4
$\begingroup$

I can comprehend some but not all of the proofs. I do not understand how the limit definition of a derivative is derived in this context, and those are highlighted in $\color{#009900}{\text{green}}$.

This is from PDE Evans, 2nd edition, pages 127-128.

Theorem 5 (Solving the Hamilton-Jacobi equation). Suppose $x \in \mathbb{R}^n$, $t > 0$, and $u$ defined by the Hopf-Lax formula $$u(x,t)=\min_{y\in\mathbb{R}^n} \left\{tL\left(\frac{x-y}{t} \right) + g(y) \right\}$$ is differentiable at a point $(x,t) \in \mathbb{R}^n \times (0,\infty)$. Then $$u_t(x,t)+H(Du(x,t))=0.$$

Proof. 1.) Fix $v \in \mathbb{R}^n, h > 0$. Owing to Lemma 1, \begin{align} u(x+hv,t+h)&=\min_{y \in \mathbb{R}^n} \left\{ hL\left(\frac{x+hv-y} {h}\right)+u(y,t)\right\} \\ &\le hL(v)+u(x,t). \end{align} Hence, $$\frac{u(x+hv,t+h)-u(x,t)}{h} \le L(v).$$ Let $h \rightarrow 0^+$, to compute $$\underbrace{v \cdot Du(x,t)+u_t(x,t)}_{\color{#009900}{\text{How is this expression obtained?}}} \le L(v).$$ This inequality is valid for all $v \in \mathbb{R}^n$, and so $$u_t(x,t)+H(Du(x,t))=u_t(x,t)+\max_{v \in \mathbb{R}^n} \{v \cdot Du(x,t)-L(v) \} \le 0. \tag{31}$$ The first equality holds since $H = L^*:=\max_{v \in \mathbb{R}^n} \{v \cdot Du(x,t)-L(v) \}$, by convex duality of Hamiltonian and Lagrangian (page 121 of the book).

2.) Now chose $z$ such that $u(x,t)=tL(\frac{x-z}{t})+g(z)$. Fix $h > 0$ and set $s=t-h,y=\frac st x+(1- \frac st)z$. Then $\frac{x-z}{t}=\frac{y-z}{s}$, and thus \begin{align} u(x,t)-u(y,s) &\ge tL\left(\frac{y-z}{s} \right) + g(z) - \left[sL\left(\frac{y-z}{s} \right)+g(z) \right] \\ &= (t-s)L\left(\frac{y-z}{s} \right) \\ &=hL\left(\frac{y-z}{s} \right). \end{align} That is, $$\frac{u(x,t)-u((1-\frac ht)x+\frac htz,t-h)}{h} \ge L\left(\frac{x-z}{t} \right).$$ Let $h \rightarrow 0^+$, to see that $$\underbrace{\frac{x-z}{t} \cdot Du(x,t)+u_t(x,t)}_{\color{#009900}{\text{How is the limit definition of derivative applied here exactly?}}} \ge L\left(\frac{x-z}{t} \right).$$ Consequently,
\begin{align} u_t(x,t)+H(Du(x,t))&=u_t(x,t)+\max_{v\in\mathbb{R}^n} \{v \cdot Du(x,t)-L(v) \} \\ &\ge u_t(x,t)+\frac{x-z}{t} \cdot Du(x,t)-L\left(\frac{x-z}{t} \right) \\ &\ge 0 \end{align} This inequality and $\text{(31)}$ complete the proof.

$\endgroup$
1
$\begingroup$

For $f:\mathbb{R}^n \times \mathbb{R}^+ \rightarrow \mathbb{R}$, the derivative at a point $(x,t)$ is a linear operator mapping $(v,s) \in \mathbb{R}^n \times \mathbb{R}^+$ into $\hat{D}f(x,t)\cdot (v,s) \in \mathbb{R}$ such that

$$|f(x+v,t +s)-f(x,t) -\hat{D}f(x,t)\cdot (v,s)| = o(\|(v,s)\|),$$

as $\|(v,s)\| \rightarrow 0.$

The $1 \times (n+1)$ matrix of this operator with respect to the canonical basis has the partial derivatives as components:

$$[Df(x,t)]= (f_{x_1},f_{x_2},\ldots,f_{x_n},f_{t}).$$

This implies for $s=h > 0$

$$\lim_{h \rightarrow 0} \frac{f(x+hv,t +h)-f(x,t)}{h} =\hat{D}f(x,t)\cdot (v,1),$$

or

$$\lim_{h \rightarrow 0} \frac{f(x+hv,t +h)-f(x,t)}{h} ={D}f(x,t)\cdot v + f_{t}(x,t),$$

where the $1 \times n$ matrix of $D$ has the partial derivatives with respect to $x_1,\ldots,x_n$ as components.

$\endgroup$
  • $\begingroup$ Basically in this case $D$ is the gradient operator. $\endgroup$ – RRL Jun 18 '14 at 2:58
  • $\begingroup$ "...the derivative at a point $(x,t)$ is a linear operator mapping $(v,s) \in \mathbb{R}^n \times \mathbb{R}^+$ into $\hat{D}f(x,t)\cdot (v,s) \in \mathbb{R}$..." I understand most of your answer, but I have one small question: I am not too familiar with functional analysis yet; why do we get the byproduct of $(v,s)$? Is this simply the result from the definition of taking a derivative of a function $f(x,t)$ in $\mathbb{R}^n \times\mathbb{R}^+$? $\endgroup$ – Cookie Jun 18 '14 at 6:33
  • 1
    $\begingroup$ I made some changes to make it clearer. The definition of a multivariate derivative means if you add an incremental vector $(v,s)$ to $(x,t)$ then $f(x+v,t+s)$ has a linear(affine) approximation with an error that goes to zero as the norm of $(v,s)$ goes to 0. In your problem you have $(hv,h)$ as the increment which goes to zero in norm iff $h \rightarrow 0.$ Dividing the linear approximation by $h$, using linearity $\hat{D}(x,t)(hv,h) = h\hat{D}(x,t)(v,1)$, and taking the limit as $h \rightarrow 0$ you get the result you want -- since the error term goes to 0. $\endgroup$ – RRL Jun 18 '14 at 6:57
  • $\begingroup$ It's analagous to $f(x+h) - f(x) - f'(x)h = e(h)$ where $e(h) \rightarrow 0$ as $h \rightarrow 0$ in the single variable case. This is equivalent to $\lim [f(x+h)-f(x)]/h = f'(x).$ $\endgroup$ – RRL Jun 18 '14 at 7:04
  • 1
    $\begingroup$ Not quite. $u(x+hv,t+h) = u(x,t) + L\cdot(hv,h) + ho(h) = u(x,t) +hL\cdot(v,1) + ho(h)$ where $L = \hat{D}(x,t)$ is the linear operator. In matrix terms $L\cdot(v,1) = (u_{x_1},\ldots,u_{x_n},u_t)\cdot(v_1,\dots,v_n,1)$. Then $\lim[u(x+hv,t+h)-u(x,t)]/h = L\cdot(v,1)$ $\endgroup$ – RRL Jun 18 '14 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.