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Hamilton Jacobi Equations and Viscosity Solutions — Examples

1 · Prerequisites

2 · Summary

These companions compute the viscosity theory of the main page on explicit equations and mark its scope boundaries. The eikonal equation is tested at its ridge: the positive norm has no upper contact at the tip but fails the supersolution test, while the negative norm has upper contacts with slopes in $[- 1, 1]$ and no lower contact; distance to the boundary of the unit ball is verified as the nonsmooth solution of the unit eikonal Dirichlet problem, and the counterexample of three distinct eikonal solutions on an interval shows that omitting endpoint data destroys uniqueness. The quadratic Hopf--Lax formula is identified with an infimal convolution and the quadratic Moreau envelope is computed separately for the unbounded datum. Two scope counterexamples show that nonconvexity breaks the Hopf--Lax equation and that nonsuperlinearity restricts the Lagrangian to a bounded velocity set, and that minima of subsolutions of a proper equation need not be subsolutions. A smooth datum whose characteristics cross at $t = 1 / 2$ produces a Hopf--Lax solution with a forming corner, and the Cole--Hopf family for $\sin$ is shown to converge to the Hopf--Lax solution uniformly on finite time strips. None of the examples uses a choice principle.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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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.

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The quadratic Hopf--Lax formula as an infimal convolution

Example

Let $n \geq 1$ and $H (p) = \frac{1}{2} ∣ p ∣^{2}$ . Its Legendre transform is $L (v) = \frac{1}{2} ∣ v ∣^{2}$ . For every bounded uniformly continuous $u_{0} : R^{n} \to R$ , the Hopf--Lax operator of The Hopf--Lax operator and the Hopf--Lax formula is, for $t > 0$ , $Q_{t} u_{0} (x) = \inf_{y \in R^{n}} {u_{0} (y) + \frac{∣ x - y ∣^{2}}{2 t}},$ the infimal convolution with the quadratic kernel $q_{t} (z) = ∣ z ∣^{2} / (2 t)$ . The infimum is attained, and for every minimiser $y$ the Euler relation $D u_{0} (y) = (x - y) / t$ holds whenever $u_{0}$ is differentiable at $y$ (A Hopf--Lax minimiser satisfies the characteristic Euler relation at differentiability points). Separately, the quadratic datum $w_{0} (y) = \frac{1}{2} ∣ y ∣^{2}$ is unbounded and so is outside the datum class in The Hopf--Lax operator and the Hopf--Lax formula. Its algebraic infimal convolution $I_{t} (x) : = \inf_{y \in R^{n}} {\frac{1}{2} ∣ y ∣^{2} + \frac{∣ x - y ∣^{2}}{2 t}}$ has the unique minimiser $y = x / (1 + t)$ and value $I_{t} (x) = ∣ x ∣^{2} / (2 (1 + t))$ ; this separate calculation is the Moreau envelope of the quadratic function and does not apply the bounded-data Hopf--Lax theorem to $w_{0}$ .

Verification

Given: The Hamiltonian $H (p) = \frac{1}{2} ∣ p ∣^{2}$ on $R^{n}$ , its Legendre transform $L$ , a bounded uniformly continuous datum $u_{0}$ , the operators $Q_{t}$ of The Hopf--Lax operator and the Hopf--Lax formula, and the unbounded quadratic datum $w_{0} (y) = \frac{1}{2} ∣ y ∣^{2}$ .

[F1] $L (v) = \sup_{p \in R^{n}} (p \cdot v - H (p))$ , the supremum taken in $\overline{R}$ (The Legendre transform of a finite-valued convex Hamiltonian).

[F2] For $t > 0$ , $Q_{t} u_{0} (x) = \inf_{y} {u_{0} (y) + t L ((x - y) / t)}$ , and under convexity and superlinearity of $H$ the infimum is finite and attained for bounded uniformly continuous $u_{0}$ (The Hopf--Lax operator and the Hopf--Lax formula, Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers).

[F3] At a minimiser $y$ of $u_{0} (y) + t L ((x - y) / t)$ , if $u_{0}$ is differentiable at $y$ and $L$ at $(x - y) / t$ , then $D u_{0} (y) = D L ((x - y) / t)$ (A Hopf--Lax minimiser satisfies the characteristic Euler relation at differentiability points).

Proof technique: compute the quadratic conjugate, use the in-class Hopf--Lax suppliers only for bounded uniformly continuous data, and evaluate the separate quadratic infimum by completing the square.

1.1F1algebra

The conjugate of the quadratic Hamiltonian. Fix $v \in R^{n}$ and complete the square: $p \cdot v - \frac{1}{2} ∣ p ∣^{2} = \frac{1}{2} ∣ v ∣^{2} - \frac{1}{2} ∣ p - v ∣^{2}$ , whose supremum over $p$ is attained at $p = v$ with value $\frac{1}{2} ∣ v ∣^{2}$ . Hence $L (v) = \frac{1}{2} ∣ v ∣^{2}$ by [F1]; in particular $L$ is finite, convex and superlinear.

2.1step 1.1F2F3algebra

The formula, attainment and the Euler relation. Substituting $L (v) = ∣ v ∣^{2} / 2$ into the definition of $Q_{t}$ gives the displayed infimal convolution. The infimum is attained by [F2], and for every minimiser $y$ the conditional Euler relation $D u_{0} (y) = D L ((x - y) / t) = (x - y) / t$ is [F3]; both suppliers use only the bounded uniformly continuous data class.

3.1step 1.1algebra∎

The separate quadratic infimum. For $t > 0$ and all $x, y$ , completing the square gives $\frac{1}{2} ∣ y ∣^{2} + \frac{∣ x - y ∣^{2}}{2 t} = \frac{1 + t}{2 t} ∣ y - \frac{x}{1 + t} ∣^{2} + \frac{∣ x ∣^{2}}{2 (1 + t)}$ . Since the coefficient $(1 + t) / (2 t)$ is positive, the infimum over $y$ is attained uniquely at $y = x / (1 + t)$ with value $∣ x ∣^{2} / (2 (1 + t))$ . This is a direct computation for the unbounded datum $w_{0}$ and makes no assertion that $w_{0}$ lies in the domain of the Hopf--Lax operator.

