In Stokes Flow, the Stream Function Associated with the Velocity of The

Home , Stokes flow, Stream function

In Stokes flow, the stream function associated with the velocity of the fluid satisfies the biharmonic equation. The detailed behavior of solutions to the biharmonic equation on regions with corners has been historically difficult to characterize. The problem was first examined by Lord Rayleigh in 1920; in 1973, the existence of infinite oscillations in the domain Green’s function was proven in the case of the right angle by S. Osher. In this paper, we observe that, when the biharmonic equation is formulated as a boundary integral equation, the solutions are representable by rapidly convergent P µj µj series of the form j(cjt sin (βj log (t)) + djt cos (βj log (t))), where t is the distance from the corner and the parameters µj, βj are real, and are determined via an explicit formula depending on the angle at the corner. In addition to being analytically perspicuous, these representations lend themselves to the construction of highly accurate and efficient numerical discretizations, significantly reducing the number of degrees of freedom required for the solution of the corresponding integral equations. The results are illustrated by several numerical examples.

On the solution of Stokes equation on regions with corners

M. Rachh‡ , K. Serkh† , Technical Report YALEU/DCS/TR-1536 October 2, 2017

This author’s work was supported in part by ONR N00014-14-1-0797/16-1-2123 This author’s work was supported by the NSF Mathematical Sciences Postdoctoral Research Fellowship (award number 1606262)

‡ Dept. of Mathematics, Yale University, New Haven CT 06511 † Courant Institute of Mathematical Sciences, New York University, New York NY 10012 Approved for public release: distribution is unlimited. Keywords: Integral equations, Stokes ﬂow, Polygonal domains, Biharmonic equation, Corners

1 On the solution of Stokes equation on regions with corners

Manas Rachh, Kirill Serkh

October 2, 2017

Contents

1 Introduction and problem formulation 4

2 Preliminaries 4 2.1 Velocity boundary value problem ...... 5 2.2 Integral equation formulation ...... 6 2.3 Integral equation in tangential and normal coordinates ...... 7 2.4 Integral equations on the wedge ...... 7

3 Analytical Apparatus 10

4 Analysis of the integral equation 14 4.1 Tangential odd, normal even case ...... 15 4.1.1 The values of p, q, and z in (96) ...... 15 4.1.2 Invertibility of B in (131) ...... 20 4.2 Tangential even, normal odd case ...... 22 4.2.1 The values of p, q, and z in (100) ...... 22 4.2.2 Invertibility of B in (166) ...... 25

5 Numerical Results 26

6 Conclusions and extensions 26 6.1 Inﬁnite oscillations of the biharmonic Green’s function ...... 27 6.2 Curved boundaries ...... 28 6.3 Generalization to three dimensions ...... 28 6.4 Other boundary conditions ...... 28 6.5 Modiﬁed biharmonic equation ...... 28

7 Acknowledgements 28

8 Appendix A 28

9 Appendix B 28 9.1 Tangential odd, normal even case ...... 29 9.2 Tangential even, normal odd case ...... 31

2 1. Introduction and problem formulation

In classical potential theory, solutions to elliptic partial differential equations are represented by potentials on the boundaries of the regions. Over the last four decades, several well-conditioned boundary integral representations have been developed for the biharmonic equation with various boundary conditions. These integral representations have been studied extensively when the boundaries of the regions are approximated by a smooth curve (see, [1, 2, 3, 4, 5, 6, 7, 8], for example). In all of these representations, the kernels of the integral equation are at worst weakly singular and, in same cases, even smooth. The corresponding solutions to the integral equations tend to be as smooth as the boundaries of the domains and the incoming data. However, when the boundary of the region has corners, the solutions to both the differential equation and the corresponding integral equations are known to develop singularities. The solutions to the differential equation have been studied extensively on regions with corners for both Dirichlet and gradient boundary conditions. In particular, the behavior of solutions to the biharmonic equation on a wedge enclosed by straight boundaries θ = 0 and θ = α, and a circular arc r = r0, where (r, θ) are polar coordinates, has received much attention over the years. To the best of our knowledge, this particular problem was first studied by Lord Rayleigh in 1920 [9], and over the decades, by A. Dixon [10], Dean and Montagnon [11], Szegö[12], and Moffat [13], to name a few. In 1973, S. Osher showed that the Green’s function for the biharmonic equation on a right angle wedge has infinitely many oscillations in the vicinity of the corner on all but finitely many rays [14]. The more complicated structure of the Green’s function for the biharmonic equation is explained, in part, by the fact that the biharmonic equation does not have a maximum principle associated with it, while, for example, Laplace’s equation does. While much has been published about the solutions to the differential equation, the solutions of the corresponding integral equation have been studied much less exhaustively. Recently, a detailed analysis of solutions to integral equations corresponding to the Laplace and Helmholtz equations on regions with corners was carried out by the second author and V. Rokhlin [15, 16, 17]. They observed that the solutions to these integral equations can be expressed as rapidly convergent series of singular powers in the Laplace case and Bessel functions of non-integer order in the Helmholtz case. In this paper, we investigate the solutions to a standard integral equation corresponding to the velocity boundary value problem for Stokes equation. For the velocity boundary value problem, the stream function associated with the velocity field satisfies the biharmonic equation with gradient boundary conditions (it turns out that the same integral equation can be used for Dirichlet boundary conditions, see, for example [8]). We show that, if the boundary data is smooth on each side of the corner, then the solutions of this integral µj equation can be expressed a rapidly convergent series of elementary functions of the form t cos (βj log t ) µj | | and t sin (βj log t ), where the parameters µj, βj can be computed explicitly by a simple formula depending only on the angle| at| the corner. Furthermore, we prove that, for any N, there exists a linear combination of the first N of these basis functions which satisfies the integral equation with error O( t N ), where t is the distance from the corner. | | The detailed information about the analytical behavior of the solution in the vicinity of corners, discussed in this paper, allows for the construction of purpose-made discretizations of the integral equation. These discretizations accurately represent the solutions near corners using far fewer degrees of freedom than graded meshes, which are commonly used in such environments; thereby leading to highly efficient numerical solvers. The rest of the paper is organized as follows. In Section 2, we discuss the mathematical preliminaries for the governing equation and it’s reformulation as an integral equation. In Section 3, we derive several analytical results using techniques required for the principal results which are derived in Section 4. We illustrate the performance of the numerical scheme which utilizes the explicit knowledge of the structure of the solution to the integral equation in Section 5. In Section 6, we present generalizations and extensions of the apparatus of this paper. The proof of several results in Section 4 are technical and are presented in Sections 8 and 9.

2. Preliminaries

In this paper, vector-valued quantities are denoted by bold, lower-case letters (e.g. h), while tensor- valued quantities are bold and upper-case (e.g. T). Subscript indices of non-bold characters (e.g. hj or Tjkl) are used to denote the entries within a vector (h) or tensor (T). We use the standard Einstein summation

3 convention; in other words, there is an implied sum taken over the repeated indices of any term (e.g. the P k symbol ajbj is used to represent the sum j ajbj). Let denote the space of functions which have k continuous derivatives. C Suppose now that Ω is a simply connected open subset of R2. Let Γ denote the boundary of Ω and 2 suppose that Γ is a simple closed curve of length L with nc corners. Let γ : [0,L] R denote an arc length parameterization of Γ in the counter-clockwise direction, and suppose that the location→ of the corners are

given by γ(sj), j = 1, 2, . . . nc with 0 = s1 < s2 . . . < snc < snc+1 = L. Furthermore, suppose that γ is analytic on the intervals (sj, sj+1) for each j = 1, 2, . . . nc. Let τ (x) and ν(x) denote the positively-oriented unit tangent and the outward unit normal respectively, for x Γ. Let hτ = hjτj and hν = hjνj denote the tangential and normal components of the vector h respectively,∈ see Figure 1.

Figure 1: A sample domain Ω with three corners

2.1. Velocity boundary value problem The equations of incompressible Stokes flow with velocity boundary conditions on a domain Ω with boundary Γ are ∆u + p = 0 in Ω, (1) − ∇ u = 0 in Ω, (2) ∇ · u = h on Γ, (3) where u is the velocity of the fluid, p is the fluid pressure and h is the prescribed velocity on the boundary. For any h which satisfies Z h ν dS = 0 , (4) Γ · there exists a unique velocity field u and a pressure p, defined uniquely up to a constant, that satisfy (1) – (3). We summarize the result in the following lemma (see [18] for a proof). 2 R Lemma 1. Suppose h L (Γ) and satisfies Γ h ν dS = 0 , then there exists a unique velocity u and a pressure p which is unique∈ up to a constant which satisfy· the velocity boundary value problem (1) – (3). Remark 2. The Stokes equation with velocity boundary conditions can be reformulated as a biharmonic 2 equation with gradient boundary conditions. First, we represent the velocity u:Ω R as u = ⊥w, where → ∇ w :Ω R is the stream function associated with the velocity field and ⊥ is the operator given by → ∇ ∂w ∂x2 ⊥w = −∂w . (5) ∇ ∂x1

Next, we observe that u = ⊥w automatically satisﬁes the divergence free condition (2). Finally, taking ∇ the dot product of ⊥ with (1), we observe that w satisﬁes the biharmonic equation with gradient boundary conditions given by∇

∆2w = 0 in Ω (6)

⊥w = h on Γ . (7) ∇

4 2.2. Integral equation formulation Following the treatment of [19, 5], the fundamental solution to the Stokes equations (the Stokeslet) is given by " # 1 (xj yj)(xk yk) Gj,k (x, y) = log x y δij + − − , j, k 1, 2 , (8) 4π − | − | x y 2 ∈ | − | 2 for x, y R , and x = y, where δij is the Kronecker delta function. The stress tensor Tj,k,` (x, y) associated with the∈ Green’s function,6 or the stresslet, is given by

1 (xj yj)(xk yk)(x` y`) Tj,k,` (x, y) = − − − j, k, ` 1, 2 , (9) −π x y 4 ∈ | − | x, y R2 and x = y. The stresslet T is roughly analogous to a dipole in electrostatics. The double layer Stokes∈ potential is6 the velocity field due to a surface density of stresslets µ and is defined by Z ( Γ[µ](x))j = Tk,j,` (y, x) µk (y) ν`(y) dSy , j, k, ` 1, 2 , (10) D Γ ∈ 2 for x R . Clearly, Γ[µ](x) satisfies Stokes equation for x Ω. The∈ following lemmaD describes the behavior of the double layer∈ Stokes potential as x x where x Γ. → 0 0 ∈ 2 Lemma 3. Suppose that µ:Γ R and let Γ[µ](x) denote a double layer Stokes potential (10). Then [µ](x) satisfies the jump relation:→ D DΓ 1 Z lim Γ[µ](x) = µ(x0) + p.v. K0(x0, y)µ(y)dSy , (11) x x0 D −2 x→Ω Γ ∈ where x , y Γ, and K is given by 0 ∈ 0 2 1 (y x) ν(y) (y1 x1) (y1 x1)(y2 x2) K0(x, y) = − · 4 − − −2 (12) −π x y (y1 x1)(y2 x2)(y2 x2) | − | − − − for x, y Γ and p.v. R denotes a principal value integral. ∈ The following lemma states that the kernel K0 is smooth if the boundary Γ is smooth. k 2 k Lemma 4. The kernel K is − if γ is with limiting values 0 C C 2 1 γ10 (t) γ10 (t)γ20 (t) lim K0(γ(t), γ(s)) = κ(γ(t)) 2 , (13) t s γ0 (t)γ0 (t) γ0 (t) → −π 1 2 2

where κ(γ(t)) is the curvature at γ(t). Furthermore, K0 is analytic if γ is analytic. The following theorem reduces the velocity boundary value problem (1) – (3) to an integral equation on the boundary by representing u as double layer Stokes potential with unknown density µ, i.e. u(x) = [µ](x) x Ω. (14) DΓ ∈ 2 R 2 Lemma 5. Suppose h L (Γ) and that Γ h ν dS = 0. Then there exists a unique solution µ L (Γ) which satisﬁes ∈ · ∈ 1 Z µ(x) + p.v. K0(x, y)µ(y) dSy = h(x) , x Γ , (15) −2 Γ ∈ R and Γ µ ν dS = 0. Furthermore, u(x) = Γ[µ](x) satisﬁes Stokes equations (1), (2), along with the boundary· conditions u(x) = h(x) for x Γ. D ∈ Proof. See, for example, [5] for a proof.

