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5.3 Solution Curves / Phase Portraits

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

\(\lambda\)'s are real and distinct

\[\vec{x}' = \begin{bmatrix} -3 & \sqrt{2} \\ \sqrt{2} & -2 \end{bmatrix} \vec{x}\]
\[\lambda = -1, -4\]\[\vec{v} = \begin{bmatrix} 1 \\ \sqrt{2} \end{bmatrix}, \begin{bmatrix} -\sqrt{2} \\ 1 \end{bmatrix}\]
A phase portrait on a Cartesian coordinate system with axes x_1 and x_2. The origin is a nodal sink. Two straight-line trajectories correspond to the 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 other trajectories are curved and approach the origin. Arrows on all trajectories point toward the origin, indicating that as time t increases, solutions decay to zero. The curves are tangent to the eigenvector associated with the dominant eigenvalue \lambda = -1 as they approach the origin.
Phase portrait showing a nodal sink at the origin.

Which one is asymptote into origin?

Origin is \(t = \infty\) because \(e^{\lambda t} \to 0\) as \(t \to \infty\).

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

So, the eigenvector w/ \(\lambda = -1\) is more important \(\to\) asymptote.

The origin is a nodal sink

  • point: the origin
  • solutions: go into origin
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Phase Portrait Analysis: Improper Nodal Source

Often also called improper nodal sink.

Follow asymptote into/out of origin

System of Equations

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

Eigenvalues and Eigenvectors:

\[ \lambda = 2, 4 \]\[ \vec{v} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix} \]
A phase portrait on a Cartesian coordinate system with axes x_1 and x_2. The origin is an improper nodal source. Multiple green solution curves originate from the origin and move outward into all four quadrants. Two straight-line trajectories are visible corresponding to the eigenvectors: one along the line defined by the vector \begin{bmatrix} 1 \\ 3 \end{bmatrix} with eigenvalue \lambda = 2, and another along the line defined by the vector \begin{bmatrix} 1 \\ 1 \end{bmatrix} with eigenvalue \lambda = 4. Arrows on the curves indicate movement away from the origin as time t increases.
Phase portrait showing trajectories leaving the origin.

\( \begin{bmatrix} 1 \\ 3 \end{bmatrix}, \lambda = 2 \)

\( \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \lambda = 4 \)

Origin: \( t = -\infty \)

\( e^{2t} > e^{4t} \)

Solutions follow \( \begin{bmatrix} 1 \\ 3 \end{bmatrix} \) (\( \lambda = 2 \)) when they leave origin.

Origin is an improper nodal source

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Eigenvalues are Real and Have Opposite Signs

Consider the linear system of differential equations:

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

The eigenvalues and corresponding eigenvectors for this system are:

\[ \lambda = 3, -1 \]\[ \vec{v} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 1 \\ -2 \end{bmatrix} \]

Phase Portrait Analysis

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 corresponds to the eigenvector \begin{bmatrix} 1 \\ 2 \end{bmatrix} with eigenvalue \lambda = 3; arrows on this line point away from the origin, indicating instability. The other line corresponds to the eigenvector \begin{bmatrix} 1 \\ -2 \end{bmatrix} with eigenvalue \lambda = -1; arrows on this line point toward the origin, indicating stability in that direction. Hyperbolic trajectories fill the regions between these lines, curving toward the unstable eigenvector as time increases.

Conclusion:

The origin is a saddle point.

  • For \(\lambda = 3\), the eigenvector is \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\).
  • For \(\lambda = -1\), the eigenvector is \(\begin{bmatrix} 1 \\ -2 \end{bmatrix}\).
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\(\lambda\)'s are real and repeated

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

\(\lambda = 1, 1\)

\(\vec{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\)

enough eigenvectors

origin is a proper nodal source

no preferred asymptote

A phase portrait of a proper nodal source in a 2D Cartesian coordinate system with axes x_1 and x_2. Numerous straight-line trajectories radiate outward from the origin in all directions, indicated by arrows pointing away from the center. Two specific vectors are labeled: the horizontal vector is labeled with the eigenvector \begin{bmatrix} 1 \\ 0 \end{bmatrix} and eigenvalue \lambda = 1, and the vertical vector is labeled with the eigenvector \begin{bmatrix} 0 \\ 1 \end{bmatrix} and eigenvalue \lambda = 1. The symmetry of the outward flow shows no preferred direction or asymptote.
Phase portrait showing a proper nodal source where all trajectories radiate from the origin.

both \(\lambda\)'s are the same, so both eigenvectors are equally important \(\rightarrow\) solutions are along linear combos of them

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Phase Portrait Analysis: Improper Nodal Sink

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

Eigenvalues and Eigenvectors:

\[ \lambda = -3, -3 \]

\[ \vec{v} = \begin{bmatrix} 3 \\ 2 \end{bmatrix} \]

Defect of one

A phase portrait on a Cartesian coordinate system with axes x1 and x2. A single straight-line solution is shown along the eigenvector  \vec{v} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}  for  \lambda = -3 . All other trajectories are curved in an 'S' shape, resembling an incomplete spiral, and all arrows point toward the origin (0,0). This represents an improper nodal sink.
Figure 1: Phase portrait for a system with a repeated negative eigenvalue and a defect of one.

