Linear Systems Slides Phase Plane Analysis
Linear Systems & Stability
Graduate Seminar in Differential Equations
\(\dot{x} = Ax\)
The Central Question
How can we predict the global behavior of a dynamical system by looking at its critical points?
For a linear system \(\dot{\mathbf{x}} = A\mathbf{x}\), the origin is typically the unique equilibrium. Its nature is entirely determined by the eigenvalues of \(A\).
"The eigenvalues are the DNA of the trajectory."
\(\det(A - \lambda I) = 0\)
The Trace-Determinant Plane
Characteristic Equation for \(2 \times 2\) matrix \(A\):
\(\lambda^2 - \tau \lambda + \Delta = 0\)
\(\tau = \text{tr}(A) = \lambda_1 + \lambda_2\)
\(\Delta = \det(A) = \lambda_1 \lambda_2\)
Discriminant: \(D = \tau^2 - 4\Delta\)
\(\tau\)
\(\Delta\)
Unstable (\(\tau > 0\))
Stable (\(\tau < 0\))
Qualitative Regions
Classification Taxonomy
Eigenvalues \(\lambda_1, \lambda_2\) Stability Type Geometrical Name Real, distinct, negative Asymptotically Stable Stable Node Real, distinct, positive Unstable Unstable Node Real, opposite signs Unstable Saddle Point Complex, \(\text{Re}(\lambda) < 0\) Asymptotically Stable Stable Spiral Purely Imaginary (\(\text{Re} = 0\)) Stable (Neutrally) Center
Sketching Strategy
1
Find Eigenspaces
The eigenvectors define the invariant lines (manifolds). These are the backbones of the flow.
2
Determine Direction
Analyze the sign of the eigenvalues. Do trajectories move toward or away from the origin?
3
Interpolate Flow
Fill in the trajectories between eigenspaces, ensuring they don't cross (uniqueness theorem).
"What happens at \(\Delta = 0\)?"
When the determinant is zero, we have a non-isolated equilibrium or a line of critical points.
In this case, the system is singular. The linear approximation is robust for hyperbolic points, but what if we perturb a non-hyperbolic point?
Next: Nonlinear Linearization
Phase Plane Worksheet Phase Plane Diagnostic
Dossier: Linear Stability Analysis
Researcher:
Timestamp:
1
Classification of Planar Systems
Consider the following system \(\dot{\mathbf{x}} = A\mathbf{x}\). For each matrix \(A\), determine the trace (\(\tau\)), determinant (\(\Delta\)), eigenvalues (\(\lambda_1, \lambda_2\)), and classify the critical point at the origin.
\(A = \begin{pmatrix} 1 & -2 \\ 3 & -4 \end{pmatrix}\)
Trace (\(\tau\)):
Determinant (\(\Delta\)):
Eigenvalues:
Classification:
\(A = \begin{pmatrix} -1 & -1 \\ 1 & -1 \end{pmatrix}\)
Trace (\(\tau\)):
Determinant (\(\Delta\)):
Eigenvalues:
Classification:
2
Phase Portrait Construction
For the system \(\dot{x} = 2x + y, \dot{y} = x + 2y\), calculate the eigenvectors and sketch the phase portrait below. Be sure to indicate the directions of flow on all invariant lines.
Work Space: Eigenvectors
Analytical Conclusions
PHASE PLANE (x-y)
Label axes and identify the type of equilibrium point.
3
The Singular Case
Investigate the system where \(\Delta = 0\): \(\dot{x} = x + y, \dot{y} = x + y\). Describe the set of equilibrium points and sketch the resulting phase portrait. How does this differ from the hyperbolic cases discussed in lecture?
