How to find the general solution of homogeneous systems with constant coefficients

📚 Introduction to Homogeneous Systems

A homogeneous system of linear differential equations with constant coefficients is a system of the form:

$\mathbf{x}'(t) = A \mathbf{x}(t)$

where $\mathbf{x}(t)$ is a vector of unknown functions, $\mathbf{x}'(t)$ is its derivative, and $A$ is a constant matrix.

📜 Historical Background

The study of linear differential equations dates back to the 17th century with the development of calculus by Isaac Newton and Gottfried Wilhelm Leibniz. The techniques for solving these systems evolved over centuries, with significant contributions from mathematicians such as Euler, Lagrange, and Cauchy. The use of linear algebra, particularly eigenvalues and eigenvectors, provided a powerful framework for finding general solutions, solidifying in the 20th century.

🔑 Key Principles for Finding General Solutions

• 🔢 Eigenvalues and Eigenvectors: Find the eigenvalues $\lambda$ of the matrix $A$ by solving the characteristic equation $\det(A - \lambda I) = 0$, where $I$ is the identity matrix. For each eigenvalue, find the corresponding eigenvector $\mathbf{v}$ by solving $(A - \lambda I)\mathbf{v} = 0$.
• 📈 Real and Distinct Eigenvalues: If all eigenvalues $\lambda_1, \lambda_2, ..., \lambda_n$ are real and distinct, the general solution is given by: $\mathbf{x}(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} + ... + c_n \mathbf{v}_n e^{\lambda_n t}$, where $c_i$ are arbitrary constants and $\mathbf{v}_i$ are the corresponding eigenvectors.
• 👯 Repeated Eigenvalues: If an eigenvalue $\lambda$ has algebraic multiplicity $m > 1$, then there may be fewer than $m$ linearly independent eigenvectors associated with $\lambda$. In this case, we need to find generalized eigenvectors to form a complete set of linearly independent solutions.
• 👻 Complex Eigenvalues: If $A$ is a real matrix and $\lambda = a + bi$ is a complex eigenvalue with eigenvector $\mathbf{v} = \mathbf{u} + i\mathbf{w}$, then $\overline{\lambda} = a - bi$ is also an eigenvalue with eigenvector $\overline{\mathbf{v}} = \mathbf{u} - i\mathbf{w}$. Two linearly independent real solutions are given by: $e^{at}(\mathbf{u}\cos(bt) - \mathbf{w}\sin(bt))$ and $e^{at}(\mathbf{u}\sin(bt) + \mathbf{w}\cos(bt))$.
• ✍️ General Solution: The general solution is a linear combination of $n$ linearly independent solutions, where $n$ is the dimension of the system.

🏢 Real-world Examples

Example 1: Simple 2x2 System

Consider the system $\mathbf{x}' = A\mathbf{x}$, where $A = \begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix}$.

1. Find Eigenvalues: $\det(A - \lambda I) = (1-\lambda)^2 - 4 = \lambda^2 - 2\lambda - 3 = (\lambda - 3)(\lambda + 1) = 0$. So, $\lambda_1 = 3$ and $\lambda_2 = -1$.
2. Find Eigenvectors:
   • For $\lambda_1 = 3$: $(A - 3I)\mathbf{v}_1 = \begin{bmatrix} -2 & 1 \\ 4 & -2 \end{bmatrix} \mathbf{v}_1 = 0$. An eigenvector is $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$.
   • For $\lambda_2 = -1$: $(A + I)\mathbf{v}_2 = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix} \mathbf{v}_2 = 0$. An eigenvector is $\mathbf{v}_2 = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$.
3. General Solution: $\mathbf{x}(t) = c_1 \begin{bmatrix} 1 \\ 2 \end{bmatrix} e^{3t} + c_2 \begin{bmatrix} 1 \\ -2 \end{bmatrix} e^{-t}$.

Example 2: Complex Eigenvalues

Consider the system $\mathbf{x}' = A\mathbf{x}$, where $A = \begin{bmatrix} -1 & 2 \\ -1 & -1 \end{bmatrix}$.

1. Find Eigenvalues: $\det(A - \lambda I) = (-1-\lambda)^2 + 2 = \lambda^2 + 2\lambda + 3 = 0$. So, $\lambda = -1 \pm i\sqrt{2}$.
2. Find Eigenvectors:
   • For $\lambda = -1 + i\sqrt{2}$: $(A - (-1 + i\sqrt{2})I)\mathbf{v} = \begin{bmatrix} -i\sqrt{2} & 2 \\ -1 & -i\sqrt{2} \end{bmatrix} \mathbf{v} = 0$. An eigenvector is $\mathbf{v} = \begin{bmatrix} -i\sqrt{2} \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} + i \begin{bmatrix} -\sqrt{2} \\ 0 \end{bmatrix}$.
3. General Solution: $\mathbf{x}(t) = c_1 e^{-t} \left( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \cos(\sqrt{2}t) - \begin{bmatrix} -\sqrt{2} \\ 0 \end{bmatrix} \sin(\sqrt{2}t) \right) + c_2 e^{-t} \left( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \sin(\sqrt{2}t) + \begin{bmatrix} -\sqrt{2} \\ 0 \end{bmatrix} \cos(\sqrt{2}t) \right)$.

Conclusion

Finding the general solution of homogeneous systems with constant coefficients involves understanding eigenvalues, eigenvectors, and how to handle different cases, including real distinct, repeated, and complex eigenvalues. This knowledge provides a powerful tool for analyzing a variety of dynamical systems.