THE DIRICHLET PROBLEM

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ANNALES DE

L’INSTITUT FOURIER

LesAnnales de l’institut Fouriersont membres du

Wolfgang Arendt & A. F. M. ter Elst

The Dirichlet problem without the maximum principle Tome 69, n^o2 (2019), p. 763-782.

<http://aif.centre-mersenne.org/item/AIF_2019__69_2_763_0>

Cet article est mis à disposition selon les termes de la licence Creative Commons attribution – pas de modification 3.0 France.

http://creativecommons.org/licenses/by-nd/3.0/fr/

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THE DIRICHLET PROBLEM

WITHOUT THE MAXIMUM PRINCIPLE

by Wolfgang ARENDT & A. F. M. TER ELST (*)

Abstract. — Consider the Dirichlet problem with respect to an elliptic operator

A=−

d

X

k,l=1

∂kakl∂l−

d

X

k=1

∂kbk+

d

X

k=1

ck∂k+c0

on a bounded Wiener regular open set Ω ⊂ R^d, whereakl, ck ∈ L∞(Ω,R) and bk, c0∈L∞(Ω,C). Suppose that the associated operator onL2(Ω) with Dirichlet boundary conditions is invertible. Then we show that for allϕ∈C(∂Ω) there exists a uniqueu∈C(Ω)∩H_loc¹ (Ω) such thatu|∂Ω=ϕandAu= 0.

In the case when Ω has a Lipschitz boundary andϕ∈C(Ω)∩H^1/2(Ω), then we show thatucoincides with the variational solution inH¹(Ω).

Résumé. — Considérons le problème de Dirichlet par rapport à un opérateur elliptique

A=−

d

X

k,l=1

∂kakl∂l−

d

X

k=1

∂kbk+

d

X

k=1

ck∂k+c0

sur un ensemble ouvert régulier de Wiener borné Ω⊂R^d, oùakl, ck∈L∞(Ω,R) et bk, c0 ∈ L∞(Ω,C). Supposons que 0 n’est pas une valeur propre de Aavec conditions aux limites Dirichlet. Alors nous montrons que pour toutϕ∈C(∂Ω) il existe un uniqueu∈C(Ω)∩H_loc¹ (Ω) tel queu|_∂Ω=ϕetAu= 0.

Dans le cas où Ω a une frontière Lipschitz etϕ∈C(Ω)∩H^1/2(Ω), nous montrons queucoïncide avec la solution variationnelle dansH¹(Ω).

Keywords:Dirichlet problem, Wiener regular, holomorphic semigroup.

2010Mathematics Subject Classification:31C25, 35J05, 31B05.

(*) The second-named author is most grateful for the hospitality extended to him during a fruitful stay at the University of Ulm. He wishes to thank the University of Ulm for financial support. Part of this work is supported by an NZ-EU IRSES counterpart fund and the Marsden Fund Council from Government funding, administered by the Royal Society of New Zealand. Part of this work is supported by the EU Marie Curie IRSES program, project “AOS”, No. 318910.

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1. Introduction

Let Ω ⊂ R^d be an open bounded set with boundary Γ. Throughout this paper we assume that d > 2. The classical Dirichlet problem is to find for each ϕ ∈ C(Γ) a function u ∈ C(Ω) such that u|Γ = ϕ and

∆u= 0 as distribution on Ω. The set Ω is calledWiener regularif for every ϕ∈C(Γ) there exists a uniqueu∈C(Ω) such thatu|Γ =ϕand ∆u= 0 as distribution on Ω.

The Dirichlet problem has been extended naturally to more general second-order operators. For all k, l ∈ {1, . . . , d} let a_kl: Ω → R be a bounded measurable function and suppose that there exists aµ >0 such that

(1.1) Re

d

X

k,l=1

akl(x)ξkξl>µ|ξ|²

for allx∈Ω andξ∈C^d. Further, for allk∈ {1, . . . , d}letb_k, c_k, c₀: Ω→C be bounded and measurable. Define the mapA:H_loc¹ (Ω)→ D⁰(Ω) by

hAu, vi_D⁰_{(Ω)×D(Ω)}=

d

X

k,l=1

Z

Ω

a_kl(∂_ku)∂_lv+

d

X

k=1

Z

Ω

b_ku ∂_kv

+

d

X

k=1

Z

Ω

ck(∂ku)v+ Z

Ω

c0u v for allu∈H_loc¹ (Ω) andv∈C_c^∞(Ω). Givenϕ∈C(Γ), by aclassical solution of the Dirichlet problem we understand a function u ∈ C(Ω)∩H_loc¹ (Ω) satisfyingAu= 0 andu|Γ =ϕ. For the pure second-order case (that isbk= ck = c0 = 0) Littman–Stampacchia–Weinberger [11] proved that for all ϕ∈C(Γ) there exists a unique classical solutionu. Then Stampacchia [13, Théorème 10.2] added real valued lower order terms, under the condition (see [13], (9.2’)) that there exists aµ⁰>0 such that

