# Lipschitz Continuity for the Solutions of Triharmonic Equation

Authors: Zhou Yu; Xiao Bing
Journal: Mathematical Problems in Engineering
 (2019)

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## Abstract

LetD be the unit disk in the complex plane C and denote T=∂D. Write Hom+T,∂Ω for the class of all sense-preserving homeomorphism of T onto the boundary of a C2 convex Jordan domain Ω. In this paper, five equivalent conditions for the solutions of triharmonic equations ∂z∂z¯3ω=ff∈CD¯ with Dirichlet boundary value conditions ωzz¯zz¯T=γ2∈CT,ωzz¯T=γ1∈CT and ωT=γ0∈Hom+T,∂Ω to be Lipschitz continuous are presented.


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## Body

## 1. Introduction

LetD be the unit disk in the complex plane C and denote T=∂D. Let D and D′ be subdomains of the complex plane C. Write CnD for the set of all complex-valued n-times continuously differentiable functions from D into C. A real-valued function u, defined in an open subset D of the complex plane C, is harmonic if it satisfies Laplace’s equation:(1)Δuz=∂2u∂x2z+∂2u∂y2z=0.A complex-valued functionw=u+iv is harmonic if both u and v are real harmonic. For a harmonic mapping f in D, it can be represented as(2)fz=hz+gz¯,g0=0,where h and g are analytic in D and unique determined by f. By a result of Lewy [1], the Jacobian of a sense-preserving locally univalent harmonic f in D satisfies(3)Jfz=fzz2−fz¯z2=h′z2−g′z2>0.See [2] for more properties of harmonic mappings in the plane. We say that a function u:D⟼ℝ is absolutely continuous on line in the region D if for every closed rectangle R⊂D with sides parallel to the axes x and y, u is absolutely continuous on almost every horizontal line and almost every vertical line in R. Such a function has, of course, partial derivatives ux and uy everywhere in D. A sense-preserving topological mapping f=u+iv:D⟶D′ is said to be K,K′-quasiconformal if it satisfies:a f is absolutely continuous on lines in D.b There are constants K≥1 and K′≥0, such that(4)Lf2≤KLflf+K′,a.e.inD,where Lfz=fzz+fz¯z, lfz=fzz−fz¯z, fz=1/2fx−ify and fz¯=1/2fx+ify. If K′=0, then f is called a K-quasiconformal mapping. It is well known that a quasiconformal mapping f from D onto a Jordan domain Ω has a homeomorphic extension f∗ to the closure D¯; see [3].A mappingh from D onto D′ is said to be Lipschitz if there exists constant L>0, such that the following inequality(5)hz1−hz2≤L⋅z1−z2,holds for every z1,z2∈D.Letf∈LpD;ℂ2<p and ω∈C6D which satisfies the following triharmonic equation:(6)∂z∂z¯3ω=f,inD,with the following associated Dirichlet boundary value condition:(7)ω=γ0,inT,ωzz¯=γ1,inT,ωzz¯zz¯=γ2,inT,where γ0,γ1,γ2∈CT;ℂ. By [[4], Theorem 3.2.2], it is known that all solutions to equation (6) satisfying the condition (7) are given by(8)ωz=Pγ0z+G1γ1z+G2γ2z−G3fz,where(9)Pγ0z=12π∫∂Dg1z,ζγ0ζdζζ,(10)G1γ1z=12π∫∂Dg^2z,ζγ1ζdζζ,(11)G2γ2z=12π∫∂Dg^3z,ζγ2ζdζζ,(12)G3fz=1π∫DG^3z,ζfζdξdη,(13)g1z,ζ=1−z21−zζ¯2,g^2z,ζ=1−z21zζ¯log1−zζ¯+1z¯ζlog1−z¯ζ+1,g^3z,ζ=1−z2∑k=2∞1k2zζ¯k−1+z¯ζk−1+1−1−z42∑k=2∞1kk+1zζ¯k−1+z¯ζk−1+12,G^3z,ζ=14ζ−z4log1−zζ¯ζ−z2+141−z21−ζ2zζ¯+z¯ζ−4−1−z41−ζ44log1−zζ¯zζ¯2+log1−z¯ζz¯ζ2+1zζ¯+1z¯ζ+1−z21−ζ2z2+ζ22log1−zζ¯zζ¯+log1−z¯ζz¯ζ+1−z21−ζ2∑k=1∞1k2zζ¯k−1+z¯ζk−1.


