160 Mr. P. H. Ling on a Certain 



Now <3> need not involve H : for if it does, we may 

 solve the equation connecting them for p x and get an 

 identity 



%(<lu 92, q3,Pi,P2,P3)= zS P(q 1 , q 2 , qs,K ,p 2 ,p s ). 



By replacing H by H, we get ^(H), an integral of the 

 problem. Then <J> — ^(H) is also an integral. Call it 

 /^'(H). Then ^ ' may not involve H . If it does, repeat 

 the process. Then ultimately we arrive at a function not 

 involving H . 



The condition that <I> should be an integral is that the 

 Poisson-bracket (H, <J>) should vanish. And we have 



(H,*) = (H** o )+/*{(H ,* 1 ) + (H 1> *,)} + . ... (6) 



Since <3> is a periodic function, we may write, by Fourier's 

 theorem, 



where the A's are functions of the jo's only, and the summa- 

 tion extend? over all integral values of m 1? m 2 , m 3 . Then 

 since H does not involve q u q 2i q^ or p 39 



(H , <S>o) = - * |p 2m, A . e l(mqi + "*» + mm) 

 _ l O^o ^ m J± e *0»iS' + »»2f 2 + Mask) 



Bj»2 2 



= (identically). 



for all integral values of m l} m 2 , m s . 



Now the quantity in the bracket cannot vanish unless the 

 Hessian does. 



Hence A mi , m2t mz vanishes, except Aqo^. It follows that 

 ^> cannot involve q l9 q 2 , but may involve q^. 



NOW let Hi= 2 ^m l m 2 m- ii el{miqi + mm + mm) ' 



m x , m 2 , m 3 



<J> = V g*(«» 1 Si + »t 2 S2 + W*3S3) 



m 1} w»2> w*3 



