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A 2007 Honours thesis in applied mathematics, supervised by John Sader. It explains the Wiener-Hopf technique (product and sum decompositions), then follows Atkinson and de Lara's analysis of an oscillating clamped plate in viscous fluid. It extends this to flexural and torsional blade vibrations, derives hydrodynamic functions, and compares them with exact results. Appendices cover large-argument expansions and contour integrals. It sits in Phil's Stakgold Chapter 7 support folder as a reference.

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The University of Melbourne Department of Mathematics and Statistics The Wiener-Hopf Method and Its Applications in Fluids Juwen Ho Supervisor : Associate Professor John Sader Applied Mathematics Honours Thesis November 28, 2007 Acknowledgements I would like to thank my supervisor John Sader for giving me th e opportunity to explore the very interesting world of ﬂuids. His encourag ement, passion and patience has proven to be invaluable to me throughout the year. A special thanks to Assoc. Prof. Barry Hughes and Paul Pearce f or helping me with the numerous questions I have in complex analysis. Thank you to the other Honours students in G69 for making this year a more enjoyable one and a year to treasure. i Contents Acknowledgements i 1 Introduction 1 2 The Wiener-Hopf Technique 5 2.1 The Wiener-Hopf Method . . . . . . . . . . . . . . . . . . . . 5 2.2 Product Decomposition of A(ζ) . . . . . . . . . . . . . . . . . 8 2.3 Sum Decomposition of B(ζ)A+(ζ) . . . . . . . . . . . . . . . 11 3 Oscillating Plate with Clamped End in Viscous Fluid 14 3.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 3.2 Scaling of Governing Equations . . . . . . . . . . . . . . . . . 17 3.3 Solving for the ﬂuid reaction . . . . . . . . . . . . . . . . . . 18 3.3.1 Inviscid Solution . . . . . . . . . . . . . . . . . . . . . 18 3.3.2 Correction to the Inviscid Solution . . . . . . . . . . . 23 3.4 Equation of Motion of the Plate . . . . . . . . . . . . . . . . . 34 4 Flexural and Torsional Vibrations 44 4.1 Normal Oscillations in a Viscous Fluid . . . . . . . . . . . . . 45 4.1.1 Inviscid Solution . . . . . . . . . . . . . . . . . . . . . 46 4.1.2 Correction to the Inviscid Solution . . . . . . . . . . . 50 ii 4.1.3 Hydrodynamic Function Γn(β) . . . . . . . . . . . . . 52 4.2 Torsional Oscillations in a Viscous Fluid . . . . . . . . . . . . 54 4.2.1 Inviscid Solution . . . . . . . . . . . . . . . . . . . . . 54 4.2.2 Correction to the Inviscid Solution . . . . . . . . . . . 57 4.2.3 Hydrodynamic Function Γt(β) . . . . . . . . . . . . . 59 5 Results and Discussion 63 5.1 Comparison with Exact Computed results . . . . . . . . . . . 63 6 Conclusions 71 6.1 Future research . . . . . . . . . . . . . . . . . . . . . . . . . . 72 A Product Decomposition of H(ζ) 73 A.1 Large ζexpansion of H+(ζ) . . . . . . . . . . . . . . . . . . . 74 A.2 Large ζexpansion of H−(ζ) . . . . . . . . . . . . . . . . . . . 76 B Sum decomposition of S(ζ) 78 B.1 Large ζexpansion of S−(ζ) . . . . . . . . . . . . . . . . . . . 78 B.2 Large ζexpansion of S+(ζ) . . . . . . . . . . . . . . . . . . . 80 C Contour integrals 81 C.1 Integral in Equation (3.18) . . . . . . . . . . . . . . . . . . . 82 C.2 Integral in Equation (3.37) . . . . . . . . . . . . . . . . . . . 83 C.3 Integral in Equation (4.14) . . . . . . . . . . . . . . . . . . . 84 C.4 Integral in Equation (4.19) and (4.32) . . . . . . . . . . . . . 84 C.5 Integral in Equation (4.25) . . . . . . . . . . . . . . . . . . . 85 C.6 Integral in Equation (4.29) . . . . . . . . . . . . . . . . . . . 86 Bibliography 87 iii iv Chapter 1 Introduction The geometry of an object submerged in ﬂuid and its interacti ons with the ﬂuid has long been an open area of research. Since the time of S tokes, who studied the ﬂow of a viscous ﬂuid around a sphere, it was perce ived that the structure of the object in the ﬂuid is important. Demonstrat ed by his study on spheres in ﬂuid, Stokes came up with the well known formula of the drag force on a sphere, which includes variables that are linked t o the geometry of the sphere. Flat plates or blades in ﬂuids have been studied since the tim e of Bla- sius and used considerably throughout the scientiﬁc commun ity particularly in engineering. These plates have been implemented to desig n better and smaller diagnostic tools and to be able to measure some physi cal quantity such as density or viscosity of a ﬂuid. Such devices can be mad e in the order of microns, which is sometimes lumped into the categor y of a micro- electrical mechanical system (MEMS) or even to the scale of n anometres and categorized as a nano-electrical mechanical system (NEM S). Arguably, the atomic force microscope (AFM), which is a device which co nsists of a clamped cantilever beam with a weighted tip at the free end, has been 1 seen as one of the greatest uses of plates/blades. Not only is it be able to measure physical quantities, but it can also provide informa tion about the surface of a material down to nanometre resolution, for test ing of nanoma- terials and the list goes on. Typically, the AFM undergoes os cillations in ﬂuid and therefore understanding how the ﬂuid and the cantil ever interact is important to ensure accuracy and precision when measuring . Quite a number of people have modelled the AFM cantilever imm ersed in an inviscid ﬂuid, that is, a ﬂuid having no viscosity. But, ev en though most regions of the problem can be modeled by an inviscid ﬂuid, vis cous eﬀects are signiﬁcant and penetrate a small distance away from its surf ace. As length scales decrease, viscous eﬀects cannot be neglected since t he thickness of the viscous penetration depth becomes comparable with the leng th scale. This is especially true when modelling small objects such as the A FM cantilever and thus we must consider viscous eﬀects on the cantilever. It is known from the study of semi-inﬁnite plates in a two dimen sional viscous ﬂuid ﬂow that the edge of a plate is of particular impo rtance. The edge of a blade gives a signiﬁcant contribution due to viscou s eﬀects that may be diﬀerent from regions far away from the edge. Thus, a bl ade with 2 edges would be make it all the more important to be able to acc ount for the viscous eﬀects and subsequently the ﬂuid’s eﬀect on the b lade. Since we are considering blades with very large aspect ratio where th e length greatly exceeds the width, we need to account for the edges at the side s of the blade only, since it can be approximated by a blade with inﬁnite length and ﬁnite width. Tuck [17] has done a numerical analysis on a blade with neglig ible thick- ness oscillating parallel to its surface orientation by sol ving integral equa- tions. He found that the coeﬃcient due to the damping force as the blade 2 oscillates behaves like C/√β∼3.2/√β, whereβis the Reynolds number associated with this problem. The Reynolds number used in Tu ck’s analysis was β=ρωL2 4µ, whereρandµare the density and the viscosity of the ﬂuid and Lis the length of the blade. He proposed that to obtain a ﬁrst non-zero damping coeﬃcient to correct for the potential (or inviscid) ﬂow, on e would have to carry out an asymptotic expansion for large β. The method he proposed to use was the Wiener-Hopf technique to be able to obtain the da mping coeﬃcient. Recently, Atkinson and de Lara [3] have used the Wiener-Hopf m ethod to solve a ﬂuid ﬂow problem on an oscillating plate that is sim iliar to the problem considered by Tuck. Their problem consists of an inﬁ nitely wide plate but ﬁnite length, clamped at one end and the free end und ergoes oscillations. They have also solved the problem of a ﬂat disk vibrating in a viscous ﬂuid [2]. Armed with this knowledge we can apply thei r methodology to carry out the analysis suggested by Tuck. In Chapter 2 of this thesis, the Wiener-Hopf method is discuss ed and in particular how to decompose functions since it is the crux of the method. Chapter 3 focuses on expounding Atkinson and de Lara’s analy sis, which will serve as the basis for the other problems considered her e. In Chapter 4, the Wiener-Hopf method is applied again to the problem cons idered by Tuck and a problem where the blade undergoes torsional oscil lations about the middle of the blade. The forces and moments experienced b y the blade are calculated and the hydrodynamic function is obtained. T his function will give the damping coeﬃcient to be able to compare with res ults from Tuck. 3 In Chapter 5, the forces and moments obtained in the previous chap- ter are compared with their corresponding exact solutions o btained by Van Eysden and Sader [6]. A comparison is also made with the ﬁndin gs of Sader [15] for the problem of Tuck and the ﬁndings of Green and Sader [7] for the problem of torsional oscillations of a blade. 4 Chapter 2 The Wiener-Hopf Technique The Wiener-Hopf technique was originally formulated to solv e certain classes of integral equations. Since then, this technique has been w idely used to solve certain types of partial diﬀerential equations (PDEs ) particularly par- ticularly mixed boundary value problems. This has been prov en very useful use when solving radiation or diﬀraction problems [13] wher e there are ob- stacles that generate reﬂected and transmitted waves [9]. I t is also quite closely related to the Riemann-Hilbert problem. In this chap ter, we shall review the main ideas on how to solve problems using this tech nique. 2.1 The Wiener-Hopf Method Suppose we have a boundary value problem with a PDE and the bou ndary conditions (BCs) are speciﬁed only on the semi-inﬁnite domai n. The prob- lem can be either the half plane or the whole of R2and the boundary data is typically given on the real line. An example would be the Lapl ace equation 5 on the upper half plane. ∇2φ= 0 BCs:  φ= 0, y = 0 andx>0 ∂φ ∂y= 0, y= 0 andx<0 We typically apply Fourier transforms to the PDE and its boun dary con- ditions and we assume that we can solve the Fourier transform ed equation. The Fourier transform of the spatial variable, x, used throughout this thesis is deﬁned as F(ζ) =/integraldisplay∞ −∞f(x)eiζxdx . Because the boundary conditions are speciﬁed over the semi-i nﬁnite do- main, we would normally ﬁnd that there will be two unknown fun ctions corresponding to the unspeciﬁed part of the domain in the BCs . One would be analytic in the upper half plane ( Im(ζ)>τ+) and we shall call this a plus function. The other would be analytic in the lower half plane (Im(ζ)<τ−) and we shall call this a minus function. After applying the Fo urier trans- formed boundary conditions to the PDE, we would typically ge t an equation of the form φ+(ζ) =ψ−(ζ) A(ζ)+B(ζ), (2.1) whereφ+andψ−are the unknown functions. This equation has a common region of analyticity if τ+< Im (ζ)< τ− (see Fig. 2.1). In the most strict case, this analytic strip w ould only be a line and we can view this as the limit as τ−→0+andτ+→0−, where τ−approaches 0 from above and τ+approaches 0 from below. We will now 6 τ+τ− Re(ζ)Im(ζ) Figure 2.1: The shaded area represents the strip where the (2 .1) is deﬁned want to perform a product decomposition of the form A(ζ) =A−(ζ)A+(ζ), (2.2) whereA−(ζ) is analytic in the lower half plane and A+(ζ) is analytic in the upper half plane. Substituting (2.2) and multiplying both s ides of equation (2.1) byA+(ζ), we obtain A+(ζ)φ+(ζ) =ψ−(ζ) A−(ζ)+B(ζ)A+(ζ). (2.3) We perform another decomposition on the function B(ζ)A+(ζ) into a sum of two functions, i.e. B(ζ)A+(ζ) =C+(ζ) +C−(ζ). (2.4) As usual,C−(ζ) is analytic in the lower half plane and C+(ζ) is analytic in the upper half plane. Substituting (2.4) into (2.3) and re arranging the 7 terms so that the plus functions are on one side of the equatio n and the minus functions are on the other side of the equation to give A+(ζ)φ+(ζ)−C+(ζ) =ψ−(ζ) A−(ζ)+C−(ζ)≡G(ζ), (2.5) whereG(ζ) is deﬁned as as above. The equation that is deﬁned by G(ζ) is analytic in the strip τ+<Im(ζ)<τ−. But the ﬁrst part of (2.5) is analytic in the upper half plane and the second part of the equation is an alytic in the lower half plane. Therefore, by analytic continuation, we can conclude that the function G(ζ) is analytic over the whole complex plane. Suppose that we can show that |A+(ζ)φ+(ζ)−C+(ζ)|<|ζ|pas|ζ| → ∞, Im (ζ)>τ−, /vextendsingle/vextendsingle/vextendsingle/vextendsingleψ−(ζ) A−(ζ)+C−(ζ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle<|ζ|qas|ζ| → ∞, Im (ζ)<τ+. Then by the generalised Liouville’s theorem, the function G(ζ) is a poly- nomialP(ζ) of degree less than equal to the intergral part of min(p,q). Therefore we can solve for the two functions φ+(ζ) andψ−(ζ) separately, i.e. φ+(ζ) =P(ζ) +C+(ζ) A+(ζ), ψ−(ζ) =A−(ζ)(P(ζ)−C−(ζ)). Upon solving the two unknown functions separately, we can no w solve the original boundary value problem by taking inverse Fouri er transforms. 