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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!

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  • $\begingroup$ 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 $|q|>1$? $\endgroup$ – Ian Aug 11 '17 at 17:09
  • $\begingroup$ 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. $\endgroup$ – Merkh Aug 11 '17 at 17:29
  • $\begingroup$ 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. $\endgroup$ – Ian Aug 11 '17 at 17:32
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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.

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