Quasiconformal harmonic mappings between two doubly connected domains in the plane
aa r X i v : . [ m a t h . C V ] F e b QUASICONFORMAL HARMONIC MAPPINGS BETWEEN TWODOUBLY CONNECTED DOMAINS IN THE PLANE
DAVID KALAJA
BSTRACT . It is known for some time that there exists an energy-minimal dif-feomorphism between two doubly-connected domains Ω and D provided that Mod(Ω) ≤ Mod D and that diffeomorphism is harmonic [4]. In this note wegive a short proof of the fact that for given annuli Ω and D satisfying the con-dition Mod(Ω) ≤ Mod( D ) there exist a K − quasiconformal harmonic diffeo-morphism f : Ω onto −→ D , where K = K ( τ ) , τ = Mod(Ω) / Mod( D ) and lim τ → K ( τ ) = 1 .
1. I
NTRODUCTION
Minimizers of Dirichlet energy.
It is well-known the Riemann mapping the-orem which assert that for given two Jordan domains Ω and D in the complex plane C there is a conformal diffeomorphism f : Ω onto −→ D . In the case of doubly con-nected domains, such a statement is not true in general. However for any givendoubly connected domain D there is a positive real number r < and a conformaldiffeomorphism between the circular annulus A ( r, def == { z ∈ C : r < | z | < } and D . The number r is unique, and the expression Mod( D ) := log r is calledthe conformal modulus of Ω . The Dirichlet energy integral of a smooth mapping f : Ω → D is defined by E [ f ] = Z Ω ( | f z | + | f ¯ z | ) dxdy. If f is a sense preserving diffeomorphism, then E [ f ] ≥ Z Ω ( | f z | − | f ¯ z | ) dxdy = µ ( D ) , with the equality if and only if f ¯ z ≡ , or what is the same if and only if f isconformal. On the other hand, if the domains Ω and D are not conformally equiv-alent, then the minimizer of the Dirichlet energy is not a conformal mapping. Theauthors of [4] have considered this problem and proved the following result. If thedoubly-connected domains Ω and D are so that Mod(Ω) ≤ Mod( D ) , then thereexists a minimizer of the Dirichlet energy throughout the class of diffeomorphismsbetween Ω and D . That diffeomorphism is a harmonic diffeomorphism betweenthem. A more general result has been obtained by the author in [5]. Mathematics Subject Classification.
Primary 31A05; Secondary 42B30 .
Key words and phrases.
Harmonic mappings, Quasiconformal mappings, Annuli. If Ω = A ( r, and D = A ( R, , then the minimiser of Dirichlet energy is acertain radial harmonic mapping of the form f ( z ) = az + b/ ¯ z , provided that theNitsche condition R ≤ r/ (1 + r ) is satisfied ([1]).Let us also mention the following result about the existence of harmonic diffeo-morphisms. If the image annulus has Dini’s smooth boundary, then Kovalev andLi in [7] have proved that there exist a harmonic diffeomorphism between D and Ω provided that Mod( D ) ≤ Mod(Ω) ≤ Mod( D ) + ǫ, where ǫ = ǫ (Ω) .1.2. Mappings of finite distortion.
A homeomorphism w = f ( z ) between planardomains Ω and D has finite distortion if a) f lies in the Sobolev space W , loc (Ω , D ) of functions whose first derivatives are locally integrable, and b) f satisfies thedistortion inequality | f ¯ z | ≤ ω ( z ) | f z | , ≤ ω ( z ) < almost everywhere in Ω . Such mappings are generalizations ofquasiconformal homeomorphisms where one works with the stronger assumption ω ( z ) ≤ k < . Quasiconformal mapping are usually called K − quasiconformalmappings, where K = (1 + k ) / (1 − k ) . The Jacobian determinant of a mapping f of finite distortion is non-negative almost everywhere, since J f ( z ) = | f z | − | f ¯ z | = (1 − ω ( z ) ) | f z | ≥ . The distortion function is defined by the rule(1.1) K ( z, f ) = | f z | + | f ¯ z | | f z | − | f ¯ z | = k Df ( z ) k J f ( z ) if J f ( z ) > . Here k A k = 12 Tr( A T A ) is the square of mean Hilbert-Schmidt norm. We conveniently set K ( z, f ) = 1 if f z = f ¯ z = 0 . Notice that then K ( z, f ) = 1 and we have the equality K ( z, f ) = 1 if and only if f is conformal, by the Looman Menchoff theorem. The classicalformulations of the extremal Gr¨otsch and Teichm¨uller problems are concerned withfinding mappings Ω → D in some class (for instance, with free or prescribedboundary values) which have smallest L ∞ -norm of the distortion function, thus”extremal quasiconformal mappings”. Another extremal problem is to minimizeintegral means(1.2) K [ f ] = Z D K ( z, f ) dxdy, z = x + iy, where f : D → Ω is a mapping of finite distortion. It has been shown in [1] forEuclidean metric and in [6] for some radial metrics that the minimisation problemfor Dirichlet energy of diffeomorphism is equivalent with the minimization prob-lem of integral (1.2). If Ω = A ( R, and D = A ( r, are circular annuli, then theminimizer of integral (1.2) is f if and only if g = f − is a minimizer of Dirich-let energy which is a radial harmonic diffeomorphism. A similar statement holds .C. HARMONIC MAPPINGS BETWEEN DOUBLY CONNECTED DOMAINS IN THE PLANE 3 for arbitrary domains. As it is said before, the minimizer of Dirichlet energy is acertain harmonic diffeomorphism, provided that Mod(Ω) ≤ Mod( D ) . This im-plies that, the minimizer of the integral (1.2) is the inverse of a harmonic mapping,provided that Mod(Ω) ≤ Mod( D ) . Question 1.1. If Ω = A ( r, and D = A ( R, , then the extremal quasiconfromalmapping mapping the inner boundary onto the inner boundary defined by f ( z ) = | z | log(1 /R ) / log(1 /r ) − z is unique up to the rotation and thus it is not harmonic for R = r . The same conclusion can be made for arbitrary doubly-connected domains.The following question arises however, for given doubly-connected domains Ω and D whether there exist a quasiconformal harmonic mapping of Ω onto D (which isnot extremal quasiconformal). We solve this question for the case
Mod(Ω) ≤ Mod( D ) . In this paper we givea simply proof of the following theorem. Theorem 1.2.
