Differential and Integral Equations
Volume 25, Numbers 11-12 (2012) , 1037–1052
ON THE EQUATION det ∇ϕ = f PRESCRIBING ϕ = 0
ON THE BOUNDARY
Olivier Kneuss
Department of Mathematics, University of California, Berkeley, CA
(Submitted by: Fabrice Bethuel)
Abstract. We discuss the existence of a regular map ϕ satisfying

det ∇ϕ = f in Ω
ϕ=0
on ∂Ω,
whereRΩ is a bounded smooth domain and f is a regular function satisfying Ω f = 0.
1. Introduction
In this article we discuss the existence of ϕ : Ω ⊂ Rn → Rn such that
det ∇ϕ = f in Ω
(1.1)
ϕ=0
on ∂Ω,
where Ω is a bounded smooth domain and
R f : Ω → R. Using the divergence
theorem one immediately deduces that Ω f = 0 is a necessary condition to
solve (1.1) (see Lemma 15 for a more precise statement). The main results
of this article can be summarized as follows (see Theorems 2, 4, and 6 for a
more general statement).
Theorem
1. Let Ω ⊂ Rn be the unit ball and f ∈ C 1 (Ω) be such that
R
Ω f = 0.
(i) If |f | > 0 on ∂Ω, then there exists ϕ ∈ C 1 (Ω; Rn ) ∩ C 0,1/n (Ω; Rn )
satisfying (1.1).
(ii) If f is of the form dist(·, ∂Ω)n−1 near ∂Ω, then there exists ϕ ∈
1
C (Ω; Rn ) satisfying (1.1).
(iii) If supp f ⊂ Ω, then
inf
ϕ∈C0∞ (Ω;Rn )
k det ∇ϕ − f kC 1 (Ω) = 0.
Accepted for publication: July 2012.
AMS Subject Classifications: 35F30.
1037
1038
Olivier Kneuss
This article uses similar
techniques as in [2] where the existence of a ϕ as
R
regular as f (where Ω f = meas Ω) satisfying
det ∇ϕ = f in Ω
and
ϕ = id on ∂Ω
was proved (see also [1] for a complete overview): roughly speaking we look
for a solution ϕ of the form ϕ = φ ◦ ψ where ψ is a diffeormophism and ψ a
radial solution, i.e., a solution of the form α(x)x/|x| with α a function. However, contrary to the case where we prescribe the identity on the boundary,
these techniques seem not to allow one to solve (1.1) in its full generality.
Finally, note that,
R contrary to the case ϕ = id on the boundary, the
integral condition Ω f = 0 is not sufficient for the existence of a solution
C k , k ≥ 1, up to the boundary of (1.1). Indeed (cf. Lemma 17):
- f has to vanish identically on the boundary in order to have a solution
C 1 up to the boundary.
- f has to be of the form dist(·, ∂Ω)n−1 h(·) near ∂Ω for a continuous
function h if we want the solution to be C 2 up to the boundary.
2. Notation
We gather here the main notation that will be used throughout the article.
Let Ω, O ⊂ Rn be bounded open sets.
- Balls in Rn are denoted by B (x) := {y ∈ Rn : |y − x| < }. When x = 0
we simply write B instead of B (0); similarly we write B instead of B1 .
- For g ∈ C 0 (Rn ) and ϕ ∈ C 1 (Ω; Rn ) we let
ϕ∗ (g) := g(ϕ) det ∇ϕ.
We will use several times the following elementary property: for ϕ1 ∈
C 1 (Ω; O) and ϕ2 ∈ C 1 (O; Rn ) we have (ϕ2 ◦ ϕ1 )∗ (g) = ϕ∗1 (ϕ∗2 (g)).
- The set of diffeormorphisms of class k, k ≥ 1 an integer, is denoted by
Diff k (Ω; O) := {ϕ ∈ C k (Ω; O) : ϕ−1 ∈ C k (O; Ω)}.
We say that Ω is C k -diffeomorphic to B if there exists ϕ ∈ Diff k (B; Ω).
- C0k (Ω) stands for the set of C k (Ω) functions with compact support in Ω.
- For k ≥ 0 an integer and g a function, ∇k g stands for the set of derivatives of order k of g with the convention ∇0 g = g.
- dist(·, ∂Ω) is the usual distance function to the set ∂Ω, i.e.,
dist(x, ∂Ω) = inf |y − x|.
y∈∂Ω
On the equation det ∇ϕ = f prescribing ϕ = 0 on the boundary
1039
3. Main results
In this section we state the three main results of this article, which will
be proved in the next section. We start with a result giving a solution with
interior regularity.
Theorem 2. Let n ≥ 2 and k ≥ 1 be two integers and Ω ⊂ Rn be an open
set such that Ω is C k+1 -diffeomorphic to B. Let also V be a neighborhood of
∂Ω, gR ∈ C k (Rn ) and f ∈ C k (Ω) be such that inf Rn |g| > 0, |f | > 0 in V ∩ Ω,
and Ω f = 0. Then there exists ϕ ∈ C k (Ω; Rn ) ∩ C 0,1/n (Ω; Rn ) such that
∗
ϕ (g) = f in Ω
ϕ=0
on ∂Ω.
