434 Mr. 0. Heaviside on the 



if R is the steady-flow resistance of line (both conductors), 

 and K is the conductance of the insulator, both per unit length 

 of line ; 



? =(K/»,)B =-KR %, 



if E = effective steady-flow resistance at the z = terminals, 

 and 



_gi sin gli— KR 2 cos gli 

 gi cos gli + KR : sin gli y 



if Jii — effective steady-flow resistance at the z = l terminals. 



The expression on the right side of (176) is, of course, real in 

 the exponential form, and the steady distribution of V is got by 



KY = -dCJdz. 



Using the thus obtained expressions, we reach the (172) form 

 of C solution, and the corresponding 



Y=Y o —t(-p4>')- 1 u( i 0wdz.# t . . . (176a) 



The value of <£' here, got by differentiation with respect to 

 p, may be written in many ways, of which one of the most 

 useful, for expansions in Fourier series, is the following. Let 



w = (1 + q 2 )^ cos(mz + 6) ; 

 then 



#_ m d_ C , r/m _Jri_\ ,1 

 dp" &"coa 2 dpi™ 11 {^"(?n/S") 2 + Z 1 Zj ml J 



I dm 



{ COs2ml d^T) (W (m/V'Az) - 1 }' M 



2S"cos 2 dp 

 Corresponding to this, 



^ ml =w { m jsylXz ■ • ■ (178 > 



finds the angles ml; it is got by the union of 



tan0 = S"Z o /m, tan (wiZ + 0) = 8%/ro, . . (179) 



which are equivalent to (160 b). 



For example, if we take R" = R, constant, thus abolishing 

 inertia, and S" = Sjo, no leakage, and S constant (R and S not 

 containing p, that is to say), the expansion of Y an arbitrary 

 function of z is 



1 Y $m(mz + 6)dz 

 V.-SrinQns + g) , v f- d m Zi _^ y (180) 



2 t C ° S ,M d(mi) $p (m/Spf + ZxZ J 



