On discreteness of spectrum of Schrödinger operator with bounded potential

Let's consider a complete noncompact Riemannian manifold $M$ without boundary which is representable as $K\cup D$, where $K$ is a compact set and $D$ is isometric to the product $\mathbf{ R_0} \times \mathrm{S}_1\times \mathrm{S}_2\times\cdots\times \mathrm{S}_k$ (where $\mathbf{ R_0}=(r_0,+\infty)$, and $\mathrm{S}_i$ are compact Riemannian manifolds without boundary) with metric
$$ds^2=dr^2+q_1^2(r)d\theta_1^2+\cdots+q_k^2(r)d\theta_k^2,$$
where $d\theta_i^2$ is the metric on $\mathbb{S}_i$ and $q_i(r)$ is a smooth positive function on $\mathrm{R}_0$. We assume $\dim\mathbb{S}_i=n_i$ and denote $s(r)=q_1^{n_1}(r)\cdots q_k^{n_k}(r)$. The manifold $M$ is called a manifold with end. Since its end $D$ is a simple warped product, $M$ is the simplest case of a quasimodel manifold.
On the manifold $M$ we study the Laplace–Beltrami operator
$$-\Delta=-\mathrm{div}\nabla$$
and the Schrödinger operator
$$-\Delta=-\mathrm{div}\nabla+c(r,\theta).$$
We denote
$$F(r)=\left(\frac{s'(r)}{2s(r)}\right)' +\left(\frac{s'(r)}{2s(r)}\right)^2.$$

Theorem 1. Let's $c(r,\theta)\geq 0$. The spectrum of the Schrödinger operator $L$ on the manifold $M$ is discrete if one of the following conditions is satisfied:
$$ V(D)<\infty\quad \text{ and }\quad \lim\limits_{r\to \infty}\frac{V(D\setminus B(r))}{\mathrm{cap }(B(1),B(r))}=0,$$
or
$$\mathrm{cap }\, B(1)>0\quad \text{ and }\quad\lim\limits_{r\to \infty}\frac{V(B(r))}{\mathrm{cap }\, B(r)}=0.$$

We can note that the conditions of the theorem 1 are not just sufficient, but necessary for discreteness of the Laplacian spectrum.
Theorem 2. If there is a function $\tilde{c}(r)$ on manifold $M$ such that $c(r,\theta)\geq \tilde{c}(r)$ and $\tilde{c}(r)+F(r)>-C \ (C=\mathrm{const}>0)$, then the spectrum of the Schrödinger operator $L$ on the manifold $M$ is discrete if
$$\forall \omega>0\quad\lim_{r\to\infty}\int\limits_r^{r+\omega}\left(\tilde{c}(r)+F(r)\right)dr=+\infty.$$