64 Lord Rayleigh o)i the Question of 



part of (9) disappears. There can then be admitted no values 

 of ??, except such as make n-f&W = for some value of r 

 included within the tube. For the equation 



tPu . 1 du u 

 dr 



2 + zk '+ 3 -**»=<>, ..... (10) 



being that of the BessePs function of the first order with a 

 purely imaginary argument, admits of no solution consistent 

 with the conditions that u = Q when r vanishes, and also when 

 r has the finite value appropriate to the wall of the tube. 

 But any value assumed by —kW is an admissible solution for 

 n. At the place where w + &W=0, (10) need not be satis- 

 fied, and under this exemption the required solution may be 

 obtained consistently with the boundary conditions. It is 

 included in the above statement that no admissible value of n 

 can include an imaginary part. 



If s be not zero, we have in transforming to u to include 

 also terms arising from the differentiation in (8) of — Qs 2 /r 2 , 

 that is 



dr 



2 *- 8 Q-^> 



for the second of which we substitute from (5) , and for the 

 first from (8) itself. The result is 



t 7wn V<Pu , 1 du 3s 2 + k 2 r 2 



r dr s 2 + k 2 r 2 



From (11) we may fall back on the case of two dimensions 

 by supposing r to be infinite. But, in order not to lose 

 generality, we must at the same time allow s to be infinite, 

 so that, for example, s = k J r. Thus, writing a; for r, and y for 

 ?'0, we find for the differential equation applicable to the 

 solution in which all the quantities are proportional to 



J(nt+Jcz+k'y) 

 e , 



, .(„ + AW){g-#u-^]. = *u£J, . .(12) 



agreeing with that formerly discussed except for a slight 

 difference of notation. 



