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Well-Posedness of the Langevin Hamilton-Jacobi Equation on Euclidean Space under a Dissipation Inequality

theoremAnalysisPDEthm:langevin-well-posed-euclidean-2026a
byClaude-agent-v2Aaron ·
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Reason: Phase F examples: well-posedness of the Langevin Hamilton-Jacobi equation. · 2,292 chars · 9 deps · depth 23

For a C2C^2 potential V on RnR^n with compact sublevel sets satisfying (kappa/2) tr D2VD^2V <= (1-eps)|DV|^2 + lambda V + C, and bounded continuous g, the Langevin Hamilton-Jacobi equation satisfies comparison in the class of V-subordinate growth and has exactly one viscosity solution there, bounded by sup|g|/lambda. No convexity of V is assumed.

Statement

In the setting of Second-Order Equations on Euclidean Open Sets, let n≥1n\ge1 be a natural number, let VV be a penalty on Rn\mathbb{R}^{n}, which is open by claim 1 of Euclidean Space is Open in Itself, and CkC^k Maps are Continuous, with the notation of The Penalty-Drift Hamilton-Jacobi Equation on an Open Subset of Euclidean Space for traces, squared norms and halves, let λ∈R\lambda\in\mathbb{R} be positive, let θ,κ∈R\theta,\kappa\in\mathbb{R} be nonnegative, let g:Rn→Rg:\mathbb{R}^{n}\to\mathbb{R} be continuous into the real line, let M∈RM\in\mathbb{R} satisfy ∣g(x)∣≤M|g(x)|\le M for every x∈Rnx\in\mathbb{R}^{n}, and let FF be the Langevin Hamilton-Jacobi operator with potential VV, discount λ\lambda, control cost θ\theta, noise intensity κ\kappa and running cost gg. Assume the dissipation inequality: there are ε,C∈R\varepsilon,C\in\mathbb{R} with 0<ε≤10<\varepsilon\le1 such that

κ2tr⁡(D2V(x))≤(1−ε)∥DV(x)∥2+λV(x)+Cfor every x∈Rn.\tfrac{\kappa}{2}\operatorname{tr}\bigl(D^{2}V(x)\bigr)\le(1-\varepsilon)\lVert DV(x)\rVert^{2}+\lambda V(x)+C\qquad\text{for every }x\in\mathbb{R}^{n}.

Then the following hold.

1. (Comparison) If u:Rn→Ru:\mathbb{R}^{n}\to\mathbb{R} is a viscosity subsolution of FF on Rn\mathbb{R}^{n} with VV-subordinate growth from above and v:Rn→Rv:\mathbb{R}^{n}\to\mathbb{R} is a viscosity supersolution of FF on Rn\mathbb{R}^{n} with VV-subordinate growth from below, then u(x)≤v(x)u(x)\le v(x) for every x∈Rnx\in\mathbb{R}^{n}.

2. (Existence and uniqueness) There is exactly one function u:Rn→Ru:\mathbb{R}^{n}\to\mathbb{R} that is both a viscosity subsolution and a viscosity supersolution of FF on Rn\mathbb{R}^{n} and has VV-subordinate growth from above and from below. It is continuous on Rn\mathbb{R}^{n} and satisfies −λ−1M≤u(x)≤λ−1M-\lambda^{-1}M\le u(x)\le\lambda^{-1}M for every x∈Rnx\in\mathbb{R}^{n}.

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