Method of Riemann surfaces in the study of supercavitating flow around two hydrofoils in a channel

Transcription

1 Physica D 235 ( Method of Riemann surfaces in the study of supercaitating flow around two hydrofoils in a channel Y.A. Antipo a,, V.V. Silestro b a Department of Mathematics, Louisiana State Uniersity, Baton Rouge, LA 70803, USA b Department of Mathematics, Gubkin Russian State Uniersity of Oil and Gas, Moscow , Russia Aailable online 2 August 2007 Abstract In the framework of the Tulin model of supercaitating flow, the problem of reconstructing the free surface of a channel and the shapes of the caities behind two hydrofoils placed in an ideal fluid is soled in closed form. The conformal map that transforms a parametric plane with three cuts along the real axis into the triple-connected flow domain is found by quadratures. The use of the theory of Riemann surfaces (the Schottky doubles enables the non-linear model problem to be reduced to two separate Riemann Hilbert problems on a hyperelliptic surface of genus two. The solution to the first problem is a rational function with certain zeros and poles on a Riemann surface. The second problem is soled in terms of singular integrals with the Weierstrass kernel. The essential singularities of the solution at the infinite points of the surface due to a pole of the kernel are remoed by soling a real analogue of the Jacobi inersion problem on the surface. The unknown parameters of the conformal map are recoered from a system of certain algebraic and transcendental equations. c 2007 Elseier B.V. All rights resered. Keywords: Supercaitating flow; Conformal map; Riemann surfaces; Riemann Hilbert problem 1. Introduction When an obstacle is placed stationary in a moing fluid the flow breaks away from the barrier along separating streamlines, and a wake (a dead-zone forms behind the obstacle. In the case of high-speed flow the wake becomes a apor-filled caity. For water under atmospheric pressure for example, this occurs when the speed is 100 f/s or more (lower speeds under reduced pressure [1]. Caitating flow has been studied intensiely since marine engineers at the end of the 19th century became aware of the serious problem due to caitation: an increase in pressure causes the caities to collapse and release energy resulting in a force which may damage submarine propellers when they are operating at certain depths. Caitation has long been of interest not only in the field of shipbuilding and hydraulic machinery, but also in chemical processing, nuclear physics and medicine (potential bioeffects of ultrasound caused by acoustic caitation in blood essels. Corresponding author. Tel.: ; fax: address: antipo@math.lsu.edu (Y.A. Antipo. Modeling of caitation is based on the results by Brillouin (1911 (see, e.g. [1] who proed that the maximum elocity must be attained on the free surface and also that the boundary of a caity is conex. Good references to the theory of caitating flow around obstacles are [1 3]. Tulin [4] proposed a model of the caitating flow which admits the presence of the singularity of the solution at the point say, C, the two streamlines along the caity attempt to close it, namely, log(dw/dz K (w w 0 1/2, z C, K = const, (1.1 w 0 = w(c, w = w(z is the complex potential of the motion. This condition extends the class of solutions to the goerning boundary-alue problem which makes possible to reconstruct a flow that meets the condition dz = 0, (1.2 L and is therefore single-alued. Here L is the boundary of the caity combined with the boundary of the hydrofoil /$ - see front matter c 2007 Elseier B.V. All rights resered. doi: /j.physd

2 Y.A. Antipo, V.V. Silestro / Physica D 235 ( For model problems of caitating flow, numerical and analytical methods are employed. The former technique requires the solution of associated integral equations with the contour to be recoered. The analytical methods for the study of caitating flow in simply- and double-connected domains are well deeloped [1,3,5]. These methods are based on the use of a Schwarz Christoffel transformation to map a circle, a quadrant, or a plane with a cut (simply-connected flow, or an annulus, a rectangular, or a plane with two cuts on the real axis (doubleconnected flow into the flow domain. A closed-form solution to the model problem can then be found in terms of elementary and elliptic functions for a simply- and double-connected flow, respectiely. In the case of free boundary problems in a tripleconnected flow domain, the problem of fluid mechanics can be formulated [6] as the Hilbert boundary-alue problem with a piecewise constant coefficient on three cuts along the real axis. The actual problem soled in [6] concerned non-caitating flow of a fluid around a single foil in a half-plane