Office of Naval Research Department of the Navy Contract ~onr-220(28)
VISCOUS EFFECT ON SURFACE WAVES GENERATED BY STEADY DISTURBANCES
T. Yao-tsu Wu and R. E. Messick
Reproduction in whole or in part is permitted for any purpose of the UnitedStates Government
Engineering Division California Institute of Technology Pasadena, California
Report No. 85-8 Approved February, 1958 M. S. Plesset Abstract
A linearized theory is applied here to investigate the viscous effect on water waves generated and maintained by a system of ex- ternal disturbances which is distributed over the free surface of an otherwise uniform flow. The flow is taken to be in the steady state
configuration. The analysis is carried out to yield the asymptotic expressions for the surface wave when the Reynolds number of the flow is either large or small. 1. Introduction
The problem of the decay of a train of free-running simple waves over a water surface due to the viscous effect has been treated by Lamb 1 2 and Basset . Since in the general practice water waves are usually generated by some localized external disturbances, this consideration sug- gested to the present authors the investigation of the viscous attenuation, in space, of the water waves generated and maintained by a system of ex- ternal normal stress and shearing stress distributed over the free surface of an otherwise uniform flow. The scope of the present investigation will be limited to the case when the external forcing functions are independent of the time and the motion has been maintained for a long duration such that only the steady state solution of the problem will be of interest. Since the viscous attenuation turns out to depend significantly on the wave length, e specially when the wave length is small, both the gravitdtional and sur - face tension effect are here taken into account.
It is well known that by neglecting the viscous effect, one is led in general to the so-called singular perturbation problem; because in doing so, the order of the differential equation governing the flow momentum is re- duced by one and consequently certain boundary condition or conditions have to be relaxed. In the case of a fluid flow passing over a solid body, the condition that the velocity vanishes at the solid surface in the presence of viscosity is replaced by one that the velocity is tangential to the surface when the viscosity is neglected. In the case of a free surface flow, such as the present problem in question, the condition that states the balance of the viscous part of the total stress at the free surface must be omitted with the omission of the viscosity. One of the direct consequences that arises in such problems of steady, inviscid water waves is that the mathematical solu- tion becomes, strictly speaking, indeterminate. More precisely, it is found to be not possible in general to obtain the unique mathematical solution with
the desired physical properties by imp0 sing only the boundedness conditions at infinity. Within the framework of the potential theory for the steady sur- face wave problems, some stronger radiation conditions are actually nece s-
sary; for example, in the absence of the surface tension effect, the steady
surface wave is required to exist only on the downstream side of the external
disturbance. The necessity of imposing such radiation conditions has been 3 a topic of some dispute. To avoid such conditions, Rayleigh introduced a
small dissipative force, proportional to the relative flow velocity; thereby
the irrotationality property of the flow is preserved. Recently, it has been 6 pointed out by different authors ((3reen4, stoker5, De Prima and Wu ) that
the steady state solution can be obtained uniquely if the solution is interpreted
as the limit of a corresponding initial value problem, as the time tends to
infinity. Incidently, it was revealed in Ref. 6 that the coefficient of
Rayleigh' s force is actually a time- limiting factor. Now, with the additional
viscous effect taken into account, it is shown in the present work that the
mathematical solution becomes determinate, and bears automatically the
desired physical properties (as required by the radiation conditions) and
also exhibits the actual attenuation that takes place.
The outline of this paper is as follows. By assuming that the result-
ing motion is a small perturbation of the basic uniform flow, the same linear -
ized theory as used by ~amb'is applied to treat the present problem.
Oseen's equation is used to describe the momentum conservation; and the flow
is further decomposed into two parts, one being irrotational and the other solenoidal. After the integral representation of the surface wave is obtained, by using the Fourier transform method, the analysis is carried out further to yield the asymptotic solutions In closed form for two special cases when the Reynolds number is large or small. The Reynolds number of this prob- lem is defined as Re = (U/v)(r/gf'2 where U is the free stream velocity, v the kinematic viscosity of the fluid, g the gravitational constant and cr the ratio of the surface tension to the fluid density. For large values of
Re (small viscous effect) and U> cm ( = (4grf/4 ) , the result shows that a train of gravity waves exists downstream, and capillary waves upstream of the disturbance. Both of these waves are attenuated by the viscosity; the smaller the wave length, the faster is the space rate of attenuation. In the critical case of U = c the viscosity is found to play a significant role to m' assure the existence of the steady solution. For small values of Re (large viscous effect) and moderate values of U, it is shown that a train of vis- cous surface waves of large wave length propagates only on the downstream side, with its amplitude and wave length greatly affected by the viscosity.
