7.5 Derivatives of Functionals
417
Making use of the computing result of the third term, the former integral of the
fourth term can be written as
−
x 1
x 0
ηη
2 d
dx
F y y y dx =
x 1
x 0
η
1
2
d
dx
d
dx
F y y y
d
dx
η
2 dx
Using integration by parts to the later integral of the fourth term, we obtain
−
x 1
x 0
F y y y 2ηη
η
dx = −
x 1
x 0
F y y y (η
2
)
dη
= −2F y y y ηη
2
x 1
x 0
+
x 1
x 0
η
d
dx
F y y y
d
dx
η
2
dx
=
x 1
x 0
d
dx
F y y y
d
dx
η
2
dη
=
d
dx
F y y y
d
dx
η
2
η
x 1
x 0
−
x 1
x 0
η
d
2
dx 2
F y y y
d
dx
η
2
dx
= −
x 1
x 0
η
d
2
dx 2
F y y y
d
dx
η
2 dx
Substituting the second, third, fourth terms into the expression of the third
variation, we obtain
δ
3 J =
ε
3
3!
x 1
x 0
(F yyy η
3
+ 3F yyy η
2
η
+ 3F yy y ηη
2
+ F y y y η
3
)dx
=
ε
3
3!
x 1
x 0
η
F yyy +
3
2
F yyy
d
dx
−
3
2
d
dx
F yy y
d
dx
+
1
2
d
dx
d
dx
F y y y
d
dx
−
d
2
dx 2
F y y y
d
dx
η
2 dx
=
ε
3
3!
[η, T (y)η
2
] =
ε
3
3!
[T (y)η
2
, η]
Thus, the third derivative of the functional J [y] is obtained
J
[y] = T (y) = F yyy +
3
2
F yyy
d
dx
−
3
2
d
dx
F yy y
d
dx
+
1
2
d
dx
d
dx
F y y y
d
dx
−
d
2
dx 2
F y y y
d
dx
(7.5.5)
It can be seen from the expression (7.3.5) that the third derivative of the functional
J [y] is still an operator.