16—Calculus of Variations 386plus terms of higher order inδyandδy′.
Put this into Eq. (16.6), andδI=
∫badx
[
∂F
∂y
δy+
∂F
∂y′
δy′
]
(16.7)
For example, LetF=x^2 +y^2 +y′^2 on the interval 0 ≤x≤ 1. Take a base path to be a straight
line from(0,0)to(1,1). Choose for the change in the pathδy(x) =x(1−x). This is simple and it
satisfies the boundary conditions.
I[y]=
∫ 1
0dx
[
x^2 +y^2 +y′^2
]
=
∫ 1
0dx
[
x^2 +x^2 + 1^2
]
=
5
3
I[y+δy]=
∫ 1
0[
x^2 +
(
x+x(1−x)
) 2
+
(
1 +(1− 2 x)
) 2 ]
=
5
3
+
1
6
+
11
30
^2
(16.8)
The value of Eq. (16.7) is
δI=
∫ 1
0dx
[
2 yδy+ 2y′δy′
]
=
∫ 1
0dx[2xx(1−x) + 2(1− 2 x)] =
1
6
Return to the general case of Eq. (16.7) and you will see that I’ve explicitly used onlypartof theassumption that the endpoint of the path hasn’t moved,∆a= ∆b= 0. There’s nothing in the body
of the integral itself that constrains the change in they-direction, and I had to choose the functionδy
by hand so that this constraint held. In order to use the equationsδy(a) =δy(b) = 0more generally,
there is a standard trick: integrate by parts. You’llalwaysintegrate by parts in these calculations.
∫badx
∂F
∂y′
δy′=
∫badx
∂F
∂y′
dδy
dx
=
∂F
∂y′
δy
∣
∣
∣∣
ba−
∫badx
d
dx
(
∂F
∂y′
)
δy(x)
This expression allows you to use the information that the path hasn’t moved at its endpoints in they
direction either. The boundary term from this partial integration is
∂F
∂y′
δy
∣
∣∣
∣
ba=
∂F
∂y′
(
b,y(b)
)
δy(b)−
∂F
∂y′
(
a,y(a)
)
δy(a) = 0
Put the last two equations back into the expression forδI, Eq. (16.7) and the result is
δI=
∫badx
[
∂F
∂y
−
d
dx
(
∂F
∂y′
)]
δy (16.9)
Use this expression for the same exampleF=x^2 +y^2 +y′^2 withy(x) =xand you have
δI=
∫ 1
0dx
[
2 y−
d
dx
2 y′
]
δy=
∫ 1
0dx[2x−0]x(1−x) =
1
6
This is sort of like Eq. (8.16),