with a free surface. A method of Riemann surfaces for an elasticity problem [7] on a system of cracks along an interface with mixed boundary conditions was extended for the problem of caitating flow in the whole plane around three foils [8]. In this paper we model the steady flow of an ideal fluid in a channel with a free surface when two plates are held stationary. The main mathematical tool used is the method of the Riemann Hilbert problem on a Riemann surface. This technique was applied and deeloped [9 11] for the solution of model problems of acoustic and electromagnetic scattering from a perforated sandwich panel and an anisotropic impedance half-plane. In Section 2 we formulate the problem using the non-linear model of caitating flow by Tulin. Section 3 maps the problem into two Hilbert problems on three cuts on the real axis. One of the cuts is the image of the boundary of the channel (a portion of it is a free boundary and therefore is unknown. The other two cuts are the images of the caities whose boundaries formed by the hydrofoils and the unknown conex contours. The deriatie of the conformal map z = f (ζ is represented as a quotient of two functions, dw/dz and dw/dζ, w is a complex potential of the flow. In Section 4 the function dw/dζ is found as a rational function with certain zeros and poles on a hyperelliptic surface of genus two. To define the function ω(ζ = log(v 1dw(z/dz (V is the speed at infinity, in Section 5 we reduce the Hilbert problem on the three cuts with a piecewise constant coefficient to the Riemann Hilbert problem on the Riemann surface introduced in Section 4. Its solution is found by quadratures in terms of singular integrals with the Weierstrass kernel. Initially, it has an inadmissible exponential growth at infinity. The conditions which make the solution bounded at infinity are written as the Jacobi inersion problem for hyperelliptic integrals. This non-linear problem requires finding two points on the surface and four integers. The solution of the Jacobi problem is found in closed form by reducing it to a system of two algebraic equations with the right-hand side expressed through the Riemann θ-function of the surface. Section 6 writes down additional conditions to be satisfied in order to fix 21 Fig. 1. Flow domain D. unknown real parameters. The system of equations for the unknowns consists of 8 linear and 13 non-linear relations which are algebraic and transcendental equations. Finally, equations of the free surface of the channel and the boundaries of the two caities are found by quadratures. 2. Formulation Let two hydrofoils B 1 D 1 and B 2 D 2 (Fig. 1 be placed in an incompressible graity-free fluid which is moing steadily and irrotationally in a channel. The bottom { < x <, y = 0} of the channel is solid, and its upper boundary is a free surface. Far away from the hydrofoils, the flow is uniform with elocity = (V, 0 across the channel of depth h. It is assumed that at the ends B j and D j ( j = 1, 2 the jets break away from the hydrofoils, and caities B j C j D j ( j = 1, 2 form behind the foils. The caities are conex and bounded but not closed. The unknown boundaries of the caities are streamlines. In the framework of the model considered, the elocity ector is constant and prescribed on the boundaries of the caities (the constants are not necessarily the same. The loops A j B j C j D j A j are smooth in a neighborhood of the points B j and D j. At the stagnation points A j (unknown a priori, the flow branches and the elocity ector anishes. Under these assumptions, the model problem of fluid mechanics is reduced to that of finding a complex potential of the motion w(z = φ + iψ in the 3-connected domain say, D, occupied by the fluid (the physical domain, together with boundary conditions of the form { W ± Im w(z = 0, z E± 1 E± 2 j = 1, 2, W j, z L j, { dw dz = V, z E + 1 E+ 2 j = 1, 2, V j, z B j C j D j, arg dw 0, z E dz = 1 E 2 α j, z A j B j j = 1, 2. (2.1 π α j, z A j D j, Here W ± 0 and W j ( j = 1, 2 are some constants, dw/dz = x i y, x and y are the elocity components, V j are positie constants defined by the Bernoulli equation 1 2 (V j 2 V 2 + p j p = 0, j = 1, 2, (2.2 ρ p and p j are the pressure at infinity and in the caities,