(1.2)

Z

Ω

c0v+

d

X

k=1

Z

Ω

bk∂kv>µ⁰ Z

Ω

v

for allv∈C_c^∞(Ω)⁺. Gilbarg–Trudinger [10, Theorem 8.31] merely assume that

(1.3)

Z

Ω

c0v+

d

X

k=1

Z

Ω

bk∂kv>0

for allv∈C_c^∞(Ω)⁺ in order to obtain the same conclusion. A consequence of these assumptions is a weak maximum principle, which implies that

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kuk_C(Ω) 6 kϕkC(Γ) for all u ∈ H_loc¹ (Ω)∩C(Ω) satisfying Au = 0 and u|Γ =ϕ. We may consider (1.3) as a kind of submarkov condition since it is equivalent to−A1Ω60 inD⁰(Ω).

The aim of this paper is to show that the positivity condition (1.3) and the maximum principle are not needed for the well-posedness of the Dirichlet problem. In addition we allow thebkandc0to be complex valued.

In order to state the main results of this paper in a more precise way we need a few definitions. Define the forma:H¹(Ω)×H¹(Ω)→Cby

(1.4) a(u, v) =

d

X

k,l=1

Z

Ω

akl(∂ku)∂lv+

d

X

k=1

Z

Ω

bku ∂kv

+

d

X

k=1

Z

Ω

ck(∂ku)v+ Z

Ω

c₀u v.

LetA^D be the operator inL2(Ω) associated with the forma|_H¹

0(Ω)×H₀¹(Ω). In other words,A^D is the realisation of the elliptic operator A in L₂(Ω) with Dirichlet boundary conditions. This operator has a compact resolvent.

Moreover, if (1.3) is valid, then kerA^D={0}by [10, Corollary 8.2]. Instead of (1.3) we assume the condition kerA^D={0}, which is equivalent to the uniqueness of the Dirichlet problem (cf. Proposition 2.3 below).

The main result of this paper is the following well-posedness result for the Dirichlet problem.

Theorem 1.1. — Let Ω ⊂ R^d be an open bounded Wiener regular set with d > 2. For all k, l ∈ {1, . . . , d} let akl: Ω → R be a bounded measurable function and suppose that there exists aµ >0such that

Re

d

X

k,l=1

akl(x)ξkξl>µ|ξ|²

for allx∈Ωand ξ∈C^d. Further, for all k∈ {1, . . . , d} letbk, c0: Ω→C andc_k: Ω→Rbe bounded and measurable. LetA^Dbe as above. Suppose 06∈σ(A^D). Then for allϕ∈C(Γ)there exists a uniqueu∈C(Ω)∩H_loc¹ (Ω) such thatu|Γ =ϕandAu= 0.

Moreover, there exists a constant c >0such that kuk_C(Ω)6ckϕk_C(Γ)

for all ϕ ∈ C(Γ), where u ∈ C(Ω)∩H_loc¹ (Ω) is such that u|_Γ = ϕ and Au= 0.

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Instead of the homogeneous equation Au = 0 one can also consider the inhomogeneous equation Au= f0+Pd

k=1∂kfk. We shall do that in Theorem 2.13.

Adopt the notation and assumptions of Theorem 1.1. DefineP:C(Γ)→ C(Ω) by P ϕ = u, where u ∈ C(Ω)∩H_loc¹ (Ω) is such that u|_Γ = ϕ and Au= 0. Note thatP ϕis theclassical solutionof the Dirichlet problem.

If Ω has even a Lipschitz boundary (which implies Wiener regularity), then there is also a variational solution of the Dirichlet problem that we describe next. Denote by Tr :H¹(Ω) → L2(Γ) the trace operator. Again leta_kl, b_k, c_k, c₀ ∈L_∞(Ω) and suppose that the ellipticity condition (1.1) is satisfied. Further suppose that 06∈σ(A^D). Then for eachϕ∈TrH¹(Ω) there exists a uniqueu∈H¹(Ω), called thevariational solution, such that Au= 0 and Tru=ϕ(cf. Lemma 2.1). Defineγ: TrH¹(Ω)→ H¹(Ω) by settingγϕ=u.

The second result of this paper says that the variational solution and the classical solution coincide, if both are defined.

Theorem 1.2. — Adopt the notation and assumptions of Theorem 1.1.

Suppose thatΩhas a Lipschitz boundary. Letϕ∈C(Γ)∩TrH¹(Ω). Then P ϕ=γϕalmost everywhere onΩ.

The last main result of this paper concerns a parabolic equation. LetA_c denote the part of the operatorA^D inC0(Ω). So

D(Ac) ={u∈D(A^D)∩C0(Ω) :A^Du∈C0(Ω)}

andA_c=A^D|_D(A_c₎.

