3.5 Two dimensional systems and their vector fields

Let us take a moment to talk about constant coefficient linear homogeneous systems in the plane. Much intuition can be obtained by studying this simple case. Suppose we use coordinates $(x,y)$ for the plane as usual, and suppose $P=\left[\begin{array}{cc}{a}&{b}\\{c}&{d}\end{array}\right]$ is a $2\, \times \, 2$ matrix. Consider the system

$\label{pln:eq} \begin{bmatrix} x \\ y \end{bmatrix} ' = P \begin{bmatrix} x \\ y \end{bmatrix} \qquad \text{or} \qquad \begin{bmatrix} x \\ y \end{bmatrix} ' = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}.$

(3.5.1)

The system is autonomous (compare this section to Section 1.6) and so we can draw a vector field (see end of Section 3.1). We will be able to visually tell what the vector field looks like and how the solutions behave, once we find the eigenvalues and eigenvectors of the matrix $P$ . For this section, we assume that $P$ has two eigenvalues and two corresponding eigenvectors.

Six trajectories, each marked with a dot where it starts, wind outward from the origin in the same direction of rotation, looping steadily wider until they leave the frame — an unstable spiral, with a further dot at the origin marking the equilibrium. The two sliders produce every other case in the section's table: raising the lower-left entry uncoils the spirals into a node whose curves run out along straight directions, then into a saddle, while lowering the diagonal entry reverses the drift into closed loops and finally an inward spiral. Adjustable parameters: Diagonal entry a — shifts both eigenvalues (stability) (a) = 1, Lower-left entry c — sign of the discriminant (spiral vs node) (c) = -4. Viewing window: x from -6.33 to 6.33, y from -3.92 to 3.92. — Lebl's summary table lists six behaviors; this one family produces all six. For x′ = Ax with A = [[a, 1], [c, a]] the eigenvalues are exactly λ = a ± √c, so the two sliders are the two things that decide the picture: a shifts both eigenvalues (stability), and the sign of c decides complex versus real. It opens on Lebl's own spiral source [[1, 1], [−4, 1]], λ = 1 ± 2i. Drag c up from −4 and the spiral uncoils into an unstable node as the discriminant passes zero — at c = 0 exactly you get the defective [[a, 1], [0, a]], the shape of §3.7's [[3, 1], [0, 3]]. At c = a² one eigenvalue is 0, the singular case this section's last paragraph sets aside; past it the two real eigenvalues have opposite signs and the node becomes a saddle. Then put c back to −4 and drag a down through 0: spiral source → Lebl's own center [[0, 1], [−4, 0]], λ = ±2i → spiral sink, λ = −1 ± 2i.

1

Suppose that the eigenvalues of $P$ are real and positive. We find two corresponding eigenvectors and plot them in the plane. For example, take the matrix $\begin{bmatrix} 1&1 \\ 0&2 \end{bmatrix}$ . The eigenvalues are 1 and 2 and corresponding eigenvectors are $\begin{bmatrix} 1\\0 \end{bmatrix}$ and $\begin{bmatrix} 1 \\1 \end{bmatrix}$ . See Figure $\PageIndex{1}$ .

Plot on a grid from -3 to 3 showing only two red eigenvector segments from the origin: one along the 45-degree diagonal to (1, 1), one along the horizontal axis to (1, 0). — Figure $\PageIndex{1}$ : Eigenvectors of $P$ .

Now suppose that $x$ and $y$ are on the line determined by an eigenvector $\vec{v}$ for an eigenvalue $\lambda$ . That is, $\begin{bmatrix} x \\ y \end{bmatrix} = a \vec{v}$ for some scalar $a$ . Then

$\begin{bmatrix} x\\y \end{bmatrix}' = P \begin{bmatrix} x\\y \end{bmatrix} = P(a \vec{v}) = a(P \vec{v}) = a \lambda \vec{v} \nonumber$

The derivative is a multiple of $\vec{v}$ and hence points along the line determined by $\vec{v}$ . As $\lambda > 0$ , the derivative points in the direction of $vec{v}$ when $\alpha$ is positive and in the opposite direction when $\alpha$ is negative. Let us draw the lines determined by the eigenvectors, and let us draw arrows on the lines to indicate the directions. See Figure $\PageIndex{2}$ .

