Let Ω be an exterior Lipschitz domain in R2. It is proved that the Helmholtz decomposition of the vector fields in Lp(Ω;R2) exists if p satisfies |1/p−1/2|<1/4+ε with some constant ε=ε(Ω)∈(0,1/4], where it is allowed to take ε=1/4 if ∂Ω∈C1.
Citation: Keiichi Watanabe. The Helmholtz decomposition of vector fields for two-dimensional exterior Lipschitz domains[J]. AIMS Mathematics, 2024, 9(7): 17886-17900. doi: 10.3934/math.2024870
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Let Ω be an exterior Lipschitz domain in R2. It is proved that the Helmholtz decomposition of the vector fields in Lp(Ω;R2) exists if p satisfies |1/p−1/2|<1/4+ε with some constant ε=ε(Ω)∈(0,1/4], where it is allowed to take ε=1/4 if ∂Ω∈C1.
Let Ω be an exterior Lipschitz domain in R2, i.e., the complement of a bounded planar Lipschitz domain. The aim of this paper is to show the existence of the Helmholtz decomposition of the vector fields in Lp(Ω;R2) provided that p satisfies
|1p−12|<14+ε | (1.1) |
with some constant ε=ε(Ω)∈(0,1/4], where it is allowed to take ε=1/4 if ∂Ω∈C1. Let Lp,σ(Ω) be the subspace of functions f in Lp(Ω;R2) such that ∫Ωf⋅∇φdx=0 for every φ∈˙H1p(Ω). Notice that Lp,σ(Ω) is the closure of C∞c,σ(Ω):={ϕ∈C∞c(Ω;R2):divϕ=0} in Lp(Ω;R2), see [4,Proposition 2.5]. In addition, let Gp(Ω):=Im∇p=∇p˙H1p(Ω) denote the space of gradients in Lp(Ω;R2), where ∇p:˙H1p(Ω)→Lp(Ω;R2) is the linear map. We aim to prove the following theorem.
Theorem 1.1. Let Ω⊂R2 be an exterior Lipschitz domain. Then there exists ε=ε(Ω)∈(0,1/4] such that for every p subject to (1.1) the Helmholtz decomposition
Lp(Ω;R2)=Lp,σ(Ω)⊕Gp(Ω) | (1.2) |
holds, where the direct sum is topological. In particular, if ∂Ω∈C1, it is allowed to take ε=1/4, i.e., the Helmholtz decomposition (1.2) is valid for every 1<p<∞.
The Helmholtz decomposition (1.2), a useful tool in the study of the incompressible Navier-Stokes equations, is well-known in the case that the boundary ∂Ω is smooth (see e.g., [9,Theorem 1.6] and [11,Theorem 1.4]). In the case of exterior domains with lower regularity (locally Lipschitz), it was proved by Lang and Méndez [7,Theorem 6.1] as well as Tolksdorf and the author [12,Proposition 2.3] that the Helmholtz decomposition (1.2) holds for all p satisfying |1/p−1/2|<1/6+ε with some constant ε(Ω)>0 provided that Ω⊂Rd with d≥3. However, to the best of the author's knowledge, there seems to be no result on the Helmholtz decomposition (1.2) in the case of exterior planar Lipschitz domains.
It is well-known (cf. [2,Lemma III.1.2]) that the existence of the Helmholtz decomposition (1.2) is equivalent to the unique solvability of the following weak Neumann problem: Given f∈Lp(Ω;R2), consider the weak Neumann problem
⟨∇u,∇φ⟩Ω=⟨f,∇φ⟩Ωfor every φ∈˙H1p′(Ω). | (1.3) |
The present paper aims to prove the following theorem.
Theorem 1.2. Let Ω⊂R2 be an exterior Lipschitz domain. Then there exists ε=ε(Ω)∈(0,1/4] having the following property: If p satisfies (1.1), for every f∈Lp(Ω) Problem (1.3) admits a solution u∈˙H1p(Ω) subject to the estimate
‖∇u‖Lp(Ω;R2)≤C‖f‖Lp(Ω;R2) |
with some positive constant C>0 which depends only on Ω and p. In particular, the solution is unique in ˙H1p(Ω;R2) up to an additive constant. Furthermore, if ∂Ω∈C1, then it is allowed to take ε=1/4.
Remark 1.3. Let us make some comments on Theorem 1.2.
(1) The constant ε appearing in Theorem 1.2 arises from analyses of elliptic systems on bounded Lipschitz domains (i.e., analyses near the boundary ∂Ω). More precisely, ε is given by ε=εNeu∈(0,1/4], where εNeu is a constant arising in the result of the Neumann-Laplacian, see Lemma 3.3 below. This is due to the fact that we will construct a solution to (1.3) via a cut-off procedure so that, roughly speaking, a solution to (1.3) may be given as a sum of the solution to the weak Neumann problem in a bounded Lipschitz domain and the fundamental solution of the Laplace equation. Namely, the restriction (1.1) on p stems essentially from the roughness of the boundary ∂Ω but the unboundedness of the domain does not restrict the range of p, which is entirely different from the strategy in [7]. A similar observation for the large-time behavior of the three-dimensional Navier-Stokes flow in an exterior Lipschitz domain was made by the author [13].