Then −Ac generates a holomorphic C₀-semigroup on C₀(Ω). Moreover, e^−tA^cu= e^−tA^Dufor allu∈C₀(Ω)andt >0.

In Section 2 we prove Theorem 1.1 via an iteration argument. Section 3 is devoted to the comparison of the classical and the variational solutions of the Dirichlet problem. Theorem 1.2 is proved there with the help of a deep result of Dahlberg [7]. We consider the semigroup on C0(Ω) in Section 4 and prove Theorem 1.3.

2. The Dirichlet problem

In this section we prove Theorem 1.1 on the well-posedness of the Dirich- let problem. The technique is a reduction to the Stampacchia result mentioned in the introduction. For this reason we introduce the following two forms and operators.

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Adopt the notation and assumptions of Theorem 1.1. For allλ∈Rdefine the formsaλ,bλ:H¹(Ω)×H¹(Ω)→Cby

aλ(u, v) =a(u, v) +λ(u, v)_L₂_(Ω) and b_λ(u, v) =

d

X

k,l=1

Z

Ω

a_kl(∂_ku)∂_lv+

d

X

k=1

Z

Ω

c_k(∂_ku)v+λ Z

Ω

u v,

where a is as in (1.4). Define similarly Aλ,Bλ: H_loc¹ (Ω) → D⁰(Ω) and let B^D be the operator associated with the sesquilinear formb₀|_H1

0(Ω)×H₀¹(Ω). It follows from ellipticity that there exists aλ0>0 such that

µ

2kvk²_H1(Ω)6Rea_λ₀(v) and µ

2 kvk²_H1(Ω)6Reb_λ₀(v)

for all v ∈ H¹(Ω). Note that Bλ satisfies the submarkovian condition

−Bλ1^Ω 60, that is (1.3), and even Stampacchia’s condition (1.2) for all λ > 0. So we can and will apply Stampacchia’s result (in the proof of Lemma 2.8).

We first investigate the operatorA^DinL2(Ω). Note thatf0+Pd

k=1∂kfk∈ D⁰(Ω) for allf₀, f₁, . . . , f_d∈L₁(Ω). The next lemma is also valid if thea_kl andck are complex valued.

Lemma 2.1. — Let f1, . . . , fd ∈ L2(Ω). Let p˜ ∈ (1,∞) be such that p˜> d+2^2d . Further let f₀ ∈ L_p_˜(Ω). Then there exists a unique u∈ H₀¹(Ω) such thatAu=f₀+Pd

k=1∂kfk.

Proof. — There exists a uniqueT ∈ L(H₀¹(Ω)) such that (T u, v)_H1 0(Ω)= a(u, v) for all u, v ∈ H₀¹(Ω). Then T is injective because kerA^D = {0}.

Moreover, the inclusion H₀¹(Ω) ,→ L2(Ω) is compact. Hence the opera- torT is invertible by the Fredholm–Lax–Milgram lemma, [5, Lemma 4.1].

Clearlyv 7→Pd

k=1(fk, ∂kv)_L₂_(Ω) is continuous fromH₀¹(Ω) into C. Define F:C_c^∞(Ω)→CbyF(v) =hf0, vi_D⁰_{(Ω)×D(Ω)}. We claim thatF extends to a continuous function fromH₀¹(Ω) into C. Ifd>3, then H₀¹(Ω)⊂Lr(Ω), wherer=_d−2^2d . SoH₀¹(Ω)⊂L_q(Ω), whereqis the dual exponent of ˜p. The last inclusion is also valid ifd= 2. So in any case the mapF extends to a continuous function fromH₀¹(Ω) intoC. Then the lemma follows.

The next lemma is valid for a general bounded open set Ω and does not use the condition 06∈σ(A^D). It is an extension of [1, Lemma 4.2].

Lemma 2.2. — Letu ∈ C0(Ω)∩H_loc¹ (Ω) and f1, . . . , fd ∈ L2(Ω). Let

˜

p∈ (1,∞) be such that p˜ > d+2^2d . Further let f0 ∈ Lp˜(Ω). Suppose that Au=f0+Pd

k=1∂kfk. Thenu∈H₀¹(Ω).

