On Riemann’s mapping theorem. AshotVagharshakyan Institute of Mathematics Armenian National Academy of Sciences, Bagramian 24-b, 1 1 Yerevan, Armenia. 0 e-mail: [email protected] 2 n a Abstract. InthispaperwegiveanewproofofRiemann’swell knownmapping J theorem. The suggested method permits to prove an analog of that theorem for 4 three dimensional case. ] V Keywords: Quasi - conformal mapping, Riemann’s theorem. C . h t 1 Introduction a m [ By Liouville’s theorem, see [2], p. 130, in three dimensional case, only superpos- 2 ition of isometric, dilatation and inverse transformations are conformal. To get v an analog for Riemann’s mapping theorem, one introduce a family of mappings 4 named quasi - conformal. This family is wider, nevertheless we do not have a 6 4 natural analog of conformal mappings like of two dimensional case. In this paper 5 we introduce a new family of mappings, named weak - conformal and obtain more . 5 natural generalization of Riemann’s theorem. 0 0 The proof of the main result of the present paper is interesting for two dimen- 1 sional case too. Actually, we give a new proof of Riemann’s classical theorem, : v where the specific properties of complex analysis do not used. This permits us to i X prove a similar theorem in three dimensional case. r a 2 Classes of mappings For any matrix M = {a } with eigenvalues λ , ...,λ denote by ij 1 n n n n |M|2 = a2 = |λ |2, ij i i=1 j=1 i=1 XX X n n tr(M) = a = λ ii i i=1 i=1 X X 1 and n det(M) = λ . i i=1 Y Let ϕ(x,y,z) = (A,B,C) be a continuously differentiable mapping. Denote by A′ A′ A′ x y z J = B′ B′ B′  x y z  C′ C′ C′ x y z   the Jacobi matrix. Let G = J∗J. We have |ϕ(~x+∆~x)−ϕ(~x)|2 = (J∆~x,J∆~x)+o(|∆~x|2) = (∆~x,G∆~x)+o(|∆~x|2). Definition 1. A continuously differentiable one to one mapping ϕ : Ω → Ω 1 2 of the domain Ω ⊂ R3 on Ω ⊂ R3 is conformal if for each point ~x ∈ Ω there is 1 2 1 a number M(~x) such that |ϕ(~x+∆~x)−ϕ(~x)| = M(~x)|∆~x|+o(|∆~x|). Lemma 1. Let ϕ be a continuously differentiable mapping with the Jacobi matrix J and G = J∗J. Then ϕ is conformal if and only if 27det(G) = tr3(G). Proof. The eigenvalues λ ,λ ,λ of the matrix G are nonnegative. The 1 2 3 lemma’s condition means that λ +λ +λ 3 1 2 3 = λ λ λ 1 2 3 3 (cid:18) (cid:19) This equality is valid only if all eigenvalues are the same, i. e. λ = λ = λ . 1 2 3 Example. Letusconsidertheinversetransformation,whichthepoint(x,y,z) 6= (0,0,0) maps to (A,B,C), where x y z (A,B,C) = , , , x2 +y2 +z2 x2 +y2+z2 x2 +y2 +z2 (cid:18) (cid:19) We have A′ A′ A′ x y z J = B′ B′ B′ =  x y z  C′ C′ C′ x y z   2 −x2 +y2 +z2 −2xy −2xz 1 = −2xy x2 −y2 +z2 −2yz (x2 +y2 +z2)2   −2xz −2yz x2 +y2 −z2   Consequently, 1 0 0 1 G = 0 1 0 (x2 +y2+z2)2   0 0 1   So, the condition of lemma 1 is satisfy and hence this mapping is conformal. Definition 2. A quasi-conformal mapping is a continuously differentiable homeomorphism ϕ : Ω → Ω 1 2 for which the ball B(~x,r) = {~y; |~x−~y| < r} maps to {ϕ(~y); ~y ∈ B(~x,r)} = {ϕ(~x)+J(~y −~x); ~y ∈ B(~x,r)}+o(r2) and the ratio of the main diagonals of the ellipsoid {J(~z); |~z| = r)} are uniformly bounded for all points ~x ∈ Ω. In this paper we introduce a new family of mappings, which