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in Evans' book (pag. 135) there's the following example (already asked

in this site, but never answered):

Consider the initial-value problem \begin{cases} u_t+\frac 12|Du|^2=0 & \text{in }\mathbb{R}^n \times (0,\infty) \\ \qquad \qquad \, \, \, \, u = |x| & \text{on } \mathbb{R}^n \times \{t=0\} \tag{49} \end{cases} The Hopf-Lax formula for the unique weak solution of $(49)$ is $$u(x,t)=\min_{y \in \mathbb{R}^n} \left\{\frac{|x-y|^2}{2t}+|y| \right\} \tag{50}$$ Assume $|x| > t$. Then \begin{align}D_y \left(\frac{|x-y|^2}{2t} + |y| \right) = \frac{y-x}{t}+\frac{y}{|y|} & (y \not=0)\end{align} and this expression equals $0$ if $x=y+\frac{y}{|y|}t, y=(|x|-t)\frac{x}{|x|} \not=0$.

Thus $u(x,t)=|x|-\frac t2$ if $|x| > t$. If $|x| > t$, the minimum in $(50)$ is attained at $y=0$. Consequently $$u(x,t)=\begin{cases}|x|-\frac{t}{2} & \text{if }|x| > t \\ \qquad\frac{|x|^2}{2t} & \text{if }|x| \le t \end{cases}$$

The thoery behind that formula is clear, but I really can't understand some steps.

  • I can find $x=y+\frac{y}{|y|}t$ but how can I derive \begin{align} y=(|x|-t)\frac{x}{|x|} \end{align} I can't find it.

  • How can I see a priori that I have to distinguish the cases $|x|\leq t$ and $|x| >t$ in order to find the minimum?

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1 Answer 1

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The expression for $y$ is only possible with the posed restrictions $|x|>t$ and $t>0$. So, the expresion for $x$ has to be read:

$$x=y+\frac{y}{|y|}t= \begin{cases} y+t&\text{if}&y\geq 0 \land(x<-t\lor x>t)\\ y-t&\text{if}&y<0\land(x<-t\lor x>t) \end{cases} $$

Considering that $x<-t\implies y<0$ and $x>t\implies y\geq0$, we can remove each one from its corresponding slot:

$$x= \begin{cases} y+t&\text{if}&y\geq 0 \land x>t\\ y-t&\text{if}&y<0\land x<-t \end{cases} $$

Now, as $x<-t\implies x<0$ and$x>t\implies x>0$, we can add the redundant condition to each slot

$$x= \begin{cases} y+t&\text{if}&y\geq 0 \land x>t\land x>0\\ y-t&\text{if}&y<0\land x<-t\land x<0 \end{cases} $$

We see now a function that can be inverted as it is injective. We can isolate $y$ in the expressio for each branch:

$$y= \begin{cases} x-t&\text{if}&y\geq 0 \land x>t\land x>0\\ x+t&\text{if}&y<0\land x<-t\land x<0 \end{cases} $$

remove the redundant conditions:

$$y= \begin{cases} x-t&\text{if}&x>t\land x>0\\ x+t&\text{if}&x<-t\land x<0 \end{cases} $$

And sumarizing, with the restrictions implicitly granted:

$y=x-\dfrac{x}{|x|}t=(|x|-t)\dfrac{x}{|x|}$

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