Week 6 notes. dz = 0, it follows by taking limit that f has no zeros in B δ/2, contradicting the hypothesis that f(z 0 ) = 0.

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1 Week 6 notes. Proof of Riemann Mapping Theorem Remark.. As far as the Riemann-mapping theorem and properties of the map near the boundary, I am following the approach in O. Costin notes, though I have avoided introducing concepts that we don t absolutely need. We will do some preliminary Lemmas that will help us prove the existence of a analytic homeomorphism f : D D (unit disk with as the center), where D is simply connected. Using a combination of explicit inversion, translation and dilatation, there is no loss of generality in restricting to D D, where D contains. First, we will need some preliminary lemmas. Lemma.. (Hurwitz Theorem) If f n are analytic and f n in a domain D C and f n f uniformly on compact sets, then if the limiting f is not identically zero, then f in D either. Proof. First note that the limiting function f must be analytic. For an arbitrary contour closed contour C contained in an arbirary compact subset of D, we have C f n(z)dz = and on taking the limit, the same is true for C f(z)dz. Moerara s theorem implies analyticity of f. Assume there exists z D where f(z ) =. Since zeros of a nontrivial analytic f are isolated, there exists a neighborhood B δ D around z so that f on B δ \ {z }. Using Cauchy integral formula, both f n and f n converge uniformly on B δ/ ; therefore, f n/f n f /f on B δ/ uniformly. Therefore, since f n B δ/ fn dz =, it follows by taking limit that f has no zeros in B δ/, contradicting the hypothesis that f(z ) =. Lemma.3. Assume F is a family of analytic functions f : D D. Then for any sequence {f n } n= F, there exists a subsequence that converges uniformly in an arbitrary compact subset K of D. The limiting function f is analytic in D. Proof. Take any compact set K D. Define d = dist{k, D}. Take another compact subset K so that K K D with dist{k, K } d/. Then, since for z K, for any f F, f(z), and therefore, for z K, (.) f (z) = K f(z )dz (z z) 4L d, where L is the length of K. Hence the family F is equi-continuous. From Arzela- Ascoli Theorem, there exists a subsequence {f nj } j= which converges uniformly in the compact set K. Uniform convergence implies that the limiting function f is analytic, as argued from using Moerara s theorem as in the last Lemma. Definition.4. Assume D D and contains. Define S to be the set of analytic functions f : D D with f() = and f () > so that f : f(d) D exists as a single valued function (which is analytic). Remark.5. Note that f S need not be an analytic homeomorphism between D and D since we only have f(d) D. However, any analytic homeomorphism between D and D belongs to S. Also, note that S F. Notice that the set is not empty since the identity map is in S.

2 Proof of Riemann mapping theorem Define (.) M = sup{f () : f S} Clearly M since for identity map f () =. This means there exists a sequence of function f n S so that f n () = and lim n f n () = M. Since {f n} n= F, it follows from previous lemma that there exists a subsequence { } f nj with the j= propertythatf nj F uniformly incompactsubsetsofd. ThislimitingF, satisfies F() =, F () = M, the latter because f n() = f n(z)dz z =δ z and uniform convergence allows one to take limit inside. Now, take any z D and consider g(z) = F(z) F(z ) and the sequence of functions g nj (z) = f nj (z) f nj (z ). We know that for any compact K D \{z }, g nj converges uniformly to g, and since f nj is one-one, g nj (z) for z K. From Hurwitz Theorem, g(z) for z K since g is not identically zero, since F () = M. Since z D is arbitrary, this shows that the limiting analytic function F is one-one between D and F(D). This implies that F S. Assume contrary to what needs to be proved, there exists a D, so that F(z) a for z D. Consider the map F(z)+a (.3) h(z) = +āf(z) Calculation shows (.4) h () = M( a )/( a) Since F(z) D, it follows from properties of Mobius map that F(z)+a +āf(z) <, and since F(z) a, it follows that h is one-one between D and h(d). We notice h() = b := a. Now, consider the following Mobius map that moves b to zero followed by a rotation to ensure that the map is in S. (.5) h (z) = We note that h () = and ( h(z) b bh(z) ) h () h () (.6) h () = M( a ) a ( a ) = M a (+ a ) > M since a <. Sinceh isamobiusmapofhandh(z)isone-one, itfollowsthath (z) is also one-one. Therefore h S. However, h () > M contradicts the definition of M in (.). This contradicts the assumption that there exists a D for which F(z)+a for all z D, implying F : D D is an analytic homeomorphism between D and D. We proved uniqueness in an earlier Lemma.. Boundary Properties of Conformal Map Remark.. From Riemann mapping theorem, there exists - analytic mapping (i.e. analytic homeomorphism) f : D D for any simply connected domain D. We now seek to understand how f behaves near the boundary. The following Lemmas address the issue of boundary behavior of f.

