Harmonic Mappings and Shear Construction. References. Introduction - Definitions

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1 Harmonic Mappings and Shear Construction KAUS 212 Stockholm, Sweden Tri Quach Department of Mathematics and Systems Analysis Aalto University School of Science Joint work with S. Ponnusamy and A. Rasila January 27, 212 Contents References Introduction Preliminaries Shear Construction Minimal Surfaces Examples Simple One Polygonal Mappings Polygonal Mappings - ω(z) = z n Polygonal Mappings - ω(z) = z 2n Polygonal Mappings - ω(z) = z 2 Strip Mappings Strip Mappings - ω(z) = z 4 Harmonic Mappings and Shear Construction 2/45 References [CluShe] J. Clunie, T. Sheil-Small, Harmonic univalent functions, Ann. Acad. Sci. Fenn. Ser. A I Math. vol. 9, pp. 3 25, [DriDur] K. Driver, P. Duren, Harmonic shears of regular polygons by hypergeometric functions, J. Math. Anal. Appl. vol. 239, no. 1, pp , [Duren] P. Duren, Harmonic Mappings in the Plane, Cambridge Tracts in Mathematics, 156. Cambridge University Press, 24. [PonQuaRas] S. Ponnusamy, T. Quach, A. Rasila, Harmonic Shears of Slit and Polygonal Mappings, preprint, arxiv: , 212. Introduction - Definitions A real-valued function u(x, y) is harmonic it satisfies the Laplace equation u = 2 x 2 u + 2 y 2 u =. A complex-valued function f (x + iy) = u(x, y) + iv(x, y) from a domain f : D C is harmonic if the two coordinate functions u, v are harmonic. A complex-valued harmonic function is a harmonic mapping if it is univalent (one-to-one). Harmonic Mappings and Shear Construction 3/45 Harmonic Mappings and Shear Construction 4/45

2 Introduction Compositions of Complex-Valued Harmonic Functions Introduction Example 1 If f : Ω C, g : f (Ω) C are harmonic functions, g f is not necessarily harmonic. If f : Ω C is analytic and g : f (Ω) C is harmonic, then g f is harmonic. If f : Ω C is harmonic and g : f (Ω) C is analytic, then g f is not necessarily harmonic. In fact, even a square or the reciprocal of a harmonic function need not be harmonic. Inverse of a harmonic mapping need not be harmonic. Square may not be a harmonic mapping: f (x, y) = x 2 y 2 + ix. Im f 2 = 16x. Simple harmonic mapping that is not necessarily conformal is an affine mapping: f (z) = αz + γ + β z, In fact, we have: α β. If f is harmonic, then αf + γ + β f, is harmonic as well. Harmonic Mappings and Shear Construction 5/45 Harmonic Mappings and Shear Construction 6/45 Introduction Example 2 (1/2) Introduction Example 2 (2/2) Consider a mapping f (z) = z 1 2 z2, which maps the open D onto a hypocycloid with three cusps. To verify its univalence, suppose f (z 1 ) = f (z 2 ) for z 1, z 2 D. Then 2(z 2 z 1 ) = z 2 1 z 2 1 = ( z 1 + z 2 )( z 1 z 2 ). By taking absolute values on both sides. We see that this is impossible unless z 1 = z 2, because z 1 + z 2 < 2. A similar argument shows that f (z) = z 1 n zn is univalent for each n 2. (a) n = 2 (b) n = 5 Figure: f (z) = z 1 n zn. Harmonic Mappings and Shear Construction 7/45 Harmonic Mappings and Shear Construction 8/45

3 Preliminaries Canonical Representation Theorem In a simply connected domain Ω C, a complex-valued harmonic function f has a representation f = h + ḡ, where h and g are analytic in Ω. The representation is unique up to an additive constant. For a harmonic mapping f of the unit disk D, it is convenient to choose the additive constant so that g() =. Then the representation f = h + ḡ is unique and is called the canonical representation of f. Preliminaries / z and / z Definition z = 1 2 where z = x + iy. ( ) x i, y z = 1 ( ) 2 x + i, y Note that f / z = is another way to write the Cauchy-Riemann equations. Laplacian of f is f = 4 2 f z z. Function f with continuous second partial derivatives is harmonic if and only if f / z is analytic. Harmonic Mappings and Shear Construction 9/45 Harmonic Mappings and Shear Construction 1/45 Notations and Jacobian Connection between the differential operators ( ) f = f z z. Suppose that f = h + ḡ, then we have following notations: f z = f / z = h/ z = h, f z = f / z = g/ z = g. Jacobian of f is defined by J f (z) = h (z) 2 g (z) 2. Function f is locally univalent and sense-preserving if J f (z) >. Shear Construction Definition A domain Ω C is said to be convex in the horizontal direction (CHD) if its intersection with each horizontal line is connected (or empty). Shear construction is based on the following Theorem: Theorem Let f = h + g be a harmonic and locally univalent in the unit disk D. Then f is univalent in D and its range is a CHD domain if and only if h g is a conformal mapping of D onto a CHD domain. A way to construct harmonic mappings from analytic functions. Introduced by Clunie and Sheil-Small in 1984 [CluShe]. Harmonic Mappings and Shear Construction 11/45 Harmonic Mappings and Shear Construction 12/45

