Coupled Linear Systems in Matrix Form
A coupled system of two first-order ordinary differential equations has the general form:
An equilibrium point occurs where both rates of change are simultaneously zero ($\mathbf{x}' = \mathbf{0}$), which for $\det(\mathbf{A}) \ne 0$ is the origin $(0, 0)$.
Eigenvalues, Eigenvectors & General Solutions
Assuming solutions of the form $\mathbf{x}(t) = \mathbf{v}e^{\lambda t}$ leads to the eigenvalue equation:
Phase Plane Trajectories & Stability Classification
The trajectories plotted in the $(x, y)$ plane describe the geometric evolution of the system over time:
Classification by Eigenvalues:
- Stable Node (Attractor): Both eigenvalues are real and negative ($\lambda_1 < \lambda_2 < 0$). All trajectories flow into the origin as $t \to \infty$.
- Unstable Node (Repeller): Both eigenvalues are real and positive ($\lambda_1 > \lambda_2 > 0$). All trajectories flow outward away from the origin.
- Saddle Point (Unstable): Real eigenvalues with opposite signs ($\lambda_1 < 0 < \lambda_2$). Trajectories approach along $\mathbf{v}_1$ and diverge along $\mathbf{v}_2$.
- Stable Spiral (Focus): Complex $\lambda = p \pm iq$ with $p < 0$. Trajectories spiral inward toward origin.
- Unstable Spiral (Focus): Complex $\lambda = p \pm iq$ with $p > 0$. Trajectories spiral outward.
- Center (Neutrally Stable): Purely imaginary $\lambda = \pm iq$ ($p = 0$). Trajectories form concentric closed elliptical orbits (e.g. ideal predator-prey cycles).
Write the coupled system $\frac{dx}{dt} = 4x - 2y$ and $\frac{dy}{dt} = x + y$ in matrix form $\mathbf{x}' = \mathbf{A}\mathbf{x}$.
Solution:
Let $\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}$. [A1]
$\frac{d}{dt}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$. [A2]
Find the eigenvalues of matrix $\mathbf{A} = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}$.
Solution:
$\det(\mathbf{A} - \lambda\mathbf{I}) = \det\begin{pmatrix} 4 - \lambda & -2 \\ 1 & 1 - \lambda \end{pmatrix} = 0$. [M1]
$(4 - \lambda)(1 - \lambda) - (-2)(1) = 0 \implies 4 - 5\lambda + \lambda^2 + 2 = 0$. [M1]
$\lambda^2 - 5\lambda + 6 = 0 \implies (\lambda - 2)(\lambda - 3) = 0$. [M1]
$\lambda_1 = 2, \quad \lambda_2 = 3$. [A1]
For the system in Exercise 2 with $\lambda_1 = 2$ and $\lambda_2 = 3$, find the corresponding eigenvectors.
Solution:
For $\lambda_1 = 2$: $(\mathbf{A} - 2\mathbf{I})\mathbf{v}_1 = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies x = y \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. [M1][A1]
For $\lambda_2 = 3$: $(\mathbf{A} - 3\mathbf{I})\mathbf{v}_2 = \begin{pmatrix} 1 & -2 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies x - 2y = 0 \implies \mathbf{v}_2 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$. [M1][A1]
Classify the equilibrium point $(0, 0)$ for the system in Exercises 2 and 3 and state its stability.
Solution:
Both eigenvalues are real and strictly positive: $\lambda_1 = 2 > 0$ and $\lambda_2 = 3 > 0$. [R1]
Classification: **Unstable node** (or nodal source / repeller). [A1]
All non-zero trajectories diverge away from the origin as $t \to \infty$. [R1]
A system has eigenvalues $\lambda_1 = -4$ and $\lambda_2 = 3$. Classify the equilibrium point $(0, 0)$ and sketch or describe the behavior of trajectories in the phase plane.
Solution:
The eigenvalues are real with opposite signs ($\lambda_1 < 0 < \lambda_2$). [R1]
Classification: **Saddle point** (unstable). [A1]
Behavior: Trajectories approach the origin along the line corresponding to eigenvector $\mathbf{v}_1$, but diverge away from the origin along the line of eigenvector $\mathbf{v}_2$. Almost all trajectories bend away, exhibiting hyperbolic geometry. [A2]
Given the system matrix $\mathbf{A} = \begin{pmatrix} -1 & 5 \\ -2 & -3 \end{pmatrix}$, find the eigenvalues and determine if the origin is a stable or unstable spiral.
Solution:
$\det(\mathbf{A} - \lambda\mathbf{I}) = (-1 - \lambda)(-3 - \lambda) - 5(-2) = 0 \implies \lambda^2 + 4\lambda + 3 + 10 = 0$. [M1]
$\lambda^2 + 4\lambda + 13 = 0 \implies \lambda = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3i$. [M1][A1]
The real part is $p = -2 < 0$. [R1]
Therefore, the origin is a **stable spiral** (attractor). Trajectories spiral inward toward $(0, 0)$ as $t \to \infty$. [A1]
Explain why a system with eigenvalues $\lambda = \pm 4i$ produces closed periodic orbits (centers) in the phase plane.