3 Definition.. A set of points of {z n } n= D is said to approach the boundary D, if for any compact set K D, there exists n so that z n / K for n > n. A curve Γ D, parameterized by Γ : {z : z = γ(t) : a < t < b}, is said to approach D if there exists t (a,b) so that z = γ(t) / K for t > t. Lemma.3. If {z n } n= or Γ approaches D, then f(z n) (or f(γ(t))) approaches D. Proof. Let K D be an arbitrarycompact set. Then, sincef is continuous(since it is analytic), K = f (K ) D is compact. There exists n so that z n / K for n > n. Therefore, because f is -, f(z n ) / f(k) = K. Therefore {f(z n )} n= approaches D. Similar proof goes for f(γ(t)). Corollary.4. If f : D D is an analytic homeomorphism, then if {z n } D approaches D, then f(z n ) as {z n } approaches D. Proof. Choose K,n = { ζ : ζ n}. From previous Lemma, there exists for each n, an integer m n so that f(z j ) / K,n, i.e. > f(z j ) > n for j > m n. This gives the desired result. Lemma.5. If D is a simple closed curve that is continuous (i.e., the graph of a continous - function from [a,b] to D, except at the end points, where γ(a) = γ(b)), and h : D D is an analytic homeomorphism, then h extends to a continous homeomorphism between D and D. Further if D is analytic, then h can be analytically continued across D. Proof. We will only outline of the proof and that too for the analytic case. Let S : D H + be an explicit fractional linear map that maps the circle into the upper-half plane. Suppose the boundary is analytic. Then there exists an analytic function g : [a,b] D where g is analytic and - except at the end points, where g(a) = g(b). The composition S h g is analytic at points on one side of the real axis close to (a,b) and approaches real value as (a,b) is approached, from previous Lemma. From reflection principle this composition is analytic on (a, b). Since S and g are locally invertible analytic functions, so must be h. 3. Theodorsen s integral equation for explicit construction of analytic homeomorphism Riemann mapping theorem, though general, is not constructive and provides no clues about the analytic homeomorphism f : D D. This is not the case in the Theodorsen s formulation, though it is restricted to D being near circular. Using sequence of explicit maps, many other domains can be mapped to D, and so the near circular limitation is not as serious as it appears. We start by developing some preliminary notions that relate real and imaginary parts of an analytic function in the disk D. 3.. Hilbert Transform and properties. Definition 3.. For π periodic C function φ, we define Hilbert Transform H so that (3.7) H[φ](ν) = π ( ) ν ν π cot φ(ν )dν 3

