1911]
HYDRODYNAMICAL NOTES
19
When we substitute in the second and fourth of these equations the values of A arid B, derived from the first and third, there results
0(1 - cos 2a) + D sin 2a = 0, Osin 2a- D (1 - cos 2a) = 0;
and these can only be harmonized when cos 2a = 1, or a = sir, where s is an integer. In physical problems, a is thus limited to the values TT and 27T. To these cases (4) is applicable with C and D arbitrary, provided that we make
0=0, B + (I~-~}D = O................(5 bis)
n
= 0?'n/« JCOB
20
cos
n6
+ DrB/* jsinf-
20 - 1
sin -
.(6)
Thus
making
When s = l, a = 77-, the corner disappears and we have simply a straight boundary (fig. 1). In this case n = I gives a nugatory result. When n = 2, we have
f = O2(l-cos26>) = 2Cy) ..................... (8)
Fig. 1. Fig. 2.
= 4 (- - l] r~2+n/s \Gcos (— - 20) + Z>sin (~ - 20}\. ...(7)
Vo/ / VQ / VO /I '
\* / \. \ o / \ o 1 )
and V2-^ = 40. When n - 3,
^ = (7r3 (cos 0 - cos 30) + Dr3 (sin 0 - i sin 30),
In rectangular coordinates
(9) (10)
(11)
solutions which obviously satisfy the required conditions.
When s = 2, a = 27r, the boundary consists of a straight wall extending from the origin in one direction (fig. 2). In this case (6) and (7) give
TJT = 6Vn [cos (Jn0 - 20) - cos ^16}
+ Dr^ j sin QK 0-20)- (l -^sin^0, ...... (12)
V^ = (2n - 4) r^
cos
20) + D sin ( Jn0 - 20)}. . . .(13) 2—2 represents an irrotational motion.