Giga Y. Surface Evolution Equations: A Level Set Approach

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148 Chapter 3. Comparison principle

Corollary 3.5.4. Assume that F = F (x, t, r, p, X) is independent of x and r and

satisﬁes (F1), (F2). Assume that F is geometric. Assume that

X → F (t, p, X) is concave on S

for all p ∈ R

\{0}, t ∈ [0,T].Letu ∈ C(R

× [0,T]) be an F-solution of

(3.1.1) in R

× (0,T) satisfying (3.5.2) with some L>0.Ifu(x, 0) is convex,

then x → u(x, t) is convex for all t ∈ [0,T].

This immediately follows from Theorem 3.5.2 by replacing u by −u since the

concavity of X → F (t, p, X)isequivalenttothatofX → F (t, p, −X).

Remark 3.5.5 (Applicability). The concavity (resp. convexity) of X → F

(p, X)

is fulﬁlled if and only if f in (1.6.1) is convex (resp. concave) in ∇n,sinceF

deﬁned by (1.6.4). Our Theorem 3.5.2 and Corollary 3.5.4 apply to the level set

equations of (1.5.2), (1.5.4), (1.5.6) provided that conditions in §3.1.3 are fulﬁlled.

For the aﬃne curvature ﬂow equation (1.5.14) or its generalization V =

(κ

)

, α>0 one should take the orientation n inward for convex curves so that

the equation is not a trivial one V ≡ 0. For the convex initial curves Γ

let u

be a

function representing Γ

, i.e., Γ

= ∂{ u

> 0} = ∂{u

< 0} with n = −∇u/|∇u|.

One can take u

so that it is convex; however it is impossible to take concave u

The convexity of the solution u of the level set equation of V =(κ

)

is preserved

if α ≥ 1, by Corollary 3.5.4.

Since

[det(

X + Y

)]

1/m

≥

((det X)

1/m

+(det Y )

1/m

)

for X ≥ O, Y ≥ O, X, Y ∈ S

(see e.g. D. S. Mintrinovi´c (1970)), for the

equation V =(κ

···κ

N−1

)

1/(N−1)

in R

the right-hand side is concave in ∇n

so that F

(p, X)isconvexinX. Unfortunately, we again have to consider convex

initial data u

for convex initial hypersurface Γ

so neither Theorem 3.5.2 nor

Corollary 3.5.4 apply to conclude that solution u is convex in x. So it is unlikely

that Theorem 3.5.2 and Corollary 3.5.4 apply for surface evolution equations by

principal curvatures like (1.6.22), (1.6.23).

3.6 Spatially inhomogeneous equations

We present a few versions of comparison principles for (3.1.1) when F also depends

on the spatial variable x. We restrict ourselves to the case when Ω is bounded,

since the results for unbounded Ω are very complicated to state. We ﬁrst study

the case when inhomogeneity appears only in the ﬁrst order term.

3.6.1 Inhomogeneity in ﬁrst order perturbation

The next result applies when F is of the form

F (x, t, r, p, X)=F

(t, r, p, X)+F

(x, t, p)(3.6.1)

3.6. Spatially inhomogeneous equations 149

with F

satisfying the assumptions on F of Theorem 3.1.1 (including (F3)) and

∈ C(Ω × [0,T] × R

)thatsatisﬁes

(x, t, p) − F

(y,t,q)|≤C(1 + |p|)|x − y| (3.6.2)

with some constant C>0 independent of x, y ∈

Ω, p ∈ R

. In particular, the

level set equation of (1.5.2) applies even when the constancy property of c in §3.1.3

is relaxed as a uniform Lipschitz continuity in x:

|c(x, t) − c(y, t)|≤C|x − y|.