2.2 Product Decomposition of A(ζ) In performing the Wiener-Hopf method, we need to decompose a f unction into a product of two functions, one analytic in the lower hal f plane and the 8 ×× ×× Re(ζ)Im(ζ) τ+τ− C1 C2 C3C4 Figure 2.2: Contour integration for product decomposition . The crosses are the zeros of H(ζ) other in the upper half plane. In some simple cases, this can b e done by inspection, but most of the time doing this decomposition is not immediately obvious. Here we shall see a general systematic way in perfor ming this decomposition by using contour integration. We will consider the function A(ζ) to be analytic in the strip τ+< Im(ζ)<τ−. We may also consider that the function may have some zeros inside the strip of analyticity as well. We shall choose a rec tangular contour that is below all of the zeros inside the strip. This strip wil l also contain the inversion contour of the inverse Fourier transform since th e contour can be anywhere within the analytic strip (see Fig. 2.2). The function A(ζ) is analytic inside this rectangle comprises of the con- tours Γ =C1+C2+C3+C4and there are no zeros inside this rectangle. We also require that the function A(ζ)→1 as|ζ| → ∞ to ensure that the contributions from C2andC4vanish as |ζ| → ∞ . In order to express A(ζ) 9 as a product A+(ζ)A−(ζ), we take the logarithm of equation (2.2) to get ln(A(ζ)) = ln(A+(ζ)A−(ζ)) = ln(A+(ζ)) + ln(A−(ζ)). We also note that ln( A(ζ)) is also analytic in the rectangle because it has no branch points since there are no zeros there, so we can appl y the Cauchy integral formula. ln(A(ζ)) =1 2πi/integraldisplay Γln(A(z)) z−ζdz =1 2πi/bracketleftbigg/integraldisplay C1ln(A(z)) z−ζdz+/integraldisplay C3ln(A(z)) z−ζdz+/integraldisplay C3+C4ln(A(z)) z−ζdz/bracketrightbigg . So far we have no guarantee that the contributions from the co ntours of C2andC4cancel out due to the multivaluedness of the logarithm as we g o around the contour. We may have in the limit as |ζ| → ∞ that ln(A(ζ))→ln(e2πin) = 2πin, n ∈Z. But for the purposes of this thesis, we will only consider the case whenA(ζ) is even and the inversion contour for the inverse Fourier tra nsform on the real axis. It can be shown that given these conditions, the co ntributions from the contour C2andC4will cancel each other. For a more general method to deal with cases where the function is not even and no t necessarily analytic on the real axis, we would refer to some texts regard ing these types of decomposition [11]. Therefore we have ln(A+(ζ)) + ln(A−(ζ)) =1 2πi/integraldisplay C1ln(A(z)) z−ζdz+1 2πi/integraldisplay C3ln(A(z)) z−ζdz . In the ﬁrst integral, we may let ζbe anywhere below C1whereC1is ar- 10 bitrarily close to Im(ζ) =τ+from above. We can therefore conclude that that the integral 1 2πi/integraldisplay C1ln(A(z)) z−ζdz, Im (ζ)<τ+ is analytic in the lower half plane and can be identiﬁed as a mi nus function. Similiarly for the second integral, we may let ζbe anywhere above C3where C3is arbitrarily close to Im(ζ) =τ+from above. We can now conclude that the integral 1 2πi/integraldisplay C3ln(A(z)) z−ζdz, Im (ζ)>τ+ is analytic in the upper half plane and can be identiﬁed as a pl us function. Since we have analyticity on the real axis, we therefore have thatC1andC3 on the real axis. So we now obtain the following expressions ln(A+(ζ)) =1 2πi/integraldisplay∞ −∞ln(A(z)) z−ζdz, Im (ζ)>0 (2.6) ln(A−(ζ)) =−1 2πi/integraldisplay−∞ ∞ln(A(z)) z−ζdz =−1 2πi/integraldisplay∞ −∞ln(A(z)) z−ζdz, Im (ζ)<0 (2.7) Exponentiating equation (2.6) and (2.7) we obtain A+(ζ) = exp/parenleftbigg1 2πi/integraldisplay∞ −∞ln(A(z)) z−ζdz/parenrightbigg , A−(ζ) = exp/parenleftbigg −1 2πi/integraldisplay∞ −∞ln(A(z)) z−ζdz/parenrightbigg .(2.8) 2.3 Sum Decomposition of B(ζ)A+(ζ) Similiarly for the sum decompostion, for some cases the deco mposition can be done by inspection but most of the time it requires a more ge neral ap- 11 proach by using contour integration again. Re(ζ)Im(ζ) τ− τ+C1 C2 C3C4 Figure 2.3: Contour integration for sum decomposition We again have that B(ζ)A+(ζ) is analytic in the strip τ+<Im(ζ)<τ− and so we can apply the Cauchy integral formula to obtain B(ζ)A+(ζ) =1 2πi/integraldisplay ΓB(z)A+(z) z−ζdz , where Γ =C1+C2+C3+C4(see Fig. 2.3). We require that |B(ζ)A+(ζ)| →0 so that the contributions of the contours C2andC4will vanish as |ζ| → ∞ . Therefore, we have that B(ζ)A+(ζ) =1 2πi/integraldisplay ΓB(z)A+(z) z−ζdz =1 2πi/integraldisplay C1B(z)A+(z) z−ζdz+1 2πi/integraldisplay C3B(z)A+(z) z−ζdz . (2.9) The ﬁrst integral in (2.9) is analytic for Im(ζ)<τ−and therefore a minus function while the second integral is analytic for Im(ζ)>τ+and therefore 12 a plus function. Putting together this fact with (2.4), we ge t C+(ζ) =1 2πi/integraldisplay∞+iτ+ −∞+iτ+B(z)A+(z) z−ζdz C−(ζ) =1 2πi/integraldisplay−∞+iτ− ∞+iτ−B(z)A+(z) z−ζdz , =−1 2πi/integraldisplay∞+iτ− −∞+iτ−B(z)A+(z) z−ζdz . As before in the previous section, we will consider the case w henτ+→0− andτ−→0+. Hence, we now obtain the expression for the sum decompo- sitions as C+(ζ) =1 2πi/integraldisplay∞ −∞B(z)A+(z) z−ζdz, Im (ζ)>0, C−(ζ) =−1 2πi/integraldisplay∞ −∞B(z)A+(z) z−ζdz, Im (ζ)<0.(2.10) 13 Chapter 3 Oscillating Plate with Clamped End in Viscous Fluid We will now review the method proposed by Atkinson and de Lara [3] to solve the problem of obtaining the frequency response on a re ctangular plate oscilating in a viscous ﬂuid. The model used here is a plate th at has a length ofL, a width of Wand a thickness of Twhere it is assume that T≪L,W andL≈WorL < W . The plate is clamped along one edge while the other edges are free to vibrate at a frequency of ω(see Fig. 3.1). The motivation for using such a model was to design sensors to be able to measure ﬂuid viscosity and density. This would be implement ed in a MEMS where typically the length and the width are in the order of mi limetres and the thickness is in the order of microns. Since the model considered here is wider than it is long, the a uthors assumed that all longitudinal cross sections behave the sam e with negligible eﬀects on the 2 parallel edges with the third edge having the d ominant 14 LWT Figure 3.1: Plate clamped at one end with length L, widthWand thickness TandT≪W,LandL<W eﬀect on the plate. In other words, they have considered an in ﬁnitely wide plate clamped on one edge and the other free to vibrate and thu s giving the main contribution to the viscous ﬂuid ﬂow. Therefore, the 3-d imensional model can now be reduced to a problem involving planar ﬂow alo ng one longitudinal cross section. The approach that was taken by Atkinson and de Lara was to solv e for the ﬂuid ﬂow surrounding the plate and obtain the ﬂuid pressu re exerted on the vibrating plate. This is done in two steps: ﬁrst, the case where the ﬂuid is inviscid is solved and then obtain a correction to the invi scid solution by allowing viscous eﬀects via the Wiener-Hopf technique menti oned in Chap. 2. This is thought of as the leading order in a matched asympto tic expansion between the region of inviscid ﬂow and the region of viscoud ﬂ ow near the surface. Once the ﬂuid pressure is found, the equation of mot ion of the thin plate is solve by balancing the forces experienced by the pla te. 3.1 Assumptions 6 main assumptions were made in describing the problem: 1. The width of the plate is far greater than the length of the p late. 15 2. The ﬂuid is considered to be inﬁnite with no wall eﬀects. 3. The amplitude of oscillation is smaller than any length sc ale in the problem. 4. The ﬂuid is incompressible with density ρ. 5. The thickness of the plate is negligible. 6. The ﬂuid is Newtonian with a viscosity of µ. Assumption 1 means that the dominant length scale in our prob lem is just the length L. A consequence of assumption 3 is that the non-linear con- vective term ( u·∇u) that appears in the Navier-Stokes equation which gov- erns viscous ﬂuid ﬂow can now be removed. This is because u∼amplitude and so, for suﬃciently high frequencies but not too high such that the ﬂuid becomes compressible, O(amplitude ) =∂u ∂t≫u· ∇u=O(amplitude2). From assumption 4, the continuity equation can also be reduc ed to a simpler form. For most practical purposes of modeling small mechani cal systems, the ﬂuid velocity doesn’t exceed the speed of sound and the wa ve length of the speed of sound is far greater than the length L. We have assumption 5 so that we can use the theory of thin plates when solving for th e equation of motion for the plate. Also, the eﬀects of the ﬂuid coming from the thickness of the plate are neglected. Therefore, the governing equati on for this problem would be the linearized Navier-Stokes equations: ρ∂u ∂t=−∇p+µ∇2u, ∇ ·u= 0.(3.1) 16 3.2 Scaling of Governing Equations Since we are considering oscillations of the plate, it would be easier to work in terms of the Fourier transformed version of (3.1) where the F ourier transform of the time variable is ˜X=/integraldisplay∞ −∞Xeiωtdt . Applying the Fourier transform to (3.1), we get −iω˜u=−∇˜p+µ∇2˜u, ∇ ·˜u= 0.(3.2) From (3.2) we make all the variables dimensionless by deﬁnin g ˜p=ωρLU ˆp, ˜u=Uˆuand the coordinates x,yandzare scaled by dividing by the length of the plate L. We also deﬁne a dimensionless quantity β=ρωL2 µ, where this quantity is commonly known as the Reynolds number [4]. Al- though the Reynolds number is associated with the non-linear convective term, this is not the case here. The governing equation now be comes −iˆu=−ˆ∇ˆp+1 βˆ∇2ˆu, (3.3a) ˆ∇ ·ˆu= 0. (3.3b) From here on, we will drop the hatted signs for simplicity and note that all the results are presented in the frequency domain and that al l variables are dimensionless. 17 3.3 Solving for the ﬂuid reaction 3.3.1 Inviscid Solution yz 1 0 Figure 3.2: The plate representing by the line going from 0 (c lamped end) to 1 (free end) in the scaled coordinate system In the ﬁgure, the plate is located in the z= 0 plane along the y-axis betweeny= 0 (the clamped end) and y= 1 (the free end) (see Fig. 3.2). Using the assumptions stated in Sec. 3.1 together with the vi scosity of the ﬂuid being zero ( β→ ∞), we set the velocity and the pressure of the inviscid ﬂuid to be u=∇φ (3.4a) p=iφ , (3.4b) whereφis the velocity potential so that ∇2φ= 0. With this choice of u andp, it solves (3.3). We will also deﬁne the pressure diﬀerentia l [p] about z= 0 as [p] =p|z=0+−p|z=0−. (3.5) The velocity potential satisﬁes Laplace’s equation after s ubstituting equa- tion (3.4a) into (3.3b). Laplace’s equation is given to be ∇2φ= 0. 