14 Mr. 0. Heaviside on the 



That of a rectangle will be given later in the course of the 

 following subsidence solution. 



Consider the subsidence from the initial state of steady flow 

 to zero, when the impressed force that supported the current 

 is removed, in a prism of rectangular section. Let 2a and 

 2b be its sides, parallel to x and y respectively, the origin 

 being taken at the centre. Let Hj and H 2 be the x and y 

 components of the magnetic force at the time t. Let E be 

 the intensity of the magnetic-force vector E, which is parallel 

 to z; then the two equations of induction ((6), (7), Part I.), 

 or 



curl H = 47jT, — curl E = /jlTL, 

 are reduced to 



~d^=^' S =lA " • • • • ( 15a ) 



S-f^* 4 ^ • • • • ^) 



if r is the current density, k the conductivity, //, the induc- 

 tivity. [I speak of the intensity of a " force " and of the 

 ei density " of a flux, believing a distinction desirable.] The 

 equation of F is therefore 



^ + jf) r =^ kr > ■ ■ ■ ■ ^ 



of which an elementary solution is 



T = cos mx cos ny e pt , (18a) 



if 



47T/^= — (m 2 4-?i 2 ) (19a) 



At the boundary we have, during the subsidence, E = 0, or 

 r = 0; therefore 



cos mx cos ny — O at the boundary, 

 or 



cos ma = 0, cos7ifr = 0, (20a) 



or ma = ^7r, |7r, §tt, &c, nb = ditto. The general solution is 

 therefore the double summation over m and n, 



r = 2S A cos mx cos ny e pt , 



if we find A to make the right member represent the initial 

 state. This has to be r = r o _, a constant. 

 Now 



1 = S(2 /ma) sin ma cos mx, from x = — a to + a, 



1 = X(2/nb) sin nb cos ny, from y = — b to + 6. 



