research40 min studyLesson 846 in Mathematics

Invariant measures and ergodicity

Construct stationary Gaussian laws from an infinite-time covariance trace and distinguish existence of a law from contraction toward it.

Before you begin

On this page

The question for today

What you’ll understand

Check the trace condition for a stationary Ornstein–Uhlenbeck law.
Use same-noise contraction to bound convergence in Wasserstein distance.
Identify an undamped zero mode that prevents an invariant probability law.

Random forcing can balance dissipation and produce a stationary distribution, but stability of the deterministic equation alone does not guarantee a probability measure in the desired infinite-dimensional space. The accumulated covariance must have finite trace there. Existence and convergence are also separate questions: an invariant law can coexist with undamped deterministic directions, while a contraction estimate can establish uniqueness and convergence once a compatible law exists. A diagonal Ornstein–Uhlenbeck system displays these distinctions without relying on a nonlinear ergodicity theorem.

Accumulate covariance over the whole past

Consider a linear evolution $dX=AXdt+B\,dW$ on a real separable Hilbert space, with cylindrical W and a bounded deterministic amplitude B. Suppose the semigroup satisfies $\|S(t)\|\le e^{-\omega t}$ for some ω>0 and the integral over positive times of $\|S(t)B\|_{\mathrm{HS}}^2$ is finite. Then its infinite-time covariance is a positive trace-class operator. The centered Gaussian law with that covariance is an H-valued invariant probability measure.

The covariance integral is interpreted through its positive quadratic form, with trace convergence supplied by the Hilbert–Schmidt integral. Its invariance follows by splitting the infinite past into an older contribution evolved by the semigroup and a fresh independent convolution. Merely checking finite-time covariance does not control the accumulation over an infinite past. Conversely, writing formal stationary coordinate variances does not define an H-valued law if their sum diverges. The state space is part of the invariant-measure claim.

Q_\infty=\int_0^\infty S(r)BB^*S(r)^*dr,\qquad \operatorname{Tr}Q_\infty=\int_0^\infty\|S(r)B\|_{\mathrm{HS}}^2dr<\infty

Here B is a noise amplitude and Q_infinity is a covariance. The assumed exponential stability and finite trace give a genuine stationary Gaussian law on H.↔ Scroll wide equations horizontally.

Diagonal balance and coupling give quantitative conclusions

For a diagonal generator $Ae_k=-\lambda_ke_k$ with λk bounded below by a positive constant, and bounded diagonal amplitude $Be_k=b_ke_k$ , stationary coordinate variances are $b_k^2/(2\lambda_k)$ . Their sum must be finite. The scalar balance is the covariance form of the Lyapunov equation. With an unbounded generator, a full operator Lyapunov identity is interpreted weakly on adjoint-domain test vectors; products involving A and the covariance cannot be assumed bounded without a domain argument.

Couple two solutions with the same future noise. Their difference is the deterministic semigroup applied to their initial difference, so it contracts exponentially. Taking an optimal initial coupling gives the same contraction in quadratic Wasserstein distance between laws with finite second moments. The stationary Gaussian law has such a moment by its trace condition, hence this proves uniqueness among finite-second-moment invariant laws and exponential attraction in that distance. Stronger conclusions for arbitrary initial laws or nonlinear dynamics need their own arguments.

q_k^\infty=\frac{b_k^2}{2\lambda_k},\qquad W_2(P_t^*\mu,P_t^*\nu)\le e^{-\omega t}W_2(\mu,\nu)

The variance formula is diagonal. The Wasserstein estimate uses the stated semigroup contraction and finite second moments of both initial laws.↔ Scroll wide equations horizontally.

Zero modes and infinite-dimensional tightness matter

On a periodic domain, the Laplacian has a constant zero mode. If white forcing acts on that mode, its coordinate is an undamped Brownian motion and has no invariant probability distribution on the real line. Adding a positive mass term or projecting both dynamics and noise to mean-zero functions can remove that specific obstruction. An unforced zero mode instead retains its initial value and can support many invariant distributions, so deleting noise there does not create uniqueness by itself.

For nonlinear equations, averaging time-evolved laws can help construct invariant measures, but an infinite-dimensional tightness argument usually needs compactness beyond a bound on the H-norm. Closed bounded balls need not be compact. Existence of an invariant law also does not automatically imply mixing, a spectral gap or a quantitative rate. These distinctions are easy to miss when every finite-dimensional approximation has a smooth stationary density. Uniform covariance tails or stronger energy-space compactness are what permit passage to a state-space probability measure.

Work it through

Different noise strengths can yield equal stationary variances

Take two independent coordinates with drift rates one and four and noise amplitudes one and two. Find the stationary covariance, its trace, and the finite-time covariance from zero initial data.

The first stationary variance is 1/(2×1)=1/2. The second is 2²/(2×4)=1/2. Independence and diagonal dynamics make the cross covariance zero.
The stationary covariance is diag(1/2,1/2), with trace one. From zero initial data, the two variances are (1−exp(−2t))/2 and (1−exp(−8t))/2, respectively.
The stronger second noise is balanced by stronger damping, so its equilibrium variance matches the first while its approach is faster. Same-noise coupling contracts the full Euclidean difference norm at least as exp(−t), determined by the slowest drift direction.

The result

Stationary variance depends on the ratio of squared noise amplitude to damping; the slowest decay direction controls this simple coupling rate.

The living notebook

Explore the model

Interactive lab

Interactive lab · markov

Stationarity and finite-time convergence are separate readouts

Pause and Reset, then set both controls to p=0.25 and q=0.5. The stationary probability of state zero is 2/3; after two Steps, corresponding to two transitions, predict 11/16 from initial state zero. Reset, set p=0.5 and q=0.25, and take two Steps; predict stationary probability 1/3 and two-transition probability 3/8. This exact two-state chain illustrates equilibrium versus approach. It does not compute an SPDE covariance or prove infinite-dimensional tightness.

P(state 0)P(state 1)Stationary distribution

P(state 1) at step 0: 0%
Stationary P(state 1): 60%
Chain behavior: Monotone convergence

Transition p · 0 → 10.3Transition q · 1 → 00.2

t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. 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 forAn equilibrium formula and a convergence claim answer different questions, even in a fully solvable model.

Pause & reason

Check your understanding

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Read the checkpoint explanation

Summable coordinate variances produce an H-valued Gaussian law. Without that trace condition, finite-dimensional stationary distributions may exist at every cutoff while no such invariant Gaussian probability exists on the chosen H.

Your turn

Try it yourself

On ell² let drift eigenvalues be lambda_k=k^r for r>0 and use unit cylindrical noise amplitude. Determine when the diagonal stationary law is ell²-valued. Include r=1 and r=2, and explain why finite cutoffs cannot settle the answer.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The stationary covariance eigenvalues are 1/(2k^r), so apply the p-series criterion.

Compare with the solution

The stationary trace is one-half the sum of k^(−r), finite exactly when r>1. At r=1 the harmonic divergence rules out an ell²-valued Gaussian stationary law; at r=2 the trace is pi²/12. Every finite cutoff has finite trace for every r>0, so its stationary law alone does not establish an infinite-dimensional limit. In this diagonal system the same high-frequency obstruction also affects the finite-time ell² convolution when r≤1.

Keep the idea

Take it with you

Stationary Gaussian laws require an infinite-time covariance with finite state-space trace.
Same-noise contraction yields a precise Wasserstein convergence statement.
Zero modes and covariance tails can obstruct existence or uniqueness despite finite-cutoff success.

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