are generalizations of conformal mappings. For those mappings, which we name weak - conformal, we have an analog of Riemann’s mapping theorem. Definition 3. A weak-conformal mapping is a continuously differentiable ho- momorphism ϕ : Ω → Ω 1 2 for which the ball B(~x,r) = {~y; |~x−~y| < r} maps to {ϕ(~y); ~y ∈ B(~x,r)} = {ϕ(~x)+J(~y −~x); ~y ∈ B(~x,r)}+o(r2) and the the main diagonals of the ellipsoid {J(~z); |~z| = r)} form geometric progression for all points ~x ∈ Ω. 3 Lemma 2. Let ϕ be a continuously differentiable mapping with Jacobi matrix J. Then it is weak - conformal if and only if tr2(G)−|G|2 3 = 8det(G) tr3(G), where G = J∗J. (cid:0) (cid:1) Proof. In terms of eigenvalues of the matrix G = J∗J, this condition one can write as follows λ λ λ (λ +λ +λ )3 = (λ λ +λ λ +λ λ )3. 1 2 3 1 2 3 1 2 3 1 3 2 So, (λ λ λ −λ3)(λ +λ +λ )3 −(λ λ +λ λ +λ λ )3 + λ λ +λ λ +λ2 3 = 0. 1 2 3 3 1 2 3 1 2 3 1 3 2 1 3 2 3 3 After simple transformations we get (cid:0) (cid:1) (λ λ −λ2) (λ λ +λ λ +λ λ ) λ2 +λ2 +λ λ −λ λ (λ +λ +λ )2 = 0. 1 2 3 1 2 3 1 3 2 1 2 1 2 1 2 1 2 3 The last cond(cid:0)ition is equivalent to (cid:0)the following on(cid:1)e (cid:1) λ λ −λ2 λ λ −λ2 λ λ −λ2 = 0. 1 2 3 1 3 2 3 2 1 Consequently, our c(cid:0)ondition m(cid:1)e(cid:0)ans that e(cid:1)ig(cid:0)envalues o(cid:1)f the matrix J∗J form a geometric progression. 3 Green’s function in R2 In this section we introduce Green function and prove some of its properties. Definition 4. Let Ω be a domain in R2. A function G(~x,~y), ~x 6= ~y ∈ Ω is called Green function for the domain Ω, if it satisfies the following conditions: 1. G(~x,~y) is continuous from below and G(~x,~y) > 0, ~x 6= ~y ∈ Ω; 2. for each fixed point ~y ∈ Ω there is a harmonic function h(~x,~y), ~x ∈ Ω such that 1 1 G(~x,~y) = log +h(~x,~y); 2π |~x−~y| 3. if u(~x) is an arbitrary harmonic function defined in Ω and satisfying the condition u(~x) ≤ G(~x,~y), ~x ∈ Ω\{~y}, 4 then u(~x) ≤ 0, ~x ∈ Ω. In three dimension case, if Ω ⊂ R3, the Green’s function can be defined by the same way replacing the second condition to the following one 1 G(~x,~y) = +h(~x,~y), ~x ∈ Ω\{~y}. 4π|~x−~y| It iswell known, that iftheboundaryofadomainΩ 6= R2 haspositive capacity, thenithasuniqueGreenfunction, see[4], p. 138. Inparticular, ifΩ 6= R2 issimply connected then it has Green function. For a fixed point ~y ∈ Ω and for an arbitrary number 0 < t < +∞ let us denote Ω = {~x; G(~x,~y) > t}. t Note that for arbitrary value of t > 0 the domain Ω is connected and its Green t function is G(~x,~y)−t. Lemma 3. Let u(~x), ~x ∈ Ω, be a harmonic function and {~x; |~x−~x | ≤ r} ⊂ Ω. 0 If for some point ~x, |~x−~x | = r, we have 0 u(~x) = inf{u(~y); |~y −~x | < r} 0 then 1 |∇u(~x)| ≥ (u(~x )−u(~x)). 2r 0 Proof. For arbitrary 0 ≤ ϕ < 2π and 0 < t < 1 we have r2 −(rt)2 r2 −(rt)2 1−t ≥ ≥ . |reiϕ −t(~x−~x )|2 (r +rt)2 2 0 So, u(~x +t(~x−~x ))−u(~x) 0 0 = |~x +t(~x−~x )−~x| 0 0 1 2π r2 −(rt)2 = (u(~x +reiϕ)−u(~x))dϕ ≥ 2π(r−rt) |reiϕ −t(~x−~x )|2 0 Z0 0 1 2π 1 ≥ (u(~x +reiϕ)−u(~x))dϕ = (u(~x )−u(~x)). 