4 Shear Construction Shear Construction Information we have: Write f = h + ḡ. Denote ϕ = h g. Dilatation ω = g /h, such that ω(z) < 1, for z D. Solve h, g from differential equations: { h g = ϕ, ωh g =. We have h(z) = g(z) = ϕ (ζ) 1 ω(ζ) dζ, Observe that [ f (z) = h(z) + g(z) = 2 Re ϕ (ζ) ω(ζ) 1 ω(ζ) dζ. ϕ ] (ζ) 1 ω(ζ) dζ ϕ(z). Harmonic Mappings and Shear Construction 13/45 Harmonic Mappings and Shear Construction 14/45 Minimal Surfaces Shear Construction & Minimal Surface Minimimal surface is a surface which minimizes the area with a fixed curve as its boundary. Harmonic mapping f = h + ḡ can be lifted to a minimal surface if and only if the dilatation is a square of an analytic function. Suppose that ω = q 2. Then the corresponding minimal surface has the form where {u, v, w} = {Re f, Im f, 2 Im ψ}, ψ(z) = ϕ(ζ) q(ζ) 1 ω(ζ) dζ. Algorithm 1. Supervisor gives you conformal mappings ϕ. 2. Come up with easy dilatations ω. 3. Compute h and g. 4. Construct f = h + ḡ. 5. Compute ψ if possible for minimal surface. 6. Plot f and minimal surface. 7. Profit. Harmonic Mappings and Shear Construction 15/45 Harmonic Mappings and Shear Construction 16/45

5 Example - Simple [Duren, pp. 39-4] (1/4) Example - Simple [Duren, pp. 39-4] (2/4) Suppose ϕ(z) = z, thus ϕ (z) = 1. Let ω(z) = z 2. We have: dζ h(z) = 1 ζ 2 = 1 2 log 1 + z 1 z, Therefore: g(z) = f (z) = h(z) + g(z) = z + 2 Re ζ 2 1 ζ 2 dζ = z log 1 + z 1 z. [ log 1 + z ] 1 z = z + log 1 + z 1 z Figure: Shearing of the identity map with the dilatation ω(z) = z 2. Harmonic Mappings and Shear Construction 17/45 Harmonic Mappings and Shear Construction 18/45 Example - Simple (3/4) Example - Simple (4/4) For the minimal surface, let us compute ψ(z) = ζ 1 ζ 2 dζ = 1 2 log(1 z2 ). Then we have [ ] ( ) u(z) = Re z + log 1 + z 2 Re z 1 z = arctan 1 + z 2 Re z, [ ] v(z) = Im z + log 1 + z 1 z = Im z, [ w(z) = 2 Im 1 ] 2 log(1 z2 ) = Arg (1 z 2 ). Figure: Minimal surface of the identity map with the dilatation ω(z) = z 2. Harmonic Mappings and Shear Construction 19/45 Harmonic Mappings and Shear Construction 2/45

6 Example - Polygonal Mappings (1/2) Example - Polygonal Mappings (2/2) Lemma The function ϕ defined by ϕ(z) = (1 ζ n ) 2/n dζ maps the unit disk D onto regular n-gons Dilatations considered: [DriDur]: z, z n, z n/2. In the last dilatation n is even. [PonQuaRas]: z 2n, z 2. In the latter dilatation n is odd. 1. (a) n = 3 1. Figure: Polygonal mapping ϕ. (b) n = 4 Harmonic Mappings and Shear Construction 21/45 Harmonic Mappings and Shear Construction 22/45 Polygonal Mappings - Hypergeometric Functions Polygonal Mappings - Hypergeometric Functions Gaussian hypergeometric function: (a) n (b) n F (a, b; c; z) = 1 + z n, z < 1, n!(c) n n=1 where (α) n is Pochhammer symbol defined as follows: (α) n = α(α + 1) (α + n 1) = Euler integral presentation: For Re c > Re b > : F (a, b, c; z) = Γ(c) Γ(b)Γ(c b) Γ(α + n), α C. Γ(α) t b 1 (1 t) c b 1 (1 zt) a dt. Appel s hypergeometric function: F 1 (a, b 1, b 2 ; c; x, y) = k= l= (a) k+l (b 1 ) k (b 2 ) l x k y l, (c) k+l k! l! where (α) n is Pochhammer symbol. Euler integral presentation: For Re c > Re a > : F 1 (a, b 1, b 2 ; c; x, y) = C where C = Γ(c) Γ(a)Γ(c a). t a 1 (1 t) c a 1 (1 xt) b 1 (1 yt) b 2 dt, Harmonic Mappings and Shear Construction 23/45 Harmonic Mappings and Shear Construction 24/45

