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The Helmholtz decomposition of vector fields for two-dimensional exterior Lipschitz domains

  • 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/p1/2|<1/4+ε with some constant ε=ε(Ω)(0,1/4], where it is allowed to take ε=1/4 if ΩC1.

    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/p1/2|<1/4+ε with some constant ε=ε(Ω)(0,1/4], where it is allowed to take ε=1/4 if ΩC1.



    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

    |1p12|<14+ε (1.1)

    with some constant ε=ε(Ω)(0,1/4], where it is allowed to take ε=1/4 if ΩC1. Let Lp,σ(Ω) be the subspace of functions f in Lp(Ω;R2) such that Ωfφdx=0 for every φ˙H1p(Ω). Notice that Lp,σ(Ω) is the closure of Cc,σ(Ω):={ϕCc(Ω;R2):divϕ=0} in Lp(Ω;R2), see [4,Proposition 2.5]. In addition, let Gp(Ω):=Imp=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 ΩR2 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 ΩC1, 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/p1/2|<1/6+ε with some constant ε(Ω)>0 provided that ΩRd with d3. 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 fLp(Ω;R2), consider the weak Neumann problem

    u,φΩ=f,φΩfor every φ˙H1p(Ω). (1.3)

    The present paper aims to prove the following theorem.

    Theorem 1.2. Let ΩR2 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 fLp(Ω) Problem (1.3) admits a solution u˙H1p(Ω) subject to the estimate

    uLp(Ω;R2)CfLp(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 ΩC1, 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, ΩRd (d3), 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/p1/2|<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 GR2, let u,vG=Gu(x)v(x)dx. For vector-valued functions U=(U1,U2) and V=(V1,V2) defined on GR2, let U,VG=j=1,2GUj(x)Vj(x)dx. For R>0, let BR(0)={xR2|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 Cc(G;X) be the set of all C-functions on R2 whose supports are compact and contained in DR2. For 1p and GR2, let Lp(G;X) be the Lebesgue space equipped with the norm Lp(G;X). For 1p< and sR, 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 uD(G;X)=(Cc(G;X)) such that there exists vHsp(G;X) with v|G=u. Furthermore, the norm Hsp(G;X) is defined by the usual quotient norm:

    uHsp(G;X)=infvHsp(R2;X)

    where the infimum is taken over all vHsp(R2;X) such that its restriction v|G to G coincides in D(G;X) with u. In particular, if uHsp(G;X) vanishes on the boundary G then the space will be attached with the subscript 0, i.e., Hsp,0(G;X). For 1p< and sR, let ˙Hsp(R2;X) be the homogeneous Sobolev space equipped with the norm Hsp(R2;X). For 1p< and GR2, we also define

    ˙H1p(G):={[u]=u+R:uLp,loc(G)  and  uLp(G;R2)}

    with the norm ˙H1p(G)=Lp(G), where uLp,loc(G) means uLp(G) for any bounded domain G with GG. If X=R, we often write Lp(G)=Lp(G;R), Hsp(R2)=Hsp(R2;R), and ˙Hsp(R2)=˙H1p(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,uν=0on Σ,u=0on Γ. (3.1)

    Here, DR2 is a bounded Lipschitz domain with Ω=ΣΓ with Σ=DΓ and Σ. We suppose that there exist some constants 0<R1<R2 such that ΣBR1(0) and ΓR2¯BR2(0). Let

    Lp,0(D):={fLp(D):f,1D=0}.

    Then the result for (3.1) reads as follows.

    Theorem 3.1. Let DR2 be a bounded Lipschitz domain. Suppose that there exist some constants 0<R1<R2 such that ΣBR1(0) and ΓR2¯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 fLp,0(D) Problem (3.1) admits a solution uH1+sp(D) subject to the estimate

    uH1+sp(D)CfLp(D)

    with some positive constant C>0 which depends only on D, p, and s. Furthermore, if DC1, 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)={uH1+sp,0(D):ΔuLp(D)},ΔDp,s,wu=Δu (3.2)

    with 0s<1/p. Here, ΔuLp(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)={uH1+sp(D):vLp(D) s.t. φH1+sp,0(D):u,φD=v,φD},ΔNp,s,wu=v (3.3)

    with 0s<1/p. Notice that the Neumann boundary condition u/ν=0 on D is interpreted in the sense that

    Δu,φD=u,φDfor any φH1+sp(D).