### 1.1. Background

For a harmonic homeomorphismf of the unit disk onto itself, it is interesting to study the conditions which can imply f to be quasiconformal. It was Martio [5] who was the first one to pose and study for this problem. Let ψ∈C1ℝ be the boundary function of f, and he proved that the conditions satisfy minψ′>0 and the Dini condition(14)∫0πϖttdt<∞,where ϖ is the modulus of continuity of ψ,(15)ϖt=supψ′x−ψ′y:x−y<t,can guarantee to imply the quasiconformality of f. Pavlović [6] considered the above question to replace the Dini condition by the Hilbert transformation of ψ′ belong to L∞. In [7], Kalaj and Mateljević gave the equivalent conditions for a harmonic orientation preserving diffeomorphism between two plane Jordan domains Ω and D with C2 boundaries to be K,K′-quasiconformal. Their result was generalized to the case for K,K′-quasiconformal mappings satisfying Poisson’s equation by Li et al. [8]. In [9], Chen and Chen obtained several equivalent conditions for a harmonic mapping of upper half-plane ℍ onto itself to be K,K′-quasiconformal mapping. Recently, after characterizing the boundary conditions for Lipschitz continuity of α¯-harmonic mappings, Chen [10] presented four equivalent conditions for the K,K′-quasiconformal solutions of the α¯-Poisson equation with a nonhomogeneous term to be Lipschitz continuous. His result is a kind of generalization of the result obtained by Pavlović [6].In what follows, we writeHom+T,∂Ω for the class of all sense-preserving homeomorphism of T=∂D onto the boundary of a C2 convex Jordan domain Ω. The aim of this paper is to study for the solutions of triharmonic equations ∂z∂z¯3ω=ff∈CD¯ with Dirichlet boundary value condition ωzz¯zz¯T=γ2∈CT,ωzz¯T=γ2∈CT,ωT=γ0∈Hom+T,∂Ω to be Lipschitz continuous.The organization at the rest of this paper is as follows: In Section2, we will make some estimations which will be used for proving the main result (Theorem 1). In Section 3, the proof of Theorem 1 is given. At the last section, an example is given to show that the family of solution DD⟶Ωf,γ2,γ1,γ0 (see this definition in Section 3) is nonempty.


## 1.1. Background

For a harmonic homeomorphismf of the unit disk onto itself, it is interesting to study the conditions which can imply f to be quasiconformal. It was Martio [5] who was the first one to pose and study for this problem. Let ψ∈C1ℝ be the boundary function of f, and he proved that the conditions satisfy minψ′>0 and the Dini condition(14)∫0πϖttdt<∞,where ϖ is the modulus of continuity of ψ,(15)ϖt=supψ′x−ψ′y:x−y<t,can guarantee to imply the quasiconformality of f. Pavlović [6] considered the above question to replace the Dini condition by the Hilbert transformation of ψ′ belong to L∞. In [7], Kalaj and Mateljević gave the equivalent conditions for a harmonic orientation preserving diffeomorphism between two plane Jordan domains Ω and D with C2 boundaries to be K,K′-quasiconformal. Their result was generalized to the case for K,K′-quasiconformal mappings satisfying Poisson’s equation by Li et al. [8]. In [9], Chen and Chen obtained several equivalent conditions for a harmonic mapping of upper half-plane ℍ onto itself to be K,K′-quasiconformal mapping. Recently, after characterizing the boundary conditions for Lipschitz continuity of α¯-harmonic mappings, Chen [10] presented four equivalent conditions for the K,K′-quasiconformal solutions of the α¯-Poisson equation with a nonhomogeneous term to be Lipschitz continuous. His result is a kind of generalization of the result obtained by Pavlović [6].In what follows, we writeHom+T,∂Ω for the class of all sense-preserving homeomorphism of T=∂D onto the boundary of a C2 convex Jordan domain Ω. The aim of this paper is to study for the solutions of triharmonic equations ∂z∂z¯3ω=ff∈CD¯ with Dirichlet boundary value condition ωzz¯zz¯T=γ2∈CT,ωzz¯T=γ2∈CT,ωT=γ0∈Hom+T,∂Ω to be Lipschitz continuous.The organization at the rest of this paper is as follows: In Section2, we will make some estimations which will be used for proving the main result (Theorem 1). In Section 3, the proof of Theorem 1 is given. At the last section, an example is given to show that the family of solution DD⟶Ωf,γ2,γ1,γ0 (see this definition in Section 3) is nonempty.