7 Polygonal Mappings - ω(z) = z n (1/5) Idea is to rewrite integrals as Gaussian hypergeometric functions. Chosen dilatation ω(z) = z n leads to integrals: h(z) = (1 ζ n ) 1 2/n dζ, g(z) = By substituting u = ζ n, we have h(z) = 1 n n ζ n (1 ζ n ) 1 2/n dζ. (1 u) 1 2/n u 1/n 1 du. Polygonal Mappings - ω(z) = z n (2/5) Substitution u = z n t leads to h(z) = 1 n (1 z n t) 1 2/n (z n t) 1/n 1 z n }{{} =zt 1/n 1 = z t 1/n 1 (1 z n t) 1 2/n dt n = zf (1 + 2n, 1n ; 1 + 1n ) ; zn. By similar procedure, we have g(z) = zn+1 n + 1 F dt (1 + 2n, 1 + 1n ; 2 + 1n ; zn ). Harmonic Mappings and Shear Construction 25/45 Harmonic Mappings and Shear Construction 26/45 Polygonal Mappings - ω(z) = z n (3/5) Polygonal Mappings - ω(z) = z n (4/5) (a) n = 3 (b) n = 4 Figure: Harmonic shears of polygonal mappings with a dilatation ω(z) = z n. For minimal surfaces, we assume that n is even, say n = 2m. Then q(z) = z m and h (z) = (1 z 2m ) 1 1/m. Therefore ψ(z) = = zm+1 2m ζ m (1 ζ 2m ) 1 1/m dζ = 2 z1+n/2 n + 2 F t 1/2m 1/2 (1 z 2m t) 1 1/m dt (1 + 2n, n ; n ; zn ). Harmonic Mappings and Shear Construction 27/45 Harmonic Mappings and Shear Construction 28/45

8 Polygonal Mappings - ω(z) = z n (5/5) Polygonal Mappings - ω(z) = z 2n (1/5) Figure: Minimal surface of the polygonal mappings with the dilatation ω(z) = z n, for n = 4. Chosen dilatation ω(z) = z 2n leads to integrals: h(z) = (1 ζ n ) 2/n (1 ζ 2n ) 1 dζ, g(z) = ζ 2n (1 ζ n ) 2/n (1 ζ 2n ) 1 dζ. This can be rewritten as: h(z) = (1 ζ n ) 1 2/n (1 + ζ n ) 1 dζ, g(z) = ζ 2n (1 ζ n ) 1 2/n (1 + ζ n ) 1 dζ. Harmonic Mappings and Shear Construction 29/45 Harmonic Mappings and Shear Construction 3/45 Polygonal Mappings - ω(z) = z 2n (2/5) Polygonal Mappings - ω(z) = z 2n (3/5) By the change of variable ζ = zt 1/n, we obtain h(z) = z n (1 z n t) 1 2/n (1 + z n t) 1 t 1/n 1 dt. This can be rewritten by using Appel s hypergeometric functions as follows: ( 1 h(z) = zf 1 n, n, 1; ) n ; zn, z n Again in the same fashion, we have g(z) = z2n+1 2n + 1 F 1 (2 + 1n, 1 + 2n, 1; 3 + 1n ; zn, z n ). (a) n = 3 (b) n = 4 Figure: Harmonic shears of polygonal mappings with a dilatation ω(z) = z 2n. Harmonic Mappings and Shear Construction 31/45 Harmonic Mappings and Shear Construction 32/45

