Colloquium publications.

2

CAMBRIDGE COLLOQUIUM.

(22) over a three dimensional region, and reducing the right
hand member by an integration by parts.
What conditions must soi, s2 satisfy in order to make possible
the equation (22)? A short analysis shows that a necessary
and sufficient condition is given by equation (17), which may be
deduced directly, starting from this point of view.
The analysis follows. In the first place, by making use of the
identities (see (8) page 34)
W, = V a V W W + V2 W. W1, W12' = Viao'W2' + V2 1, W2'
we have
W1 X Wf' = (V1 X V2){(a. W) (O W2') - (. W1) (a W2')}
whence, from the definition (21),
H1,Pl = - ( W1) (a W2') + (a -FV1) (/. W2')
By the condition of isogeneity (10"), we get the other new form
mentioned for H,,,,, as well as additional ones of the same character.
The equation just written, by comparison with (22), gives
us the equation
k[(aW T1) ( ' W2') - (. Wl) (a. W2')] = W1" V1 + W2"'V.2
In the right hand member the following substitutions may be
made, to make the two members similar in construction:
W'. Vi = VL1 (V1f - V2a) W2' = (V1 V1i) (. fW2')
- (V2'V.1) (C' W2')
W2' V.2 = V2' (Via + V2y) )W2' = (VI V2) (a'W2')
+ (V2 -V2) (O. W2').
Hence the equation becomes
(V.1 V1 + V<p2 V2 - ka W1) (. 7W2') +
+ ( V 10 - Vvl V2+ k*. Wi) (a W2') = 0.
By hypothesis, 0l1, (2, k are independent of W1', W2'; hence
by putting W2' = V2 and Vi successively the following equations
are deduced
kac W1 = V' 1 V + Vs2 'V2
k. Wi = Vrl V2 - VP2' V