3 74 Y.A. Antipo, V.V. Silestro / Physica D 235 ( The boundary conditions (2.1 on the boundary of the flow transform into ones on the sides of the cuts. The free surface boundary condition becomes Fig. 2. Parametric ζ -plane with the cuts l 0, l 1, and l 2. respectiely, α j are the angles the foils make with the real axis, L 0 = E 1 E 2 E+ 2 E+ 1 is the boundary of the channel, E j = + i0, E + j = + ih, E + 1 E+ 2 is a free surface. The first equation in (2.1 reflects the fact that the stream function ψ is piecewise constant on the boundary of the flow that consists of L 0 and two lops L j = A j B j C j D j A j ( j = 1, 2. The complex potential w is an analytic function in the flow domain D. It becomes a single-alued function in the domain D cut along the two stream lines P j A j joining the critical points A j ( j = 1, 2 with the infinite point + iy, 0 y h. The problem of interest is non-linear first, for the shapes of the free surface of the channel, the boundaries of the caities, and the position of the stagnation points A j being unknown and second, for the boundary conditions (2.1 being non-linear. 3. Conformal mapping Let z = f (ζ be a conformal mapping of a parametric ζ -plane cut along three segments of the real axis l 1 = [1/k, 1], l 0 = [k 1, k 2 ], and l 2 = [1, 1/k] (0 < k < 1, 1 < k 1 < k 2 < 1 onto the 3-connected flow domain D. Such a map always exists [12]. It becomes unique if, in addition, it is required that the segments l j are transformed into the loops L j ( j = 0, 1, 2, and the images of the points a j, b j, c j, d j ( j = 1, 2, e 1, and e 2 are the points of the physical domain A j, B j, C j, D j, + iy, and + iy (0 y h, respectiely. None of the points of the real axis k 1, k 2, k, or the points a j, b j, c j, d j, and e j ( j = 1, 2 can be prescribed arbitrary. They hae to be found as a part of the solution to the problem. What is known howeer, is that these points lie on either side of the cuts: a j, b j, c j, d j l j and e 1, e 2 l 0, and they follow each other in clockwise direction (the same direction is chosen for the contours L j in the flow domain D. The image pattern of the flow domain by the inerse transform ζ = f 1 (z is shown in Fig. 2. The solid line corresponds to the solid boundary of the flow, and the broken line is the image of the unknown cures. It will be conenient to introduce a new function ω(ζ, ω(ζ = log dw(z, z = f (ζ. (3.1 V dz Then ω(ζ = log V V iθ, (3.2 V = dw/dz is the speed, and θ = arg dw/dz is the angle that the elocity ector makes with the x-axis. The function ω( f 1 (z is defined in the flow domain D cut along the streamlines joining the point + iy (0 y h with the points A 1 and A 2. Re ω(ζ = { 0, ζ e2 e 1 log(v j /V, ζ b j c j d j, j = 1, 2, (3.3 and the boundary condition on the solid part of the boundary can be written as 0, ζ e 1 e 2 Im ω(ζ = α j, ζ a j b j j = 1, 2. (3.4 π α j, ζ d j a j, Since dw/dz = 0 at z = A j ( j = 1, 2, the new function ω(ζ has the logarithmic singularities at the points ζ = a j. In a neighborhood of the points c j, the function ω(ζ has a singularity [4,5] ω(ζ M j (w w j 1/2, z C j, M j 0, (3.5 w j = w(c j, and a single branch of the square root is considered in the w-plane cut along the axis arg(w w j = π. A neighborhood w w j < ε of the point w = w j cut along the line arg(w w j = π corresponds to a neighborhood of the point z = C j which locally is a semi-disc. Therefore, ww j N j (z C j 2, z C j, N j 0. Next, z C j N j (ζ c j, z C j, N j 0. It follows from (3.5 now that the function ω(ζ has simple poles at the points ζ = c j ω(ζ M j (ζ c j 1, ζ c j. (3.6 Here M j are non-zero real constants and j = 1, 2. The recoering of the mapping function z = f (ζ is the key step in the definition of the free surfaces. Its deriatie can be found from the relation dw dζ = dw d f dz dζ (3.7 proided the deriaties dw/dζ and dw/dz are known. Because of the link dw dz = V e ω(ζ, (3.8 to find the deriatie dw/dz, one needs to sole the Hilbert problem (3.3 and (3.4 for the function ω(ζ. As for the function dw/dζ, since Im w( f (ζ is a piecewise constant function on the sides of the cuts l j ( j = 0, 1, 2, the imaginary part of the deriatie dw/dζ is equal to zero on l j Im dw dζ = 0, ζ l 0 l 1 l 2. (3.9 At the points a j and c j ( j = 1, 2, the function dw/dζ has simple zeros if these points do not coincide with the ending points of the cuts. Indeed, according to the model, Im w(z has constant alues W j on the cures P j A j, A j B j C j, A j D j C j, and C j Q j. These lines are the images of the lines p j a j, the corresponding parts of the cuts l j, and the lines c j q j in the parametric ζ -plane (Fig. 2. Since the cures P j A j and C j Q j are orthogonal to the cures A j B j C j and A j D j C j, the cures

