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Research article Special Issues

Some results on free boundary problems of incompressible ideal magnetohydrodynamics equations

  • Received: 31 August 2021 Revised: 01 November 2021 Accepted: 06 November 2021 Published: 18 January 2022
  • We survey some recent results related to free boundary problems of incompressible ideal magnetohydrodynamics equations, and present the main ideas in the proofs of the ill-posedness in 2D when the Taylor sign condition is violated given [1], and the well-posedness of a linearized problem given in [2] in general n-dimensions (n2) when the Taylor sign condition is satisfied and the free boundaries are closed.

    Citation: Chengchun Hao, Tao Luo. Some results on free boundary problems of incompressible ideal magnetohydrodynamics equations[J]. Electronic Research Archive, 2022, 30(2): 404-424. doi: 10.3934/era.2022021

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  • We survey some recent results related to free boundary problems of incompressible ideal magnetohydrodynamics equations, and present the main ideas in the proofs of the ill-posedness in 2D when the Taylor sign condition is violated given [1], and the well-posedness of a linearized problem given in [2] in general n-dimensions (n2) when the Taylor sign condition is satisfied and the free boundaries are closed.



    Let C denote the complex plane and Cn the n-dimensional complex Euclidean space with an inner product defined as z,w=nj=1zj¯wj. Let B(a,r)={zCn:|za|<r} be the open ball of Cn. In particular, the open unit ball is defined as B=B(0,1).

    Let H(B) denote the set of all holomorphic functions on B and S(B) the set of all holomorphic self-mappings of B. For given φS(B) and uH(B), the weighted composition operator on or between some subspaces of H(B) is defined by

    Wu,φf(z)=u(z)f(φ(z)).

    If u1, then Wu,φ is reduced to the composition operator usually denoted by Cφ. If φ(z)=z, then Wu,φ is reduced to the multiplication operator usually denoted by Mu. Since Wu,φ=MuCφ, Wu,φ can be regarded as the product of Mu and Cφ.

    If n=1, B becomes the open unit disk in C usually denoted by D. Let Dm be the mth differentiation operator on H(D), that is,

    Dmf(z)=f(m)(z),

    where f(0)=f. D1 denotes the classical differentiation operator denoted by D. As expected, there has been some considerable interest in investigating products of differentiation and other related operators. For example, the most common products DCφ and CφD were extensively studied in [1,10,11,12,13,23,25,26], and the products

    MuCφD,CφMuD,MuDCφ,CφDMu,DMuCφ,DCφMu (1.1)

    were also extensively studied in [14,18,22,27]. Following the study of the operators in (1.1), people naturally extend to study the operators (see [5,6,30])

    MuCφDm,CφMuDm,MuDmCφ,CφDmMu,DmMuCφ,DmCφMu.

    Other examples of products involving differentiation operators can be found in [7,8,19,32] and the related references.

    As studying on the unit disk becomes more mature, people begin to become interested in exploring related properties on the unit ball. One method for extending the differentiation operator to Cn is the radial derivative operator

    f(z)=nj=1zjfzj(z).

    Naturally, replacing D by in (1.1), we obtain the following operators

    MuCφ,CφMu,MuCφ,CφMu,MuCφ,CφMu. (1.2)

    Recently, these operators have been studied in [31]. Other operators involving radial derivative operators have been studied in [21,33,34].

    Interestingly, the radial derivative operator can be defined iteratively, namely, mf can be defined as mf=(m1f). Similarly, using the radial derivative operator can yield the related operators

    MuCφm,CφMum,MumCφ,CφmMu,mMuCφ,mCφMu. (1.3)

    Clearly, the operators in (1.3) are more complex than those in (1.2). Since CφMum=MuφCφm, the operator MuCφm can be regarded as the simplest one in (1.3) which was first studied and denoted as mu,φ in [24]. Recently, it has been studied again because people need to obtain more properties about spaces to characterize its properties (see [29]).