(2) Concerning the higher-dimensional case, say, Ω⊂Rd (d≥3), that was discovered by Lang and Méndez [7,Theorem 6.1] as well as Tolksdorf and the author [12,Proposition 2.3], in their papers, it was proved that the Helmholtz decomposition (1.2) exists provided that p satisfies |1/p−1/2|<1/6+ε with some constant ε(Ω)>0. Although the present paper is restricted to dealing with the two-dimensional case, it is easy to extend to the higher-dimensional case and obtain the same result as in [7,12]. Nevertheless, the present paper provides a simpler proof than that of [7] (notice that the proof in [12] relied on the result obtained in [7]). It should be emphasized that it seems to be difficult to extend the approach of Lang and Méndez [7] to the two-dimensional case since their approach heavily relied on potential theory, which causes several difficulties in the two-dimensional case due to a logarithm singularity of the fundamental solution to the Laplace equation.
This paper is organized as follows: In the next section, we introduce some notation and the function spaces that will be used throughout this paper. Section 3 is concerned with the solvability result of some elliptic systems in bounded domains. Then, in the last section, we give the proof of Theorem 1.2 via a cut-off technique introduced by Shibata [10]. It should be noted that some modifications are required in contrast to Shibata's argument [10] since we may not expect the H2p(Ω)-regularity for the solution to elliptic systems due to the lack of the smoothness of the boundary. Note that, for the same reason, we may not use the approaches of Miyakawa [9] as well as Simader and Sohr [11] to prove Theorem 1.2.
As usual, let N and R be the set of all natural and real numbers, respectively. For scalar-valued functions u and v defined on G⊂R2, let ⟨u,v⟩G=∫Gu(x)v(x)dx. For vector-valued functions U=(U1,U2) and V=(V1,V2) defined on G⊂R2, let ⟨U,V⟩G=∑j=1,2∫GUj(x)Vj(x)dx. For R>0, let BR(0)={x∈R2∣|x|<R}. For Banach spaces X and Y, let L(X,Y) be all bounded linear operators from X into Y, where we will write L(X)=L(X,X) to simplify the notation. For a Banach space X, denote by X′ the dual space of X. Throughout this paper, the letter C stands for a generic constant that does not depend on the quantities whenever there is no confusion.
Let X be a complex Banach space and let R2 be endowed with the Lebesgue measure. Let C∞c(G;X) be the set of all C∞-functions on R2 whose supports are compact and contained in D⊂R2. For 1≤p≤∞ and G⊂R2, let Lp(G;X) be the Lebesgue space equipped with the norm ‖⋅‖Lp(G;X). For 1≤p<∞ and s∈R, let Hsp(R2;X) be the inhomogeneous Sobolev space endowed with the norm ‖⋅‖Hsp(R2;X). The inhomogeneous Sobolev space on G is defined by the collection of all u∈D′(G;X)=(C∞c(G;X))′ such that there exists v∈Hsp(G;X) with v|G=u. Furthermore, the norm ‖⋅‖Hsp(G;X) is defined by the usual quotient norm:
‖u‖Hsp(G;X)=inf‖v‖Hsp(R2;X) |
where the infimum is taken over all v∈Hsp(R2;X) such that its restriction v|G to G coincides in D′(G;X) with u. In particular, if u∈Hsp(G;X) vanishes on the boundary ∂G then the space will be attached with the subscript 0, i.e., Hsp,0(G;X). For 1≤p<∞ and s∈R, let ˙Hsp(R2;X) be the homogeneous Sobolev space equipped with the norm ‖⋅‖Hsp(R2;X). For 1≤p<∞ and G⊂R2, we also define
˙H1p(G):={[u]=u+R:u∈Lp,loc(G) and ∇u∈Lp(G;R2)} |
with the norm ‖⋅‖˙H1p(G)=‖∇⋅‖Lp(G), where u∈Lp,loc(G) means u∈Lp(G′) for any bounded domain G′ with G′⊂G. If X=R, we often write Lp(G)=Lp(G;R), Hsp(R2)=Hsp(R2;R), and ˙Hsp(R2)=˙H−1p(R2;R) for short. The Hölder conjugate exponent of p is denoted by p′.
In this section, we give the solvability result for elliptic systems in the case of bounded domains. Consider the following elliptic system:
{−Δu=fin D,∂u∂ν=0on Σ,u=0on Γ. | (3.1) |
Here, D⊂R2 is a bounded Lipschitz domain with ∂Ω=Σ∪Γ with Σ=∂D∖Γ and Σ≠∅. We suppose that there exist some constants 0<R1<R2 such that Σ⊊BR1(0) and Γ⊂R2∖¯BR2(0). Let
Lp,0(D):={f∈Lp(D):⟨f,1⟩D=0}. |
Then the result for (3.1) reads as follows.
