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I'm searching for lower bounds of bilinear forms arising in FEM for elliptic second order PDEs with mixed boundaries.

I did some research and found:

$$\max_{v_{h}\in\mathcal{V}_h(\mathcal{\Omega})}a(v_{h}, v_{h}) \geq Ch^{d-2}\, ,$$

where $d$ is the Dimension of the domain, $C$ a positive constant and $h$ the mesh size.

The bilinear form is defined as $a(v_h,v_h):= \int_\Omega \nabla v_h A \nabla v_h \, d\Omega$ and $A$ is an elliptic operator.

I found this bound quite often but not a single proof to it. Does anyone know some literature or how the proof works?

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  • $\begingroup$ Is $A$ an operator or a matrix? Also, is $\|y\|_2$ the $L_2$ norm? $\endgroup$ – Wolfgang Bangerth Nov 19 '18 at 18:32
  • $\begingroup$ $A$ is an elliptic operator. The original PDE reads: $-div(A \nabla u) = f$ with mixed boundary conditions. The norm is the Euclidian norm. The inequality appears during the calculation of condition number of the stiffness matrix. It is used to get a lower bound for it. But I can not find a proof for it. $\endgroup$ – user29088 Nov 19 '18 at 18:51
  • $\begingroup$ Then what are domain and range of $A$? I'm actually pretty sure that you want $A$ to be a symmetric and positive definite $d\times d$ matrix. $\endgroup$ – Wolfgang Bangerth Nov 19 '18 at 23:48
  • $\begingroup$ I don't think your lower bound makes sense -- what is $y$? As written, the $\max$ is not attained (you can take $y$ arbitrarily close to zero). Should this be $v_h$, too? Are you sure the constant is independent of the choice of $V_h$? $\endgroup$ – Christian Clason Nov 20 '18 at 7:37
  • $\begingroup$ @ChristianClason you are right. I did a big mistake but fixed it now. Sorry for the circumstance $\endgroup$ – user29088 Nov 20 '18 at 9:48
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I agree that $A$ should be a symmetric positive matrix, whose smallest eigenvalue is bounded away from 0. Let $K_A$ denote your resulting matrix and $K$ denote the FEM matrix for the Laplacian on the same mesh. Then you write $$ \max_{ v \neq 0 } \frac{ v^T K_A v }{ v^T v } \geq \lambda_{min, A} \max_{ v \neq 0 } \frac{ v^T K v }{ v^T v } \geq \lambda_{min, A} \max_{ v \neq 0 } \frac{ v^T k_{ele} v }{ v^T v } $$ where $k_{ele}$ is one arbitrary element matrix.

Whe largest eigenvalue of one arbitrary element matrix is $O(h^{d-2})$.

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