    To reconsider the operator CφmMu, people find the fact

    CφmMu=mi=0Cimi(miu)φ,φ. (1.4)

    Motivated by (1.4), people directly studied the sum operator (see [2,28])

    Smu,φ=mi=0MuiCφi,

    where uiH(B), i=¯0,m, and φS(B). Particularly, if we set u0um10 and um=u, then Smu,φ=MuCφm; if we set u0um10 and um=uφ, then Smu,φ=CφMum. In [28], Stević et al. studied the operators Smu,φ from Hardy spaces to weighted-type spaces on the unit ball and obtained the following results.

    Theorem A. Let mN, ujH(B), j=¯0,m, φS(B), and μ a weight function on B. Then, the operator Smu,φ:HpHμ is bounded and

    supzBμ(z)|uj(φ(z))||φ(z)|<+,j=¯1,m, (1.5)

    if and only if

    I0=supzBμ(z)|u0(z)|(1|φ(z)|2)np<+

    and

    Ij=supzBμ(z)|uj(z)||φ(z)|(1|φ(z)|2)np+j<+,j=¯1,m.

    Theorem B. Let mN, ujH(B), j=¯0,m, φS(B), and μ a weight function on B. Then, the operator Smu,φ:HpHμ is compact if and only if it is bounded,

    lim|φ(z)|1μ(z)|u0(z)|(1|φ(z)|2)np=0

    and

    lim|φ(z)|1μ(z)|uj(z)||φ(z)|(1|φ(z)|2)np+j=0,j=¯1,m.

    It must be mentioned that we find that the necessity of Theorem A requires (1.5) to hold. Inspired by [2,28], here we use a new method and technique without (1.5) to study the sum operator Smu,φ from logarithmic Bergman-type space to weighted-type space on the unit ball. To this end, we need to introduce the well-known Bell polynomial (see [3])

    Bm,k(x1,x2,,xmk+1)=m!mk1i=1ji!mk1i=1(xii!)ji,

    where all non-negative integer sequences j1, j2,,jmk+1 satisfy

    mk+1i=1ji=kandmk+1i=1iji=m.

    In particular, when k=0, one can get B0,0=1 and Bm,0=0 for any mN. When k=1, one can get Bi,1=xi. When m=k=i, Bi,i=xi1 holds.

    In this section, we need to introduce logarithmic Bergman-type space and weighted-type space. Here, a bounded positive continuous function on B is called a weight. For a weight μ, the weighted-type space Hμ consists of all fH(B) such that

    fHμ=supzBμ(z)|f(z)|<+.

    With the norm Hμ, Hμ becomes a Banach space. In particular, if μ(z)=(1|z|2)σ(σ>0), the space Hμ is called classical weighted-type space usually denoted by Hσ. If μ1, then space Hμ becomes the bounded holomorphic function space usually denoted by H.

    Next, we need to present the logarithmic Bergman-type space on B (see [4] for the unit disk case). Let dv be the standardized Lebesgue measure on B. The logarithmic Bergman-type space Apwγ,δ consists of all fH(B) such that

    fpApwγ,δ=B|f(z)|pwγ,δ(z)dv(z)<+,

    where 1<γ<+, δ0, 0<p<+ and wγ,δ(z) is defined by

    wγ,δ(z)=(log1|z|)γ[log(11log|z|)]δ.

    When p1, Apwγ,δ is a Banach space. While 0<p<1, it is a Fréchet space with the translation invariant metric ρ(f,g)=fgpApωγ,δ.

    Let φS(B), 0r<1, 0γ<, δ0, and aB{φ(0)}. The generalized counting functions are defined as

    Nφ,γ,δ(r,a)=zj(a)φ1(a)wγ,δ(zj(a)r)

    where |zj(a)|<r, counting multiplicities, and

    Nφ,γ,δ(a)=Nφ,γ,δ(1,a)=zj(a)φ1(a)wγ,δ(zj(a)).

    If φS(D), then the function Nφ,γ,δ has the integral expression: For 1γ<+ and δ0, there is a positive function F(t) satisfying

    Nφ,γ,δ(r,u)=r0F(t)Nφ,1(t,u)dt,r(0,1),uφ(0).