18 Laplace’s equation has the solution φ= arctan(z y). If we shift our origin from y= 0 toy=s, φ= arctan(z y−s) is also a solution to Laplace’s equation. Using superpositi on overs, we get the general solution for the velocity potential to be φ=/integraldisplay1 0f(s)arctan(z y−s)ds , (3.6) where the function f(s) is to be determined from the boundary conditions. We deﬁne the standard arctangent function to take the values arctan(z y−s) =  πifz= 0+,y<s 0 ifz= 0+,y>s −πifz= 0−,y<s 0 ifz= 0−,y>s(3.7) The boundary conditions are imposed on the plane z= 0 since that is where the plate is located. •The vertical velocity, denoted by uzis continuous everywhere and 19 known for the plate as a function of ybut unknown outside the plate uz|z=0+=uz|z=0−=  unknown function y<0 uz(y) 0 <y< 1 unknown function y>1(3.8) •The pressure diﬀerential is unknown for the plate but equal t o zero outside the plate for y >1. On the other side of the plate at y <0, we do not expect the pressure diﬀerential to be zero but ﬁnite . This represents the clamping of the plate at y= 0 [p] =  ﬁnite y<0 unknown function to be determined 0 <y< 1 0 y>1(3.9) Atkinson and de Lara quoted the additional condition that th e tangential velocity, denoted by uyis continuous and equal to zero at z= 0 for the inviscid solution. This condition is the no-slip boundary co ndition and is a condition only for the viscous case since inviscid ﬂuids can have a non-zero tangential velocity and not a condition used to solve Laplac e’s equation. This, however, is used as a check for the inviscid solution on w hether it is satisﬁed even though it is an inviscid ﬂuid. We now have two bo undary conditions for Laplace’s equation and therefore we are able to solve it. The vertical velocity can be obtained by diﬀerentiating (3. 6) with respect toz: uz=/integraldisplay1 0f(s)1/parenleftBig 1 + (z y−s)2/parenrightBig1 y−sds . (3.10) Applying the boundary condition (3.8) to (3.10), we get the s ingular integral 20 equation uz(y) =/integraldisplay1 0f(s) y−sds,0<y< 1 (3.11) where the integral is deﬁned as a Cauchy principal value and s ubsequently all integrals are deﬁned as Cauchy principal value. We refer to some texts [1, 11, 12] on obtaining the general solution of (3.11). This is given to be f(s) =1 π2/radicalbig s(1−s)/integraldisplay1 0uz(y′)/radicalbig y′(1−y′) y′−sdy′+A/radicalbig s(1−s) =1 π2/radicalbig s(1−s)/integraldisplay1 0uz(y′)/radicalBigg 1−y′ y′(1 +O(s))dy′+A/radicalbig s(1−s) =1 π2/radicalbig s(1−s)/parenleftBigg A+1 π2/integraldisplay1 0uz(y′)/radicalBigg 1−y′ y′dy′/parenrightBigg +O(√s),(3.12) where we have done a binomial expansion for small sin the integral. Ap- plying the boundary condition (3.9) to (3.12) requires that ats= 0 the functionf(s) is ﬁnite and therefore the coeﬃcient of the 1 /√shas to be zero: A+1 π2/integraldisplay1 0uz(y′)/radicalBigg 1−y′ y′dy′= 0 ⇒A=−1 π2/integraldisplay1 0uz(y′)/radicalBigg 1−y′ y′dy′. (3.13) We can now re-write the function f(s) to be f(s) =1 π2/radicalbig s(1−s)/integraldisplay1 0uz(y′)/parenleftBigg/radicalbig y′(1−y′) y′−s−/radicalBigg 1−y′ y′/parenrightBigg dy′ =1 π2/radicalbig s(1−s)/integraldisplay1 0uz(y′)/parenleftbiggy′√1−y′−(y′−s)√1−y′ √y′(y′−s)/parenrightbigg dy′ =1 π2/radicalbiggs 1−s/integraldisplay1 0uz(y′) y′−s/radicalBigg 1−y′ y′dy′. (3.14) 21 The function f(s) has now been determined and we now proceed to ﬁnding an expression for the tangential velocity at z= 0. Atz= 0+andz= 0−, (3.6) becomes φ|z=0+=π/integraldisplay1 yf(s)ds φ|z=0−=−π/integraldisplay1 yf(s)ds(3.15) by using the values of the arctangent speciﬁed in (3.7). By di ﬀerentiating (3.15) with respect to yusing (3.4a), we get the tangential velocity uyat z= 0+to be uy=−πf(y). We observe a singularity at the edge of the plate where it has t he following leading order behaviour uy|y→1−≈ −π π2√1−y/integraldisplay1 0uz(y′) y′−1/radicalBigg 1−y′ y′dy′ =1 π√1−y/integraldisplay1 0uz(y′)/radicalbig y′(1−y′)dy′ =B√1−y, (3.16) where the constant Bis a weighted average of the vertical velocity over the plate: B=1 π/integraldisplay1 0uz(y′)/radicalbig y′(1−y′)dy′. (3.17) Thus we need to correct for the tangential velocity since it h as a square root singularity at the edge of the plate. Apart from this, we can a lso calculate the vertical velocity outside the plate using (3.11) uz=/integraldisplay1 0f(s) y−sds, y> 1. 22 Substituting (3.14) into the above equation and swapping th e order of inte- gration we get uz=1 π2/integraldisplay1 0uz(y′)/radicalBigg 1−y′ y′/parenleftbigg/integraldisplay1 01 (y−s)(y′−s)/radicalbiggs 1−sds/parenrightbigg dy′. Using contour integral techniques in App. C.1, we obtain the expression for the leading order behaviour as y→1+ uz=1 π2/integraldisplay1 0uz(y′)/radicalBigg 1−y′ y′/parenleftbigg π/radicalbiggy y−11 y′−y/parenrightbigg dy′ ⇒uz|y→1+≈ −1 π√y−1/integraldisplay1 0uz(y′)1/radicalbig y′(1−y′)dy′ =−B√y−1. (3.18) We also see that the vertical velocity also has a square root s ingularity at the edge and this will also need to be corrected. The pressure diﬀerential for the inviscid case can be calcul ated by using (3.4b), (3.5) and (3.15) to obtain the following expression : [p] = 2πi/integraldisplay1 yf(s)ds . (3.19) 3.3.2 Correction to the Inviscid Solution As seen in the previous section, both the tangential and vert ical velocity need to be corrected. This means that we need to include visco sity into our governing equations for the ﬂuid ﬂow. Noting that we are solv ing a 2-D ﬂow 23 in theyandzcoordinates, the component form of (3.3) is as follows: −iuy=−∂p ∂y+1 β∇2uy (3.20a) −iuz=−∂p ∂z+1 β∇2uz (3.20b) ∂uy ∂y+∂uz ∂z= 0 (3.20c) with ∇2=∂2 ∂y2+∂2 ∂z2. We introduce the stream function ψ uy=−∂ψ ∂z uz=∂ψ ∂y(3.21) such that (3.20c) is automatically satisﬁed. Substituting (3.21) into (3.20a,b) and eliminating the pressure term by diﬀerentiating (3.20a ) with respect to z and diﬀerentiating (3.20b) with respect to yand subtracting both equations, we obtain the following expression: ∇2(∇2+iβ)ψ= 0. (3.22) As we can see, excluding the second term inside the brackets i s equivalent to the inviscid case where it satisﬁes Laplace’s equation ( ∇2φ= 0) and therefore it is a ﬁrst order approximation. In this correcti on, we propose a local coordinate system at the edge of the plate, given by y= 1 +ǫY z=ǫZ ,(3.23) 24 whereǫis some small but positive dimensionless number. In this new co- ordinate system, the origin is now at the edge of the plate and the plate is located in the negative X-axis (see Fig. 3.3). YZ 0−1/ǫ Figure 3.3: Coordinate system speciﬁed by (3.23) where the o rigin is at the edge of the plate. When β→ ∞, the clamped end of the plate ( Y=−1/ǫ) will tend to −∞ Under the new coordinate system, we get the expression for th e Laplacian operator ( ∇2) to be ∇2=1 ǫ2/parenleftbigg∂2 ∂Y2+∂2 ∂Z2/parenrightbigg . We are interested in the case when βis large and so in order to make the two terms inside the bracket of (3.22) comparable in the new c oordinate system, we require 1 ǫ2=β ⇒ǫ=1√β We now take the Fourier transform of Yon (3.22) and we obtain the follow- ing diﬀerential equation in Z. ∂4ψ ∂Z4+ (i−2ζ2)∂2ψ ∂Z2+ (ζ4−iζ2)ψ= 0 25 The 4 independent solutions obtained from the diﬀerential e quation are e|ζ|Z, e−|ζ|Z,e√ ζ2−iZande−√ ζ2−iZ. We seek solutions that tend to zero at inﬁnity since the ﬂuid is not aﬀected by the oscillations of the plate far away. Hence we construct the solution as ψ=  A1(ζ)e−√ ζ2−iZ−iB1(ζ)e−|ζ|Z, Z > 0 A2(ζ)e√ ζ2−iZ−iB2(ζ)e|ζ|Z, Z < 0(3.24) where/radicalbig ζ2−iand|ζ|are deﬁned to have positive real part and the functions A1,A2,B1andB2are to be determined from the boundary conditions. Using the expression for ψ, the expressions for uyanduzcan be easily obtained from the Fourier transform of Yof (3.21) and once those have been found, the pressure pas a function of Zcan be obtained by taking the Fourier transform of (3.20b): uy=  A1(ζ)/radicalbig ζ2−ie−√ ζ2−iZ−iB1(ζ)|ζ|e−|ζ|Z, Z > 0 −A2(ζ)/radicalbig ζ2−ie√ ζ2−iZ+iB2(ζ)|ζ|e|ζ|Z, Z < 0(3.25a) uz=  −iζA1(ζ)e−√ ζ2−iZ−ζB1(ζ)e−|ζ|Z, Z > 0 −iζA2(ζ)e√ ζ2−iZ−ζB2(ζ)e|ζ|Z, Z < 0(3.25b) p=  iǫ|ζ| ζB1(ζ)e−|ζ|Z, Z > 0 −iǫ|ζ| ζB2(ζ)e|ζ|Z, Z > 0(3.25c) We now list down the boundary conditions at Z= 0 for the new solution. •The velocity components are continuous across the plane Z= 0 with 26 the tangential component equal to zero uy|Z=0+=uy|Z=0−= 0 uz|Z=0+=uz|Z=0−(3.26) •The tangential velocity outside the plate is zero and on the p late, it must cancel the velocity obtained from (3.16) and so uy=  −B√1−x=−B√ −ǫY, Y < 0 0, Y > 0(3.27) •The vertical velocity is zero on the plate, as it is already kn own and has been accounted for in the inviscid solution, and unknown outside the plate: uz=  0, Y < 0 unknown function , Y > 0(3.28) •The pressure diﬀerential on the plate is unknown and zero out side: [p] =  unknown function , Y < 0 0, Y > 0(3.29) The ﬁrst thing to notice from the boundary conditions is that A1(ζ) =A2(ζ) andB1(ζ) =B2(ζ). This comes directly from the boundary condition (3.26). Then the pressure diﬀerential from (3.25c) will have the exp ression [p] =p|Z=0+−p|Z=0− = 2iǫ|ζ| ζB1(ζ). (3.30) 27 Note that in Atkinson and de Lara’s paper, the factor of ǫis missing. We will now use the Wiener-Hopf technique to ﬁnd an expression for the pressure diﬀerential correction. We take the Fourier tr ansform of the boundary conditions (3.27), (3.28) and (3.29). The half Fou rier transform F−=/integraldisplay0 −∞FeiζYdY where the integral converges for Im(ζ)<0 and therefore F−is a minus function since it is analytic in the lower half plane Im(ζ)<0 and denoted with a ’-’ subscript. Similiarly, the half Fourier transform F+=/integraldisplay∞ 0FeiζYdY where the integral converges for Im(ζ)>0 and therefore analytic in the upper half plane Im(ζ)>0.F+is called a plus function and is indicated with a ’+’ subscript. The Fourier transform of the boundary conditions simpiﬁes t o half Fourier transforms that are either analytic in the upper or lower hal f plane. There- fore, we can classify the velocity components and the pressu re diﬀerential into plus and minus functions depending on where the region i s non-zero. For the tangential velocity and the pressure diﬀerential, i t is non-zero for Y <0 and therefore uyand[p] are minus functions and shall be denoted byuy=uy−and[p] =[p]−respectively. For the vertical velocity, it is non-zero for Y >0 and therefore it is a plus function and shall be denoted asuz=uz+. So, forZ= 0, we have three equations which are (3.25a),(3.25b) and (3.30) involving 4 unknowns [p]−,uz+,A1(ζ) andB1(ζ). Note that we have speciﬁed the boundary conditions for tangential velocity a nd so we know 28 the form of uy−. The explicit form of uywill be considered as derived later for convenience. We can express A1(ζ) andB1(ζ) using (3.30) and (3.25a) to be in terms of [p]−anduy−. A1(ζ) =ζ 2ǫ/radicalbig ζ2−i[p]−+uy−/radicalbig ζ2−i B1(ζ) =ζ 2iǫ|ζ|[p]−. Substituting into (3.25b) we obtain the following function al equation in terms of the two remaining unknowns uz+and[p]− uz+=iζ2 2ǫ[p]−/bracketleftBigg 1 |ζ|−1/radicalbig ζ2−i/bracketrightBigg −iζ/radicalbig ζ2−iuy− =iζ2 2ǫ|ζ|/radicalbig ζ2−i[p]−(/radicalbig ζ2−i− |ζ|)−iζ/radicalbig ζ2−iuy− =iζ2(ζ2−i− |ζ|2) 4ǫ|ζ|(ζ2−i)/parenleftBigg 2/radicalbig ζ2−i |ζ|+/radicalbig ζ2−i/parenrightBigg [p]−−iζ/radicalbig ζ2−iuy−(3.31) where in going from the second line to the third line, the ﬁrst term on the RHS is multiplied by 2/radicalbig ζ2−i(|ζ|+/radicalbig ζ2−i) on the numerator and denominator. We see that this is of the form stated in the prev ious chapter in (2.1) with uy−known. We now deﬁne a function H(ζ) where H(ζ) =|ζ|+/radicalbig ζ2−i 2/radicalbig ζ2−i. We see that H(ζ) is an even function and as |ζ| → ∞ thatH(ζ)→1 and it has no zeros on the real axis, thus containing the inversion c ontour for the inverse Fourier transform. This satisﬁes the condition to p erform a product decomposition H(ζ) =H+(ζ)H−(ζ) as discussed in Sec. 2.3. The function |ζ|is not an analytic function and thus we need another representation to make it analytic. Since (3.31) is deﬁned o n the real axis, 29 we can represent |ζ|as |ζ|= lim δ→0/radicalbig ζ+iδ/radicalbig ζ−iδ =/radicalbig ζ+/radicalbig ζ−(3.32) since the branch point for√ζ+iδis in the lower half plane and√ζ−iδis in the upper half plane. Choosing both the branch cuts to be pa rallel to the real axis gives that√ζ+= lim δ→0√ζ+iδis analytic in the upper half plane and therefore it is a plus function and similiarly√ζ−= lim δ→0√ζ−iδis analytic in the lower half plane and therefore it is a minus fu nction. Thus, the function H(ζ) becomes H(ζ) =√ζ+√ζ−+/radicalbig ζ2−i 2/radicalbig ζ2−i. We shall refer to App. A on simplifying the expression for H+(ζ) andH−(ζ) and obtain their respective expansions as |ζ| → ∞ . Since (3.31) is deﬁned only on the real axis, we note that in th e ﬁrst term on the RHS1of the equation that ( ζ2−i− |ζ|2) is now equal to −i since|ζ|2=ζ2here and so (3.31) now becomes uz+=ζ2 4ǫ√ζ+√ζ−(ζ−eiπ/4)(ζ+eiπ/4)/parenleftbigg1 H+(ζ)H−(ζ)/parenrightbigg [p]−−iζ/radicalbig ζ2−iuy− Rearranging the ﬁrst term on the RHS so that the [p]−term purely consists 1RHS=right hand side 30 of minus functions gives uz+(ζ+eiπ/4)H+(ζ)√ζ+ ζ =ζ 4ǫ√ζ−(ζ−eiπ/4)H−(ζ)[p]−−i/radicalBigg ζ+eiπ/4 ζ−eiπ/4H+(ζ)/radicalbig ζ+uy−(3.33) We would not be able to proceed further until we know somethin g about the form of uy−. From the Fourier transform of (3.27), we get uy−=−e−iπ/4B√ǫ√π√ζ−. Therefore (3.33) becomes uz+(ζ+eiπ/4)H+(ζ)√ζ+ ζ =ζ 4ǫ√ζ−(ζ−eiπ/4)H−(ζ)[p]−+B√ǫ√πeiπ/4√ζ+√ζ−/radicalBigg ζ+eiπ/4 ζ−eiπ/4H+(ζ) =ζ 4ǫ√ζ−(ζ−eiπ/4)H−(ζ)[p]−+B√ǫ√πeiπ/4 √ζ+√ζ−/radicalBigg ζ+eiπ/4 ζ−eiπ/4−1 H+(ζ) +B√ǫ√πeiπ/4H+(ζ). We now deﬁne a function S(ζ) such that S(ζ) =S+(ζ) +S−(ζ) where S(ζ) = √ζ+√ζ−/radicalBigg ζ+eiπ/4 ζ−eiπ/4−1 H+(ζ). Note that as |ζ| → ∞ we have that S(ζ)→0 and therefore it satisﬁes the condition discussed in Sec.2.3. Thus we can perform a sum dec omposition onS(ζ). We shall refer to App. B to simplify the expressions for S+(ζ) andS−(ζ) and to obtain their respective expansions as |ζ| → ∞ . Once we have done the sum decomposition, we rearrange (3.33) so that all the plus 31 functions are on one side of the equality and the minus functi ons are on the other side: uz+(ζ+eiπ/4)H+(ζ)√ζ+ ζ−eiπ/4B√π√ǫ(S+(ζ) +H+(ζ)) =ζ 4ǫ√ζ−(ζ−eiπ/4)H−(ζ)[p]−+eiπ/4B√π√ǫS−(ζ)≡G(ζ).