PERTURBATION THEORY FOR EXPONENTIAL M CURVE 187 
The characteristic values D, are then obtained as 
the roots of the infinite determinant. 
D, — D;° + aBi, aBi2, Bis, 
aBo, Dy — De® + aBo2, aes, 
aBs1, aB32, D, — D3° + aBss , 
ms CO DOPE EOS SCE 2 Ae RCEC (30) 
Having determined D, from equation (30), the 
Ajm are obtained by solving the system of linear 
equations (28). 
EVALUATION OF Bnm() 
AS AN INDEFINITE INTEGRAL 
The primary task in the perturbation method is 
the evaluation of the exchange integrals Bam(A) 
defined in equation (29). We shall accomplish this 
by proving that Bpm(A), as a function of A, satisfies 
a differential equation of the first order for which 
an explicit solution can be given. For this purpose 
we shall study the function 
#(z) = U,°@) Un) ; (31) 
where 
Un%(z) + [z+ Dn] Unr(2) = 0, (32) 
U,%(z) + lz + D,) Un%(z) = 0. (33) 
By multiplying equation (32) by U,%(z), equation 
(83) by U,,°(2) and subtracting, we obtain 
— U,° U 2) = — 0,9 — D.) Un Un; 
(34) 
= — (D,° — po fo m(z) U,°(x) dx . (35) 
Now it can be verified by direct substitution that 
F + 2F (D, +.Dm + 22) + 2F 
= (Dn® — Dz) (Un Un? — Un® Un!) 
= (Dr) Dao [ Fe) dz . (36)° 
From equation (36) it follows that 
P= 2 Qe + Deo + DOF +5 ?'| 
aS F (Dp? — D,®)? i F(a) dz. (37) 
0 
We may also note that 
F() = 2U,,°0) U,(0) = —2, (38) 
€This is the first occasion in the author’s experience where 
use is made of the fact that the product of two functions, 
each of which is a solution of a distinct second order ordinary 
linear differential equation, satisfies a fourth order linear 
differential equation.45? 
o o 
/ e-™ Fdz = e™ (F + F)| +» I e™ Fdz 
0 0 
0 
2 
= ve fe  Fdz . (39) 
z 
[- ty af F(a) dx = — pe | F(a) dx 
+3 fern waif re dz. (40) 
o 
iy Wad! 
fe F(z) dz = aN 
We now substitute equation ah in the integrand 
of equation (29) and obtain 
e- F(z) dz. (41) 
o 
Bum 0) =| re eM dem | ede x 
0 
2) (2 + Dj + Dio) F +A Fi 
dz| “* m + Dy) +5 
+ ¥ (Dy? — Dy’)? i Faas} 
1 ao 
= — Dat — Dar} | Nz 
= 2 | e* FG) de 
+[eet rato sri lo| 
rf [Ge + Dit + Dad) P+ 5 Fa 
= je / o* Fe) 
[ 22 + Dp "+ Dao) +5 M+ (Dat — D4) |e 
= 1 — 2) Bam (d) 
= eh 
+ Ban(X) po. °+ Da) +e +2 @,0 - 0,9] 
(42) 
It follows that the exchange integral B,m(A) satisfies 
the first order differential equation 
dg) ae 
The solution of equation (48) is 
» eS Wane) 
Bam (A) = we ee j 
r 
dr _ 20 +9) -2 a tet 1 (pe - Do)? (44) 
AV att i ; 
0 
