Regularity of the free boundary in the obstacle problem
for the fractional Laplacian.
Begoña Barrios Barrera
University of Texas at Austin.
Universidad de La Laguna.
2 Julio 2015
B. Barrios.
(UT)
Regularity of free boundary.
1 / 24
1
Introduction.
2
Principal Result.
The key ingredient: Non degeneracy condition
A useful transformation
3
Sketch of the proof
B. Barrios.
(UT)
Regularity of free boundary.
2 / 24
Introduction.
Scheme
1
Introduction.
2
Principal Result.
The key ingredient: Non degeneracy condition
A useful transformation
3
Sketch of the proof
B. Barrios.
(UT)
Regularity of free boundary.
3 / 24
Introduction.
Our problem
Given a smooth function ϕ : Rn → R, the obstacle problem for the
fractional Laplacian can be written as
min u − ϕ, (−∆)s u = 0 in Rn ,
(1) =
lim|x|→∞ u(x) = 0,
where
ˆ
s
u(x)−u(x+z)
(−∆) u(x) = cn,s PV
Rn
dz
,
|z|n+2s
s ∈ (0, 1),
is the fractional Laplacian. FL
Some notations:
We will denote the free boundary as Γ(u) := ∂{u = ϕ} ⊂ Rn .
x0 ∈ Γ(u) is called singular if the contact set {u = ϕ} has zero density
at x0 , that is, if
{u = ϕ} ∩ Br (x0 )
lim
= 0.
r ↓0
|Br (x0 )|
B. Barrios.
(UT)
Regularity of free boundary.
4 / 24
Introduction.
Our problem
Given a smooth function ϕ : Rn → R, the obstacle problem for the
fractional Laplacian can be written as
min u − ϕ, (−∆)s u = 0 in Rn ,
(1) =
lim|x|→∞ u(x) = 0,
where
ˆ
s
u(x)−u(x+z)
(−∆) u(x) = cn,s PV
Rn
dz
,
|z|n+2s
s ∈ (0, 1),
is the fractional Laplacian. FL
Some notations:
We will denote the free boundary as Γ(u) := ∂{u = ϕ} ⊂ Rn .
x0 ∈ Γ(u) is called singular if the contact set {u = ϕ} has zero density
at x0 , that is, if
{u = ϕ} ∩ Br (x0 )
lim
= 0.
r ↓0
|Br (x0 )|
B. Barrios.
(UT)
Regularity of free boundary.
4 / 24
Introduction.
Lower dimensional obstacle problem
Using the extension method (e
u (x, 0) = u(x)) and doing an even reflection
e(x, −y )) the previous problem is equivalent to the following
(e
u (x, y ) = u
global lower dimensional obstacle problem:

e(x, 0) ≥
u



e =
La u
(1) =
e ≥
u
L

a


e(x, y ) →
u
ϕ(x)
0
0
0
in
in
in
as
Rn ,
lower obstacle problem
n+1
R
\ {(x, 0) : u(x) = ϕ(x)} ,
Rn+1 ,
|(x, y )| → ∞,
e := − div |y |a ∇x,y u
e ,
La u
x,y
a := 1 − 2s.
? s = 1/2 then La = −∆x,y so we recover the Signorini problem in Rn+1 .
The free boundary is not well understood in bounded domain Ω ⊆ Rn+1
symmetric with respect to the hyperplane {y = 0}.
B. Barrios.
(UT)
Regularity of free boundary.
5 / 24
Introduction.
Lower dimensional obstacle problem
Using the extension method (e
u (x, 0) = u(x)) and doing an even reflection
e(x, −y )) the previous problem is equivalent to the following
(e
u (x, y ) = u
global lower dimensional obstacle problem:

e(x, 0) ≥
u



e =
La u
(1) =
e ≥
u
L

a


e(x, y ) →
u
ϕ(x)
0
0
0
in
in
in
as
Rn ,
lower obstacle problem
n+1
R
\ {(x, 0) : u(x) = ϕ(x)} ,
Rn+1 ,
|(x, y )| → ∞,
e := − div |y |a ∇x,y u
e ,
La u
x,y
a := 1 − 2s.
? s = 1/2 then La = −∆x,y so we recover the Signorini problem in Rn+1 .
The free boundary is not well understood in bounded domain Ω ⊆ Rn+1
symmetric with respect to the hyperplane {y = 0}.
B. Barrios.
(UT)
Regularity of free boundary.
5 / 24
Introduction.
Lower dimensional obstacle problem
Using the extension method (e
u (x, 0) = u(x)) and doing an even reflection
e(x, −y )) the previous problem is equivalent to the following
(e
u (x, y ) = u
global lower dimensional obstacle problem:

e(x, 0) ≥
u



e =
La u
(1) =
e ≥
u
L

a


e(x, y ) →
u
ϕ(x)
0
0
0
in
in
in
as
Rn ,
lower obstacle problem
n+1
R
\ {(x, 0) : u(x) = ϕ(x)} ,
Rn+1 ,
|(x, y )| → ∞,
e := − div |y |a ∇x,y u
e ,
La u
x,y
a := 1 − 2s.
? s = 1/2 then La = −∆x,y so we recover the Signorini problem in Rn+1 .
The free boundary is not well understood in bounded domain Ω ⊆ Rn+1
symmetric with respect to the hyperplane {y = 0}.
B. Barrios.
(UT)
Regularity of free boundary.
5 / 24
Introduction.
Lower dimensional obstacle problem
Let be
Ω = B1 the unit ball in Rn+1 ,
B1 := B1 ∩ {y = 0},
an obstacle ϕ : B1 → R such that ϕ|∂B1 < 0 and
e : B1 → R
u
the solution of the following

e(x, 0) ≥
u



e =
La u
(2) =
e ≥
L
u

a


e =
u
local lower dimensional obstacle problem:
ϕ(x)
0
0
0
on B1 ∩ {y = 0}
in B1 \ {y = 0} ∩ {u = ϕ}
on B1
on ∂B1 ,
The main questions for both problems (1) and (2) are
? the regularity of u,
? the regularity of the free boundary.
B. Barrios.
(UT)
Regularity of free boundary.
6 / 24
Introduction.
Lower dimensional obstacle problem
Let be
Ω = B1 the unit ball in Rn+1 ,
B1 := B1 ∩ {y = 0},
an obstacle ϕ : B1 → R such that ϕ|∂B1 < 0 and
e : B1 → R
u
the solution of the following