Remark 3. (i) Recall that “Ω is C k+1 -diffeomorphic to B” means that there
exists ψ ∈ Diff k+1 (B;RΩ).
(ii) The condition Ω f = 0 is always necessary (cf. Lemma 15).
(iii) In the particular case |f | > 0 on ∂Ω, the obtained solution belongs
to ∩1≤p<n/(n−1) W 1,p (Ω, Rn ) but not to W 1,n/(n−1) (Ω, Rn ); moreover, it does
not belong to C 0,β (Ω; Rn ) for 1/n < β ≤ 1 (cf. Proposition 13 (i)).
(iv) With a more involved proof we can show that for every > 0
there exists ϕ satisfying all the conclusions of the theorem with in addition ϕ (Ω) ⊂ B . This can be seen as an optimal control of the size of the
image of the solution.
Our second result gives a sufficient condition (and somehow partially necessary; cf. the following remark) to have a solution regular up to the boundary.
Theorem 4. Let n ≥ 2 be an integer, Ω ⊂ Rn be an open set such that Ω is
3
1
n
1
C
R -diffeomorphic to B, g ∈ C (R ) and f ∈ C (Ω) be such that inf Rn |g| > 0,
B f = 0, and
f (x) = dist(x, ∂Ω)m c(x) near ∂Ω,
where m ≥ n − 1 and where c ∈ C 1 (Ω) satisfies |c| > 0 on ∂Ω. Then there
exists ϕ ∈ C 1 (Ω; Rn ) such that
∗
ϕ (g) = f in Ω
(3.1)
ϕ=0
on ∂Ω.
R
Remark 5. (i) The condition Ω f = 0 is always necessary (cf. Lemma 15)
and Remark 3 (iv) is also verified.
(ii) Surprisingly enough, even if g, f, c, and Ω are smooth, the proof does
not give, in general, ϕ ∈ C 2 (Ω; Rn ). For example, taking n = 2, Ω = B,
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Olivier Kneuss
g ≡ 1, and f = (1 − |x|)|x| near the boundary, the solution constructed in
the proof is not C 2 up to the boundary (cf. Remark 14 (i)).
(iii) However, with exactly the same proof (using Proposition 13 (iii)),
one can show the following: let Ω = B, k ≥ 2 be an integer, g ∈ C k (Rn ) and
f ∈ C k (B) be such that inf Rn |g| > 0,
Z
1
f = 0, and f (x) = (1 − |x|)ns−1 n−1 near ∂B,
|x|
B
where s ≥ 1 is an integer; then there exists ϕ ∈ C k (B; Rn ) satisfying (3.1).
(iv) Using Proposition 17, f has to be of the form
f (x) = dist(x; ∂Ω)n−1 h(x)
near the the boundary where h is continuous if we want the solution to be
C 2 (Ω; Rn ). This gives a condition of the behavior of f near the boundary:
indeed, in order to have a solution of (3.1) C 2 up to the boundary, f has to
vanish on ∂Ω at least as fast as dist(·; ∂Ω)n−1 .
Our last theorem is an approximation result.
Theorem 6. Let n ≥ 2 and k ≥ 0 be two integers, and let Ω ⊂ Rn be an
open set such that Ω is C k+1 -diffeomorphic to B. Let also g ∈ C k (Rn ) and
f ∈ C k (Ω) be such that inf Rn |g| > 0,
∇s f = 0
and
R
B
on ∂Ω, for every 0 ≤ s ≤ k
f = 0. Then
inf
ϕ∈C0∞ (Ω;Rn )
kϕ∗ (g) − f kC k (Ω) = 0.
Remark 7. The previous theorem
raises the following interesting open quesR
tion: given f ∈ C0k (Ω) with Ω f = 0, does there exist ϕ ∈ C0k (Ω; Rn ) such
that
det ∇ϕ = f in Ω?
Note that the above result says that we can always uniformly approximate
any such f by the Jacobian of smooth maps with compact support.
4. Proof of the main results
In this section we prove the three results stated in the last section. We
start by proving an elementary lemma which will allow us to assume g ≡ 1.
On the equation det ∇ϕ = f prescribing ϕ = 0 on the boundary
1041
Lemma 8. Let n ≥ 2 and k ≥ 1 be two integers, Ω ⊂ Rn be a bounded open
set, g ∈ C k (Rn ) with inf Rn |g| > 0, f ∈ C k (Ω), and ψ ∈ C k (Ω; Rn ) be such
that
∗
ψ (1) = f in Ω
ψ=0
on ∂Ω.
Then there exists ϕ ∈ C k (Ω; Rn ) satisfying
∗
ϕ (g) = f in Ω
ϕ=0
on ∂Ω.
