dissipative operator

A dissipative operator has the following properties:Engel and Nagel Proposition II.3.14

• From the inequality given above, we see that for any x in the domain of A, if ‖x‖ ≠ 0 then $\backslash |(\backslash lambda\; I-A)x\backslash |\backslash ne\; 0,$ 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 $\backslash |(\backslash lambda\; I-A)x\backslash |>\backslash lambda\backslash |x\backslash |$ for all non-null x in the domain, then, by the triangle inequality, $\backslash |\backslash lambda\; x\backslash |+\backslash |Ax\backslash |\backslash ge\backslash |(\backslash lambda\; I-A)x\backslash |>\backslash lambda\backslash |x\backslash |,$ which implies that A itself has an inverse.) We may then state that

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

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

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

By the Hahn–Banach theorem this set is nonempty.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'. In the Hilbert space case (using the canonical duality between a Hilbert space and its dual) it consists of the single element x.Engel and Nagel Exercise II.3.25i More generally, if X is a Banach space with a strictly convex dual, then J(x) consists of a single element.Engel and Nagel Example II.3.26

Using this notation, A is dissipative if and only ifEngel and Nagel Proposition II.3.23 for all x ∈ D(A) there exists a x' ∈ J(x) such that

:$\{\backslash rm\; Re\}\backslash langle\; Ax,x\text{'}\backslash rangle\backslash leq0.$

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

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

:$\backslash therefore\backslash |x-Ax\backslash |\backslash geq\backslash |x+Ax\backslash |$

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

::$x\; \backslash cdot\; A\; x\; =\; x\; \backslash cdot\; (-x)\; =\; -\; \backslash |\; x\; \backslash |^\{2\}\; \backslash leq\; 0,$

: 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 $x^\{*\}Ax,$ which must be nonpositive for any x, is $x^\{*\}\backslash frac\{A+A^\{*\}\}2x.$ 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 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 $\backslash lambda,\; \backslash lambda-A$ has an inverse and the operator $(\backslash lambda+A)(\backslash lambda-A)^\{-1\}$ 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 $H^1([0,\backslash ;1];\backslash ;\backslash mathbf\{R\})$ with u(1) = 0. D(A) is dense in L²([0, 1]; R). Moreover, for every u in D(A), using integration by parts,

::$\backslash langle\; u,\; A\; u\; \backslash rangle\; =\; \backslash int\_\{0\}^\{1\}\; u(x)\; u\text{'}(x)\; \backslash ,\; \backslash mathrm\{d\}\; x\; =\; -\; \backslash frac1\{2\}\; u(0)^\{2\}\; \backslash leq\; 0.$

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

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

: so the Laplacian is a dissipative operator.

{{reflist|2}}

• {{ Cite book | last1=Engel| first1=Klaus-Jochen| last2=Nagel| first2=Rainer | title=One-parameter semigroups for linear evolution equations | year=2000| publisher=Springer}}
• {{cite book

|author1=Renardy, Michael |author2=Rogers, Robert C.

| title = An introduction to partial differential equations

| series = Texts in Applied Mathematics 13

| edition = Second

|publisher = Springer-Verlag

| location = New York

| year = 2004

| pages = 356

| isbn = 0-387-00444-0

}} (Definition 12.25)

Category:Operator theory

Category:Linear operators