202 MR JOHN DOUGALL ON AN ANALYTICAL THEORY OF 
where E, = som ,- Work diffce. from 0= $= x(p) —32" log p + fp,’ log p, 
or E, = neon . work diffce. from 96=p~=logp, the part omitted being merely 
a constant. | 
Now in the system 6==log p 
u=(a+1)/p | pp = — 2u(a + 1)/p? 
w=0 p= 
Hence the solution 
u= —(a+1)p/a? | pp= 2u(a — 7)/a” 
w= — 2(a—3)z/a? pz=0 
« 
will, taken along with 6=p=log p, give u and pz=0 at p=a. The balanced solution 
gives pp = —16u/a? , w=2(3—a)z/a’, at the edge. 
Hence 
2 fh (7 
[oes i She Hiay = Ns \ Be 
2 ean) aelog me ey 
and 
U(py 5 %1) = Py . ) 
ll {Z 
= 3} a 20 (3 —a)e+ du, } de. 
w(p, 5 4)= 24 i Bah] —» | Qu 
(u) Transitory extensional modes. 
The solution is given by (100) with 
= Goxp cosh xz 
JokPy work diffce. from the system a cosh 2Q«h - 
nie Srp(a+ 1)K2h(cosh 2xh + 1)" 
In the system mentioned 
u=«G, Kp{(cosh 2«h +a) cosh «z+ 2«z sinh xz} 
w=xGykp{(cosh 2xh —a) sinh xz+2«z cosh xz} 
rere 
a =P =G,xp{(cosh 2xh +1) sinh xz+ 2«z cosh xz} 
ra 
=| 
= 
4 
of (Ds Goxp{(cosh 2«h +3) cosh Kz+ 2«z sinh xz} 
ae Gy «p{ (cosh 2xh +a) cosh xz + 2xz sinh xz} - 
Kp 
- fier Gee? hy é 
The balancing system for u and zp at p=a is therefore 
G) Ka 
JQ ka 
p= Joxp cosh «xz, @= cosh 2xh¢p. 
In the balanced system, at the edge 
{ (cosh 2xh — a)sinh «z+ 2«z cosh Kz \ 
Wi 
ad y Ka 
IPP ea ety ae { (cosh 2 kh + 3)cosh «z+ 2«z sinh xz \ 
2p. ad) Ka 
Hence for the free solution with edge values v=, , Lee 
7 al 
é 1 J kp, if On (cosh 2«h — a sinh «z+ 2«z cosh xz) EP 
K, = 7 Sar i) Oy mL: Tie 
Z (cosh 2xh - —.—., ; 
2a Cg ay ne) + Ku,(cosh 2xh + 3 cosh «z+ 2«z sinh xz 
