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A Nash Type Result for Divergence Parabolic Equation Related to H?rmander’s Vector Fields

2021-01-15 03:22:46HOULinglingandNIUPengcheng
Journal of Partial Differential Equations 2020年4期

HOU Lingling and NIU Pengcheng

1 Institute for Mathematical Sciences,Renmin University of China,Beijing 100872,China.

2 Department of Applied Mathematics,Northwestern Polytechnical University, Xi’an 710129,China.

Abstract. In this paper we consider the divergence parabolic equation with bounded and measurable coefficients related to H?rmander’s vector fields and establish a Nash type result, i.e., the local H?lder regularity for weak solutions. After deriving the parabolic Sobolev inequality, (1,1) type Poincaré inequality of H?rmander’s vector fields and a De Giorgi type Lemma,the H?lder regularity of weak solutions to the equation is proved based on the estimates of oscillations of solutions and the isomorphism between parabolic Campanato space and parabolic H?lder space. As a consequence, we give the Harnack inequality of weak solutions by showing an extension property of positivity for functions in the De Giorgi class.

Key Words: H?rmander’s vector fields; divergence parabolic equation; weak solution; H?lder regularity;Harnack inequality.

1 Introduction

Schauder theory for the solutions to linear elliptic and parabolic equations withCαcoefficients or VMO coefficients has been completed. De Girogi has followed the local H?lder continuity for the solutions to the divergence elliptic equation with bounded and measurable coefficients

and given the a priori estimate of H?lder norm (see [1]). Nash in [2] derived independently the similar result for the solutions to the parabolic equation with a different approach from[1]. Hereafter Moser in[3]developed a new method(nowadays it is called the Moser iteration method) and proved again results above-mentioned to elliptic and parabolic equations. These important works break a new path for the study of regularity for weak solutions to partial differential equations.

In[4],Fabes and Stroock proved the Harnack inequality for linear parabolic equations by going back to Nash’s original technique in[2]. A very interesting approach has been raised by De Benedetto(see[5])for proving a Harnack inequality of functions belonging to parabolic De Giorgi classes.The approach was used to derive the H?lder continuity of solutions to linear second order parabolic equations with bounded and measurable coefficients(see[6]). Giusti[7]applied the approach to give a proof of the Harnack inequality in the elliptic setting.

Square sum operators constructed by vector fields satisfying the finite rank condition were introduced by H?rmander(see[8]), who deduced that such operators are hypoelliptic. Many authors carried on researches to such operators and acquired numerous important results([9-14]). Nagel,Stein and Wainger([15])concluded the deep properties of balls and metrics defined by vector fields of this type. Many other authors obtained very appreciable results related to H?rmander’s square sum operators,for instance fundamental solutions ([16]), the Poincaré inequality ([17]), potential estimates ([18]), also see[4,19-21],etc. All these motivate the study to degenerate elliptic and parabolic equations formed from H?rmander’s vector fields. Schauder estimates to degenerate elliptic and parabolic equations related to noncommutative vector fields have been handled in [22-24], etc. Bramanti and Brandolini in [25] investigated Schauder estimates to the following H?rmander type nondivergence parabolic operator

where coefficientsaij(t,x),bi(t,x)andc(t,x)areCα.

In this paper,we are concerned with the divergence parabolic equation with bounded and measurable coefficients related to H?rmander’s vector fields and try to establish a Nash type result for weak solutions which will play a crucial role for corresponding nonlinear problems.

where

is the smooth vector field,is the adjoint ofXj,j=1,···,q. The summation conventions in(1.1)are omitted.Denote

then(1.1)is written by

Throughout this paper we make the following assumptions:

(C1)aij(x,t)∈L∞(QT)(i,j=1,···q)and there exists Λ>0 such that

(C2) form>(Q+2)/2,Qis the local homogeneous dimension relative to ?,bi(x,t)(i=1,···,q)andc(x,t)satisfy

(C3)

Let

we call thatuis a weak sub-solution(super-solution)to the equation(1.1),ifu∈V2(QT)satisfies

We callbelongs to the De Giorgi class

Now we state the main results of this paper.

Theorem 1.1(H?lder Regularity).withbe the weak solutionto(1.1)with(C1),(C2)and(C3),then forQ??QT,there exist C=C(Q,η,Λ,δ,dP)≥1,andβ,satisfyingsuch that

whereis the parabolic metric(see(2.6)below)and

Theorem 1.2(Harnack Inequality).Letbe the weak solutionto(1.1)with(C1),(C2)and(C3),andthen

where C depends on Q,Λ,η,and(a′?1)?1.