Theorem 3.6.1. Assume that Ω is bounded. Assume that (F1)–(F3) and (F4)

hold. Assume that for R>0 there is a modulus ω

such that

|F (x, t, r, p, X) − F (y,t,r, p,X)|≤ω

(|x − y|(|p| + 1)) (3.6.3)

holds for x, y ∈

Ω, t ∈ [0,T], |r|≤R, p ∈ R

\{0}, X, Y ∈ S

. Then the

comparison principle (BCP) is valid. If we assume (F3



) instead of (F3),then

(BCP) is still valid (by replacing solutions by F

Ω

-solutions) provided that

(i) F

Ω

is invariant under positive multiplication;

(ii) there exists f ∈F

Ω

such that f



(ρ)ρ/f(ρ) is bounded on [0, 1].

Corollary 3.6.2. Assume that Ω is bounded and that F satisﬁes (F1), (F2) and

(3.6.3). Assume that F satisﬁes

|F (x, t, p, ±I)|≤c

|p|

−η

on Ω × [0,T] × (B

(0) \{0})(3.6.4)

for some r

> 0, c

> 0, η>0.Then(BCP) holds if F is geometric.

From the proof of Lemma 3.1.3 it is clear that the assumption on the existence

of c

guarantees the existence of f ∈F

Ω

satisfying (ii) of Theorem 3.6.1. The

assumption (i) of Theorem 3.6.1 is automatically fulﬁlled since F is geometric as

proved in Lemma 3.1.3. Thus, Corollary 3.6.2 follows from Theorem 3.6.1.

Proof of Theorem 3.6.1. We argue in the same way as in the proof of Theorem

3.1.1 right before formula (3.4.12), where x-dependence should be taken into ac-

count. Instead of (3.4.13) we obtain

2γ + F (x

, ˆu, ˆϕ

,X) − F (y

, ˆu, ˆϕ

,X) ≤ 0. (3.6.5)

We send β →∞to get t

αβ

− s

αβ

→ 0. If x

αβ

− y

αβ

→ 0asβ →∞,wegeta

contradiction as in the Case B of the proof of Theorem 3.1.1. Thus we may assume

that x

αβ

→ x

, y

αβ

→ y

with x

= y

. We may also assume that

αβ

→

t, s

αβ

→

t, ˆu → u

∈ R,

ˆϕ

→ q

:= αf



(|p

|)p

/|p

| with p

= x

− y

;

X → X

∈ S

(by (3.4.11))

150 Chapter 3. Comparison principle

as β →∞by taking a subsequence. Thus

2γ + F (x

t, u

) − F (y

t, u

) ≤ 0. (3.6.6)

By (3.6.3) with R =sup

|u| + 1 this inequality yields

2γ − ω

(|x

− y

|(|q

| +1))≤ 0. (3.6.7)

Since M

→ sup

w (as σ →∞by Lemma 3.4.1), we observe that αf (|x

−y

|) →

0asα →∞. By the assumption (ii),

− y

||q

| = αf



(|p

|)|p

|≤Cα f(|p

with some C independent of α.Thus|x

− y

||q

|→0asα →∞.Since|p

|→0

as α →∞, (3.6.7) now yields 2γ ≤ 0 which is a contradiction. 

Remark 3.6.3. (i) The assumption (3.6.4) is fulﬁlled for F = F

of (1.6.19)

provided that g is ˆe

or ˆe



(m>). If F

in (3.6.1) is equal to such an F

, F

of (3.6.1) satisﬁes the assumption of Corollary 3.6.2 provided that F

is geometric

and satisﬁes (3.6.2). For example, Corollary 3.6.2 guarantees that (BCP) holds for

the level set equation of

V = K + c(x, t)

if c satisﬁes the uniform (in t) Lipschitz continuity in x provided that the Gaussian

curvature K is interpreted as its modiﬁed form ˆe

N−1

(κ

,...,κ

N−1

(ii) It is of course possible to prove (CP) when Ω is unbounded. The proof

is similar to that of Theorem 3.1.4. Since we shall send α →∞,wemodifythe

classiﬁcation Case 1, Case 2 as follows. The details are left to the interested readers.

Case 1. There is a sequence α

→∞such that for each r ∈ (0,r

∗

)thereis

→∞such that θ(β

)=µ

(r, β

)asr → 0.