e(x, 0) ≥
u



e =
La u
(2) =
e ≥
L
u

a


e =
u
local lower dimensional obstacle problem:
ϕ(x)
0
0
0
on B1 ∩ {y = 0}
in B1 \ {y = 0} ∩ {u = ϕ}
on B1
on ∂B1 ,
The main questions for both problems (1) and (2) are
? the regularity of u,
? the regularity of the free boundary.
B. Barrios.
(UT)
Regularity of free boundary.
6 / 24
Introduction.
Known results
?The classical case s = 1 ⇒ Well understood.
L. Caffarelli, The regularity of free boundaries in higher dimensions. Acta Math. 139 (1977), no. 3-4, 155-184.
L. Caffarelli, The obstacle problem revisited, J. Fourier Anal. Appl. 4 (1998), 383-402.
A. Petrosyan, H. Shahgholian, N. Uraltseva. Regularity of free boundaries in obstacle-type problems, volume 136 of
Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, (2012).
The main results establish that
The solution u ∈ C 1,1 .
Whenever ∆ϕ ≤ −c0 < 0 then,
1
2
3
4
the blow-up of u at any free boundary point is a unique homogeneous
polynomial of degree 2;
the free boundary splits into the union of a set of regular points and a
set of singular points;
the set of regular points is an open subset of the free boundary of class
C 1,α , α > 0;
singular points are either isolated or locally contained inside a C 1
submanifold.
B. Barrios.
(UT)
Regularity of free boundary.
7 / 24
Introduction.
Known results
?The classical case s = 1 ⇒ Well understood.
L. Caffarelli, The regularity of free boundaries in higher dimensions. Acta Math. 139 (1977), no. 3-4, 155-184.
L. Caffarelli, The obstacle problem revisited, J. Fourier Anal. Appl. 4 (1998), 383-402.
A. Petrosyan, H. Shahgholian, N. Uraltseva. Regularity of free boundaries in obstacle-type problems, volume 136 of
Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, (2012).
The main results establish that
The solution u ∈ C 1,1 .
Whenever ∆ϕ ≤ −c0 < 0 then,
1
2
3
4
the blow-up of u at any free boundary point is a unique homogeneous
polynomial of degree 2;
the free boundary splits into the union of a set of regular points and a
set of singular points;
the set of regular points is an open subset of the free boundary of class
C 1,α , α > 0;
singular points are either isolated or locally contained inside a C 1
submanifold.
B. Barrios.
(UT)
Regularity of free boundary.
7 / 24
Introduction.
Known results
? The fractional case s 6= 1.
Regularity of the solution
I. Athanasopoulos, L. Caffarelli, Optimal regularity of lower dimensional obstacle problems, Zap. Nauchn. Sem.
S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 310 (2004), 49-66.
For s = 1/2 and ϕ ≡ 0: u ∈ C 1,1/2 (optimal regularity).
L. Caffarelli, S. Salsa, L. Silvestre, Regularity estimates for the solution and the free boundary of the obstacle problem for
the fractional Laplacian, Invent. Math. 171 (2008), 425-461.
L. Silvestre, Regularity of the obstacle problem for a fractional power of the Laplace operator, Comm. Pure Appl. Math.
60 (2007), 67-112.
For 0 < s < 1 and ϕ 6= 0: u ∈ C 1,s (optimal regularity).
Study of the free boundary
I. Athanasopoulos, L. Caffarelli, S. Salsa, The structure of the free boundary for lower dimensional obstacle problems,
Amer. J. Math. 130 (2008) 485-498.
For s = 1/2 and ϕ ≡ 0: the set of regular points is an open subset of the free
boundary of class C 1,α , α > 0.
L. Caffarelli, S. Salsa, L. Silvestre, Regularity estimates for the solution and the free boundary of the obstacle problem for
the fractional Laplacian, Invent. Math. 171 (2008), 425-461.
For 0 < s < 1 and ϕ 6= 0: the set of regular points is an open subset of the
free boundary of class C 1,α , α > 0.
B. Barrios.
(UT)
Regularity of free boundary.
8 / 24
Introduction.
Known results
? The fractional case s 6= 1.
Regularity of the solution
I. Athanasopoulos, L. Caffarelli, Optimal regularity of lower dimensional obstacle problems, Zap. Nauchn. Sem.
S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 310 (2004), 49-66.
For s = 1/2 and ϕ ≡ 0: u ∈ C 1,1/2 (optimal regularity).
L. Caffarelli, S. Salsa, L. Silvestre, Regularity estimates for the solution and the free boundary of the obstacle problem for
the fractional Laplacian, Invent. Math. 171 (2008), 425-461.
L. Silvestre, Regularity of the obstacle problem for a fractional power of the Laplace operator, Comm. Pure Appl. Math.
60 (2007), 67-112.
For 0 < s < 1 and ϕ 6= 0: u ∈ C 1,s (optimal regularity).
Study of the free boundary
I. Athanasopoulos, L. Caffarelli, S. Salsa, The structure of the free boundary for lower dimensional obstacle problems,
Amer. J. Math. 130 (2008) 485-498.
For s = 1/2 and ϕ ≡ 0: the set of regular points is an open subset of the free
boundary of class C 1,α , α > 0.
L. Caffarelli, S. Salsa, L. Silvestre, Regularity estimates for the solution and the free boundary of the obstacle problem for
the fractional Laplacian, Invent. Math. 171 (2008), 425-461.
For 0 < s < 1 and ϕ 6= 0: the set of regular points is an open subset of the
free boundary of class C 1,α , α > 0.
B. Barrios.
(UT)
Regularity of free boundary.
8 / 24
Introduction.
Known results
? The fractional case s 6= 1.
Regularity of the solution
I. Athanasopoulos, L. Caffarelli, Optimal regularity of lower dimensional obstacle problems, Zap. Nauchn. Sem.
S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 310 (2004), 49-66.
For s = 1/2 and ϕ ≡ 0: u ∈ C 1,1/2 (optimal regularity).
L. Caffarelli, S. Salsa, L. Silvestre, Regularity estimates for the solution and the free boundary of the obstacle problem for
the fractional Laplacian, Invent. Math. 171 (2008), 425-461.
L. Silvestre, Regularity of the obstacle problem for a fractional power of the Laplace operator, Comm. Pure Appl. Math.
60 (2007), 67-112.
For 0 < s < 1 and ϕ 6= 0: u ∈ C 1,s (optimal regularity).
Study of the free boundary
I. Athanasopoulos, L. Caffarelli, S. Salsa, The structure of the free boundary for lower dimensional obstacle problems,
Amer. J. Math. 130 (2008) 485-498.
For s = 1/2 and ϕ ≡ 0: the set of regular points is an open subset of the free
boundary of class C 1,α , α > 0.
L. Caffarelli, S. Salsa, L. Silvestre, Regularity estimates for the solution and the free boundary of the obstacle problem for
the fractional Laplacian, Invent. Math. 171 (2008), 425-461.
For 0 < s < 1 and ϕ 6= 0: the set of regular points is an open subset of the
free boundary of class C 1,α , α > 0.
B. Barrios.
(UT)
Regularity of free boundary.
8 / 24
Introduction.
Known results
? The fractional case s 6= 1.
Regularity of the solution
I. Athanasopoulos, L. Caffarelli, Optimal regularity of lower dimensional obstacle problems, Zap. Nauchn. Sem.
S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 310 (2004), 49-66.
For s = 1/2 and ϕ ≡ 0: u ∈ C 1,1/2 (optimal regularity).
L. Caffarelli, S. Salsa, L. Silvestre, Regularity estimates for the solution and the free boundary of the obstacle problem for
the fractional Laplacian, Invent. Math. 171 (2008), 425-461.
L. Silvestre, Regularity of the obstacle problem for a fractional power of the Laplace operator, Comm. Pure Appl. Math.