Assume that A and B are two doubly-connected domains in com-plex plane so that Mod( A ) ≤ Mod( B ) . Then there exists a quasiconformal har-monic diffeomorphism f of A onto B . Moreover its distortion function µ f ( z ) = f ¯ z ( z ) /f z ( z ) is a constant function. Remark 1.3. If H = Mod( B )Mod( A ) = m √ m , then from the proof of Theorem 1.2, theproduced mapping f is K = √ m − quasiconformal. So it is not extremalquasiconformal mapping, because H < K . It can be also proved, for circular an-nuli that the produced harmonic mapping in Theorem 1.2 is not radial harmonic,and therefore is not minimizer of the Dirichlet energy (for example they have dif-ferent distortion function).The case
Mod( A ) ≥ Mod( B ) is much harder. If we consider the most beatifulsituation of concentric annuli A ( r, and A ( R, , then a harmonic diffeomor-phism between A ( r, and A ( R, exist if and only if R ≤ r r . This J.C. C.Nitsche conjecture has been solved by Iwaniec, Kovalev and Onninen in [3] . In thelimiting case, i.e. when R = 2 r/ (1 + r ) , then such harmonic mapping is uniqueup to a rotation and it is not quasiconformal [3, Theorem 1.4] . A nice consequenceof Nitsche inequality has been given by Feng and Tang in [2] .
2. A
N AUXILIARY RESULT
Lemma 2.1.
Let X be a doubly connected domain in R and assume that Φ ∈ C ( X ) . Assume that Y = { ( x, y, Φ( x, y )) : ( x, y ) ∈ X } , is an annulus in R . Assume that C = sup ( x,y ) ∈ X max { x ( x, y ) , y ( x, y ) } q x ( x, y ) + Φ y ( x, y ) .C ′ = inf ( x,y ) ∈ X min { x ( x, y ) , y ( x, y ) } q x ( x, y ) + Φ y ( x, y ) . .C. HARMONIC MAPPINGS BETWEEN DOUBLY CONNECTED DOMAINS IN THE PLANE 4 Then C ′ Mod( Y ) ≤ Mod( X ) ≤ C Mod( Y ) . In particular if Φ( x, y ) = m ( x + y ) , then Mod( X ) = 1 + m √ m Mod( Y ) . Proof.
