Legendre Transform Article Index for
Legendre 		Website Links For
Legendre 		 

Information About

Legendre Transform
  CATEGORIES ABOUT LEGENDRE TRANSFORMATION
   
transforms
   
duality theories
   
fundamental physics concepts
   
APPAREL
BABY
BEAUTY
BOOKS
CAR TOYS
CELL PHONES
DVD'S
ELECTRONICS
GOURMET FOOD
GROCERIES
HEALTH & PERSONAL
HOME & GARDEN
JEWELRY
MUSIC
MUSIC INSTRUMENTS
OFFICE PRODUCTS
SOFTWARE
SPORTING GOODS
TOOLS & HARDWARE
TOYS
VIDEO GAMES
SHOPPING HOME
MORE SHOPPING...



In Mathematics , it is often desirable to express a functional relationship f(x)\, as a different function, whose argument is the derivative of ''f'' , rather than ''x'' . If we let ''p'' = ''df''/''dx''  be the argument of this new function, then this new function is written f^\star(p)\, and is called the Legendre transform of the original function, after Adrien-Marie Legendre .

The Legendre transform f^\star of a function f\, is defined as follows:

:f^\star(p) = \mathrm{max}_x(px-f(x)).

The notation \max_x indicates the maximization of the expression with respect to the variable x while ''p'' is held constant. The Legendre transform is its own inverse. Like the familiar Fourier Transform , the Legendre transform takes a function f(x) and produces a function of a different variable p. However, while the Fourier transform consists of an integration with a Kernel , the Legendre transform uses maximization as the transformation procedure. The transform is well behaved only if f(x) is a Convex Function :

: rac{d^2f}{dx^2}> 0

The Legendre transformation is an application of the Duality relationship between points and lines. The functional relationship specified by ''f''(''x'') can be represented equally well as a set of (''x'', ''y'') points, or as a set of tangent lines specified by their slope and intercept values.

The Legendre transformation can be generalized to the Legendre-Fenchel Transformation . It is commonly used in Thermodynamics and in the hamiltonian formulation of classical mechanics.


DEFINITIONS

The definition of the Legendre transform can be made more explicit. To maximize px-f(x) with respect to x, we set its derivative equal to zero:

:: rac{d}{dx} \left(px-f(x) ight) = p-{df(x) \over dx} = 0. \quad \quad (1)\,

Thus, the expression is maximized when
:p = {df(x) \over dx} \quad \quad \quad \quad \quad \quad (2).
This is a maximum because the second derivative is negative:

:{d^2 \over dx^2}(xp-f(x)) = -{d^2f(x) \over dx^2} < 0,

since f was assumed convex. Next we invert (2) to obtain x as a function of p and plug this into (1) , which gives the more useful form,

::f^\star(p) = p \,\, x(p) - f(x(p)).

This definition gives the conventional procedure for calculating the Legendre transform of f(x): find p = {df \over dx}, invert for x and substitute into the expression xp-f(x). This definition makes clear the following interpretation: the Legendre transform produces a new function, in which the independent variable x is replaced by p = {df \over dx}, which is the derivative of the original function with respect to x.


Another definition

There is a third definition of the Legendre transform: f\, and f^\star are said to be Legendre transforms of each other if their first Derivative s are Inverse Function s of each other:

:Df = \left( Df^\star ight)^{-1}.

We can see this by taking derivative of f^\star:

:{df^\star(p) \over dp} = {d \over dp}(xp-f(x)) = x.

Combining this equation with the maximization condition results in the following pair of reciprocal equations:

:p = {df \over dx}(x),

:x = {df^\star \over dp}(p).

We see that Df and Df^\star are inverses, as promised. They are unique up to an additive constant which is fixed by the additional requirement that

:f(x) + f^\star(y) = x\,y.

Although in some cases (e.g. thermodynamic potentials) a non-standard requirement is used:

:f(x) - f^\star(y) = x\,y.

The standard constraint will be considered in this article unless otherwise noted. The Legendre transformation is its own inverse, and is related to Integration By Parts .


APPLICATIONS



Thermodynamics


The strategy behind the use of Legendre transforms is to shift the dependence of a function from one Independent Variable to another (the derivative of the original function with regard to this independent variable) by taking the difference between the original function and their product. They are used to transform among the various Thermodynamic Potential s. For example, while the Internal Energy is an explicit function of the '' Extensive Variables '', Entropy , Volume (and Chemical Composition )

: U = U(S,V,\{N_i\})\,

the Enthalpy , the (non standard) Legendre transform of ''U'' with respect to −''PV''

: H = U + PV \, = H(S,P,\{N_i\})\,
: P=\, -\left( rac{\partial U}{\partial V} ight)_S\,

becomes a function of the entropy and the '' Intensive Quantity '', Pressure , as natural variables, and is useful when the (external) ''P'' is constant. The Free Energies ( Helmholtz and Gibbs ), are obtained through further Legendre transforms, by subtracting ''TS'' (from ''U'' and ''H'' respectively), shift dependence from the entropy ''S'' to its conjugate intensive variable Temperature ''T'', and are useful when it is constant.


