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7.3 The Eigenvalue Method for Linear Systems

Solve constant coefficient homogeneous systems

\[ \vec{x}' = A \vec{x} \quad A: \text{constant matrix} \]

First graphical interpretation of solutions

\[ \vec{x}' = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \vec{x} \quad \vec{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \]

\[ \begin{bmatrix} x_1' \\ x_2' \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \]

Pick \( x_1, x_2 \)

The system gives vector tangent to solution at \( x_1, x_2 \) we picked.

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For example, if we picked \( \vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \), then \( \vec{x}' = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \)

A vector field plot on x1, x2 axes with blue tangent vectors and green hyperbolic solution curves.

\[ \vec{x} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \vec{x}' = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \]

\[ \vec{x} = \begin{bmatrix} 0 \\ -1 \end{bmatrix}, \vec{x}' = \begin{bmatrix} -1 \\ 0 \end{bmatrix} \]

\[ \vec{x} = \begin{bmatrix} -1 \\ 0 \end{bmatrix}, \vec{x}' = \begin{bmatrix} 0 \\ -1 \end{bmatrix} \]

Repeat w/ other points to fill up the space

\[ \vec{x} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \vec{x}' = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

These are vectors tangent to solution curves

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Vector Field Visualization

The following figure displays a vector field on a Cartesian coordinate system, showing the direction and relative magnitude of vectors at various points in the plane.

A vector field plot on a grid from -1 to 1 on both axes, showing arrows pointing away from the origin along the line y=x and towards the origin along y=-x.
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Notice we see two straight line solutions.

A hand-drawn phase portrait with axes x1 and x2, showing two straight-line trajectories and vector arrows.

They are: \( \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \end{bmatrix} \)

These can be found from the matrix \( A \) w/o plotting the arrows.

\[ \vec{x}' = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \vec{x} \quad A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \]

Eigenvalues and Eigenvectors of \( A \):

\[ \det(A - \lambda I) = 0 \quad \begin{vmatrix} -\lambda & 1 \\ 1 & -\lambda \end{vmatrix} = 0 \quad \lambda^2 = 1 \implies \lambda = 1, -1 \]

\( \lambda = 1 \) solve \( (A - \lambda I)\vec{v} = \vec{0} \)

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Straight Line Solutions and Eigenvalues

\[ \begin{bmatrix} -1 & 1 & 0 \\ 1 & -1 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \dots \vec{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]

This is one of the straight line solutions. Its eigenvalue \( \lambda = 1 \) is positive, so the arrows go away from the origin.

Solving for \( \lambda = -1 \)

Solve \( (A - \lambda I) \vec{v} = \vec{0} \)

\[ \vdots \] \[ \vec{v} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \]

\( \lambda < 0 \) so they go into the origin.

General Solution to \( \vec{x}' = A \vec{x} \)

(as shown last time)

\[ \vec{x} = c_1 e^{\lambda_1 t} \vec{v}_1 + c_2 e^{\lambda_2 t} \vec{v}_2 \]

\( \lambda_1, \vec{v}_1 \) and \( \lambda_2, \vec{v}_2 \) are the eigenvalue/eigenvector pairs.

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System Solution and Phase Portrait

For this system, \( \vec{x}' = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \vec{x} \)

Solution is \( \vec{x} = c_1 e^t \begin{bmatrix} 1 \\ 1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \)

Straight Line Solutions Behavior

The straight line solutions show up:

  • \( \lim_{t \to \infty} \vec{x} \Rightarrow e^{-t} \to 0 \) so solutions end up following \( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \)
  • \( \lim_{t \to -\infty} \vec{x} \Rightarrow e^t \to 0 \) so solutions follow \( \begin{bmatrix} 1 \\ -1 \end{bmatrix} \)

Phase Portrait

Phase portrait showing hyperbolic trajectories on x1-x2 axes with arrows indicating flow direction.
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Phase Portrait Analysis

The following figure illustrates a phase portrait for a system of linear differential equations, showing the vector field and trajectories in the xy-plane.

