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Consider the Hamilton Jacobi equation $$\begin{cases} u_t + |\nabla u|^2 = 0 & x \in \mathbb{R}^n, t > 0\\ u = g(x) & t = 0 \end{cases}$$ Suppose $g$ is Lipschitz and $|g(x)| \leq \frac{M}{|x|},$ where $M > 0.$ Show that $u(x,t) \to 0$ as $t \to \infty.$


I know the solution is given by $$ u(x,t) = \min_{y \in \mathbb{R}^n} \left\{ \frac{|x-y|^2}{4t} + g(y) \right\}$$ which is just the Hopf-Lax formula with the Lagrangian $L(q) = \frac{|q|^2}{4}.$ However, I am completely stumped on showing that $u \to 0$ as $t \to \infty.$ One thing we do know is that for $g$ Lipschitz, we have $$ |u(x,t) - g(x)| \leq Ct$$ for some $C, t > 0$ A proof of this can be found in Evans chapter 3.

I feel like this latter fact should be enough for one to show that $u \to 0,$ but I've been stuck on it for some time now. Any sugggestions? Thanks in advance.

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There's a missing assumption $g\ge 0$; without it the statement is false since $u(x,t)\le g(x)$ for all $t\ge 0$.

An intuitive way to think of the minimum of $\frac{|x-y|^2}{4t} + g(y)$ is that it's the smallest number $C$ for which the equation $g(y) = C- \frac{|x-y|^2}{4t}$ has a solution. Well this is not quite intuitive yet, but think of the family $C- \frac{|x-y|^2}{4t}$ parameterized by $C$ as an upside-down paraboloid rising from $C=0$ until it meets the graph of $g$. Then the claim is: when the paraboloid is very flat ($t$ is large), it can't rise far before hitting the graph of $g$. Then the statement becomes obvious because $g$ decays at infinity: that far-away, low part of the graph of $g$ will get in the way of the rising paraboloid.

But "it's obvious" still isn't a proof, so let's fix $x$ and suppose $u(x,t) \ge \epsilon$ for some large $t$. Then $g(y) + |x-y|^2/(4t) \ge \epsilon $ for all $y$. Pick $y$ such that $g(y) \le \epsilon/2$ and conclude that $$ t\le \frac{|x-y|^2}{2\epsilon} \tag{1} $$ So, for every $\epsilon>0$ there exist $t_0$ such that $u(x,t)<\epsilon$ when $t>t_0$. The very definition of $\lim_{t\to\infty}u(x,t) = 0$.

And the above can be made uniform with respect to $x$, using the assumption $g(x)\le \frac{M}{|x|}$. Indeed, any $y$ with $|y|\ge 2M/\epsilon$ will do now, so in (1) we can be assured to have $|x-y|\le 2M/\epsilon$ no matter what $x$ is, leading to
$$ t\le \frac{2M^2 }{\epsilon^3} \tag{2} $$

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  • $\begingroup$ Wow! Impressive explaination, the intuitive explaination really changed the way I look at these minimums! Thanks! $\endgroup$ – Merkh Aug 18 '17 at 17:29

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