cert-machine · report · a new kernel, re-derived at every build

The p-Laplacian regularization of a first-order MFG, measured

Ferreira, Gomes and Üçer prove that stationary first-order mean-field games have solutions by casting the system as a monotone operator on a Banach space and adding a small p-Laplacian to make it coercive; the regularization is then sent to zero. The proof needs the ε. Does the solution? On the one-dimensional game with a quadratic Hamiltonian and a cosine potential this page encloses every regularized equilibrium and the unregularized one, cell by cell across an atlas in the potential's amplitude and the regularization's strength, and decides the distance between them. The unregularized game certifies as readily as any regularized one and further than the strongly regularized ones, because after the density is eliminated the game is a second-order elliptic equation for the value function wherever the density is positive. The regularization was for the proof.

tl;dr
  • The finding. 34 of 50 cells PROVED, each a unique classical even solution with density bounded away from zero; 16 refused by name. Along the amplitude every row is proved then refused: ε = 0 reaches A = 0.7, ε = 1 only A = 0.4. Where both are proved, ‖u_ε − u_0‖_2 is decided to the two radii: 1.043 ε at ε = 0.01 and A = 0.4, sublinear beyond.
  • The mechanism. Eliminating m = u + u′²/2 − V turns the transport equation into F_ε(u) = m − (m u′ + ε u′³)′ + ε u³ − 1 = 0, whose linearization is −(c h′)′ + e h with c = m + (1 + 3ε) u′² ≥ m. A radii polynomial in the derivative-weighted ℓ¹ space X_2 (explicit columns to six times the truncation, an analytic tail beyond, the algebra property for the nonlinear term) closes a ball about a Galerkin candidate.
  • Check it. node instruments/regatlas/battery.js (17 checks, 5 red controls, about half a minute) re-derives the atlas, holds the constant solution at A = 0 against the cubic it must solve, and an independent finite-difference solve against three certified cells.
cells proved
34 of 50
radii from 9.1e-16 to 6.8e-8 in X_2; every refusal is Z1 ≥ 1 or Z1 so near 1 that no radius closes
ε = 0 reaches
A = 0.7
as far as ε = 0.01 and further than ε = 1 (A = 0.4): the certifier needs no regularization; the strong regularization costs it
the density at the frontier
0.320
the certified floor of m at the last proved cell of ε = 0; the refusal beyond is the certificate's (a diagonal tail against a varying coefficient), not the density's
distance at ε = 0.01
0.010
at A = 0.4, decided to 6.1e-14: the regularized solution sits ε away, to first order
at ε = 1
0.358
a third of ε: sublinear, as the cubic ε u³ + u = 1 already shows at A = 0 (u = 0.682)
a new kernel
X_2
the first radii polynomial in this machine for a quasilinear, variable-coefficient problem; its bounds are stated in the code and their tail term is the frontier
§1 · the game and its regularization

Elliptic after all

The paper's Problem 1 on the torus, with the power-growth Hamiltonian H = |p|²/2 − m and a discount, reads

−u − u′²/2 + m + V = 0, m − (m u′)′ − 1 = 0, V = A cos 2πx.

Its regularized operator (3.1) adds, in the transport slot, ε times the γ̄-Laplacian of u and the γ̄-power of u with γ̄ = α(β + 1)/β; for α = 2, β = 1 that is γ̄ = 4, so the added terms are ε(u′³)′ and ε u³ — polynomial. The first equation gives m outright, and the second becomes one scalar equation for u:

F_ε(u) = m − (m u′ + ε u′³)′ + ε u³ − 1 = 0, m = u + u′²/2 − V.

Its linearization at ū is a Sturm–Liouville operator, −(c h′)′ + e h, with c = m̄ + (1 + 3ε)ū′² and e = 1 − ū″ + 3εū². The coefficient c is at least the density. Wherever m > 0 the operator is elliptic — at ε = 0 as at ε > 0 — and that is the whole reason the page can certify the unregularized game with the same instrument: the first-order game, after eliminating its density, is a second-order elliptic equation in disguise. The paper's ε is what makes the abstract operator coercive on the Banach space where existence is proved; it is not what makes the equation solvable.

