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IB Mathematics AI AHL Coupled differential equations MAI Study Notes - New Syllabus

IB Mathematics AI AHL Coupled differential equations MAI Study Notes

LEARNING OBJECTIVE

  • Phase portrait for the solutions of coupled differential equations

Key Concepts: 

  • Phase Portraits in Coupled Differential Equations
  • Eigenvalues and Eigenvectors

MAI HL and SL Notes – All topics

PHASE PORTRAITS IN COUPLED DIFFERENTIAL EQUATIONS

Phase Portraits in Coupled Differential Equations

A phase portrait is a graphical representation that shows the trajectories of a dynamical system in the phase plane, illustrating how the system evolves over time from different initial conditions.

Consider the system of coupled first-order differential equations:

\(\frac{dx}{dt} = ax + by\)
\(\frac{dy}{dt} = cx + dy\)

Here, \(a\), \(b\), \(c\), and \(d\) are constants, and the system describes the rate of change of two variables \(x\) and \(y\) with respect to time \(t\).

Phase portraits help visualize the behavior of solutions for all initial points \((x_0, y_0)\) by plotting trajectories (paths) followed by the system in the plane.

Example 

Sketch some trajectories for the system: \[ \begin{aligned} x_1′ &= x_1 + 2x_2 \\ x_2′ &= 3x_1 + 2x_2 \end{aligned} \quad \Rightarrow \quad \vec{x}’ = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} \vec{x} \]

▶️ Answer/Explanation

We choose some points in the phase plane and compute the direction vectors by multiplying with the system matrix:

\[ \vec{x} = \begin{pmatrix} -1 \\ 1 \end{pmatrix} \Rightarrow \vec{x}’ = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \] \[ \vec{x} = \begin{pmatrix} 2 \\ 0 \end{pmatrix} \Rightarrow \vec{x}’ = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 6 \end{pmatrix} \] \[ \vec{x} = \begin{pmatrix} -3 \\ -2 \end{pmatrix} \Rightarrow \vec{x}’ = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} \begin{pmatrix} -3 \\ -2 \end{pmatrix} = \begin{pmatrix} -7 \\ -13 \end{pmatrix} \]                                                      Interpretation:

  • At point \( (-1, 1) \), the direction vector is \( \langle 1, -1 \rangle \).
  • At point \( (2, 0) \), the vector points in direction \( \langle 2, 6 \rangle \).
  • At point \( (-3, -2) \), the direction is \( \langle -7, -13 \rangle \).

This helps sketch the flow of solutions (trajectories) in the phase portrait.

Key Features of Phase Portraits

Phase portraits consist of trajectories, also called orbits, representing solutions to the system for different initial conditions.

  • Equilibrium points: Points where \(\frac{dx}{dt} = 0\) and \(\frac{dy}{dt} = 0\). The system remains at rest here.
  • Stability: Determines whether trajectories move towards or away from equilibrium points.
  • Types of equilibrium points: Nodes, saddles, spirals, centers, depending on system parameters.
  • Direction fields: Vector fields showing instantaneous direction of movement at any point.

Applications of Phase Portrait Analysis

Phase portraits are used widely in physics, biology, economics, and engineering to study systems such as:

  • Mechanical oscillators (e.g., pendulums, springs)
  • Electrical circuits (e.g., RLC circuits)
  • Population dynamics (predator-prey models)
  • Control systems stability
  • Neural networks and biological rhythms

They provide visual insights into system stability, oscillations, and possible bifurcations.

EIGENVALUES AND THEIR IMPACT ON SOLUTIONS

Eigenvalues and Their Impact on Solutions

To analyze the system, we write it in matrix form:

\(\frac{d}{dt} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}\)

The eigenvalues \(\lambda_1, \lambda_2\) of the coefficient matrix determine the nature of the equilibrium point at the origin:

Real and distinct eigenvalues

•If both $\lambda_1, \lambda_2 > 0$: unstable node
•If both $\lambda_1, \lambda_2 < 0$: stable node
•If $\lambda_1 < 0 < \lambda_2$: saddle point (unstable)

Real and repeated eigenvalues

• If there are enough linearly independent eigenvectors, we get a star node.
• If not, the system is an improper node (trajectories curve into/out of the same direction).
• Stability is determined by the sign of the repeated eigenvalue.

Complex eigenvalues $\lambda = a \pm bi$

Trajectories are spirals or centers.

• If $a > 0$: spiral out (unstable focus)
• If $a < 0$: spiral in (stable focus)
• If $a = 0$: center (neutral stability)

Zero eigenvalues

• Suggests a line of equilibrium points or degenerate behavior.
• System may not be asymptotically stable even if other eigenvalues are negative.
• Requires deeper qualitative or nonlinear analysis.

Example 

Find the eigenvalues of $ A = \begin{bmatrix} 3 & 4 \\ 2 & 1 \end{bmatrix} $ and interpret what they imply about the system behavior.

