Skip to main content
MathOverflow

Return to Question

Notice removed Draw attention by CommunityBot
Bounty Ended with no winning answer by CommunityBot
Notice added Draw attention by lasik43
Bounty Started worth 50 reputation by lasik43
added 33 characters in body
Source Link
lasik43
lasik43
  • 61
  • 4

Let $-\Delta \colon H^2(\mathbb R^d) \to \mathbb R^d$ be the (negative) Laplacian on the full space and $-\Delta_L$ the Laplacian acting on $L^2([-L,L]^d)$ with some boundary conditions making it self-adjoint, say Dirichlet. Furthermore, let $\Omega \subset \mathbb R^d$ be a fixed bounded set (smooth if needed), $h\colon [0,1] \to [0,1]$ continuouscontinuously differentiable with $h(0) = 0$ and $E>0$. I am interested whether it is true that $$\lim_{L\to \infty}\operatorname{tr}h(1_\Omega 1_{]-\infty,E[}(-\Delta_L) 1_\Omega) = \operatorname{tr} h(1_\Omega 1_{]-\infty,E[}(-\Delta) 1_\Omega).$$ Here $1_\Omega$ means the corresponding multiplication operator. Are there any known results?

Let $-\Delta \colon H^2(\mathbb R^d) \to \mathbb R^d$ be the (negative) Laplacian on the full space and $-\Delta_L$ the Laplacian acting on $L^2([-L,L]^d)$ with some boundary conditions making it self-adjoint, say Dirichlet. Furthermore, let $\Omega \subset \mathbb R^d$ be a fixed bounded set (smooth if needed), $h\colon [0,1] \to [0,1]$ continuous and $E>0$. I am interested whether it is true that $$\lim_{L\to \infty}\operatorname{tr}h(1_\Omega 1_{]-\infty,E[}(-\Delta_L) 1_\Omega) = \operatorname{tr} h(1_\Omega 1_{]-\infty,E[}(-\Delta) 1_\Omega).$$ Here $1_\Omega$ means the corresponding multiplication operator. Are there any known results?

Let $-\Delta \colon H^2(\mathbb R^d) \to \mathbb R^d$ be the (negative) Laplacian on the full space and $-\Delta_L$ the Laplacian acting on $L^2([-L,L]^d)$ with some boundary conditions making it self-adjoint, say Dirichlet. Furthermore, let $\Omega \subset \mathbb R^d$ be a fixed bounded set (smooth if needed), $h\colon [0,1] \to [0,1]$ continuously differentiable with $h(0) = 0$ and $E>0$. I am interested whether it is true that $$\lim_{L\to \infty}\operatorname{tr}h(1_\Omega 1_{]-\infty,E[}(-\Delta_L) 1_\Omega) = \operatorname{tr} h(1_\Omega 1_{]-\infty,E[}(-\Delta) 1_\Omega).$$ Here $1_\Omega$ means the corresponding multiplication operator. Are there any known results?

added 68 characters in body
Source Link
lasik43
lasik43
  • 61
  • 4

Let $-\Delta \colon H^2(\mathbb R^d) \to \mathbb R^d$ be the (negative) Laplacian on the full space and $-\Delta_L$ the Laplacian acting on $L^2([-L,L]^d)$ with some boundary conditions making it self-adjoint, say Dirichlet. Furthermore, let $\Omega \subset \mathbb R^d$ be a fixed bounded set (smooth if needed), $h\colon [0,1] \to [0,1]$ continuous and $E>0$. I am interested whether it is true that $$\lim_{L\to \infty}\operatorname{tr}h(1_\Omega 1_{]-\infty,E[}(-\Delta_L) 1_\Omega) = \operatorname{tr} h(1_\Omega 1_{]-\infty,E[}(-\Delta) 1_\Omega).$$ AreHere $1_\Omega$ means the corresponding multiplication operator. Are there any known results?

Let $-\Delta \colon H^2(\mathbb R^d) \to \mathbb R^d$ be the (negative) Laplacian on the full space and $-\Delta_L$ the Laplacian acting on $L^2([-L,L]^d)$ with some boundary conditions making it self-adjoint, say Dirichlet. Furthermore, let $\Omega \subset \mathbb R^d$ be a fixed bounded set (smooth if needed), $h\colon [0,1] \to [0,1]$ continuous and $E>0$. I am interested whether it is true that $$\lim_{L\to \infty}\operatorname{tr}h(1_\Omega 1_{]-\infty,E[}(-\Delta_L) 1_\Omega) = \operatorname{tr} h(1_\Omega 1_{]-\infty,E[}(-\Delta) 1_\Omega).$$ Are there any known results?

Let $-\Delta \colon H^2(\mathbb R^d) \to \mathbb R^d$ be the (negative) Laplacian on the full space and $-\Delta_L$ the Laplacian acting on $L^2([-L,L]^d)$ with some boundary conditions making it self-adjoint, say Dirichlet. Furthermore, let $\Omega \subset \mathbb R^d$ be a fixed bounded set (smooth if needed), $h\colon [0,1] \to [0,1]$ continuous and $E>0$. I am interested whether it is true that $$\lim_{L\to \infty}\operatorname{tr}h(1_\Omega 1_{]-\infty,E[}(-\Delta_L) 1_\Omega) = \operatorname{tr} h(1_\Omega 1_{]-\infty,E[}(-\Delta) 1_\Omega).$$ Here $1_\Omega$ means the corresponding multiplication operator. Are there any known results?

Source Link
lasik43
lasik43
  • 61
  • 4

Trace type convergence of the Laplacian on the box to the Laplacian on $\mathbb R^d$

Let $-\Delta \colon H^2(\mathbb R^d) \to \mathbb R^d$ be the (negative) Laplacian on the full space and $-\Delta_L$ the Laplacian acting on $L^2([-L,L]^d)$ with some boundary conditions making it self-adjoint, say Dirichlet. Furthermore, let $\Omega \subset \mathbb R^d$ be a fixed bounded set (smooth if needed), $h\colon [0,1] \to [0,1]$ continuous and $E>0$. I am interested whether it is true that $$\lim_{L\to \infty}\operatorname{tr}h(1_\Omega 1_{]-\infty,E[}(-\Delta_L) 1_\Omega) = \operatorname{tr} h(1_\Omega 1_{]-\infty,E[}(-\Delta) 1_\Omega).$$ Are there any known results?