derivation of Euler-Lagrange differential equation (elementary)

derivation of Euler-Lagrange differential equation (elementary)

Let

[

e

,

c

]

be a finite

subinterval

of

(

a

,

b

)

. Let the function

h

:

ℝ

→

ℝ

be chosen so that a)

h

is twice

differentiable

, b)

h

⁢

(

t

)

=

0

when

t

∉

[

e

,

c

]

, c)

h

⁢

(

t

)

>

0

when

t

∈

(

e

,

c

)

, and d)

∫

e

c

h

⁢

(

t

)

⁢

𝑑

t

=

1

.

Choose

f

⁢

(

λ

,

t

)

=

q

⁢

(

t

)

+

λ

⁢

h

⁢

(

t

)

. It is easy to see that this function satisfies the requirements for

f

laid out in the main entry. Then, we can write

g

⁢

(

λ

,

x

)

=

∫

a

b

L

⁢

(

t

,

q

⁢

(

t

)

+

λ

⁢

h

⁢

(

t

)

,

q

˙

⁢

(

t

)

+

λ

⁢

h

˙

⁢

(

t

)

)

⁢

𝑑

t

Let us split the integration into three parts — the

integral

from

a

to

e

, the integral from

e

to

c

, and the integral from

c

to

b

. By the way

h

was chosen, the integrand reduces to

L

⁢

(

t

,

q

⁢

(

t

)

⁢

(

t

)

,

q

˙

⁢

(

t

)

)

when

t

∈

(

a

,

e

)

or

t

∈

(

c

,

b

)

. Hence the pieces of the integral over the

intervals

(

a

,

e

)

and

(

c

,

b

)

do not depend on

λ

and we have

d

⁢

g

d

⁢

λ

=

d

d

⁢

λ

⁢

∫

e

c

L

⁢

(

t

,

q

⁢

(

t

)

+

λ

⁢

h

⁢

(

t

)

,

q

˙

⁢

(

t

)

+

λ

⁢

h

˙

⁢

(

t

)

)

⁢

𝑑

t

Since

[

e

,

c

]

is closed and

bounded

, it is

compact

. By our assumption, the

derivative

of the integrand is

continuous

. Since

continuous functions

on compact sets are

uniformly continuous

, the derivative of the integrand is uniformly continuous. This imples that it is permissible to interchange

differentiation

and integration:

d

⁢

g

d

⁢

λ

=

∫

e

c

d

d

⁢

λ

⁢

L

⁢

(

t

,

q

⁢

(

t

)

+

λ

⁢

h

⁢

(

t

)

,

q

˙

⁢

(

t

)

+

λ

⁢

h

˙

⁢

(

t

)

)

⁢

𝑑

t

Using the

chain rule

(several variables) and setting

λ

=

0

, we have

d

⁢

g

d

⁢

λ

|

λ

=

0

=

∫

e

c

h

⁢

(

t

)

⁢

∂

⁡

L

∂

⁡

q

⁢

(

t

,

q

⁢

(

t

)

,

q

˙

⁢

(

t

)

)

+

h

˙

⁢

(

t

)

⁢

∂

⁡

L

∂

⁡

q

˙

⁢

(

t

,

q

⁢

(

t

)

,

q

˙

⁢

(

t

)

)

d

⁢

t

Integrating by parts and using the fact that

h

was chosen so as to vanish at the endpoints

e

qnd

c

, we find that

d

⁢

g

d

⁢

λ

|

λ

=

0

=

∫

e

c

h

⁢

(

t

)

⁢

[

∂

⁡

L

∂

⁡

q

⁢

(

t

,

q

⁢

(

t

)

,

q

˙

⁢

(

t

)

)

-

d

d

⁢

t

⁢

(

∂

⁡

L

∂

⁡

q

˙

⁢

(

t

,

q

⁢

(

t

)

,

q

˙

⁢

(

t

)

)

)

]

d

⁢

t

=

∫

e

c

h

⁢

(

t

)

⁢

E

⁢

L

⁢

(

t

)

⁢

𝑑

t

(The last equals sign defines

E

⁢

L

as the quantity in the brackets in the first integral.)

I claim that requiring

d

⁢

g

/

d

⁢

λ

=

0

for all finite intervals

[

e

,

c

]

⊂

(

a

,

b

)

implies that the

E

⁢

L

⁢

(

t

)

must equal zero for all

t

∈

[

a

,

b

]

. By our assumptions,

E

⁢

L

is a continuous function. Hence, for every

t

0

∈

(

a

,

b

)

and every

ϵ

>

0

, there must exist and

[

e

,

c

]

⊂

(

a

,

b

)

such that

t

0

∈

[

e

,

c

]

and

t

1

∈

[

e

,

c

]

implies that

|

E

⁢

L

⁢

(

t

0

)

-

E

⁢

L

⁢

(

t

1

)

|

<

ϵ

. Therefore,

|

d

⁢

g

d

⁢

λ

|

λ

=

0

-

E

L

(

t

0

)

|

=

|

∫

e

c

h

(

t

)

(

E

L

(

t

)

-

E

L

(

t

0

)

)

d

t

|

<

ϵ

|

∫

e

c

h

(

t

)

d

t

|

=

ϵ

Since this must be true for all

ϵ

>

0

, it follows that

E

⁢

L

⁢

(

t

0

)

=

0

for all

t

0

∈

(

a

,

b

)

. In other words, q satisfies the Euler-Lagrange equation.

Title

derivation

of

Euler-Lagrange differential equation

(elementary)

Canonical name

DerivationOfEulerLagrangeDifferentialEquationelementary

Date of creation

2013-03-22 14:45:35

Last modified on

2013-03-22 14:45:35

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

16

Author

rspuzio (6075)

Entry type

Derivation

Classification

msc 47A60

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