Remarks

What is and is not applied. The bounded-data statements are applied only to bounded uniformly continuous $u_{0}$ ; the quadratic datum is treated by the displayed algebraic computation, which is the Moreau envelope of $\frac{1}{2} ∣ \cdot ∣^{2}$ and does not claim a Hopf--Lax solution for it.

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A Hopf--Lax solution with a forming corner from smooth data

Example

Let $H (p) = p^{2} / 2$ , $L (v) = v^{2} / 2$ , and $u_{0} (x) = e^{- x^{2}} - 1$ . Then $u_{0}$ is smooth, bounded and uniformly continuous, with $p_{0} (y) : = u_{0}^{'} (y) = - 2 y e^{- y^{2}}$ and $u_{0}^{''} (y) = (4 y^{2} - 2) e^{- y^{2}}$ . The characteristic projection is $X_{t} (y) = y + t p_{0} (y) = y - 2 t y e^{- y^{2}}$ , and its lifted value is $Z_{t} (y) = u_{0} (y) + \frac{t}{2} p_{0} (y)^{2}$ . For $0 \leq t < \frac{1}{2}$ , $X_{t}$ is a diffeomorphism of $R$ , and $u_{c l} (X_{t} (y), t) = Z_{t} (y)$ is the classical characteristic solution. At $t = \frac{1}{2}$ , $X_{t}^{'} (0) = 0$ ; for $t > \frac{1}{2}$ , putting $s_{t} = \sqrt{\log (2 t)}$ gives $X_{t} (- s_{t}) = X_{t} (0) = X_{t} (s_{t}) = 0$ , so the characteristic projection is no longer injective and its single-valued classical graph breaks down. The Hopf--Lax formula $u (x, t) = \inf_{y \in R} {e^{- y^{2}} - 1 + \frac{∣ x - y ∣^{2}}{2 t}}, t > 0,$ is finite, satisfies $- 1 \leq u (x, t) \leq 0$ , is uniformly continuous in $x$ , and is a viscosity solution of $u_{t} + \frac{1}{2} u_{x}^{2} = 0$ with initial datum $u_{0}$ (The Hopf--Lax formula solves the Hamilton--Jacobi Cauchy problem, The Hopf--Lax operator preserves a modulus of continuity). For each fixed $t > \frac{1}{2}$ , the minimisers at $x = 0$ are exactly $y = \pm s_{t}$ , and $u (0, t) = \frac{1 + \log (2 t)}{2 t} - 1.$ For $x > 0$ sufficiently close to $0$ , the unique minimiser tends to $s_{t}$ as $x ↓ 0$ ; for $x < 0$ sufficiently close to $0$ , it tends to $- s_{t}$ as $x ↑ 0$ . Thus the one-sided spatial derivatives tend to $- s_{t} / t$ from the right and $s_{t} / t$ from the left, so $u (\cdot, t)$ is continuous but has a corner at $x = 0$ .

Verification

Given: The Hamiltonian $H (p) = p^{2} / 2$ with Lagrangian $L (v) = v^{2} / 2$ , the datum $u_{0} (y) = e^{- y^{2}} - 1$ , its derivatives $p_{0} = u_{0}^{'}$ , $u_{0}^{''}$ , the characteristic data $X_{t} (y) = y + t p_{0} (y)$ , $Z_{t} (y) = u_{0} (y) + \frac{t}{2} p_{0} (y)^{2}$ , and the Hopf--Lax function $u = Q_{t} u_{0}$ (The Hopf--Lax operator and the Hopf--Lax formula, Characteristic crossing and caustic for a first-order PDE).

[F1] $Q_{t} u_{0}$ satisfies the bounds $- 1 \leq Q_{t} u_{0} \leq 0$ , is spatially uniformly continuous, and is a viscosity solution with datum $u_{0}$ (The Hopf--Lax operator preserves a modulus of continuity, The Hopf--Lax formula solves the Hamilton--Jacobi Cauchy problem); the defining infimum is attained (Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers).

[F2] Differentiation rules for the elementary functions give $p_{0} (y) = - 2 y e^{- y^{2}}$ , $p_{0}^{'} (y) = (4 y^{2} - 2) e^{- y^{2}}$ , and $X_{t}^{'} (y) = 1 + t p_{0}^{'} (y) = 1 - 2 t e^{- y^{2}} + 4 t y^{2} e^{- y^{2}}$ (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 $U \subseteq R^{m} \to R^{n}$ ).

[F3] A continuous scalar function takes every value between its endpoint values (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ), and for a differentiable scalar function on an interval there is a mean-value point (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ). The bound $e^{y^{2}} \geq 1 + y^{2}$ from $1 + x \leq \exp (x)$ for every real $x$ , hence $(1 - p)^{m} \leq \exp (- m p)$ gives $∣ y ∣ e^{- y^{2}} \leq ∣ y ∣ / (1 + y^{2}) \to 0$ . The exponential is smooth and the elementary chain and algebra rules apply (The exponential function is smooth and $(\exp)^{'} = \exp$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

Proof technique: explicit characteristic and minimiser computations.

1.1F2F3algebra

Classical solution before the first singular time. For $0 \leq t < 1 / 2$ , [F2] gives $X_{t}^{'} (y) \geq 1 - 2 t > 0$ . The mean value theorem [F3] makes $X_{t}$ strictly increasing, and $X_{t} (y) - y = - 2 t y e^{- y^{2}} \to 0$ at both infinities, so the intermediate value theorem gives a unique inverse $Y (x, t)$ . This inverse is continuous jointly: near any fixed $(x, t)$ , $X_{t}^{'}$ has a positive lower bound, and the mean value theorem bounds changes in $Y$ by changes in $x$ and in $X_{t} (Y)$ . Differentiating the identity $x = Y + t p_{0} (Y)$ by difference quotients then gives $Y_{x} = (1 + t p_{0}^{'} (Y))^{- 1}$ and $Y_{t} = - p_{0} (Y) / (1 + t p_{0}^{'} (Y))$ , continuously. Thus $u_{c l} (x, t) = u_{0} (Y) + (t / 2) p_{0} (Y)^{2}$ is $C^{1}$ , with $u_{c l, x} = p_{0} (Y)$ and $u_{c l, t} = - p_{0} (Y)^{2} / 2$ , by substitution of these derivatives. It solves the equation and has the initial datum $u_{0}$ .