Solution:
The eigenvalues have a real part of zero ($p = 0$). [R1]
Because there is no real exponential decay ($e^{-pt}$) or growth ($e^{pt}$), the solution components consist purely of sines and cosines: $c_1 \cos(4t) + c_2 \sin(4t)$. [R1]
This creates perpetual closed concentric periodic cycles (center equilibrium) with period $T = \frac{2\pi}{4} = \frac{\pi}{2}$. [A1]
In a simplified predator-prey model with prey population $x$ and predator population $y$:
$$\frac{dx}{dt} = 0.4x - 0.02xy, \qquad \frac{dy}{dt} = -0.3y + 0.01xy$$Find the non-zero equilibrium point $(\bar{x}, \bar{y})$ where both populations remain stationary.
Solution:
Set $\frac{dx}{dt} = 0 \implies x(0.4 - 0.02y) = 0$. For $x \ne 0$: $0.02y = 0.4 \implies y = \frac{0.4}{0.02} = 20$. [M1][A1]
Set $\frac{dy}{dt} = 0 \implies y(-0.3 + 0.01x) = 0$. For $y \ne 0$: $0.01x = 0.3 \implies x = \frac{0.3}{0.01} = 30$. [M1][A1]
Non-zero equilibrium point: $(\bar{x}, \bar{y}) = (30, 20)$.
For a linear system with solution $\mathbf{x}(t) = c_1 e^{-2t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 e^{-5t}\begin{pmatrix} -1 \\ 1 \end{pmatrix}$, find the particular solution given the initial condition $\mathbf{x}(0) = \begin{pmatrix} 1 \\ 5 \end{pmatrix}$.
Solution:
At $t = 0$: $c_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}$. [M1]
System: $c_1 - c_2 = 1$ and $2c_1 + c_2 = 5$. [M1]
Add the equations: $3c_1 = 6 \implies c_1 = 2$. [A1]
$c_2 = c_1 - 1 = 2 - 1 = 1$. [A1]
$\mathbf{x}(t) = 2e^{-2t}\begin{pmatrix} 1 \\ 2 \end{pmatrix} + e^{-5t}\begin{pmatrix} -1 \\ 1 \end{pmatrix}$.
In the solution of Exercise 9, which term dominates the trajectory as $t \to \infty$?
Solution:
Comparing decay rates: $e^{-5t}$ decays significantly faster than $e^{-2t}$. [R1]
As $t$ increases, $e^{-5t} \to 0$ much quicker, meaning the trajectory becomes tangent to the slow eigenvector direction $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$. [R1]
The term $2e^{-2t}\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ dominates. [A1]
Core Formulas (Topic 5 AHL)
| Matrix System Form | $$\frac{d\mathbf{x}}{dt} = \mathbf{A}\mathbf{x}$$ |
| Characteristic Equation | $$\det(\mathbf{A} - \lambda\mathbf{I}) = \lambda^2 - \text{tr}(\mathbf{A})\lambda + \det(\mathbf{A}) = 0$$ |
| General Real Solution | $$\mathbf{x}(t) = c_1 e^{\lambda_1 t}\mathbf{v}_1 + c_2 e^{\lambda_2 t}\mathbf{v}_2$$ |
| Complex Eigenvalues $\lambda = p \pm iq$ | $$p < 0 \implies \text{Stable Spiral}, \quad p > 0 \implies \text{Unstable Spiral}, \quad p = 0 \implies \text{Center}$$ |
Top 5 IB Exam Traps to Avoid
- Trap 1: Confusing Node and Saddle: Both nodes have eigenvalues of the *same sign* ($++, --$); saddles have eigenvalues of *opposite signs* ($+-$).
- Trap 2: Forgetting to find eigenvectors: Stating the eigenvalues alone is not sufficient to write down the general solution vector $\mathbf{x}(t)$.
- Trap 3: Inverting spiral stability: If $\lambda = -3 \pm 2i$, the negative real part $-3$ creates exponential decay $e^{-3t} \implies$ stable inward spiral.
- Trap 4: Missing arrow directions on phase portraits: An IB phase portrait requires arrows indicating the direction of trajectory movement as time $t$ increases.
- Trap 5: Non-zero equilibrium in predator-prey: Don't assume $(0, 0)$ is the only equilibrium point; non-linear models have interior co-existence equilibria.
5-Question Concept Check
Score: - / 51. If a $2 \times 2$ system has eigenvalues $\lambda_1 = -2$ and $\lambda_2 = -5$, what is the classification of $(0,0)$?
2. What type of equilibrium occurs when eigenvalues are purely imaginary ($\lambda = \pm 3i$)?
3. For eigenvalues $\lambda_1 = -3$ and $\lambda_2 = 4$, the origin is a:
4. What is the characteristic equation for matrix $\mathbf{A} = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$?
5. If $\lambda = 2 \pm 5i$, the equilibrium point is an: