Appl. Math. J. Chinese Univ. Ser. B
2006, 21(3): 343-349
REGULARITY FOR VERY WEAK SOLUTIONS TO
A-HARMONIC EQUATION
Liu Lin Gao Hongya
Abstract. In this paper, the following result is given by using Hodge decomposition and weak
reverse Hölder inequality: For every r1 with
³
2
³
p − 2n+1 100n p 23+n/(p−1) + 1
´ ´−1
b
a
< r1 < p ,
there exists the exponent r2 = r2 (n, r1 , p) > p, such that for every very weak solution u ∈
Wr11 ,loc (Ω ) to A-harmonic equation, u also belongs to Wr12 ,loc (Ω ) . In particular, u is the weak
solution to A-harmonic equation in the usual sense.
§1 Introduction
Let Ω be a bounded regular domain in Rn (n ≥ 2). By regular domain we understand any
domain of finite measure for which the estimates for Hodge decomposition in (2.1) and (2.2 )
are justified. See [5], [7], [4]. A Lipschitz domain, for example, is regular.
We denote by Lp (Ω ) (1 ≤ p < ∞) the space of functions defined on Ω , such that |f (x)|p is
integrable with respect to the measure dx with the norm
µZ
k f kp,Ω = kf kLp (Ω) =
¶1/p
| f (x) |p dx
.
Ω
The space Lploc (Ω ) consists of functions which belong to Lp (F ) for every compact subset F ⊂ Ω .
¡ 1
¢
The symbol Wp1 (Ω ) Wp,
loc (Ω ) , 1 ≤ p < ∞ stands for the class of functions which belong to
Lp (Ω ) (Lploc (Ω )) and whose weak partial derivatives exist and also belong to Lp (Ω ) (Lploc (Ω )).
We consider solutions u of a quasilinear second order equation
divA(x, ∇u(x)) = 0,
(1.1)
where A : Ω ×Rn −→ Rn satisfies the usual measuability conditions (Caratheodory conditions)
and that for some 1 < p < ∞, the following conditions hold:
Received:2004-06-28, Revised:2005-07-10.
MR Subject Classification:35J65.
Keywords: A-harmonic equation, very weak solution, Hodge decomposition, weak reverse Hölder inequality.
344
Appl. Math. J. Chinese Univ. Ser. B
Vol. 21, No. 3
(I) the Lipschitz type condition
p−2
|A(x, ξ) − A(x, ζ)| ≤ b |ξ − ζ| (|ξ| + |ζ|)
;
(II) the monotonicity inequality
2
p−2
hA(x, ξ) − A(x, ζ), ξ − ζi ≥ a |ξ − ζ| (|ξ| + |ζ|)
;
(III) the homogeneity condition
A(x, λ ξ) = |λ|p−2 λ A(x, ξ)
for almost every x ∈ Ω and all ξ, ζ ∈ Rn , 0 < a ≤ b < ∞, λ ∈ R.
Remark 1. The mapping A(x, ξ) = |ξ|p−2 ξ, which generates the p-harmonic equation
³
´
p−2
div | ∇u(x) |
∇u(x) = 0,
satisfies the assumptions (I) (II) (III).
We need the following definition.
1
Definition 1. A function u ∈ Wp,
loc (Ω ) is said to be a weak solution to (1.1) if
Z
hA(x, ∇u), ∇φidx = 0
(1.2)
Ω
1
(Ω ).
holds for all φ ∈ Wp,0
Recall that the original test function for weak solution is that φ ∈ C0∞ (Ω ) ,and then by an
1
approximation argument it is extended to all φ ∈ Wp,0
(Ω ).
By the Lipschitz type inequality, it is clear that in order to guarantee the integrability
of the integrand of (1.2) with φ ∈ C0∞ (Ω ), it is only necessary that u ∈ Wr,1 loc (Ω ), where
max{1, p − 1} ≤ r. For this reason we give the following definition.