2. Formulation of the Problem
The problem in question concerns the water waves generated by a pressure distribution and a viscous shearing stress applied on the free sur- face of a deep water which has a uniform free stream velocity U. We shall restrict our selves to the two-dimensional motion in an xy-plane, with the y-axis pointing vertically upward and the x-axis coinciding with the undis- turbed free surface, directed along U. The liquid medium is taken to be incompressible, of constant density p, and viscous, of kinematic viscosity v. The resulting motion is assumed to be a small perturbation so that a linear theory can be applied. Furthermore, we are interested only in the steady state solution; the time-dependence of the flow will thereby be omitted.
It will be seen later that, in contrast with the nonviscous case, the viscous steady state problem is uniquely determinate and hence no radiation condi- tion need be imposed.
With the total flow velocity denoted by U + $= ( U+u, v), the per- turbation velocity < of the incompressible flow satisfies the continuity e qua tio n
4 div q=u +v = 0, 1x1' m, YCO, (1) x Y where the subscripts denote the partial derivatives. By assuming that
1;) 1;) << U, the Navier-Stokes equations for the steady flow in the region 1x1 c oo, y< 0, may be linearized to give the Oseen equations
where p is the pressure, g the gravitational constant, and v2 the
Laplacian. Equations (1) - (3) are the basic equations governing the flow motion. On the displaced free surface, y = ~(x),the kinematic boundary condition that the flow velocity is tangential to the displacement ~(x),may be expressed, after linearization,. as
When both the gravity and surface tension are considered, the linearized boundary conditions for the stresses at the frke surface are (cf. Ref. 1) : In the above, P denotes the external pressure (or the normal stress acting in the y-direction with the sign reversed, to be more general), Q the ex- ternal shearing stress acting in the x-direction, p= p v, and (pcr) is the surface tension of the liquid-air interface. Equation (5) expresses the con- dition that the normal stress component, (2pvy- P), has a jump across the free surface by an amount pu'Cxx due to the surface tension effect; while
Eq. (6) states the condition that the shearing stress is continuous at the free surface. The arbitrary functions ~(x)and ~(x)are assumed to be abso- lutely integrable:
We further require that
'C, Zx be absolutely integrable; and u, v be absolutely integrable with respect to x for any fixed y50 . (8)
This condition is suggested by the argument that when P and Q are sub- ject to condition (7), the resulting disturbance will be damped out at infinity by the viscosity. This completes the statement of the problem.
It is known1' that the linearized flow (c p) governed by Eqs. (1)- (3) can be uniquely decomposed into two parts, one is irrotational (Gl, p) and the other is solenoidal (G, 0) .
--+ + -C q = TI+G, curl ql =O, div q2 = 0, such that
It follows from (9) that Tl has a velocity potential ,
and < in this two-dimensional flow is conveniently expressed in terms of a stream function $ ,
Since also satisfies (1), cfl satisfies the Lapkce equation 71
(P,+'PYY = O for ~xleoo,yc~. (13)
Substituting (12) in the second equation of (10) and integrating, we obtain
Furthermore, the first equation of (10) can be integrated to yield
Now p may be eliminated from (5) and (15). The boundary conditions then become
Urx=q), -Ix on y=~; (16)
Uvx+ gZ - 0 cxx- 2v (rPxx+ Y,) = - ~(S/Pon y=~; (17)
v(z(P,+ lyy-"1 = Q(X)/P on y=~. (18)
Our problem now becomes to first solve for v , $ and Z from (13) and (14) with conditions (16)-(18); the velocity is then obtained from (9), (1 1) and (l2), and the pressure p, from (15).
Next we introduce the Fourier transform of f(x) and its inversion, de- fined as
Y -ikx ikx f(k) = e f(x) dx , f(x) = - e f(k) dk .
-00 -00
The existence of the Fourier transform of , $, < etc. is actually implied by conditions (7) and (8). The transform of (13) and (14) under condition
(8) are
they have the solution of the form
2 1/2 provided (k +iUk/v) is defined to have its real part positive for k real.