The following lemma extends Lemma 5 to the case where the boundary Γ is an open arc. Lemma 6. Suppose h L2(Γ) then there exists a unique solution µ L2(Γ) which satisﬁes ∈ ∈ 1 Z µ(x) + p.v. K0(x, y)µ(y) dSy = h(x) x Γ , (16) −2 Γ ∈

where K0 is deﬁned by (12).

5 2.3. Integral equation in tangential and normal coordinates It turns out that it is convenient to represent both the velocity on the boundary h and the solution of the integral equation µ in terms of their tangential and normal coordinates, denoted by h = (hτ , hν ) and µ = (µτ , µν ) respectively, as opposed to their Cartesian coordinates. In this section, we discuss the representation in the tangential and normal coordinates of the double layer Stokes potential, and the corresponding integral equation for the velocity boundary value problem. Let R(x) denote the unitary transformation that converts vectors expressed in Cartesian coordinates to vectors expressed in tangential and normal components, i.e. τ (x) τ (x) R(x) = 1 2 x Γ. (17) n1(x) n2(x) ∈

Let R∗(x) denote the adjoint of R(x). Suppose u(x) = (u1(x), u2(x)) is the double layer Stokes potential with density µ(x) = (µτ (x), µν (x)), given by u(x) = ˜ [µ](x) x Ω, (18) DΓ ∈ where Z µ (y) ˜ [µ](x) = T (y, x) R (y) τ ν (y)dS , j, k, ` = 1, 2 , (19) Γ k,j,` ∗ y ` y D j Γ · µν ( ) k where x Ω. The following theorem reduces the velocity boundary value problem (1) – (3) to an integral equation∈ on the boundary in the rotated frame.

2 R Lemma 7. Suppose h = (hτ , hν ) L (Γ) and that Γ hν dS = 0. Then there exists a unique solution 2 ∈ µ = (µτ , µν ) L (Γ) which satisﬁes ∈ Z 1 µτ (x) µτ (y) hτ (x) + p.v. K(x, y) dSy = , x Γ , (20) −2 µν (x) Γ µν (y) hν (x) ∈ R along with Γ µν dS = 0, where

K(x, y) = R(x)K (x, y)R∗(y) , x, y Γ. (21) 0 ∈ Furthermore, u(x) = ˜Γ[µ](x) satisﬁes Stokes equations (1), (2), along with the boundary conditions u(x) = h(x) for x Γ. D ∈ Proof. The result is a straightforward consequence of Lemma 5.

The following lemma extends Lemma 7 to the case where the boundary Γ is an open arc.

2 2 Lemma 8. Suppose h = (hτ , hν ) L (Γ) then there exists a unique solution µ = (µτ , µν ) L (Γ) which satisﬁes ∈ ∈ Z 1 µτ (x) µτ (y) hτ (x) + p.v. K(x, y) dSy = , x Γ , (22) −2 µν (x) Γ µν (y) hν (x) ∈ where K is deﬁned by (21).

2.4. Integral equations on the wedge Suppose γ(t):[ 1, 1] R2 is a wedge with interior angle πθ and side length 1 on either side of the corner, parametrized− by arc-length→ (see, Figure 2). In a slight misuse of notation, let µ (t) denote µ (γ(t)) for all 1 < t < 1. Likewise, let µ (t), h (t), τ τ − ν τ and hν (t) denote µν (γ(t)), hτ (γ(t)) and hν (γ(t)), respectively. The integral equation (22) for the velocity boundary value problem is then given by

1 µ (t) Z 1 µ (s) h (t) τ + p.v. K(γ(t), γ(s)) τ ds = τ , 1 < t < 1 , (23) −2 µν (t) 1 µν (s) hν (t) − − where K is deﬁned by (21). In this case, the kernel K has a simple form which is given by the following lemma.

6 Figure 2: A wedge with interior angle πθ

Lemma 9. Suppose γ :[ 1, 1] R2 is deﬁned by the formula − → ( t (cos(πθ), sin(πθ)) if 1 < t < 0 , γ(t) = − · − (24) (t, 0) if 0 < t < 1 .

Suppose further that kj(s, t), j = 1, 2, 3, 4, are deﬁned by t sin(πθ) k1(s, t) = (s + t cos (πθ))(s cos (πθ) + t) , (25) π (s2 + t2 + 2st cos(πθ))2 t sin(πθ) k2(s, t) = t sin (πθ)(t + s cos (πθ)) , (26) π (s2 + t2 + 2st cos(πθ))2 t sin(πθ) k3(s, t) = (s + t cos (πθ))s sin (πθ) , (27) −π (s2 + t2 + 2st cos(πθ))2

t sin(πθ) 2 k4(s, t) = st sin (πθ) . (28) −π (s2 + t2 + 2st cos(πθ))2 for all 1 < s, t < 1. If 0 < t < 1, then − " #  k1(s, t) k2(s, t)  if 1 < s < 0 ,  k3(s, t) k4(s, t) − K(γ(t), γ(s)) = " # (29)  0 0  if 0 < s < 1 .  0 0

Likewise, if 1 < t < 0, then − " #  0 0  if 1 < s < 0 ,  0 0 − K(γ(t), γ(s)) = " # (30)  k1(s, t) k2(s, t)  − if 0 < s < 1 .  k (s, t) k (s, t) 3 − 4 In the following theorem, we show that, when Γ is a wedge, the integral equation (23) decouples into two independent integral equations on the interval [0, 1].

2 Theorem 10. Suppose µτ , µν are functions in L [ 1, 1]. Let hτ , hν be deﬁned by (23). We denote the odd − and the even parts of a function f by fo and fe respectively, where 1 1 f (s) = (f(s) f( s)) , and , f (s) = (f(s) + f( s)) , (31) o 2 − − e 2 − for 1 < s < 1. Then for 0 < t < 1, − h (t) 1 µ (t) Z 1 k (s, t) k (s, t) µ (s) τ,e = τ,e 1,1 1,2 τ,e ds , (32) hν,o(t) −2 µν,o(t) − 0 k2,1(s, t) k2,2(s, t) µν,o(s)

7 and h (t) 1 µ (t) Z 1 k (s, t) k (s, t) µ (s) τ,o = τ,o + 1,1 1,2 τ,o ds , (33) hν,e(t) −2 µν,e(t) 0 k2,1(s, t) k2,2(s, t) µν,e(s) where kj,`(s, t), j, ` = 1, 2, are given by

1 t sin (πθ)(s t cos (πθ))(t s cos (πθ)) k (s, t) = (k (s, t) k ( s, t)) = − − , (34) 1,1 2 1 − − 1 − π(s2 + t2 2st cos (πθ))2 − 1 t2 sin2 (πθ)(t s cos (πθ)) k (s, t) = ( k (s, t) + k ( s, t)) = − , (35) 1,2 2 − 2 − 2 − π(s2 + t2 2st cos (πθ))2 − 1 st sin2 (πθ)(s t cos (πθ)) k (s, t) = (k (s, t) k ( s, t)) = − , (36) 2,1 2 3 − − 3 − π(s2 + t2 2st cos (πθ))2 − 1 st2 sin3 (πθ) k (s, t) = ( k (s, t) + k ( s, t)) = , (37) 2,2 2 − 4 − 4 − π(s2 + t2 2st cos (πθ))2 − for all 0 < s, t < 1. Proof. Substituting the expression for the kernel K given by (29), (30) in (23), we get

h (t) 1 µ (t) Z 0 k (s, t) k (s, t) µ (s) τ = τ + 1 2 τ ds , (38) hν (t) −2 µν (t) 1 k3(s, t) k4(s, t) µν (s) − for 0 < t < 1 and h (t) 1 µ (t) Z 1 k (s, t) k (s, t) µ (s) τ = τ + − 1 2 τ ds , (39) hν (t) −2 µν (t) k3(s, t) k4(s, t) µν (s) 0 − for 1 < t < 0. Then, making the change of variable s s in (38) and the change of variable t t in (39),− we get → − → − h (t) 1 µ (t) Z 1 k ( s, t) k ( s, t) µ ( s) τ = τ + 1 − 2 − τ − ds , (40) hν (t) −2 µν (t) k3( s, t) k4( s, t) µν ( s) 0 − − − for 0 < t < 1 and

h ( t) 1 µ ( t) Z 1 k (s, t) k (s, t) µ (s) τ − = τ − + − 1 − 2 − τ ds , (41) hν ( t) −2 µν ( t) k3(s, t) k4(s, t) µν (s) − − 0 − − − for 0 < t < 1. Finally, combining (40), (41), we get

h (t) 1 µ (t) Z 1 k (s, t) k (s, t) µ (s) τ,e = τ,e 1,1 1,2 τ,e ds , (42) hν,o(t) −2 µν,o(t) − 0 k2,1(s, t) k2,2(s, t) µν,o(s) h (t) 1 µ (t) Z 1 k (s, t) k (s, t) µ (s) τ,o = τ,o + 1,1 1,2 τ,o ds . (43) hν,e(t) −2 µν,e(t) 0 k2,1(s, t) k2,2(s, t) µν,e(s)

Thus, the integral equation (23) clearly splits into two cases: Tangential odd, normal even: the tangential components of both the velocity ﬁeld h and the density µ, • hτ (t) and µτ (t), are odd functions of t and the normal components hν (t) and µν (t) are even functions of t. Tangential even, normal odd: the tangential components of both the velocity ﬁeld h and the density µ, • hτ (t) and µτ (t), are even functions of t and the normal components hν (t) and µν (t) are odd functions of t.