Case 2. For suﬃciently large α,sayα>α

,thereisr

∈ (0,r

∗

) such that

θ(β) >µ

,β) for all β>β

with some β

3.6.2 Inhomogeneity in higher order terms

Unfortunately, results in §3.6.1 do not apply to (3.1.1) when a spatial imho-

mogeneity appears in higher order terms. For example, it does not apply to

F (x, t, p)=−a(x)traceX unless a ≥ 0 is a constant. The next result applies

to such an F .

Theorem 3.6.4. Assume that Ω is bounded. Assume that (F1)–(F3) and (F4)

hold. Assume that for R>0 there is a modulus ω

such that

F (x, t, r, p, X) − F (y, t, r, p, Y ) ≥−ω

(|x − y|(|p| +1)+µ|x − y|

)(3.6.8)

3.6. Spatially inhomogeneous equations 151

for x, y ∈

Ω, t ∈ [0,T], p ∈ R

\{0}, X,Y ∈ S

, |r|≤R, µ>0 whenever

−3µI ≤



O −Y



≤ 3µJ, J =



I −I

−II



. (3.6.9)

Then (BCP) holds.

Remark 3.6.5. The assumption (F2) follows from (3.6.8)–(3.6.9) so it is redun-

dant. Indeed,

ξXξ −

ηY η =

(ξ − η)X(ξ − η)+2

ηX(ξ − η)+

η(X − Y )η

≤|X||ξ − η|

+2|X||ξ − η||η| +

η(X − Y )η

≤|X|(1 + ε

−1

)|ξ − η|

+ ε|X||η|

η(X − Y )η

for ξ, η ∈ R

, ε>0. If X ≤ Y , this implies



O −Y



≤ (1 + ε

−1

)|X|J + ε|X|I. (3.6.10)

(By the way estimate (3.6.10) where ε|X| replaced by |X − Y | + ε|X| in front of

I holds for arbitrary X, Y ∈ S

.) This evidently implies



O −Y



≤ (1 + ε

−1

)|X|J

with X

= X − ε|X|I, Y

= Y + ε|X|I.ThusX

, Y

satisﬁes (3.6.9) by a suitable

choice of µ = µ

. By (3.6.8) we have

F (x, t, r, p, X

) − F (x, t, r, p, Y

) ≥ 0.

Sending ε to zero yields (F2).

We give a typical example of F to which Theorem 3.6.4 applies.

Corollary 3.6.6. Assume that Ω is bounded. Assume that F is of the form (3.6.1)

with F

∈ C(Ω × [0,T] × R

) satisfying (3.6.2). Assume that F

is of the form

(x, t, r, p, X)=−trace(

A(x, t, p)XA(x, t, p)) (3.6.11)

and A(x, t, p) are S

-valued bounded continuous functions in Ω×[0,T]×(R

\{0}).

Assume that A ≥ O and satisﬁes

|A(x, t, p) − A(y,t,p)|≤C|x − y| (3.6.12)

with C>0 independent of p ∈ R

\{0}, x, y ∈ Ω, t ∈ [0,T].Then(BCP) holds

for (3.1.1).

152 Chapter 3. Comparison principle

Proof. Since (F3) and (F4) are clearly fulﬁlled, it suﬃces to prove that (3.6.9)

implies (3.6.8). Such a property is closed under addition of F .Thus,wemay

assume that F = F

, since the assertion is clear for F

From (3.6.9) we see X ≤ Y , more precisely,

ξXξ −

ηY η ≤ 3µ|ξ − η|

.This

implies

(x, t, r, p, X) − F

(y,t,r,p,Y ) ≥−3µ||A(x, t, p) − A(y, t,p)||

where || · ||

denotes the Hilbert–Schmidt norm. By (3.6.12) the estimate (3.6.8)

is fulﬁlled by choosing ω

(s)=3NC

s. 