## 2. Some Estimations

The following lemma was obtained in [11].Lemma 1 [[11], Lemma 2.1].
Suppose thatγ1∈CT and G1γ1 are defined in (10). Then, the following inequality(16)max∂∂zG1γ1z,∂∂z¯G1γ1z≤γ1∞maxx∈0,1hx+π23−11/2z,holds for every z∈D, where γ1∞=supz∈Tγ1z and(17)hx=1−x∑n=2∞n−12n2xn−21/2≤1−x1/2,x∈0,1.In order to give the complete proof of Theorem 1, we also need the following two lemmas.Lemma 2.
Suppose thatγ2∈CT and G2γ2 are defined in (11). Then, the following inequalities(18)∂∂zG2γ2z≤3+2z−2z2+2z3−z4π4/45−11/2⋅γ2∞2,∂∂z¯G2γ2z≤3+2z−2z2+2z3−z4π4/45−11/2⋅γ2∞2,hold for every z∈D, where γ2∞=supz∈Tγ2z.Proof.
Here we only calculate the estimation of∂/∂zG2γ2z because the calculation for another one is similar. Let(19)L1z,t=A1z,t+A2z,t,where(20)A1z,t=−z¯∑k=2∞ze−itk−1+z¯eitk−1k2+1+1−z2e−it∑k=1∞kze−itk−1k+12,A2z,t=−zz¯2∑k=2∞ze−itk−1+z¯eitk−1kk+1+12+1−z4e−it2∑k=1∞kze−itk−1k+1k+2.
By virtue of Hölder’s inequality, we get(21)12π∫02πA1z,tγ2eitdt≤z⋅γ2∞12π∫02π∑k=2∞ze−itk−1+z¯eitk−1k2+1dt+1−z2γ2∞12π∫02π∑k=1∞kze−itk−1k+12dt≤z⋅γ2∞12π∫02π∑k=2∞ze−itk−1+z¯eitk−1k2+12dt1/2+1−z2γ2∞12π∫02π∑k=1∞kze−itk−1k+122dt1/2=z⋅γ2∞2∑k=2∞z2k−1k4+11/2+1−z2γ2∞∑k=1∞k2z2k−1k+141/2≤z⋅γ2∞2∑k=1∞1k4−11/2+1−z2γ2∞2∑k=1∞1k4−11/2=π445−11/21+z−z2⋅γ2∞,12π∫02πA2z,tγ2eitdt≤z3⋅γ2∞12π∫02π∑k=2∞ze−itk−1+z¯eitk−1kk+1+12dt+1−z42γ2∞12π∫02π∑k=1∞kze−itk−1k+1k+2dt≤z3⋅γ2∞12π∫02π∑k=2∞ze−itk−1+z¯eitk−1kk+1+122dt1/2+1−z42γ2∞12π∫02π∑k=1∞kze−itk−1k+1k+22dt1/2=z3⋅γ2∞2∑k=2∞z2k−1k2k+12+141/2+1−z42γ2∞∑k=1∞k2z2k−1k+12k+221/2≤z3⋅γ2∞2∑k=2∞1k4+141/2+1−z42γ2∞∑k=1∞1k+121/2=2π4/45−7/41/2⋅z3+π2/6−11/21−z42⋅γ2∞.
Hence,(22)∂∂zG2γ2z=12π∫02πL1z,tγ2eitdt≤12π∫02πA1z,tγ2eitdt+12π∫02πA2z,tγ2eitdt≤π445−11/21+z−z2⋅γ2∞+2π4/45−7/41/2⋅z3+π2/6−11/21−z42⋅γ2∞≤3+2z−2z2+2z3−z4π4/45−11/2⋅γ2∞2.
This finishes the proof.Lemma 3.
