Operational Methods in Mathematical Pliysics. 409 

 we obtain the solution of (5) in the form 



ru = S mh(gr/c) 



c sinh q v ' 



because ru must vanish with r, and u — Gt at the surface 

 r = c. 



It is quite easy to interpret (7) by the aid of equation (2)* ; 

 but it seems unnecessary to write the result out at length, as 

 the observations refer to the mean temperature w, which we 

 shall proceed to calculate in symbolic form. 



Thus we have the mean temperature 



u=~\ r 2 u dr = 3 ( coth q A Gt, 



c\\ \q q 2 ) 



(8) 



and this has now to be interpreted by means of equation (2). 

 In the first place we know that 



qcothq = l+p 2 --~q*+... 



and so the expansion corresponding to (3) is 



3(icoth ? -I) = l-i f/+ ... = i_ 1 1 5 ^ + ... 



Thus here ^^ Ni= _^ /fl , (9) 



Again in (8) we can write 



¥(p) = o(q cosh q— sinh q)lq z , 



A(p) = sinh^/g, 



because each of these functions can be expressed in positive 

 integral powers of p. The roots a of AQt?) =0 are given by 



q = n7rt, p=z—n 2 'rr 2 a 2 /c 2 = a, . . . (10) 



where n is any positive integer f- 

 Then we find that X 



A' (a)— - -^cosh q— -=r- cosh a 

 q dp zol L 



and by (6) w \ ' 3 , oa 2 , 



* It will be found that the operator in equation (7) really involves 

 only q 2 and so can be expressed as the quotient of two series in p • but 

 the form used in (7) is more compact and is easier to work with. 



t Negative integers do not give any fresh values for p : and so should 

 not be included because we require only the complete set of values of u. 



X Since A(a)=0 it is only necessary to differentiate sinhy, in order 

 to evaluate A'(x). 



