ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The eikonal equation as a viscosity equation at a tip

Example

Let $n \geq 1$ and consider the eikonal equation $∣ D u ∣ = 1$ on $R^{n}$ , i.e. the first-order equation $u_{t} + H (D u) = 0$ with $H (p) = ∣ p ∣ - 1$ read on functions independent of $t$ . The Euclidean norm $u (x) = ∣ x ∣$ is a classical solution on $R^{n} ∖ {0}$ , and at its tip $a = 0$ it exhibits the difference between the subsolution inequality $∣ D ϕ ∣ \leq 1$ and the reverse supersolution inequality $∣ D ϕ ∣ \geq 1$ . There is no $C^{1}$ test function $ϕ$ with $u - ϕ$ having a local maximum at $0$ : such a contact would force $ϕ (x) \geq ϕ (0) + ∣ x ∣$ near $0$ , hence $⟨ D ϕ (0), h ⟩ \geq ∣ h ∣$ for every direction $h$ , which is impossible for a linear functional. Writing $x_{0}$ for the coordinate indexed by $0 < n$ , the $C^{1}$ functions $ϕ (x) = ε x_{0}$ with $0 < ε < 1$ satisfy $ϕ \leq ∣ x ∣$ near $0$ with equality at $0$ , so $u - ϕ$ has a local minimum at $0$ and the supersolution test would require $∣ D ϕ (0) ∣ = ε \geq 1$ , which fails. Hence $u$ is a subsolution of $∣ D u ∣ = 1$ but not a supersolution at the tip: it satisfies the viscosity subsolution inequality $∣ D u ∣ \leq 1$ at the tip and is a viscosity solution of $∣ D u ∣ = 1$ on $R^{n} ∖ {0}$ .

Verification

Given: The tip $a = 0 \in R^{n}$ , the function $u (x) = ∣ x ∣$ , the test-function definition of viscosity sub- and supersolutions for $u_{t} + H (D u) = 0$ with $H (p) = ∣ p ∣ - 1$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem), and the Euclidean inner product $⟨ \cdot, \cdot ⟩$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ) with gradient as in The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case.

[F1] $u - ϕ$ has a local maximum (respectively minimum) at $0$ for $ϕ \in C^{1}$ precisely when $u (x) - ϕ (x) \leq u (0) - ϕ (0)$ (respectively $\geq$ ) for all $x$ near $0$ , and the viscosity inequalities are tested at such contacts (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem). For the stationary extension $U (x, t) = u (x)$ on $R^{n} \times (0, 1)$ , a space--time contact with $Φ$ has $Φ_{t} = 0$ : restrict to the time line, where the differentiable function $t \mapsto Φ (x, t)$ has a local extremum. Restricting to the spatial slice therefore gives the inequalities $∣ D ϕ ∣ \leq 1$ and $∣ D ϕ ∣ \geq 1$ used here; conversely, a spatial test extends to a time-independent space--time test. The norm is continuous by [F2], so the required semicontinuity holds.

[F2] The Euclidean norm satisfies $∣ t h ∣ = ∣ t ∣ ∣ h ∣$ for $t \in R$ , and for $x \neq 0$ it is differentiable at $x$ with gradient $x / ∣ x ∣$ , of norm $1$ : The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ gives $∣ x + t h ∣^{2} = ∣ x ∣^{2} + 2 t ⟨ x, h ⟩ + t^{2} ∣ h ∣^{2}$ , the norm axioms and the triangle inequality in clause 2 of Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation give $∣ x + t h ∣ \geq ∣ x ∣ - ∣ t ∣ ∣ h ∣$ and $∣ ∣ x + t h ∣ - ∣ x ∣ ∣ \leq ∣ t ∣ ∣ h ∣$ , Cauchy--Schwarz (clause 1 there) gives $∣ ⟨ x, h ⟩ ∣ \leq ∣ x ∣ ∣ h ∣$ , and since $∣ x + t h ∣ + ∣ x ∣ \geq ∣ x ∣ > 0$ the identity $∣ x + t h ∣ - ∣ x ∣ = (2 t ⟨ x, h ⟩ + t^{2} ∣ h ∣^{2}) / (∣ x + t h ∣ + ∣ x ∣)$ differs from $t ⟨ x, h ⟩ / ∣ x ∣$ by at most $2 t^{2} ∣ h ∣^{2} / ∣ x ∣$ , so the gradient is $x / ∣ x ∣$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case).