60 (2007), 67-112.
For 0 < s < 1 and ϕ 6= 0: u ∈ C 1,s (optimal regularity).
Study of the free boundary
I. Athanasopoulos, L. Caffarelli, S. Salsa, The structure of the free boundary for lower dimensional obstacle problems,
Amer. J. Math. 130 (2008) 485-498.
For s = 1/2 and ϕ ≡ 0: the set of regular points is an open subset of the free
boundary of class C 1,α , α > 0.
L. Caffarelli, S. Salsa, L. Silvestre, Regularity estimates for the solution and the free boundary of the obstacle problem for
the fractional Laplacian, Invent. Math. 171 (2008), 425-461.
For 0 < s < 1 and ϕ 6= 0: the set of regular points is an open subset of the
free boundary of class C 1,α , α > 0.
B. Barrios.
(UT)
Regularity of free boundary.
8 / 24
Introduction.
Known results
N. Garofalo, A. Petrosyan, Some new monotonicity formulas and the singular set in the lower dimensional obstacle
problem, Invent. Math. 177 (2009), 415-461.
For s = 21 proved a stratification results for singular points in terms of
the homogeneity of the blow-ups of u.
Therefore...
Even for s = 1/2 the full structure of the free boundary was far from being
understood:
? Whether the free boundary had Hausdorff dimension n − 1.
? The definitions of regular and singular points do not exhaust all
possible free boundary points.
? A priori the union of regular and singular points may consists of a
very small fraction of the whole free boundary.
B. Barrios.
(UT)
Regularity of free boundary.
9 / 24
Introduction.
Known results
N. Garofalo, A. Petrosyan, Some new monotonicity formulas and the singular set in the lower dimensional obstacle
problem, Invent. Math. 177 (2009), 415-461.
For s = 21 proved a stratification results for singular points in terms of
the homogeneity of the blow-ups of u.
Therefore...
Even for s = 1/2 the full structure of the free boundary was far from being
understood:
? Whether the free boundary had Hausdorff dimension n − 1.
? The definitions of regular and singular points do not exhaust all
possible free boundary points.
? A priori the union of regular and singular points may consists of a
very small fraction of the whole free boundary.
B. Barrios.
(UT)
Regularity of free boundary.
9 / 24
Principal Result.
Scheme
1
Introduction.
2
Principal Result.
The key ingredient: Non degeneracy condition
A useful transformation
3
Sketch of the proof
B. Barrios.
(UT)
Regularity of free boundary.
10 / 24
Principal Result.
The result.
Theorem: [B. B, A. Figalli, X. Ros-Oton]: Let u be:
(H1 ) either a solution to (1) with ϕ : Rn → R satisfying ϕ ∈ C 3,γ (Rn ),
∆ϕ ≤ 0 in {ϕ > 0},
∅=
6 {ϕ > 0} ⊂⊂ Rn , for some γ > 0.
(H2 ) or a solution to (2) with ϕ : B1 → R satisfying ϕ ∈ C 3,γ (B1 ),
∆ϕ ≤ −c0 < 0 in {ϕ > 0},
∅=
6 {ϕ > 0} ⊂⊂ B1 , for some c0 , γ > 0.
Then,
Γ(u) = Γ1+s (u) ∪ Γ2 (u), where Γ1+s (u) (resp. Γ2 (u)) is a open (resp.
closed) subset of Γ(u). Moreover
Γ1+s (u) is a (n − 1)-dimensional manifold of class C 1,α .
k
k
Γ2 (u) = ∪n−1
k=0 Γ2 (u), where Γ2 (u) is contained in a k-dimensional
1
manifold of class C .
At every singular point, the blow-up of u is a homogeneous polynomial of
degree 2.
B. Barrios.
(UT)
Regularity of free boundary.
11 / 24
Principal Result.
The result.
Theorem: [B. B, A. Figalli, X. Ros-Oton]: Let u be:
(H1 ) either a solution to (1) with ϕ : Rn → R satisfying ϕ ∈ C 3,γ (Rn ),
∆ϕ ≤ 0 in {ϕ > 0},
∅=
6 {ϕ > 0} ⊂⊂ Rn , for some γ > 0.
(H2 ) or a solution to (2) with ϕ : B1 → R satisfying ϕ ∈ C 3,γ (B1 ),
∆ϕ ≤ −c0 < 0 in {ϕ > 0},
∅=
6 {ϕ > 0} ⊂⊂ B1 , for some c0 , γ > 0.
Then,
Γ(u) = Γ1+s (u) ∪ Γ2 (u), where Γ1+s (u) (resp. Γ2 (u)) is a open (resp.
closed) subset of Γ(u). Moreover
Γ1+s (u) is a (n − 1)-dimensional manifold of class C 1,α .
k
k
Γ2 (u) = ∪n−1
k=0 Γ2 (u), where Γ2 (u) is contained in a k-dimensional
1
manifold of class C .
At every singular point, the blow-up of u is a homogeneous polynomial of
degree 2.
B. Barrios.
(UT)
Regularity of free boundary.
11 / 24
Principal Result.
The result.
Theorem: [B. B, A. Figalli, X. Ros-Oton]: Let u be:
(H1 ) either a solution to (1) with ϕ : Rn → R satisfying ϕ ∈ C 3,γ (Rn ),
∆ϕ ≤ 0 in {ϕ > 0},
∅=
6 {ϕ > 0} ⊂⊂ Rn , for some γ > 0.
(H2 ) or a solution to (2) with ϕ : B1 → R satisfying ϕ ∈ C 3,γ (B1 ),
∆ϕ ≤ −c0 < 0 in {ϕ > 0},
∅=
6 {ϕ > 0} ⊂⊂ B1 , for some c0 , γ > 0.
Then,
Γ(u) = Γ1+s (u) ∪ Γ2 (u), where Γ1+s (u) (resp. Γ2 (u)) is a open (resp.
closed) subset of Γ(u). Moreover
Γ1+s (u) is a (n − 1)-dimensional manifold of class C 1,α .
k
k
Γ2 (u) = ∪n−1
k=0 Γ2 (u), where Γ2 (u) is contained in a k-dimensional
1
manifold of class C .
At every singular point, the blow-up of u is a homogeneous polynomial of
degree 2.
B. Barrios.
(UT)
Regularity of free boundary.
11 / 24
Principal Result.
Our
The result.
objective is to show that, for (1) and (2), if ∆ϕ ≤ −c0 < 0 then:
? Regular and singular points do exhaust all possible free boundary points.
? Singular points are either isolated or locally contained inside a C 1
submanifold.
That is,
we obtain the same structure result on the free boundary
as in the case s = 1.
B. Barrios.
(UT)
Regularity of free boundary.
12 / 24
Principal Result.
Our
The result.
objective is to show that, for (1) and (2), if ∆ϕ ≤ −c0 < 0 then:
? Regular and singular points do exhaust all possible free boundary points.
? Singular points are either isolated or locally contained inside a C 1
submanifold.
That is,
we obtain the same structure result on the free boundary
as in the case s = 1.
B. Barrios.
(UT)
Regularity of free boundary.
12 / 24
Principal Result.
The key ingredient: Non degeneracy condition
A key ingredient in the proof of the main Theorem is the non-degeneracy
condition:
Lemma
Let u solve the obstacle problem (1) with ϕ satisfying (H1 ). Then there
exist constants c1 , r1 > 0 such that the following holds: for any x0 ∈ Γ(u)
we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
Let be 0 ≤ w := (−∆)s u 6= 0 ⇒ (−∆)u = (−∆)1−s w < 0, {u > ϕ}.
Let be u1 := u − ϕ ⇒ ∆u1 ≥ C > 0 in Λ ∩ Br (x1 ), Λ := {u1 > 0}.
That is u2 (x) := u1 (x) − C /2n|x − x1 |2 is sub-harmonic in Λ ∩ Br (x1 ).
Then 0 < u1 (x1 ) ≤ supΛ∩Br (x1 ) u2 = sup∂(Λ∩Br (x1 )) u2 .
Moreover, since, u2 < 0 on ∂Λ ∩ Br (x1 ) we deduce that
C
0 < sup u2 ≤ sup u1 − c1 r 2 ,
c1 :=
.
2n
Λ∩∂Br (x1 )
∂Br (x1 )
The result follows by letting x1 → x0 .
B. Barrios.
(UT)
Regularity of free boundary.
13 / 24
Principal Result.
The key ingredient: Non degeneracy condition
A key ingredient in the proof of the main Theorem is the non-degeneracy
condition:
Lemma
Let u solve the obstacle problem (1) with ϕ satisfying (H1 ). Then there
exist constants c1 , r1 > 0 such that the following holds: for any x0 ∈ Γ(u)
we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
Let be 0 ≤ w := (−∆)s u 6= 0 ⇒ (−∆)u = (−∆)1−s w < 0, {u > ϕ}.
Let be u1 := u − ϕ ⇒ ∆u1 ≥ C > 0 in Λ ∩ Br (x1 ), Λ := {u1 > 0}.
That is u2 (x) := u1 (x) − C /2n|x − x1 |2 is sub-harmonic in Λ ∩ Br (x1 ).
Then 0 < u1 (x1 ) ≤ supΛ∩Br (x1 ) u2 = sup∂(Λ∩Br (x1 )) u2 .
Moreover, since, u2 < 0 on ∂Λ ∩ Br (x1 ) we deduce that
