Consider the elliptic eigenvalue problem $$ \begin{cases} \int_{B(r)} A(x) \nabla u \cdot \nabla \phi \, dx &= \ \ \frac{\lambda_1}{r^2}\int_{B(r)} u \phi \, dx \\ \qquad \qquad \qquad \quad u&= \ \ \ 0\quad\text{on} \quad \partial B(r), \end{cases} $$ where $B(r)$ is a ball of radius $r$ and $\phi \in W_0^{1,2}(B(r))$ is a standard test function. By the classical theory it is known that the principal eigenfunction $u$ (solving the above equation) is non-negative and minimizes the Rayleigh quotient. Instead of merely non-negativity, I am interested in finding a quantitative lower bound in $\overline B(r)$.

I was able to find out that if the coefficient matrix $A$ is assumed to be uniformly elliptic, $C^1$-function of $x$, the principal eigenfunction behaves like the distant function to the boundary: \begin{equation}\tag 1 u \ge c(1-\sigma)r \quad \text{in} \quad \overline B(\sigma r), \quad \text{for} \quad 0<\sigma \le 1. \end{equation} Obviously, due to the scale invariance of the equation, the constant $c$ must depend on $u$. Now my question is threefold:

  1. Does equation (1) hold if we merely assume the coefficient matrix to be "bounded and measurable"? With this I mean that $A$ is measurable function of $x$ and the following hold:

    • $A(x) p \cdot p \ge \lambda|p|^2, \quad x, p \in \mathbb{R}^n, \quad\lambda \in \mathbb{R}_+,$
    • $|A(x)p| \le \Lambda |p|, \quad \Lambda \in \mathbb{R}_+$.
  2. If yes, what would be a good reference for the result in this generality? If not, what are the minimal assumptions which are needed for $A$ to satisfy in order to equation (1) to hold?

  3. What is the exact behavior of the constant $c$ in (1) and what would be a good reference for this? In particular is it possible to prove for sufficiently smooth $A$ that $$\tag 2 \frac{1}{|B(r)|}\int_{B(r)} u \, dx \le \frac{C}{(1-\sigma)^\gamma}\inf_{B(\sigma r)} u $$ for some constants $C$ and $\gamma$ which only depend on the a priori information and, in particular, do not depend on $u$ or $r$.

  • $\begingroup$ The keyword to look for is symmetrization, I think, as in this paper. $\endgroup$
    – username
    Feb 1, 2014 at 21:57
  • $\begingroup$ Thanks for the paper. There seems to be some interesting references there. However, these papers seem overly complicated for my purposes. Shouldn't this kind of results for eigenfunctions be almost classical? The equation is linear after all. Yet, it seems difficult to find a reference for the results. $\endgroup$ Feb 4, 2014 at 8:18
  • $\begingroup$ Oh, sorry about that. I forgot the eigenvalue in the problem definition. It's now corrected. :P $\endgroup$ Feb 5, 2014 at 13:10
  • $\begingroup$ That's what people usually do, I guess. The problem with this scaling argument is, however, that then it is not immediate anymore what is the scaling behavior of the estimates and, in particular, whether they are scale invariant or not. So I usually prefer keeping the r in the estimates so that it is easier to see how they transform under scalings. $\endgroup$ Feb 6, 2014 at 12:59
  • $\begingroup$ One more remark: If we know that the principal eigenfunction attains its maximum in a point with a quantified distance to the boundary, then it is possible to show the estimate (2) by using the Harnack inequality - at least for $\sigma \ge \sigma_0$ with a quantitatively determinable $\sigma_0 <1$. Then we can probably adjust the constants to make it true for any $\sigma$. Moreover, I think this kind of result should be true as the structure of the matrix $A$ should give an upper bound for the speed of diffusion which determines where the maximum is attained. $\endgroup$ Feb 7, 2014 at 9:15

1 Answer 1


Heavily edited answer, closer to what the question asks.