A phase portrait in the xy-plane from -2 to 2, showing a saddle point at the origin with blue vector flow lines.
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Example: Linear System of ODEs

Example

\[ \vec{x}' = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \vec{x} \quad \leftrightarrow \quad \begin{cases} x_1' = x_1 + 2x_2 \\ x_2' = 2x_1 + x_2 \end{cases} \]

Solution:

\[ \vec{x} = C_1 e^{\lambda_1 t} \vec{v}_1 + C_2 e^{\lambda_2 t} \vec{v}_2 \]

To find the eigenvalues, we solve the characteristic equation:

\[ \det(A - \lambda I) = 0 \implies \begin{vmatrix} 1 - \lambda & 2 \\ 2 & 1 - \lambda \end{vmatrix} = 0 \]
\[ (1 - \lambda)^2 - 4 = 0 \]
\[ (1 - \lambda)^2 = 4 \]
\[ 1 - \lambda = \pm 2 \]
\[ \lambda = -1, \quad \lambda = 3 \]

\( \lambda = -1 \)

Straight line for this goes into origin

\( \lambda = 3 \)

Goes away from origin

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Solving for Eigenvectors and General Solution

Case 1: \(\lambda = -1\)

Solve \((A - \lambda I)\vec{v} = \vec{0}\):

\[\begin{bmatrix} 2 & 2 & 0 \\ 2 & 2 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\]

The corresponding eigenvector is:

\[\vec{v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\]

Case 2: \(\lambda = 3\)

\[\begin{bmatrix} -2 & 2 & 0 \\ 2 & -2 & 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\]

The corresponding eigenvector is:

\[\vec{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

General Solution

The general solution is given by:

\[\vec{x} = c_1 e^{-t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\]
  • 1st row: \(x_1(t) = c_1 e^{-t} + c_2 e^{3t}\)
  • 2nd row: \(x_2(t) = -c_1 e^{-t} + c_2 e^{3t}\)
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Phase Portrait Analysis

Quick sketch of phase portrait

Phase portrait on x1-x2 axes showing hyperbolic trajectories and eigenvectors [1, 1] and [1, -1].

The diagram shows the behavior of the system in the phase plane:

  • For the eigenvector \(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\), \(\lambda > 0\) (unstable direction).
  • For the eigenvector \(\begin{bmatrix} 1 \\ -1 \end{bmatrix}\), \(\lambda < 0\) (stable direction).

Solution if this is the initial condition

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Example: System of Linear Differential Equations

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

\[\begin{aligned} x_1' &= 4x_1 + 3x_2 + 3x_3 \\ x_2' &= -8x_1 - 7x_2 - 3x_3 \\ x_3' &= 8x_1 + 8x_2 + 4x_3 \end{aligned}\]

This system can be written in matrix form as:

\[\vec{x}' = \begin{bmatrix} 4 & 3 & 3 \\ -8 & -7 & -3 \\ 8 & 8 & 4 \end{bmatrix} \vec{x}\]

Since the matrix \(A\) is \(3 \times 3\), there are 3 eigenvalue/eigenvector pairs.

Finding Eigenvalues

We solve the characteristic equation \(\det(A - \lambda I) = 0\):

\[\begin{vmatrix} 4 - \lambda & 3 & 3 \\ -8 & -7 - \lambda & -3 \\ 8 & 8 & 4 - \lambda \end{vmatrix} = 0 \implies \lambda = -4, 4, 1\]
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Corresponding Eigenvectors

The eigenvectors corresponding to each eigenvalue are:

\[\vec{v} = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} (\lambda = -4), \quad \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} (\lambda = 4), \quad \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} (\lambda = 1)\]

General Solution

The solution is a linear combination of \(e^{\lambda t} \vec{v}\):

\[\vec{x} = c_1 e^{-4t} \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} + c_2 e^{4t} \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} + c_3 e^t \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}\]

This is how we solve \(\vec{x}' = A\vec{x}\) if \(\lambda\)'s are real and distinct.

Next time: complex \(\lambda\)'s