Let $H: \mathbf{R}^n \rightarrow \mathbf{R}$ be a bounded continuous function. Set
$$\tag{1}
\int\_{\mathbf{R}^n}\left\{|\nabla \xi|^2+H(x) \xi^2\right\} \mathrm{d} x \geqslant 0, \quad \forall \xi \in C\_c^{\infty}\left(\mathbf{R}^n\right)
$$
Then by Theorem 8.38 in Gilbarg & Trudinger's book, in a ball $B\_R$ we should have $$-\Delta \phi\_R +
H(x)\phi\_R=\lambda\_R\phi\_R,$$
where $\lambda\_R\ge0$ is the first nonzero eigenvalue, with $\phi\_R>0$.


1. If there is no $H(x)$, it will be simple to prove that $\lambda\_R$ is a decreasing function when $R$ becomes greater, so there will be a limit for $\lambda\_R$ when $R \rightarrow +\infty$.  
 But here we have $H(x)$, so we can't simply let $\phi(x)=\Phi(xR/a)$ to make the ball bigger (with radius $aR$): do we still have the property that the first nonzero eigenvalue is a decreasing function with respect to $R$?
2. What is the value of $\lim\_{R\rightarrow+\infty}\lambda\_R=\lambda\_0$, is it 0?
3. If we let the $R$ tend to $+\infty$, can we also get a limit function $\Phi$, which satisfies that $$-\Delta \Phi +
H(x)\Phi=\lambda\_0\Phi,
$$
and what type is this convergence? Do we need local convergence or?


All these questions have a vague answer in my mind, but I don't know how to figure out the details, if you have any references I would be very glad!



Given an eigenvalue equation (elliptic PDE) in a ball $B_R$, prove the convergence of the first nonzero $\lambda_R$ and its eigenfunction $\phi_R$
For simplicity let me take $H$ smooth or at least $C^\alpha$ so that I don't have to worry about elliptic estimates:


1. Limit as $R\to\infty$:


Use
$$
\lambda\_R = \inf\_{\phi \in C^\infty\_c(B\_R)}\frac{\int |\nabla \phi|^2 + H \phi^2}{\int \phi^2} 
$$
Since $C^\infty\_c(B\_R) \subset C^\infty\_c(B\_{R+s})$ for all $s>0$ we see that $\lambda\_R$ is non-increasing.


2. The limit $\lambda\_\infty$ (which exists since $\lambda\_R$ is non-increasing and $\geq 0$) depends on $H$. For example if $H=0$ then then since we know the first Dirichlet eigenvalue of a ball we can see that $\lambda\_\infty=0$. On the other hand, if $H=1$ then
$$
\int\_{B\_R} |\nabla \phi|^2 + \phi^2 \geq \int \phi^2
$$
so $\lambda\_\infty \geq 1$ (actually we should be able to prove $\lambda\_\infty=1$).
3. Yes. Choose the first eigenfunction $\phi\_R$ and normalize so that $\phi\_R(0) = 1$. Then, the Harnack inequality says that for $K\Subset B\_R$ there's $C=C(K,H)$ so that $C^{-1}\leq \phi\_R \leq C$ on $K$. Bootstrapping to $C^{2,\alpha}$ we can take a subsequence to find $\phi\_\infty$ solving
$$
-\Delta \phi\_\infty+H\phi\_\infty = \lambda\_\infty\phi\_\infty.
$$
Note that, $\phi\_\infty$ will not necessarily (never?) be in $L^2(\mathbb{R}^n)$. (For example, when $H$ is constant, we get $\phi\_\infty = 1$.)


