Sloshing problem in a half-plane covered by a dock with two equal gaps

Transcription

1 Sloshing prolem in a half-plane covered y a dock with two equal gaps O. V. Motygin N. G. Kuznetsov Institute of Prolems in Mech Engineering Russian Academy of Sciences St.Petersurg, Russia

2 STATEMENT OF THE PROBLEM y x z Ansatz: u(x, y, z, t) = u(x, y) sin mz cos ωt Prolems for symmetric u (+) and antisymmetric u ( ) mode u (±) = m 2 u (±), (x, y) Q x u (+) = 0, u ( ) = 0, x = 0, y < 0 y u (±) = 0, on docks y u (±) ν (±) u (±) = 0, (x, y) F u (±) 2 dx dy + u (±) 2 dx < Q F 1 y x + 1 F Q

3 REFERENCES For the case m = 0, = 0 A.M.J. Davis (1970) P. Henrici, B.A. Troesch, L. Wuytack (1970) J.W. Miles (1972) B.A. Troesch, H.R. Troesch (1972) B.A. Troesch, H.R. Troesch (1974) D.W. Fox, J.R. Kuttler (1983)

4 LOCAL ASYMPTOTICS NEAR DOCK TIPS Polar coordinates (ρ, θ), (z = ρe iθ, z = x + iy): y u = 0 (, 0) θ y u = νu (ρ, θ) By Wigley (1964), Nazarov, Plamenevsky (1994) u P (ρ, log ρ), where P is a polynomial function Local asymptotics for the potential u (±) (ρ, θ) = c (±){ π/ν (±) + ρ ( cos θ log ρ + (π θ) sin θ )} + d (±) ρ cos θ + ψ (±) (ρ, θ) and for a complex conjugate stream function ϕ (±) (ρ, θ) = (±) c (±) ρ [ sin θ log ρ + (θ π) cos θ ] d (±) ρ sin θ + φ (±) (ρ, θ) with some coefficients (±), c (±) and d (±) As ρ 0, ψ (±), φ (±) = O(ρ 1+δ ), ψ (±) = φ (±) = O(ρ δ ) where δ > 0

5 ANTISYMMETRIC MODES. INTEGRAL EQUATION. Completing Miles (1972) scheme (ased on Fourier transform) For m = 0 u ( ) (x) = ν( ) π +1 u ( ) (ξ) log x + ξ x ξ dξ, x (, + 1) In the fluid: u ( ) (x, y) = 1 2π For m 0 +1 u ( ) (ξ) log (x + ξ)2 + y 2 (x ξ) 2 + y 2 dξ u ( ) (x) = ν( ) π +1 Restoring the potential in the fluid y u ( ) (x, y) = 1 2π +1 (K 0 is the modified Bessel function) [ K0 ( m(x + ξ) ) K0 ( m x ξ )] u ( ) (ξ) dξ, ( u ( ) (ξ) [K 0 m (x + ξ)2 + y 2) x (, + 1) K 0 ( m (x ξ)2 + y 2)] dξ Consequences: (i) For any 0, there exist a sequence of eigenvalues 0 < 1 < 2... ν n ( )..., ν n ( ) () as n and u ( ) 1 (x; ) 0 for x [0, 1] (ii) The values of ν n ( ) () are monotonically decaying in

6 NUMERICAL RESULTS, ANTISYMMETRIC MODES Solid lines = 0.05, dashed lines = 1.0 m Solid lines m = 0, dashed lines ( ) m = 3.25, (- - -) m = 5.0

7 SYMMETRIC MODES, INTEGRAL EQUATION, m = 0 Fourier transform u (+) (x, y; ) = 2 π 0 f(k) e ky cos kx dk Applying y and the inverse transform for y = 0 k f(k) = 0 y u (+) (ξ, 0; ) cos kξ dξ = ν (+) () +1 u (+) (ξ, 0; ) cos kξ dξ Then, u (+) (x, 0; ) = ν (+) () 2 π 0 k 1 cos kx dk +1 u (+) (ξ, 0; ) [ cos kξ 1 ] dξ (since +1 Integral equation u (+) (x, 0) dx = 0 y zero flux condition) +1 [ ] u (+) (x) = ν (+) π 1 log(x + ξ) + log x ξ u (+) (ξ) dξ, The integral equation is not correct! (it has no solutions orthogonal to constant) Assuming the contrary, restoring the potential in the fluid y u (+) (x, y) = 1 2π +1 x (, + 1) u (+) (ξ) log [ (x + ξ) 2 + y 2)( (x ξ) 2 + y 2)] dξ u (+) (x, y) = O( z 2 ) as z. Such decaying is not allowed.

