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Assume that $A=A(r,1)=\{x: r<||x||<1\} \subset R^n$ is an annulus.

Whether is known the constant of Poincare inequality for A or some its estimation (w.r.t. $L^2$): the constant $C$ in the inequality $||f||_ {L^2(A)}\le C ||\nabla f||_{L^2(A)},$ where $ f\in W^{1,2}_0(A) $

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A quick look on Wikipedia tells me that the Poincare constant depends on the domain and on the exponent $p$ (when one is looking at the $L^p$ norm of the gradient). Are you interested in estimates of the Poincare constant for certain $p$ in particular, or just any $p$? (My guess is that the $p=2$ case would be easier than the others.) –  Yemon Choi Feb 5 '12 at 8:33
    
Yes, I have had in mind the case p=2! –  Marijan Feb 5 '12 at 9:12
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2 Answers

The constant $C^{-2}$, that is, the infimum of the Raleigh quotient $$\min _ { f \in W^{1,2} _0(A) } \frac {\int_A|\nabla f|^2 dx}{ \int_A|f|^2 dx } $$ is the first eigenvalue of the Laplacian on $A$ with Dirichlet boundary conditions. As a general fact, a positive solution of $-\Delta f=\lambda f$ with Dirichlet boundary conditions is necessarily the first eigenfunction. Here, this allow to find it just by solving the corresponding ODE, where radial simmetry is assumed. The subject is of course classic. However, I do not have access to MatSciNet in this moment, but I'm pretty sure that a search on "first eigenvalue of the Laplacian on an annulus" should give you useful results; you may also like to write and solve the ODE by yourself, and to compute the corresponding Raleigh quotient.

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This is almost certainly solved in many physics textbooks, either in electromagnetism or maybe thermodynamics (steady heat flow). –  Deane Yang Feb 5 '12 at 12:18
    
Indeed ... –  Pietro Majer Feb 5 '12 at 12:45
    
Probably has to do with zero of Bessel function... –  Marijan Feb 5 '12 at 13:27
    
@Deane Yang, I would be grateful if you give a reference –  Marijan Feb 5 '12 at 13:50
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The relevant topic is "separation of variables". Try googling "separation of variables eigenvalues laplacian annulus". –  Deane Yang Feb 5 '12 at 19:16
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If you apply Poincare Inequality on the ball $B\supset A$, thus

$|\vert f-\frac{1}{|B|}\int_A f|\vert_{L^2(B)}\le C |\vert\nabla f|\vert_{L^{2}(B)}$

and as we know $f=0$ outside $A$, thus

$||\frac{1}{|B|}\int f||\le\frac{|A|}{|B|}||f||=\alpha ||f||$

By triangular inequality,

$|\vert f-\frac{1}{|B|}\int f|\vert_{L^2}\ge (1-\alpha)||f||_{L^2}$.

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