110 Proceedings of the Royal Society of Edinburgh. [Sess. 
constants l x and l 2 (since these depend only on a ± and a 2 ) and therefore does 
not affect the constant of the integral (5) or the constant of energy : this 
shows that all the orbits, which differ from each other only in having 
different values of the constant e v have the same values for the constant of 
the integral (5) and the constant of energy : and hence that the infinitesimal 
transformation which corresponds to the integral (5) transforms these 
orbits into each other : that is to say, the integral (5) is the adelphic 
integral of the dynamical system. 
§ 7. Determination of the adelphic integral in Case II . — We now pro- 
ceed to the discussion of “Case II,” in which the ratio sjs 2 is a rational 
number (say equal to m/n), but no terrain cos (np 1 — mp 2 ) is present among 
the third-order terms in the Hamiltonian function H. Certain terms of 
the series (5) now contain in their denominators the factor (ns 1 — ms. 2 ), 
which vanishes since sjs. 2 = m/n: and therefore the series (5) as it stands 
cannot converge in Case II, unless the terms which have zero denominators 
have numerators which also vanish. We have here come upon the real 
root of the principal difficulty of Celestial Mechanics : by removing it here, 
so as to obtain a valid adelphic integral in Cases II and III, we shall be 
enabled to remove it from the whole subject. 
To fix ideas, we shall suppose that s 1 = 2, s 2 = l, so that sfs 2 has the 
rational value 2, and the denominator (gj — 2s a ), which occurs frequently 
in the series (5), is zero. 
In this case the equation for (p 3 becomes 
2 a< />3 , _ 2 aH 3 
dp 1 dp 2 dp 1 
& 
dpf 
and indeed the equation for any one of the functions 0 3 , </> 4 , </> 5 , . . . takes 
the form 
2 0 ^ + ^ = a known sum of terms of the type qf m qf l sin (kp 1 + lp 2 )> 
op i 3p 2 
Now in integrating the differential equations for </> 3 , </> 4 , . . . in § 4, we 
used only the “ particular integral,” which corresponds term-by-term to the 
known function on the right-hand side of the equation : so that, e.g., the 
integral of the equation 
0^3 
l dpi 
— - ” 1 “ S 2 
d ih: 
Op . 2 
qp sin p l 
would be taken to be 
<t>3 = ~ “ C0S Ik- 
S 1 
The reason for this was that the “ complementary function,” or arbitrary 
part of the solution of the differential equation, is a function of {s 2 p 1 ^s 1 p 2 ), 
