Game Theoretical Methods in PDEsTutorial
Marta Lewicka
University of Pittsburgh
1
Linear PDEs (∆) and probability
• Initial position of token: x0 ∈Ω⊂ R2.
• Moves: random increments of length ε.
• uε(x0): probability of hitting Oε(Γ1),
the first time Oε(∂Ω) is hit.
THM uε ⇒ v∈ C (Ω) as ε→ 0, and v is a viscosity solution to ∆v = 0:
∆φ(x0)≤ 0 ∆φ(x0)≥ 0
In fact, v is the classical solution to:∆v = 0 in Ω
v = 1 on Γ1 and v = 0 on ∂Ω\Γ1
2
Proof that v is a viscosity solution
uε(x) = 14(uε(x+ εe1)+uε(x− εe1)+uε(x+ εe2)+uε(x− εe2))
⇔ 0 = (uε(x+ εe1)−uε(x))+(uε(x− εe1)−uε(x))
+(uε(x+ εe2)−uε(x))+(uε(x− εe2)−uε(x))
• uε ⇒ v in Ω as ε→ 0 (can be proven rigorously).
Take φ smooth, (v−φ)(x0) = 0≤ (v−φ)(x) ∀x.
Find xε→ x0: (uε−φ)(xε)≤ (uε−φ)(x)+ε3 ∀x
⇔ uε(x)−uε(xε)≥ φ(x)−φ(xε)− ε3
• 0≥ (φ(xε+ εe1)−φ(xε))+(φ(xε− εe1)−φ(xε))
+(φ(xε+ εe2)−φ(xε))+(φ(xε− εe2)−φ(xε))−4ε3
(Taylor) = ε2∂11φ(xε)+ ε2∂22φ(xε)+o(ε2) = ε2∆φ(xε)+o(ε2)
• Conclude: 0≥ ∆φ(xε)+o(1). Hence: ∆φ(x0)≤ 0.3
Non-linear PDEs (∆p)
∆pu = div(|∇u|p−2∇u) = |∇u|p−2∆u+⟨
∇(|∇u|p−2) : ∇u⟩
= |∇u|p−2(
∆u+(p−2)〈∇2u : ∇u|∇u|⊗
∇u|∇u|〉
)= |∇u|p−2(∆u+(p−2) ∆∞u)
• Taylor expand: u(x) = u(x0)+ 〈∇u(x0),x− x0〉+ 1
2〈∇2u(x0) : (x− x0)⊗ (x− x0)〉+o(ε2)
Average:ffl
Bε(x0)u = u(x0)+
12〈∇
2u(x0) :ffl(x− x0)⊗ (x− x0)〉
= u(x0)+12〈∇
2u(x0) : ε2
4 Id〉= u(x0)+ε2
8 ∆u+o(ε2)[• Thm (Manfredi-Parviainen-Rossi 2012)
Let u ∈ C (Ω) satisfy: ∀x0 ∈Ω u(x0) =ffl
Bε(x0)u+o(ε2). Then ∆u = 0 in Ω.
]• 1
2(supBε(x0)u+ infBε(x0)
u)≈ 12
(u(x0+ ε
∇u(x0)|∇u(x0)|
)+u(x0− ε∇u(x0)|∇u(x0)|
))
= u(x0)+ε2
2
⟨∇2u(x0) : ∇u(x0)
|∇u(x0)|⊗ ∇u(x0)|∇u(x0)|
⟩+o(ε2)
= u(x0)+ε2
2 ∆∞u(x0)+o(ε2)4
Non-linear PDEs (∆p)
∆pu = div(|∇u|p−2∇u) = |∇u|p−2∆u+⟨
∇(|∇u|p−2) : ∇u⟩
= |∇u|p−2(
∆u+(p−2)〈∇2u : ∇u|∇u|⊗
∇u|∇u|〉
)= |∇u|p−2(∆u+(p−2) ∆∞u)
• We have:ffl
Bε(x0)u = u(x0)+
ε2
8 ∆u+o(ε2)
• 12(supBε(x0)
u+ infBε(x0)u) = u(x0)+
ε2
2 ∆∞u(x0)+o(ε2)
Hence: 4p+2
fflBε(x0)
u+ p−2p+2
12(supBε(x0)
u+ infBε(x0)u)
= u(x0)+ε2
2(p+2)(∆u+(p−2)∆∞u)+o(ε2)
Concluding: If ∆pu = 0 then:
u(x0) =α2 supBε(x0)
u+ α2 infBε(x0)
u+βffl
Bε(x0)u+o(ε2)
(Ω⊂ R2, 2≤ p < ∞) (α = p−2p+2, α+β = 1)
5
Tug-of-War game for ∆p
X = Ω∪Γε. Rules of the game: • Token initially at x0 ∈Ω.