Remark 9. With exactly the same proof we show that if ψ ∈ C k (Ω; Rn ) ∩
C 0,1/n (Ω; Rn ), then ϕ ∈ C k (Ω; Rn ) ∩ C 0,1/n (Ω; Rn ).
Proof. Since inf Rn |g| > 0, we can define h : Rn → R by
Z h(x)
g(x1 , . . . , xn−1 , t) dt = xn .
0
C k (Rn ).
Note that h ∈
Then φ defined by φ(x) = (x1 , . . . , xn−1 , h(x))
k
n
n
belongs to Diff (R ; φ(R )) and satisfies
φ∗ (g) = 1 in Rn
and φ(0) = 0.
It is then easily checked that ϕ = φ ◦ ψ has all the wished-for properties.
We now prove the first main result.
Proof of Theorem 2. Step 1 (simplification). Using Lemma 8, we can
assume with no loss of generality that g ≡ 1. We also claim that we can
assume that Ω = B. Indeed, by hypothesis there exists ψ ∈ Diff k+1 (B; Ω).
Note that fe := ψ ∗ (f ) belongs to C k (B) and satisfies
Z
fe = 0 and |fe| > 0 in B \ B1− ,
B
for some > 0 small enough. If φ ∈ C k (B; Rn ) ∩ C 0,1/n (B; Rn ) solves
φ∗ (1) = fe in B
then it is easily seen that ϕ =
φ◦ψ −1
ϕ∗ (1) = f in Ω
and φ = 0 on ∂B,
∈ C k (Ω; Rn )∩C 0,1/n (Ω; Rn ) and solves
and
ϕ = 0 on ∂Ω,
which proves the assertion. We also claim that we can assume f < 0 in
B \ B1− . Indeed, if f > 0 in B \ B1− and if ϕ satisfies
ϕ∗ (1) = −f in B
and ϕ = 0 on ∂B,
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Olivier Kneuss
then ψ = (−ϕ1 , ϕ2 , . . . , ϕn ) satisfies
ψ ∗ (1) = f in B
and ψ = 0 on ∂B.
Step 2. Applying Lemma 10 there exists ϕ1 ∈ Diff k (B; B) such that
supp(ϕ1 − id) ⊂ B, ϕ∗1 (f ) ∈ C k (B), ϕ∗1 (f ) ≡ 1 in a neighborhood of 0, and
Z r
x
sn−1 ϕ∗1 (f )(s ) ds > 0 for every r ∈ (0, 1) and every x 6= 0
|x|
0
Z 1
x
sn−1 ϕ∗1 (f )(s ) ds = 0 for every x 6= 0.
|x|
0
Step 3. Let α : B → R be defined by α(0) = 0 and, for x ∈ B \ {0},
Z |x|
1/n
x
α(x) = n
sn−1 ϕ∗1 (f )(s ) ds
|x|
0
x
. Using Lemma 13 we have
and ϕ2 : B → Rn be defined by ϕ2 (x) = α(x) |x|
ϕ2 ∈ C k (B; Rn ) ∩ C 0,1/n (B; Rn ) and
∗
ϕ2 (1) = ϕ∗1 (f ) in B
ϕ2 = 0
on ∂B.
Then ϕ = ϕ2 ◦ (ϕ1 )−1 belongs to C k (B; Rn ) ∩ C 0,1/n (B; Rn ) and satisfies
∗
ϕ (1) = f in B
ϕ=0
on ∂B,
which concludes the proof.
We now prove the second main result.
Proof of Theorem 4. Step 1. Reasoning exactly as in Steps 1 and 2 of
the previous proof (using in particular Lemma 18) we can assume Ω = B,
c < 0 on ∂Ω, and the
g ≡ 1, f = (1 − |x|)m e
c(x) with e
c ∈ C 1 (B) such that e
1
existence of ϕ1 ∈ Diff (B; B) such that supp(ϕ1 − id) ⊂ B, ϕ∗1 (f ) ∈ C 1 (B),
ϕ∗1 (f ) ≡ 1 in a neighborhood of 0, and
Z r
x
sn−1 ϕ∗1 (f )(s ) ds > 0 for every r ∈ (0, 1) and every x 6= 0
|x|
0
Z 1
x
sn−1 ϕ∗1 (f )(s ) ds = 0 for every x 6= 0.
|x|
0
Step 2. Let α : B → R be defined by α(0) = 0 and, for x ∈ B \ {0},
1/n
Z |x|
x
,
α(x) = n
sn−1 ϕ∗1 (f )(s ) ds
|x|
0
On the equation det ∇ϕ = f prescribing ϕ = 0 on the boundary
1043
x
and let ϕ2 : B → Rn be defined by ϕ2 (x) = α(x) |x|
. Since ϕ∗1 (f ) = f =
(1 − |x|)m e
c(x) near ∂B we get, using Lemma 13 (ii), that ϕ2 belongs to
C 1 (B; Rn ) and satisfies
∗
ϕ2 (1) = ϕ∗1 (f ) in B
ϕ2 = 0
on ∂B.