4 4 It is readily checked from explicit integration that H[] = and that for k Z +, we get () (3.8) H[cos(kν)] = sin(kν), H[sin(kν)] = cos(kν) This implies for any k Z \, H [ e ikν] = isgn(k)e ikν. The above properties allow extension of operator H to π-periodic function that are L [,π], henceforth denoted as L. Assume (3.9) φ(ν) = k Z Then, we may define (3.) H[φ](ν) = k Z\{} φ k e ikν isgn(k)φ k e ikν From Parseval s equality, it follows that H[φ] L φ L, and that the equality holds iff φ =. Further, it also follows that H H = I on L functions with zero average, i.e. φ = ; this may be readily checked from (3.). Lemma 3.. Assume g to be analytic in D and g C ( D ). Suppose g(re iν ) = φ(r,ν)+iψ(r,ν), then on ζ = e iν, (3.) ψ(,ν) = ψ()+h[φ(, )](ν) Proof. Recall from Posson integral formula relating a Harmonic function in the unit circle to the conjugate harmonic function on the boundary (see Theorem 3.4, page, week notes), we have for ζ = re iν, (3.) ψ(r,ν) = ψ()+ π π rsin(ν θ)φ(,θ)dθ rcos(ν θ)+r Now, if we had φ(,θ) =, we know that the harmonic φ(r,ν) =, implying from C-R conditions that ψ(r,ν) = const.. However, if ψ() =, then ψ(r,ν) =. This implies (3.3) = π rsin(ν θ)dθ rcos(ν θ)+r Subtracting from (3.) a multiple of (3.3), it follows (3.4) ψ(r,ν) = ψ()+ π π rsin(ν θ)[φ(,θ) φ(,ν)]dθ rcos(ν θ)+r φ(,θ) φ(,ν) From assumed regularity of g on D, lim θ ν θ ν taking the limit r in (3.4) results in (3.5) ψ(,ν) = ψ()+ π π sin(ν θ) Using cos(ν θ) = cot ν θ, and using π definition of H. sin(θ ν)[φ(,θ) φ(,ν)]dθ cos(ν θ) exists. Therefore, on cot ν θ =, the Lemma follows on () Using H[] =, it is useful to express it as π π cot ν ν [cos(kν ) cos(kν)]dν, use simplifications of cosa cosb in terms of sin before integration can be done. Similarly, with sin(kν).

5 5 Remark 3.3. We could use the formulae (3.8) to extend Lemma 3. to analytic g in D with L valued boundary values. This is left as an exercise. 3.. Derivation of Theodorsen Integral equation. Assume we have a simply connected bounded domain D whose boundary is described by (3.6) D := { z C : z = R(φ)e iφ, φ π } where R is a π periodic smooth function with R > and R small. Consider the map f : D D where f() = and f() = R. For ζ D, we use the parametrization (3.7) ζ = e iν, with ν π Then, the boundary angle correspondence φ = φ(ν) is determined from f ( e iν) = R(φ)e iφ. From orientation preserving nature of f, note φ is an increasing function while φ(ν) ν is a π periodic function. Now consider the analytic function ( ) f(ζ) (3.8) g(ζ) = log ζ For ζ = e iν D, we have (3.9) Re g(ζ) = logr(φ(ν)) and (3.) Im g(ζ) = φ(ν) ν Using Lemma 3. π (3.) φ(ν) ν = π ( ) ν ν cot log(r(φ(ν )) Equation (3.) is the Theodorsen integral equation for the boundary correspondence function φ(ν). It will be proved that this equation has solution provided the boundary is regular with small enough R, i.e. near circle. Once this is established, and we may determine Fourier coefficients in the form (3.) φ(ν) ν = b j sin(jν +δ j ). Then (3.) implies ( ) f(ζ) (3.3) log = logr()+ c j ζ j, where c j = b j e iδj ζ j= j= wheretheconditionf() = R()fixesthe realconstantlogr()appearingin(3.3). Thus the analytic homeomorphism f : D D is determined as a a series Solution to Theodorsen s integral equation. It is useful to define (3.4) χ(ν) = φ(ν) ν We note that while φ is not π periodic function of ν, χ is. Then, we may write Theodorsen integral equation in (3.) abstractly as (3.5) χ(ν) = H[ρ( )](ν) =: M[χ](ν) where (3.6) ρ(ν) := log[r(ν +χ(ν))]

