Problem 15 - Differential Equations and Linear Algebra

Question

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

Step-by-Step Solution

Answer

The short answer to the initial-value problem is: $$ \mathbf{x}(t) = \begin{bmatrix} \frac{1}{2}(e^{-t} - e^{-7t}) \\ -\frac{1}{10}e^{-t} + \frac{1}{2}e^{-7t} \end{bmatrix} $$

Step 1: Find eigenvalues and eigenvectors of A

First, we'll find the eigenvalues and eigenvectors of the matrix A. Given: $$ A = \begin{bmatrix} -2 & -1 \\ 1 & -4 \end{bmatrix} $$ Let's find the eigenvalues by finding the determinant of A - λI, where I is the identity matrix, and λ are the eigenvalues: $$ \begin{vmatrix} -2 - \lambda & -1 \\ 1 & -4 - \lambda \end{vmatrix} = (\lambda + 2)(\lambda + 4) - (-1)(1) = \lambda^2 + 6\lambda + 7 $$ Now, we'll find the roots of this polynomial: $$ \lambda^2 + 6\lambda + 7 = (\lambda + 1)(\lambda + 7) = 0 $$ So, the eigenvalues are λ_1 = -1 and λ_2 = -7. Next, we'll find the eigenvectors associated with the eigenvalues. For λ_1 = -1: $$ (A - (-1) I) v_1 = 0 \Rightarrow \begin{bmatrix} -1 & -1 \\ 1 & -3 \end{bmatrix} \begin{bmatrix} v_{11} \\ v_{12} \end{bmatrix} = 0 $$ Without loss of generality, let v_12 = 1, then we have v_11 = -1. Thus, v_1 = [-1, 1]^T. Similarly, for λ_2 = -7: $$ (A - (-7) I) v_2 = 0 \Rightarrow \begin{bmatrix} 5 & -1 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} v_{21} \\ v_{22} \end{bmatrix} = 0 $$ Let v_22 = 1, then v_21 = 1/5. Thus, v_2 = [1/5, 1]^T.

Step 2: Construct matrix exponential of A

The general form of the matrix exponential e^{At} can be written as: $$ e^{At} = P e^{\Lambda t} P^{-1} $$ where P is matrix formed by eigenvectors, Λ is the diagonal matrix formed by eigenvalues, and P^{-1} is the inverse of matrix P. $$ P = \begin{bmatrix} -1 & 1/5 \\ 1 & 1 \end{bmatrix}, \quad \Lambda = \begin{bmatrix} -1 & 0 \\ 0 & -7 \end{bmatrix} $$ Now, we'll find the inverse of matrix P: $$ P^{-1} = \frac{1}{\det(P)} \begin{bmatrix} 1 & -1/5 \\ -1 & -1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & -1/5 \\ -1 & -1 \end{bmatrix} $$ Now the matrix exponential is: $$ e^{At} = Pe^{\Lambda t}P^{-1} = \begin{bmatrix} -1 & 1/5 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} e^{-t} & 0 \\ 0 & e^{-7t} \end{bmatrix} \frac{1}{2} \begin{bmatrix} 1 & -1/5 \\ -1 & -1 \end{bmatrix} $$

Step 3: Compute matrix exponential at time t

After the matrix exponential multiplication, we have: $$ e^{At} = \frac{1}{2} \begin{bmatrix} -1 + \frac{1}{5}e^{-7t} & -e^{-t} + e^{-7t} \\ e^{-t} - e^{-7t} & \frac{1}{5}e^{-t} - e^{-7t} \end{bmatrix} $$

Step 4: Obtain the state vector x(t)

Now, we can obtain the state vector at time t by multiplying the matrix exponential with the initial state vector x_0, which is given as: $$ \mathbf{x}_0 = \begin{bmatrix} 0 \\ -1 \end{bmatrix} $$ Thus, we have: $$ \mathbf{x}(t) = e^{At}\mathbf{x}_0 = \frac{1}{2}\begin{bmatrix} -1 + \frac{1}{5}e^{-7t} & -e^{-t} + e^{-7t} \\ e^{-t} - e^{-7t} & \frac{1}{5}e^{-t} - e^{-7t} \end{bmatrix} \begin{bmatrix} 0 \\ -1 \end{bmatrix} $$ So the final solution is: $$ \mathbf{x}(t) = \begin{bmatrix} \frac{1}{2}(e^{-t} - e^{-7t}) \\ -\frac{1}{10}e^{-t} + \frac{1}{2}e^{-7t} \end{bmatrix} $$

Key Concepts

Eigenvalues and Eigenvectors

The concept of eigenvalues and eigenvectors is central to understanding many problems in linear algebra and differential equations, including initial-value problems.

In the given exercise, the first step involves finding the eigenvalues \( \lambda \) of the matrix \( A \) by solving the characteristic equation \( \det(A - \lambda I) = 0 \). The eigenvalues are the scalars that, when substituted in place of \( \lambda \) in this equation, will make the determinant of the matrix \( A - \lambda I \) equal to zero.

After finding the eigenvalues, the next step is to determine the eigenvectors. An eigenvector is a non-zero vector \( v \) that satisfies the equation \( A v = \lambda v \). This means that when the matrix \( A \) acts on the eigenvector \( v \) it simply scales it by the factor of the corresponding eigenvalue \( \lambda \).

Identifying the eigenvalues and eigenvectors is crucial because they provide the foundation for understanding the behavior of systems described by matrices. They reveal the directions in which the system evolves without changing and the rate of expansion or contraction in these directions.

Matrix Exponential

The matrix exponential \( e^{At} \) is a key tool for solving systems of linear differential equations. In the context of our initial-value problem, this represents the evolution of the state vector over time.

To construct the matrix exponential, we first form a diagonal matrix \( \Lambda \) from the eigenvalues of \( A \) and a matrix \( P \) from the corresponding eigenvectors. The matrix exponential is then calculated using the formula \( e^{At} = P e^{\Lambda t} P^{-1} \).

The exponentiation of the matrix \( \Lambda \) is straightforward since \( \Lambda \) is diagonal; it simply involves exponentiating the eigenvalues on the diagonal. The challenging part is to find the inverse of the eigenvector matrix \( P \) and to perform the multiplication \( P e^{\Lambda t} P^{-1} \).

Understanding the matrix exponential is essential because it directly relates to how the system changes over time, capturing both the growth and decay processes governed by the system's dynamics. It determines how initial conditions propagate through the system, making it fundamental for prediction and control.

State Vector

In the study of differential equations and linear dynamical systems, the concept of a 'state vector' represents the state of a system at any given time.

In the given exercise, the initial state vector \( \mathbf{x}_0 \) is specified, setting the initial conditions for our system. The state vector \( \mathbf{x}(t) \) evolves over time and is given by multiplying the matrix exponential \( e^{At} \) by the initial state vector \( \mathbf{x}_0 \).

The state vector captures all the essential information needed to describe the system at any moment. It is composed of state variables that can change over time as the system evolves. For a given initial value problem like ours, the state vector \( \mathbf{x}(t) \) can be interpreted as the position of our system in the state space at any time \( t \) after taking into account both the natural dynamics of the system (represented by the matrix \( A \) and its exponential \( e^{At} \) ) and any external forces or conditions indicated by the initial state \( \mathbf{x}_0 \).

Through this representation, we can track the trajectory of the system, making the state vector an invaluable tool for understanding and predicting the behavior of dynamic systems.