C
0 < sup u2 ≤ sup u1 − c1 r 2 ,
c1 :=
.
2n
Λ∩∂Br (x1 )
∂Br (x1 )
The result follows by letting x1 → x0 .
B. Barrios.
(UT)
Regularity of free boundary.
13 / 24
Principal Result.
The key ingredient: Non degeneracy condition
A key ingredient in the proof of the main Theorem is the non-degeneracy
condition:
Lemma
Let u solve the obstacle problem (1) with ϕ satisfying (H1 ). Then there
exist constants c1 , r1 > 0 such that the following holds: for any x0 ∈ Γ(u)
we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
Let be 0 ≤ w := (−∆)s u 6= 0 ⇒ (−∆)u = (−∆)1−s w < 0, {u > ϕ}.
Let be u1 := u − ϕ ⇒ ∆u1 ≥ C > 0 in Λ ∩ Br (x1 ), Λ := {u1 > 0}.
That is u2 (x) := u1 (x) − C /2n|x − x1 |2 is sub-harmonic in Λ ∩ Br (x1 ).
Then 0 < u1 (x1 ) ≤ supΛ∩Br (x1 ) u2 = sup∂(Λ∩Br (x1 )) u2 .
Moreover, since, u2 < 0 on ∂Λ ∩ Br (x1 ) we deduce that
C
0 < sup u2 ≤ sup u1 − c1 r 2 ,
c1 :=
.
2n
Λ∩∂Br (x1 )
∂Br (x1 )
The result follows by letting x1 → x0 .
B. Barrios.
(UT)
Regularity of free boundary.
13 / 24
Principal Result.
The key ingredient: Non degeneracy condition
A key ingredient in the proof of the main Theorem is the non-degeneracy
condition:
Lemma
Let u solve the obstacle problem (1) with ϕ satisfying (H1 ). Then there
exist constants c1 , r1 > 0 such that the following holds: for any x0 ∈ Γ(u)
we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
Let be 0 ≤ w := (−∆)s u 6= 0 ⇒ (−∆)u = (−∆)1−s w < 0, {u > ϕ}.
Let be u1 := u − ϕ ⇒ ∆u1 ≥ C > 0 in Λ ∩ Br (x1 ), Λ := {u1 > 0}.
That is u2 (x) := u1 (x) − C /2n|x − x1 |2 is sub-harmonic in Λ ∩ Br (x1 ).
Then 0 < u1 (x1 ) ≤ supΛ∩Br (x1 ) u2 = sup∂(Λ∩Br (x1 )) u2 .
Moreover, since, u2 < 0 on ∂Λ ∩ Br (x1 ) we deduce that
C
0 < sup u2 ≤ sup u1 − c1 r 2 ,
c1 :=
.
2n
Λ∩∂Br (x1 )
∂Br (x1 )
The result follows by letting x1 → x0 .
B. Barrios.
(UT)
Regularity of free boundary.
13 / 24
Principal Result.
The key ingredient: Non degeneracy condition
A key ingredient in the proof of the main Theorem is the non-degeneracy
condition:
Lemma
Let u solve the obstacle problem (1) with ϕ satisfying (H1 ). Then there
exist constants c1 , r1 > 0 such that the following holds: for any x0 ∈ Γ(u)
we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
Let be 0 ≤ w := (−∆)s u 6= 0 ⇒ (−∆)u = (−∆)1−s w < 0, {u > ϕ}.
Let be u1 := u − ϕ ⇒ ∆u1 ≥ C > 0 in Λ ∩ Br (x1 ), Λ := {u1 > 0}.
That is u2 (x) := u1 (x) − C /2n|x − x1 |2 is sub-harmonic in Λ ∩ Br (x1 ).
Then 0 < u1 (x1 ) ≤ supΛ∩Br (x1 ) u2 = sup∂(Λ∩Br (x1 )) u2 .
Moreover, since, u2 < 0 on ∂Λ ∩ Br (x1 ) we deduce that
C
0 < sup u2 ≤ sup u1 − c1 r 2 ,
c1 :=
.
2n
Λ∩∂Br (x1 )
∂Br (x1 )
The result follows by letting x1 → x0 .
B. Barrios.
(UT)
Regularity of free boundary.
13 / 24
Principal Result.
The key ingredient: Non degeneracy condition
A key ingredient in the proof of the main Theorem is the non-degeneracy
condition:
Lemma
Let u solve the obstacle problem (1) with ϕ satisfying (H1 ). Then there
exist constants c1 , r1 > 0 such that the following holds: for any x0 ∈ Γ(u)
we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
Let be 0 ≤ w := (−∆)s u 6= 0 ⇒ (−∆)u = (−∆)1−s w < 0, {u > ϕ}.
Let be u1 := u − ϕ ⇒ ∆u1 ≥ C > 0 in Λ ∩ Br (x1 ), Λ := {u1 > 0}.
That is u2 (x) := u1 (x) − C /2n|x − x1 |2 is sub-harmonic in Λ ∩ Br (x1 ).
Then 0 < u1 (x1 ) ≤ supΛ∩Br (x1 ) u2 = sup∂(Λ∩Br (x1 )) u2 .
Moreover, since, u2 < 0 on ∂Λ ∩ Br (x1 ) we deduce that
C
0 < sup u2 ≤ sup u1 − c1 r 2 ,
c1 :=
.
2n
Λ∩∂Br (x1 )
∂Br (x1 )
The result follows by letting x1 → x0 .
B. Barrios.
(UT)
Regularity of free boundary.
13 / 24
Principal Result.
The key ingredient: Non degeneracy condition
A key ingredient in the proof of the main Theorem is the non-degeneracy
condition:
Lemma
Let u solve the obstacle problem (1) with ϕ satisfying (H1 ). Then there
exist constants c1 , r1 > 0 such that the following holds: for any x0 ∈ Γ(u)
we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
Let be 0 ≤ w := (−∆)s u 6= 0 ⇒ (−∆)u = (−∆)1−s w < 0, {u > ϕ}.
Let be u1 := u − ϕ ⇒ ∆u1 ≥ C > 0 in Λ ∩ Br (x1 ), Λ := {u1 > 0}.
That is u2 (x) := u1 (x) − C /2n|x − x1 |2 is sub-harmonic in Λ ∩ Br (x1 ).
Then 0 < u1 (x1 ) ≤ supΛ∩Br (x1 ) u2 = sup∂(Λ∩Br (x1 )) u2 .
Moreover, since, u2 < 0 on ∂Λ ∩ Br (x1 ) we deduce that
C
0 < sup u2 ≤ sup u1 − c1 r 2 ,
c1 :=
.
2n
Λ∩∂Br (x1 )
∂Br (x1 )
The result follows by letting x1 → x0 .
B. Barrios.
(UT)
Regularity of free boundary.
13 / 24
Principal Result.
The key ingredient: Non degeneracy condition
A key ingredient in the proof of the main Theorem is the non-degeneracy
condition:
Lemma
Let u solve the obstacle problem (1) with ϕ satisfying (H1 ). Then there
exist constants c1 , r1 > 0 such that the following holds: for any x0 ∈ Γ(u)
we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
Let be 0 ≤ w := (−∆)s u 6= 0 ⇒ (−∆)u = (−∆)1−s w < 0, {u > ϕ}.
Let be u1 := u − ϕ ⇒ ∆u1 ≥ C > 0 in Λ ∩ Br (x1 ), Λ := {u1 > 0}.
That is u2 (x) := u1 (x) − C /2n|x − x1 |2 is sub-harmonic in Λ ∩ Br (x1 ).
Then 0 < u1 (x1 ) ≤ supΛ∩Br (x1 ) u2 = sup∂(Λ∩Br (x1 )) u2 .
Moreover, since, u2 < 0 on ∂Λ ∩ Br (x1 ) we deduce that
C
0 < sup u2 ≤ sup u1 − c1 r 2 ,
c1 :=
.
2n
Λ∩∂Br (x1 )
∂Br (x1 )
The result follows by letting x1 → x0 .
B. Barrios.
(UT)
Regularity of free boundary.
13 / 24
Principal Result.
The key ingredient: Non degeneracy condition
Lemma
Let u solve the obstacle problem (2) with ϕ satisfying (H2 ) with constant
c0 . Then there exist constants c1 , r1 > 0 such that the following holds: for
any x0 ∈ Γ(u) we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
c0
e(x, y ) − ϕ(x) − 2n+2(1+a)
Let be w (x, y ) := u
|x − x1 |2 + y 2 , where
u(x1 ) > ϕ(x1 ) > 0. Let us take r > 0 small enough so that ϕ > 0 in
Br (x1 ). Therefore,
e(x, y ) − |y |a ∆ϕ(x) + c0 ≥ 0 in U,
−La w (x, y ) = −La u
with U := Br (x1 , 0) \ {y = 0} ∩ {u = ϕ} ⊂⊂ B1 .
Then
0 < w (x1 , 0) ≤ supU w = sup∂U w < sup∂Br (x1 ,0) w ≤ supBr (x1 ,0) w .
c0
Letting x1 → x0 we get supBr (x0 ,0) (e
u − ϕ) ≥ 2n+2(1+a)
r 2.
Y This supremum is attained on {y = 0}.
B. Barrios.
(UT)
Regularity of free boundary.
14 / 24
Principal Result.
The key ingredient: Non degeneracy condition
Lemma
Let u solve the obstacle problem (2) with ϕ satisfying (H2 ) with constant
c0 . Then there exist constants c1 , r1 > 0 such that the following holds: for
any x0 ∈ Γ(u) we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
c0
e(x, y ) − ϕ(x) − 2n+2(1+a)
Let be w (x, y ) := u
|x − x1 |2 + y 2 , where
u(x1 ) > ϕ(x1 ) > 0. Let us take r > 0 small enough so that ϕ > 0 in
Br (x1 ). Therefore,
e(x, y ) − |y |a ∆ϕ(x) + c0 ≥ 0 in U,
−La w (x, y ) = −La u
with U := Br (x1 , 0) \ {y = 0} ∩ {u = ϕ} ⊂⊂ B1 .
Then
0 < w (x1 , 0) ≤ supU w = sup∂U w < sup∂Br (x1 ,0) w ≤ supBr (x1 ,0) w .
c0
Letting x1 → x0 we get supBr (x0 ,0) (e
u − ϕ) ≥ 2n+2(1+a)
r 2.
Y This supremum is attained on {y = 0}.
B. Barrios.
(UT)
Regularity of free boundary.
14 / 24
Principal Result.
The key ingredient: Non degeneracy condition
Lemma
Let u solve the obstacle problem (2) with ϕ satisfying (H2 ) with constant