Theorem 3.1. Let D⊂R2 be a bounded Lipschitz domain. Suppose that there exist some constants 0<R1<R2 such that Σ⊊BR1(0) and Γ⊂R2∖¯BR2(0). Then there exists ε=ε(D)∈(0,1/4] having the following property: If p satisfies (1.1), for every s∈[0,1/p) and for every f∈Lp,0(D) Problem (3.1) admits a solution u∈H1+sp(D) subject to the estimate
‖u‖H1+sp(D)≤C‖f‖Lp(D) |
with some positive constant C>0 which depends only on D, p, and s. Furthermore, if ∂D∈C1, then it is allowed to take ε=1/4.
To prove this theorem, we define the weak Dirichlet-Laplacian ΔDp,s,w on Lp(D) as
D(ΔDp,s,w)={u∈H1+sp,0(D):Δu∈Lp(D)},ΔDp,s,wu=Δu | (3.2) |
with 0≤s<1/p. Here, Δu∈Lp(D) is understood in the sense of distributions. We also define the weak Neumann-Laplacian ΔNp,s,w on Lp(D) as
D(ΔNp,s,w)={u∈H1+sp(D):∃v∈Lp(D) s.t. ∀φ∈H1+sp′,0(D):⟨∇u,∇φ⟩D=⟨v,φ⟩D},ΔNp,s,wu=v | (3.3) |
with 0≤s<1/p. Notice that the Neumann boundary condition ∂u/∂ν=0 on ∂D is interpreted in the sense that
⟨Δu,φ⟩D=−⟨∇u,∇φ⟩Dfor any φ∈H1+sp′(D). |
Mimicking the argument as in Sections 3–5 in [15], we may prove the following results.
Lemma 3.2. Let D⊂R2 be a bounded Lipschitz domain. Then there exists εDir=εDir(D)∈(0,1/4] having the following property: If p satisfies (1.1), for every s∈[0,1/p) the operator ΔDp,s,w defined by (3.2) generates a C0-semigroup of contractions on Lp(D). Furthermore, if ∂D∈C1, then it is allowed to take εDir=1/4.
Lemma 3.3. Let D⊂R2 be a bounded Lipschitz domain. Then there exists εNeu=εNeu(D)∈(0,1/4] having the following property: If p satisfies (1.1), for every s∈[0,1/p) the operator ΔNp,s,w defined by (3.3) generates a C0-semigroup of contractions on Lp(D). Furthermore, if ∂D∈C1, then it is allowed to take εNeu=1/4.
Remark 3.4. Let us make a few comments on Lemmas 3.2 and 3.3.
(1) If s=0, Lemma 3.2 was obtained by Wood [15,Proposition 4.1]. However, it is easy to extend its result to the case of 0<s<1/p due to [6,Theorem 1.3].
(2) In the case of D⊂Rd with d≥3, Lemma 3.3 with s=0 was proved by Wood [15,Theorem 5.6]. It is not difficult to extend the result of [15,Theorem 5.6] to the case of D⊂R2 if one replaces [15,Theorem 2.6] by [8,Corollary 4.2].
For every λ>0, consider the resolvent problem for the Laplacian with Dirichlet boundary condition
{λu−Δu=fin D,u=0on ∂D | (3.4) |
as well as the resolvent problem for the Laplacian with Neumann boundary condition
{λu−Δu=fin D,∂u∂ν=0on ∂D. | (3.5) |
By the Hille-Yosida theorem, we infer from Lemma 3.2 that for every λ>0, there exists a unique solution u satisfying
λ‖u‖Lp(D)≤‖f‖Lp(D). |
Let p and s be the same numbers as in Lemma 3.2. From the invertibility results in [8,Corollary 4.2], we have
‖u‖H1+sp(D)≤C‖−λu+f‖H−1+sp(D)≤C‖−λu+f‖Lp(D)≤C(λ‖u‖Lp(D)+‖f‖Lp(D))≤C‖f‖Lp(D) |
since H−1+sp(D)↩Lp(D) due to −1+s<0. Hence, the solution u∈H1+sp,0(D) to (3.4) verifies
λ‖u‖Lp(D)+‖u‖H1+sp(D)≤C‖f‖Lp(D)for all λ>0 |
provided that f∈Lp(D). Likewise, we infer from Lemma 3.3 and the invertibility results in [8,Corollary 4.2] that there exists a unique solution u∈H1+sp(D) satisfying
λ‖u‖Lp(D)+‖u‖H1+sp(D)≤C‖f‖Lp(D)for all λ>0 |
provided that f∈Lp,0(D), where p and s are the same numbers as in Lemma 3.3. With the aforementioned preliminaries, we are in a position to prove Theorem 3.1.
