348 Mr. 0. Heaviside on the 



Therefore we must have 



i _ v m 2 /Sj? z 



Simplified, it makes this theorem 



4,{0)- z \ p dp) ' 



if the p's are the roots of (f>(p)=Q. This is correct. 



To determine the effect of longitudinal impressed force, 

 keeping to the case of uniform intensity over the cross section 

 of either conductor. Let a steady impressed force of integral 

 amount e be introduced in the line at distance z x ; it may be 

 partly in one and partly in the other conductor, as in Part II. 

 By elementary methods, we can find the steady state of V, C 

 it will set up. If, then, we remove e , we can, by the prece- 

 ding, find the transient state that will result. Let Y be the 

 steady state of V set up, and Y 1 what it becomes at time t 

 after removal of e ; then V — Y x represents the state at time 

 t after e is put on. So, if %Au represent the V set up by the 

 unit impressed force at z l7 



Y=Y -e $AueP t 



will give the distribution of Y at time t after e Q is put on, 

 being zero when £ = 0, and V when t = co . No zero value of 

 p is admissible here. 



From this we deduce that the effect of e lasting from t=ti 

 to t=t 1 + dt 1 at the later time t is 



— %Aupe dt 1 ePV- t i> ; 



therefore, by time integration, the effect due to an impressed 

 force e at one spot, variable with the time, starting at time 



t is 



= —LAupeP t \ , 

 J t 



e e-P*idt h 



in which e is a function of t v 



By integrating along the line, we find the effect of a con- 

 tinuously distributed impressed force, e per unit length, to be 



Jo . ^0 



wherein e is a function of both z x and t 1} and starts at time ^ ; 

 whilst A is a function of z v the position of the elementary 

 impressed force edz v 