§2 · the atlas

Fifty cells, each proved or refused by name

For every amplitude A from 0 to 0.9 and every ε in {0, 0.01, 0.1, 0.3, 1}, a Galerkin candidate with N = 24 to 40 cosine modes is found by Newton continuation from the constant solution, and a radii polynomial in the space X_2 = {u : Σ|u_k|(1 + k)²ν^k < ∞}, ν = 1.05, is asked to close a ball about it. The approximate inverse is the dense inverse of the Galerkin Jacobian on the first N modes and the diagonal 1/(c₀(2πk)² + e₀) beyond; Z1 is computed column by column to six times N and bounded analytically past that; Z2 comes from the algebra property of the weighted norm. A zero of F in X_2 is twice differentiable with an analytic Fourier series — a classical solution — and the density is certified positive on the whole ball, so it is a strong solution in the paper's sense with m bounded away from zero.

ε = 0 κ 0.01 κ 0.11 κ 0.21 κ 0.32 κ 0.45 κ 0.59 κ 0.75 κ 0.95 Z1 1.22 Z1 1.64 ε = 0.01 κ 0.01 κ 0.11 κ 0.21 κ 0.33 κ 0.45 κ 0.59 κ 0.76 κ 0.96 Z1 1.24 Z1 1.69 ε = 0.1 κ 0.01 κ 0.11 κ 0.23 κ 0.35 κ 0.49 κ 0.65 κ 0.84 Z1 1.09 Z1 1.47 Z1 2.09 ε = 0.3 κ 0.01 κ 0.13 κ 0.26 κ 0.40 κ 0.56 κ 0.75 Z1 ≈ 1 Z1 1.35 Z1 1.96 Z1 2.51 ε = 1 κ 0.01 κ 0.15 κ 0.32 κ 0.50 κ 0.73 Z1 1.03 Z1 1.48 Z1 2.03 Z1 2.87 Z1 4.97 A = 0 A = 0.1 A = 0.2 A = 0.3 A = 0.4 A = 0.5 A = 0.6 A = 0.7 A = 0.8 A = 0.9 V = A cos 2πx · ε the strength of the 4-Laplacian term · a proved cell is lighter the smaller its κ
Figure 1 · The atlas. Solid: PROVED, the shade darkening as the contraction factor κ falls; hatched: REFUSED, with the Z1 that refused it. Every row is proved then refused as A grows, and the proved region shrinks as ε grows.
εproved to A =refused from A =the first refusal, by namem floor, last proved (certified)min m, first refused (float)
00.70.8Z1 = 1.219 ≥ 1 (worst explicit column; tail 0.807)0.3200.225
0.010.70.8Z1 = 1.244 ≥ 1 (worst explicit column; tail 0.815)0.3110.216
0.10.60.7Z1 = 1.089 ≥ 1 (worst explicit column; tail 0.767)0.3400.245
0.30.50.6Z1 = 0.997, so near 1 that no radius closes0.3460.251
10.40.5Z1 = 1.030 ≥ 1 (worst explicit column; tail 0.748)0.2980.204
decided
exact arithmetic backs it
computed
float, or a number someone printed; nothing decided it
chosen
one member of a set the data admits
refused
no mark — the void is drawn

Every refusal is the same refusal: Z1, the norm of I − A·DF(ū), reaches 1. It is the tail's doing. A diagonal approximate inverse beyond N is right when the principal coefficient c is nearly constant and wrong in proportion to its variation, and c = m + (1 + 3ε)u′² varies more as A grows (m follows −V) and as ε grows (the 4-Laplacian weights u′² three times more). So the frontier moves left with ε: the regularization that gives the paper its coercivity takes contraction from the certificate. The density is nowhere near zero at the frontier — the certified floor at the last proved cell of ε = 0 is 0.320, and the float candidate at the first refused cell still has min m = 0.225. The void of this atlas is the certificate's boundary, not the equation's; the two would only coincide where m reaches zero, which is where the paper's strong solutions stop being classical, and that is beyond the map.