c0 . Then there exist constants c1 , r1 > 0 such that the following holds: for
any x0 ∈ Γ(u) we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
c0
e(x, y ) − ϕ(x) − 2n+2(1+a)
Let be w (x, y ) := u
|x − x1 |2 + y 2 , where
u(x1 ) > ϕ(x1 ) > 0. Let us take r > 0 small enough so that ϕ > 0 in
Br (x1 ). Therefore,
e(x, y ) − |y |a ∆ϕ(x) + c0 ≥ 0 in U,
−La w (x, y ) = −La u
with U := Br (x1 , 0) \ {y = 0} ∩ {u = ϕ} ⊂⊂ B1 .
Then
0 < w (x1 , 0) ≤ supU w = sup∂U w < sup∂Br (x1 ,0) w ≤ supBr (x1 ,0) w .
c0
Letting x1 → x0 we get supBr (x0 ,0) (e
u − ϕ) ≥ 2n+2(1+a)
r 2.
Y This supremum is attained on {y = 0}.
B. Barrios.
(UT)
Regularity of free boundary.
14 / 24
Principal Result.
The key ingredient: Non degeneracy condition
Lemma
Let u solve the obstacle problem (2) with ϕ satisfying (H2 ) with constant
c0 . Then there exist constants c1 , r1 > 0 such that the following holds: for
any x0 ∈ Γ(u) we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
c0
e(x, y ) − ϕ(x) − 2n+2(1+a)
Let be w (x, y ) := u
|x − x1 |2 + y 2 , where
u(x1 ) > ϕ(x1 ) > 0. Let us take r > 0 small enough so that ϕ > 0 in
Br (x1 ). Therefore,
e(x, y ) − |y |a ∆ϕ(x) + c0 ≥ 0 in U,
−La w (x, y ) = −La u
with U := Br (x1 , 0) \ {y = 0} ∩ {u = ϕ} ⊂⊂ B1 .
Then
0 < w (x1 , 0) ≤ supU w = sup∂U w < sup∂Br (x1 ,0) w ≤ supBr (x1 ,0) w .
c0
Letting x1 → x0 we get supBr (x0 ,0) (e
u − ϕ) ≥ 2n+2(1+a)
r 2.
Y This supremum is attained on {y = 0}.
B. Barrios.
(UT)
Regularity of free boundary.
14 / 24
Principal Result.
The key ingredient: Non degeneracy condition
Lemma
Let u solve the obstacle problem (2) with ϕ satisfying (H2 ) with constant
c0 . Then there exist constants c1 , r1 > 0 such that the following holds: for
any x0 ∈ Γ(u) we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
c0
e(x, y ) − ϕ(x) − 2n+2(1+a)
Let be w (x, y ) := u
|x − x1 |2 + y 2 , where
u(x1 ) > ϕ(x1 ) > 0. Let us take r > 0 small enough so that ϕ > 0 in
Br (x1 ). Therefore,
e(x, y ) − |y |a ∆ϕ(x) + c0 ≥ 0 in U,
−La w (x, y ) = −La u
with U := Br (x1 , 0) \ {y = 0} ∩ {u = ϕ} ⊂⊂ B1 .
Then
0 < w (x1 , 0) ≤ supU w = sup∂U w < sup∂Br (x1 ,0) w ≤ supBr (x1 ,0) w .
c0
Letting x1 → x0 we get supBr (x0 ,0) (e
u − ϕ) ≥ 2n+2(1+a)
r 2.
Y This supremum is attained on {y = 0}.
B. Barrios.
(UT)
Regularity of free boundary.
14 / 24
Principal Result.
The key ingredient: Non degeneracy condition
Lemma
Let u solve the obstacle problem (2) with ϕ satisfying (H2 ) with constant
c0 . Then there exist constants c1 , r1 > 0 such that the following holds: for
any x0 ∈ Γ(u) we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
c0
e(x, y ) − ϕ(x) − 2n+2(1+a)
Let be w (x, y ) := u
|x − x1 |2 + y 2 , where
u(x1 ) > ϕ(x1 ) > 0. Let us take r > 0 small enough so that ϕ > 0 in
Br (x1 ). Therefore,
e(x, y ) − |y |a ∆ϕ(x) + c0 ≥ 0 in U,
−La w (x, y ) = −La u
with U := Br (x1 , 0) \ {y = 0} ∩ {u = ϕ} ⊂⊂ B1 .
Then
0 < w (x1 , 0) ≤ supU w = sup∂U w < sup∂Br (x1 ,0) w ≤ supBr (x1 ,0) w .
c0
Letting x1 → x0 we get supBr (x0 ,0) (e
u − ϕ) ≥ 2n+2(1+a)
r 2.
Y This supremum is attained on {y = 0}.
B. Barrios.
(UT)
Regularity of free boundary.
14 / 24
Principal Result.
The key ingredient: Non degeneracy condition
Lemma
Let u solve the obstacle problem (2) with ϕ satisfying (H2 ) with constant
c0 . Then there exist constants c1 , r1 > 0 such that the following holds: for
any x0 ∈ Γ(u) we have
sup (u − ϕ) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Br (x0 )
Sketch of the proof:
c0
e(x, y ) − ϕ(x) − 2n+2(1+a)
Let be w (x, y ) := u
|x − x1 |2 + y 2 , where
u(x1 ) > ϕ(x1 ) > 0. Let us take r > 0 small enough so that ϕ > 0 in
Br (x1 ). Therefore,
e(x, y ) − |y |a ∆ϕ(x) + c0 ≥ 0 in U,
−La w (x, y ) = −La u
with U := Br (x1 , 0) \ {y = 0} ∩ {u = ϕ} ⊂⊂ B1 .
Then
0 < w (x1 , 0) ≤ supU w = sup∂U w < sup∂Br (x1 ,0) w ≤ supBr (x1 ,0) w .
c0
Letting x1 → x0 we get supBr (x0 ,0) (e
u − ϕ) ≥ 2n+2(1+a)
r 2.
Y This supremum is attained on {y = 0}.
B. Barrios.
(UT)
Regularity of free boundary.
14 / 24
Principal Result.
A useful transformation
A useful transformation:
e a solution of (1) or (2) and x0 ∈ Γ(u). We define
Let be u
e(x, y )−ϕ(x)+
v x0 (x, y ) := u
1
∆ϕ(x0 ) y 2 + (∇∆ϕ)(x0 ) · (x − x0 ) y 2
2(1 + a)
that satisfy
? v x0 (x, 0) = u(x) − ϕ(x) ≥ 0, x ∈ Rn .
?|La v x0 (x, y )| ≤ C |y |a |x − x0 |1+γ , (x, y ) ∈ Rn+1 \ {(x, 0) : v x0 (x, 0) = 0} .
? v x0 depends continuously on x0 .
We get that
supBr (x0 ) v x0 (x, 0) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Moreover (using a weak Harnack inequality)
ˆ
|y |a |v x0 (x, y )|2 dx dy ≥ c2 r n+a+5
∀ r ∈ (0, r2 ).
Br (x0 ,0)
B. Barrios.
(UT)
Regularity of free boundary.
15 / 24
Principal Result.
A useful transformation
A useful transformation:
e a solution of (1) or (2) and x0 ∈ Γ(u). We define
Let be u
e(x, y )−ϕ(x)+
v x0 (x, y ) := u
1
∆ϕ(x0 ) y 2 + (∇∆ϕ)(x0 ) · (x − x0 ) y 2
2(1 + a)
that satisfy
? v x0 (x, 0) = u(x) − ϕ(x) ≥ 0, x ∈ Rn .
?|La v x0 (x, y )| ≤ C |y |a |x − x0 |1+γ , (x, y ) ∈ Rn+1 \ {(x, 0) : v x0 (x, 0) = 0} .
? v x0 depends continuously on x0 .
We get that
supBr (x0 ) v x0 (x, 0) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Moreover (using a weak Harnack inequality)
ˆ
|y |a |v x0 (x, y )|2 dx dy ≥ c2 r n+a+5
∀ r ∈ (0, r2 ).
Br (x0 ,0)
B. Barrios.
(UT)
Regularity of free boundary.
15 / 24
Principal Result.
A useful transformation
A useful transformation:
e a solution of (1) or (2) and x0 ∈ Γ(u). We define
Let be u
e(x, y )−ϕ(x)+
v x0 (x, y ) := u
1
∆ϕ(x0 ) y 2 + (∇∆ϕ)(x0 ) · (x − x0 ) y 2
2(1 + a)
that satisfy
? v x0 (x, 0) = u(x) − ϕ(x) ≥ 0, x ∈ Rn .
?|La v x0 (x, y )| ≤ C |y |a |x − x0 |1+γ , (x, y ) ∈ Rn+1 \ {(x, 0) : v x0 (x, 0) = 0} .
? v x0 depends continuously on x0 .
We get that
supBr (x0 ) v x0 (x, 0) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Moreover (using a weak Harnack inequality)
ˆ
|y |a |v x0 (x, y )|2 dx dy ≥ c2 r n+a+5
∀ r ∈ (0, r2 ).
Br (x0 ,0)
B. Barrios.
(UT)
Regularity of free boundary.
15 / 24
Principal Result.
A useful transformation
A useful transformation:
e a solution of (1) or (2) and x0 ∈ Γ(u). We define
Let be u
e(x, y )−ϕ(x)+
v x0 (x, y ) := u
1
∆ϕ(x0 ) y 2 + (∇∆ϕ)(x0 ) · (x − x0 ) y 2
2(1 + a)
that satisfy
? v x0 (x, 0) = u(x) − ϕ(x) ≥ 0, x ∈ Rn .
?|La v x0 (x, y )| ≤ C |y |a |x − x0 |1+γ , (x, y ) ∈ Rn+1 \ {(x, 0) : v x0 (x, 0) = 0} .
? v x0 depends continuously on x0 .
We get that
supBr (x0 ) v x0 (x, 0) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Moreover (using a weak Harnack inequality)
ˆ
|y |a |v x0 (x, y )|2 dx dy ≥ c2 r n+a+5
∀ r ∈ (0, r2 ).
Br (x0 ,0)
B. Barrios.
(UT)
Regularity of free boundary.
15 / 24
Principal Result.
A useful transformation
A useful transformation:
e a solution of (1) or (2) and x0 ∈ Γ(u). We define
Let be u
e(x, y )−ϕ(x)+
v x0 (x, y ) := u
1
∆ϕ(x0 ) y 2 + (∇∆ϕ)(x0 ) · (x − x0 ) y 2
2(1 + a)
that satisfy
? v x0 (x, 0) = u(x) − ϕ(x) ≥ 0, x ∈ Rn .
?|La v x0 (x, y )| ≤ C |y |a |x − x0 |1+γ , (x, y ) ∈ Rn+1 \ {(x, 0) : v x0 (x, 0) = 0} .
? v x0 depends continuously on x0 .
We get that