(3.34) We have now re-expressed (3.31) into a function that consists of plus func- tions only on one side of the equality and a function that cons ists of minus functions only on the other side of the equality. On the LHS2of (3.34), the function is analytic in the lower half plane and the RHS of (3.34), the function is analytic in the upper half plane. Since we have th at (3.34) holds on the real axis, by analytic continuation, we have that the f unctionG(ζ) is analytic in the whole complex plane. The next step in the Wi ener-Hopf method is to be able to use the generalised Liouville’s theor em to be able to say what does the function G(ζ) is. In this case, we will be using the standard Liouville’s theorem instead of the generalised ve rsion. To be able to do this we require some knowledge of the unknown functions [p]−and uz+. The pressure diﬀerential at Y= 0 can be singular but has to be inte- grable to give a ﬁnite force on the plate. More precisely, we e xpect a square root singularity in the pressure diﬀerential at the edge and therefore the Fourier transform of the pressure diﬀerential, [p]−∼1/√ζ−. At the same time, we also require that the vertical velocity outside the plate must cor- rect the singular behaviour obtained in the inviscid soluti on. Therefore the Fourier transform of the vertical velocity outside the plat e,uz+∼1/√ζ+. Using these two facts together with the fact that the H+(ζ),H−(ζ)→1 and 2LHS=left hand side 32 S+(ζ),S−(ζ)→0 as|ζ| → ∞ , we have shown that the LHS of (3.34) goes to a constant and the RHS of (3.34) goes to zero as |ζ| → ∞ . By the standard Liouville’s theorem, we have that the function G(ζ) is an entire function which is identically zero. We now have two equations with two unknowns, namely uz+(ζ+eiπ/4)H+(ζ)√ζ+ ζ−eiπ/4B√π√ǫ(S+(ζ) +H+(ζ)) = 0, ζ 4ǫ√ζ−(ζ−eiπ/4)H−(ζ)[p]−+eiπ/4B√π√ǫS−(ζ) = 0.(3.35) We are now able to solve for [p]−anduz+using (3.35) to obtain the cor- rection for the pressure diﬀerential. Since the correction is signiﬁcant near the edge of the plate, we would use the expansion of the functi onsH+(ζ), H−(ζ),S+(ζ) andS−(ζ) as|ζ| → ∞ to get the leading order correction. Rearranging (3.35) and using binomial expansions, we get uz+|ζ→∞≈eiπ/4B√π ǫ1√ζ+, [p]−|ζ→∞≈ −2iBǫ√ 2√π√ǫ1√ζ−.(3.36) Taking the inverse Fourier transform of the above equations and re-writing the equations in terms of the original variable y, we obtain uz|y→1+≈B√y−1, [p]|y→1−≈2√ 2e−iπ/4ǫB1√1−y. We see that the vertical velocity obtained from this correct ion cancels out the singularity obtained from the inviscid solution in (3.1 8) just outside the plate. As for the pressure diﬀerential, the correction has a n added mass 33 contribution out of phase with the velocity of the plate. Thi s is clearly seen as the real part of the pressure diﬀerential. It also has a dam ping component which is in phase with the velocity of the plate which can be id entiﬁed as the imaginary part of the term. 3.4 Equation of Motion of the Plate Since we now have an expression for the force exerted by the ﬂu id on the plate, mainly the pressure diﬀerential, we can now use it to s olve the equation of motion for the vibrating plate. The force exerted by the ﬂu id,fﬂuid, is obtained by superposing the solution of the inviscid case an d the correction to the invisicid case. Atkinson and de Lara have made a simpliﬁcation to calculatin g the pres- sure diﬀerential for the inviscid case. The correction to th e inviscid case is mainly important near the edge at y= 1, the inviscid pressure diﬀerential is approximated by the pressure diﬀerential at the other end y= 0. Using (3.19) and setting y= 0, swapping order of integration and using contour integration techniques (App. C.2), we obtain the pressure d iﬀerential at the clamped end to be [p] = 2πi/integraldisplay1 0f(s)ds = 2πi/integraldisplay1 0/integraldisplay1 01 π2/radicalbiggs 1−suz(y′) y′−s/radicalBigg 1−y′ y′dsdy′ = 2πi1 π2/integraldisplay1 0(−π)uz(y′)/radicalBigg 1−y′ y′dy′ =−2π2i1 π2/integraldisplay1 0uz(y′)/radicalBigg 1−y′ y′dy′ = 2π2iA , (3.37) 34 whereAis given by (3.13). Therefore the full expression of the pres sure diﬀerential by superposing the inviscid with the correctio n is [p] = 2π2iA+ 2√ 2e−iπ/4ǫB1√1−y. (3.38) yzFz=0+ Fz=0−ˆjˆk Figure 3.4: Forces acting on the plate The ﬂuid force can be calculated via the stress vector. The fo rce per unit area exerted by the ﬂuid on the top plate, z= 0+, isn·Twhere nis the normal of the top plate, ˆk, andTis the stress tensor (see Fig. 3.4). The force per unit area in the z-direction on the top plate is Fz=0+=n·T·ˆk =Tzz =−p|z=0++∂uz ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=0+. Using the continuity equation ∇ ·u=∂uy ∂y+∂uz ∂z= 0 and the no-slip condition on the surface of the plate, we ﬁnd th at the second term in the continuity equation vanishes since there is no ho rizontal velocity 35 at the surface, thus giving the result ∂uz ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=0+=∂uz ∂z/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=0−= 0. Therefore, the force on the top plate has been reduced to Fz=0+=−p|z=0+. Similiarly, the force on the bottom plate has the same form as the force on the top plate except for the minus sign. This comes from the fa ct that the normal of the bottom plate is −ˆk. Putting all of these forces together, the net force of the ﬂuid per unit area is the z-direction is Fz=−p|z=0++p|z=0−=−[p], (3.39) where this is the dimensionless force per unit area. We unscale back to the original pressure unit by multiplying the pressure diﬀe rential byωρLU . Therefore, ﬂuid force per unit area is fﬂuid=−ωρLU/parenleftbigg 2π2iA+ 2√ 2e−iπ/4ǫB1√1−y/parenrightbigg . Apart from the ﬂuid force, we can also have a driving force for the plate as well. For simplicity, we model the driving force, fdrive, as a sinusoid with frequency ωand with strength equal to one on a line parallel to the clamping at a distance yfaway from the clamping and it is expressed in units of pressure fdrive=δ(y−yf). Note that this diﬀers from Atkinson and de Lara by a factor of e−iωt. This 36 is because the driving force that they used is in the time doma in and not the frequency domain as stated here. Both can be made equivalent by taking the Fourier transform of their driving force although they seem to have changed their deﬁnition for the Fourier transform to e−iωtinstead ofeiωtas used in solving the ﬂuid force. This change of deﬁnition does not aﬀe ct the rest of their analysis. Using the expression for modelling a thin plate with a small d eﬂection in thez-direction with an elastic restoring force while still assum ing that all longitudinal cross sections act the same way [16], the equat ion of motioin of the plate can be reduced to ρshL∂2w ∂t2+Eh3 12∂4w ∂y4=ft ﬂuid+ft drive, (3.40) whereρsis the density of the plate, wis the dimensionless deﬂection of the plate andhis the dimensionless thickness of the plate and Eis the Young’s modulus of the plate. wandhhave both been scaled with respect to L. The forces are in the time domain as well. Taking the Fourier t ransform of (3.40), we get Eh3 12∂4w(y,ω) ∂y4−ρshL2ω2w(y,ω) =δ(y−yf)−ωρLU [p] ∂4w(y,ω) ∂y4−12 Eh2ρsL2ω2w(y,ω) =12 Eh3δ(y−yf)−12 Eh3ωρLU [p] ∂4w(y,ω) ∂y4−c(ω)4w(y,ω) =12 Eh3δ(y−yf)−12 Eh3ωρLU [p], (3.41) 37 wherec(ω) andg(y,ω) take the following expressions: c(ω) =/parenleftbigg3ρs E/parenrightbigg1/4/parenleftbigg2ωL h/parenrightbigg1/2 , g(y,ω) =12 Eh3δ(y−yf)−12 Eh3ωρLU [p]. [p] used here is the full expression for the pressure diﬀerenti al, given by (3.38). The boundary conditions for the plate are: •Displacement is zero at y= 0 w|y=0= 0. •The ﬁrst derivative is zero at y= 0 which represents clamping at the end ∂w ∂y|y=0= 0. •There is no bending moments at the free end y= 1 which can be written as ∂2w ∂y2|y=1= 0. •There are no shearing forces at the free end y= 1 which can be written as ∂3w ∂y3|y=1= 0. The constants AandBthat appear in the RHS of (3.41) are weighted av- erages of the vertical velocity uz. We can re-write this as the ﬁrst derivative ofwwith respect to time. After scaling and taking the Fourier tr ansform of 38 the time variable, we ﬁnd that uz(y) =−iωL Uw(y,ω). (3.42) Atkinson and de Lara solved the equation of motion by constru cting a solution to the equation of motion of the form w(y,ω) =wp(y,ω) +C1sinh(c(ω)y) +C2cosh(c(ω)y) +C3sin(c(ω)y) +C4cos(c(ω)y), (3.43) wherewp(y,ω) is constructed using the Green’s function, G(y′,y′), for equa- tion (3.41) wp(y,ω) =/integraldisplay1 0G(y,y′)g(y′,ω)dy′. Their approach is very tedious since will we end up with an int egral equation where the unknown w(y,ω) appears on both sides of the equation of (3.43) sincewp(y,ω) containsw(y,ω) in the constants AandBing(y,ω). Hence, the method of Green’s functions is not appropriate in this ca se. Instead, we will use an eigenfunction expansion to solve the equation of motion. The homogeneous problem will be denoted as ∂4φ ∂y4−c4 0φ= 0, (3.44) wherec0is related to the resonant frequencies of the plate. The gene ral solution to the homogenous problem is φ=C1sinh(c0y) +C2cosh(c0y) +C3sin(c0y) +C4cos(c0y). Applying the boundary conditions to the solution of the homo geneous prob- 39 lem, we need to solve a homogenous linear system. Requiring t hat we must have non-trivial solutions, the determinant of the associat ed matrix of the linear system must be zero. We then ﬁnd that there are certain values ofc0 can only be attained for non-trivial solutions and they satis fy the relation 1 + cosh(c0,n)cos(c0,n) = 0, n∈N. We can see that there are an inﬁnite number of values of c0,nsuch that the above equation is satisﬁed. Using the boundary conditions, we can eliminate 3 of the coeﬃcients of the general solution and the function φcan be written as an inﬁnite sum φ=∞/summationdisplay n=1Cn/bracketleftbigg cosh(c0,ny)−cos(c0,ny) +cosh(c0,n) + cos(c0,n) sinh(c0,n) + sin(c0,n)(sin(c0,ny)−sinh(c0,ny))/bracketrightbigg ⇒φ=∞/summationdisplay n=1Cnφn, whereφnare the basis functions for the solution and Cnare coeﬃcients that depend on n. Each of the φnsatisﬁes (3.44) and so we see that φis a superposition of all the basis functions. Returning to sol ving (3.41), we expand the displacement in terms of the basis functions w(y,ω) =∞/summationdisplay n=1wmφn(y,ω) (3.45) and substitute it into (3.41) to get ∞/summationdisplay n=1wn/parenleftBig φ(iv) n(y,ω)−c(ω)4φn(y,ω)/parenrightBig =g(y,ω), (3.46) 40 where g(y,ω) =12 Eh3δ(y−yf)−12 Eh3ωρLU/parenleftbigg 2π2iA+ 2√ 2e−iπ/4ǫB1√1−y/parenrightbigg =12 Eh3δ(y−yf) +24 Eh3ρω2L2/parenleftBigg∞/summationdisplay n=1wn/bracketleftBigg/integraldisplay1 0φn(y′)/radicalBigg 1−y′ y′dy′ +√ 2 πeiπ/4ǫ1√1−y/integraldisplay1 0φn(y′)1/radicalbig y′(1−y′)dy′/bracketrightBigg/parenrightBigg , where we have used the expression for AandBfrom (3.13) and (3.17) and have substituted the relation between uz(y) with the displacement of the platew(y) from (3.42). We have the property of the basis functions tha t they are orthogonal to each other [8] and we can scale the basi s functions so that /integraldisplay1 0φn(y)φm(y)dy=  1,ifn=m 0,ifn/negationslash=m Multiplying φm(y,ω) (3.46) and integrating from 0 to 1 and using the or- thogonality relations above we get the following expressio n: ∞/summationdisplay n=1wnδm,n(c4 0,n−c4)/integraldisplay1 0φm(y)φn(y)dy=/integraldisplay1 0φm(y)g(y)dy , (3.47) where the RHS is /integraldisplay1 0φm(y)g(y)dy=12 Eh3φm(yf) +24 Eh3ρω2L2× /parenleftBigg∞/summationdisplay n=1wn/bracketleftBigg/integraldisplay1 0/integraldisplay1 0φm(y)φn(y′)/radicalBigg 1−y′ y′dy′dy +√ 2 πeiπ/4ǫ/integraldisplay1 0/integraldisplay1 01√1−yφm(y)φn(y′)1/radicalbig y′(1−y′)dy′dy/bracketrightBigg/parenrightBigg . 