Cfrgtkky

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 $$u(x,t)=min_{yinmathbb{R}^n} left{tLleft(frac{x-y}{t} right) + g(y) right}$$ 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{ hLleft(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 tLleft(frac{y-z}{s} right) + g(z) - left[sLleft(frac{y-z}{s} right)+g(z) right] \ &= (t-s)Lleft(frac{y-z}{s} right) \ &=hLleft(frac{y-z}{s} right).
end{align}
That is, $$frac{u(x,t)-u((1-frac ht)x+frac htz,t-h)}{h} ge Lleft(frac{x-z}{t} right).$$
Let $h rightarrow 0^+$, to see that $$underbrace{frac{x-z}{t} cdot Du(x,t)+u_t(x,t)}_{color{#009900}{text{How is the limit definition of derivative applied here exactly?}}} ge Lleft(frac{x-z}{t} right).$$
Consequently,

begin{align}
u_t(x,t)+H(Du(x,t))&=u_t(x,t)+max_{vinmathbb{R}^n} {v cdot Du(x,t)-L(v) } \
&ge u_t(x,t)+frac{x-z}{t} cdot Du(x,t)-Lleft(frac{x-z}{t} right) \ &ge 0
end{align}
This inequality and $text{(31)}$ complete the proof.

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.