536  Royal  Society: — 
<p  from  (1), 
ff^  =  —  —(V  +PW  d^ 
dy  dtdyK   °*    }     dx~' 
-«*$=        d*   (V    \-T>\     d^   \ 
dx  dtdxK   "^    }      dy\ 
A  solution  of  these  equations  is 
ff0=—  ^-(Po  +  P),     ^=  constant (10) 
Substituting  the  value  of  0  in  terms  of  P,  as  given  in  equation  (4), 
<r   dP       d  /TJ    .  TJN 
ss-a^ <n> 
The  quantity  -£-  is  evidently  a  velocity;  let  us"  therefore  for  con- 
ciseness  call  it  R,  then 
S+S+3-*  • w 
24.  Let  P0'  be  the  value  of  P0  at  the  time  t—  r,  and  at  a  point 
on  the  negative  side  of  the  sheet,  whose  coordinates  area?,  y,  (z  —  Rr), 
and  let 
Q=rp»'* C3) 
At  the  upper  limit  when  r  is  infinite  P0'  vanishes.     Hence  at  the 
lower  limit,  when  r=0  and  P0'=P0,  we  must  have 
P°-^  +  R^' <l4> 
but  by  equation  (12) 
dp0_    dv   ^d?  :1(rt 
Hence  the  equation  will  be  satisfied  if  we  make 
p=-f ->-■»£*•* ci6) 
25.  This,  then,  is  a  solution  of  the  problem.  Any  other  solution 
must  differ  from  this  by  a  system  of  closed  currents,  depending  on 
the  initial  state  of  the  sheet,  not  due  to  any  external  cause,  and 
which  therefore  must  decay  rapidly.  Hence,  since  we  assume  an 
eternity  of  past  time,  this  is  the  only  solution  of  the  problem. 
This  solution  expresses  P,  a  function  due  to  the  action  of  the 
induced  current,  in  terms  of  P0',  and  through  this  of  P0,  a  function 
of  the  same  kind  due  to  the  external  magnetic  system.  By  dif- 
ferentiating P  and  P0  with  respect  to  z,  we  obtain  the  magnetic  poten- 
tial, and  by  differentiating  them  with  respect  to  t,  we  obtain,  by 
equation  (10),  the  current-function.  Hence  the  relation  between 
P  and  P0,  as  expressed  by  equation  (16),  is  similar  to  the  relation 
