5The second variation
IB Variational Principles
5.1 The second variation
So far, we have only looked at the “first derivatives” of functionals. We can
identify stationary points, but we don’t know if it is a maximum, minimum or a
saddle. To distinguish between these, we have to look at the “second derivatives”,
or the second variation.
Suppose x(t) = x
0
(t) is a solution of
δF [x]
δy(x)
= 0,
i.e. F [x] is stationary at y = y
0
.
To determine what type of stationary point it is, we need to expand
F
[
x
+
δx
]
to second order in
δx
. For convenience, let
δx
(
t
) =
εξ
(
t
) with constant
ε
1.
We will also only consider functionals of the form
F [x] =
Z
β
α
f(x, ˙x, t) dt
with fixed-end boundary conditions, i.e.
ξ
(
α
) =
ξ
(
β
) = 0. We will use both dots
( ˙x) and dashes (x
0
) to denote derivatives.
We consider a variation x 7→ x + δx and expand the integrand to obtain
f(x + εξ, ˙x + ε
˙
ξ, t) − f(x, ˙x, t)
= ε
ξ
∂f
∂x
+
˙
ξ
∂f
∂ ˙x
+
ε
2
2
ξ
2
∂
2
f
∂x
2
+ 2ξ
˙
ξ
∂
2
f
∂x∂ ˙x
+
˙
ξ
2
∂
2
f
∂ ˙x
2
+ O(ε
3
)
Noting that 2ξ
˙
ξ = (ξ
2
)
0
and integrating by parts, we obtain
= εξ
∂f
∂x
−
d
dt
∂f
∂ ˙x
+
ε
2
2
ξ
2
∂
2
f
∂x
2
−
d
dt
∂
2
f
∂x∂ ˙x
+
˙
ξ
2
∂f
∂ ˙x
2
.
plus some boundary terms which vanish. So
F [x + εξ] − F [x] =
Z
β
α
εξ
∂f
∂x
−
d
dt
∂f
∂ ˙x
dt +
ε
2
2
δ
2
F [x, ξ] + O(ε
3
),
where
δ
2
F [x, ξ] =
Z
β
α
ξ
2
∂
2
f
∂x
2
−
d
dt
∂
2
f
∂x∂ ˙x
+
˙
ξ
2
∂
2
f
∂ ˙x
2
dt
is a functional of both x(t) and ξ(t). This is analogous to the term
δx
T
H(x)δx
appearing in the expansion of a regular function
f
(
x
). In the case of normal
functions, if
H
(
x
) is positive,
f
(
x
) is convex for all
x
, and the stationary point
is hence a global minimum. A similar result holds for functionals.
In this case, if
δ
2
F
[
x, ξ
]
>
0 for all non-zero
ξ
and all allowed
x
, then a
solution x
0
(t) of
δF
δx
= 0 is an absolute minimum.
Example
(Geodesics in the plane)
.
We previously shown that a straight line is
a stationary point for the curve-length functional, but we didn’t show it is in
fact the shortest distance! Maybe it is a maximum, and we can get the shortest
distance by routing to the moon and back.
Recall that f =
p
1 + (y
0
)
2
. Then
∂f
∂y
= 0,
∂f
∂y
0
=
y
0
p
1 + (y
0
)
2
,
∂
2
f
∂y
02
=
1
p
1 + (y
0
)
2
3
,
with the other second derivatives zero. So we have
δ
2
F [y, ξ] =
Z
β
α
˙
ξ
2
(1 + (y
0
)
2
)
3/2
dx > 0
So if we have a stationary function satisfying the boundary conditions, it is an
absolute minimum. Since the straight line is a stationary function, it is indeed
the minimum.
However, not all functions are convex
[citation needed]
. We can still ask whether
a solution
x
0
(
t
) of the Euler-Lagrange equation is a local minimum. For these,
we need to consider
δ
2
F [x
0
, ξ] =
Z
β
α
(ρ(t)
˙
ξ
2
+ σ(t)ξ
2
) dt,
where
ρ(t) =
∂
2
f
∂ ˙x
2
x=x
0
, σ(t) =
∂
2
f
∂x
2
−
d
dt
∂
2
f
∂x∂ ˙x
x=x
0
.
This is of the same form as the Sturm-Liouville problem. For
x
0
to minimize
F [x] locally, we need δ
2
F [x
0
, ξ] > 0. A necessary condition for this is
ρ(t) ≥ 0,
which is the Legendre condition.
The intuition behind this necessary condition is as follows: suppose that
ρ
(
t
) is negative in some interval
I ⊆
[
α, β
]. Then we can find a
ξ
(
t
) that makes
δ
2
F
[
x
0
, ξ
] negative. We simply have to make
ξ
zero outside
I
, and small but
crazily oscillating inside
I
. Then inside
I
,
˙x
2
wiill be very large while
ξ
2
is kept
tiny. So we can make δ
2
F [y, ξ] arbitrarily negative.
Turning the intuition into a formal proof is not difficult but is tedious and
will be omitted.
However, this is not a sufficient condition. Even if we had a strict inequality
ρ(t) > 0 for all α < t < β, it is still not sufficient.
Of course, a sufficient (but not necessary) condition is
ρ
(
t
)
>
0
, σ
(
t
)
≥
0, but
this is not too interesting.
Example. In the Branchistochrone problem, we have
T [x] ∝
Z
β
α
r
1 + ˙x
2
x
dt.
Then
ρ(t) =
∂
2
f
∂ ˙x
2
x
0
> 0
σ(t) =
1
2x
2
p
x(1 + ˙x
2
)
> 0.
So the cycloid does minimize the time T .