# Nonlinear Schrödinger equation with discrete Laplacian

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 $\Delta_{\text{disc}}u(u) = u(n+1) - 2u(n) + u(n-1)$?

• 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). – Mateusz Kwaśnicki Aug 5 '18 at 23:08
• 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. – user101142 Aug 7 '18 at 17:58
• @VeikkoApell: 11 edits in 2 days might be a bit too much. Why don't you give this question some time to settle? – Alex M. Aug 7 '18 at 20:24

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.
• @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?". – Mateusz Kwaśnicki Aug 7 '18 at 16:59
• (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? – Mateusz Kwaśnicki Aug 7 '18 at 16:59
• @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). – Mateusz Kwaśnicki Aug 7 '18 at 17:24