Definition 2.[7] A function u ∈ Wr,1 loc (Ω ) with max{1, p − 1} < r < p is called a very weak
solution to (1.1), if (1.2) holds for all φ ∈ W 1 r ,0 (Ω ).
r−p+1
In [7], Iwaniec and Sbordone proved the following regularity result for very weak solutions
to (1.1).
Lemma 1. There exist exponents
1 < r1 = r1 (n, p, a, b) < p < r2 = r2 (n, p, a, b) < ∞,
such that every very weak solution u ∈ Wr11 ,loc (Ω ) to A-harmonic equation belongs to Wr12 ,loc (Ω ).
Lemma 2. Let BR denotes the open ball of radius R. BσR stands for the ball of radius σR
with the same center as in BR . If u ∈ Wp1 (BR ), 1 ≤ p < ∞, then for any 0 < σ ≤ 1,
k u − uBσR kLp (BR ) ≤
³ 2 ´n/p
σ
diamBR k ∇u kLp (BR )
R
R
where uBσR = −BσR u(x)dx = |B1σR | BσR u(x)dx is the average value of u(x) over BσR .
Lemma 2 is from Lemma 1.5 in [1].
(1.3)
Liu Lin, et al.
Lemma 3.
REGULARITY FOR VERY WEAK SOLUTIONS TO A-HARMONIC ...
345
1
If u ∈ Wp,
loc (Ω ), then for any ball BR ⊂⊂ Ω ,
ku − uBR kLp (BR ) ≤ 21+n/p Rk∇ukLp (BR ) .
(1.4)
Proof. (1.4) is obtained by letting σ = 1 in Lemma 2.
1
Lemma 4.[8] Let 1 ≤ p < n. If u ∈ Wp,
loc (Ω ), then for any ball BR ⊂⊂ Ω ,
ku − uBR k
np
L n−p
(BR )
≤ C(n)
p ³ p ´(n−p)/np
k∇ukLp (BR ) .
n−p p−1
(1.5)
See Lemma 7.16 and 7.12 in [3] for Lemma 4.
The following lemma ([2], Chap.5, Proposition 2.1) gives a relationship between the weak
reverse Hölder inequality and the local Lp integrability.
Lemma 5. Let 0 < R < R0 = dist(a, ∂Ω ), ∀a ∈ Ω . If for functions g(x) ∈ Lploc (Ω ), 1 < p <
∞, the following weak reverse Hölder inequality holds:
Z
Z
´ ps
³Z
− |g(x)|p dx ≤ θ − |g(x)|p dx + C − |g(x)|s dx
(1.6)
BR
,
BR
BR
2
where 1 ≤ s < p and 0 ≤ θ < 1, then there exists an exponent p0 = p0 (θ, p, n, C) > p such that
0
g(x) ∈ Lploc (Ω ), and the following inequality holds:
³Z
´ 10
³Z
´ p1
p
p0
− |g(x)| dx
(1.7)
≤ C1 − |g(x)|p dx
BR
BR
2
,
where C1 depends only on n, C, p, θ, R0 .
It is a typical situation in P.D.E. and also in the calculus of variations that one wants to
build up the regularity theory for very weak solutions or minimizers of a variational integral
under the least possible hypotheses. Therefore we consider the very weak solutions to (1.1) in
this paper. Our primary result is the following regularity theorem with 1 < p < n + 1.
³
³
´ ´−1
2
< r1 < p, there exists
Theorem. For every r1 with p − 2n+1 100n p 23+n/(p−1) + 1 ab
the exponent r2 = r2 (n, r1 , p) > p, such that for every very weak solution u ∈ Wr11 ,loc (Ω ) to
A-harmonic equation, u also belongs to Wr12 ,loc (Ω ) . In particular, u is the weak solution to
(1.1) in the usual sense.
Remark 2. Lemma 1 shows that in order to obtain the regularity of very weak solutions to A1
1
harmonic equation, it is not necessary that u ∈ Wp,
loc (Ω ), a weaker space u ∈ Wr1 ,loc (Ω ) (r1 < p)
is sufficient. The theorem improves Lemma 1 and obtains the estimate for r1 .
