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PAGE 1

5.3 Solution Curves / Phase Portraits

Summary of \(\vec{x}' = A\vec{x}\)

\(\lambda\)'s real and distinct

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

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

\(\vec{v} = \begin{bmatrix} 1 \\ \sqrt{2} \end{bmatrix}, \begin{bmatrix} -\sqrt{2} \\ 1 \end{bmatrix}\) ← straight line solutions

A phase portrait on a Cartesian coordinate system with axes x_1 and x_2. The origin is a sink. Two straight-line solutions are shown corresponding to eigenvectors: one along the vector \begin{bmatrix} 1 \\ \sqrt{2} \end{bmatrix} with eigenvalue \lambda = -1, and another along \begin{bmatrix} -\sqrt{2} \\ 1 \end{bmatrix} with eigenvalue \lambda = -4. All trajectories (solution curves) are directed toward the origin. The curves approach the origin tangent to the eigenvector associated with \lambda = -1 (the dominant eigenvalue as t goes to infinity).
Phase portrait showing an improper nodal sink.

Origin: \(t \to \infty\) (\(e^{\lambda t} \to 0\) as \(t \to \infty\) if \(\lambda < 0\))

At \(t \to \infty\), \(e^{-t} > e^{-4t}\)

So near origin, solution curves follow the eigenvector of \(\lambda = -1\)

Origin is an equilibrium

Also an improper nodal sink

  • Solutions go into origin
  • "point"
  • along an asymptote into/out of origin
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Phase Portrait Analysis: Improper Nodal Source

Given the system of linear differential equations:

\[ \vec{x}' = \begin{bmatrix} 5 & -1 \\ 3 & 1 \end{bmatrix} \vec{x} \]

The eigenvalues and corresponding eigenvectors are calculated as:

\[ \lambda = 2, 4 \]\[ \vec{v} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]
A phase portrait on a Cartesian coordinate system with axes x_1 and x_2. Two straight-line trajectories are shown corresponding to the eigenvectors: one along the line through \begin{bmatrix} 1 \\ 3 \end{bmatrix} for \lambda = 2, and another along the line through \begin{bmatrix} 1 \\ 1 \end{bmatrix} for \lambda = 4. Multiple curved trajectories originate from the origin and move outward into the four quadrants, indicating an unstable source. The curves are tangent to the eigenvector \begin{bmatrix} 1 \\ 3 \end{bmatrix} as they leave the origin.
Figure 1: Phase portrait showing trajectories for the system with eigenvalues \( \lambda = 2, 4 \).

Analysis of Behavior

  • Origin: \( t = -\infty \)
  • Dominant Term near Origin: Since \( e^{2t} > e^{4t} \) as \( t \to -\infty \), the trajectory is dominated by the smaller eigenvalue near the origin.

Observation:

\( \vec{v} = \begin{bmatrix} 1 \\ 3 \end{bmatrix} \) leave along this near origin.

Origin is an improper nodal source

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Linear Systems: Real Eigenvalues with Opposite Signs

\(\lambda\)'s real and opposite in signs

\[ \vec{x}' = \begin{bmatrix} 1 & 1 \\ 4 & 1 \end{bmatrix} \vec{x} \]

\[ \lambda = 3, -1 \]

\[ \vec{v} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 1 \\ -2 \end{bmatrix} \]

A phase portrait on a Cartesian coordinate system with axes x1 and x2. The origin is a saddle point. Two straight-line trajectories (eigenvectors) pass through the origin. One line, corresponding to \lambda = 3 and eigenvector \begin{bmatrix} 1 \\ 2 \end{bmatrix}, has arrows pointing away from the origin (unstable). The other line, corresponding to \lambda = -1 and eigenvector \begin{bmatrix} 1 \\ -2 \end{bmatrix}, has arrows pointing toward the origin (stable). Hyperbolic trajectories in the four regions between these lines curve toward the unstable eigenvector and away from the stable one.
Figure 1: Phase portrait showing a saddle point at the origin.