▶️ Answer/Explanation

Solve the characteristic equation:
$ \det(A – \lambda I) = \lambda^2 – 4\lambda – 5 = 0 $ Eigenvalues:
\(\lambda_1 = 5\) (positive), \(\lambda_2 = -1\) (negative).
Positive eigenvalue → exponential growth along eigenvector.
Negative eigenvalue → exponential decay along eigenvector.
This indicates a saddle point behavior.

Sketching Trajectories

Trajectories are curves in the phase plane that represent solution paths. To sketch these:

  1. Find equilibrium points by solving \(ax + by = 0\) and \(cx + dy = 0\).
  2. Calculate eigenvalues and eigenvectors of the matrix \(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\).
  3. Plot eigenvectors through the equilibrium point; these form the directions of trajectories.
  4. Determine behavior near equilibrium using eigenvalues’ sign and type.
  5. Draw curves moving along or spiraling around eigenvectors accordingly.

Exact Solutions for Real Distinct Eigenvalues

When eigenvalues \(\lambda_1 \neq \lambda_2\) are real and distinct, the general solution is:

$\begin{bmatrix} x \\ y \end{bmatrix} = C_1 \mathbf{v}_1 e^{\lambda_1 t} + C_2 \mathbf{v}_2 e^{\lambda_2 t} $

where \(\mathbf{v}_1, \mathbf{v}_2\) are eigenvectors associated with \(\lambda_1, \lambda_2\) and \(C_1, C_2\) are constants determined by initial conditions.

These solutions trace out trajectories in the phase plane shaped by the exponential growth/decay along eigenvectors.

Example

Given eigenvalues \(\lambda_1=5\), \(\lambda_2=-1\) with eigenvectors $ \mathbf{v}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} $ Write the general solution of the system.

▶️ Answer/Explanation

The general solution is:
$ \mathbf{x}(t) = c_1 e^{5t} \begin{bmatrix} 2 \\ 1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} $ where \(c_1, c_2\) are arbitrary constants determined by initial conditions.

CONNECTION TO EIGENVALUES AND EIGENVECTORS

Connection to Eigenvalues and Eigenvectors

The solutions to the system $ \frac{dx}{dt} = ax + by, \quad \frac{dy}{dt} = cx + dy $ can be expressed in terms of the eigenvalues and eigenvectors of the system matrix $ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}. $

Specifically, each eigenvalue \(\lambda\) corresponds to an eigenvector \(\mathbf{v}\), and the general solution is a linear combination of terms of the form $ \mathbf{x}(t) = e^{\lambda t} \mathbf{v}. $

The eigenvalues determine the growth, decay, or oscillatory behavior of the solutions, while the eigenvectors define the direction of these trajectories in the phase plane.

The eigenvalues are found by solving $ \det(A – \lambda I) = 0, $ and eigenvectors satisfy $ (A – \lambda I) \mathbf{v} = \mathbf{0}. $

The eigenvalues and eigenvectors fully characterize the system’s behavior near equilibrium:

  • Eigenvalues’ real parts indicate stability (negative = stable, positive = unstable).
  • Imaginary parts cause oscillations (complex conjugate eigenvalues).
  • Eigenvectors determine directions of fastest growth or decay.

Understanding this connection simplifies sketching phase portraits and predicting long-term system behavior.

Example 

Explain how eigenvalues and eigenvectors relate to solutions of $ \frac{d\mathbf{x}}{dt} = A \mathbf{x} $ where \(A\) is a 2×2 matrix.

▶️ Answer/Explanation

Eigenvalues \(\lambda\) solve \(\det(A – \lambda I) = 0\).
Corresponding eigenvectors \(\mathbf{v}\) satisfy \((A – \lambda I)\mathbf{v} = \mathbf{0}\).
Each eigenvalue-eigenvector pair gives a solution of form:
$ \mathbf{x}(t) = e^{\lambda t} \mathbf{v} $ The general solution is a linear combination of these solutions.

Application 

  • Population dynamics: Modeling predator-prey relationships and competing species.
  • Mechanical systems: Analyzing vibrations and stability of coupled oscillators.
  • Electrical circuits: Studying behavior of RLC circuits and feedback systems.
  • Control systems: Determining system stability and response to inputs.
  • Economics: Modeling interacting economic variables or markets over time.

Example

Use the system $\frac{dx}{dt} = 3x + 4y, \quad \frac{dy}{dt} = 2x + y $ to model the interaction between two species populations.

Discuss how eigenvalues help in understanding the stability.

▶️ Answer/Explanation

The eigenvalues (\(5\) and \(-1\)) show one population component grows exponentially, and the other decays.
This indicates unstable coexistence with one species dominating.
The phase portrait with saddle behavior confirms this instability.
Such analysis helps ecologists predict long-term outcomes in predator-prey or competing species models.

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