Proof of Theorem 3.1. Let ε∈(0,1/4] be ε=min{εDir,εNeu}, where εDir and εNeu are the same numbers as in Lemmas 3.2 and 3.3, respectively. In addition, let p and s satisfy (1.1) and 0≤s<1/p, respectively. Assume that f∈Lp,0(D). Then, for every λ>0, there exists a unique solution uDir∈H1+sp,0(D) to
{λuDir−ΔuDir=fin D,uDir=0on Σ∪Γ. |
In addition, u satisfies the estimate
λ‖uDir‖Lp(D)+‖uDir‖H1+sp(D)≤C‖f‖Lp(D)for all λ>0. | (3.6) |
By complex interpolation, we observe that for every t∈(0,s), there holds
‖uDir‖H1+tp(D)≤Cλs−t1+s‖f‖Lp(D)for all λ>0. |
Similarly, for every λ>0, there exists a unique solution uNeu∈H1+sp(D) to
{λuNeu−ΔuNeu=fin D,∂uNeu∂ν=0on Σ∪Γ. |
Moreover, u satisfies the estimate
λ‖uNeu‖Lp(D)+‖uNeu‖H1+sp(D)≤C‖f‖Lp(D)for all λ>0. |
By complex interpolation, we observe that for every t∈(0,s), there holds
‖uNeu‖H1+tp(D)≤Cλs−t1+s‖f‖Lp(D)for all λ>0. | (3.7) |
Let ζ∈C∞c(D;[0,1]) be a cut-off function such that ζ=1 if |x|≤R1+ϵ and ζ=0 if |x|≥R2−ϵ, where ϵ:=(R1+R2)/3. Set v=ζuDir+(1−ζ)uNeu. We see that v solves
{λv−Δv=f+R0fin D,∂v∂ν=0,on Σ,v=0on Γ, |
where we have set
R0f:=Δζ(uNeu−uDir)+2∇ζ⋅∇(uNeu−uDir). |
From (3.6) and (3.7), there exists λ0≥1 such that for all λ≥λ0, we have
‖R0f‖Lp(D)≤12‖f‖Lp(D), |
which together with a Neumann series argument implies that the operator I+R0:Lp,0(D)→Lp(D) is invertible. Thus, we see that
{λu−Δu=fin D,∂u∂ν=0on Σ,u=0on Γ, | (3.8) |
is solvable for all λ≥λ0. Notice that the uniqueness of the solution follows from the duality argument. Indeed, suppose that u∈H1+sp(D) solves (3.8) with f vanishing in D. For any λ≥λ0 and ϕ∈C∞c(D), consider
{λu0−Δu0=ϕin D,∂u0∂ν=0on Σ,u0=0on Γ. |
Then we infer from the divergence theorem that
⟨u,ϕ⟩D=⟨u,λu0−Δu0⟩D=⟨λu−Δu,u0⟩D=0. |
Since ϕ∈C∞c(D) is arbitrary, we deduce that u=0 in D, which gives the uniqueness assertion. In the following, let Ap,s be the operator defined by
D(Ap,s)={u∈H1+sp(D):∂u∂ν=0on Σ and u=0on Γ}Ap,su=Δu |
with 0≤s<1/p. From the aforementioned argument, we see that the resolvent set ρ(Ap,s) of Ap,s contains [λ0,∞).
We next deal with the case 0≤λ<λ0. Since (3.8) may be written as (λI−Ap,s)u=f, we find that (3.8) and
(I+(λ−2λ0)Rp,s(λ0))(2λ0I−Ap,s)u=f,Rp,s(λ0):=(2λ0I−Ap,s)−1 |
are equivalent. Hence, to prove that [0,λ0) is contained in ρ(Ap,s), it suffices to show the invertibility of the operator I+(λ−2λ0)Rp,s(λ0). Since it follows from the Rellich-Kondrachov theorem (cf. [1,Theorem 6.3]) that Rp,s(λ0) is a compact operator from H1+sp(D) into Lp(D), we observe that [0,λ0)⊂ρ(Ap,s) follows from the Fredholm alternative theorem and the injection of I+(λ−2λ0)Rp,s(λ0). To see this, for any λ∈[0,λ0), take u1∈Ker(I+(λ−2λ0)Rp,s(λ0)), i.e.,
(I+(λ−2λ0)Rp,s(λ0))u1=0for any u1∈Lp,0(D). |
From the definition of Rp,s(λ0), we see that u1∈D(Ap,s) and w solves
{λu1−Δu1=0in D,∂u1∂ν=0on Σ,u1=0on Γ. | (3.9) |
We first deal with the case 2≤p<∞. In this case, we may assume u1∈D(A2,s) due to the boundedness of D. Using the divergence theorem, we deduce from (3.9) that
0=λ‖u1‖2L2(D)+12‖D(u1)‖2L2(D), |
where we have set D(u1):=∇u1+[∇u1]⊤. By the Korn inequality (cf. [14,Theorem A.4]), we obtain
λ‖u1‖2L2(D)+1C‖∇u1‖2L2(D)≤0 |
with some constant C>0 depending only on D. Since λ∈[0,λ0), we see that u1=0 in D due to the Dirichlet boundary condition u1=0 on Γ. This completes the proof of [0,λ0)⊂ρ(Ap,s) in the case that p satisfies 2≤p<∞. The remaining case 1<p<2 follows from the duality [5,Corollary A.4.3].