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Proof. — As at the end of the previous proof there exists anM₀>0 such that|R

Ωf0v|6M0kvk_H1(Ω)for allv∈H₀¹(Ω). SetM =M0+Pd

k=1kfkk2. Let ε > 0. Setv_ε = (Reu−ε)⁺. Then suppv_ε ⊂Ω is compact. Hence there exists an open Ω1 ⊂ R^d such that suppvε ⊂ Ω1 ⊂ Ω1 ⊂ Ω. Then vε∈H₀¹(Ω₁). Moreover,

d

X

k,l=1

Z

Ω1

akl(∂ku)∂lv+

d

X

k=1

Z

Ω1

bku ∂kv+

d

X

k=1

Z

Ω1

ck(∂ku)v+ Z

Ω1

c0u v

= Z

Ω1

f₀v+

d

X

k=1

Z

Ω1

fk∂kv (2.1)

for allv ∈ C_c^∞(Ω₁). Since u|_Ω₁ ∈H¹(Ω₁) it follows that (2.1) is valid for allv∈H₀¹(Ω1). Choosingv=vε gives

d

X

k,l=1

Z

Ω

akl(∂ku)∂lvε+

d

X

k=1

Z

Ω

bku ∂kvε+

d

X

k=1

Z

Ω

ck(∂ku)vε+ Z

Ω

c0u vε

6M0kvεkH¹(Ω)+

d

X

k=1

kfkk2k∂kvεk26MkvεkH¹(Ω). On the other hand, ∂_kv_ε = ∂_k((Reu−ε)⁺) = 1[Reu>ε]∂_kReu for all k∈ {1, . . . , d}by [10, Lemma 7.6]. Therefore

Re

d

X

k,l=1

Z

Ω

akl(∂ku)∂lvε+ Re

d

X

k=1

Z

Ω

bku ∂kvε

+ Re

d

X

k=1

Z

Ω

ck(∂ku)vε+ Re Z

Ω

c0u vε

=

d

X

k,l=1

Z

Ω

akl(∂kvε)∂lvε+ Re

d

X

k=1

Z

Ω

bku ∂kvε

+

d

X

k=1

Z

Ω

c_k(∂_kReu)v_ε+ Re Z

Ω

c₀u v_ε

= Rea(vε) +ε

d

X

k=1

Z

Ω

(Rebk)∂kvε−

d

X

k=1

Z

Ω

(Imbk) (Imu)∂kvε

+ε Z

Ω

(Rec0)vε− Z

Ω

(Imc0) (Imu)vε

>µ

2 kvεk²_H1(Ω)−λ₀kvεk²₂−ε M⁰|Ω|^1/2kvεkH¹(Ω)−M⁰kuk2kvεkH¹(Ω),

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whereM⁰ =kc0k∞+Pd

k=1kbkk∞. Sincekvεk2=k(Reu−ε)⁺k26kuk26

|Ω|^1/2kuk_C₀_(Ω), it follows that µ

2k(Reu−ε)⁺k²_H1(Ω)6M⁰⁰k(Reu−ε)⁺k_H1(Ω)+λ0|Ω| kuk²_C

0(Ω)

for allε∈(0,1], whereM⁰⁰=M+M⁰|Ω|^1/2(kuk_C₀_(Ω)+ 1).

Therefore the sequence ((Reu−2⁻ⁿ)⁺)_n∈N₀ is bounded inH₀¹(Ω). Pass- ing to a subsequence if necessary, we may assume without loss of generality that there exists aw∈H₀¹(Ω) such that lim(Reu−2⁻ⁿ)⁺=w weakly in H₀¹(Ω). Then lim(Reu−2⁻ⁿ)⁺ = w in L2(Ω). But lim(Reu−2⁻ⁿ)⁺ = (Reu)⁺ in L₂(Ω). So (Reu)⁺ = w ∈ H₀¹(Ω). Similarly one proves that (Reu)⁻,(Imu)⁺,(Imu)⁻ ∈H₀¹(Ω). Sou∈H₀¹(Ω).

Lemma 2.2 together with the condition 06∈σ(A^D) gives the uniqueness in Theorem 1.1.

Proposition 2.3. — For all ϕ ∈ C(Γ) there exists at most one u ∈ C(Ω)∩H_loc¹ (Ω) such thatu|Γ=ϕand Au= 0.

Proof. — Letu∈C(Ω)∩H_loc¹ (Ω) and suppose thatu|Γ= 0 andAu= 0.

Thenu∈C0(Ω). Henceu∈H₀¹(Ω) by Lemma 2.2. AlsoAu= 0. Therefore u∈D(A^D) andA^Du= 0. But 06∈σ(A^D). Sou= 0.

In the next proposition we use that Ω is Wiener regular.

Proposition 2.4. — Letλ > λ₀ andp∈(d,∞]. Letf₀∈L_p/2(Ω) and f1, . . . , fd ∈ Lp(Ω). Then there exists a unique u∈ H₀¹(Ω)∩C0(Ω) such thatBλu=f₀+Pd

k=1∂kfk.

Proof. — Sinceakl andckare real valued for allk, l∈ {1, . . . , d}we may assume thatf₀, . . . , f_dare real valued. By [10, Theorem 8.31] there exists a uniqueu∈C(Ω)∩H_loc¹ (Ω) such thatB_λu=f₀+Pd

k=1∂_kf_k andu|_Γ= 0.

Thenu∈C0(Ω) and the existence follows from Lemma 2.2. The uniqueness

follows from Proposition 2.3.