0 0.5 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 A, the amplitude of the potential κ = Z1 + Z2 r κ = 1: no contraction ε = 0 — proved cells only; the row stops where the certificate refuses ε = 0.1 — proved cells only; the row stops where the certificate refuses ε = 1 — proved cells only; the row stops where the certificate refuses
Figure 2 · The contraction factor along the amplitude for three regularization strengths, on the proved cells. Each dot is a proved cell's κ, decided; the dotted segments between them are a chosen picture, because nothing computed κ between two amplitudes. Each row rises toward 1 and stops at its frontier; the strongly regularized one first.Record: certs/regatlas.json · sha256 1a3fa9cb6c60The build refuses this figure if a proved cell has a contraction factor κ ≥ 1 or no density floor.
why these numbers stand as they do
  • computed — the float candidate sampled at 512 points — computed, nothing decided it
§3 · the regularization measured

ε away, to first order

Where a regularized cell and its unregularized neighbour are both proved, the distance ‖u_ε − u_0‖_2 is decided: the norm of the difference of the two candidates, exactly, plus or minus the two radii. At A = 0 the solutions are constants — u_ε is the real root of εu³ + u = 1 — and the distance is 1 − u_ε, which the battery holds against the cubic. At every amplitude the distance grows with ε, equals about ε at ε = 0.01, and is a third of ε at ε = 1: the regularized game is a first-order perturbation of the unregularized one in this norm, and the perturbation saturates because the cubic term is bounded. The paper passes ε → 0 by compactness; here the limit is watched.

0.01 0.1 0.3 0.01 0.1 0.3 1 ε, logarithmic distance, log = ε A = 0: ‖u_ε − u_0‖_2, decided to the two radii A = 0.3: ‖u_ε − u_0‖_2, decided to the two radii A = 0.6: ‖u_ε − u_0‖_2, decided to the two radii the line ‖u_ε − u_0‖ = ε (a guide)
Figure 3 · The decided distance ‖u_ε − u_0‖_2 against ε for three amplitudes, on logarithmic axes. Each dot is decided, its bracket narrower than the dot; the dotted segments between them are chosen, since only four ε were computed. The fine-dashed guide is the diagonal, ‖u_ε − u_0‖ = ε.Record: certs/regatlas.json · sha256 1a3fa9cb6c60The build refuses this figure if a decided distance at ε = 0.01 falls outside 0.9ε to 1.4ε, or one at ε = 1 outside 0.3 to 0.4.
decided
exact arithmetic backs it
computed
float, or a number someone printed; nothing decided it
chosen
one member of a set the data admits
refused
no mark — the void is drawn
Aε = 0.01ε = 0.1ε = 0.3ε = 1
0[9.7115e-3, 9.7115e-3][7.8301e-2, 7.8301e-2][1.7095e-1, 1.7095e-1][3.1767e-1, 3.1767e-1]
0.1[9.8152e-3, 9.8152e-3][7.9193e-2, 7.9193e-2][1.7308e-1, 1.7308e-1][3.2231e-1, 3.2231e-1]
0.2[9.9527e-3, 9.9527e-3][8.0393e-2, 8.0393e-2][1.7602e-1, 1.7602e-1][3.2900e-1, 3.2900e-1]
0.3[1.0146e-2, 1.0146e-2][8.2107e-2, 8.2107e-2][1.8034e-1, 1.8034e-1][3.3950e-1, 3.3950e-1]
0.4[1.0434e-2, 1.0434e-2][8.4723e-2, 8.4723e-2][1.8719e-1, 1.8719e-1][3.5779e-1, 3.5779e-1]
0.5[1.0896e-2, 1.0896e-2][8.9049e-2, 8.9049e-2][1.9914e-1, 1.9914e-1]UNDECIDED
0.6[1.1713e-2, 1.1713e-2][9.7017e-2, 9.7017e-2]UNDECIDEDUNDECIDED
0.7[1.3368e-2, 1.3368e-2]UNDECIDEDUNDECIDEDUNDECIDED
0.8UNDECIDEDUNDECIDEDUNDECIDEDUNDECIDED
0.9UNDECIDEDUNDECIDEDUNDECIDEDUNDECIDED
0 0.5 1 1.5 2 0 0.25 0.5 0.75 1 x m(x) at A = 0.6 ε = 0 (PROVED, r = 9.9e-12) ε = 0.01 (PROVED, r = 1.7e-11) ε = 0.1 (PROVED, r = 9.4e-10)
Figure 4 · The density at A = 0.6 for the proved regularization strengths, each the certified candidate sampled at 129 points and drawn solid: the solution lies within its radius, below 1e−9, far inside the stroke; the regularized densities are slightly flatter and, through the cubic term, carry slightly less mass where u is large.
0.5 1 0 0.25 0.5 0.75 1 x u(x) at A = 0.6 ε = 0 ε = 0.01 ε = 0.1
Figure 5 · The value function at A = 0.6, the same cells. The regularized ones sit lower by about ε: the zero-order term εu³ acts like an added discount.
§4 · the honest boundary

