I am currently studying eigenvalue problems.

I already worked through the Minimax-principles, seen why $\lambda_{h, m} \geq \lambda_m$, when comparing the eigenvalues of a discretization and the continuous ones.

The next lemma tries to establish conditions when we can get convergence between these two, finally arriving at the theorem that eigenvalues converge twice as fast as eigenvectors do.

The lemma states the following with $V$ being the space we are looking for eigenvectors in, $V_h$ being a discretation of that and $V_m$ being the space which is spanned by the first $m$ eigenvectors:

Define $$C_{h,m} := \inf_{v \in V_m} \frac{\| P_h v \|}{\|v\|}, \quad 1 \leq m \leq N.$$

If there exists a $m$ such that $C_{h,m} > 0$, we get

$$\lambda_m \leq \lambda_{h,m} \leq C_{h,m}^{-2} \, \lambda_m.$$

There is a remark stating that the $1-C^{-1}_{h,m}$ is a measure for how close $V_m$ and $P_h V_m$ are and that it is the norm of the inverse of the operator $P_h$. Can you help me how to arrive at this conclusion?

My thoughts so far were that the $C_{h,m}$ are measuring the minimal $\cos$ between $V_m$ and $P_h V_m$. As we want to get $C_{h,m} \to 1$ later on for $h \to 0$, it means that $P_h V_m$ would become parallel to $V_m$. What is the connection to the Inverse of the operator $P_h$?

  • $\begingroup$ What is $\sigma_{h,m}$ in your description? $\endgroup$ Jun 17 '18 at 5:16
  • $\begingroup$ Also, as a projection operator, $P_h$ obviously doesn't have an inverse. $\endgroup$ Jun 17 '18 at 5:19
  • $\begingroup$ @WolfgangBangerth The assumption that there exists an $m$ such that $C_{h,m} > 0$ excludes the case that there is a $v \in \ker{P_h}$, the restricted operator, so that $P_h V_m$ is $m$-dimensional, hence the minimax principle applicable. $\endgroup$
    – mdot
    Jun 17 '18 at 8:27
  • $\begingroup$ @WolfgangBangerth Sorry, the $\sigma_{h,m}$ was a typo. It is simply a $C_{h,m}$. It is only implicitly stated that Ritz-projector $P_h$ is restricted to $P_h \colon V_m \to P_h V_m$. The $C_{h,m}$ assumption should therefore guarantee that one can invert this restricted operator. $\endgroup$
    – mdot
    Jun 17 '18 at 10:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.