Self-induction of Wires \ 433 



which, besides, seem to give us some view of the inner 

 meaning of the expansions and of the operations producing 

 them that is wanting in the treatment of a special problem 

 on its own merits by the easiest way that presents itself. 



Leaving, now, the question of variable electrical constants, 

 let the line be homogeneous from beginning to end, so that 

 H" and &' are functions of p, but not of z. The normal 

 current- functions are then simply 



X = cosms, Y = sin mz, 



where m is the function of p given by — m^B/'S", so that 



w — cos mz + q sin mz, \ ,.. _ . . 



u— (m/&") (sin mz — q cos mz). J 



Let there be a single impressed force e 2 at z = z 2 ; then the 

 differential equations of the currents on the left and right sides 

 of the same, corresponding to (162), will be 



n f , N cos mz a + q 1 sin mz 2 -\ 



■ . cos mz2 + q sin mz 2 \ 



C^icosmz + q^mmz) (m/Sff)(go _ gi) *, ) 



where q and q x are given by 



S" tj _ (m/S n ) sin ?nl—Z 1 cos ml n rnj , 



2°-~m °> ^"(m/S'OcosmZ + ZiSinmr ' ^ W(?) 



As before, in the case of an arbitrary distribution of e we are 

 led to the solution (165), wherein for w (and for u in the 

 corresponding V formula) use the expressions (174), in which 

 q is to be the common value of the q Q and q x of (1606), and 



^=(m/S")( ?0 -?i)=0 (175) 



is the determinantal equation of the ^'s. 



Use (170) to find the final steady current distribution. 

 Thus, now, 



C = | (cos mz + q x sin mz) j (cos mz + q Q sin mz)edz 



+ (cos mz + q Q sin mz) I (cos mz + q ± sin mz)edz}-+--^ (q —qi) } (176) 



in which m, q , q u and S" have the p = values. They are, 

 iH=(-l)*, 



S" = K, m=(-RK)§=^say> 



