Phase Portraits

When two quantities change together, like two competing populations, you can follow them as a single moving point $(x, y)$. A phase portrait is a picture of the paths (trajectories) that point can take. For linear systems $\dfrac{dx}{dt} = ax + by$, $\dfrac{dy}{dt} = cx + dy$, the eigenvalues and eigenvectors of the matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ tell you everything: whether paths head into the origin or away from it, whether they spiral, and which way they leave. This page covers the IB Mathematics AI HL scope: exact solutions for real eigenvalues, and qualitative sketches for every case.

Key ideas

The system in matrix form

Write the system as

$$\frac{d}{dt}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}, \qquad \text{or} \qquad \dfrac{d\vec{x}}{dt} = M\vec{x}$$

The origin $(0, 0)$ is an equilibrium point: there both rates are zero, so a point that starts there stays there. In the IB course the eigenvalues are always distinct and non-zero, so the origin is the only equilibrium. In a population model, $x$ and $y$ often measure how far each population is from its stable population values, so the origin represents the equilibrium state.

The eigenvalues are the roots of the characteristic equation

$$\lambda^2 - (a + d)\lambda + (ad - bc) = 0$$

Exact solution for distinct real eigenvalues

If $M$ has distinct real eigenvalues $\lambda_1$ and $\lambda_2$ with eigenvectors $\vec{p}_1$ and $\vec{p}_2$, then the general solution is

$$\begin{pmatrix} x \\ y \end{pmatrix} = A e^{\lambda_1 t}\vec{p}_1 + B e^{\lambda_2 t}\vec{p}_2$$

where the constants $A$ and $B$ come from the initial conditions. (Check: each term satisfies $\dfrac{d\vec{x}}{dt} = M\vec{x}$ because $M\vec{p}_1 = \lambda_1\vec{p}_1$.) The IB guide only requires exact solutions in this case.

Each eigenvector gives a straight-line trajectory through the origin: if you start on the line through $\vec{p}_1$, you stay on it, moving away if $\lambda_1 \gt 0$ and toward the origin if $\lambda_1 \lt 0$. As $t \to \infty$ the term with the larger eigenvalue dominates, which tells you the long-term direction.

The five cases

Eigenvalues	Name	What trajectories do
Both real and positive	Unstable node	All move away from the origin
Both real and negative	Stable node	All move toward the origin
Real, opposite signs	Saddle point	Come in along the negative eigenvalue’s eigenvector, leave along the positive one’s; all others eventually move away
Complex, $p \pm qi$	Spiral	Spiral out if $p \gt 0$ (unstable), in if $p \lt 0$ (stable)
Imaginary, $\pm qi$	Centre	Closed circles or ellipses around the origin: periodic motion

“Stable” means trajectories that start near the origin approach it as $t \to \infty$.

Sketching a phase portrait

1. Find the eigenvalues and classify the origin.
2. Real eigenvalues: draw the eigenvector lines through the origin, with arrows (in for negative $\lambda$, out for positive $\lambda$). Other trajectories curve between them. For a node they leave (or arrive at) the origin tangent to the eigenvector of the eigenvalue closer to zero, and far from the origin run parallel to the other one.
3. Complex or imaginary eigenvalues: find the direction of rotation by checking one point. At $(1, 0)$ the velocity is $\begin{pmatrix} a \\ c \end{pmatrix}$, so if $c \gt 0$ the point is moving up: anticlockwise. If $c \lt 0$: clockwise.
4. Check a few velocity vectors $\left(\dfrac{dx}{dt}, \dfrac{dy}{dt}\right)$ at points like $(1, 0)$, $(0, 1)$ and $(1, 1)$. Trajectories are vertical where $\dfrac{dx}{dt} = 0$ and horizontal where $\dfrac{dy}{dt} = 0$.
5. Add arrows to every trajectory to show the direction of time.

Second-order equations

The equation $\dfrac{d^2x}{dt^2} + a\dfrac{dx}{dt} + bx = 0$ becomes a linear system by letting $y = \dfrac{dx}{dt}$ (as on the Euler’s method for systems page):

$$\frac{dx}{dt} = y, \qquad \frac{dy}{dt} = -bx - ay, \qquad M = \begin{pmatrix} 0 & 1 \\ -b & -a \end{pmatrix}$$

So you can describe the motion of a spring with its phase portrait: a centre means it oscillates forever, a stable spiral means the oscillations die down, and a stable node means it returns to rest without oscillating.

