5The second variation
IB Variational Principles
5.2 Jacobi condition for local minima of F [x]
Legendre tried to prove that
ρ >
0 is a sufficient condition for
δ
2
F >
0. This is
known as the strong Legendre condition. However, he obviously failed, since it is
indeed not a sufficient condition. Yet, it turns out that he was close.
Before we get to the actual sufficient condition, we first try to understand
why thinking ρ > 0 is sufficient isn’t as crazy as it first sounds.
If
ρ >
0 and
σ <
0, we would want to create a negative
δ
2
F
[
x
0
, ξ
] by choosing
ξ
to be large but slowly varying. Then we will have a very negative
σ
(
t
)
ξ
2
while
a small positive ρ(t)
˙
ξ
2
.
The problem is that
ξ
has to be 0 at the end points
α
and
β
. For
ξ
to take a
large value, it must reach the value from 0, and this requires some variation of
ξ
,
thereby inducing some
˙
ξ
. This is not a problem if
α
and
β
are far apart - we
simply slowly climb up to a large value of
ξ
and then slowly rappel back down,
maintaining a low
˙
ξ
throughout the process. However, it is not unreasonable to
assume that as we make the distance
β −α
smaller and smaller, eventually all
ξ
will lead to a positive
δ
2
F
[
x
0
, ξ
], since we cannot reach large values of
ξ
without
having large
˙
ξ.
It turns out that the intuition is correct. As long as
α
and
β
are sufficiently
close,
δ
2
F
[
x
0
, ξ
] will be positive. The derivation of this result is, however, rather
roundabout, involving a number of algebraic tricks.
For a solution x
0
to the Euler Lagrange equation, we have
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
.
Assume
ρ
(
t
)
>
0 for
α < t < β
(the strong Legendre condition) and assume
boundary conditions ξ(α) = ξ(β) = 0. When is this sufficient for δ
2
F > 0?
First of all, notice that for any smooth function w(x), we have
0 =
Z
β
α
(wξ
2
)
0
dt
since this is a total derivative and evaluates to
wξ
(
α
)
− wξ
(
β
) = 0. So we have
0 =
Z
β
α
(2wξ
˙
ξ + ˙wξ
2
) dt.
This allows us to rewrite δ
2
F as
δ
2
F =
Z
β
α
ρ
˙
ξ
2
+ 2wξ
˙
ξ + (σ + ˙w)ξ
2
dt.
Now complete the square in ξ and
˙
ξ. So
δ
2
F =
Z
β
α
"
ρ
˙
ξ +
w
ρ
ξ
2
+
σ + ˙w −
w
2
ρ
ξ
2
#
dt
This is non-negative if
w
2
= ρ(σ + ˙w). (∗)
So as long as we can find a solution to this equation, we know that
δ
2
F
is
non-negative. Could it be that
δ
2
F
= 0? Turns out not. If it were, then
˙
ξ = −
w
ρ
ξ. We can solve this to obtain
ξ(x) = C exp
−
Z
x
α
w(s)
ρ(s)
ds
.
We know that ξ(α) = 0. But ξ(α) = Ce
0
. So C = 0. Hence equality holds only
for ξ = 0.
So all we need to do is to find a solution to (
∗
), and we are sure that
δ
2
F >
0.
Note that this is non-linear in
w
. We can convert this into a linear equation
by defining w in terms of a new function u by w = −ρ ˙u/u. Then (∗) becomes
ρ
˙u
u
2
= σ −
ρ ˙u
u
0
= σ −
(ρ ˙u)
0
u
+ ρ
˙u
u
2
.
We see that the left and right terms cancel. So we have
−(ρ ˙u)
0
+ σu = 0.
This is the Jacobi accessory equation, a second-order linear ODE.
There is a caveat here. Not every solution
u
will do. Recall that
u
is used
to produce
w
via
w
=
−ρ ˙u/u
. Hence within [
α, β
], we cannot have
u
= 0 since
we cannot divide by zero. If we can find a non-zero
u
(
x
) satisfying the Jacobi
accessory equation, then
δ
2
F >
0 for
ξ 6
= 0, and hence
y
0
is a local minimum of
F .
A suitable solution will always exists for sufficiently small
β − α
, but may
not exist if β − α is too large, as stated at the beginning of the chapter.
Example
(Geodesics on unit sphere)
.
For any curve
C
on the sphere, we have
L =
Z
C
q
dθ
2
+ sin
2
θ dφ
2
.
If θ is a good parameter of the curve, then
L[φ] =
Z
θ
2
θ
1
q
1 + sin
2
θ(φ
0
)
2
dθ.
Alternatively, if φ is a good parameter, we have
L[θ] =
Z
φ
2
φ
1
q
(θ
0
)
2
+ sin
2
θ dφ.
We will look at the second case.
We have
f(θ, θ
0
) =
q
(θ
0
)
2
+ sin
2
θ.
So
∂f
∂θ
=
sin θ cos θ
q
(θ
0
)
2
+ sin
2
θ
,
∂f
∂θ
0
=
θ
0
q
(θ
0
)
2
+ sin
2
θ
.
Since
∂f
∂φ
= 0, we have the first integral
const = f − θ
0
∂f
∂θ
0
=
sin
2
θ
q
(θ
0
)
2
+ sin
2
θ
So a solution is
c sin
2
θ =
q
(θ
0
)
2
+ sin
2
θ.
Here we need c ≥ 1 for the equation to make sense.
We will consider the case where
c
= 1 (in fact, we can show that we can
always orient our axes such that
c
= 1). This occurs when
θ
0
= 0, i.e.
θ
is a
constant. Then our first integral gives
sin
2
θ
=
sin θ
. So
sin θ
= 1 and
θ
=
π/
2.
This corresponds to a curve on the equator. (we ignore the case
sin θ
= 0 that
gives θ = 0, which is a rather silly solution)
There are two equatorial solutions to the Euler-Lagrange equations. Which,
if any, minimizes L[θ]?
We have
∂
2
f
∂(θ
0
)
2
θ=π/2
= 1
and
∂
2
f
∂θ∂θ
0
= −1,
∂
2
∂θ∂θ
0
= 0.
So ρ(x) = 1 and σ(x) = −1. So
δ
2
F =
Z
φ
2
φ
1
((ξ
0
)
2
− ξ
2
) dφ.
The Jacobi accessory equation is
u
00
+
u
= 0. So the general solution is
u ∝
sin φ − γ cos φ. This is equal to zero if tan φ = γ.
Looking at the graph of
tan φ
, we see that
tan
has a zero every
π
radians.
Hence if the domain
φ
2
− φ
1
is greater than
π
(i.e. we go the long way from
the first point to the second), it will always contain some values for which
tan φ
is zero. So we cannot conclude that the longer path is a local minimum (it
is obviously not a global minimum, by definition of longer) (we also cannot
conclude that it is not a local minimum, since we tested with a sufficient and
not necessary condition). On the other hand, if
φ
2
− φ
1
is less than
π
, then we
will be able to pick a γ such that u is non-zero in the domain.