Then ϕ = ϕ2 ◦ (ϕ1 )−1 has all the desired properties, which concludes the
proof.
Finally, we prove the last main result.
Proof of Theorem 6. Step 1. As in Step 1 of the proof of Theorem 2,
we can assume g ≡ 1 and Ω = B. Using that ∇s f = 0 on ∂B for every
0 ≤ s ≤ k, we deduce that
inf
h∈C0∞ (B)
kh − f kC k (B) = 0,
and hence we can also assume that f ∈ C0∞ (B). Then, in particular, there
exists σ > 0 small enough such that f ≡ 0 in B \ B1−3σ . For every δ > 0
small enough let fδ ∈ C0∞ (B) be such that
(
f
in (B \ B1−σ ) ∪ B1−3σ
n
fδ (x) =
e−1/(1−σ−|x|)
−δ |x|n−1 (1−σ−|x|)n+1 in B1−σ \ B1−2σ
Z
fδ = 0 and lim kfδ − f kC k (B) = 0.
B\B1−3σ
δ→0
Using the last equation the theorem will be proved if we show, for every
δ > 0 small, the existence of a ϕδ ∈ C0∞ (B; Rn ) such that ϕ∗δ (1) = fδ . This
will be done in Step 2.
Step 2. Let δ > 0 be small enough. Note that
Z
fδ = 0 and fδ < 0 in B1−σ \ B1−2σ .
B1−σ
Hence, using Lemma 10 (applied to B1−σ ) there exists ϕ1 ∈ Diff ∞ (B; B)
such that supp(ϕ1 − id) ⊂ B1−σ , ϕ∗1 (fδ ) ∈ C ∞ (B), ϕ∗1 (fδ ) ≡ 1 in a neighborhood of 0, and
Z r
x
sn−1 ϕ∗1 (fδ )(s ) ds > 0 for every r ∈ (0, 1 − σ) and every x 6= 0
|x|
0
Z 1−σ
x
sn−1 ϕ∗1 (fδ )(s ) ds = 0 for every x 6= 0.
|x|
0
1044
Olivier Kneuss
Define α : B → R by α(0) = 0 and, for x ∈ B \ {0},
Z |x|
1/n
x
sn−1 ϕ∗1 (fδ )(s ) ds
α(x) = n
|x|
0
x
and ϕ2 : B → Rn by ϕ2 (x) = α(x) |x|
. Since ϕ∗1 (fδ ) = fδ near ∂B1−σ , we get,
using Lemma 13 (iv), that ϕ2 ∈ C0∞ (B; Rn ) and satisfies ϕ∗2 (1) = ϕ∗1 (fδ ) in
B. Finally, ϕ = ϕ2 ◦ (ϕ1 )−1 belongs to C0∞ (B; Rn ) and satisfies ϕ∗ (1) = fδ
in B, which concludes the proof.
5. Positive radial integration
In this sectionR we show how to modify the mass distribution of some
f ∈ C k (B) with B f = 0 in order to have strictly positive integrals on every
radius starting from 0 and going to points in B and vanishing integrals on
every radius starting from 0 and going to points on ∂B.
Lemma 10. Let n ≥ 2 and k ≥ 1 be two integers, > 0, and f ∈ C k (B) be
such that
Z
f = 0 and f < 0 in B \ B1− .
B
Then there exists ϕ ∈ Diff k (B; B) such that supp(ϕ − id) ⊂ B, ϕ∗ (f ) ∈
C k (B), ϕ∗ (f ) ≡ 1 in a neighborhood of 0, and
Z r
x
sn−1 ϕ∗ (f )(s ) ds > 0 for every r ∈ (0, 1) and every x 6= 0
|x|
0
Z 1
x
sn−1 ϕ∗ (f )(s ) ds = 0 for every x 6= 0.
|x|
0
R
Proof. Step 1. Note that B1− f > 0. Hence, applying Lemma 11.21 of [1]
to f and B1− , there exists ϕ1 ∈ Diff ∞ (B; B) such that supp(ϕ1 −id) ⊂ B1−
and such that, writing f1 = ϕ∗1 (f ), we have f1 (0) > 0 and
Z r
x
sn−1 f1 (s ) ds > 0, for every x 6= 0 and every r ∈ (0, 1 − ].
|x|
0
R
Note that B f1 = 0.
Step 2. Choose η > 0 small enough so that f1 > 0 in B η and
Z r
x
sn−1 f1 (s ) ds > 0, for every x 6= 0 and every r ∈ (η, 1 − ].
|x|
η
On the equation det ∇ϕ = f prescribing ϕ = 0 on the boundary
1045
Let f2 ∈ C k (B) be such that supp(f2 − f1 ) ⊂ Bη , f2 > 0 in B η , f2 ≡ 1 in a
neighborhood of 0, and
Z
Z
f2 =
f1 .