4 Y.A. Antipo, V.V. Silestro / Physica D 235 ( p j a j and q j c j are orthogonal to the cuts l j ( j = 1, 2. Therefore, in a neighborhood of the points ζ = a j and c j, w( f (ζ = w(a j + K j (ζ a j 2 +, ζ a j, K j 0, w( f (ζ = w(c j + K j (ζ c j 2 +, ζ c j, K j 0, (3.10 and dw/dζ = 0 at ζ = a j and ζ = c j, j = 1, 2. When the points a j or c j coincide with one of the branch points, the function dw/dζ is bounded and not equal to zero. The point ζ = corresponds to a finite point of the flow domain. Therefore, in a neighborhood of the point ζ =, w = w 0 + cζ 1 +, c = const. Clearly, the function dw/dζ has a zero of the second order at the infinite point of the parametric ζ -plane. Analyze next the behaior of the function dw/dζ at the points e 1 and e 2. The function z = f (ζ maps the ζ -plane with three cuts onto the domain D. At z + iy (0 y h, the flow domain D looks like a strip of width h. Therefore, locally, f (ζ = (1 j1 h log(ζ e j + O (1, ζ e j, σ π j = 1, 2, (3.11 σ = 1 if e j does not coincide with the ends of the cut l 0, and σ = 2 otherwise. Then the function dw/dζ has simple poles at the points e 1 and e 2, dw dζ = dw d f dz dζ = (1 j1 hv 1 + O (1, σ π ζ e j ζ e j, j = 1, 2. (3.12 In a neighborhood of any ending point (call it ˆζ of the cuts, if it coincides with none of the critical points a j, c j, or e j, f (ζ K (ζ ˆζ 1/2 (K 0. Therefore, in this case, dw/dζ K (ζ ˆζ 1/2, ζ ˆζ, K is a non-zero constant. Thus, the functions dw/dζ and ω(ζ sole the two Hilbert boundary-alue problems (3.9, (3.3 and (3.4 in the classes of functions prescribed. The solution of these problems will be found in Sections 4 and 5 by a method that exploits the symmetry of the parametric ζ -plane, the fact that the imaginary part of the unknown function is equal to zero on the whole boundary (in the case of dw/dζ or on its portion (in the case of ω(ζ and the theory of Riemann surfaces of algebraic functions. After the solution of the problems the mapping function z = f (ζ is found by integration of the deriatie d f/dζ. 4. Function dw/dζ Let R be the hyperelliptic surface gien by the algebraic equation u 2 = p(ζ, p(ζ = (ζ 2 1(k 2 ζ 2 1(ζ k 1 (ζ k 2. (4.1 The surface is formed by gluing two copies C 1 and C 2 of the extended complex ζ -plane C { } cut along the segments l j, ( j = 0, 1, 2. The positie sides l + j of the cuts l j C 1 are glued with the negatie sides l j of the cuts l j C 2, and the sides l j C 1 are glued with l + j C 2. The function u(ζ defined by (4.1 is single-alued on R: { p 1/2 (ζ, ζ C u = 1 p 1/2 (4.2 (ζ, ζ C 2. Here p 1/2 (ζ is a branch fixed by the condition p 1/2 (ζ kζ 3, ζ. (4.3 The pair (ζ, p 1/2 (ζ denotes a point on C 1 with affix ζ, whilst the notation (ζ, p 1/2 (ζ will be used for its counterpart on the second sheet. Notice that the sides of the cuts l j ( j = 0, 1, 2 form the symmetry line for the surface R which splits the surface into two symmetrical hales, and the surface itself is therefore the Schottky double. Since Im dw/dζ = 0 on l j ( j = 0, 1, 2 the function dw/dζ can be analytically continued by symmetry on the whole Riemann surface. Then (dw/dζ + (dw/dζ = 0, (ζ, u (l 0 l 1 l 2 R. (4.4 By the generalized Liouille theorem, the solution of this simplest form of the Riemann Hilbert problem is a rational function say, R(ζ, u, on the Riemann surface R. If the points a j, c j do not coincide with the branch points of the surface R, this function has simple zeros at the points (a j, u(a j and (c j, u(c j which simultaneously belong to the first and second sheet of the surface. Notice that the function dw/dζ is bounded and not equal to zero at any point of the set {a 1, a 2, c 1, c 2 } if that point coincides with one of the branch points of the surface. The rational function R(ζ, u on the surface has zeros of the second order at the infinite points (, + and (, of the surface R. At the points e j (but not at the points ē j ( j = 1, 2 this function has simple poles and res R(ζ, u = (1 j1 hv ζ =e j σ π. (4.5 Notice that the number of poles and zeros (counting multiplicities of this function is the same. Consider the most interesting and realistic case when none of the points a j, c j, e j ( j = 1, 2 coincide with the branch points of the surface. Then σ = 1 and the function R(ζ, u must hae simple zeros at the points (a j, u(a j and (c j, u(c j. The most general form of the rational function R(ζ, u on the surface R is R(ζ, u = R 1 (ζ + R 2(ζ, (4.6 u the functions R j (ζ hae simple poles at the points e 1 and e 2. At infinity, R 1 (ζ = O(ζ 2 and R 2 (ζ = O(ζ. The functions with these properties hae the form R 1 (ζ = ( 1 ζ e 1 1 ζ e 2 N 0,

5 76 Y.A. Antipo, V.V. Silestro / Physica D 235 ( R 2 (ζ = in 1 + in 2 ζ + N 3 ζ e 1 + N 4 ζ e 2, (4.7 N j ( j = 0, 1,..., 4 are arbitrary constants. It is necessary to impose the following four conditions [ R 1 (ζ + R ] 2(ζ = (1 j1 hv u(ζ σ π. res ζ =e j res ζ =ē j [ R 1 (ζ + R 2(ζ u(ζ which define the constants N 0, N 3, and N 4 N 0 = hv 2π, ] = 0, (4.8 N 3 = hv 2π u(e 1, N 4 = hv 2π u(e 2. (4.9 Substitute now (4.7 and (4.9 into (4.6 and take (ζ, u as a point on the first sheet C 1. This defines the expression of the deriatie dw/dζ on the parametric ζ -plane dw dζ = i[n 1 + N 2 ζ + r(ζ ] p 1/2, ζ C, (4.10 (ζ r(ζ = ihv 2π ( p 1/2 (ζ + p 1/2 (e 1 ζ e 1 p1/2 (ζ + p 1/2 (e 2 ζ e 2. (4.11 Now, the function dw/dζ has to