Without loss of generality, we may assume that the origin of the coordinates is located interior to the complement of the domain Ω. Then, we may take R0>0 so large that there holds R2∖Ω⊂BR(0) for any R>R0. Set DR:=Ω∩B8R(0) and AR:={x∈R2∣2R≤|x|≤7R}.
Step 1: Construction of U1
Following the argument of Shibata [10,Section 3], we introduce cut-off functions as follows. Let ψ0,χ0∈C∞c(R2;[0,1]) be the cut-off functions satisfying
ψ0(x)={1for |x|≤4R,0for |x|≥5R,χ0(x)={1for |x|≤6R,0for |x|≥7R, |
respectively. In addition, let ψ∞,χ∞∈C∞(R2;[0,1]) be smooth functions defined by
ψ∞=1−ψ0(x),χ∞(x)={χ0(9R−|x|)for |x|≤9R,1for |x|>9R, |
respectively. Clearly, there holds ψ∞(x)=0 for |x|≤4R and ψ∞(x)=1 for |x|≥5R, while χ∞(x)=0 for |x|≤2R and χ∞(x)=1 for |x|≥3R. Let U0 be the inverse of −Ap,s and U∞ be the solution operator to
⟨∇u,∇φ⟩R2=⟨F,∇φ⟩R2for every φ∈˙H1p′(R2), | (4.1) |
i.e., the solution u to this equation may be written as u=U∞(F), where F has been assumed to be F∈Lp(R2;R2). Since the solution u to (4.1) is unique up to an additive constant, there is no loss of generality in assuming ⟨U∞(F),1⟩D=0. Notice that the Fourier multiplier theorem and the Poincaré-Wirtinger inequality yield
‖∇U∞(F)‖Lp(R2;R2)+‖U∞(F)‖Lp(G)≤C‖F‖Lp(R2;R2) |
for every bounded Lipschitz domain G⊂R2. Given f∈Lp(Ω;R2), define the linear operator U1 by
U1(f)=χ0U0(ψ0f)+(1−χ0)U∞(ψ∞f). |
Here, ψ∞f may be regarded as a function defined on R2 since ψ∞f vanishes in BR(0). In the following, for all smooth functions h defined on R2, the term hU0(ψ0f) is regarded as a function that is extended by zero to all of R2. Then we see that U1(f) satisfies
⟨∇U1(f),∇φ⟩Ω=⟨f,∇φ⟩Ω+⟨R1(f),φ⟩Ωfor every φ∈˙H1p′(Ω) |
with
R1(f):=−2(∇χ0)⋅∇(U0(ψ0f)−U∞(ψ∞f))−(Δχ0)(U0(ψ0f)−U∞(ψ∞f)). |
Here, we have used the identity
⟨ψ0f,∇(χ0φ)⟩Ω+⟨ψ∞f,∇(χ∞φ)⟩Ω=⟨(ψ0+ψ∞)f,∇φ⟩Ω=⟨f,∇φ⟩Ω, |
which follows from ψ0+ψ∞=1 and the fact that χ0=1 on supp(ψ0) and χ∞=1 on supp(ψ∞). Since we may write
R1(f)=−(∇χ0)⋅∇(U0(ψ0f)−U∞(ψ∞f))−div((∇χ0)(U0(ψ0f)−U∞(ψ∞f))), |
we have
⟨R1(f),φ⟩R2=⟨R1(f),∇φ⟩R2for every φ∈˙H1p′(R2) | (4.2) |
with
R1(f):=−ψ0f+(∇χ0)U0(ψ0f)+χ0∇U0(ψ0f)−(∇χ0)U∞(ψ∞f)−χ0∇U∞(ψ∞f). | (4.3) |
In fact, it follows that
⟨R1(f),φ⟩R2=−⟨∇(U0(ψ0f)−U∞(ψ∞f)),(∇χ0)φ⟩R2+⟨(∇χ0)(U0(ψ0f)−U∞(ψ∞f)),∇φ⟩DR=−⟨∇U0(ψ0f),∇(χ0φ)⟩DR+⟨∇U∞(ψ∞f),∇(χ0φ)⟩R2+⟨χ0∇(U0(ψ0f)−U∞(ψ0f)),∇φ⟩DR+⟨(∇χ0)(U0(ψ0f)−U∞(ψ∞f)),∇φ⟩DR |
for any φ∈˙H1p′(R2). Here, we have used the identity
−⟨div((∇χ0)(U0(ψ0f)−U∞(ψ∞f))),φ⟩R2=⟨(∇χ0)(U0(ψ0f)−U∞(ψ∞f)),∇φ⟩DR, |
which may be justified since supp(∇χ0) is contained in an annulus (i.e., a bounded domain). Since χ0=1 on supp(ψ0) and χ0=0 on supp(ψ∞), we deduce that
⟨∇U0(ψ0f),∇(χ0φ)⟩DR=⟨ψ0f,∇φ⟩Ω,⟨∇U∞(ψ∞f),∇(χ0φ)⟩R2=0, |
which yields the representation (4.3). Clearly, for any f∈Lp(Ω;R2), we have
‖R1(f)‖Lp(R2)≤C‖f‖Lp(Ω;R2),‖R1(f)‖Lp(R2;R2)≤C‖f‖Lp(Ω;R2). |
We also infer from (4.2) that
‖R1(f)‖˙H−1p(R2)≤C‖f‖Lp(Ω;R2). |
In addition, we infer from 1∈˙H1p(R2) and (4.2) that ⟨R1(f),1⟩R2=⟨R1(f),0⟩R2=0.
