Self-induction of Wires. 351 



The first summation cancels the second at the first moment, 

 and ultimately vanishes, leaving the second part to represent 

 the final periodic solution. Take u=0 ; and use the u, iv, M 

 expressions of (126) and (ICO), and let <j> m stand for 

 m 2 + $p(R' m + y m p), so that <f> m — gives the p's for a par- 

 ticular m 2 . Then we obtain, (with V = at both ends), 



v=4s 



cos mz\ cos mz\ . e Q dz x . (p sin nt + n cos nt) 



dz ¥ ^* { f^) 



dp 

 \mz 

 (d \. 



cos mz 1 cos mz 1 . e sin nt . dz 1 



=£? 2 — •% zm>. — ' • • • (138) 



dp 

 because d 2 /dt 2 = —n 2 . But, if e ='2,e m , the equation of V m is 



de 

 — <f>nNm= -^ sin nt, 



(by (60) and (63), Part II.), so that 



\j jL—\®^m ® v e m smnt /-iqqn 



Ym= -*™llz- =Z -Tz^ ld__ \df • (139) 



[dt p )dp 



by a well-known algebraical theorem, the summation being 

 with respect to them's, which are the roots of <£ TO = 0, con- 

 sidered as algebraic. We have also 



2 C l 



e = j- 2 cos mz I coamzie Q dzi, . . . (140) 

 1 Jo 



the summation being with respect to m. 



Uniting (139) and (140), there results the previous equation 

 (138), in which the summation is with respect to all the p's 

 belonging to all the m's. In the case m = 0, the 2/1 must be 

 halved. In the form of a summation with respect to m, 

 similar to (77) for C, the corresponding V solution is 



v _ 2V ^ m sin mz j(L' TO — m 2 /$n 2 )n sin nt + R' m cos nt\ 

 -~SS" i - R^ + (L M -m 2 /S» 2 )V, 



the impressed force being V sinw£, at z=0. This, on the 

 assumption R' m = R/, U m = U, will be found to be the expan- 

 sion of the form (80), Part II. 



