Explicit viscosity solution of the crystalline curvature flow
with fattening
Norbert Požár
Faculty of Mathematics and Physics
Institute of Science and Engineering, Kanazawa University
Abstract
We give a simple explicit example of a solution of the level set formulation of the crystalline curvature flow in two dimensions whose one level set fattens, leading to nonuniquenessof the flow, and prove that it is the unique viscosity solution. The motivation of this note is
to illustrate the notion of viscosity solutions introduced recently by Y. Giga and the author.

1

Introduction

The purpose of this short note is to illustrate the notion of viscosity solutions of the crystalline
mean curvature flow introduced by Y. Giga and the author in [13, 14]. We will construct an
explicit viscosity solution with a polygonal figure 8 initial data as an example of fattening inthis problem. Fattening refers to the essential nonuniqueness present in the mean curvature
flow where multiple evolutions are possible. A well-known example of this phenomenon in thecontext of the standard mean curvature flow in two dimensions is the initially connected, selfintersecting curve resembling the figure 8. During the evolution, the curve either splits at the
self-intersection into two disconnected components that contract to two distinct points, or it
separates at the self-intersection and evolves as a simple closed curve that contracts to a single
point; see [7, 8].
The level set formulation of the mean curvature flow has a unique solution even when theevolution of the curve is not unique. The level set describing the evolution of the curve however
develops a nonempty interior. That is, the level set fattens. Examples of fattening are discussed
for instance in [7, 8, 15, 17, 18], but none of them are given by an explicit formula.
On the other hand, the solutions of the crystalline curvature flow in two dimensions, reviewed
in Section 2, are in general polygonal curves whenever the initial curve is polygonal, and so we can
expect that the crystalline algorithm [22, 23] for tracking such evolutions will give an explicit
solution. One such solution is constructed in Section 3, and the proof that it is the uniqueviscosity solution of the level set formulation is presented in Section 3.2.

2

The crystalline mean curvature flow

The same way that one can understand the standard curvature flow as the formal gradient flowof the curve length, one can consider the formal gradient flow t 7→ Et of the anisotropic perimeter∫(2.1)
Pσ (E) :=σ(ν) dH1∂∗E

defined for sets of finite perimeter E ⊂ R2 , where ν is the outer unit normal, σ : S 1 → (0, ∞)expresses the anisotropy of the perimeter, and H1 is the one-dimensional Hausdorff measure.
1This formally yields the geometric evolution
V = −κσ ,(2.2)

where V is the normal velocity of the evolving boundary ∂Et . The term κσ is the anisotropic
mean curvature of ∂Et corresponding to the anisotropy σ. It is the first variation of theanisotropic perimeter Pσ [2, 4]. When the anisotropy σ is nonsmooth so that the Wulff shape,that is, the set with the smallest anisotropic perimeter for a given volume, is a convex polytope,
it is referred to as the crystalline mean curvature flow. This problem was introduced by Angenent and Gurtin [3] and Taylor [21] as a model of growth of crystals. Solutions of the flow in
two dimensions are often evolving polygonal curves. Furthermore, the velocity of each segment
can be found quite easily as a constant inversely proportional to the length of the segment; see(3.5). This motivates the crystalline algorithm [21–23]. It is a first order ODE system for the
positions of the individual segments. This algorithm provides an efficient way to find the evolution before the onset of singularities. When singularities occur, the evolution can sometimes becontinued in a natural way [1], but there are many examples where a much more care must be
taken as in the case of collisions and topological changes [16, 19].
The viscosity solutions [10–14] provide one way to select a unique continuation when no
fattening occurs. It is also known that the crystalline algorithm generates a viscosity solution
when no singularities occur [9, 11]. Another notion of generalized solutions was developed in[5, 6]. It is defined in terms of the signed distance function to the evolving curve (surface) andrelies for existence on the minimizing movements algorithm [2]. For a more detailed discussionsee [6, 13] and the reference therein.
The level set equation for the crystalline curvature flow can be formally written as
ut = |∇u| div ∇σ(∇u)

in R2 × (0, ∞),(2.3)

so that every level set of the auxiliary function u defines a curve that evolves under the geometric flow (2.2) [12]. The term div ∇σ(∇u) must be properly understood in the sense of a
subdifferential of the anisotropic total variation energy∫
E(ψ) :=σ(∇ψ) dx.
R2