For later, we introduce the function space
Hp(R2)={g∈Lp(R2)∩˙H−1p(R2):supp(g)⊂AR,⟨g,1⟩R2=0}. |
Clearly, from the aforementioned argument, we see that R1(f)∈Hp(R2) for any f∈Lp(Ω;R2).
Step 2: Construction of U2
Given g∈Hp(R2), we intend to construct the solution operator to
⟨∇u2,∇φ⟩Ω=⟨g,φ⟩Ωfor every φ∈˙H1p′(Ω) | (4.4) |
possessing the estimate
‖∇u2‖Lp(Ω;R2)≤C‖g‖Lp(R2). | (4.5) |
Notice that, by using this operator U2, it follows that U1(f)−U2(R1(f)) solves (1.3).
In the following, let g∈Hp(R2). Let V∞ be the operator defined by
(V∞f)(x)=−∫R2E(x−y)g(y)dy, |
where E(x−y) stands for the fundamental solution of the Laplace equation:
E(x−y)=(2π)−1log|x−y|. |
By [10,(82)], the formula
⟨∇V∞(g),∇φ⟩R2=⟨g,φ⟩R2for every φ∈˙H1p′(R2) |
may be justified. Furthermore, for every g∈Hp(R2), we have
‖V∞(g)‖Lp(B9R(0))+sup|x|≥9R|x||V∞(g)|≤C‖g‖Lp(R2),‖∇V∞(g)‖Lp(R2;R2)+sup|x|≥9R|x|2|V∞(g)|≤C‖g‖Lp(R2),‖∇2V∞(g)‖Lp(R2;R4)≤C‖g‖Lp(R2), | (4.6) |
see [10,(80)].
We next consider the following elliptic problem:
{−Δu3=g|DRin D_R ,∂u3∂ν=0on ∂Ω,u3=0on ∂B8R(0). | (4.7) |
By Theorem 3.1, we know that (4.7) admits a unique solution u3∈H1+sp(DR) provided that p satisfies (1.1) with some ε∈(0,1/4] and s satisfies 0≤s<1/p. Notice that we may take ε=1/4 if ∂Ω∈C1. Denote by V0 the solution operator to (4.7), i.e., u3=V0(g). From Theorem 3.1, we see that V0(g) solves
⟨∇V0(g),∇φ⟩DR=⟨g,φ⟩DRfor every φ∈˙H1p′(DR) |
possessing the estimate
‖V0(g)‖H1+sp(DR)≤C‖f‖Lp(DR). | (4.8) |
Let V∞,0 be V∞,0(g)=V∞(g)+cg with a constant cg such that
∫6R≤|x|≤7R(V∞(g)+cg)dx=0. |
Furthermore, V∞,0(g) verifies
‖V∞,0(g)‖Lp(B9R(0))≤CR‖g‖Lp(R2;R2). |
For every g∈Hp(R2), let U2(f)=ψ0V0(g)+ψ∞V∞,0(g). As noted before, for all smooth functions h defined on R2, the term hV0(g) is regarded as a function that is extended by zero to all of R2. Clearly, there holds
∫6R≤|x|≤7RU2(g)dx=0 | (4.9) |
since there holds U2(g)=V∞,0(g) for 6R≤|x|≤7R. Note that it follows from (4.6) and (4.8) that
‖U2(g)‖Lp(DR)+‖∇U2(g)‖Hsp(Ω)≤C‖g‖Lp(Ω). | (4.10) |
We also see that U2(g) satisfies
⟨∇U2(g),∇φ⟩Ω=⟨g+R2(g),φ⟩Ωfor every φ∈˙H1p′(Ω) | (4.11) |
with
R2(g):=−2(∇ψ0)⋅∇(V0(g)−V∞,0(g))−(Δψ0)(V0(g)−V∞,0(g)). | (4.12) |
Similarly to (4.2), we set
R2(g):=∇(ψ∞V0(g))+∇(ψ0V∞,0(g)), | (4.13) |
so that there holds
⟨R2(g),φ⟩R2=⟨g,φ⟩R2−⟨R2(g),∇φ⟩R2for every φ∈˙H1p′(R2). | (4.14) |
Indeed, by
R2(g)=2(∇ψ∞)⋅∇V0(g)+(Δψ∞)V0(g)+2(∇ψ0)⋅∇V∞,0(g)+(Δψ0)V∞,0(g)=div((∇ψ∞)V0(g)+ψ∞∇V0(g)+(∇ψ0)V∞,0(g)+ψ0∇V∞,0(g))−ψ∞ΔV0(g)−ψ0ΔV∞,0(g), |
we have
⟨R2(g),φ⟩R2=⟨g,φ⟩R2−⟨(∇ψ∞)V0(g)+ψ∞∇V0(g),∇φ⟩DR+⟨(∇ψ0)V∞,0(g)+ψ0∇V∞,0(g),∇φ⟩B8R(0), |
which yields the representation of R2(g). Since 1∈˙H1p′(R2) and g∈Hp(R2), the relation (4.14) gives ⟨R2(g),1⟩R2=0. In addition, we also deduce from (4.13) and (4.14) that there holds R2(g)∈Hp(R2). Thus, to construct the solution operator to (4.4), it remains to verify the invertibility of I+R2 on Hp(R2).
