579 Internat. J. Math. & Math. Sci. VOL. 13 NO. 3 (1990) 579-590 THREE DIMENSIONAL GREEN’S FUNCTION FOR SHIP MOTION AT FORWARD SPEED MATIUR RAHMAN Department of Applied Mmthematics Technical University of Nova Scotia P.O. BOX I000 Halifax, Nova Scotia Canada B3J 2X4 (Received December 12, 1988) ABSTRACT. The Green’s function formulation for ship motion at forward speed contains double integrals with singularities in the path of integrations with respect to the wave number. In this the study, double have integrals analysis provides an efficient way of computing been replaced by single It has been found that this integrals with the use of complex exponential integrals. the resistance wave for three dimensional potential problem of ship motion with forward speed. KEY WORDS AND PHRASES. Ship motion, Green’s function, Hydrodynamics, Wave Resistance and Wave Responses. I. INTRODUCTION. In ship hydrodynamics, Green’s functions play a very important role in predicting the wave resistance, wave induced responses at zero forward speed, and the motions of a vessel advancing in waves. The Green’s function formulation for ship motions at forward speed is the most difficult part of the problem partly because it contains double integrals and partly because of the presence of the singularities in the path of integrations with respect to the wave numbers. been paid to evaluate Nowadays, considerable interest has the three dimensional Green’s function for ship motions at forward speed. Many authors including Haskind [I] and Havelock [2] have expressed the Green’s function having a constant forward speed as a double integral. This form of Green’s function is not suitable for numerical analysis because the detailed computation of the double integral is very expensive. Therefore, in the present study, we have replaced the double integral by a single integral (see Wu and Taylor [3]) involving a complex exponential integral, and it is found that it is more efficient to calculate the Green’s function numerically. 2. A FORM OF THE GREEN’S FUNCTION. U which is moving at constant forward speed Consider the coordinate system oxyz (see surface free upwards from the mean along the x axis and z measured positive Figure I). 580 M. RAHMAN Z Figure 1" The Coodinate System It is assumed that a ship is travelling at a constant forward speed U along the Ox direction and oscillating at a frequency Ialtone [4] - Wehausen and as in the form of e 1960 have shown that the Green’s function which satisfies the exact in free surface condition can be written as =I G(x y,z;a,b c) +_K 7 7/2 d6 L 2 /(x-a)2 + (y-b)2 ’ dO o o F(0,k)dk + 2_ 7 7/2 F(8,k)dk d8 L + (z-c) 2 Rankine source located at (a,b,c) 2 2 /(x-a) 2 + (y-b) + (z+c) Image about the mean free surface R g 7 F(6,k)dk where R 1+2g R k e k[(z+c) + i(x-a)cSe]cos[k(y-b)sine| gk-( w+kUcos e) 2 (x,y,z) is the field point and (a,b,c) is the source distribution point. parameters in Equation (2.1) are defined by o cos T (a,b,-c) acceleration due to gravity F(,k) where at (-) if J is called the Strouhal number g The contours L and L 2 are defined as follows: The other < (2.3) t (2.4) if T -I (2.2) 581 SPEED THREE DIMENSIONAL GREEN’S FUNCTION FOR SHIP MOTION AT FORWARD kl k2 k3 k4 and two singular points in the L There are two singular points in L Equation (2.1). 