    When φS(D) and δ=0, the generalized counting functions become the common counting functions. Namely,

    Nφ,γ(r,a)=zφ1(a),|z|<r(logr|z|)γ,

    and

    Nφ,γ(a)=Nφ,γ(1,a)=zφ1(a)(log1|z|)γ.

    In [17], Shapiro used the function Nφ,γ(1,a) to characterize the compact composition operators on the weighted Bergman space.

    Let X and Y be two topological spaces induced by the translation invariant metrics dX and dY, respectively. A linear operator T:XY is called bounded if there is a positive number K such that

    dY(Tf,0)KdX(f,0)

    for all fX. The operator T:XY is called compact if it maps bounded sets into relatively compact sets.

    In this paper, j=¯k,l is used to represent j=k,...,l, where k,lN0 and kl. Positive numbers are denoted by C, and they may vary in different situations. The notation ab (resp. ab) means that there is a positive number C such that aCb (resp. aCb). When ab and ba, we write ab.

    In this section, we obtain some properties on the logarithmic Bergman-type space. First, we have the following point-evaluation estimate for the functions in the space.

    Theorem 3.1. Let 1<γ<+, δ0, 0<p<+ and 0<r<1. Then, there exists a positive number C=C(γ,δ,p,r) independent of zK={zB:|z|>r} and fApwγ,δ such that

    |f(z)|C(1|z|2)γ+n+1p[log(11log|z|)]δpfApwγ,δ. (3.1)

    Proof. Let zB. By applying the subharmonicity of the function |f|p to Euclidean ball B(z,r) and using Lemma 1.23 in [35], we have

    |f(z)|p1v(B(z,r))B(z,r)|f(w)|pdv(w)C1,r(1|z|2)n+1B(z,r)|f(w)|pdv(w). (3.2)

    Since r<|z|<1 and 1|w|21|z|2, we have

    log1|w|1|w|1|z|log1|z| (3.3)

    and

    log(1log1|w|)log(1log1|z|). (3.4)

    From (3.3) and (3.4), it follows that there is a positive constant C2,r such that wγ,δ(z)C2,rwγ,δ(w) for all wB(z,r). From this and (3.2), we have

    |f(z)|pC1,rC2,r(1|z|2)n+1wγ,δ(z)B(z,r)|f(w)|pwγ,δ(w)dv(w)C1,rC2,r(1|z|2)n+1wγ,δ(z)fpApwγ,δ. (3.5)

    From (3.5) and the fact log1|z|1|z|1|z|2, the following inequality is right with a fixed constant C3,r

    |f(z)|pC1,rC2,rC3,r(1|z|2)n+1+γ[log(11log|z|)]δfpApwγ,δ.

    Let C=C1,rC2,rC3,rp. Then the proof is end.

    Theorem 3.2. Let mN, 1<γ<+, δ0, 0<p<+ and 0<r<1. Then, there exists a positive constant Cm=C(γ,δ,p,r,m) independent of zK and fApwγ,δ such that

    |mf(z)zi1zi2zim|Cm(1|z|2)γ+n+1p+m[log(11log|z|)]δpfApwγ,δ. (3.6)

    Proof. First, we prove the case of m=1. By the definition of the gradient and the Cauchy's inequality, we get

    |f(z)zi||f(z)|˜C1supwB(z,q(1|z|))|f(w)|1|z|, (3.7)

    where i=¯1,n. By using the relations

    1|z|1|z|22(1|z|),
    (1q)(1|z|)1|w|(q+1)(1|z|),

    and

    log(11log|z|)log(11log|w|),

    we obtain the following formula

    |f(w)|˘C1(1|z|2)γ+n+1p[log(11log|z|)]δpfApwγ,δ

    for any wB(z,q(1|z|)). Then,

    supwB(z,q(1|z|))|f(w)|˘C1(1|z|2)γ+n+1p[log(11log|z|)]δpfApwγ,δ.

    From (3.1) and (3.2), it follows that

    |f(z)zi|ˆC1(1|z|2)γ+n+1p+1[log(11log|z|)]δpfApwγ,δ. (3.8)

    Hence, the proof is completed for the case of m=1.