Worked examples

Example 1: A saddle point

Consider $\dfrac{dx}{dt} = x + y$, $\dfrac{dy}{dt} = 4x + y$, with $x(0) = 2$ and $y(0) = 0$.

(a) Find the eigenvalues and eigenvectors, and classify the equilibrium.

(b) Find the exact solution, and describe the long-term behaviour.

Solution.

(a) $M = \begin{pmatrix} 1 & 1 \\ 4 & 1 \end{pmatrix}$. The characteristic equation is

$$\lambda^2 - 2\lambda + (1 - 4) = \lambda^2 - 2\lambda - 3 = (\lambda - 3)(\lambda + 1) = 0$$

so $\lambda = 3$ or $\lambda = -1$.

• $\lambda = 3$: $\begin{pmatrix} -2 & 1 \\ 4 & -2 \end{pmatrix}\begin{pmatrix} p \\ q \end{pmatrix} = \vec{0}$ gives $q = 2p$, so $\vec{p}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$.
• $\lambda = -1$: $\begin{pmatrix} 2 & 1 \\ 4 & 2 \end{pmatrix}\begin{pmatrix} p \\ q \end{pmatrix} = \vec{0}$ gives $q = -2p$, so $\vec{p}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$.

The eigenvalues are real with opposite signs, so the origin is a saddle point.

(b) The general solution is

$$\begin{pmatrix} x \\ y \end{pmatrix} = Ae^{3t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} + Be^{-t}\begin{pmatrix} 1 \\ -2 \end{pmatrix}$$

At $t = 0$: $A + B = 2$ and $2A - 2B = 0$, so $A = B = 1$:

$$x = e^{3t} + e^{-t}, \qquad y = 2e^{3t} - 2e^{-t}$$

As $t \to \infty$, $e^{-t} \to 0$ and $e^{3t}$ grows, so $x$ and $y$ both increase without bound with $\dfrac{y}{x} \to 2$: the trajectory moves away from the origin, approaching the direction of the line $y = 2x$.

Example 2: A stable node

Consider $\dfrac{dx}{dt} = -3x + y$, $\dfrac{dy}{dt} = x - 3y$, with $x(0) = 3$ and $y(0) = 1$. Find the exact solution and describe the phase portrait.

Solution. $M = \begin{pmatrix} -3 & 1 \\ 1 & -3 \end{pmatrix}$, with characteristic equation $\lambda^2 + 6\lambda + 8 = (\lambda + 2)(\lambda + 4) = 0$, so $\lambda = -2$ or $\lambda = -4$.

• $\lambda = -2$: $\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}$ gives $q = p$, so $\vec{p}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.
• $\lambda = -4$: $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ gives $q = -p$, so $\vec{p}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

General solution: $\begin{pmatrix} x \\ y \end{pmatrix} = Ae^{-2t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} + Be^{-4t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$. At $t = 0$: $A + B = 3$ and $A - B = 1$, so $A = 2$, $B = 1$:

$$x = 2e^{-2t} + e^{-4t}, \qquad y = 2e^{-2t} - e^{-4t}$$

Both eigenvalues are negative, so the origin is a stable node: every trajectory approaches $(0, 0)$. The $e^{-4t}$ term dies away faster, so near the origin trajectories come in along the line $y = x$ (the eigenvector of $\lambda = -2$, the eigenvalue closer to zero).

Left: the saddle point of Example 1, with eigenvector lines $y = 2x$ (out, $\lambda = 3$) and $y = -2x$ (in, $\lambda = -1$). Right: the stable node of Example 2.

Example 3: A spiral in a population model

Two species live on an island. Their populations differ from their stable values by $x$ and $y$ (in hundreds), where

$$\frac{dx}{dt} = -x - 2y, \qquad \frac{dy}{dt} = 2x - y$$

Classify the equilibrium, sketch a typical trajectory starting from $(3, 0)$, and describe what happens to the populations.

Solution. $M = \begin{pmatrix} -1 & -2 \\ 2 & -1 \end{pmatrix}$, with characteristic equation $\lambda^2 + 2\lambda + 5 = 0$:

$$\lambda = \frac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i$$

The eigenvalues are complex with negative real part, so the origin is a stable spiral. Direction: at $(3, 0)$ the velocity is $\begin{pmatrix} -3 \\ 6 \end{pmatrix}$, so the point is moving up and to the left: the spiral turns anticlockwise.

