Dissipative operator explained

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)

\|(λI-A)x\|\geqλ\|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]

\|(λI-A)x\|\ne0,

so the kernel of λIA is just the zero vector and λIA is therefore injective and has an inverse for all λ > 0. (If we have the strict inequality

\|(λI-A)x\|>λ\|x\|

for all non-null x in the domain, then, by the triangle inequality,

\|λx\|+\|Ax\|\ge\|(λI-A)x\|>λ\|x\|,

which implies that A itself has an inverse.) We may then state that

\|(λI-A)-1z\|\leq

1
λ

\|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=(λI-A)x.

(We could equally well write these as

\|(I-\kappaA)-1z\|\leq\|z\|or\|(I-\kappaA)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

2=\langle
X':\|x'\|
X

x',x\rangle\right\}.

By the Hahn–Banach theorem this set is nonempty.[3] In the Hilbert space case (using the canonical duality between a Hilbert space and its dual) it consists of the single element x.[4] More generally, if X is a Banach space with a strictly convex dual, then J(x) consists of a single element.[5] Using this notation, A is dissipative if and only if[6] for all xD(A) there exists a x' ∈ J(x) such that

{\rmRe}\langleAx,x'\rangle\leq0.

In the case of Hilbert spaces, this becomes

{\rmRe}\langleAx,x\rangle\leq0

for all x in D(A). Since this is non-positive, we have

\|x-Ax\|2=\|x\|2+\|Ax\|2-2{\rmRe}\langleAx,x\rangle\geq\|x\|2+\|Ax\|2+2{\rmRe}\langleAx,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,

(λI+A)(λI-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

xAx=x(-x)=-\|x\|2\leq0,

so A is a dissipative operator.

x*Ax,

which must be nonpositive for any x, is

x*

A+A*
2x.
The eigenvalues of this quadratic form must therefore be nonpositive. (The fact that the real part of

x*Ax,

must be nonpositive implies that the real parts of the eigenvalues of A must be nonpositive, but this is not sufficient. For example, if

A=\begin{pmatrix}-1&3\\0&-1\end{pmatrix}

then its eigenvalues are negative, but the eigenvalues of A+A* are −5 and 1, so A is not dissipative.) An equivalent condition is that for some (and hence any) positive

λ,λ-A

has an inverse and the operator

(λ+A)(λ-A)-1

is a contraction (that is, it either diminishes or leaves unchanged the norm of its operand). If the time derivative of a point x in the space is given by Ax, then the time evolution is governed by a contraction semigroup that constantly decreases the norm (or at least doesn't allow it to increase). (Note however that if the domain of A is a proper subspace, then A cannot be maximally dissipative because the range will not have a high enough dimensionality.)

H1([0, 1];R)

with u(1) = 0. D(A) is dense in L2([0, 1]; R). Moreover, for every u in D(A), using integration by parts,

\langleu,Au\rangle=

1
\int
0

u(x)u'(x)dx=-

1{2}
u(0)

2\leq0.

Hence, A is a dissipative operator. Furthermore, since there is a solution (almost everywhere) in D to

uu'=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.

\langleu,\Deltau\rangle=\int\Omegau(x)\Deltau(x)dx=-\int\Omega|\nablau(x)|2dx=-\|\nablau

2
\|
L2(\Omega;R)

\leq0,

so the Laplacian is a dissipative operator.

References

Notes and References

  1. Encyclopedia: 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 Example II.3.26
  6. Engel and Nagel Proposition II.3.23