• probability α2 → player I moves by at most ε
• probability α2 → player II moves by at most ε
• probability β→ computer
moves randomly by at most ε
Game stops, when token position xn ∈ Γε
Then player II pays F(xn) to player I. Boundary data: F : Γε→ R.
Strategies: σI = σkI : Xk+1→ X∞k=1, σII = σk
II : Xk+1→ X∞k=1Borel functions, σk
I,II(x0, . . .xk) ∈ Bε(xk),
and σkI,II(x0, . . .xk) = xk if xk ∈ Γε.
.6
Tug-of-War game for ∆p
X = Ω∪Γε. Strategies: σI,II = σkI,II : Xk+1→ X∞k=1,
Probability measures:
γk[x0, . . .xk] =α2δ
σkI (x0,...xk)
+ α2δ
σkII(x0,...xk)
+β
LbBε(xk)|Bε| if xk ∈Ω
δxk if xk ∈ Γε
• Probabilities Px0,kσI,σII :
Px0,kσI,σII(A1× . . .Ak) =
´A1
. . .´
Ak1 dγk−1[x0, . . .xk−1] . . . dγ0[x0].
• They generate (Kolmogoroff extension) probability Px0σI,σII
on sets of histories (x0,x1, . . .) ∈ X∞.
• Define the expectation: Ex0σI,σII[Fτ] =
´X∞ F(xτ) dPx0
σI,σII
where τ = mink; xk ∈ Γε and F(xτ) =
F(xk) if τ = k+∞ if τ =+∞
7
Tug-of-War game for ∆p
minimum gain of player I: uI(x0) = supσIinfσII E
x0σI,σII[Fτ]
maximum loss of player II: uII(x0) = infσII supσIEx0
σI,σII[Fτ]
THM 1 (Peres-Schramm-Sheffield-Wilson 2008 (p = ∞))(Peres-Sheffield 2008 (1 < p < ∞))(Manfredi-Parviainen-Rossi 2012 (2≤ p < ∞))
uI = uII = uε := the unique p-harmonious function: u(x) =α
2sup
Bε(x)u+
α
2inf
Bε(x)u+β
Bε(x)
u in Ω
u = F in Γε
THM 2 When ε→ 0 then uε ⇒ v the unique viscosity solution to:∆pv = 0 in Ω
u = F on ∂Ω
8
Plan for the lecture
• complete proofs of Theorem 1 and Theorem 2
• proof of the Harnack inequality for ∆pv = 0,
using tug-of-war game interpretation
• limiting cases p→ ∞ and p→ 1
• p = ∞: infinity Laplacian ∆∞v = 0
and absolutely minimizing Lipschitz extensions
• p = 1: motion by mean curvature vt−|∇v|div( ∇v|∇v|) = 0
and “geometric” equations, Kohn-Serfaty game
9
Basic facts on ∆pv = div(|∇v|p−2∇v), 1 < p < ∞
• p = 2: Laplacian, p > 2: degenerate elliptic, 1 < p < 2: singular
• Euler-Lagrange eqn for: Ip(v) = 1p´
Ω|∇v|p on v ∈W 1,p(Ω)
⇒ existence and uniqueness of a weak solution to:
∆pv = 0v|∂Ω = g
• weak solutions are C 1,αloc , in general not C 2 when ∇v(x) = 0
(Uraltseva: p > 2, Lewis, DiBenedetto: 1 < p < 2),
• a continuous v is called a viscosity solution if:
∇φ(x0) 6= 0⇒ ∆pφ(x0)≤ 0
∇φ(x0) 6= 0⇒ ∆pφ(x0)≥ 0
• weak solutions = viscosity solutions (Juutinen-Lindqvist-Manfredi 2001)
• Open problems: If v|Bρ= 0 v≡ 0 in Ω
?⇒(true for n = 2 by complex analysis methods: Manfredi 1988)
10
Proof of Thm 1 (β 6= 0): existence of p-harmonious u(after Luiro, Parviainen, Saksman 2013)
Let u : X → R bounded, Borel. Look for fixed points of the operator:
(Tu)(x) =
α2 supBε(x)u+ α
2 infBε(x)u+βffl
Bε(x)u in Ω
u(x) in Γε
• Tu is also bounded, Borel. • Let u0 =
infΓε
F in Ω
F in Γε
• Then u1 = Tu0 ≥ u0 and u2 = Tu1 ≥ Tu0 = u1 and ... u j+1 ≥ u j
Hence u j u. In fact u j ⇒ u in Ω.