Key Observations

  • No second straight line solution visible.
  • "S" shape is typical ("incomplete" spiral).
Improper nodal sink
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Complex \(\lambda\)'s

\[ \vec{x}' = \begin{bmatrix} 2 & -5 \\ 1 & -2 \end{bmatrix} \vec{x} \]

Eigenvalues:

\[ \lambda = i, -i \]

Eigenvectors:

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

purely imaginary : ovals

how to orient them?

Two small sketches of ellipses on Cartesian axes. The first ellipse is oriented with its major axis along the line y = x, and the second is oriented with its major axis along the line y = -x. They are separated by the word 'or'.

Pick an eigenvector:

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

The vectors \(\begin{bmatrix} 2 \\ 1 \end{bmatrix}\) and \(\begin{bmatrix} 1 \\ 0 \end{bmatrix}\) tell us the direction of the major axis.

* incomplete see next page

A phase portrait showing concentric elliptical trajectories centered at the origin in the x_1, x_2 plane. The ellipses are elongated along an axis in the first and third quadrants. Green arrows on the trajectories indicate a counter-clockwise (CCW) direction. Two vectors are drawn from the origin: one labeled \begin{bmatrix} 2 \\ 1 \end{bmatrix} and another labeled \begin{bmatrix} 1 \\ 0 \end{bmatrix}. A red vector labeled \begin{bmatrix} 2 \\ 1 \end{bmatrix} points from the tip of the horizontal vector.

Direction:

pick \(\vec{x}\), look at \(\vec{x}'\)

if \(\vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\), \(\vec{x}' = \begin{bmatrix} 2 \\ 1 \end{bmatrix}\)

right, up so CCW (counter-clockwise)

origin is a center

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Ellipse Orientation

The eigenvectors are given by:

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

These vectors form a parallelogram with the real and imaginary components:

\[ \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \quad \pm \begin{bmatrix} 1 \\ 0 \end{bmatrix} \]
A Cartesian coordinate system with axes labeled x1 and x2. A red parallelogram is centered at the origin. The vertices and sides 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 a vertex in the first quadrant, and  \begin{bmatrix} 1 \\ 0 \end{bmatrix}  and its negative  -\begin{bmatrix} 1 \\ 0 \end{bmatrix}  define horizontal segments. A small black ellipse is sketched inside this parallelogram.
Figure 1: Parallelogram formed by the real and imaginary parts of the eigenvectors on the \(x_1-x_2\) plane.

The ellipse is inside the parallelogram.

A conceptual diagram showing a blue ellipse perfectly inscribed within a red tilted parallelogram, illustrating the geometric relationship between the eigenvector components and the resulting trajectory shape.
Figure 2: Detail of the ellipse inscribed within the parallelogram.
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Linear Systems: Complex Eigenvalues and Spiral Sinks

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

\[ \lambda = -1 + i, \quad -1 - i \]

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

Orient the "ovals" the same way as in the previous case.

This time, the real part of \( \lambda \) is \( -1 \), so the \( e^{-t} \) factor drives solutions into the origin. The ovals shrink as they go.

A phase portrait on a Cartesian coordinate system with axes labeled  x_1  and  x_2 . A green spiral curve starts from the outer regions and spirals inward toward the origin in a clockwise direction. Arrows on the spiral indicate the direction of motion toward the center, representing a spiral sink or stable spiral.

Spiral source/sink

The trajectory shown is a spiral sink because the real part of the eigenvalue is negative, causing the magnitude to decay over time.

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One Unusual Case: One Eigenvalue is 0

Consider the system of differential equations:

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

The eigenvalues and corresponding eigenvectors for this system are:

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

General Solution

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

Note on the first term:

Everything on this line is a solution to \( \vec{x}' = A\vec{x} \) (a whole line of equilibria).

Phase Portrait

A phase portrait on a Cartesian coordinate system with axes x1 and x2. A line of equilibria is shown along the eigenvector  \begin{bmatrix} -2 \\ 1 \end{bmatrix}  corresponding to  \lambda = 0 . Other trajectories are straight lines parallel to the eigenvector  \begin{bmatrix} 1 \\ 2 \end{bmatrix}  corresponding to  \lambda = 5 . Since  \lambda = 5  is positive, arrows on these trajectories point away from the line of equilibria, indicating an unstable growth direction.
Figure 1: Phase portrait showing a line of equilibria for the case where one eigenvalue is zero.