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Volume 2010, Article ID 879821, 12 pagesdoi:10.1155/2010/879821 Research Article Global Optimal Regularity for the Parabolic Polyharmonic Equations Fengping Yao Department of Mathematics

Volume 2010, Article ID 879821, 12 pages

doi:10.1155/2010/879821

Research Article

Global Optimal Regularity for

the Parabolic Polyharmonic Equations

Fengping Yao

Department of Mathematics, Shanghai University, Shanghai 200436, China

Correspondence should be addressed to Fengping Yao,yfp1123@math.pku.edu.cn

Received 21 February 2010; Accepted 3 June 2010

Academic Editor: Vicentiu D Radulescu

Copyrightq 2010 Fengping Yao This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

We show the global regularity estimates for the following parabolic polyharmonic equation

u t −Δm

u  f inRn × 0, ∞, m ∈ Zunder proper conditions Moreover, it will be verified

that these conditions are necessary for the simplest heat equation u t − Δu  f inRn × 0, ∞.

1 Introduction

Regularity theory in PDE plays an important role in the development of second-order elliptic and parabolic equations Classical regularity estimates for elliptic and parabolic

the theory of partial differential equations, are two fundamental estimates for elliptic and parabolic equations and the basis for the existence, uniqueness, and regularity of solutions

regularity estimates in Orlicz spaces, for the following parabolic polyharmonic problems:

where x  x1, , x n, Δ  n

i1∂2/∂x i2 and m is a positive integer Since the 1960s, the

need to use wider spaces of functions than Sobolev spaces arose out of various practical problems Orlicz spaces have been studied as the generalization of Sobolev spaces since they

We denote the distance inRn1as

δ z1, z2  max|x1− x2|, |t1− t2|1/2m

for z1 x1, t1, z2 x2, t2 1.3

and the cylinders inRn1as

Q R  B R×−R 2m , R 2m

D x ν u ∂ |ν| u

∂ ν1

x1· · · ∂ ν n

x n

where ν  ν1, ν2, , ν n  is a multiple index, ν i ≥ 0 i  1, 2, , n, and |ν| n

i1|ν i| For

parabolic equations with the help of fundamental solutions and Green functions Moreover,



Q 1/6

φ

D 2m u 2

dz



Q 1/6

φ

|u t|2

dz ≤ C



Q 1/2

φ f 2

dz



Q 1/2

φ

|u|2

for

φ x  |x| p/2 with p > 2,1.6 is reduced to the local L pestimates In fact, we can replace 2 of

φ| · |2 in 1.6 by the power of p1for any p1> 1.

we are interested in the estimate like



Rn ×0,∞ φ D 2m u

dz



Rn ×0,∞ φ |u t |dz ≤ C



Rn ×0,∞ φ f dz, 1.8

framework and iteration-covering procedure as in6,12 , more complicated analysis should

be carefully carried out with a proper dilation of the unbounded domain

Here for the reader’s convenience, we will give some definitions on the general Orlicz spaces

Definition 1.1 A convex function φ :R → Ris said to be a Young function if

Definition 1.2 A Young function φ is said to satisfy the globalΔ2condition, denoted by φ ∈

Δ2, if there exists a positive constant K such that for every s > 0,

there exists a number a > 1 such that for every s > 0,

φ s ≤ φ as

Example 1.3 i φ1s  1  |s| log1  |s| − |s| ∈ Δ2, but φ1s /∈ ∇2

ii φ2s  e |s| − |s| − 1 ∈ ∇2, but φ2s /∈ Δ2

iii φ3s  |s| α 1  | log |s|| ∈ Δ2∩ ∇2, α > 1.

