Problem 23 - Differential Equations and Linear Algebra

Question

Let \(A\) be a \(2 \times 2\) nondefective matrix. If all eigenvalues of \(A\) have negative real part, prove that every solution to \(\mathbf{x}^{\prime}=A \mathbf{x}\) satisfies $$ \lim _{t \rightarrow \infty} \mathbf{x}(t)=\mathbf{0} $$

Step-by-Step Solution

Answer

The general solution of the system of linear differential equations \(\mathbf{x}^{\prime}=A \mathbf{x}\) is given by \(\mathbf{x}(t)=c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2\), where \(\lambda_1\) and \(\lambda_2\) are the eigenvalues of A with negative real parts, and \(\mathbf{v}_1\) and \(\mathbf{v}_2\) are the corresponding eigenvectors. Since the real parts of \(\lambda_1\) and \(\lambda_2\) are negative, the exponential terms \(e^{\lambda_1 t}\) and \(e^{\lambda_2 t}\) will tend to zero as \(t\) goes to infinity. Therefore, we have \(\lim _{t \rightarrow \infty} \mathbf{x}(t) = \mathbf{0}\), which proves that every solution to the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) satisfies \(\lim_{t \rightarrow \infty} \mathbf{x}(t)=\mathbf{0}\) given that all eigenvalues have negative real parts.

Step 1: Define the given matrix and system of equations

The given matrix A is a \(2 \times 2\) nondefective matrix, so it has two linearly independent eigenvectors, and all eigenvalues have negative real parts. We have the system of linear differential equations, \[ \mathbf{x}^{\prime}=A \mathbf{x}, \] where \(\mathbf{x}\) is the vector of variables that depend on t.

Step 2: Find the eigenvalues of A

In general, find the eigenvalues (\(\lambda_1\) and \(\lambda_2\)) of matrix A by solving the following characteristic equation: \[ det(A - \lambda I)=0, \] where \(I\) is the identity matrix. Since A has negative real parts of eigenvalues, let's consider: \[ \lambda_1=\alpha_1-i\beta_1 \quad \text{and} \quad \lambda_2=\alpha_1+i\beta_1, \] where \(\alpha_1<0\) and \(\beta_1 \in \mathbb{R}\).

Step 3: Find the eigenvectors

For each eigenvalue (\(\lambda_1\) and \(\lambda_2\)), find the corresponding eigenvector \(\mathbf{v}_1\) and \(\mathbf{v}_2\) by solving the following system of linear equations: \[ (A - \lambda I)\mathbf{v} = \mathbf{0}. \]

Step 4: Express the solution in terms of eigenvalues and eigenvectors

The general solution of the system of linear differential equations \(\mathbf{x}^{\prime}=A \mathbf{x}\) can be written using the eigenvalues and eigenvectors as: \[ \mathbf{x}(t)=c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2, \] where \(c_1\) and \(c_2\) are constants that depend on the initial conditions.

Step 5: Analyze the limit of the solution as t goes to infinity

Now, we need to show that \(\lim_{t \rightarrow \infty} \mathbf{x}(t)=\mathbf{0}\). Using the general solution we got in Step 4: \[ \lim _{t \rightarrow \infty} \mathbf{x}(t)=\lim _{t \rightarrow \infty} c_1 e^{\lambda_1 t} \mathbf{v}_1+\lim _{t \rightarrow \infty} c_2 e^{\lambda_2 t} \mathbf{v}_2. \] Since the real parts of \(\lambda_1\) and \(\lambda_2\) are negative, the exponential terms \(e^{\lambda_1 t}\) and \(e^{\lambda_2 t}\) will tend to zero as \(t\) goes to infinity. Therefore: \[ \lim _{t \rightarrow \infty} \mathbf{x}(t) = \mathbf{0}. \] This proves that every solution to the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) satisfies \(\lim_{t \rightarrow \infty} \mathbf{x}(t)=\mathbf{0}\), given that all eigenvalues of the nondefective matrix A have negative real parts.

Key Concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are foundational concepts in linear algebra with widespread applications in differential equations, stability analysis, and beyond. When we talk about an eigenvalue \( \lambda \) of a matrix \( A \), we refer to a scalar for which there exists a non-zero vector \( \textbf{v} \), called the eigenvector, such that \( A\textbf{v} = \lambda\textbf{v} \). The eigenvector \( \textbf{v} \) is therefore a vector that is only scaled by \( A \) and not rotated.

These concepts are crucial in solving systems of linear differential equations, \( \textbf{x}^\prime = A\textbf{x} \), as they allow us to find solutions that grow or decay exponentially over time. In the given exercise, eigenvalues with negative real parts imply that the corresponding solutions will tend to zero as time \( t \) approaches infinity. This characteristic leads to a deeper understanding of a system's long-term behavior and its stability.

Matrix Exponential

The matrix exponential, denoted as \( e^{At} \), is an important function in the context of solving linear systems of differential equations. It is used to express the solution of a system \( \textbf{x}^\prime = A\textbf{x} \) where \( A \) is a square matrix. The matrix exponential is analogous to the scalar exponential function and shares similar properties, such as facilitating the calculation of the system's state at any time \( t \).

When \( A \) is diagonalizable, the matrix exponential can be computed using the eigenvalues and eigenvectors, greatly simplifying the task. In practice, this involves computing terms of the form \( e^{\lambda t} \) where \( \lambda \) is an eigenvalue. As discussed in the solution, if the eigenvalues are negative, these terms will decay to zero as \( t \) approaches infinity, confirming the system's tendency towards the zero vector.

System Stability

System stability is a pivotal concept in the study of differential equations and control theory. It determines whether a system's response will remain bounded over time or not. For linear systems of the form \( \textbf{x}^\prime = A\textbf{x} \), the eigenvalues of matrix \( A \) are key indicators of the system's stability.

When all eigenvalues have negative real parts, as stated in the exercise, the system is said to be stable because solutions to the system decay to zero over time. This means that regardless of the initial state of the system, the system's response will eventually stabilize at the equilibrium point, which is the origin in this case. Stable systems are desirable in many applications because they recover from disturbances and return to equilibrium without external intervention.

Nondefective Matrix

A nondefective matrix, also known as a diagonalizable matrix, is a square matrix that can be represented as the product of a diagonal matrix \( D \) and two invertible matrices \( P \) and \( P^{-1} \), where \( P \) contains all the linearly independent eigenvectors and \( D \) contains the corresponding eigenvalues on the diagonal. Mathematically, \( A = PDP^{-1} \).

The nondefectiveness of the matrix \( A \) in the exercise guarantees that the matrix has a full set of linearly independent eigenvectors, which simplifies the process of solving the system and analyzing its behavior. When a matrix is nondefective, we can foresee the system's evolution through time more clearly, as the general solution involves distinct exponential terms related to each eigenvalue and eigenvector pair. This distinctness ensures that each solution component decays independently based on the eigenvalues' negativity, leading to the zero vector as \( t \) becomes large.