# Solving Hamilton Jacobi equation using Lagrangians

Consider the Hamilton Jacobi equation, $$\begin{cases} u_t + H(Du) = 0 & x \in \mathbb{R}^n, t > 0\\ u(x,0) = \max(|x|^2 - 1, 0) \end{cases}$$ 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?

• Why is L so hard to compute? The expression is clearly maximized along the angular coordinates when p is proportional to q, since the Hamiltonian is radial, and then you have just a 1D problem of maximizing $k(|q|^2-|q|)$ over $k \geq 0$. Is the problem that this sup is $+\infty$ for $|q|>1$? – Ian Aug 11 '17 at 17:09
• Perhaps its my poor understanding of how the transform works. It seems to me that $L(q) = \infty$ always, and then $u(x,t) = \infty,$ obviously incorrect. If you have a better understanding of what is going on, feel free to write an answer. – Merkh Aug 11 '17 at 17:29
• Why would $L(q)=\infty$ always? Fix $q$ and freeze $|p|$ for the moment. Then the first term is biggest when $p=|p| \frac{q}{|q|}$. So the supremum is the supremum over $|p|$ of $|p||q|-|p|=|p|(|q|-1)$. This is $0$ at $p=0$ if $|q| \leq 1$ and then $+\infty$ otherwise. Then your formula always allows you to choose some $y$ with $L((x-y)/t)=0$: just choose $y$ proportional to $x$ and not too large. I don't know if this deals with the rest of the problem so I don't feel comfortable writing an answer. – Ian Aug 11 '17 at 17:32

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.