8 Figure 3: A keyhole contour

3. Analytical Apparatus

In this section, we investigate integrals of the form Z 1 z kj,`(s, t)s ds , j, ` = 1, 2 , (44) 0 for 0 < t < 1, where z C with Re(z) > 1, and kj,` are defined in (34) – (37). We use the principal branch of log for the definition∈ of sz, i.e., Arg(s)− [0, 2π). The principal result of this section is Theorem 15. ∈ On inspecting the kernels kj,`(s, t), j, ` = 1, 2, we observe that it suffices to evaluate integrals of the form

1 Z 1 sz I(z, θ, t) = ds , for 0 < t < 1 , (45) π (s2 + t2 2st cos (πθ))2 0 − where Re(z) > 1. Using standard techniques in complex analysis, we ﬁrst derive an expression for the above integral from− 0 to in the following lemma. ∞ Lemma 11. Suppose z C, 1 < Re(z) < 3, z = 0, 1, 2, and θ C. Then ∈ − 6 ∈ Z z 1 ∞ s z 3 I1(z, θ, t) = ds = a(z, θ)t − , (46) 2 2 2 π 0 (s + t 2st cos (πθ)) − for 0 < t < 1 where z sin (2πθ) cos (π(1 θ)z) + 2 sin (π(1 θ)z) 1 z sin2 (πθ) a(z, θ) = − − − . (47) 4 sin (πz) sin3 (πθ) Proof. Let Γ = Γ Γ Γ Γ denote a keyhole contour where key ε ∪ 1 ∪ R ∪ 2 n π π o n p o Γ = εeix , < x < , Γ = x + iε , 0 x R2 ε2 , ε − − 2 2 1 ≤ ≤ − n p o Γ = Reix , x < x < 2π x , Γ = x iε , 0 x R2 ε2 , R 0 − 0 2 − ≤ ≤ − where x0 = arctan ε/R, see Figure 3. Using Cauchy’s integral theorem, Z sz Z Z Z Z sz ds = + + + ds , (48) 2 2 2 iπθ 2 iπ(2 θ)2 Γkey (s + t 2st cos (πθ)) Γε Γ1 ΓR Γ2 (s te ) s te − − ! −! −! d sz d sz = 2πi + , (49) ds iπ(2 θ)2 ds iπθ 2 s te iπθ (s te ) iπ(2−θ) − − s=te − s=te z 3 2πt − iπθz 2πiz iπθz iπθ(z 2) 2πiz iπθ(z 2) = (z 2) e e e− z e − e e− − , (50) −sin3 (πθ) − − − −

9 for 0 < t < 1. Suppose 0 < t < 1, then there exists a constant M < such that 1 ∞

z s Re(z) 2 M1ε , (51) (s2 + t2 2st cos (πθ)) ≤ − for all s Γ . Taking the limit as ε 0, we get ∈ ε → Z z s Re(z)+1 lim ds lim πM1ε = 0 , (52) ε 0 2 2 2 ε 0 Γε (s + t 2st cos (πθ)) ≤ → − → since Re(z) > 1. Similarly, for 0 < t < 1, there exists a constant M < such that − 2 ∞

sz M 2 , (53) 2 2 2 4 Re(z) (s + t 2st cos (πθ)) ≤ R − − for all s Γ . Taking the limit as R , we get ∈ R → ∞

Z sz 2πM lim ds lim = 0 , (54) R 2 2 2 R 3 Re(z) ΓR (s + t 2st cos (πθ)) ≤ R − →∞ − →∞ since Re(z) < 3. On Γ and Γ , taking the limits as ε 0 and R , we get 1 2 → → ∞ Z sz lim lim ds = I1(z, θ, t) , (55) ε 0 R 2 2 2 → →∞ Γ1 (s + t 2st cos (πθ)) Z − z s 2πiz lim lim ds = e I1(z, θ, t) . (56) 2 2 2 ε 0 R Γ (s + t 2st cos (πθ)) − → →∞ 2 − The result follows by taking the limit ε 0 and R in (50), and using (52) and (54) to (56). → → ∞

Suppose now that I2(z, θ, t) is deﬁned by z Z ∞ s I (z, θ) = ds , (57) 2 (s2 + t2 2st cos (πθ))2 1 − for 0 < t < 1. Clearly, I(z, θ, t) = I (z, θ, t) I (z, θ, t) , (58) 1 − 2 where I(z, θ, t) is given by (45) and I1(z, θ, t) by (46). In the following lemma, we compute an expression for I2(z, θ, t) by deriving a Taylor expansion for 1 f(s, t) = , for s > 1, t < 1 . (s2 + t2 2st cos (πθ))2 | | | | − Lemma 12. . Suppose that θ C. Then for all t < 1 and s > 1 ∈ | | | |

1 X∞ n f(s, t) = = an(s)t , (59) (s2 + t2 2st cos (πθ))2 − n=0 where 1 an(s) = ((n + 3) sin ((n + 1)πθ) (n + 1) sin ((n + 3)πθ)) . (60) 4sn+4 sin3 (πθ) − Furthermore, for 1 < Re(z) < 3, and z = 0, 1, 2, − 6 Z 1 ∞ X∞ I (z, θ, t) = szf(s, t) ds = F (n, z, θ)tn , (61) 2 π 1 n=0 where (n + 1) sin ((n + 3)πθ) (n + 3) sin((n + 1)πθ) F (n, z, θ) = − . (62) 4π( z + n + 3) sin3 (πθ) −

10 Proof. For fixed s > 1, f(s, t) is analytic in t < 1. Using Cauchy’s integral formula, the coefficients of the Taylor series are| | given by | | Z Z 2π 1 f(s, ξ) 1 inx ix an(s) = n+1 dξ = e− f(s, e ) dx , (63) 2πi ξ =1 ξ 2π 0 | | 1 Z 2π e inx = − dx , (64) 2π (s ei(πθ+x))2(s ei( πθ+x))2 0 − − − Z 2π inx 1 e− = 4 2 2 dx . (65) 2πs i(πθ+x) i(−πθ+x) 0 1 e 1 e − s − s Using the Taylor expansion of 1/(1 x)2 given by − 1 X∞ = (n + 1)xn , (66) (1 x)2 − n=0 for 0 < x < 1, equation (65) simplifies to,

Z 2π i(`+m)x+(` m)πθ 1 1 inx X∞ X∞ e − a (s) = e− (` + 1)(m + 1) dx (67) n 2π s4 s`+m 0 `=0 m=0 Due to the orthogonality of the Fourier basis einx in L2[0, 2π], the only terms that contribute to the integral in (67) are when ` + m = n. Thus, { }

inπθ n 1 X i(` m)πθ e− X 2i`πθ a (s) = (` + 1)(m + 1)e − = (` + 1)(n ` + 1)e . (68) n sn+4 sn+4 − `+m=n `=0 The result for the Taylor series for f(s, t) then follows by using n X dp 1 ei(n+1)x `pei`x = − p = 0, 1, 2 . (69) dxp 1 eix `=0 −

Given the Taylor expansion for f(s, t), the integral I2(z, θ, t) can be computed by switching the order of summation and integration and using Z ∞ z s an(s) ds = F (n, z, θ) , (70) 1 which concludes the proof.

Using Lemmas 11 and 12, we compute I(z, θ, t) in the following theorem. Theorem 13. Suppose 1 < Re(z) < 3, and z = 0, 1, 2, then − 6 Z 1 z s z 3 X∞ n I(z, θ, t) = = a(z, θ)t − F (n, z, θ)t , (71) (s2 + t2 2st cos (πθ))2 − 0 − n=1 for 0 < t < 1. We observe that both the expressions on the left and right of (71) are analytic functions of z for Re(z) > 1 and z not an integer. In the following theorem, we extend the deﬁnition of I(z, θ, t) to Re(z) > 1 and −z not an integer. − Theorem 14. Suppose that z C, Re(z) > 1, z = 0, 1, 2 ..., and θ C. Then ∈ − 6 ∈ Z 1 z s z 3 X∞ n I(z, θ, t) = = a(z, θ)t − F (n, z, θ)t , (72) (s2 + t2 2st cos (πθ))2 − 0 − n=1 for 0 < t < 1.

11 Proof. The result follows from (71) and using analytic continuation.

We now present the principal result of this section in the following theorem.

Theorem 15. Suppose z C, Re(z) > 1, z = 0, 1, 2,..., and θ C. Recall that, ∈ − 6 ∈ t sin (πθ)(s t cos (πθ))(t s cos (πθ)) k (s, t) = − − , (73) 1,1 π(s2 + t2 2st cos (πθ))2 − t2 sin2 (πθ)(t s cos (πθ)) k (s, t) = − , (74) 1,2 π(s2 + t2 2st cos (πθ))2 − st sin2 (πθ)(s t cos (πθ)) k (s, t) = − , (75) 2,1 π(s2 + t2 2st cos (πθ))2 − st2 sin3 (πθ) k (s, t) = , (76) 2,2 π(s2 + t2 2st cos (πθ))2 − for 0 < s, t < 1. Then

1 Z X∞ k (s, t)szds = a (z, θ) tz + F (n, z, θ) tn , (77) 1,1 1,1 · 1,1 · 0 n=1 1 Z X∞ k (s, t)szds = a (z, θ) tz + F (n, z, θ) tn , (78) 1,2 1,2 · 1,2 · 0 n=2 1 Z X∞ k (s, t)szds = a (z, θ) tz + F (n, z, θ) tn , (79) 2,2 2,2 · 2,1 · 0 n=1 1 Z X∞ k (s, t)szds = a (z, θ) tz + F (n, z, θ) tn , (80) 2,2 2,2 · 2,2 · 0 n=2 for 0 < t < 1, where 1 a (z, θ) = [(z + 1) sin (πθ) cos (πz(1 θ)) sin (π(z(1 θ) + θ))] , (81) 1,1 2 sin (πz) − − − 1 a (z, θ) = (z 1) sin (πθ) sin (πz(1 θ)) , (82) 1,2 −2 sin (πz) − − 1 a (z, θ) = (z + 1) sin (πθ) sin (πz(1 θ)) , (83) 2,1 2 sin (πz) − 1 a (z, θ) = [(z + 1) sin (πθ) cos (πz(1 θ)) + sin (π(z(1 θ) θ))] , (84) 2,2 2 sin (πz) − − − and n sin (πθ) cos (nπθ) + cos (πθ) sin (nπθ) F (n, z, θ) = , (85) 1,1 2π(n z) − (n 1) sin (πθ) sin (nπθ) F (n, z, θ) = − , (86) 1,2 2π(n z) − (n + 1) sin (πθ) sin (nπθ) F (n, z, θ) = , (87) 2,1 − 2π(n z) − n sin (πθ) cos (nπθ) cos (πθ) sin (nπθ) F (n, z, θ) = − . (88) 2,2 2π(n z) −

12 Proof. The result follows from observing that Z 1 z sin (2πθ) 3 k1,1(s, t)s ds = t I(z + 2, θ, t) + t I(z, θ, t) (89) 0 − 2 · · + t2(1 + cos2 (πθ)) I(z + 1, θ, t) , · Z 1 z 2 2 k1,2(s, t)s ds = t sin (πθ)(t I(z, θ, t) cos (πθ)I(z + 1, θ, t)) , (90) 0 · − Z 1 z 2 k2,1(s, t)s ds = t sin (πθ)(I(z + 2, θ, t) t cos (πθ) I(z + 1, θ, t)) , (91) 0 − · Z 1 z 2 3 k2,2(s, t)s ds = t sin (πθ) I(z + 1, θ, t) , (92) 0 · and using the formula for I(z, θ, t) derived in Theorem 14.