Here σ : R2 → [0, ∞) is the positively one-homogeneous extension of σ in (2.1) given as σ(p) =|p|σ(p/|p|), p ̸= 0, and it is assumed to be convex. Any such function we shall call an anisotropy.
If σ is furthermore given as the maximum of a finite number of linear functions,
σ(p) := max xi · p,iwe shall call it a crystalline anisotropy. In this case the vector field ∇σ(∇u) is in general notcontinuous even if the level sets of u are smooth curves and therefore one cannot expect to beable to express div ∇σ(∇u) as a function. In fact, this operator becomes nonlocal on flat parts
parallel to the sides of the Wulff shape.
In this note we will focus on the crystalline anisotropy given by the ℓ1 norm: σ(p) := ∥p∥1 =|p1 | + |p2 |. Clearly we can take xi to be the four points (1, 1), (−1, 1), (1, −1) and (−1, −1).
These happen to also be the vertices of the Wulff shape W corresponding to σ,
W = {x : σ ◦ (x) ≤ 1} = {x : ∥x∥∞ ≤ 1},
where σ ◦ : R2 → [0, ∞) is the convex polarσ ◦ (x) := max {x · p : σ(p) ≤ 1} = ∥x∥∞ ,
2

x2
2

x2
u0 > 0
u0 < 0

2

1

1
1

2

x1

1

2

x1

Figure 1: Left: Level sets {u0 = θ} of the initial data for θ = −1 (two points), − 12 (two squares),
0 (thick figure ‘8’), and 12 . Right: Level sets {u(·, t) = 0} (gray region) and {u(·, t) = 21 } att = 0.45.
see [20]. The Wulff shape is the set with the smallest anisotropic perimeter given the volume
and in this case it happens to be a square.
It is important to understand for what kind of curves we can define the crystalline curvature.
In the case of the ℓ1 anisotropy, we can for instance consider the following class: A curve is regular
for the anisotropy σ = ∥·∥1 if it is a Lipschitz regular boundary of an open set and consists onlyof axes-aligned segments with positive length; see also [23].

3The example of fattening

We consider the initial data)(u0 (x) = min ∥x − (1, 1)∥∞ − 1, ∥x + (1, 1)∥∞ − 1, 1 ,x ∈ R2 ,for the level set equation (2.3). Note that −1 ≤ u0 ≤ 1. The zero level set {u0 = 0} of u0 is
the boundary of the union of two axes-aligned squares of side-length 2 touching at the origin,
centered at (1, 1) and (−1, −1), respectively; see Figure 1. This is the polygonal figure ‘8’ shape.
Other level sets {u0 = θ} for −1 < θ < 1 are boundaries of unions of two squares of sidelength 2 + 2θ with the same centers as above, but the squares are disjoint for −1 < θ < 0, while
the intersection has a nonempty interior for θ > 0.
The evolution of the initial polygonal curves {u0 = θ} is therefore quite clear for θ ∈ Θ :=(−1, 0) ∪ (0, 1) by the crystalline algorithm since these initial curves are regular, see above, withrespect to the anisotropy σ = ∥·∥1 . From these evolving curves, we will construct a level set
function u : R2 × [0, ∞) → R and show that it is the unique viscosity solution of (2.3) withinitial data u0 .

3.1

Crystalline algorithm

We will express the evolving curves as the boundaries of evolving open sets Etθ ⊂ R2 for θ ∈ Θ,with initial data E0θ = {u0 < θ}.