2 integral of These singular points can be obtained as follows: cose (2.5) + /i-4 COS8 (2.6) /1-4 2 /gkl /gk3 /g--72K cos-----m /gk--4 The alternative forms of these singularities k I, k 2, k and k can be written as 4 3 k for /2 e 2 k 2 cosS) : /1-4 cos8 v 2 e 2 2 (I w; and where v g It should be noted here that the ranges indicated. 0 ( e (2.8) T2cos kl=k 3 and k2=k 4. These singularities are real in It is, however, worth mentioning here that in the range y, the singularities k and k 2 become complex quantities and are either given by +/- /gk2’ /gkl i/4T cos 2T cos0 (2.9) or k2 ’kl 2z cos0 2 +/- i/4 cos-I z2cos2 0 (2.10) v Thus the integrand in the integral F(8,k)dk]de O O contains no real singular points in the path of integration from 0 to 3. (R). EVALUATION OF INTEGRALS. The Green’s function given in equation (2. I) is difficult to numerically because as we have seen in the previous section, the contours L have singularities at k l,k2,k 3 and k 4. 2 This difficulty can be overcome by introducing the Cauchy Principal Value (PV) integrals. k.i o- F integrate and L k2 ’k > M. RAHMAN 582 The first contour integral along the path L G 2-K L can be rewritten as 712 F(0,k)dk d0 L F( 0,k)dk 0, equation (3.1) can be written as When z 2_g GL1 where f dO y the evaluate f o (V.V.) f F( 0,k)dk o (P.V.) To 7/2 F(0,k)dk k k 2 f f fy dO {+ F( O,k)dk F(O,k)dk (3.2) 0 (3.3) deformationsandf the along integral F(0,k) in terms of its singularities. F(O,k) 7/2 kl- k2-e f+ f + f 0 k1+e k2+e Lim integral 2g / we decompose We write exp{k[(z+c) + i(x-a)cosO]} cos(k(y-b)sin0) g II-4 zcos O [Z_k Z_f the (3.4) which can be put in the following compact form kl F(8,k) 2g,/1-4"rcosO k2 [[2kl k-k2] [exp(kXl) + exp(kx2)] (3.5) where xl= w+ (z+c) + iw+, x 2 (z+c) + i w. (x-a)cosO + (y-b)sin0, w_ (x-a)cosO- (y-b)sln O We know that (z+c)0 so we can redefine x x {(z+c) iw+}, x and x {(z+c) 2 2 (3.6) as follows (3.7) iw_}. Thus, equation (3.5) can be rewritten as F(0,k) 2g,/I-4 z cos {exp(-kXl) + exp(-kx2) [kl o k2 {exp(-kXl) + exp(-kx2) k-k2 k-kl (3.8) The integration along the deformatlons.and fIn equation (3.2) can be obtained according to the residue theorem. f+ f 2g,/I-4 zcos O Thus F(8,k)dk {kl(exp(-klX I) Thus, equation (3.9) reduces to + exp(-klX2)) + k2(exp(-k2x I) + exp(-k2x2))} (3.9) THREE DIMENSIONAL GREEN’S FUNCTION FOR SHIP MOTION AT FORWARD SPEED =2__K 12 / d0 Gel / (P.V.) /2 i ,/I-4 os F(e,k)dk 0 y + 583 e {kl(e-klXl + e -k IX2) + k 2 (e-k 2Xl + e-k2x2)} de (3.10) In a similar manner the second contour integral along path L 2 equation (2.1) can be obtained GL 2 2g /2f de F(0,k)dk 0 {k3(k3Xl ,/1-4 cos e /2 f (P.V.) k4(e-k4Xl e-k3x2) + + e-k4x2)} (3.11) Therefore the Greens function in equation (2.1) can be rewritten as follows: a,b,c) Gl(x,y,z; R R f d0 0f F(0,k)dk + 0 /2 + 2g f+ de (P.V.) x /2 /2 i ,/I-4 cos F( e,k)dk {k (e -klXl + e -k x2) + y -k {k3(e /2 J e f 0 3 Xl+ e -k 3 x2) k4(e k2(e -k x 4 + e -k -k4 x2 2 Xl+ e-k2x2)} de (3.12) de} or + G + G + G + iG 3 5 2 4 G G (3.13) where G1 G2 ’ =2g f R/2 Y G3 de 0 f F(e,k)dk 0 /2 G5 f +f /2 ,/1-4 cos O ,/I-4 cos e {k G f + /2f e 4 f F(e,k)dk (P.V.) 0 y {kl(e-klXl+ e-klX2) + 3(e-k3xl+ e-k3x2) k k2(e -k 2 Xl+ e -k2x 2 )} 4(e-k4xl+ e-k4x2) de de (3.14) There are two cases to be considered. Case I y= 0 if and in this case G 0. 