4. Analysis of the integral equation

Suppose that γ(t):[ 1, 1] R2 is a wedge with interior angle πθ and side length 1 on either side of the corner, parametrized− by arc→ length (see Figure 2). Suppose further that the odd and the even parts of a function f are denoted by fo and fe respectively (see (31)). In Section 2.4, we observed that the integral equation 1 µ (t) Z 1 µ (s) h (t) τ + p.v. K(γ(t), γ(s)) τ ds = τ , 1 < t < 1 , (93) −2 µν (t) 1 µν (s) hν (t) − − simplifies into two uncoupled integral equations on the interval [0,1], given by 1 µ (t) Z 1 k (s, t) k (s, t) µ (s) h (t) τ,o + 1,1 1,2 τ,o ds = τ,o , (94) −2 µν,e(t) 0 k2,1(s, t) k2,2(s, t) µν,e(s) hν,e(t) for 0 < t < 1, and 1 µ (t) Z 1 k (s, t) k (s, t) µ (s) h (t) τ,e 1,1 1,2 τ,e ds = τ,e , (95) −2 µν,o(t) − 0 k2,1(s, t) k2,2(s, t) µν,o(s) hν,o(t) for 0 < t < 1, where K is defined in (21) and the kernels kj,`(s, t), j, ` = 1, 2, are defined in (34) – (37). As in Section 2.4, we refer to (94) as the tangential odd, normal even case and to (95) as the tangential even, normal odd case. In Section 4.1, we investigate equation (94), i.e., the tangential odd, normal even case. We observe that, if µτ,o(t) and µν,e(t) are of the form µ (t) = p t z sgn(t) , and µ (t) = q t z , (96) τ,o · | | ν,e · | | for 1 < t < 1, where p, q, z C, then for certain values of p, q, and z depending on the angle πθ, the − ∈ corresponding components of the velocity hτ,o(t) and hν,e(t), defined by (94), are smooth. We also prove the converse. Suppose that N is a positive integer. Then for all but countably many 0 < θ < 2, there exist pn,j, qn,j, zn,j C, n = 1, 2,...N and j = 1, 2, such that the following holds. Suppose αn, βn C, ∈ ∈ n = 0, 1,...N, and hτ,o(t) and hν,e(t) are given by N ! N X X h (t) = α t n sgn(t) , and h (t) = β t n , (97) τ,o n| | ν,e n| | n=0 n=0 for 1 < t < 1. Then, there exist unique numbers cn, dn C, n = 0, 1,...N, such that, if − ∈ N ! X µ (t) = c + c p t zn,1 + d p t zn,2 sgn(t) , (98) τ,o 0 n n,1| | n n,2| | n=1 N X µ (t) = d + c q t zn,1 + d q t zn,2 , (99) ν,e 0 n n,1| | n n,2| | n=1

13 N+1 for 1 < t < 1, then µτ,e(t), µν,o(t) satisfy equation (94) with error O( t ). We prove this result in Theorem− 27. | | Similarly, in Section 4.2, we investigate equation (95), i.e., the tangential even, normal odd case. We observe that, if µτ,e(t) and µν,o(t) are of the form

µ (t) = p t z , and µ (t) = q t z sgn(t) , (100) τ,e · | | ν,o · | | for 1 < t < 1, where p, q, z C, then for certain values of p, q, and z depending on the angle πθ, the − ∈ corresponding components of the velocity hτ,e(t) and hν,o(t), deﬁned by (95), are smooth. We also prove the converse in the following sense. Suppose that N is a positive integer. Then for all but countably many 0 < θ < 2, there exist pn,j, qn,j, zn,j C, n = 1, 2,...N and j = 1, 2, such that the following holds. Suppose ∈ αn, βn C, n = 0, 1,...N, and hτ,e(t) and hν,o(t) are given by ∈ N N ! X X h (t) = α t n , and h (t) = β t n sgn(t) (101) τ,e n| | ν,o n| | n=0 n=0

for 1 < t < 1. Then, there exist unique numbers cn, dn C for n = 0, 1,...N, such that, if − ∈ N X µ (t) = c + c p t zn,1 + d p t zn,2 (102) τ,e 0 n n,1| | n n,2| | n=1 N ! X µ (t) = d + c q t zn,1 + d q t zn,2 sgn(t) , (103) ν,o 0 n n,1| | n n,2| | n=1

4.1. Tangential odd, normal even case In this section, we investigate the tangential odd, normal even case (see equation (94)). In Section 4.1.1, we investigate the values of p, q and z in (96) for which the resulting components of the velocity are smooth functions. In Section 4.1.2, we show that, for every h of the form (97), there exists a density µ of the form (98), (99), which satisﬁes the integral equation (94) to order N.

4.1.1. The values of p, q, and z in (96) Suppose that µτ,o(t) and µν,e(t) are given by

µ (t) = p t z sgn(t) , and µ (t) = q t z , (104) τ,o · | | ν,e · | |

for 1 < t < 1, where p, q, z C. In this section, we determine the values of p, q and z such that hτ,o(t) and − ∈ hν,e(t) deﬁned by

h (t) 1 µ (t) Z 1 k (s, t) k (s, t) µ (s) τ,o = τ,o + 1,1 1,2 τ,o ds , (105) hν,e(t) −2 µν,e(t) 0 k2,1(s, t) k2,2(s, t) µν,e(s)

are smooth functions of t for 0 < t < 1, where the kernels kj,`(s, t), j, ` = 1, 2, are defined in (34) – (37). The principal result of this section is Theorem 24. The following lemma describes sufficient conditions for p, q, and z such that, if µ is defined by (104), then the velocity h given by (105) is smooth.

14 Theorem 17. Suppose that θ (0, 2), z is not an integer, and z satisﬁes det A(z, θ) = 0, where ∈ 1 a (z, θ) a (z, θ) A(z, θ) = I + 1,1 1,2 , (106) −2 a2,1(z, θ) a2,2(z, θ)

I is the 2 2 identity matrix, and aj,`(z, θ), j, ` = 1, 2, are given by (81) – (84). Furthermore, suppose that (p, q) ×A(z, θ) , where A denotes the null space of the matrix A. Suppose ﬁnally that ∈ N { } N{ } µ (t) = p tz , and µ (t) = q tz , (107) τ,o · ν,e ·

for 0 < t < 1. Then hτ,o(t) and hν,e(t) deﬁned by (105) satisfy

h (t) X∞ p τ,o = F(n, z, θ) tn , (108) hν,e(t) q · n=1 for 0 < t < 1, where F (n, z, θ) F (n, z, θ) F(n, z, θ) = 1,1 1,2 , (109) F2,1(n, z, θ) F2,2(n, z, θ)

and Fj,`(n, z, θ), j, ` = 1, 2, are given by (85) – (88). Proof. Substituting µ (t) = p tz and µ (t) = q tz in (105) and using Theorem 15, we get τ,o · ν,e · " z R 1 z # hτ,o(t) 1/2p t + 0 (p k1,1(s, t) + q k1,2(s, t)) s ds = − · R 1 · · , (110) hν,e(t) 1/2q tz + (p k (s, t) + q k (s, t)) sz ds − · 0 · 2,1 · 2,2 p X∞ p = A(z, θ) tz + F(n, z, θ) tn , (111) q q · n=1

Since (p, q) A(z, θ) , we note that A(z, θ) (p, q) = (0, 0) and thus ∈ N { } · h (t) X∞ F (n, z, θ) F (n, z, θ) p τ,o = 1,1 1,2 tn . (112) hν,e(t) F2,1(n, z, θ) F2,2(n, z, θ) q · n=1

A straightforward calculation shows that

(z sin (πθ) sin (πzθ)) (z sin (πθ) sin (πz(2 θ))) det A(z, θ) = − − − (113) 4 sin2 (πz)

Thus, if z is not an integer and either

z sin (πθ) sin (πz(2 θ)) = 0 , (114) − − or z sin (πθ) sin (πzθ) = 0 , (115) − then det A(z, θ) = 0. Remark 18. If z satisﬁes the implicit relations (114), (115), we have that det A(z, θ) = 0. It is then straightforward to determine p and q via an explicit formula in terms of the entries of A(z, θ), since (p, q) (A(z, θ)). ∈ N In the following theorem, we prove the existence of the implicit functions z(θ) deﬁned by (114), (115) on the interval θ (0, 2). ∈

15 Figure 4: An illustrative domain V for the case N = 2, where θ1, θ2, . . . θ5 are the combined branch points of the functions z1,1(θ), z2,1(θ), and z2,2(θ). The large dashes are used to denote the interval (0, 2) for reference. The thin dashes are the locations of the branch cuts at θ1, θ2 . . . θ5.

Theorem 19. Suppose that N 2 is an integer. Then there exists 6N 7 real numbers θ1, θ2, . . . θ6N 7 (0, 2) such that the following holds.≥ Suppose that D is the strip in the upper− half plane with 0 < Re(θ)−< ∈2 6N 7 that includes the interval (0, 2) θ − , i.e. \{ j}j=1 6N 7 D = θ C : Re(θ) (0, 2) , 0 Im(θ) < θj − . (116) { ∈ ∈ ≤ ∞} \ { }j=1 Then, there exists a simply connected open set D V C and analytic functions zn,1(θ): V C, n = 1, 2 ...N, which satisfy ⊂ ⊂ →

z sin (πθ) sin (πz(2 θ)) = 0 , z(1) = n , (117) − − for θ V , and analytic functions zn,2(θ): V C, n = 2, 3,...N, which satisfy ∈ → z sin (πθ) sin (πzθ) = 0 , z(1) = n , (118) −

for θ V (see Figure 4 for an illustrative domain V ). Moreover, the functions zn,1(θ), n = 1, 2 ...N, do not∈ take integer values for all θ V 1 , and satisfy det A(z (θ), θ) = 0, n = 1, 2 ...N, for all ∈ \{ } n,1 θ V (see (106), (113)). Similarly, the functions zn,2(θ), n = 2, 3,...N, do not take integer values for all θ ∈ V 1 , and satisfy det A(z (θ), θ) = 0, n = 2, 3 ...N, for all θ V (see (106), (113)). ∈ \{ } n,2 ∈ Proof. The proof is technical and is contained in Section 8.

Remark 20. In fact, the real numbers θ1, θ2 . . . θ6N 7 are the combined branch points of the functions − zn,1(θ), n = 1, 2,...N, and zn,2(θ), n = 2, 3,...N. (see Section 8).

Remark 21. As shown in Section 8, the functions zn,1(θ), n = 2 ...N, and zn,2(θ), n = 3,...N, have exactly three branch singularities each, and the functions z1,1(θ) and z2,2(θ) have exactly one branch singularity each. We plot z2,1(θ), z2,2(θ), z3,1(θ), and z3,2(θ) in Figure 5.

Suppose now that, as in Theorem 19, zn,1(θ), n = 1, 2,...N, are analytic functions which satisfy det A(zn,1(θ), θ) = 0, and zn,2(θ), n = 2, 3,...N, are analytic functions which satisfy det A(zn,2(θ), θ) = 0, n = 2, 3,...N. We observed in Remark 18 that pn,1(θ) and qn,1(θ) are determined explicitly by zn,1(θ) for n = 1, 2 ...N, and, similarly pn,2(θ) and qn,2(θ) are determined explicitly by zn,2(θ) for n = 2, 3 ...N. We recall that, if µ (t) = p t zn,j , and µ (t) = q t zn,j , (119) τ,o n,j · | | ν,e n,j · | | then the corresponding components of the velocity hτ,o(t) and hν,e(t) deﬁned by (105) satisfy hτ,o(t) X∞ pn,j m = F(m, zn,j, θ) t , (120) hν,e(t) qn,j · m=1 for 0 < t < 1, n = 1, 2,...N when j = 1, and n = 2, 3 ...N when j = 2, where F is deﬁned by (109) (see Theorem 17).

16 z2,2(θ) z2,1(θ) 40

20 30

10 20

0 10

10 0 − 0 0.5 1 1.5 2 0 0.5 1 1.5 2 θ θ

z3,1(θ) z3,2(θ) 40 20 30

20 10

0 0

10 − 0 0.5 1 1.5 2 0 0.5 1 1.5 2 θ θ

Figure 5: Plots for the real and imaginary parts of the functions z2,1(θ) (top left), z2,2(θ) (top right), z3,1(θ) (bottom left), and z3,2(θ) (bottom right) for 0 < θ < 2. The solid lines represent the real part of z, the dashed line represents the imaginary part of z, and the vertical dotted lines indicate the locations of the branch points.

