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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.