2 The penalized problem
Let
Ω
δ
:=
{
x
∈
Ω
:
dist
(
x
,
∂
Ω
)
<
δ
}
and
V
δ
:=
{
𝐮
∈
W
1
,
p
(
Ω
;
ℝ
m
)
:
u
i
≥
0
,
Δ
p
u
i
≥
0
,
Δ
p
u
i
=
0
in
Ω
δ
,
u
i
=
φ
i
on
∂
Ω
}
.
Furthermore, we set
V
:=
⋃
δ
>
0
V
δ
.
Observe that the above definition is consistent due to the assumption
φ
i
>
0
on
∂
Ω
. Also, by
Δ
p
u
i
≥
0
we mean that for any test function
ζ
∈
C
c
∞
(
Ω
)
with
ζ
≥
0
we have
-
∫
Ω
∇
ζ
⋅
|
∇
u
i
|
p
-
2
∇
u
i
d
x
≥
0
.
This implies that there is a Radon measure
μ
i
such that for any
test function
ζ
∈
C
c
∞
(
Ω
)
we have
∫
Ω
ζ
𝑑
μ
i
=
-
∫
Ω
∇
ζ
⋅
|
∇
u
i
|
p
-
2
∇
u
i
d
x
.
To simplify the notation, we denote
μ
i
by
Δ
p
u
i
,
and
d
μ
i
by
Δ
p
u
i
d
x
. (It should be noted that
this notation is not meant to imply
μ
i
is absolutely continuous
with respect to the Lebesgue measure. In fact, for the minimizer,
the two measures are mutually singular as we will see in Theorem 3.5.) It is also worth noting that
(2.1)
u
i
>
0
in
Ω
δ
by the strong maximum principle, since
u
i
is p-harmonic in
Ω
δ
, and while it is positive on
∂
Ω
, it is nonnegative everywhere.
Let
f
ε
:
ℝ
→
ℝ
be
f
ε
(
t
)
:=
{
1
+
1
ε
(
t
-
1
)
,
t
≥
1
,
1
+
ε
(
t
-
1
)
,
t
<
1
.
We are interested in minimizing the penalized functional
J
ε
(
𝐮
)
:=
∫
∂
Ω
Γ
(
x
,
A
ν
𝐮
(
x
)
)
d
σ
+
f
ε
(
|
{
|
𝐮
|
>
0
}
|
)
over V. The significance of the above penalization is that it forces the volume
|
{
|
𝐮
|
>
0
}
|
to be 1 for small enough ε; see Theorem 4.3. (Notice that the components of
𝐮
∈
V
are p-harmonic
near
∂
Ω
; therefore they are smooth enough near the boundary, and it makes sense to compute their derivatives along
∂
Ω
.)
We first consider the minimizer of
J
ε
over
V
δ
.
Lemma 2.1.
Let
u
∈
V
. Then we have
(2.2)
∫
Ω
ψ
i
Δ
p
u
i
𝑑
x
+
∫
Ω
∇
ψ
i
⋅
A
[
u
i
]
𝑑
x
=
∫
∂
Ω
ψ
i
A
ν
u
i
𝑑
σ
,
where the
ψ
i
are
C
1
functions.
Proof.
Let
ϕ
k
∈
C
∞
(
Ω
)
be such that
ϕ
k
≡
1
on
Ω
~
k
:=
Ω
-
Ω
1
/
k
and
ϕ
k
≡
0
on
∂
Ω
.
We know that
𝐮
∈
V
δ
for some δ. Suppose
k is large enough so that
1
k
<
δ
, and thus
Ω
1
/
k
⊂
Ω
δ
.
Then we have
∫
Ω
∇
(
ϕ
k
ψ
i
)
⋅
A
[
u
i
]
𝑑
x
=
∫
Ω
1
/
k
∇
(
ϕ
k
ψ
i
)
⋅
A
[
u
i
]
𝑑
x
+
∫
Ω
~
k
∇
(
ϕ
k
ψ
i
)
⋅
A
[
u
i
]
𝑑
x
=
∫
Ω
1
/
k
∇
(
ϕ
k
ψ
i
)
⋅
A
[
u
i
]
𝑑
x
+
∫
Ω
~
k
∇
ψ
i
⋅
A
[
u
i
]
𝑑
x
.
Now, noting that
∂
Ω
1
/
k
=
∂
Ω
~
k
∪
∂
Ω
,
and by using the integration by parts formula proved in [6],
we get
∫
Ω
1
/
k
∇
(
ϕ
k
ψ
i
)
⋅
A
[
u
i
]
𝑑
x
=
∫
Ω
1
/
k
∇
(
ϕ
k
ψ
i
)
⋅
A
[
u
i
]
+
ϕ
k
ψ
i
Δ
p
u
i
d
x
=
-
∫
∂
Ω
~
k
ϕ
k
ψ
i
A
ν
u
i
𝑑
σ
+
∫
∂
Ω
ϕ
k
ψ
i
A
ν
u
i
𝑑
σ
=
-
∫
∂
Ω
~
k
ψ
i
A
ν
u
i
𝑑
σ
→
k
→
∞
-
∫
∂
Ω
ψ
i
A
ν
u
i
𝑑
σ
.
In addition, we have
∫
Ω
~
k
∇
ψ
i
⋅
A
[
u
i
]
𝑑
x
→
k
→
∞
∫
Ω
∇
ψ
i
⋅
A
[
u
i
]
𝑑
x
,
and
∫
Ω
∇
(
ϕ
k
ψ
i
)
⋅
A
[
u
i
]
𝑑
x
=
-
∫
Ω
ϕ
k
ψ
i
Δ
p
u
i
𝑑
x
→
k
→
∞
-
∫
Ω
ψ
i
Δ
p
u
i
𝑑
x
,
which together give the desired result.
∎
We can similarly show that
∫
Ω
Δ
p
u
i
𝑑
x
=
∫
∂
Ω
A
ν
u
i
𝑑
σ
.
In addition, note that
∫
Ω
u
i
Δ
p
u
i
𝑑
x
is meaningful
(since
u
i
-
φ
i
∈
W
0
1
,
p
while
Δ
p
u
i
∈
W
-
1
,
p
/
(
p
-
1
)
and
φ
i
is continuous) and we can similarly show that
(2.3)
∫
Ω
u
i
Δ
p
u
i
+
|
∇
u
i
|
p
d
x
=
∫
∂
Ω
φ
i
A
ν
u
i
𝑑
σ
.
Note that
u
i
=
φ
i
on
∂
Ω
.
Lemma 2.2.
For
u
∈
V
we have
∑
i
=
1
m
∫
Ω
|
∇
u
i
|
p
𝑑
x
≤
C
+
C
∫
∂
Ω
∑
i
≤
m
ψ
i
(
x
)
A
ν
u
i
d
σ
.
Remark.
As we will see, the above inequality actually holds for each summand.
Furthermore, with a slight modification of the last part of the proof
we obtain that
∫
Ω
Δ
p
u
i
𝑑
x
≤
C
+
C
∫
∂
Ω
ψ
i
(
x
)
A
ν
u
i
𝑑
σ
.
Proof.
Let
𝐡
0
be the vector-valued function in Ω satisfying
Δ
p
h
0
i
=
0
, and taking the boundary values
𝝋
on
∂
Ω
. Note that
h
0
i
is
C
1
and we can plug
it in (2.2). By subtracting the resulting relation
from (2.3) we get
∫
Ω
(
u
i
-
h
0
i
)
Δ
p
u
i
𝑑
x
+
∫
Ω
∇
(
u
i
-
h
0
i
)
⋅
A
[
u
i
]
𝑑
x
=
0
.
Hence
∫
Ω
|
∇
u
i
|
p
𝑑
x
=
∫
Ω
∇
u
i
⋅
A
[
u
i
]
𝑑
x
=
∫
Ω
(
h
0
i
-
u
i
)
Δ
p
u
i
𝑑
x
+
∫
Ω
∇
h
0
i
⋅
A
[
u
i
]
𝑑
x
≤
∫
Ω
h
0
i
Δ
p
u
i
𝑑
x
+
C
∫
Ω
|
∇
h
0
i
|
p
𝑑
x
+
1
2
∫
Ω
|
A
[
u
i
]
|
p
p
-
1
𝑑
x
≤
C
∫
Ω
Δ
p
u
i
𝑑
x
+
C
+
1
2
∫
Ω
|
∇
u
i
|
p
𝑑
x
,
where we have used the facts that
u
i
,
Δ
p
u
i
≥
0
and
|
A
[
u
i
]
|
p
p
-
1
=
|
∇
u
i
|
p
. Thus we have
∫
Ω
|
∇
u
i
|
p
𝑑
x
≤
C
∫
Ω
Δ
p
u
i
𝑑
x
.
But since
ψ
i
,
Δ
p
u
i
≥
0
we get
∫
Ω
|
∇
u
i
|
p
𝑑
x
≤
C
∫
Ω
Δ
p
u
i
𝑑
x
≤
C
C
i
∫
Ω
ψ
i
Δ
p
u
i
𝑑
x
,
where
C
i
=
max
Ω
¯
1
ψ
i
>
0
. Hence by (2.2)
we get
∫
Ω
|
∇
u
i
|
p
𝑑
x
≤
C
∫
Ω
ψ
i
Δ
p
u
i
𝑑
x
=
-
C
∫
Ω
∇
ψ
i
⋅
A
[
u
i
]
𝑑
x
+
C
∫
∂
Ω
ψ
i
A
ν
u
i
𝑑
σ
≤
C
~
∫
Ω
|
∇
ψ
i
|
p
𝑑
x
+
1
2
∫
Ω
|
A
[
u
i
]
|
p
p
-
1
𝑑
x
+
C
∫
∂
Ω
ψ
i
A
ν
u
i
𝑑
σ
≤
C
+
1
2
∫
Ω
|
∇
u
i
|
p
𝑑
x
+
C
∫
∂
Ω
ψ
i
A
ν
u
i
𝑑
σ
,
which gives the desired.
∎
Theorem 2.3.
There exists a minimizer
u
ε
δ
∈
V
δ
for
J
ε
.
Proof.
Let
{
𝐮
k
}
⊂
V
δ
be a minimizing sequence.
Then by the above lemma and (1.2) we have
∑
i
=
1
m
∫
Ω
|
∇
u
k
i
|
p
𝑑
x
≤
C
+
C
∫
∂
Ω
∑
i
≤
m
ψ
i
A
ν
u
k
i
d
σ
≤
C
+
C
∫
∂
Ω
Γ
(
x
,
A
ν
u
k
1
,
…
,
A
ν
u
k
m
)
+
C
d
σ
≤
C
+
C
J
ε
(
𝐮
k
)
.
Hence
∥
∇
𝐮
k
∥
L
p
is bounded. In addition,
for the dual exponent
q
=
p
p
-
1
we can see that
∥
A
[
𝐮
k
]
∥
L
q
=
∥
∇
𝐮
k
∥
L
p
p
-
1
is also bounded. Hence, up to a subsequence, we can assume that
∇
u
k
i
⇀
∇
u
i
in
L
p
,
A
[
u
k
i
]
⇀
A
[
u
i
]
in
L
q
,
and
𝐮
k
→
𝐮
a.e. in Ω. Thus we have
u
i
≥
0
.
Also,
u
i
=
φ
i
on
∂
Ω
, since
u
k
i
-
φ
i
∈
W
0
1
,
p
(
Ω
)
,
which is a closed and convex set, hence weakly closed.
Finally, to see that
Δ
p
u
i
has the desired properties,
notice that for an appropriate test function ϕ we have
∫
Ω
∇
ϕ
⋅
A
[
u
i
]
𝑑
x
=
lim
k
→
∞
∫
Ω
∇
ϕ
⋅
A
[
u
k
i
]
𝑑
x
due to the weak convergence of
A
[
𝐮
k
]
. Therefore
𝐮
∈
V
δ
.
Now we can repeat the proof of [16, Lemma 3.3]
to deduce the weak lower semicontinuity of
J
ε
with
respect to this sequence, and conclude the proof (the convexity of Γ is needed here).
∎
Although Hopf’s lemmas for p-harmonic functions are well known (see for example [15]), we include the proof of the following version as we need a specific form for the constant.
Lemma 2.4 (Hopf’s lemma for p-harmonic functions).
Suppose h is a p-harmonic function on
B
1
(
0
)
with nonnegative
boundary values on
∂
B
1
. Then we have
h
(
x
)
≥
c
(
n
,
p
)
dist
(
x
,
∂
B
1
)
sup
B
1
/
2
h
.
Proof.
Consider the function
g
(
x
)
=
e
-
λ
|
x
|
2
-
e
-
λ
for some
λ
>
0
. Note
that
g
=
0
on
∂
B
1
, and
0
<
g
<
1
on
B
1
. We also
have
∂
i
g
=
-
2
λ
x
i
e
-
λ
|
x
|
2
,
∂
i
j
g
=
(
4
λ
2
x
i
x
j
-
2
λ
δ
i
j
)
e
-
λ
|
x
|
2
.
Now we have
Δ
g
=
(
4
λ
2
|
x
|
2
-
2
n
λ
)
e
-
λ
|
x
|
2
,
and
Δ
∞
g
:=
∑
i
,
j
∂
i
g
∂
j
g
∂
i
j
g
=
∑
i
,
j
4
λ
2
(
4
λ
2
x
i
2
x
j
2
-
2
λ
δ
i
j
x
i
x
j
)
e
-
3
λ
|
x
|
2
=
4
λ
2
(
4
λ
2
|
x
|
4
-
2
λ
|
x
|
2
)
e
-
3
λ
|
x
|
2
.
Therefore
Δ
p
g
=
div
(
|
∇
g
|
p
-
2
∇
g
)
=
|
∇
g
|
p
-
4
(
|
∇
g
|
2
Δ
g
+
(
p
-
2
)
Δ
∞
g
)
=
(
2
λ
|
x
|
)
p
-
4
(
4
λ
2
|
x
|
2
(
4
λ
2
|
x
|
2
-
2
n
λ
)
+
(
p
-
2
)
4
λ
2
(
4
λ
2
|
x
|
4
-
2
λ
|
x
|
2
)
)
e
-
(
p
-
1
)
λ
|
x
|
2
=
(
2
λ
)
p
-
1
|
x
|
p
-
2
(
2
λ
|
x
|
2
-
n
+
(
p
-
2
)
(
2
λ
|
x
|
2
-
1
)
)
e
-
(
p
-
1
)
λ
|
x
|
2
=
(
2
λ
)
p
-
1
|
x
|
p
-
2
(
2
(
p
-
1
)
λ
|
x
|
2
-
n
-
p
+
2
)
e
-
(
p
-
1
)
λ
|
x
|
2
.
