A sharp Hölder estimate for elliptic equations in
two variables
Tonia Ricciardi∗
Dipartimento di Matematica e Applicazioni
Università di Napoli Federico II
Via Cintia, 80126 Naples, Italy
fax: +39 081 675665
tonia.ricciardi@unina.it
Abstract
We prove a sharp Hölder estimate for solutions of linear two-dimensional, divergence form elliptic equations with measurable coefficients, such
that the matrix of the coefficients is symmetric and has unit determinant.
Our result extends some previous work by Piccinini and Spagnolo [7]. The
proof relies on a sharp Wirtinger type inequality.
Key Words: linear elliptic equation, measurable coefficients, Hölder regularity
MSC 2000 Subject Classification: 35J60
1
Introduction
1
Let Ω be a bounded open subset of Rn and let u ∈ Hloc
(Ω) be a weak solution
to the linear elliptic equation in divergence form
(1)
(aij uxi )xj = 0,
where aij , i, j = 1, 2, . . . , n, are bounded measurable functions in Ω satisfying
the ellipticity condition
(2)
λ|ξ|2 ≤ aij (x)ξi ξj ≤ Λ|ξ|2 ,
for all ξ ∈ Rn and for some 0 < λ ≤ Λ. It is well known (see, e.g., [1, 2, 3, 6] and
references therein) that solutions to (1) are Hölder continuous. More precisely,
there exists some 0 < α < 1 such that for every compact subset K ⊂⊂ Ω there
holds
|u(x) − u(y)|
sup
< +∞.
|x − y|α
x,y∈K,x6=y
∗ Partially supported by Regione Campania L.R. 5/02 and by the MIUR National Project
Variational Methods and Nonlinear Differential Equations
1
Moreover, estimates for the Hölder exponent α may be obtained, depending on
the ellipticity constant L = Λ/λ only. In this note we consider the case n = 2,
which unless otherwise stated we assume henceforth. For this case, sharp Hölder
estimates were obtained by Piccinini and Spagnolo in [7] under the symmetry
assumption
(3)
aij (x) = aji (x).
We collect in the following theorem some results from [7] which are relevant to
our considerations.
1
Theorem 1 (Piccinini and Spagnolo [7]). Let u ∈ Hloc
(Ω) be a weak solution to (1).
(i) If the coefficients aij satisfy (2)–(3), then α = L−1/2 .
(ii) If the coefficients aij satisfy (2)–(3) and if furthermore aij = a(x)δij for
some measurable function a = a(x) satisfying λ ≤ a(x) ≤ Λ for all x ∈ Ω,
then α = π4 arctan L−1/2 .
These values of α are sharp.
In particular, Theorem 1–(ii) implies that the optimal value of α increases by
restricting the matrices A to the class of isotropic matrices A = a(x)I. Thus,
it is natural to seek other classes of matrices A for which the corresponding
optimal value of α may be improved. Our main result in this note is to show
that if A = (aij ) satisfies (2)–(3) and if furthermore A has unit determinant,
namely
(4)
det A(x) = a11 (x)a22 (x) − a12 (x)a21 (x) ≡ 1,
then a more accurate integral characterization of α may be obtained, pointwise
in Ω. Moreover, our result is sharp within the class of matrices satisfying (4).
It should be mentioned that condition (4) has attracted much interest recently,
see [4].
Theorem 2 (Main result). Let A = (aij ) satisfy (2)–(3), let x0 ∈ Ω and let
A satisfy (4) in a ball Br0 (x0 ) ⊂ Ω. Then there holds
(5)
sup
0<|x−x0 |<r0
|u(x) − u(x0 )|
< +∞
|x − x0 |α
with
!−1
Z
(6)
α = α(x0 , r0 ) = 2π ess sup
aij (x0 + rξ)ξi ξj dσ
0<r<r0
.
|ξ|=1
We note that under assumption (4), we may choose λ = 1/Λ in (2) and
therefore the ellipticity constant takes the value L = Λ2 . Hence, the PiccininiSpagnolo estimate in Theorem 1–(i) yields in this case α = Λ−1 . On the other
hand, recalling that Λ = supx∈B sup|ξ|=1 aij (x)ξi ξj , it is clear that α(x0 , r0 ) ≥
Λ−1 .
