where G is defined in Equation [24] and A(s) is the same determinant which was obtained 
in ®quation [25]. Therefore the poles of this problem are exactly the same as the poles of 
the former problem and the inverse of the Laplace transform is equal to the sum of the 
residues. 
For the pole at the origin p = zp, g= 0, and@=Q. A(0) is defined in Equation [29] 
and the other terms have the values Q? 
2 p2 
2 Kip + Ky 
p2 + pw oe 2 [51] 
y2 Tour) 1 @r) Loa @ Iotur) J (ur) _ ard, (ur) [52] 
p77) Jae), | 1@E)| ¥ 2 UG 2 Jy (ud) 
}2 P a 
qa D,, (gt, 9a)+ — Dy, Gr, ¢a)=— (1+ — log— (53] 
Ko Ko tr 
At the other poles of the integrand where A (A_) = 0; p =ta@, and g=7f8, which were 
determined in the former problem. The determinant A (\,) is defined in Equation [36] as a 
function of N, (8,2), N, (8, a), and P/x, but it may also be defined in terms of new functions 
M, ‘B,, 5), M, (8, 5), and G/x, as 
G 
A(,)=—B, 5 M(B, 5) +—* Mg (B, 6) = 0 [54] 
2 
where 
My (B,7) = ia, 4 Dy (ia, 7, te, 2) + = Doo (itn? ia, 2) 
2 [55] 
=F [250m T Ct, nad + = Coo (a 59| 
iM. (ples = =H 22 eae (a, 7 0, 2) fale Coa, 78, 0)| (56] 
K 
2 
If Equations [54] and [36] are combined, it may be shown that 
My (8B, 5) No (8, 2) = -1 [57] 
If #*(b) is written for 
* (« Bb 2. 2 no lng 7 
p*(6) = - a [58] 
P-@Q se 2 has = 
Ky b 
13 