The arbitrary functions A(k) and ~(k)can be determined from the trans- form of Eqs. (16)-(18), that is, after elimination of 'g , from
rc* With A and B determined from the above equations, 'g is then given by the
transform of (16) , Substituting the value of A and B in the above equation and applying the inverse transform, one obtains, after some manipulation, the following integral representation of c (x) ,
00 2 1/2 , 1 ~(x)= - - ikx ~(k)+~-~[(2vk+i~)k/~kl-2v(k+iUk/v) ) (k) dk . 2 22 2 1/2 I, (gtrk ) -1kl (u- ~ivk)~-4vk (k +iuk/v)
In the remaining part of this paper, we shall confine ourselves primarily to
the discussion of c .
First, it is convenient to define the fundamental solutions of <(x) :
12 Z(x) = H&x) when P(x) = p U B(x), Q(x) = 0 ; (21)
12 G(x) = HQ(x) when (X) = 0 Q(x) = Tp U a(x) ,
where b(x) denotes the Dirac delta function. Thus, Hp(x) and HQ(x)
are the surface displacement due to respectively a concentrated normal pres-
sure and a concentrated shearing stress at the origin. Since the problem is
linear, the solution of G(x) for arbitrary ~(x)and ~(x)can be deduced
by super position:
N 12 The integral representation of Hp is obtained simply by setting P(k)= ~p U
N and ~(k)= 0 in (20). In terms of the following nondimensional quantities,
xl=k x, kt = k/km , , u = u/c,, a = v km/U, (234 m 0
where Hp(x) is expressed, after the @me is dropped for kt, as
2 2 2 1/2 with f(k; uo, a) = (k t 1) - 2uo k(1- 2i a k)': 8u0 a k2 [ a k (a k t i)] (25)
Similarly,
In (23), cm is the minimum phase velocity of a train of simple harmonic surface waves in a nonviscous medium and k the corresponding wave m ' 1 number ; - is thus a characteristic length of this problem. The quantity km a may be regarded as the inverse of the Reynold number Re= U/(V km) ; it provides a measure of the relative importance between the viscous effect and the inertia effect. When the viscosity is neglected, a=0, the de- nominator of the integrand in (24) and (26) becomes
which has two simple zeros on the positive real k-axis at R and K for u0> 1 (or Us cm), or a double zero at k=1 for uo = 1, or two complex conjugate zeros for uo< 1. Hence, with a = 0 and u > 1, the problem, 0- strictly speaking, is indeterminate. The steady nonviscous problem may be made determinate either by imposing a proper radiation condition at 3 x=f oo, or by adopting some artifice such as that due to Rayleigh , or by considering anew the corresponding initial value problem4* 5*6. However, with the viscosity of the medium taken into account, a>O, the integrand of
(24) and (26) can be shown to have no singularity on the positive real k-axis
(the path of integration). Since for k>O, the imaginary part of f(k; uo , a)
can be written
the quantity inside the curly bracket decreases monotonically from unity at
k=0 to 1/2 at k=m. Furthermore, the real part of f(k; uo, a) does not
vanish at k=0. This proves the above statement. Therefore, the con-
sideration of the viscous effect not only will exhibit the physical attenuation
of the surface wave that takes place, but also makes the problem mathemati-
cally determinate.
3. Analytic Behavior of the Integral Representation
In order to facilitate the evaluation of integrals (24) and (26), we in-
vestigate first the analytic behavior of their integrands with uo and a re-
garded as two positive parameters. The following discussion will be limited
to that of (24), while the discussion of (26) follows almost in parallel. The
function f(k; uo, a) of (25) will be abbreviated simply as f(k) whenever it
is unnecessary to mention its dependence on u or a. 0
If k is taken to be a complex variable, f(k) has two branch points
at k = 0 and k = - i/a. Wi th the introducti~nof a branch cut connecting
these two points along the imaginary k-axis, f(k) is then an analytic function of k in the cut plane. As the theorem of residues will be applied to
evaluate integrals (24) and (26), we proceed to find the poles of the integrand, or the zeros of f(k). We next show that f(k) has exactly two simple zeros in the cutk-plane for allpositive, realvaluesof u and a, provided that 0 4 112 2 2a is different from (uo t 1) + u0 .
