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5.2 The Eigenvalue Method for Homogeneous Systems

We want to solve the following system of first-order differential equations:

\[\begin{cases} x_1'(t) = ax_1 + bx_2 \\ x_2'(t) = cx_1 + dx_2 \end{cases}\]

Goal: \[\begin{aligned} x_1(t) &= ? \\ x_2(t) &= ? \end{aligned}\]

Matrix Representation

The system can be written as a matrix equation:

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

This takes the general form:

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

where \(a, b, c, d\) are constants.

Solutions

We look for solutions of the form:

\[\vec{x} = \vec{v}e^{\lambda t}\]

(For an \(n \times n\) matrix \(A\), there are \(n\) of these solutions).

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

General Solution

The general solution is a linear combination of all individual solutions:

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

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Eigenvalue and Eigenvector Problems

To solve the equation \( A\vec{v} = \lambda\vec{v} \) for \( \lambda \) and \( \vec{v} \):

Step 1:

Solve \( \det(A - \lambda I) = 0 \) for \( \lambda \)

Then solve for \( \vec{v} \):

Solve \( (A - \lambda I)\vec{v} = \vec{0} \) using the \( \lambda \)'s found 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:

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

To find the eigenvalues, solve \( \det(A - \lambda I) = 0 \):

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

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

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Finding Eigenvectors

Starting with the characteristic equation:

\[ \lambda^2 - 3\lambda - 4 = 0 \quad \text{characteristic eq.} \]

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

\[ \lambda = -1, \quad \lambda = 4 \]

Calculating Corresponding Eigenvectors

Now the corresponding eigenvectors. Solve \( (A - \lambda I)\vec{v} = \vec{0} \) for \( \vec{v} \).

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

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

Using row reduction on the augmented matrix:

\[ \begin{bmatrix} 2 & 2 & 0 \\ 3 & 3 & 0 \end{bmatrix} \quad \text{row reduction} \]

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

From the reduced matrix, we see that \( b \) is free, and the first row gives the equation:

\[ a + b = 0 \]

Thus, the eigenvector is of the form:

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

Choose any \( b \neq 0 \), e.g., \( b = -1 \)

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Eigenvalues and General Solutions

First Eigenpair:

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

Similarly, the other pair is:

\[ \vec{v} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}, \quad \lambda = 4 \]

General Solution

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

Note: The solution follows the form \( \vec{v} e^{\lambda t} \).

Initial Conditions

\( c_1, c_2 \) come from initial conditions.

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...\[ c_1 = -\frac{3}{5}, \quad c_2 = \frac{1}{5} \]

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Phase Diagrams and Phase Portraits

Graphing \( x_1 \) vs \( x_2 \) leads to Phase Diagrams or Phase Portraits.

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

As \( t \to \infty \), \( \vec{x} \to c_2 e^{4t} \begin{bmatrix} 2 \\ 3 \end{bmatrix} \), which is parallel to the vector \( \begin{bmatrix} 2 \\ 3 \end{bmatrix} \).

As \( t \to -\infty \), \( \vec{x} \to c_1 e^{-t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \), which is parallel to the vector \( \begin{bmatrix} 1 \\ -1 \end{bmatrix} \).

Eigenvectors act like asymptotes to solution curves.

A phase portrait on a Cartesian coordinate system with axes x_1 and x_2 . Two straight lines representing eigenvectors pass through the origin. The first line corresponds to the vector \begin{bmatrix} 2 \\ 3 \end{bmatrix} with eigenvalue \lambda = 4 ; arrows on this line point away from the origin. The second line corresponds to the vector \begin{bmatrix} 1 \\ -1 \end{bmatrix} with eigenvalue \lambda = -1 ; arrows on this line point toward the origin. Red hyperbolic solution curves are shown in the regions between the eigenvectors, flowing towards the origin along the stable direction and away along the unstable direction. — Figure: A phase portrait on a Cartesian coordinate system with axes \( x_1 \) and \( x_2 \). Two straight lines representing eigenvectors pass through the origin. The first line corresponds to the vector \( \begin{bmatrix} 2 \\ 3 \end{bmatrix} \) with eigenvalue \( \lambda = 4 \); arrows on this line point away from the origin. The second line corresponds to the vector \( \begin{bmatrix} 1 \\ -1 \end{bmatrix} \) with eigenvalue \( \lambda = -1 \); arrows on this line point toward the origin. Red hyperbolic solution curves are shown in the regions between the eigenvectors, flowing towards the origin along the stable direction and away along the unstable direction.

\( \begin{bmatrix} 2 \\ 3 \end{bmatrix}, \lambda = 4 \to \lambda > 0 \) moving away from origin

\( \begin{bmatrix} 1 \\ -1 \end{bmatrix}, \lambda = -1 \to \lambda < 0 \) moving towards 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:

\(\lambda = -4\), \(\vec{v} = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\)
\(\lambda = -2\), \(\vec{v} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\)

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

Asymptotes: \(\begin{bmatrix} 1 \\ -1 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix}\). Solutions go toward the origin because \(\lambda < 0\) for both eigenvalues.

When \(t > 0\), \(e^{-2t} > e^{-4t}\).

So, solutions follow the eigenvector of \(\lambda = -2 \rightarrow \begin{bmatrix} 1 \\ 1 \end{bmatrix}\).

For \(t < 0\), the opposite situation occurs \(\rightarrow\) solutions follow \(\begin{bmatrix} 1 \\ -1 \end{bmatrix}\).

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Origin is \( t \) at \( \infty \) since both \( \lambda \)'s are negative.

Solutions start with \( t < 0 \) then increase to \( t = \infty \).

\( \rightarrow \) follow

\( \begin{bmatrix} 1 \\ -1 \end{bmatrix} \)

then

\( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \)

A phase portrait on a Cartesian coordinate system with axes x_1 and x_2 . Two straight-line trajectories (eigenvectors) are shown: one along the line y = x corresponding to the vector \begin{bmatrix} 1 \\ 1 \end{bmatrix} and another along the line y = -x corresponding to the vector \begin{bmatrix} 1 \\ -1 \end{bmatrix} . All trajectories, including the curved ones in red, have arrows pointing toward the origin, indicating a stable node where both eigenvalues ( \lambda ) are negative. The curved trajectories initially follow the direction of the eigenvector \begin{bmatrix} 1 \\ -1 \end{bmatrix} before bending to approach the origin along the direction of the eigenvector \begin{bmatrix} 1 \\ 1 \end{bmatrix} . — Figure: A phase portrait on a Cartesian coordinate system with axes \( x_1 \) and \( x_2 \). Two straight-line trajectories (eigenvectors) are shown: one along the line \( y = x \) corresponding to the vector \( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \) and another along the line \( y = -x \) corresponding to the vector \( \begin{bmatrix} 1 \\ -1 \end{bmatrix} \). All trajectories, including the curved ones in red, have arrows pointing toward the origin, indicating a stable node where both eigenvalues (\( \lambda \)) are negative. The curved trajectories initially follow the direction of the eigenvector \( \begin{bmatrix} 1 \\ -1 \end{bmatrix} \) before bending to approach the origin along the direction of the eigenvector \( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \).