128 Mr. 0. Heaviside on the 



the terminal conditions. Also, let 



—s 2 = 4:7rfj,kp+/jLcp 2 + m 2 (15) 



Then (14) becomes 



d 1 rl 



. . . (16) 



dr r ar 



which is the equation of the J\(sr) and its complementary 

 function, which call K 1 (sr). Thus, for reference, 



JoO) = l 



JiO)=- ^7Z7y J o(sr) = i8r(l-± ^ + i ^-. . .y 



2 2 + 2 2 4 2 



.2^,2 



d(sr 



K («r) = - J -^- + Ji («•) log jw 



K (sr) = J (sr) . log «r + -gg- - (1 + i) gajg + 

 K 1 (,r) = - 



' We have therefore the following sets of solutions, in the 

 wire, dielectric, and sheath respectively, the A's and B's being 

 constants : — 



R 1 =A 1 J 1 (sir) cos (mz + 6)6?*, 

 47T7 1 = A 1 J l (5^)771 sin (mz + 6)e pt , 

 4-771^ = A 1 J (5 1 r)s 1 cos (mz -f 6)6?* , 



H 2 = { A 2 J \(s 2 r) + BaKx^r)} cos (mz + 0)eP*. 

 ±7ry 2 = {A 2 J 1 (s 2 r) + BgK,^)}™ sin ( ms + 0) eP S } 

 4ttT 2 = {A 2 J (s 2 r) + B 2 K (s 2 r)\s 2 cos (mz + 0)6?' , 



H 3 = {A 3 J 1 ( 5s r) + B 3 K 1 (s 3 r)} cos (mz + 0)eP l , 

 4-7773= {A 3 Ji(s 3 r) + BaK^Sg^Jm sin (wis + #)e p ', 

 47rr 3 = { A 3 J (^) + B3Ko(jj 3 r)}*3 cos (mz + 6)ei>K 



To harmonize these, we have the boundary conditions of 

 continuity of tangential electric and magnetic forces, and of 

 normal electric and magnetic currents (or of magnetic induc- 

 tion). Thus y 1 = y 2 and p i T 1 =p 2 T 2 , at r=a 1 , give us 

 A 2 /A 1 (J 1 K -J Ki)(5 2 a 1 )=J 1 (5 1 a 1 )K (5 2 a 1 ) 



— (/W/V2) JoK«i)K, (s 2 a 2 ), 

 B 2 IA 1 (J 1 K —J K 1 )(s 2 a 1 )=(p 1 sJp 2 s 2 )Jo(s 1 a 1 )J 1 (s 2 a 1 ) 



— J 1 (s l a 1 )J (s 2 a l ). 



> (17) 



(18) 



(19) 



