LagrangeMultiplierAppliedToTheLegendreTransform 2009-04-02 09:14:44 2009-04-02 21:45:30 Example example of using Lagrange multipliers % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here 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.\\\\ \textbf{THE LEGENDRE-LAGRANGE PROBLEM}\\ Let $f(\overline{x})$ be a function of the real n-vector $\overline{x}$ and $g(\overline{p})$ its Legendre transform defined by \begin{align} g(\overline{p}) \;=\;\overline{p}.\overline{x}-f(\overline{x})\quad\quad p_i\;=\;\frac{\partial f}{\partial x_i} \end{align} The vector $\overline{p}$ is a function of $\overline{x}$, or conversely, $\overline{x}$ is a function of $\overline{p}$, so that $g(\overline{p})$ is a function of $\overline{p}$ alone. The Legendre transform is also alternatively defined as the maximum of the function $\overline{p}.\overline{x}-f(\overline{x})$. This maximum is reached for $\overline{x}$ 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: \begin{align} h(\overline{x})\;=\;0 \end{align} This equation defines one of the components, $x_\alpha$ for example, as a function of all the others. Putting this function into (1) would give us the transform $g(\overline{p})$ as a function of the vector $\overline{p}$ with n-1 components. It would be nice if we could instead use a Lagrange multiplier $k$ and compute the maximum of the function $$\overline{p}.\overline{x}-f(\overline{x})+kh(\overline{x})$$ with all its n components. But then, how does the constraint (2) on $\overline{x}$ translate to $\overline{p}$? The answer is amazingly simple, as we shall see next.\\\\ \textbf{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 $x_\alpha$ as a function of the other components: \begin{align} p_i=\frac{df}{dx_i}=\frac{\partial f}{\partial x_i}+\frac{\partial f}{\partial x_\alpha}\frac{\partial x_\alpha}{\partial x_i}\quad\quad \frac{dh}{dx_i}=\frac{\partial h}{\partial x_i}+\frac{\partial h}{\partial x_\alpha}\frac{\partial x_\alpha}{\partial x_i}=0\quad\quad i\neq\alpha \end{align} Taking the value of \Large $\frac{\partial x_\alpha}{\partial x_i}$ \normalsize from the second set of equations (3) and putting it into the first set, we get: \begin{align} p_i\;=\;\frac{\partial f}{\partial x_i}-\Phi\frac{\partial h}{\partial x_i}\quad \mbox{with} \quad \Phi\;=\;\frac{\frac{\partial f}{\partial x_\alpha}}{\frac{\partial h}{\partial x_\alpha}}\quad\mbox{and}\quad i\neq\alpha \end{align} From definition (1), the Legendre transform $g$ is therefore: \begin{align} g(\overline{p})\;=\;\sum_{i\neq\alpha} x_i\left( \frac{\partial f}{\partial x_i} - \Phi\frac{\partial h}{\partial x_i}\right)-f(\overline{x}) \end{align}\\ \textbf{BACK TO THE LAGRANGE MULTIPLIER}\\ In the first equation (4), if we set $i=\alpha$ we get $p_\alpha = 0$ and conversely, getting the value of $\Phi$. So, in (5), we may remove the condition $i\neq\alpha$ and sum over all the n components of $\overline{p}$. The constraint $p_\alpha = 0$ reduces the number of independent components to n-1. In fact, we are back to the traditional Lagrange method with the multiplier $\Phi$ 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 $\overline{x}$, according to which component $x_\alpha$ 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 $\Phi_1, \Phi_2...\Phi_m$. They are computed by equating to zero any set of m components from $\overline{p}$. \begin{thebibliography}{1} \bibitem {A} \PMlinkexternal{Fersanz at PM - Legendre Transform}{http://planetmath.org/encyclopedia/LegendreTransform.html}\\ This link is actually broken but, hopefully, should be operative soon. \bibitem {B} \PMlinkexternal{Wikipedia - Legendre transformation}{http://en.wikipedia.org/wiki/Legendre_transformation} \end{thebibliography}