Hamilton-Lagrange mechanics


A Legendre transform is used in Classical Mechanics to derive the Hamiltonian Formulation from the Lagrangian One , and conversely. While the Lagrangian is an explicit function of the positional Coordinates ''q''''j'' and generalized Velocities d''q''''j'' /d''t'' (and time), the Hamiltonian shifts the functional dependence to the positions and '' Momenta '',defined as p_j= rac{\partial L}{\partial \dot{q}_j}

:H\left(q_j,p_j,t ight) = \sum_i \dot{q}_i p_i - L(q_j,\dot{q}_j,t) \,.

Each of the two formulations has its own applicability, both in the Theoretical foundations of the subject, and in practice, depending on the ease of Calculation for a particular problem. The coordinates are not necessarily Rectilinear , but can also be Angle s, etc. An optimum choice takes advantage of the actual physical Symmetries .


An example - variable capacitor

As another example from Physics , consider a parallel-plate Capacitor whose plates can approach or recede from one another, exchanging work with external mechanical Force s which maintain the plate separation — analogous to a Gas in a Cylinder with a Piston . We want the attractive force f between the plates as a function of the variable separation '''x'''. (The two Vectors point in opposite directions.) If the '' Charges '' on the plates remain constant as they move, the force is the negative Gradient of the Electrostatic energy

: U (Q, \mathbf{x} ) = \begin{matrix} rac{1}{2} \end{matrix} QV \,.

However, if the '' Volt age'' between the plates ''V'' is maintained constant by connection to a Battery , which is a reservoir for charge at constant potential difference, the force now becomes the negative gradient of the Legendre transform

: U - QV = -\begin{matrix} rac{1}{2} \end{matrix} QV \,.

The two functions happen to be negatives only because of the Linear ity of the Capacitance . Of course, for given charge, voltage and distance, the Static force must be the same by either calculation since the plates cannot "know" what might be held constant as they move.


EXAMPLES


The Exponential Function e''x'' has  ''x'' ln ''x'' − ''x''  as a Legendre transform since the respective first derivatives e''x'' and ln ''x'' are inverse to each other. This example shows that the respective Domain s of a function and its Legendre transform need not agree.

Similarly, the Quadratic Form

: f(x) = \begin{matrix} rac{1}{2} \end{matrix} \, x^t \, A \, x

with ''A'' a Symmetric Invertible ''n''-by-''n''- Matrix has

: f^\star(y) = \begin{matrix} rac{1}{2} \end{matrix} \, y^t \, A^{-1} \, y

as a Legendre transform.


LEGENDRE TRANSFORMATION IN ONE DIMENSION


In one dimension, a Legendre transform to a function ''f'' : R → R with an invertible first derivative may be found using the formula

: f^\star(y) = y \, x - f(x), \, x = \dot{f}^{-1}(y)

This can be seen by integrating both sides of the defining condition restricted to one-dimension

: \dot{f}(x) = \dot{f}^{\star-1}(x)

from ''x''0 to ''x''1, making use of the Fundamental Theorem Of Calculus on the left hand side and Substituting

: y = \dot{f}^{\star-1}(x)

on the right hand side to find

: f(x_1) - f(x_0) = \int_{y_0}^{y_1} y \, \ddot{f}^\star(y) \, dy

  • ''′(''y''0) = ''x''0, ''f---''′(''y''1) = ''x''1. Using Integration By Parts the last integral simplifies to


: y_1 \, \dot{f}^\star(y_1) - y_0 \, \dot{f}^\star(y_0) - \int_{y_0}^{y_1} \dot{f}^\star(y) \, dy
= y_1 \, x_1 - y_0 \, x_0 - f^\star(y_1) + f^\star(y_0).

Therefore,

: f(x_1) + f^\star(y_1) - y_1 \, x_1 = f(x_0) + f^\star(y_0) - y_0 \, x_0.

Since the left hand side of this equation does only depend on x1 and the right hand side only on x0, they have to evaluate to the same constant.

: f(x) + f^\star(y) - y \, x = C,\, x = \dot{f}^\star(y) = \dot{f}^{-1}(y).