Proof of Theorem 3.6.4. We take f(η)=η

∈Fand argue as in the proof of

Theorem 3.6.1 right before the formula (3.6.5); we invoke (F3) to handle Case 1

and Case A. Instead of (3.6.5) we obtain

2γ + F (x

, ˆu, ˆϕ

,X) − F (y

, ˆu, ˆϕ

, −Y ) ≤ 0

without using X ≤ Y .Wesendβ →∞and obtain

2γ + F (x

t, u

) − F (y

t, u

, −Y

) ≤ 0(3.6.13)

instead of (3.6.6), where q

= f



(|p

|)p

/|p

|(= 4|p

), p

= x

− y

In the estimate (3.4.11) we may take λ

, λ

−1

= αK

(ρ), K

(ρ)=f



(ρ)/ρ +



(ρ)withp = x

− y

to get a bound

−3αK

(ρ

)I ≤



O −Y



≤ 3αK

(ρ

with ρ

= |x

− y

|;seetheestimateof



σδ

−O −Y

σδ



in the proof of Theorem 3.1.4, Case 2. Thus X

, Y

∈ S

fulﬁlls (3.6.9) with

µ = αK

(ρ

). Applying (3.6.8) to (3.6.13) we obtain

2γ − ω

(|x

− y

|)(|q

| +1)+αK

(ρ

)|x

− y

) ≤ 0(3.6.14)

by taking R =sup

|u| + 1. We have observed that αf(|x

− y

|)=α|ρ

→ 0

as α →∞in the proof of Theorem 3.6.1. Thus, the quantity in ω

tends to zero

as α →∞. Sending α →∞in (3.6.14) now yields a contradiction 2γ ≤ 0. Thus

(BCP) has been proved. 

Example. By Corollary 3.6.2 we observe that (BCP) holds for the level set

equation of (1.5.2) provided that β>0 is continuous and γ in (1.3.7) is convex

and C

except the origin and that a ≥ 0andc ∈ C(Ω×[0,T]) satisﬁes the Lipschitz

continuity in x uniformly in t, i.e.,

|c(x, t) − c(y, t)|≤C|x − y|.

3.6. Spatially inhomogeneous equations 153

We seek a suﬃcient condition for (BCP) when β and γ also depend on x.We

ﬁrst recall the level set equation of

β(x, n)V = −a div

ξ(x, n)+c(x, t, n). (3.6.15)

where ξ = ∇

γ, p → γ(x, p) is convex, positively homogeneous of degree 1. If we

write its level set equation in the form of (3.1.1),

F (x, t, p, X)

= −

β(x, −

|p|

)

{trace∇

γ(x, −p)X +trace∇

∇

γ(x, −p)}|p|−c|p|,

(3.6.16)

(x, p, X)=−

β(x, −

|p|

)

(trace∇

γ(x, −p)X))|p|, (3.6.17)

(x, t, p, X)=−

β(x, −

|p|

)

(trace∇

∇

γ(x, −p))|p|−c|p|. (3.6.18)

The condition (3.6.2) for F

in (3.6.18) is fulﬁlled if (and only if)

M(x, t, p)=−a

trace∇

∇

γ(x, −p)

β(x, −p/|p|)

− c (x, t, −p/|p|)(3.6.19)

is Lipschitz continuous in x uniformly in t, p, i.e.,

|M(x, t, p) − M(y,t,p)|≤C|x − y| (3.6.20)

with C independent of x, y ∈

Ω, t ∈ [0,T], p ∈ R

\{0}. To apply Corollary 3.6.6

we shall ﬁnd a condition that

A(x, p)=



(∇

γ(x, −p)



|p|))

1/2

(3.6.21)

satisﬁes (3.6.12) so that F

in (3.6.17) is written in the form (3.6.11) with (3.6.12).

Since A is positively homogeneous of degree zero in p, it suﬃces to check the

condition that (3.6.12) for p ∈ S

N−1

. The next result gives a suﬃcient condition

for (3.6.12).