Thus for
1
2
≤
|
x
|
≤
1
and large enough λ we have
Δ
p
g
≥
2
λ
p
-
1
(
(
p
-
1
)
λ
2
-
n
-
p
+
2
)
e
-
(
p
-
1
)
λ
>
0
.
Now we have
h
≥
inf
B
¯
1
/
2
h
>
(
inf
B
¯
1
/
2
h
)
g
on
B
¯
1
/
2
(note that h is positive on
B
1
by
maximum principle), and on
B
1
-
B
¯
1
/
2
we have
Δ
p
h
=
0
<
Δ
p
g
.
Also on
∂
B
1
we have
h
≥
0
=
g
. Hence by the maximum
principle we have
h
(
x
)
≥
g
(
x
)
(
inf
B
¯
1
/
2
h
)
for
x
∈
B
1
. But by the Harnack’s inequality we have
inf
B
¯
1
/
2
h
≥
C
sup
B
¯
1
/
2
h
for some constant C which does not depend on h. Hence we obtain
h
(
x
)
≥
C
g
(
x
)
sup
B
¯
1
/
2
h
.
On the other hand note that
g
(
x
)
=
g
(
x
)
-
g
(
x
/
|
x
|
)
=
∫
1
|
x
|
1
d
d
t
g
(
t
x
)
𝑑
t
=
∫
1
|
x
|
1
x
⋅
∇
g
(
t
x
)
𝑑
t
=
∫
1
|
x
|
1
-
2
λ
t
|
x
|
2
e
-
λ
t
2
|
x
|
2
d
t
=
2
λ
|
x
|
2
∫
1
1
|
x
|
t
e
-
λ
t
2
|
x
|
2
𝑑
t
≥
2
λ
|
x
|
2
∫
1
1
|
x
|
t
e
-
λ
𝑑
t
=
λ
e
-
λ
|
x
|
2
(
1
|
x
|
2
-
1
)
=
λ
e
-
λ
(
1
-
|
x
|
2
)
≥
λ
e
-
λ
(
1
-
|
x
|
)
=
λ
e
-
λ
dist
(
x
,
∂
B
1
)
,
which gives the desired.
∎
If h is a p-harmonic function on
B
r
(
x
0
)
, then
h
~
(
x
)
:=
h
(
x
0
+
r
x
)
is a p-harmonic function on
B
1
(
0
)
. Hence we have
h
(
x
0
+
r
x
)
=
h
~
(
x
)
≥
c
(
n
,
p
)
dist
(
x
,
∂
B
1
)
sup
B
1
/
2
h
~
=
c
(
n
,
p
)
(
1
-
|
x
|
)
sup
B
1
/
2
h
~
=
c
(
n
,
p
)
r
-
r
|
x
|
r
sup
B
r
/
2
(
x
0
)
h
=
c
(
n
,
p
)
dist
(
x
0
+
r
x
,
∂
B
r
(
x
0
)
)
1
r
sup
B
r
/
2
(
x
0
)
h
.
Lemma 2.5.
Let
w
∈
W
1
,
p
(
Ω
)
be a nonnegative function. Then there exists
c
>
0
, depending only on p and the
dimension, such that for any ball
B
¯
r
(
x
0
)
⊂
Ω
we have
(
1
r
sup
B
r
/
2
(
x
0
)
h
)
p
⋅
|
B
r
(
x
0
)
∩
{
w
=
0
}
|
≤
c
∫
B
r
(
x
0
)
|
∇
(
w
-
h
)
|
p
d
y
,
where h satisfies
Δ
p
h
=
0
in
B
r
(
x
0
)
taking boundary
values equal to w on
∂
B
r
(
x
0
)
.
Proof.
Let
τ
∈
(
0
,
1
)
be fixed. For ξ with
|
ξ
|
=
1
we set
t
ξ
:=
inf
{
t
∈
[
τ
r
,
r
]
:
w
(
x
0
+
t
ξ
)
=
0
}
provided that this set is nonempty. Otherwise we set
t
ξ
:=
r
.
Now note that
w
-
h
and w are absolutely continuous in almost
every direction ξ; in particular we have
w
(
x
0
+
t
ξ
ξ
)
=
0
(note that this will not be necessarily true if we allow τ to
be zero). Also
w
-
h
is
ℋ
n
-
1
-a.e. zero on
∂
B
r
(
x
0
)
as its trace is zero there, so
(
w
-
h
)
(
x
0
+
r
ξ
)
=
0
. Thus for almost
every ξ for which
t
ξ
<
r
we have
h
(
x
0
+
t
ξ
ξ
)
=
(
w
-
h
)
(
x
0
+
r
ξ
)
-
(
w
-
h
)
(
x
0
+
t
ξ
ξ
)
=
∫
t
ξ
r
d
d
t
(
(
w
-
h
)
(
x
0
+
t
ξ
)
)
𝑑
t
=
∫
t
ξ
r
∇
ξ
(
w
-
h
)
(
x
0
+
t
ξ
)
𝑑
t
≤
(
r
-
t
ξ
)
p
-
1
p
(
∫
t
ξ
r
|
∇
(
w
-
h
)
(
x
0
+
t
ξ
)
|
p
𝑑
t
)
1
p
.
On the other hand, using Hopf’s lemma we get
h
(
x
0
+
t
ξ
ξ
)
≥
c
(
n
,
p
)
dist
(
x
0
+
t
ξ
ξ
,
∂
B
r
(
x
0
)
)
1
r
sup
B
r
/
2
(
x
0
)
h
=
c
(
n
,
p
)
(
r
-
t
ξ
)
1
r
sup
B
r
/
2
(
x
0
)
h
.
Hence we obtain
(
r
-
t
ξ
)
(
1
r
sup
B
r
/
2
(
x
0
)
h
)
p
≤
C
(
n
,
p
)
∫
t
ξ
r
|
∇
(
w
-
h
)
(
x
0
+
t
ξ
)
|
p
𝑑
t
.
Note that this inequality is trivially satisfied if
t
ξ
=
r
.
Now by integrating with respect to
d
ξ
we get
C
(
n
,
p
)
∫
B
r
(
x
0
)
|
∇
(
w
-
h
)
(
x
)
|
p
𝑑
x
≥
C
(
n
,
p
)
∫
∂
B
1
(
0
)
∫
t
ξ
r
|
∇
(
w
-
h
)
(
x
0
+
t
ξ
)
|
p
𝑑
t
𝑑
ξ
≥
(
1
r
sup
B
r
/
2
(
x
0
)
h
)
p
∫
∂
B
1
(
0
)
(
r
-
t
ξ
)
𝑑
ξ
=
(
1
r
sup
B
r
/
2
(
x
0
)
h
)
p
∫
∂
B
1
(
0
)
∫
t
ξ
r
1
𝑑
t
𝑑
ξ
≥
(
1
r
sup
B
r
/
2
(
x
0
)
h
)
p
∫
B
r
(
x
0
)
-
B
τ
r
(
x
0
)
χ
{
w
=
0
}
𝑑
x
,
where the last inequality follows from the definition of
t
ξ
.
Finally, we get the desired by letting
τ
→
0
.
∎
Lemma 2.6.
Let
u
=
u
ε
δ
be a minimizer of
J
ε
over
V
δ
, and let
B
⊂
Ω
be a ball. Then there exists a unique
v
i
∈
W
1
,
p
(
Ω
)
that
minimizes the functional
∫
Ω
|
∇
v
i
|
p
𝑑
x
among all functions with
v
i
=
φ
i
on
∂
Ω
and
v
i
≤
0
on
{
u
i
=
0
}
-
B
. The functions
v
i
also satisfy:
v
i
=
0
on
{
u
i
=
0
}
-
B
,
𝐯
=
(
v
1
,
…
,
v
m
)
∈
V
δ
,
0
≤
u
i
≤
v
i
≤
C
0
=
max
∂
Ω
|
𝝋
|
,
∫
Ω
v
i
Δ
p
v
i
𝑑
x
=
0
.
Remark.
Instead of a ball B, we can also use other open subsets of Ω
in the above lemma.
Essentially, all we need is that the p-energy functional has a minimum over the corresponding set K in the following proof; so no regularity assumption is actually needed regarding such open sets.
Proof.
It is easy to see that
K
:=
{
v
∈
W
1
,
p
(
Ω
)
:
v
=
φ
i
on
∂
Ω
and
v
≤
0
on
{
u
i
=
0
}
-
B
}
is a closed convex subset of
W
1
,
p
(
Ω
)
. It is nonempty too
as
u
i
∈
K
. So there exists a unique
v
i
∈
K
minimizing
the strictly convex and coercive functional
∫
Ω
|
∇
v
|
p
𝑑
x
.
Then for every
v
∈
K
we have
d
d
t
|
t
=
0
∫
Ω
|
∇
(
v
i
+
t
(
v
-
v
i
)
)
|
p
𝑑
x
≥
0
,
and hence
v
i
satisfies the variational inequality
(2.4)
∫
Ω
|
∇
v
i
|
p
-
2
∇
v
i
⋅
∇
(
v
-
v
i
)
d
x
≥
0
.
Now note that
v
=
v
i
-
ζ
∈
K
for any test function
ζ
∈
C
c
∞
(
Ω
)
with
ζ
≥
0
. Therefore
-
∫
Ω
|
∇
v
i
|
p
-
2
∇
v
i
⋅
∇
ζ
d
x
≥
0
,
which means
Δ
p
v
i
≥
0
. As a result,
v
i
≤
max
∂
Ω
φ
i
≤
C
0
by the maximum principle.
Next note that if
spt
ζ
does not intersect
{
u
i
=
0
}
-
B
,
then we also have
v
i
+
ζ
∈
K
. Thus we also get
∫
Ω
|
∇
v
i
|
p
-
2
∇
v
i
⋅
∇
ζ
d
x
≥
0
,
which together with the previous inequality implies
-
∫
Ω
|
∇
v
i
|
p
-
2
∇
v
i
⋅
∇
ζ
d
x
=
0
.
Therefore
Δ
p
v
i
=
0
in the interior of
Ω
-
(
{
u
i
=
0
}
-
B
)
=
(
Ω
-
{
u
i
=
0
}
)
∪
B
.
In particular,
Δ
p
v
i
=
0
in
Ω
δ
since
u
i
>
0
in
Ω
δ
by (2.1).
In addition, for
ϵ
>
0
we have
v
=
max
(
v
i
,
-
ϵ
)
∈
K
.
By plugging this test function in (2.4) we get
0
≤
∫
Ω
|
∇
v
i
|
p
-
2
∇
v
i
⋅
∇
(
v
-
v
i
)
d
x
=
∫
{
v
i
<
-
ϵ
}
|
∇
v
i
|
p
-
2
∇
v
i
⋅
∇
(
-
ϵ
-
v
i
)
d
x
=
-
∫
{
v
i
<
-
ϵ
}
|
∇
v
i
|
p
𝑑
x
.
By letting
ϵ
→
0
we obtain
∫
{
v
i
<
0
}
|
∇
v
i
|
p
𝑑
x
=
0
,
and hence
v
i
≥
0
. In particular, we must have
v
i
=
0
on
{
u
i
=
0
}
-
B
as
v
i
is assumed to be nonpositive there. Furthermore,
note that we have so far shown
𝐯
=
(
v
1
,
…
,
v
m
)
∈
V
δ
.
Next, since
Δ
p
v
i
=
0
in the exterior of
{
u
i
=
0
}
-
B
,
and
Δ
p
u
i
≥
0
, the maximum principle implies that
u
i
≤
v
i
(note that
u
i
,
v
i
have the same boundary values in the exterior of
{
u
i
=
0
}
-
B
).
Finally, for
ζ
∈
C
c
∞
(
Ω
)
with
ζ
≥
0
and small enough ϵ we have
v
i
±
ϵ
ζ
v
i
=
(
1
±
ϵ
ζ
)
v
i
∈
K
.
By plugging this test function in (2.4) we get
0
≤
±
ϵ
∫
Ω
|
∇
v
i
|
p
-
2
∇
v
i
⋅
∇
(
ζ
v
i
)
d
x
⟹
∫
Ω
|
∇
v
i
|
p
-
2
∇
v
i
⋅
∇
(
ζ
v
i
)
d
x
=
0
.
In other words
∫
Ω
ζ
v
i
Δ
p
v
i
𝑑
x
=
0
.
By letting
ζ
→
1
we obtain
∫
Ω
v
i
Δ
p
v
i
𝑑
x
=
0
,
as desired. Alternatively, we can take ζ to be 1 over a neighborhood of
{
u
i
=
0
}
-
B
. From this and that
Δ
p
v
i
=
0
in the exterior of
{
u
i
=
0
}
-
B
,
we obtain
∫
Ω
v
i
Δ
p
v
i
𝑑
x
=
0
.
∎
Theorem 2.7.
Let
u
=
u
ε
δ
be a minimizer of
J
ε
over
V
δ
. There exists
a constant
M
=
M
ε
, independent of δ, such that
if for some j we have
1
r
sup
B
r
/
2
(
x
)
u
j
≥
M
,
then
B
r
(
x
)
⊂
{
|
u
|
>
0
}
, and
Δ
p
u
i
=
0
in
B
r
(
x
)
for every i.
Proof.
Let
𝐯
∈
V
δ
be the function given by Lemma 2.6
for
B
r
(
x
)
. Then we have
J
ε
(
𝐮
)
≤
J
ε
(
𝐯
)
.
Let
𝐡
0
be the vector-valued function in Ω satisfying
Δ
p
h
0
i
=
0
, and taking the boundary values
𝝋
on
∂
Ω
. Since
0
≤
u
i
≤
v
i
≤
h
0
i
, for each
z
∈
∂
Ω
we have
∂
ν
h
0
i
(
z
)
≤
∂
ν
v
i
(
z
)
≤
∂
ν
u
i
(
z
)
≤
0
.
Then by using the fact that
𝐮
,
𝐯
,
𝐡
0
take the same boundary values and therefore have equal tangential
derivatives on
∂
Ω
, we deduce that
a
≤
A
ν
h
0
i
(
z
)
≤
A
ν
v
i
(
z
)
≤
A
ν
u
i
(
z
)
,
where a is a lower bound for
A
ν
h
0
i
(note that a
does not depend on δ).
