Coupled differential equations and phase portraits

IB Mathematics Applications and Interpretation HL· U4.16· 12 min read

1. Structure of Linear Coupled 2D Systems★★★★☆HL only⏱ 15 min

All linear homogeneous 2D coupled systems follow the standard matrix form, where the rates of change of variables x and y are linear combinations of x and y themselves. These systems are widely used to model interacting populations, chemical reaction rates, and compartmental epidemic flows.

📘 Definition

Standard coupled system form

\begin{bmatrix} \frac{dx}{dt} \\ \frac{dy}{dt} \end{bmatrix} = A \begin{bmatrix} x \\ y \end{bmatrix}

A is a constant 2x2 coefficient matrix that fully defines the dynamic behaviour of the system

Example:

For two competing species, A may contain positive growth terms and negative inter-species interaction terms

📐 Worked Example

Write the following coupled system in standard matrix form: dx/dt = 3x - 2y, dy/dt = x + 4y

1
Identify the coefficients of x and y in the dx/dt equation
2
These are 3 and -2, forming the first row of matrix A
3
Identify the coefficients of x and y in the dy/dt equation
4
These are 1 and 4, forming the second row of matrix A
5
$A = \begin{bmatrix} 3 & -2 \\ 1 & 4 \end{bmatrix}$

✓ Quick check

Test your understanding of system structure

Which of the following is a valid linear coupled 2D system?
- dx/dt = 2xy, dy/dt = x + y
- dx/dt = 3x - y, dy/dt = -x + 2y
- dx/dt = t + x, dy/dt = y
Reveal answer
dx/dt = 3x - y, dy/dt = -x + 2y —
All terms are linear in x and y, with no t dependence or product terms

2. Classifying Equilibrium Points via Eigenvalues★★★★☆HL only⏱ 18 min

🔬 Derivation

Goal:

Find the condition for non-trivial equilibrium points

Starting from:

A \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}

1
The trivial equilibrium at (0,0) always exists for homogeneous systems
2
Non-trivial equilibria exist only if det(A) = 0, otherwise (0,0) is the only equilibrium
3
Calculate eigenvalues of A by solving det(A - \lambda I) = 0

Result:

The sign and type of eigenvalues (real, complex, repeated) fully define the equilibrium classification

Eigenvalue type	Equilibrium classification	Stability
Two distinct positive real	Unstable node	All trajectories move away
Two distinct negative real	Stable node	All trajectories converge
One positive, one negative real	Saddle point	Unstable, only 2 trajectories converge
Complex with positive real part	Unstable spiral	Trajectories spiral outwards
Complex with negative real part	Stable spiral	Trajectories spiral inwards
Pure imaginary	Centre	Closed periodic orbits, neutral stability

📐 Worked Example

Classify the equilibrium at (0,0) for system with A = \begin{bmatrix} 1 & 2 \ 2 & 1 \end{bmatrix}

1
Solve characteristic equation det(A - \lambda I) = 0
2
$(1-\lambda)^2 - 4 = 0 \implies \lambda^2 - 2\lambda -3 =0$
3
Factor to get eigenvalues \lambda = 3 and \lambda = -1
4
One positive, one negative real eigenvalue means this is a saddle point, unstable

3. Sketching Fully Labelled Phase Portraits★★★★★HL only⏱ 20 min

📐 Worked Example

Sketch the phase portrait for a stable node with eigenvalues -1 and -2, eigenvectors (1,0) and (0,1)

1
Draw x and y axes, mark the equilibrium point at (0,0)
2
Draw the eigenvector lines along the x and y axes, add arrows pointing towards (0,0)
3
Add curved trajectories that approach the slower eigenvector (eigenvalue -1, x axis) as t \to \infty
4
Label the equilibrium as stable node, add direction arrows to all trajectories

4. Interpreting Phase Portraits for Applied Models★★★★☆HL only⏱ 12 min

For real-world models such as SIR epidemic compartments or competing species populations, phase portraits let you predict long-term outcomes without solving the full analytical solution. You can identify threshold conditions that lead to population extinction, coexistence, or epidemic fade-out.

📐 Worked Example

For a competing species model with a stable node at (200, 150), interpret the long-term behaviour

1
The stable node means all initial population values will converge to this equilibrium point
2
This tells you the two species will coexist at steady state populations of 200 and 150 respectively
3
No matter the initial non-zero population counts, neither species will go extinct

5. Second Order Differential Equations via Coupling★★★★★HL only⏱ 15 min

A second order differential equation of the form d²x/dt² = f(x, dx/dt, t) can always be rewritten as a system of two coupled first order equations. This lets you apply the same numerical and phase portrait techniques to second order models, and it is a recurring IB AI HL Paper 3 task, especially for mechanical systems like a damped oscillator d²x/dt² + a dx/dt + bx = 0.

