204 Mr Ghree, On thin rotating isotropic disks. [May 4, 



Putting the right-hand side of (10) zero, it is easily verified 

 that this is a solution. We thus have 



8 = A + G (z 2 - if 2 ) - ift)Vr 2 /(m +n) (11). 



Noticing that 



d ( (du dw\] _ d<$ d 2 w d I dw^ 

 dr\ [dz dr)} dz dz 2 dr\ dr , 

 we may transform (9) into 



d 2 w 1 dw d 2 w m d8 2m ri ... _. 



t? + -j-+TT = 7T = Gz ( 12 )- 



dr r dr dz n dz n 



Of this a particular solution is 



lu = -™Cz\ 

 on 



A complementary solution is easily obtained in ascending powers 

 of r and z. Of this we require for our present purpose only the 

 terms of the first and third degrees, which separately satisfy (12) 

 when the right-hand side is zero. Thus we get 



07) 



w = 0L 1 z+e 1 (2z 3 -ozr 2 )- (> - Cz 3 (13), 



oil 



where a x and e x are new constants. 



Employing (1) and (11), we in like manner easily transform (8) 

 into 



d 2 u 1 du u d 2 u _ m d8 <o 2 pr 



dr* r dr r 2 dz* n dr n ' 



A particular solution is 





= r[G ™-^P-) (14). 



n m + nj 



711 ft) p 



c- 



n in + n 



For the complementary solution we require only the terms in odd 

 powers of r and z up to the third degree. Terms in negative 

 powers of z are of course inadmissible, and r~ l is the only negative 

 power of r which satisfies the differential equation. Thus the 

 complementary solution is 



u = - + ar + e (z 3 - 2zr 2 ) + £ {^z 2 r - r 3 ), 



where D, a, e, £ are new constants whose coefficients separately 

 satisfy (14) when its right-hand side is 0. Thus for our complete 

 solution of (14) we get 



u = 5 + ar + e(z 3 - 2zr 2 ) + Wz 2 r -r 3 ) + l (c - - -^-) r 3 . . .(15). 

 r 8 V n m + nj 