8 NON-EXISTENCE OF SYMMETRIC MODES DECAYING AT INFINITY (m = 0) If Q ( u (+) 2 + u (+) 2) dx dy <, then u (+) 0 in Q The proof follows M. McIver (1999) (the 14th IWWWFB) Introduce the stream function ϕ complex conjugate to u = u (+) By the assumption u, ϕ 0 as z 0 y x ϕ = 0 ϕ = 0 ϕ = 0 Q Green s formula for harmonic functions u 2 ϕ 2 and x {( u 2 ϕ 2) ( n x x n u 2 ϕ 2)} ds = 0 Q + oundary conditions 0 u 2 dy + +1 x y ( u 2 ϕ 2) dx = 0 ( By Cauchy Riemann conditions y u 2 ϕ 2) ( ) = x 2uϕ and ( (+ oundary conditions) in the gap 2uϕ = 2 ν (+) y uϕ = 1 ) ν (+) x ϕ 2 0 u 2 dy = 1 ν (+) +1 ( ) x ϕ 2 dx = 1 ν [ϕ(x, x=+1 (+) 0)]2 = 0 x= u (+) (0, y) = x u (+) (0, y) = 0 u (+) 0 in Q

9 SYMMETRIC MODES, CORRECT INTEGRAL EQUATION Following Davis (1970), introduce { W (z; ξ) = 1 8(z ξ) log π (1 2z)(1 2ξ) 1 + 2z ( ) 1 + 2z log 2 1 2z 1 + 2ξ ( ) } 1 + 2ξ log ξ πi(z + ξ) For one gap ( 1, 1) all modes can e found from u(x, 0) = ν 1 1 u(ξ, 0) Re W (x, ξ) dξ The symmetric two-gap prolem is equivalent to the spectral prolem +1 u (+) (x) = ν (+) G(x, 0; ξ) u (+) (ξ) dξ, x (, + 1) Green s function: G(x, y; ξ) = 1 2 Re{ W (z ; ξ ) + W (z 1 2 ; ξ 1 2 ) where + W (z ; ξ ) + W (z 1 2 ; ξ 1 2 )} g 0 () g 0 () = 1 [ 2 2 log(2) + 2(1 + 2 ) log[2(1 + )] (1 + 2) 2 log(1 + 2) ] 2π and +1 G(x, 0; ξ) dx = For y = 0 y [ G(x, y, ξ) + 1 π log z ξ ] = { +1 G(x, 0; ξ) dξ = 0 0 for x (0, ) ( + 1, + ) 1 for x (, + 1)

10 Restoring the potential in the fluid y u (+) (x, y) = +1 u (+) (ξ) G(x, y; ξ) dξ, (x, y) Q Hence, u (+) (x, y) = c + O( z 2 ) as z, c = const 0 Another way to the integral equation For m 0 the kernel y Fourier transform G 0 (x, ξ) = π [ ( ) ( )] 1 K 0 m(x + ξ) + K0 m x ξ will not deliver solutions satisfying zero flux condition y u (+) (x, 0) dx = ν (+) u (+) (x, 0) dx = 0 The correct kernel orthogonal to constant is as follows G(x, ξ) = G 0 (x, ξ) f(x) f(ξ) + C 0 where f(x) = +1 G 0 (x, ξ) dξ, C 0 = +1 f(ξ) dξ Integral equation +1 u (+) (x) = ν (+) G(x, 0; ξ) u (+) (ξ) dξ, x (, + 1)