6 6 It is to be noted from properties of R(φ) that that ρ is π periodic in ν. We will now prove that the operator M is contractive in the space L provided we assume R is small enough. To keep the notation from getting clumsy, we define (3.7) ρ (ν) = logr(ν +χ (ν)), ρ (ν) = logr(ν +χ (ν)) Lemma 3.4. Assume R(φ) > R > and R (φ) < ǫ. Then, for sufficiently small ǫ, the operator M : L L is contactive. and therefore there exists unique solution χ L satisfying (3.5). Proof. It is readily checked from (3.7) and applying mean value theorem that (3.8) ρ (ν) ρ (ν) ǫ χ (ν) χ (ν) R Therefore, (3.9) ρ ρ L ǫ χ χ L R Since (3.3) H[ρ ] H[ρ ] L = H[ρ ρ ] L ρ ρ L It follows that (3.3) M[χ ] M[χ ] ǫ R χ χ L and therefore M : L L is contractive for small enough ǫ. If we define (3.3) χ (ν) = M[](ν), it follows from (3.5) that if χ B L, a ball of sized χ L about the origin, then it follows for small enough ǫ that (3.33) M[χ] L M[] L + M[χ] M[] L χ L Hence M : B B. Therefore, from contraction mapping theorem, there exists unique solution χ satisfying (3.5) Remark 3.5. Note that while unique solution has been proved in a small ball around the origin, the solution is globally unique from Riemann mapping theorem for domains D whose boundary D = f( D ) are images of L valued functions. Also, regularity of φ(ν) can be proved if we assume R to be smooth by using higher order Sobolev spaces Ḣk of π periodic functions. 4. Introduction to Elliptic functions Remark 4.. The treatment here follows the classic Whittaker & Watson s Modern analysis book. Definition 4.. A function f is doubly periodic if there exists two nonzero ω,ω C, whose ratio is not purely real, such that (4.34) f(z + ω ) = f(z) ; f(z + ω ) = f(z) for all values of z where f(z) exists. Definition 4.3. A doubly-periodic meromorphic function (functions with only pole singularities) is called an elliptic function.

7 7 Remark 4.4. The periods ω and ω play the same role in the elliptic function theory as a single period does in theory of circular (trigonometric) functions. Definition 4.5. The parallelogram obtained by joining, ω, ω + ω, ω and is called a fundamental period-parallelogram there is no point ω, except for the vertices, with the property that f(z + ω) = f(z). ω ω + ω ω Figure. Fundamental Period Parallelogram Remark 4.6. It is clear that translations of the fundamental period-parallelogram (in Fig. ) cover C (complex-plane) in a mesh, with vertices at m ω + n ω, m, n Z (integers). Doubly periodic property of an elliptic function implies that the same set of values is taken in each period parallelogram. Remark 4.7. For integration purposes, it it is not convenient to deal with actual meshes if they have singularities of the function on the boundaries of the period- parallelogram. Because of periodicity, there is no loss of generality in choosing a contour, not necessarily coinciding with a period-parallelogram, but a mere-translation of it that guarantees no singularities on the boundary. Such a parallelogram will be referred to as a cell. Definition 4.8. The set of poles (or zeros) of an elliptic function in any given cell is called an irreducible set; all other poles (or zeros) of the function are congruent to it, i.e. differ by an additive factor of m ω + n ω for Z. Lemma 4.9. The number of poles of an elliptic function f(z) in any cell must be finite. Proof. Suppose otherwise. Then, there must be a limit point of the set of poles within the finite cell. The limit point is a non-isolated singularity, which contradicts the definition of elliptic function. Lemma 4.. The number of zeros of an elliptic function f(z) in any cell must be finite as well. Proof. Note /f is an elliptic function as well, with zeros of f becoming poles of its reciprocal. Applying 4.9, the proof follows.