    Mimicking the argument as in Sections 3–5 in [15], we may prove the following results.

    Lemma 3.2. Let DR2 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 DC1, then it is allowed to take εDir=1/4.

    Lemma 3.3. Let DR2 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 DC1, 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 DRd with d3, 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 DR2 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

    {λuΔu=fin D,u=0on D (3.4)

    as well as the resolvent problem for the Laplacian with Neumann boundary condition

    {λuΔu=fin D,uν=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

    λuLp(D)fLp(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

    uH1+sp(D)Cλu+fH1+sp(D)Cλu+fLp(D)C(λuLp(D)+fLp(D))CfLp(D)

    since H1+sp(D)Lp(D) due to 1+s<0. Hence, the solution uH1+sp,0(D) to (3.4) verifies

    λuLp(D)+uH1+sp(D)CfLp(D)for all λ>0

    provided that fLp(D). Likewise, we infer from Lemma 3.3 and the invertibility results in [8,Corollary 4.2] that there exists a unique solution uH1+sp(D) satisfying

    λuLp(D)+uH1+sp(D)CfLp(D)for all λ>0

    provided that fLp,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 0s<1/p, respectively. Assume that fLp,0(D). Then, for every λ>0, there exists a unique solution uDirH1+sp,0(D) to

    {λuDirΔuDir=fin D,uDir=0on ΣΓ.

    In addition, u satisfies the estimate

    λuDirLp(D)+uDirH1+sp(D)CfLp(D)for all λ>0. (3.6)

    By complex interpolation, we observe that for every t(0,s), there holds

    uDirH1+tp(D)Cλst1+sfLp(D)for all λ>0.

    Similarly, for every λ>0, there exists a unique solution uNeuH1+sp(D) to

    {λuNeuΔuNeu=fin D,uNeuν=0on ΣΓ.

    Moreover, u satisfies the estimate

    λuNeuLp(D)+uNeuH1+sp(D)CfLp(D)for all λ>0.

    By complex interpolation, we observe that for every t(0,s), there holds

    uNeuH1+tp(D)Cλst1+sfLp(D)for all λ>0. (3.7)

    Let ζCc(D;[0,1]) be a cut-off function such that ζ=1 if |x|R1+ϵ and ζ=0 if |x|R2ϵ, where ϵ:=(R1+R2)/3. Set v=ζuDir+(1ζ)uNeu. We see that v solves

    {λvΔv=f+R0fin D,vν=0,on Σ,v=0on Γ,

    where we have set

    R0f:=Δζ(uNeuuDir)+2ζ(uNeuuDir).

    From (3.6) and (3.7), there exists λ01 such that for all λλ0, we have

    R0fLp(D)12fLp(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

    {λuΔu=fin D,uν=0on Σ,u=0on Γ, (3.8)

    is solvable for all λλ0. Notice that the uniqueness of the solution follows from the duality argument. Indeed, suppose that uH1+sp(D) solves (3.8) with f vanishing in D. For any λλ0 and ϕCc(D), consider

    {λu0Δu0=ϕin D,u0ν=0on Σ,u0=0on Γ.

    Then we infer from the divergence theorem that

    u,ϕD=u,λu0Δu0D=λuΔu,u0D=0.

    Since ϕCc(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)={uH1+sp(D):uν=0on   Σ   and   u=0on   Γ}Ap,su=Δu

    with 0s<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 0λ<λ0. Since (3.8) may be written as (λIAp,s)u=f, we find that (3.8) and

    (I+(λ2λ0)Rp,s(λ0))(2λ0IAp,s)u=f,Rp,s(λ0):=(2λ0IAp,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+(λ2λ0)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+(λ2λ0)Rp,s(λ0). To see this, for any λ[0,λ0), take u1Ker(I+(λ2λ0)Rp,s(λ0)), i.e.,

    (I+(λ2λ0)Rp,s(λ0))u1=0for any u1Lp,0(D).