Step 3: Invertibility of I+R2
To show the invertibility of I+R2 on Hp(R2), we first note that R2 is a compact operator on Hp(R2). In fact, we see that supp(R2(g))⊂AR and
‖R2(g)‖Hsp(R2)≤C‖g‖Lp(R2), |
where the Kato-Ponce type inequality has been applied (cf. [3,Theorem 1]). Let (gj)j∈N be a bounded sequence in Hp(R2). Then we infer from the Rellich-Kondrachov theorem (cf. [1,Theorem 6.3]) that Hsp(R2) is compactly embedded into Lp(AR), and thus the operator R2 may be regarded as a compact operator from Hp(R2) into Lp(AR). Namely, there exists a subsequence (R2(gj(k)))k∈N⊂(R2(gj))j∈N such that
limk→∞‖R2(gj(k))−Rg‖Lp(AR)=0 | (4.15) |
with some Rg∈Lp(AR). Let ˜Rg be the zero extension of Rg to R2. Since supp(R2(gj(k)))⊂AR, it follows from the Poincaré-Wirtinger inequality that
|⟨R2(gj(k))−˜Rg,φ⟩R2|≤‖R2(gj(k))−Rg‖Lp(AR)‖φ−|AR|−1∫ARφdx‖Lp′(AR)≤‖R2(gj(k))−Rg‖Lp(AR)‖∇φ‖Lp′(R2;R2) | (4.16) |
for every φ∈˙H1p′(R2). Together with (4.15), we deduce that
limk→∞‖R2(gj(k))−˜Rg‖Lp(R2)∩˙H−1p(R2)=0. |
Notice that, similarly to (4.16), we have ˜Rg∈˙H−1p(R2). In addition, there holds
⟨˜Rg,1⟩R2=⟨Rg,1⟩AR=limk→∞⟨R2(gj(k)),1⟩AR=0 |
due to the construction of R2. Since supp(˜Rg)⊂AR, we observe ˜Rg∈Hp(R2). Therefore, the operator R2 is a compact operator from Hp(R2) into itself.
We next verify the following lemma.
Lemma 4.1. Assume that Ω⊂R2 is an exterior Lipschitz domain and p satisfies (1.1) with some 0<ε≤1/4 as well as p≥2. Let Hp(R2), R2, and U2 be as above. If g∈Hp(R2) satisfies (I+R2)g=0, then U2(g)=0 in Ω.
Proof. Let ω∈C∞c(R2;[0,1]) be a function such that ω(x)=1 for |x|≤1 and ω(x)=0 for |x|≥2. Set ωL(x)=ω(x/L). By (4.10), we have U2(g)∈H1+sp,loc(Ω), which implies ωLU2(g)∈˙H1p(Ω). Here, u∈H1+sp,loc(Ω) means u∈H1+sp(Ω′) for any bounded domain Ω′ with Ω′⊂Ω. Hence, the formula (4.11) together with (I+R2)g=0 yields
0=⟨(I+R2)g,ωLU2(g)⟩Ω=⟨∇U2(g),∇(ωLU2(g))⟩Ω=⟨ωL∇U2(g),∇U2(g)⟩Ω+⟨(∇ωL)⋅∇U2(g),U2(g)⟩Ω. | (4.17) |
Let L>10R. Then there holds supp(∇ψ∞)∩supp(∇ωL)=∅. Recalling the definition of U2, by (4.6), we obtain
|⟨(∇ωL)⋅∇U2(g),U2(g)⟩Ω|≤|⟨(∇ωL)⋅∇V∞(g),cg⟩R2|+|⟨(∇ωL)⋅∇V∞(g),V∞(g)⟩R2| |
≤CRLsupx∈R2|∇ω(x)|∫L≤|x|≤2L‖g‖Lp(R2)(|cg|+‖g‖Lp(R2)|x|−1)|x|−2dx, |
where a constant CR may depend on R but is independent of L. We then observe
limL→∞⟨(∇ωL)⋅∇U2(g),U2(g)⟩Ω=0. |
Thus, letting L→∞ in (4.17) implies that ‖∇U2(g)‖L2(Ω;R2)=0. Hence, U2(g) is a constant. In particular, we infer from (4.9) that U2(g)=0.
