1911-12.] Singular Solutions of Partial Differential Equations. 155 
second degree in p and q to have a singular onefold. For, by what has 
already been said, it must be completely represented by 
tiifzP +fxY + tyJJ# +f x ) (f z q +f y ) + ^ 3 (f z q +f y ) 2 = 0 . . (6) 
where y]s v \/s 2 , and are any functions of x, y , and 0. But this may 
evidently be written in reduced form 
«Al fzP + (&} ± x/^2 2 ~ 'I'MfzV '= - «Al /* - (^2 ± n/^2 2 - 'I'Mfv ■ • 00 
Hence Lagrange’s Subsidiary System gives 
dx dy dz 
tl fz («A 2 ± ^2 -^2 )fz - ^lfx - (<A 2 ± )/y 
_ +//&/ +//^ 
0 
Hence one integral is 
/(^) - c x . 
Suppose 
u{xyz) — c 2 
is another integral when the upper sign is taken, and 
= C 3 
when the lower sign is taken with the square root, then the complete 
integrals of the two reduced parts are 
f(xyz) + Au = B, 
f{xyz) + A'v = B\ 
Hence in this case the so-called singular onefold — and it satisfies 
Cauchy’s conditions — is the infinity of surfaces common to both reduced 
parts. 
As an example we may take the case already quoted, Example 1 — 
(xp + yq - z) 2 -p 2 -q 2 = —EL 
{x 2 + y A - 1) 
which, being of the second degree in p and q, reduces to 
p(x 2 - 1) + q(xy ± Jx 2 + y 2 - 1) = xz ± --===. 
six 1 + y A - 1 
Example 4. — 
(px - 2 z) 2 + qz(px + qy - 2z) = 0 
<f> p = 0 gives 2 x(px - 2 z) + xzq = 0 
<£<z = 0 gives z(px + qy - 2 z) + yzq = 0 
which are equivalent to 
2 px + qz- 4s = 0, z = 0, 
and 
2px + qz- 42 = 0, and px + qy -2z — 0. 