3

x2

Gθt

Ftθ

b(t)
θ
x1

−θ

Figure 2: The boundary of Etθ for θ ∈ (0, 1) and 0 ≤ t < t♯ .
Case −1 < θ < 0
E0θ := {u0 < θ} is the union of two disjoint squares
E0θ = Q1+θ (1, 1) ∪ Q1+θ (−1, −1),

(3.1)

where Qr (c) := {x : ∥x − c∥∞ < r}. Each square can be evolved independently. If we denotethe half-length of their sides as a = a(t), a(0) = 1 + θ, the crystalline curvature on the facet isgiven as
κσ =

1
.
a

This follows from a general formula κσ = (length of Wulff shape facet)/(length of facet), which
is rigorously written in (3.5) below. The velocity in the outer normal direction of each side of
the square happens to be the derivative a′ . Therefore under the crystalline algorithm, a satisfies a′ = − 1 ,t > 0,a
a(0) = 1 + θ,√2∗∗,θ is the extinction
with exact solution a(t) = (1 + θ)2 − 2t for t < t∗ := (1+θ)
2 , where t = t∗time, that is, the time such that a(t) ↘ 0 as t ↗ t . We continue the solution as a(t) = 0 fort ≥ t∗ .
Remark 3.1. In two dimensions, the flow with σ = ∥·∥1 reduces the area of each connected componentof the interior of the curve at a constant rate −8. Since the initial area of each component in this case is4(1 + θ)2 , the area becomes 0 exactly at t∗ .

We defineEtθ := Qa(t) (1, 1) ∪ Qa(t) (−1, −1),

−1 < θ < 0, t ≥ 0.

Note that Etθ = ∅ for t ≥ t∗ .
Case 0 < θ < 1
E0θ = {u0 < θ} is again a union of two squares as in (3.1), however, this time they are notdisjoint. In fact E0θ has 8 facets; see Figure 1. By the symmetry with respect to both diagonals,
we need to only consider the two of them in {(x1 , x2 ) : x2 > |x1 |}, namely
F0θ := {(−θ, x2 ) : θ < x2 < 1 + θ},Gθ0 := {(x1 , 1 + θ) : −θ < x1 < 1 + θ},
4see Figure 2. The crystalline curvature is 0 on F0θ by (3.5) and a positive constant dependingon the length of the facet on Gθ0 . Therefore F0θ does not move, only shortens, while Gθ0 moves
in the direction of E0θ . We introduce b = b(t) to be the distance to the x1 -axis of the facet Gθt ,Gθt := {(x1 , b(t)) : −θ < x1 < b(t)}.

Ftθ := {(−θ, x2 ) : θ < x2 < b(t)},We define the time t♯ = t♯,θ as the time when b(t♯ ) = θ, and Ftθ♯ vanishes. At this time Etθ♯ is thesquare Qθ (0, 0). After that the evolution is self-similar (homothetic) as in the previous case.
Let us find an expression for b(t) = bθ (t). The crystalline curvature on Gθt is
κσ =

2
.
b(t) + θ

As before, b′ is the normal velocity of the facet Gθt and therefore b is the solution of b′ = − 2 , t > 0,b+θb(0) = 2 + θ,with formulab(t) =√(2 + 2θ)2 − 4t − θ.

We have b(t♯ ) = θ at t♯ = t♯,θ := 1 + 2θ. After this time, Etθ is the square
Etθ = Qc(t) (0),where c(t) = cθ (t) =vanishes.
We have√θ2 − 2(t − t♯ ), t♯ ≤ t < t♯ + θ2 . At t∗ := t♯ + θ2 = 1 + 2θ + 12 θ2 the set
2

2


♯
θ
θ

At ∪ (−At ), 0 ≤ t < t = 1 + 2θ,θ
Et = Qc(t) (0),t♯ ≤ t < t∗ := t♯ + 12 θ2 = 1 + 2θ + 21 θ2 ,∅,t ≥ t∗ ,with Aθt = {(x1 , x2 ) : −θ < x1 , x2 < b(t)}. As in Remark 3.1, we can find t♯ and t∗ by considering
the area of E0θ and the fact that the area of Etθ decreases with a constant rate −8.
We clearly have the following monotonicity:Etη ⊂ Etθ ⊂ Esθ

for η ≤ θ, 0 ≤ s ≤ t,η, θ ∈ Θ.