• So: u ⇔ u j+1 = Tu j ⇒ Tu, so Tu = u and u = F on Γε.
.11
Proof of Thm 1 (β 6= 0): existence of p-harmonious u(after Luiro, Parviainen, Saksman 2013)
Let u : X → R bounded, Borel. Look for fixed points of the operator:
(Tu)(x) =
α2 supBε(x)u+ α
2 infBε(x)u+βffl
Bε(x)u in Ω
u(x) in Γε
• Tu is also bounded, Borel. • Let u0 =
infΓε
F in Ω
F in Γε
• Then u1 = Tu0 ≥ u0 and u2 = Tu1 ≥ Tu0 = u1 and ... u j+1 ≥ u j
Hence u j u. In fact u j ⇒ u in Ω.
• So: u ⇔ u j+1 = Tu j ⇒ Tu, so Tu = u and u = F on Γε.
• Uniqueness: If u 6= v are two solutions, call M = sup(u− v)> 0.
∀x u(x)− v(x)= α2(supBε(x)u−supBε(x)v)+α
2(infBε(x)u−infBε(x)v)
+βffl
Bε(x)(u−v)≤αsupBε(x)(u−v)+βffl
Bε(x)(u−v)
≤ αM+βffl
Bε(x)(u− v)
.12
Proof of Thm 1 (β 6= 0): existence of p-harmonious u(after Luiro, Parviainen, Saksman 2013)
Let u : X → R bounded, Borel. Look for fixed points of the operator:
(Tu)(x) =
α2 supBε(x)u+ α
2 infBε(x)u+βffl
Bε(x)u in Ω
u(x) in Γε
• Tu is also bounded, Borel. • Let u0 =
infΓε
F in Ω
F in Γε
• Then u1 = Tu0 ≥ u0 and u2 = Tu1 ≥ Tu0 = u1 and ... u j+1 ≥ u j
Hence u j u. In fact u j ⇒ u in Ω.
• So: u ⇔ u j+1 = Tu j ⇒ Tu, so Tu = u and u = F on Γε.
• Uniqueness: If u 6= v are two solutions, call M = sup(u− v)> 0.
∀x u(x)− v(x)≤ αM+βffl
Bε(x)(u− v).
If u(xn)−v(xn)→M and xn→ x0∈X then: M ≤ αM+βffl
Bε(x0)(u− v).
But β = 1−α, so: M ≤ffl
Bε(x0)(u− v). Hence u− v≡M in Bε(x0).
By extension: u− v≡M in X , contradicting u = v on Γε.
13
Proof of Thm 1: u is the game value
Lemma Let u : Ω→ R be bounded and Borel. Fix ε,δ > 0.
There exists suboptimal selections Ssup,Sin f : Ω→Ω, such that:
• Ssup,Sin f are Borel functions, and ∀x ∈Ω Ssup(x),Sin f (x)∈ Bε(x)
• ∀x ∈Ω u(Ssup(x))≥ supBε(x)u−δ, u(Sin f (x))≤ infBε(x)u+δ
Example Ssup may not exist with closed balls Bε(x) !
Let A⊂R3 be bounded, Borel, and such that: A+ B1(0) is not Borel.
Let u = χA. Assume that: ∀x u(Ssup(x))≥ supB1(x)u−12.
Then: (Ssup)−1(A) = x ∈ R3; supB1(x)u = 1= A+ B1(0)
Therefore Ssup cannot be a Borel function.
14
Proof of Thm 1: u is the game value
Recall: uI(x0) = supσIinfσII E
x0σI,σII[Fτ],
uII(x0) = infσII supσIEx0
σI,σII[Fτ].Clearly: uI ≤ uII.