Corollary 2.5. — Letλ > λ0 andp∈(d,∞]. Let f0∈ L_p/2(Ω) and f₁, . . . , f_d∈L_p(Ω). Letu∈H₀¹(Ω)and suppose thatBλu=f₀+Pd

k=1∂_kf_k. Thenu∈C0(Ω).

Proof. — By Proposition 2.4 there exists a ˜u∈H₀¹(Ω)∩C0(Ω) such that Bλu˜=f0+Pd

k=1∂kfk. ThenBλ(u−u) = 0. So˜ bλ(u−u, v) = 0 first for˜ allv∈C_c^∞(Ω) and then by density for allv ∈H₀¹(Ω). Choosev =u−u.˜ Then ^µ₂ku−uk˜ ²_H1(Ω)6Reb_λ(u−u) = 0. So˜ u= ˜u∈C₀(Ω).

We next wish to add the other lower order terms.

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Proposition 2.6. — There exists ac >0such that for allΦ∈C¹(R^d) there exists a uniqueu∈H¹(Ω)∩C(Ω)such that u|Γ= Φ|Γ andAu= 0.

Moreover,

kuk_C(Ω)6ckΦ|Γk_C(Γ).

For the proof we need some lemmas. In the next lemma we introduce a parameterδin order to avoid duplication of the proof.

Lemma 2.7. — Fixδ∈[0, λ0+ 1].

(1) For allf ∈L2(Ω)andλ > λ0there exists a uniqueu∈H₀¹(Ω)such that

(2.2) bλ(u, v) =

d

X

k=1

(bkf, ∂kv)_L₂_(Ω)+ ((c0−δ1^Ω)f, v)_L₂_(Ω) for allv∈H₀¹(Ω).

For allλ > λ0 defineRλ:L2(Ω)→L2(Ω) byRλf =u, whereu∈H₀¹(Ω) is as in(2.2).

(2) There exists ac₁>0such that

kRλfkL_q(Ω)6c₁(λ−λ₀)^−1/4kfkL₂(Ω)

for allλ > λ0andf ∈L2(Ω), where ¹_q =¹₂−_4d¹. (3) There exists ac2>1such that

kRλfk_L_q_(Ω)6c2kfk_L_p_(Ω)

for allλ∈[λ0+ 1,∞),p, q∈[2,∞]andf ∈Lp(Ω)with ¹_q = ¹_p−_4d¹. (4) Ifλ > λ₀,p∈(d,∞]andf ∈L_p(Ω), then R_λf ∈C₀(Ω).

Proof.

(1). This follows from the Lax–Milgram theorem.

(2). Define M =kc0−δ1^Ωk_L_∞_(Ω)+Pd

k=1kbkk_L_∞_(Ω). Let λ > λ0, f ∈ L₂(Ω) and setu=R_λf. Then

µ

2 kuk²_H1(Ω)+ (λ−λ0)kuk²_L₂_(Ω)

6Rebλ₀(u) + (λ−λ0)kuk²_L

2(Ω)

= Rebλ(u)

= Re

d

X

k=1

(bkf, ∂ku)_L₂_(Ω)+ Re((c0−δ1^Ω)f, u)_L₂_(Ω) 6Mkfk_L₂_(Ω)kuk_H1(Ω).

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SokukH¹(Ω)62µ⁻¹MkfkL₂(Ω) and

kRλfk_L₂_(Ω)=kuk_L₂_(Ω)6

s 2

µ(λ−λ₀)Mkfk_L₂_(Ω).

The Sobolev embedding theorem implies that there exists a c₁ >0 such thatkvk_L_q

1(Ω)6c1kvk_H1(Ω) for allv ∈H₀¹(Ω), where _q¹

1 = ¹₂ −_2d¹. (The extra factor 2 is to avoid a separate case ford= 2.) ThenkRλfk_L_q

1(Ω)6 2µ⁻¹c1Mkfk_L₂_(Ω). Hence

kRλfkL_q(Ω)6kRλfk^1/2_L

2(Ω)kRλfk^1/2_L

q1(Ω)6c2(λ−λ0)^−1/4kfkL₂(Ω), wherec2= (2/µ)^3/4c^1/2₁ M.

(3). Apply Corollary 2.5 with p = 4d and λ = λ₀+ 1. It follows that Rλ₀+1f ∈C0(Ω) for allf ∈Lp(Ω). Clearly the mapRλ₀+1|_L_p_(Ω):Lp(Ω)→ C₀(Ω) has a closed graph. Hence it is continuous. In particular, there exists ac3>0 such thatkRλ₀+1fkL_∞(Ω)=kRλ₀+1fkC₀(Ω) 6c3kfkL_p(Ω) for all f ∈Lp(Ω).