8 8 Lemma 4.. The sum of residues at all poles of an elliptic function in a cell is zero. Proof. Let f be an elliptic function. Let C be the the boundary of a cell with vertices t, t + ω, t + ω + ω and t + ω. Then f(z)dz = { t+ω t+ω+ω t+ω t } f(z)dz C t t+ω t+ω +ω t+ω () The first and third integrals, as well as second and fourth integrals, can be combined as: t + ω t [f(z) f(z +ω )]dz t+ω t [f(z) f(z +ω )]dz and each of these integrands vanish due to periodicity. So, the lemma follows. Remark 4.. There cannot be an elliptic function with only one simple pole in a cell, as otherwise the sum of residues cannot be zero. Lemma 4.3. An elliptic function with no poles in a cell is merely a constant. Proof. If elliptic f is analytic everywherein a cell, it follows f(z) K (constant) for z on or within the cell, since the cell is a compact set. From periodicity, f is bounded everywhere. From Liouville s theorem, f must be a constant. Definition 4.4. The order of an elliptic function is the number of roots in a cell of the equation (4.35) f(z) = c Remark 4.5. As will be shown, this is independent of the value of c chosen, but only depends on the elliptic function under consideration. Lemma 4.6. The order of an elliptic function f is the same as the number of poles (including multiplicities) of f in a cell. Proof. Consider (4.36) C f (z) f(z) c dz Without loss of generality, the cell boundary C is chosen so that it not only avoids poles of f(z), but also zeros of f(z) c. Breaking up the integral (4.36) into four parts, as in the proof of Lemma 4. and using periodicity of the integrand as before, we conclude that the expression in (4.36) is zero. However, (4.36) is the merely the difference between the number of zeros of f(z) c and the number of poles of f(z) c (the same as f(z)), within the cell. Thus, the Lemma follows. Remark 4.7. Note that the number of poles of f is not dependent on c and hence the order of an elliptic function, as defined in definition 4.4, is independent of c, as claimed above. Lemma 4.8. The sum of the irreducible zeros (including multiplicity) is congruent to the sum of poles (including multiplicity).

9 9 Proof. Let C denote the boundary of a cell, that will be chosen, without loss of generality, to avoid any zeros or poles of f. Then, (4.37) zf (z) C f(z) dz = = { t+ω t t+ω t + + t+ω t+ω +ω t+ω t+ω+ω + } z f (z) f(z) t+ω { zf (z) t f(z) (z +ω ) f } (z +ω ) dz f(z +ω ) t+ω { zf (z) t f(z) (z +ω )f } (z +ω ) dz f(z +ω ) = t+ω f { ω (z) t+ω t f(z) dz +ω f } (z) t f(z) dz = { } ω [lnf(z)] t+ω t +ω [ln f(z)] t+ω t = mω +nω, for Z. However, we note that zf (z)/f(z) has a simple pole with residue z at a simple zero (z being the location of the zero) and has a simple pole with residue z p at a simple pole z p. For an higher order zero or poles, the residues are multiplied by the order of zero or pole. Thus, from residue theory, and relation (4.37), it follows that z j z pk = mω + n ω j k where the multiplicity of zeros or poles are included in the summation (i.e. if there is a second order zero at a point, then two of the indices j in the summation correspond to that point). Hence the Lemma follows. dz 5. Construction of Weirstrass elliptic function P(z) Definition 5.. We define the Weirstrass elliptic function as: (5.38) P(z) = { } z + (z Ω ) Ω where (5.39) Ω = mω + nω and the summation in (5.38) extends to all Z, except for () = (,). Remark 5.. When Ω, is large, it is clear that the general term of the series in (5.38) is O( Ω 3 ), and so the double sum over m and n converges absolutely and uniformly in any compact set that excludes the set of points Ω. Thus P(z) as defined above is indeed an analytic function, except for double poles at Ω. To show periodicity and other properties of P(z), it is easier to first consider the properties of P (z). Because of uniform convergence, it follows that (5.4) P (z) = (z Ω ) 3 On rearranging the shifting the index m in the above summation, it is clear that (5.4) P (z + ω ) = P (z)