§2 Proof of the Theorem
Let η ∈ C0∞ (BR ) , 0 ≤ η ≤ 1 , |∇η| ≤ R4 and η ≡ 1 in B R be a test function. First, we
2
1
suppose that u ∈ Wp−ε,
(0 < ε < 1/2) is a very weak solution to (1.1). Consider the
loc (Ω )
Hodge decomposition
|∇(ηu)|−ε ∇(ηu) = ∇φ + H,
p−ε
1
where φ ∈ W p−ε
(Ω ) ; H ∈ L 1−ε (Ω ) is a divergence free vector field and the following estimates
,0
1−ε
hold[4] :
k∇φk p−ε ≤ C(n, p) k∇(ηu)k1−ε
p−ε ,
1−ε
(2.1)
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Appl. Math. J. Chinese Univ. Ser. B
Vol. 21, No. 3
2
kHk p−ε ≤ 100n 2 p ε k∇(ηu)k1−ε
p−ε .
(2.2)
1−ε
We introduce the Hodge decomposition so that the factor ε is included in the right hand
side of (2.2). This will play a crucial role in our derivation.
Let
E(η, u) =| ∇(ηu) |−ε ∇(ηu)− | η∇u |−ε η∇u,
then by an elementary inequality [6,(4.1)]
2ε (1 + ε)
| X − Y |1−ε
1−ε
| |X |−ε X− | Y |−ε Y |≤
0 < ε < 1,
X, Y ∈ Rn
we have
2ε (1 + ε)
| u∇η |1−ε .
(2.3)
1−ε
A useful technique in the following calculus is to use ∇φ in Hodge decomposition as a test
function in (1.2). We have
Z
Z
Z
−ε
hA(x, ∇u), E(η, u)idx +
hA(x, ∇u), |η∇u| η∇uidx =
hA(x, ∇u), Hidx,
| E(η, u) |≤
BR
BR
namely,
Z
BR
Z
−ε
hA(x, ∇u), |η∇u|
Z
η∇uidx =
hA(x, ∇u), Hidx −
BR
BR
hA(x, ∇u), E(η, u)idx =
BR
I1 + I2 ,
(2.4)
we have by the monotonicity inequality that
Z
Z
hA(x, ∇u), | η∇u |−ε η∇uidx =
η 1−ε | ∇u |−ε hA(x , ∇u), ∇uidx ≥
BR
BR
Z
Z
a
η
1−ε
−ε
| ∇u |
p
| ∇u |p−ε dx.
| ∇u | dx ≥ a
BR
(2.5)
BR
2
By the Lipschitz type inequality, Hölder’s inequality and (2.2) we have
¯ Z
¯Z
¯
¯
¯
|I1 | = ¯
hA(x, ∇u), Hidx¯¯ ≤
|A(x, ∇u)| |H|dx ≤
BR
BR
Z
p−1
b
|∇u|
BR
p−1
p−1
|H| dx ≤ b k∇ukp−ε kHk p−ε ≤
1−ε
2
1−ε
b k∇ukp−ε 100n 2 p ε k∇(ηu)kp−ε =
2
p−1
1−ε
2 b 100n p ε k∇ukp−ε k∇(ηu)kp−ε .
Notice that adding a constant to u will not change ∇u and the A-harmonic equation (1.1).
Therefore we can assume that the average value of u on BR is equal to zero. Then the Poincaré
inequality (1.4) yields
³
´1−ε
1−ε
1−ε
k∇(ηu)kp−ε = ku∇η + η∇ukp−ε ≤ ku∇ηkp−ε + kη∇ukp−ε
≤
REGULARITY FOR VERY WEAK SOLUTIONS TO A-HARMONIC ...
Liu Lin, et al.
µ
4
kukp−ε + k∇ukp−ε
R
Hence
¶1−ε
µ
4 1+n/(p−ε)
·2
R k∇ukp−ε + k∇ukp−ε
R
³
´
1−ε
23+n/(p−1) + 1 k∇ukp−ε .