Hence by property (2) of Γ we have
∫
∂
Ω
Γ
(
x
,
A
ν
𝐮
(
x
)
)
-
Γ
(
x
,
A
ν
𝐯
(
x
)
)
d
σ
=
∑
i
=
1
m
∫
∂
Ω
Γ
(
x
,
A
ν
u
1
,
…
,
A
ν
u
i
-
1
,
A
ν
u
i
,
A
ν
v
i
+
1
,
…
,
A
ν
v
m
)
(2.5)
-
Γ
(
x
,
A
ν
u
1
,
…
,
A
ν
u
i
-
1
,
A
ν
v
i
,
A
ν
v
i
+
1
,
…
,
A
ν
v
m
)
d
σ
≥
C
a
∑
i
=
1
m
∫
∂
Ω
A
ν
u
i
-
A
ν
v
i
d
σ
,
where
C
a
>
0
is the lower bound of the
∂
ξ
i
Γ
on the set
{
(
x
,
ξ
)
:
ξ
i
≥
a
}
. On the other hand, using the
identity (2.3) we get
C
0
∫
∂
Ω
A
ν
u
i
-
A
ν
v
i
d
σ
≥
∫
∂
Ω
φ
i
(
A
ν
u
i
-
A
ν
v
i
)
𝑑
σ
(2.6)
=
∫
Ω
u
i
Δ
p
u
i
+
|
∇
u
i
|
p
d
y
-
∫
Ω
v
i
Δ
p
v
i
+
|
∇
v
i
|
p
d
y
≥
∫
Ω
|
∇
u
i
|
p
𝑑
y
-
∫
Ω
|
∇
v
i
|
p
𝑑
y
,
where
C
0
=
max
∂
Ω
|
𝝋
|
, and in the
last line we used the facts that
∫
Ω
v
i
Δ
p
v
i
𝑑
y
=
0
and
u
i
,
Δ
p
u
i
≥
0
. Now consider the function
h
i
in
B
r
(
x
)
satisfying
Δ
p
h
i
=
0
, and taking boundary
values equal to
u
i
. We extend
h
i
to be equal to
u
i
outside of
B
r
(
x
)
. Then we have
𝐡
=
(
h
1
,
…
,
h
m
)
∈
V
δ
.
In addition,
h
i
=
u
i
=
φ
i
on
∂
Ω
and
h
i
=
u
i
=
0
on
{
u
i
=
0
}
-
B
r
(
x
)
. Hence due to the minimality property of
v
i
given by Lemma 2.6 we have
∫
Ω
|
∇
v
i
|
p
𝑑
y
≤
∫
Ω
|
∇
h
i
|
p
𝑑
y
.
Combining this with the above inequality we get
C
0
∫
∂
Ω
A
ν
u
i
-
A
ν
v
i
d
σ
≥
∫
Ω
|
∇
u
i
|
p
-
|
∇
v
i
|
p
d
y
≥
∫
Ω
|
∇
u
i
|
p
-
|
∇
h
i
|
p
d
y
≥
C
∫
B
r
(
x
)
|
∇
(
u
i
-
h
i
)
|
p
𝑑
y
,
where the last inequality can be proved similarly to the proof of [9, Lemma
3.1]. (Note that in
the last line we have also used the fact that
u
i
=
h
i
outside
B
r
(
x
)
.)
Summing the above inequality for each i, and using the facts that
J
ε
(
𝐮
)
≤
J
ε
(
𝐯
)
, and
f
ε
has Lipschitz constant equal to
1
ε
,
we get
C
a
C
0
∑
i
≤
m
∫
B
r
(
x
)
|
∇
(
u
i
-
h
i
)
|
p
𝑑
y
≤
C
a
∫
∂
Ω
∑
i
≤
m
(
A
ν
u
i
-
A
ν
v
i
)
d
σ
≤
∫
∂
Ω
Γ
(
x
,
A
ν
𝐮
(
x
)
)
-
Γ
(
x
,
A
ν
𝐯
(
x
)
)
d
σ
≤
f
ε
(
|
{
|
𝐯
|
>
0
}
|
)
-
f
ε
(
|
{
|
𝐮
|
>
0
}
|
)
≤
1
ε
|
B
r
(
x
)
∩
{
|
𝐮
|
=
0
}
|
,
since
0
≤
u
i
≤
v
i
, and outside of
B
r
(
x
)
,
|
𝐮
|
=
0
implies
|
𝐯
|
=
0
. Therefore by Lemma 2.5 applied
to
u
j
we obtain
|
B
r
(
x
)
∩
{
|
𝐮
|
=
0
}
|
≥
ε
C
a
C
0
∑
i
≤
m
∫
B
r
(
x
)
|
∇
(
u
i
-
h
i
)
|
p
d
y
≥
ε
C
a
C
0
∫
B
r
(
x
)
|
∇
(
u
j
-
h
j
)
|
p
𝑑
y
≥
ε
C
a
c
C
0
(
1
r
sup
B
r
/
2
(
x
)
h
j
)
p
⋅
|
B
r
(
x
)
∩
{
u
j
=
0
}
|
≥
ε
C
a
c
C
0
(
1
r
sup
B
r
/
2
(
x
)
u
j
)
p
⋅
|
B
r
(
x
)
∩
{
u
j
=
0
}
|
≥
ε
C
a
M
p
c
C
0
|
B
r
(
x
)
∩
{
|
𝐮
|
=
0
}
|
,
since
|
𝐮
|
=
0
implies
u
j
=
0
, and
h
j
≥
u
j
as
u
j
is p-subharmonic. Hence if
M
>
(
c
C
0
ε
C
a
)
1
p
,
then
|
B
r
(
x
)
∩
{
|
𝐮
|
=
0
}
|
must be zero, as
desired. Note that in this case the above inequality also implies
that
u
i
=
h
i
in
B
r
(
x
)
for each i; so
u
i
satisfies
the equation in
B
r
(
x
)
.
∎
Corollary 2.8.
All minimizers
u
ε
δ
are Lipschitz, and for every
Ω
′
⊂
⊂
Ω
there exists
a constant
K
ε
=
K
ε
(
Ω
′
)
, independent
of δ, such that
∥
𝐮
ε
δ
|
Ω
′
∥
Lip
≤
K
ε
.
In addition,
Δ
p
(
u
ε
δ
)
i
=
0
in the
open set
{
|
u
ε
δ
|
>
0
}
.
Proof.
For simplicity we set
𝐮
=
𝐮
ε
δ
.
First let us show that
{
|
𝐮
|
>
0
}
is an open set. Suppose
x
∈
{
|
𝐮
|
>
0
}
. Then
u
j
(
x
)
>
0
for some j. Then
for small enough r we must have
1
r
sup
B
r
/
2
(
x
)
u
j
≥
1
r
u
j
(
x
)
≥
M
.
Hence the previous theorem implies that
B
r
(
x
)
⊂
{
|
𝐮
|
>
0
}
and we have
Δ
p
u
i
=
0
in
B
r
(
x
)
.
Next note that
∇
𝐮
=
0
a.e. in
{
|
𝐮
|
=
0
}
. So suppose
x
∈
{
|
𝐮
|
>
0
}
∩
Ω
′
. Let
Ω
′
⊂
⊂
Ω
~
⊂
⊂
Ω
,
and
B
=
B
d
(
x
)
, where
d
=
dist
(
x
,
∂
(
{
|
𝐮
|
>
0
}
∩
Ω
~
)
)
.
If
∂
B
touches
∂
{
|
𝐮
|
=
0
}
then
B
d
+
d
′
(
x
)
intersects
{
|
𝐮
|
=
0
}
, and by previous theorem we have
1
d
+
d
′
sup
B
(
d
+
d
′
)
/
2
(
x
)
u
i
≤
M
for every i. Hence in the limit
d
′
→
0
we get
1
d
sup
B
d
/
2
(
x
)
u
i
≤
M
.
Now since the
u
i
are p-harmonic in B, as shown in the proof
of [7, Lemma 3.1], we have
|
∇
u
i
(
x
)
|
≤
C
1
d
sup
B
d
/
2
(
x
)
u
i
≤
C
M
,
where the constant C depends only on p and the dimension n.
On the other hand, if
∂
B
touches
∂
Ω
~
then,
by the interior derivative estimate of [11],
we obtain (the dependence on d follows from the proof of this estimate; see [11, equation (3.4)])
|
∇
u
i
(
x
)
|
≤
C
(
n
,
p
)
d
n
∥
𝐮
∥
W
1
,
p
≤
C
,
since
d
≥
dist
(
Ω
′
,
∂
Ω
~
)
, and
∥
𝐮
∥
W
1
,
p
is bounded independently of δ
as will be shown now. Let
Ω
′
⊂
⊂
Ω
be a smooth open set
with
|
Ω
-
Ω
′
|
=
1
. Let
𝐮
0
be a vector-valued function
on
Ω
-
Ω
′
that satisfies the equation
Δ
p
u
0
i
=
0
, and
takes the boundary values
𝝋
on
∂
Ω
and 0 on
∂
Ω
′
. Then for every small enough δ
we have
𝐮
0
∈
V
δ
. Hence (remember that
𝐮
=
𝐮
ε
δ
)
C
=
J
ε
(
𝐮
0
)
≥
J
ε
(
𝐮
ε
δ
)
≥
∫
∂
Ω
Γ
(
x
,
A
ν
𝐮
ε
δ
(
x
)
)
𝑑
σ
≥
∫
∂
Ω
∑
i
=
1
m
ψ
i
(
x
)
A
ν
(
u
ε
δ
)
i
-
C
d
σ
,
where we used (1.2) in the last line. Thus by
Lemma 2.2 the
∥
∇
𝐮
ε
δ
∥
L
p
(
Ω
;
ℝ
m
)
is bounded as
δ
→
0
, and the boundedness of
∥
𝐮
ε
δ
∥
W
1
,
p
follows from Poincaré inequality and the fact that all of
𝐮
ε
δ
’s have the same boundary values.
Finally, to see that
𝐮
is Lipschitz continuous on all
of Ω, note that
𝐮
has p-harmonic components near
the smooth boundary
∂
Ω
, attaining smooth boundary conditions
𝝋
; hence the gradient of
𝐮
is bounded
near the boundary too.
∎
Lemma 2.9.
There exists
δ
0
=
δ
0
(
ε
)
>
0
such that for
every δ we have
|
u
ε
δ
|
>
0
in
Ω
δ
0
.
Remark.
Note that as a consequence,
Δ
p
(
u
ε
δ
)
i
=
0
on
Ω
δ
0
for every δ (by Theorem 2.7).
In other words,
𝐮
ε
δ
∈
V
δ
0
for every δ.
Proof.
Suppose to the contrary that there is a sequence
𝐮
k
=
𝐮
ε
δ
k
for which we have
2
d
k
:=
dist
(
{
|
𝐮
k
|
=
0
}
,
∂
Ω
)
→
0
.
Then the midpoint of the closest points on
{
|
𝐮
k
|
=
0
}
and
∂
Ω
, which we call
x
k
, has distance
d
k
from
both of these sets. So the boundary of the ball
B
d
k
(
x
k
)
touches both of these sets. In addition, by Theorem 2.7,
for every
t
>
0
we must have
1
d
k
sup
B
d
k
/
2
(
x
k
+
t
ν
k
)
u
k
i
≤
M
ε
for every i (here
ν
k
is the direction of the line segment
from
x
k
to its closest point on
{
|
𝐮
k
|
=
0
}
).
So in the limit
t
→
0
we get
(2.7)
sup
B
d
k
/
2
(
x
k
)
u
k
i
≤
M
ε
d
k
.
We also have
sup
B
d
k
(
x
k
)
|
𝐮
k
|
≥
c
0
,
where
c
0
=
min
i
min
∂
Ω
φ
i
>
0
. Because at the point
y
k
∈
∂
B
d
k
(
x
k
)
∩
∂
Ω
we have
u
k
i
(
y
k
)
=
φ
i
(
y
k
)
≥
c
0
(note that
u
k
i
is continuous up to the boundary).
Next consider the functions
𝐮
^
k
(
x
)
:=
𝐮
(
x
k
+
d
k
x
)
sup
B
d
k
(
x
k
)
|
𝐮
k
|
on
B
1
. Then
u
^
k
i
is positive and p-harmonic
on
B
1
, and we have
sup
B
1
|
𝐮
^
k
|
=
1
.
In addition,
by (2.7) we have
sup
B
1
/
2
u
^
k
i
=
sup
B
d
k
/
2
(
x
k
)
u
k
i
sup
B
d
k
(
x
k
)
|
𝐮
k
|
≤
M
ε
d
k
c
0
→
k
→
∞
0
.
Furthermore, note that
u
^
k
i
is a uniformly bounded sequence
of p-harmonic functions on
B
1
, so there is
α
>
0
such
that for all
r
<
1
the Hölder norms
∥
u
^
k
i
∥
C
0
,
α
(
B
¯
r
)
are uniformly bounded (see [10, p. 251]). Hence,
by a diagonal argument, we can construct a subsequence of
u
^
k
i
,
which we still denote by
u
^
k
i
, that locally uniformly
converges to a nonnegative p-harmonic function
u
^
∞
i
on
B
1
. In addition,
u
^
∞
i
must vanish on
B
1
/
2
by the above estimate. Thus by the strong maximum principle we must
have
𝐮
^
∞
≡
0
on
B
1
.
Now for
y
k
∈
∂
B
d
k
(
x
k
)
∩
∂
Ω
and
r
<
d
k
we have
osc
B
r
(
y
k
)
∩
Ω
u
k
i
≤
C
(
n
,
p
)
(
r
α
+
osc
B
r
(
y
k
)
∩
∂
Ω
φ
i
)
≤
C
r
α
for some
α
∈
(
0
,
1
)
. This estimate holds by [14, Theorem 4.19] when
1
<
p
≤
n
. And when
p
>
n
this estimate
holds due to the uniform Hölder continuity of
u
k
i
on
Ω
¯
,
since
∥
𝐮
k
∥
W
1
,
p
(
Ω
)
is uniformly bounded as we
have seen in the proof of Corollary 2.8. Hence for
r
=
d
k
/
2
we have
min
B
d
k
/
2
(
y
k
)
∩
B
d
k
(
x
k
)
u
k
i
≥
min
B
d
k
/
2
(
y
k
)
∩
Ω
u
k
i
≥
1
2
c
0
,
where
c
0
=
min
i
min
∂
Ω
φ
i
. Therefore for
y
^
k
=
1
d
k
(
y
k
-
x
k
)
∈
∂
B
1
we have
min
B
1
/
2
(
y
^
k
)
∩
B
1
u
^
k
i
=
1
sup
B
d
k
(
x
k
)
|
𝐮
k
|
min
B
d
k
/
2
(
y
k
)
∩
B
d
k
(
x
k
)
u
k
i
≥
c
>
0
,
since
sup
B
d
k
(
x
k
)
|
𝐮
k
|
≤
m
C
0
where
C
0
=
max
∂
Ω
|
𝝋
|
.