Starting from $(3, 0)$, the trajectory winds anticlockwise around the origin, getting closer each turn. So the populations oscillate above and below their stable values, with smaller and smaller swings, and settle at the equilibrium in the long run.

Example 4: A centre from a spring

An undamped spring satisfies $\dfrac{d^2x}{dt^2} + 4x = 0$.

(a) Write this as a system with $y = \dfrac{dx}{dt}$, and classify the equilibrium.

(b) Show that every trajectory lies on an ellipse $4x^2 + y^2 = k$, and describe the trajectory through $(1, 0)$.

Solution.

(a) $\dfrac{dx}{dt} = y$, $\dfrac{dy}{dt} = -4x$, so $M = \begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$. The characteristic equation is $\lambda^2 + 4 = 0$, so $\lambda = \pm 2i$: purely imaginary. The origin is a centre. At $(1, 0)$ the velocity is $\begin{pmatrix} 0 \\ -4 \end{pmatrix}$ (moving down), so trajectories go clockwise.

(b) Differentiate $4x^2 + y^2$ with respect to $t$:

$$\frac{d}{dt}\left(4x^2 + y^2\right) = 8x\frac{dx}{dt} + 2y\frac{dy}{dt} = 8xy + 2y(-4x) = 0$$

So $4x^2 + y^2$ stays constant along every trajectory. Through $(1, 0)$, $k = 4$: the ellipse $4x^2 + y^2 = 4$, which crosses the axes at $x = \pm 1$ and $y = \pm 2$. The spring oscillates between displacements $-1$ and $1$ forever, with maximum speed $2$ as it passes $x = 0$.

Left: the stable spiral of Example 3 ($\lambda = -1 \pm 2i$), turning anticlockwise into the origin. Right: the centre of Example 4 ($\lambda = \pm 2i$), with clockwise ellipses $4x^2 + y^2 = k$.

Common mistakes

Classifying from the matrix entries instead of the eigenvalues. Negative numbers in the matrix don’t make a system stable. For example, $\begin{pmatrix} -1 & 3 \\ 2 & -1 \end{pmatrix}$ has negative diagonal entries, but its eigenvalues are $-1 \pm \sqrt{6}$, about $1.45$ and $-3.45$, so the origin is a saddle. Always find the eigenvalues.

Mixing up the eigenvectors when writing the solution. Each eigenvector must be paired with its own eigenvalue: $Ae^{\lambda_1 t}\vec{p}_1 + Be^{\lambda_2 t}\vec{p}_2$. Check by confirming $M\vec{p}_1 = \lambda_1\vec{p}_1$.

Calling a saddle point stable. A saddle has one direction in, but every trajectory not exactly on that line eventually moves away. In real situations it’s unstable.

Guessing the direction of a spiral or centre. Complex eigenvalues don’t tell you clockwise or anticlockwise. Check the velocity at one point, such as $(1, 0)$, where it is $\begin{pmatrix} a \\ c \end{pmatrix}$.

Drawing trajectories that cross. Trajectories of these systems never cross (except at the equilibrium, which they only approach). If your sketch has two paths crossing, one of them is wrong.

Forgetting arrows, or confusing a spiral with a centre. Without arrows a phase portrait doesn’t say whether paths go in or out. A real part of zero gives closed loops; any non-zero real part turns them into spirals.

Practice

1. (Warm-up) Find the eigenvalues of each matrix $M$ and classify the equilibrium of $\dfrac{d\vec{x}}{dt} = M\vec{x}$.

• (a) $M = \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}$
• (b) $M = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$
• (c) $M = \begin{pmatrix} 0 & -9 \\ 1 & 0 \end{pmatrix}$

Solution

(a) The matrix is diagonal, so $\lambda = 2, 5$: both positive, an unstable node.

(b) $\lambda^2 - 2\lambda + (1 - 4) = \lambda^2 - 2\lambda - 3 = 0$, so $\lambda = 3, -1$: opposite signs, a saddle point.

(c) $\lambda^2 - 0\lambda + (0 + 9) = \lambda^2 + 9 = 0$, so $\lambda = \pm 3i$: imaginary, a centre.

2. (Warm-up) For $\dfrac{dx}{dt} = 2x + y$, $\dfrac{dy}{dt} = x + 2y$, find the eigenvalues and eigenvectors, write down the general solution, and classify the equilibrium.