Theorem 2 is sharp, in the sense of the following
2
Example 1 (Sharpness). Let Ω = B the unit ball in R2 and let
A = p(θ)−1 I + p(θ) − p(θ)−1
(7)
x⊗x
|x|2
in B \ {0}.
where θ = arg x and p : R → R+ is 2π-periodic measurable function bounded
from above and away from 0. Then det A(x) ≡ 1. Moreover, for every ξ ∈ R2
such that |ξ| = 1 and for every 0 < r < 1 we have aij (rξ)ξi ξj ≡ p(θ) and
R
−1
2π
therefore formula (6) yields α(0, r) = 2π 0 p(θ) dθ
=: α0 for every 0 <
r < 1. On the other hand the function u ∈ H 1 (B) defined by
Z
u(x) = |x|α0 cos α0
(8)
arg x
p
0
satisfies equation (1) with A given by (7), and its Hölder exponent is exactly
α0 .
For the reader’s convenience we have included a verification of Example 1 in
the Appendix.
Theorem 2 readily implies the following
Corollary 1. Let A = (aij ) satisfy (2)–(3), let x0 ∈ Ω and suppose A satisfies
(4) in Ω. Then the greatest lower bound of the admissible values of α in (5) is
given by
!−1
Z
α(x0 ) = 2π
inf
aij (x0 + rξ)ξi ξj dσ
ess sup
0<r0 <dist(x0 ,∂Ω)
0<r<r0
.
|ξ|=1
The remaining part of this note is devoted to the proof of Theorem 2.
Notation
Let x1 = ρ cos θ, x2 = ρ sin θ be the polar coordinate transformation. We
denote uxi = ∂u/∂xi , uρ = ∂u/∂ρ, uθ = ∂u/∂θ. Following the notation in [7],
we denote ∇u = (ux1 , ux2 ) and ∇u = (uρ , uρθ ). We denote by J(θ) the rotation
matrix
cos θ
− sin θ
J(θ) =
.
sin θ
cos θ
In this notation, we have
(9)
∇u = J(θ)∇u.
Finally, we denote by h·, ·i the Euclidean scalar product in R2 .
2
Proof of Theorem 2
In the spirit of [7], a key ingredient in the proof of Theorem 2 is a sharp Wirtinger
type inequality.
3
Lemma 1 (sharp Wirtinger type inequality). Let a : R → R be a 2πperiodic measurable function which is bounded from above and away from 0.
Then
Z 2π
Z 2π 2 Z 2π
1
1 0 2
2
aw ≤
(10)
a
(w )
2π
a
0
0
0
1
for every w ∈ Hloc
(R) such that w is 2π-periodic and satisfies
Equality in (10) is attained if and only if w is of the form
!
Z θ
2π
a+ϕ
(11)
w(θ) = C cos R 2π
a 0
0
R 2π
0
aw = 0.
for some C 6= 0 and ϕ ∈ R.
Proof. Let
1
X = w ∈ Hloc
(R) : w is 2π−periodic
and consider the functional I defined by
R 2π
(w0 )2
1
I(w) = 0R 2πa
0
aw2
for all w ∈ X \ {0}. We equivalently have to show that
2π
Z
inf I(w) : w ∈ X \ {0},
aw = 0 =
0
!2
2π
R 2π
0
a
and that the infimum is attained exactly on functions of the form (11). By
Sobolev embeddings and standard compactness arguments, there exists w ∈
R 2π
X \ {0} satisfying 0 aw = 0 and such that
Z
I(w) = inf I(w) : w ∈ X,
2π
aw = 0 > 0.
0
R 2π By the minimum property of w we have that I 0 (w) φ − (2π)−1 0 aφ = 0 for
all φ ∈ X. By properties of w it follows that
Z
0
2π
1 0 0
w φ − I(w)
a
Z
2π
∀φ ∈ X.
awφ = 0
0
R
Rθ
2π
1
Let Θ = 0 a, w(θ) = W (Θ), φ(θ) = Φ(Θ). Then W, Φ ∈ Hloc
(R) are 0 a periodic and
Z
R 2π
0
a
0
Z
0
R 2π
0
W (Θ)Φ (Θ) dΘ = I(w)
0
W (Θ)Φ(Θ) dΘ.
0
Equivalently, W is a weak solution for
−(W 0 )0 = I(w)W.
4
a
It follows that I(w) = (2π/
R 2π
0
a)2 and
2π
W (Θ) = C cos( R 2π Θ + ϕ)
a
0
for some C 6= 0 and ϕ ∈ R. Recalling the definition of W and Θ, we conclude
the proof.