Thus
u
^
k
i
has a uniform positive lower bound on a subset
of
B
1
with positive volume (where the volume is independent
of k). So no subsequence of
𝐮
^
k
can converge
locally uniformly to
𝐮
^
∞
≡
0
, because
otherwise they will uniformly converge to 0 outside a set of small
volume, contradicting the uniform boundedness from below.
∎
Now we can find a minimizer for
J
ε
over V.
Theorem 2.10.
There exists a minimizer
u
ε
∈
V
for
J
ε
.
Moreover,
u
ε
is a Lipschitz function, and
Δ
p
u
ε
i
=
0
in the open set
{
|
u
ε
|
>
0
}
.
Remark.
As we will see in the following proof,
𝐮
ε
δ
∈
V
δ
0
for
δ
0
=
δ
0
(
ε
)
given by the above lemma. So in fact
𝐮
ε
is a minimizer of
J
ε
over some
V
δ
, and therefore it has all the properties of
𝐮
ε
δ
’s that we have proved so far. In particular, we have
|
𝐮
ε
|
>
0
on
Ω
δ
0
.
Proof.
As we have shown in the proof of Corollary 2.8,
∥
∇
𝐮
ε
δ
∥
L
p
(
Ω
;
ℝ
m
)
is bounded as
δ
→
0
. Hence there is a subsequence such that
𝐮
ε
δ
⇀
𝐮
ε
weakly in
W
1
,
p
(and also a.e.) with
A
(
∇
(
u
ε
δ
)
i
)
⇀
A
(
∇
u
ε
i
)
in
L
q
as
δ
→
0
. So, in particular,
u
ε
i
≥
0
,
u
ε
i
is p-subharmonic, and attains the boundary
condition
φ
i
. Furthermore, by Corollary 2.8,
𝐮
ε
δ
→
𝐮
ε
uniformly on compact subsets of Ω. Hence for each ball
B
¯
⊂
{
|
𝐮
ε
|
>
0
}
and all small enough δ we have
B
¯
⊂
{
|
𝐮
ε
δ
|
>
0
}
.
Therefore by using test functions with support in B together with
A
(
∇
(
u
ε
δ
)
i
)
⇀
A
(
∇
u
ε
i
)
we can conclude that
u
ε
i
is p-harmonic in B.
The same reasoning applied to test functions with support in
Ω
δ
0
, for
δ
0
given by the previous lemma, implies that
u
ε
i
is p-harmonic in
Ω
δ
0
, and thus
𝐮
ε
∈
V
δ
0
⊂
V
.
In particular,
u
ε
i
is p-harmonic near the smooth
boundary
∂
Ω
, attaining smooth boundary conditions
φ
i
,
so it is Lipschitz near
∂
Ω
. Moreover,
𝐮
ε
is Lipschitz inside Ω away from its boundary, because it is the
uniform limit of a sequence of Lipschitz functions with uniformly
bounded Lipschitz constants. Hence
𝐮
ε
is
Lipschitz on all of Ω.
Finally, note that
𝐮
ε
minimizes
J
ε
over V, since for every
𝐰
∈
V
we have
𝐰
∈
V
δ
for some δ. Thus we obtain
J
ε
(
𝐮
ε
δ
)
≤
J
ε
(
𝐰
)
.
However,
𝐮
ε
δ
→
𝐮
ε
,
so we get
J
ε
(
𝐮
ε
)
≤
J
ε
(
𝐰
)
due to the semicontinuity of
J
ε
.
∎
3 Regularity of solutions to the penalized problem
To simplify the notation, throughout this section we will suppress the index ε in
𝐮
ε
.
Theorem 3.1.
For
τ
∈
(
0
,
1
4
)
there exists
m
ε
(
τ
)
such that if for each i we have
1
r
sup
B
r
/
2
(
x
)
u
i
≤
m
ε
(
τ
)
,
then
B
τ
r
(
x
)
⊂
{
|
u
|
=
0
}
.
Proof.
Similarly to Lemma 2.6, we can show that there
is
v
i
∈
W
1
,
p
(
Ω
)
that minimizes the functional
∫
Ω
|
∇
v
i
|
p
𝑑
x
among all functions with
v
i
=
φ
i
on
∂
Ω
and
v
i
≤
0
on
{
u
i
=
0
}
∪
B
¯
τ
r
(
x
)
. The function
v
i
also satisfies
Δ
p
v
i
≥
0
,
∫
Ω
v
i
Δ
p
v
i
𝑑
x
=
0
,
u
i
≥
v
i
≥
0
(to see this, note that
Δ
p
v
i
≥
Δ
p
u
i
on
Ω
-
(
{
u
i
=
0
}
∪
B
¯
τ
r
(
x
)
)
⊂
{
|
𝐮
|
>
0
}
,
and
v
i
-
u
i
≤
0
on
{
u
i
=
0
}
∪
B
¯
τ
r
(
x
)
or
∂
Ω
). In addition, we have
𝐯
=
(
v
1
,
…
,
v
m
)
∈
V
δ
1
⊂
V
(where
δ
1
is small enough so that
B
¯
τ
r
(
x
)
⊂
Ω
-
Ω
δ
1
).
Thus
J
ε
(
𝐮
)
≤
J
ε
(
𝐯
)
.
Let us assume that
δ
1
is small enough so that
B
¯
r
(
x
)
⊂
Ω
-
Ω
δ
1
and
𝐮
∈
V
δ
1
. Let
𝐰
be a vector-valued
p-harmonic function in
Ω
δ
1
with boundary values
equal to
𝝋
on
∂
Ω
and equal to 0 on
∂
Ω
δ
1
-
∂
Ω
. Then we have
u
i
≥
v
i
≥
w
i
≥
0
(since
𝐮
,
𝐯
are also p-harmonic on
Ω
δ
1
,
and nonnegative everywhere). Thus for each
z
∈
∂
Ω
we have
0
≥
∂
ν
w
i
(
z
)
≥
∂
ν
v
i
(
z
)
≥
∂
ν
u
i
(
z
)
.
Next using the fact that
𝐮
,
𝐯
,
𝐰
take
the same boundary values on
∂
Ω
, and therefore have equal
tangential derivatives on
∂
Ω
, we deduce that
0
≥
A
ν
w
i
(
z
)
≥
A
ν
v
i
(
z
)
≥
A
ν
u
i
(
z
)
.
Now similar to (2) we can show that
∫
∂
Ω
Γ
(
x
,
A
ν
𝐯
(
x
)
)
-
Γ
(
x
,
A
ν
𝐮
(
x
)
)
d
σ
≤
C
1
∑
i
=
1
m
∫
∂
Ω
A
ν
v
i
-
A
ν
u
i
d
σ
,
where
C
1
>
0
is the upper bound of
∂
ξ
i
Γ
’s
on the set
{
(
x
,
ξ
)
:
ξ
i
≤
0
}
. On the other hand, using the
identity (2.3) we obtain (using the notation
c
0
=
min
i
min
∂
Ω
φ
i
)
c
0
∫
∂
Ω
A
ν
v
i
-
A
ν
u
i
d
σ
≤
∫
∂
Ω
φ
i
(
A
ν
v
i
-
A
ν
u
i
)
𝑑
σ
=
∫
Ω
v
i
Δ
p
v
i
+
|
∇
v
i
|
p
d
y
-
∫
Ω
u
i
Δ
p
u
i
+
|
∇
u
i
|
p
d
y
(3.1)
=
∫
Ω
|
∇
v
i
|
p
𝑑
y
-
∫
Ω
|
∇
u
i
|
p
𝑑
y
,
where in the last line we used the facts that
∫
Ω
v
i
Δ
p
v
i
𝑑
y
=
0
,
and
Δ
p
u
i
=
0
on
{
u
i
≠
0
}
⊂
{
|
𝐮
|
>
0
}
.
Summing the above inequality for each i, and using the facts that
J
ε
(
𝐮
)
≤
J
ε
(
𝐯
)
, and
the derivative of
f
ε
is bounded below by ε,
we get
C
1
c
0
∑
i
≤
m
∫
Ω
|
∇
v
i
|
p
-
|
∇
u
i
|
p
d
y
≥
C
1
∫
∂
Ω
∑
i
≤
m
(
A
ν
v
i
-
A
ν
u
i
)
d
σ
≥
∫
∂
Ω
Γ
(
x
,
A
ν
𝐯
(
x
)
)
-
Γ
(
x
,
A
ν
𝐮
(
x
)
)
d
σ
≥
f
ε
(
|
{
|
𝐮
|
>
0
}
|
)
-
f
ε
(
|
{
|
𝐯
|
>
0
}
|
)
≥
ε
|
{
|
𝐮
|
>
0
}
∩
{
|
𝐯
|
=
0
}
|
(3.2)
≥
ε
|
{
|
𝐮
|
>
0
}
∩
B
τ
r
(
x
)
|
,
since
u
i
≥
v
i
≥
0
and
v
i
=
0
in
B
τ
r
(
x
)
.
Next we define
g
:
(
0
,
∞
)
→
ℝ
by
g
(
t
)
:=
{
t
p
-
n
p
-
1
-
(
τ
r
)
p
-
n
p
-
1
,
p
>
n
,
log
t
-
log
(
τ
r
)
,
p
=
n
,
(
τ
r
)
p
-
n
p
-
1
-
t
p
-
n
p
-
1
,
p
<
n
.
Note that g is an increasing function that vanishes at
t
=
τ
r
,
and is negative for
t
<
τ
r
. In addition,
g
(
|
x
|
)
is a p-harmonic
function in
ℝ
n
-
{
0
}
, which is negative on
B
τ
r
(
x
)
and vanishes on
∂
B
τ
r
(
x
)
. Now let us define
h
i
:
B
τ
r
(
x
)
→
ℝ
by
h
i
(
y
)
:=
min
{
u
i
(
y
)
,
s
i
g
(
τ
r
)
(
g
(
|
y
-
x
|
)
)
+
}
,
where
s
i
:=
max
B
¯
τ
r
(
x
)
u
i
. We extend
h
i
by
u
i
outside of
B
τ
r
(
x
)
. Note that
we have
h
i
=
0
on
{
u
i
=
0
}
∩
B
¯
τ
r
(
x
)
and
h
i
=
u
i
=
φ
i
on
∂
Ω
. Hence
h
i
competes
with
v
i
, and we have
∫
Ω
|
∇
v
i
|
p
𝑑
x
≤
∫
Ω
|
∇
h
i
|
p
𝑑
x
.
Therefore we can exchange
v
i
by
h
i
in inequality (3.2)
to get
ε
c
0
C
1
|
{
|
𝐮
|
>
0
}
∩
B
τ
r
(
x
)
|
≤
∑
i
≤
m
∫
B
τ
r
(
x
)
|
∇
h
i
|
p
-
|
∇
u
i
|
p
d
y
.
Now since
h
i
=
0
on
B
τ
r
(
x
)
, we can rewrite the above
inequality as
(3.3)
ε
c
0
C
1
|
{
|
𝐮
|
>
0
}
∩
B
τ
r
(
x
)
|
+
∑
i
≤
m
∫
B
τ
r
(
x
)
|
∇
u
i
|
p
d
y
≤
∑
i
≤
m
∫
B
τ
r
(
x
)
-
B
τ
r
(
x
)
|
∇
h
i
|
p
-
|
∇
u
i
|
p
d
y
.
But
|
∇
h
i
|
p
-
|
∇
u
i
|
p
≤
-
p
|
∇
h
i
|
p
-
2
∇
h
i
⋅
∇
(
u
i
-
h
i
)
,
since for two vectors
a
,
b
we have
|
a
|
p
-
|
b
|
p
≤
-
p
|
a
|
p
-
2
a
⋅
(
b
-
a
)
due to the convexity of the function
⋅
↦
|
⋅
|
p
(see for example [12]).
So we can estimate the right-hand side of (3.3) as follows
(using integration by parts, and the facts that
Δ
p
h
i
=
0
on
{
u
i
>
h
i
}
,
h
i
=
0
on
∂
B
τ
r
(
x
)
, and
h
i
=
u
i
on
∂
B
τ
r
(
x
)
):
∫
B
τ
r
(
x
)
-
B
τ
r
(
x
)
|
∇
h
i
|
p
-
|
∇
u
i
|
p
d
y
≤
-
p
∫
B
τ
r
(
x
)
-
B
τ
r
(
x
)
|
∇
h
i
|
p
-
2
∇
(
u
i
-
h
i
)
⋅
∇
h
i
d
y
=
p
∫
∂
B
τ
r
(
x
)
(
u
i
-
h
i
)
|
∇
h
i
|
p
-
2
∇
h
i
⋅
ν
d
σ
-
p
∫
∂
B
τ
r
(
x
)
(
u
i
-
h
i
)
|
∇
h
i
|
p
-
2
∇
h
i
⋅
ν
d
σ
=
p
∫
∂
B
τ
r
(
x
)
u
i
|
∇
h
i
|
p
-
2
∇
h
i
⋅
ν
d
σ
=
C
(
n
,
p
,
τ
)
s
i
p
-
1
r
p
-
1
∫
∂
B
τ
r
(
x
)
u
i
𝑑
σ
,
where the last equality is calculated using the fact
h
i
(
y
)
=
s
i
g
(
τ
r
)
(
g
(
|
y
-
x
|
)
)
+
=
0
on
B
¯
τ
r
(
x
)
; hence on
∂
B
τ
r
(
x
)
we have
∇
h
i
=
C
(
τ
)
s
i
r
n
-
p
p
-
1
{
2
|
p
-
n
|
p
-
1
|
y
-
x
|
2
-
n
-
p
p
-
1
(
y
-
x
)
,
p
≠
n
,
|
y
-
x
|
-
2
(
y
-
x
)
,
p
=
n
,
and thus
|
∇
h
i
|
p
-
2
∇
h
i
⋅
ν
=
C
(
n
,
p
,
τ
)
s
i
p
-
1
r
n
-
p
|
y
-
x
|
2
-
n
-
p
|
y
-
x
|
p
-
2
(
y
-
x
)
⋅
(
y
-
x
)
τ
r
=
C
(
n
,
p
,
τ
)
s
i
p
-
1
r
n
-
p
|
y
-
x
|
-
n
+
2
1
τ
r
=
C
(
n
,
p
,
τ
)
s
i
p
-
1
r
p
-
1
.