    From the definition of Rp,s(λ0), we see that u1D(Ap,s) and w solves

    {λu1Δu1=0in D,u1ν=0on Σ,u1=0on Γ. (3.9)

    We first deal with the case 2p<. In this case, we may assume u1D(A2,s) due to the boundedness of D. Using the divergence theorem, we deduce from (3.9) that

    0=λu12L2(D)+12D(u1)2L2(D),

    where we have set D(u1):=u1+[u1]. By the Korn inequality (cf. [14,Theorem A.4]), we obtain

    λu12L2(D)+1Cu12L2(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 2p<. 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 R2ΩBR(0) for any R>R0. Set DR:=ΩB8R(0) and AR:={xR22R|x|7R}.

    Step 1: Construction of U1

    Following the argument of Shibata [10,Section 3], we introduce cut-off functions as follows. Let ψ0,χ0Cc(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,χC(R2;[0,1]) be smooth functions defined by

    ψ=1ψ0(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,φR2=F,φR2for 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 FLp(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),1D=0. Notice that the Fourier multiplier theorem and the Poincaré-Wirtinger inequality yield

    U(F)Lp(R2;R2)+U(F)Lp(G)CFLp(R2;R2)

    for every bounded Lipschitz domain GR2. Given fLp(Ω;R2), define the linear operator U1 by

    U1(f)=χ0U0(ψ0f)+(1χ0)U(ψf).

    Here, ψf may be regarded as a function defined on R2 since ψf 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),φΩ=f,φΩ+R1(f),φΩfor every φ˙H1p(Ω)

    with

    R1(f):=2(χ0)(U0(ψ0f)U(ψf))(Δχ0)(U0(ψ0f)U(ψf)).

    Here, we have used the identity

    ψ0f,(χ0φ)Ω+ψf,(χφ)Ω=(ψ0+ψ)f,φΩ=f,φΩ,

    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(ψf))div((χ0)(U0(ψ0f)U(ψf))),

    we have

    R1(f),φR2=R1(f),φR2for every φ˙H1p(R2) (4.2)

    with

    R1(f):=ψ0f+(χ0)U0(ψ0f)+χ0U0(ψ0f)(χ0)U(ψf)χ0U(ψf). (4.3)

    In fact, it follows that

    R1(f),φR2=(U0(ψ0f)U(ψf)),(χ0)φR2+(χ0)(U0(ψ0f)U(ψf)),φDR=U0(ψ0f),(χ0φ)DR+U(ψf),(χ0φ)R2+χ0(U0(ψ0f)U(ψ0f)),φDR+(χ0)(U0(ψ0f)U(ψf)),φDR

    for any φ˙H1p(R2). Here, we have used the identity

    div((χ0)(U0(ψ0f)U(ψf))),φR2=(χ0)(U0(ψ0f)U(ψf)),φDR,

    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,φΩ,U(ψf),(χ0φ)R2=0,

    which yields the representation (4.3). Clearly, for any fLp(Ω;R2), we have

    R1(f)Lp(R2)CfLp(Ω;R2),R1(f)Lp(R2;R2)CfLp(Ω;R2).

    We also infer from (4.2) that

    R1(f)˙H1p(R2)CfLp(Ω;R2).

    In addition, we infer from 1˙H1p(R2) and (4.2) that R1(f),1R2=R1(f),0R2=0.

    For later, we introduce the function space

    Hp(R2)={gLp(R2)˙H1p(R2):supp(g)AR,g,1R2=0}.

    Clearly, from the aforementioned argument, we see that R1(f)Hp(R2) for any fLp(Ω;R2).

    Step 2: Construction of U2

    Given gHp(R2), we intend to construct the solution operator to

    u2,φΩ=g,φΩfor every φ˙H1p(Ω) (4.4)

    possessing the estimate

    u2Lp(Ω;R2)CgLp(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 gHp(R2). Let V be the operator defined by

    (Vf)(x)=R2E(xy)g(y)dy,

    where E(xy) stands for the fundamental solution of the Laplace equation:

    E(xy)=(2π)1log|xy|.

    By [10,(82)], the formula

    V(g),φR2=g,φR2for every φ˙H1p(R2)

    may be justified. Furthermore, for every gHp(R2), we have

    V(g)Lp(B9R(0))+sup|x|9R|x||V(g)|CgLp(R2),V(g)Lp(R2;R2)+sup|x|9R|x|2|V(g)|CgLp(R2),2V(g)Lp(R2;R4)CgLp(R2), (4.6)

    see [10,(80)].