Solution

$\lambda^2 - 4\lambda + 3 = (\lambda - 1)(\lambda - 3) = 0$, so $\lambda = 1$ or $3$.

• $\lambda = 3$: $\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}$ gives $\vec{p} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.
• $\lambda = 1$: $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ gives $\vec{p} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

$$\begin{pmatrix} x \\ y \end{pmatrix} = Ae^{3t}\begin{pmatrix} 1 \\ 1 \end{pmatrix} + Be^{t}\begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

Both eigenvalues are positive: an unstable node. All trajectories move away from the origin.

3. (Warm-up) For $\dfrac{dx}{dt} = x - 2y$, $\dfrac{dy}{dt} = 3x - 4y$:

• (a) Find the velocity vector at $(1, 0)$ and at $(0, 1)$.
• (b) Find the eigenvalues and classify the equilibrium.

Solution

(a) At $(1, 0)$: $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$. At $(0, 1)$: $\begin{pmatrix} -2 \\ -4 \end{pmatrix}$.

(b) $\lambda^2 - (1 - 4)\lambda + (-4 + 6) = \lambda^2 + 3\lambda + 2 = (\lambda + 1)(\lambda + 2) = 0$, so $\lambda = -1, -2$: both negative, a stable node.

4. (Core) Solve $\dfrac{dx}{dt} = 4x - 2y$, $\dfrac{dy}{dt} = x + y$, with $x(0) = 3$ and $y(0) = 2$. Describe the long-term behaviour of the trajectory.

Solution

$\lambda^2 - 5\lambda + (4 + 2) = (\lambda - 2)(\lambda - 3) = 0$.

• $\lambda = 2$: $\begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}$ gives $\vec{p}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.
• $\lambda = 3$: $\begin{pmatrix} 1 & -2 \\ 1 & -2 \end{pmatrix}$ gives $\vec{p}_2 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

At $t = 0$: $A\begin{pmatrix} 1 \\ 1 \end{pmatrix} + B\begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$, so $A + 2B = 3$ and $A + B = 2$, giving $B = 1$, $A = 1$:

$$x = e^{2t} + 2e^{3t}, \qquad y = e^{2t} + e^{3t}$$

Both eigenvalues are positive (unstable node), so the trajectory moves away from the origin. The $e^{3t}$ terms dominate, so $\dfrac{y}{x} \to \dfrac{1}{2}$: far from the origin the path runs parallel to the direction $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

5. (Core) Consider $\dfrac{dx}{dt} = 2x + 3y$, $\dfrac{dy}{dt} = 2x + y$.

• (a) Show that the eigenvalues are $4$ and $-1$, find corresponding eigenvectors, and classify the equilibrium.
• (b) Find the solution with $x(0) = 1$, $y(0) = -1$, and describe it.
• (c) Find the solution with $x(0) = 4$, $y(0) = 1$, and describe its long-term behaviour.

Solution

(a) $\lambda^2 - 3\lambda + (2 - 6) = \lambda^2 - 3\lambda - 4 = (\lambda - 4)(\lambda + 1) = 0$. For $\lambda = 4$: $\begin{pmatrix} -2 & 3 \\ 2 & -3 \end{pmatrix}$ gives $\vec{p}_1 = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$. For $\lambda = -1$: $\begin{pmatrix} 3 & 3 \\ 2 & 2 \end{pmatrix}$ gives $\vec{p}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$. Opposite signs: a saddle point.

(b) $(1, -1)$ is on the eigenvector line for $\lambda = -1$, so $A = 0$, $B = 1$: $x = e^{-t}$, $y = -e^{-t}$. The point moves along the line $y = -x$ straight into the origin.

(c) $3A + B = 4$ and $2A - B = 1$. Adding: $5A = 5$, so $A = 1$, $B = 1$:

$$x = 3e^{4t} + e^{-t}, \qquad y = 2e^{4t} - e^{-t}$$

The $e^{4t}$ terms dominate, so the trajectory moves away from the origin with $\dfrac{y}{x} \to \dfrac{2}{3}$, approaching the direction $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$.

6. (Core) For $\dfrac{dx}{dt} = x - 5y$, $\dfrac{dy}{dt} = x - 3y$, find the eigenvalues, classify the equilibrium, and state the direction in which trajectories turn.