4πr 0 2r 0 Z0 5 Passing to the limit if t → 1−0 we get the required result. Remark. The analogous result is true in R3. Theorem 1. Let G(~x,~y) be a Green function for a simply connected domain Ω in R2. Then |∇G(~x,~y)| =6 0, ~x ∈ Ω\{~y}. Proof. Let us assume, that at some point ~x ∈ Ω we have 0 |∇G(~x ,~y)| = 0. 0 Denote Ω+ = {~x, G(~x,~y) > G(~x ,~y)} 0 and − Ω = {~x, G(~x,~y) < G(~x ,~y)}. 0 Note that the domain Ω+ is connected. For sufficiently small number ǫ > 0, such that the following condition B(~x ,ǫ) ⊂ Ω\{~y}, 0 holds, we consider the open set A = B(~x ,ǫ)\{~x; G(~x,~y) = G(~x ,~y)}. 0 0 The set A consists of an even number of components. Otherwise, we could find a point ~x ∈ B(~x ,ǫ)∩{~x; G(~x,~y) = G(~x ,~y)} 1 0 0 in some neighborhood of which the function G(~x,~y) − G(~x ,~y) would preserve 0 its sign. So, ~x would be the point of local extremum, which is impossible for 1 the nonconstant harmonic function G(~x,~y), ~x ∈ B(~x ,ǫ). Moreover, the set A 0 cannot have only two components. Indeed if it had two components then the boundary ∂Ω+ ∩B(~x ,ǫ) would be smooth. Consequently, by lemma 1 we would 0 have |∇G(~x ,~y)| > 0. 0 Thus, the domain A has at least four connected components. This implies that the open set Ω− consists of more than two connected components. Since our domain is simply connected, one of those components has the boundary, completely laying inside of ∂Ω+. On that connected component the function G(~x,~y) is identically constant end equal G(~x ,~y). This is a contradiction. 0 Note that in theorem 1, the condition ”simply connected”, is essential. Indeed, for the domain {~x; 1 < |~x| < 2} theorem 1 does not valid. 6 4 New proof of Riemann’s theorem In this section we give a new proof of Riemann’s well known theorem on conformal mapping. In this proof we do not use methods of complex analysis. Let us denote D(~y,r) = {~x; |~x−~y| < r}. Theorem 2. Let Ω be a simply connected domain in R2. If Ω 6= R2 then there is a one to one conformal mapping ϕ : Ω → D of the domain Ω on the unit disk D = D(~0,1). Proof. Let us fix a point ~y ∈ Ω and G(~x,~y) be Green function of the domain Ω. Let us consider the following dynamical system in Ω\{~y} d~x(t) ∇G(~x(t),~y) = − e2πG(~x(t),~y), 0 < t < 1. (1) dt 2π|∇G(~x(t),~y)|2 For an arbitrary solution of this equation we have d d~x(t) e−2πG(~x(t),~y) = −2πe−2πG(~x(t),~y) ∇G(~x(t),~y), = 1. dt dt (cid:18) (cid:19) (cid:0) (cid:1) Consequently, 1 1 G(~x(t),~y) = ln , 0 < t < 1. 2π t In the neighborhood of each point ~x ∈ Ω\{~y} the equation (1) has a unique solution passing through the point ~x, see [1] p. 19. In the neighborhood of the point ~y the equation (1) may be written in the following form d~x(t) ~x(t)−~y = exp{2πh(~y,~y)}+o(t), t → 0. dt |~x(t)−~y| So, for each solution of our equation we have ~x(t) = ~y +~atexp{2πh(~y,~y)}+o(t), t → 0, where ~a is a vector with norm one. 7 Consequently, for each point x ∈ Ω\{~y} we can find a unique vector~a =~a(~x), such that there is a solution ~x(t) of our equation which passes through the point ~x and at the same time in the neighborhood of the point ~y satisfies the condition ~x(t)−~y lim =~aexp{2πh(~y,~y)}. t→0 t Let us define the mapping ϕ : Ω → D as follows, ϕ(~y) = 0 and for the arbitrary point ~x ∈ Ω\{~y} we put ϕ(~x) =~a(~x)e−2πG(~x,~y). It is obvious, that ϕ(~x) is