Skip to content

Stability and Phase Plane Analysis

For x"=f(x)\mathbf{x}" = \mathbf{f}(\mathbf{x})A critical point x\mathbf{x}^* satisfies f(x)=0\mathbf{f}(\mathbf{x}^*) = \mathbf{0}.

Let A=Jf(x)A = J\mathbf{f}(\mathbf{x}^*) be the Jacobian at the critical point. The eigenvalues of AA Determine the local stability:

Eigenvalues of AATypeStability
Both real, negativeStable nodeAsymptotically stable
Both real, positiveUnstable nodeUnstable
Real, opposite signsSaddle pointUnstable
Complex, Re(λ)<0\mathrm{Re}(\lambda) \lt 0Stable spiralAsymptotically stable
Complex, Re(λ)>0\mathrm{Re}(\lambda) > 0Unstable spiralUnstable
Purely imaginaryCenter(Marginally) stable

Definition. A critical point x\mathbf{x}^* is stable if for every ε>0\varepsilon > 0There Exists δ>0\delta > 0 such that x(0)x<δ\|\mathbf{x}(0) - \mathbf{x}^*\| \lt \delta implies x(t)x<ε\|\mathbf{x}(t) - \mathbf{x}^*\| \lt \varepsilon for all t>0t > 0.

It is asymptotically stable if it is stable and x(t)x\mathbf{x}(t) \to \mathbf{x}^* as tt \to \infty.

Theorem 9.1 (Lyapunov). If there exists a continuously differentiable function VV (a Lyapunov Function) such that V(x)=0V(\mathbf{x}^*) = 0, V(x)>0V(\mathbf{x}) > 0 for xx\mathbf{x} \neq \mathbf{x}^* And V˙0\dot{V} \leq 0 in a neighbourhood of x\mathbf{x}^* Then x\mathbf{x}^* is stable. If V˙<0\dot{V} \lt 0 for xx\mathbf{x} \neq \mathbf{x}^* Then x\mathbf{x}^* is asymptotically stable.

Problem. Find and classify the critical points of x=xyx' = x - y, y=x2+y21y' = x^2 + y^2 - 1.

Solution

Solution. Set x=0x' = 0 and y=0y' = 0:

xy=0    y=xx - y = 0 \implies y = x

x2+x21=0    2x2=1    x=±1/2x^2 + x^2 - 1 = 0 \implies 2x^2 = 1 \implies x = \pm 1/\sqrt{2}

Critical points: (1/2,1/2)(1/\sqrt{2}, 1/\sqrt{2}) and (1/2,1/2)(-1/\sqrt{2}, -1/\sqrt{2}).

The Jacobian is J=(112x2y)J = \begin{pmatrix} 1 & -1 \\ 2x & 2y \end{pmatrix}.

At (1/2,1/2)(1/\sqrt{2}, 1/\sqrt{2}): J=(1122)J = \begin{pmatrix} 1 & -1 \\ \sqrt{2} & \sqrt{2} \end{pmatrix}.

tr(J)=1+2>0\mathrm{tr}(J) = 1 + \sqrt{2} > 0, det(J)=2+2=22>0\det(J) = \sqrt{2} + \sqrt{2} = 2\sqrt{2} > 0.

τ24Δ=(1+2)282=3+2282=362<0\tau^2 - 4\Delta = (1 + \sqrt{2})^2 - 8\sqrt{2} = 3 + 2\sqrt{2} - 8\sqrt{2} = 3 - 6\sqrt{2} \lt 0.

Complex eigenvalues with positive real part: unstable spiral.

At (1/2,1/2)(-1/\sqrt{2}, -1/\sqrt{2}): J=(1122)J = \begin{pmatrix} 1 & -1 \\ -\sqrt{2} & -\sqrt{2} \end{pmatrix}.

tr(J)=12<0\mathrm{tr}(J) = 1 - \sqrt{2} \lt 0, det(J)=2+2=0\det(J) = -\sqrt{2} + \sqrt{2} = 0.

Wait, det(J)=(1)(2)(1)(2)=22=22<0\det(J) = (1)(-\sqrt{2}) - (-1)(-\sqrt{2}) = -\sqrt{2} - \sqrt{2} = -2\sqrt{2} \lt 0.

Negative determinant: saddle point (unstable). \blacksquare

9.5 Phase Portraits for 2D Nonlinear Systems

Section titled “9.5 Phase Portraits for 2D Nonlinear Systems”

For the nonlinear system x=f(x)\mathbf{x}' = \mathbf{f}(\mathbf{x})The Hartman-Grobman theorem States that near a hyperbolic critical point (one where the Jacobian has no eigenvalues on the Imaginary axis), the nonlinear phase portrait is topologically equivalent to the linearized one.

Procedure for sketching phase portraits:

  1. Find all critical points by solving f(x)=0\mathbf{f}(\mathbf{x}) = \mathbf{0}.
  2. Compute the Jacobian JJ at each critical point.
  3. Classify each critical point using the eigenvalue analysis from Section 4.9.
  4. Sketch the local behaviour near each critical point.
  5. Connect the local pictures using nullclines (x=0x' = 0 and y=0y' = 0 curves).

A limit cycle is an isolated closed periodic orbit. Limit cycles are inherently nonlinear Phenomena --- linear systems cannot have isolated closed orbits.