41 We will deﬁne the quantities Am,n=δm,n(c4 0,m−c4)/integraldisplay1 0φm(y)φn(y)dy Bm,n=/integraldisplay1 0/integraldisplay1 0φm(y)φn(y′)/radicalBigg 1−y′ y′dy′dy Cm,n=/integraldisplay1 0/integraldisplay1 01√1−yφm(y)φn(y′)1/radicalbig y′(1−y′)dy′dy . Therefore, (3.47) becomes ∞/summationdisplay n=1Am,nwm=12 Eh3φm(yf) +24 Eh3ρω2L2/parenleftBigg∞/summationdisplay n=1/bracketleftBigg Bm,n+√ 2 πeiπ/4ǫCm,n/bracketrightBigg wn/parenrightBigg ⇒∞/summationdisplay n=1/bracketleftBigg Am,n−24 Eh3ρω2L2/parenleftBigg Bm,n+√ 2 πeiπ/4ǫCm,n/parenrightBigg/bracketrightBigg wn=12 Eh3φm(yf). This can now be written as a linear system Dw=r, (3.48) where the matrix Dhas entries Dm,n=Am,n−24 Eh3ρω2L2/parenleftBigg Bm,n+√ 2 πeiπ/4ǫCm,n/parenrightBigg and the vector rhas entries speciﬁed by rm=12 Eh3φm(yf). To solve this, we would have to truncate the series in (3.48) t o include the ﬁrstkterms and this will give a k×kmatrixDwhere the entries will have to be computed numerically. Solving the linear sys tem in (3.48) by using methods such as Gauss elimination (since in general the matrix 42 Dis dense) gives the coeﬃcients for the wm. Thus we can calculate the displacement of the plate subjected to the driving force, th e ﬂuid force and the elastic restoring force of the plate from (3.45) for a giv enω. We can also increase the accuracy of the solution by including more term s in the series and subsequently the matrix. 43 Chapter 4 Flexural and Torsional Vibrations We would like to be able to test the method that was proposed by Atkinson and de Lara in using the Wiener-Hopf technique to obtain the le ading or- der behaviour of a plate vibrating in a viscous ﬂuid. Recentl y, Van Eysden and Sader [5] have solved the linearized Navier-Stokes equat ion exactly on a vibrating blade having normal and torsional oscillations for all values of Reynolds number. The solution comes from solving a system of linear equa- tions where the entries of the matrix consist of Meijer-G func tions. The authors have computed a quantity which is called the normali zed hydrody- namic function, Γ(Re), where Re is the Reynolds number for bo th cases. Γn(Re) is for the case with normal oscillations and this is rela ted to the force per unit length ,fﬂuid, exerted by the ﬂuid on the blade while Γt(Re) is for the case with torsional oscillations and it is related to the moment per unitlength ,mﬂuidexerted by the ﬂuid on the blade. Note that the forces that were calculated by Atkinson and de Lara [3] were in units of force per unitarea. There is an extra term κwhich signiﬁes the wave number in the 44 z-direction but it can be shown that for κ= 0, the problem becomes a 2-D problem where it has inﬁnite length in the x-direction. The general form of fﬂuidandmﬂuidfor this geometry can be expressed as fﬂuid=π 4ρω2L2Γn(β)w(x), (4.1a) mﬂuid=−π 8ρω2L4Γt(β)Φ(x). (4.1b) To be consistent with the previous analysis, the Reynolds nu mber will be denoted by βand note that it is the same deﬁnition used by the Atkinson and de Lara and Las the length across the blade. The functions w(x) is the displacement of the blade as a function of xwhile Φ(x) is the angle of deﬂection of the blade as a function of xand is deﬁned in both the papers by Van Eysden and Sader [5, 6]. This is because the displaceme nt and the angle of the blade is the same for a speciﬁed value of x. We will attempt to use the Wiener-Hopf method to capture the as ymp- totic behaviour of both cases for large values of βand will compare it with the tables that have been calculated from the exact solution . 4.1 Normal Oscillations in a Viscous Fluid We will proceed with the same steps used in the previous chapt er to calculate the inviscid solution and the correction to the inviscid sol ution and later obtain the force per unit length to be able to ﬁnd out the hydro dynamic function Γn(β). The superscript nin all the variables and constants here is to denote that it is for the normal oscillation case (see Fi g. 4.1) and not raised to the power n. 45 z y −1 21 2Directionof motion Figure 4.1: Blade performing normal oscillations 4.1.1 Inviscid Solution In this problem, we have that the blade is centred at the origi n and its edges are at y=±1/2. The governing equation for the inviscid ﬂuid is the same, namely Laplace’s equation, where the velocity and the pressure can be determined by (3.4). The general solution for Laplace’s e quation for this case is superposing the fundamental solution with a functio nfn(s) which is to be determined from the boundary conditions over the lengt h across the blade, which gives φn(y) =/integraldisplay1/2 −1/2fn(s)arctan/parenleftbiggz y−s/parenrightbigg ds , (4.2) wherefn(s) is to denote normal oscillations. The arctangent function still carries the same deﬁnition as in (3.7). We now list the bounda ry conditions on the plane z= 0 needed to specify the solution of Laplace’s equation. •The vertical velocity is continuous everywhere and known fo r the blade but unknown outside the blade. Also, we note that it is also sy mmetric about the origin: un z|z=0+=un z|z=0−=  unknown function , y>/vextendsingle/vextendsingle1 2/vextendsingle/vextendsingle uz(y), y</vextendsingle/vextendsingle1 2/vextendsingle/vextendsingle un z(y) =un z(−y).(4.3) 46 •The pressure diﬀerential is unknown for the blade but equal to zero outside the blade. Also, due to the symmetry of the problem, i t is also symmetric about the origin: [p]n=  unknown function to be determined , y</vextendsingle/vextendsingle1 2/vextendsingle/vextendsingle 0, y>/vextendsingle/vextendsingle1 2/vextendsingle/vextendsingle [p]n(y) = [p]n(−y).(4.4) We will also use the no-slip condition as a check for the invisc id solution as in the previous chapter. The vertical velocity is obtained b y diﬀerentiating (4.2) with respect to zand setting z= 0 and using the boundary condition in (4.3), we obtain the following expression un z(y) =/integraldisplay1/2 −1/2fn(s) y−sds, y</vextendsingle/vextendsingle/vextendsingle/vextendsingle1 2/vextendsingle/vextendsingle/vextendsingle/vextendsingle. (4.5) Using the same technique in the referred texts, the general s olution for the functionfn(s) is fn(s) =1 π2√ 1−4s2/integraldisplay1/2 −1/2un z(y′)/radicalbig 1−4y′2 y′−sdy′+A√ 1−4s2 =2s π2√ 1−4s2/integraldisplay1/2 0un z(y′)/radicalbig 1−4y′2 y′2−s2dy′+A√ 1−4s2,(4.6) where in going from the ﬁrst line to the second line we used the symmetry condition on the vertical velocity. At z= 0+andz= 0−, (4.2) becomes φn z=0+=π/integraldisplay1/2 yfn(s)ds , φn z=0−=−π/integraldisplay1/2 yfn(s)ds .(4.7) 47 Therefore the pressure diﬀerential is [p] = 2πi/integraldisplay1/2 yfn(s)ds . (4.8) We now use the boundary condition (4.4) to set the constant Ain (4.6). In other words, we require that [ p]n= 0 aty=−1/2 ⇒0 =/integraldisplay1/2 −1/2fn(s)ds (4.9) ⇒0 =/integraldisplay1/2 −1/2/integraldisplay1/2 0un z(y′)2s π2√ 1−4s2/radicalbig 1−4y′2 y′2−s2dy′ds+/integraldisplay1/2 −1/2A√ 1−4s2ds ⇒A= 0, since in the ﬁrst term in the second line equals to zero becaus e we are integrating over an odd function over a symmetric domain and thusA= 0. Therefore the function fn(s) is now fn(s) =2s π2√ 1−4s2/integraldisplay1/2 0un z(y′)/radicalbig 1−4y′2 y′2−s2dy′. (4.10) We now examine the behaviour of the tangential velocity of th e inviscid solution. By taking the derivative with respect to yof (4.7), we have uy|z=0+=−πfn(y). (4.11) 48 Asy→ ±1/2, un y|z=0+≈  2√ 2 π√1−2y/integraldisplay1/2 0un z(y′)/radicalbig 1−4y′2dy′,asy→/parenleftbig1 2/parenrightbig − −2√ 2 π√1 + 2y/integraldisplay1/2 0un z(y′)/radicalbig 1−4y′2dy′,asy→/parenleftbig −1 2/parenrightbig + =  Bn √1−2y,asy→/parenleftbig1 2/parenrightbig− −Bn √1 + 2y,asy→/parenleftbig −1 2/parenrightbig+, (4.12) whereBnis a weighted average of the vertical velocity, Bn=2√ 2 π/integraldisplay1/2 0un z(y′)/radicalbig 1−4y′2dy′. (4.13) There is a singularity in the tangential velocity that needs to be corrected by including viscous eﬀects. Also, we can use (4.5) to determ ine the vertical velocity outside the plate: un z(y) =/integraldisplay1/2 −1/2fn(s) y−sds, y>/vextendsingle/vextendsingle/vextendsingle/vextendsingle1 2/vextendsingle/vextendsingle/vextendsingle/vextendsingle. Substituting (4.10) into the equation above and swapping th e order of inte- gration, we get un z(y) =/integraldisplay1/2 0/integraldisplay1/2 −1/22s π2√ 1−4s2un z(y′)/radicalbig 1−4y′2 y′2−s21 y−sdsdy′.(4.14) 49 Using contour integration techniques (App. C.3), we obtain un z(y) =/integraldisplay1/2 02y π2/radicalbig 1−4y2un z(y′)/radicalbig 1−4y′2 y′2−y2(πi)dy′ ≈  −4i π/radicalbig 2(1−2y)/integraldisplay1/2 0un z(y′)/radicalbig 1−4y′2dy′,asy→/parenleftbig1 2/parenrightbig+ 4i π/radicalbig 2(1 + 2y)/integraldisplay1/2 0un z(y′)/radicalbig 1−4y′2dy′,asy→/parenleftbig −1 2/parenrightbig− =  −Bn √2y−1,asy→/parenleftbig1 2/parenrightbig+ Bni√2y+ 1,asy→/parenleftbig −1 2/parenrightbig−. We see that the vertical velocity also has a singular behavio ur and needs to be corrected as well. 4.1.2 Correction to the Inviscid Solution We will concentrate on retrieving the correction near the y= 1/2 edge. The analysis done in Sec.3.3.2 is the same here except instead of using the change of variables stated in (3.23), we will use y=1 2+ǫY z=ǫZ ,(4.15) whereǫ= 1/√βas usual. The governing equation (3.22) is the same in this case and so taking the Fourier transform of this equatio n and solving it, gives the same expression for ψnin (3.24) and therefore the expressions foruyn,uznand[p]nare the same as in (3.25). The boundary conditions (3.26), (3.28) and (3.29) are all the same except for the cond ition on the tangential velocity in (3.27). The tangential velocity on the blade must cancel the singula r behaviour 50 obtained in the inviscid solution in (4.12) and it is zero out side the plate. Therefore, we have un y=  −Bn √1−2y=−Bn √ 21√ −ǫY=−B′n √ −ǫY, Y < 0 0, Y >0 where we have used (4.15) and B′n=Bn/√ 2. Notice that the analysis from here onwards is the same as in Sec.3.3.2 except we just replac eBwithB′n. Hence we can immediately write down the expressions for uznand[p]nas |ζ| → ∞ using (3.36) uzn +||ζ|→∞≈eiπ/4B′n√π ǫ1√ζ+ [p]n −||ζ|→∞≈ −2iB′nǫ√ 2√π√ǫ1√ζ− Taking the inverse Fourier transform and re-writing it in the original vari- ablesy, we get un z≈B′n √ ǫY=Bn √ 2ǫY=Bn √2y−1(4.16a) [p]n≈2√ 2B′nǫe−iπ/4 √ −ǫY=2√ 2Bnǫe−iπ/4 √ −2ǫY=2√ 2Bnǫe−iπ/4 √1−2y(4.16b) We have now obtained the correction for the pressure diﬀeren tial at the edgey= 1/2. We can repeat a rather similiar analysis for the edge y=−1/2, performing the same steps as described in Sec. 3.3.2. Howe ver, a simple argument will help shorten the computation to obtai n the pres- sure diﬀerential at the edge y=−1/2. For the blade that performs normal oscillations, the pressure is symmetric about the origin. T herefore the cor- rection to the pressure diﬀerential near y=−1/2 would have the same form 51 as (4.16b) except with the with the term√1−2yreplaced with√1 + 2y. The total correction for the pressure diﬀerential, [ p]n c, is just summing the contributions from each end of the blade giving [p]n c= 2√ 2Bnǫe−iπ/4/parenleftbigg1√1−2y+1√1 + 2y/parenrightbigg . From the above equation, it is clear that it is symmetric in th e region −1/2< y<1/2. When repeating the analysis for the edge y=−1/2, it can be shown that the vertical velocity cancels the singular behaviour o utside the blade as well. 4.1.3 Hydrodynamic Function Γn(β) We proceed to calculate the force per unit length exerted by t he ﬂuid on the blade. The pressure diﬀerential is the sum of the inviscid co ntribution and the total correction of the inviscid case [p]n= 2πi/integraldisplay1/2 yfn(s)ds+2√ 2Bnǫe−iπ/4/parenleftbigg1√1−2y+1√1 + 2y/parenrightbigg .