1.2F2algebra

Breakdown of the projection. At $t = \frac{1}{2}$ we have $X_{t}^{'} (0) = 1 - 2 t = 0$ . For $t > \frac{1}{2}$ put $s_{t} = \sqrt{\log (2 t)} > 0$ ; then $e^{- s_{t}^{2}} = 1 / (2 t)$ , so $X_{t} (\pm s_{t}) = \pm s_{t} - 2 t (\pm s_{t}) / (2 t) = 0 = X_{t} (0)$ , and the projection is not injective; it is locally decreasing near $y = 0$ because $X_{t}^{'} (0) = 1 - 2 t < 0$ . The loss of rank at $t = 1 / 2$ , $y = 0$ , is the caustic of Characteristic crossing and caustic for a first-order PDE; the later equal projections show global folding, without asserting local noninjectivity at $y = 0$ for $t > 1 / 2$ .

2.1step 1.1step 1.2F1F2F3algebra∎

Bounds, minimisers at $x = 0$ , and the corner. By [F1] the Hopf--Lax function is finite, $- 1 \leq u \leq 0$ and uniformly continuous in $x$ . For $g_{t} (y) : = e^{- y^{2}} - 1 + y^{2} / (2 t)$ we have $g_{t}^{'} (y) = y (1 / t - 2 e^{- y^{2}})$ , so for $t > \frac{1}{2}$ the critical points are $0, \pm s_{t}$ , with $g_{t}^{''} (0) = 1 / t - 2 < 0$ and $g_{t}^{''} (\pm s_{t}) = 2 s_{t}^{2} / t > 0$ ; since $g_{t} (y) \to \infty$ as $∣ y ∣ \to \infty$ and $g_{t} (\pm s_{t}) = - 1 + (1 + \log (2 t)) / (2 t) < 0 = g_{t} (0)$ (since $g_{t}^{'} (y) < 0$ for $0 < y < s_{t}$ and [F3] makes $g_{t}$ strictly decreasing there), the global minimisers of $g_{t}$ are exactly $\pm s_{t}$ , and the value is $(1 + \log (2 t)) / (2 t) - 1$ . For $G_{t} (x, y) : = u_{0} (y) + (x - y)^{2} / (2 t)$ any minimiser obeys $∣ x - y ∣ \leq \sqrt{2 t}$ , so all minimisers for $∣ x ∣ \leq 1$ lie in a fixed compact interval. For every neighbourhood of ${- s_{t}, s_{t}}$ , the complement in this interval has a positive gap above $\min g_{t}$ by continuity and compact attainment [F1]; uniform convergence $G_{t} (x, \cdot) \to g_{t}$ on the interval forces every minimiser into that neighbourhood for all sufficiently small $∣ x ∣$ . since $G_{t} (x, y) - G_{t} (x, - y) = - 2 x y / t$ , a minimiser cannot be negative when $x > 0$ nor positive when $x < 0$ , and $y = 0$ is not a minimiser for $x \neq 0$ because $\partial_{y} G_{t} (x, 0) = - x / t \neq 0$ . Hence all minimisers have the sign of $x$ and, as $x ↓ 0$ , they converge to $s_{t}$ (and to $- s_{t}$ as $x ↑ 0$ ). The stationarity equation is $x = F_{t} (y) : = y - 2 t y e^{- y^{2}}$ with $F_{t}^{'} (\pm s_{t}) = 2 s_{t}^{2} > 0$ , so on small intervals around $\pm s_{t}$ , $F_{t}^{'}$ is bounded below by a positive constant. The mean value and intermediate value theorems [F3] give a unique local inverse there, and its difference quotient has derivative $1 / F_{t}^{'} (y)$ , which is continuous. Since every minimiser is on the corresponding interval for small $∣ x ∣$ , this inverse is the unique $C^{1}$ minimising branch on each punctured side. Along a branch the envelope derivative is $u_{x} = (x - y (x)) / t$ , whose one-sided limits are $- s_{t} / t$ (from the right) and $s_{t} / t$ (from the left); these unequal finite limits show that $u (\cdot, t)$ has a corner at $x = 0$ while remaining continuous.

Remarks

What is claimed. The computation identifies the minimisers and the one-sided derivatives at the corner; it does not assert local noninjectivity of $X_{t}$ near $y = 0$ , where $X_{t}^{'} (0) < 0$ and the map is locally decreasing, and it does not claim that the classical solution extends past $t = \frac{1}{2}$ .

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The negative absolute value solves the eikonal equation in the viscosity sense

Example

Let $u (x) = - ∣ x ∣$ on $R$ . Then $∣ u^{'} ∣ = 1$ for $x \neq 0$ , and $u$ is a viscosity solution of the stationary eikonal equation $∣ u^{'} ∣ - 1 = 0$ . Equivalently, for any $T > 0$ its evolutionary extension $U (x, t) : = - ∣ x ∣ - t$ is a viscosity solution of $U_{t} + ∣ U_{x} ∣ = 0$ on $R \times (0, T)$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem). At every point $x \neq 0$ the equation holds classically. At the cusp $(0, t_{0})$ , every $C^{1}$ upper test has $ϕ_{t} (0, t_{0}) = - 1$ and satisfies $∣ ϕ_{x} (0, t_{0}) ∣ \leq 1$ , so the subsolution inequality holds, and there is no $C^{1}$ test function for which $U - ϕ$ has a local minimum at $(0, t_{0})$ ; hence the supersolution test is vacuous. This is the complementary cusp to The eikonal equation as a viscosity equation at a tip, where the positive absolute value fails the supersolution test because it has lower tests with slopes of modulus less than one.

Verification

Given: The function $U (x, t) = - ∣ x ∣ - t$ on $U = R \times (0, T)$ , the equation $U_{t} + ∣ U_{x} ∣ = 0$ , and the test-function definition of viscosity sub- and supersolutions.