a one to one mapping and ϕ(Ω) = D. Recall some facts about the constructed mapping, which permit to assert that it is conformal. Let us take two solutions ~x(t), ~x (t) 1 of the equation (1). We denote by α the angle between the vectors ~a(~x(t)) and ~a(~x (t)). For arbitrary numbers 0 < t < t < 1 denote by U the domain bounded 1 0 1 by the curves γ = {~x(t); t < t < t }, γ = {~x (t); t < t < t } 1 0 1 2 1 0 1 and ~ ~ γ = {~x; G(~x,~y) = G(x(t ),~y)}, γ = {~x; G(~x,~y) = G(x(t ),~y)}. 3 0 4 1 Let m~ (~x) be the unite outer normal to the boundary of the domain U at the point ~x ∈ ∂U. For an arbitrary point ~x ∈ γ ∪γ we have 1 2 d~x(t) ,m~ (~x(t)) = 0, t < t < t . dt 0 1 (cid:18) (cid:19) Consequently, ∂G(~x,~y) (∇G(~x,~y),m~ (~x)) = = 0. ∂m~ If ~x ∈ γ then we have m~ (~x) = −~n(~x), where ~n(~x) is the outer normal to the 3 boundary of the domain {~x; G(~x,~y) > t }. If ~x ∈ γ then we have m~ (~x) = ~n(~x), 0 4 where ~n(~x) is the outer normal to the domain {~x; G(~x,~y) > t}. Therefore, ∂G(~x,~y) ∂G(~x,~y) ds = ds. ∂~n ∂~n Zγ3 Zγ4 8 Passing to the limit we get ∂G(~x,~y) ∂G(~x,~y) α = 2π lim ds = 2π ds. t0→+0Zγ3 ∂~n Zγ4 ∂~n From definition of the mapping ϕ(~x) we have |ϕ(~x(t))−ϕ(~x (t)| = |t(~x)||~a(~x(t))−~a(~x (t))| = 1 1 ∂G(~x,~y) = |t(~x)| 2π ds = ∂~n (cid:12) Zγ4 (cid:12) (cid:12) (cid:12) ∂G(~x(t),~y) (cid:12) (cid:12) = 2π exp{−2πG((cid:12)~x(t),~y)}|~x(t)−~x(cid:12)(t)|+o(|~x(t)−~x (t)|). 1 1 ∂~n Further on we can write |ϕ(~x(t+∆t))−ϕ(~x(t))|+o(|∆t|) = |∆t|+o(|∆t|) = d~x(t) −1 = |~x(t+∆t)−~x(t)| = dt (cid:12) (cid:12) (cid:12) (cid:12) ∂G(~x(t),~y) (cid:12) (cid:12) = 2π exp{−2πG(~x(t),~y)}(cid:12)|~x(t+(cid:12)∆t)−~x(t)|+o(|∆t|). ∂~n Let r > 0 be sufficiently small. We choose ∆t and~x (t) such that the equalities 1 |~x(t+∆t)−~x(t)| = |~x(t)−~x (t)| = r 1 hold. The vectors ~x(t+∆t)−~x(t) and ~x(t)−~x (t) 1 are orthogonal. Consequently, the image of the disk D(~x(t),r) is a circle, as a first approximation, once if the orthogonal vectors ϕ(~x(t+∆t))−ϕ(~x(t)) and ϕ(~x(t))−ϕ(~x (t)) 1 satisfy the condition |ϕ(~x(t+∆t))−ϕ(~x(t))| = |ϕ(~x(t))−ϕ(~x (t))|+o(r). 1 9 The last condition holds since ∂G(~x(t),~y) |ϕ(~x(t+∆t))−ϕ(~x(t))| = 2π exp{−2πG(~x(t),~y)}r+o(r) ∂~n and ∂G(~x(t),~y) |ϕ(~x(t))−ϕ(~x (t)| = 2π exp{−2πG(~x(t),~y)}r +o(r). 1 ∂~n Remark. For constructed mapping at the points ~x ∈ Ω\{~y} we have ′ |ϕ(~x)| = 2π|∇G(~x,~y)|exp{−2πG(~x,~y)}. At the point ~y we have ′ |ϕ(~y)| = 2πexp{−2πh(~y,~y)}. 5 Green’s function in R3 Definition 5. We say that a domain Ω ⊂ R3 is simply connected if 1. for an arbitrary bounded domain Ω ⊂ R3 if we have ∂Ω ⊂ Ω then it 1 1 follows Ω ⊂ Ω; 1 2. anarbitraryclosedcurvelayingindomainΩpermitscontinuousdeformation in domain Ω to the point. Lemma 4. Let Ω be a simply connected domain in R3. Let Ω be a bounded domain with smooth boundary and G(~x,~y) is its Green function. Then ∇G(~x,~y) 6= 0, ~x ∈ Ω\{~y}. Proof. Since the boundary of our domain is smooth so, we have ∇G(~x,~y) 6= 0, ~x ∈ ∂Ω. Let us assume that {~x; ~x ∈ Ω, ∇G(~x,~y) = 0} =6 ∅. Let 0 < t < ∞ be the biggest number for which there is a point x ∈ Ω such that 0 0 G(~x ,~y) = t and 0 0 ∇G(~x ,~y) = 0. 0 Denote Ω+ = {~x, G(~x,~y) > t } 0 10