Letλ>λ₀+ 1 andf ∈L₂(Ω). Writeu=R_λf and u₀=R_λ₀₊₁f. Then bλ(u, v) =bλ₀+1(u0, v) andbλ(u−u0, v) =−(λ−λ0−1) (u, v)_L₂_(Ω)for all v∈H₀¹(Ω). Henceu−u₀∈D(B^D) and (B^D+λ I)(u−u₀) =−(λ−λ₀−1)u₀. Consequently

Rλ= I−(λ−λ0−1) (B^D+λ I)⁻¹ Rλ₀+1

for all λ>λ0+ 1. Since the semigroup generated by −B^D has Gaussian bounds, there exists ac4>1 such thatk(B^D+λ I)⁻¹k_∞→∞6c4λ⁻¹ for allλ>λ₀+ 1. ThenkRλfkL∞(Ω)62c₃c₄kfkL_p(Ω) for allλ>λ₀+ 1 and f ∈Lp(Ω).

Finally let p⁰ ∈ (2,4d) and let q⁰ ∈ (2,∞) be such that _q¹0 = _p¹0 − _4d¹. There exists aθ ∈(0,1) such that _p¹0 = ^1−θ₂ + ^θ_p. Then _q¹0 = ^1−θ_q , where

1

q = ¹₂−_4d¹. Letc₁>0 be as in Statement (2). The operatorR_λis bounded from L₂(Ω) into L_q(Ω) with norm at most c₁ by Statement (2), and we just proved that the operatorRλis bounded fromLp(Ω) intoL∞(Ω) with norm at most 2c₃c₄. Hence by interpolation the operator R_λ is bounded fromLp⁰(Ω) intoLq⁰(Ω) with norm bounded byc^1−θ₁ (2c3c4)^θ6c1+ 2c3c4, which gives Statement (3).

(4). This is a special case of Corollary 2.5.

The main step in the proof of Proposition 2.6 is the next lemma.

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Lemma 2.8. — There exist λ > λ₀ and c > 0 such that for all Φ ∈ C¹(Ω)∩H¹(Ω)there exists a uniqueu∈H¹(Ω)∩C(Ω)such thatu|Γ= Φ|Γ

andA_λu= 0. Moreover,

kuk_C(Ω)6ckΦ|Γk_C(Γ).

Proof. — Chooseδ= 0 in Lemma 2.7. Letc1andc2be as in Lemma 2.7.

Letλ∈(λ₀+ 1,∞) be such that c₁c^2d−1₂ (λ−λ₀)^−1/4(1 +|Ω|)6 ¹2. Let Rλ be as in Lemma 2.7. Setϕ= Φ|Γ.

There exist unique w,w˜ ∈ H₀¹(Ω) such that a_λ(w, v) = a_λ(Φ, v) and bλ( ˜w, v) = bλ(Φ, v) for all v ∈H₀¹(Ω). Then ˜w∈C0(Ω) by Corollary 2.5.

Defineu= Φ−wand ˜u= Φ−w. Then ˜˜ u∈H¹(Ω)∩C(Ω) and ˜u|_Γ =ϕ.

Moreover,aλ(u, v) = 0 andbλ(˜u, v) = 0 for allv∈H₀¹(Ω), andkuk˜ _C(Ω)6 kϕk_C(Γ)by the result of Stampacchia mentioned in the introduction ([13, Théorème 3.8]).

Letv∈H₀¹(Ω). Then

bλ(˜u−u, v) =

d

X

k=1

(bku, ∂kv)_L₂_(Ω)+ (c0u, v)_L₂_(Ω)

and ˜u−u=R_λuby the definition ofR_λ.

For all n ∈ {0, . . . ,2d} define pn = _2d−n^4d . Then p0 = 2, p_2d−1 = 4d, p_2d =∞and _p¹

n = _p¹

n−1 −_4d¹ for alln∈ {1, . . . ,2d}. So k˜u−ukL_pn(Ω)6 c2kuk_L_pn−1_(Ω)for alln∈ {2, . . . ,2d}and

k˜u−uk_L_p

1(Ω)6c1(λ−λ0)^−1/4kuk_L₂_(Ω) by Lemma 2.7 (3) and (2). Then

kuk_L_p

1(Ω)6c₁(λ−λ₀)^−1/4kuk_L₂_(Ω)+ (1 +|Ω|)k˜uk_L_∞_(Ω) and

kuk_L_pn_(Ω)6c₂kuk_L_pn−1_(Ω)+ (1 +|Ω|)k˜uk_L_∞_(Ω) for alln∈ {2, . . . ,2d}. It follows by induction tonthat

kuk_L_pn_(Ω)6c₁cⁿ⁻¹₂ (λ−λ₀)^−1/4kuk_L₂_(Ω)+ (1 +|Ω|)

n−1

X

k=0

c^k₂k˜uk_L_∞_(Ω)

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for alln ∈ {2, . . . ,2d}. So u ∈ L_p_2d−1(Ω) = L_4d(Ω) and ˜u−u= R_λu ∈ C₀(Ω) by Lemma 2.7 (4). In particularu∈C(Ω). Moreover,

kukL∞(Ω)