Classification

Origin is a saddle point

For \(\lambda = 3\): \(\vec{v} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}\)

For \(\lambda = -1\): \(\vec{v} = \begin{bmatrix} 1 \\ -2 \end{bmatrix}\)

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λ's are repeated

\[ \vec{x}' = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \vec{x} \]
\( \lambda = 1, 1 \)
\( \vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix} \)

(matrix is complete)

→ not defective

A phase portrait of a system with repeated eigenvalues \lambda = 1 . It shows a Cartesian coordinate system with axes labeled x_1 and x_2 . Numerous straight-line trajectories radiate outward from the origin in all directions, forming a star-like pattern. Each trajectory has an arrow pointing away from the origin, indicating a source. The horizontal axis is labeled with the eigenvector \begin{bmatrix} 1 \\ 0 \end{bmatrix} and \lambda = 1 . The vertical axis is labeled with the eigenvector \begin{bmatrix} 0 \\ 1 \end{bmatrix} and \lambda = 1 .

Phase portrait showing a star node (proper nodal source).

eigenvalues are equal, so neither eigenvector is more important → solutions are linear combos of both

origin is sometimes called a "star node" and is a proper nodal source

↳

no preferred asymptote

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Phase Portrait of a Linear System with Repeated Eigenvalues

\[ \vec{x}' = \begin{bmatrix} 1 & -4 \\ 4 & -7 \end{bmatrix} \vec{x} \]

Eigenvalues and Eigenvectors:

  • \( \lambda = -3, -3 \)
  • \( \vec{v} = \begin{bmatrix} 3 \\ 2 \end{bmatrix} \)
  • Defect of one
A phase portrait in the x_1-x_2 plane showing an improper nodal sink. A single straight-line solution exists along the eigenvector \vec{v} = \begin{bmatrix} 3 \\ 2 \end{bmatrix} for \lambda = -3, with arrows pointing toward the origin. Other trajectories are curved, approaching the origin and becoming tangent to the straight-line solution. All arrows indicate motion toward the origin, signifying stability.
Figure 1: Phase portrait for the system with a repeated eigenvalue and a defect of one.

One straight line solution visible.

Origin is an improper nodal sink

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Complex \(\lambda\)'s

\[ \vec{x}' = \begin{bmatrix} 2 & -5 \\ 1 & -2 \end{bmatrix} \vec{x} \quad \lambda = i, -i \quad \vec{v} = \begin{bmatrix} 2+i \\ 1 \end{bmatrix}, \begin{bmatrix} 2-i \\ 1 \end{bmatrix} \]

\(\lambda\)'s are purely imaginary, so solution curves are ovals.

Orientation?

Two small sketches of Cartesian axes with ellipses. The first ellipse is tilted and has no direction indicated. The second ellipse is also tilted and has no direction indicated. They represent the two possible rotational orientations (clockwise vs counter-clockwise).

* incomplete see next page

To do that, look at a \(\vec{v}\):

\[ \vec{v} = \begin{bmatrix} 2+i \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} + i \begin{bmatrix} 1 \\ 0 \end{bmatrix} \]

Sum of them gives us direction of the major axis

A phase portrait on a Cartesian coordinate system with axes labeled x_1 and x_2. Several concentric green ellipses (ovals) are centered at the origin. Arrows on the curves indicate a counter-clockwise orientation. A vector labeled \begin{bmatrix} 2 \\ 1 \end{bmatrix} is drawn from the origin to the boundary of one of the ellipses, representing the major axis direction.
Origin is a center

Direction:

  • Pick \(\vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\)
  • Look at \(\vec{x}' = \begin{bmatrix} 2 \\ 1 \end{bmatrix}\)
  • right, up
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Ellipse Orientation

Eigenvectors:

\[ \vec{v} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} + i \begin{bmatrix} 1 \\ 0 \end{bmatrix} , \quad \begin{bmatrix} 2 \\ 1 \end{bmatrix} - i \begin{bmatrix} 1 \\ 0 \end{bmatrix} \]