Suppose thatf∈CD¯ and G3f are defined in (12). Then, the following inequalities(23)∂∂zG3fz≤M2z+M3z+M4z+M5z+M6z⋅f∞,∂∂z¯G3fz≤M2z+M3z+M4z+M5z+M6z⋅f∞,hold for every z∈D, where f∞=supz∈Dfz and(24)M2z=16+z2+11z4−5z6+z848,M3z=1+4z+z24,M4z=4π2/6−11/21−z215+π2z65,M5z=23πz+21−z21/21+z2224,M6z=22π2/6−11/2z3+1+z21−z21/224.Proof.
Here we only calculate the estimation of∂/∂zG3fz because the calculation for another one is similar. In order to do this, we introduce following functions:(25)L2z=1π∫Dζ−zζ¯−z¯24⋅−2log1−zζ¯ζ−z2+1−ζ21−zζ¯fζdσζ,L3z=1π∫D1−ζ24⋅−z¯zζ¯+z¯ζ−4+1−z2ζ¯fζdσζ,L4z=A5z+A6z,L5z=A7z+A8z+A9z,L6z=A10z+A11z,where(26)A5z=1π∫D−z¯1−ζ2∑k=1∞zζ¯k−1+z¯ζk−1k2fζdσζ,A6z=1π∫D1−ζ21−z2∑k=1∞kζ¯zζ¯k−1k+12fζdσζ,A7z=12π∫Dz¯1−ζ2z2+ζ2∑n=1∞zζ¯n−1+z¯ζn−1nfζdσζ,A8z=12π∫D−z¯1−z21−ζ2∑n=1∞zζ¯n−1+z¯ζn−1nfζdσζ,A9z=12π∫D−1−z21−ζ2z2+ζ2∑n=1∞nζ¯zζ¯n−1n+1fζdσζ,A10z=12π∫Dzz¯21−ζ4∑n=1∞zζ¯n−1+z¯ζn−1n+1fζdσζ,A11z=14π∫D1−z41−ζ4−ζ¯⋅∑n=1∞nzζ¯n−1n+2fζdσζ.
Firstly, we give the estimation ofL2z. Set(27)w=z−ζ1−z¯ζ.
Then,(28)Jζw=1−z221−z¯w4,ζ−z=wz2−11−z¯w,1−ζ2=1−w21−z21−z¯w2.
Therefore,(29)L2z=1π∫Dζ−zζ¯−z¯24⋅−2log1−zζ¯ζ−z2+1−ζ21−zζ¯fζdσζ≤1−z25f∞4π∫Dw31−z¯w72log1w2+1−w21−z¯wdσw=1−z25f∞2A3z+A4z,where(30)A3z=12π∫Dw3log1/w21−z¯w7dσw,A4z=12π∫Dw31−w21−z¯w8dσw.
Since(31)12π∫02πdφ1−zρe−iφ7≤12π∫02πdφ1−zρe−iφ61/2⋅12π∫02πdφ1−zρe−iφ81/2≤∑n=0∞n+12n+224z2nρ2n1/2⋅∑n=0∞n+12n+22n+3236z2nρ2n1/2≤∑n=0∞n+12n+22n+3236z2nρ2n,we get (by letting w=ρeiφ)(32)A3z=2∫01ρ4log1ρ212π∫02πdφ1−zρe−iφ7dρ≤∑n=0∞n+12n+22n+3218z2n∫01ρ2n+4log1ρ2dρ=∑n=0∞n+12n+22n+3292n+52z2n≤∑n=0∞n+12n+22n+3236n+1n+3z2n≤∑n=0∞n+2436z2n=16+z2+11z4−5z6+z8361−z25,A4z=∫01ρ41−ρ212π∫02πdφ1−zρe−iφ8dρ=∑n=0∞n+12n+22n+3236z2n∫01ρ2n+41−ρ2dρ=∑n=0∞n+12n+22n+32182n+52n+7z2n≤∑n=0∞n+12n+22n+32182n+22n+6z2n=∑n=0∞n+2472z2n=16+z2+11z4−5z6+z8721−z25,which implies that(33)L2z≤1−z25f∞2A3z+A4z≤16+z2+11z4−5z6+z848⋅f∞.
