Dissipative operator

On a Hilbert space, an operator $A$ is dissipative when $\mathrm{Re}(Ax,x)\le 0$ for every $x\in D(A)$. This is the infinitesimal form of norm contraction.

For a closed densely defined dissipative operator, the estimate $\Vert (\alpha I-Z)f\Vert \ge \alpha \Vert f\Vert$ makes its range closed and its kernel trivial. Thus a positive spectral value means this range is proper, which is equivalent to a nonzero vector in $\ker (\alpha I-Z^{\ast })$. Dissipativity is essential: a real multiplication operator may have positive continuous spectrum with no adjoint eigenvector.

// Target: analysis.bigb

If $\mathrm{Re}⟨Lv,v⟩\le 0$ in a finite-dimensional Hilbert space, $I-hL/2$ is invertible for every positive $h$. The relation $v_{+}-v_{-}=(h/2)L(v_{+}+v_{-})$ gives $\Vert v_{+}{\Vert }^2-\Vert v_{-}{\Vert }^2=(h/2)\mathrm{Re}⟨L(v_{+}+v_{-}),v_{+}+v_{-}⟩\le 0$. Thus the Crank-Nicolson method contracts for such a linear system even when $L$ is a non-normal matrix. This is a direct energy result, not an unqualified inequality by a scalar rational function evaluated at a logarithmic norm.

A maximal dissipative operator is dissipative and has no proper dissipative extension. Equivalently, for a densely defined dissipative operator, $\lambda I-A$ is onto for some positive $\lambda$.

The Lumer-Phillips theorem says that a densely defined operator generates a contraction C0-semigroup exactly when it is maximal dissipative.