Topic 4 of 6

Systems and phase portraits

Step 20 · Differential equations topics

Write linked differential equations as x′ = Ax, solve them with eigenvalues and eigenvectors, and read the long-run behavior from a phase portrait.

Systems as matrix equations

Two quantities that change together, like two populations or two tanks, give a system: x′ = ax + by and y′ = cx + dy. Write it as one matrix equation, x′ = Ax.

\begin{pmatrix} x \\ y \end{pmatrix}' = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}

Example: Writing a system

Write x′ = 3x + y, y′ = x + 3y as a matrix equation.

The coefficients form the rows.

Answerx′ = Ax with A = $\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$

Example: A second-order equation as a system

Write y″ + 3y′ + 2y = 0 as a system.

Let x₁ = y and x₂ = y′.
Then x₁′ = x₂ and x₂′ = −2x₁ − 3x₂.

Answerx′ = Ax with A = $\begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}$

Eigenvalue solutions

If Av = λv, then $\mathbf{x} = e^{\lambda t}\mathbf{v}$ solves x′ = Ax: it moves along the line through v, growing or shrinking. With two independent eigenvectors, the general solution combines two of these.

\mathbf{x}(t) = c_1e^{\lambda_1 t}\mathbf{v}_1 + c_2e^{\lambda_2 t}\mathbf{v}_2

Example: Two real eigenvalues

Solve x′ = Ax with A = $\begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$ .

The eigenvalues are 4 and 2, with eigenvectors ⟨1, 1⟩ and ⟨1, −1⟩.
Each gives a straight-line solution.

Answer $\mathbf{x} = c_1e^{4t}\langle 1, 1\rangle + c_2e^{2t}\langle 1, -1\rangle$

Example: Complex eigenvalues

What do solutions of x′ = Ax look like for A = $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ ?

The eigenvalues are ±i.
x = cos t, y = sin t is one solution: it runs round the unit circle.

AnswerCircles around the origin

Phase portraits

A phase portrait draws solution paths in the xy-plane, with the direction of motion. The eigenvalues of A decide the picture around the origin, where the system is at rest.

Eigenvalues	Picture	Stable?
Both negative	Node: all paths flow in	Yes
Both positive	Node: all paths flow out	No
Opposite signs	Saddle	No
Complex, real part negative	Spiral in	Yes
Complex, real part positive	Spiral out	No
Purely imaginary	Center: closed loops	Neither

A saddle, for x′ = x and y′ = −y: paths come in along the y-axis and leave along the x-axis, bending away from the origin.

Example: Classifying

Classify the origin for x′ = Ax with A = $\begin{pmatrix} -2 & 0 \\ 0 & -5 \end{pmatrix}$ .

The eigenvalues are −2 and −5, both negative.

AnswerA stable node: every path flows into the origin

Example: A spiral

Classify the origin for A = $\begin{pmatrix} -1 & -2 \\ 2 & -1 \end{pmatrix}$ .

The eigenvalues are −1 ± 2i.
Their real part, −1, is negative.

AnswerA stable spiral: paths wind into the origin

Try it yourself

Work each one on paper first, then open the answer to check.

Write x′ = 2x − y, y′ = 4x + y as x′ = Ax.
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$\begin{pmatrix} 2 & -1 \\ 4 & 1 \end{pmatrix}$
What are the eigenvalues of $\begin{pmatrix} 2 & 0 \\ 0 & -3 \end{pmatrix}$ , and what kind of point is the origin?
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2 and −3: a saddle
The eigenvalues are −1 and −4. Is the origin stable?
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Yes: a stable node
The eigenvalues are 2 ± 3i. What does the phase portrait show?
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An unstable spiral
Av = 5v. Which solution of x′ = Ax does v give?
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$e^{5t}\mathbf{v}$
The eigenvalues are ±4i. What do the paths look like?
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Closed loops around the origin: a center