be equal to zero at the point a j and c j ( j = 1, 2. The first two conditions define uniquely the real constants N 1 and N 2 N 1 = a 2r(a 1 a 1 r(a 2 a 1 a 2, N 2 = r(a 2 r(a 1 a 1 a 2. (4.12 The other two conditions dw/dζ(c j = 0 ( j = 1, 2 gie two real non-linear equations for the parameters a j, c j, and e j ( j = 1, 2 N 1 + N 2 c j + r(c j = 0, j = 1, 2. ( Riemann Hilbert problem on a Riemann surface 5.1. Formulation The Hilbert problem (3.3 and (3.4 for the function ω(ζ may be formulated as a boundary-alue problem on the Riemann surface of the algebraic function (4.1. Introduce the function { ω(ζ, (ζ, u C1 Φ(ζ, u = (5.1 ω( ζ, (ζ, u C 2. This function satisfies the symmetry condition Φ(ζ, u = Φ(ζ, u, (5.2 (ζ, u = ( ζ, u( ζ is the point symmetrical to a point (ζ, u(ζ with respect to the lines l 0 l 1 l 2 along which the two hales of the surface are glued. Thus, if (ζ, u C 1, then (ζ, u C 2. On the symmetry lines, the boundary alues of the function Φ(ζ, u are Φ + (ξ, = ω(ξ, Φ (ξ, = ω(ξ, (ξ, l 1 l 0 l 2, = u(ξ. (5.3 It is clear from inspection of the boundary conditions (3.3 and (3.4 that the function Φ(ζ, u is continuous through the cure e 1 e 2 (a part of the contour l 0 and is discontinuous through the contours l 1, l 2 and the rest e 2 e 1 of the contour l 0. Thus, the function Φ(ζ, u soles the following Riemann Hilbert problem on the Riemann surface R. Find all the functions Φ(ζ, u analytic in R \ (l 0 l 1 l 2, Hölder-continuous up to the boundary l 0 l 1 l 2 with boundary alues satisfying the relations 0, (ξ, e 1 e 2 Φ + (ξ, Φ (ξ, = 2iα j, (ξ, a j b j 2i(π α j, (ξ, d j a j, j = 1, 2, Φ + (ξ, + Φ (ξ, = { 0, (ξ, e2 e 1 2 log(v j /V, (ξ, b j c j d j, j = 1, 2, (5.4 and the symmetry condition (5.2. The function Φ(ζ, u has logarithmic singularities at the points (a j, u(a j and poles of the first order at the points (c j, u(c j ( j = 1, 2. It is bounded at the points (b j, u(b j, (d j, u(d j, both infinite points (, ±, and Φ(e j, u(e j = 0, j = 1, Factorization Let L = b 1 c 1 d 1 e 2 e 1 b 2 c 2 d 2. To sole the problem (5.4, consider first the homogeneous Riemann Hilbert problem X + (ξ, = X (ξ,, (ξ, L. (5.5 By the Sokhotski Plemelj formulas the function { } 1 χ 0 (ζ, u = exp log(1dw 2πi L [ ] (ζ d1 (ζ d 2 (ζ e 1 1/4 = (ζ b 1 (ζ b 2 (ζ e 2 { } u dξ exp 4 L (ξ ζ (5.6 satisfies the boundary condition (5.5. Here dw is the Weierstrass kernel dw = 1 ( 1 + u dξ, 2 ξ ζ u = u(ζ, = u(ξ. (5.7 Also, it is directly erified that the function χ 0 (ζ, u meets the symmetry condition (5.2. Howeer, at the points (, ± it has essential singularities. This is because of the second order

6 Y.A. Antipo, V.V. Silestro / Physica D 235 ( It will be conenient to take the solution of the factorization problem (5.5 as Fig. 3. Canonical cross-sections a j and b j ( j = 1, 2. pole of the kernel dw at both infinite points of the surface. To remoe these essential singularities, analyze the function χ 0 (ζ, uχ 1 (ζ, uχ 1 (ζ, u, (5.8 χ 1 (ζ, u = exp { ( γ j +m j a j +n j b j dw }. (5.9 Here a j and b j ( j = 1, 2 form a system of canonical crosssections of the surface R. The contours a 1 and a 2 are smooth loops which lie on both sheets of the surface and coincide with the banks of the cuts l 0 and l 2, respectiely (Fig. 3. The positie direction is that which leaes the sheet C 1 on the left. The loop b 1 consists of the segments [k 1, 1] C 1 and [1, k 1 ] C 2. The contour b 2 consists of the segments [1, 1] C 1 and [1, 1] C 2. In Fig. 3, the part of the contours b j which lies on the first sheet is shown as the solid lines. The broken line corresponds to the part lying on C 2. The loops a j and b ν (ν j do not intersect. The cross-section a j intersects b j from left to the right, and there is only one point of intersection. The contours γ 1 and γ 2 are continuous cures with the same starting point (1/k, 0. The ending points q j = (ζ j, u j R (u j = u(ζ j are to be determined. These points may lie on either sheet of the surface. The contours γ j cannot intersect the canonical cross-sections. The parameters m j and n j are integers and will be found later on. Now, formulas (5.6 and (5.9 can be combined to gie an expression for the function (5.8 χ 0 (ζ, uχ 1 (ζ, uχ 1 (ζ, u [ (ζ d1 (ζ d 2 (ζ e 1 = (ζ b 1 (ζ b 2 (ζ e 2 ] 1/4 (ζ + 1/k 2 χ(ζ, u (ζ ζ1 (ζ ζ 1 (ζ ζ 2 (ζ ζ 2, (5.10 χ(ζ, u = exp { u 1 dξ 4 L ξ ζ u + 2m j a + j dξ p 1/2 (ξ(ξ ζ ( } 1 dξ Γ j 2 ξ ζ. (5.11 Here the contours Γ j are cures symmetrical with respect to the real line, lying on one of the sheets of the surface, starting at the points (ζ j, u j and ending at (ζ j, u j. The contours a + 1 and a + 2 are the upper banks of the cuts l 0 and l 2, respectiely. Notice that the integrals oer the loops b j anish in formula (5.11 because of the symmetry. X (ζ, u = (ζ d 1 1/4 (ζ d 2 