[F1] The subsolution inequality is tested at local maxima of $U - ϕ$ and the supersolution inequality at local minima, for $C^{1}$ test functions $ϕ$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

[F2] At a local extremum of a differentiable function of two variables both partial derivatives vanish (Fermat's theorem: an interior differentiable local extremum has zero gradient); away from $x = 0$ the function $U$ is $C^{1}$ with $U_{t} = - 1$ and $U_{x} = \mp 1$ , so it solves the equation classically there and is a viscosity solution on the open set $R ∖ {0} \times (0, T)$ by Classical solutions are viscosity solutions and differentiable viscosity solutions solve the equation pointwise.

Proof technique: direct contact computation at the cusp.

1.1F2

Away from the cusp. On the open set ${x \neq 0}$ the function $U$ is $C^{1}$ with $U_{t} = - 1$ and $∣ U_{x} ∣ = 1$ , so it is a viscosity solution of the equation there by [F2].

1.2F1F2algebra

Upper contacts at the cusp. Fix $t_{0}$ and let $ϕ \in C^{1}$ with $U - ϕ$ having a local maximum at $(0, t_{0})$ ; normalize $ϕ (0, t_{0}) = U (0, t_{0}) = - t_{0}$ . Restricting to the line $x = 0$ and using [F2] gives $ϕ_{t} (0, t_{0}) = - 1$ . Writing $p : = ϕ_{x} (0, t_{0})$ and testing $x = h$ , $x = - h$ with $h ↓ 0$ in the inequality $- ∣ x ∣ \leq ϕ (x, t_{0}) - ϕ (0, t_{0}) = p x + o (∣ x ∣)$ gives $p \geq - 1$ and $p \leq 1$ , that is $∣ p ∣ \leq 1$ . Hence $ϕ_{t} + ∣ ϕ_{x} ∣ = - 1 + ∣ p ∣ \leq 0$ , which is the subsolution inequality.

1.3F1F2algebra

No lower contact at the cusp. If $U - ϕ$ had a local minimum at $(0, t_{0})$ , the same computation with the inequality reversed would give $ϕ_{t} (0, t_{0}) = - 1$ and $p \leq - 1$ (from $h > 0$ ) together with $p \geq 1$ (from $h < 0$ ), an impossibility; hence the set of lower contacts at the cusp is empty and the supersolution inequality holds vacuously.

2.1step 1.1step 1.2step 1.3∎

Conclusion. Steps 1.2 and 1.3 show that $U$ is a subsolution everywhere on $R \times (0, T)$ and a supersolution everywhere, hence a viscosity solution; step 1.1 identifies the classical region. The cusp supports the subsolution inequality but admits no lower test, which is the complementary behaviour to the positive absolute value at its ridge point.

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Vanishing viscosity selects the Hopf--Lax solution for bounded data

Example

Let $u_{0} (x) = \sin x$ on $R$ , a bounded uniformly continuous datum, and let $H (p) = p^{2} / 2$ . For $ε > 0$ and $t > 0$ define $u^{ε} (x, t) = - 2 ε \log ⁣ [(4 π ε t)^{- 1 / 2} \int_{R} \exp ⁣ (- \frac{\sin y + (x - y)^{2} / (2 t)}{2 ε}) d y],$ and set $u^{ε} (x, 0) = \sin x$ . Then $u^{ε}$ is a bounded classical solution for $t > 0$ of $u_{t}^{ε} + \frac{1}{2} ∣ u_{x}^{ε} ∣^{2} = ε u_{x x}^{ε},$ and it attains $u_{0}$ uniformly as $t ↓ 0$ . The Hopf--Lax function $Q_{t} u_{0} (x) : = \inf_{y \in R} {\sin y + \frac{(x - y)^{2}}{2 t}}, Q_{0} u_{0} = u_{0},$ is a bounded uniformly continuous viscosity solution of $u_{t} + \frac{1}{2} ∣ u_{x} ∣^{2} = 0$ with initial trace $u_{0}$ , and for every $T < \infty$ one has $\sup_{x \in R, 0 \leq t \leq T} ∣ u^{ε} (x, t) - Q_{t} u_{0} (x) ∣ \leq \max {ε \log (1 + T), 2 ε \log (\sqrt{\frac{8}{π ε}} + \sqrt{2})} .$ Both terms on the right tend to zero as $ε ↓ 0$ . Thus the viscous solutions converge uniformly on every finite time strip. The estimate permits an $O (ε ∣ \log ε ∣)$ error and does not assert a uniform $O (ε)$ rate.

Verification

Given: The datum $u_{0} (x) = \sin x$ , the Hamiltonian $H (p) = p^{2} / 2$ with Legendre transform $L (v) = v^{2} / 2$ , all displayed integrals interpreted as absolutely convergent improper integrals of continuous functions, the heat kernel $Γ (z, s) = (4 π s)^{- 1 / 2} e^{- z^{2} / (4 s)}$ (The heat kernel on $R^{n}$ and its causal extension), the viscous equations and the Hopf--Lax function $Q_{t} u_{0}$ (The Hopf--Lax operator and the Hopf--Lax formula).

[F1] $\sin^{'} = \cos$ , $∣ \cos ∣ \leq 1$ , $∣ \sin x - \sin y ∣ \leq ∣ x - y ∣$ , and $\sin$ is bounded (The derivatives of sine and cosine are cosine and minus sine, Sine and cosine are $1$ -Lipschitz on $R$ , Parity and the Pythagorean identity for sine and cosine).

[F2] The Gaussian integral and change of variable for improper integrals are The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ and Change of variable in an improper integral. The logarithm derivative is $1 / x$ for $x > 0$ (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t), and the kernel is the explicit function of The heat kernel on $R^{n}$ and its causal extension; the exponential derivative, chain and algebra rules are The exponential function is smooth and $(\exp)^{'} = \exp$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ and Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ . On a compact $(x, t)$ -neighbourhood with $t > 0$ , every required kernel derivative is bounded by a fixed polynomial-times-Gaussian function of $y$ , integrable by comparison with a Gaussian. On each compact $y$ -interval its difference quotients converge uniformly; the mean value theorem bounds their tails by that same integrable majorant. Splitting into this interval and its tail justifies differentiation under the improper integral without a Lebesgue change-of-variable theorem or a choice assumption (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[F3] The quadratic conjugate is $L (v) = v^{2} / 2$ by completing the square in The Legendre transform of a finite-valued convex Hamiltonian. For the bounded uniformly continuous datum $\sin$ , the infimum is attained (Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers) and The Hopf--Lax formula solves the Hamilton--Jacobi Cauchy problem supplies the viscosity solution, finite-strip bounded uniform continuity, initial trace and uniqueness. At a differentiable minimizer the derivative is zero by Fermat's theorem: an interior differentiable local extremum has zero gradient. The semigroup is The Hopf--Lax operators form a semigroup (dynamic programming).

