Modern Arabic mathematical notation

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)

$\Vert (\lambda I-A)x\Vert \ge \lambda \Vert x\Vert .$

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.

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

• λI − A is injective for all λ > 0 and

$\Vert (\lambda I-A{)}^{-1}z\Vert \le \frac{1}{\lambda }\Vert z\Vert$

for all z in the range of λI − A.

• 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.

Equivalent characterization

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

$J(x):=\{x^{\prime }\in X^{\prime }:\Vert x^{\prime }{\Vert }_{X^{\prime }}^2=\Vert x{\Vert }_X^2=⟨x^{\prime },x⟩\}.$

By the Hahn–Banach theorem this set is nonempty. If X is reflexive, then J(x) consists of a single elementPotter or Ceramic Artist Truman Bedell from Rexton, has interests which include ceramics, best property developers in singapore developers in singapore and scrabble. Was especially enthused after visiting Alejandro de Humboldt National Park.. In the Hilbert space case (using the canonical duality between a Hilbert space and its dual) it consists of the single element x. Using this notation, A is dissipative if and only if for all x ∈ D(A) there exists a x' ∈ J(x) such that

$\mathrm{R}\mathrm{e}⟨Ax,x^{\prime }⟩\le 0.$

Examples

• 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

$x\cdot Ax=x\cdot (-x)=-\Vert x{\Vert }^2\le 0,$

so A is a dissipative operator.

• So long as the domain of an operator A is the whole Euclidean space, then it is dissipative if and only if it does not have any eigenvalue with positive real part, and (consequently) all such operators are maximally dissipative. An equivalent condition is that for some (and hence any) positive $\lambda ,\lambda -A$ has an inverse and the operator $(\lambda +A)(\lambda -A{)}^{-1}$ is a contraction (that is, it diminishes 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. (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′ with domain D(A) equal to those functions u in the Sobolev space 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,

$⟨u,Au⟩=\underset{0}{\overset{1}{\int }}u(x)u^{\prime }(x)\mathrm{d}x=-\frac{1}{2}u(0{)}^2\le 0.$

Hence, A is a dissipative operator. Furthermore, since there is a solution in D to $u-\lambda u^{\prime }=f$ 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) 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,

$⟨u,\mathrm{\Delta }u⟩=\underset{\mathrm{\Omega }}{\int }u(x)\mathrm{\Delta }u(x)\mathrm{d}x=-\underset{\mathrm{\Omega }}{\int }|\mathrm{\nabla }u(x){|}^2\mathrm{d}x=-\Vert \mathrm{\nabla }u{\Vert }_{L^2(\mathrm{\Omega };\mathbf{R})}\le 0,$

so the Laplacian is a dissipative operator. However in this case it is not maximally dissipative because there is no compactly supported solution to $u-\lambda \mathrm{\Delta }u=f$ if f is not compactly supported.

Notes

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References

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• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534 (Definition 12.25)