Hence (3.3) becomes
(3.4)
ε
c
0
C
1
|
{
|
𝐮
|
>
0
}
∩
B
τ
r
(
x
)
|
+
∑
i
≤
m
∫
B
τ
r
(
x
)
|
∇
u
i
|
p
d
y
≤
C
(
n
,
p
,
τ
)
∑
i
≤
m
s
i
p
-
1
r
p
-
1
∫
∂
B
τ
r
(
x
)
u
i
d
σ
.
On the other hand we have
∫
∂
B
τ
r
(
x
)
u
i
𝑑
σ
≤
c
(
n
,
τ
)
(
∫
B
τ
r
(
x
)
u
i
𝑑
y
+
∫
B
τ
r
(
x
)
|
∇
u
i
|
𝑑
y
)
(3.5)
≤
c
(
n
,
τ
)
(
(
s
i
+
1
)
⋅
|
{
|
𝐮
|
>
0
}
∩
B
τ
r
(
x
)
|
+
∫
B
τ
r
(
x
)
|
∇
u
i
|
p
d
y
)
,
where in the last line we estimated
u
i
,
|
∇
u
i
|
from above
by
s
i
,
1
+
|
∇
u
i
|
p
on the set
{
u
i
>
0
}
⊂
{
|
𝐮
|
>
0
}
.
Next note that
(3.6)
s
i
=
max
B
¯
τ
r
(
x
)
u
i
≤
sup
B
r
/
2
(
x
)
u
i
≤
r
m
ε
(
τ
)
,
since
τ
<
1
2
. Combining inequalities (3.4),
(3.5), and (3.6), we get
ε
c
0
C
1
|
{
|
𝐮
|
>
0
}
∩
B
τ
r
(
x
)
|
+
∑
i
≤
m
∫
B
τ
r
(
x
)
|
∇
u
i
|
p
d
y
≤
c
C
∑
i
≤
m
s
i
p
-
1
r
p
-
1
(
(
s
i
+
1
)
⋅
|
{
|
𝐮
|
>
0
}
∩
B
τ
r
(
x
)
|
+
∫
B
τ
r
(
x
)
|
∇
u
i
|
p
d
y
)
≤
c
C
m
ε
p
-
1
(
τ
)
(
|
{
|
𝐮
|
>
0
}
∩
B
τ
r
(
x
)
|
∑
i
≤
m
(
s
i
+
1
)
+
∑
i
≤
m
∫
B
τ
r
(
x
)
|
∇
u
i
|
p
d
y
)
.
Now if
m
ε
(
τ
)
is small enough, we must necessarily
have
|
𝐮
|
=
0
on
B
τ
r
(
x
)
, as desired.
∎
Now let us set
U
:=
{
x
∈
Ω
:
|
𝐮
(
x
)
|
>
0
}
,
E
:=
{
x
∈
Ω
:
|
𝐮
(
x
)
|
=
0
}
.
Lemma 3.2.
For every i we have
U
=
{
x
∈
Ω
:
u
i
(
x
)
>
0
}
,
E
=
{
x
∈
Ω
:
u
i
(
x
)
=
0
}
.
Proof.
By Theorem 2.10, each
u
i
is p-harmonic in
the open set U. So in each component of U either
u
i
>
0
or
u
i
≡
0
(by the strong maximum principle). Now consider
a component of U, say
U
1
. If
∂
U
1
does not intersect
∂
Ω
, then it must be a subset of E. Therefore every
u
i
vanishes on
∂
U
1
, and hence every
u
i
vanishes
on
U
1
by the maximum principle. So we would have
U
1
⊂
E
,
which is a contradiction. Thus
∂
U
1
must intersect
∂
Ω
.
Hence each
u
i
>
0
on
U
1
, since they are positive on
∂
Ω
.
Therefore each
u
i
is positive on every component of U, as desired.
∎
Corollary 3.3.
There are
c
,
C
>
0
such that for
x
∈
U
near
∂
E
we have
c
⋅
dist
(
x
,
∂
E
)
≤
|
𝐮
(
x
)
|
≤
C
⋅
dist
(
x
,
∂
E
)
.
Proof.
The right-hand side inequality holds according to the Lipschitz regularity
of the solutions, Theorem 2.10. To see the left-hand side inequality, we argue indirectly. Assume to the contrary that there exists a sequence
x
k
∈
U
such that
(3.7)
|
𝐮
(
x
k
)
|
≤
1
k
dist
(
x
k
,
∂
E
)
.
Let
r
k
=
dist
(
x
k
,
∂
E
)
and define
𝐮
k
(
x
)
=
𝐮
(
x
k
+
r
k
x
)
r
k
.
The sequence
𝐮
k
is uniformly bounded and uniformly Lipschitz in
B
1
due to Lipschitz regularity
of
𝐮
and assumption (3.7).
Recall that
Δ
p
u
k
i
=
0
in U, then we
may choose a converging subsequence
𝐮
k
→
𝐮
0
such
that
u
0
i
is also p-harmonic. Furthermore, by Theorem 3.1
we get that
sup
B
1
/
2
(
0
)
|
𝐮
0
|
=
lim
k
→
∞
sup
B
1
/
2
(
0
)
|
𝐮
k
|
≥
m
ε
>
0
,
since
|
𝐮
k
(
0
)
|
>
0
. Also, (3.7) yields that
𝐮
0
(
0
)
=
0
, which
contradicts the maximum (minimum) principle; remember that each component
of
𝐮
0
is nonnegative.
∎
Corollary 3.4.
There exists
c
=
c
ε
∈
(
0
,
1
)
such that for any
x
∈
∂
U
and small enough r we have
(3.8)
c
≤
|
E
∩
B
r
(
x
)
|
|
B
r
(
x
)
|
≤
1
-
c
.
Proof.
The proof is similar to the proof of [9, Theorem 4.2].
By Theorem 3.1, there exists
z
∈
B
r
/
2
(
x
)
such
that
|
𝐮
(
z
)
|
≥
m
ε
r
>
0
. Now for any
y
∈
B
τ
r
(
z
)
we have
|
𝐮
(
y
)
-
𝐮
(
z
)
|
≤
Lip
(
𝐮
)
|
y
-
z
|
<
Lip
(
𝐮
)
τ
r
<
m
ε
r
2
,
provided that τ is small enough. Hence we must have
|
𝐮
(
y
)
|
>
m
ε
r
2
>
0
.
This gives the upper estimate in (3.8).
To prove the estimate from below, suppose to the contrary that there
exists a sequence of points
x
k
∈
∂
U
and radii
r
k
→
0
such that
|
{
|
𝐮
|
=
0
}
∩
B
r
k
(
x
k
)
|
<
1
k
|
B
r
k
(
x
k
)
|
=
1
k
r
k
n
|
B
1
|
.
Now let us define
𝐮
k
(
x
)
=
𝐮
(
x
k
+
r
k
x
)
r
k
.
Note that
𝐮
k
(
0
)
=
𝐮
(
x
k
)
=
0
, and thus
𝐮
k
is uniformly bounded and uniformly Lipschitz in
B
1
=
B
1
(
0
)
due to Lipschitz regularity of
𝐮
. Also
|
{
|
𝐮
k
|
=
0
}
∩
B
1
|
=
1
r
k
n
|
{
|
𝐮
|
=
0
}
∩
B
r
k
(
x
k
)
|
→
k
→
∞
0
.
Let
v
k
i
be a p-harmonic function in
B
1
/
2
with boundary
data
v
k
i
=
u
k
i
on
∂
B
1
/
2
. Then
h
k
i
(
x
)
=
r
k
v
k
i
(
x
-
x
k
r
k
)
is a p-harmonic function in
B
r
k
/
2
(
x
k
)
with boundary
data
h
k
i
=
u
i
on
∂
B
r
k
/
2
(
x
k
)
. Now, similarly
to the proof of Theorem 2.7, we can show that
(3.9)
∫
B
1
/
2
|
∇
(
u
k
i
-
v
k
i
)
|
p
𝑑
x
=
1
r
k
n
∫
B
r
k
/
2
(
x
k
)
|
∇
(
u
i
-
h
k
i
)
|
p
𝑑
x
≤
C
ε
1
r
k
n
|
{
|
𝐮
|
=
0
}
∩
B
r
k
(
x
k
)
|
→
k
→
∞
0
.
(Note that the constant C does not depend on the radius
r
k
or the point
x
k
.)
Since
u
k
i
and therefore
v
k
i
are uniformly Lipschitz
in
B
1
/
4
, we may assume that
u
k
i
→
u
0
i
and
v
k
i
→
v
0
i
uniformly in
B
1
/
4
. Observe that
Δ
p
v
0
i
=
0
, and
(3.9) implies that
u
0
i
=
v
0
i
+
C
for
some constant C. Thus
Δ
p
u
0
i
=
0
in
B
1
/
4
and
from the strong minimum principle it follows
u
0
i
≡
0
in
B
1
/
4
, since
u
0
i
≥
0
and
u
0
i
(
0
)
=
lim
u
k
i
(
0
)
=
0
.
On the other hand the nondegeneracy property, Theorem 3.1,
implies that (since
x
k
is not in the interior of
{
|
𝐮
|
=
0
}
)
∥
𝐮
k
∥
L
∞
(
B
1
/
4
)
=
1
r
k
∥
𝐮
∥
L
∞
(
B
r
k
/
4
(
x
k
)
)
≥
m
ε
2
>
0
.
Therefore we get
∥
𝐮
0
∥
L
∞
(
B
1
/
4
)
≥
m
ε
/
2
,
which is a contradiction.
∎
Hence we can apply the results in [4, Section 4]
and in [5, Section 3] to conclude (see also [9, Sections 5 and 6])
Theorem 3.5.
Let
u
=
u
ε
be a minimizer of
J
ε
over V. Then we have:
The
(
n
-
1
)
-
dimensional Hausdorff measure of
∂
E
is locally
finite, i.e.
ℋ
n
-
1
(
Ω
′
∩
∂
E
)
<
∞
for
every
Ω
′
⊂
⊂
Ω
. Moreover, there exist positive constants
c
ε
,
C
ε
, depending on
n
,
p
,
Ω
,
Ω
′
,
ε
,
such that for each ball
B
r
(
x
)
⊂
Ω
′
with
x
∈
∂
E
we have
c
ε
r
n
-
1
≤
ℋ
n
-
1
(
B
r
(
x
)
∩
∂
E
)
≤
C
ε
r
n
-
1
.
There exist Borel functions
q
i
=
q
ε
i
such that
Δ
p
u
i
=
q
i
ℋ
n
-
1
⌞
∂
E
,
that is, for any
ζ
∈
C
0
∞
(
Ω
)
we have
-
∫
Ω
A
[
u
i
]
⋅
∇
ζ
d
y
=
∫
∂
E
ζ
q
i
𝑑
ℋ
n
-
1
.
For
ℋ
n
-
1
-
a.e. points
x
∈
∂
E
we have
c
ε
≤
∑
i
=
1
m
q
i
(
x
)
≤
C
ε
.
For
ℋ
n
-
1
-
a.e. points
x
∈
∂
E
an outward unit normal
ν
=
ν
E
(
x
)
is defined, and
u
i
(
x
+
y
)
=
(
q
i
(
x
)
)
1
p
-
1
(
y
⋅
ν
)
+
+
o
(
|
y
|
)
,
which allows us to define
A
ν
u
i
(
x
)
=
q
i
(
x
)
at those points.
The reduced boundary
∂
red
E
satisfies
ℋ
n
-
1
(
∂
E
-
∂
red
E
)
=
0
.
4 The original problem
In this section we will show that for
ε
>
0
small enough,
a minimizer of
J
ε
over V satisfies
|
{
|
𝐮
ε
|
>
0
}
|
=
1
,
and hence it can be regarded as a solution to our original problem
(1.1). Remember that
U
=
U
ε
=
{
|
𝐮
ε
|
>
0
}
,
E
=
E
ε
=
{
|
𝐮
ε
|
=
0
}
.
Note that by Lemma 2.9, the free boundary
∂
E
has a positive distance from the fixed boundary
∂
Ω
.
We say
x
∈
∂
E
is a regular point of the free boundary
if it satisfies (3) and (4) in Theorem 3.5. The set of such regular points of the free boundary will be denoted
by
ℛ
=
ℛ
ε
; Theorem 3.5
shows that
ℋ
n
-
1
(
∂
E
-
ℛ
)
=
0
.
Lemma 4.1.
There is a constant
C
>
0
, independent of ε,
such that
inf
ℛ
ε
(
∑
i
≤
m
q
ε
i
)
≤
C
.
Remark.
Note that
∑
i
≤
m
q
ε
i
≥
c
ε
>
0
by Theorem 3.5.
Proof.
Let
Ω
′
⊂
⊂
Ω
be a smooth open set with
|
Ω
-
Ω
′
|
=
1
.
Let
𝐮
0
be a vector-valued function on
Ω
-
Ω
′
that satisfies the equation
Δ
p
u
0
i
=
0
, and takes the
boundary values
𝝋
on
∂
Ω
and 0
on
∂
Ω
′
. Then for some small enough
δ
0
we
have
𝐮
0
∈
V
δ
0
⊂
V
; hence
C
=
∫
∂
Ω
Γ
(
x
,
A
ν
𝐮
0
)
𝑑
σ
+
1
=
J
ε
(
𝐮
0
)
≥
J
ε
(
𝐮
ε
)
=
∫
∂
Ω
Γ
(
x
,
A
ν
𝐮
ε
)
d
σ
+
f
ε
(
|
{
|
𝐮
ε
|
>
0
}
|
)
≥
∫
∂
Ω
∑
i
=
1
m
ψ
i
A
ν
u
ε
i
-
C
d
σ
+
f
ε
(
|
{
|
𝐮
ε
|
>
0
}
|
)
≥
C
∑
i
=
1
m
∫
Ω
|
∇
u
ε
i
|
p
d
x
-
C
+
f
ε
(
|
{
|
𝐮
ε
|
>
0
}
|
)
≥
-
C
+
1
ε
(
|
{
|
𝐮
ε
|
>
0
}
|
-
1
)
,
where we have used (1.2) and Lemma 2.2.
Thus we get the bound
|
U
|
=
|
{
|
𝐮
ε
|
>
0
}
|
≤
1
+
C
ε
.
Note that
J
ε
(
𝐮
0
)
, and thus C, does
not depend on ε due to the definition of
f
ε
.
As a result, we have a lower bound for the volume of E. Hence,
by the isoperimetric inequality, we have a lower bound for
ℋ
n
-
1
(
∂
E
)
,
independent of ε. Now note that (keep in mind that
ν
E
points to the interior of U)
∫
∂
Ω
A
ν
u
ε
i
𝑑
σ
-
∫
∂
E
A
ν
u
ε
i
𝑑
ℋ
n
-
1
=
∫
U
Δ
p
u
ε
i
𝑑
x
=
0
.
