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Minima of viscosity subsolutions need not be subsolutions

Statement refuted

False claim: a finite minimum of finitely many viscosity subsolutions of a first-order equation is again a viscosity subsolution.

Take the stationary proper equation $F (x, u, u^{'}) : = u + 1 - (u^{'})^{2} = 0 on Ω = (- 1, 1),$ whose operator $F (r, p) = r + 1 - p^{2}$ is continuous and strictly increasing in $r$ (proper), and the two affine functions $u_{1} (x) = 1 - 2 x$ and $u_{2} (x) = 1 + 2 x$ . Both are classical subsolutions: $F (u_{i}, u_{i}^{'}) = u_{i} + 1 - 4 = u_{i} - 3 \leq 0$ on $Ω$ because $u_{i} \leq 3$ there. Their pointwise minimum is $u (x) = 1 - 2 ∣ x ∣$ , a peak at $x_{0} = 0$ with value $1$ . The $C^{1}$ test function $ϕ \equiv 1$ satisfies $ϕ \geq u$ on $Ω$ with equality at $0$ , so $u - ϕ$ has a local maximum at $0$ ; but $F (0, u (0), ϕ^{'} (0)) = 1 + 1 - 0 = 2 > 0$ , so $u$ fails the subsolution test at the corner. Hence the finite-minimum operation is not admissible for subsolutions of a proper equation, while the finite maximum of subsolutions and the finite minimum of supersolutions are, by the same active-index contact argument as Finite maxima of subsolutions and finite minima of supersolutions: the active function has the same value and the same test at the contact. The zero-order term contributes to this residual but is not essential to failure of the minimum operation: for $F (p) = 1 - p^{2}$ the same two affine functions have residual $- 3$ , while their minimum has the upper-test residual $F (0) = 1 > 0$ .

Facts & Assumptions

Given: The interval $Ω = (- 1, 1)$ , the proper operator $F (r, p) = r + 1 - p^{2}$ , the functions $u_{1} = 1 - 2 x$ , $u_{2} = 1 + 2 x$ , their pointwise minimum $u = \min (u_{1}, u_{2}) = 1 - 2 ∣ x ∣$ , and the constant test $ϕ \equiv 1$ .

[F1]

For a proper operator $F (x, r, p)$ the viscosity subsolution inequality is $F (x_{0}, u (x_{0}), D ϕ (x_{0})) \leq 0$ at every point $x_{0}$ where $u - ϕ$ has a local maximum, $ϕ \in C^{1}$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem for the test-function scheme; here this displayed inequality defines the zero-order stationary adaptation; the cited item itself treats only the evolutionary operator with no zero-order term).

[F2]

A real-valued function has a local maximum at $x_{0}$ when its value near $x_{0}$ is at most its value at $x_{0}$ ; the total derivative of a $C^{1}$ function is computed from its partial derivatives; at a differentiable local extremum the derivative vanishes (Fermat's theorem: an interior differentiable local extremum has zero gradient, Local and strict local extrema for scalar fields on Euclidean open sets, The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ).

Counterexample

technique · explicit touching test at the corner of the minimum

1.1F1F2algebra

The two affine functions are subsolutions. For $i = 1, 2$ and $x \in (- 1, 1)$ we have $u_{i} (x) = 1 \mp 2 x \leq 3$ , and $u_{i}$ is $C^{1}$ with $u_{i}^{'} = \mp 2$ , so $F (u_{i}, u_{i}^{'}) = u_{i} + 1 - 4 = u_{i} - 3 \leq 0$ pointwise on $Ω$ . Since at a local maximum of $u_{i} - ϕ$ the differentiability of $u_{i}$ and $ϕ$ forces $D ϕ (x_{0}) = D u_{i} (x_{0})$ by [F2], we get $F (x_{0}, u_{i} (x_{0}), D ϕ (x_{0})) = u_{i} (x_{0}) - 3 \leq 0$ ; hence $u_{1}$ and $u_{2}$ are viscosity subsolutions.

2.1step 1.1F1F2algebra

The minimum fails the test. The pointwise minimum is $u = 1 - 2 ∣ x ∣$ with $u (0) = 1$ and $u (x) < 1$ for $x \neq 0$ ; the constant $ϕ \equiv 1$ is $C^{1}$ with $ϕ^{'} = 0$ and satisfies $ϕ \geq u$ on $Ω$ with equality exactly at $0$ , so $u - ϕ$ has a local maximum at $0$ by [F2]. The subsolution test of [F1] at $x_{0} = 0$ would require $F (0, u (0), ϕ^{'} (0)) = 1 + 1 - 0 = 2 \leq 0$ , which is false. Hence the finite minimum of the two subsolutions is not a subsolution.

3.1step 1.1step 2.1∎

Conclusion. Step 1.1 exhibits two viscosity subsolutions of the proper equation $u + 1 - (u^{'})^{2} = 0$ and step 2.1 shows their pointwise minimum fails the subsolution test at the peak; For this stationary operator, the finite-maximum and dual finite-minimum rules follow directly by choosing an active index at the contact: its function value is unchanged there and its test is the same. This is the argument of Finite maxima of subsolutions and finite minima of supersolutions, whose stated operator has no zero-order term. The computation $F (p) = 1 - p^{2}$ with the same slopes and constant upper test also shows failure without a zero-order term.

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Sources

Michael G. Crandall, Hitoshi Ishii and Pierre-Louis Lions, User's guide to viscosity solutions of second order partial differential equations, Bulletin of the American Mathematical Society 27 (1992), 1--67 (complete article) (standard reference, not scraped)
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)