y r= « (1 — y) cos x = — | a2 + a cos as - ^a2 cos 2# ; ............... (4)
= - ^a2 + a (1 + -fa2) cos x - £a2 cos 20 + f a3 cos 30, ......... (5)
which is correct to a3 inclusive, ft being of order a4. In calculating (2) to the approximation now intended we omit the term in ay. In association with a/3 and 7 we take e~sv = 1 ; in association with ft, e~-v = 1 — 2y; while
a2e-^ = a2(l-22/ + 22/2-t2/3)-Thus on substitution for yz and y3 from (o)
a2g-2!/ = a2 {1 — 2y -t- a3 - 4a3 cos x + a2 cos 2* — £ a3 cos 30}. In like manner
2/3e~22/ cos 20 = 2/3 cos 2a? - 2«/3 (cos a; + cos 30).
Since the terms in cos 0 are of the fifth order, we may replace a cos 0 by y, and we get
. Z72 - 20y = 1 + a2 + a4 + 2y (1 - £> - a2 - 2a4 + /3)
+ (a4 + 2/3) cos 2x + (- |a5 + 47 - 2a/3) cos 3« ....... (6)
The constancy of (6) requires the annulment of the coefficients of y and of cos 2# and cos 3tv, so that
and #=l-a2--fa4 .................................. (8)
The value of g in (8) differs from that expressed in equation (11) of my former paper. The cause is to be found in the difference of suppositions with respect to ty. Here we have taken ty = 0 at the free surface, which leads to a constant term in the expression for y, as seen in (5), while formerly the constant term was made to disappear by a different choice of -\|r.
There is no essential difficulty in carrying the approximation to y two stages further than is attained in (5). If S, e are of the 6th and 7th order, they do not appear. The longest part of the work is the expression of e~y as a function of as. We get
. (3a2 125a4 J 2a3 0 125a4 + cos Zx j-j- + -^g- - /3j - -g- cos 30 + -j-gg- cos 4c, . . .(9)
and thence from (1)
9a3 625aB
125a5 K , x
os5« ........................................... (10) 2gy = 1 + 2 (1 - g) y + cPe'*" + 2/3e~^ cos 2a?