}(0) = \boldsymbol{x_0} = \begin{bmatrix} 3 \\ 0 \end{bmatrix}\): \[\boldsymbol{x}(t) = e^{At}\boldsymbol{x_0} = \begin{bmatrix} 3e^t - e^{-5t} & 2e^t + 2e^{-5t} \\ -2e^t - e^{-5t} & -e^t + 3e^{-5t} \end{bmatrix}\begin{bmatrix} 3 \\ 0 \end{bmatrix} = \begin{bmatrix} 9e^t - 3e^{-5t} \\ -6e^t \end{bmatrix}\] Thus, the solution to the initial-value problem is: \[\boldsymbol{x}(t) = \begin{bmatrix} 9e^t - 3e^{-5t} \\ -6e^t \end{bmatrix}\]

Key Concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is central to solving differential equations, especially when dealing with systems of linear equations represented in matrix form. Given a square matrix \( A \), its eigenvalues \( \lambda \) are values that satisfy the equation \( \det(A - \lambda I) = 0 \), where \( I \) is the identity matrix. When you solve this polynomial, you find the eigenvalues, which tell you the scaling factors along the directions defined by their associated eigenvectors.

In practical terms:

• Eigenvalues reveal the nature of the system dynamics, such as stability and oscillations.
• Eigenvectors provide directions that remain unchanged under the transformation represented by \( A \).
• The approach to find them typically involves solving the characteristic polynomial.

In this exercise, the eigenvalues are \( 1 \) and \( -5 \), found by solving \( \lambda^2 + 4\lambda - 5 = 0 \). Each eigenvalue is then used to find its corresponding eigenvector by solving the equation \((A - \lambda I) \mathbf{v} = 0\). This meticulous process is crucial for analyzing and solving the given initial-value problem.

Matrix Exponential

The matrix exponential \( e^{At} \) is an essential tool in solving systems of linear differential equations. It allows us to express the evolution of a system over time, given an initial condition. For a matrix \( A \), the exponential is defined using its eigenvectors and eigenvalues.

The process involves:

• Computing \( \mathbf{P} \), a matrix whose columns are eigenvectors of \( A \).
• Finding \( \mathbf{D} \), a diagonal matrix with eigenvalues of \( A \) as its diagonal elements.
• Calculating \( \mathbf{P}^{-1} \).
• Constructing \( e^{\mathbf{D}t} \), a diagonal matrix with exponential terms \( e^{\lambda t} \) on its diagonal.
• Combining these to form \( e^{At} = \mathbf{P} e^{\mathbf{D}t} \mathbf{P}^{-1} \).

This matrix exponential captures the system's behavior over time, allowing the construction of the solution for given initial conditions. In the provided exercise, we used this method to compute the matrix exponential which was then applied to the initial condition.

Initial-Value Problem

In differential equations, an initial-value problem consists of finding a solution to a differential equation that satisfies certain conditions at a given time, often \( t = 0 \). Specifically, we seek a function \( \mathbf{x}(t) \) such that \( \mathbf{x}(0) = \mathbf{x_0} \) and the differential equation \( \mathbf{x}' = A\mathbf{x} \) is satisfied.

Key elements of handling initial-value problems include:

• Finding a suitable solution form, such as the matrix exponential in linear cases.
• Applying initial conditions to solve any constants or vectors determined by the conditions.
• Ensuring the solution fits both the equation and the initial conditions.

In this exercise, the initial condition was \( \mathbf{x}(0) = \begin{bmatrix} 3 \ 0 \end{bmatrix} \). We used the matrix exponential derived from \( A \) and applied it to this initial vector to find the explicit solution \( \mathbf{x}(t) \). This ensures the dynamic system described by the differential equation starts with the given initial state.

Linear Algebra

Linear algebra underpins much of the process involved in solving differential equations, particularly through matrices, eigenvalues, and eigenvectors. Linear algebra provides the tools for:

• Breaking down complex systems into understandable components (using matrices).
• Identifying invariant directions and scaling behaviours (eigenvectors and eigenvalues).
• Decomposing matrices to simplify problem solving (as with diagonalization).

In this exercise, linear algebra facilitated the solving of the system \( \mathbf{x}' = A\mathbf{x} \) by allowing us to express \( A \) in terms suited for easy computation of \( e^{At} \). It also helped in forming the solution by applying a structured approach to comprehend system behaviours. Applying linear algebra concepts lets us manipulate systems of equations effectively, bringing clarity to complex dynamics in a systematic manner.