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I am looking for some help to determine the large-time behavior of the unique solution for the equation in $\mathbb R^+ \times \mathbb R$ $$u_t+\vert\nabla u\vert^\frac{2}{3}=0,\ \ \ \ u(0,x)=-\cos x$$ More specifically, I am thinking about how to determine the behavior of u at $(2015,0)$ or $(n,0)$ for any large integer $n$, but I have no idea that where to start.

Many thanks to the help in advance.

EditPiAf
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Sam Wong
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1 Answers1

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Long comment: The Hamilton-Jacobi PDE is derived from a canonical transformation involving a type-2 generating function $u(t,x)$ which makes vanish the new Hamiltonian $K=H(u_x) + u_t$. Here, $H(p)=|p|^{2/3}$ is the original Hamiltonian.

For convex Hamiltonians $H$ which grow faster than the identity, the unique continuous viscosity solution can be obtained via the Lax-Hopf formula $$ u(t,x) = \inf_{y\in \Bbb R} \left\lbrace t L \big(\tfrac{x-y}{t}\big) - \cos y\right\rbrace $$ where the Lagrangian $L$ is the Legendre-Fenchel transform of $H$: $$ L(q) = \sup_{p\in\Bbb R} \lbrace qp - H(p)\rbrace = \frac{4}{27}|q|^3 . $$ However, the Hamiltonian is nonconvex here, which invalidates the previous approach. One may have a look at this post for complements.

EditPiAf
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  • Hi Harry, if we want to determine the value of $u$ at $(2015,0)$, then shall we need to know the period? Or can $2015$ be regarded as a enough large time and so that $u$ vanishes at $(2015,0)$? Thank you:) – Sam Wong Dec 10 '18 at 02:27
  • Thank you so much. It's clear to me now :) – Sam Wong Dec 10 '18 at 09:21
  • Hi Harry. I checked your another post yesterday. Does your another post imply that $u(2018,0)=0$ since $u(n,0)$ is closed to $0$ when $n\ge 5$ by your another post? But my friend told me that $u(2018,0)=-1$. Is he wrong? – Sam Wong Dec 10 '18 at 22:39
  • Thanks for your hint, I will reconsider this problem :) – Sam Wong Dec 11 '18 at 10:08
  • Thank you so much:) – Sam Wong Dec 11 '18 at 14:03
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    The Hamiltonian is not convex...This answer does not seem correct. – Jeff Dec 19 '18 at 16:50