📂Partial Differential Equations

Legendre Transformation

• Regarding x and p, when emphasizing that they are variables of a partial differential equation, it is indicated with regular font $x,p \in \mathbb{R}^{n}$, and when emphasizing that it is a function of $s$, it is indicated with bold font $\mathbf{x}, \mathbf{p} \in \mathbb{R}^{n}$.

Definition ¹

For simplicity, let’s say the Lagrangian is a function of variable $v\in \mathbb{R}^{n}$ only.

$$L(v) = L : \mathbb{R}^{n} \to \mathbb{R}$$

Let’s say the Lagrangian $L$ satisfies the following two conditions

• (a) $L$ is a convex function.

  $$\lambda L(v_{1}) + (1-\lambda)L(v_{2}) \le L\big( \lambda v_{1} +(1-\lambda)v_{2} \big) \quad \forall\ v_{1},v_{2}\in \mathbb{R}^{n},\quad \forall\ 0\le \lambda \le 1$$

• (b) $\lim \limits_{ |v|\to \infty} \dfrac{ L(v) }{ |v| }=+\infty$

Then the Legendre transform $L^{\ast} : \mathbb{R}^{n} \to \mathbb{R}$ of $L$ is defined as follows.

$$L^{\ast} (p) := \sup \limits_{v \in \mathbb{R}^{n}} \left\{ p\cdot v -L(v) \right\} \quad \forall \ p \in \mathbb{R}^{n}$$

It is also called the Fenchel transform.

Explanation

At this time, $p$ is the same as the variable $\mathbf{p}= D_{v}L\big( \dot{\mathbf{x}},\ \mathbf{x}\big)$ of the Hamiltonian. That $L^{\ast}$ is well defined can be proved as follows. Also, although it is defined as $\sup$, it can also be shown to actually be the same as $\max$.

Theorem

The Legendre transform $L^{\ast}(p)$ is well defined. Also, $\sup \left\{ p\cdot v -L(v) \right\} = \max \left\{ p\cdot v -L(v) \right\}$ holds.

Proof

Well-definedness

The fact that $L^{\ast}(p)$ has a real value is proved by reductio ad absurdum. The fact that $L$ is continuous due to condition (a) is trivial. And by the definition of the Legendre transform$(\sup)$, $-\infty$ cannot be a value.

$$L^{\ast}(p) = \sup \left\{ p\cdot v -L(v) \right\} \in (-\infty, \infty]$$

Now, if we assume $L^{\ast}(p)=\infty$ and show a contradiction, it is proved to have a real value.

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$L^{\ast}(p)=\infty$ Let’s assume. Then there exists a sequence $\left\{ v_{k} \right\}_{k=1}^\infty$ that satisfies the conditions below.

$$\begin{equation} a_{k}:= p\cdot v_{k} -L(v_{k}) \to \infty \quad \text{as } k \to \infty \label{eq1} \end{equation}$$

And let’s assume $v_{k} \ne 0$ for all $k$. The reason this assumption is valid is because there can only be a finite number of cases where $v_{k}=0$ due to $a_{k} \to \infty$, and a subsequence excluding them can be re-assigned as $v_{k}$. Now there are two cases where $\left\{ v_{k} \right\}$ is bounded or not. Showing that a contradiction arises in both cases ends the proof.

• Case 1. If $\left\{ v_{k} \right\}$ is bounded

  Since $\left\{ v_{k} \right\}$ is bounded, there exists a subsequence converging to some point, let’s call it again $v_{k}$. Then there exists a $\left\{ v_{k} \right\}$ like below.

  $$v_{k} \to v_{0} \quad \text{as } k \to \infty,\quad \mathrm{for\ some}\ v_{0}\in \mathbb{R}^{n}$$

  Since $L$ is continuous, $a_{k} \to p \cdot v_{0} -L(v_{0}) \in \mathbb{R}^{n} \quad \text{as } k \to \infty$ and this is a contradiction by $\eqref{eq1}$.

• Case 2. If $\left\{ v_{k} \right\}$ is unbounded

  Since it is unbounded, it can be assumed as follows.

  $$|v_{k}| \to \infty \quad \text{as } k \to \infty$$

  By the assumption, since $|v_{k}| \ne 0$, dividing both sides of $a_{k}=p \cdot v_{k}-L(v_{k})$ by $|v_{k}|$ gives the following.

  $$\dfrac{a_{k}}{|v_{k}|}=\dfrac{p\cdot v_{k}}{|v_{k}|}-\dfrac{L(v_{k})}{|v_{k}|}$$

  Here, applying the Cauchy-Schwarz inequality to the first term on the right-hand side gives the following.

  $$\left| \dfrac{p\cdot v_{k}}{|v_{k}|} \right| \le |p| \left| \dfrac{v_{k}}{|v_{k}|}\right| = | p |$$

  Then, taking the limit where $k \to \infty$, by (b) the following is obtained.

  $$\lim \limits_{ |v_{k}|\to \infty} \dfrac{a_{k}}{|v_{k}|} \le \lim \limits_{ |v_{k}|\to \infty} |p| -\dfrac{L(v_{k})}{|v_{k}|} = -\infty$$ At this time, by the assumption that $|v_{k}|=\infty$, $a_{k}$ must diverge faster to $-\infty$. However, this contradicts $\eqref{eq1}$.

Since a contradiction is found in both possible cases, the assumption $L^{\ast}(p)=\infty$ is wrong. Therefore, the Legendre transform is well defined.

$$L^{\ast}(p) \in \mathbb{R}$$

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$\sup=\max$

This is equivalent to showing that there exists $v_{p} \in \mathbb{R}^{n}$ satisfying the following conditions.

$$L^{\ast}(p) = p\cdot v_{p}-L(v_{p})$$

First, by the definition of the Legendre transform$(\sup)$, there exists a sequence $\left\{ v_{k} \right\}$ satisfying the conditions below.

$$\begin{equation} a_{k} := p\cdot v_{k}-L(v_{k}) \to L^{\ast}(p) \quad \text{as } k \to \infty \label{eq2} \end{equation}$$

First, let’s assume that $\left\{ v_{k} \right\}$ is not bounded. Then $|v_{k}| \to \infty$ and as shown above, $a_{k} \to -\infty$ and this is a contradiction. Therefore, $\left\{ v_{k} \right\}$ is bounded. Since $\left\{ v_{k} \right\}$ is bounded, there exists a subsequence converging to $v_{k} \to v_{p}$. Therefore, the following holds.

$$p\cdot v_{k} - L(v_{k}) \to p \cdot v_{p} -L(v_{p}) \quad \text{as } k \to \infty$$

But since it was $p\cdot v_{k}-L(v_{k}) \to L^{\ast}(p)$ from $\eqref{eq2}$, the following holds.

$$p \cdot v_{p} -L(v_{p})=L^{\ast}(p)$$

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