    We will use the mathematical induction to complete the proof. Assume that (3.6) holds for m<a. For convenience, let g(z)=a1f(z)zi1zi2zia1. By applying (3.7) to the function g, we obtain

    |g(z)zi|˜C1supwB(z,q(1|z|))|g(w)|1|z|. (3.9)

    According to the assumption, the function g satisfies

    |g(z)|ˆCa1(1|z|2)γ+n+1p+a1[log(11log|z|)]δpfApwγ,δ.

    By using (3.8), the following formula is also obtained

    |g(z)zi|ˆCa(1|z|2)γ+n+1p+a[log(11log|z|)]δpfApwγ,δ.

    This shows that (3.6) holds for m=a. The proof is end.

    As an application of Theorems 3.1 and 3.2, we give the estimate in z=0 for the functions in Apωγ,δ.

    Corollary 3.1. Let 1<γ<+, δ0, 0<p<+, and 0<r<2/3. Then, for all fApwγ,δ, it follows that

    |f(0)|C(1r2)γ+n+1p[log(11logr)]δpfApwγ,δ, (3.10)

    and

    |mf(0)zl1zlm|Cm(1r2)γ+n+1p+m[log(11logr)]δpfApwγ,δ, (3.11)

    where constants C and Cm are defined in Theorems 3.1 and 3.2, respectively.

    Proof. For fApwγ,δ, from Theorem 3.1 and the maximum module theorem, we have

    |f(0)|max|z|=r|f(z)|C(1r2)γ+n+1p[log(11logr)]δpfApwγ,δ,

    which implies that (3.10) holds. By using the similar method, we also have that (3.11) holds.

    Next, we give an equivalent norm in Apwγ,δ, which extends Lemma 3.2 in [4] to B.

    Theorem 3.3. Let r0[0,1). Then, for every fApwγ,δ, it follows that

    fpApwγ,δBr0B|f(z)|pwγ,δ(z)dv(z). (3.12)

    Proof. If r0=0, then it is obvious. So, we assume that r0(0,1). Integration in polar coordinates, we have

    fpApwγ,δ=2n10wγ,δ(r)r2n1drS|f(rζ)|pdσ(ζ).

    Put

    A(r)=wγ,δ(r)r2n1andM(r,f)=S|f(rζ)|pdσ(ζ).

    Then it is represented that

    fpApwγ,δr00+1r0M(r,f)A(r)dr. (3.13)

    Since M(r,f) is increasing, A(r) is positive and continuous in r on (0,1) and

    limr0A(r)=limx+xγ[log(1+1x)]δe(2n1)x=limx+xγδe(2n1)x=0,

    that is, there is a constant ε>0(ε<r0) such that A(r)<A(ε) for r(0,ε). Then we have

    r00M(r,f)A(r)dr2r01r0maxεrr0A(r)1+r02r0M(r,f)dr2r01r0maxεrr0A(r)minr0r1+r02A(r)1+r02r0M(r,f)A(r)dr1r0M(r,f)A(r)dr. (3.14)

    From (3.13) and (3.14), we obtain the inequality

    fpApwγ,δ1r0M(r,f)A(r)dr.

    The inequality reverse to this is obvious. The asymptotic relationship (3.12) follows, as desired.

    The following integral estimate is an extension of Lemma 3.4 in [4]. The proof is similar, but we still present it for completeness.

    Lemma 3.1. Let 1<γ<+, δ0, β>γδ and 0<r<1. Then, for each fixed wB with |w|>r,

    Bωγ,δ(z)|1z,w|n+β+1dv(z)1(1|w|)βγ[log(11log|w|)]δ.