11 NUMERICAL SCHEME, SYMMETRIC MODES (m = 0) Change the coordinate system y 1 x to seek solutions to the integral equation in the form u (+) (x) = ν (+) 1 1 u (+) (x) = G(x, 0; ξ) u (+) (ξ) dξ, x ( 1, 1) (2k + 1) 1 2 ck P k (x) k=1 where P k are the Legendre functions of the first kind Since the system {P k } is orthogonal, the integral equation turns to 2c m + ν (2n + 1) 1 1 { 2 (2m + 1) 2 Imn + I π mn} cn = 0, n, m = 1, 2,... where k=1 I mn = I mn = P m (x) P n (ξ) log x ξ dx dξ, P m (x) P n (ξ) log(x + ξ ) dx dξ By Davis (1970) I mn = 0 when m n is odd; for even m n Besides, I mn = 2 2n + 1 I mn = (m + n)(2 + m + n) [ (m n) 2 1 ] P m (ξ) { Q n+1 ( ξ 4 2) Q n 1 ( ξ 4 2) } dξ where Q k are the Legendre functions of the second kind

12 NUMERICAL RESULTS AND ASYMPTOTICS m = ν ,..., 5 (dashed lines) and ν (+) 1,..., ν (+) 5 (solid lines) Asymptotics for large lim ν( ) n+1() = lim ν n (+) () = 2λ n where λ n are eigenvalues of the prolem with one gap ( 1, 1) λ 2n+1 = n+1(0), λ 2n = ν (+) n (0), n = 0, 1, 2,... For m = 0, ν (+) 0 = 0 corresponds to trivial mode u (+) 0 = const; For m 0 the mode u (+) 0 is not trivial and ν (+) 0 0 For m = 0, as, π 1 1 w ( ) 1 (x) = ν( ) 1 π = log(2) [ O ( 1/ log 2 ) (w ( ) 1 0, + O (1/ log ), ] + x log x + (1 x) log(1 x) 1 0 w ( ) 1 (x) dx = 1)

13 DERIVATIVE OF EIGENVALUES AS FUNCTIONS OF SPACING BETWEEN GAPS ɛ = > 0 : u n, ν n : u n, ν n ɛ y x F 0 F Q By Green s formula over Q, 0 = {u n y u n u n } y u n dx + F F 0 Antisymmetric modes { } u n x u n u n y u n dy By the condition on F and on F 0, [ ν ν ] F u n u n dx = F 0 u n x u n dy Since u n(x, y) = ɛ x u n(θ(y), y), 0 θ(y) ɛ, when x [0, ɛ], [ ν ν ] u n (x, 0) u n(x, 0) dx = ɛ x u n(θ(y), y) x u n (0, y) dy F F 0 As ɛ 0, this gives d n d = 0 +1 x u ( ) n (0, y) 2 dy ( ) u (x, 0) 2 dx n

14 Symmetric modes dν (+) n d = 0 y [ u (+) n (0, y) 2 + m 2 (+) u n (0, y) 2] dy +1 (+) u (x, 0) 2 dx n Interesting to compare with the Rayleigh quotient [ (±) ν n (±) R u = 2 n (x, y) 2 + m 2 (±) u n (x, y) 2] dx dy 2 +1 (±) u n (x, 0) 2 dx Example of another configuration (symmetric in x) for which the method is applicale y x

15 SUMMARY: Sloshing prolem is considered for an inviscid incomressile heavy fluid that occupies a channel with rigid walls and is covered y a rigid dock with two equal, strip-like apertures. The prolem is reduced to spectral prolems for integral equations (different for symmetric and antisymmetric modes). It is proved that the antisymmetric (symmetric) sloshing eigenvalues are monotonically decreasing (increasing) functions of spacing etween gaps and formulae for their derivatives are otained. Simplicity of eigenvalues is shown and their limit properties when spacing tends to zero and infinity are studied. Numerical scheme is developed and computations for eigenvalues are made. Asymptotics of solutions to the prolem near dock tips are otained (m = 0). Non-existence of symmetric modes decaying at infinity is proved (m = 0).