Wasserstein gradient flows and JKO time steps
Derive a variational implicit transport step, separate an exact quadratic toy from entropy-driven diffusion, and state a valid convergence setting.
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The question for today
What you’ll understand
- Define the JKO minimization and derive its one-step energy decrease.
- Identify entropy and confinement as the energy behind a Fokker–Planck flow.
- Compute an exact pure-potential step and distinguish it from explicit gradient descent.
Gradient descent can be formulated without subtracting vectors: choose a new state that lowers an energy while paying for its squared distance from the old state. Using Wasserstein distance makes the states probability measures. The JKO scheme applies this variational idea one time step at a time. Its power includes recovering diffusion equations from energy minimization, but a finite-dimensional quadratic example illustrates only the implicit stepping mechanism, not the full analytic convergence theorem or entropy-driven spreading.
An energy decrease follows from a feasible competitor
Let E be an energy on P₂, choose a positive time step τ, and suppose the displayed minimization has a solution with finite old energy. The next measure minimizes energy plus W₂ squared from the old measure divided by 2τ. The old measure itself is a competitor with zero transport penalty. Therefore the new energy plus its transport penalty is at most the old energy, yielding both an energy decrease and a quantitative bound on one-step movement.
This inequality does not guarantee minimizers exist for arbitrary energies. Properness, lower semicontinuity, and coercivity or compactness conditions enter the direct method, while suitable displacement convexity can supply uniqueness and stable limiting behavior. Nor does energy decrease alone prove that discrete trajectories converge to a particular PDE. One must control the accumulated transport movement and identify a limit through the energy’s variation and the conservation structure.
Entropy produces diffusion rather than simple contraction
For an absolutely continuous probability density r, use the entropy convention ∫r log r, which is negative Shannon differential entropy. Add the confining potential energy ∫|x|²r/2. Assign infinite entropy to singular measures. Starting from a P₂ probability density with finite energy, the classical JKO setting for this smooth quadratic confinement gives convergence of suitable time interpolants as τ tends to zero to the Fokker–Planck evolution. This is a statement about the vanishing-step sequence, not exact equality after one finite minimization.
Formally the energy variation is log r + 1 + |x|²/2. Transport down its gradient gives velocity −∇log r − x; inserting this into mass conservation produces Δr + div(xr). The logarithmic term spreads mass while confinement pulls it inward. The formal differentiation explains the equation, while existence and convergence require the weak analytic theorem. Gaussian stationary density is consistent with this balance because its logarithmic gradient is −x.
Remove entropy to isolate an exactly solvable implicit step
For the different energy E(ρ) = one-half its second moment, atoms are allowed and there is no diffusion term. Couple an old position x with a candidate new position y. The one-step objective integrates y²/2 + |y − x|²/(2τ). Minimizing this strictly convex quadratic pointwise gives y = x/(1 + τ). The resulting deterministic pushforward achieves the lower bound for all couplings, proving the exact measure update.
Iterating this pure-potential step contracts each position by (1 + τ) to the negative k-th power. At fixed elapsed time kτ, letting τ shrink gives exponential contraction e to the minus time, matching the corresponding nondiffusive transport flow. A finite τ update remains implicit Euler, whereas explicit gradient descent uses factor 1 − τ and may overshoot. The entropy-free calculation cannot be used to claim that a Gaussian’s variance collapses under the entropy-plus-confinement Fokker–Planck equation.
Work it through
One implicit step for two atoms
Use only E(ρ) = (1/2)∫x² dρ, take τ = 1/2, and start from equal atoms at −2 and 2. Find the next measure and verify the variational energy bound.
The pointwise minimizer sends x to x/(1 + τ) = 2x/3. Hence the next measure has equal atoms at −4/3 and 4/3; this is the pure-potential problem, where singular measures are permitted.
The old energy is two. The new second moment is 16/9, giving new energy 8/9. Each particle moves distance 2/3, so W₂ squared is 4/9; the increasing map is optimal on the line.
Because 2τ = 1, the transport penalty is 4/9. The full one-step objective is 8/9 + 4/9 = 4/3, which is at most the old-energy competitor value two.
The implicit step contracts the atoms to ±4/3, and its objective 4/3 verifies the energy-decrease inequality without modeling diffusion.
The living notebook
Explore the model
Interactive lab · optimization
An exact numerical match for a different step rule
Pause and Reset, then set both step size η = 0.5 and initial position x₀ = 2. Predict position 1 after one Step and 0.5 after two. These explicit quadratic-descent values equal the pure-potential JKO positions for τ = 1. In a fresh run Reset, then set η = 1 and x₀ = 2; one Step gives 0, which differs from the τ = 1 JKO result 1. The lab is explicit scalar descent, not an entropy or transport minimizer.
- Position x0
- 2
- Objective x²/2
- 2
- Iteration behavior
- Stable step
Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. 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 forA selected contraction factor can agree across two algorithms without making their time-step parameters or general equations identical.
Pause & reason
Check your understanding
Your turn
Try it yourself
For the same pure-potential energy, start from δ₂ and take τ = 1. Compare the minimizing next atom with the candidate δ₀, compute each objective, and identify the continuous contraction limit.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Minimize y²/2 + (y − 2)²/2. Keep this implicit objective separate from an explicit gradient-descent formula.
Compare with the solution
The derivative is 2y − 2, so the minimizing position is y = 1, with positive second derivative two. Its energy is 1/2 and its transport penalty is 1/2, totaling one. The candidate y = 0 has zero energy but penalty four divided by two, totaling two, so it is worse. The old atom has objective two as well. Explicit descent with step one would jump to zero and is therefore not this implicit minimizer. Repeated JKO steps of size τ give position 2(1 + τ)^(−k); with kτ approaching t and τ tending to zero, the limit is 2e^(−t).
Keep the idea
Take it with you
- JKO minimizes energy plus a Wasserstein movement penalty and gives a one-step dissipation bound.
- Entropy-plus-confinement flow has a vanishing-step convergence theorem beyond a formal variation.
- An entropy-free implicit contraction is a useful exact example with different dynamics from diffusion.
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