In words, the sets are nonincreasing in time for each θ, and they form a monotonically increasing family in θ. This monotonicity of Etθ is clear from the explicit formulas, but it is also a
consequence of the comparison principle for the crystalline algorithm. In fact, it is rather easy
to see that
dist(Etη , (Etθ )c ) ≥ θ − η = dist(E0η , (E0θ )c )whenever η ≤ θ, η, θ ∈ Θ.

(3.2)

Moreover, due to the semigroup property we can always compare the evolving set with thehomothetic solution starting from Qδ (c) centered at any point c.
Lemma 3.2. Whenever for some c ∈ R2 , t ≥ 0 and δ > 0 we have x ∈ Etθ for all x ∈ R2 with2∥x − c∥∞ < δ, we can conclude that c ∈ Esθ for t ≤ s < t + δ2 .
5

We will now define the level set function({)}
u(x, t) := inf θ ∈ (−1, 0) ∪ (0, 1) : x ∈ Etθ ∪ {1} .

(3.3)

It is easy to see that −1 ≤ u ≤ 1 on R2 ×[0, ∞), u(x, t) = 1 for ∥x∥∞ ≥ 3 or t ≥ limθ→1− t∗ = 3.5,
and u is nondecreasing in time. Finally, u is Lipschitz in space with Lipschitz constant 1 by
(3.2).
In fact, we can write a rather explicit formula for u. To simplify the situation, we again takeadvantage of the symmetries of u guaranteed by the symmetries of the initial data and give thevalues of u(·, t) only on the set {x2 ≥ |x1 |}. Let us define{√
x22 + 2t + 2 − 2,
) x2 < (t − 1)/2,(√
ξ(x2 , t) := 1
4(x2 − 1)2 + 12t − 4 , otherwise.
3 x2 +
It is not difficult to check that ξ is 1-Lipschitz continuous √since at x2 = (t − 1)/2 both branches
are equal to x2 . Note also that ξ(x2 , t) < 0 for 0 ≤ x2 < 2 1 − t and 0 ≤ t < 1 = t♯,0 .
For x2 < (t − 1)/2, θ = ξ(x2 , t) is the solution of bθ (t) = x2 , that is, the point x might lie onthe edge Gθt . Otherwise θ = ξ(x2 , t) is a solution of cθ (t) = x2 , in which case x lies on the topedge of Qc(t) . Let us combine this with the possibility that x lies on the edge Ftθ , which does
not move:()
v(x, t) := min max(−x1 , ξ(x2 , t)), 1 .
On the other hand, x can lie on the boundary of Qa(t) (1, 1) or in the fattened zero set, which
we express by)(√w(x, t) := min∥x − (1, 1)∥2∞ + 2t − 1, 0 .
Then we can write u for x2 ≥ |x1 | as{v(x, t), v(x, t) > 0,u(x, t) =w(x, t), otherwise.

(3.4)

u can be extended to the rest of R2 by symmetry.

3.2

Viscosity solution

We claim that u is the unique viscosity solution of (2.3) with initial data u0 .
The definition of viscosity solutions introduced in [13, 14] can be written in n = 2 with
σ(p) = ∥p∥1 = |p1 | + |p2 | in the following simplified way. First we introduce admissible support
functions as the Lipschitz functions on R whose zero level set {ψ = 0} is the union of finite
number of bounded closed intervals of positive length. Here we should note that the definitionof admissible functions in [14] is potentially more general but the important point is that the classconsidered here still gives the density result [14, Theorem 1.3]: For every Lipschitz continuousfunction ψ on R with {ψ = 0} compact and any ρ > 0 there exists an admissible support function
ζ with sign ψ(x) ≤ sign ζ(x) ≤ sup|y−x|<ρ sign ψ(y). Note that sign s := −1, 0, 1 whenever s < 0,
s = 0, s > 0, respectively. This result is a simple consequence of the structure of open subsetsof the real line.
For such an admissible support function ψ, we will defineΛ[ψ](x) :=

