This is not anHeavily edited answer to the question, just an explicit calculationcloser to see what shows up in a discontinuous casethe question asks.
Let us look at a one dimensional piecewise constant case, namely the domainThere is an answer to the simpler question: $$ C (\sigma)^{\gamma} \int_{B_1} u^2 dx \leq \int_{B_{\sigma}} u^2 dx $$ where $(-r,r)$$C$ and $\gamma$ depend only on $\lambda,\Lambda$, and the coefficientdimension. Such estimates are available for a general not necessarily symmetric $A(x)=a$ when$A$ bounded and measurable in dimension 2 $x<0$(see Alessandrini 2010), and for $A$ Lipschitz continuous in dimension $A(x)=b$ when$3$ or more $x>0$(see Garofalo & Lin 1986 and Alessandrini, Rondi, Rosset and Vessela).
Let us now consider These results are general unique continuation results, not specific to the first eigenvalue/eigenvector of problem; in that context, the problem $$ -\frac{d}{dx}(A(x)\frac{d}{dx} u) = \frac{\lambda}{r^2} u \mbox{ in }H^1_0(-r,r), $$ normalised by $u(0)=1$. A simple comparison usinghypothesis on the Raleigh quotient shows thatregularity of $\lambda=\frac{\pi^2}{4}\nu$ with$A$ is probably optimal. $$ \min(a,b)\leq \nu \leq \max(a,b) $$
An explicit computation shows that $u,\nu$ are given byA result slightly closer to what you want is for example
$$
u=\frac{\displaystyle\cos\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)\sin\left(\frac{\pi\sqrt{\nu}x}{2\sqrt{a}r}\right)}{\displaystyle\sin\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)}+\cos\left(\frac{\pi\sqrt{\nu}x}{2\sqrt{a}r}\right)\mbox{ when }x<0,
$$$$
C (\frac{\sigma}{\tau})^{\gamma} \int_{\partial B_{\tau r}} u^2 dx \leq \int_{\partial B_{\sigma r}} u^2 dx
$$
andfor $\tau\leq 1/2$, which is proved in the same papers. Both result follow from the the doubling inequality
$$
u=-\frac{\displaystyle\cos\left(\frac{\pi\sqrt{\nu}}{2\sqrt{b}}\right)\sin\left(\frac{\pi\sqrt{\nu}x}{2\sqrt{b}r}\right)}{\displaystyle\sin\left(\frac{\pi\sqrt{\nu}}{2\sqrt{b}}\right)}+\cos\left(\frac{\pi\sqrt{\nu}x}{2\sqrt{b}r}\right)\mbox{ when }x>0,
$$$$
C\int_{\partial B_{2\sigma r}} u^2 \leq \int_{\partial B_{\sigma r}} u^2.
$$
andIn the other direction $\nu$ is(bound from above), you have the smallest positive solution ofthree sphere's inequality
$$
\tan\left(\frac{\pi\sqrt{\nu}}{2\sqrt{b}}\right)\sqrt{\nu}\sqrt{a}=-\tan\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)\sqrt{\nu}\sqrt{b}.
$$$$
\|u\|_{L^2(B_{r_2})} \leq \|u\|^{\alpha}_{L^2(B_{r_1})}\|u\|^{1-\alpha}_{L^2(B_{r_3})}
$$
ItFor 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 easynot $\min u$.
Below is an attempt to seedecide what to expect in terms of exponents in dimension 1.
What you would like is that for the solution to stay positive, it must stay fairly close tobehaves reasonably nicely near the boundary $\min(a,b)$$\partial B_r$. For example, when $b/a \gg1$ A counter-example, $\nu\approx 4a$in dimension 1, aswould be if the solution concentrateswas like $u\approx \exp(-\sigma r/n) - \exp(-r/n)$ because in that case the half domain whereconstant $c$ in your estimate would depend in $\sigma$ and $r$ in the conductivity iswrong way.
Let us consider the lowestinterval $(-1,1)$ (you can always rescale the argument), and becomes close tothe eigenvalue problem $\sin(\frac{\pi}{R}(x+R))$$$ - \frac{1}{n^2}(a u')' = \lambda u, \quad u(-1)=u(1)=0,$$ with $1\leq a\leq2$ for example. Near the boundaries: we have atBy changing variables to $-r$$v=u/\sqrt{a}$, you obtain $$ u\approx \frac{\cos^2\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)}{\sin\left(\frac{\pi\sqrt{\nu}}{2\sqrt{a}}\right)}(1+\frac{x}{r}) $$$$ - \frac{1}{n^2}v'' + q v = \lambda r v, $$ and atwith $r$ $$ u\approx \frac{\cos^2\left(\frac{\pi\sqrt{\nu}}{2\sqrt{b}}\right)}{\sin\left(\frac{\pi\sqrt{\nu}}{2\sqrt{b}}\right)}(1-\frac{x}{r}) $$$q= \sqrt{a}''/\sqrt{a}$ and $r=1/\sqrt{a}$.
Now, choose $q,r$ to be discontinuous, and periodic (editedof period $1$) So the scaling infor $x>0$ 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 dependenceboundary 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 stillan excellent candidate for the oneDirichlet 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.