Solution to Homework 4, Math 7651, Tanveer 1. Show that the function w(z) = 1 2

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1 Solution to Homework 4, Math 7651, Tanveer 1. Show that the function w(z) = 1 (z + 1/z) maps the exterior of a unit circle centered around the origin in the z-plane to the exterior of a straight line cut from 1 to 1 in the w-pane. What is the inverse of this mapping. Be specific on choice of branch if the inverse function involves branches. Solution: If we express z = re iθ, then the domain of interest in the z-plane corresponds to r > 1. We note than that w = ξ + iη = 1 ( r + 1 ) cos θ + 1 ( r 1 ) sin θ (1) r r and minor axis So, r = const > 1 corresponds ellipses with major axis r + 1 r r 1 r. We note that as r, w 1 z and as r 1+, w cos θ, and therefore the boundary z = 1 is mapped to the slit [ 1, 1] in the w-plane. We note that since zw = z + 1, solving the quadratic and seeking the root that matches asymptotic behavior w z, as z, we have z = w + w 1 () where we define [ i w 1 = w 1 exp arg(w 1) + i ] arg(w + 1), with arg (w±1) ( π, π] (3). Solve the potential problem φ = in the strip region < y < 1 with boundary condition φ(x, 1) = 1, φ(x, ) =. What a priori assumption is needed on the growth of φ as x ± to assure that there is only one solution to this problem. Solution: Clearly on inspection φ(x, y) = y is a solution of φ = and satisfies stated boundary conditions. We now address the question fo uniqueness. If φ is another solution, we then have for Φ = φ φ satisfy Φ = with homogenous boundary conditions Φ(x, ) = Φ(x, 1) =. Uniqueness revolves around question of use of maximimum principle. Let Ψ be the harmonic conjugate of Φ. Then, consider the analytic function χ = e Φ+iΨ, (4) We note that maximimum principle applied to χ and 1/χ relate to maximimum principle and minimum principles for Φ. Now, note with z = x + iy, ζ = ζ(z) := e πz (5) maps the strip in the z-plane to the upper-half plane, with {z : Iz = } mapped to {ζ : Iζ =, Rζ > } and {z : Iz = 1} mapped to {ζ : Iζ =, Rζ < }. In order to apply the Phragmen-Lindeloff principle to χ, we have to assume χ is not singular at ζ = (corresponding to z = and that that χ e C ζ β = e Ceβπx (6) 1

2 for some β [, 1). This means that we have to assume we are in the class of functions for which φ(x) Ce βπx as x (7) Now consider the a priori condition on Φ + iψ that rules out a singularity at ζ =. Claim it is enough to require a priori that Φ + iψ is weakly singular at ζ =. To prove this consider γ(ζ) = iζ(φ + iψ). it is analytic in the half-plane and continuous upto the real axis, including ζ =. Since Iγ =, Schwartz reflection principle implies that γ is an analytic function in the lower-half plane as well, and in particular analytic at ζ =. But from a priori condition on weak singularity of Φ + iψ, it follows that γ() =, and therefore Φ + iψ is analytic at ζ =. Now, we claim that the a priori weak singularity hypothesis of Φ + iψ is implied by requiring Φy Φ x, Ce βπx as x (8) for β [, 1). To prove this we note that from Cauchy Riemann condition d dz (Φ + iψ) = Φ x iφ y. Therefore, using ζ = e πz, ζ (z) = e πx, (8) may be replaced by d dζ (Φ + iψ) C ζ β+1 (9) near ζ = (i.e. x = ). On integration, we obtain Φ + iψ C ζ β (1) which gives the a priori weak singularity condition. Thus the a priori condition on the class of solution φ to the originally posed problem should include φy φ x, Ce βπx as x (11) for some β [, 1), in addition to (7). Note: It is possible to replace (8) by Φ Ce βπx, by using a more refined argument. This is expected from the symmetry of the problem between x = and x = + and the result (7) 3. Solve the potential problem for φ(x, y) φ = in y > with φ(x, ) = 1 for x > 1, φ(x, ) = 1 for x < 1 and φ (x, ) = for x < 1 y State and prove conditions that will make the solution you find unique. Hint: Think of mapping to a semi-infinite strip and Schwarz Reflection Principle.

3 Solution: We want to map the upper-half z = x + iy plane H into the semiinfinite half-strip S := {ζ : Iζ > : Rζ ( 1, 1)} We already know from class that the map from H to the rectangular domain S c := {ζ : c > Iζ > : Rζ ( 1, 1)} with z = ±1 corresponding to ζ = ±1 and z = ±1+ic corresponding to ζ = ± 1 k is given from Scwartz reflection principle to 1 z ζ = A dz 1 z 1 k z and that in the limit k +, c corresponding to the semi-infinite strip S. Therefore, for a strip. ζ = A sin 1 z. Or z = sin z A. In order for z = ±1 to correspond to ζ = ±1, A = πζ π, or z = g(ζ) = sin maps S into H. We can check in particular that real axis segement [ 1, 1] is mapped to real axis segment [ 1, 1] This map is conformal (i.e. preserves angles) except at ζ = ±1 where g =. Therefore, the boundary condition y φ(x, ) = for x ( 1, 1) becomes in the ζ = ξ + iη plane the following equation η Φ(ξ, ) = for ξ ( 1, 1) (1) where Φ(ξ, η) = φ (x(ξ, η), y(ξ, η)) where x(ξ, η) and y(ξ, η) is defined by x+iy = g(ξ + iη). Translating the conditions on φ on different parts of H, the other boundary conditions on Φ(ξ, η) in S are Φ( 1, η) = 1, Φ(1, η) = 1 for η > (13) We note that Φ(ξ, η) = ξ solves ξ,η Φ = as well as satisfy boundary conditions (1), (13). This implies φ(x, y) = ξ(x, y) = Rg 1 (x + iy) = π R { sin 1 (x + iy) } (14) To explore whether or not this is the only solution take the difference between two solutions Φ in S. Then, define analytic function W so that W (ξ + iη) = Φ(ξ, η) + i Ψ(ξ, η) Therefore, if we assume that the solution we are seeking has continuous first derivatives as η + for ξ ( 1, 1), then from Cauchy Riemann conditions for ξ ( 1, 1) η Φ(ξ, ) = ξ Ψ(ξ, ) 1 Variables z and ζ are switched here from class notes 3