To show this, it is convenient first to remove the branch cut by the
transformation
which maps the entire cut k-plane conformally into the region ( t 12 1 of
the complex t-plane. Then
4 2 3 2 2 f(k(t)) = (16a2t3f1 { t (t- i) t 16a t t 4i auo (t- i12(t3 t it t t - i)
It is thus obvious that f has a triple pole at t=0 and five zeros in the
entire t-plane. Now on the unit circle 1 t 1 = 1, let t = eiQ. then
Hence on t = eiQ, f = 0 has two particular solutions for a> 0 :
2 2 4 1/2 Q = 0 with 4a t 4a uo - 1 = 0 (or 2a = (uo t 1) - ut) ; (314
2 4 112 2 Q = n with 4u2-4auo - 1 =0 (or 20 = (uo t 1) t u0 ) .
The relationships between a and uo shown in the above parentheses are
taken to keep a positive, real: As uo increases from zero to infinity,
a of (31a) decreases from 1/2 to zero, whilst a of (31b) increases from 1/2 to infinity. By expanding f near the point t =+ 1, one finds that the above two particular solutions are two simple zeros of f. In the original cut k-plane, these two particular simple zeros of f lie on the two sides of the cut
i 4 2 at k=--to with 2a = (uot 1) -uo 2a , (324
i 4 1/2 2 and at k=--0 with 2a=(uotl) tu 2a o . (32b)
The physical significance of these particular solutions of f =O is as yet not
clear to the authors. Their contribution to the final solution, however, will
be made explicit later for the special cases of interest.
In the general case when a and u are not restricted by the relations 0 in (31), f then has no zero on the unit circle ( t ( = 1, or on the cut in the
k-plane. Furthermore, Eq. (30) shows that on t = eio , f is a periodic
function of 0 with period 2rr, and Im f is non-negative with a> 0. Hence
1 1 df 1 f. 3 dt = arg f = 0 -2s 10.0
8 which, from the theory of functions , shows that inside I t 1 = 1, the number
of zeros of f is equal to the number of poles of f, a zero or a pole of
order m being counted m times. But the only pole of f (see Eq. 29) is
at t=0 and is of order 3, hence the number of f inside I t( = 1 is also
3. Since f is known to have five zeros in the entire t-plane, this leaves
f to have two zeros outside I t ( = 1. Therefore, it follows from the con-
formal transformation that f has exactly two zeros in the entire cut
k-plane provided a and uo are not restricted by the relations in (31).
This completes our proof. -b Having known the number of zeros of f(k), we can easily determine
these zeros asymptotically for small or large values of a as follows.
f 3. 1. Small values of a.
When a is taken to be small, the two zeros of f(k), say kl and k2, will lie near their corresponding nonviscous value YG and K (see Eq. 27) because k 1 and k2 are seen to be continuous functions of a. The ex- pansion of f(k; u ,a), obtained from (25), for a small suggests that k 0 1 and k2 may be expanded as
Upon substitution of these expansions into that of f(k) = 0, we find
where IC and rC2 are given by (27). Thus for u >1 and a small, 0 1 lies in the first quadrant whilst k2, in the fourth quadrant of the k-plane.
The above expansion also holds valid for uoc 1 ; it breaks down, however,
as uo approaches unity. At u = 1, we have the different expansion: 0 so kl and k2 still lie separately in the first and fourth quadrant of the k-plane.
For a given small a, there exists a large uo such that the parti- cular relation in (32a) is satisfied. In that case, f(k) has an additional - 1 zero at k = -i(2a) . But since this zero of f(k) lies on the imaginary axis at a large distance from the origin for a << 1, its contribution to the motion can be seen to be rapidly damped 6ut away from the origin and will thus be neglected in this case of small a.
3. 2. A critical value of a.
Suppose that f(k) has a simple zero at k=a t ib (and hence a simple
pole of the integrand of Eq. 24). If the theorem of residues is applied with an appropriate construction of the integration contour to enclose this pole,
the contribution from the residue at this pole will therefore contain a term exp(-)bx'l tiax) so that the real part a of the pole gives the wave number and the imaginary part b gives the attenuation of the surface wave. The
surface ceases to have the wavy form when a vanishes; this condition will be referred to as the critical condition.
On the positive imaginary t-axis, we let t = iq , q>0. Then from (29)
3 3 2- Now tq2-qt 1 =q t (q-1) tq>O for "$0, so that Im f = 0 at q = 1 only.