Let Γ be the family of curves that separate the boundary components of Y .Let be ρ a positive function defined in Y so that A Y ( ρ ) := Z Y ρ ( z ) dµ ( z ) < ∞ . Denote all such so called allowable weights ρ by Π . Here µ is two-dimensionalLebesgue measure on Y . Now let L Y (Γ) be the extremal length of Γ , i.e. L Y ( ρ ) = inf γ ∈ Γ L Y ( ρ ) , where L Y ( ρ ) = Z γ ρ ( w ) | dw | . The modulus of annulus Y is now given by the formula Mod( Y ) = inf ρ ∈ Π A Y ( ρ ) L Y ( ρ ) . Let Γ be the family of curves that separate the boundary components of X .Then γ ∈ Γ if and only if γ = P r R ( γ ) ∈ Γ . Moreover if Π is the set of allallowable weights in X , then ρ ∈ Π , if and only if ρ q x + Φ y ∈ Π . Now if ρ ∈ Π , is so that L Y ( ρ ) ≥ , then for ρ ( x, y ) = ρ ( x, y, Φ( x, y )) q x + Φ y ∈ Π . Moreover L X ( ρ ) ≥ N = 1 / √ C. Namely if γ ( t ) = ( x ( t ) , y ( t )) : [0 , → X is in Γ , then γ ( t ) = ( x ( t ) , y ( t ) , Φ( x ( t ) , y ( t ))) :[0 , → Y is in Γ . Then we have N ≤ N L γ ( ρ ) = N Z ρ ( γ ( t )) | dγ ( t ) | = N Z ρ ( γ ( t )) q ˙ x (1 + Φ x ) + ˙ y (1 + Φ y ) dt ≤ Z ρ ( γ ( t )) q x + Φ y p ˙ x + ˙ y dt = Z γ ρ ( w ) | dw | = L γ ( ρ ) . (2.1) .C. HARMONIC MAPPINGS BETWEEN DOUBLY CONNECTED DOMAINS IN THE PLANE 5 Therefore L γ ( ρ ) ≥ N .Further we have A Y ( ρ ) = Z Y ρ ( ζ ) dS ( ζ )= Z X ρ ( x, y, Φ( x, y )) q x + Φ y dxdy = Z X ρ ( x, y ) dxdy = A X ( ρ ) . (2.2)Therefore A X ( ρ ) L X ( ρ ) ≤ N A Y ( ρ ) L Y ( ρ ) We obtain that
Mod( X ) ≤ C Mod( Y ) , where N = 1 /C . Similarly we get theinequality from the other side. (cid:3)
3. P
ROOF OF T HEOREM
Proof.
Assume without loss of generality that
Mod( A ) < Mod( B ) and let m besuch a positive real number so that m √ m = Mod( B )Mod( A ) . Let Φ( x, y ) = m ( x + y ) .Then consider B = { ( x, y, Φ( x, y )) : ( x, y ) ∈ B } . Then B is a planar domain inthe plane z = m ( x + y ) . Denote that plane by Π and assume that e and e is anorthonormal basis of Π . Then e = ( α , β , γ ) , e = ( α, β, γ ) . Then α α + β β + γ γ = α + β + γ − α + β + γ − and γ = m ( α + β ) , γ = m ( α + β ) . One of particular solution of the above system is α = 1 √ m + 2 m + m √ m + 2 m ,β = − m √ m + 2 m , γ = m √ m + 2 m α = 0 , β = − √ m , γ = − m √ m . Assume that the outer boundary ∂ B of B is the boundary of the Jordan domain B . Denote its lift on Π (w.r.t. Φ ) by B ⊃ B and assume that ϕ : U
7→ B is aconformal diffeomorphism, which exists in virtue of Riemann mapping theorem.Assume that its representation in the basis e and e is ϕ ( z ) = u ( z ) e + v ( z ) e .Then its projection φ to R × { } ∼ = C is a harmonic diffeomorphism of the unitdisk U onto a Jordan domain Ω ⊂ C . More precisely .C. HARMONIC MAPPINGS BETWEEN DOUBLY CONNECTED DOMAINS IN THE PLANE 6 φ ( z ) = α u + αv + i ( β u + βv ) . So(3.1) φ ( z ) = 12 ( α + β + i ( β − α )) ϕ ◦ + 12 ( α − β + i ( α + β )) ϕ ◦ . Here ϕ ◦ ( z ) = u ( z ) + iv ( z ) is conformal.Then φ ( z ) is a harmonic diffeomorphism of U onto Ω . If we put the abovevalues of parameters α, . . . , γ we get the following transformation φ ( z ) = U ϕ ◦ ( z ) + V ϕ ◦ ( z ) , where U = − (1 + i ) m √ m + 2 m − i (cid:18) √ m + 1 √ m + 2 m (cid:19) and V = (1 + i ) m √ m + 2 m − i (cid:18) √ m − √ m + 2 m (cid:19) . Let P = φ − ( B ) . Since ϕ ( P ) = B , in view of Lemma 2.1, it follows that Mod( P ) = Mod( B ) = √ m m Mod( B ) . It follows that
Mod( A ) = Mod( P ) , so there is a conformal diffeomorphism a between A and P . The desired harmonic diffeomorphism between A and B isgiven by f ( z ) = φ ( a ( z )) . Then that harmonic mapping is also K = √ m − quasiconformal mapping of A onto B . Namely VU = − − (1 − i ) m + √ m − i ) m + √ m from where it follows that | φ ¯ z || φ z | = m m + √ m . The rest of the proof is a simple matter. It should be noted the following, allother solutions give the same constant of quasiconformality, because we have aprojection from the same plane. (cid:3)
By following the same lines of the proof of Theorem 1.2, we can also prove thefollowing theorem.
Theorem 3.1.
Assume that f is a harmonic homeomophism between double con-nected domains A and B . Then, for every real number c > Mod( B ) , there existsa double-connected domain C , with the conformal modulus c , together with a har-monic diffeomorphism F = af + bf of A onto C . Here a and b are some constants. .C. HARMONIC MAPPINGS BETWEEN DOUBLY CONNECTED DOMAINS IN THE PLANE 7 R EFERENCES [1] A
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