    Proof. Fix |w| with |w|>r0 (0<r0<1). It is easy to see that

    log1r1rforr0r<1. (3.15)

    By applying Theorem 3.3 with

    fw(z)=1(1z,w)n+β+1

    and using (3.15), the formula of integration in polar coordinates gives

    B1|1z,w|n+β+1ωγ,δ(z)dv(z)1r0M(r,fw)(1r)γ[log(11logr)]δr2n1dr. (3.16)

    By Proposition 1.4.10 in [15], we have

    M(r,fw)1(1r2|w|2)β+1. (3.17)

    From (3.16) and (3.17), we have

    B1|1z,w|β+2nωγ,δ(z)dv(z)1r01(1r2|w|2)β+1(1r)γ[log(11logr)]δr2n1dr1r01(1r|w|)β+1(1r)γ[log(11logr)]δr2n1dr|w|r01(1r|w|)β+1(1r)γ[log(11logr)]δr2n1dr+1|w|1(1r|w|)β+1(1r)γ[log(11logr)]δr2n1dr=I1+I2.

    Since [log(11logr)]δ is decreasing in r on [|w|,1], we have

    I2=1|w|1(1r|w|)β+1(1r)γ[log(11logr)]δr2n1dr1(1|w|)β+1[log(11log|w|)]δ1|w|(1r)γdr1(1|w|)βγ[log(11log|w|)]δ. (3.18)

    On the other hand, we obtain

    I1=|w|r01(1r|w|)β+1(1r)γ[log(11logr)]δr2n1dr|w|r0(1r)γβ1(log21r)δdr.

    If δ=0 and β>γ, then we have

    I1(0)(1|w|)γβ.

    If δ0, then integration by parts gives

    I1(δ)=1γβ(1|w|)γβ(log21|w|)δ+1γβ(1r0)γβ(log21r0)δ+δγβI1(δ1).

    Since δ<0, γβ<0 and

    (log21r)δ1(log21r)δforr0<r<|w|<1,

    we have

    I1(δ)1γβ(1|w|)γβ(log21|w|)δ+δγβI1(δ)

    and from this follows

    I1(δ)(1|w|)γβ(log21|w|)δ(1|w|)γβ[log(11log|w|)]δ

    provided γβδ<0. The proof is finished.

    The following gives an important test function in Apwγ,δ.

    Theorem 3.4. Let 1<γ<+, δ0, 0<p<+ and 0<r<1. Then, for each t0 and wB with |w|>r, the following function is in Apwγ,δ

    fw,t(z)=[log(11log|w|)]δp(1|w|2)δp+t+1(1z,w)γδ+n+1p+t+1.

    Moreover,

    sup{wB:|w|>r}fw,tApwγ,δ1.

    Proof. By Lemma 3.1 and a direct calculation, we have

    fw,tpApwγ,δ=B|[log(11log|w|)]δp(1|w|2)δp+t+1(1z,w)γδ+n+1p+t+1|pwγ,δ(z)dA(z)=(1|w|2)p(t+1)δ[log(11log|w|)]δ×B1|1z,w|γδ+p(t+1)+n+1wγ,δ(z)dA(z)1.

    The proof is finished.

    In this section, for simplicity, we define

    Bi,j(φ(z))=Bi,j(φ(z),φ(z),,φ(z)).

    In order to characterize the compactness of the operator Smu,φ:Apwγ,δHμ, we need the following lemma. It can be proved similar to that in [16], so we omit here.

    Lemma 4.1. Let 1<γ<+, δ0, 0<p<+, mN, ujH(B), j=¯0,m, and φS(B). Then, the bounded operator Smu,φ:Apwγ,δHμ is compact if and only if for every bounded sequence {fk}kN in Apwγ,δ such that fk0 uniformly on any compact subset of B as k, it follows that

    limkSmu,φfkHμ=0.

    The following result was obtained in [24].

    Lemma 4.2. Let s0, wB and

    gw,s(z)=1(1z,w)s,zB.

    Then,

    kgw,s(z)=sPk(z,w)(1z,w)s+k,

    where Pk(w)=sk1wk+p(k)k1(s)wk1+...+p(k)2(s)w2+w, and p(k)j(s), j=¯2,k1, are nonnegative polynomials for s.

    We also need the following result obtained in [20].

    Lemma 4.3. Let s>0, wB and

    gw,s(z)=1(1z,w)s,zB.