By Lemma 4.1 and the Fredholm alternative theorem, we may show the existence of the inverse of I+R2 on L(Hp(R2)) provided that p satisfies (1.1). Namely, we may show the following lemma.
Lemma 4.2. Assume that Ω⊂R2 is an exterior Lipschitz domain and p satisfies (1.1) with some 0<ε≤1/4. Let R2 be the operator defined by (4.12). Then the inverse of I+R2 on L(Hp(R2)) exists.
Proof. It suffices to show that the kernel of I+R2 is trivial. To this end, let g be an element of Hp(R2) such that (I+R2)g=0.
We first deal with the case p≥2. By Lemma 4.1, we have U2(g)=0. Recalling the definition of U2, there holds
ψ0V0(g)+(1−ψ0)V∞,0(g)=0in Ω. | (4.18) |
Notice that it follows from the definition of ψ0 that V∞,0(g)=0 for |x|≥5R and V0(g)=0 for x∈Ω∩B4R(0). Let
V={V0(g)for 4R<|x|≤8R,0for 0≤|x|≤4R. |
Since V0(g) is a solution to (4.7) and satisfies (4.8), it is clear that V solves
{−ΔV=g|B8R(0)in B8R(0),V=0on ∂B8R(0). | (4.19) |
Since V∞,0(g)=0 for |x|≥5R, we observe V∞,0(g)∈H1+sp(B8R(0)) as follows from (4.6). In addition, V∞,0(g) also solves (4.19). Thus, by the uniqueness of the solution to (4.19), we obtain V=V∞,0(g) in B8R(0), i.e., V0(g)=V∞,0(g) in Ω∩B8R.
By virtue of the relation (4.18), we have V∞,0(g)=0, and thus there holds g=ΔV∞,0(g)=0 in Ω. Recalling that supp(g)⊂AR, it is necessary to have g=0 in R2.
Concerning the remaining case p<2, we note that if u2∈˙H1p(Ω) satisfies (4.4) with g=0, then u2 is a constant. In fact, for any f∈Lp′(Ω;R2), let U2∈˙H1p′(Ω) be a solution to
⟨∇U2,∇φ⟩Ω=⟨f,∇φ⟩Ωfor every φ∈˙H1p(Ω). |
Then, the aforementioned argument ensures the existence of U2 since the inverse of I+R2 on L(Hp(R2)) exists. From the assumption, we know that u2∈˙H1p(Ω) fulfills ⟨∇u2,∇φ⟩Ω=0 for every φ∈˙H1p′(Ω), and hence there holds 0=⟨∇U2,∇u2⟩Ω=⟨f,∇u2⟩Ω. Then we deduce that u2 is a constant since f∈Lp′(Ω;R2) is arbitrary.
Since (I+R2)g=0, it follows from (4.11) that U2(g)∈˙H1p(Ω) solves
⟨∇U2(g),∇φ⟩Ω=0for every φ∈˙H1p′(Ω). |
As seen before, we deduce that U2(g) is a constant. In particular, U2(g)=0 as follows from (4.9). Then, mimicking the argument as in the case 2≤p, we may show the invertibility of I+R2 on L(Hp(R2)).
From Lemma 4.2, we may construct the solution operator to (4.4) with the desired estimate (4.5). Then, as noted before (cf. Step 2 in this section), we see that U1(f)−U2(R1(f)) satisfies (3.5) with the desired estimate. Hence, it remains to verify the uniqueness of the solution to (3.5), but this may be proved along the same line as in the latter part of the proof of Lemma 4.2. Thus, the proof of Theorem 1.2 is complete. Namely, Theorem 1.1 has been proved.
For an exterior Lipschitz domain Ω⊂R2, we have proved the Helmholtz decomposition of the vector fields in Lp(Ω;R2) provided that p satisfies |1/p−1/2|<1/4+ε with some constant ε=ε(Ω)∈(0,1/4]. In particular, it is allowed to take ε=1/4 if ∂Ω∈C1. We have presented a new proof of the Helmholtz decomposition of the vector fields for two-dimensional exterior domains, which is different from the previous approaches of Miyakawa [9] as well as Simader and Sohr [11].
The author declares he has not used Artificial Intelligence (AI) tools in the creation of this article.
The author was partially supported by JSPS KAKENHI Grant Number 21K13826.
The author declares that he has no conflict of interest.
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