Proof technique: Cole--Hopf representation, an approximate-identity bound, and the variational reduction of the limit.

1.1F1F2algebra

The Cole--Hopf family solves the viscous equation. Write $W^{ε} (x, t) : = \int_{R} Γ (x - y, ε t) e^{- \sin y / (2 ε)} d y$ . The substitution $z = 2 \sqrt{ε t} r$ and $\int e^{- r^{2}} d r = \sqrt{π}$ show $\int Γ (z, ε t) d z = 1$ , and the same substitution with the Gaussian tail shows that $Γ (\cdot, ε t)$ is an approximate identity as $t ↓ 0$ . Differentiating the kernel gives $\partial_{t} Γ (\cdot, ε t) = ε \partial_{x x} Γ (\cdot, ε t)$ , so by [F2] $W_{t}^{ε} = ε W_{x x}^{ε}$ , $W^{ε} > 0$ , and $u^{ε} = - 2 ε \log W^{ε}$ satisfies $u_{t}^{ε} = ε u_{x x}^{ε} - ∣ u_{x}^{ε} ∣^{2} / 2$ , that is $u_{t}^{ε} + ∣ u_{x}^{ε} ∣^{2} / 2 = ε u_{x x}^{ε}$ , on $t > 0$ . The approximate identity applied to the bounded uniformly continuous function $e^{- \sin (\cdot) / (2 ε)}$ gives $W^{ε} \to e^{- \sin x / (2 ε)}$ uniformly as $t ↓ 0$ , hence $u^{ε} (x, t) \to \sin x$ uniformly; and $- 1 \leq u^{ε} \leq 1$ because $e^{- 1 / (2 ε)} \leq e^{- \sin y / (2 ε)} \leq e^{1 / (2 ε)}$ and the kernel has unit mass.

1.2F1F2F3algebra

Uniform comparison with Hopf--Lax. Fix $x$ , $t > 0$ , put $g (y) = \sin y + (x - y)^{2} / (2 t)$ and $m = \min g = Q_{t} u_{0} (x)$ , and take one minimiser $y_{*}$ . Then $- 1 \leq m \leq 1$ , $g^{'} (y_{*}) = 0$ , and $g^{''} (y) \leq A : = 1 + 1 / t$ . Applying the mean value theorem [F2] to $g^{'}$ shows that the derivative of $g (y_{*} + r) - m - A r^{2} / 2$ is nonpositive for $r > 0$ and nonnegative for $r < 0$ ; a second application gives $g (y_{*} + r) \leq m + A r^{2} / 2$ . The full Gaussian integral therefore gives $W^{ε} (x, t) \geq e^{- m / (2 ε)} / \sqrt{1 + t}$ , hence $u^{ε} - m \leq ε \log (1 + t)$ . Conversely, $g - m \geq 0$ everywhere, and $g - m \geq (x - y)^{2} / (2 t) - 2$ . If $∣ y - x ∣ \geq \sqrt{8 t}$ , the latter is at least $(x - y)^{2} / (4 t)$ . Splitting the integral at this radius and using [F2] bounds its normalized ratio by $e^{m / (2 ε)} W^{ε} (x, t) \leq (4 π ε t)^{- 1 / 2} (2 \sqrt{8 t} + \int_{R} e^{- (x - y)^{2} / (8 ε t)} d y) = \sqrt{\frac{8}{π ε}} + \sqrt{2} .$ Thus $u^{ε} - m \geq - 2 ε \log (\sqrt{8 / (π ε)} + \sqrt{2})$ . These two bounds hold for every $x$ and $t > 0$ ; at $t = 0$ the functions agree. Taking their maximum for $0 \leq t \leq T$ gives the stated absolute error bound. Its right-hand side tends to zero: writing $s = 1 / ε \geq 1$ , the integral formula for the logarithm in [F2] gives $\log s = \int_{1}^{s} d r / r \leq \int_{1}^{s} d r / \sqrt{r} = 2 (\sqrt{s} - 1)$ , hence $(\log s) / s \to 0$ .

2.1step 1.1step 1.2F1F3∎

The limit is the viscosity solution. By [F3] the function $Q_{t} u_{0}$ is a viscosity solution of $u_{t} + ∣ u_{x} ∣^{2} / 2 = 0$ ; independently, since $u_{0}$ is 1-Lipschitz with $∣ u_{0} ∣ \leq 1$ , the bounds $u_{0} (x) - t / 2 \leq Q_{t} u_{0} (x) \leq u_{0} (x)$ hold (competitor $y = x$ and the Lipschitz bound for $\sin$ ), so $Q_{t} u_{0}$ has the initial trace $u_{0}$ and is bounded and uniformly continuous. The uniform estimate of step 1.2, whose right-hand side tends to zero, then gives locally uniform convergence of the viscous family to this viscosity solution without invoking a general vanishing-viscosity theorem and without a momentum-Lipschitz hypothesis.

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Nonconvexity can break the equation; nonsuperlinearity can limit the Lagrangian domain

Statement refuted

False claim: the Hopf--Lax formula of The Hopf--Lax operator and the Hopf--Lax formula solves the equation for every finite superlinear Hamiltonian, and the Legendre transform of a finite Hamiltonian is real-valued on all of $R^{n}$ .

