📂Partial Differential Equations

Proof that the Hopf-Lax Formula Satisfies the Hamilton-Jacobi Equation

Theorem ¹

Hopf-Lax Formula

$$ u(x,t) = \min \limits_{y \in \mathbb{R}^n} \left\{ tL\left( \dfrac{x-y}{t} \right) +g(y) \right\} $$

Let’s denote $x \in \mathbb{R}^n$ and $t>0$. And suppose that $u$ defined by the Hopf-Lax formula is differentiable at point $(x,t)$. Then, $u$ satisfies the Hamilton-Jacobi Equation.

$$ u_{t}(x, t) + H\big( Du(x, t) \big) =0 $$

Proof

Lemma: Generalization of Hopf-Lax Formula

Let’s denote $t>0$. Then, for any given $x \in \mathbb{R}^n$ and $0< s< t$, the following holds:

$$ u(x, t) = \min \limits_{y \in \mathbb{R}^n} \left\{ (t-s) L \left( \dfrac{x-y}{t-s} \right) +u(y, s) \right\} $$

It is equal to the Hopf-Lax Formula when $s=0$.

• Part 1.

  Suppose fixed $v \in \mathbb{R}^n$ and $h>0$ are given. Then, by the lemma, the following holds:

  $$ \begin{align*} u(x+hv, t+h) &= \min \limits_{y \in \mathbb{R}^n} \left\{ hL\left( \dfrac{x+hv-y}{h} \right) +u(y, t) \right\} \\ & \le hL(v)+u(x, t) \end{align*} $$

  This is obtained by substituting $t+h$ for $t$, and $t$ for $s$ in the lemma’s formula. The inequality holds because for all $y \in \mathbb{R}^n$, it is the minimum, so naturally substituting any $x$ for $y$ will yield a value equal or greater. By moving $u$ to the other side and dividing by $h$, we obtain the following:

  $$ \dfrac{u(x+hv, t+h)-u(x, t)}{h} \le L(v) $$

  The left-hand side can be rewritten as below:

  $$ \dfrac{ u\big( (x, t)+h(v, 1) \big) -u(x, t) }{h} $$

  It’s rearranged to a form suitable for differentiation. Taking the limit as $h^+ \rightarrow 0$, we obtain:

  $$ \begin{align} && (v, 1) \cdot \big( Du(x, t), u_{t}(x, t) \big) &\le L(v) \nonumber \\ \implies && v\cdot Du(x, t)+ u_{t}(x, t) &\le L(v) \nonumber \\ \implies && u_{t}(x, t) + v\cdot Du(x, t)- L(v) &\le 0 \label{eq1} \end{align} $$

  This holds for all $v \in \mathbb{R}^n$, and since $H(p)=L^{\ast}(p)=\max\limits_{v \in \mathbb{R}^n} \left\{ p\cdot v-L(v) \right\}$, by $\eqref{eq1}$, the following holds:

  $$ u_{t}(x, t) + H\big( Du(x, t) \big) = u_{t}(x, t)+\max_{v \in \mathbb{R}^n} \left\{ v \cdot Du(x, t)-L(v) \right\} \le 0 $$

• Part 2.

  By the Hopf-Lax formula, for fixed $x$, $t$, there exists a $z=z_{z, t} \in \mathbb{R}^n$ that satisfies:

  $$ \begin{equation} u(x, t)=tL\left( \dfrac{x-z}{t} \right) + g(z) \label{eq2} \end{equation} $$

  $y$ is set as the minimum value among all $y$. And let there be a fixed $0 < h <t$, and let’s denote $s=t-h$, $y=\dfrac{s}{t}x+\left( 1-\dfrac{s}{t} \right)z$. Then, the following holds:

  $$ \begin{equation} 0<s<t, \quad 0<\dfrac{s}{t}<1, \quad \dfrac{x-z}{t}=\dfrac{y-z}{s} \label{eq3} \end{equation} $$

  Thus:

  $$ \begin{align*} u(x, t)-u(y, s) & \ge tL\left( \dfrac{x-z}{t} \right) +g(z) -\left[ sL\left(\frac{y-z}{s}\right) + g(z) \right] \\ &= (t-s)L\left( \dfrac{x-z}{t} \right) \end{align*} $$

  The first line holds by $\eqref{eq2}$ and the Hopf-Lax formula, and the second line by $\eqref{eq3}$. Since $s=t-h$, the following holds:

  $$ y=\dfrac{s}{t}x+\left( 1-\dfrac{s}{t} \right)z=\left( 1-\dfrac{h}{t}\right )x + \dfrac{h}{t}z $$

  Hence, the inequality above is as follows:

  $$ \begin{align*} u(x, t)-u(y, s) &= u(x, t)-u\bigg( \left( 1-\dfrac{h}{t}\right )x + \dfrac{h}{t}z, t-h \bigg) \\ & \ge & hL\left(\dfrac{x-z}{t}\right) =(t-s)L\left( \dfrac{x-z}{t} \right) \end{align*} $$

  Dividing both sides by $h$ and rearranging the numerator and denominator of the left-hand side by multiplying by $-1$ to prepare for differentiation yields:

  $$ \dfrac{ u\left( (x, t)-h\left( \dfrac{x-z}{t}, 1 \right) \right) - u(x, t) }{-h} \ge L\left(\dfrac{x-z}{t}\right) $$

  Taking the limit of the above inequality as $h \rightarrow 0^+$ results in:

  $$ \begin{align} &&\left( \dfrac{x-z}{t}, 1 \right) \cdot \left( Du(x, t), u_{t}(x, t) \right) = \dfrac{x-z}{t}\cdot Du(x, t) + u_{t}(x, t) &\ge L\left( \dfrac{x-z}{t} \right) \nonumber \\ \implies && \dfrac{x-z}{t}\cdot Du(x, t) + u_{t}(x, t) - L\left( \dfrac{x-z}{t} \right) &\ge 0 \label{eq4} \end{align} $$

  Consequently, the following holds:

  $$ \begin{align*} u_{t}(x, t)+H(Du) &= u_{t}(x, t)+\max_{v \in \mathbb{R}^n} \left\{ v \cdot Du(x, t)-L(v) \right\} \\ &\ge u_{t}(x, t)+\dfrac{x-z}{t}\cdot Du(x, t)-L\left( \dfrac{x-z}{t} \right) \\ & \ge 0 \end{align*} $$

  The second line holds by choosing $v=\dfrac{x-z}{t}$, and the last line by $\eqref{eq4}$.

Part 1. and Part 2. prove that the following holds:

$$ u(x, t) = \min \limits_{y \in \mathbb{R}^n} \left\{ (t-s) L \left( \dfrac{x-y}{t-s} \right) +u(y, s) \right\} $$

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