supBr (x0 ) v x0 (x, 0) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Moreover (using a weak Harnack inequality)
ˆ
|y |a |v x0 (x, y )|2 dx dy ≥ c2 r n+a+5
∀ r ∈ (0, r2 ).
Br (x0 ,0)
B. Barrios.
(UT)
Regularity of free boundary.
15 / 24
Principal Result.
A useful transformation
A useful transformation:
e a solution of (1) or (2) and x0 ∈ Γ(u). We define
Let be u
e(x, y )−ϕ(x)+
v x0 (x, y ) := u
1
∆ϕ(x0 ) y 2 + (∇∆ϕ)(x0 ) · (x − x0 ) y 2
2(1 + a)
that satisfy
? v x0 (x, 0) = u(x) − ϕ(x) ≥ 0, x ∈ Rn .
?|La v x0 (x, y )| ≤ C |y |a |x − x0 |1+γ , (x, y ) ∈ Rn+1 \ {(x, 0) : v x0 (x, 0) = 0} .
? v x0 depends continuously on x0 .
We get that
supBr (x0 ) v x0 (x, 0) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Moreover (using a weak Harnack inequality)
ˆ
|y |a |v x0 (x, y )|2 dx dy ≥ c2 r n+a+5
∀ r ∈ (0, r2 ).
Br (x0 ,0)
B. Barrios.
(UT)
Regularity of free boundary.
15 / 24
Principal Result.
A useful transformation
A useful transformation:
e a solution of (1) or (2) and x0 ∈ Γ(u). We define
Let be u
e(x, y )−ϕ(x)+
v x0 (x, y ) := u
1
∆ϕ(x0 ) y 2 + (∇∆ϕ)(x0 ) · (x − x0 ) y 2
2(1 + a)
that satisfy
? v x0 (x, 0) = u(x) − ϕ(x) ≥ 0, x ∈ Rn .
?|La v x0 (x, y )| ≤ C |y |a |x − x0 |1+γ , (x, y ) ∈ Rn+1 \ {(x, 0) : v x0 (x, 0) = 0} .
? v x0 depends continuously on x0 .
We get that
supBr (x0 ) v x0 (x, 0) ≥ c1 r 2
∀ r ∈ (0, r1 ).
Moreover (using a weak Harnack inequality)
ˆ
|y |a |v x0 (x, y )|2 dx dy ≥ c2 r n+a+5
∀ r ∈ (0, r2 ).
Br (x0 ,0)
B. Barrios.
(UT)
Regularity of free boundary.
15 / 24
Sketch of the proof
Scheme
1
Introduction.
2
Principal Result.
The key ingredient: Non degeneracy condition
A useful transformation
3
Sketch of the proof
B. Barrios.
(UT)
Regularity of free boundary.
16 / 24
Sketch of the proof
Step1: Frequency formula and study of the blow ups
Let u solve the obstacle problem (1) (resp. (2)) with ϕ satisfying (H1 )
(resp. (H2 )). Let x0 ∈ Γ(u) be and set
ˆ
x0
x0
H (r , v ) :=
|y |a (v x0 )2 .
(3.1)
∂Br (x0 ,0)
Then
there exist constants C0 , r0 > 0, independent of x0 , such that the
function
FF
r 7→ Φx0 (r , v x0 ) := r + C0 r 2
d
log max Hx0 (r , v x0 ), r n+a+4+2γ ,
dr
is monotone nondecreasing on (0, r0 ). In particular, the limit
lı́m Φx0 (r , v x0 ) := Φx0 (0+ , v x0 )
r ↓0
exists.
B. Barrios.
(UT)
Regularity of free boundary.
17 / 24
Sketch of the proof
Look!
If H(r , v ) =
´
∂Br
Step1: Frequency formula and study of the blow ups
|y |a v 2 ≥ r n+a+4+2γ then
´
Φ(r , v ) = (1 + C0 r ) (n + a) + 2 N (r , v ) − 2r
Br
v La v
H(r , v )
!
,
where
N (r , v ) :=
r
´
|y |a |∇v |2
´Br
a 2
∂Br |y | v
is the classic Almgren’s frequency function.
Let m be such that Φ(0+ , v ) = n + a + 2m. Then
H(r , v ) ≤ C̄ r n+a+2m
∀ r ∈ (0, r0 ).
Moreover, thanks to the non-degeneracy condition,
m ≤ 2,
∀ ε, H(r , v ) ≥ r n+a+2m+ε ∀ r ∈ (0, rε,x0 ) ⇒ m = 2 or m = 1 + s.
Also
Φ(0+ , v ) = n + a + 2 N (0+ , v ).
B. Barrios.
(UT)
Regularity of free boundary.
18 / 24
Sketch of the proof
Look!
If H(r , v ) =
´
∂Br
Step1: Frequency formula and study of the blow ups
|y |a v 2 ≥ r n+a+4+2γ then
´
Φ(r , v ) = (1 + C0 r ) (n + a) + 2 N (r , v ) − 2r
Br
v La v
H(r , v )
!
,
where
N (r , v ) :=
r
´
|y |a |∇v |2
´Br
a 2
∂Br |y | v
is the classic Almgren’s frequency function.
Let m be such that Φ(0+ , v ) = n + a + 2m. Then
H(r , v ) ≤ C̄ r n+a+2m
∀ r ∈ (0, r0 ).
Moreover, thanks to the non-degeneracy condition,
m ≤ 2,
∀ ε, H(r , v ) ≥ r n+a+2m+ε ∀ r ∈ (0, rε,x0 ) ⇒ m = 2 or m = 1 + s.
Also
Φ(0+ , v ) = n + a + 2 N (0+ , v ).
B. Barrios.
(UT)
Regularity of free boundary.
18 / 24
Sketch of the proof
Look!
If H(r , v ) =
´
∂Br
Step1: Frequency formula and study of the blow ups
|y |a v 2 ≥ r n+a+4+2γ then
´
Φ(r , v ) = (1 + C0 r ) (n + a) + 2 N (r , v ) − 2r
Br
v La v
H(r , v )
!
,
where
N (r , v ) :=
r
´
|y |a |∇v |2
´Br
a 2
∂Br |y | v
is the classic Almgren’s frequency function.
Let m be such that Φ(0+ , v ) = n + a + 2m. Then
H(r , v ) ≤ C̄ r n+a+2m
∀ r ∈ (0, r0 ).
Moreover, thanks to the non-degeneracy condition,
m ≤ 2,
∀ ε, H(r , v ) ≥ r n+a+2m+ε ∀ r ∈ (0, rε,x0 ) ⇒ m = 2 or m = 1 + s.
Also
Φ(0+ , v ) = n + a + 2 N (0+ , v ).
B. Barrios.
(UT)
Regularity of free boundary.
18 / 24
Sketch of the proof
Look!
If H(r , v ) =
´
∂Br
Step1: Frequency formula and study of the blow ups
|y |a v 2 ≥ r n+a+4+2γ then
´
Φ(r , v ) = (1 + C0 r ) (n + a) + 2 N (r , v ) − 2r
Br
v La v
H(r , v )
!
,
where
N (r , v ) :=
r
´
|y |a |∇v |2
´Br
a 2
∂Br |y | v
is the classic Almgren’s frequency function.
Let m be such that Φ(0+ , v ) = n + a + 2m. Then
H(r , v ) ≤ C̄ r n+a+2m
∀ r ∈ (0, r0 ).
Moreover, thanks to the non-degeneracy condition,
m ≤ 2,
∀ ε, H(r , v ) ≥ r n+a+2m+ε ∀ r ∈ (0, rε,x0 ) ⇒ m = 2 or m = 1 + s.
Also
Φ(0+ , v ) = n + a + 2 N (0+ , v ).
B. Barrios.
(UT)
Regularity of free boundary.
18 / 24
Sketch of the proof
Look!
If H(r , v ) =
´
∂Br
Step1: Frequency formula and study of the blow ups
|y |a v 2 ≥ r n+a+4+2γ then
´
Φ(r , v ) = (1 + C0 r ) (n + a) + 2 N (r , v ) − 2r
Br
v La v
H(r , v )
!
,
where
N (r , v ) :=
r
´
|y |a |∇v |2
´Br
a 2
∂Br |y | v
is the classic Almgren’s frequency function.
Let m be such that Φ(0+ , v ) = n + a + 2m. Then
H(r , v ) ≤ C̄ r n+a+2m
∀ r ∈ (0, r0 ).
Moreover, thanks to the non-degeneracy condition,
m ≤ 2,
∀ ε, H(r , v ) ≥ r n+a+2m+ε ∀ r ∈ (0, rε,x0 ) ⇒ m = 2 or m = 1 + s.
Also
Φ(0+ , v ) = n + a + 2 N (0+ , v ).
B. Barrios.
(UT)
Regularity of free boundary.
18 / 24
Sketch of the proof
Look!
If H(r , v ) =
´
∂Br
Step1: Frequency formula and study of the blow ups
|y |a v 2 ≥ r n+a+4+2γ then
´
Φ(r , v ) = (1 + C0 r ) (n + a) + 2 N (r , v ) − 2r
Br
v La v
H(r , v )
!
,
where
N (r , v ) :=
r
´
|y |a |∇v |2
´Br
a 2
∂Br |y | v
is the classic Almgren’s frequency function.
Let m be such that Φ(0+ , v ) = n + a + 2m. Then
H(r , v ) ≤ C̄ r n+a+2m
∀ r ∈ (0, r0 ).
Moreover, thanks to the non-degeneracy condition,
m ≤ 2,
∀ ε, H(r , v ) ≥ r n+a+2m+ε ∀ r ∈ (0, rε,x0 ) ⇒ m = 2 or m = 1 + s.
Also
Φ(0+ , v ) = n + a + 2 N (0+ , v ).
B. Barrios.
(UT)
Regularity of free boundary.
18 / 24
Sketch of the proof
Look!
If H(r , v ) =
´
∂Br
Step1: Frequency formula and study of the blow ups
|y |a v 2 ≥ r n+a+4+2γ then
´
Φ(r , v ) = (1 + C0 r ) (n + a) + 2 N (r , v ) − 2r
Br
v La v
H(r , v )
!
,
where
N (r , v ) :=
r
´
|y |a |∇v |2
´Br
a 2
∂Br |y | v
is the classic Almgren’s frequency function.
Let m be such that Φ(0+ , v ) = n + a + 2m. Then
H(r , v ) ≤ C̄ r n+a+2m
∀ r ∈ (0, r0 ).
Moreover, thanks to the non-degeneracy condition,
m ≤ 2,
∀ ε, H(r , v ) ≥ r n+a+2m+ε ∀ r ∈ (0, rε,x0 ) ⇒ m = 2 or m = 1 + s.
Also
Φ(0+ , v ) = n + a + 2 N (0+ , v ).
B. Barrios.
(UT)
Regularity of free boundary.
18 / 24
Sketch of the proof
Step1: Frequency formula and study of the blow ups
Let set now
m :=
Φx0 (0+ , v ) − n − a ⇒ m = 2 or m = 1 + s,
2
and let
vrx0 (x, y ) :=
v x0 (x0 + rx, ry )
,
dr
be a blow-up sequence with
1/2
.
dr := r −n−a Hx0 (r , v )
Since, Hx0 (r , v ) ≥ r n+a+4+2γ using the increasing behavior of Φx0 (r , v )
and some regularity properties, we can pass to the limit r → 0 getting that