Secondly, we derive the estimation ofL3z. This can easily been obtained by(34)L3z≤f∞4π∫Dz2z+4+1−z2dσζ=z2+4z+1f∞4.
Thirdly, we calculate the upper bound ofL44. By letting ζ=ρeit, we get(35)A5z≤2z⋅f∞∫01ρ1−ρ212π∫02π∑k=1∞zρe−itk−1+z¯ρeitk−1k2dtdρ≤2z⋅f∞∫01ρ1−ρ212π∫02π∑k=1∞zρe−itk−1+z¯ρeitk−1k22dt1/2dρ≤2z⋅f∞2∑k=1∞1k41/2∫01ρ1−ρ2dρ=π2z⋅f∞65,A6z≤21−z2f∞∫01ρ21−ρ212π∫02π∑k=1∞kzρe−itk−1k+12dtdρ≤21−z2f∞∫01ρ21−ρ212π∫02π∑k=1∞kzρe−itk−1k+122dt1/2dρ≤21−z2f∞∑k=1∞k2k+141/2∫01ρ21−ρ2dρ≤21−z2f∞∑k=2∞1k21/2∫01ρ21−ρ2dρ=41−z2π2/6−11/2f∞15.Then,(36)L4z≤A5z+A6z≤4π2/6−11/21−z215+π2z65⋅f∞.
Next, we estimate the upper bound ofL5z. By letting ζ=ρeit, we get(37)A7z≤zz2+1f∞⋅∫01ρ1−ρ2∫02π∑n=1∞zρe−itn−1+z¯ρeitn−1ndtdρ≤zz2+1⋅f∞⋅∫01ρ1−ρ212π∫02π∑n=1∞zρe−itn−1+z¯ρeitn−1n2dt1/2dρ=zz2+1⋅f∞⋅∫01ρ1−ρ22∑n=1∞z2n−1ρ2n−1n21/2dρ≤zz2+1⋅f∞⋅∫01ρ1−ρ22∑n=1∞1n21/2dρ=3πzz2+1⋅f∞12,A8z≤z⋅1−z2f∞2π∫01ρ1−ρ2∫02π∑n=1∞zρe−itn−1+z¯ρeitn−1ndtdρ≤z⋅1−z2⋅f∞⋅∫01ρ1−ρ212π∫02π∑n=1∞zρe−itn−1+z¯ρeitn−1n2dt1/2dρ=z⋅1−z2⋅f∞⋅∫01ρ1−ρ22∑n=1∞z2n−1ρ2n−1n21/2dρ=3πz1−z2⋅f∞12,A9z≤1−z21+z22f∞8π∫01ρ2∫02π∑n=1∞nzρe−itn−1n+1dtdρ≤1−z21+z22f∞4∫01ρ212π∫02π∑n=1∞nzρe−itn−1n+12dt1/2dρ=1−z21+z22f∞4∫01ρ2∑n=1∞n2z2n−1ρ2n−1n+1221/2dρ≤1−z21+z22f∞4∫01ρ2∑n=1∞z2n−11/2dρ=1−z21/21+z22f∞12.
Therefore,(38)L5z≤A7z+A8z+A9z≤3π1+z2z+3πz1−z2+1−z1/21+z2212⋅f∞=23πz+1−z1/21+z2212⋅f∞.
At last, we calculate the upper bound ofL6z. By letting ζ=ρeit, we get(39)A10z≤z3f∞2π∫01ρ1−ρ4∫02π∑n=1∞zρe−itn−1+z¯ρeitn−1n+1dtdρ≤z3f∞∫01ρ1−ρ412π∫02π∑n=1∞zρe−itn−1+z¯ρeitn−1n+12dt1/2dρ≤z3f∞∫01ρ1−ρ42∑n=1∞1n+121/2dρ=2π2/6−11/2z3f∞3,A11z≤1−z4f∞4π∫01ρ21−ρ4∫02π∑n=1∞nzρe−itn−1n+2dtdρ≤1−z4f∞2∫01ρ21−ρ412π∫02π∑n=1∞nzρe−itn−1n+22dt1/2dρ≤1−z4f∞2∫01ρ1−ρ4∑n=1∞z2n−11/2dρ=1+z21−z21/2f∞6.Hence,(40)L6z≤A10z+A11z≤22π2/6−11/2z3+1+z21−z21/26⋅f∞.
Combining with above estimations fromL2z to L6z, we get(41)∂∂zG3fz≤∑k=26Lkz≤∑k=26Mkz⋅f∞,where Mkzk=2,3,…,6 are defined in (15). This finished the proof.