    We next consider the following elliptic problem:

    {Δu3=g|DRin D_R ,u3ν=0on Ω,u3=0on B8R(0). (4.7)

    By Theorem 3.1, we know that (4.7) admits a unique solution u3H1+sp(DR) provided that p satisfies (1.1) with some ε(0,1/4] and s satisfies 0s<1/p. Notice that we may take ε=1/4 if ΩC1. 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),φDR=g,φDRfor every φ˙H1p(DR)

    possessing the estimate

    V0(g)H1+sp(DR)CfLp(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))CRgLp(R2;R2).

    For every gHp(R2), let U2(f)=ψ0V0(g)+ψV,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)CgLp(Ω). (4.10)

    We also see that U2(g) satisfies

    U2(g),φΩ=g+R2(g),φΩfor 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):=(ψV0(g))+(ψ0V,0(g)), (4.13)

    so that there holds

    R2(g),φR2=g,φR2R2(g),φR2for 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)+ψV0(g)+(ψ0)V,0(g)+ψ0V,0(g))ψΔV0(g)ψ0ΔV,0(g),

    we have

    R2(g),φR2=g,φR2(ψ)V0(g)+ψV0(g),φDR+(ψ0)V,0(g)+ψ0V,0(g),φB8R(0),

    which yields the representation of R2(g). Since 1˙H1p(R2) and gHp(R2), the relation (4.14) gives R2(g),1R2=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)CgLp(R2),

    where the Kato-Ponce type inequality has been applied (cf. [3,Theorem 1]). Let (gj)jN 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)))kN(R2(gj))jN such that

    limkR2(gj(k))RgLp(AR)=0 (4.15)

    with some RgLp(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,φR2|R2(gj(k))RgLp(AR)φ|AR|1ARφdxLp(AR)R2(gj(k))RgLp(AR)φLp(R2;R2) (4.16)

    for every φ˙H1p(R2). Together with (4.15), we deduce that

    limkR2(gj(k))˜RgLp(R2)˙H1p(R2)=0.

    Notice that, similarly to (4.16), we have ˜Rg˙H1p(R2). In addition, there holds

    ˜Rg,1R2=Rg,1AR=limkR2(gj(k)),1AR=0

    due to the construction of R2. Since supp(˜Rg)AR, we observe ˜RgHp(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 ΩR2 is an exterior Lipschitz domain and p satisfies (1.1) with some 0<ε1/4 as well as p2. Let Hp(R2), R2, and U2 be as above. If gHp(R2) satisfies (I+R2)g=0, then U2(g)=0 in Ω.

    Proof. Let ωCc(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, uH1+sp,loc(Ω) means uH1+sp(Ω) for any bounded domain Ω with ΩΩ. Hence, the formula (4.11) together with (I+R2)g=0 yields

    0=(I+R2)g,ωLU2(g)Ω=U2(g),(ωLU2(g))Ω=ωLU2(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),cgR2|+|(ωL)V(g),V(g)R2|
    CRLsupxR2|ω(x)|L|x|2LgLp(R2)(|cg|+gLp(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 ΩR2 is an exterior Lipschitz domain and p satisfies (1.1) with some 0<ε1/4. 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 p2. By Lemma 4.1, we have U2(g)=0. Recalling the definition of U2, there holds

    ψ0V0(g)+(1ψ0)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 xΩB4R(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 ΩB8R.

    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 fLp(Ω;R2), let U2˙H1p(Ω) be a solution to

    U2,φΩ=f,φΩfor 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,φΩ=0 for every φ˙H1p(Ω), and hence there holds 0=U2,u2Ω=f,u2Ω. Then we deduce that u2 is a constant since fLp(Ω;R2) is arbitrary.

    Since (I+R2)g=0, it follows from (4.11) that U2(g)˙H1p(Ω) solves

    U2(g),φΩ=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 2p, 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 ΩR2, we have proved the Helmholtz decomposition of the vector fields in Lp(Ω;R2) provided that p satisfies |1/p1/2|<1/4+ε with some constant ε=ε(Ω)(0,1/4]. In particular, it is allowed to take ε=1/4 if ΩC1. 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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