Qualitative Theory of Ordinary Differential Equations — Advanced Dynamical Systems
Linear Systems Answer Key Answer Key & Teacher Guide
Phase Plane Analysis of Linear Systems
INSTRUCTOR COPY
1. Classification of Planar Systems
Matrix A: \(\begin{pmatrix} 1 & -2 \\ 3 & -4 \end{pmatrix}\)
Trace (\(\tau\)): \(1 - 4 = -3\)
Determinant (\(\Delta\)): \((1)(-4) - (-2)(3) = -4 + 6 = 2\)
Characteristic Eq: \(\lambda^2 + 3\lambda + 2 = 0 \implies (\lambda+1)(\lambda+2)=0\)
Eigenvalues: \(\lambda_1 = -1, \lambda_2 = -2\)
Classification: Stable Node (Sink)
Matrix B: \(\begin{pmatrix} -1 & -1 \\ 1 & -1 \end{pmatrix}\)
Trace (\(\tau\)): \(-1 - 1 = -2\)
Determinant (\(\Delta\)): \((-1)(-1) - (-1)(1) = 1 + 1 = 2\)
Characteristic Eq: \(\lambda^2 + 2\lambda + 2 = 0\)
Eigenvalues: \(\lambda = -1 \pm i\)
Classification: Stable Spiral (Focus)
2. Phase Portrait Construction
Calculation Pathway:
Matrix: \(\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\)
\(\tau = 4, \Delta = 3 \implies \lambda^2 - 4\lambda + 3 = 0\)
\(\lambda_1 = 3, \lambda_2 = 1\) (Unstable Node)
For \(\lambda = 3\): \(\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\mathbf{v} = 0 \implies \mathbf{v}_1 = (1, 1)^T\)
For \(\lambda = 1\): \(\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\mathbf{v} = 0 \implies \mathbf{v}_2 = (1, -1)^T\)
Visual Expectation
Source
Trajectories emerge from the origin, becoming parallel to the \(\lambda=3\) eigenvector (\(y=x\)) as \(t \to \infty\).
Instructional Considerations
Critical Concept: Emphasize that in planar systems, the fast eigenvalue dominates the long-term behavior (\(t \to \infty\)), while the slow eigenvalue determines the direction from which trajectories approach the origin (\(t \to -\infty\)).
Common Pitfall: Students often swap the stability of spirals. Remind them: if \(\tau < 0\), the "exponential part" \(e^{\text{Re}(\lambda)t}\) decays, drawing the spiral inward.
The Singular Case (Problem 3): This system has \(\lambda = 2, 0\). The entire line \(x+y=0\) consists of non-isolated equilibrium points. Trajectories are parallel lines moving away from this "equilibrium line" since the other eigenvalue is positive.
Nonlinear Linearization Slides Nonlinear Linearization
Local Stability Analysis
Hartman-Grobman and the Power of the Jacobian
The Nonlinear Reality
Most dynamical systems are of the form \(\dot{\mathbf{x}} = \mathbf{f}(\mathbf{x})\), where \(\mathbf{f}\) is nonlinear.
Direct solution is usually impossible. We must look at local behavior near fixed points \(\mathbf{x}^*\).
Taylor Expansion:
\(\mathbf{f}(\mathbf{x}) \approx \mathbf{f}(\mathbf{x}^*) + D\mathbf{f}(\mathbf{x}^*)(\mathbf{x}-\mathbf{x}^*)\)
Since \(\mathbf{f}(\mathbf{x}^*) = 0\), the linear part dominates nearby.
The Jacobian Matrix
The Jacobian \(J\) represents the best linear approximation to the vector field \(\mathbf{f}\) at a point.
\(J = \begin{pmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \end{pmatrix}\)
Workflow:
Find fixed points \(\mathbf{f}(\mathbf{x}^*) = 0\).
Compute the general Jacobian matrix \(J(\mathbf{x})\).
Evaluate \(J\) at each fixed point \(\mathbf{x}^*\).
Find eigenvalues of \(J(\mathbf{x}^*)\).
The Theorem
Hartman-Grobman
Near a hyperbolic equilibrium point, the nonlinear flow is topologically conjugate to its linearization.
What is "Hyperbolic"?
An equilibrium point is hyperbolic if all eigenvalues of its Jacobian have non-zero real parts.
Sinks, Sources, Saddles
Centers, Non-isolated points
The Danger Zone: Non-Hyperbolic Points
If \(\text{Re}(\lambda) = 0\), the linearization fails to guarantee the local behavior.
Higher-order terms (the ones we threw away in the Taylor expansion) suddenly matter. A "linear center" could actually be a stable spiral or an unstable spiral in the full nonlinear system.
Center Problem
Linearization says "Center", but reality could be anything.
Local vs. Global
Linearization tells us what happens infinitesimally close to a point. It does not determine the global geometry of the phase plane.
Workshop Challenge:
Apply linearization to the Simple Pendulum system. Where does the linearization break down, and why is that physically significant?