The proofs of Theorems are based on the readjustment of De Giorgi’s approach,and some new ingredients applying to our setting are replenished. It is worth emphasizing that we do not impose any artifical condition to the measure of metric ball induced by H?rmander’s vector fields.

We note that(1.1)involves the special case

The rest of the paper is organized as follows: Section 2 explains H?rmander’s vector fields,the parabolic Campanato space and parabolic H?lder space;several new preliminary results including the parabolic Sobolev inequality,(1,1)type Poincaré inequality of H?rmander’s vector fields and a De Giorgi type Lemma are inferred. In section 3 we prove that weak solutions to(1.1)are actually in the De Giorgi classDG(QT)and derive some properties of functions inDG(QT). Section 4 is devoted to proofs of main results.Theorem 1.1 is proved based on the estimates of oscillations of solutions and the isomorphism between the parabolic Campanato space and parabolic H?lder space. Theorem 1.2 is followed by using Theorem 1.1 and an extension property of positivity for functions in the De Giorgi class.

2 Preliminaries

LetX1,···,Xq(q≤n) areC∞vector fields inN.Throughout this paper, we always suppose that these vector fields satisfy the finite rank condition [8], i.e., there exists a positive integerssuch thatspansNat every point.

Definition 2.1(Carnot-Carathéodory distance, [26]).An absolutely continuous curve γ:[0,T]→?is said sub-unitary,if it is Lipschitz continuous and satisfies that for every N anda.e.t∈[0,T],

LetΦ(x,y)be the class of sub-unitary curves connected x and y,we define the Carnot-Carathéodory distance(C-C distance)by

The C-C metric ball is defined by

and the Lebesgue measure of metric ballBR(x0) by |BR(x0)| . A fundamental doubling property with respect to the metric balls was showed in [15], namely, there are positive constantsC1>1 andR0,such that forx0∈? and 0<R<R0,

whereQ=log2C1,Qacts as a dimension and is called the local homogeneous dimension relative to ?. It is easy to see from(2.1)that for any 0<R≤R0andθ∈(0,1),

whereC1andR0are constants in(2.1).

The gradient ofu∈C1(?)is denoted byand the norm ofXuis of the formThe Sobolev spacerelated to vector fieldsX1,···,Xqis the completion ofunder the norm

Definition 2.2.The parabolic Sobolev spaceon vector fields X1,···,Xq is the set of allfunctions u satisfying Xu∈L2(QT)andThe norm on V2(QT)is

In the sequel,we also need spacesis the set of functions inV2(QT)satisfying

Obviously,we have

respectively,where?? is the boundary of ?.

Let Q??QT,we define the parabolic metricdP:

Definition 2.3(H?lder space).For α∈(0,1], let Cα(Q)be the set of functions u:satisfying

its norm is

Definition 2.4(Campanato space).For1≤p<+∞and λ≥0,if u∈Lp(Q)satisfies

wherethen we say that u belongs to the Campanato spaceLp,λ(Q)with the norm

Let us state two useful cut-off functionsandη(t)([27,28])which satisfyoutsideBR)and

The following result is well known.

Lemma 2.1(Sobolev inequality,[29-32]).For1≤p<Q,there exist C>0and R0>0such thatfor any x∈?and0<R≤R0,we have for any

where1≤k≤Q/(Q?p). In particular,letthen

In the light of(2.8)we can prove

Theorem 2.1(Parabolic Sobolev inequality).Forit follows u∈L2(Q+2)/Q(QT)and

whereQR=BR(x0)×(t0?R2,t0]?QT,BR=BR(x0)and

Proof.Using the H?lder inequality and(2.8)withp=1,it sees

Consequently,

Integrating it with respect tot,(2.9)is derived.

We arrive at by(2.9)and the Young inequality that

and(2.10)is proved.

Lemma 2.2.For0<θ<1and κ>1,there exists a positive constant Cκ such that

Proof.Denote

Then it can be verified that

Let

we have thatF(θ)is continuous and uniformly bounded on[0,1],which impliesf(θ)≤Cκfor someCκ>0.

Theorem 2.2((1,1)type Poincaré inequality).Let u∈W1,1(BR)and

If|E0|>0,then

where C>0relies only on Q.

Proof.Consider firstu∈Lips(B(x,R)),then by the result of Jerison[17],

whereudx. Since the above inequality has the self-improvement property(see[33]),i.e. there existsκ>1,such that

it follows

Applying

it arrives at

and then

Noting

we have from(2.2)that

Using it and the H?lder inequality,it follows

Now(2.11)is obtained by combining it and the density ofLip(BR)inW1,1(BR).

Theorem 2.3(De Giorgi type lemma).Let u∈W1,1(BR)and

Then we have

where C>0relies only on Q.