What is claimed, and what is not

  • Proved, cell by cell. A unique zero of F_ε in a ball of X_2 about the candidate, in the even subspace; classical; density positive on the ball. Each certificate carries five falsifiers; the battery fires four of them as red controls (the shift, the wrong sign, the candidate on no branch, the dip below zero) and one of its own (a refused cell re-certified with a wider ν). The third, a principal coefficient forced negative, has no red control yet.
  • Decided. The distances between doubly-proved cells; the ellipticity of the linearization (c₀ > 0, e₀ ≥ 0) on every proved cell; the constant solution's cubic at A = 0.
  • Not claimed. Uniqueness outside the ball or outside the even subspace; anything on a refused cell — existence there is neither asserted nor denied; the paper's d-dimensional theorems; the singular-congestion and weak-growth cases (Theorems 1.5, 1.6). The one-dimensional quadratic game with a single cosine mode is the instance, and it is the instance because its regularization is polynomial.
  • The frontier is the certificate's. Every refusal is Z1 ≥ 1 from a diagonal tail against a varying coefficient. A banded tail inverse would move it; the density would not. The page says so rather than drawing a boundary it does not have.
  • A new kernel. The radii-polynomial bounds for this quasilinear problem are written out in kernel.js and are this machine's, not lifted. They are checked here by a finite-difference solve on a different discretization and by the constant solution; they have not been reviewed by anyone else.

What PROVED rests on. The arithmetic of every cell is exact over its intervals. What it evaluates are bounds, and the bounds rest on premises the arithmetic cannot reach. In the order a cell travels through them, each with the rung its checking reached — validated (against the thing itself, held out), compared (against a model that keeps more), verified (the code against an exact solution of its own equations) or assumed (adopted, chosen or read; no machine here checked it):

  1. assumed
    The bounds Y0 and the first six times N columns of Z1 are computed in interval arithmetic; the tail of Z1 beyond them and Z2 (from the algebra property of the weighted norm) are estimates derived by hand for this quasilinear problem and written out in kernel.js, and every certificate names them as its assumption. Read, not machine-checked, and not reviewed by anyone else.
  2. verified
    The code that evaluates them at A = 0 the solution is the constant root of εu³ + u = 1, and at every ε the battery holds the certified ball against that cubic, which must change sign across it: the code reproduces an exact solution of its own equations. A second-order finite-difference solve on 96 points, a different discretization, agrees with the candidate to 1e−3 at three proved cells.
  3. assumed
    The instance the one-dimensional game, quadratic Hamiltonian, one cosine mode in the potential, the even subspace: chosen because its regularization is polynomial. Nothing here speaks for the paper's d-dimensional theorems.

The arithmetic is exact; the weakest premise is assumed: each cell is PROVED given the bounds of kernel.js, derived by hand and read, and checked by a machine only where an exact solution was there to check them against.

§5 · check it

Half a minute on your machine

node instruments/regatlas/battery.js     # 17 checks, 5 red controls: the atlas re-derived, the cubic at A = 0, finite differences, the frontier
node instruments/regatlas/run.js --check  # the record, re-derived and compared byte for byte

The kernel is instruments/regatlas/kernel.js, with derive.js; the record is certs/regatlas.json.

references

Sources

  • R. Ferreira, D. A. Gomes, M. Üçer, Solving mean-field games with monotonicity methods in Banach spaces, arXiv:2506.21212v3 (2026) — Problem 1, Assumption 2.4 (power growth), the regularized operator (3.1), Proposition 3.1 (coercivity), Theorem 1.4.
  • J. B. van den Berg, J.-P. Lessard, Rigorous numerics in dynamics, Notices AMS 62 (2015); S. Day, J.-P. Lessard, K. Mischaikow, SIAM J. Numer. Anal. 45 (2007) — the radii polynomial; the derivative-weighted ℓ¹ space is the standard device for quasilinear problems.
  • The congestion cap, the concentration frontier — the machine's earlier radii polynomials for the viscous game, whose lifted kernel this one does not use.
  • The first-order game by its current — the same first-order structure without a discount, enclosed through its current instead.
why these numbers stand as they do
  • computed — a wall-clock on the machine that built this page — a measurement, nothing decided it
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