• We show that uII ≤ u. Symetrically u≤ uI. Then uI = uII = u.
• Fix η > 0. Choose σII suboptimal selection, so that:
u(σkII(x0 . . .xk))≤ infBε(xk)
u+ η
2k+1
• Let σI be any strategy for player I.
uII(x0)≤ supσIEx0
σI,σII[Fτ]≤ supσI
Ex0σI,σII
[u(xτ)+
η
2τ
]≤ supσI
Ex0σI,σII
[u(x0)+
η
20
]= u(x0)+η
⇑
because
u(xk)+η
2k, (Borel σ-algebra)k
is a supermartingale.
15
Proof of Thm 1: u is the game value
Lemma
u(xk)+η
2k, F x0k
is a supermartingale.
Proof Ex0σI,σII
u(xk)+
η
2k | Fx0
k−1
(x0 . . .xk−1) =
=´
X(u+η
2k) dγk−1[x0 . . .xk−1]
= α2 u(σI(x0 . . .xk−1))+
α2 u(σII(x0 . . .xk−1))+β
fflBε(xk−1)
u dy+ η
2k
≤ α2 supBε(xk−1)
u+ α2
(infBε(xk−1)
u+ η
2k
)+β
fflBε(xk−1)
u dy+ η
2k
= u(xk−1)+(α2 +1) η
2k ≤ u(xk−1)+η
2k−1
Theorem (Doob’s optional stopping)
If fk,Fk is a supermartingale and τ1,τ2 are bounded stopping times,
then: τ1 ≤ τ2 ⇒ E[ fτ2]≤ E[ fτ1].
16
p-harmonious u and the martingale calculation
Similarly, one can prove:
Lemma Let σI, σII be any two fixed strategies. Then:
infσII Ex0σI,σII
[u(xτ∗)]≤ u(x0)≤ supσIEx0
σI,σII[u(xτ∗)]
for every stopping time τ∗ ≤ τ.
Proof Use the suboptimal selection strategy σII:
infσII Ex0σI,σII
[u(xτ∗)]≤ Ex0σI,σII
[u(xτ∗)]≤ Ex0σI,σII
[u(xτ∗)+
η
2τ∗]
≤ Ex0σI,σII
[u(x0)+
η
20
]= u(x0)+η
because
u(xk)+η
2k, F x0k
is a supermartingale.
17
Proof of Thm 2: uε ⇒ v continuous(after Manfredi, Parviainen, Rossi 2012)
Lemma (∼ Ascoli-Arzela theorem for discontinuous functions)
Let uε : Ω→ R be a sequence of functions such that:
(i) ∃C ∀x ∈ Ω ∀ε |uε(x)| ≤C
(ii) ∀η > 0 ∃r0 ∃ε0 ∀x,y ∈ Ω|x− y|< r0
∀ε < ε0 |uε(x)−uε(y)|< η
Then uε ⇒ v in Ω and v is a continuous function.
• Take uε : Ω∪Γε→ R to be the unique p-harmonious functions
with parameter ε and boundary data F in Γε
• uε is equibounded. Need to verify condition (ii).
• Case 1: x ∈Ω, y ∈ ∂Ω ⇒ Case 2: x,y ∈Ω.
18
Proof of Thm 2: verification of (ii) for x0 ∈Ω, y ∈ ∂Ω
(after Manfredi, Parviainen, Rossi 2012)
|uε(x0)−uε(y)|= |uε(x0)−F(y)| ≤ |uε(x0)−F(z)|+Lδ
= |supσI
infσII
Ex0σI,σII[F(xτ)−F(z)]|+Lδ≤ Lsup
σIinfσII
Ex0σI,σII[|xτ− z|]+Lδ
Take any σI and σII = σzII “pulling towards z”
σzII(xk) =
xk+ ε
z−xk|z−xk|
if xk 6∈ Bδ(z)xk if xk ∈ Bδ(z)
Compute: Ex0σI,σ
zII
|xk− z|−Cε2k | F x0
k−1
(x0 . . .xk−1)
≤ α2(|xk−1− z|+ε)+ α
2(|xk−1− z|−ε)+βffl
Bε(xk−1)|s− z| ds−Cε2k
= α|xk−1− z|+β
(|xk−1− z|+ ε2
8 ∆(|s− z|)+o(ε2))−Cε2k
≤ |xk−1− z|−Cε2(k−1) for C >> 1.