x
 v0 0 (x, 0) ≥ 0 in Rn ,
x0
La v0 (x, y ) = 0 in Rn+1 \ {(x, 0) : v0x0 (x, 0) = 0},
(a) =

La v0x0 (x, y ) ≥ 0 in Rn+1 .
with vrx0 → v0x0 and kv0x0 kL2 (∂B1 ,|y |a ) = 1.
(b) N (ρ, v0x0 ) = limr →0+ N (ρ, vrx0 ) = limr →0+ N (r ρ, v x0 ) = m.
B. Barrios.
(UT)
Regularity of free boundary.
19 / 24
Sketch of the proof
Step1: Frequency formula and study of the blow ups
Let set now
m :=
Φx0 (0+ , v ) − n − a ⇒ m = 2 or m = 1 + s,
2
and let
vrx0 (x, y ) :=
v x0 (x0 + rx, ry )
,
dr
be a blow-up sequence with
1/2
.
dr := r −n−a Hx0 (r , v )
Since, Hx0 (r , v ) ≥ r n+a+4+2γ using the increasing behavior of Φx0 (r , v )
and some regularity properties, we can pass to the limit r → 0 getting that

x
 v0 0 (x, 0) ≥ 0 in Rn ,
x0
La v0 (x, y ) = 0 in Rn+1 \ {(x, 0) : v0x0 (x, 0) = 0},
(a) =

La v0x0 (x, y ) ≥ 0 in Rn+1 .
with vrx0 → v0x0 and kv0x0 kL2 (∂B1 ,|y |a ) = 1.
(b) N (ρ, v0x0 ) = limr →0+ N (ρ, vrx0 ) = limr →0+ N (r ρ, v x0 ) = m.
B. Barrios.
(UT)
Regularity of free boundary.
19 / 24
Sketch of the proof
Step1: Frequency formula and study of the blow ups
Let set now
m :=
Φx0 (0+ , v ) − n − a ⇒ m = 2 or m = 1 + s,
2
and let
vrx0 (x, y ) :=
v x0 (x0 + rx, ry )
,
dr
be a blow-up sequence with
1/2
.
dr := r −n−a Hx0 (r , v )
Since, Hx0 (r , v ) ≥ r n+a+4+2γ using the increasing behavior of Φx0 (r , v )
and some regularity properties, we can pass to the limit r → 0 getting that

x
 v0 0 (x, 0) ≥ 0 in Rn ,
x0
La v0 (x, y ) = 0 in Rn+1 \ {(x, 0) : v0x0 (x, 0) = 0},
(a) =