Lemma 3.6.7. Let Ω be an open set in R

. Assume that γ = γ(x, p), x ∈ Ω,

p ∈ R

is convex and positively homogeneous of degree 1 on R

as a function

of p. Assume that γ ∈ C(

Ω × R

) is C

as a function of p except at the origin

and Γ=∇

γ ∈ C(Ω × (R

\{0}). Then the square root Γ

1/2

of Γ is Lipschitz

continuous on

Ω × S

N−1

(respectively, Lipschitz continuous in x ∈ Ω uniformly in

p ∈ S

N−1

) if one of following conditions holds.

(i) Γ is C

and its derivative ∇

Γ, ∇

Γ (resp. ∇

Γ) is Lipschitz continuous on

Ω × S

N−1

(resp. Lipschitz continuous in x ∈ Ω uniformly in p ∈ S

N−1

154 Chapter 3. Comparison principle

(ii) All eigenvalues of Γ(x, p)+p ⊗ p are uniformly away from 0 in

Ω × S

N−1

and Γ has the corresponding Lipschitz regularity.

Proof. If (i) holds, Γ has the Lipschitz property in

Ω × U where U is some tubular

neighborhood of S

N−1

.SinceΓ≥ 0, this implies that Γ

1/2

has the desired Lipschitz

continuity; see M. G. Crandall, H. Ishii, P.-L. Lions (1992) [Example 3.6].

By homogeneity the matrix Γ(x, p) has a zero eigenvalue with eigenvector

p ∈ S

N−1

. If (ii) holds, (Γ + p ⊗ p)

1/2

is positive so it has the desired Lipschitz

property. Since

(Γ(x, p))

1/2

= Q

((Γ(x, p)+p ⊗ p)

1/2

),p∈ S

N−1

with Q

(X)=R

, R

= I − p ⊗ p/|p|

, this yields the Lipschitz property of

(Γ(x, p))

1/2

. 

Summarizing these results, we conclude that F in (3.6.16) satisﬁes the as-

sumptions of Corollary 3.6.6 if (i) or (ii) of Lemma 3.6.7 is fulﬁlled for Γ and β,

∇

γ(x, −p)andc are continuous in Ω × [0,T] × S

N−1

and Lipschitz continuous

in x uniformly in p ∈ S

N−1

and t ∈ [0,T] provided that β ≥ β

> 0onΩ × S

N−1

with some constant β

Remark 3.6.8. It seems to be nontrivial to allow spatially inhomogeneity in the

second order term for (BCP) if the singularity at ∇u = 0 in (3.1.1) is very strong

so that (F3) is violated. We consider the level set equation of V =((a(x))

where a ∈ C

∞

(Ω) such that inf

Ω

a>0. It is of the form

+ F (x, ∇u, ∇

u)=0

with

F (x, p, X)=− (G(x, p, X))

/|p|

,G(x, p, X)=trace((aI)R

X(aI)).

We argue as in the proof of Theorem 3.6.4 for f (η)=η

∈Fto obtain (3.6.13)

with the estimate

−3µI ≤



O −Y



≤ 3µJ, µ = αK

(ρ

This relation yields

G(y

) − G(x

) ≥−Nµ|a(x

) − a(y

It also yields

|G(x

)|, |G(y

)|≤C

(ρ

Since

F (x, p, X) − F (y, p,Y )

|p|

{(G(x, p, X))

+ G(x, p, X)G(y, p, Y )+(G(y,p,Y ))

}

×{G(y,p,Y ) − G(x, p, X)},

3.7. Boundary value problems 155

plugging in x

yields

F (x

) − F (y

) ≥−αC

(ρ

)

(ρ

)



(ρ

))

(3.6.22)

if a is Lipschitz continuous. Here C

and C

are constants independent of α.By

the form of f we see f



(η)η

/30 = f



(η)η/6=f(η)soαK

(ρ

)ρ

→ 0asα →∞

since αf (ρ

) → 0. However, there is no control on K

(ρ

)/q

nor 1/ρ

as α →∞

so it is not clear that the right-hand side of (3.6.22) tends to zero as α →∞.