We note that the implicit functions z (θ), satisfying (118), are defined for n 2, as opposed to the n,2 ≥ implicit functions zn,1(θ), satisfying (117), which are defined for n 1. We observe that the function z1,2(θ), defined by z (θ) 1, satisfies (118), since when z = 1, ≥ 1,2 ≡ z sin (πθ) sin (πzθ) = sin (πθ) sin (πθ) = 0 , (121) − − for all θ. In the following lemma, we compute the velocity field when (µ (t), µ (t)) = (0, 1) t. τ,o ν,e · Lemma 22. Suppose that θ C, µτ,o(t) = 0 and µν,e(t) = t, for 0 < t < 1. Then hτ,o(t) and hν,e(t) defined by (105) satisfy ∈ hτ,o(t) X∞ 0 n = F1(θ)t + F(n, 1, θ) t , (122) hν,e(t) 1 · n=2 for 0 < t < 1, where F is defined in (109) and 1 sin2 (πθ) F1(θ) = − . (123) −2π πθ sin (πθ) cos (πθ) −

17 z Proof. Substituting µτ,o(t) = 0 and µν,e(t) = t in (105), where z is not an integer, the corresponding components on the boundary hτ,o(t) and hν,e(t) using Theorem 15 are given by

h (t) a (z, θ) X∞ F (n, z, θ) τ,o = 1,2 tz + 1,2 tn , (124) hν,e(t) 1/2 + a2,2(z, θ) F2,2(n, z, θ) − n=2 for 0 < t < 1 , where a1,2(z, θ), a2,2(z, θ) are defined in (82), (84) respectively, and F1,2(n, z, θ),F2,2(n, z, θ) are defined in (86), (88) respectively. The result then follows from taking the limit as z 1. → It is clear from (120), (122) that there is no constant term in the Taylor series of the components of the velocity hτ,o(t) and hν,e(t). The following lemma computes the velocity field when the components of the density µτ,o(t) and µν,e(t) are constants.

Lemma 23. Suppose that θ C, µτ,o(t) = p0 and µν,e(t) = q0, where p0, q0 are constants. Then hτ,o(t) and ∈ hν,e(t) defined by (105) satisfy hτ,o(t) p0 X∞ p0 n = F0(θ) + F(n, 0, θ) t , (125) hν,e(t) q0 q0 · n=1 for 0 < t < 1, where F is defined in (109) and 1 π sin (πθ) + π(1 θ) cos (πθ) π(1 θ) sin (πθ) F0(θ) = − − − − . (126) −2π π(1 θ) sin (πθ) π sin (πθ) π(1 θ) cos (πθ) − − − − − Proof. The result follows from taking the limit z 0 in (111). → In the following theorem, we describe the matrix B(θ) that maps the coefficients of the basis functions (p t zn,j , q t zn,j ) to the Taylor expansion coefficients of the corresponding velocity field. n,j| | n,j| | Theorem 24. Suppose N 2 is an integer. Suppose further that, as in Theorem 19, θ1, θ2, . . . θ6N 7 ≥ − are real numbers on the interval (0, 2), and that zn,1(θ), n = 1, 2,...N, are analytic functions satisfying det A(zn,1(θ), θ) = 0 for θ V C, where V is a simply connected open set containing the strip D with ∈ ⊂ 6N 7 Re(θ) (0, 2) and the interval (0, 2) θj j=1− . Similarly, suppose that zn,2(θ), n = 2, 3 ...N, are analytic functions∈ satisfying det A(z (θ), θ)\{ = 0 }for θ V . Let (p , q ) A(z (θ), θ) , n = 1, 2,...N, and n,2 ∈ n,1 n,1 ∈ N { n,1 } (pn,2, qn,2) A(zn,2(θ), θ) , n = 2, 3,...N. Suppose that z1,2(θ) 1, p1,2 = 0, and q1,2 = 1. Finally, suppose that∈ N { } ≡

N ! X µ (t) = c + c p t zn,1 + d p t zn,2 sgn(t) , (127) τ,o 0 n n,1| | n n,2| | n=1 N ! X µ (t) = d + c q t zn,1 + d q t zn,2 , (128) ν,e 0 n n,1| | n n,2| | n=1 for 1 < t < 1, where cj, dj C, j = 0, 1 ...N. Then − ∈ N ! X h (t) = α t n sgn(t) + O( t N+1) (129) τ,o n| | | | n=0 N ! X h (t) = β t n + O( t N+1) , (130) ν,e n| | | | n=0 for 1 < t < 1, where −     α0 c0  β0   d0       .   .   .  = B(θ)  .  , (131)     αN  cN  βN dN

18 B(θ) is a (2N + 2) (2N + 2) matrix, and θ V . The 2 2 block of B(θ) which maps cn, dn to α`, β` is given by × ∈ × pn,1(θ) pn,2(θ) B`,n(θ) = F(`, zn,1(θ), θ) F(`, zn,2(θ), θ) , (132) qn,1(θ) qn,2(θ) for `, n = 1, 2,...N, where F is deﬁned in (109), except for the case ` = n = 1. In the case ` = n = 1, the matrix B1,1(θ) is given by p1,1(θ) B1,1(θ) = F(1, z1,1(θ), θ) F1(θ) , (133) q1,1(θ)

where F is deﬁned in (109), and F1 is deﬁned in (123). Finally, if either ` = 0 or n = 0, then the matrices B`,n(θ) are given by

0 0 B (θ) = , (134) `,0 0 0

B0,0(θ) = F0(θ) , (135)

B0,n(θ) = F(n, 0, θ) , (136)

for `, n = 1, 2,...N, where F is deﬁned in (109), and F0 is deﬁned in (126). Proof. It follows from Theorem 19 that for θ V , det A(z (θ), θ) = 0, n 1, 2,...N, and det A(z (θ), θ) = ∈ n,1 ∈ n,2 0, n = 2, 3 ...N. Moreover, zn,1(θ), n = 1, 2 ...N, and zn,2(θ), n = 2, 3 ...N, are not integers for θ V 1 . Since (p , q ) A(z , θ) , n = 1, 2,...N, and, (p , q ) A(z , θ) , n = 2, 3,...N∈, we\{ ob-} n,1 n,1 ∈ N { n,1 } n,2 n,2 ∈ N { n,2 } serve that pn,1, qn,1, and zn,1, n = 1, 2 ...N and pn,2, qn,2, and zn,2, n = 2, 3 ...N satisfy the conditions of Theorem 17 for θ V 1 and the corresponding entries of the matrix B(θ) can be derived from (108) in Theorem 17. ∈ \{ } Furthermore, we observe that the density corresponding to p1,2, q1,2 and z1,2 satisfy the conditions for Lemma 22, and thus the corresponding entries of the matrix B(θ) can be derived from (122) in Lemma 22. Finally, the entries of B(θ) corresponding to µτ,o(t), µν,e(t) = (c0, d0) can be obtained from Lemma 23, with which the result follows for θ V 1 . The result for θ = 1 follows∈ by\{ taking} the limit θ 1 in (131) and from the observation that the limit B(θ) as θ 1 exists (see Theorem 38 in Section 9). → →

4.1.2. Invertibility of B in (131) zn,j zn,j The matrix B(θ) is a mapping from coefficients of the basis functions (pn,j t , qn,j t ) to the Taylor expansion coefficients of the corresponding velocity field (see (131)). In this| section,| we| | observe that B(θ) is invertible for all θ (0, 2) except for countably many values of θ. We then use this result to derive a converse of Theorem 24.∈ The principal result of this section is Theorem 27. In the following lemma, we show that B(θ) is invertible for all θ (0, 2) except for countably many values of θ. ∈

Lemma 25. Suppose that N 2 is an integer. There exists a countable set φ ∞ (0, 2), such that ≥ { m}m=1 ⊂ B(θ) is an invertible matrix for θ (0, 2) φm m∞=1. Moreover, the limit points of φm m∞=1 are a subset of 6N 7 ∈ \{ } { } θ − 0, 2 , where θ , j = 1, 2 ... 6N 7, are the branch points of the functions z (θ), n = 1, 2 ...N, { j}j=1 ∪ { } j − n,1 and zn,2(θ), n = 2, 3 ...,N, deﬁned in Theorem 19.

6N 7 Proof. Suppose that V is as deﬁned in Theorem 19. Recall that the interval (0, 2) θj j=1− V . Clearly, the entries of B(θ) are analytic functions of θ V 1 . Furthermore, using Theorem\{ } 38 in⊂ Section 9, we ∈ \{ }

19 observe that B(θ) is analytic at θ = 1 as well, since " # 1/2 0  − ` = j = 0   0 1/2  − " #  0 1/2  − ` = j = 2m = 0  1/2 0 6 − lim B`,j(θ) = " # . (137) θ 1  1/2 0 →  − ` = j = 2m + 1   0 1/2  − " #  0 0  otherwise  0 0

Thus, det B(θ) is an analytic function for θ V . Moreover, using (137), det(B(1)) = 0. Using standard results in complex analysis, since det B(θ) is∈ not identically zero, we conclude that6 the matrix B(θ) is invertible for all θ (0, 2) except for a countable set of values of θ = φm, m = 1, 2,.... Moreover, the set of limit points of det∈B(θ) = 0, i.e. the values of θ for which B(θ) is not invertible, is a subset of ∂V (0, 2), 6N 7 ∩ where ∂V is the boundary of the set V . Clearly, ∂V (0, 2) = θ − 0, 2 . ∩ { j}j=1 ∪ { }

Remark 26. In Lemma 25, we show that det B(θ) has countably many zeros on the interval (0, 2). In fact, it is possible to show that there are finitely many{ zeros} of det B(θ) on the interval (0, 2). On inspecting the form of the entries of B(θ), we note that det B(θ) is a linear{ combination} of T (θ)/P (θ) where T (θ) is a trigonometric polynomial of degree less than{ or equal} to N and P (θ) is a finite product of functions (zj,`(θ) kj,`) for some integer kj,`. A detailed analysis shows that the functions zj,`(θ) are essentially non-oscillatory− for θ (0, 2). Since both the functions T (θ) and P (θ) are band-limited functions, det B(θ) cannot have infinitely∈ many zeros for θ (0, 2). { } ∈ The following theorem is the principal result of this section. Theorem 27. Suppose that N 2 is an integer. Then for each θ (0, 2) except for countably many values, ≥ ∈ there exist pn,j, qn,j, zn,j C, n = 1, 2 ...N and j = 1, 2, such that the following holds. Suppose αn, βn C, ∈ ∈ n = 0, 1,...N, and hτ,o(t) and hν,e(t) are given by

N ! N ! X X h (t) = α t n sgn(t) , and h (t) = β t n , (138) τ,o n| | ν,e n| | n=0 n=0 for 1 < t < 1. Then there exist unique numbers cn, dn C, n = 0, 1,...N, such that, if µτ,o(t) and µν,e(t) deﬁned− by ∈

N ! X µ (t) = c + c p t zn,1 + d p t zn,2 sgn(t) , (139) τ,o 0 n n,1| | n n,2| | n=1 N X µ (t) = d + c q t zn,1 + d q t zn,2 , ν,e 0 n n,1| | n n,2| | n=1 for 1 < t < 1, then µ (t) and µ (t) satisfy − τ,o ν,e h (t) 1 µ (t) Z 1 k (s, t) k (s, t) µ (s) τ,o = τ,o + 1,1 1,2 τ,o ds , (140) hν,e(t) −2 µν,e(t) 0 k2,1(s, t) k2,2(s, t) µν,e(s) for 1 < t < 1 with error O( t N+1), where k (s, t), j, ` = 1, 2, are deﬁned in (34) – (37). − | | j,`

20 Proof. Suppose zn,1(θ), n = 1, 2,...N, and zn,2(θ), n = 2, 3 ...N, are the implicit functions that satisfy det A(z, θ) = 0 as deﬁned in Theorem 19. Let (p , q ) A(z (θ), θ) , n = 1, 2 ...N, and n,1 n,1 ∈ N { n,1 } (pn,2, qn,2) A(zn,2(θ), θ) . Let z1,2 = 1, p1,2 = 0, and q1,2 = 1. Given pn,j, qn,j, and zn,j, n = 1, 2 ...N and j = 1,∈2, N let { B(θ) be the} (2N + 2) (2N + 2) matrix deﬁned in Theorem 24. Suppose further that × φ ∞ (0, 2) are the values of θ for which B(θ) is not invertible (see Lemma 25). Finally, since B(θ) is { m}m=1 ⊂ invertible for all θ (0, 2) φ ∞ , let ∈ \{ m}m=1     c0 α0  d0   β0       .  1  .   .  = B− (θ)  .  . (141)     cN  αN  dN βN The result then follows from using Theorem 24.

