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Proof that the Hopf-Lax Formula Satisfies the Hamilton-Jacobi Equation 📂Partial Differential Equations

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

Theorem 1

Hopf-Lax Formula

u(x,t)=min⁡y∈Rn{tL(x−yt)+g(y)} 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∈Rnx \in \mathbb{R}^n and t>0t>0. And suppose that uu defined by the Hopf-Lax formula is differentiable at point (x,t)(x,t). Then, uu satisfies the Hamilton-Jacobi Equation.

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

Proof

Lemma: Generalization of Hopf-Lax Formula

Let’s denote t>0t>0. Then, for any given x∈Rnx \in \mathbb{R}^n and 0<s<t0< s< t, the following holds:

u(x,t)=min⁡y∈Rn{(t−s)L(x−yt−s)+u(y,s)} 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=0s=0.

  • Part 1.

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

    u(x+hv,t+h)=min⁡y∈Rn{hL(x+hv−yh)+u(y,t)}≤hL(v)+u(x,t) \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+ht+h for tt, and tt for ss in the lemma’s formula. The inequality holds because for all y∈Rny \in \mathbb{R}^n, it is the minimum, so naturally substituting any xx for yy will yield a value equal or greater. By moving uu to the other side and dividing by hh, we obtain the following:

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

    The left-hand side can be rewritten as below:

    u((x,t)+h(v,1))−u(x,t)h \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+→0h^+ \rightarrow 0, we obtain:

    (v,1)⋅(Du(x,t),ut(x,t))≤L(v)  ⟹  v⋅Du(x,t)+ut(x,t)≤L(v)  ⟹  ut(x,t)+v⋅Du(x,t)−L(v)≤0 \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∈Rnv \in \mathbb{R}^n, and since H(p)=L∗(p)=max⁡v∈Rn{p⋅v−L(v)}H(p)=L^{\ast}(p)=\max\limits_{v \in \mathbb{R}^n} \left\{ p\cdot v-L(v) \right\}, by (eq1)\eqref{eq1}, the following holds:

    ut(x,t)+H(Du(x,t))=ut(x,t)+max⁡v∈Rn{v⋅Du(x,t)−L(v)}≤0 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 xx, tt, there exists a z=zz,t∈Rnz=z_{z, t} \in \mathbb{R}^n that satisfies:

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

    yy is set as the minimum value among all yy. And let there be a fixed 0<h<t0 < h <t, and let’s denote s=t−hs=t-h, y=stx+(1−st)zy=\dfrac{s}{t}x+\left( 1-\dfrac{s}{t} \right)z. Then, the following holds:

    0<s<t,0<st<1,x−zt=y−zs \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:

    u(x,t)−u(y,s)≥tL(x−zt)+g(z)−[sL(y−zs)+g(z)]=(t−s)L(x−zt) \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 (eq2)\eqref{eq2} and the Hopf-Lax formula, and the second line by (eq3)\eqref{eq3}. Since s=t−hs=t-h, the following holds:

    y=stx+(1−st)z=(1−ht)x+htz 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:

    u(x,t)−u(y,s)=u(x,t)−u((1−ht)x+htz,t−h)≥hL(x−zt)=(t−s)L(x−zt) \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 hh and rearranging the numerator and denominator of the left-hand side by multiplying by −1-1 to prepare for differentiation yields:

    u((x,t)−h(x−zt,1))−u(x,t)−h≥L(x−zt) \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→0+h \rightarrow 0^+ results in:

    (x−zt,1)⋅(Du(x,t),ut(x,t))=x−zt⋅Du(x,t)+ut(x,t)≥L(x−zt)  ⟹  x−zt⋅Du(x,t)+ut(x,t)−L(x−zt)≥0 \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:

    ut(x,t)+H(Du)=ut(x,t)+max⁡v∈Rn{v⋅Du(x,t)−L(v)}≥ut(x,t)+x−zt⋅Du(x,t)−L(x−zt)≥0 \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=x−ztv=\dfrac{x-z}{t}, and the last line by (eq4)\eqref{eq4}.

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

u(x,t)=min⁡y∈Rn{(t−s)L(x−yt−s)+u(y,s)} 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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  1. Lawrence C. Evans, Partial Differential Equations (2nd Edition, 2010), p127-128 ↩︎