Remark 1.4 If a function φ satisfies1.10 and 1.11, then

φ θ1s  ≤ Kθ α1

1 φ s, φ θ2s  ≤ 2aθ α2

for every s > 0 and 0 < θ2≤ 1 ≤ θ1< ∞, where α1 log2K and α2 loga2 1

Remark 1.5 Under condition1.12, it is easy to check that φ satisfies

s→ 0

φ s

s→ ∞

s

Definition 1.6 Assume that φ is a Young function Then the Orlicz class K φRn is the set of



Rn

φ g dx < ∞. 1.14

The Orlicz space L φRn  is the linear hull of K φRn

Lemma 1.7 see 2  Assume that φ ∈ Δ2∩ ∇2and g ∈ L φ Ω Then

1 K φ Ω  L φ Ω,

0 Ω is dense in L φ Ω,

3



Ωφ g dx∞

0

x∈ Ω : g > μ d

φ

μ

Now let us state the main results of this work

Theorem 1.8 Assume that φ is a Young function and u satisfies

u t x, t − Δux, t  fx, t in R n × 0, ∞,

Then if the following inequality holds



Rn ×0,∞ φ D2u

dz



Rn ×0,∞ φ |u t |dz ≤ C



Rn ×0,∞ φ f dz, 1.17

One has

Theorem 1.9 Assume that φ ∈ Δ2∩∇2 If u is the solution of1.1-1.2 with f ∈ L φRn ×0, ∞,

then1.8 holds.

Remark 1.10 We would like to point out that theΔ2condition is necessary In fact, if the local

we have



Q 1/6



Q 1/6 φ

⎛

⎝

∂2u

∂x1∂x2

2⎞

⎠dz

≤ C



Q 1/2

φ f 2

dz



Q 1/2

φ

|u|2

dz

≤ C



Q

φ sdz,

1.20

which implies that

In this section we show that φ satisfies the global∇2condition if u satisfies1.16 and estimate

1.17 is true

Proof Now we consider the special case in1.16 when

0Rn1 is a cutoff function satisfying

u x, t 

t

0

1

4πt − s n/2



Rn

e −|x−y|2/4 t−s f

y, s

dy ds. 2.3

It follows from1.17, 2.1, and 2.2 that



Rn ×0,∞ φ |u t |dz ≤ C



Rn ×0,∞ φ f dz ≤ C1φ

ρ

4π n/2

t

0

1

t − s n2/2



B2

 x − y 2

n

2



e −|x−y|2/4 t−s f

y, s

dy ds. 2.5

Define

D:z  x, t ∈ R n × 0, ∞ : |x| > 4, |x| ≥ 4√nt

x − y 2

|x|2

x − y ≥ |x| − y ≥ |x| − |x|

2· 4π n/2

t

0

1

t − s n2/2



B1

e −|x−y|2/4 t−s dy ds

2· 4π n/2

t

0

1

t − s



y ∈B1

e −|ξ|2/4 dξ ds ξ 



x − y

t − s

2· 4π n/2

t

0

t − s n−2/2

2|x| n



y ∈B1

|ξ| n e −|ξ|2/4 dξ ds

≥ Cρ|x| −nt

0

t − s n−2/2 ds ≥ C2ρ |x| −n t n/2

2.9



D

φ

C2ρ |x| −n t n/2

dx dt ≤ C1φ

ρ

which implies that

1

1/n

4 √

n

φ

C2ρr −n

r n−1dr dt ≤ C1φ

ρ

By changing variable we conclude that, for any ρ > 0,

αρ

0

φ σ

σ2 dσ≤ C3φ

ρ

where α  C24−n n −n/2 Let ρ2≥ ρ1and 0 < ε ≤ α/2 Then we conclude from 2.12 that

φ

ρ2

ρ2 ≥ 1

C3

αρ2

0

φ σ

σ2 dσ≥ 1

C3

αρ1

ερ1

φ σ

σ2 dσ



ερ1

C3

 1

ερ1 − 1

αρ1 ≥ φ



ερ1

2C3ερ1.