1/4 (ζ b 1 3/4 (ζ b 2 3/4 (ζ e 1 1/4 (ζ e 2 3/4 χ(ζ, u. (ζ ζ1 (ζ ζ 1 (ζ ζ 2 (ζ ζ 2 (5.12 Let β be one of the points b j, d j, or e j. Analysis of the singular integrals in (5.11 in a neighborhood of that point indicates that X (ζ, u Λ β (ζ β 1/2, ζ β, Λ β = const 0, (5.13 while at the point ζ = β that lies on the opposite side of the corresponding cut, Λ b j (ζ b j, ζ b j Λ X (ζ, u d j, ζ d j Λ e1, ζ ē 1 Λ e2 (ζ e 2, ζ ē 2, Λ β = const 0, j = 1, 2. (5.14 At the points q j = (ζ j, u j, the factorization function X (ζ, u has simple poles and it is bounded at the points (ζ j, u j. At present, the function χ(ζ, u and therefore the function X (ζ, u hae an inappropriate exponential growth at infinity caused by the pole of the second order at infinity of the kernel dw. In order to get a solution of a finite order at infinity, expand the function χ(ζ, u as ζ, X (ζ, u = ζ exp { ( ν=1 u ζ ν ξ ν1 dξ Γ j 2 [ 1 ξ ν1 dξ 4 L + 2m j a + j ]} ξ ν1 dξ p 1/2 (ξ + O(ζ, ζ. (5.15 Because of the symmetry of the contours Γ j, the requirement of the canonical function to hae a finite order at infinity enables us to write the following system of non-linear equations Im { 1 ξ ν1 dξ 4 L + 2m j a + j [ ξ ν1 dξ p 1/2 (ξ ]} ξ ν1 dξ γ j = 0, ν = 1, 2. (5.16 This system presents a real analogue of the Jacobi inersion problem on the Riemann surface R. To reduce the system (5.16 to the classical inersion problem, notice that ( Re L 2 a + j ξ ν1 dξ ξ ν1 dξ p 1/2 (ξ = = 0, a j dω ν = A νj, Re A νj = 0, b j dω ν = B νj, Im B νj = 0, (5.17

7 78 Y.A. Antipo, V.V. Silestro / Physica D 235 ( dω ν are abelian differentials of the first kind of the surface R: dω ν = ξ ν1 dξ, ν = 1, 2, (5.18 and A νj and B νj are the A- and B-periods of the abelian integrals ω ν = ω ν (ζ, u = (ζ,u (1/k,0 ξ ν1 dξ, ν = 1, 2. (5.19 Now, the imaginary part of the system ( 1 ξ ν1 dξ ξ ν1 dξ + m j A νj + n j B νj = 0, 4 L γ j ν = 1, 2 (5.20 coincides with the system (5.16. On the other hand, it can be written as the classical Jacobi problem [ ] ων (q j + m j A νj + n j B νj = ην, ν = 1, 2, (5.21 η ν = 1 4 L ξ ν1 dξ. (5.22 The solution to the inersion problem (5.21 with respect to the two points q j = (ζ j, u j R and the four integers m j and n j ( j = 1, 2 (the function χ(ζ, u in (5.11 is independent of the constants n j makes the function χ(ζ, u bounded at infinity Jacobi inersion problem Following [9,11] normalize first the basis of abelian integrals ˆω = T ω, T = A 1, (5.23 A = {A νj } (ν, j = 1, 2. The A- and B-periods of the canonical basis ˆω are A νj = d ˆω ν = δ νj, a j B νj = d ˆω ν = T νm B mj. (5.24 b j m=1 Here δ νj is the Kronecker symbol. The matrix B = {B νj } (ν, j = 1, 2 is symmetric, and Im B is positie definite. For the canonical basis, the inersion problem (5.21 becomes ˆω ν (q j = g ν κ ν n j B νj m ν g ν κ ν, (modulo the periods, ν = 1, 2, (5.25 g ν = T νj η j + κ ν, ν = 1, 2, (5.26 and κ 1 and κ 2 are the Riemann constants for the chosen system of a and b cross-sections computed in [11] κ ν = 1 2 (ν + B ν1 + B ν2. (5.27 The affixes ζ 1 and ζ 2 of the points q 1 and q 2 on the surface R coincide with the two zeros of the Riemann θ-function F(q = θ( ˆω 1 (q g 1, ˆω 2 (q g 2 { = exp πi B µν t µ t ν t 1,t 2 = + 2πi µ=1 ν=1 } t ν [ ˆω ν (q g ν ]. (5.28 ν=1 Since the matrix Im B is positie definite the double series (5.28 conerges exponentially. The affixes of the points q 1 and q 2 sole the system of two equations [13] ζ ν 1 + ζ ν 2 = ε ν, ν = 1, 2, (5.29 ε ν = 2 m=1 T jm a + j τ m+ν1 dτ p 1/2 (τ ζ ν F (q res. q= µ C µ F(q µ=1 (5.30 Here µ = (, (1 µ1 C µ (µ = 1, 2 are the two infinite points of the Riemann surface. By soling (5.29, ζ j = ε 1 (1 j 2ε 2 ε1 2, j = 1, 2. ( Since Re B νj = 0, the integers m ν are gien by [13,11] m ν = Re (ĝ ν, and the integers n ν equations ĝ ν = g ν κ ν ˆω ν (q j, (5.32 sole the system of linear algebraic n j Im (B νj = Im (ĝ ν, ν = 1, 2. (5.33 This completes the solution to the Jacobi problem (5.25 and therefore (5.21. We emphasize that the function χ(ζ, u is independent of the integers n ν Function dw/dz The definition of the function dw/dz requires soling the Riemann Hilbert problem (5.4. Using the factorization (5.5 write the boundary condition (5.4 as Φ + (ξ, X + (ξ, Ψ + (ξ, = Φ (ξ, X (ξ, Ψ (ξ,, (ξ, L, (5.34