Remark 28. In Remark 16, we noted that the components of the density µ can be expressed in terms of functions of the form t α cos (β log t ) and t α sin (β log t ), as opposed to t z, where z = α+iβ. The precise statement is as follows.| | We observe| | that, when| | θ is real,| | if det A(z, θ) = 0|,| then det A(¯z, θ) = 0. Moreover, if (p, q) A(z, θ) , then (¯p, q¯) A(¯z, θ) . Thus, the numbers p, q, and z occur in complex conjugates. Results∈ analogous N { to} Theorem 24 and∈ N Theorem{ } 27 can be derived for the case when the components of µ are given by

N X µ (t) = c + c (r t αn cos (β log t ) s t αn sin (β log t )) (142) τ,o 0 n n| | n | | − n| | n | | n=1 N X + d (s t αn cos (β log t ) + r t αn sin (β log t )) sgn(t) n n| | n | | n| | n | | n=1 N X µ (t) = d + c (v t αn cos (β log t ) w t αn sin (β log t )) (143) ν,e 0 n n| | n | | − n| | n | | n=1 N X + d (w t αn cos (β log t ) + v t αn sin (β log t )) , n n| | n | | n| | n | | n=1 for 1 < t < 1, where z = α + iβ , p = r + is , and q = v + iw . The advantage for using the − n n n n n n n n n representation (142), (143) for the density µ is the following. If the components of the velocity hτ,o(t) and hν,e(t) are real, then the solution cn, dn, n = 0, 1 ...N when µ deﬁned by (142), (143) which satisﬁes (105) to order N accuracy, is also real.

4.2. Tangential even, normal odd case In this section, we investigate tangential even, normal odd case (see equation (95)). In Section 4.2.1, we investigate the values of p, q and z in (96) for which the resulting components of the velocity are smooth functions. In Section 4.2.2, we show that, for every h of the form (101), there exists a density µ of the form (102), (103), which satisﬁes the integral equation (95) to order N. The proofs of the results in this section are essentially identical to the corresponding proofs in Section 4.1. For brevity, we present the statements of the theorems without proof.

4.2.1. The values of p, q, and z in (100) Suppose that µτ,e(t) and µν,o(t) are given by µ (t) = p t z , and µ (t) = q t z sgn(t) , (144) τ,e · | | ν,o · | | for 1 < t < 1, where p, q, z C. In this section, we determine the values of p, q and z such that hτ,e(t) and − ∈ hν,o(t) deﬁned by h (t) 1 µ (t) Z 1 k (s, t) k (s, t) µ (s) τ,e = τ,e 1,1 1,2 τ,e ds , (145) hν,o(t) −2 µν,o(t) − 0 k2,1(s, t) k2,2(s, t) µν,o(s)

21 are smooth functions of t for 0 < t < 1. The principal result of this section is Theorem 33. The following lemma describes sufficient conditions for p, q, and z such that, if µ is defined by (144), then the velocity h given by (145) is smooth. Theorem 29. Suppose that θ (0, 2), z is not an integer, and z satisfies det A(z, θ) = 0 where ∈ 1 a (z, θ) a (z, θ) A(z, θ) = I 1,1 1,2 , (146) −2 − a2,1(z, θ) a2,2(z, θ)

I is the 2 2 identity matrix and aj,`(z, θ), j, ` = 1, 2, are given by (81) – (84). Furthermore, suppose that (p, q) ×A(z, θ) , where A denotes the null space of the matrix A. Suppose ﬁnally that ∈ N { } N{ } µ (t) = p tz , and µ (t) = q tz , (147) τ,e · ν,o · for 0 < t < 1. Then hτ,e(t) and hν,o(t) deﬁned by (145) satisfy

h (t) X∞ p τ,o = F(n, z, θ) tn , (148) hν,e(t) − q · n=1 for 0 < t < 1, where F (n, z, θ) F (n, z, θ) F(n, z, θ) = 1,1 1,2 , (149) F2,1(n, z, θ) F2,2(n, z, θ) and Fj,`(n, z, θ), j, ` = 1, 2, are given by (85) – (88). A straightforward calculation shows that (z sin (πθ) + sin (πzθ)) (z sin (πθ) + sin (πz(2 θ))) det A(z, θ) = − (150) 4 sin2 (πz) Thus, if z is not an integer and either

z sin (πθ) + sin (πz(2 θ)) = 0 , (151) − or z sin (πθ) + sin (πzθ) = 0 , (152) then det A(z, , θ) = 0. In the following theorem, we prove the existence of the implicit functions z(θ) deﬁned by (151), (152) on the interval (0, 2).

Theorem 30. Suppose that N 2 is an integer. Then there exists 6N 7 real numbers θ1, θ2, . . . θ6N 7 (0, 2) such that the following holds.≥ Suppose that D is the strip in the upper− half plane with 0 < Re(θ)−< ∈2 6N 7 that includes the interval (0, 2) θ − , i.e. \{ j}j=1 6N 7 D = θ C : Re(θ) (0, 2) , 0 Im(θ) < θj − . (153) { ∈ ∈ ≤ ∞} \ { }j=1 Then, there exists a simply connected open set D V C and analytic functions zn,1(θ): V C, n = 2, 3 ...N, which satisfy ⊂ ⊂ →

z sin (πθ) + sin (πz(2 θ)) = 0 , z(1) = n , (154) − for θ V , and analytic functions zn,2(θ): V C, n = 1, 2,...N, which satisfy ∈ → z sin (πθ) + sin (πzθ) = 0 , z(1) = n , (155) for θ V (see Figure 4 for an illustrative domain V ). Moreover, the functions zn,1(θ), n = 2, 3 ...N, do not take integer∈ values for all θ V 1 and satisfy det A(z (θ), θ) = 0, n = 2, 3 ...N, for all θ V (see (146), ∈ \{ } n,1 ∈ (150)). Similarly, the functions zn,2(θ), n = 1, 2,...N, do not take integer values for all θ V 1 and satisfy det A(z (θ), θ) = 0, n = 1, 2 ...N, for all θ V (see (146), (150)). ∈ \{ } n,2 ∈

22 We note that the implicit functions z (θ), satisfying (154), are defined for n 2, as opposed to, the n,1 ≥ implicit functions zn,2(θ), satisfying (155), which are defined for n 1. We observe that the function z1,1(θ) defined by z (θ) 1, satisfies (154), since when z = 1, ≥ 1,1 ≡ z sin (πθ) + sin (πz(2 θ)) = sin (πθ) sin (πθ) = 0 , (156) − −

for all θ. In the following lemma, we compute the velocity ﬁeld when (µτ,e(t), µν,o(t)) = (0, 1)t.

Lemma 31. Suppose that θ C, µτ,e(t) = 0 and µν,o(t) = t, for 0 < t < 1. Then hτ,e(t) and hν,o(t) deﬁned by (145) satisfy ∈ hτ,e(t) X∞ 0 n = F1(θ)t F(n, 1, θ) t , (157) hν,o(t) − 1 · n=2 for 0 < t < 1, where F is deﬁned in (109) and

1 sin2 (πθ) F (θ) = . (158) 1 −2π π(2 θ) + sin (πθ) cos (πθ) − It is clear from (148), (157) that there is no constant term in the Taylor series of the components of the velocity hτ,e and hν,o. The following lemma computes the velocity ﬁeld when the components of the density µτ,e and µν,o are constants.

Lemma 32. Suppose that θ C, µτ,e(t) = p0 and µν,o(t) = q0, where p0, q0 are constants. Then hτ,e(t) and ∈ hν,o(t) deﬁned by (145) satisfy hτ,e(t) p0 X∞ p0 n = F0(θ) F(n, 0, θ) t , (159) hν,o(t) q0 − q0 · n=1 for 0 < t < 1, where F is deﬁned in (149) and

1 π + sin (πθ) π(1 θ) cos (πθ) π(1 θ) sin (πθ) F0(θ) = − − − . (160) −2π π(1 θ) sin (πθ) π + sin (πθ) + π(1 θ) cos (πθ) − − In the following theorem, we describe the matrix B(θ) that maps the coefficients of the basis functions (p t zn,j , q t zn,j ) to the Taylor expansion coefficients of the corresponding velocity field. n,j| | n,j| | Theorem 33. Suppose N 2 is an integer. Suppose further that, as in Theorem 30, θ1, θ2, . . . θ6N 7 are real ≥ − numbers on the interval (0, 2) zn,1(θ), n = 2, 3,...N, are analytic functions satisfying det A(zn,1(θ), θ) = 0 for θ V C, where V is a simply connected open set containing the strip D with Re(θ) (0, 2) and the ∈ ⊂ 6N 7 ∈ interval (0, 2) θ − . Similarly, suppose that z (θ), n = 1, 2,...N, are analytic functions satisfying \{ j}j=1 n,2 det A(zn,2(θ), θ) = 0 for θ V . Let (pn,1, qn,1) A(zn,1(θ), θ) , n = 2, 3,...N, and (pn,2, qn,2) A(z (θ), θ) , n = 1, 2,...N∈ . Suppose that z (∈θ) N {1, p = 0, and} q = 1. Finally, suppose that ∈ N{ n,2 } 1,1 ≡ 1,1 1,1 N ! X µ (t) = c + c p t zn,1 + d p t zn,2 , (161) τ,e 0 n n,1| | n n,2| | n=1 N ! X µ (t) = d + c q t zn,1 + d q t zn,2 sgn(t) , (162) ν,o 0 n n,1| | n n,2| | n=1 (163)

for 1 < t < 1, where cj, dj C, j = 0, 1 ...N. Then − ∈ N ! X h (t) = α t n + O( t N+1) (164) τ,o n| | | | n=0 N ! X h (t) = β t n sgn(t) + O( t N+1) , (165) ν,e n| | | | n=0

23 for 1 < t < 1, where −     α0 c0  β0   d0       .   .   .  = B(θ)  .  , (166)     αN  cN  βN dN

B(θ) is a (2N + 2) (2N + 2) matrix, and θ V . The 2 2 block of B(θ) which maps cn, dn to α`, β` is given by × ∈ × pn,1(θ) pn,2(θ) B`,n(θ) = F(`, zn,1(θ), θ) F(`, zn,2(θ), θ) , (167) − qn,1(θ) qn,2(θ) for `, n = 1, 2,...N, where F is defined in (109), except for the case ` = n = 1. In the case ` = n = 1, the matrix B1,1(θ) is given by p1,2(θ) B1,1(θ) = F1(θ) F(1, z1,2(θ), θ) , (168) − q1,2(θ) where F is defined in (109), and F1 is defined in (158). Finally, if either ` = 0 or n = 0, then the matrices B`,n(θ) are given by 0 0 B (θ) = , (169) `,0 0 0

B0,0(θ) = F0(θ) , (170) B (θ) = F(n, 0, θ) , (171) 0,n − for `, n = 1, 2,...N, where F is deﬁned in (109), and F0 is deﬁned in (160).