Bη
Bη
Using Theorem 11 there exists ϕ2 ∈ Diff k (B; B) such that
ϕ∗2 (f1 ) = f2 in B
and
supp(ϕ2 − id) ⊂ Bη .
In particular, we have f2 = f1 = f < 0 in B \ B1− and
Z r
x
sn−1 f2 (s ) ds > 0, for every x 6= 0 and every r ∈ (0, 1 − ]. (5.1)
|x|
0
R
Note that B f2 = 0.
Step 3. We claim the existence of f3 ∈ C k (B) such that supp(f3 − f2 ) ⊂
B \ B 1− , f3 < 0 in B \ B1− and
Z 1
x
(5.2)
sn−1 f3 (s ) ds = 0, for every x 6= 0.
|x|
0
Using (5.1), there exists δ > 0 small enough so that, for every x 6= 0,
Z 1
Z 1−+2δ
x
x
n−1
sn−1 f2 (s ) ds > 0.
(5.3)
s
f2 (s ) ds +
|x|
|x|
1−2δ
0
Let ρ ∈ C ∞ ([0, 1]; [0, 1]) be such that
1 in [1 − + 2δ, 1 − 2δ]
ρ=
0 in [0, 1 − + δ] ∪ [1 − δ, 1].
Define h ∈ C k (Rn \ {0}) by
R1
R 1−
x
x
− 1− sn−1 (1 − ρ(s))f2 (s |x|
) ds − 0 sn−1 f2 (s |x|
) ds
h(x) =
.
R1
n−1 ρ(s) ds
1− s
Using (5.3), the definition of ρ, and the fact that f2 < 0 in B \ B1− , we get
h < 0. Note also that h(λx) = h(x) for every x 6= 0 and every λ > 0. It is
easily seen that f3 defined by f3 (x) = (1 − ρ(|x|))f2 (x) + ρ(|x|)h(x) has all
the required properties.
R
Step 4. Integrating (5.2)
over
the
unit
sphere
we
get
that
B f3 = 0, and
R
therefore, recalling that B f2 = 0 and that supp(f2 − f3 ) ⊂ B \ B1− ,
Z
Z
f2 =
f3 .
B1−
B1−
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Olivier Kneuss
Hence, recalling also that f2 , f3 < 0 in B \ B1− , we can apply Theorem 11
and find ϕ3 ∈ Diff k (B; B) such that
ϕ∗3 (f2 ) = f3 in B
supp(ϕ3 − id) ⊂ B \ B 1− .
and
Step 5. Note that f3 ∈ C k (B) satisfies f3 ≡ 1 in a neighborhood of 0
and
Z r
x
sn−1 f3 (s ) ds > 0 for every r ∈ (0, 1) and every x 6= 0
|x|
0
Z 1
x
sn−1 f3 (s ) ds = 0 for every x 6= 0.
|x|
0
Let ϕ = ϕ1 ◦ ϕ2 ◦ ϕ3 . Noticing that ϕ ∈ Diff k (B; B) and satisfies supp(ϕ −
id) ⊂ B and ϕ∗ (f ) = f3 , the proof is finished.
In the previous proof we have used a classical result about the Jacobian
(see [1] Theorem 10.11 or [3] for a proof).
Theorem 11. Let k ≥ 1 be an integer, Ω be a bounded connected open set
in Rn , and f, g ∈ C k (Ω) be such that
Z
Z
f · g > 0 in Ω,
f=
g, and supp(f − g) ⊂ Ω.
Ω
Ω
k
Then there exists ϕ ∈ Diff (Ω; Ω) such that
∗
ϕ (g) = f in Ω
supp(ϕ − id) ⊂ Ω.
6. Radial solutions
In this section, we investigate some properties of radial solutions, i.e.
solutions of the form α(x)x/|x| where α is a function. We start with an
elementary lemma.
Lemma 12. Let n ≥ 2 be an integer and f ∈ C 0,1 (B) be such that
Z r
x
sn−1 f (s ) ds ≥ 0, for every r ∈ (0, 1] and every x 6= 0.
|x|
0
Let α : B → R be defined by α(0) = 0 and, for x ∈ B \ {0},
Z |x|
1/n
x
α(x) = n
sn−1 f (s ) ds
.
|x|
0
Then α ∈ C 0,1/n (B).
On the equation det ∇ϕ = f prescribing ϕ = 0 on the boundary
Proof. It is enough to note that αn ∈ C 0,1 (B).
1047
Proposition 13. Let n ≥ 2 and k ≥ 1 be two integers and let f ∈ C k (B)
be such that f ≡ 1 in a neighborhood of 0 and
Z r
x
sn−1 f (s ) ds > 0, for every r ∈ (0, 1) and every x 6= 0
|x|
0
Z 1
x
sn−1 f (s ) ds = 0, for every x 6= 0.
|x|
0
Let α : B → R be defined by α(0) = 0 and, for x ∈ B \ {0},
Z |x|
1/n
x
α(x) = n
sn−1 f (s ) ds
,
|x|
0
x
and let ϕ : B → Rn be defined by ϕ(x) = α(x) |x|
. Then ϕ ∈ C k (B; Rn ) ∩
C 0,1/n (B; Rn ) and
ϕ∗ (1) = f in B
ϕ=0
on ∂B.