=kuk_L_p

2d(Ω)

6c1c^2d−1₂ (λ−λ0)^−1/4kuk_L₂_(Ω)+ 2d(1 +|Ω|)c^2d−1₂ k˜uk_L_∞_(Ω)

6c1c^2d−1₂ (λ−λ0)^−1/4(1 +|Ω|)kuk_L_∞_(Ω)+ 2d(1 +|Ω|)c^2d−1₂ k˜uk_L_∞_(Ω) 6 1

2kuk_L_∞_(Ω)+ 2d(1 +|Ω|)c^2d−1₂ kuk˜ _L_∞_(Ω) by the choice ofλ. So

kuk_L_∞_(Ω)64d(1 +|Ω|)c^2d−1₂ k˜uk_L_∞_(Ω)64d(1 +|Ω|)c^2d−1₂ kϕk_C(Γ)

and the proof of the lemma is complete.

We next wish to remove the λ in Lemma 2.8. For future purposes, we consider the full inhomogeneous problem.

Proposition 2.9. — Letp∈(d,∞],f0 ∈L_p/2(Ω) and letf1, . . . , fd ∈ L_p(Ω). Letu∈H₀¹(Ω)be such thatAu=f₀+Pd

k=1∂_kf_k. Thenu∈C₀(Ω).

Proof. — Without loss of generality we may assume that p ∈ (d,4d).

Chooseλ=δ =λ0+ 1 in Lemma 2.7 and in Proposition 2.4. By Propo- sition 2.4 there exists a unique ˜u ∈ H₀¹(Ω) ∩C0(Ω) such that Bλu˜ = f0+Pd

k=1∂kfk. Ifv∈C_c^∞(Ω), then bλ(˜u, v) =hf0+

d

X

k=1

∂kfk, vi_D⁰_{(Ω)×D(Ω)}

=a(u, v)

=b_λ(u, v) +

d

X

k=1

(b_ku, ∂_kv)_L₂_(Ω)+ ((c₀−δ1Ω)u, v)_L₂_(Ω). So

bλ(˜u−u, v) =

d

X

k=1

(bku, ∂kv)_L₂_(Ω)+ ((c0−δ1^Ω)u, v)_L₂_(Ω)

and by density for allv ∈H₀¹(Ω). Henceu−u˜ =R_λu, whereR_λ is as in Lemma 2.7. For alln ∈ {0, . . . ,2d−1} define pn = _2d−n^4d . Then u−˜u∈ L₂(Ω) = L_p₀(Ω). It follows by induction to n that u ∈ L_p_n−1(Ω) and u−u˜∈Lp_n(Ω) for alln∈ {1, . . . ,2d−1}, where the last part follows from Lemma 2.7 (3). Henceu−u˜ ∈L_p_2d−1(Ω) = L_4d(Ω) and u∈L_p(Ω). Then Lemma 2.7 (4) givesu−u˜=Rλu∈C0(Ω) and thereforeu∈C0(Ω).

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Corollary 2.10. — Letp∈(d,∞]. Then(A^D)⁻¹(L_p(Ω))⊂C₀(Ω).

Corollary 2.11. — There exists ac⁰ >0such thatk(A^D)⁻¹fkL∞(Ω)6 c⁰kfk_L_∞_(Ω)for allf ∈L_∞(Ω).

Proof. — Closed graph theorem.

Proof of Proposition 2.6. — Let c, λ > 0 be as in Lemma 2.8 and let c⁰ > 0 be as in Corollary 2.11. By Lemma 2.8 there exists a unique ˜u ∈ H¹(Ω)∩C(Ω) such that ˜u|Γ = Φ|Γ and Aλu˜ = 0. By Lemma 2.1 there exists a uniquew∈H₀¹(Ω) such thata(w, v) =a(Φ|_Ω, v) for allv∈H₀¹(Ω).

Setu= Φ|Ω−wand ˜w= Φ|Ω−u. Then˜

a(w, v) =a(Φ|Ω, v) =aλ(Φ|Ω, v)−λ(Φ, v)_L₂_(Ω)=aλ( ˜w, v)−λ(Φ, v)_L₂_(Ω)

=a( ˜w, v) +λ( ˜w, v)_L₂_(Ω)−λ(Φ, v)_L₂_(Ω)=a( ˜w, v)−λ(˜u, v)_L₂_(Ω) for allv∈H₀¹(Ω). So

a(˜u−u, v) =a(w−w, v) =˜ −λ(˜u, v)_L₂_(Ω).

Since ˜u−u ∈ H₀¹(Ω) it follows that A^D(˜u−u) = −λu. Consequently,˜ u= ˜u+λ(A^D)⁻¹u˜∈C0(Ω) by Corollary 2.10. Moreover,

kuk_C(Ω)=kuk_L_∞_(Ω)6k˜uk_L_∞_(Ω)+λk(A^D)⁻¹uk˜ _L_∞_(Ω) 6(1 +c⁰λ)k˜uk_L_∞_(Ω)6(1 +c⁰λ)ckΦ|Γk_C(Γ)

and the proof of Proposition 2.6 is complete.

