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3.1 Derivation of The Euler-Lagrange Equation

This document derives the Euler-Lagrange equation, which provides a necessary condition for a function y(x) to be an extremal of a given functional I and minimize or maximize I. It considers small variations of y(x) and uses Taylor series expansion to show that the extremals y(x) must satisfy the Euler-Lagrange equation, a second-order ordinary differential equation involving the partial derivatives of F with respect to y and y'. A fundamental lemma of calculus of variations is also stated, establishing that if an integral of a function η times a continuous function g is zero for all η, then g must be zero everywhere.

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574 views2 pages

3.1 Derivation of The Euler-Lagrange Equation

This document derives the Euler-Lagrange equation, which provides a necessary condition for a function y(x) to be an extremal of a given functional I and minimize or maximize I. It considers small variations of y(x) and uses Taylor series expansion to show that the extremals y(x) must satisfy the Euler-Lagrange equation, a second-order ordinary differential equation involving the partial derivatives of F with respect to y and y'. A fundamental lemma of calculus of variations is also stated, establishing that if an integral of a function η times a continuous function g is zero for all η, then g must be zero everywhere.

Uploaded by

Akhilrajscribd
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© © All Rights Reserved
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MATH2650

3.1 Derivation of the Euler-Lagrange Equation


The classical problem in Calculus of Variations is to find the functions y(x) which extremize a given
functional
Z x2
F (x, y, y ) dx
I[y] =
x1

subject to the boundary conditions y(x1 ) = y1 , y(x2 ) = y2 . The solutions y(x) of this problem are
called the extremals of I (or of F ), and the corresponding values of I are the extrema. We derive the
Euler-Lagrange equation, which provides a necessary condition for y(x) to be an extremal of I.
Suppose y(x) is an extremal, i.e. a particular1 function which extremizes I. We consider small
variations of the form
y(x) = y(x) + (x) ,
where is a small parameter and (x) is any2 function satisfying the boundary conditions (x1 ) = 0,
(x2 ) = 0 (so that the variation y(x) satisfies the same boundary conditions as the extremal y(x)).
Since y(x) is an extremal, the functional I should be stationary with respect to all such variations,
i.e. we must have



dI
I[
y ] I[y]
=0
()
= lim
d =0 0

for every possible choice of (x). Let I = I[


y ] I[y]. Then
Z
Z x2
Z x2
Z x2

F (x, y + , y + ) dx
F (x, y, y ) dx =
F (x, y, y ) dx
I =
x1

x1

x1

x2

F (x, y, y ) dx .

x1

We expand2 the first integrand as a Taylor series, keeping only the leading terms:

Z x2 
Z x2
F
F

I =
F
(x, y, y ) +
+ + O( ) dx
F
(x,y,
) dx

y
y
x1
x1

Z x2 
F
F
+ dx + O(2) .
=

y
y
x1
Now, integrating the second term by parts and applying the boundary conditions on ,

x


Z x2 
F
d F
F 2

dx + O(2) .
+
I =

y x1
y
dx y

x1

Thus the requirement () becomes:



 Z x2 


I
F
d F
lim

dx = 0 ,
0

y
dx y
x1
and, because this must hold for every 2 choice of (x), we deduce3 the Euler-Lagrange equation:


F
d F

=0
x [x1 , x2 ] .
y
dx y
1 This

is a slight abuse of notation, as originally y(x) was general. If you prefer, call the particular extremal y0 (x) or similar.
the derivation to be rigorous, should be continuously differentiable and F should have continuous partial derivatives.
3 Formally, this step requires a lemma (see overleaf) sometimes known as the Fundamental Lemma of Calculus of Variations.

2 For

MATH2650
For a particular function F (x, y, y ), the Euler-Lagrange equation is a 2nd-order ODE which the
extremals y(x) must satisfy.

For a fully rigorous approach, the final step in the preceding derivation of the Euler-Lagrange equation requires the following
lemma, known as the Fundamental Lemma of Calculus of Variations (which is non-examinable this year).
Lemma: Let g(x) be continuous on [x1 , x2 ] and suppose that
Z

x2

(x)g(x) dx = 0
x1

for every continuously differentiable function (x) satisfying (x1 ) = (x2 ) = 0. Then g(x) = 0 x [x1 , x2 ].
Proof: Suppose there exists a point (x1 , x2 ) such that g() 6= 0, WLOG g() > 0. Then because g is continuous on the
interval [x1 , x2 ], there exists a subinterval (1 , 2 ) containing on which g(x) > 0. Consider (x) defined as follows:
8
<(x 1 )2 (x 2 )2
(x) =
:0

if x [1 , 2 ] ,
if x
/ [1 , 2 ] .

Then (x) is continuously differentiable on [x1 , x2 ] and (x1 ) = (x2 ) = 0. But


Z

x2

(x)g(x) dx =
x1

(x 1 )2 (x 2 )2 g(x) dx > 0 ,

since both (x) and g(x) are strictly positive on (1 , 2 ); thus we have established a contradiction, and we deduce that there
can be no such point . Therefore g(x) = 0 on (x1 , x2 ) and, by continuity, g must also be zero at the endpoints, hence g(x) = 0
x [x1 , x2 ].

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