Lagrange multiplier applied to the Legendre transform

Lagrange multiplier applied to the Legendre transform

Since the

Legendre transform

is a

stationary point

of a function,

Lagrange multipliers

should be the natural choice for handling constraints. However, there is a problem due to the fact that we are mostly interested in its functional dependence on the transform parameter, and not on its value.

THE LEGENDRE-LAGRANGE PROBLEM

Let

f

⁢

(

x

¯

)

be a function of the real

n-vector

x

¯

and

g

⁢

(

p

¯

)

its Legendre transform defined by

g

⁢

(

p

¯

)

=

p

¯

.

x

¯

-

f

⁢

(

x

¯

)

p

i

=

∂

⁡

f

∂

⁡

x

i

(1)

The vector

p

¯

is a function of

x

¯

, or conversely,

x

¯

is a function of

p

¯

, so that

g

⁢

(

p

¯

)

is a function of

p

¯

alone.
The Legendre transform is also alternatively defined as the

maximum

of the function

p

¯

.

x

¯

-

f

⁢

(

x

¯

)

. This maximum is reached for

x

¯

satifying the second set of equations in (1).
Suppose now that the

components

x

i

are not all independent, but are linked by the constraint:

h

⁢

(

x

¯

)

=

0

(2)

This equation defines one of the components,

x

α

for example, as a function of all the others. Putting this function into (1) would give us the transform

g

⁢

(

p

¯

)

as a function of the vector

p

¯

with n-1 components. It would be nice if we could instead use a Lagrange multiplier

k

and compute the maximum of the function

p

¯

.

x

¯

-

f

⁢

(

x

¯

)

+

k

⁢

h

⁢

(

x

¯

)

with all its n components. But then, how does the constraint (2) on

x

¯

translate

to

p

¯

? The answer is amazingly simple, as we shall see next.

DIRECT COMPUTATION OF THE TRANSFORM

We are going first to compute the Legendre transform the hard way, without the help of a multiplier, by considering

x

α

as a function of the other components:

p

i

=

d

⁢

f

d

⁢

x

i

=

∂

⁡

f

∂

⁡

x

i

+

∂

⁡

f

∂

⁡

x

α

⁢

∂

⁡

x

α

∂

⁡

x

i

d

⁢

h

d

⁢

x

i

=

∂

⁡

h

∂

⁡

x

i

+

∂

⁡

h

∂

⁡

x

α

⁢

∂

⁡

x

α

∂

⁡

x

i

=

0

i

≠

α

(3)

Taking the value of

∂

⁡

x

α

∂

⁡

x

i

from the second set of equations (3) and putting it into the first set, we get:

p

i

=

∂

⁡

f

∂

⁡

x

i

-

Φ

⁢

∂

⁡

h

∂

⁡

x

i

with

Φ

=

∂

⁡

f

∂

⁡

x

α

∂

⁡

h

∂

⁡

x

α

and

i

≠

α

(4)

From

definition

(1), the Legendre transform

g

is therefore:

g

⁢

(

p

¯

)

=

∑

i

≠

α

x

i

⁢

(

∂

⁡

f

∂

⁡

x

i

-

Φ

⁢

∂

⁡

h

∂

⁡

x

i

)

-

f

⁢

(

x

¯

)

(5)

BACK TO THE LAGRANGE MULTIPLIER

In the first equation (4), if we set

i

=

α

we get

p

α

=

0

and conversely, getting the value of

Φ

. So, in (5), we may remove the condition

i

≠

α

and sum over all the n components of

p

¯

. The constraint

p

α

=

0

reduces the number of independent components to n-1. In fact, we are back to the traditional Lagrange method with the multiplier

Φ

and an additional constraint. We even have the choice between n such constraints. They generate up to n functionally different transforms corresponding to the n possible forms of the function

x

¯

, according to which component

x

α

we choose to eliminate.
The method extends easily to the case of more than one constraint:

h

1

=

h

2

=

…

⁢

h

m

=

0

. We use m multipliers

Φ

1

,

Φ

2

⁢

…

⁢

Φ

m

. They are computed by equating to zero any set of m components from

p

¯

.

References

1

http://planetmath.org/encyclopedia/LegendreTransform.html

Fersanz at PM - Legendre Transform

This link is actually broken but, hopefully, should be operative soon.

2

http://en.wikipedia.org/wiki/Legendre_transformation

Wikipedia -

Legendre transformation

Title

Lagrange multiplier applied to the Legendre transform

Canonical

name

LagrangeMultiplierAppliedToTheLegendreTransform

Date of creation

2013-03-22 18:52:47

Last modified on

2013-03-22 18:52:47

Owner

dh2718 (16929)

Last modified by

dh2718 (16929)

Numerical id

7

Author

dh2718 (16929)

Entry type

Example

Classification

msc 26B10

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