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5.2 The Eigenvalue Method for Homogeneous Systems

want to solve

\[ \begin{aligned} x_1' &= ax_1 + bx_2 \\ x_2' &= cx_1 + dx_2 \end{aligned} \]

goal: find \( x_1(t) = ? \)

\( x_2(t) = ? \)

rewrite as a matrix equation

\[ \begin{bmatrix} x_1' \\ x_2' \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \]

\( a, b, c, d \) constants

\[ \vec{x}' = A \vec{x} \]

solutions are of the form \( e^{\lambda t} \vec{v} \)

\( \lambda \) : eigenvalue of \( A \)
\( \vec{v} \) : corresponding eigenvector

(\( n \) of these for \( n \times n \) \( A \))

general solution: \( \vec{x} = c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2 + \dots \)

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Eigenvalues and Eigenvectors

General Procedure

To find the eigenvalues \(\lambda\):

\[ \det(A - \lambda I) = 0 \]

Note: \(I\) is the identity matrix.

To find the eigenvectors \(\vec{v}\):

\[ \text{solve } (A - \lambda I)\vec{v} = \vec{0} \text{ using } \lambda\text{'s above} \]

Example

Given the system of differential equations:

\[ \begin{aligned} x_1' &= x_1 + 2x_2 \\ x_2' &= 3x_1 + 2x_2 \end{aligned} \]

This can be written in matrix form as:

\[ \vec{x}' = \underbrace{\begin{bmatrix} 1 & 2 \\ 3 & 2 \end{bmatrix}}_{A} \vec{x} \]

Find \(\lambda\):

Set the determinant of \(A - \lambda I\) to zero:

\[ \begin{vmatrix} 1 - \lambda & 2 \\ 3 & 2 - \lambda \end{vmatrix} = 0 \]

\((1 - \lambda)(2 - \lambda) - 6 = 0\)

\(\lambda^2 - 3\lambda - 4 = 0\)

characteristic eq.

\((\lambda - 4)(\lambda + 1) = 0\)

\(\lambda = -1, \quad \lambda = 4\)

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Solving for Eigenvectors

Now solve \((A - \lambda I)\vec{v} = \vec{0}\) using those \(\lambda\) for \(\vec{v}\).

Case: \(\lambda = -1\)

\[\begin{bmatrix} 2 & 2 \\ 3 & 3 \end{bmatrix} \vec{v} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}, \quad \vec{v} = \begin{bmatrix} a \\ b \end{bmatrix}\]

Augmented matrix:

\[\begin{bmatrix} 2 & 2 & 0 \\ 3 & 3 & 0 \end{bmatrix}\]

Then row reductions:

\[\rightarrow \begin{bmatrix} 1 & 1 & 0 \\ 3 & 3 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\]

\(a = -b\)

\(b\) is free

\[\vec{v} = \begin{bmatrix} -b \\ b \end{bmatrix}\]

Choose any \(b \neq 0\), let \(b = -1\):

\[\vec{v} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}, \quad \lambda = -1\]\(\rightarrow\)\[e^{-t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}\]

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Differential Equations: General Solution and Initial Conditions

Similarly, for \(\lambda = 4, \vec{v} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}\) \(\rightarrow e^{4t} \begin{bmatrix} 2 \\ 3 \end{bmatrix}\)

General solution:

\[ \vec{x} = c_1 e^{-t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 e^{4t} \begin{bmatrix} 2 \\ 3 \end{bmatrix} \]

\(c_1, c_2\) come from initial condition: \(x_1(0) = ?\) and \(x_2(0) = ?\)

For example, if \(x_1(0) = 1\) and \(x_2(0) = 0\)

Then, at \(t = 0\):

\[ \begin{bmatrix} 1 \\ 0 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 \begin{bmatrix} 2 \\ 3 \end{bmatrix} \]

Solve for \(c_1, c_2 \dots\) \(c_1 = \frac{3}{5}, c_2 = \frac{1}{5}\)

Construct the phase diagram / portrait

\(\rightarrow\) graph of \(x_1\) vs \(x_2\)

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Phase Portrait of a Linear System

The general solution for the system is given by:

\[ \vec{x} = c_1 e^{-t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 e^{4t} \begin{bmatrix} 2 \\ 3 \end{bmatrix} \]

Solution curves follow the eigenvectors as asymptotes.