(4.17) As in the previous chapter in (3.39), we have found that the ne t force excerted by the ﬂuid on the blade in the z-direction is Fz=−[p]n. This gives the dimensionless force per unit area. To obtain t he force per unit length, we have to integrate the above expression over t he length of the blade and multiplying by ωρLU and thus fﬂuid=−ωρLU/integraldisplay1/2 −1/2[p]ndy . (4.18) 52 Substituting (4.17) into (4.18), and swapping order of inte gration we get, fﬂuid=−ωρLU/bracketleftBigg 2πi/integraldisplay1/2 −1/2/integraldisplayy −1/2fn(s)dyds +2√ 2Bnǫe−iπ/4/integraldisplay1/2 −1/2/parenleftbigg1√1−2y+1√1 + 2y/parenrightbigg dy/bracketrightBigg =−ωρLU/bracketleftBigg 2πi/integraldisplay1/2 −1/2(s+ 1/2)fn(s)ds+ 2√ 2Bnǫe−iπ/4(2√ 2)/bracketrightBigg =−ωρLU/bracketleftBigg 2πi/integraldisplay1/2 −1/2sf(s)ds+ 8Bnǫe−iπ/4/bracketrightBigg (4.19) since/integraltext1/2 −1/2fn(s)ds= 0 from (4.9). Substituting fn(s) from (4.10) into the ﬁrst term and swapping the order of integration, the integra l can be reduced using contour integration techniques (App. C.4) to give fﬂuid=−ωρLU/bracketleftBigg 2πi/integraldisplay1/2 0/parenleftBig −π 2/parenrightBig2 π2un z(y′)/radicalbig 1−4y′2dy′ +16√ 2 πǫe−iπ/4/integraldisplay1/2 0un z(y′)/radicalbig 1−4y′2dy′/bracketrightBigg , (4.20) where we have used the expression for Bnfrom (4.13). We can re-write the vertical velocity in terms of the ﬁrst derivative of the disp lacement of the blade with respect to time, t, noting that the displacement is a function ofxand noty. Therefore, after taking the Fourier transform of the time derivative and scaling, we get un z(y) =−iωL Uw(x). Substituting un z(y) into (4.20), re-arranging the terms and performing the 53 integrals which are elementary, we obtain fﬂuid=π 4ρω2L2/bracketleftBigg 1 +16√ 2 πǫeiπ/4/bracketrightBigg w(x). We can now identify that the hydrodynamic function from (4.1 a) for the case of normal oscillations is Γn(β) = 1 +16√ 2 π√βeiπ/4. (4.21) 4.2 Torsional Oscillations in a Viscous Fluid We shift our attention to use the method on to a blade which und ergoes torsional oscillations (see Fig. 4.2). The inviscid soluti on is ﬁrst obtained and subsequently the correction to the inviscid solution. T hese are to be used later to compute the moment per unit length to determine the form of the hydrodynamic function Γt(β). The superscript tin all the variables and constants here is to denote case of the torsional oscillatio n and not raised to the powert. z y −1 21 2Direction of motion Figure 4.2: Blade performing torsional oscillations 4.2.1 Inviscid Solution As for the normal oscillation case, the governing equation f or the inviscid ﬂuid is Laplace’s equation and as usual, the velocity and pre ssure proﬁle 54 can be determined using (3.4). The potential φtis still of the form in (4.2) whereft(s) is denoted for torsional oscillations is to be determined b y the boundary conditions. The boundary conditions are the same as in (4.3) and (4.4) exc ept that instead of having the vertical velocity and the pressure diﬀ erential to be symmetric, we require that they are antisymmetric: ut z(y) =−ut z(−y), [p]t(y) =−[p]t(−y).(4.22) The general form of the function ft(s) is the same as in the previous case, that is ft(s) =1 π2√ 1−4s2/integraldisplay1/2 −1/2ut z(y′)/radicalbig 1−4y′2 y′−sdy′+A√ 1−4s2. Using the antisymmetric condition for the vertical velocit y in (4.22), ft(s) =2 π2√ 1−4s2/integraldisplay1/2 0ut z(y′)y′/radicalbig 1−4y′2 y′2−s2dy′+A√ 1−4s2.(4.23) The pressure diﬀerential is still has the same form as in (4.8 ). Using the boundary condition in (4.4) that the pressure diﬀerential i s zero aty=−1/2, we have that/integraldisplay1/2 −1/2ft(s)ds= 0. (4.24) Substituting ft(s) from (4.23) into the above equation and swapping order of integration, /integraldisplay1/2 0/integraldisplay1/2 −1/22 π2√ 1−4s2ut z(y′)y′/radicalbig 1−4y′2 y′2−s2dsdy′+/integraldisplay1/2 −1/2A√ 1−4s2ds= 0. (4.25) 55 Using contour integration techniques (App. C.5) on the ﬁrst term, we ﬁnd that the term vanishes and thus we ﬁnd that A= 0. The function f(s) is now ft(s) =2 π2√ 1−4s2/integraldisplay1/2 0ut z(y′)y′/radicalbig 1−4y′2 y′2−s2dy′. (4.26) Using the the expression for the tangential velocity from (4 .11) forz= 0+, ut y|z=0+=−2π π2/radicalbig 1−4y2/integraldisplay1/2 0ut z(y′)y′/radicalbig 1−4y′2 y′2−y2dy′ ≈8 π/radicalbig 1−4y2/integraldisplay1/2 0ut z(y′)y′ /radicalbig 1−4y′2dy′ =  4√ 2 π√1−2y/integraldisplay1/2 0ut z(y′)y′ /radicalbig 1−4y′2dy′, y→/parenleftbig1 2/parenrightbig− 4√ 2 π√1 + 2y/integraldisplay1/2 0ut z(y′)y′ /radicalbig 1−4y′2dy′, y→/parenleftbig −1 2/parenrightbig+ =  Bt √1−2y, y→/parenleftbig1 2/parenrightbig− Bt √1 + 2y, y→/parenleftbig −1 2/parenrightbig+, (4.27) whereBtis a weighted average of the vertical velocity Bt=4√ 2 π/integraldisplay1/2 0ut z(y′)y′ /radicalbig 1−4y′2dy′. (4.28) We see that there is a singular behaviour near the edges of the plate and this will have to be corrected Using the expresion in (4.5) to determine the vertical veloc ity outside the blade, ut z(y) =/integraldisplay1/2 −1/2ft(s) y−sds, y>/vextendsingle/vextendsingle/vextendsingle/vextendsingle1 2/vextendsingle/vextendsingle/vextendsingle/vextendsingle. Substituting (4.26) into the above equation and swapping or der of integra- 56 tion gives ut z(y) =/integraldisplay1/2 0/integraldisplay1/2 −1/22 π2√ 1−4s2ut z(y′) y−sy′/radicalbig 1−4y′2 y′2−s2dsdy′. (4.29) Using contour integration techniques (App. C.6), the verti cal velocity will now have the form ut z(y) =/integraldisplay1/2 0(πi)ut z(y′)2y′ π2/radicalbig 1−4y2/radicalbig 1−4y′2 y′2−y2dy′ ≈ −8i π/radicalbig 1−4y2/integraldisplay1/2 0ut z(y′)y′ /radicalbig 1−4y′2dy′ =  −4√ 2 π√2y−1/integraldisplay1/2 0ut z(y′)y′ /radicalbig 1−4y′2dy′, y→/parenleftbig1 2/parenrightbig− −4i√ 2 π√2y+ 1/integraldisplay1/2 0ut z(y′)y′ /radicalbig 1−4y′2dy′, y→/parenleftbig −1 2/parenrightbig+ =  −Bt √2y−1, y→/parenleftbig1 2/parenrightbig− −Bti√2y+ 1, y→/parenleftbig −1 2/parenrightbig+. The vertical velocity outside the plate needs to be correcte d for the singular behaviour outside the plate. 4.2.2 Correction to the Inviscid Solution As with the previous section, we shall start by looking at the edgey= 1/2. The change of variables to a local coordinate system at the ed ge is the same in the previous section in (4.15). The governing equation is still the same as in the previous chapter and we have the same expressions for uyt,uztand [p]tin (3.25). As usual, the boundary conditions are the same exc ept for the tangential velocity. The tangential velocity on the blade has to cancel the singul ar behaviour obtained in the inviscid solution in (4.27) and outside the b lade it is zero. 57 Thus, the boundary condition for the tangential velocity is given as uyt=  −Bt √1−2y=−Bt √ −2ǫY=−B′t √ −ǫY, Y < 0 0, Y > 0, where we have used the local coordinate system in the previou s section and B′t=Bt/√ 2 whereBis deﬁned in (4.28) We have again that this is the same case as before in Sec. 3.3.2 except withBreplaced by B′t. Thus we can straight away write down the expres- sion for the vertical velocity and pressure diﬀerential in t erms of the original variableybut with the deﬁnition of Btin (4.28). ut z≈B′t √ ǫY=Bt √ 2ǫY=Bt √2y−1(4.30a) [p]t≈2√ 2B′tǫe−iπ/4 √ −ǫY=2√ 2Btǫe−iπ/4 √ −2ǫY=2√ 2Btǫe−iπ/4 √1−2y.(4.30b) As usual, the singular behaviour for the vertical velocity h as been canceled and have a square root singularity for the pressure diﬀerent ial. We can repeat the procedure for the edge y=−1/2 but we shall use a similiar argument in the previous section to obtain the corr ection. Noting that, just as the vertical velocity is antisymmetric about t he origin, so too the pressure diﬀerential is antisymmetric about the origin as well. Therefore, the correction to the pressure diﬀerential at the edge y=−1/2 is the same as in (4.30b) except that it has a minus sign and the term√1−2yis replaced with√1 + 2y. Hence, the total correction to the pressure diﬀerential, [ p]t c, is [p]t c= 2√ 2Bǫe−iπ/4/parenleftbigg1√1−2y−1√1 + 2y/parenrightbigg . 58 4.2.3 Hydrodynamic Function Γt(β) The total pressure diﬀerential is the sum of the inviscid par t and the cor- rection [p]t= 2πi/integraldisplay1/2 yft(s)ds+ 2√ 2Btǫe−iπ/4/parenleftbigg1√1−2y−1√1 + 2y/parenrightbigg .(4.31) It not appropriate to consider the force per unit length on a b lade performing torsional oscillations since the pressure diﬀerential is a ntisymmetric and so the total force per unit length on the blade is zero. Instead, we calculate the moment per unit length of the blade. yz ˆjˆk /Bullet ˆimz=0+ mz=0− Figure 4.3: Moments experienced by the blade. The ˆi-direction is pointing out of the page The moment caused by the force acting on the top plate about th e origin is force on the top plate multiplied by the perpendicular dis tance from the origin: mz=0+=−yFz=0+ since in a right-handed coordinate system, the moment is posi tive if it is acting to rotate in the anti-clockwise direction. The moment caused by the force on the top blade is acting to rotate in the clockwise dir ection, hence the minus sign (see Fig. 4.3). Similiarly, the moment caused by the force 59 on the bottom blade about the origin is mz=0−=yFz=0−. So, the dimensionless net moment per unit area experienced by the blade about the origin is mnet=−yFz. The moment per unit length can be found by integrating the net moment across the length of the blade mﬂuid=−L/integraldisplay1/2 −1/2yFzdy =ωρL2U/integraldisplay1/2 −1/2y[p]tdy , whereFzis the net force in the z-direction. Substituting (4.31) into the above equation and swapping the order of integration, mﬂuid=ωρL2U/bracketleftBigg 2πi/integraldisplay1/2 −1/2/integraldisplays −1/2yft(s)dyds +2√ 2Btǫe−iπ/4/integraldisplay1/2 −1/2y√1−2y−y√1 + 2ydy/bracketrightBigg =ωρL2U/bracketleftBigg πi/integraldisplay1/2 −1/2(s2−1 4)ft(s)ds+ 2√ 2Bǫe−iπ/4/parenleftbigg2√ 3/parenrightbigg/bracketrightBigg =ωρL2U/bracketleftBigg πi/integraldisplay1/2 −1/2s2ft(s)ds+ 2√ 2Btǫe−iπ/4/parenleftbigg2√ 3/parenrightbigg/bracketrightBigg since/integraltext1/2 −1/2ft(s)ds= 0 from (4.24). Substituting the expression for ft(s) from (4.26) and Btfrom (4.28), swapping the order of integration and per- 60 forming contour integration (App. C.4) mﬂuid=ωρL2U/bracketleftBigg πi/integraldisplay1/2 0/integraldisplay1/2 −1/22s2 π2√ 1−4s2y′/radicalbig 1−4y′2 y′2−s2ut z(y′)dsdy′ +16√ 2 3πǫe−iπ/4/integraldisplay1/2 0y′ /radicalbig 1−4y′2ut z(y′)dy′/bracketrightBigg =ωρL2U/bracketleftBigg πi/integraldisplay1/2 0/parenleftBig −π 2/parenrightBig2y′/radicalbig 1−4y′2 π2ut z(y′)dy′ +16√ 2 3πǫe−iπ/4/integraldisplay1/2 0y′ /radicalbig 1−4y′2ut z(y′)dy′/bracketrightBigg =ωρL2U/bracketleftBigg −i/integraldisplay1/2 0y′ut z(y′)/radicalbig 1−4y′2dy′ +16√ 2 3πǫe−iπ/4/integraldisplay1/2 0y′ /radicalbig 1−4y′2ut z(y′)dy′/bracketrightBigg . (4.32) Displacement yz Φ(x) Figure 4.4: Displacement and the function Φ( x) for torsional oscillations We can replace the vertical velocity as the ﬁrst derivative w ith respect to time of the displacement of the blade. Because it is execut ing torsional oscillations, the displacement is deﬁned by the arc traced b y the edges with respect to the z= 0 plane (see Fig. 4.4). Taking the Fourier transform of the ﬁrst derivative with respect to time and unscaling, ut z(y) =−iωL2 UyΦ(x), 61 where Φ(x)1is the angle of the blade with respect to z= 0. Substituting the vertical velocity into (4.32) and performing the integr als which again are elementary, mﬂuid=−ρω2L4/bracketleftBigg/integraldisplay1/2 0y′2/radicalbig 1−4y′2dy′+16√ 2 3πǫeiπ/4/integraldisplay1/2 0y′2 /radicalbig 1−4y′2dy′/bracketrightBigg Φ(x) =−ρω2L4/bracketleftBigg π 128+16√ 2 3πǫeiπ/4π 32/bracketrightBigg Φ(x) =−π 8ρω2L4/bracketleftBigg 1 16+4√ 2 3πǫeiπ/4/bracketrightBigg Φ(x). We can