Stable and unstable manifolds
Replace linear eigenspaces by nonlinear invariant geometry, derive a curved stable graph, and distinguish local sheets from their global continuations.
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The question for today
What you’ll understand
- State the local stable-manifold theorem at a smooth hyperbolic equilibrium.
- Derive an invariance equation for a graph through a saddle.
- Explain exponential sensitivity transverse to a stable manifold.
A saddle has trajectories that approach it, even though most nearby trajectories eventually leave. Linearization identifies the tangent directions of those special approaches. Nonlinear terms bend the directions into invariant manifolds, separating precise long-time outcomes. Finding such a curve is more informative than drawing a few simulated trajectories near an unstable point: it describes exactly which initial conditions cancel the growing component and why a tiny transverse perturbation eventually spoils that cancellation.
The tangent decomposition becomes invariant geometry
For a vector field, with integer r at least one, let p be a hyperbolic equilibrium. The stable and unstable generalized eigenspaces of its Jacobian split the full tangent space. The local stable-manifold theorem supplies a embedded local manifold through p, tangent to the stable eigenspace, consisting of nearby states whose forward evolution stays suitably near p and converges to it. The unstable counterpart uses backward evolution and is tangent to the unstable eigenspace.
The dimensions match the linear spaces, but the manifolds need not be straight. Hyperbolicity gives exponential approach along the stable manifold locally, with rates bounded relative to the spectral gap; it does not force every nearby trajectory to converge. Forward and backward descriptions also require the corresponding solutions to exist for the times used. Global stable and unstable sets are obtained by continuing local pieces along the flow and can be immersed or folded rather than one simple graph over a fixed coordinate axis.
A graph must carry its own velocity
Consider a planar system with a possible invariant graph y=h(x). Along that graph, the chain rule requires y dot to equal h prime of x times x dot. Equating this expression with the system's second component gives an invariance equation for h. The tangent condition at the equilibrium selects the appropriate stable or unstable branch. Solving only an instantaneous nullcline, where one component of velocity vanishes, generally finds a different object: a trajectory can cross a nullcline rather than follow it.
For x dot = −2x and y dot = y+5x², the linear stable space is the x-axis. Trying a quadratic graph gives h(x)=−x², and substitution verifies it exactly: its chain-rule velocity is 4x², matching h(x)+5x². On that curve x decays exponentially and y follows its square. The x-axis itself is not invariant except at the origin, because at a nonzero point with y zero the second component equals 5x² and pushes the trajectory away from the axis.
Transverse errors grow while tangent motion shrinks
The triangular example can be solved completely. Its vertical solution is an exponentially growing term with coefficient y(0)+x(0)² plus a decaying quadratic term. Exactly on the stable graph the growing coefficient vanishes. A vertical perturbation epsilon changes that coefficient to epsilon, yielding eventual departure proportional to epsilon times exp(t). Forward numerical integration therefore has difficulty tracing a stable manifold all the way into a saddle: tiny rounding or truncation errors can excite its unstable direction.
The unstable manifold in this example is the y-axis, because x must vanish to approach the origin in backward time. Stable and unstable manifolds can meet away from their equilibria in more complicated systems. An orbit joining distinct equilibria is heteroclinic; one leaving and returning to the same equilibrium is homoclinic. These global connections require a separate analysis. The local theorem neither guarantees their existence nor determines whether their intersections are transverse, a distinction that later becomes central to horseshoes.
Work it through
Choose the unique stable vertical coordinate
For x dot = −2x, y dot = y+5x² with x(0)=2, find the initial y that approaches the origin and calculate the state at time (log 2)/2.
First x(t)=2 exp(−2t). Solving the forced linear y equation gives y(t)=exp(t)(y(0)+4)−4 exp(−4t).
Convergence requires cancellation of the growing coefficient, so y(0)=−4. The resulting trajectory lies on y=−x², the stable graph derived from invariance.
At t=(log 2)/2, exp(−2t)=1/2 and exp(−4t)=1/4. Thus x=1 and y=−1. If the initial y were −4+epsilon instead, an extra epsilon exp(t) term would remain.
The stable initial point is (2,−4); its later point is (1,−1). The stable graph is curved but tangent to the horizontal stable eigenspace at the origin.
The living notebook
Explore the model
Interactive lab · fields
Compare full stable and unstable planes in a simpler family
Pause and reset, then set both controls to q = −0.2 and Ω = 0. Predict radial inward arrows: interpreting this smooth field as an ODE gives a sink whose local stable manifold has dimension two. For a fresh comparison, pause and reset, then set q = 0.2 and Ω = 0; the arrows reverse and the origin has a two-dimensional local unstable manifold. The field panel does not draw the lesson's saddle graph y = −x² or compute invariant manifolds; it illustrates the dimensional meaning in a family with no mixed stable/unstable directions.
- Fx at (1, 0)
- 0.833
- Fy at (1, 0)
- 0.5
- Field magnitude there
- 0.972
The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forStable-manifold dimension is determined locally by contracting directions, while its nonlinear shape needs an invariance calculation.
Pause & reason
Check your understanding
Your turn
Try it yourself
For x dot = −x and y dot = 3y+10x², find a stable graph y=ax² and determine the stable initial y when x(0)=1. Explain the effect of an added vertical error epsilon.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Set h'(x)(−x)=3h(x)+10x², then isolate the homogeneous growing term in y.
Compare with the solution
The graph equation gives −2a=3a+10, hence a=−2. Thus the stable initial coordinate is y(0)=−2. With x(t)=exp(−t), the exact vertical solution is y(t)=exp(3t)(y(0)+2)−2 exp(−2t). The chosen coefficient cancels the growing term and gives approach to the origin. Perturbing it by epsilon adds epsilon exp(3t), so an arbitrarily small transverse error eventually grows. The graph remains tangent to the x-axis at zero, while the y-axis is the unstable manifold.
Keep the idea
Take it with you
- A hyperbolic equilibrium has smooth local stable and unstable manifolds tangent to its spectral subspaces.
- An invariant graph satisfies a chain-rule equation rather than a nullcline equation.
- The local theorem controls nearby geometry; global connections and numerical sensitivity need additional arguments.
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