Proof.Denoting the function

This proves(2.12).

3 Several auxiliary lemmas

For the weak sub-solution(super-solution)to(1.1),we have

Lemma 3.1.Letbe the bounded weak sub-solution(or super-solution)to(1.1)with(C1),(C2)and(C3)then

where p>Q+2,η=min{p,2m},γ(·)relies only on Q, Λand p,

Proof.We only prove the conclusion for the weak sub-solution and the proof for the weak super-solution is similar. Multiplying the test functionζ2(u?k)+to(1.1)and integrating onBρ×(t0,t),it yields

Consequently,

wheret0<t≤t0+τ, 0<τ<1. NoticingXiu=Xi(u?k)+and

we obtain from(3.2)that

Denote the third,fourth and fifth term in the right hand side of(3.3)respectively by

Applying(C1)toI1gives

In virtue of the Cauchy inequality,[u>k]={(x,t)|u>k}and

we see that

and

Substituting these estimates into(3.3)and using(C1)to the second term in the left hand side of(3.3),it is not difficult to derive

Choosingwe have

Denote the third and fourth term in the right hand side of(3.4)by

Employing(C2),(3.1)and the H?lder inequality,it shows

whereit knows by the Interpolation inequality and the Young inequality that for 0<θ<1,

It implies from it and(2.10)that

Using(2.10)toII2,it gets

Putting estimates forII1andII2into(3.4)and choosingit leads to

Taking?small enough andλ0=Λ?1,we conclude

and the proof is completed.

Remark 3.1.Lemma 3.1 indicates that the weak solution to(1.1)belongs to the De Giorgi class.

Next we give several useful properties for functions in the De Giorgi class.

Lemma 3.2.Let u∈DG+(QT),

then for any positive integer s,either

or

where C relies only on Q,λ0,η,δ and γ(·),for0<ρ<2R.

Proof.Denote

Thenklis increasing,AR(kl)is decreasing,and|AR(k,t)|≤(1?σ)|BR|by(3.5). Applying Theorem 2.3 and(3.5),we have fort0≤t≤t0+aR2,

Integrating it intover(t0,t0+aR2)and notingand

we obtain

Lettingξ(x)be the cut-off function betweenBR(x0)andB2R(x0)and using

and(2.2),we obtain from(1.3)(kis changed tokl)that

If(3.6)is invalid,i.e.,there existsl, 0≤l≤s?1,such thatthen we noteto arrive at from the previous estimate(3.9)that

Taking it into(3.8),it obtains

Squaring both sides and summing with respect tolfrom 0 tos?1, we have by usingthat

which implies(3.7).

Lemma 3.3.Let u∈DG+(QT),

then there exists a positive integer s0=s0(σ)≥1relying on Q,λ0,η,δ,σ and γ(·),such that either

or

Proof.Suppose thatξ(x)is a cut-off function betweenBβR(x0)(0<β<1)andBR(x0),such thatand denote

We observeu?k≤μ?k=Hand apply(1.3)onto see

On the other hand,for any integers1≥1,

it follows

and obtain from(3.13)and(3.14)that

where the factR2|BR|=|QR|is used.Obviously,from 1?σ≤1 and

Choosingβ∈(0,1)such that

Sinceγ(?)is decreasing,we know thatis strictly increasing and its inverse function?=?(q)satisfies?(q)→0 asq→0. Usingand choosing

it follows

and so(3.17)becomes

As?=?(q)is increasing andit implies

Pickinga=a(σ)>0 satisfying

With it and(3.18),it derives

Similarly,denoting

and repeating the process above,we have for a larges2≥1,

Lettings0=max{s1,s2},it shows by combining(3.19)and(3.20)that

which is(3.12).

Lemma 3.4.Letthere exists θ∈(0,1)depending on Q,λ0,η,δ,σ and γ(·),such that for k<μ,if

then

To prove Lemma 3.4,we recall a known result.

Lemma 3.5([34]).Given a non-negative sequence{yh}(h=0,1,2,···)satisfying the recursion relation

where b>1and ?>0,if

then

Proof of Lemma 3.4.Denote

thenRmis decreasing,is increasing and

Take a cut-off functionbetweenandthen

Using

Denotingit followsfromApplying it into(3.25),we have by(2.10),

it shows

Substitutinginto(3.27),we derive from(3.26)that

Consequently,

Let

Then(3.29)becomes

we have

which gives(3.23).