La v0x0 (x, y ) ≥ 0 in Rn+1 .
with vrx0 → v0x0 and kv0x0 kL2 (∂B1 ,|y |a ) = 1.
(b) N (ρ, v0x0 ) = limr →0+ N (ρ, vrx0 ) = limr →0+ N (r ρ, v x0 ) = m.
B. Barrios.
(UT)
Regularity of free boundary.
19 / 24
Sketch of the proof
Step1: Frequency formula and study of the blow ups
Then (a) + (b) imply that
v0x0 is an homogeneous function of degree m = 1 + s or m = 2 in B1/2 .
n−1 .
If m = 1 + s ⇒ v0x0 (x) = c((x − x0 ) · ν)1+s
+ , for some ν ∈ S
x0
If m = 2 ⇒ x0 is a singular point and v0 is an homogeneous
polynomial of degree 2.
⇓
Γ(u) = Γ2 (u) ∪ Γ1+s (u)
where
Γ2 (u) := x0 ∈ Γ(u) : Φx0 (0+ , v ) = n + a + 4 .
is a closed set consists of singular points and
√√
Γ1+s (u) := x0 ∈ Γ(u) : Φx0 (0+ , v ) = n + a + 2(1 + s)
is an open set consists of regular points.
B. Barrios.
(UT)
Regularity of free boundary.
20 / 24
Sketch of the proof
Step1: Frequency formula and study of the blow ups
Then (a) + (b) imply that
v0x0 is an homogeneous function of degree m = 1 + s or m = 2 in B1/2 .
n−1 .
If m = 1 + s ⇒ v0x0 (x) = c((x − x0 ) · ν)1+s
+ , for some ν ∈ S
x0
If m = 2 ⇒ x0 is a singular point and v0 is an homogeneous
polynomial of degree 2.
⇓
Γ(u) = Γ2 (u) ∪ Γ1+s (u)
where
Γ2 (u) := x0 ∈ Γ(u) : Φx0 (0+ , v ) = n + a + 4 .
is a closed set consists of singular points and
√√
Γ1+s (u) := x0 ∈ Γ(u) : Φx0 (0+ , v ) = n + a + 2(1 + s)
is an open set consists of regular points.
B. Barrios.
(UT)
Regularity of free boundary.
20 / 24
Sketch of the proof
Step1: Frequency formula and study of the blow ups
Then (a) + (b) imply that
v0x0 is an homogeneous function of degree m = 1 + s or m = 2 in B1/2 .
n−1 .
If m = 1 + s ⇒ v0x0 (x) = c((x − x0 ) · ν)1+s
+ , for some ν ∈ S
x0
If m = 2 ⇒ x0 is a singular point and v0 is an homogeneous
polynomial of degree 2.
⇓
Γ(u) = Γ2 (u) ∪ Γ1+s (u)
where
Γ2 (u) := x0 ∈ Γ(u) : Φx0 (0+ , v ) = n + a + 4 .
is a closed set consists of singular points and
√√
Γ1+s (u) := x0 ∈ Γ(u) : Φx0 (0+ , v ) = n + a + 2(1 + s)
is an open set consists of regular points.
B. Barrios.
(UT)
Regularity of free boundary.
20 / 24
Sketch of the proof
Step1: Frequency formula and study of the blow ups
Then (a) + (b) imply that
v0x0 is an homogeneous function of degree m = 1 + s or m = 2 in B1/2 .
n−1 .
If m = 1 + s ⇒ v0x0 (x) = c((x − x0 ) · ν)1+s
+ , for some ν ∈ S
x0
If m = 2 ⇒ x0 is a singular point and v0 is an homogeneous
polynomial of degree 2.
⇓
Γ(u) = Γ2 (u) ∪ Γ1+s (u)
where
Γ2 (u) := x0 ∈ Γ(u) : Φx0 (0+ , v ) = n + a + 4 .
is a closed set consists of singular points and
√√
Γ1+s (u) := x0 ∈ Γ(u) : Φx0 (0+ , v ) = n + a + 2(1 + s)
is an open set consists of regular points.
B. Barrios.
(UT)
Regularity of free boundary.
20 / 24
Sketch of the proof
Step2:
Finally, using
a Monneau-type monotonicity formula
and the
nondegeneracy property,
we obtain the
Uniqueness of blow-ups for points in Γ2 .
Continuity of the blow-up profiles with respect to x0 ∈ Γ2 .
That is v0x0 = p2x0 (x) = 12 hAx0 x, xi. Therefore we can define
Γk2 (u) := {x0 ∈ Γ2 (u) : dim(ker Ax0 ) = k},
k = 0, . . . , n − 1.
Using a Whitney’s extension theorem and the implicit function theorem
For any x0 ∈ Γk2 (u) there exists r = rx0 > 0 such that
Γk2 (u) ∩ Br (x0 ) is contained in a connected C 1 k-dimensional manifold.
B. Barrios.
(UT)
Regularity of free boundary.
21 / 24
Sketch of the proof
Step2:
Finally, using
a Monneau-type monotonicity formula
and the
nondegeneracy property,
we obtain the
Uniqueness of blow-ups for points in Γ2 .
Continuity of the blow-up profiles with respect to x0 ∈ Γ2 .
That is v0x0 = p2x0 (x) = 12 hAx0 x, xi. Therefore we can define
Γk2 (u) := {x0 ∈ Γ2 (u) : dim(ker Ax0 ) = k},
k = 0, . . . , n − 1.
Using a Whitney’s extension theorem and the implicit function theorem
For any x0 ∈ Γk2 (u) there exists r = rx0 > 0 such that
Γk2 (u) ∩ Br (x0 ) is contained in a connected C 1 k-dimensional manifold.
B. Barrios.
(UT)
Regularity of free boundary.
21 / 24
Sketch of the proof
Step2:
Finally, using
a Monneau-type monotonicity formula
and the
nondegeneracy property,
we obtain the
Uniqueness of blow-ups for points in Γ2 .
Continuity of the blow-up profiles with respect to x0 ∈ Γ2 .
That is v0x0 = p2x0 (x) = 12 hAx0 x, xi. Therefore we can define
Γk2 (u) := {x0 ∈ Γ2 (u) : dim(ker Ax0 ) = k},
k = 0, . . . , n − 1.
Using a Whitney’s extension theorem and the implicit function theorem
For any x0 ∈ Γk2 (u) there exists r = rx0 > 0 such that
Γk2 (u) ∩ Br (x0 ) is contained in a connected C 1 k-dimensional manifold.
B. Barrios.
(UT)
Regularity of free boundary.
21 / 24
Sketch of the proof
Step2:
Thank you for your attention.
B. Barrios.
(UT)
Regularity of free boundary.
22 / 24
Sketch of the proof
Step2:
FL
ˆ
s
(−∆) u(x) = cn,s P.V
Rn
u(x) − u(y )
b(ξ), 0 < s < 1.
dy = |ξ|2s u
|x − y |n+2s
Some basic properties:
Well define in a pointwise way for suitable functions in
C 2s+β (Rn ) ∩ L∞ (Rn ).
(−∆)s (u(λx)) = λ2s ((−∆)s u) (λx).
Forms a semigrup: (−∆)s1 ◦ (−∆)s2 = (−∆)s1 +s2 .
(−∆)0 = Id,
(−∆)1 = (−∆).
Fundamental solution Φ(x) = |x|2s−N .
Maximum principle, Hopf’s Lemma, Mean value property...
B. Barrios.
(UT)
Regularity of free boundary.
23 / 24
Sketch of the proof
Step2:
FL
ˆ
s
(−∆) u(x) = cn,s P.V
Rn
u(x) − u(y )
b(ξ), 0 < s < 1.
dy = |ξ|2s u
|x − y |n+2s
Some basic properties:
Well define in a pointwise way for suitable functions in
C 2s+β (Rn ) ∩ L∞ (Rn ).
(−∆)s (u(λx)) = λ2s ((−∆)s u) (λx).
Forms a semigrup: (−∆)s1 ◦ (−∆)s2 = (−∆)s1 +s2 .
(−∆)0 = Id,
(−∆)1 = (−∆).
Fundamental solution Φ(x) = |x|2s−N .
Maximum principle, Hopf’s Lemma, Mean value property...
B. Barrios.
(UT)
Regularity of free boundary.
23 / 24
Sketch of the proof
Step2:
FL
ˆ
s
(−∆) u(x) = cn,s P.V
Rn
u(x) − u(y )
b(ξ), 0 < s < 1.
dy = |ξ|2s u
|x − y |n+2s
Some basic properties:
Well define in a pointwise way for suitable functions in
C 2s+β (Rn ) ∩ L∞ (Rn ).
(−∆)s (u(λx)) = λ2s ((−∆)s u) (λx).
Forms a semigrup: (−∆)s1 ◦ (−∆)s2 = (−∆)s1 +s2 .
(−∆)0 = Id,
(−∆)1 = (−∆).
Fundamental solution Φ(x) = |x|2s−N .
Maximum principle, Hopf’s Lemma, Mean value property...
B. Barrios.
(UT)
Regularity of free boundary.
23 / 24
Sketch of the proof
Step2:
FL
ˆ
s
(−∆) u(x) = cn,s P.V
Rn
u(x) − u(y )
b(ξ), 0 < s < 1.
dy = |ξ|2s u
|x − y |n+2s
Some basic properties:
Well define in a pointwise way for suitable functions in
C 2s+β (Rn ) ∩ L∞ (Rn ).
(−∆)s (u(λx)) = λ2s ((−∆)s u) (λx).
Forms a semigrup: (−∆)s1 ◦ (−∆)s2 = (−∆)s1 +s2 .
(−∆)0 = Id,
(−∆)1 = (−∆).
Fundamental solution Φ(x) = |x|2s−N .
Maximum principle, Hopf’s Lemma, Mean value property...
B. Barrios.
(UT)
Regularity of free boundary.
23 / 24
Sketch of the proof
Step2:
FL
ˆ
s
(−∆) u(x) = cn,s P.V
Rn
u(x) − u(y )
b(ξ), 0 < s < 1.
dy = |ξ|2s u
|x − y |n+2s
Some basic properties:
Well define in a pointwise way for suitable functions in
C 2s+β (Rn ) ∩ L∞ (Rn ).
(−∆)s (u(λx)) = λ2s ((−∆)s u) (λx).
Forms a semigrup: (−∆)s1 ◦ (−∆)s2 = (−∆)s1 +s2 .
(−∆)0 = Id,
(−∆)1 = (−∆).
Fundamental solution Φ(x) = |x|2s−N .
Maximum principle, Hopf’s Lemma, Mean value property...
B. Barrios.
(UT)
Regularity of free boundary.
23 / 24
Sketch of the proof
Step2:
FL
ˆ
s
(−∆) u(x) = cn,s P.V
Rn
u(x) − u(y )
b(ξ), 0 < s < 1.
dy = |ξ|2s u
|x − y |n+2s
Some basic properties:
Well define in a pointwise way for suitable functions in
C 2s+β (Rn ) ∩ L∞ (Rn ).
(−∆)s (u(λx)) = λ2s ((−∆)s u) (λx).
Forms a semigrup: (−∆)s1 ◦ (−∆)s2 = (−∆)s1 +s2 .
(−∆)0 = Id,
(−∆)1 = (−∆).
Fundamental solution Φ(x) = |x|2s−N .
Maximum principle, Hopf’s Lemma, Mean value property...
B. Barrios.
(UT)
Regularity of free boundary.
23 / 24
Sketch of the proof
Step2:
L. Caffarelli, L. Silvestre, An extension problem related to the fractional Laplacian. CPDE (2007).
in Rn+1
+ , a = 1 − 2s,

 La (w (x, y )) := − div(y a ∇w ) = 0

in Rn .
w (x, 0) = u(x)
− div(y a ∇w ) = −y a (∆x w + wyy +
{z
|
1 − 2s
wy ) = 0.
y
}
Laplacian in dim n + (1 − 2s) + 1, (radial way)
ˆ
w (x, y ) = (Ps (·, y ) ∗ u(·))(x) = dn,s
Rn
y 2s
(y 2 + |x − η|2 )
n+2s
2
u(η)dη,
−κs lim+ y 1−2s wy (x, y ) = (−∆)s u(x), x ∈ Rn .
y →0
FL
B. Barrios.
(UT)
Regularity of free boundary.
24 / 24
Sketch of the proof
Step2:
L. Caffarelli, L. Silvestre, An extension problem related to the fractional Laplacian. CPDE (2007).
in Rn+1
+ , a = 1 − 2s,

 La (w (x, y )) := − div(y a ∇w ) = 0

in Rn .
w (x, 0) = u(x)
− div(y a ∇w ) = −y a (∆x w + wyy +
{z
|
1 − 2s
wy ) = 0.
y
}
Laplacian in dim n + (1 − 2s) + 1, (radial way)
ˆ
w (x, y ) = (Ps (·, y ) ∗ u(·))(x) = dn,s
Rn
y 2s
(y 2 + |x − η|2 )
n+2s
2
u(η)dη,
−κs lim+ y 1−2s wy (x, y ) = (−∆)s u(x), x ∈ Rn .
y →0
FL
B. Barrios.
(UT)
Regularity of free boundary.
24 / 24
Sketch of the proof
Step2:
L. Caffarelli, L. Silvestre, An extension problem related to the fractional Laplacian. CPDE (2007).
in Rn+1
+ , a = 1 − 2s,

 La (w (x, y )) := − div(y a ∇w ) = 0

in Rn .
w (x, 0) = u(x)
− div(y a ∇w ) = −y a (∆x w + wyy +
{z
|
1 − 2s
wy ) = 0.
y
}
Laplacian in dim n + (1 − 2s) + 1, (radial way)
ˆ
w (x, y ) = (Ps (·, y ) ∗ u(·))(x) = dn,s
Rn
y 2s
(y 2 + |x − η|2 )
n+2s
2
u(η)dη,
−κs lim+ y 1−2s wy (x, y ) = (−∆)s u(x), x ∈ Rn .
y →0
FL
B. Barrios.
(UT)
Regularity of free boundary.
24 / 24