170 THEORY OF NEWTONIAN FORCES. [PT. I. CH. IV. 



The condition for a minimum is that 



J(v)< J(v + hs\ 

 for all values of h. 



-nj du dv , ds 



Now = + h x , etc. 



dx dx ox 



*.:.. (/ay. AV . 



dy 



9 ds dv ds dv ds 



[das dx dy dy dz dz 



Integrating 



/ \ ?/ \ r/ \ z.o r/ \ OL [[[ fdvds dv ds dv ds 

 (3) J(u) = J(v) + h 2 J(s) + 2h U- 5- 



JJJ \9^9^ 



Now in order that J"(v) may be a minimum, we must have 



j, 2 r/ \ oi, rrr^ ds , ^ a5 8v ^^ ^ 



(4) h*J(s) + 2h L ^-+^-^-+^ 5- dr>0, 



JJJ \9^ 8^c 9/ d dz dzj 



for aW values of h, positive or negative. But as 5 is as yet un- 

 limited, we may take h so small that the absolute value of the 

 term in h is greater than that of the term in h 2 , and if we choose 

 the sign of h opposite to that of the integral, making the product 

 negative, the whole will be negative. 



The only way to leave the sum always positive is to have the 

 integral vanish. (It will be observed that the above is exactly 

 the process of the calculus of variations. We might put Sv instead 

 of hs.) 



The condition for a minimum is then 



dvds dvds dv ds 



, N 



(5) o-5-5~5- 



JJJ(9#9# dydy 



But by Green's theorem, this is equal to 



Now at the surface the function is given, hence u and v must 

 have the same values, and 5 = 0. 