We fill in the rest of the arrows for the vector field and we also draw a few solutions. See Figure $\PageIndex{3}$ . Notice that the picture looks like a source with arrows coming out from the origin. Hence we call this type of picture a source or sometimes an unstable node.

Arrows point away from the origin everywhere. Two straight guide lines mark the eigendirections: a rising diagonal drawn dashed, and a horizontal one along the axis, where the solid axis fills the dashes so it reads solid. A family of trajectories curves outward between them, six marked with a dot where they start and a seventh dot at the origin marking the equilibrium. A readout gives the eigenvalues and calls the origin an unstable node; dragging the second eigenvalue below zero switches it to a saddle. Adjustable parameter: Second eigenvalue b — Lebl's case 1 is b = 2; below 0 the source becomes a saddle (b) = 2. Viewing window: x from -6.33 to 6.33, y from -3.92 to 3.92. — Lebl's own case 1, x′ = Ax with A = [[1, 1], [0, 2]]: the eigenvalues are 1 and 2 and the eigenvectors are (1, 0) and (1, 1), exactly as the text computes, so the two straight rays you see are those eigendirections. The slider carries the second eigenvalue. Because A stays upper triangular its eigenvalues are always 1 and b, and the second eigenvector is always (1, b − 1) — so dragging b does two things at once, tilting that ray and changing how fast solutions run out along it. Note the existing six-pictures explorer cannot draw this matrix: its family [[a, 1], [c, a]] forces both diagonal entries equal.

Plot on a grid from -3 to 3 with the eigenvector lines through the origin drawn in black, the diagonal y equals x and the horizontal axis, and red arrows to (1, 1) and (1, 0) marking direction. — Figure $\PageIndex{2}$ : Eigenvectors of $P$ with directions.

Vector field of blue arrows on a square from -3 to 3, all pointing away from the origin, with red trajectories running outward along the horizontal axis and the diagonal y equals x: a source. — Figure $\PageIndex{3}$ : Example source vector field with eigenvectors and solutions.

2

Suppose both eigenvalues were negative. For example, take the negation of the matrix in case 1, $\begin{bmatrix} -1 & -1\\ 0 & -2 \end{bmatrix}$ . The eigenvalues are -1 and -2 and corresponding eigenvectors are the same, $\begin{bmatrix} 1\\0 \end{bmatrix}$ and $\begin{bmatrix}1\\1\end{bmatrix}$ . The calculation and the picture are almost the same. The only difference is that the eigenvalues are negative and hence all arrows are reversed. We get the picture in Figure $\PageIndex{4}$ . We call this kind of picture a sink or sometimes a stable node.

Interactive figureSink: the source matrix negated, with the sign of the eigenvalues on a sliderDrag the Overall sign s of s·[[1,1],[0,2]] slider from -1.5 to 1.5.

XYZ Graph · viewer build 5edf91b

This section builds the sink by negating case 1: A = [[−1, −1], [0, −2]], eigenvalues −1 and −2, the same eigenvectors (1, 0) and (1, 1). "The only difference is that the eigenvalues are negative and hence all arrows are reversed" — so here that sentence is the slider. The matrix is s·[[1, 1], [0, 2]], and s = −1 is Lebl's sink exactly. Drag s up to +1 and you land on Figure 3's source, arrow for arrow; pass through s = 0 and the matrix is zero, every point is an equilibrium and the trajectories stop dead. Scaling a matrix scales its eigenvalues but never moves its eigenvectors, which is why the two straight rays hold still through the whole drag.

