# Compute the Hopf-Lax formula for a given PDE

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?

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:

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