## 3. Main Results

The Hilbert transform of a functionf∈L1T is defined by the formula(42)Hfφ=−1π∫0+πfφ+t−fφ−t2tant/2dt,for a.e. φ∈0,2π. See [[12], Chapter VII]. In what follows, we always say that Ω is a convex Jordan domain with C2 boundary. Let DD⟶Ωf,γ2,γ1,γ0 denote the family of solutions ω:D⟶Ω ω∈C6D of the triharmonic equation ∂z∂z¯3ω=f∈CD¯ in D with the boundary value conditions ωzz¯zz¯T=γ2∈CT and ωzz¯T=γ1∈CT, ωT∈Hom+T,∂Ω, and each ω is a sense-preserving diffeomorphism. The following theorem gives a boundary characterizations of K,K′-quasiconformal mappings from D onto Ω satisfying triharmonic equations to be Lipschitz continuous.Theorem 1.
Suppose thatω∈DD⟶Ωf,γ2,γ1,γ0 with the representation (8). Then, the following conditions are equivalent:(a)
ω is a K,K′-quasiconformal mapping and ∂Pγ0reiθ/∂r≤L in D, where L>0 is a constant(b)
ω is Lipschitz with respect to Euclidean metric(c)
Pγ0 is Lipschitz with respect to Euclidean metric(d)
Pγ0 is a K,K′-quasiconformal mapping(e)
γ0 is absolutely continuous on T, γ0′∈L∞T and Hγ0′∈L∞TProof.
SinceωT=γ0∈Hom+T,∂Ω, by a Theorem of Radó [13], Kneser [14], and Choquet [15], Pγ0 is sense-preserving harmonic diffeomorphism between Dand Ω.
We now first prove thatb⇔c holds. By Lemmas 1–3, we have(43)∇G1γ1z=∂G1γ1z∂z+∂G1γ1z∂z¯≤γ1∞2maxx∈0,1hx+π23−11/2z≤γ1∞21−z1/2+π23−11/2z≤6γ1∞,(44)∇G2γ2z=∂G2γ2z∂z+∂G2γ2z∂z¯≤3+2z−2z2+2z3−z4π445−11/2⋅γ2∞≤7π445−11/2⋅γ2∞,(45)∇G3fz=∂G3fz∂z+∂G3fz∂z¯≤2∑k=26Mkz⋅f∞≤210148+4π4/45−11/215+π265+3π+412+2π2/6−11/2+112⋅f∞.
Hence, the proof of equivalence of (b) and (c) is derived by (43)-(45) and inequalities (46)(46)∇ω≤∇Pγ0+∇G1γ1+∇G2γ2+∇G3f,∇Pγ0≤∇ω+∇G1γ1+∇G2γ2+∇G3f.
Secondly, we prove thata⟹b holds. By the K,K′-quasiconformality of ω, we get(47)∇ω≤Klωz+K2lωz2+4K′2≤Klωz+K′.
Letz=reiθ∈D. Thus,(48)∂ωreiθ∂r=eiθωz+e−iθωz¯≥lωz.
Hence,(49)∇ω≤K∂ωreiθ∂r+K′.
By inequalities (43)–(45), we obtain(50)∂G1γ1reiθ∂r=eiθ⋅∂G1γ1reiθ∂z+e−iθ⋅∂G1γ1reiθ∂z¯≤∂G1γ1z∂z+∂G1γ1z∂z¯≤6γ1∞,(51)∂G2γ2reiθ∂r=eiθ⋅∂G2γ2reiθ∂z+e−iθ⋅∂G2γ2reiθ∂z¯≤∂G2γ2z∂z+∂G2γ2z∂z¯≤7π445−11/2⋅γ2∞,(52)∂G3freiθ∂r=eiθ⋅∂G3freiθ∂z+e−iθ⋅∂G3freiθ∂z¯≤∂G3fz∂z+∂G3fz∂z¯≤210148+4π4/45−11/215+π265+3π+412+2π2/6−11/2+112⋅f∞.
Therefore, by (50)–(52), we get(53)∇ω≤K∂ωreiθ∂r+K′≤K∂Pγ0reiθ∂r+∂G1γ1reiθ∂r+∂G2γ2reiθ∂r+∂G3fz∂r+K′<∞.
This finishes the implication froma to b.
Thirdly, we prove thatb⟹a holds. Since ω is Lipschitz and b is equivalent to c, Pγ0 is Lipschitz. Thus, there are two constants M and L such that ∇ω≤M and ∇Pγ0≤L hold for every z∈D. Therefore,(54)∇ω2≤Jω+∇ω2≤Jω+M2,which says that w is a 1,M2-quasiconformal mapping. Moreover,(55)∂Pγ0reiθ∂r≤∇Pγ0≤L.
Thus, the proof of implication fromb to a is finished.
At last, sincePγ0 is a sense-preserving harmonic diffeomorphism, the equivalence for c,d, and e follows from [[7], Corollary 1.3]. Hence, the proof is finished.Remark 1.
The reason to suppose that∂Pγ0reiθ/∂r≤L in part a of Theorem 1 can be seen [10, Proposition 1.1].