Define ||| · |||: H_loc¹ (Ω)→[0,∞] by

|||u|||= sup

δ>0

sup

Ω₀⊂Ω open d(Ω₀,Γ)=δ

δ Z

Ω₀

|∇u|² 1/2

.

Finally we need the following Caccioppoli inequality.

Proposition 2.12. — There exists a c⁰ > 1 such that |||u||| 6 c⁰kuk_L₂_(Ω) for allu∈H¹(Ω)such thatAu= 0.

Proof. — See [9, Theorem 4.4].

Now we are able to prove Theorem 1.1.

Proof of Theorem 1.1. — The uniqueness is already proved in Proposi- tion 2.3.

Letc >0 andc⁰>1 be as in Propositions 2.6 and 2.12. Let Φ∈C¹(R^d)∩

H¹(R^d). By Proposition 2.6 there exists a unique u∈H¹(Ω)∩C(Ω) such

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thatu|Γ= Φ|Γ andAu= 0. Moreover,

kuk_C(Ω)+|||u|||6kuk_C(Ω)+c⁰kukL₂(Ω)

6(2 +|Ω|)c⁰kuk_C(Ω) 6(2 +|Ω|)c c⁰kΦ|ΓkC(Γ). (2.3)

It follows from (2.3) that we can define a linear mapF: {Φ|_Γ: Φ∈C¹(R^d)∩

H¹(R^d)} → H¹(Ω)∩C(Ω) by F(Φ|Γ) = u, whereu ∈ H¹(Ω)∩C(Ω) is such that u|Γ = Φ|Γ and Au = 0. Now let ϕ ∈ C(Γ). By the Stone–

Weierstraß theorem there are Φ1,Φ2, . . . ∈ C¹(R^d)∩H¹(R^d) such that lim Φn|Γ = ϕ in C(Γ). Set un = F(Φn|Γ) for all n ∈ N. Then it follows from (2.3) that (u_n)_n∈_N is a Cauchy sequence in C(Ω). Let u = limu_n in C(Ω). Also (un)_n∈N is a Cauchy sequence in H_loc¹ (Ω) by (2.3). So u∈ H_loc¹ (Ω). SinceAu_n= 0 for alln∈N, one deduces thatAu= 0. Moreover, u|Γ = limun|Γ= lim Φn|Γ=ϕ. This proves existence. Finally,

kuk_C(Ω)= limkunk_C(Ω)

6lim(2 +|Ω|)c c⁰kΦn|Γk_C(Γ)= (2 +|Ω|)c c⁰kϕk_C(Γ).

This completes the proof of Theorem 1.1.

Theorem 1.1 has the following extension.

Let ϕ ∈ C(Γ), p ∈ (d,∞], f0 ∈ Lp/2(Ω) and let f1, . . . , fd ∈ Lp(Ω).

Then there exists a unique u ∈ C(Ω)∩H_loc¹ (Ω) such that u|Γ = ϕ and Au=f₀+Pd

k=1∂_kf_k.

Proof. — The uniqueness follows as in the proof of Proposition 2.3.

By Lemma 2.1 there exists au0∈H₀¹(Ω) such thatAu0=f0+Pd

k=1∂kfk. Thenu₀ ∈C₀(Ω) by Proposition 2.9. By Theorem 1.1 there exists au₁ ∈ C(Ω)∩H_loc¹ (Ω) such thatu1|Γ=ϕandAu1= 0. Defineu=u0+u1. Then u∈C(Ω)∩H_loc¹ (Ω). Moreover, u|_Γ =ϕandAu=f₀+Pd

k=1∂_kf_k. We conclude this section with some results for the classical solution.

They will be used in Section 3 and are of independent interest. Recall that P:C(Γ) → C(Ω) is given by P ϕ = u, where u∈ C(Ω)∩H_loc¹ (Ω) is the classical solution, sou|Γ=ϕandAu= 0.

Proposition 2.14. — LetΦ∈C(Ω)∩H_loc¹ (Ω). Suppose there exists a w∈H₀¹(Ω) such thatAΦ =Aw. Then w∈C(Ω)andP(Φ|Γ) = Φ−w.

Proof. — Write ˜w= Φ−P(Φ|Γ). Then ˜w∈C₀(Ω)∩H_loc¹ (Ω) andAw˜= AΦ =Aw=f0+Pd

k=1∂kfk, where f0=c0w+Pd

l=1cl∂lw∈L2(Ω) and fk =−Pd

l=1alk∂lw−bkw∈L2(Ω) for allk∈ {1, . . . , d}. So ˜w∈H₀¹(Ω)