Therefore we get
∫
∂
E
A
ν
u
ε
i
𝑑
ℋ
n
-
1
=
∫
∂
Ω
A
ν
u
ε
i
𝑑
σ
=
∫
Ω
Δ
p
u
ε
i
𝑑
x
≤
C
+
C
∫
∂
Ω
ψ
i
A
ν
u
ε
i
𝑑
σ
,
where the last inequality follows from the remark below Lemma 2.2.
Thus we have
inf
ℛ
ε
(
∑
i
≤
m
A
ν
u
ε
i
)
ℋ
n
-
1
(
∂
E
)
≤
∫
∂
E
∑
i
≤
m
A
ν
u
ε
i
d
ℋ
n
-
1
≤
C
+
C
∫
∂
Ω
∑
i
≤
m
ψ
i
A
ν
u
ε
i
d
σ
≤
C
+
C
∫
∂
Ω
Γ
(
x
,
A
ν
𝐮
ε
)
𝑑
σ
(by (1.2))
≤
C
+
C
J
ε
(
𝐮
ε
)
≤
C
+
C
J
ε
(
𝐮
0
)
≤
C
,
which gives the desired (noting that
q
i
=
A
ν
u
ε
i
by Theorem 3.5).
∎
Lemma 4.2.
For small enough ε we have
|
{
|
𝐮
ε
|
>
0
}
|
≥
1
.
Proof.
Consider a point
z
0
∈
Ω
c
which has distance
δ
0
from
∂
Ω
. Then the ball
B
δ
0
(
z
0
)
is
an exterior tangent ball to
∂
Ω
. Let
t
=
t
(
ε
)
be the first time at which
∂
B
δ
0
+
t
(
z
0
)
intersects
∂
{
|
𝐮
ε
|
=
0
}
, at a point
x
0
=
x
0
(
ε
)
.
Now let v be a p-harmonic function in
B
δ
0
+
t
(
z
0
)
-
B
¯
δ
0
(
z
0
)
with boundary values 0 on
∂
B
δ
0
+
t
(
z
0
)
and
c
0
on
∂
B
δ
0
(
z
0
)
, where
c
0
=
min
i
min
∂
Ω
φ
i
>
0
.
Then on
∂
(
Ω
∩
B
δ
0
+
t
(
z
0
)
)
we
have
v
≤
u
i
; so by the maximum principle we have
v
≤
u
i
in
Ω
∩
B
δ
0
+
t
(
z
0
)
. However, by an easy modification
of the proof of Hopf’s lemma (Lemma 2.4), we can see
that
v
(
x
)
≥
c
c
0
dist
(
x
,
∂
B
δ
0
+
t
(
z
0
)
)
,
where the constant c only depends on
n
,
p
,
δ
0
. Therefore,
for points x in the line segment between
x
0
,
z
0
we have
u
i
(
x
)
≥
v
(
x
)
≥
c
c
0
dist
(
x
,
∂
B
δ
0
+
t
(
z
0
)
)
=
c
c
0
|
x
-
x
0
|
.
Now consider the ball
B
r
(
x
0
)
for small enough r. Then
we have
1
r
sup
B
r
/
2
(
x
0
)
u
i
≥
1
r
c
c
0
r
2
=
c
c
0
2
,
independently of ε.
Let
𝐡
be the vector-valued function which satisfies
Δ
p
h
i
=
0
in
B
r
(
x
0
)
, and is equal to
𝐮
in
Ω
-
B
r
(
x
0
)
.
By Lemma 2.5 and the fact that
h
i
≥
u
i
we have
∫
B
r
(
x
0
)
|
∇
(
u
i
-
h
i
)
|
p
d
y
≥
C
(
1
r
sup
B
r
/
2
(
x
0
)
h
i
)
p
⋅
|
B
r
(
x
0
)
∩
{
u
i
=
0
}
|
≥
C
(
1
r
sup
B
r
/
2
(
x
0
)
u
i
)
p
⋅
|
B
r
(
x
0
)
∩
{
u
i
=
0
}
|
≥
C
|
B
r
(
x
0
)
∩
{
u
i
=
0
}
|
≥
C
|
B
r
(
x
0
)
∩
{
|
𝐮
|
=
0
}
|
.
Next let
𝐯
be the function given by Lemma 2.6
for
B
r
(
x
0
)
. We know that
J
ε
(
𝐮
)
≤
J
ε
(
𝐯
)
.
Then similarly to the proof of Theorem 2.7 we can see
that
C
∑
i
≤
m
∫
B
r
(
x
0
)
|
∇
(
u
i
-
h
i
)
|
p
𝑑
y
≤
∫
∂
Ω
Γ
(
x
,
A
ν
𝐮
)
-
Γ
(
x
,
A
ν
𝐯
)
d
σ
≤
f
ε
(
|
{
|
𝐯
|
>
0
}
|
)
-
f
ε
(
|
{
|
𝐮
|
>
0
}
|
)
.
A closer inspection of the proof of Theorem 2.7 reveals
that the constant C in the above estimate only depends on
n
,
p
,
Ω
,
𝝋
,
Γ
.
Now suppose to the contrary that
|
{
|
𝐮
|
>
0
}
|
<
1
.
Then, since
0
≤
u
i
≤
v
i
, and outside of
B
r
(
x
0
)
,
|
𝐮
|
=
0
implies
|
𝐯
|
=
0
, we have
|
{
|
𝐯
|
>
0
}
|
≤
|
{
|
𝐮
|
>
0
}
|
+
|
B
r
(
x
0
)
∩
{
|
𝐮
|
=
0
}
|
<
1
for small enough r. Hence, using the monotonicity of
f
ε
,
we have
f
ε
(
|
{
|
𝐯
|
>
0
}
|
)
-
f
ε
(
|
{
|
𝐮
|
>
0
}
|
)
≤
f
ε
(
|
{
|
𝐮
|
>
0
}
|
+
|
B
r
(
x
0
)
∩
{
|
𝐮
|
=
0
}
|
)
-
f
ε
(
|
{
|
𝐮
|
>
0
}
|
)
=
ε
|
B
r
(
x
0
)
∩
{
|
𝐮
|
=
0
}
|
.
Combining this estimate with the estimates of the above paragraph,
and using (3.8), we obtain
0
<
C
|
B
r
(
x
0
)
∩
{
|
𝐮
|
=
0
}
|
≤
ε
|
B
r
(
x
0
)
∩
{
|
𝐮
|
=
0
}
|
,
which gives a positive lower bound for ε, and results
in a contradiction.
∎
Theorem 4.3.
When ε is small enough, we have
|
{
|
𝐮
ε
|
>
0
}
|
=
1
.
Proof.
By the above lemma we only need to show that
|
{
|
𝐮
ε
|
>
0
}
|
≤
1
.
To this end, we will compare
𝐮
ε
with a suitable
perturbation of itself. Let
x
0
∈
ℛ
, and let
ρ
:
ℝ
→
ℝ
be a nonnegative smooth function supported in
(
0
,
1
)
. For small
enough
r
,
λ
>
0
we consider the vector field
T
r
(
x
)
:=
{
x
+
r
λ
ρ
(
|
x
-
x
0
|
r
)
ν
(
x
0
)
if
x
∈
B
r
(
x
0
)
,
x
elsewhere
.
Here,
ν
(
x
0
)
is the outward normal vector provided in (4) of
Theorem 3.5. We can easily see that for x
in
B
r
(
x
0
)
we have
(4.1)
D
T
r
(
x
)
⋅
=
I
⋅
+
λ
ρ
′
(
|
x
-
x
0
|
r
)
〈
x
-
x
0
,
⋅
〉
|
x
-
x
0
|
ν
(
x
0
)
,
where I is the identity matrix. Hence, if λ is small enough,
T
r
is a diffeomorphism that maps
B
r
(
x
0
)
onto itself.
Now consider
𝐯
r
(
x
)
:=
𝐮
(
T
r
-
1
(
x
)
)
for
r
>
0
small enough. Similarly to the proof of Theorem 3.1,
we consider the vector-valued function
𝐰
whose components
minimize the Dirichlet p-energy subject to the condition
w
i
≤
0
on
{
𝐮
=
0
}
∪
(
B
¯
r
(
x
0
)
∩
{
𝐯
r
=
0
}
)
.
With a calculation similar to (3) and
(2) we get
0
≤
J
ε
(
𝐰
)
-
J
ε
(
𝐮
)
≤
C
∑
i
=
1
m
∫
Ω
|
∇
w
i
|
p
-
|
∇
u
i
|
p
d
x
+
f
ε
(
|
{
|
𝐰
|
>
0
}
|
)
-
f
ε
(
|
{
|
𝐮
|
>
0
}
|
)
(4.2)
≤
C
∑
i
=
1
m
∫
B
r
(
x
0
)
|
∇
v
r
i
|
p
-
|
∇
u
i
|
p
d
x
+
f
ε
(
|
{
|
𝐰
|
>
0
}
|
)
-
f
ε
(
|
{
|
𝐮
|
>
0
}
|
)
,
where in the last inequality we have compared the Dirichlet p-energy
of
𝐰
with that of
𝐯
r
χ
B
r
(
x
0
)
+
𝐮
χ
Ω
-
B
r
(
x
0
)
.
Now notice that
∫
B
r
(
x
0
)
|
∇
v
r
i
|
p
𝑑
x
=
∫
B
r
(
x
0
)
|
D
T
r
(
T
r
-
1
(
x
)
)
-
1
∇
u
i
(
T
r
-
1
(
x
)
)
|
p
𝑑
x
=
∫
B
r
(
x
0
)
|
D
T
r
(
y
)
-
1
∇
u
i
(
y
)
|
p
|
det
D
T
r
(
y
)
|
d
y
=
r
n
∫
B
1
|
D
T
r
(
y
)
-
1
∇
u
i
(
y
)
|
p
|
det
D
T
r
(
y
)
|
d
z
,
z
=
y
-
x
0
r
.
From (4.1), for small enough λ we can write
D
T
r
(
y
)
-
1
=
I
+
(
∑
k
=
1
∞
(
-
1
)
k
λ
k
ρ
′
(
|
z
|
)
k
〈
z
,
ν
〉
k
-
1
|
z
|
k
-
1
)
〈
z
,
⋅
〉
|
z
|
ν
(
x
0
)
(4.3)
=
I
-
λ
ρ
′
(
|
z
|
)
〈
z
,
⋅
〉
|
z
|
ν
(
x
0
)
+
λ
2
g
(
λ
,
z
)
〈
z
,
⋅
〉
|
z
|
ν
(
x
0
)
for some g. Hence we have
D
T
r
(
y
)
-
1
∇
u
i
(
y
)
=
∇
u
i
(
y
)
-
λ
ρ
′
(
|
z
|
)
〈
z
,
∇
u
i
(
y
)
〉
|
z
|
ν
(
x
0
)
+
O
(
λ
2
)
.
Thus
|
D
T
r
(
y
)
-
1
∇
u
i
(
y
)
|
2
=
|
∇
u
i
(
y
)
|
2
-
2
λ
ρ
′
(
|
z
|
)
〈
z
,
∇
u
i
(
y
)
〉
|
z
|
〈
ν
(
x
0
)
,
∇
u
i
(
y
)
〉
+
O
(
λ
2
)
,
and therefore
|
D
T
r
(
y
)
-
1
∇
u
i
(
y
)
|
p
=
|
∇
u
i
(
y
)
|
p
(
1
-
p
λ
ρ
′
(
|
z
|
)
〈
z
,
∇
u
i
(
y
)
〉
|
z
|
|
∇
u
i
(
y
)
|
2
〈
ν
(
x
0
)
,
∇
u
i
(
y
)
〉
)
+
O
(
λ
2
)
.
Also, we have (noting that
D
T
r
is the identity matrix plus a
rank 1 matrix)
|
det
D
T
r
(
y
)
|
=
1
+
λ
ρ
′
(
|
z
|
)
〈
z
,
ν
(
x
0
)
〉
|
z
|
.
All these together, we obtain (remember that
y
=
x
0
+
r
z
)
r
-
n
∫
B
r
(
x
0
)
|
∇
v
r
i
|
p
-
|
∇
u
i
|
p
d
x
=
λ
∫
B
1
|
∇
u
i
(
y
)
|
p
ρ
′
(
|
z
|
)
(
〈
z
,
ν
(
x
0
)
〉
|
z
|
-
p
〈
z
,
∇
u
i
(
y
)
〉
〈
∇
u
i
(
y
)
,
ν
(
x
0
)
〉
|
z
|
|
∇
u
i
(
y
)
|
2
)
𝑑
z
+
O
(
λ
2
)
.
Now consider the blowup sequence
𝐮
r
(
z
)
:=
1
r
𝐮
(
x
0
+
r
z
)
.
We know that as
r
→
0
(see [5])
{
u
r
i
>
0
}
∩
B
1
→
{
z
:
z
⋅
ν
(
x
0
)
>
0
}
∩
B
1
,
∇
u
i
(
y
)
=
∇
u
r
i
(
z
)
→
(
q
i
(
x
0
)
)
1
p
-
1
ν
(
x
0
)
χ
{
z
⋅
ν
(
x
0
)
>
0
}
a.e. in
B
1
.
Therefore we get
r
-
n
∫
B
r
(
x
0
)
|
∇
v
r
i
|
p
-
|
∇
u
i
|
p
d
x
→
r
→
0
-
(
p
-
1
)
λ
|
q
i
(
x
0
)
|
p
p
-
1
∫
B
1
∩
{
z
⋅
ν
(
x
0
)
>
0
}
ρ
′
(
|
z
|
)
〈
z
,
ν
(
x
0
)
〉
|
z
|
𝑑
z
+
O
(
λ
2
)
.
Note that formula (4.3) for
(
D
T
r
)
-
1
does
not depend on r, and the function
⋅
↦
|
⋅
|
p
is
continuous; so the
O
(
λ
2
)
term converges to an
O
(
λ
2
)
term as
r
→
0
. Next note that
div
(
ρ
(
|
z
|
)
ν
)
=
ρ
′
(
|
z
|
)
|
z
|
〈
z
,
ν
〉
.