    Then,

    kgw,s(z)=kt=1a(k)t(t1j=0(s+j))z,wt(1z,w)s+t,

    where the sequences (a(k)t)t¯1,k, kN, are defined by the relations

    a(k)k=a(k)1=1

    for kN and

    a(k)t=ta(k1)t+a(k1)t1

    for 2tk1,k3.

    The final lemma of this section was obtained in [24].

    Lemma 4.4. If a>0, then

    Dn(a)=|111aa+1a+n1a(a+1)(a+1)(a+2)(a+n1)(a+n)n2k=0(a+k)n2k=0(a+k+1)n2k=0(a+k+n1)|=n1k=1k!.

    Theorem 4.1. Let 1<γ<+, δ0, 0<p<+, mN, ujH(B), j=¯0,m, and φS(B). Then, the operator Smu,φ:Apwγ,δHμ is bounded if and only if

    M0:=supzBμ(z)|u0(z)|(1|φ(z)|2)γ+n+1p[log(11log|φ(z)|)]δp<+ (4.1)

    and

    Mj:=supzBμ(z)|mi=jui(z)Bi,j(φ(z))|(1|φ(z)|2)γ+n+1p+j[log(11log|φ(z)|)]δp<+ (4.2)

    for j=¯1,m.

    Moreover, if the operator Smu,φ:Apwγ,δHμ is bounded, then

    Smu,φApwγ,δHμmj=0Mj. (4.3)

    Proof. Suppose that (4.1) and (4.2) hold. From Theorem 3.1, Theorem 3.2, and some easy calculations, it follows that

    μ(z)|mi=0ui(z)if(φ(z))|μ(z)mi=0|ui(z)||if(φ(z))|=μ(z)|u0(z)||f(φ(z))|+μ(z)|mi=1ij=1(ui(z)nl1=1nlj=1(jfzl1zl2zlj(φ(z))k1,,kjC(i)k1,,kjjt=1φlt(z)))|=μ(z)|u0(z)f(φ(z))|+μ(z)|mj=1mi=j(ui(z)nl1=1nlj=1(jfzl1zl2zlj(φ(z))k1,,kjC(i)k1,,kjjt=1φlt(z)))|μ(z)|u0(z)|(1|φ(z)|2)γ+n+1p[log(11log|φ(z)|)]δpfApwγ,δ+mj=1μ(z)|mi=jui(z)Bi,j(φ(z))|(1|φ(z)|2)γ+n+1p+j[log(11log|φ(z)|)]δpfApwγ,δ=M0fApwγ,δ+mj=1MjfApwγ,δ. (4.4)

    By taking the supremum in inequality (4.4) over the unit ball in the space Apwγ,δ, and using (4.1) and (4.2), we obtain that the operator Smu,φ:Apwγ,δHμ is bounded. Moreover, we have

    Smu,φApwγ,δHμCmj=0Mj, (4.5)

    where C is a positive constant.

    Assume that the operator Smu,φ:Apwγ,δHμ is bounded. Then there exists a positive constant C such that

    Smu,φfHμCfApwγ,δ (4.6)

    for any fApwγ,δ. First, we can take f(z)=1Apwγ,δ, then one has that

    supzBμ(z)|u0(z)|<+. (4.7)

    Similarly, take fk(z)=zjkApwγ,δ, k=¯1,n and j=¯1,m, by (4.7), then

    μ(z)|u0(z)φk(z)j+mi=j(ui(z)Bi,j(φk(z))))|<+ (4.8)

    for any j{1,2,,m}. Since φ(z)B, we have |φ(z)|1. So, one can use the triangle inequality (4.7) and (4.8), the following inequality is true

    supzBμ(z)|mi=jui(z)Bi,j(φ(z))|<+. (4.9)

    Let wB and dk=γ+n+1p+k. For any j{1,2,,m} and constants ck=c(j)k, k=¯0,m, let

    h(j)w(z)=mk=0c(j)kfw,k(z), (4.10)

    where fw,k is defined in Theorem 3.4. Then, by Theorem 3.4, we have

    Lj=supwBh(j)wApwγ,δ<+. (4.11)

