# Finding mimimum of the unique, weak solution to an IVP

PDE Evans, 2nd edition, page 134

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

Since $(50)$ calls for minimums, we get the $u(x,t)$ to be the formula above. My question pertains only the top formula of the bracket.

If $x > t$, then we are supposed to get $|x|-\frac{t}{2}$. My book doesn't show this, and I have trouble deriving that top formula. How can I verify this? (I understand getting the bottom formula though.) All I have so far:

Substitute $y$ with $(|x|-t)\frac{x}{|x|}$ into $(50)$. Then for all $|x| > t$ \begin{align}u(x,t)&=\min_{y\in\mathbb{R}^n} \left\{\frac{|x-y|^2}{2t} + |y| \right\}\\&=\min_{y\in\mathbb{R}^n} \left\{\frac t2 + (x-t) \right\}\\&=\min_{y\in\mathbb{R}^n}\left\{\frac t2 + |x|-t \right\}\\&=\min_{y\in\mathbb{R}^n}\left\{|x|-\frac t2 \right\} \\&=|x|-\frac t2 \end{align} where $x=|x|$ because $|x|>t\ge0$, and $x-t=|x-t|$ because $|x-t|\ge|x|-|t|>t-|t|=0$.