We shall also need the following result, which is a variant of a well known
sufficient condition for Hölder continuity. See, e.g., [2, 3, 5]. We state it for
arbitrary n ≥ 2 and p > 1. For the reader’s convenience we have included a
proof in the Appendix.
1,p
Lemma 2. Let B be the unit ball in Rn , n ≥ 2. Let u ∈ Wloc
(B) for some
p > 1 and suppose that for all 0 < r < 1 there holds the inequality
Z
(12)
|∇u|p dx ≤ K(r)rn−p+pα
Br
for some nondecreasing function K, where Br is the ball of radius r centered at
0. Then
|u(x) − u(0)|
< +∞.
sup
|x|α
x∈B\{0}
We have included a proof of Lemma 2 in the Appendix.
Proof of Theorem 2. Let 0 < r < r0 . For every 0 < r < r0 let
Z
g(r) =
aij uxi uxj dx.
|x−x0 |<r
In view of Lemma 2 with n = p = 2, a sufficient condition for (5) is that the
function G(r) = r−2α g(r) be bounded near 0. Setting P = J ∗ AJ = (pij ), we
have by (9)
aij uxi uxj =hA∇u, ∇ui = hAJ∇u, J∇ui = hP ∇u, ∇ui
2
uθ
uθ
=p11 u2ρ + (p12 + p21 ) uρ
+ p22
.
ρ
ρ
Note that det P = 1 and P = P ∗ . Therefore, denoting p11 = p, p12 = p21 = q,
p22 = (1 + q 2 )/p, we can write
2
1 + q 2 uθ
uθ
aij uxi uxj = pu2ρ + 2quρ
+
.
ρ
p
ρ
In view of (1) for every k ∈ R we have aij uxi uxj = ((u − k)aij uxi )xj and
therefore by the divergence theorem and an approximation argument (see [7])
Z
g(r) =
(u − k)aij uxi nj dσ
Sr
where Sr is the circle of center x0 and radius r and n is the outward normal to
Sr . Since n = (cos θ, sin θ) = Je1 we have
aij uxi nj = hA∇u, ni = hAJ∇u, Je1 i = hP ∇u, e1 i = puρ + q
5
uθ
ρ
and therefore by Hölder’s inequality
Z
Z
uθ
q uθ
√
√
g(r) =
(u − k) puρ + q
p(u − k)
puρ + √
dσ =
dσ
ρ
p ρ
Sr
Sr
#1/2
"Z
2
Z q uθ
√
2
puρ + √
dσ
.
≤
p(u − k) dσ
p ρ
Sr
Sr
R
Choosing k = |Sr |−1 Sr pu dσ, rescaling and using (10) with a(θ) = p(r, θ),
w(θ) = u(r, θ) − k, we obtain the inequality
2 Z
Z
Z
1 2
p(u − k)2 dσ ≤ |Sr |−1
p dσ
uθ dσ.
Sr
Sr
Sr p
It follows that
−1
Z
Z
g(r) ≤ |Sr |
p dσ
Sr
Sr
1 2
u dσ
p θ
1/2 "Z
√
Sr
q uθ
puρ + √
p ρ
#1/2
2
dσ
.
Equivalently, we may write
Z
−1
p dσ r
g(r) ≤ |Sr |
Sr
"Z
×
Sr
1
p
uθ
ρ
#1/2 "Z
2
dσ
Sr
√
q uθ
puρ + √
p ρ
#1/2
2
dσ
.
√
By the inequality ab ≤ 12 (a + b) for every a, b > 0 we obtain for a.e. 0 < r < r0
that
Z " 2 2 #
Z
1 uθ
1
q uθ
√
−1
p dσ r
(13) g(r) ≤
|Sr |
+
puρ + √
dσ
2
ρ
p ρ
Sr p
Sr
Z "
2 #
Z
1
uθ
1 + q2
uθ
2
−1
puρ + 2quρ
=
|Sr |
p dσ r
+
dσ
2
ρ
p
ρ
Sr
Sr
Z
Z
1
−1
|Sr |
p dσ r
hP ∇u, ∇ui
=
2
S
S
Z r
Z r
1
=
|Sr |−1
p dσ r
hA∇u, ∇ui
2
Sr
S
Z r Z
1
≤ ess sup
|Sr |−1
p dσ r
hA∇u, ∇ui.
2
0<r<r0
Sr
Sr
Finally, we note that for all ξ ∈ R2 , |ξ| = 1, we have ξ = Je1 and therefore
(14)
p(rξ) = hP (rξ)e1 , e1 i = hJ ∗ A(rξ)Je1 , e1 i = hA(rξ)ξ, ξi = aij (rξ)ξi ξj .
