200 MR JOHN DOUGALL ON AN ANALYTICAL THEORY OF 
sinh xh = 0, but equal to—sinh xz sinh xz, when cosh xh =0, we may transform (1) if we 
put h+z2=C, h+z,=G, into 
pees OL + LE Seen xG, cosh K6J'xp,Gy Kp, (k a pos. imag. root of sinh 2«h) . 2 aa) 
Srph p  2rph** 
For 
— 
v=G, «xp, we have when p=a, pw= —  (2G,/xa + KaGoKa) . 
Hence, when the transitory part of v in (iii) is balanced, the solution becomes (p> p,) 
v= ede a uli cosh xf, cosh KlJq'xp, (Gi’xp - 
Pi 2G, Ka+ KaGoKa, , ) (iv) 
Srph p = 2Arph'e ; ; 
J 
25, Ka + Kad Ka tae 
erate 
This gives at p =a, 
bgt ed 
~ cosh xf, cosh KJ q Kp, ( — 
Qarph 25,'Ka + Kad \Ka 
Hence, for the free displacement at (p,, z,) under symmetrical transverse traction 2, on 
p=, Bettis Theorem gives (omitting the rigid body rotation) 
a , l 2h 
Pies - Jo Kp, cosh Ke( - averse) cosh KlQ,dl_. : : (v) 
From this 
i 2) Kp + KpJ Kp ae 
5 $ Q,d 
(pe) => z= Tea Nea ca cosh ral (cosh KL 
The series passes continuously, as p increases to a, into the limit 
— cosh xf, | “cosh «6Q,d¢, provided this latter series converges. 
By Fourier’s Theorem we know that it does, namely, to the value 2,, it being noted 
that { Od =0. The solution is thus verified. Of course, it could easily be obtained 
by the Fourier method ab initio. 
The series (iv) converges very rapidly unless p and p, are nearly equal. By an 
application of the Residue Calculus, it may be transformed into a series in which the 
functions of p,p, are the fluctuating functions, and the functions of ¢, G the con- 
vergence factors. For consider the function of «, 
1 cosh «(2h —€) cosh xé, J 
Tp sinh 2«h 
2Gy Ka + KaGyKa zy! ) 
0 Kpi( Gy Kp = 2S) Ka + Kad Ka 0 Kp 
It is easy to see that log « disappears from the last factor, and that the whole function 
is a uniform, odd function of «; also that if (>G,p>p,, the function vanishes at 
infinity in such a way as to make the total sum of its residues equal to zero. 
The poles of the function are «=0, the (pure imaginary) zeroes of sinh 2«h, and 
the (real) zeroes of 2J,/ka+«aJ.ea. The function being odd, we have (series of residues 
at pos. imag. roots of sinh 2«h)+(series of residues at pos. roots of 2J)xa+«aJ ca) 
+4 residue at («=0), equal to zero. 
