Consider the Hamilton Jacobi equation, \begin{equation} \begin{cases} u_t + H(Du) = 0 & x \in \mathbb{R}^n, t > 0\\ u(x,0) = \max(|x|^2 - 1, 0) \end{cases} \end{equation} Show that for $H(p) = |p|,$ then $u(x,t) = 0$ when $t = |x| - 1.$
My idea for this problem was to use the Hopf Cole formula to get the exact solution for $u,$ and then hopefully it would be clear that $u(x, |x| - 1) = 0$. The Hopf Cole formula is given by $$ u(x,t) = \min_{y \in \mathbb{R}^n} \bigg\{ t L\Big(\frac{x-y}{t} \Big) + \max(|y|^2 - 1,0) \bigg\}, $$ where $L$ is the Lagrangian associated with the Hamiltonian $H(p).$ The Legendre transform is used to determine $L,$ $$ L(q) = \sup_{p \in \mathbb{R}^n} \{ p\cdot q - H(p) \} = \sup_{p \in \mathbb{R}^n} \{ p\cdot q - |p|\}.$$
Normally, we have that $H$ grows superlinearly at least, so we can solve that $q(p) = \frac{dH}{dp}$, meaning that doing the Legendre transform is easy enough. However, this Hamiltonian does not grow superlinearly, and I am stuck on the possibly easy problem of determining $L(q).$ My qyestions are:
How to determine the Lagrangian in this situation? Does this overall seem like the most logical way to go about this problem, or are other methods more reasonable?
Thanks in advance for any comments, suggestions, or calculations!
The Lagrangian that you wrote down is given by
$$L(q)=\begin{cases} 0 & |q| \leq 1 \\ +\infty & |q|>1 \end{cases}.$$
Consequently, assuming everything else you wrote is correct, $u(x,t)=\max \{ r(x,t)^2-1,0 \}$ where $r(x,t)$ is the minimum value of $|y|$ such that $\left | \frac{x-y}{t} \right | \leq 1$. I guess this is $\max \{ |x|-t,0 \}$, which gives the desired result.