In the paper "Global existence and scattering for rough solutions of a nonlinear Schrödinger equation on $\mathbb{R}^3$" by Colliander, Keel, Staffilani, Takaoka and Tao it is argued in the beginning that the NLS $$i\partial_t\phi(x,t)+\Delta\phi(x,t)=\vert\phi(x,t)\vert^2\phi(x,t)$$ with initial datum $\phi_0\in H^s(\mathbb{R}^3)$ is globally-in-time well-posed for $s\geq 1$.

And the precise argument has been discussed in this thread.

I am wondering whether the discrete NLS is also globally well-posed $$i\partial_t u(t,n) = -\Delta_{\text{disc}} u(t,n) + \lvert{u(t,n)}\rvert^2 u(t,n), \qquad t\in \mathbb{R},\ n\in\mathbb{Z},$$ where $\Delta_{\text{disc}}u(u) = u(n+1) - 2u(n) + u(n-1)$?

  • $\begingroup$ I do not see a reason to call this a Schrödinger equation: it is a first-order ODE with coefficients that depend on the parameter $x$. I am not an expert in ODEs, but I suppose ODE methods should work here (we have $|\phi(t, x)| = |\phi_0(x)|$, so if $s$ is large enough, then $\phi$ is bounded; in this case the solution is a smooth function of the initial condition, and I guess it is possible to deduce smooth dependence on the coefficient $\sin(x)$ as well). $\endgroup$ Commented Aug 5, 2018 at 23:08
  • $\begingroup$ Rather than the techniques relying on the presence of a continuous Laplacian, I would say that when you replace the Laplacian with something else, you have a different animal, and the right/interesting questions to ask about it are not the same. $\endgroup$
    – user101142
    Commented Aug 7, 2018 at 17:58
  • $\begingroup$ @VeikkoApell: 11 edits in 2 days might be a bit too much. Why don't you give this question some time to settle? $\endgroup$
    – Alex M.
    Commented Aug 7, 2018 at 20:24

1 Answer 1


My impression is, that in the discrete case the interesting questions and tools are sometimes different from the continuous case.

Studying the discrete Laplacion also means that one is interested in discrete solution. For example, in one dimension people are often investigating the discrete nonlinear Schrödinger equation $$i\partial_t u(t,n) = -\Delta_{\text{disc}} u(t,n) + \lvert{u(t,n)}\rvert^2 u(t,n), \qquad t\in \mathbb{R},\ n\in\mathbb{Z},$$ where $\Delta_{\text{disc}}u(u) = u(n+1) - 2u(n) + u(n-1)$. As you have suspected, the global-in-time well-posedness theory is much easier than in the continuous case. For example, in the space $\ell^2(\mathbb{Z})$, the existence and uniqueness of local solutions to the above equations follows from the classical Picard-Lindelöf theorem (regard the equation as an ODE with $C^1$-right-hand side in the Hilbert space $\ell^2(\mathbb{Z})$). The conservation of the $\ell^2$-norm (check $\frac{d}{dt} \lVert u(t)\rVert_2^2 = 0$) shows that there is no blow-up and hence the solution exists globally. Similar arguments work in weighted $\ell^2$-spaces, see e.g. Lemma 2 in the paper Asymptotic stability of small bound states in the discrete nonlinear Schrödinger equation by Kevrekidis, Pelinovsky and Stefanov.

You say that tools such as classical Strichartz estimates are gone in the discrete case. Such questions are also studied in the literature, see for example the paper Dispersion estimates for one-dimensional discrete Schrödinger and wave equations by Egorova, Kopylova, and Teschl. They prove dispersive estimates (and deduce Strichartz estimates) in the discrete setting.

  • $\begingroup$ The question I tried to answer is: On which space is the discrete NLS globally well-posed? $\endgroup$
    – gsa
    Commented Aug 7, 2018 at 16:41
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    $\begingroup$ @VeikkoApell: (1/2) I was about to write a similar (but much less detailed) answer, but right before posting it I realised what you really mean. I think your question is quite misleading, if not ill-posed at all. You can replace the Laplacian by whatever operator you like, but in each case the answer will be diffferent. If you omit $\Delta$ completely, or replace it by the discrete Laplacian $\Delta_{\operatorname{disc}}$, a natural answer is "this equation has nothing to do with $H^s$, why on Earth would you consider this?". $\endgroup$ Commented Aug 7, 2018 at 16:59
  • $\begingroup$ (2/2) If you use a non-local "elliptic" operator, like the fractional Laplace operator $(-\Delta)^s$, the answer might be: "there are similar tools, and quite likely you can follow the same approach to some extent". If you choose a "non-elliptic" operator, such as $(-\Delta)^{-s}$, the answer should still be positive, but it may require completely different methods. Finally, if you change $\Delta$ to something like $-i \phi$, the answer can well change to a negative one. I think what we are missing is the exact context of your question. Are you willing to consider a particular equation? $\endgroup$ Commented Aug 7, 2018 at 16:59
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    $\begingroup$ @VeikkoApell: One more though: if you are really interested in the discrete Laplacian and $H^s(\mathbb{R}^d)$, then still gsa's answer is the first step, and should not be downvoted: writing $x = z + r$ with $z \in \mathbb{Z}^3$ and $r \in [0,1)^3$, the solution of the continuous problem can be written as $\phi(x,t) = u_r(z,t)$, where $r$ is a parameter and $u_r$ solves gsa's variant. The question becomes: does the solution of the gsa's $\ell^2$-valued ODE depend on the initial condition in a sufficiently smooth way (at least for $s$ large enough). $\endgroup$ Commented Aug 7, 2018 at 17:24
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    $\begingroup$ @VeikkoApell: For numerical solutions, finite element methods seem more appropriate. If you insist on the discrete Laplace operator as an approximation of the continuous one, it still seems reasonable to consider discrete variable and measure smoothness with the help of discrete Sobolev spaces, in a way which is uniform with respect to the mesh size. I am not working in this area, but already the first reference suggested by Google seems to provide the right toolkit. $\endgroup$ Commented Aug 7, 2018 at 22:54

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