📘 Definition

Reduction to a coupled first order system

\text{Let } y = \frac{dx}{dt}, \text{ then } \frac{dx}{dt} = y, \quad \frac{dy}{dt} = f(x, y, t)

Introduce a new variable y equal to the first derivative dx/dt. The single second order equation then becomes two first order equations in x and y, which you can solve numerically or analyse with eigenvalues.

Example:

For \frac{d^2x}{dt^2} + 3\frac{dx}{dt} + 2x = 0, set y = dx/dt to get \frac{dx}{dt} = y and \frac{dy}{dt} = -3y - 2x

📐 Worked Example

Write \frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 3x = 0 as a system of coupled first order equations, and state its coefficient matrix.

1
Let y = dx/dt, so the first equation is simply dx/dt = y
2
Rearrange the original equation for the highest derivative: d²x/dt² = -4(dx/dt) - 3x
3
Replace dx/dt with y and d²x/dt² with dy/dt to get the second equation
4
$\frac{dx}{dt} = y, \quad \frac{dy}{dt} = -3x - 4y$
5
In matrix form the coefficient matrix is:
6
$A = \begin{bmatrix} 0 & 1 \\ -3 & -4 \end{bmatrix}$

📐 Worked Example

Use Euler's method with step size h = 0.1 to estimate x(0.2) for \frac{d^2x}{dt^2} = -x, given x(0) = 1 and \frac{dx}{dt}(0) = 0. Compare with the exact solution x = \cos t.

1
Reduce to a system: let y = dx/dt, then dx/dt = y and dy/dt = -x
2
Euler update equations: x_{n+1} = x_n + h,y_n and y_{n+1} = y_n + h(-x_n)
3
Step 1 (t = 0 to 0.1): x_1 = 1 + 0.1(0) = 1, y_1 = 0 + 0.1(-1) = -0.1
4
Step 2 (t = 0.1 to 0.2): x_2 = 1 + 0.1(-0.1) = 0.99, y_2 = -0.1 + 0.1(-1) = -0.2
5
$x(0.2) \approx 0.99$
6
The exact solution gives \cos(0.2) \approx 0.980, so Euler's method slightly overestimates here; a smaller step size h would reduce the error.

6. Common Pitfalls

Wrong move:

Classifying a saddle point as unstable node

Why:

Saddle points have one positive and one negative eigenvalue, not two positive eigenvalues

Correct move:

Always check the sign of both eigenvalues before classification

Wrong move:

Drawing trajectories that cross each other in the phase portrait

Why:

Uniqueness theorem for ODEs guarantees no two distinct trajectories can intersect

Correct move:

Ensure all trajectories never meet except at equilibrium points

Wrong move:

Adding direction arrows pointing away from a stable spiral equilibrium

Why:

Negative real part of complex eigenvalues means trajectories must spiral inwards over time

Correct move:

Verify eigenvalue signs before adding direction arrows

Wrong move:

Forgetting to label eigenvector lines in the phase portrait

Why:

IB exam mark schemes explicitly award marks for correctly drawn and labelled eigenvector lines

Correct move:

Mark and label all eigenvector lines before drawing curved trajectories

Wrong move:

Treating a centre equilibrium as a stable spiral

Why:

Pure imaginary eigenvalues produce closed periodic orbits, not inward spiralling trajectories

Correct move:

Check that the real part of complex eigenvalues is exactly zero before drawing closed orbits

7. Quick Reference Cheatsheet

Eigenvalue condition	Equilibrium type	Stability	Trajectory shape
Two distinct real, both negative	Stable node	Asymptotically stable	Curves converge to equilibrium, tangent to slow eigenvector
Two distinct real, both positive	Unstable node	Unstable	Curves diverge away from equilibrium
Real, opposite signs	Saddle point	Unstable	Two trajectories converge, all others diverge
Complex, negative real part	Stable spiral	Asymptotically stable	Trajectories spiral inwards
Complex, positive real part	Unstable spiral	Unstable	Trajectories spiral outwards
Pure imaginary	Centre	Neutrally stable	Closed elliptical periodic orbits

What's Next

Mastering linear coupled systems and phase portraits gives you the foundation to tackle far more realistic non-linear dynamic systems that appear frequently in IB AI HL Paper 3 problem sets. You will learn to analyse predator-prey Lotka-Volterra models, which produce closed periodic orbits representing natural population cycles, and extend your understanding to non-linear equilibrium classification via Jacobian matrices. You will also connect these concepts to numerical differential equation solving methods like Euler's method, which you can use to approximate trajectories for systems with no analytical closed-form solution. These skills are heavily weighted for the 20% Paper 3 investigation component of your final exam.