Theorem 9.2 (Poincaré-Bendixson). If a trajectory of a C1C^1 planar system is confined to a Closed bounded region RR that contains no critical points, then the trajectory approaches a closed Periodic orbit as tt \to \infty.

Remark. The Poincaré-Bendixson theorem is specific to two dimensions. In three or more Dimensions, much more complex behaviour (chaos) is possible.

Example: Van der Pol oscillator. The equation

x+μ(x21)x+x=0x'' + \mu(x^2 - 1)x' + x = 0

With μ>0\mu > 0 has a unique stable limit cycle. This system models electrical circuits with Nonlinear resistance and arises in biology (cardiac rhythms, neuron firing).

9.7 Worked Example: Lotka-Volterra Analysis

Section titled “9.7 Worked Example: Lotka-Volterra Analysis”

Problem. Analyze the stability of the Lotka-Volterra system x=x(2y)x' = x(2 - y), y=y(x1)y' = y(x - 1).

Solution

Solution. Critical points: (0,0)(0, 0) and (1,2)(1, 2).

Jacobian: J=(2yxyx1)J = \begin{pmatrix} 2 - y & -x \\ y & x - 1 \end{pmatrix}.

At (0,0)(0, 0): J=(2001)J = \begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}. Eigenvalues 22 and 1-1: saddle point (unstable).

At (1,2)(1, 2): J=(0120)J = \begin{pmatrix} 0 & -1 \\ 2 & 0 \end{pmatrix}. det(J)=2>0\det(J) = 2 > 0, tr(J)=0\mathrm{tr}(J) = 0. Eigenvalues ±i2\pm i\sqrt{2}: center.

Remark. For the linearized system, the center is (marginally) stable. However, for the Nonlinear Lotka-Volterra system, the trajectories are actually closed orbits surrounding (1,2)(1, 2). This can be verified using the first integral H=xlnx+2lnyyH = x - \ln x + 2\ln y - yWhich is constant Along trajectories. \blacksquare

9.8 Intuition: What Do Phase Portraits Reveal?

Section titled “9.8 Intuition: What Do Phase Portraits Reveal?”

Phase portraits are flow maps for dynamical systems. Each point in the plane represents a state of the system, and the trajectory through that point shows how the state evolves over time. Critical points are states where nothing changes (the system is in equilibrium). The eigenvalues of the Jacobian at a critical point tell you what happens to nearby trajectories: negative real parts mean trajectories spiral or flow inward (stability), positive real parts mean they spiral or flow outward (instability), and purely imaginary eigenvalues mean they orbit the critical point (center).

The Hartman-Grobman theorem guarantees that near a hyperbolic critical point (no eigenvalues on the imaginary axis), the nonlinear system looks qualitatively the same as its linearisation. This is why eigenvalue analysis is so powerful: it tells you the local geometry of the flow without solving the equations. Limit cycles, which are isolated periodic orbits, are inherently nonlinear phenomena. The Poincare-Bendixson theorem says that in two dimensions, a trajectory trapped in a bounded region with no critical points must approach a periodic orbit, which means two-dimensional systems cannot exhibit chaos. In three or more dimensions, chaos becomes possible because trajectories have more room to avoid intersecting themselves.

The competing species model is:

x=x(r1a11xa12y),y=y(r2a21xa22y)x' = x(r_1 - a_{11}x - a_{12}y), \quad y' = y(r_2 - a_{21}x - a_{22}y)

Where ri>0r_i > 0 are growth rates and aij>0a_{ij} > 0 are competition coefficients. The four critical Points are (0,0)(0, 0), (r1/a11,0)(r_1/a_{11}, 0), (0,r2/a22)(0, r_2/a_{22}) And the coexistence point (x,y)(x^*, y^*) where both xx' and yy' vanish.

The stability of the coexistence point determines whether both species survive. If a11a22>a12a21a_{11}a_{22} > a_{12}a_{21}Coexistence is stable; otherwise, one species drives the other To extinction (competitive exclusion).

Mistake 1: Assuming that linearization always determines stability. Linearization determines the local stability of hyperbolic critical points (where no eigenvalue has zero real part). For non-hyperbolic critical points (e.g., purely imaginary eigenvalues), linearization is inconclusive and nonlinear analysis is required. Do not assume that linearization always works.

Mistake 2: Confusing stability with asymptotic stability. A critical point is stable if nearby trajectories stay nearby, and asymptotically stable if they converge to the critical point. A center is stable but not asymptotically stable. Do not confuse the two concepts.

Mistake 3: Forgetting that the Jacobian must be evaluated at the critical point. The Jacobian matrix Jf(x)J\mathbf{f}(\mathbf{x}^*) must be evaluated at the critical point x\mathbf{x}^*, not at an arbitrary point. Evaluating at the wrong point leads to incorrect eigenvalues and wrong stability conclusions.

Mistake 4: Assuming that all critical points are isolated. Critical points can be isolated or form a continuum (e.g., a line of critical points). Do not assume that all critical points are isolated; check the structure of the vector field.

Mistake 5: Confusing the phase plane with the phase portrait. The phase plane is the coordinate space (x,y)(x, y), while the phase portrait is the collection of trajectories in the phase plane. Do not confuse the two; the phase portrait includes the trajectories, not just the coordinate axes.

flowchart TD
A[9_Stability And Phase Plane Analysis] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]