8 Ψ ± (ξ, are the boundary alues of the function Ψ(ζ, u = Ψ 0 (ζ + uψ 1 (ζ, ( Ψ ν (ζ = 1 α j 2 π + log(v j/v πi d j a j b j + b j c j d j d j a j Y.A. Antipo, V.V. Silestro / Physica D 235 ( dξ X + (ξ, (ξ ζ ν, ν = 0, 1. (5.35 By the generalized Liouille theorem, the general solution of the problem (5.34 has the form Φ(ζ, u = X (ζ, u{ψ 0 (ζ + Ω 0 (ζ + u[ψ 1 (ζ + Ω 1 (ζ ]}, (5.36 Ω j (ζ ( j = 0, 1 are certain rational functions on the complex ζ -plane. The function X (ξ, u has simple zeros at the points b j ( j = 1, 2, ē 2 and the solution itself has simple poles at the points c 1 and c 2. The functions X (ζ, u and u(ζ grow at infinity as ζ and ζ 3, respectiely. Therefore, the functions Ω 0 (ζ and Ω 1 (ζ are gien by ( im Ω 0 (ζ = M j ζ j j+2 u(c j + im j+4u(b j ζ c j=0 j ζ b j im 7u(e 2, ζ e 2 Ω 1 (ζ = ( im j+2 ζ c j + im j+4 ζ b j + im 7 ζ e 2, (5.37 M j ( j = 0, 1,..., 7 are arbitrary real constants. The factorization function X (ζ, u has simple poles at the points q j = (ζ j, u j. To make these points remoable singularities for the function Φ(ζ, u, the following two complex conditions need to be satisfied Ψ 0 (ζ j + Ω 0 (ζ j + u j [Ψ 1 (ζ j + Ω 1 (ζ j ] = 0, j = 1, 2. (5.38 At infinity, after the solution of the Jacobi inersion problem, since the factorization function X (ζ, u has a simple pole, the function Φ(ζ, u has a pole of the third order. To write down the conditions which eliminate these singularities, expand the functions u(ζ, Ω 0 (ζ, Ω 1 (ζ, and Ψ 1 (ζ at infinity u(ζ = kζ 3 + K 1 ζ 2 + K 2 ζ +, Ω 0 (ζ = M 2 ζ 2 + M 1 ζ + M 0 +, Ω 1 (ζ = Ω 10 ζ Ψ 1 (ζ = Ψ 10 ζ + Ω 11 ζ 2 + Ω 12 ζ 3 +, + Ψ 11 ζ 2 + Ψ 12 ζ 3 +, (5.39 ( K 1 = k 2 (k 1 + k 2, K 2 = k k k k 2 k 1k 2, Ω 1ν = i (c ν j M j+2 + b ν j M j+4 + im 7 e2 ν, ν = 0, 1, 2, Ψ 1ν = 1 2 ( α j π + log(v j /V πi d j a j b j + b j c j d j d j a j ξ ν dξ X + (ξ,, ν = 0, 1, 2. (5.40 The solution to the Riemann Hilbert problem (5.4 is finite at infinity proided the following three complex conditions are met M 2 + k(ψ 10 + Ω 10 = 0, M 1 + K 1 (Ψ 10 + Ω 10 + k(ψ 11 + Ω 11 = 0, M 0 + K 2 (Ψ 10 + Ω 10 + K 1 (Ψ 11 + Ω 11 + k(ψ 12 + Ω 12 = 0. (5.41 At the points ζ = e j ( j = 1, 2 the function Φ(ζ, u anishes as required. It follows from this fact that the speed of the fluid far away from the foils approaches the gien alue V. The function Φ(ζ, u is bounded at the points b j and d j, and it has the logarithmic singularity at the points ζ = a j. 6. Definition of the elocity and the free boundaries The solution of the Riemann Hilbert problem (5.4 defines the elocity of the fluid = ( x, y by x i y = dw dz = V exp{φ(ζ, u}, (ζ, u C 1. (6.1 The solution found possesses 21 unknown real constants. Among them, there are ten images a j, b j, c j, d j, and e j ( j = 1, 2 of the points A j, B j, C j, D j, and E ± j, respectiely. The three parameters k, k 1, and k 2 of the conformal map z = f (ζ are also to be determined. The last group of the unknowns are the eight real constants M j ( j = 0, 1,..., 7 in the representation (5.36 of the solution to the Riemann Hilbert problem (5.4. For their definition, we hae already written two real conditions (4.13, two complex equations (5.38 and three complex conditions (5.41. Thus, in total, there are 12 real equations. We should also add the following fie real conditions which specify the geometry of the problem. First, d f dζ dζ = λ j, j = 1, 2, (6.2 d j a j b j which define the prescribed lengths λ 1 and λ 2 of the foils. Here the deriatie d f/dζ is expressed through the functions dw/dζ and dw/dz as d f dζ = dw/dζ dw/dz. (6.3 Next, we specify the two distances µ 1 and µ 2 from the ending points D j to the bottom of the channel by Im d j e 0 d f dζ dζ = µ j, j = 1, 2, (6.4

9 80 Y.A. Antipo, V.V. Silestro / Physica D 235 ( e 0 is an arbitrary fixed point on the portion e 1 e 2 of the cut l 0. If it turns out that k 2 e 1 e 2, then k 2 can be chosen as the point e 0. The fifth condition describes the distance µ between the two ending points D 1 and D 2 of the hydrofoils Re d2 d 1 d f dζ = µ. (6.5 dζ We emphasize that, in general, the conformal map z = f (ζ is a multi-alued function. It becomes a one-to-one map if d f dζ = 0, j = 1, 2. (6.6 l j dζ These two complex equations add four real conditions for the unknown parameters. Thus, in total we hae 21 real conditions for the same number of real constants. Note that the system of these equations can be split into a set of eight linear equations for the constants M j ( j = 0, 1,..., 7 and a system of 13 non-linear (algebraic and transcendental equations for the parameters a j, b j, c j, d j, e j ( j = 1, 2, k, k 1, and k 2. These non-linear equations can be soled numerically. Finally, we determine the equations for the unknown a priori boundaries of the caities behind the hydrofoils and the free surface of the channel. Integrating the function (6.3 gies the boundaries of the caities τ dw( f (ζ /dζ z(τ = dζ + B j, τ b j c j, z B j C j, dw/dz and b j j = 1, 2, (6.7 z(τ = τ d j dw( f (ζ /dζ dζ + D j, τ c j d j, z C j D j, dw/dz j = 1, 2, (6.8 whilst letting τ e 2 e 1 by integration we recoer the free surface of the channel τ dw( f (ζ /dζ z(τ = dζ + B, τ e 2 e 1, z E + b dw/dz 2 E+ 1. (6.9 The integrals (6.7 and (6.8 are taken either oer the banks of the corresponding cuts, or oer smooth cures on the cut