## 4. An Example

Following example says that the family of solutionDD⟶Ωf,γ2,γ1,γ0 is nonempty.Example 1.
Letω=zz6+2z,z∈D. Then,
a
1 ω satisfies the triharmonic equation ∂z∂z¯3ω=144z in D and Dirichlet boundary value condition:(56)ωT=3eit,ωzz¯T=12eit,ωzz¯zz¯T=72eit.
a
2 ω is a 1,54-quasiconformal mapping of D onto Ω=z:z<3 and ∂Pγ0/∂r≤3969 in D.
a
3 ω is Lipschitz continuous.Proof.
Letz=reit,t∈0,2π. Obviously, ωT=3eit≔γ0, ωz=4z6+2, ωz¯=3z2z4, ωzz¯T=12eit≔γ1, ωzz¯zz¯T=72eit≔γ2, and ∂z∂z¯3ω=144z≔f. It follows that(57)Jw=wz2−wz2=4z6+22−9z12=7z12+16z6+4>0,∇w2−Jw=4z6+2+3z62−7z12+8z6+4=42z12+12z6≤54.
Thus,ω is a sense-preserving 1,54-quasiconformal homeomorphism from D onto Ω=z:z<3. In addition, by (8) and (50)–(52), we get(58)∂Pγ0∂r≤∂w∂r+∂G1γ1reiθ∂r+∂G2γ2reiθ∂r+∂G3fz∂r≤7r6+2r+6γ1∞+14γ2∞+20f∞≤3969.Hence, we have proved that w satisfies a1 and a2.
Further,(59)∇ω=4z6+2+3z6≤9,which says that ω is Lipschitz continuous. Hence, the proof of a3 is finished.



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*Source: 3261254-2019-09-17.xml*