10 Similarly, ω in the above, can be replaced by ω. Thus, it is clear that P (z) is a doubly periodic analytic function (elliptic function) with periods ω and ω, and order three. Further, (5.4) P ( z) = = (z +Ω ) 3 However, we can replace Ω in the above by Ω, since the set of values attained by Ω is the same. Therefore (5.43) P ( z) = = (z Ω ) 3 = P(z) In the same manner, it follows that (5.44) P( z) = { z + (z +Ω ) } Ω = P(z) On integrating (5.4), it follows that (5.45) P(z + ω ) = P(z) + A for some constant A. However, if we now substitute z = ω into (5.44), we obtain P( ω ) = P(ω ) + A Using (5.45), it follows that A =. Hence, from (5.45), we obtain (5.46) P(z +ω ) = P(z) Similarly (5.47) P(z +ω ) = P(z) Thus P(z) is a second order elliptic function, with fundamental periods ω and ω. Notice it cannot possibly have a smaller period, because P(z) has a pole at z = and there are no other poles in the period parallelogram, with vertices, ω, ω + ω and ω. Lemma 5.3. P(z) satisfies the differential equation: (5.48) P (z) = 4 P (z) g P(z) g 3 where (5.49) g = 6 Ω 4, g 3 = 4 Ω 6 Proof. We first note that P(z) z is a regular function in a neighborhood of z = and has the Taylor series expansion: (5.5) P(z) z = g z + 8 g 3 z 4 + O(z 6 ) where g and g are as defined by (5.49). Further on differentiating, it follows that (5.5) P (z) = z 3 + g z + 7 g 3 z 3 + O(z 5 ) Cubing and and squaring relations (5.5) and (5.5), respectively, we get (5.5) P 3 (z) = z g z g 3 + O(z )

11 (5.53) P (z) = 4 z 6 5 g z 4 7 g 3 + O(z ) Hence (5.54) P (z) 4 P 3 (z) = g z g 3 + O(z ) Therefore, in a neighborhood of z =, (5.55) P (z) 4 P 3 (z) + g P(z) + g 3 = O(z ) Since, the expression on the left hand side of (5.55) is clearly an elliptic function with possible singularities only at singularities of P and P, i.e at z = and all points congruent to it, it follows from (5.55) that it has no singularities anywhere (since it cannot haveoneat ). Thus, the left hand side of(5.55) must be aconstant as it is entire and bounded. Equation (5.55) implies that the constant must be zero, since the right hand side is at z =. Thus, (5.48) follows. Remark 5.4. Conversely, given a differential equation in the form: ( ) dy (5.56) = 4 y 3 g y g 3 dz with given g and g 3, if numbers ω and ω can be determined so that (5.49) is valid, then the general solution of the differential equation is given by (5.57) y(z) = P(±z + α), where α is some constant of integration. This can be proved by simply transforming the dependent variable y through the transformation y = P(u). It is then seen that the differential equation (5.56) becomes ( ) du = dz Remark 5.5. The differential equation (5.56) arises naturally in studying the travelling wave solution to the Korteweg Devries equation for u(x,t) of the form: (5.58) u t + 6u u x u xxx = For a travelling wave solution u(x,t) = F(x c t). Then (5.59) c F + 6 F F F = Integrating once, (5.6) c F + 3 F F = d for some constant d. Multiplying (5.6) by F and integrating again, we get (5.6) F = c F + F 3 d F + b for some constant b. By shifting F by aconstant, i.e. introducing F = G + constant and choosing constant suitably, we get an equation for G of the form (5.6) G = G 3 g G g 3 This can be further transformed into (5.57) by merely scaling both the dependent and independent variables by constants. Thus, the elliptic function is useful in studying this integrable partial differential equation.