ζ -plane C \ (l 0 l 1 l 2. The contour of integration bτ in (6.9 is a smooth cure on C \ (l 0 l 1 l 2 which do not intersect the images of the cuts P j A j (the complex potential w is a single-alued analytic function in the flow domain D cut along the streamlines P j A j joining the stagnation points A j and the infinite point + iy, 0 y h, and b is the image of a point B D. In the case shown in Fig. 1 B 2 can be taken as the the point B. 7. Conclusions In the present paper, in the framework of the Tulin s model of caitating flow the steady streaming flow in a channel with a free surface around two hydrofoils (plates with caities of unknown shape behind them has been analyzed. By a conformal map of a parametric complex plane cut along three cuts on the real axis onto the triple-connected flow domain, the model problem has been reduced to two Riemann Hilbert problems on a Schottky double which is a hyperelliptic surface of genus two. The solution of these problems was found to be necessary in order to recoer the actual form of the conformal map. The solution to the first problem turned out to be a rational function on the Riemann surface and was constructed explicitly without quadratures. The second problem required the use of singular integrals with the Weierstrass kernel. Due to the second order pole of the kernel, in general, the solution had an essential singularity at the infinite points of the surface. The procedure of elimination of the essential singularities of the solution led to the classical Jacobi inersion problem of genus two which was soled explicitly. Final equations for the elocity, the free surface of the channel, caities and the conformal mapping itself possessed 21 unknown real parameters. To recoer them the same number of real conditions (13 non-linear and 8 linear equations hae been found. The non-linear system of algebraic and transcendental equations can be separated from the linear system. It is worth pointing out some limitations and possible extensions to the work done in the paper. First, for the method presented it is essential to be able to map a parametric plane cut along n segments on the same straight line into the n-connected flow domain. If 1 n 3, then such a map always exists [12]. If n 4, in general, it does not. For this more general case, it is hoped to deelop another technique based on the theory of the Riemann Hilbert problem for symmetric automorphic functions. Second, the paper concerns the supercaitating steady flow around hydrofoils. In the transient case for not too rapidly arying flows, it is possible (at least approximately to assume that the free boundaries of the caities are streamlines [14] and modify the procedure of the paper accordingly. Acknowledgments This work was partly funded by NSF through grant DMS , Louisiana Board of Regents through grant LEQSF( ENH-TR-09 and Russian Foundation for Basic Research through grant References [1] G. Birkhoff, E.H. Zarantello, Jets, Wakes, and Caities, Academic Press, New York, [2] P. Eisenberg, M.P. Tulin, Caitation, in: V.L. Streeter (Ed., Handbook of Fluid Dynamics, McGraw-Hill, New York, 1961, pp [3] M.I. Gureich, The Theory of Jets in an Ideal Fluid, Pergamon Press, Oxford, [4] M.P. Tulin, Supercaitating flows small perturbation theory, J. Ship Res. 7 (3 ( [5] A.G. Terent e, Non-linear theory of caitational flow around obstacles, Fluid Dynam. 11 ( [6] G.P. Cherepano, Flow of an ideal fluid with free surface in doubly connected and triply connected regions, J. Appl. Math. Mech. 27 ( [7] E.I. Zeroich, A mixed problem of elasticity theory for the plane with cuts that lie on the real axis, in: Proc. Symp. Continuum Mechanics

10 Y.A. Antipo, V.V. Silestro / Physica D 235 ( and Related Problems of Analysis, ol. 1, Mecniereba, Tbilisi, 1973, pp (in Russian. [8] V.V. Silestro, Analytical solution of the problem of the caitating flow around a system of plates by the method of Riemann surfaces, in: Dynamics of Continua with Free Boundaries, Cheboksary, 1996, pp (in Russian. [9] Y.A. Antipo, V.V. Silestro, Factorization on a Riemann surface in scattering theory, Quart. J. Mech. Appl. Math. 55 ( [10] Y.A. Antipo, V.V. Silestro, Vector functional-difference equation in electromagnetic scattering, IMA J. Appl. Math. 69 ( [11] Y.A. Antipo, V.V. Silestro, Electromagnetic scattering from an anisotropic impedance half-plane at oblique incidence: The exact solution, Quart. J. Mech. Appl. Math. 59 ( [12] R. Courant, Dirichlet s Principle, Conformal Mapping, and Minimal Surfaces, Interscience Publishers, New York, [13] E.I. Zeroich, Boundary alue problems in the theory of analytic functions in Hölder classes on Riemann surfaces, Russian Math. Sureys 26 ( [14] D. Gilbarg, Unsteady flows with free boundaries, Z. Angew. Math. Phys. 3 (