Hence:|xk− z|−Cε2k,F x0
k
is a supermartingale.
19
Proof of Thm 2: verification of (ii) for x0 ∈Ω, y ∈ ∂Ω
(after Manfredi, Parviainen, Rossi 2012)
By Doob’s Thm: Ex0σI,σ
zII[|xτ− z|−Cε2τ]≤ Ex0
σI,σzII[|x0− z|] = |x0− z|
Hence: Ex0σI,σ
zII[|xτ− z|]≤ |x0− z|+Cε2Ex0
σI,σzII[τ]
Consider:∆g =−2(n+2) in BR(z)\ Bδ(z)g = 0 on ∂Bδ(z)∂g∂~n = 0 on ∂BR(z)
i.e: g(s) = g(|s− z|) and g(r) =−ar2−b logr+ c
g satisfies: ∀Bε(s0)⊂ annulusffl
Bε(s0)g(s) ds = g(s0)− ε2
• Compute: Ex0σI,σ
zII
g(|xk− z|)+Cε2k | F x0
k−1
(x0 . . .xk−1)
≤ α2g(|xk−1−z|+ε)+α
2g(|xk−1−z|−ε)+β(g(|xk−1− z|)− ε2)+Cε2k
≤ g(|xk−1−z|)+Cε2(k−1) because g is increasing and concave.
20
Proof of Thm 2: verification of (ii) for x0 ∈Ω, y ∈ ∂Ω
(after Manfredi, Parviainen, Rossi 2012)
Hence, Doob’s Thm yields: Ex0σI,σ
zII[g(|xτ∗−z|)+Cε2τ∗]≤ g(|x0−z|)
where τ∗ is the exit time into the inner ball Bδ(z). Clearly: τ≤ τ∗
Hence:
Ex0σI,σ
zII[Cε2τ]≤ Ex0
σI,σzII[Cε2τ∗]
≤ g(|x0− z|)−Ex0σI,σ
zII[g(|xτ∗− z|)]
≤ Lg dist(x0,∂Bδ(z))+Lgε≤ Lg(|x0− z|+ ε)
Recall: Ex0σI,σ
zII[|xτ− z|]≤ |x0− z|+Cε2Ex0
σI,σzII[τ] ≤Cδ(|x0−z|+ε)
• Concluding: |uε(x0)−uε(y)| ≤Cδ,Ω (|x0− y|+ ε)+CFδ
which gives exactly condition (ii)!
21
Proof of Thm 2: the limit satisfies ∆pv = 0(after Manfredi, Parviainen, Rossi 2012)
We proved that: uε ⇒ v in Ω. Clearly v = F on ∂Ω
Verification that v is a viscosity solution:Find xε→ x0 so that: (uε−φ)(xε)≤ (uε−φ)(x)+ ε3 ∀x
• uε(xε) =α2 supBε(xε)uε+
α2 infBε(xε)uε+β
fflBε(xε)
uε
• φ(xε) =α2 supBε(xε)φ+ α
2 infBε(xε)φ+βffl
Bε(xε)φ
−βε2
8 (∆φ(xε)+(p−2)∆∞φ(uε))+o(ε2)
Hence: (uε−φ)(xε)=α2(supuε−supφ)+α
2(infuε− infφ)+βffl(uε−φ)
+ βε2
81
|∇φ|p−2∆pφ(xε)+o(ε2)
≥ (uε−φ)(xε)+βε2
81
|∇φ|p−2∆pφ(xε)+o(ε2).
Consequently: 1|∇φ|p−2∆pφ(xε)≤ o(1).
Since ∇φ(x0) 6= 0, we get: ∆pφ(x0)≤ 0.
22
Harnack’s inequality for p-harmonious functions (2≤ p < ∞)(after Luiro, Parviainen, Saksman 2013)
THM 3 For each ε, let uε be a p-harmonious function in Ω. Then:
∀Bρ(z0)⊂ Br(z0)B2r(z0)⊂Ω
∃Cp ∀ε < ρ osc(uε,Bρ)≤ Cpρ
r osc(uε,Br)
Corollary Let uε be a positive p-harmonious function in B2r(z0).