Two separate scope boundaries occur. (a) Convexity is essential if the Hopf--Lax formula is claimed to solve the equation for the original Hamiltonian: for $H (p) = p^{2} - 2 ∣ p ∣$ on $R$ , which is finite and superlinear but not convex, the formula applied to $u_{0} \equiv 0$ produces $u (x, t) = t$ , and at every interior point the smooth function itself is an upper test whose residual $u_{t} + H (u_{x}) = 1 + H (0) = 1 > 0$ violates the viscosity subsolution inequality (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem). (b) Superlinearity is needed for the full-domain real-valued Lagrangian conclusion: for the finite convex Hamiltonian $H (p) = \sqrt{1 + p^{2}}$ , which is not superlinear, the Legendre transform equals $- \sqrt{1 - v^{2}}$ for $∣ v ∣ \leq 1$ and $+ \infty$ for $∣ v ∣ > 1$ , so finite action is available only when $∣ x - y ∣ \leq t$ ; this is exactly the full-domain finiteness conclusion of Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers that fails without superlinearity. The second clause does not show failure of the value or the solution property: for $u_{0} \equiv 0$ the formula is $Q_{t} u_{0} \equiv - t$ , whose residual is $- 1 + H (0) = 0$ .

Facts & Assumptions

Given: The Hamiltonians $H (p) = p^{2} - 2 ∣ p ∣$ and $H (p) = \sqrt{1 + p^{2}}$ on $R$ , their Legendre transforms $L$ , the bounded uniformly continuous datum $u_{0} \equiv 0$ , and the extended infimum formula $Q_{t}$ defined in [F2].

[F1]

$L (v) = \sup_{p \in R} (p v - H (p))$ for a finite Hamiltonian on $R$ , the supremum being taken in $\overline{R}$ , and $L (v) = + \infty$ is possible (The Legendre transform of a finite-valued convex Hamiltonian).

[F2]

For the two Hamiltonians considered here, define the extended infimum formula $Q_{t} u_{0} (x) : = \inf_{y \in R} {u_{0} (y) + t L ((x - y) / t)}$ for $t > 0$ and $Q_{0} u_{0} : = u_{0}$ . This extends the same expression in The Hopf--Lax operator and the Hopf--Lax formula beyond that definition's convex-superlinear hypotheses. Here $L (v) \geq - H (0)$ and $u_{0} = 0$ , so the infimum is well defined in $R \cup {+ \infty}$ , with $t (+ \infty) = + \infty$ . The finiteness and attainment theorem Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers applies under convexity and superlinearity only; it is not invoked for these two Hamiltonians.

[F3]

A function $u$ is a viscosity subsolution of $u_{t} + H (u_{x}) = 0$ only if $ϕ_{t} + H (ϕ_{x}) \leq 0$ at every local maximum of $u - ϕ$ with $ϕ \in C^{1}$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

Counterexample

technique · split each conjugate supremum at the sign of $p$ and evaluate the resulting formula

1.1F1algebra

Conjugate of the nonconvex Hamiltonian. For $H (p) = p^{2} - 2 ∣ p ∣$ and $p \geq 0$ one has $p v - H (p) = p (v + 2) - p^{2}$ , whose supremum over $p \geq 0$ is $(v + 2)_{+}^{2} / 4$ ; for $p \leq 0$ , writing $p = - q$ with $q \geq 0$ gives $p v - H (p) = - q v - q^{2} + 2 q = q (2 - v) - q^{2}$ , whose supremum is $(2 - v)_{+}^{2} / 4$ . Hence $L (v) = \max {(v + 2)_{+}^{2} / 4, (2 - v)_{+}^{2} / 4}$ , where $r_{+} = \max (r, 0)$ ; in particular $L (0) = 1$ , $L (v) = (2 + ∣ v ∣)^{2} / 4 \geq 1$ for every $v$ and $L (v) \geq 4$ for $∣ v ∣ \geq 2$ , so $\inf_{v} L (v) = 1$ , attained at $v = 0$ .

1.2F1F2algebra

Conjugate of the non-superlinear Hamiltonian. For $H (p) = \sqrt{1 + p^{2}}$ and $∣ v ∣ < 1$ , the function $p \mapsto p v - \sqrt{1 + p^{2}}$ has derivative $v - p / \sqrt{1 + p^{2}}$ , which vanishes exactly at $p = v / \sqrt{1 - v^{2}}$ , where the value is $- \sqrt{1 - v^{2}}$ ; at $v = \pm 1$ the supremum is $0$ , approached along the infinite tail $p \to \pm \infty$ ; and for $∣ v ∣ > 1$ the expression tends to $+ \infty$ along $p = sgn (v) r$ as $r \to \infty$ . Hence $L (v) = - \sqrt{1 - v^{2}}$ for $∣ v ∣ \leq 1$ and $L (v) = + \infty$ for $∣ v ∣ > 1$ : the Legendre transform is not real-valued on all of $R$ , finite action being available only when $∣ x - y ∣ / t \leq 1$ , that is $∣ x - y ∣ \leq t$ . For $u_{0} \equiv 0$ the formula is $Q_{t} u_{0} = t \inf_{v} L (v) = - t$ ; the residual of $u (x, t) = - t$ is $u_{t} + H (u_{x}) = - 1 + H (0) = - 1 + 1 = 0$ , so as a formal expression it does satisfy the equation, and no failure of the solution property is claimed in this clause.

2.1step 1.1F2F3algebra

The formula is not a solution for the nonconvex Hamiltonian. With $u_{0} \equiv 0$ , the Hopf--Lax formula of [F2] is $Q_{t} u_{0} (x) = t \inf_{v \in R} L (v) = t$ for every $x$ and $t > 0$ (the substitution $v = (x - y) / t$ turns the infimum over $y$ into the infimum over $v$ ). The function $u (x, t) = t$ is smooth, and for a smooth function every point is both an upper and a lower contact with the test function itself; as an upper test, [F3] requires $u_{t} + H (u_{x}) \leq 0$ , but $u_{t} = 1$ and $u_{x} = 0$ give $1 + H (0) = 1 > 0$ . Hence the Hopf--Lax output is not a viscosity subsolution, therefore not a viscosity solution, of $u_{t} + H (u_{x}) = 0$ .

3.1step 1.1step 1.2step 2.1∎

Conclusion. Part (a) exhibits a finite superlinear nonconvex Hamiltonian whose Hopf--Lax output fails the subsolution inequality pointwise, and part (b) exhibits a finite convex non-superlinear Hamiltonian whose Legendre transform is finite only on a bounded velocity set, so superlinearity cannot be dropped from the full-domain finiteness conclusion. The two failures are of different kinds: convexity is needed for the equation to hold, superlinearity for the Lagrangian to be finite everywhere.

