Given an eigenvalue equation (elliptic PDE) in a ball $B_R$, prove the convergence of the first nonzero $\lambda_R$ and its eigenfunction $\phi_R$

Question

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!

Answer 1

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 $\phi_\infty = 1$.)