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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.

EditPiAf
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Cookie
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1 Answers1

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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.

RRL
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  • Basically in this case $D$ is the gradient operator. – RRL Jun 18 '14 at 02:58
  • "...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 $\mathbb{R}^n \times\mathbb{R}^+$?

     – Cookie Jun 18 '14 at 06:33
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    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. – RRL Jun 18 '14 at 06:57
  • 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 $\lim [f(x+h)-f(x)]/h = f'(x).$ – RRL Jun 18 '14 at 07:04
  • I will try to apply your explanation to the book's proof (and to see if I understand this on my part as well): So, when we look at $u(x+hv,t+h)$ in part 1 of Evans' proof, this expression can be written as $$u(x+hv,t+h)=u(x,t)+u(hv,h)$$ where $u(hv,h)$ is the incremental vector, as you are describing. And we can also say $u(hv,h)=h\cdot u(v,1)$ because this transformation--from $u(x,t)$ to $u(x+hv,t+h)$--is linear (or affine). – Cookie Jun 18 '14 at 07:16
  • If this recapitulation seems okay to you, then I think I will understand the whole thing better, as the rest follows in your answer (in which you apply the differential operator to $f(x,t) \cdot (v,1)$, and in finding the derivative, the product rule was employed). – Cookie Jun 18 '14 at 07:25
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    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 $\lim[u(x+hv,t+h)-u(x,t)]/h = L\cdot(v,1)$ – RRL Jun 18 '14 at 07:30