of Edinburgh, Session 1881-82. 
349 
Developing the second member we get 
nffyX = cj)Y{y(fi'X ) + g\Yycf>'\ + ^>YyA.] + g 2 YyX . 
If we substitute for n g its expression (22), and identify the co- 
efficients of the same powers of g in both members, we get the two 
conditions : 
(23) 
( mffyX = <f>.Y(y(j>X) 
( mfJ y\ = Y (ycfi'X) + cj>(Y yX) . 
From these two relations we eliminate Y(ycjdX) by treating the 
second equation by <£. Thus we get : 
or 
mffyX - m.fiYyX = - f 2 YyX 
(■ 4 > 2 — m 2 cf> + mfYyX = £YyX = 0 . 
Now VyA. represents any vector perpendicular to y. Calling it p 0 , 
we have 
(24) JforSw ,- 0 
( ip 0 = (4> 2 ~ m 2 < h + m i)Po = 0 • 
From this equation we may also draw the inverse function 
(25) 
% < / , 'Vo = ( m 2 -#>(>> 
when Syp 0 = 0 . 
To these equations (24), (25) correspond similar equations for the 
conjugate <£', namely, for Sy l p 1 = 0 we have : 
(26) 
( f P 1 = (c/)' 2 - mf>' + = 0 
l %0'“Vh = ( w 2-0')/V 
It is evident by the expressions (12) and (12 bis) of m 1 and m 2 that 
the coefficients in the relation (26) are the same as in the relation 
(24). 
§ 4. The equations (24) and (26) may be applied in order to find 
the directions of the vectors which coincide with the direction of 
the corresponding function <f> or </>'. 
Calling cr the vector, which satisfies 
V <x<£cr = 0 , 
or 
(27) (<f>-h)<r = 0, 
we see that a must satisfy 
Sycr = 0 , 
