Dissipative operator

In linear theory, dissipative operators are linear operators that are defined on real or complex Banach spaces and that satisfy certain norm estimates. Through the Lumer-Phillips theorem , they play an important role in the consideration of strongly continuous semigroups .

definition

Be a Banach space and . A linear operator with ${\displaystyle X}$ ${\displaystyle D(A)\subset X}$ ${\displaystyle A:D(A)\rightarrow X}$

${\displaystyle \|(\lambda -A)x\|\geq \lambda \|x\|}$

for everyone and is called dissipative . This name goes back to Ralph Phillips . ${\displaystyle \lambda >0}$ ${\displaystyle x\in D(A)}$

If a linear operator and is dissipative, it is called accretive . This designation was introduced by Tosio Kato and Kurt Friedrichs . ${\displaystyle A}$ ${\displaystyle -A}$ ${\displaystyle A}$

Hilbert dream

If is a Hilbert space , then a linear operator is dissipative if and only if ${\displaystyle X}$ ${\displaystyle A:D(A)\rightarrow X}$

${\displaystyle \operatorname {Re} \,\langle Ax,x\rangle \leq 0}$

applies to all , where denotes the real part . ${\displaystyle x\in D(A)}$ ${\displaystyle \operatorname {Re} }$

Inferences

Let be a dissipative operator on a Banach space . ${\displaystyle (A,D(A))}$ ${\displaystyle X}$

• ${\displaystyle \lambda -A}$is for a surjective if and only if is for all surjective. Then m-dissipative and produces a strongly continuous operator half-group. ${\displaystyle \lambda >0}$ ${\displaystyle \lambda -A}$ ${\displaystyle \lambda >0}$ ${\displaystyle (A,D(A))}$
• ${\displaystyle A}$is complete exactly when the image of for one is complete . ${\displaystyle \lambda -A}$ ${\displaystyle \lambda >0}$

example

Considering on a limited area of the Laplacian with Dirichlet boundary condition on (see -space ), so we have: ${\displaystyle \Omega \subset \mathbb {R} ^{n}}$ ${\displaystyle \Delta }$ ${\displaystyle L^{2}(\Omega )}$ ${\displaystyle L^{p}}$ ${\displaystyle D(\Delta )=H^{2}(\Omega )\cap H_{0}^{1}(\Omega )}$

${\displaystyle \langle \Delta u,u\rangle =-\langle \nabla u,\nabla u\rangle =-\|\nabla u\|^{2}\leq 0}$.

The set of Lax-Milgram demonstrates that is m-dissipative and thus creates a strong continuous operator semigroup. ${\displaystyle \Delta :D(\Delta )\rightarrow L^{2}(\Omega )}$