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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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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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The eikonal equation on an interval has many solutions when endpoint data are omitted

Statement refuted

False claim: the stationary eikonal equation $∣ u^{'} ∣ = 1$ on $Ω = (0, 1)$ has a unique viscosity solution when no boundary data are prescribed.

The functions $u_{1} (x) = x$ , $u_{2} (x) = 1 - x$ and $u_{3} (x) = \frac{1}{2} - ∣ x - \frac{1}{2} ∣ = \min {x, 1 - x}$ are viscosity solutions of $∣ u^{'} ∣ = 1$ on $(0, 1)$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem, Viscosity testing by first-order jets, and closure of the jet inequality). They are pairwise distinct, and their continuous boundary traces are respectively $(0, 1)$ , $(1, 0)$ and $(0, 0)$ ; in particular the equation without prescribed endpoint data does not select a unique solution, and the examples are distinguished by their boundary traces.

Facts & Assumptions

Given: The interval $Ω = (0, 1)$ , the equation $∣ u^{'} ∣ - 1 = 0$ written as $U_{t} + F (U_{x}) = 0$ on $U = Ω \times (0, \infty)$ with $F (p) = ∣ p ∣ - 1$ , the profiles $u_{1} (x) = x$ , $u_{2} (x) = 1 - x$ , $u_{3} (x) = \min {x, 1 - x}$ , and their time-independent lifts $U_{i} (x, t) = u_{i} (x)$ .

[F1]

A viscosity subsolution is defined by $ϕ_{t} + F (ϕ_{x}) \leq 0$ at every local maximum of $U - ϕ$ , and a supersolution by $ϕ_{t} + F (ϕ_{x}) \geq 0$ at every local minimum, for $C^{1}$ test functions $ϕ$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

[F2]

A function $U$ is a viscosity subsolution if and only if every $p = (p_{x}, p_{t}) \in D^{+} U$ satisfies $p_{t} + F (p_{x}) \leq 0$ , and a supersolution if and only if every $p \in D^{-} U$ satisfies $p_{t} + F (p_{x}) \geq 0$ (Viscosity testing by first-order jets, and closure of the jet inequality).

Counterexample

technique · direct test-function verification at the peak and explicit evaluation of the boundary traces

1.1F2algebra

Time-independent profiles have zero test time-derivative. Let $U (x, t) = u (x)$ for a continuous $u : (0, 1) \to R$ and let $ϕ \in C^{1} (U)$ with $U - ϕ$ having a local extremum at $(x_{0}, t_{0})$ . Then, testing along the time line $x = x_{0}$ and along the space line $t = t_{0}$ , the jet characterisation [F2] gives $p_{t} = 0$ for every $p \in D^{+} U (x_{0}, t_{0})$ or $p \in D^{-} U (x_{0}, t_{0})$ : the inequality $u (x_{0}) - ϕ (x_{0}, t) \leq u (x_{0}) - ϕ (x_{0}, t_{0})$ near $t_{0}$ forces $\partial_{t} ϕ (x_{0}, t_{0}) = 0$ in the upper case (and dually in the lower case), and the remaining spatial inequality is the one-sided differentiability statement for $u$ at $x_{0}$ . Consequently the viscosity conditions for $U$ reduce to: every upper-test spatial slope $p_{x}$ satisfies $F (p_{x}) \leq 0$ , and every lower-test spatial slope $p_{x}$ satisfies $F (p_{x}) \geq 0$ .

2.1step 1.1F1F2algebra

The three profiles are solutions. For $x_{0} \neq \frac{1}{2}$ each $u_{i}$ is $C^{1}$ near $x_{0}$ with $∣ u_{i}^{'} ∣ = 1$ , so at any upper or lower test the spatial slope equals $u_{i}^{'} (x_{0}) = \pm 1$ , and [F1] gives $F (\pm 1) = 0$ in both directions. At $x_{0} = \frac{1}{2}$ , only $u_{3}$ is non-differentiable: writing $h = x - \frac{1}{2}$ we have $u_{3} (\frac{1}{2} + h) = \frac{1}{2} - ∣ h ∣$ . If $ϕ$ is an upper test with slope $p = ϕ_{x} (\frac{1}{2}, t_{0})$ , then $u_{3} \leq ϕ$ near the point gives for $h > 0$ the inequality $- h + o (h) \leq p h + o (h)$ , hence $p \geq - 1$ , and for $h < 0$ the inequality $h + o (∣ h ∣) \leq p h + o (∣ h ∣)$ , hence $p \leq 1$ ; thus every upper-test slope lies in $[- 1, 1]$ and $F (p) \leq 0$ . If $ϕ$ were a lower test, $u_{3} \geq ϕ$ would give for $h > 0$ that $p \leq - 1$ and for $h < 0$ that $p \geq 1$ , which is impossible; hence there is no lower test at the peak and the supersolution condition is vacuous. By step 1.1 the same reduction applies to $u_{1}$ and $u_{2}$ at every point. Hence all three profiles are viscosity solutions of $∣ u^{'} ∣ = 1$ on $(0, 1)$ .

3.1step 2.1algebra∎

Distinctness and boundary traces. At $x = \frac{1}{4}$ the values are $(u_{1}, u_{2}, u_{3}) = (\frac{1}{4}, \frac{3}{4}, \frac{1}{4})$ and at $x = \frac{3}{4}$ they are $(\frac{3}{4}, \frac{1}{4}, \frac{1}{4})$ , which distinguishes every pair; also $u_{1} - u_{2} = 2 x - 1$ changes sign on $(0, 1)$ . The continuous extensions to $[0, 1]$ have boundary values $u_{1} (0) = 0, u_{1} (1) = 1$ ; $u_{2} (0) = 1, u_{2} (1) = 0$ ; and $u_{3} (0) = u_{3} (1) = 0$ . Thus the equation with no prescribed boundary data admits at least three distinct viscosity solutions, and the displayed candidates carry different endpoint traces.

Remarks

What this shows. Comparison and uniqueness on a bounded domain require the boundary condition to be imposed; without it the solution class is not a singleton even for the simplest non-smooth first-order equation. This is the reason the page states comparison and uniqueness on $R^{n}$ or, on a bounded cylinder, with boundary data on the parabolic boundary.

