Essential Uses of
Probability in Analysis
Part III. Robin problem
in fractal domains
Krzysztof Burdzy
University of Washington
Inspiration – physical phenomena
• physiology
• heterogeneous catalysis
• electrochemistry
Partial differential equation
B
⎧Δu ( x) = 0, x ∈ D \ B,
⎪ ∂u
⎪
⎨ ( x) = c u ( x), x ∈ ∂D,
⎪ ∂n
⎪⎩u ( x) = 1, x ∈ ∂B.
D
Robin boundary
condition
Remarks
(i) Robin boundary condition is not Robin’s
(Gustafson and Abe (1998))
u ( x) > 0, x ∈ D \ B
(iii) In the rest of the paper c = 1
(ii)
Robin condition:
∂u
( x) = u ( x)
∂n
Is the whole surface active?
Definition. We say that the whole surface
of D is active if
∂u
inf x∈∂D
( x) > 0
∂n
inf x∈D \ B u ( x) > 0
Problem. Give a geometric characterization
of domains whose whole surface is active.
Simple observations
If ∂D is smooth or even Lipschitz then
the whole surface is active.
If the area of the boundary of D
is infinite then it is not the case
that the whole surface is active.
Domains with infinite surfaces
Suppose that inf x∈D \ B u ( x) > 0 and | ∂D |= ∞
Then
inf x∈∂D
∂u
( x) > 0
∂n
and the flow outside the domain is
∂u
∫∂D ∂n ( x) dσ ( x) = ∞
The flow inside the domain through ∂B
is finite – a contradiction.
Example I
y=x
α
α >1
Is the whole surface active?
α ∈ (1,2) ⇒ yes
α ≥ 2 ⇒ no
Example II
2
2
2
2
− kα
− kβ
− kα
− kβ
β > 1, β > α
dim = 2 ⇒ α > 1
dim ≥ 3 ⇒ α > 0
2
Is the whole surface active?
β < 2α ⇒
β ≥ 2α ⇒
yes
no
− kα
2
− kβ
Example III
ρ
k
dim ≥ 3
ρ < 1/ 5
Is the whole
surface active?
ρ kβ
β >1
d = dim
Yes, if
β < (d − 1) /(d − 2)
Otherwise no.
Main result
D1
Dn
z
rn
Is the whole surface active?
n
sup ∑ | ∂Dn ∩ ∂D | ∑ rk2− d < ∞ ⇒ yes
z∈∂D n ≥1
∃z ∈ ∂D :
k =1
n
d −1
2− d
r
r
∑ n ∑ k =∞⇒
n ≥1
k =1
no
Green function estimate
D1
B
Dn
γn
x ∈γ k
y ∈γ m
rn
G ( x, y ) - Green function with Neumann
boundary conditions on ∂D and Dirichlet
boundary conditions on ∂B
k
k
c1 ∑ r j ≤ G ( x, y ) ≤ c2 ∑ r j
j =1
2− d
j =1
2− d
m>k
D1
B
γn
x ∈γ k
Dn
y ∈γ m
m>k
rn
G(x, y) - Green function with Neumann
boundary conditions on ∂D and Dirichlet
boundary conditions on ∂B
k
G(x, y) ≈ ∑r j
2−d
j =1
~
G ( x, y ) - Green function with Dirichlet
boundary conditions on ∂D ∪ ∂B
~
log G ( x, y ) ≈ f ( x, y )
Robin
Green function asymptotics
x
y→∞
• Neumann – linear growth of G ( x, y )
• Dirichlet, Robin – exponential decay
~
of G ( x, y )
Feynman-Kac formula
x
B
Brownian motion reflected on
at time
∂B
τ
L(t )
- Local time on
∂D
D
and killed on
∂D
u ( x) = E x exp(− L(τ ∂B ))
Ma and Song (1990), Papanicolau (1990)
∂B
u ( x) = E x exp(− L(τ ∂B ))
Problem: inf x∈D \ B u ( x) > 0 ?
u ( x) = 0 ?
u ( x) = 0 ⇔ P ( L(τ ∂B ) = ∞) = 1
E x L(τ ∂B ) < ∞ ⇒ Px ( L(τ ∂B ) = ∞) = 0
E x L(τ ∂B ) = ∞ ⇒
/ Px ( L(τ ∂B ) = ∞) = 1
x
L1
L2
L3
L4
L = L1 + L2 + K + Ln
The Law of Large Numbers
and the Central Limit Theorem are
applicable when L1 , L2 , L3 , K
are “nearly” independent.
In fact, Lk +1 = Lk (1 + o(1))
space
time
T
x
x
T
L
- amount of time spent at level x
before T
Ray-Knight Theorem
space
time
T
x
x
T
L - amount of time spent at level x
before T
x→L
x
T
is a diffusion with an explicit
distribution.
x
L1
L2
L3
L4
L = L1 + L2 + K + Ln
The joint distribution of L1 , L2 , L3 , K
is given (approximately) by the Ray-Knight
Theorem.
γ1
A
y
v x
Assume:
Prove:
γ2
Challenge
z
B
ω x (dz ) < ω y (dz ) ∀z ∈ γ 2
ω x (dv) > ω y (dv) ∀v ∈ γ 1
P (TA < TB ) > P (TA < TB )
x
y
References
• Bass, B and Chen (2005) “On the Robin problem in
fractal domains” (preprint)
• Gustafson and Abe (1998) “The third boundary
condition-was it Robin's?” Math. Intelligencer 20, 63-71.
• Ma and Song (1990) “Probabilistic methods in
Schrodinger equations” Seminar on Stochastic
Processes 1989, 135-164, Progr. Probab., 18,
Birkhauser, Boston.
• Papanicolaou (1990) “The probabilistic solution of the
third boundary value problem for second order elliptic
equations” Probab. Theory Rel. Fields 87 27-77.