Mr. Integral
Math notes/Differential Equations/Systems and Phase Planes

Nonlinear Systems and Local Linearization

2 MIN READ · PRACTISE AT YOUR OWN PACE   ·   FREE

LESSON 5 OF 5See the unit map ↗

Use a Jacobian near an equilibrium and recognize inconclusive eigenvalues.

Builds on Forced Systems and Variation of Constants

The bigger question: How do coupled variables move together?

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The idea

For x′=f(x)\mathbf x'=\mathbf f(\mathbf x), an equilibrium satisfies f(x∗)=0\mathbf f(\mathbf x_*)=0. Write a small displacement u=x−x∗\mathbf u=\mathbf x-\mathbf x_*. The first-order approximation is u′=Ju\mathbf u'=J\mathbf u, where J=Df(x∗)J=D\mathbf f(\mathbf x_*).

Visual guide

VISUAL GUIDEThe sign of the nonlinear term can decide stability
Both y′ = −y³ and y′ = y³ have zero linear derivative at equilibrium. In the slope-versus-state graph, −y³ points solutions toward zero and y³ points them away. The linearized equation alone misses this difference.-1.3-2-0.65-1000.6511.32state yslope y′
  • Stable: −y³
  • Unstable: y³
Both y′ = −y³ and y′ = y³ have zero linear derivative at equilibrium. In the slope-versus-state graph, −y³ points solutions toward zero and y³ points them away. The linearized equation alone misses this difference.

Method and assumptions

Compute the Jacobian before substituting the equilibrium. Strictly negative eigenvalue real parts imply local asymptotic stability; a positive real part implies instability. If every eigenvalue has nonzero real part, the equilibrium is hyperbolic and the linearized phase behavior is locally robust. Zero real parts make this test inconclusive.

Worked example: a nonlinear stable state

For x′=−x+x2,y′=−2yx'=-x+x^2,y'=-2y, the origin has Jacobian diag⁡(−1,−2)\operatorname{diag}(-1,-2) and is locally asymptotically stable. At (1,0)(1,0) the Jacobian is diag⁡(1,−2)\operatorname{diag}(1,-2), a saddle. The nonlinear system has different behavior near different equilibria.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

The Jacobian at a nonlinear equilibrium has a zero eigenvalue. Can linearization alone always decide stability?

Hint 1 · Find a starting point

A zero eigenvalue is not a hyperbolic direction.

Hint 2 · Take the next step

Compare scalar equations y′=−y³ and y′=y³ at zero.

Show the reasoning

Answer: No; nonlinear terms may decide.

Both have linearization y′=0, but opposite stability, so more analysis is needed.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: the same linearization, opposite outcomes

Both scalar equations x′=−x3x'=-x^3 and x′=x3x'=x^3 have derivative zero at the origin. In the first, arrows point inward and solutions approach zero; in the second they point outward. A zero linearization cannot distinguish them.

Interpreting the result

Local stability is not a global claim about every starting point. A stable equilibrium may have a limited basin of attraction, and a linear approximation loses accuracy far from its expansion point.

Practice

  1. Find equilibria of x′=x−x3x'=x-x^3.
  2. Classify them using derivatives.
  3. What should you do when an eigenvalue has zero real part?
Show worked solutions
  1. x=−1,0,1x=-1,0,1.
  2. f′=1−3x2f'=1-3x^2: the outer equilibria attract and zero repels.
  3. Use nonlinear terms, phase-line signs or another suitable argument; the linear test is inconclusive.

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

Pause before the next idea.

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