4.3 Variational Problems of Functionals with Higher Order Derivatives
277
In formula (4.3.20), starting with the second term under the integral sign, perform
integration by parts k times respectively, here k = 1, 2, . . . , n, and take notice that
when the left endpoint is fixed, there are δy| x=x 0 = 0, δy
x=x 0
= 0, δy
x=x 0
= 0,
…, δy
(n−1)
x=x 0
= 0 when the function (4.3.15) obtains extremum, there should be
δ J = 0, that is
δ J =
x1
x0
F y +
n
k=1
(−1)
k
d k F y (k)
dx k
δydx + F| x=x1 δx 1 +
F y δy + F y δy
−
d F y
dx
δy + · · ·
+ F y (k) δy
(k−1) −
d F y (k)
dx
δy
(k−2) +
d 2 F y (k)
dx 2 δy
(k−3) − · · · + (−1)
(k−1)
d k−1 F y (k)
dx k−1 δy + · · ·
+ F y (n) δy
(n−1) −
d F y (n)
dx
δy
(n−2) +
d 2 F y (n)
dx 2 δy
(n−3) − · · · +(−1)
(n−1)
d n−1 F y (n)
dx n−1 δy
x=x1
=
Fδx +
F y +
n−1
k=1
(−1)
k
d k F y (k+1)
dx k
δy+
F y +
n−2
k=1
(−1)
k
d k F y (k+2)
dx k
δy
+ · · ·
+
⎡
⎣ F y ( j) +
n− j
k=1
(−1)
k
d k F y (k+ j)
dx k
⎤
⎦ δy
( j−1) + · · · +
F y (n−1) −
d F y (n)
dx
δy
(n−2) + F y (n) δy
(n−1)
x=x1
+
x1
x0
F y +
n
k=1
(−1)
k
d k F y (k)
dx k
δydx = 0
(4.3.21)
It can be seen that the integral terms in the above formula should be zero, therefore
the formula (4.2.16) holds.
The variations δx 1 , δy, δy
, …, δy
(n−1) are not independent respectively, but there
is the following relation
δy| x=x 1 = δy 1 − y
(x 1 )δx 1
(4.3.22)
Applying the above relation to δy
, we have
δy
x=x 1
= δy
1 − y
(x 1 )δx 1
(4.3.23)
Applying the above relation to δy
(k) , we have
δy
(k)
x=x 1
= δy
(k)
1 − y
(k+1)
(x 1 )δx 1 (k ≥ 0)
(4.3.24)
The formula (4.3.24) has contained the formula (4.3.22) and formula (4.3.23).
Substituting the relations expressed by the formula (4.3.24) into the formula (4.3.21),
after management we obtain
δ J =
F − y
F y +
n−1
k=1
(−1)
k
d k F y (k+1)
dx k
− y
F y +
n−2
k=1
(−1)
k
d k F y (k+2)
dx k
− · · ·