Thus (noting that
ρ
(
|
z
|
)
is zero near
∂
B
1
)
∫
B
1
∩
{
z
⋅
ν
(
x
0
)
>
0
}
ρ
′
(
|
z
|
)
〈
z
,
ν
(
x
0
)
〉
|
z
|
𝑑
z
=
-
∫
B
1
∩
{
z
⋅
ν
(
x
0
)
=
0
}
ρ
(
|
z
|
)
𝑑
z
=
-
ω
n
-
1
∫
0
1
ρ
(
t
)
t
n
-
1
𝑑
t
=
-
C
ρ
ω
n
-
1
,
where
ω
n
-
1
is the volume of the
(
n
-
1
)
-dimensional ball
of radius 1, and
C
ρ
depends only on ρ. Hence we
can write
∫
B
r
(
x
0
)
|
∇
v
r
i
|
p
-
|
∇
u
i
|
p
d
x
=
[
(
p
-
1
)
λ
C
ρ
ω
n
-
1
|
q
i
(
x
0
)
|
p
p
-
1
+
O
(
λ
2
)
]
r
n
+
o
(
r
n
)
.
On the other hand,
lim
r
→
0
r
-
n
|
B
r
(
x
0
)
∩
{
|
𝐯
r
|
>
0
}
|
=
lim
r
→
0
r
-
n
∫
{
|
𝐯
r
|
>
0
}
∩
B
r
(
x
0
)
𝑑
x
=
lim
r
→
0
r
-
n
∫
{
|
𝐮
|
>
0
}
∩
B
r
(
x
0
)
|
det
D
T
r
(
y
)
|
d
y
=
∫
B
1
∩
{
z
⋅
ν
(
x
0
)
>
0
}
1
+
λ
ρ
′
(
|
z
|
)
〈
z
,
ν
(
x
0
)
〉
|
z
|
d
z
=
1
2
ω
n
-
λ
ω
n
-
1
∫
0
1
ρ
(
t
)
t
n
-
1
𝑑
t
=
1
2
ω
n
-
λ
C
ρ
ω
n
-
1
.
Thus for
A
0
:=
(
{
|
𝐮
|
>
0
}
-
B
r
(
x
0
)
)
∪
(
{
|
𝐯
r
|
>
0
}
∩
B
r
(
x
0
)
)
we have
|
A
0
|
-
|
{
|
𝐮
|
>
0
}
|
=
|
B
r
(
x
0
)
∩
{
|
𝐯
r
|
>
0
}
|
-
|
B
r
(
x
0
)
∩
{
|
𝐮
|
>
0
}
|
=
-
λ
C
ρ
ω
n
-
1
r
n
+
o
(
r
n
)
.
In addition, it is easy to see that
{
|
𝐰
|
>
0
}
⊂
A
0
.
Now suppose to the contrary that
|
{
|
𝐮
|
>
0
}
|
>
1
.
Then we can choose r small enough so that
|
A
0
|
=
|
{
|
𝐮
|
>
0
}
|
-
λ
C
ρ
ω
n
-
1
r
n
+
o
(
r
n
)
>
1
.
Therefore, using the monotonicity of
f
ε
we get
f
ε
(
|
{
|
𝐰
|
>
0
}
|
)
-
f
ε
(
|
{
|
𝐮
|
>
0
}
|
)
≤
f
ε
(
|
A
0
|
)
-
f
ε
(
|
{
|
𝐮
|
>
0
}
|
)
=
1
ε
(
|
A
0
|
-
|
{
|
𝐮
|
>
0
}
|
)
=
-
1
ε
λ
C
ρ
ω
n
-
1
r
n
+
o
(
r
n
)
.
Finally, by putting all these estimates in (4.2),
we obtain
0
≤
C
∑
i
=
1
m
∫
B
r
(
x
0
)
|
∇
v
r
i
|
p
-
|
∇
u
i
|
p
d
x
+
f
ε
(
|
A
0
|
)
-
f
ε
(
|
{
|
𝐮
|
>
0
}
|
)
=
[
(
p
-
1
)
λ
C
ρ
ω
n
-
1
∑
i
=
1
m
|
q
i
(
x
0
)
|
p
p
-
1
+
O
(
λ
2
)
]
r
n
-
1
ε
λ
C
ρ
ω
n
-
1
r
n
+
o
(
r
n
)
.
Dividing by
r
n
and letting
r
→
0
, and then dividing by λ
and letting
λ
→
0
, we get
1
ε
≤
(
p
-
1
)
∑
i
=
1
m
|
q
i
(
x
0
)
|
p
p
-
1
.
Now if we choose
x
0
such that
∑
i
≤
m
q
i
(
x
0
)
≤
inf
ℛ
ε
(
∑
i
≤
m
q
i
)
+
1
,
then by Lemma 4.1 (and the equivalence of all norms on
the finite-dimensional space
ℝ
m
) we have
∑
i
≤
m
|
q
i
(
x
0
)
|
p
p
-
1
≤
C
,
independently of ε. However, this implies that ε
has a positive lower bound, which is a contradiction.
∎
5 Regularity of the free boundary (case
p
=
2
)
We are going to show that
ℛ
is an analytic hypersurface when
p
=
2
. To see this, we first derive the free boundary condition, also known as the optimality condition, in the following lemma. We perturb the optimal set Ω and compute the first variation of the energy functional
J
ε
. To perform this computation, it is
crucial to ensure that the p-harmonic solution within the perturbed domain is differentiable with respect to the perturbation parameter. When
p
=
2
, this can be established through the implicit function theorem . However, it is noteworthy that for
p
≠
2
the following proof breaks down, primarily due to the ill-posedness of the derivative of the map
u
↦
Δ
p
u
. Nevertheless, we believe a different approach may give a direct proof of the smoothness of the free boundary. This is left to future investigations.
Lemma 5.1.
Let
u
be a solution of the minimization problem
(1.1) for
p
=
2
. Let
h
i
be the solution of
{
Δ
h
i
=
0
in
Ω
-
E
,
h
i
=
0
on
E
,
h
i
=
∂
ξ
i
Γ
(
x
,
∂
ν
𝐮
)
on
∂
Ω
.
Then, on the regular part of the free boundary, we have
(5.1)
∑
i
=
1
m
∂
ν
h
i
∂
ν
u
i
=
C
for some positive constant C.
Proof.
Let
x
1
and
x
2
be two regular points in
ℛ
with corresponding unit normal vectors
ν
(
x
1
)
and
ν
(
x
2
)
.
Also, let
ρ
:
ℝ
→
ℝ
be a nonnegative smooth
function supported in
(
0
,
1
)
. Similarly to the proof of Theorem
4.3 we define the vector field
T
r
,
λ
(
x
)
:=
{
x
-
r
λ
ρ
(
|
x
-
x
1
|
r
)
ν
(
x
1
)
if
x
∈
B
r
(
x
1
)
,
x
+
r
λ
ρ
(
|
x
-
x
2
|
r
)
ν
(
x
2
)
if
x
∈
B
r
(
x
2
)
,
x
elsewhere
,
for small enough
r
,
λ
>
0
(which makes
T
r
,
λ
a diffeomorphism from
B
r
(
x
a
)
onto itself for
a
=
1
,
2
).
Now for some fixed
r
>
0
let
E
λ
=
T
r
,
λ
-
1
(
E
)
, and assume that
𝐰
λ
solves
{
Δ
w
λ
i
=
0
in
Ω
-
E
λ
,
w
λ
i
=
φ
i
on
∂
Ω
,
w
λ
i
=
0
on
∂
E
λ
.
Define
𝐯
λ
(
y
)
:=
𝐰
λ
(
T
r
,
λ
-
1
(
y
)
)
.
We are going to show that
λ
↦
𝐯
λ
is
a
C
1
map from a neighborhood of
λ
=
0
into
W
1
,
2
(
Ω
-
E
)
.
We know that each
v
λ
i
satisfies an elliptic PDE of
the form
F
[
v
,
λ
]
=
F
(
D
y
2
v
,
∇
y
v
,
y
,
λ
)
=
0
in
U
=
Ω
-
E
.
We also know that
F
=
Δ
when
y
∉
B
r
(
x
1
)
∪
B
r
(
x
2
)
or when
λ
=
0
. In addition, we can consider F as a
C
1
map
F
:
W
1
,
2
(
U
)
×
ℝ
→
W
-
1
,
2
(
U
)
,
(
v
,
λ
)
↦
F
[
v
,
λ
]
,
where
U
=
Ω
-
E
.
Now we employ the implicit function theorem to show that
λ
↦
𝐯
λ
is
C
1
. This can be readily deduced from the fact that
∂
v
F
|
λ
=
0
:
W
0
1
,
2
(
U
)
→
W
-
1
,
2
(
U
)
is invertible, since we have
∂
v
F
|
λ
=
0
⋅
=
d
d
s
|
s
=
0
F
[
v
+
s
⋅
,
0
]
=
d
d
s
|
s
=
0
Δ
(
v
+
s
⋅
)
=
Δ
⋅
.
Therefore,
𝐯
λ
=
𝐮
+
λ
𝐮
0
+
o
(
λ
)
in
W
1
,
2
(
U
)
, where
𝐮
0
∈
W
0
1
,
2
(
U
)
solves
0
=
d
d
λ
|
λ
=
0
F
[
v
λ
i
,
λ
]
=
∂
v
F
u
0
i
+
∂
λ
F
.
In other words
Δ
u
0
i
=
-
∂
λ
F
|
v
=
u
i
,
λ
=
0
.
Note that we also have
∇
𝐯
λ
=
∇
𝐮
+
λ
∇
𝐮
0
+
o
(
λ
)
,
since
λ
↦
𝐯
λ
is a
C
1
map into
W
1
,
2
(
U
)
; so
λ
↦
∇
𝐯
λ
is a
C
1
map into
L
2
(
U
)
.
Now let
h
i
be the solution of
Δ
h
i
=
0
in
U
=
Ω
-
E
with boundary data
h
i
=
∂
ξ
i
Γ
(
x
,
∂
ν
𝐮
)
on
∂
Ω
and
h
i
=
0
on
∂
E
. Then for small
λ
>
0
we have (note that for
p
=
2
we have
A
ν
=
∂
ν
)
∫
∂
Ω
Γ
(
x
,
∂
ν
𝐯
λ
)
-
Γ
(
x
,
∂
ν
𝐮
)
d
σ
=
∫
∂
Ω
∑
i
∂
i
Γ
(
x
,
∂
ν
𝐮
)
(
∂
ν
v
λ
i
-
∂
ν
u
i
)
d
σ
+
o
(
λ
)
=
λ
∫
∂
Ω
∑
i
∂
i
Γ
(
x
,
∂
ν
𝐮
)
∂
ν
u
0
i
d
σ
+
o
(
λ
)
=
λ
∫
∂
Ω
∑
i
h
i
∂
ν
u
0
i
d
σ
+
o
(
λ
)
=
λ
∑
i
∫
U
∇
h
i
⋅
∇
u
0
i
+
h
i
Δ
u
0
i
d
x
+
o
(
λ
)
=
-
λ
∑
i
∫
(
B
r
(
x
1
)
∪
B
r
(
x
2
)
)
-
E
h
i
∂
λ
F
|
v
=
u
i
,
λ
=
0
d
x
+
o
(
λ
)
.
Note that in the last line we have used the facts that
Δ
h
i
=
0
in U and
u
0
i
=
0
on
∂
U
=
∂
Ω
∪
∂
E
.
Also, we have
∂
λ
F
|
v
=
u
i
,
λ
=
0
=
0
outside
B
r
(
x
1
)
∪
B
r
(
x
2
)
,
because in that region
F
=
Δ
for all λ.
Now let us extend
𝐰
λ
to all of Ω by setting
it equal to 0 on
E
λ
. Note that
w
λ
i
is
positive on
Ω
-
E
λ
by the maximum principle. Hence
{
|
𝐰
λ
|
>
0
}
=
Ω
-
E
λ
.
Furthermore, similarly to the proof of Theorem 4.3,
we obtain
f
ε
(
|
{
|
𝐰
λ
|
>
0
}
|
)
-
f
ε
(
|
{
|
𝐮
|
>
0
}
|
)
≤
1
ε
(
|
E
|
-
|
E
λ
|
)
=
λ
ε
(
∫
B
r
(
x
2
)
∩
{
|
𝐮
|
>
0
}
ρ
′
(
|
x
-
x
2
|
)
〈
x
-
x
2
,
ν
(
x
2
)
〉
|
x
-
x
2
|
d
x
-
∫
B
r
(
x
1
)
∩
{
|
𝐮
|
>
0
}
ρ
′
(
|
x
-
x
1
|
)
〈
x
-
x
1
,
ν
(
x
1
)
〉
|
x
-
x
1
|
d
x
)
=
λ
ε
o
(
r
n
)
.
Therefore if we compare the energy of
𝐮
with
𝐰
λ
(it is easy to see that
𝐰
λ
∈
V
) we get (in the
second equality below we use the fact that
𝐯
λ
=
𝐰
λ
near
∂
Ω
)
0
≤
J
ε
(
𝐰
λ
)
-
J
ε
(
𝐮
)
=
∫
∂
Ω
Γ
(
x
,
∂
ν
𝐰
λ
)
-
Γ
(
x
,
∂
ν
𝐮
)
d
σ
+
f
ε
(
|
{
|
𝐰
λ
|
>
0
}
|
)
-
f
ε
(
|
{
|
𝐮
|
>
0
}
|
)
=
∫
∂
Ω
Γ
(
x
,
∂
ν
𝐯
λ
)
-
Γ
(
x
,
∂
ν
𝐮
)
d
σ
+
λ
ε
o
(
r
n
)
=
-
λ
∑
i
∫
(
B
r
(
x
1
)
∪
B
r
(
x
2
)
)
-
E
h
i
∂
λ
F
|
v
=
u
i
,
λ
=
0
d
x
+
o
(
λ
)
+
λ
ε
o
(
r
n
)
.
Hence if we divide by λ and let
λ
→
0
we obtain
(5.2)
0
≤
-
∑
i
∫
(
B
r
(
x
1
)
∪
B
r
(
x
2
)
)
-
E
h
i
∂
λ
F
|
v
=
u
i
,
λ
=
0
d
x
+
o
(
r
n
)
.
So we need to compute
∂
λ
F
|
v
=
u
i
,
λ
=
0
.
Next let us compute F explicitly. Set
x
=
T
r
,
λ
-
1
(
y
)
so that
y
=
T
r
,
λ
(
x
)
. To simplify the notation we suppress
the λ or r in the indexes. We have
v
i
(
T
(
x
)
)
=
v
i
(
y
)
=
w
i
(
x
)
.
Hence
∂
x
k
w
i
=
∑
j
∂
y
j
v
i
∂
x
k
T
j
,
∂
x
k
x
k
2
w
i
=
∑
j
∂
x
k
(
∂
y
j
v
i
∂
x
k
T
j
)
=
∑
j
,
ℓ
∂
y
j
y
ℓ
2
v
i
∂
x
k
T
j
∂
x
k
T
ℓ
+
∑
j
∂
y
j
v
i
∂
x
k
x
k
2
T
j
.
