2Euler-Lagrange equation
IB Variational Principles
2.1 Functional derivatives
Definition
(Functional)
.
A functional is a function that takes in another real-
valued function as an argument. We usually write them as
F
[
x
] (square brackets),
where
x
=
x
(
t
) :
R → R
is a real function. We say that
F
[
x
] is a functional of
the function x(t).
Of course, we can also have functionals of many functions, e.g.
F
[
x, y
]
∈ R
for x, y : R → R. We can also have functionals of a function of many variables.
Example.
Given a medium with refractive index
n
(
x
), the time taken by a
path x(t) from x
0
to x
1
is given by the functional
T [x] =
Z
x
1
x
0
n(x) dt.
While this is a very general definition, in reality, there is just one particular
class of functionals we care about. Given a function
x
(
t
) defined for
α ≤ t ≤ β
,
we study functional of the form
F [x] =
Z
β
α
f(x, ˙x, t) dt
for some function f.
Our objective is to find a stationary point of the functional
F
[
x
]. To do so,
suppose we vary
x
(
t
) by a small amount
δx
(
t
). Then the corresponding change
δF [x] of F [x] is
δF [x] = F [x + δx] − F [x]
=
Z
β
α
f(x + δx, ˙x + δ ˙x, t) − f (x, ˙x, t)
dt
Taylor expand to obtain
=
Z
β
α
δx
∂f
∂x
+ δ ˙x
∂f
∂ ˙x
dt + O(δx
2
)
Integrate the second term by parts to obtain
δF [x] =
Z
β
α
δx
∂f
∂x
−
d
dt
∂f
∂ ˙x
dt +
δx
∂f
∂ ˙x
β
α
.
This doesn’t seem like a very helpful equation. Life would be much easier if
the last term (known as the boundary term)
h
δx
∂f
∂ ˙x
i
β
α
vanishes. Fortunately, for
most of the cases we care about, the boundary conditions mandate that the
boundary term does indeed vanish. Most of the time, we are told that
x
is fixed
at
t
=
α, β
. So
δx
(
α
) =
δx
(
β
) = 0. But regardless of what we do, we always
choose boundary conditions such that the boundary term is 0. Then
δF [x] =
Z
β
α
δx
δF [x]
δx(t)
dt
where
Definition (Functional derivative).
δF [x]
δx
=
∂f
∂x
−
d
dt
∂f
∂ ˙x
is the functional derivative of F[x].
If we want to find a stationary point of F , then we need
δF [x]
δx
= 0. So
Definition (Euler-Lagrange equation). The Euler-Lagrange equation is
∂f
∂x
−
d
dt
∂f
∂ ˙x
= 0
for α ≤ t ≤ β.
There is an obvious generalization to functionals F [x] for x(t) ∈ R
n
:
∂f
∂x
i
−
d
dt
∂f
∂ ˙x
i
= 0 for all i.
Example
(Geodesics of a plane)
.
What is the curve
C
of minimal length between
two points A, B in the Euclidean plane? The length is
L =
Z
C
d`
where d` =
p
dx
2
+ dy
2
.
There are two ways we can do this:
(i)
We restrict to curves for which
x
(or
y
) is a good parameter, i.e.
y
can be
made a function of x. Then
d` =
p
1 + (y
0
)
2
dx.
Then
L[y] =
Z
β
α
p
1 + (y
0
)
2
dx.
Since there is no explicit dependence on y, we know that
∂f
∂y
= 0
So the Euler-Lagrange equation says that
d
dx
∂f
∂y
0
= 0
We can integrate once to obtain
∂f
∂y
0
= constant
This is known as a first integral, which will be studied more in detail later.
Plugging in our value of f , we obtain
y
0
p
1 + (y
0
)
2
= constant
This shows that y
0
must be constant. So y must be a straight line.
(ii)
We can get around the restriction to “good” curves by choosing an arbitrary
parameterization
r
= (
x
(
t
)
, y
(
t
)) for
t ∈
[0
,
1] such that
r
(0) =
A
,
r
(1) =
B
.
So
d` =
p
˙x
2
+ ˙y
2
dt.
Then
L[x, y] =
Z
1
0
p
˙x
2
+ ˙y
2
dt.
We have, again
∂f
∂x
=
∂f
∂y
= 0.
So we are left to solve
d
dt
∂f
∂ ˙x
=
d
dt
∂f
∂ ˙y
= 0.
So we obtain
˙x
p
˙x
2
+ ˙y
2
= c,
˙y
p
˙x
2
+ ˙y
2
= s
where
c
and
s
are constants. While we have two constants, they are not
independent. We must have
c
2
+
s
2
= 1. So we let
c
=
cos θ
,
s
=
sin θ
.
Then the two conditions are both equivalent to
( ˙x sin θ)
2
= ( ˙y cos θ)
2
.
Hence
˙x sin θ = ±˙y cos θ.
We can choose a θ such that we have a positive sign. So
y cos θ = x sin θ + A
for a constant A. This is a straight line with slope tan θ.