χ(x)
L(x)

x ∈ {ψ = 0},

6

(3.5)

where L(x) is the length of the connected component of {ψ = 0} that contains x, and χ(x) is
the sum of the signs of the two boundary points of the connected component. We define thesign of the boundary point as 1 if ψ ≥ 0 in its neighborhood, and −1 otherwise. Therefore thepossible values of χ(x) are −2, 0 and 2.
Definition 3.3. We say that an upper semi-continuous function u is a viscosity subsolution of(2.3) whenever the following conditions are satisfied:(i1 ) If φ(x, t) = ψ(x1 ) + f (x2 ) + g(t), where ψ is an admissible support function and f ,g ∈ C 1 (R), f ′ (x̂2 ) ̸= 0, and u − φ(· − he1 ) has a global maximum at (x̂, t̂) for all h ∈ R,
|h| small, and x̂1 ∈ int {ψ = 0}, theng ′ (t̂) − |f ′ (x̂2 )|Λ[ψ](x̂1 ) ≤ 0.
(i2 ) Same as (i1 ) but with subscripts 1 and 2 exchanged.
(ii) If φ(x, t) = ψ(x) + g(t), where ψ ∈ Lip(R2 ) and g ∈ C 1 (R), and u − φ(· − h) has a global
maximum at (x̂, t̂) for all h ∈ R2 , |h| small, and x̂ ∈ int {ψ = 0}, theng ′ (t̂) ≤ 0.
(iii) If φ(x, t) = f (x) + g(t), where f, g ∈ C 1 (R), and u − φ has a global maximum at (x̂, t̂),and ∂x1 f (x̂) ̸= 0 ̸= ∂x2 f (x̂), theng ′ (t̂) ≤ 0.
A viscosity supersolution is defined analogously by replacing upper semi-continuous, maximum and ≤ by lower semi-continuous, minimum and ≥, respectively.
A function is a viscosity solution if it is both a viscosity subsolution and a viscosity supersolution.
Intuitively, u − φ having a maximum at (x̂, t̂) can be interpreted as the graph of φ touching
the graph of u from above. In the language of the level sets, the sub-level set of φ touches theboundary of the sub-level set of u from inside. Note that in two dimensions we do not need toconsider ess inf or ess sup in tests (i1 ) and (i2 ) as written in [13, 14] because Λ[ψ] is constant oneach connected component of {ψ = 0}; see also [10, 12].
Theorem 3.4. The function u defined in (3.3) and given explicitly in (3.4) is the viscosity
solution of the crystalline curvature flow with initial data u0 .
Proof. Let us show that u is a viscosity subsolution since this is more difficult. Indeed, u isnondecreasing in time and hence tests (ii) and (iii) are automatic for a supersolution.
Let us get the simpler tests (ii) and (iii) out of the way first.
Test (ii). Suppose that φ, ψ, g and (x̂, t̂) are as in Definition 3.3(ii). We need to show thatg ′ (t̂) ≤ 0. Suppose therefore that g ′ (t̂) > 0. We can find δ > 0 such that ψ(x) = 0 = ψ(x̂) for∥x − x̂∥∞ < δ and g(t) < g(t̂) for t ∈ (t̂ − δ, t̂).
Let us set θ̂ := u(x̂, t̂). By the maximality of u − φ at (x̂, t̂), we getu(x, t) ≤ θ̂ − φ(x̂, t̂) + φ(x, t) = θ̂ + g(t) − g(t̂) < θ̂,for ∥x − x̂∥∞ < δ, t ∈ (t̂−δ, t̂). This can be interpreted in the following way: for every t ∈ (t̂−δ, t̂)and any θ ∈ (θ̂ + g(t) − g(t̂), θ̂) ∩ Θ ̸= ∅ we have x ∈ Etθ but x ∈/ Et̂θ for ∥x − x̂∥∞ < δ. But
7