Then there exists C (independent of ε) so that:
∀Bρ(x0)⊂ Br(z0) supBρuε ≤ C infBρ
uε
Corollary Harnack’s inequality for p-harmonic v: supBρv≤ C infBρ
v
Prior proofs of Harnack’s inequality:
• Ladyzhenskaya-Uraltseva 1968, Serrin 1967
• Nash-Moser iteration, DeGiorgi classes (DiBenedetto-Trudinger 1984)
• Here: completely different proof!
23
Proof of Thm 3: Harnack’s inequality(after Luiro, Parviainen, Saksman 2013)
Let x0,y0 ∈ Br(z0). Let z ∈ B2r: |x0− z|= mε = |y0− z|, m = b|x0−y0|εc.
Define the strategy σzII:
• player II cancels the earliest move of player I
• otherwise player II “pulls towards z” (by ε)
Position of token:xk+1 = x0+∑i∈Jk
Iwi+∑i∈Jk
IIwi+∑i∈Jk
randomwi
Define the stopping time τ∗II = mink; (i),(ii),(iii):(i) player II has won m times more than player I: |Jk
II|= |JkI |+m
(ii) player I has won d2rεe times more than player II: |Jk
I |= |JkII|+d
2rεe
(iii) total random moves are large: |∑i∈Jkrandom
wi| ≥ 2r.
In particular: xτ∗II∈ B6r(z0). Also: τ∗II ≤ τ, so (by lemma):
u(x0)−u(y0)≤ supσI,σII
(Ex0
σI,σzII[uτ∗II
]−Ey0σ
zI,σII
[uτ∗I]
)24
Proof of Thm 3: Harnack’s inequality(after Luiro, Parviainen, Saksman 2013)
Fix σI,σII. Then: Ex0σI,σ
zII[uτ∗II
]−Ey0σ
zI,σII
[uτ∗I]
= Ex0σI,σ
zII[uτ∗II
χgame endsby (i)
]+Ex0σI,σ
zII[uτ∗II
χgame endsby (ii) or (iii)
]
−Ey0σ
zI,σII
[uτ∗Iχ
game endsby (i)
]−Ey0σ
zI,σII
[uτ∗Iχ
game endsby (ii) or (iii)
]
= Ex0σI,σ
zII[uτ∗II
χgame endsby (ii) or (iii)
]−Ey0σ
zI,σII
[uτ∗Iχ
game endsby (ii) or (iii)
]
≤ P ( supB6r(z0)
u− infB6r(z0)
u)≤ Cp|x0−y0|
r ( supB6r(z0)
u− infB6r(z0)
u)
where P = probability that game ends by (ii) or (iii).
• Concluding: |u(x0)−u(y0)| ≤Cp|x0−y0|
r ( supB6r(z0)
u− infB6r(z0)
u)
⇒ osc(u,Bρ)≤Cpρ
6r(sup
B6r
u− infB6r
u)
25
Proof of Thm 3: P≤Cp|x0−y0|
r(after Luiro, Parviainen, Saksman 2013)
To estimate P, couple the game with a cylinder walk:
Rules of the walk:
• token initially at (z,mε)
• probability α2 → player I moves by (0,ε)
• probability α2 → player II moves by (0,−ε)
• probability β→ random move by (y,0), y ∈ Bε(0)
Lemma P = probability (the walk does not exit the cylinderthrough its bottom)
≤Cpmε+ ε
2r
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The limiting case p→ ∞ (β = 0)
Theorem (Bhattacharya-DiBenedetto-Manfredi 1989, Jensen 1993)
Let vp be the unique solution to: ∆pvp = 0, vp|∂Ω= F.
Then vp ⇒ v∞ in Ω as p→∞, and v∞ is the unique viscosity solution to:
∆∞v∞ = 0, v∞|∂Ω = F
∇φ(x0) 6= 0⇒
〈∇2φ : ∇φ
|∇φ|⊗∇φ
|∇φ|〉(x0)≤ 0
∇φ(x0) 6= 0⇒ ∆∞φ(x0)≥ 0
Theorem (Aronsson, Aronsson-Crandall-Juutinen 1967, 2004)
v∞ ∈ Lip(Ω,R) has the uniquely defining property:
∀ open V ⊂Ω ∀v : V → Rv|∂V = v∞|∂V
Lip(v∞,V )≤ Lip(v,V )
• v∞ is AMLE (Absolutely Minimizing Lipschitz Extension)
“ ∆∞v = 0 is the Euler-Lagrange equation of ‖∇v‖∞”27
The limiting case p = ∞
• ∆∞v = 0 is a fully nonlinear equation,
only viscosity solution definition valid (no integration by parts)
• n = 2: v∞ ∈ C 1,α (Savin, Savin-Evans 2008),
n > 2: v∞ differentiable everywhere (Evans-Smart 2011)
Thm (Peres-Schramm-Sheffield-Wilson 2008)
uε ⇒ v∞ as ε→ 0 where uε are values of the Tug-of-War game:uε(x) = 1
2 supBε(x)uε+12 infBε(x)uε in Ω
uε = F on Γε
An AMLE extension
is inifinity harmonic!