3.7 Boundary value problems

We consider the boundary value problem

+ F (x, t, u, ∇u, ∇

u)=0 in Q =Ω× (0,T), (3.7.1)

B(x, ∇u)=0 on ∂Ω × (0,T). (3.7.2)

When Ω is convex with C

boundary ∂Ω and (3.7.2) is the homogeneous Neumann

boundary condition, i.e., B(x, p)=ν(x),p,whereν is the outward unit normal

of ∂Ω, it is easy to state a condition to guarantee the comparison principle:

(BCPB) Let u and v be sub- and supersolutions of (3.7.1)–(3.7.2) in

Ω × (0,T),

where Ω is bounded. If −∞<u

∗

≤v

∗

<∞ at t =0,thenu

∗

≤v

∗

in Q.

Theorem 3.7.1. Assume that Ω is a convex domain with C

boundary in R

Assume that F is independent of x. Assume that (F1)–(F3) and (F4) holds. Then

the comparison principle (BCPB) holds. If we assume (F3



) instead of (F3),then

(BCPB) is still valid (by replacing solutions by F

Ω

-solutions) provided that F

Ω

invariant under positive multiplication.

By Lemma 3.1.3 this result yields:

Corollary 3.7.2. Assume that Ω is a convex domain with C

boundary in R

.As-

sume that F is independent of x and r. Assume that F :[0,T]×(R

\{0})×S

→

R is continuous and degenerate elliptic. Then (BCPB) holds if F is geometric.

Proof of Theorem 3.7.1. To simplify the presentation we give a proof when F is

also independent of r.

We ﬁrst construct a ‘barrier’ of the boundary. Let d(x, ∂Ω) be the distance of

x and the boundary ∂Ω. As well known, d(x, ∂Ω) is C

near ∂Ωforx ∈ Ω; see e.g.

D. Gilbarg and N. S. Trudinger (1998) [Lemma 14.16]. We set b(x)=−d(x, ∂Ω)

near ∂Ω and extend to Ω in a suitable way so that b ∈ C

(Ω) and b<0inΩ.

Clearly, b fulﬁlls ν(x)=∇b(x). We set g(x)=b(x) − inf

Ω

b so that g ≥ 0inΩ.

We argue in the same way as in the proof of Theorem 3.1.1. Let z

) ∈ Q × Q be a maximizer of w − ϕ

. We invoke the assumption

that u ≤ v at t = 0 and observe that t

> 0, s

> 0forlargeσ,sayα>α

156 Chapter 3. Comparison principle

β>β

. The main diﬀerence from the proof of Theorem 3.1.1 is that x

and y

may be on ∂Ω.

We divide the situation into two cases following Remark 3.4.3. We ﬁx α>α

and suppress the subscript α.

Case 1. For each r>0thereisβ

→∞(as r → 0) and a maximizer z

w − ϕ

in Q

∗∗

× Q

∗∗

such that |x

− y

|≤r,wherez

=(x

)and

∗∗

= Ω × (0,T].

We ﬁrst ﬁx r and denote β

simply by β.

(w − ϕ

)(x, t, y

) ≤ (w − ϕ

)(x

)

for (x, t) ∈ Q

∗∗

,wesee

max

∗∗

(u − ϕ

)=(u − ϕ

)(x

)

(x, t)=αf(|x − y

|)+β(t − s

)

+ γt.

Our goal is to prove

)+F (t

,α∇

f(|p|),α∇

f(|p|)) ≤ 0atp = p

(3.7.3)

if p

:= x

− y

=0and

)=2β(t

− s

)+γ ≤ 0(3.7.4)

if p

=0.Sinceu is a subsolution in Q,thisisclearifx

∈ Ω. So we may assume

that x

∈ ∂Ω. We modify ϕ

˜ϕ

= ϕ

+ f(|x − x

|)+|t − t

so that u − ˜ϕ

takes a strict maximum at x = x

.Wesetϕ

=˜ϕ

+ δb.Bythe

convergence of maximum points (Lemma 2.2.5), there is a maximizer (x

)of

u − ϕ

in Q converging to (x

)asδ ↓ 0.