Lemma 3.6.Let u∈DG+(QT)(u∈DG?(QT)),

If forand0<σ<1,u satisfies

then there exists s=s(σ)≥1depending on Q,λ0,η,δ,σ and γ(·)such that either

or

where

Proof.Lets0be the constant in Lemma 3.3. If(3.32)is invalid fors≥s0,then from Lemma 3.3 and(3.31),

Employing Lemma 3.2(sandHare changed intos?s0?1 andrespectively),it follows

whereCrelies onQ,λ0,η,andγ(·). Letθbe the constant in Lemma 3.4 and chooseslarge enough satisfying

It implies(3.33)from Lemma 3.4.

4 Proofs of main results

We first prove an oscillation estimate for the weak solution to(1.1).

Lemma 4.1.Letbe the weak solution to(1.1)with(C1),(C2)and(C3),

then for any R∈(0,R0],there exists β,such that

whererelies on Q,η,Λ,δ and

Proof.Denoteand one of the following two inequalities holds:

If(4.2)is valid,then Lemma 3.6 implies that for

and there existss1=s1(1/2)≥1,such that one of the following two inequalities holds:

It sees that by(4.4),

and by(4.5),

If (4.3) is valid, then by Lemma 3.6 there existss2=s2(1/2)≥1, such that one of the following two inequalities holds:

It shows that from(4.8),

and from(4.9),

Let us takes0=max{s1,s2},we derive by(4.6)and(4.10)that

and by(4.7)and(4.11)that

Combining(4.12)and(4.13),it follows that forR∈(0,R0], 0<R0≤1,

ReplacingRin (4.14) byR0and denotingandit impliesand from(4.14)and the above estimate that

(1)If?≥υ,it givesand thenChoosingwe derive from(4.15)that

where we used

(2)If?<υ,thenandChoosingit leads to from(4.15)that

where

Selectβ=min{β1,β2},we prove(4.1)by combining estimates in cases(1)and(2).

The following is an isomorphism between Lp′,λ(Q)andCα(Q).

Lemma 4.2.LetQ??QT,we have for0<λ≤p′,p′≥1,

where

Proof.Supposeu∈Cα(Q).For anyZ=(x,t)∈Q,denote

we have for

and

and

On the contrary,ifwe have for any

and

Sinceis continuous onZ,it knows by(4.19)thatis continuous in Q. In addition,Lebesgue’s Theorem assures

hencea.e.Z∈Q andis independent ofR.Now we see from(4.19)that

which shows

Substituting it into(4.21),it happens

and notingR=dP(Y1,Y2),we arrive at

Combing(4.16)we get the conclusion.

Proof of Theorem 1.1.For anyZ0=(x0,t0)∈Q,denote

By employing Lemma 4.1,we derive for

which impliesand soby Lemma 4.2.

Now let us estimate [u]β;Q. For anyZ0=(x0,t0),Z1=(x1,t1)∈Q, without losing of generality,supposet0≥t1. IfthenZ1∈QR(Z0)?QT,R=dP(Z0,Z1),and from(4.22),

ifthen

It follows(1.4)by combining(4.23)and(4.24).

To prove Theorem 1.2, we need an extension property of positivity for functions in the De Giorgi class.

Lemma 4.3.Let u∈DG?(QT)and u≥0in

For ?∈(0,1),if

then there exist R0>0and a positive integer s≥1relying on Q,λ0,η and γ(·), such that for BR=BR(x0), 0<R≤R0,either

or

Proof.Let?≤1/8,we have from(4.25)and(2.2)that

and then

As a result of Lemma 3.6,there existss>1,such that either

(Min(3.32) is changed tok; actually, it is suitable from the proofs of lemmas in Section 3),or

where we usedH=kwhich is followed fromH:=k?infuandu≥0. If(4.28)holds,we chooseR0satisfying

and for 0<R≤R0,it yields by(2.2)that

Notingit implies from(4.28)that

This proves(4.26).

If(4.29)holds,choose a suitable?>0 such thatis an integer,and arrive at by employing Lemma 3.6 withNtimes that

which is(4.27).

Proof of Theorem 1.2.For simplicity,we assumea′=2 and denote

Consider two functions

wheresis determined in Lemma 4.3,and letr0(r0<R)be the largest root to the equationm(r)=μ(r). Sincem(r)→∞asr→R?0, anduis continuous and bounded init sees thatr0is well defined,m(r)>μ(r) forr0<r≤R, and there exists (x1,t1)∈Qr0=such that

Now introduce

and

wherethen we haveNoting the meaning ofr0,it follows

From Lemma 4.1,(2.2),(4.31)andwe obtain

and

Choosing?so small thatC?β<1 andit yields

Forx∈BR/2(x0),we have

and so

Employing this result and(4.25)in Lemma 4.3 on the domainB6R(x1)×(t1,t0+2R2],we can obtain

Acknowledgement

This work is supported by the National Natural Science Foundation of China(No.1177-1354).


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