There is an answer to the simpler question: $$ C (\sigma)^{\gamma} \int_{B_1} u^2 dx \leq \int_{B_{\sigma}} u^2 dx $$ where $C$ and $\gamma$ depend only on $\lambda,\Lambda$, and the dimension. Such estimates are available for a general not necessarily symmetric $A$ bounded and measurable in dimension 2 (see Alessandrini 2010), and for $A$ Lipschitz continuous in dimension $3$ or more (see Garofalo & Lin 1986 and Alessandrini, Rondi, Rosset and Vessela). These results are general unique continuation results, not specific to the eigenvalue problem; in that context, the hypothesis on the regularity of $A$ is probably optimal. A result slightly closer to what you want is for example $$ C (\frac{\sigma}{\tau})^{\gamma} \int_{\partial B_{\tau r}} u^2 dx \leq \int_{\partial B_{\sigma r}} u^2 dx $$ for $\tau\leq 1/2$, which is proved in the same papers. Both result follow from the the doubling inequality $$ C\int_{\partial B_{2\sigma r}} u^2 \leq \int_{\partial B_{\sigma r}} u^2. $$
In the other direction (bound from above), you have the three sphere's inequality $$ \|u\|_{L^2(B_{r_2})} \leq \|u\|^{\alpha}_{L^2(B_{r_1})}\|u\|^{1-\alpha}_{L^2(B_{r_3})} $$ For every $r_1<r_2<r_3<R$ in $B_R$, with the same assumptions on $A$. But that's not exactly what you wanted, as this is not $\min u$.

Below is an attempt to decide what to expect in terms of exponents in dimension 1.

What you would like is that the solution behaves reasonably nicely near the boundary $\partial B_r$. A counter-example, in dimension 1, would be if the solution was like $u\approx \exp(-\sigma r/n) - \exp(-r/n)$ because in that case the constant $c$ in your estimate would depend in $\sigma$ and $r$ in the wrong way.

Let us consider the interval $(-1,1)$ (you can always rescale the argument), and the eigenvalue problem $$ - \frac{1}{n^2}(a u')' = \lambda u, \quad u(-1)=u(1)=0,$$ with $1\leq a\leq2$ for example. By changing variables to $v=u/\sqrt{a}$, you obtain $$ - \frac{1}{n^2}v'' + q v = \lambda r v, $$ with $q= \sqrt{a}''/\sqrt{a}$ and $r=1/\sqrt{a}$.

Now, choose $q,r$ to be discontinuous, and periodic (of period $1$) for $x>0$ and $x<0$. More precisely, choose $q$ to be $q(x)= q_0(nx)$ for $x<0$, and $q(x) = q_0(nx + t)$ for $x>0$ and $r$ of a similar form.

Now, if you try to construct a rescaled 'full space' solution (forgetting the boundary constraint), you can solve it using Floquet Theory, which tells you that the solution will look like $$ exp(\theta_{\pm}nx) \psi_{\pm}(nx) \mbox{ for } \pm x >0, \mbox{ with } \psi_{\pm} \mbox{ periodic}. $$ Playing with the parameter $t$ (try numerically, e.g. with Matthieu's equations for $q_0$ and $r_0$), you can cook-up a case where $\theta_{-}>0$ and $\theta_{+}<0$. Since this 'full space' solution decays exponentially, it is an excellent candidate for the Dirichlet problem, and therefore the real solution will be exponentially close to that, and you are in the bad case previously described.

It is not a true counter-example, as it uses a very small lower bound for $a$. But I think it tells you that at best with respect to (3) you cannot hope for more than a logarithmic dependence on $(1-\sigma)$ in terms of the a priori information.

  • $\begingroup$ Thanks for the answer. I did not do the calculations, but I think your end result is exactly of the correct form I am looking for, isn't it? Your solutions is basically $$u=C(1-x/R)$$ at the boundary, i.e. it behaves like the distance function, right? Naturally, the scaling depends on the normalization and when you normalize $u(0)=1$, the scaling in $R$ changes correspondingly. $\endgroup$ Feb 5, 2014 at 13:24

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