Form a parallelogram with:

\[ \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \pm \begin{bmatrix} 1 \\ 0 \end{bmatrix} \]
A Cartesian coordinate system with axes x1 and x2. A red parallelogram is centered at the origin. The vertices of the parallelogram are defined by the vectors \begin{bmatrix} 2 \\ 1 \end{bmatrix} and \begin{bmatrix} 1 \\ 0 \end{bmatrix} . Specifically, the vector \begin{bmatrix} 2 \\ 1 \end{bmatrix} points to the upper right vertex, and \begin{bmatrix} 1 \\ 0 \end{bmatrix} points to the right along the x1-axis. The opposite vectors -\begin{bmatrix} 2 \\ 1 \end{bmatrix} and -\begin{bmatrix} 1 \\ 0 \end{bmatrix} are also indicated, completing the parallelogram shape.
Figure 1: Parallelogram formed by the real and imaginary parts of the eigenvectors.

Ellipse is inside the parallelogram

A diagram showing a blue ellipse inscribed within a red parallelogram. The ellipse is tangent to the midpoints of the parallelogram's sides, illustrating how the eigenvector components constrain the orientation and size of the resulting elliptical trajectory.
Figure 2: Ellipse inscribed within the bounding parallelogram.
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If the real part of \( \lambda \) is not zero, then the ovals grow or shrink as they go around.

\[ \vec{x}' = \begin{bmatrix} 1 & -1 \\ 5 & -3 \end{bmatrix} \vec{x} \]

\( \lambda = -1 + i, \quad -1 - i \)

\( \vec{v} = \begin{bmatrix} 2 + i \\ 5 \end{bmatrix}, \quad \begin{bmatrix} 2 - i \\ 5 \end{bmatrix} \)

Figure out orientation and direction as before.

Here, the real part of \( \lambda \) is negative, so the "oval" shrinks as they go.

A phase portrait on a Cartesian coordinate system with axes labeled x_1 and x_2 . A green spiral curve originates from the outer region and spirals inward toward the origin (0,0) in a clockwise direction. Arrows on the spiral indicate the inward flow. This visual represents a spiral sink, where trajectories converge to the equilibrium point at the origin due to the negative real part of the eigenvalues ( \text{Re}(\lambda) < 0 ).

Origin is a spiral sink

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An Unusual Case: One Eigenvalue is Zero

Consider the system of differential equations:

\[ \vec{x}' = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} \vec{x} \]

The eigenvalues and corresponding eigenvectors are:

\[ \lambda = 0, \quad \lambda = 5 \] \[ \vec{v} = \begin{bmatrix} -2 \\ 1 \end{bmatrix}, \quad \vec{v} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} \]

General Solution

\[ \vec{x} = c_1 \begin{bmatrix} -2 \\ 1 \end{bmatrix} + c_2 e^{5t} \begin{bmatrix} 1 \\ 2 \end{bmatrix} \]

Phase Portrait Analysis

A phase portrait on a Cartesian coordinate system with axes x_1 and x_2. A line through the origin with a negative slope represents the eigenvector \begin{bmatrix} -2 \\ 1 \end{bmatrix} corresponding to eigenvalue \lambda = 0. A series of parallel lines with a positive slope of 2 (parallel to the eigenvector \begin{bmatrix} 1 \\ 2 \end{bmatrix}) fill the plane. Arrows on these lines point away from the line \lambda = 0, indicating growth due to the positive eigenvalue \lambda = 5. The line \lambda = 0 is marked as a line of equilibrium points where every point satisfies \vec{x}' = \vec{0}.
Phase portrait showing trajectories for a system with one zero eigenvalue.

Observations for \(\lambda = 0\):

  • The line defined by the eigenvector \(\begin{bmatrix} -2 \\ 1 \end{bmatrix}\) consists entirely of equilibrium points.
  • At every point on this line, \(\vec{x}' = \vec{0}\).
  • There are no directions (no movement) along this line.