2taking t > t̂ − δ2 , we reach a contradiction with Lemma 3.2 that states that the neighborhoodof x̂ cannot disappear from Etθ that fast. Therefore g ′ (t̂) ≤ 0.
Test (iii). Suppose now that φ, f and g is as in Definition 3.3(iii). There is ε > 0 suchthat {x : f (x) = f (x̂), |x − x̂| < ε} is a C 1 curve with normal vector p̂ := ∇f (x̂) ̸= 0 at x̂.
Furthermore, as u(·, t̂) − f has a global maximum at x̂, we conclude thatu(x, t̂) ≤ u(x̂, t̂) + f (x) − f (x̂)

in R2 ,and so{x : f (x) < f (x̂), |x − x̂| < ε} ⊂ {u(·, t̂) < θ̂} =: Ut̂θ̂ ,(3.6)

and x̂ ∈ ∂Ut̂θ̂ . We can express the sub-level sets as
Ut̂θ̂ :=∪

Et̂θ .

θ∈Θ, θ<θ̂

See Figure 3 for the boundaries of Utθ̂ for 0 < t < t∗,θ̂ .
In the following, we say that an edge of A ⊂ R2 , that is, a maximal line segment Γ of∂A, is convex if A is convex in its neighborhood ( ). More precisely, there exists a convex
neighborhood C of the edge Γ such that A ∩ C is convex. Similarly, the edge Γ is concave if
Ac ∩ C is convex for some convex neighborhood C ( ). Finally, if the edge is neither, we saythat it is indeterminate ( ). Similarly we will talk about convex ( ) or concave ( ) corners.
Therefore by (3.6) and the fact that p̂ is not parallel to either of the axes, x̂ can be locatedonly on the concave corners of the boundary of Ut̂θ̂ . More precisely, θ̂ ∈ (0, 1], x̂1 = −x̂2 and0 < t̂ < t♯,θ̂ = 1 + 2θ̂. But then u(x̂, t) = u(x̂, t̂) for t near t̂ because this corner does not moveand hence g ′ (t̂) = 0.
Test (i1 ). It is time to tackle the more challenging test (i1 ). We will see that here we test thehorizontal facets, that is, the one dimensional edges parallel to the x1 -axis. Suppose that φ,ψ, f , g and (x̂, t̂) are as in Definition 3.3(i1 ). We need to estimate the quantity Λ[ψ](x̂1 ) frombelow, which means controlling the length of the component of {ψ = 0} containing x̂1 and the
behavior of ψ near its boundary points.
We again set θ̂ := u(x̂, t̂). The maximality of u − φ at (x̂, t̂) is equivalent toψ(x̂1 + v1 ) + f (x̂2 + v2 ) + g(t̂) = φ(x̂ + v, t̂) ≥ u(x̂ + v, t̂) − θ̂ + φ(x̂, t̂),for v ∈ R2 , which after rearrangement using ψ(x̂1 ) = 0 reads
ψ(x̂1 + v1 ) ≥ u(x̂ + v) − θ̂ + f (x̂2 ) − f (x̂2 + v2 ).

(3.7)

We immediately haveψ(x1 ) ≥ 0

whenever (x1 , x̂2 ) ∈/ Ut̂θ̂ := {u(·, t̂) < θ̂}.

(3.8)

Let s = f ′ (x̂2 )/|f ′ (x̂2 )| be the sign of f ′ (x̂2 ) ̸= 0. There is δ > 0 such that f (x̂2 )−f (x̂2 −hs) >0 for h ∈ (0, δ). Therefore by (3.7) we get
ψ>0

on {x1 : (x1 , x̂2 − hs) ∈/ Ut̂θ̂ for some h ∈ (0, δ)}.