The following extra assertions are also satisfied:
(i) If f < 0 on ∂B, then ϕ ∈ ∩1≤p<n/(n−1) W 1,p (B; Rn ),
ϕ∈
/ W 1,n/(n−1) (B; Rn ), and ϕ ∈
/ C 0,β (B; Rn ) for every 1/n < β ≤ 1.
m
(ii) If f (x) = c(x)(1 − |x|) near ∂B, where m ≥ n − 1 and where c ∈
1
C (B) is such that c < 0 on ∂B, then ϕ ∈ C 1 (B; Rn ).
ns−1
(iii) If f (x) = − (1−|x|)
near ∂B, where s ≥ 1 is an integer, then
|x|n−1
ϕ ∈ C k (B; Rn ).
(iv) If f ∈ C ∞ (B) and
n
e−1/(1−|x|)
f (x) = − n−1
|x|
(1 − |x|)n+1
near ∂B,
then ϕ ∈ C ∞ (B; Rn ) and satisfies
∇l ϕ = 0
on ∂B for every integer l ≥ 0.
In particular, ϕ extended by 0 outside B belongs to C0∞ (Rn ; Rn ).
Remark 14. (i) The conclusion concerning the regularity of the third extra
statement is no longer true in general if f (x) is only assumed to be of the
form c(x)(1 − |x|)ns−1 near the boundary for some c ∈ C ∞ (B) with c < 0
on ∂B. Indeed, taking n = 2, s = 1, and c(x) = −|x| near ∂B, a simple
calculation gives that ϕ ∈
/ C 2 (B; Rn ).
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Olivier Kneuss
(ii) In the third extra statement, if f is of the form f (x) = −(1 −
|x|)m /|x|n−1 near ∂B, then ϕ(x) = (n/(k + 1))1/n (1 − |x|)(m+1)/n x/|x| near
∂B. This shows in particular that, if we want the solution to be smooth
near the boundary, m has to be of the form ns − 1 with s ≥ 1 an integer.
Proof. Step 1. Using the implicit function theorem, the fact that f ≡ 1
in a neighborhood of 0, and the previous lemma, we easily obtain that ϕ ∈
C k (B; Rn ) ∩ C 0,1/n (B; Rn ) and ϕ = 0 on ∂B. Finally (cf. Step 2.2 in the
proof of Lemma 11.10 in [1]) we get ϕ∗ (1) = f in B, which proves the main
assertion.
Step 2. Let us show the first extra assertion. Since f < 0 on ∂B, there
exists > 0 small enough such that
−1/ < f < − in B \ B1− .
(6.1)
W 1,p (B; Rn ).
Step 2.1. We start by showing that ϕ ∈ ∩1≤p<n/(n−1)
Since
k
n
1,p
n
ϕ ∈ C (B; R ), it is enough to prove that ϕ ∈ W (B \ B1− ; R ) for every
1 ≤ p < n/(n − 1), which is equivalent to proving that α ∈ W 1,p (B \ B1− )
for every 1 ≤ p < n/(n − 1). Note that
Z 1
1/n 1/n
Z |x|
x
x
n−1
α(x) = n
s
f (s ) ds
= −n
sn−1 f (s ) ds
.
|x|
|x|
0
|x|
Fix 1 ≤ i ≤ n. Differentiating the previous equation with respect to xi , we
get that, for every x ∈ B, x 6= 0,
|x|δij − xi xj R1
P
|x|
xi
x
ds
n|x|n−1 f (x) |x|
− n nj=1 |x| sn fxj (s |x|
)
|x|2
αxi (x) =
, (6.2)
(n−1)/n
R1
x
− n |x| sn−1 f (s |x|
) ds
where δij = 1 if i = j and 0 otherwise. Using (6.1) we get that
Z
Z
1
|αxi (x)|p dx ≤ K
p(n−1)/n dx
R
1
x
B\B1−
B\B1−
n−1
− n |x| s
f (s |x| ) ds
Z
K
1
≤ p(n−1)/n
dx
n )p(n−1)/n
(1
−
|x|
B\B1−
Z
K
1
≤ p(n−1)/n
dx.
p(n−1)/n
B\B1− (1 − |x|)
Since the previous integral is finite when 1 ≤ p < n/(n − 1), we have proved
that ϕ ∈ ∩1≤p<n/(n−1) W 1,p (B; Rn ).
