63-64] EXAMPLES. 79 



we see that the lines = 0, 6 = irjn are parts of the same stream 

 line. Hence if we put n = TT/, we obtain the required solution in 

 the form 



- ft -ft 



( f&amp;gt; = Ar a cos ) ^ = Ar a sm ............... (3). 



The component velocities along and perpendicular to r, are 



A IT a&quot; 1 ^ JJ A TT a 1 1*6 



A-r cos , and A -r sin ; 

 a a a a 



and are therefore zero, finite, or infinite at the origin, according as 

 a. is less than, equal to, or greater than TT. 



64. We take next some cases of cyclic functions. 

 1. The assumption 



W= /JL\OgZ ........................... (1) 



gives &amp;lt; = -^logr, ^ = -^0 ..................... (2). 



The velocity at a distance r from the origin is fi/r } this point 

 must therefore be isolated by drawing a small closed curve 

 round it. 



If we take the radii 6 const, as the stream-lines we get the 

 case of a (two-dimensional) source at the origin. (See Art. 60.) 



If the circles r = const, be taken as stream-lines we get the 

 case of Art. 28 ; the motion is now cyclic, the circulation in any 

 circuit embracing the origin being 2 7r/i. C 



J ~/ 



2. Let us take * 



(3). 



If we denote by r 1} r 2 the distances of any point in the plane xy 

 from the points (+ a, 0), and by lt 6 2 , the angles which these 

 distances make with positive direction of the axis of #, we have 



z a = r^ty z 4- a = r 2 e^ 2 , 

 whence &amp;lt;/&amp;gt; = -//,logn/r 2 , -&amp;gt;|r = - /z (6^ - 2 ) ............ (4). 



The curves &amp;lt;/&amp;gt; = const., ty const, form two orthogonal systems of 

 coaxal circles. 