9 Polygonal Mappings - ω(z) = z 2n (4/5) Polygonal Mappings - ω(z) = z 2n (5/5) The minimal surface is determined by the integral ψ(z) = ζ n (1 ζ n ) 1 2/n (1 + ζ n ) 1 dζ = zn+1 t 1/n (1 z n t) 1 2/n (1 + z n t) 1 dt n ( = z n+1 F n, n ; 1; ) n ; zn, z n. Figure: Minimal surface of the polygonal mappings with the dilatation ω(z) = z 2n, for n = 4. Harmonic Mappings and Shear Construction 33/45 Harmonic Mappings and Shear Construction 34/45 Polygonal Mappings - ω(z) = z 2 (1/5) In this example, we assume that n is odd. This time we have following integrals: h(z) = (1 ζ n ) 2/n (1 ζ 2 ) 1 dζ, g(z) = ζ 2 (1 ζ n ) 2/n (1 ζ 2 ) 1 dζ. Recall the following identities: { 1 + ζ n = (1 + ζ)(1 ζ + ζ n 2 + ζ n 1 ), n is odd, 1 ζ n = (1 ζ)(1 + ζ + + ζ n 2 + ζ n 1 ), From above, we have 1 + ζ n 1 ζ n 1 + ζ 1 ζ = 1 + ζ2 + + ζ 2(n 1). Polygonal Mappings - ω(z) = z 2 (2/5) h(z) = The above identity leads to 1 + ζ ζ 2(n 1) n 1 (1 ζ n ) 1+2/n (1 + ζ n ) dζ = By computation, we have h(z) = g(z) = n 1 k= n k=1 z 2k+1 ( 2k + 1 2k + 1 F 1 n ( 2k + 1 z 2k+1 2k + 1 F 1 n k=, n, n ζ 2k (1 ζ n ) 1+2/n (1 + ζ n ) dζ. ) 2k + 1, 1; 1 +, z n, z n, n ) 2k + 1, 1; 1 +, z n, z n. n Harmonic Mappings and Shear Construction 35/45 Harmonic Mappings and Shear Construction 36/45

10 Polygonal Mappings - ω(z) = z 2 (3/5) Polygonal Mappings - ω(z) = z 2 (4/5) The minimal surface is determined by the integral (a) n = 3 (b) n = 5 Figure: Harmonic shears of polygonal mappings with a dilatation ω(z) = z 2. ψ(z) = n 1 = = ζ(1 ζ n ) 2/n (1 ζ 2 ) 1 dζ k= n 1 z 2(k+1) k= n 1 = k= n ζ 2k+1 (1 ζ n ) 1 2/n (1 + ζ n ) 1 dζ t 2(k+1)/n 1 (1 z n t) 1 2/n (1 + z n t) 1 dt z 2(k+1) ( 2(k + 1) 2(k + 1) F 1, n n ) 2(k + 1) ; 1; 1 + ; z n, z n. n Harmonic Mappings and Shear Construction 37/45 Harmonic Mappings and Shear Construction 38/45 Polygonal Mappings - ω(z) = z 2 (5/5) Figure: Minimal surface of the polygonal mappings with the dilatation ω(z) = z 2, for n = 5. Strip Mappings (1/2) Lemma The function ϕ defined by ϕ(z) = 1 + ζ 4 1 ζ 4 dζ maps the unit disk D onto a CHD domain Ω. Ω = Ω 1 Ω 2, where { [ ]} e iπ/4 Γ(5/4) Ω 1 = x + iy : x R, y Re, Γ(1/4) { [ ] } e iπ/4 Γ(5/4) Ω 2 = x + iy : x Re, y R. Γ(1/4) Harmonic Mappings and Shear Construction 39/45 Harmonic Mappings and Shear Construction 4/45

11 Strip Mappings (2/2) Strip Mappings - ω(z) = z 4 (1/4) Figure: Strip Mappings Chosen dilatation ω(z) = z 4 leads to integrals: (1 + ζ 4 ) 1/2 h(z) = (1 ζ 4 ) 2 dζ. g(z) = By the substitution ζ = zt 1/4, we have ζ 4 (1 + ζ4 ) 1/2 (1 ζ 4 ) 2 dζ. h(z) = zf 1 ( 1 4, 2, 1 2 ; 5 4 ; z4, z 4 ), ( g(z) = z5 5 5 F 1 4, 2, 1 2 ; 9 ) 4 ; z4, z 4. Harmonic Mappings and Shear Construction 41/45 Harmonic Mappings and Shear Construction 42/45 Strip Mappings - ω(z) = z 4 (2/4) Strip Mappings - ω(z) = z 4 (3/4) Figure: Harmonic shear of the strip mapping with dilatation ω(z) = z 4. For minimal surface, we compute ψ(z) = The substitution ζ = zt 1/4 leads to ψ(z) = z3 4 = z3 3 F 1 ζ 2 (1 ζ 4 ) 2 (1 + ζ 4 ) 1/2 dζ. t 1/4 (1 z 4 t) 2 (1 + z 4 t) 1/2 dt ( 3 4, 2, 1 2 ; 7 ) 4 ; z4, z 4. Harmonic Mappings and Shear Construction 43/45 Harmonic Mappings and Shear Construction 44/45

12 Strip Mappings - ω(z) = z 4 (4/4) Figure: Minimal surface associated with the harmonic shear of the strip mapping with dilatation ω(z) = z 4. Harmonic Mappings and Shear Construction 45/45