But since Re ff0 at r) = 1 with ado, f has no zero on the positive
imaginary t -axis. On the negative imaginary t -axis, let t = -iq , q l0 ;
the n 3 2 The factor (q .-r) -q- 1) in the imaginary part of f can be shown to have only one real zero at
for which Re f also vanishes provided
We denote this zero of f by
t2 = - (1. 8393) i with a = 1.0957 = ac say , (35) which is independent of the value of u . Transforming back to the k-plane 0 by (28), we may then assert that f is free from zeros on the positive imaginary k-axis, and has only one zero on the negative imaginary k-axis at
k =-i for a=a =1.0957andforallreal u>0. 2 C 0 (36)
The above is the only critical value of a for which the zero of f is ac purely imaginary.
3.3 Large values of a.
For ( t 1 > 1 and a>> 1, the leading terms in the expansion of f 2 2 obtained from (29) start with f 2 1 tiu t (4a)-l This suggests that the 0 . two zeros of f outside I t I = 1 may be expanded in the form
1 tl- ii- a1/2ein/4ta3tb3a-1'2tc a - t... 3 , 0
- 1 2h28/4 ta4tb4a-1/2tc4a t... . 16
Substituting these expressions in (29) and setting f = 0, we find
In order to have the above asymptotic expansions valid for large a and at the same time to retain I t ll 1 1, I t21 > 1, uo must be bounded away - 1 2 from zero and must not be large, or more precisely, O(a ) < uo < a . When uo2 = O(aml) and a 21 1, the first and the last term in the curly bracket of (29) must be considered together; in this manner one obtains
2 valid for P z lauo = O(l), a>>~l. On the other hand, when uo = 0(a) and a>> 1, the last term in the curly bracket of (29) becomes predominant. 3 2 Now this term has a double zero at t = i and the factor (t + it + t -i) can be shown to have two zeros inside I t 1 = 1 and the third zero at t = - (1.8393)i (see Eq. 35). By expanding f about t = i and - (1.8393)i. one finds the two zeros of f outside I t 1 = 1 given by
which hold valid for y u /a of the order unity or greater, and a 1. 0 >>
It may be noted that all the above diffesent expansions become invalid for u = O(al'). In this range of u the last two terms in the curly 0 0 bracket of (29) are of the same order of magnitude and must be considered together; conse~uentlythe evaluation of the two zeros of f outside I t 1 = 1 becomes in this case so complicdted that it does not seem worthwhile to get the explicit determination. However, since the two zeros of f, tl(uo,a) and t2(uo, a), are seen to be continuous functions of u and a, the 0 qualitative behavior of tl and t2 (and hence the behavior of the solution E ) when u = and a >> 1 may always be obtained by "interpolatingu 0 O(b/') their behavior at u = O(1) and u = O(a). It may also be pointed out that 0 0 in this range of u = O(a , a particular solution of f = 0 exists 0 (see Eq. 31b):
t =-1 for ~~=at~(a-~)and a>>l. 3 0 (40)
Finally, transforming the above results back to the original k-plane
by (28), we obtain, for a>> 1, the zeros of f(k) given by:
t ~(a-~), (41)
valid for P 5 4a uf of the order unity or smaller;
and for y 1 u /a of the order unity or greater, 0 As U--. 0, then uo+ 0 and a- m such that p= 4au:--r 0, hence from Eq. (4 1) kl- i and k2 - - i, which are the corresponding nonviscous solution. Thus, we can foresee that the viscous effect becomes insignifi- cant at small values of U. Physically this is because the variation in velocity becomes everywhere too small for the viscosity to be effective.
. By summing up the results (33), (34), (36), (41)-(43), we obtain the loci of the two zeros k and k2 of f(k) for some specific ranges of u 1 0 and a, as depicted in Fig. 1. The curves in the upper half plane are the
loci of kl(uo,a) and those in the lower half plane, the loci of k2(uo, a).
As a- 0, kl and k approach their nonviscous value 1( and 2 e2 given by (27). When a increases, k has its real part decreasing and 1 converges toward the origin (except when U+ 0), as a- ao, the loci of
being bounded within the first quadrant; whilst k2 moves across the kl point k = - i at a = 1.0957 for all values of uo into the third quadrant
and then approaches the origin (again except when U-- 0) as a- m.
