Dissipative operator

In mathematics, a dissipative operator is a linear operator A defined on a linear subspace D(A) of Banach space X, taking values in X such that for all λ > 0 and all xD(A)

\|(\lambda I-A)x\|\geq\lambda\|x\|.

A couple of equivalent definitions are given below. A dissipative operator is called maximally dissipative if it is dissipative and for all λ > 0 the operator λIA is surjective, meaning that the range when applied to the domain D is the whole of the space X.

An operator that obeys a similar condition but with a plus sign instead of a minus sign (that is, the negation of a dissipative operator) is called an accretive operator.[1]

The main importance of dissipative operators is their appearance in the Lumer–Phillips theorem which characterizes maximally dissipative operators as the generators of contraction semigroups.

Properties

A dissipative operator has the following properties[2]

\|(\lambda I-A)^{-1}z\|\leq\frac{1}{\lambda}\|z\|
for all z in the range of λIA. This is the same inequality as that given at the beginning of this article, with z=(\lambda I-A)x. (We could equally well write these as \|(I-\kappa A)^{-1}z\|\leq\|z\|\text{ or }\|(I-\kappa A)x\|\geq\|x\| which must hold for any positive κ.)

Equivalent characterizations

Define the duality set of xX, a subset of the dual space X' of X, by

J(x):=\left\{x'\in X':\|x'\|_{X'}^2=\|x\|_{X}^2=\langle x',x\rangle \right\}.

By the Hahn–Banach theorem this set is nonempty.[3] If X is reflexive, then J(x) consists of a single element. In the Hilbert space case (using the canonical duality between a Hilbert space and its dual) it consists of the single element x.[4] Using this notation, A is dissipative if and only if[5] for all xD(A) there exists a x'J(x) such that

{\rm Re}\langle Ax,x'\rangle\leq0.

In the case of Hilbert spaces, this becomes {\rm Re}\langle Ax,x\rangle\leq0 for all x in D(A). Since this is non-positive, we have

\|x-Ax\|^2=\|x\|^2+\|Ax\|^2-2{\rm Re}\langle Ax,x\rangle\geq\|x\|^2+\|Ax\|^2+2{\rm Re}\langle Ax,x\rangle=\|x+Ax\|^2
\therefore\|x-Ax\|\geq\|x+Ax\|

Since I−A has an inverse, this implies that (I+A)(I-A)^{-1} is a contraction, and more generally, (\lambda I+A)(\lambda I-A)^{-1} is a contraction for any positive λ. The utility of this formulation is that if this operator is a contraction for some positive λ then A is dissipative. It is not necessary to show that it is a contraction for all positive λ (though this is true), in contrast to (λI−A)−1 which must be proved to be a contraction for all positive values of λ.

Examples

x \cdot A x = x \cdot (-x) = - \| x \|^{2} \leq 0,
so A is a dissipative operator.
\langle u, A u \rangle = \int_{0}^{1} u(x) u'(x) \, \mathrm{d} x = - \frac1{2} u(0)^{2} \leq 0.
Hence, A is a dissipative operator. Furthermore, since there is a solution (almost everywhere) in D to u-\lambda u'=f for any f in H, the operator A is maximally dissipative. Note that in a case of infinite dimensionality like this, the range can be the whole Banach space even though the domain is only a proper subspace thereof.
\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.

Notes

  1. "Dissipative operator". Encyclopedia of Mathematics.
  2. Engel and Nagel Proposition II.3.14
  3. The theorem implies that for a given x there exists a continuous linear functional φ with the property that φ(x)=‖x‖, with the norm of φ equal to 1. We identify ‖x‖φ with x'.
  4. Engel and Nagel Exercise II.3.25i
  5. Engel and Nagel Proposition II.3.23

References

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