Vector field of blue arrows on a square from -3 to 3, all pointing toward the origin, with red trajectories flattening onto the horizontal axis and sliding in without circling: a sink. — Figure $\PageIndex{4}$ : Example sink vector field with eigenvectors and solutions.

3

Suppose one eigenvalue is positive and one is negative. For example the matrix $\left[\begin{array}{cc}{1}&{1}\\{0}&{-2}\end{array}\right]$ . The eigenvalues are $1$ and $−2$ and corresponding eigenvectors are $\left[\begin{array}{c}{1}\\{0}\end{array}\right]$ and $\left[\begin{array}{c}{1}\\{-3}\end{array}\right]$ .

We reverse the arrows on one line (corresponding to the negative eigenvalue) and we obtain the picture in Figure $\PageIndex{5}$ . We call this picture a saddle point.

Arrows and a family of trajectories make an X through the origin, sweeping in along one direction and leaving along the other; six are marked with a dot where they start, and a seventh dot at the origin marks the equilibrium. Two guide lines mark the eigendirections: a steep one drawn dashed, and a horizontal one along the axis, where the solid axis fills the dashes so it reads solid. A readout gives the eigenvalues and calls the origin a saddle; dragging the unstable eigenvalue below zero switches it to a stable node. Adjustable parameter: Unstable eigenvalue p — Lebl's saddle is p = 1; below 0 both are negative and it is a sink (p) = 1. Viewing window: x from -6.33 to 6.33, y from -3.92 to 3.92. — The section's saddle is A = [[1, 1], [0, −2]] — eigenvalues 1 and −2, eigenvectors (1, 0) and (1, −3), which is the steep incoming line you can see in the raster this replaces. Here the positive eigenvalue is the slider while the negative one is pinned at −2, so the incoming direction never moves and you are watching only the outgoing one. At p = 1 this is Lebl's picture. Drag p down through 0 and the outgoing arm loses its push, then reverses: with both eigenvalues negative every trajectory now falls into the origin and the saddle has become a sink. A saddle is exactly the case where the two eigenvalues straddle zero.

Phase portrait of a saddle on axes from -3 to 3: red trajectories in a blue vector field sweep in along a steep eigenvector line of slope about -3, then out along the horizontal axis. — Figure $\PageIndex{5}$ : Example saddle vector field with eigenvectors and solutions.

For the next three cases we will assume the eigenvalues are complex. In this case the eigenvectors are also complex and we cannot just plot them in the plane.

4

Suppose the eigenvalues are purely imaginary. That is, suppose the eigenvalues are $\pm ib$ . For example, let $P = \begin{bmatrix} 0&1\\-4&0\end{bmatrix}$ . The eigenvalues turn out to be $\pm 2i$ and eigenvectors are $\begin{bmatrix} 1 \\ 2i \end{bmatrix}$ and $\begin{bmatrix} 1 \\ -2i \end{bmatrix}$ . Consider the eigenvalue $2i$ and its eigenvector $\begin{bmatrix} 1\\ 2i \end{bmatrix}$ . The real and imaginary parts of $\vec{v} e^{i 2t}$ are

$Re \begin{bmatrix} 1\\2i \end{bmatrix} e^{i2t} = \begin{bmatrix} cos(2t)\\-2sin(2t) \end{bmatrix},\quad Im \begin{bmatrix} 1\\2i \end{bmatrix} e^{i2t} = \begin{bmatrix} sin(2t) \\2cos(2t) \end{bmatrix} \nonumber$

We can take any linear combination of them to get other solutions, which one we take depends on the initial conditions. Now note that the real part is a parametric equation for an ellipse. Same with the imaginary part and in fact any linear combination of the two. This is what happens in general when the eigenvalues are purely imaginary. So when the eigenvalues are purely imaginary, we get ellipses for the solutions. This type of picture is sometimes called a center. See Figure $\PageIndex{6}$ .