4. Solution for Small Values of a (~ar~eReynolds Number)
The integral (24) will now be calculated by applying the theorem of
residues. For x > 0, we cons~ructa closed contour r consisting of the
original path along the real k-axis from k = 0 to k = R, a circular arc of
large radius ) kt = R in the first quadrant and a return along the imagi-
nary axis from k=iR back to the origin. Then F encloses one simple
pole at k=kl, given by (33) of (34) for OKa< the contribution along the circular arc vanishes. Hence, application of the k - plane Figure I. theorem of residues to (24) yields, for x>O , where the integration is performed along the imaginary axis, and Res. (kl) denotes the residue of the integrand at k=kl, given by ik x' 1 ikxl - Res. (kl) = lim (k - kl) e /f(k) = e [ df(kl)/dk] . (444 k+ kl The integral ~(x)converges even when a = 0, hence to obtain the first approximation of L(x), the a appearing in its integrand may be neglected for 0 where K , K are given by (27) and ei(z) denotes the exponential integral 9 From the known asymptotic expansions of ei(z) for I z 1 large or small , 20 one readily deduces that for IdlxJ>>1 , which holds valid for all u; and for 1 lC2xI << 1 , 2 u 0 4 -2 22 12 2 (x (u- 1 log u x t O(x , a) for uo> 1 , - -2 0 2 U - 1/2 1-2(1-u:)m cos-'u2-- o 2012u xt~(x 2,a) for uO 11 2 2 -- ~xtO(x ,a) for u -1. TI 0 Since L(x)-+O like x-' as x-tm, L(x) merely represents the local elevation which is important only in a neighborhood of the origin. By making use of (25), (33), (34) in calculating Res. (k ), we finally obtain for 1 x>O, O 2 2 2 u 4a uo I( XI 0 ~inr(~x~+~(x~)tO(a)for uo>l, 2 u 2 = exp [-( Jq- 8% a) XI cos u xl t ~(xf)tO(r) for uo< 1 , -,* 0 (I-uO 41 (48) for u = 1 , 0 in which L(xl) is given by (45) and may be approximated as shown in (46). (47) - For x< 0, we take the closed contour r to be the boundary of the fourth quadrant of the k-plane, with the branch cut on the imaginary axis lying just outside of r . It was shLwn in the lhst section that for 0 < a << 1, I? encloses only one simple pole at k=k given by (33) or (34). Hence, 2' applying again the theorem of residues, we obtain u 0 for x proportional to exp(x1 /2a) for x<0, which is negligibly small since a< sult will then be accurate up to O(a) (since this is the lowest order that a appears in the integrand). Carrying out the details as the previous case for x >0, we obtain here for x < 0, 0 < a << 1 , 2 u o 4au:~,"xl Hp(x) = - 7 7sin /C 2xt + L(JxlJ)+ O(a) for u0 > 1, (uo - 1) expr (l] 2 u 2 -*2 - -*2 exp[(J1-,4+8au:)xlo cosu x~+~(lx~l)+~(a)foruo for u = 1 0 where ~(lxll)represents the same function defined by (45) with x now replaced by I x ( . Some important features of the viscous effect may be pointed out here. First, with nonvanishing viscosity the problem is mathematically deter - tllinate so that for u > 1 and a small, the gravity wave sin IC x auto- 0 1 matically appears only downstream (x > 0) and the capillary wave sin rCZx, only upstream. These waves are attenuated by the viscosity at the rate shown in their respective exponents; the viscous attenuation is more marked for the short capillary wave than for the gravity wave since IC 6 2> 1' For u0> 1, we may define an attenuation length for waves of wave number to be (in the physical units) kmrG - 1 which corresponds to an attentuation factor e at x= . Now from (27), where h = A,/& is the wave length and Am , the wave length at K. = 1. Then 1 This expression is probably accurate for A > 3Am or X (since Hp has < -3 m a different expression near u = 1, or near X= X ) For a water surface, 0 m 2 Hence when X = Xm/3, C = 5.05 cm., 'rn = 1. 73 cm. and (cm/4v km) = 44 cm. P X = 3Xm, C = 410 cm., and when A= loohm, & = 3 kilometers. Thus, the viscous damping on very long waves is hardly noticeable. Furthermore, it is of interest to note that for u0> 1, the viscous effect only attenuates the wave amplitude without affecting,sup to O(a),d the wave length, its velocity or its phase, provided uo is not too close to unity. For u0< 1, the attenuation of the wave is due to both the dispersion A 1/2 (given by the term (1- u4) ' in the exponent) and the viscosity; the for- 0 mer effect becomes more marked than the la-tter as the velocity u becomes 0 smaller. Finally, when u = 1, the viscous effect then modifies slightly 0 the wave length from the nonviscous value km to [ 1 + 2(v km/cm) and gives an attenuation length, in the physical units (which is equal to 3.48 cm. for a water surface). Furthermore, the vis- cosity now greatly affects the wave amplitude; in fact, the amplitude has no limit as a+ 0. Therefore, to get an intelligible result in this case it is necessary to retain the viscosity. Or, what is equivalent, when viscosity is neglected, it has been shown in a previous work6 that no steady state solution exists at u = 1. 