2.13

φ

ρ

ρ ≥ 1

C3

αρ

ερ

φ σ

σ

1

σ dσ≥ φ



ε2ρ

2C2

3ε2ρln α

where we choose that ρ1 ερ, ρ2 σ in 2.13 Set a  1/ε2 Then we have

φ

ρ



α√

a

2C2 3

aφ ρ

a

a

completes our proof

3 Proof of the Main Result

use the following two lemmas The first lemma is the following integral inequality

Lemma 3.1 see 6  Let φ ∈ Δ2∩∇2, g ∈ L φRn1, and p ∈ 1, α2, where α2is defined in1.12.

Then for any b1, b2> 0 one has

0

1

μ p



{z∈R n1 :|g|>b 1μ}

g p

dz d

φ

b2μ

≤ Cb1, b2, φ 

Rn1φ g dz. 3.1 Moreover, we recall the following result

Lemma 3.2 see 10, Theorem 5.5  Assume that g ∈ L pRn × 0, ∞ for p > 1 There exists a

unique solution v ∈ W 2m,1

p Rn × 0, ∞ of 1.1-1.2 with the estimate



D 2m v

L pRn ×0,∞  v t L pRn ×0,∞ ≤ Cg

L pRn ×0,∞ 3.2

Moreover, we give one important lemma, which is motivated by the iteration-covering procedure in12 To start with, let u be a solution of 1.1-1.2 Let

p 1 α2

λ p0 



Rn ×0,∞

D 2m u p

dz 1





Rn ×0,∞

f p

u λ u

λ0λ , f λ f

for any λ > 0 Then u λis still the solution of1.1-1.2 with f λ replacing f Moreover, we

write

J λ Q  −



Q

D 2m u λ p

dz1

−



Q

f λ p

E λ1 z∈ Rn × 0, ∞ : D 2m u λ > 1

Lemma 3.3 For any λ > 0, there exists a family of disjoint cylinders {Q ρ i z i}i∈Nwith z i  x i , t i ∈

E λ 1 and ρ i  ρz i , λ  > 0 such that

J λ

Q ρ i z i 1, J λ



Q ρ z i< 1 for any ρ > ρ i , 3.8

i∈N

where Q 5ρ i z i  : B 5ρ i x i  × t i − 5ρ i2m , t i  5ρ i2m Moreover, one has

Q ρ i z i ≤ 2

{z∈Q ρi z i :|D 2m u λ|p >1/4}

D 2m u λ p

dz1





{z∈Q ρi z i :|f λ|p >/4}

f λ p

dz



Proof 1 Fix any λ > 0 We first claim that

sup

w∈Rn ×0,∞sup

ρ ≥ρ0J λ

where ρ0  ρ0λ > 0 satisfies λ p |Q ρ0|  1 To prove this, fix any w ∈ R n × 0, ∞ and ρ ≥ ρ0

J λ



Q ρ w≤ 1

λ p0λ p Q ρ w



Rn ×0,∞

D 2m u p

dz1





Rn ×0,∞

f p

dz

λ p Q ρ0  1.

3.12

lim

ρ→ 0J λ

Q ρ w> 1, 3.13

which implies that there exists some ρ > 0 satisfying

J λ



Q ρ w> 1. 3.14 Therefore from3.11 we can select a radius ρ w ∈ 0, ρ0 such that

J λ

Q ρ w w 1, J λ



Q ρ w< 1 for any ρ > ρ w 3.15

Therefore, applying Vitali’s covering lemma, we can find a family of disjoint cylinders

{Q ρ i z i} such that 3.8 and 3.9 hold

−



Q ρi z i

D 2m u λ p

dz 1

−



Q ρi z i

f λ p

Therefore, by splitting the two integrals above as follows we have

Q ρ

i z i ≤

{z∈Q ρi z i :|D 2m u λ|p >1/4}

D 2m u λ p

dz

cQ ρ i z i 4





{z∈Q ρi z i :|f λ|p >/4}

f λ p

dz Q ρ i z i

3.17

Proof In the following by the elementary approximation argument as3,12 it is sufficient