On the equation det ∇ϕ = f prescribing ϕ = 0 on the boundary
1049
Step 2.2. We now show that ϕ ∈
/ W 1,n/(n−1) (B; Rn ). We start by choosing 0 < δ < small enough and choosing Ci a cone of axis R+ ei having vertex
zero and of aperture small enough so that, for every x ∈ Ci ∩ (B \ B1−δ ),
xi xj
n Z 1
X
xi
x |x|δij − |x| n−1
n
f (x)
−n
s fxj (s )
ds ≥ m > 0.
n|x|
|x|
|x|
|x|2
|x|
j=1
Combining the previous equation with (6.1) we obtain
Z
|αxi (x)|n/(n−1) dx
Ci ∩(B\B1−δ )
Z
1
n/(n−1)
dx
≥(m)
R1
n−1 f (s x ) ds
Ci ∩(B\B1−δ ) −n
s
|x|
|x|
Z
1
≥(m)n/(n−1) dx = ∞.
1
−
|x|n
Ci ∩(B\B1−δ )
Thus, ϕ ∈
/ W 1,n/(n−1) (B; Rn ).
Step 2.3. We finally prove that ϕ ∈
/ C 0,β (B; Rn ) for every 1/n < β ≤ 1.
Using (6.1), we deduce that, for x ∈ B \ B1− ,
Z 1
1/n
α(x) ≥
sn−1 ds
≥ ((1 − |x|)/n)1/n .
|x|
Hence, for |x| = 1, we deduce that, for 1 − < λ < 1,
|α(λx)|
((1 − λ)/n)1/n
|α(x) − α(λx)|
=
≥
.
|x − λx|β
|1 − λ|β
|1 − λ|β
Letting λ go to 1, we deduce that α ∈
/ C 0,β (B) for 1/n < β ≤ 1, which
proves the claim.
Step 3. We prove the second extra assertion. Using (6.2), an elementary
calculation gives that, for every x ∈ ∂B,
−nxi (−c(x))1/n if m = n − 1
lim αxi (x) =
0
if m > n − 1.
x→x,x∈B
This shows that ϕ ∈ C 1 (B; Rn ).
Step 4. The third and fourth extra statement are elementary since, near
the boundary, a direct calculation gives ϕ(x) = (1 − |x|)s x/|x| in the first
case and ϕ(x) = e−1/(1−|x|) x/|x| in the second case.
1050
Olivier Kneuss
7. Necessary conditions
In this last section we give several necessary conditions to solve our problem, namely
∗
ϕ (g) = f in Ω
ϕ=0
on ∂Ω.
We start with a lemma asserting that the total mass of f always has to
be zero in order to solve the previous problem.
Lemma 15. Let Ω ⊂ Rn be a bounded open set such that meas(∂Ω) = 0,
g ∈ C 0 (Rn ), f ∈ C 0 (Ω), and ϕ ∈ C 1 (Ω; Rn ) ∩ C 0 (Ω; Rn ) be such that |g| > 0
in Rn and
∗
ϕ (g) = f in Ω
ϕ=0
on ∂Ω.
Then
Z
f = 0.
Ω
Remark 16. If we moreover assume ϕ ∈ W 1,n (Ω; Rn ), then the previous
result can be proved using smooth approximations of ϕ and the divergence
theorem.
Proof. We extend ϕ by 0 outside of Ω. Note that this extension (still written
ϕ) satisfies ϕ ∈ C 0 (Rn ; Rn ) ∩ C 1 (Rn \ ∂Ω; Rn ). Since meas(∂Ω) = 0, ϕ is
differentiable almost everywhere. Note that det ∇ϕ ∈ L1loc (Rn ). Since ϕ is
locally Lipschitz in Rn \ ∂Ω and ϕ = 0 on ∂Ω, we deduce that
for every A ⊂ Rn such that meas(A) = 0.
meas(ϕ(A)) = 0,
Hence, we can apply Theorem 5.27 of [4] (with D = Rn , G = Ω, and v = g)
and get
Z
Z
g(ϕ) det ∇ϕ =
g(y) deg(ϕ, Ω, y) dy,
Rn
Ω
where deg is the usual topological degree (see for example [4]). Since ϕ = 0
on ∂Ω, we get, by a well-known property of the degree, that deg(ϕ, Ω, y) =
deg(0, Ω, y) = 0, for every y 6= 0. Hence
Z
Z
f=
g(ϕ) det ∇ϕ = 0,
Ω
which concludes the proof.
Ω
The next proposition gives conditions on f if we want the solution to be
regular up to the boundary.
On the equation det ∇ϕ = f prescribing ϕ = 0 on the boundary
1051
Proposition 17. Let n ≥ 2 be an integer.
(i) Let Ω be a bounded open C 1 set and ϕ ∈ C 1 (Ω; Rn ) be such that ϕ = 0
on ∂Ω. Then
det ∇ϕ = 0 on ∂Ω.
n
(ii) Let Ω ⊂ R be a bounded open C 2 set and ϕ ∈ C 2 (Ω; Rn ) be such that
ϕ = 0 on ∂Ω. Then, near ∂Ω,
det ∇ϕ(x) = dist(x, ∂Ω)n−1 h(x)
with h ∈ C 1 (Ω) ∩ C 0 (Ω).
