M2A1: A reduced form of the Euler-Lagrange equation
We have proved in the lectures that the Euler-Lagrange equation takes the form
d
fy0 = 0.
dx
fy −
(1)
For any arbitrary function1 F = F (x, y(x), y 0 (x)), the variable x appears in three
places; in the first position explicitly, and in the second and third positions implicitly.
From the chain rule, we have
δF =
∂F
∂F
∂F
δx +
δy + 0 δy 0 .
∂x
∂y
∂y
(2)
Dividing by δx and taking the limit δx → 0, we therefore conclude that the derivative
operation
d
dx
means
dF
∂F
∂F
∂F
=
+ y0
+ y 00 0 .
dx
∂x
∂y
∂y
Using the subscript notation fx =
∂f
,
∂x
(3)
etc, the EL-equation (1) can be rewritten in
a very important alternative form
fx +
d
(y 0 fy0 − f ) = 0.
dx
(4)
Proof: Form the combination in (4)
fx +
d
d
(y 0 fy0 − f ) = fx + y 00 fy0 + y 0 fy0 − (fx + y 0 fy + y 00 fy0 )
dx
dx
!
d
fy0 − fy = 0.
= y0
dx
(5)
In two special cases, we can integrate the Euler-Lagrange equation to obtain a more
amenable form:
1. If f = f (x, y, y 0 ) has x missing explicitly so f = f (y, y 0 ) then fx = 0 and we
can integrate (4) to get
y 0 fy0 − f = constant.
(6)
2. If f = f (x, y, y 0 ) has y missing explicitly so f = f (x, y 0 ) then fy = 0 and we
can integrate (1) to get
fy0 = constant.
1
I’m using F and not f because the the derivative
d
dx
in (1) operates on fy0 .
(7)