Linearization Lab Worksheet Linearization Protocol
Dossier: Local Topology Approximation
Researcher:
Linearization Summary
To analyze \(\dot{\mathbf{x}} = \mathbf{f}(\mathbf{x})\) at \(\mathbf{x}^*\):
1
The Lotka-Volterra Model
Consider a simplified predator-prey system: \[ \dot{x} = x(1-y), \quad \dot{y} = y(x-1) \] where \(x\) is prey population and \(y\) is predator population.
A. Fixed Points
Find all equilibrium points \((x^*, y^*)\).
B. Jacobian Matrix
Compute the general Jacobian \(J(x, y)\).
C. Local Analysis
Evaluate the Jacobian at the non-trivial fixed point. What are the eigenvalues? Does Hartman-Grobman apply?
2
The Damped Pendulum
The equation of motion for a damped pendulum is \(\ddot{\theta} + b\dot{\theta} + \sin\theta = 0\). Converted to a system: \(\dot{x} = y, \quad \dot{y} = -by - \sin x\).
Linearize at the origin \((0,0)\). For what values of \(b\) is the origin a stable spiral ?
Linearize at the point \((\pi, 0)\). Classify the equilibrium point.
3
Comparative Analysis
Explain the following phenomenon:
For the system \(\dot{x} = y, \dot{y} = -x + x^3\), the linearization at \((0,0)\) suggests a center. However, the energy of the system is given by \(E = \frac{1}{2}y^2 + \frac{1}{2}x^2 - \frac{1}{4}x^4\). Based on the energy conservation, can we conclude the point is truly a center? Why or why not?
Nonlinear Systems Linearization Workshop — Analytical Mechanics & Population Dynamics
Nonlinear Linearization Answer Key Instructor Guide
Linearization of Nonlinear Systems
ANSWER KEY
1. The Lotka-Volterra Model
A & B: Equilibrium & Jacobian
Fixed Points: (0,0) and (1,1).
Jacobian: \(J(x,y) = \begin{pmatrix} 1-y & -x \\ y & x-1 \end{pmatrix}\)
C: Local Analysis at (1,1)
Evaluating \(J(1,1) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\).non-hyperbolic point. Hartman-Grobman does not apply. We cannot determine stability via linearization alone.
2. The Damped Pendulum
Jacobian at origin \((0,0)\):
\(J(0,0) = \begin{pmatrix} 0 & 1 \\ -1 & -b \end{pmatrix}\). Characteristic eq: \(\lambda^2 + b\lambda + 1 = 0\).
Eigenvalues: \(\lambda = \frac{-b \pm \sqrt{b^2 - 4}}{2}\).
Conclusion: Stable spiral if \(b^2 - 4 < 0 \implies 0 < b < 2\).
Jacobian at \((\pi,0)\):
\(J(\pi,0) = \begin{pmatrix} 0 & 1 \\ 1 & -b \end{pmatrix}\). Characteristic eq: \(\lambda^2 + b\lambda - 1 = 0\).
Eigenvalues: \(\lambda = \frac{-b \pm \sqrt{b^2 + 4}}{2}\). Since \(b^2+4 > b^2\), one eigenvalue is positive and one is negative.
Conclusion: Unstable Saddle Point. Physically, this corresponds to the inverted pendulum position.
3. Comparative Analysis
The linearization gives \(\dot{x} = y, \dot{y} = -x\), which is a center. However, the full system is conservative because \(\dot{E} = y\dot{y} + x\dot{x} - x^3\dot{x} = y(-x+x^3) + x(y) - x^3(y) = 0\). Since the energy is conserved and the potential \(V(x) = \frac{1}{2}x^2 - \frac{1}{4}x^4\) has a local minimum at \(x=0\), the orbits near the origin must be closed loops. Thus, the center is robust in this specific nonlinear case, though Hartman-Grobman didn't guarantee it.
Teaching Tips
Challenge students to visualize the "Phase Portrait" for the pendulum - specifically how the saddle point at \(\pi\) connects to the next saddle at \(3\pi\).
Discuss the "Physicality" of the non-hyperbolic case: In the Lotka-Volterra model, the center is often criticized because any small perturbation (e.g., harvesting) can turn it into a spiral, changing the nature of the ecosystem entirely.
