i G. PLARR ON THE SOLUTION OF THE EQUATION Vp¢p=0. 
We shall occupy ourselves only with the octant comprised between 
+r,+u,+v, because for obvious reasons the surface will be in the other 
octants only the reproduction of what takes place in this first octant ; namely, 
the surface in the first octant will be repeated in the other octants, either in 
the position of equality of superposition or in that of symmetry. 
For E=h/,, the expression (51) gives 
O= (hy —hs) Se’ + (hg — hs) S*pe’. 
By h,>h,>h; the coefficients are both positive, and we must put 
Srxe’=0, Spe’=0, 
which gives o—r, 
The spherical conic reduces itself therefore to the point 
i 
Vv 2 (hy — hy) e 
The normal v becomes, as av=h,v , 
3 nN 
v=5hy, tge’'v=0, 
and, as h; is negative, our expression gives v directed inversely to «=v, that 
is towards the inside of the surface. 

the factor — = is positive, but for ¢ =v we have dE = 0, because 
3 
E =— Seawe’ and «2 — 1 give 
dK. = — 28dese’, and 0 = Sede’ 
dE = — 23h,Shdde'Sre, and 0 = SSrde'She’. 
Now for «=r, the last of these equations give 
0=Srde’, and this gives dE=0, so that de=0, 
namely, the tangent plane in e=¢e,.v is perpendicular to v, and e undergoes a 
minimum. 
A similar case is the one when E=h,. Then the surface reduces itself to 
the point 