3 < Therefore, equation (3.13) becomes G G + G 2 + G 4 + iG 5 (3.15) 584 M. RAHMAN Case II y cos -1 if 4T %- and in this case equation (3.13) becomes G + G G 2 + G 3 + G 4 + iG (3.16) 5 and G are highly oscillatory at large values of k 3 4 In order to calculate because of the imaginary argument of the exponential function. The double in G integrals them numerically, at minimum computer cost, these integrals must be reduced to single [5], and Inglis and Price [6]. We shall treat Case I first and evaluate the Cauchy Principal Value (P.V.) integral in G 4. integrals as suggested by Shen and Farell The term G 4 72 G4 of the Green function can be written as dO y {l-4Icos 8 where d0 7/2 I-4Tcos8 k-kl 0 13 ( for 7/2 k2exp (-kx (P.V.) dk, k-k2 0 12 dk, (P.V.) 0 14 (P’V’) 16 (P.V.) 18 (P.V.) (15+16-17-18) klexp(-kx 2) k_kl k exp(-kx (P.V.) I for y + (II +12_13_14) k exp -kx 2 2 0 k_k2 dk (3.17) (3.18) dk 7/2 O 15 (P.V.) 17 (P.V.) O k2exp (-kx i) k-k3 0 0 k4exp (-kx k_k4 dk, dk k exp -kx 3 0 0 k_k3 k4exp(-kx 2 k_k4 dk (3.19) dk ,. To obtain analytic expressions of these integrals, we consider a contour in the K k + ik’ plane as suggested by Smith et al [7] (see Figure 2) and later used by Chen et al [8]. We impose the condition that __[-KXll 1 on the o integration path 5 which makes an angle a with the real axis, argument of the exponential can be made real along the ray. Therefore, we get I [-(k + ik’) m (Iz+cl-lw+)] which simplifies to yield and a= tan -I k’ tan -I w+ Z+C i- 0 so that the THREE DIMENSIONAL GREEN’S FUNCTION FOR SHIP MOTION AT FORWARD SPEED 585 K--k+ ik 4 Figure 2" A closed contour for Thus with this value of , + w+> 0 2 w+ 0 -kV Also, we have kV K --k+ ik’ =V+c[-iw+ Integrating along the contour shown in Figure 2, I {klexp(-klXl) 2i (i) klexp(-klXl) klexp(-kx) klexp(-klX 1) klexp(-kx i dk dk k-k 5 k-kl 5 Along the path 5 0 5 kl exp(-kV) (kV) 2 klXl d(kV) exp(-u) k 0 u k Ix du exp(-klX 1) E l(-klx 1) k where EI(_Z) Therefore, for w+ > O, I k e--{- dt, [arg(z) < w exponential integral. -z exp(-klx I) {El-(klX l) It is to be noted here that for + ri} w+< 0 the contour will be as follows: (3.20) 586 M. RAHMAN K=k+ I /4 k Figure 3: Closed contour for w, <0 Thus I Also, for w+ k exp(-klXl) {El(-klXl) (3.21) ’rl} 0, klexp(-klXl) {El(-klX I) I (3.22) + i} - which is obtained using the following definitions (see Abramowitz and Stegun [9] 1965, p. 228). E such that for I w+ e (P V l(-x+lO) -E--dt i, E1(-x-lO) 0 klexp(-k2Xl) {El(-klXl)+ z’l] Similarly, we can calculate the other integrals. we get for w+ I and for w+ < ) 0 or I m k exp(-klX I) 0 or I (-k x m I k (-klX) I) < ) (P.V.) dt + J wle Thus summing up the situation, 0 {E l(-klx I) + i} 0 lexp(-k Ix I) {Z l(-k Ix I) -1) Similarly, Im(-klX2) > 0 klexp(-klX2) [El(-klX2) I], for Im(-klX2) < 0 13 k2exp(-k2x I)[E l(-k2x I) I], for Im(-k2x I) > 0 k2exp(-k2x I)[E l(-k2x I) + I], for Im(-k2x I) < 0 12 klexp(-klX2) [El(-klX2) + I], for (3.23) THREE DIMENSIONAL GREEN’S FUNCTION FOR SHIP MOTION AT FORWARD SPEED 14 k2exp(-k2x2)[El(-k2x2) k2exp(-k2x 2) [El(-k2x 2) k3exp(-k3x I) [E l(-k2x I) k3exp(-k3x I) [E l(-k3x I) k3exp(-k3x 2) [E l(-k3x2) k3exp(-k3x 2) [El(-k3x 2) k4exp(-k4x I) [E l(-k4x I) k4exp(-k4x I) [E l(-k4x I) k4exp(-k4x2) [E l(-k4x2) k4exp(-k4x 4) [El(-k4x 2) 15 16 17 18 in G Now adding the terms 4 and G 5 Im(-kzX2) for Im(-k2x2) for Im(-k3Xl for Im(-k3Xl for Im(-k3x2) for Im(-k3x2) for Im(-k4Xl for Im(-k4Xl for Im(-k4x2) for Im(-k4x2) i], for + i], + i], i], + i], i], + i], i], + i], i], given by the 587 0 ) < 0 0 < 0 ) 0 < 0 ) 0 < 0 ) 0 < 0 equations (3.24) (3.17) and (3.14), respectively, we obtain 12 G 4 {I + + iG 5 + .I_w i /2 ,/I-4cos8 [(I 5 + + i 12 13 14 16 17 18 + i (Iii + (131 + 132 112 + + 121 122)}d8 (3.25) 142)}d8 141 where