immediately identify from (4.1b) that the hydrodynam ic function for the case of a blade having torsional oscillations is give n by Γn(β) =1 16+4√ 2 3π√βeiπ/4. (4.33) 1Φ(x) has units 1 /Las deﬁned in [5] 62 Chapter 5 Results and Discussion 5.1 Comparison with Exact Computed results Having found the hydrodynamic function for both the cases of normal and torsional oscillation, we are now able to compare results wi th the exact solution that was found by Van Eysden and Sader [6]. The hydro dynamic function consists of a real and imaginary component that dep ends on the Reynolds number, β. The real component is the inertial component of the force exerted by the ﬂuid. This can be thought of as the force n eeded to move a mass of ﬂuid as the blade oscillates. The imaginary com ponent is the dissipative component of the force which acts to dampen t he oscillation of the blade or the drag force. Thus we will compare the numeri cs of the inertial and the dissipative force for a range of values for β. Since the method that was used to calculate the hydrodynamic function in this thesis is based on the assumption that the Reynolds nu mber is large, we shall compare the results for log10β >1. Note that the results that are relevant to this thesis for comparison of the exact numer ics from Van Eysden and Sader’s paper is the κ= 0 column. Using the formulas for the 63 hydrodynamic function Γn(β) and Γt(β) from (4.21) and (4.33), we can cal- culate the real and imaginary parts of the functions for vari ousβand the results are as below Normal Torsional log10βRe(Γn(β))Im(Γn(β))Re(Γt(β))Im(Γt(β)) 1.0 2.61053 1.61053 0.196711 0.134211 1.5 1.90567 0.90567 0.137973 0.0754725 2.0 1.5093 0.509296 0.104941 0.0424413 2.5 1.2864 0.286398 0.0863665 0.0238665 3.0 1.16105 0.161053 0.0759211 0.0134211 3.5 1.09057 0.090567 0.0700473 0.00754725 4.0 1.05093 0.0509296 0.0667441 0.00424413 ∞ 1 - 0.0625 - Table 5.1: Computed real and imaginary components of the hyd rodynamic function Γn(β) and Γt(β) for normal and torsional oscillations respectively as a function of βusing the Wiener-Hopf method As we can see from Table 5.1, the result from using the Wiener-H opf method follows the same trend but does not give the right nume rics as the exact solution. However, for the case when β=∞which corresponds to the inviscid case gives the correct answer when comparing wi th the exact solution. Also, it is the same result for the inviscid case in previous ﬁndings by Sader [15] for the normal oscillations as well as Green and Sader [7] for the torsional case. In the Fig. 5.1, we see that the solution obtained from the Wie ner-Hopf method has a constant slope on the log-log plot. But in all case s except Im(Γt(β)), we see that the slope decreases a bit as we increase the Rey nolds number. This indicates that there are higher order terms inﬂ uencing the results, causing it to decrease the slope but ever so slightl y. In the case 64 1 2 3 4−1.5−1−0.500.5 log10 Relog10 Re(Γf(ω))Hydrodnamic Function for normal modes 1 2 3 4−1.5−1−0.500.5 log10 Relog10 Im(Γf(ω))Hydrodnamic Function for normal modes 1 2 3 4−2.5−2−1.5−1−0.5 log10 Relog10 Re(Γt(ω))Hydrodynamic function for torsional modes 1 2 3 4−2.5−2−1.5−1−0.50 log10 Relog10 Im(Γt(ω))Hydrodynamic function for torsional modes Figure 5.1: The real and imaginary parts of Γn(β) and Γt(β) excluding the inviscid part on a log-log plot. The solid lines are the soluti ons obtained from the Wiener-Hopf method while the dashed lines are the exa ct solutions ofIm(Γt(β)), the slope seem to approach a constant value which might indicate that the higher order terms in the other cases is not aﬀecting this one. We perform a further check on whether the exact solution foll ows the behaviour of the results from the Wiener-Hopf method, namely that the hydrodynamic function is inversely proportional to the√β. If it fails, then it might mean that there is a higher order term which is non-neg ligible that needs to be included in the results of the Wiener-Hopf method. From Fig. 5.2, the exact solution doesn’t seem to be following the tren d that it is inversely proportional to√βsince it does not trace a straight line at −0.5 on the vertical axis. Moreover, it does not seem to follow any pure power of the Reynolds number which does suggest higher order terms ar e at play. From the hydrodynamic function obtained for the case of norm al oscil- 65 123456−0.51−0.5−0.49−0.48−0.47−0.46−0.45−0.44Estimated exponent for Re( Γf(ω))Estimated Power Behaviour of Re for Γf(ω) 123456−0.65−0.6−0.55−0.5−0.45Estimated exponent for Im( Γf(ω))Estimated Power Behaviour of Re for Γf(ω) 123456−0.52−0.5−0.48−0.46−0.44−0.42Estimated exponent for Re( Γt(ω))Estimated Power Behaviour of Re for Γt(ω) 123456−0.8−0.75−0.7−0.65−0.6−0.55−0.5−0.45Estimated exponent for Im( Γf(ω))Estimated Power Behaviour of Re for Γf(ω) Figure 5.2: Estimated power law behaviour for the solutions. The solid lines are the solutions from the Wiener-Hopf method and the dashed l ines are the exact solutions. A straight line indicates a power law behav iour inβ 66 lations, we can compare the damping coeﬃcient obtained here with that of Tuck’s. However, the Reynolds number used by Tuck diﬀers by a factor 4 in the denominator. So, in order to make our result comparab le with Tuck, we need to replace√βwith 2√βin the hydrodynamic function. To get the damping coeﬃcient, one needs to take the imaginary pa rt of the hydrodynamic function since it corresponds to the drag forc e: Im(Γn(β)) =8√ 2 πsin(π 4) =8 π ≈2.55. The damping coeﬃcient also falls short of the result by Tuck w hich again suggest some terms which are signiﬁcant are not accounted fo r in the Wiener- Hopf method. One explanation as to why there may be some neglected terms in the Wiener-Hopf solution is primarily due to the edges. The metho d used to solve for the ﬂuid ﬂow is to calculate the potential ﬂow ﬁrst a nd then using the information about the potential ﬂow, a correction was ob tained to in- clude viscous eﬀects. In literature, this is seen as the lead ing order term in a matched asymptotic expansion between the outer region whe re the ﬂuid is governed by the Euler equation concerning inviscid ﬂow an d the inner region where viscous eﬀect takes place. For an inﬁnite plate oscillating, the outer region consist of a ﬂuid that has zero vorticity due to t he canceling of positive and negative vorticity when oscillating [10]. H ence, it obeys the governing equation for an inviscid ﬂuid. But there is a thin l ayer where vorticity is non-zero which is the region where viscous eﬀect s have pene- trated the ﬂuid. For an inﬁnite plate, the thin boundary laye r is present 67 throughout the entire plate and has a length scale δ∼/radicalbiggµ ωρ But, for a plate that has an edge, the ﬂuid around the region of the edge ”sees” the plate as inﬁnite in one direction and thus we have n o natural length scale. If we deﬁne a local Reynolds number about the ed ge βy=y2ρω µ whereyis the distance from the edge and the corresponding length sc ale δ(y)∼y/radicalbiggµ ωρ we see that if we get closer to the edge, the local Reynolds num ber becomes small and thus the ﬂuid will be governed by Stokes ﬂow. This me ans that the assumption of a thin boundary layer breaks down near the e dges because the viscous penetration depth far away from the edges are con sidered large compared to the distance from the edge. So, we expect another contribution from the edge apart from the viscous eﬀect due to the thin boun dary layer far away from the edge that needs to be included in our solutio n. We also expect higher order terms which do not follow a pure po wer behaviour but also possibly logarithmic terms. Performing a regular per- turbation expansion for small βshows that the solution does not satisfy the condition that the ﬂuid is stationary far away from the plate . This is com- monly known in literature as Stokes paradox. For a 2-D ﬂuid ﬂow , it can be shown that the solution diverges logarithmically as we move oﬀ to inﬁnity from the edge. To correct for this, we necessarily need to hav e logarithmic 68 Region withthin boundarylayer Regionof Stokes ﬂowEdgeRegionof inviscid ﬂow Figure 5.3: Regions of inviscid ﬂow, Stokes ﬂow and thin boun dary layers occur with respect to an edge of a blade terms in the expansion and subsequently in the additional co ntribution to the solution. Further evidence of a logarithmic type behavi our comes from the asymptotics of the hydrodynamic function calculated by Sader for nor- mal oscillations and by Green and Sader for torsional oscill ations. Asβ→0, the function behaves as Γn(β)∼ −4i βln(−i√iβ) Γt(β)∼i β−3 8ln(−i/radicalbig iβ) asβ→0. So if we approach the edge, the hydrodynamic function shou ld be able to pick up the same behaviour as quoted above. Therefo re we must have logarithmic terms in the hydrodynamic function to be ab le to capture this. From the expressions of Γnand Γtabove, it is interesting to note that Stokes paradox doesn’t occur for the Im(Γt) sinceIm(βΓt) does not have anymore terms involving βwhere length scales are present. This links with the results in the corresponding ﬁgure which seem to approac h a constant slope and so we might not expect to have higher order terms cor responding to the existence of Stokes paradox for this case. 69 To properly account for the behaviour at the edges, one must d o a matched asymptotic expansion to match the region of invisci d ﬂow with the region with boundary layers and at the same time, match it with the region near the edges (see Fig. 5.3). This has been suggested by Rosenhead [14] and has been done for a few cases like ﬂow past a sphere, a c ylinder and a ﬂat plate with one edge albeit for β→0. So, any problem that has an edge in it requires one to go through these 3 steps to get the leading order behaviour. The approach taken from Atkinson and de Lar a has only completed 2 of the 3 steps mentioned above and from our result s, it seems that the third and ﬁnal step is also crucial to complete the an alysis. In their original paper, their problem is a wide clamped plate that ha s a free edge. It is likely that this edge would need to be accounted for to gi ve accurate results as well which they have not done. The problem of the vi brating disk that they have studied also have an edge and we might need to co rrect for the edges. 70 Chapter 6 Conclusions In this thesis, the main aim was to understand the Wiener-Hopf method and its application to solve the ﬂuid ﬂow problems of an inﬁni te nature. The problem studied by Atkinson and de Lara where the method w as ﬁrst applied was for an inﬁnitely wide ﬂat plate clamped on one end and free to vibrate on the other end. The method was applied to obtain the correction to the inviscid solution for the pressure diﬀerential and th e force exerted on the plate was calculated. The ﬂuid force is coupled with th e equation of motion for a thin plate. Numerous typographical errors an d some minor inconsistencies were encountered. In particular, the use o f Green’s function to solve for the equation of motion of the plate was not approp riate. Instead, the method of eigenfunction expansions was used. The Wiener-Hopf method has been utilized to solve similiar vi brational problems to which the exact solution has been obtained by Van Eysden and Sader. The goal is to extract asymptotics for large Reynolds number using this method and to compare it with the numerics from the exact solution. The method is not able to capture the full behaviour in the cas es of normal and torsional oscillations. The assumption that at the regi on near the edges 71 of the blade, the correction behaves as 1 /√βhas been suggested as the main culprit. This is because the boundary layer at the edges are n o longer thin to allow a square root behaviour. It is very likely that it is b ecause of failure of thin boundary layers at the edges that may give a non-neglig ible, higher order contributions to the asymptotics. This draws doubt on the validity of the solution obtained by A tkinson and de Lara for the wide clamped plate. The plate still have an edge in the problem and it is likely that the solution would not be able to accurately capture the true underlying behaviour. The authors have fai led to realise that they haved modeled this problem as a pure 2-D ﬂow with an ed ge and thus will need to worry about Stokes’ paradox at the edge. Nev ertheless, their method was able to give a ﬁrst approximation to the visc ous eﬀects to correct for the inviscid ﬂow. This has the behaviour for the h igh Reynolds number for regions suﬃciently far away from the edges primar ily around the origin. The same problem would also arise since they have als o solved the case of a vibrating disk which has a circular edge as well. 6.1 Future research An area for future research is to perform the ﬁnal part of the m atched asymptotic expansion for the problem as suggested by Rosenh ead. That is, to match the solution obtained in this thesis with the region near the edge where it is governed by Stokes ﬂow to obtain the term in the lea ding order behaviour. 