Therefore
0
=
Δ
w
i
=
∑
j
,
ℓ
,
k
∂
y
j
y
ℓ
2
v
i
∂
x
k
T
j
∂
x
k
T
ℓ
+
∑
j
,
k
∂
y
j
v
i
∂
x
k
x
k
2
T
j
.
It is easy to see that inside
B
r
(
x
a
)
(
a
=
1
,
2
) we have
∂
x
k
T
j
=
δ
j
k
+
(
-
1
)
a
λ
ρ
′
(
|
z
|
)
z
k
|
z
|
ν
j
(
x
a
)
,
z
=
x
-
x
a
r
,
∂
x
k
x
k
2
T
j
=
(
-
1
)
a
λ
∂
x
k
(
ρ
′
(
|
z
|
)
z
k
|
z
|
)
ν
j
(
x
a
)
.
Thus
F
[
v
,
λ
]
=
∑
j
,
ℓ
,
k
∂
y
j
y
ℓ
2
v
∂
x
k
T
j
∂
x
k
T
ℓ
+
∑
j
,
k
∂
y
j
v
∂
x
k
x
k
2
T
j
=
∑
j
,
ℓ
,
k
[
δ
j
k
+
(
-
1
)
a
λ
ρ
′
(
|
z
|
)
z
k
|
z
|
ν
j
(
x
a
)
]
[
δ
ℓ
k
+
(
-
1
)
a
λ
ρ
′
(
|
z
|
)
z
k
|
z
|
ν
ℓ
(
x
a
)
]
∂
y
j
y
ℓ
2
v
+
∑
j
,
k
[
(
-
1
)
a
λ
∂
x
k
(
ρ
′
(
|
z
|
)
z
k
|
z
|
)
ν
j
(
x
a
)
]
∂
y
j
v
in
B
r
(
x
a
)
for
a
=
1
,
2
, and
F
[
v
,
λ
]
=
Δ
v
elsewhere.
Now note that
∑
k
∂
x
k
(
ρ
′
(
|
z
|
)
z
k
|
z
|
)
=
∑
k
(
ρ
′′
(
|
z
|
)
z
k
2
r
|
z
|
2
+
ρ
′
(
|
z
|
)
1
r
|
z
|
-
ρ
′
(
|
z
|
)
z
k
2
r
|
z
|
3
)
=
1
r
ρ
′′
(
|
z
|
)
.
Hence we get
∂
λ
F
|
v
=
u
i
,
λ
=
0
=
(
-
1
)
a
(
2
ρ
′
(
|
z
|
)
∑
j
,
k
z
k
|
z
|
ν
j
(
x
a
)
∂
j
k
2
u
i
+
1
r
ρ
′′
(
|
z
|
)
∑
j
ν
j
(
x
a
)
∂
j
u
i
)
in
B
r
(
x
a
)
for
a
=
1
,
2
. Note that although a priori
z
,
u
i
in the above equation are functions of y, at
λ
=
0
we have
y
=
x
, and thus we can regard them as functions of x too.
Let
𝐮
r
(
z
)
=
1
r
𝐮
(
x
a
+
r
z
)
=
1
r
𝐮
(
x
)
and
h
r
i
(
z
)
=
1
r
h
i
(
x
a
+
r
z
)
=
1
r
h
i
(
x
)
.
Putting all these in (5.2), we get (note that in the following integration by parts the boundary term is zero, since ρ is 0 for z near
∂
B
1
and
h
i
is 0 on
∂
E
)
0
≤
-
∑
i
∫
(
B
r
(
x
1
)
∪
B
r
(
x
2
)
)
-
E
h
i
∂
λ
F
|
v
=
u
i
,
λ
=
0
d
x
+
o
(
r
n
)
=
∑
a
,
i
(
-
1
)
a
+
1
∫
B
r
(
x
a
)
-
E
h
i
(
2
ρ
′
(
|
z
|
)
∑
j
,
k
z
k
|
z
|
ν
j
(
x
a
)
∂
j
k
2
u
i
+
1
r
ρ
′′
(
|
z
|
)
∑
j
ν
j
(
x
a
)
∂
j
u
i
)
𝑑
x
+
o
(
r
n
)
=
∑
a
,
i
(
-
1
)
a
+
1
∫
B
r
(
x
a
)
-
E
(
-
2
∑
k
∂
k
[
h
i
ρ
′
(
|
z
|
)
z
k
|
z
|
]
∑
j
ν
j
(
x
a
)
∂
j
u
i
+
1
r
h
i
ρ
′′
(
|
z
|
)
∑
j
ν
j
(
x
a
)
∂
j
u
i
)
𝑑
x
+
o
(
r
n
)
=
∑
a
,
i
(
-
1
)
a
+
1
∫
B
r
(
x
a
)
-
E
(
-
2
∑
k
[
∂
k
h
i
ρ
′
(
|
z
|
)
z
k
|
z
|
+
h
i
∂
k
(
ρ
′
(
|
z
|
)
z
k
|
z
|
)
]
+
1
r
h
i
ρ
′′
(
|
z
|
)
)
∑
j
ν
j
(
x
a
)
∂
j
u
i
d
x
+
o
(
r
n
)
=
∑
a
,
i
(
-
1
)
a
+
1
∫
B
r
(
x
a
)
-
E
(
-
2
∑
k
[
∂
k
h
i
ρ
′
(
|
z
|
)
z
k
|
z
|
]
-
1
r
h
i
ρ
′′
(
|
z
|
)
)
∑
j
ν
j
(
x
a
)
∂
j
u
i
d
x
+
o
(
r
n
)
=
∑
a
,
i
(
-
1
)
a
+
1
r
n
∫
B
1
∩
{
|
𝐮
r
|
>
0
}
(
-
2
∑
k
[
∂
k
h
r
i
ρ
′
(
|
z
|
)
z
k
|
z
|
]
-
1
r
r
h
r
i
ρ
′′
(
|
z
|
)
)
∑
j
ν
j
(
x
a
)
∂
j
u
r
i
d
z
+
o
(
r
n
)
.
Now note that
∂
j
u
r
i
(
z
)
→
q
i
(
x
a
)
ν
j
(
x
a
)
=
∂
j
u
i
(
x
a
)
when
z
⋅
ν
(
x
a
)
>
0
by the results of [5].
Next note that
h
i
is Lipschitz continuous, since
u
i
is Lipschitz and we have
0
≤
h
i
≤
c
u
i
for some constant c. To see this note that the function
∂
ξ
i
Γ
(
x
,
∂
ν
𝐮
)
is positive and continuous on the compact set
∂
Ω
, so
it is bounded there, and thus for some
c
>
0
we have
h
i
=
∂
ξ
i
Γ
(
x
,
∂
ν
𝐮
)
≤
c
φ
i
=
c
u
i
on
∂
Ω
. Hence the claim follows by the maximum principle.
Therefore, by Lemma B.1 in [9], we also
have
∂
k
h
r
i
(
z
)
→
p
i
(
x
a
)
ν
k
(
x
a
)
=
∂
k
h
i
(
x
a
)
for some function
p
i
, and
h
r
i
(
z
)
→
∇
h
i
(
x
a
)
⋅
z
as
h
i
(
x
a
)
=
0
. Thus if we divide the above expression by
r
n
and let
r
→
0
we obtain
0
≤
∑
a
,
i
(
-
1
)
a
+
1
∫
B
1
∩
{
z
⋅
ν
(
x
a
)
>
0
}
(
-
2
∑
k
[
∂
k
h
i
(
x
a
)
ρ
′
(
|
z
|
)
z
k
|
z
|
]
z
)
ρ
′′
(
|
z
|
)
)
∑
j
ν
j
(
x
a
)
∂
j
u
i
(
x
a
)
d
z
=
∑
a
,
i
(
-
1
)
a
+
1
∫
B
1
∩
{
z
⋅
ν
(
x
a
)
>
0
}
(
-
2
∑
k
[
p
i
(
x
a
)
ν
k
(
x
a
)
ρ
′
(
|
z
|
)
z
k
|
z
|
]
-
p
i
(
x
a
)
(
ν
(
x
a
)
⋅
z
)
ρ
′′
(
|
z
|
)
)
∂
ν
u
i
(
x
a
)
𝑑
z
=
∑
a
,
i
(
-
1
)
a
∫
B
1
∩
{
z
⋅
ν
(
x
a
)
>
0
}
(
2
|
z
|
ρ
′
(
|
z
|
)
+
ρ
′′
(
|
z
|
)
)
(
ν
(
x
a
)
⋅
z
)
p
i
(
x
a
)
∂
ν
u
i
(
x
a
)
𝑑
z
=
∑
a
,
i
(
-
1
)
a
∂
ν
h
i
(
x
a
)
∂
ν
u
i
(
x
a
)
∫
B
1
∩
{
z
⋅
ν
(
x
a
)
>
0
}
(
2
|
z
|
ρ
′
(
|
z
|
)
+
ρ
′′
(
|
z
|
)
)
(
ν
(
x
a
)
⋅
z
)
𝑑
z
=
C
ρ
(
∑
i
∂
ν
h
i
(
x
2
)
∂
ν
u
i
(
x
2
)
-
∑
i
∂
ν
h
i
(
x
1
)
∂
ν
u
i
(
x
1
)
)
,
where
C
ρ
=
∫
B
1
∩
{
z
⋅
ν
(
x
a
)
>
0
}
(
2
|
z
|
ρ
′
(
|
z
|
)
+
ρ
′′
(
|
z
|
)
)
(
ν
(
x
a
)
⋅
z
)
𝑑
z
does not depend on
x
a
; we have also used the fact that
p
i
(
x
a
)
=
∂
ν
h
i
(
x
a
)
.
By switching the role of
x
1
,
x
2
we conclude that
∑
i
∂
ν
h
i
(
x
2
)
∂
ν
u
i
(
x
2
)
-
∑
i
∂
ν
h
i
(
x
1
)
∂
ν
u
i
(
x
1
)
must be zero, as desired.
∎
The main idea to show the regularity of the free boundary lies in
utilizing the boundary Harnack principle, which allows us to reduce the system into a scalar problem. The key tool in employing this approach is non-tangential accessibility of the domain; for the definition of non-tangentially accessible (NTA) domains we refer to [3].
Lemma 5.2.
Let
u
be a solution of the minimization problem (1.1)
for
p
=
2
. Then
U
=
{
x
:
|
u
(
x
)
|
>
0
}
is a non-tangentially
accessible domain.
Proof.
This result follows from the same analysis as of [3, Theorem 4.8]
for the function
𝒰
=
u
1
+
⋯
+
u
m
.
Note that
𝒰
is harmonic in
{
|
𝐮
|
>
0
}
=
{
𝒰
>
0
}
(these two sets are equal due to Lemma 3.2), and the
function
𝒰
is also Lipschitz continuous
and satisfies the nondegeneracy property by Corollary 3.3.
∎
Theorem 5.3.
Let
x
0
∈
R
be a regular point of the free boundary.
Then there is
r
>
0
such that
B
r
(
x
0
)
∩
∂
{
|
u
|
>
0
}
is a
C
1
,
α
hypersurface for some
α
>
0
.
Proof.
We may assume that
u
1
>
0
in
B
r
0
(
x
0
)
∩
{
|
𝐮
|
>
0
}
for some
r
0
>
0
.
First we show that for some
0
<
r
≤
r
0
there is a Hölder continuous
function g defined on
B
r
(
x
0
)
∩
∂
{
|
𝐮
|
>
0
}
,
such that in the viscosity sense we have
∂
ν
h
1
∂
ν
u
1
=
g
on
∂
E
,
where
h
1
is defined in Lemma 5.1. Since
B
r
0
(
x
0
)
∩
{
|
𝐮
|
>
0
}
is an NTA domain, the boundary Harnack inequality implies that
G
i
:=
u
i
u
1
and
H
i
=
h
i
h
1
are Hölder continuous functions in
B
r
(
x
0
)
∩
{
|
𝐮
|
>
0
}
¯
for some
0
<
r
≤
r
0
. Now if we consider a one-sided tangent ball
at some point
y
∈
B
r
(
x
0
)
∩
∂
{
|
𝐮
|
>
0
}
,
we have asymptotic developments (see [9, Lemma B.1],
noting that
h
i
is Lipschitz as we have shown in the proof of
Lemma 5.1)
u
i
(
y
+
x
)
=
q
i
(
y
)
(
x
⋅
ν
(
y
)
)
+
+
o
(
|
x
|
)
,
h
i
(
y
+
x
)
=
p
i
(
y
)
(
x
⋅
ν
(
y
)
)
+
+
o
(
|
x
|
)
.
Therefore
G
i
(
y
)
=
q
i
(
y
)
q
1
(
y
)
and
H
i
(
y
)
=
p
i
(
y
)
p
1
(
y
)
.
Thus from (5.1) we can infer that
p
1
(
y
)
q
1
(
y
)
(
1
+
∑
i
>
1
G
i
(
y
)
H
i
(
y
)
)
=
∑
i
p
i
(
y
)
q
i
(
y
)
=
∑
i
∂
ν
h
i
∂
ν
u
i
is constant for every
y
∈
B
r
(
x
0
)
∩
∂
{
|
𝐮
|
>
0
}
. Note that
G
i
,
H
i
>
0
at y as
p
i
,
q
i
>
0
. Hence by applying [13, Theorem 3.1]
we get the desired result.
∎
Corollary 5.4.
Let
u
be a solution of the minimization problem (1.1)
for
p
=
2
. Then the regular part of the free boundary,
R
,
is analytic.
Proof.
Suppose
0
∈
ℛ
and
u
1
>
0
in
B
r
∩
{
|
𝐮
|
>
0
}
.
Then we apply the hodograph-Legendre transformation
x
↦
y
=
(
x
1
,
…
,
x
n
-
1
,
u
1
)
.
Next we define the partial Legendre functions
v
1
(
y
)
:=
x
n
,
v
i
(
y
)
:=
u
i
(
x
)
for
i
=
2
,
…
,
m
,
w
i
(
y
)
:=
h
i
(
x
)
for
i
=
1
,
…
,
m
.
As
ℛ
is
C
1
,
α
, it follows that
u
i
and
h
i
are in
C
1
,
α
(
B
r
∩
{
|
𝐮
|
>
0
}
¯
)
.
So,
v
i
and
w
i
are
C
1
,
α
in a neighborhood of
the origin in
{
y
n
≥
0
}
. Now we have verified all the hypothesis
of Theorem 7.1 in [3], and through a similar argument we can obtain the analyticity of
ℛ
.
∎