4 implying Ψ(ξ, ) = without loss of generality From Schwartz reflection principle W (ζ) extends to an analytic function to the entire-strip S := {ζ : 1 < Rζ < 1} Further, since the solutions are assumed continuous upto the boundary, then repeated reflection using Schwartz reflection principle gives W to be an entire function of ζ, and therefore using Phragmen-Lindeloff principle on e W and e W, it follows that if we assume Φ(ξ, η) c 1 + c ξ + η α/, i.e. α [, 1), then the difference of two solutions Φ(ξ, η) =, and we have uniqueness. 4. a. Find a Mobius transformation that maps the unit disk centered at to the upper-half plane. b. Find a 1-1 analytic mapping of the interior of z 1 i = to the interior of the unit disk centered at i/. Solution: Consider ζ = f(z) = i z+1 z 1. It is readily checked that f(eiθ ) = cot θ maps the unit circle to the real line with z = ±1 corresponding to ζ = and ζ = respectively. This is a fractional linear map and therefore 1-1 on inspection. Further, we may check that f() = i confirming that f : D 1 H, where H is the upper-half plane. b. Find a 1-1 analytic mapping of the interior of z 1 i = to the interior of the unit disk centered at i/. Solution Note S (z) = 1 (z 1 i) maps D, which is the interior of the circle z 1 i of radius into the D 1, the unit circle centered at the origin. Further note that S 1 (w) = w i also maps the unit circle centered at i/ into D 1. Therefore, it is clear that in a 1-1 analytic manner. So f S 1 1 S : D D 1 f(z) = i + 1 (z 1 i) 5. Extend the proof of solution to Theodersen s integral equation to the space Ḣ 1, i.e. space of π-periodic functions f so that both f and f L (, π) with norm f Ḣ1 = f L + f L. Solution: Recall that if we define φ(ν) = ν + χ(ν), then χ is π periodic, and Theodorsen s integral equation becomes χ(ν) = H [ρ] (ν) (15) where It is convenient to define ρ(ν) = log [R (ν + χ(ν))] (16) ρ 1 (ν) = log [R (ν + χ 1 (ν))], ρ (ν) = log [R (ν + χ (ν))] (17) 4

5 Then, if R (ν) ɛ and R(ν) R, we note from meanvalue theorem that ρ 1 (ν) ρ (ν) ɛ χ1 (ν) χ (ν) (18) R Also, on taking derivative we have ρ (ν) = (1 + χ (ν)) [ ] d dφ log R φ=ν+χ (19) Defining (log R) (ν + χ 1 ) = P 1 (ν), (log R) (ν + χ ) = P (ν), () We note that both P 1 and P satisfy while if we define M = then from mean-value theorem, From (19), it follows that P 1, (ν) ɛ R (1) sup d φ [,π] dφ log R, () P 1 (ν) P (ν) M χ 1 (ν) χ (ν). (3) ρ (ν) ρ 1(ν) P χ (ν) χ 1(ν) + P P χ 1(ν) ɛ χ R (ν) χ 1(ν) + M (1 + χ 1 ) χ (ν) χ 1 (ν) (4) Recall Hilbert transform operator commutes with derivative and that for any ψ L, H[ψ] L ψ L. Therefore, the estimate (4) implies that for χ 1, χ in a ball of size B around the origin in the Ḣ1 space, d dν H [ρ ρ 1 ] L ɛ χ χ R 1 L + M(1 + B) χ χ 1 ɛ χ χ R 1 L + CM(1 + B) χ χ 1 Ḣ1, (5) where we used the fact that in 1-D there exists some constant so that χ C χ Ḣ1. From class notes, we already know that ρ ρ 1 L ɛ R χ χ 1 L (6) Therefore, it follows from (5)-(6) that in a ball of size B, H [ρ ] H [ρ 1 ] Ḣ1 ɛ 1 (1 + B) χ 1 χ Ḣ1, (7) 5

6 where { } ɛ ɛ 1 = max, CM R (8) If we write the Theodorsen integral equation as χ = H [ρ ] + H [ρ ρ ], (9) where ρ = log R(ν) (3) it follows from the estimate (7) that in a Ball B = ρ Ḣ1 for sufficiently small ɛ 1, we have a contractive map in the Banach space Ḣ1 and therefore the integral equation (9) has a unique solution. Note that the assumption of small ɛ 1 implies that not only does R and R exist and R has a lower limit R that stays away from, but also that R is not large. 6