0 Theintegral H of (26)maybeevaluatedinasimilarmanner. With Q the terms of O(a) and higher orders neglected, one may verify that for uo> 1 9 + ~(x') for x>O , + M(x1) for x Ha for uol 1 will, however, be omitted here. 5. Solution for Large Values of a (Small Reynolds Number) When the parameter a is large, the evaluation of integral (24) be- comes more complicated. However, when I xl is also large, we expect that some method, such as Watsonls lemma8 or the principle of steepest 9 descent , can be applied to obtain the asymptotic representation of the solution. Now, by changing the integration variable in (24) to t, with the function k(t) given by (28), the integrand then contains the term exp (ixik(t)). Consequently the saddle points, determined from dk/dt = 0, are t = i and t = - i. The path of steepest descent through these two saddle points is obtained from the conditions Im (ik(t) ) = 0 (since Im (ik) vanishes at both t = i and - i) 2nd Re (ixk)~0. Let t=c + iq , then the 2 2 steepest paths, determined from ~m(ik)=0, are F; = 0 and -5 + = 1. Hence it follows that for x > 0, the path of steepest descent is from r)= 1 to r) = +a along = 0; and for x Because of the complexity displayed in the present case of large a, we shall, for simplicity, confine ourselves in the remaining part of this work to the case uo = O(1) with a>> 1. We expect that the result of this case would also exhibit some typical features of the solution when both a and u 0 are large. On the other hand, when a is large but u =0(a-li2) or less, 0 the result reveals that the viscous effect is rather insignificant, primarily due to the small variations of the whole velocity field, and hence the in- viscid solution becomes then a good approximation. Now, for x>O, we again choose the closed contour I' to be the boun- dary of the first quadrant of the k-plane. Then, as has been shown, r will always enclose one simple pole at k=k with kl given by (41)-(43) for 1 a >> 1. Hence, by applying the theorem of residues to (24), we obtain for x>O again the same formal representation of Hp as given by (44) in which the quantities Res. (kl) and ~(x)now have to be calculated differently for a large. The residue of the integrand at kl can be calculated from (44c) by making use of (25) and (42). In this \manner, one obtains, after some manipulation, i -i}[l+~(al)]for u =0(1). Res. (kl) = - - 0 4u0 6 2u 0 Jz Next, the integral L(x) of (44b) will be approximated for both large and small values of XI. If we introduce the new variable s = - i a k in (44b), then For xt/a>> 1, the above integra1,representation is of the form to which Watson's lemma8 can be readily applied. Since most of the contribution of the integral comes from a neighborhood of s = 0, we expand the integrand for small s and large a ; then which is valid for xl>>a>>l and uo=O(l). Thus, L(X) is of the order -2 a and falls off like x-~as x +too . For small and moderate values of xt/a, we use (28) to obtain Writing the exponential function above as x ' x ' exp [- 5 @]= exp (-KTx' ) e.p(-&$ ~xP(-m) * the last factor may be expanded in a powergseries which converges rapidly, for x1/4a small, inside the range of integration 12 1. To further 2 simplify the calculation, the term with a- in the denominator may be neg- lected, for this process merely amounts to introducing a correction factor [ 1 t 0(a-')] to the final result. firtherrnore; in order to retain the con- 3 2 3 vergence of the integral at x=0, the terms 11 and i E q (1-1)' in the denominator must be grouped together. Finally, we introduce the trans- formation s = € ( 1- 1) and expand the integrand for the small parameter E = u0/2 fi , giving 3 s a t o(P') ds. 3 2 [A-(st,) (,ti) I 2 -i TT/~ If the term (s t i) is factorized as (s t e I(.