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.

Schrödingerequation: 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). $\endgroup$ – Mateusz Kwaśnicki Aug 5 '18 at 23:08