    From (4.6), (4.11), and some easy calculations, it follows that

    LjSmu,φApwγ,δHμSmu,φh(j)φ(w)Hμ=supzBμ(z)|mi=0u0(z)h(j)φ(w)(φ(z))|μ(w)|u0(w)h(j)φ(w)(φ(w))+mi=1(ui(w)ih(j)φ(w)(φ(w)))|=μ(w)|u0(w)h(j)φ(w)(φ(w))+mi=1ui(w)mk=0c(j)kfφ(w),k(φ(w))|=μ(w)|u0(w)c0+c1++cm(1|φ(z)|2)γ+n+1p+mi=1ui(w)Bi,1(φ(w)),φ(w)(d0c0++dmcm)(1|φ(w)|2)γ+n+1p+1++mi=jui(w)Bi,j(φ(w)),φ(w)j(d0dj1c0++dmdm+j1cm)(1|φ(w)|2)γ+n+1p+j++um(w)Bm,m(φ(w)),φ(w)m(d0dm1c0++dmd2m1cm)(1|φ(w)|2)γ+n+1p+m|[log(11log|φ(w)|)]δp. (4.12)

    Since , , by Lemma 4.4, we have the following linear equations

    (4.13)

    From (4.12) and (4.13), we have

    (4.14)

    On the other hand, from (4.9), we have

    (4.15)

    From (4.14) and (4.15), we get that (4.2) holds for .

    For constants , , let

    (4.16)

    By Theorem 3.4, we know that . From this, (4.12), (4.13) and Lemma 4.4, we get

    So, we have . Moreover, we have

    (4.17)

    From (4.5) and (4.17), we obtain (4.3). The proof is completed.

    From Theorem 4.1 and (1.4), we obtain the following result.

    Corollary 4.1. Let , , and is a weight function on . Then, the operator is bounded if and only if

    and

    for .

    Moreover, if the operator is bounded, then

    Theorem 4.2. Let , , , , , , and . Then, the operator is compact if and only if the operator is bounded,

    (4.18)

    for , and

    (4.19)

    Proof. Assume that the operator is compact. It is obvious that the operator is bounded.

    If , then it is clear that (4.18) and (4.19) are true. So, we suppose that . Let be a sequence in such that

    where are defined in (4.10) for a fixed . Then, it follows that uniformly on any compact subset of as . Hence, by Lemma 4.1, we have

    Then, we can find sufficiently large such that

    (4.20)

    If , then (4.20) is true.

    Now, we discuss the case of . Let , where is defined in (4.16). Then, we also have that and uniformly on any compact subset of as . Hence, by Lemma 4.1, one has that

    (4.21)

    Then, by (4.21), we know that (4.18) is true.

    Now, assume that is bounded, (4.18) and (4.19) are true. One has that

    (4.22)

    and

    (4.23)

    for any . By (4.18) and (4.19), for arbitrary , there is a , for any such that

    (4.24)

    and

    (4.25)

    Assume that is a sequence such that and uniformly on any compact subset of as . Then by Theorem 3.1, Theorem 3.2 and (4.22)–(4.25), one has that

    (4.26)

    Since uniformly on any compact subset of as . By Cauchy's estimates, we also have that uniformly on any compact subset of as . From this and using the fact that is a compact subset of , by letting in inequality (4.26), one get that

    Since is an arbitrary positive number, it follows that

    By Lemma 4.1, the operator is compact.

    As before, we also have the following result.

    Corollary 4.2. Let , , and is a weight function on . Then, the operators is compact if and only if the operator is bounded,

    and

    for .

    In this paper, we study and obtain some properties about the logarithmic Bergman-type space on the unit ball. As some applications, we completely characterized the boundedness and compactness of the operator

    from the logarithmic Bergman-type space to the weighted-type space on the unit ball. Here, one thing should be pointed out is that we use a new method and technique to characterize the boundedness of such operators without the condition (1.5), which perhaps is the special flavour in this paper.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This work was supported by Sichuan Science and Technology Program (2022ZYD0010) and the Graduate Student Innovation Foundation (Y2022193).

    The authors declare that they have no competing interests.



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