  • '' and choosing ''C'' to be zero results in the above-mentioned formula.



GEOMETRIC INTERPRETATION


For a Strictly Convex Function the Legendre-transformation can be interpreted as a mapping between the Graph of the function and the family of Tangent s of the graph. (For a function of one variable, the tangents are well-defined at all but at most Countably Many points since a convex function is Differentiable at all but at most countably many points.)

The equation of a line with Slope ''m'' and Y-intercept ''b'' is given by

:y = mx + b\,.

For this line to be tangent to the graph of a function ''f'' at the point (''x''0, ''f''(''x''0)) requires

:f\left(x_0 ight) = m x_0 + b

and

:m = \dot{f}\left(x_0 ight)

''f''' is strictly monotone as the derivative of a strictly convex function, and the second equation can be solved for ''x''0, allowing to eliminate ''x''0 from the first giving the ''y''-intercept ''b'' of the tangent as a function of its slope ''m'':

:
b = f\left(\dot{f}^{-1}\left(m ight) ight) - m \cdot \dot{f}^{-1}\left(m ight) = -f^\star(m)


  • '' denotes the Legendre transform of ''f''.


The Family of tangents of the graph of ''f'' is therefore (parameterized by ''m'') given by

:y = mx - f^\star(m)

or, written implicitly, by the solutions of the equation

:F(x,y,m) = y + f^\star(m) - mx = 0.

The graph of the original function can be reconstructed from this family of lines as the Envelope of this family by demanding

: {\partial F(x,y,m)\over\partial m} = \dot{f}^\star(m) - x = 0.

Eliminating ''m'' from these two equations gives

: y = x \cdot \dot{f}^{\star-1}(x) - f^\star\left(\dot{f}^{\star-1}(x) ight).

  • we find


: f(x) = f^{\star\star}(x).


LEGENDRE TRANSFORMATION IN MORE THAN ONE DIMENSION


For a differentiable real-valued function on an Open subset ''U'' of Rn the Legendre conjugate of the pair (''U'', ''f'') is defined to be the pair (''V'', ''g''), where ''V'' is the image of ''U'' under the Gradient mapping D''f'', and ''g'' is the function on ''V'' given by the formula

:
g(y) = \left\langle y, x ight angle - f\left(x ight), \,
x = \left(Df ight)^{-1}(y)


where

:\left\langle u,v ight angle = \sum_{k=1}^{n}u_{k} \cdot v_{k}

is the Scalar Product on R''n''.

  • ''X''''x'' with ''Y'' and T---''Y''''y'' with ''X''. If ''f'' is a real differentiable function over ''X'', then ∇''f'' is a section of the Cotangent Bundle T---''X'' and as such, we can construct a map from ''X'' to ''Y''. Similarly, if ''g'' is a real differentiable function over ''Y'', ∇''g'' defines a map from ''Y'' to ''X''. If both maps happen to be inverses of each other, we say we have a Legendre transform.



FURTHER PROPERTIES


  • .



Scaling properties

The Legendre transformation has the following scaling properties:

:
f(x) = a \cdot g(x)
\Rightarrow
f^\star(p) = a \cdot g^\star\left( rac{p}{a} ight)


:
f(x) = g(a \cdot x)
\Rightarrow
f^\star(p) = g^\star\left( rac{p}{a} ight)


It follows that if a function is Homogeneous Of Degree ''r'' then its image under the Legendre transformation is a homogeneous function of degree ''s'', where 1/''r'' + 1/''s'' = 1.


Behavior under translation


:
f(x) = g(x) + b
\Rightarrow
f^\star(p) = g^\star(p) - b


:
f(x) = g(x + y)
\Rightarrow
f^\star(p) = g^\star(p) - p \cdot y



Behavior under inversion

:
f(x) = g^{-1}(x)
\Rightarrow
f^\star(p) = - p \cdot g^\star\left( rac{1}{p} ight)



Behavior under linear transformations

Let ''A'' be a Linear Transformation from Rn to Rm. For any convex function ''f'' on Rn, one has
: \left(A f ight)^\star = f^\star A^\star

  • is the Adjoint Operator of ''A'' defined by

    • : \left \langle Ax, y^\star ight angle = \left \langle x, A^\star y^\star ight angle


A closed convex function ''f'' is symmetric with respect to a given set ''G'' of Orthogonal Linear Transformation s,
:f\left(A x ight) = f(x), \; orall x, \; orall A \in G
  • is symmetric with respect to ''G''.



Infimal convolution

The infimal convolution of two functions ''f'' and ''g'' is defined as