Solution

$\lambda^2 - (1 - 3)\lambda + (-3 + 5) = \lambda^2 + 2\lambda + 2 = 0$, so

$$\lambda = \frac{-2 \pm \sqrt{4 - 8}}{2} = -1 \pm i$$

Complex with negative real part: a stable spiral. At $(1, 0)$ the velocity is $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$, moving up, so trajectories turn anticlockwise as they spiral into the origin.

7. (Core) Each equation describes the displacement $x$ of a mass on a spring. Write each as a system with $y = \dfrac{dx}{dt}$, classify the equilibrium, and describe the motion.

• (a) $\dfrac{d^2x}{dt^2} + 3\dfrac{dx}{dt} + 2x = 0$
• (b) $\dfrac{d^2x}{dt^2} + 2\dfrac{dx}{dt} + 5x = 0$

Solution

(a) $\dfrac{dx}{dt} = y$, $\dfrac{dy}{dt} = -2x - 3y$, so $M = \begin{pmatrix} 0 & 1 \\ -2 & -3 \end{pmatrix}$ and $\lambda^2 + 3\lambda + 2 = 0$, giving $\lambda = -1, -2$. A stable node: the mass returns to rest at $x = 0$ without oscillating.

(b) $\dfrac{dx}{dt} = y$, $\dfrac{dy}{dt} = -5x - 2y$, so $\lambda^2 + 2\lambda + 5 = 0$, giving $\lambda = -1 \pm 2i$. A stable spiral: the mass oscillates back and forth with decreasing amplitude and settles at $x = 0$. (At $(1, 0)$ the velocity is $\begin{pmatrix} 0 \\ -5 \end{pmatrix}$, so the spiral is clockwise.)

8. (Challenge) Consider $\dfrac{dx}{dt} = x + ky$, $\dfrac{dy}{dt} = 2x - y$, where $k$ is a constant and $k \ne -\dfrac{1}{2}$.

• (a) Show that the eigenvalues satisfy $\lambda^2 = 1 + 2k$.
• (b) Find the values of $k$ for which the origin is a saddle point, and those for which it is a centre.

Solution

(a) $\lambda^2 - (1 - 1)\lambda + (-1 - 2k) = 0$, so $\lambda^2 = 1 + 2k$.

(b) If $k \gt -\dfrac{1}{2}$, then $1 + 2k \gt 0$ and $\lambda = \pm\sqrt{1 + 2k}$: real with opposite signs, a saddle point.

If $k \lt -\dfrac{1}{2}$, then $1 + 2k \lt 0$ and $\lambda = \pm i\sqrt{-(1 + 2k)}$: purely imaginary, a centre.

(At $k = -\dfrac{1}{2}$ both eigenvalues are $0$, which is why that value is excluded.)

9. (Challenge) The deviations $x$ and $y$ (in hundreds) of a prey and a predator population from their equilibrium values are modelled by

$$\frac{dx}{dt} = 0.2x - 0.5y, \qquad \frac{dy}{dt} = 0.5x + 0.2y$$

• (a) Explain the signs of the coefficients $-0.5$ and $0.5$ in context.
• (b) Find the eigenvalues and classify the equilibrium.
• (c) Describe the long-term behaviour of the populations, starting from $(1, 0)$, and comment on whether the model is realistic in the long run.

Solution

(a) More predators than usual ($y \gt 0$) makes the prey decrease, hence $-0.5y$ in $\dfrac{dx}{dt}$. More prey than usual ($x \gt 0$) gives predators more food, so they increase, hence $+0.5x$ in $\dfrac{dy}{dt}$.

(b) $\lambda^2 - 0.4\lambda + (0.04 + 0.25) = \lambda^2 - 0.4\lambda + 0.29 = 0$, so

$$\lambda = \frac{0.4 \pm \sqrt{0.16 - 1.16}}{2} = 0.2 \pm 0.5i$$

Complex with positive real part: an unstable spiral. At $(1, 0)$ the velocity is $\begin{pmatrix} 0.2 \\ 0.5 \end{pmatrix}$, so it turns anticlockwise.

(c) The trajectory spirals anticlockwise away from the origin. The populations oscillate around their equilibrium values with ever-growing swings: prey rise, then predators rise, then prey fall, then predators fall, each cycle larger than the last. This can’t continue forever, since eventually a deviation would make a population negative, so the linear model only describes the behaviour near the equilibrium for a limited time.