lij k i exp(-kixj) j 1,2,3,4; j 1,2. Thus, if we combine the corresponding integrands of G 4 + First Integral 1 J2 k lexp(-k Ix I) 2 klexp(-klX I__ 2 2 klexp(-klX 2) ( __l - E ,/1-4 cos8 k exp(-k x Second Integral we obtain for I ,/1-4 cos 8 1 -f [E (-k x )+2 ri ]d 8, iGs, 2 ,/I-4 zcos8 ,/1-4 zcos 8 (-klX )d 8, [E (-k Ix 2 )+2 z-i ]d e, E (-klX2)d 8, m _(-klx I) for I (-k x m for I m I) < __(-klX 2) for I (-k x m > 0 0 )0 2) < 0 588 M. RAHMAN _I Third Integral 2 k2exp(-k2x I) (-k2x J2, k2exp(-k2x I) El(-k2Xl)d --[-E ,/I-4 Tcos )+2 ri ]d O, for Im(-k2x I) ) 0 0, for Im(-k2x I) < 0 (-k2x2)+2 ]d 0, for Im(-k2x2) < 0 (-k2x 2)d O, for Im(-k2x2) < 0 0 I -- ,/I-4 Tcos /2 k 2exp(-k 2x 2) Fourth Integral _..1 Fifth Integral 1 Sixth Integral z’2 k2exp(-k2x2) i /2 ’ k exp -k x 3 3 [E 1 w 1 (-k3x ,/1-4 cos0 i/2 k3exp(-k3x2) k 3exp(-k 3x 2) [E Tcos 8 i k3exp(-k3x2) i k4exp(-k4Xl E 1 for Im(-k3x 2) < 0 for Im(-k3x 2) < 0 (-k4x )-2 rl ]d 0, [E i/2 k4exp(-k4x2 [-E (-k x )-2 ri ]d O, 4 2 Tcos )0 (-k3x 2 )+2 i ]d 0, i/2 k4exp(-k4x4) ,/1-4 (-k3x) 0 [-E k4exp (-k4x2) m Im(-k3x I) < /2 ,/I-4 zcosO ,/I-4 cos O for I for (-k3x2)d O, (-k4x ,/1-4 zcos8 Eighth Integral )+2 ri ]d O, )d 0, (-k3x E ,/I-4 zcos0 /2 ,/I-4 zcos0 Seventh Integral= E ,/1-4 cos 0 /2 ,/I-4 1 [-E ,/1-4 cos O y )d 0, (-k4x2)d 8, E O for I __(-k4x I) for I m __(-k4x I) < m (-k4x 2) for I for )0 0 )0 Im(-k4x 2) < 0 (3.26) Therefore, for Case I, we can evaluate the Green’s function given in equation (3.15). To evaluate the Green’s function for Case II given in equation (3.16), we need to express the G 3 term in exponential integrals as given below: k G 3 0 0 ek[(z+c) gk + i(x-a)cosO]cos[k(y_b)sln 8 dk (w + kUcosS) THREE DIMENSIONAL GREEN’S FUNCTION FOR SHIP MOTION AT FORWARD SPEED i where / 0 kl Jl k and k and k 2 + (J exp(-kx l)dk, I- (3.27) J3- J4 )d@ J2 4 cos 0-I kl i J2 exp(-kx 2)dk -Z-f k 2 589 2 (3.28) are the complex roots of (m + kUcos0) gk 2 0. Using the contour in Figure 2, it can be easily shown that 0 2ff] Im(-klXl) Im(-klXl) < 0 2i] Im(-klX2) Im(-klX2) < Im(-k2Xl) ) Im(-k2Xl < 0 Jl klexp(-klXl)El(-klXl klexp(-klXl) [El(-klXl) J2 klexp(-klX2) El(-klX2) klexp(-klx2) [E 1(-klx2) J3 k2exp(-k2Xl) El(-k2Xl) k2exp(-k2Xl) [E l(-k2xl) -2ri] J4 k2exp(-k2x2) [El(-k2x2) -k x + 2J. Im(-k2x2) 0 0 0 > 0 E (-k x Thus, with this information, we can evaluate the Green’s function for Case II from equation (3.16). 4. RESULTS AND CONCLUSIONS. The present form of Green’s function is equivalent to that used by Wu and Taylor, but in a different form. Chen et al [8]. However, The terms GI, G and G are all identical to those used by 3 2 and G terms in the present study, we have combined the G to correspond with the form of Wu and Taylor. 5 4 It appears that our studies are quite similar to those of Wu and Taylor, and Chen et al. The double integral arising in the evaluation of Green’s function has been replaced by a single integral with the use of complex exponential integrals. The present work has provided an alternative form but similar to that of Wu and Taylor, and has been found to be efficient for the analysis of the three dimensional potential problem of ship motion with forward speed. M. RAHMAN 590 ACKNOWLEDGEMENTS. This work has been performed under Contract No. OSC87-00549-(010) with the Defence Research Establishment Atlantic while the author was on Sabbatical Leave from the Technical University of Nova Scotia. 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