Bounded State Problems 315
− λ(ǫ; t)∇G(t,
e
φ
ǫ
(0)) · Γ(t, ǫ; −r)
o
dt
+ 2(
e
φ
ǫ
(0) − φ
0
(0)) + B
ǫ
(0) (11.3.33)
Φ(ǫ; t
1
) = λ(ǫ; t
−
1
)∇G(t,
e
φ
ǫ
(t
1
))
− 2K(ǫ)T (
e
φ
ǫ
(0),
e
φ
ǫ
(t
1
))∂
2
T (
e
φ
ǫ
(0),
e
φ
ǫ
(t
1
)) (11.3.34)
Next let ζ be an abs olutely continuous vector function such that ζ ∈ L
2
(I
0
−r
)
and ζ(t) = 0, 0 ≤ t ≤ t
1
. Using (11.3.28) we obtain
ζ(−r) ·
Z
t
1
0
[−Γ
0
(t, ǫ; −r) +
e
ψ(ǫ; t) · Γ(t, ǫ; −r)]dt
+
Z
0
−r
ζ
′
(t) ·
(
Z
t
1
0
[−Γ
0
(s, ǫ; t) +
e
ψ(ǫ; s) · Γ(s, ǫ; t)]ds
+ 2(ey
′
ǫ
− y
′
0
) + b
′
ǫ
− B
ǫ
(t)
)
dt = 0,
where
e
ψ(ǫ; t) = ψ(ǫ; t) −2(
e
φ
′
ǫ
(t) − φ
′
0
(t)).
We have seen in (11.3.30) that the vector in bra ces is constant, and thus can
be pulled out of the integral. Thus, we have
ζ(−r)·
(
Z
t
1
0
[−Γ
0
(t, ǫ; −r) +
e
ψ(ǫ; t) · Γ(t, ǫ; −r)]dt
−
Z
t
1
0
[−Γ
0
(s, ǫ; t) +
e
ψ(ǫ; s)Γ(s, ǫ; t)]ds
− 2(ey
′
ǫ
(t) − y
′
0
(t)) −(b
′