Proof. Part 1. We prove the first assertion. Since Ω is C 1 there exist
U ⊂ V a neighborhood of x and φ ∈ Diff 1 (U ; φ(U )) such that
φ(U ∩Ω) ⊂ {x ∈ Rn : xn > 0},
φ(U ∩∂Ω) ⊂ {x ∈ Rn : xn = 0},
φ(x) = 0.
In particular, there exists > 0 small enough so that ϕ◦φ−1 ∈ C 1 (B ∩{yn ≥
0}; Rn ). Note also that ϕ ◦ φ−1 = 0 on B ∩ {yn = 0}. Since ϕ ◦ φ−1 is C 1
and vanishes on yn = 0 we immediately deduce that
det ∇(ϕ ◦ φ−1 ) = 0
on B ∩ {yn = 0}.
This directly implies that det ∇ϕ(x) = 0, and hence the first assertion is
proved.
Part 2. We prove the second assertion.
Step 1. Let x ∈ ∂Ω. By hypothesis there exist U a neighborhood of x
and φ ∈ Diff 2 (U ; φ(U )) such that
φ(U ∩Ω) ⊂ {x ∈ Rn : xn > 0},
φ(U ∩∂Ω) ⊂ {x ∈ Rn : xn = 0},
ϕ◦φ−1
φ(x) = 0.
C 2 (B
In particular, there exists > 0 small enough so that
∈
∩{yn ≥
0}; Rn ). Note also that ϕ ◦ φ−1 = 0 on B ∩ {yn = 0}. We claim that, for
every y ∈ B ∩ {yn ≥ 0}, we have
det ∇(ϕ ◦ φ−1 )(y) = (yn )n−1 w(y),
where w ∈ C 1 (B ∩ {yn > 0}) ∩ C 0 (B ∩ {yn ≥ 0}). Since ϕ ◦ φ−1 = 0 on
yn = 0 and since ϕ ◦ φ−1 ∈ C 2 , we deduce that, for every 1 ≤ s ≤ n and
every 1 ≤ l ≤ n − 1,
∂(ϕ ◦ φ−1 )s )
(y) = yn wsl (y),
∂yl
where wsl ∈ C 1 (B ∩ {yn > 0}) ∩ C 0 (B ∩ {yn ≥ 0}). Recalling that
n
X
Y
∂(ϕ ◦ φ−1 )i )
det ∇(ϕ ◦ φ−1 ) =
sign(σ)
,
∂yσ(i)
σ∈Sym(n)
i=1
1052
Olivier Kneuss
we immediately get the assertion.
Step 2. Using Lemma 18 (with Ω1 = Ω and Ω2 = {xn > 0}), we
immediately get the existence of g ∈ C 1 (Ω) ∩ C 0 (Ω) such that
φn (x) = g(x) dist(x, ∂Ω),
for x ∈ Ω near ∂Ω.
Step 3 (conclusion). Let x ∈ Ω near ∂Ω. Using Steps 1 and 2 we obtain
det ∇ϕ(x) = det ∇[(ϕ ◦ φ−1 ) ◦ φ](x) = det ∇(ϕ ◦ φ−1 )(φ(x)) det ∇φ(x)
= (φn (x))n−1 w(ϕ(x)) det ∇φ(x)
= dist(x, ∂Ω)n−1 g(x)n−1 w(φ(x)) det ∇φ(x),
whence the theorem, letting h = g n−1 w(φ) det ∇φ.
In the previous proof we have used the following elementary lemma which
we do not prove.
Lemma 18. Let k ≥ 2 be an integer and Ω1 , Ω2 ⊂ Rn be two bounded open
C k sets. Let x ∈ ∂Ω1 , V be a neighborhood of x, and φ ∈ Diff k (V ; ϕ(V )) be
such that φ(∂Ω1 ∩ V ) ⊂ ∂Ω2 . Then there exist U ⊂ V a neighborhood of x
and g ∈ C k−1 (U \ ∂Ω1 ) ∩ C k−2 (U ) such that
dist(φ(x), ∂Ω2 ) = g(x) dist(x, ∂Ω1 ),
for every x ∈ U
and g ≥ m > 0 in U .
Acknowledgments. I have benefitted from interesting discussions with
B. Dacorogna and G. Csató.
References
[1] G. Csató, B. Dacorogna, and O. Kneuss, “The Pullback Equation for Differential
Forms,” Springer, Boston, 2012.
[2] G. Cupini, B. Dacorogna, and O. Kneuss, On the equation det ∇u = f with no sign
hypothesis, Calc. Var. Partial Differential Equations, 36 (2009), 251–283.
[3] B. Dacorogna and J. Moser, On a partial differential equation involving the Jacobian
determinant, Ann. Inst. H. Poincaré Anal. Non Linéaire, 7 (1990), 1–26.
[4] I. Fonseca and W. Gangbo, “Degree Theory in Analysis and Applications,” Oxford
University Press, New York, 1995.