- e-i n/4) and then the partial fraction is carried out, each term of the integrand cgn be integrated in terms of the exponential integral defined by (45a). From the known ex- 9 pansion of the exponential integral , one readily obtains valid for O~X'C$I2, a>> 1 and uo= O(1). (yE=0,5772. . . , Eulerls con- s tant) . To sum up, we combine (5 1) and (44), then for a>> 1, u = O(1) , 0 x1 u 0 - 2u0F x' H~(X)= - - e cos ( -$)[lto(a-')]t~(x) for X>O 2G 2u 0 6 where ~(x)is given by (54) and is approximated as given by (55) for OSX' Finally, for x < 0 and a large (a > 1.0957 to be precise), the path of integral (24) may be deformed to the negative imaginary k-axis since the integrand is then free from singularities inside the fourth quadrant of the k-plane. After using the transformation (28), we obtain in this case where the first integral HI comes from the integration along the right half of the unit circle ( t 1 = 1, and H2, from the integration along the imagi- nary axis, t = - ir) , r) 2 1. If we expand the integrand of H1 for small E and large a, each term can be integrated exactly. By using (1-cos 8) = 2t, HI becomes where x = xl/a. For xl<< a, H2(x) is exponentially small, hence the 0 - - value of ~(x)is well approximated by H~(X)alone, giving 4 3 4 8 a (x= -m (5)() - 12 ()- 48 t 0(d3) for x << a. (59) On the other hand, for -a<< x' _< 0, the value of H1 may be obtained either from (57) or by expanding exp(x't/a) in (57)'into a series and integrating termwise, giving which is of the order a.The integral H2 in (57) is similar in form to the integral in (54) and can be approximated for -a< HZ constitutes the most significant part of Hp (since HI is here of the 2 order a- ) Hence, u 3 0 xu2 1 Hp(x) Z----&(I-~~-~O~~)+O(~4ra a log-, a ~loga)a (60) 4Fa 2uo S;; valid for -all2< x' i; 0, a>> 1 and uo = O(1). To summarize, H~(X)is approximated as (60) for -a1/'< x1 5 0 and as (59) for xl<< -a. From the above result, namely, Eqs. (56), (59) and (60), one may note the following features of the solution: (i) The surface elevation Hp is continuous everywhere, including the origin, but dHp/dx has a logarithmic singularity at the origin. (This feature of course can also be seen from the original integral representa- tion 24.) (ii) Within the region 0 < x1<< a, there exists a train of waves of amplitude (uo/2a1") and wave length (in the physical units) which is large for large a. This wave is gradually attenuated due to vis- cosity at the rate exp (- "I ). 2uo JZa (iii) For x15 a, Hp falls off from O(a 1/2) to and diminishes - 2 like x as x--too. (iv) For XI0 Hp falls off monotonically from 0(am1I2) to ~(a-~)and event f 1. Lamb, H., Hydrodynamics, 6th ~ditidn,Dover Publications, New York, 1945. (especially Art. 349) 2. Basset, A. B. , Hydrodynamics, Cambridge Press, 1888. (Arts. 520-522) 3. Lord Rayleigh, "The Form of Standing Waves on the Surface of Running Watertt, Proc. Lond. Math. Soc. Vol. 15, p. 69, 1883. 4. Green, G. , "Waves in Deep Water Due to a Concentrated Surface Pressure", Phil. Mag. 7, Vol. 39, p. 738, 1948. 5. Stoker, J. J., "Some Remarks on Radiation Conditionstt, Wave Motion and Vibration Theory, Vol. 5, p. 97, Proc. 5th Symposium Appl. Math., Amer. Math. Soc., 1952. 6. De Prima, C.. R. and Wu, T. Y. , "On the Theory of Surface Waves in Water Generated by Moving Disturbancestt, California Institute of Technology, Engineering Division Report No. 21-23, May, 1957. 7. Lagerstrom, P. A., Cole, J. D. and Trilling, L., ItProblems in the Theory of Viscous Compressible Fluids", GALCIT Monograph, California Institute of Technology, March 1949. 8. Copson, E. T. , Theory of Functions of a Complex Variable, Oxford University Press, London, 1948. 9. Jeffreys, H. andJeffreys, B.S.. MethodsofMathematicalPhysics, 2nd Edition, Cambridge University Press, 1950. DISTRIBUTION LIST FOR UNCLASSIFIED REPORTS ON CAVITATION Contract ~onr-220(28) Chief of Naval Research i Commander Navy Department Naval Ordnance Test Station Washington 25, D. C. 3202 E. Foothill Blvd. Attn: Code 438 Pasadena, California Code 463 Attn: Head, Underwater Ord. Head, Research Div. Commanding Officer Office of Naval Re search Chief, Bureau of Ordnance Branch Office Navy De par tment The John Crerar Library Bldg. 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