In view ofLemma 3.3, given any λ > 0, we can construct a family of cylinders {Q ρ i z i}i∈N,

where z i  x i , t i  ∈ E λ 1 Fix i ≥ 1 It follows from 3.6 and 3.8 inLemma 3.3that

−



Q 10ρi z i

D 2m u λ p

dz ≤ 1, −



Q 10ρi z i

f λ p

Lemma 3.2, there exists a unique solution v ∈ W 2m,1

v t −Δ2m v  f λ inRn × 0, ∞,

with the estimate



D 2m v

L pRn ×0,∞ ≤ Cf

λ

L pRn ×0,∞ 3.20

Therefore we see that



D 2m v

L p Q 10ρi z i≤D 2m v

L pRn ×0,∞

≤ Cf

λ

L pRn ×0,∞

 Cf λ

L p Q 10ρi z i.

3.21

−



Q 10ρi z i

D 2m w p

dz≤ 2p



−



Q 10ρi z i

D 2m v p

dz −



Q 10ρi z i

D 2m v p

dz





Q 10ρi z i

f λ p

dz ≤ C.

3.23

N1> 1 such that

sup

Q 5ρi z i

D 2m w ≤ N

Set μ  λλ0 Therefore, we deduce from3.5 and 3.24 that

z ∈ Q 5ρ i z i : D 2m u > 2N

1μ

 

z ∈ Q 5ρ i z i : D 2m u λ > 2N

1

≤ 

z ∈ Q 5ρ i z i : D 2m w > N

1  

z ∈ Q 5ρ i z i : D 2m v > N

1

 

z ∈ Q 5ρ i z i : D 2m v > N

1 ≤ 1

N1p



Q 5ρi z i

D 2m v p

dz.

3.25

z ∈ Q 5ρ i z i : D 2m u > 2N

1μ

≤ C



Q z

f λ p

dz ≤ C Q 10ρ

Therefore, from3.10 inLemma 3.3we find that

z ∈ Q 5ρ i z i : D 2m u > 2N

1μ

μ p







{z∈Q ρi z i :|D 2m u|p >μ p /4}

D 2m u p

dz



{z∈Q ρi z i :|f|>μ p /4}

f p

dz



,

3.27

where C  Cn, m Recalling the fact that the cylinders {Q ρ i z i}i∈Nare disjoint,



i∈N

Q 5ρ i z i  ∪ negligible set ⊃ E λ1 z∈ Rn × 0, ∞ : D 2m u λ z > 1

z∈ Rn × 0, ∞ : D 2m u > 2N

1μ

i∈N

z ∈ Q 5ρ i z i : D 2m u > 2N

1μ

μ p







{z∈R n ×0,∞:|D 2m u|p >μ p /4}

D 2m u p

dz



{z∈R n ×0,∞:|f|>μ p /4}

f p dz



.

3.29



Rn ×0,∞ φ D 2m u

dz



0

z∈ Rn × 0, ∞ : D 2m u > 2N

1μ d

φ

2N1μ

≤ C

0

1

μ p



{z∈R n ×0,∞:|D 2m u|p >μ p /4}

D 2m u p

dz d

φ

2N1μ

 C

0

1

μ p



{z∈R n ×0,∞:|f|>μ p /4}

f p

dz d

φ

2N1μ

.

3.30



Rn ×0,∞ φ D 2m u

dz ≤ C1



Rn ×0,∞ φ D 2m u

dz  C2



Rn ×0,∞ φ f dz, 3.31

where C1  C1n, m, φ and C2 C2n, m, , φ Finally selecting a suitable  ∈ 0, 1 such that

The author wishes to thank the anonymous referee for offering valuable suggestions to

Moreover, the author wishes to thank the department of mathematics at Shanghai university

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The author wishes to thank the anonymous referee for offering valuable suggestions to

Moreover, the. .. f

for any λ > Then u λis still the solution of1.1-1.2... pRn ×0,∞ 3.20

Therefore we see that



D 2m