Proof technique: direct test-function computation at the tip and the classical-consistency proposition away from it.

1.1F1F2algebra

No upper test exists at the tip. Suppose $ϕ \in C^{1} (R^{n})$ with $u - ϕ$ having a local maximum at $0$ ; by [F1], $∣ x ∣ - ϕ (x) \leq - ϕ (0)$ near $0$ , that is $ϕ (x) \geq ϕ (0) + ∣ x ∣$ there. Substituting $x = t h$ with $∥ h ∥ = 1$ and $t ↓ 0$ and dividing by $t$ gives $⟨ D ϕ (0), h ⟩ \geq 1$ for every unit vector $h$ ; testing $h$ and $- h$ gives both $⟨ D ϕ (0), h ⟩ \geq 1$ and $- ⟨ D ϕ (0), h ⟩ \geq 1$ , an impossibility. Hence there is no upper test at the tip and the subsolution inequality holds vacuously there.

1.2F1F2algebra

A lower test with a failing supersolution inequality. Every lower contact at the tip has slope of norm at most one: if $ϕ$ is $C^{1}$ with $u - ϕ$ having a local minimum at $0$ , then after translating $ϕ$ we have $∣ x ∣ \geq ⟨ D ϕ (0), x ⟩ + o (∣ x ∣)$ ; substituting $x = t h$ with $∥ h ∥ = 1$ gives $⟨ D ϕ (0), h ⟩ \leq 1$ for $t > 0$ and $⟨ D ϕ (0), h ⟩ \geq - 1$ for $t < 0$ , hence $∣ D ϕ (0) ∣ \leq 1$ . Now take $ϕ (x) = ε x_{0}$ , where $x_{0}$ is the coordinate indexed by $0 < n$ , with $0 < ε < 1$ : then $ϕ (x) \leq ∣ x ∣$ near $0$ with equality at $0$ , so $u - ϕ$ has a local minimum at $0$ by [F1], while the supersolution condition requires $∣ D ϕ (0) ∣ = ε \geq 1$ and fails. This single lower contact shows that $u$ is not a viscosity supersolution of $∣ D u ∣ = 1$ at the tip.

1.3F2algebra

Away from the tip the equation holds classically. On the open set $R^{n} ∖ {0}$ the function $u$ is $C^{1}$ with $∣ D u ∣ = 1$ by [F2]. Its stationary extension on $(R^{n} ∖ {0}) \times (0, 1)$ extends continuously to the closed cylinder with initial datum $∣ x ∣$ , so Classical solutions are viscosity solutions and differentiable viscosity solutions solve the equation pointwise applies and it is a viscosity solution of $∣ D u ∣ = 1$ there; in particular it is both a subsolution and a supersolution at every $x \neq 0$ .

2.1step 1.1step 1.2step 1.3∎

Conclusion. By step 1.1 the subsolution test at the tip is vacuous, by step 1.2 the supersolution test fails there, and by step 1.3 both tests hold away from the tip. Hence $u$ solves the subsolution inequality $∣ D u ∣ \leq 1$ everywhere and solves $∣ D u ∣ = 1$ exactly on $R^{n} ∖ {0}$ , while it is not a viscosity solution of $∣ D u ∣ = 1$ on any neighbourhood of the tip.

Depends on

Used by

Dependency tree · two levels

37 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Hung Vinh Tran, Hamilton--Jacobi Equations: Theory and Applications, 2020 preliminary author manuscript of AMS Graduate Studies in Mathematics 213 (complete text) (standard reference, not scraped)
Alberto Bressan, Viscosity Solutions of Hamilton--Jacobi Equations and Optimal Control Problems, complete author lecture notes, Penn State University (PDF records Fall 2019 revision) (standard reference, not scraped)