Center vector field on axes from -3 to 3: blue arrows circulate clockwise and five nested red loops, taller than wide, close on themselves around the origin. — Figure $\PageIndex{6}$ : Example center vector field.

5

Now suppose the complex eigenvalues have a positive real part. That is, suppose the eigenvalues are $a \pm ib$ for some $a > 0$ . For example, let $P = \begin{bmatrix} 1&1 \\ -4&1 \end{bmatrix}$ . The eigenvalues turn out to be $1 \pm 2i$ and eigenvectors are $\begin{bmatrix}1\\2i \end{bmatrix}$ and $\begin{bmatrix} 1 \\ -2i \end{bmatrix}$ . We take $1\pm 2i$ and its eigenvector $\begin{bmatrix} 1 \\ 2i \end{bmatrix}$ and find the real and imaginary of $\vec{v}e^{(1+2i)t}$ are

$Re \begin{bmatrix} 1\\2i \end{bmatrix} e^{(1+2i)t} =e^t \begin{bmatrix} cos(2t)\\-2sin(2t) \end{bmatrix} \quad Im \begin{bmatrix} 1\\2i \end{bmatrix} e^{(1+2i)t} =e^t \begin{bmatrix} sin(2t) \\2cos(2t) \end{bmatrix} \nonumber$

Note the $e^t$ in front of the solutions. This means that the solutions grow in magnitude while spinning around the origin. Hence we get a spiral source. See Figure $\PageIndex{7}$ .

Spiral source vector field on axes from -3 to 3: red curves spiral clockwise outward from the origin, growing steadily, with blue arrows leaning outward everywhere. — Figure $\PageIndex{7}$ : Example spiral source vector field.

6

Finally suppose the complex eigenvalues have a negative real part. That is, suppose the eigenvalues are $-a \pm ib$ for some $a > 0$ . For example, let $P = \begin{bmatrix} -1& -1 \\ 4 & -1\end{bmatrix}$ . The eigenvalues turn out to be $-1 \pm 2i$ and eigenvectors are $\begin{bmatrix} 1\\ -2i \end{bmatrix}$ and $\begin{bmatrix} 1\\ 2i \end{bmatrix}$ . We take $-1-2i$ and its eigenvector $\begin{bmatrix} 1\\ 2i \end{bmatrix}$ and find the real and imaginary of $\vec{v} e^{(-1-2i)t}$ are

$Re \begin{bmatrix} 1\\2i \end{bmatrix} e^{(-1-2i)t} =e^{-t} \begin{bmatrix} cos(2t)\\ 2sin(2t) \end{bmatrix} \quad Im \begin{bmatrix} 1\\2i \end{bmatrix} e^{(-1-2i)t} =e^{-t} \begin{bmatrix} -sin(2t) \\2cos(2t) \end{bmatrix} \nonumber$

Note the $e^{-t}$ in front of the solutions. This means that the solutions shrink in magnitude while spinning around the origin. Hence we get a spiral sink. See Figure $\PageIndex{8}$ .

Spiral sink vector field on axes from -3 to 3: red trajectories wind counterclockwise inward from the edges, tightening onto the origin as the blue arrows tilt inward. — Figure $\PageIndex{8}$ : Example spiral sink vector field.

We summarize the behavior of linear homogeneous two dimensional systems given by a nonsingular matrix in Table $\PageIndex{1}$ . Systems where one of the eigenvalues is zero (the matrix is singular) come up in practice from time to time, see Example 3.1.2, and the pictures are somewhat different (simpler in a way). See the exercises.

Table $\PageIndex{1}$ : Summary of behavior of linear homogeneous two dimensional systems.
Eigenvalues	Behavior
real and both positive	source / unstable node
real and both negative	sink / stable node
real and opposite signs	saddle
purely imaginary	center point / ellipses
complex with positive real part	spiral source
complex with negative real part	spiral sink

Adapted from Differential Equations for Engineers by Jiří Lebl (https://www.jirka.org/diffyqs/), © Jiří Lebl, licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0.