347
¶1−ε
≤
<
³
´
2
p−1+1−ε
|I1 | ≤ 2 b 100n p ε 23+n/(p−1) + 1 k ∇u kp−ε
=
³
´ Z
2
p−ε
2 b 100n p 23+n/(p−1) + 1 ε
| ∇u |
dx.
(2.6)
BR
By the Lipschitz type inequality and (2.3), |I2 | can be estimated as follows:
¯Z
¯ Z
¯
¯
¯
| I2 | = ¯
hA(x, ∇u), E(η, u)idx ¯¯ ≤
| A(x, ∇u) | | E(η, u) | dx ≤
BR
BR
Z
b 2ε (1 + ε)
1−ε
p−1
| ∇u |
µ
Z
p−1
| ∇u |
BR
41−ε b 2ε (1 + ε)
R1−ε 1 − ε
|I2 | ≤
b
R1−ε
b
R1−ε
Let p00 =
yields
22−ε (1 + ε)
1−ε
22−ε (1 + ε)
1−ε
n(p−ε)
n+1−ε ,
µZ
|u|
then
n(p−ε)
n−p+1
dx
µZ
4
R
¶1−ε
1−ε
|u|
p−1
| ∇u |
1−ε
|u|
1
p0
then 1 < p0 , q 0 < ∞ ,
(p−1)p0
¶ 10 µZ
p
dx
BR
(1−ε)q 0
|u|
+
1
q0
= 1. By Hölder’s
¶ 10
dx
q
=
BR
| ∇u |
n(p−ε)
n+1−ε
n+1−ε
¶ n(p−ε)
(p−1) µZ
dx
BR
n−p+1
¶ n(p−ε)
dx.
BR
| ∇u |
=
dx =
Z
n(p−ε)
(n−p+1)(1−ε) ,
µZ
np00
n−p00
dx ≤
BR
b 2ε (1 + ε)
1−ε
n(p−ε)
Let p0 = (n+1−ε)(p−1)
, q0 =
inequality we have
1−ε
| u∇η |
|u|
n(p−ε)
n−p+1
n−p+1
¶ n(p−ε)
(1−ε)
dx
BR
n(p−ε)
n−p+1 .
Applying the Poincaré-Sobolev inequality (1.5) to p00 < n
µZ
=
|u|
BR
.
np00
n−p00
¶ n−p0000
dx
np
≤
BR
C(n)
p00
n − p00
µ
p00
00
p −1
¶(n−p00 )/np00 µZ
p00
| ∇u |
BR
¶ 100
p
dx
.
Therefore
| I2 | ≤
b
R1−ε
Ã
22−ε (1 + ε)
1−ε
p00
· C(n)
n − p00
µ
µZ
|∇u |
n(p−ε)
n+1−ε
n+1−ε
¶ n(p−ε)
(p−1)
dx
BR
p00
00
p −1
¶(n−p00 )/np00 !1−ε µZ
| ∇u |
BR
n(p−ε)
n+1−ε
n+1−ε
¶ n(p−ε)
(1−ε)
dx
,
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Appl. Math. J. Chinese Univ. Ser. B
Vol. 21, No. 3
namely,
| I2 | ≤
b
R1−ε
22−ε (1 + ε)
1−ε
Ã
p00
C(n)
n − p00
µ
p00
p00 − 1
¶(n−p00 )/np00 !1−ε µZ
| ∇u |
n(p−ε)
n+1−ε
¶ n+1−ε
n
dx
BR
.