4.2.2. Invertibility of B in (166) zn,j zn,j The matrix B(θ) is a mapping from coefficients of the basis functions (pn,j t , qn,j t ), to the corresponding Taylor expansion coefficients of the velocity field on the boundary| |h. In this| | section, we observe that B(θ) is invertible for all θ (0, 2) except for countably many values of θ. We then use this result to derive a converse of Theorem 33.∈ The following theorem is the principal result of this section. Theorem 34. Suppose that N 2 is an integer. Then for each θ (0, 2) except for countably many values, ≥ ∈ there exist pn,j, qn,j, zn,j C, n = 1, 2 ...N and j = 1, 2, such that the following holds. Suppose αn, βn C, ∈ ∈ n = 0, 1,...N, and hτ,e(t) and hν,o(t) are given by

N ! N ! X X h (t) = α t n , and h (t) = β t n sgn(t) , (172) τ,e n| | ν,e n| | n=0 n=0 for 1 < t < 1. Then there exist unique numbers cn, dn C, n = 0, 1,...N, such that, if µτ,e(t) and µν,o(t) deﬁned− by ∈

N ! X µ (t) = c + c p t zn,1 + d p t zn,2 , (173) τ,o 0 n n,1| | n n,2| | n=1 N ! X µ (t) = d + c q t zn,1 + d q t zn,2 sgn(t) , ν,e 0 n n,1| | n n,2| | n=1 for 1 < t < 1, then µ (t) and µ (t) satisfy − τ,e ν,o h (t) 1 µ (t) Z 1 k (s, t) k (s, t) µ (s) τ,e = τ,e 1,1 1,2 τ,e ds , (174) hν,o(t) −2 µν,o(t) − 0 k2,1(s, t) k2,2(s, t) µν,o(s) for 1 < t < 1 with error O( t N+1), where k , j, ` = 1, 2, are deﬁned in (34) – (37). − | | j,`

24 5. Numerical Results

To solve the integral equations (94), (95) on polygonal domains, we use the representations (139) and (173) to construct purpose made discretizations using standard techniques (see, for example [20, 21, 22]). A detailed description of this part of the procedure will be published at a later date. We illustrate the performance of the algorithm with several numerical examples. The interior velocity boundary problem was solved on each of the domains in Figures 6 to 10, where the boundary data is generated by ﬁve Stokeslets located outside the respective domains. We then compute the error E given by s P5 u (t ) u (t ) 2 E = m=1 | comp m − exact m | , (175) P5 u (t ) 2 m=1 | exact m |

where tm are targets in the interior of the domain, ucomp(t) is velocity computed numerically using the algorithm, and uexact(t) is the exact velocity at the target t. We plot the spectrum for the discretized linear systems corresponding to the associated integral equations in Figures 6 to 10. In Table 1, we report the number of discretization nodes n, the condition number of the discrete linear system κ, and the error E, for each domain.

n E κ 15 2 Γ1 220 3.0 10− 4.5 10 × 14 × 5 Γ2 285 2.1 10− 2.9 10 × 14 × 5 Γ3 489 4.8 10− 8.7 10 × 12 × 5 Γ4 968 1.1 10− 7.9 10 × 13 × 6 Γ 1343 6.7 10− 6.1 10 5 × ×

Table 1: Condition number and error for polygonal domains Γ1 –Γ5

102

101

100

1 10−

2 10−

3 10−

4 10−

Figure 6: A cone Γ1 (left) and the spectrum of the discretized linear system for Γ1 (right)

Remark 35. We note that the condition numbers reported for the boundaries Γj, j = 2, 3, 4, 5, are larger than the condition numbers of the underlying integral equations. This issue can be remedied by using a slightly more involved scaling of the discretization scheme (see, for example, [23]).

6. Conclusions and extensions

In this paper, we construct an explicit basis for the solution of a standard integral equation corresponding to the biharmonic equation with gradient boundary conditions, on polygonal domains. The explicit and detailed knowledge of the behavior of solutions to the integral equation in the vicinity of the corner was used to create purpose-made discretizations, resulting in eﬃcient numerical schemes accurate to essentially machine precision. In this section, we discuss several directions in which these results are generalizable.

25 102

101

100

1 10−

2 10−

3 10−

4 10−

Figure 7: An equilateral triangle Γ2 (left) and the spectrum of the discretized linear system for Γ2 (right)

102

101

100

1 10−

2 10−

3 10−

4 10−

Figure 8: A right angle triangle Γ3 (left) and the spectrum of the discretized linear system for Γ3 (right)

102

101

100

1 10−

2 10−

3 10−

4 10−

Figure 9: A star-shaped curve Γ4 (left) and the spectrum of the discretized linear system for Γ4 (right)

102

101

100

1 10−

2 10−

3 10−

4 10−

Figure 10: A tank-shaped curve Γ5 (left) and the spectrum of the discretized linear system for Γ5 (right)

6.1. Inﬁnite oscillations of the biharmonic Green’s function In 1973, S. Osher showed that the domain Green’s function for the biharmonic equation has inﬁnite oscillations for a right-angled wedge, and conjectured that this result holds for any domain with corners.

26 Earlier, we observed that the representations of solutions to the associated integral equations also exhibit inﬁnite oscillations near the corner (see Remark 28). The oscillatory behavior of these basis functions suggests that Osher’s conjecture may be amenable to an analysis similar to the one presented in this paper.

6.2. Curved boundaries In this paper, we derive a representation for the solutions of the integral equations associated with the biharmonic equation, on polygonal domains. In the more general case of curved boundaries with corners, the apparatus of this paper also leads to detailed representations of the solutions near corners. Speciﬁcally, the solutions to the associated integral equations are representable by rapidly convergent series of products of complex powers of t and logarithms of t, where t is the distance from the corner. This analysis closely mirrors the generalization of the authors’ analysis of Laplace’s equation on polygonal domains (see [15]) to the authors’ analysis on domains having curved boundaries with corners (see [24]).

6.3. Generalization to three dimensions The apparatus of this paper admits a straightforward generalization to surfaces with edge singularities, where the parts of the surface on either side of the edge meet at a constant angle along on the edge. The generalization to the case of edges with more complicated geometries is more involved and will be presented at a later date.

6.4. Other boundary conditions In this paper, we analyze the integral equations associated with the velocity boundary value problem for Stokes equation. The approach of this paper extends to a number of other boundary conditions, including the traction boundary value problem, and the mobility problem. In particular, the traction boundary value problem and the mobility problem can be formulated as boundary integral equations that are the adjoints of the integral equations for the velocity boundary value problem.

6.5. Modified biharmonic equation The modified biharmonic equation for a potential ψ is given by ∆2ψ α∆ψ = 0. The equation naturally arises when mixed implicit-explicit schemes are used for the time discretization− of the incompressible Navier- Stokes equation (see, for example [25]). A preliminary analysis indicates that the solutions of the integral equations associated with the modified biharmonic equation are representable as rapidly convergent series of Bessel functions of certain non-integer complex orders. The generalization of the analysis of the biharmonic equation, presented in this paper, to the analysis of the modified biharmonic equation closely mirrors the generalization of the authors’ analysis of Laplace’s equation (see [15]) to the authors’ analysis of the Helmholtz equation (see [17]).

7. Acknowledgements

The authors would like to thank Philip Greengard, Leslie Greengard, Jeremy Hoskins, and Vladimir Rokhlin for many useful discussions. M. Rachh’s work was supported in part by the Oﬃce of Naval Research award number N00014-14-1- 0797/16-1-2123. K. Serkh’s work was supported by the NSF Mathematical Sciences Postdoctoral Research Fellowship (award number 1606262).

8. Appendix A

9. Appendix B

In this section we compute the limit θ 1, of the linear transformation B(θ), which maps coefficients of singular basis functions for the solution→ of the integral equation to the Taylor expansion coefficients of the velocity field. In Section 9.1, we investigate the tangential odd, normal even case (see (94)) and, in Section 9.2, we investigate the tangential even, normal odd case (see (95)).

27 9.1. Tangential odd, normal even case Suppose that A(z, θ) is the 2 2 matrix given by (106) deﬁned in Section 4.1. Suppose N is a positive × integer. Suppose further that, as in Theorem 19, zn,1(θ), n = 1, 2,...N, are analytic functions satisfying det A(zn,1(θ), θ) = 0 for θ Vδ C, where Vδ is a neighborhood of the contour Cδ. Similarly, suppose that z (θ), n = 2, 3 ...N, are∈ analytic⊂ functions satisfying det A(z (θ), θ) = 0 for θ V . Let (p , q ) n,2 n,2 ∈ δ n,1 n,1 ∈ A(zn,1(θ), θ) , n = 1, 2,...N, and (pn,2, qn,2) A(zn,2(θ), θ) , n = 2, 3,...N. We further assume N{ } ∈ N { } 2 that the vectors (pn,1, qn,1), n = 1, 2,... and (pn,2, qn,2), n = 2, 3,..., are ` normalized. Suppose that z1,2(θ) 1, p1,2 = 0, and q1,2 = 1. Finally, suppose that B(θ) is a (2N + 2) (2N + 2) matrix deﬁned in Theorem≡ 24 for θ V . The 2 2 blocks of B(θ) are given by × ∈ δ × pn,1(θ) pn,2(θ) B`,n(θ) = F(`, zn,1(θ), θ) F(`, zn,2(θ), θ) , (176) qn,1(θ) qn,2(θ)

for `, n = 1, 2,...N, where F is deﬁned in (109), except for the case ` = n = 1. In the case ` = n = 1, the matrix B1,1(θ) is given by p1,1(θ) B1,1(θ) = F(1, z1,1(θ), θ) F1(θ) , (177) q1,1(θ)

where F is deﬁned in (109), and F1 is deﬁned in (123). Finally, if either ` = 0 or n = 0, then the matrices B`,n(θ) are given by

0 0 B (θ) = , (178) `,0 0 0

B0,0(θ) = F0(θ) , (179)

B0,n(θ) = F(n, 0, θ) , (180)

for `, n = 1, 2,...N, where F is deﬁned in (109), and F0 is deﬁned in (126). In the following lemma, we describe the behavior of zn,1(θ), pn,1(θ), and qn,1(θ), n = 1, 2,..., in the vicinity of θ = 1.

