220 Proceedings of Royal Society of Edinburgh. [sess. 
■equilibrium in which A 5 is within range of the force of A 0 , and 
A 6 is beyond it. The algebraic sum of the ordinates with their 
proper multipliers is zero, and so the diagram represents a solution 
■of equation (9). 
§ 25. In the general linear problem to find whether the 
equilibrium is stable or not for equal consecutive distances, a, let 
(as in § 4) be the work required to increase the distance 
between two atoms from D to oo . Suppose now the atoms to 
be displaced from equal distances, a, to consecutive unequal 
distances — 
.... a + u t _ 2 , a + Uj _ x , a + u it a + u i+1 , a + u i+2 , • (10). 
The equilibrium will be stable or unstable according as the 
work required to produce this displacement is, or is not, positive 
for all infinitely small values of . . . . , u i , u i+1 , . . . . Its 
amount is W 0 - W ; where W denotes the total amount of work 
required to separate all the atoms from the configuration (10) to 
infinite mutual distances. 
According to § 2 above W is given by 
• • . + + u\ + w i+x + ....). . (11); 
where 
w i =cfi(a + u i ) + <f>(2a + u i _ 1 + u i ) + 4>(3a + u t _ 2 + ^_i + u^) + . .. 
+ <j>(a +w i+1 ) + cf>(2a + u i+1 + u i+2 ) + c£(3a + u i+1 + u i+2 + u i+3 ) + . . . 
+ (12). 
Expanding each term by Taylor’s theorem as far as terms of the 
second order, and remarking that the sum of terms of the first 
order is zero for equilibrium * at equal distances, a, and putting 
</>"(D) = -/'(D), we find 
W 0 - W‘=\%{f\a){u? + u\ +l ) 
+f('2a)[(u i _i + Ui y + (u i+1 + u i+2 ) 2 ] 
+./ , (3a)[(w i _ 2 + + u { ) 2 + (u i+1 + u i+ 2 + w i+3 ) 2 ] 
+ etc. etc. etc. etc. } (13); 
* It is interesting and instructive to verify this analytically by selecting 
all the terms in H /> which contain u h and thus finding This equated 
to zero, for zero values of . . . m -\ ,% , Ui+ 1 , . . . gives equation (9) of 
the text. 