A phase portrait in the x_1-x_2 plane showing a saddle point at the origin. Two straight-line trajectories (eigenvectors) intersect at the origin. The first eigenvector, \begin{bmatrix} 1 \\ -1 \end{bmatrix} with eigenvalue \lambda = -1, has arrows pointing toward the origin (stable manifold). The second eigenvector, \begin{bmatrix} 2 \\ 3 \end{bmatrix} with eigenvalue \lambda = 4, has arrows pointing away from the origin (unstable manifold). Red hyperbolic solution curves fill the four regions between these lines, approaching the stable eigenvector as t \to -\infty and the unstable eigenvector as t \to \infty. — Figure: A phase portrait in the \(x_1-x_2\) plane showing a saddle point at the origin. Two straight-line trajectories (eigenvectors) intersect at the origin. The first eigenvector, \(\begin{bmatrix} 1 \\ -1 \end{bmatrix}\) with eigenvalue \(\lambda = -1\), has arrows pointing toward the origin (stable manifold). The second eigenvector, \(\begin{bmatrix} 2 \\ 3 \end{bmatrix}\) with eigenvalue \(\lambda = 4\), has arrows pointing away from the origin (unstable manifold). Red hyperbolic solution curves fill the four regions between these lines, approaching the stable eigenvector as \(t \to -\infty\) and the unstable eigenvector as \(t \to \infty\).

Eigenvector 1:

\[ \begin{bmatrix} 2 \\ 3 \end{bmatrix}, \lambda = 4 \text{ (away)} \]

Eigenvector 2:

\[ \begin{bmatrix} 1 \\ -1 \end{bmatrix}, \lambda = -1 \text{ (into origin)} \]

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Example: System of Linear Differential Equations

Consider the following system of first-order linear differential equations:

\[\begin{aligned} x_1' &= -3x_1 + x_2 \\ x_2' &= x_1 - 3x_2 \end{aligned}\]

This can be written in matrix form as:

\[\vec{x}' = \begin{bmatrix} -3 & 1 \\ 1 & -3 \end{bmatrix} \vec{x}\]

Eigenvalues and Eigenvectors

Suppose we found the following eigenvalues and corresponding eigenvectors:

\[\begin{aligned} \lambda_1 &= -4, & \vec{v}_1 &= \begin{bmatrix} 1 \\ -1 \end{bmatrix} \\ \lambda_2 &= -2, & \vec{v}_2 &= \begin{bmatrix} 1 \\ 1 \end{bmatrix} \end{aligned}\]

General Solution

The general solution is given by:

\[\vec{x} = c_1 e^{-4t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 e^{-2t} \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

Asymptotic Behavior

The asymptotes are defined by the eigenvectors:

\[\begin{bmatrix} 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

Both trajectories go into the origin because both eigenvalues are negative (\(\lambda < 0\)).

The origin is reached when \(t \to \infty\).
Any initial point could be from \(t = -\infty\).

When \(t \to \infty\), the term \(e^{-4t}\) decays much faster than \(e^{-2t}\) (\(e^{-4t} \ll e^{-2t}\)). Therefore, solutions follow the direction of the eigenvector associated with the slower decay rate:

Solutions follow \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\) to go into the origin, but initially follow \(\begin{bmatrix} 1 \\ -1 \end{bmatrix}\).

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Phase Portrait Analysis

The following diagram illustrates a phase portrait for a two-dimensional linear system of differential equations. The trajectories show the behavior of the system near the origin.

A phase portrait in a Cartesian coordinate system with axes labeled x_1 and x_2. The origin is a stable node (sink). Two straight-line trajectories (eigenvectors) are shown in black: one along the line x_2 = x_1 labeled with the vector \begin{bmatrix} 1 \\ 1 \end{bmatrix}, and another along the line x_2 = -x_1 labeled with the vector \begin{bmatrix} 1 \\ -1 \end{bmatrix}. Both black lines have arrows pointing toward the origin. Multiple curved trajectories in red also converge toward the origin, following the direction of the eigenvectors, specifically becoming tangent to the line x_2 = x_1 as they approach the origin, indicating that the eigenvalue associated with \begin{bmatrix} 1 \\ 1 \end{bmatrix} is more negative (dominant in the limit as t \to \infty). — Figure: A phase portrait in a Cartesian coordinate system with axes labeled \(x_1\) and \(x_2\). The origin is a stable node (sink). Two straight-line trajectories (eigenvectors) are shown in black: one along the line \(x_2 = x_1\) labeled with the vector \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\), and another along the line \(x_2 = -x_1\) labeled with the vector \(\begin{bmatrix} 1 \\ -1 \end{bmatrix}\). Both black lines have arrows pointing toward the origin. Multiple curved trajectories in red also converge toward the origin, following the direction of the eigenvectors, specifically becoming tangent to the line \(x_2 = x_1\) as they approach the origin, indicating that the eigenvalue associated with \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\) is more negative (dominant in the limit as \(t \to \infty\)).