If y

= x

,then∇ϕ

) =0forsmallδ>0. If x

∈ ∂Ω, then

ν(x

), ∇ ˜ϕ

)≥0

as Ω is convex; we suppress t

as ∇ ˜ϕ

does not depend on t.Sinceν(x), ∇b(x) =

1, this implies

ν(x

), ∇ϕ

)≥δ>0. (3.7.5)

We recall Deﬁnition 2.3.7 with ϕ = ϕ

near (x

). The estimate (3.7.5) guar-

antees that

δt

+ F (t

, ∇ϕ

, ∇

) ≤ 0at(x

)(3.7.6)

even if x

∈ ∂Ω. Sending δ → 0 we obtain (3.7.3).

3.7. Boundary value problems 157

If y

= x

, then there might be a chance that ∇ϕ

) = 0, i.e.,

(α +1)∇

f(|x − y

|)+δ∇b(x)=0 at x = x

Since b = −d(x, ∂Ω) near ∂Ω, such an x

is of the form x

= x

− C(δ)ν(x

) with

some C(δ) > 0 for small δ.Inotherwordsx

lies on the line through x

with

direction −ν(x

). Let θ be a strictly concave, increasing C

function on [0, ∞)

satisfying θ(0) = 0, θ(C(δ)) = C(δ). For a suitable choice of such θ (depending on

δ),

u − ˜ϕ

+ δθ(−b)

takes its maximum at some (ξ

,τ

) → (x

)asδ → 0and∇ ˜ϕ

) =

δ∇θ(−b)(x

)sincef is strictly convex. In a similar argument we obtain (3.7.6) at

(ξ

,τ

)evenifξ

∈ ∂Ω. Sending δ → 0 yields (3.7.4).

We observe that

min

∗∗

(v − ϕ

−

)=(v − ϕ

−

)(y

)

with ϕ

−

(y,s)=−αf(|x

− y|) − β(t

− s)

− γs. We argue in a similar way to

obtain

−

)+F (s

,α∇

f(|p|), −α∇

f(|p|)) ≥ 0atp = p

(3.7.7)

if p

=0and

−

)=2β(t

− s

) − γ ≥ 0(3.7.8)

if p

=0.Ifp

= 0, (3.7.4) and (3.7.8) yields a contradiction 2γ ≤ 0sowemay

assume that p

= 0. Subtracting (3.7.7) from (3.7.3) and sending r → 0sothat

→ 0(since|p

| <r), we obtain a contradiction 2γ ≤ 0.

Case 2. There is r

> 0 such that for suﬃciently large β any maximizer z

w − ϕ

in Q

∗∗

× Q

∗∗

satisﬁes |z

| >r

We deﬁne Ψ

βδ

(x, t, y, s)=Φ

(x, t, y, s) − δ(g(x)+g(y))

for δ>0. Let ˆz =(ˆx,

t, ˆy, ˆs) be a maximizer of Ψ

βδ

on Q × Q. Clearly, sup

βδ

↑

sup

as δ ↓ 0, so as in §3.4 for suﬃciently small δ,sayδ<δ

(β), we see that

ˆz ∈ Q

∗∗

× Q

∗∗

and |ˆx − ˆy| >r

. (Note that ˆz depends on β and δ.)

Since Ψ

βδ

= w−δ(g(x)+g(y))−ϕ

attains its maximum at ˆz over Q

∗∗

×Q

∗∗

we apply Theorem 3.3.3 with k =2,N

= N

+1,Z

= Z

= Q

∗∗

,  =2to

w − ϕ =Ψ

βδ

. Then we conclude as in the proofs of Theorem 3.1.1 and 3.1.4 (Case

2), that for each λ

> 0thereisX, Y ∈ S

such that (3.4.11) holds and

(γ +ˆϕ

, ˆϕ

+ δ∇g(ˆx),X + δ∇

g(ˆx)) ∈ P

2,+

∗∗

u(ˆx,

t), (3.7.9)

(−γ +ˆϕ

, ˆϕ

− δ∇g(ˆy),Y − δ∇

g(ˆy)) ∈ P

2,−

∗∗

v(ˆy, ˆs). (3.7.10)