Lemma 36. Suppose that zn,1(θ), n = 1, 2,..., satisfying det A(zn,1(θ), θ) = 0, be as deﬁned in Theorem 19. 2 Suppose further that (pn,1(θ), qn,1(θ)), n = 1, 2,..., be an ` normalized null vector of A(zn,1, θ). Then in the neighborhood of θ = 1,

π2n(4n2 1) z (θ) = 2n − (θ 1)3 + O( θ 1 4) (181) 2n,1 − 3 − | − | π(2n 1) p (θ) = − (θ 1) + O( θ 1 3) (182) 2n,1 − 2 − | − | q (θ) = 1 + O( θ 1 2) (183) 2n,1 | − | z (θ) = 2n + 1 + 2(2n + 1)(θ 1) + O( θ 1 2) (184) 2n+1,1 − | − | p (θ) = 1 + O( θ 1 ) (185) 2n+1,1 | − | q (θ) = O( θ 1 ) . (186) 2n+1,1 | − |

Proof. Let the superscript 0 denote derivative with respect to θ and ∂t denote the partial derivative with respect to t. Suppose that H(z, θ) is given by

H(z, θ) = z sin (πθ) sin (πz(2 θ)) (187) − −

From Theorem 19, we recall that zn,1(θ), n = 1, 2,..., satisfy H(zn,1, θ) = 0. Using the implicit function theorem ∂θH z0(θ) = = z(1 sec (πz)) (188) −∂zH − Thus, z20 n,1(1) = 0 , and z20 n+1,1(1) = 2(2n + 1) . (189)

28 Implicitly diﬀerentiating H twice we get d ∂ H z00(θ) + ∂ H + ∂ ∂ H z0(θ) + ∂ ∂ H = 0 . (190) z · dθ z z θ · θ θ

Since z20 n,1(1) = 0, we get ∂θθH 2 z200n,1(1) = = πz tan (πz) = 0 . (191) − ∂zH Proceeding in a similar fashion, we observe that

∂θθθH 2 2 2 2 z2000n,1(1) = = π z2n,1(sec (πz2n,1) z2n,1) = 2π n(1 4n ) . (192) − ∂zH − − Let p˜ (θ) = sin (πz) + 2a (z (θ), θ) , (193) n,1 − 2,2 n,1 qñ,1(θ) = 2 sin (πz)a2,1(zn,1(θ), θ) , (194) where a and a are defined in (83) and (84) respectively. Clearly, (˜p , q˜ ) (A). We then set 2,1 2,2 n,1 n,1 ∈ N pñ,1 qñ,1 pn,1(θ) = , and qn,1(θ) = (195) q 2 2 q 2 2 (˜pn,1 +q ñ,1) (˜pn,1 +q ñ,1)

The required Taylor expansions for pn,1, qn,1 are then readily obtained by using the Taylor expansions of zn,1(θ).

In the next lemma, we now describe the behavior of zn,2(θ), pn,2(θ) and qn,2(θ), n = 2, 3,..., in the vicinity of θ = 1.

Lemma 37. Suppose that zn,2(θ), n = 2, 3,..., satisfying det A(zn,2(θ), θ) = 0, be as deﬁned in Theorem 19. 2 Suppose further that (pn,2(θ), qn,2(θ)), n = 2, 3,..., be an ` normalized null vector of A(zn,2, θ). Then in the neighborhood of θ = 1, z (θ) = 2n 4n(θ 1) + O( θ 1 2) (196) 2n,2 − − | − | p (θ) = 1 + O( θ 1 ) (197) 2n,2 | − | q (θ) = 0 + O( θ 1 ) (198) 2n,2 | − | 2π2n(n + 1)(2n + 1) z (θ) = 2n + 1 + (θ 1)3 + O( θ 1 4) (199) 2n+1,1 3 − | − | p (θ) = πn(θ 1) + O( θ 1 3) (200) 2n+1,2 − − | − | q (θ) = 1 + O( θ 1 2) (201) 2n+1,2 | − | Proof. The proof proceeds in a similar manner as the proof of Lemma 36.

Combining Lemmas 36 and 37, we present the principal result of this section, which computes the limit θ 1 of the matrix B(θ) in the following theorem. → Theorem 38. Suppose that N is a positive integer and suppose further that B is given by (176) – (180). Then " # 1/2 0  − ` = j = 0   0 1/2  − " #  0 1/2  − ` = j = 2m = 0  1/2 0 6 − lim B`,j(θ) = " # , (202) θ 1  1/2 0 →  − ` = j = 2m + 1   0 1/2  − " #  0 0  otherwise  0 0

29 for all `, j = 0, 1,...N. Proof. Let F˜(n, θ) = 2πF(n, z, θ) (n z) for j, ` = 1, 2 where F is given by (109). On inspecting the entries of F˜(n, θ), we observe that · − 0 0 F˜(n, 1) = , (203) 0 0

for all n N. Since zj,`(1) = j, we conclude that ∈ ˜ pj,` F(n, θ) pj,` 0 lim F(n, zj,`, θ) = lim = , (204) θ 1 qj,` θ 1 ( z + n) qj,` 0 → → − j,` for all j = n, and ` = 1, 2. B (θ) = 0 is the zero matrix by deﬁnition and for B (θ), we have 6 n,0 0,n F˜(n, θ) 0 0 lim B0,n(θ) = lim F(n, 0, θ) = lim = , (205) θ 1 θ 1 θ 1 2πn 0 0 → → → for n = 1, 2,...N. We now turn our attention to the diagonal terms. For n = 0, it follows from a simple calculation that

1/2 0 lim B0,0(θ) = lim F0(θ) = − . (206) θ 1 θ 1 0 1/2 → → − For n 2, using Lemma 36 and Lemma 37, it follows from a rather tedious calculation that ≥ " 4 # p2n,1(θ) O( θ 1 ) ˜ 3 2 F(2n, θ) = π n(4n 1) | − 3| 4 (207) q2n,1(θ) − (θ 1) + O( θ 1 ) − 3 − | − | p (θ) (2n + 1)π(θ 1) + O( θ 1 2) F˜(2n + 1, θ) 2n+1,1 = , (208) q (θ) O(−θ 1 2) | − | 2n+1,1 | − | and p (θ) 4nπ(θ 1) + O( θ 1 2) F˜(2n, θ) 2n,2 = (209) q (θ) − O−( θ 1 2)| − | 2n,2 | − | " 4 # p2n+1,2(θ) O( θ 1 ) ˜ 3 F(2n + 1, θ) = 2π n(n+1)(2n+1) | − |3 4 . (210) q2n+1,2(θ) (θ 1) + O( θ 1 ) 3 − | − |

Finally, from the deﬁnition of F1 in (123), we note that

0 lim F1(θ) = . (211) θ 1 1/2 → − The result then follows from combining (207) – (211).

9.2. Tangential even, normal odd case Suppose that A(z, θ) is the 2 2 matrix given by (146) deﬁned in Section 4.2. Suppose N is a positive × integer. Suppose further that, as in Theorem 30, zn,1(θ), n = 2, 3,...N, are analytic functions satisfying det A(zn,1(θ), θ) = 0 for θ Vδ C, where Vδ is a neighborhood of the contour Cδ. Similarly, suppose that z (θ), n = 1, 2 ...N, are∈ analytic⊂ functions satisfying det A(z (θ), θ) = 0 for θ V . Let (p , q ) n,2 n,2 ∈ δ n,1 n,1 ∈ A(zn,1(θ), θ) , n = 2, 3,...N, and (pn,2, qn,2) A(zn,2(θ), θ) , n = 1, 2,...N. We further assume N{ } ∈ N { } 2 that the vectors (pn,1, qn,1), n = 2, 3,... and (pn,2, qn,2), n = 1, 2,..., are ` normalized. Suppose that z1,2(θ) 1, p1,2 = 0, and q1,2 = 1. Finally, suppose that B(θ) is a (2N + 2) (2N + 2) matrix deﬁned in Theorem≡ 33 for θ V . The 2 2 blocks of B(θ) are given by × ∈ δ × pn,1(θ) pn,2(θ) B`,n(θ) = F(`, zn,1(θ), θ) F(`, zn,2(θ), θ) , (212) − qn,1(θ) qn,2(θ)

30 for `, n = 1, 2,...N, where F is deﬁned in (109), except for the case ` = n = 1. In the case ` = n = 1, the matrix B1,1(θ) is given by p1,2(θ) B1,1(θ) = F1(θ) F(1, z1,2(θ), θ) , (213) − q1,2(θ)

where F is deﬁned in (109), and F1 is deﬁned in (158). Finally, if either ` = 0 or n = 0, then the matrices B`,n(θ) are given by

0 0 B (θ) = , (214) `,0 0 0

B0,0(θ) = F0(θ) , (215) B (θ) = F(n, 0, θ) , (216) 0,n −

for `, n = 1, 2,...N, where F is deﬁned in (109), and F0 is deﬁned in (160). The proofs of the results in this section are similar to the corresponding proofs in Section 9.1. For conciseness, we state the results without proof. The principal result of this section is Theorem 41. In the following lemma, we describe the behavior of zn,1(θ), pn,1(θ), and qn,1(θ), n = 2, 3,..., in the vicinity of θ = 1.

Lemma 39. Suppose that zn,1(θ), n = 2, 3,..., satisfying det A(zn,1(θ), θ) = 0, be as deﬁned in Theorem 30. 2 Suppose further that (pn,1(θ), qn,1(θ)), n = 2, 3,..., be an ` normalized null vector of A(zn,1, θ).

z (θ) = 2n + 4n(θ 1) + O( θ 1 2) (217) 2n,1 − | − | p (θ) = 1 + O( θ 1 ) (218) 2n,1 | − | q (θ) = 0 + O( θ 1 ) (219) 2n,1 | − | 2π2n(n + 1)(2n + 1) z (θ) = 2n + 1 (θ 1)3 + O( θ 1 4) (220) 2n+1,1 − 3 − | − | p (θ) = πn(θ 1) + O( θ 1 3) (221) 2n+1,1 − − | − | q (θ) = 1 + O( θ 1 2) (222) 2n+1,1 | − |

Similarly, in the following lemma, we describe the behavior of zn,2(θ), pn,2(θ), and qn,2(θ), n = 1, 2,..., in the vicinity of θ = 1.

Lemma 40. Suppose that zn,2(θ), n = 1, 2,..., satisfying det A(zn,2(θ), θ) = 0, be as deﬁned in Theorem 30. 2 Suppose further that (pn,2(θ), qn,2(θ)), n = 1, 2,..., be an ` normalized null vector of A(zn,2, θ). Then in the neighborhood of θ = 1,

π2n(4n2 1) z (θ) = 2n + − (θ 1)3 + O( θ 1 4) (223) 2n,1 3 − | − | π(2n 1) p (θ) = − (θ 1) + O( θ 1 3) (224) 2n,2 − 2 − | − | q (θ) = 1 + O( θ 1 2) (225) 2n,2 | − | z (θ) = 2n + 1 2(2n + 1)(θ 1) + O( θ 1 2) (226) 2n+1,2 − − | − | p (θ) = 1 + O( θ 1 ) (227) 2n+1,2 | − | q (θ) = 0 + O( θ 1 ) (228) 2n+1,2 | − | Finally, we present the principal result of this section in the following lemma. Theorem 41. Suppose that N is a positive integer and suppose further that B is given by (212) – (216).

31 Then " #  1/2 0  − ` = j = 2m  0 1/2  " − #  0 1/2 lim B`,j(θ) = − ` = j = 2m + 1 , (229) θ 1  1/2 0 →  − " #  0 0  otherwise  0 0 for all `, j = 0, 1, 2,...N.

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