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 x ∈ D(A)

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

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 λI − A 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.

A dissipative operator has the following properties:^[2]

• From the inequality given above, we see that for any x in the domain of A, if ‖x‖ ≠ 0 then ${\displaystyle \|(\lambda I-A)x\|\neq 0,}$ so the kernel of λI − A is just the zero vector and λI − A is therefore injective and has an inverse for all λ > 0. (If we have the strict inequality ${\displaystyle \|(\lambda I-A)x\|>\lambda \|x\|}$ for all non-null x in the domain, then, by the triangle inequality, ${\displaystyle \|\lambda x\|+\|Ax\|\geq \|(\lambda I-A)x\|>\lambda \|x\|,}$ which implies that A itself has an inverse.) We may then state that

${\displaystyle \|(\lambda I-A)^{-1}z\|\leq {\frac {1}{\lambda }}\|z\|}$

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

• λI − A is surjective for some λ > 0 if and only if it is surjective for all λ > 0. (This is the aforementioned maximally dissipative case.) In that case one has (0, ∞) ⊂ ρ(A) (the resolvent set of A).
• A is a closed operator if and only if the range of λI - A is closed for some (equivalently: for all) λ > 0.

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

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

By the Hahn–Banach theorem this set is nonempty.^[3] 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 x ∈ D(A) there exists a x' ∈ J(x) such that

${\displaystyle {\rm {Re}}\langle Ax,x'\rangle \leq 0.}$

In the case of Hilbert spaces, this becomes ${\displaystyle {\rm {Re}}\langle Ax,x\rangle \leq 0}$ for all x in D(A). Since this is non-positive, we have

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

${\displaystyle \therefore \|x-Ax\|\geq \|x+Ax\|}$

Since I−A has an inverse, this implies that ${\displaystyle (I+A)(I-A)^{-1}}$ is a contraction, and more generally, ${\displaystyle (\lambda I+A)(\lambda I-A)^{-1}}$ 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.

• For a simple finite-dimensional example, consider n-dimensional Euclidean space Rⁿ with its usual dot product. If A denotes the negative of the identity operator, defined on all of Rⁿ, then

${\displaystyle x\cdot Ax=x\cdot (-x)=-\|x\|^{2}\leq 0,}$

so A is a dissipative operator.

• So long as the domain of an operator A (a matrix) is the whole Euclidean space, then it is dissipative if and only if A+A^* (the sum of A and its adjoint) does not have any positive eigenvalue, and (consequently) all such operators are maximally dissipative. This criterion follows from the fact that the real part of ${\displaystyle x^{*}Ax,}$ which must be nonpositive for any x, is ${\displaystyle x^{*}{\frac {A+A^{*}}{2}}x.}$ The eigenvalues of this quadratic form must therefore be nonpositive. (The fact that the real part of ${\displaystyle x^{*}Ax,}$ must be nonpositive implies that the real parts of the eigenvalues of A must be nonpositive, but this is not sufficient. For example, if ${\displaystyle A={\begin{pmatrix}-1&3\\0&-1\end{pmatrix}}}$ 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 ${\displaystyle \lambda ,\lambda -A}$ has an inverse and the operator ${\displaystyle (\lambda +A)(\lambda -A)^{-1}}$ 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.)
• Consider H = L²([0, 1]; R) with its usual inner product, and let Au = u′ (in this case a weak derivative) with domain D(A) equal to those functions u in the Sobolev space ${\displaystyle H^{1}([0,\;1];\;\mathbf {R} )}$ with u(1) = 0. D(A) is dense in L²([0, 1]; R). Moreover, for every u in D(A), using integration by parts,

${\displaystyle \langle u,Au\rangle =\int _{0}^{1}u(x)u'(x)\,\mathrm {d} x=-{\frac {1}{2}}u(0)^{2}\leq 0.}$

Hence, A is a dissipative operator. Furthermore, since there is a solution (almost everywhere) in D to ${\displaystyle u-\lambda u'=f}$ 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.

• Consider H = H₀²(Ω; R) (see Sobolev space) for an open and connected domain Ω ⊆ Rⁿ and let A = Δ, the Laplace operator, defined on the dense subspace of compactly supported smooth functions on Ω. Then, using integration by parts,

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

so the Laplacian is a dissipative operator.

• Engel, Klaus-Jochen; Nagel, Rainer (2000). One-parameter semigroups for linear evolution equations. Springer.
• Renardy, Michael; Rogers, Robert C. (2004). An introduction to partial differential equations. Texts in Applied Mathematics 13 (Second ed.). New York: Springer-Verlag. p. 356. ISBN 0-387-00444-0. (Definition 12.25)