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A strict subsolution can fail the supersolution lower-test condition

Statement refuted

False claim: for a first-order equation $u_{t} + H (x, t, D u) = 0$ , passing the subsolution test at every upper contact forces the supersolution test at every lower contact, so that a viscosity subsolution which is differentiable is automatically a viscosity solution.

The claim fails for the equation $u_{t} + H (u_{x}) = 0$ on $U = R \times (0, \infty)$ with $H (p) = p$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem): the function $u (x, t) = - t$ is a viscosity subsolution but not a viscosity supersolution. The numerical residual $u_{t} + H (u_{x}) = - 1$ is the same at upper and lower contacts; what differs is the direction of the inequality that the definition requires.

Facts & Assumptions

Given: The open set $U = R \times (0, \infty)$ , the Hamiltonian $H (p) = p$ , the equation $u_{t} + H (u_{x}) = 0$ , and $u (x, t) = - t$ on $U$ .

[F1]

A viscosity subsolution is tested at local maxima of $u - ϕ$ and must satisfy $ϕ_{t} + H (z_{0}, D ϕ) \leq 0$ there; a viscosity supersolution is tested at local minima of $v - ϕ$ and must satisfy $ϕ_{t} + H (z_{0}, D ϕ) \geq 0$ there; the test function $ϕ$ is required to be $C^{1}$ , whereas $u$ and $v$ need only have their respective semicontinuity (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

[F2]

At a point where a differentiable function of several variables has a local maximum or a local minimum, its total derivative vanishes (Fermat's theorem: an interior differentiable local extremum has zero gradient).

Counterexample

technique · direct evaluation of the two contact inequalities

1.1F1F2algebra

$u$ is a viscosity subsolution. Let $ϕ \in C^{1} (U)$ and let $u - ϕ$ have a local maximum at $z_{0} = (x_{0}, t_{0}) \in U$ . The function $u - ϕ$ is differentiable with total derivative $D u - D ϕ$ , so [F2] gives $D ϕ (z_{0}) = D u (z_{0}) = (0, - 1)$ , that is $ϕ_{x} (z_{0}) = 0$ and $ϕ_{t} (z_{0}) = - 1$ . Hence $ϕ_{t} (z_{0}) + H (ϕ_{x} (z_{0})) = - 1 + 0 = - 1 \leq 0$ , which is the subsolution inequality of [F1] at the upper contact.

1.2F1algebra

$u$ is not a viscosity supersolution. Take the test function $ϕ : = u$ , which belongs to $C^{1} (U)$ . Then $u - ϕ \equiv 0$ has a local minimum at every point of $U$ , so the supersolution test of [F1] applies at, say, $z_{0} = (0, 1)$ ; but $ϕ_{t} (z_{0}) + H (ϕ_{x} (z_{0})) = - 1 + 0 = - 1 < 0$ , and the required inequality is $\geq 0$ . Hence the supersolution condition fails, and $u$ is not a viscosity solution of the equation on $U$ .

2.1step 1.1step 1.2∎

Conclusion. Step 1.1 verifies every upper contact of $u$ and step 1.2 exhibits a lower contact at which the opposite inequality fails, so the subsolution property does not imply the supersolution property; the residual is $- 1$ in both computations, and only the required direction of the inequality changes between them.

Remarks

What the example isolates. The sign asymmetry of the test-function definition is not a matter of the value of the residual but of the direction of the inequality at the two kinds of contact. This is the reason the page defines the two one-sided notions separately and why the reverse implication is false for a monotone-in-time function.

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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)