Talk:Dissipative operator

Could use some context and See also.--Cronholm¹⁴⁴ 02:47, 18 July 2007 (UTC) Gracias--Cronholm¹⁴⁴ 10:32, 18 July 2007 (UTC)

Number of elements in J(x)

It is not sufficient that X is reflexive for J(x) to consist of a single element. As a counterexample, take X a two-dimensional space equipped with the maximum norm. It is, however, sufficient if X' has strictly convex norm.

188.103.139.61 (talk) 08:33, 17 November 2011 (UTC)

Wrong spaces in last example (Laplacian)?

I am quite sure that in the last example the operator $A=\Delta$ (Laplacian) rather has to be considered on the Hilbert space of square integrable functions $L^{2}\left(\Omega ;\mathbf {R} \right)$ in stead of the Sobolev space $H_{0}^{2}\left(\Omega ;\mathbf {R} \right)$. The domain $D\left(A\right)$ could either be chosen as $D\left(A\right)=H^{2}\left(\Omega ;\mathbf {R} \right)\cap H_{0}^{1}\left(\Omega ;\mathbf {R} \right)$ or as some suitable subspace of this (e.g. compactly supported smooth functions).

Hence I want to suggest to modify this example in one of the following two ways:

• Consider $H=L^{2}\left(\Omega ;\mathbf {R} \right)$ for an open and connected domain $\Omega \subset \mathbf {R} ^{n}$ and let $A=\Delta$, the Laplace operator, defined on the dense subspace $D\left(A\right)$ of compactly supported smooth functions on $\Omega$. Then, using integration by parts,

$\langle u,\Delta u\rangle =\int _{\Omega }u(x)\Delta u(x)\,\mathrm {d} x=-\int _{\Omega }{\big |}\nabla u(x){\big |}^{2}\,\mathrm {d} x=-\|\nabla u\|_{L^{2}(\Omega ;\mathbf {R} )}\leq 0,$

so the Laplacian is a dissipative operator.

or

• Consider $H=L^{2}\left(\Omega ;\mathbf {R} \right)$ for an open and connected domain $\Omega \subset \mathbf {R} ^{n}$ and let $A=\Delta$, the Laplace operator, defined on the dense subspace $D\left(A\right)=H^{2}\left(\Omega ;\mathbf {R} \right)\cap H_{0}^{1}\left(\Omega ;\mathbf {R} \right)$. Then, using integration by parts,

$\langle u,\Delta u\rangle =\int _{\Omega }u(x)\Delta u(x)\,\mathrm {d} x=-\int _{\Omega }{\big |}\nabla u(x){\big |}^{2}\,\mathrm {d} x=-\|\nabla u\|_{L^{2}(\Omega ;\mathbf {R} )}\leq 0,$

so the Laplacian is a dissipative operator.

Could somebody please confirm this? Thank you!

DufterKunde (talk) 14:09, 11 April 2012 (UTC)

"Quasidissipative operator" listed at Redirects for discussion

The redirect Quasidissipative operator has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at Wikipedia:Redirects for discussion/Log/2023 November 23 § Quasidissipative operator until a consensus is reached. 1234qwer1234qwer4 20:09, 23 November 2023 (UTC)