(2.7)
Combining (2.4), (2.5), (2.6) and (2.7), we have
Z
³
´ Z
p−ε
n2
3+n/(p−1)
a
| ∇u |
dx ≤ 2 b 100 p 2
+1 ε
BR
p−ε
| ∇u |
dx+
BR
2
b
R1−ε
22−ε (1 + ε)
1−ε
Ã
p00
C(n)
n − p00
µ
p00
p00 − 1
¶(n−p00 )/np00 !1−ε µZ
| ∇u |
n(p−ε)
n+1−ε
¶ n+1−ε
n
dx
BR
.
n
Dividing both sides of the above inequality by | BR | = ωn R (ωn is the volume of the unit ball
in Rn ), we have
Z
³
´ Z
1
p−ε
p−ε
n2
3+n/(p−1)
a−
|
∇u
|
dx
≤
2
b
100
p
2
+
1
ε− | ∇u |
dx+
2n B R
BR
2
b
R1−ε
22−ε (1 + ε)
1−ε
n
· ( ωn R )
1−ε
n
Ã
p00
C(n)
n − p00
µ
¶(n−p00 )/np00 !1−ε
p00
00
p −1
·
µZ
¶ n+1−ε
n
n(p−ε)
n+1−ε
− | ∇u |
dx
,
BR
namely,
Z
³
´b Z
p−ε
p−ε
n+1
n2
3+n/(p−1)
dx+
dx ≤ 2
100 p 2
+1
ε − | ∇u |
− | ∇u |
a
BR
BR
2
1
R1−ε
2n+2−ε (1 + ε)
1−ε
· ( ωn R n )
1−ε
n
b
a
Ã
p00
C(n)
n − p00
µ
p00
p00 − 1
¶(n−p00 )/np00 !1−ε
·
µZ
¶ n+1−ε
n
n(p−ε)
− | ∇u | n+1−ε dx
BR
Note that
τ=
.
n(p − ε)
< p−ε
n+1−ε
and
1
R1−ε
2n+2−ε (1 + ε)
1−ε
Ã
p00
C(n)
n − p00
µ
p00
00
p −1
¶(n−p00 )/np00 !1−ε
( ωn R n )
1−ε
n
b
≤ C(n, p),
a
where C(n, p) is some constant depending only on n and p, then
Z
³
´ b Z
p−ε
p−ε
n+1
n2
3+n/(p−1)
dx ≤ 2
100 p 2
+1
− | ∇u |
ε − | ∇u |
dx+
a
BR
BR
2
REGULARITY FOR VERY WEAK SOLUTIONS TO A-HARMONIC ...
Liu Lin, et al.
µZ
¶ p−ε
τ
τ
C(n, p) − | ∇u | dx
BR
³
349
(2.8)
.
¡
¢ ´−1
2
> 0. When ε < ε0 ,
2n+1 100n p 23+n/(p−1) + 1 ab
¡ 3+n/(p−1)
¢ b
n+1
n2
0<2
100 p 2
+ 1 a ε = θ < 1 and (2.8) becomes
Let ε0 =
Z
−
p−ε
| ∇u |
BR
2
µZ
¶ p−ε
Z
τ
p−ε
τ
.
dx ≤ θ − | ∇u |
dx + C(n, p) − | ∇u | dx
BR
(2.9)
BR
Because τ < p − ε, (2.9) is a weak reverse Hölder inequality about | ∇u |. By Lemma 5 there
0
exists p0 > r1 such that | ∇u | ∈ Lploc (Ω ). Therefore it is obvious that r1 ∈ I if we assume that
I = {r ∈ [ p − ε, p ] : | ∇u | ∈ Lrloc (Ω )}. We can derive the conclusion that I is close from the
uniform estimate of (2.9). We can also obtain the conclusion that I is open from Lemma 5.
Hence I = [ p − ε, p ], namely, | ∇u | ∈ Lploc (Ω ). All the above derivation remains true for r = p.
By Lemma 5 there exists r2 = r2 (n, r1 , p) > p such that | ∇u | ∈ Lrloc2 (Ω ). Then by Sobolev
imbedding theorem we can obtain u ∈ Wr12 ,loc (Ω ). This ends the proof of the theorem .
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Applied Science School, University of Science and Technology Beijing, Beijing 100083,China.
E-mail address: liuliner@hotmail.com
College of Mathematics and Computer Science, Hebei University, Baoding 071002, China.
E-mail address: hongya-gao@sohu.com