2Euler-Lagrange equation
IB Variational Principles
2.2 First integrals
In our example above,
f
did not depend on
x
, and hence
∂f
∂x
= 0. Then the
Euler-Lagrange equations entail
d
dt
∂f
∂ ˙x
= 0.
We can integrate this to obtain
∂f
∂ ˙x
= constant.
We call this the first integral. First integrals are important in several ways. The
most immediate advantage is that it simplifies the problem a lot. We only have
to solve a first-order differential equation, instead of a second-order one. Not
needing to differentiate
∂f
∂ ˙x
also prevents a lot of mess arising from the product
and quotient rules.
This has an additional significance when applied to problems in physics.
If we have a first integral, then we get
∂f
∂ ˙x
= constant. This corresponds to
a conserved quantity of the system. When formulating physics problems as
variational problems (as we will do in Chapter 3), the conservation of energy
and momentum will arise as constants of integration from first integrals.
There is also a more complicated first integral appearing when
f
does not
(explicitly) depend on
t
. To find this out, we have to first consider the total
derivative
df
dt
. By the chain rule, we have
df
dt
=
∂f
∂t
+
dx
dt
∂f
∂x
+
d ˙x
dt
∂f
∂ ˙x
=
∂f
∂t
+ ˙x
∂f
∂x
+ ¨x
∂f
∂ ˙x
.
On the other hand, the Euler-Lagrange equation says that
∂f
∂x
=
d
dt
∂f
∂ ˙x
.
Substituting this into our equation for the total derivative gives
df
dt
=
∂f
∂t
+ ˙x
d
dt
∂f
∂ ˙x
+ ¨x
∂f
∂ ˙x
=
∂f
∂t
+
d
dt
˙x
∂f
∂ ˙x
.
Then
d
dt
f − ˙x
∂f
∂ ˙x
=
∂f
∂t
.
So if
∂f
∂t
= 0, then we have the first integral
f − ˙x
∂f
∂ ˙x
= constant.
Example.
Consider a light ray travelling in the vertical
xz
plane inside a
medium with refractive index
n
(
z
) =
√
a − bz
for positive constants
a, b
. The
phase velocity of light is v =
c
n
.
According the Fermat’s principle, the path minimizes
T =
Z
B
A
d`
v
.
This is equivalent to minimizing the optical path length
cT = P =
Z
B
A
n d`.
We specify our path by the function z(x). Then the path element is given by
d` =
p
dx
2
+ dz
2
=
p
1 + z
0
(x)
2
dx,
Then
P [z] =
Z
x
B
x
A
n(z)
p
1 + (z
0
)
2
dx.
Since this does not depend on x, we have the first integral
k = f − z
0
∂f
∂z
0
=
n(z)
p
1 + (z
0
)
2
.
for an integration constant k. Squaring and putting in the value of n gives
(z
0
)
2
=
b
k
2
(z
0
− z),
where z
0
= (a − k
2
)/b. This is integrable and we obtain
dz
√
z
0
− z
= ±
√
b
k
dx.
So
√
z −z
0
= ±
√
b
2k
(x − x
0
),
where x
0
is our second integration constant. Square it to obtain
z = z
0
−
b
4k
2
(x − x
0
)
2
,
which is a parabola.
Example
(Principle of least action)
.
Mechanics as laid out by Newton was
expressed in terms of forces and acceleration. While this is able to describe a lot
of phenomena, it is rather unsatisfactory. For one, it is messy and difficult to
scale to large systems involving many particles. It is also ugly.
As a result, mechanics is later reformulated in terms of a variational principle.
A quantity known as the action is defined for each possible path taken by
the particle, and the actual path taken is the one that minimizes the action
(technically, it is the path that is a stationary point of the action functional).
The version we will present here is an old version proposed by Maupertuis
and Euler. While it sort-of works, it is still cumbersome to work with. The
modern version is from Hamilton who again reformulated the action principle
to something that is more powerful and general. This modern version will be
discussed more in detail in Chapter 3, and for now we will work with the old
version first.
The original definition for the action, as proposed by Maupertuis, was mass
×
velocity
×
distance. This was given a more precise mathematical definition
by Euler. For a particle with constant energy,
E =
1
2
mv
2
+ U(x),
where v = |
˙
x|. So we have
mv =
p
2m(E − U(x)).
Hence we can define the action to be
A =
Z
B
A
p
2m(E − U(x)) d`,
where d` is the path length element. We minimize this to find the trajectory.
For a particle near the surface of the Earth, under the influence of gravity,
U = mgz. So we have
A[z] =
Z
B
A
p
2mE − 2m
2
gz
p
1 + (z
0
)
2
dx,
which is of exactly the same form as the optics problem we just solved. So the
result is again a parabola, as expected.
Example
(Brachistochrone)
.
The Brachistochrone problem was one of the
earliest problems in the calculus of variations. The name comes from the Greek
words br´akhistos (“shortest”) and khr´onos (“time”).
The question is as follows: suppose we have a bead sliding along a frictionless
wire, starting from rest at the origin
A
. What shape of wire minimizes the time
for the bead to travel to B?
x
y
A
B
The conservation of energy implies that
1
2
mv
2
= mgy.
So
v =
p
2gy
We want to minimize
T =
Z
d`
v
.
So
T =
1
√
2g
Z
p
dx
2
+ dy
2
√
y
=
1
√
2g
Z
s
1 + (y
0
)
2
y
dx
Since there is no explicit dependence on x, we have the first integral
f − y
0
∂f
∂y
0
=
1
p
y(1 + (y
0
)
2
)
= constant
So the solution is
y(1 + (y
0
)
2
) = c
for some positive constant c.
The solution of this ODE is, in parametric form,
x = c(θ − sin θ)
y = c(1 − cos θ).
Note that this has x = y = 0 at θ = 0. This describes a cycloid.