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Distance to the boundary solves the unit eikonal Dirichlet problem on the ball

Example

Let $n \geq 1$ and let $Ω = B (0, 1) \subseteq R^{n}$ be the open unit ball and let $d (x) : = 1 - ∣ x ∣ = dist (x, \partial Ω)$ . Then $d$ is Lipschitz with $∣ D d ∣ = 1$ at every $x \neq 0$ , continuous on $\overline{Ω}$ with $d = 0$ on $\partial Ω$ , and $d$ is a viscosity solution of the stationary eikonal equation with zero boundary data: $∣ D u ∣ = 1 in Ω, u = 0 on \partial Ω .$ At $x \neq 0$ the equation holds classically; at the centre the function has a peak, every $C^{1}$ test function $ϕ$ with $d - ϕ$ having a local maximum at $0$ satisfies $∣ D ϕ (0) ∣ \leq 1$ (the subsolution inequality), and no $C^{1}$ test function has $d - ϕ$ locally minimal at $0$ : a lower contact would require $⟨ D ϕ (0), h ⟩ \leq - ∣ h ∣$ for all $h$ , which is impossible, so no such contact exists and the supersolution test is vacuous at the centre. Hence $d$ is a viscosity solution. The example is the canonical nonsmooth boundary-value solution of the eikonal equation, obtained by cone comparisons at the centre rather than by an abstract existence theorem.

Verification

Given: $n \geq 1$ , the open unit ball $Ω = B (0, 1)$ , its boundary the unit sphere (Euclidean spheres and closed balls as subspaces of $R^{n}$ ), the distance function $d (x) = 1 - ∣ x ∣$ , the norm $∣ \cdot ∣$ (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms) and the inner product (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F1] The test-function definition of viscosity sub- and supersolutions of $∣ D u ∣ - 1 = 0$ , and its equivalent jet formulation (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem, Viscosity testing by first-order jets, and closure of the jet inequality).

[F2] For $x \neq 0$ the map $x \mapsto ∣ x ∣$ is $C^{1}$ with gradient $x / ∣ x ∣$ of norm $1$ , as computed in The eikonal equation as a viscosity equation at a tip; hence $d$ is $C^{1}$ on $Ω ∖ {0}$ with $D d (x) = - x / ∣ x ∣$ and $∣ D d (x) ∣ = 1$ ; also $d$ is Lipschitz with constant $1$ by the triangle inequality (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

Proof technique: classical verification away from the centre and cone comparisons at the centre.

1.1F2

Classical region. For $x \neq 0$ the function $d$ is $C^{1}$ near $x$ with $∣ D d ∣ = 1$ by [F2], so it is a viscosity solution of $∣ D u ∣ = 1$ on the punctured ball by the classical-consistency argument of The eikonal equation as a viscosity equation at a tip.

1.2F1F2algebra

Upper contacts at the centre. Let $ϕ \in C^{1}$ with $d - ϕ$ having a local maximum at $0$ ; normalize $ϕ (0) = d (0) = 1$ . Then $1 - ∣ x ∣ \leq ϕ (x) = 1 + ⟨ D ϕ (0), x ⟩ + o (∣ x ∣)$ near $0$ , that is $⟨ D ϕ (0), x ⟩ \geq - ∣ x ∣ + o (∣ x ∣)$ ; substitute $x = a e$ and $x = - a e$ for a unit vector $e$ , divide by $a > 0$ , and let $a ↓ 0$ to obtain $∣ ⟨ D ϕ (0), e ⟩ ∣ \leq 1$ . Taking $e$ in the direction of $D ϕ (0)$ , when this gradient is nonzero, gives $∣ D ϕ (0) ∣ \leq 1$ . Hence every upper test satisfies the subsolution inequality $∣ D ϕ (0) ∣ - 1 \leq 0$ at the centre.

1.3F1F2algebra

No lower contact at the centre. If $d - ϕ$ had a local minimum at $0$ , the reversed inequality would give $⟨ D ϕ (0), x ⟩ \leq - ∣ x ∣ + o (∣ x ∣)$ for all $x$ near $0$ ; substituting $x = a e$ and $x = - a e$ , dividing by $a > 0$ and taking $a ↓ 0$ gives $⟨ D ϕ (0), e ⟩ \leq - 1$ and $⟨ D ϕ (0), e ⟩ \geq 1$ , an impossibility. Hence no lower $C^{1}$ test exists at the centre and the supersolution inequality holds vacuously.

2.1step 1.1step 1.2step 1.3F2∎

Boundary values and conclusion. For $x \in Ω$ and $y \in \partial Ω$ , the reverse triangle inequality gives $∣ x - y ∣ \geq 1 - ∣ x ∣$ . Equality is achieved at $y = x / ∣ x ∣$ when $x \neq 0$ , and at any unit vector when $x = 0$ , so $d (x) = dist (x, \partial Ω)$ . Since $∣ x ∣ \to 1$ along sequences approaching the unit sphere, the continuous extension of $d$ to $\overline{Ω}$ vanishes exactly on $\partial Ω$ . Steps 1.2 and 1.3 give the subsolution inequality everywhere and the supersolution inequality everywhere (vacuously at the centre, classically elsewhere by step 1.1), so $d$ is a viscosity solution of the unit eikonal equation with zero boundary data on the ball.

Hamilton Jacobi Equations and Viscosity Solutions — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The eikonal equation as a viscosity equation at a tip

Example

Verification

The quadratic Hopf--Lax formula as an infimal convolution

Example

Verification

Remarks

A Hopf--Lax solution with a forming corner from smooth data

Example

Verification

Remarks

The negative absolute value solves the eikonal equation in the viscosity sense

Example

Verification

Vanishing viscosity selects the Hopf--Lax solution for bounded data

Example

Verification

Nonconvexity can break the equation; nonsuperlinearity can limit the Lagrangian domain

Statement refuted

Facts & Assumptions

Counterexample

A strict subsolution can fail the supersolution lower-test condition

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Minima of viscosity subsolutions need not be subsolutions

Statement refuted

Facts & Assumptions

Counterexample

The eikonal equation on an interval has many solutions when endpoint data are omitted

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Distance to the boundary solves the unit eikonal Dirichlet problem on the ball

Example

Verification

Sources