72 Appendix A Product Decomposition of H(ζ) We have that the function H(ζ) can be decomposed into plus and minus functions in terms of integrals over the real axis: ln(H+(ζ)) =1 2πi/integraldisplay∞ −∞ln(H(z)) z−ζdz, Im (ζ)>0 (A.1a) ln(H−(ζ)) =−1 2πi/integraldisplay∞ −∞ln(H(z)) z−ζdz Im (ζ)<0. (A.1b) whereH(ζ) is deﬁned as H(ζ) =√ζ+√ζ−+/radicalbig ζ2−i 2/radicalbig ζ2−i and√ζ+and√ζ−are deﬁned in (3.32). 73 A.1 Large ζexpansion of H+(ζ) Consider the integral/contintegraldisplay Γ1ln(H(z)) z−ζdz . There are branch points at ±iδand±e−iπ/4corresponding to the functions √ζ+√ζ−and/radicalbig ζ2−irespectively. There are no zeros for the function H(ζ) hence there are no branch points for the logarithm. So we cho ose the branch cuts such that they join between iδande−iπ/4together and similiarly between −iδand−eiπ/4. For the upper branch cut, the square root function will take the positive square root and for the lower branch cu t, the negative square root will be taken. The contour Γ 1is chosen to go along the real axis and closing in the lower half plane. This is because H+(ζ) is analytic in the upper half plane meaning that Im(ζ)>0 and so we have a pole in the upper half plane with z=ζ. × Γ1C1 C2−eiπ/4ζ −iδiδeiπ/4 Re(z)Im(z) Figure A.1: Contour used for simplifying H+(ζ) Because there are no poles but just a branch cut in the lower ha lf plane, we can analytically deform the contour Γ 1until we enclose the branch cut in the lower half plane. The contribution from the semicircl e,CR, is zero since we have that H(ζ)→1 and therefore ln( H(ζ))→0 as|ζ| → ∞ . The 74 contribution from circling the branch points can be shown to be zero. So, we have/contintegraldisplay Γln(H(z)) z−ζdz=/integraldisplay C1+C2ln(H(z)) z−ζdz . (A.2) We can take the limit as δ→0 and the RHS of the equation reduces to /integraldisplay C1+C2ln/parenleftBigg 1 2/radicalbigg z2 z2−i+1 2/parenrightBigg 1 z−ζdz =/integraldisplay−eiπ/4 0ln/parenleftbiggiz 2√ i−z2+1 2/parenrightbigg1 z−ζdz −/integraldisplay−eiπ/4 0ln/parenleftbigg−iz 2√ z2−i+1 2/parenrightbigg1 z−ζdz . (A.3) Performing a change of variables z=−reiπ/4and with the branch cut having the argument −3π/4<arg(z)<5π/4, we have /integraldisplay1 0ln/parenleftBigg −ireiπ/4+eiπ/4√ 1−r2 2eiπ/4√ 1−r2/parenrightBigg eiπ/4 reiπ/4+ζdr −/integraldisplay1 0ln/parenleftBigg reiπ/4+eiπ/4√ 1−r2 2eiπ/4√ 1−r2/parenrightBigg eiπ/4 reiπ/4+ζdr =/integraldisplay1 0ln/parenleftBigg −ir+√ 1−r2 2√ 1−r2/parenrightBigg eiπ/4 reiπ/4+ζdr −/integraldisplay1 0ln/parenleftBigg ir+√ 1−r2 2√ 1−r2/parenrightBigg eiπ/4 reiπ/4+ζdr =/integraldisplay1 0eiπ/4ln(−ir+√ 1−r2) reiπ/4+ζdr −/integraldisplay1 0eiπ/4ln(ir+√ 1−r2) reiπ/4+ζdr =−2ieiπ/4/integraldisplay1 0arcsin(r) reiπ/4+ζdz , (A.4) where from the second to the third line we can cancel the denom inator in the logarithms since they are no longer multivalued in the domain on integration and from the third to the last line, we used the id entity that 75 iarcsin(z) = ln(iz+√ 1−z2). Going back to the integral in (A.2), we have that /contintegraldisplay Γln(H(z)) z−ζdz=−2ieiπ/4/integraldisplay1 0arcsin(r) reiπ/4+ζdz /integraldisplay∞ −∞ln(H(z)) z−ζdz=−2ieiπ/4/integraldisplay1 0arcsin(r) reiπ/4+ζdz . Substituting the expression for the integral above into (A. 1a), we have ln(H+(ζ)) =−eiπ/4 π/integraldisplay1 0arcsin(r) reiπ/4+ζdz . (A.5) Expanding for |ζ| → ∞ we get ln(H+(ζ)) =−eiπ/4 πζ/integraldisplay1 0arcsin(r)dz+O/parenleftbigg1 ζ2/parenrightbigg ⇒H+(ζ)||ζ|→∞≈exp/parenleftBigg −π−2 2πeiπ/4 ζ/parenrightBigg ≈1−π−2 2πeiπ/4 ζ+··· (A.6) A.2 Large ζexpansion of H−(ζ) We still consider the same contour integral in the previous s ection except with the contour Γ 2but all its properties are the same since for H−(ζ) to be analytic in the lower half plane, there is a pole in that regio n. So, we still have the same form as (A.2) except that the contours would be e nclosing the upper branch cut: /contintegraldisplay Γ2ln(H(z)) z−ζdz=/integraldisplay C1+C2ln(H(z)) z−ζdz . (A.7) Takingδ→0, performing a change of variables z=reiπ/4and noting that the argument of the branch cut is −7π/4<arg(z)<π/4 and following 76 ×Γ1 C1C2 −eiπ/4ζ−iδiδeiπ/4 Re(z)Im(z) Figure A.2: Contour for simplifying H−(ζ) the same arguments as before, we get the RHS of (A.7) to be 2ieiπ/4/integraldisplay1 0arcsin(r) reiπ/4−ζdr . Substituting the above equation in (A.7) and substituting t he result into (A.1b), we have ln(H−(ζ)) =−eiπ/4 π/integraldisplay1 0arcsin(r) reiπ/4−ζdr . (A.8) Expanding for |ζ| → ∞ ln(H−(ζ)) =eiπ/4 πζ/integraldisplay1 0arcsin(r)dr+O/parenleftbigg1 ζ2/parenrightbigg ⇒H−(ζ)||ζ|→∞≈exp/parenleftBigg π−2 2πeiπ/4 ζ/parenrightBigg ≈1 +π−2 2πeiπ/4 ζ+··· (A.9) 77 Appendix B Sum decomposition of S(ζ) We have the function S(ζ) that can be decomposed to S+(ζ) andS−(ζ) to be S+(ζ) =1 2πi/integraldisplay∞ −∞S(z) z−ζdz Im (ζ)>0 (B.1a) S−(ζ) =−1 2πi/integraldisplay∞ −∞S(z) z−ζdz Im (ζ)<0, (B.1b) where S(ζ) = √ζ−√ζ+/radicalBigg ζ+eiπ/4 ζ−eiπ/4−1 H+(ζ). B.1 Large ζexpansion of S−(ζ) We consider the contour integral /contintegraldisplay Γ1S(z) z−ζdz =/contintegraldisplay Γ1√z−√z+/radicalBigg z+eiπ/4 z−eiπ/4H+(z) z−ζdz−/contintegraldisplay Γ1H+(z) z−ζdz =/contintegraldisplay Γ1√z−√z+/radicalBigg z+eiπ/4 z−eiπ/4H+(z) z−ζdz 78 sinceS−(ζ) is analytic in the lower half plane, so there is a pole in the l ower half plane with z=ζ. Because the contour Γ 1goes along the real axis and closes in the upper half plane and H+(ζ) is analytic in the upper half plane, the second integral vanishes by Cauchy’s theorem. The branc h points and the contour is the same as in Fig. A.2 in Appendix A.2. The cont ribution of the semi circle goes to zero since we have S(ζ)→0 as|ζ| → ∞ . So, /contintegraldisplay Γ1√z−√z+/radicalBigg z+eiπ/4 z−eiπ/4H+(z) z−ζdz=/integraldisplay C1+C2√z−√z+/radicalBigg z+eiπ/4 z−eiπ/4H+(z) z−ζdz . (B.2) Takingδ→0 and noting that the argument of the branch cut is −7π/4< arg(z)<π/4, the RHS of the equation now becomes 2i/integraldisplayeiπ/4 0/radicalBigg z+eiπ/4 eiπ/4−zH+(z) z−ζdz. Performing a change of variables z=reiπ/4, this becomes 2ieiπ/4/integraldisplay1 0/radicalbigg 1 +r 1−rH+(reiπ/4) reiπ/4−ζdr (B.3) Substituting (B.3) into (B.2) we get /integraldisplay∞ −∞S(z) z−ζdz= 2ieiπ/4/integraldisplay1 0/radicalbigg 1 +r 1−rH+(reiπ/4) reiπ/4−ζdr . (B.4) Substituting the equation above into (B.1b), S−(ζ) =−eiπ/4 π/integraldisplay1 0/radicalbigg 1 +r 1−rH+(reiπ/4) reiπ/4−ζdr , (B.5) where H+(reiπ/4) = exp/parenleftbigg −1 π/integraldisplay1 0arcsin(r′) r′+rdr′/parenrightbigg . 79 Expanding for |ζ| → ∞ for (B.5) S−(ζ)||ζ|→∞=eiπ/4 πζ/integraldisplay1 0/radicalbigg 1 +r 1−rH+(reiπ/4)dr +i πζ2/integraldisplay1 0r/radicalbigg 1 +r 1−rH+(reiπ/4)dr+O/parenleftbigg1 ζ3/parenrightbigg . Atkinson and de Lara have claimed that the following integra ls give /integraldisplay1 0/radicalbigg 1 +r 1−rH+(reiπ/4)dr=π√ 2, /integraldisplay1 0r/radicalbigg 1 +r 1−rH+(reiπ/4)dr=π 2.(B.6) It is not certain how they have evaluated these rather diﬃcul t integrals. Nonetheless, numerical integration has shown both to be acc urate to 8 deci- mal places. In the analysis, we use the values of the integral s as above since this greatly simpliﬁes the asymptotics. B.2 Large ζexpansion of S+(ζ) By using the usual method for simplifying S−(ζ), we will have to evaluate more complicated integrals. But since we are interested in t he expansion of the function as |ζ| → ∞ , we just subtract the expansion for S+(ζ) from the expansion of S(ζ) for large ζ. The expansion for S(ζ) for large ζis calculated to be S(ζ)||ζ|→∞=eiπ/4 ζ+i πζ2+··· Therefore,S+(ζ) is found to be S+(ζ)||ζ|→∞=S(ζ)−S−(ζ) =eiπ/4/parenleftbigg 1−1√ 2/parenrightbigg1 ζ+i/parenleftbigg1 π−1 2/parenrightbigg1 ζ2. (B.7) 80 Appendix C Contour integrals The main method that is being used here is calculating the res idue at inﬁnity. The way this is calculated is Res(f(z))|z=∞= lim z→∞zf(z) With this, we can make use of Cauchy residue theorem. ××× Re(z)Im(z) polesa bC1 C2CR Figure C.1: Contours for all integrals here. The contributi on going from the cut toCRcancel each other and the distance apart can be made arbitrar ily small 81 In all the integrals considered here in this thesis, there wi ll be 2 branch points on the real axis and the branch cut is chosen to join the two points through the origin. Also, we will have poles on the cut and may have poles outside as well. Because the branch cut is due to a square root function, the contribution from traversing along an arbitrarily small se micircle around the pole on the upper cut will cancel the contribution from trave rsing along an arbitrarily small semicircle around the pole on the lower cu t. Contributions from circling the branch points are also shown to be zero. So, with the contour given in the Fig. C.1, the only things that matter are poles outside the branch cut and the contour integral CR. /integraldisplay C1+C2f(z)dz= 2πi/summationdisplay Res(f(z))−/integraldisplay CRf(z)dz 2i/integraldisplay C1g(z)dz=πi/bracketleftBig/summationdisplay Res(f(z)) +Res(f(z))z→∞/bracketrightBig /integraldisplayb ag(z)dz=π/bracketleftBig/summationdisplay Res(f(z)) + lim z→∞zf(z)/bracketrightBig (C.1) since travelling along the top cut we can take out a factor of ( −1) from the square root in f(z) and it will take the value ig(z) while along the bottom cut, it will take the value −ig(z). The minus sign is removed in the second line is due to the orientation of the contour CRwith respect to the point at inﬁnity. C.1 Integral in Equation (3.18) We have the functions f(z) =1 (y−z)(y′−z)/radicalbiggz z−1, g(z) =1 (y−z)(y′−z)/radicalbiggz 1−z, 82 where there is a pole outside the cut at z=y >1 and a pole at 0 < z= y′<1 in the cut. We have that lim |z|→∞zf(z) = lim |z|→∞z (y−z)(y′−z)/radicalbiggz z−1 = 0. Therefore the contour integral in (C.1) now becomes /integraldisplay1 01 (y−s)(y′−s)/radicalbiggs 1−sds=π/parenleftbigg1 y′−y/radicalbiggy 1−y/parenrightbigg . C.2 Integral in Equation (3.37) The functions in the equation are f(z) =1 y′−z/radicalbiggz z−1, g(z) =1 y′−z/radicalbiggz 1−z, where we only have a pole at 0 <z=y′<1. We have that lim |z|→∞zf(z) = lim |z|→∞z y′−z/radicalbiggz z−1 =−1. Therefore the contour integral becomes /integraldisplay1 01 y′−z/radicalbiggz 1−zdz=−π . 83 C.3 Integral in Equation (4.14) The functions are f(z) =z√ 4z2−11 (y′2−z2)(y−z), g(z) =z√ 1−4z21 (y′2−z2)(y−z). There are poles at z=±y′on the cut where −1/2<y′<1/2 and a pole at z=y>1. The branch points are ±1/2. We have that lim |z|→∞zf(z) = lim |z|→∞z2 √ 4z2−11 (y′2−z2)(y−z) = 0. The contour integral now becomes /integraldisplay1/2 −1/2s√ 1−4s21 (y′2−s2)(y−s)ds=πiRes (f(z))|z=y =πi/parenleftBigg y/radicalbig 4y2−11 y′2−y2/parenrightBigg . C.4 Integral in Equation (4.19) and(4.32) The functions are f(z) =z2 √ 4z2−11 y′2−z2dz , g(z) =z2 √ 1−4z21 y′2−z2dz . 84 We have poles at ±y′which is on the cut and no poles outside. We have that lim |z|→∞zf(z) = lim |z|→∞z3 √ 4z2−11 y′2−z2dz =−1 2. The contour integral now reduces to /integraldisplay1/2 −1/2s2 √ 1−4s21 y′2−s2ds=−π 2. C.5 Integral in Equation (4.25) The functions are f(z) =1√ 4z2−11 y′2−z2, g(z) =1√ 1−4z21 y′2−z2. There are poles at ±y′on the cut but none outside. 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