8

(3.9)

θ̂ > 0, t̂ ≥ t♯θ̂ > 0, t̂ < t♯θ̂ ≤ 0Figure 3: Possible locations of x̂ on the boundary of Ut̂θ̂ for test (i1 ) denoted by thick line
segments.
u, ψ

u, ψ
ψ

ψ
u

x̂1

x̂1
x1

u
x1

ℓ
convex

nonconvex

Figure 4: The graphs of ψ (above) and u((·, x̂2 ), t̂) − θ̂ on a convex edge (left) vs. nonconvex
edge (right). The set {ψ = 0} is denoted by the bracketed interval.
On the other hand, by the assumption on the test function, we have ψ(x̂1 + h, t̂) = 0 for|h| < ε for some ε > 0 and therefore (3.9) yields{x̂ + v : |v1 | < ε, −δ < sv2 < 0} ⊂ Ut̂θ̂ .
This implies that x̂ must lie on one of the horizontal edges of the boundary of Ut̂θ̂ away fromthe convex corners; see Figure 3.
If x̂ is on a convex edge, both boundary points of the component of {ψ = 0} containing x̂
are positive due to (3.8) and (3.9), and hence χ(x̂) = 2; see Figure 4(left). Furthermore, thelength of the component is smaller than the length ℓ of the edge by (3.9). We conclude that
Λ[ψ](x̂) ≥ 2ℓ .
On the other hand, if x̂ does not lie on a convex edge, we can only conclude that at least
one boundary point of the component of {ψ = 0} has a positive sign and therefore χ(x̂) ≥ 0;
see Figure 4(right). This gives us only that Λ[ψ](x̂) ≥ 0.
To estimate the time derivative g ′ (t̂), we choose ξ ∈ C 1 (I, R2 ) and I a neighborhood of t̂
such that ξ(t) ∈ ∂Utθ̂ for t ∈ I. Thenu(ξ(t), t) − φ(ξ(t), t) ≤ u(x̂, t̂) − φ(x̂, t̂) = θ̂ − φ(ξ(t̂), t̂).
But u(ξ(t), t) = θ̂ and therefore t 7→ φ(ξ(t), t) has a minimum at t̂. In particular,dφ(ξ(t), t)|t=t̂ = g ′ (t̂) + f ′ (x̂2 )ξ2′ (t̂).
0 = dtNow sξ2′ (t̂) just happens to be the outer normal velocity of the horizontal edge of Ut̂θ̂ on whichx̂ lies. So if x̂ lies on a nonconvex edge, we can simply choose ξ(t) = x̂ since such edges do not
move by construction, and so we haveg ′ (t̂) = 0 ≤ |f ′ (x̂2 )|Λ[ψ](x̂1 ).
9

On the other hand, on the convex edges, we have by the crystalline algorithm sξ ′ (t̂) = − 2ℓ ,where ℓ is the length of the edge. Therefore we conclude that if x̂ is on the convex edge,2g ′ (t̂) = −|f ′ (x̂2 )|sξ ′ (t̂) = |f ′ (x̂2 )| ≤ |f ′ (x̂2 )|Λ[ψ](x̂1 ).
ℓTest (i2 ). This test can be verified as test (i1 ) by symmetry.
This concludes the proof that u defined in (3.3) and given by the explicit formula (3.4) is
a viscosity subsolution of the level set equation (2.3) in the sense of Definition 3.3. The proof
that it is a supersolution is similar in (i1 ) and (i2 ), and trivial in cases (ii) and (iii) as u isnondecreasing in time.

Acknowledgments
The author is partially supported by Japan Society for the Promotion of Science (JSPS) KAKENHI Grant No. 18K13440 (Wakate).

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Faculty of Mathematics and Physics
Institute of Science and Engineering
Kanazawa University
Kakuma town, Kanazawa, Ishikawa 920-1192
JAPANE-mail address: npozar@se.kanazawa-u.ac.jp

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