B.3 Calculus of Variations: But Briefly
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in which each curve is constrained to pass through the points (x 1 , y 1 ) and (x 2 , y 2 ).
This constraint on the path to be taken means that y(x, α) is independent of α at
those points, and hence
∂y
∂α
(x 1 ,y 1 )
=
∂y
∂α
(x 2 ,y 2 )
= 0. We are therefore able to write
Eq. (B.3.5) as
∂I
∂α
=
x 2
x 1
∂f
∂y
−
d
dx
∂f
∂y
∂y
∂α
dx .
(B.3.6)
If we now multiply Eq. (B.3.6) by the differential dα and evaluate all partial
derivatives with respect to α at α = 0, we obtain
∂I
∂α
α=0
dα =
x 2
x 1
∂f
∂y
−
d
dx
∂f
∂y
∂y
∂α
α=0
dxdα .
(B.3.7)
We shall define the quantity on the left-hand side of this equation as δI , i.e.,
∂I
∂α
α=0
dα ≡ δI ,
(B.3.8)
and call it the variation of I ; similarly, we define
∂y
∂α
α=0
dα ≡ δy, with δy thus
representing some arbitrary variation of y(x) obtained by variation of the parameter
α about α = 0. Because δy is arbitrary, we can only have
δI =
x 2
x 1
∂f
∂y
−
d
dx
∂f
∂y
dxδy = 0
(B.3.9)
if the integrand is itself zero, i.e.,
∂f
∂y
−
d
dx
∂f
∂y = 0 .
(B.3.10)
Example the shortest distance between two points in a plane.
We shall begin with the standard first year calculus expression for the differential
ds of arc length: ds =
(dx) 2 + (dy) 2 , from which we may determine the length
of an arc between two points in the (x, y)-plane as the integral
I ≡
x=x 2
x=x 1
ds =
x 2
x 1
1 +
dy
dx
2
dx .
Thus, the variational principle requires that the integral I be a minimum for the
shortest path. We have f =
1 + y 2 , from which we see that