The Hopf-Lax function I have is
$$u(t,x) = \inf_{y\in \mathbb{R}} \left\{ C|y| + \frac{(x-y)^2}{2t} \right\},$$
where $C>0$ is constant. By evaluating stationary and non-differentiable points I have found two possibilities for the infimum:
$$u_1(t,x) = \frac{x^2}{2t},\quad u_2(t,x) = C|x||1-Ct| + \frac{(x-|1-Ct|x)^2}{2t}.$$
I know the initial condition for the related Cauchy problem is given by $u(0,x) = C|x|$ which leads me to believe that the function takes the form of $u_2$ for at least small $t$. I also know that these two functions ($u_1$ and $u_2$) intersect at $t=\frac{1}{C}$, however I believe there is another intersection which is where I believe the piecewise Hopf-Lax function begins to take values according to $u_1$. I'm struggling to find a "nice" form for this point, if it exists. What is the explicit expression for the above infimum?
For fixed $(t,x)\in (0,+\infty)\times\mathbb{R}$, the function $\varphi(y) := C|y| + \frac{(x-y)^2}{2t}$ is strictly convex and superlinear, hence it has a unique minimum point $y=y(t,x)$ which is characterized by the condition $0\in [\varphi'_-(y), \varphi'_+(y)]$. It is not difficult to check that $$ y = \begin{cases} 0, & \text{if}\ |x| \leq Ct,\\ x-Ct, & \text{if}\ x > Ct,\\ x+Ct, & \text{if}\ x < -Ct, \end{cases} $$ hence $$ u(t,x) = \begin{cases} x^2/(2t), & \text{if}\ |x| \leq Ct,\\ C|x| - C^2t/2, & \text{if}\ |x| > Ct. \end{cases} $$