Consequently, we have
Z
ess sup
|Sr |−1
0<r<r0
p dσ
=(2π)−1 ess sup
0<r<r0
Sr
=α−1 (x0 , r0 ).
6
Z
aij (x0 + rξ)ξi ξj dσ
|ξ|=1
Furthermore, g is differentiable almost everywhere and
Z
g 0 (r) =
p dσ
a.e. r.
Sr
Therefore (13) yields
g(r) ≤
(15)
rg 0 (r)
.
2ᾱ
Consequently, recalling that G(r) = r−2α(x0 ,r0 ) g(r), we obtain from (15) that
(log G(r))0 ≥ 0 a.e. It follows that
Z r0
G(ρ) ≤ G(r0 ) −
G0 (r) dr ≤ G(r0 )
ρ
and finally lim supr→0 G(r) < +∞, as desired.
Appendix
We have postponed to this appendix a
Verification of Example 1. It is readily checked that u given by (8) is a weak
solution for the equation
−1 uθ
(16)
(ρp(θ)uρ )ρ + p(θ)
= 0,
ρ θ
i.e.,
Z p(θ)uρ vρ + p(θ)−1
B
uθ vθ
ρ ρ
ρ dρ dθ = 0
for all v ∈ H 1 (B) compactly supported in B. Setting
p(θ)
0
P (θ) =
0
p(θ)−1
we have, recalling (9),
Z
Z
0 = hP ∇u, ∇uiρ dρ dθ =
hP J ∗ ∇u, J ∗ ∇ui dx
B
B
Z
Z
= hJP J ∗ ∇u, ∇ui dx =
hA∇u, ∇ui dx.
B
B
Therefore, in cartesian coordinates, equation (16) takes the form (1) with A =
(aij ) given by
(17)
A =J(θ)P (θ)J ∗ (θ)
−1
p(θ) cos2 θ + p(θ)
sin2 θ
=
−1
p(θ) − p(θ)
sin θ cos θ
p(θ) − p(θ)−1 sin θ cos θ
.
p(θ) sin2 θ + p(θ)−1 cos2 θ
On the other hand, in view of (14) we have
α0 = R
2π
2π
= R 2π .
a
(rξ)ξ
ξ
ij
i
j
p
|ξ|=1
0
7
For the reader’s convenience, we include a
Proof of Lemma 2. Throughout this proof, we denote by Kj some positive constants depending on n, p and α only. RWithout loss of generality we assume that
u(0) = 0. We denote uBr = |Br |−1 Br u. By the arguments in the proof of
Theorem p. 718 in [5] we have
|u(x) − uBr | ≤ K1 K(r)rα
∀x ∈ Br .
In particular, choosing r = |x|/2, we obtain
|u(x) − uB2|x| | ≤ K2 K(2|x|)|x|α .
Therefore, it suffices to show that |uBr | ≤ K3 rα for every r > 0. To this end,
we write
Z 1
u(x) =
∇u(tx) · x dt.
0
Integrating over Br , we have
Z
Z
Z 1
Z
u=
dx
∇u(tx) · x dt =
Br
Br
Z 1
=
0
Z
∇u(z) ·
BtR
z dz
=
t tn
Z
∇u(tx) · x dx
dt
0
dt
0
1
Z
Br
1
0
dt
tn+1
Z
∇u(z) · z dz.
BtR
Therefore,
Z
Z 1
Z
dt
u ≤
|∇u(z)| |z| dz
n+1
0 t
Br
BtR
Z
1/p Z
1/p0
Z 1
dt
p
p0
|∇u(z)| dz
|z| dz
.
≤
n+1
Btr
Btr
0 t
Since
Z
Btr
we derive
Z
Z
≤
u
Br
0
|z|p =
0
nωn
(tr)p +n ,
p0 + n
1/p0
nωn
p0 +n
(tr)
p0 + n
0 t
1/p0
1/p0
Z 1
K
nωn
nωn
n+α
α−1
r
t
dt
=
rn+α .
=K
0+n
p0 + n
α
p
0
1
dt
K 1/p (tr)n/p−1+α
n+1
We conclude that
|uBr | ≤ K3 rα ,
as desired.
Acknowledgements
I would like to thank Professor Carlo Sbordone for suggesting the problem,
as well as for many interesting and stimulating discussions leading to its solution. I am grateful to Professor Nicola Fusco for many useful comments on the
manuscript.
8
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9