1911-12.] Singular Solutions of Partial Differential Equations. 157 
If 
then 
t 2 2 -'f'i'f'3 = z-x-y=0, 
p=l, 2=1, 
^ 2 + ^3 =2^ 3 , 
••• ^■ 2 =^ 3 = 1 sa y- 
(1 - Jz-x-y)p + q = 2 , 
an example discussed by Professor Chrystal in a memoir (Trans. R.S.E. 
vol. xxxvi., 1892). 
Example 6. — Suppose 
(7) gives 
Let 
= x 1 = z, 
/*=<>, fy=Q, fz= 1 , 
, AlP + (’A2 + = °- 
^-^=0 = (l-Z 2 )(x 2 + y 2 + Z 2 -l) 
z 2 — 1 satisfies our equation identically. 
x 2 + y 2 + z 2 - 1 == 0 
gives 
p=-x/z, q=-y/z, 
and the equation on substituting these values, 
^l + #2 = 0 * 
This will be satisfied by 
l/q --=X 2 + Z 2 - 1, lf/ 2 = xy. 
Hence the required equation is 
(x 2 + z 2 - l)p + [xy + (1 - z 2 ) h (x 2 + y 2 + z 2 - 1)*]^ ='0, 
an example discussed by Goursat. 
The reducible equation from which this may be supposed to be 
derived is 
(x 2 + z 2 - l)p 2 + 2xypq + (y 2 + z 2 - 1 )q 2 = 0 
On testing this for singular solutions we get 
(x 2 + z 2 - \)p + xyq = 0 
xyp + (; y 2 + z 2 - 1 )q = 
which together satisfy (9), and give for consistency 
x 2 + z 2 - 1 xy 1—0 
xy y 2 + z 2 - 1 ■ 
*1/ €/ 
(x 2 + y 2 + z 2 - 1) (1 - z 2 ) = 0, 
which also satisfies (10), and may be regarded as a singular solution. 
-° 1 . 
= 0 i 
( 9 ) 
( 10 ) 