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The limiting case p→ 1Thm (Peres-Sheffield 2008)Let 1 < p < ∞. Then the values uε of the following game converge,as ε→ 0 to v with: ∆pv = 0, v|∂Ω = F .
• token initially at x0 ∈Ω.
• the winning player (probability 12) moves token
by vector w (|w|< ε)
• token moved randomly by (p−1)−1/2|w|in the direction orthogonal to w
Game stops when token position in Γε. Player II pays F(xn) to player I.
Observe that the following structure is needed to define the game:
• p = ∞ metric (extensions to length spaces by PSSW)• 2≤ p < ∞ measure (extensions to Heisenberg groups
(Carnot-Caratheodory groups) by Ferrari-Liu-Manfredi)• 1 < p < 2 perpendicularity: Riemannian structure
(other extensions: Barron-Evans-Jenssen)29
The limiting case p→ 1Thm (Juutinen 2005, Sternberg-Williams-Ziemer 1992)
Let Ω be convex. Let vp be unique solutions to: ∆pvp = 0, vp|∂Ω=F.
Then vp ⇒ v1 in Ω as p→ 1, and v1 is the function of least gradient:
v1 ∈ BV (Ω)∩C (Ω) and v1 minimizes ‖∇v‖BV (Ω)
Moreover, v1 is a viscosity solution to ∆1:
∆1v1 = div( ∇v1|∇v1|
) = 0, v1|∂Ω = F
However, there exist viscosity solutions to ∆1, other than v1.
Motion by curvature and ∆1
Level set v = t is the image of ∂Ω under motion by curvature for time t
• curvature of the level set: κ =−div( ∇v|∇v|)
• |∇v|= |∇v ·~n|= ∂v∂~n ≈
∆t∆x
• κ = ∆x∆t ⇔ −∆1v = 1
|∇v||∇v|∆1v+1 = 0, v|∂Ω = 0
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Motion by mean curvature
The Paul/Carol game in R2: • token initially at x0 ∈ R2
• given t0 > 0, the length of game N = b t0ε2c steps
• player I (Paul) chooses wk, |wk|= 1
• player II (Carol) chooses bk =±1
• token moved to xk+1 = xk+√
2ε bkwk
After N steps, Paul pays u0(xN) to Carol.
Game value:
uε(x0, t0) = minw1
maxb1
. . . minwN
maxbN
u0(xN)
uε(x0,0) = u0(x0)
Thm (Kohn-Serfaty 2006) Let u0 ∈ Cc(R2). Then uε ⇒ v in R2× [0,∞)
locally uniformly as ε→ 0, and v is the unique viscosity solution of:vt−|∇v|∆1v = 0 in R2× (0,∞)
v(x,0) = u0(x) in R2
• existence/uniqueness: Evans-Spruck, Chen-Giga-Goto 1991.31
Conclusions
• ∃ connection between: potential theory/nonlinear PDEs,and probability theory/ stochastic tug-of-war games
• based on: harmonic/p−harmonic functions and martingalesshare a common cancellation property, via mean value property
• variety of PDEs and other problems:– Harnack inequality (Luiro-Parviainen-Saksman)– continuous uε, finite difference scheme (Armstrong-Smart)– obstacle problem (Manfredi-Rossi-Somersille, L-Manfredi)– Neumann boundary condition (Antunovic-Peres-Sheffield-Somersille)
• in fact: ∃ game interpretation for a large class of elliptic and parabolic,nonlinear equations (Kohn-Serfaty 2010)
• advantage: direct proofs by assigning “strategies”⇒ “inequalities”.Probability tools used for nonlinear equations.
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THANK YOU.
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