By Michel Cessenat

Presents the reader with uncomplicated instruments to resolve difficulties of electromagnetism in thier usual practical frameworks because of sleek mathematical tools: imperative floor equipment, and likewise semigroups, variational equipment, and so on. good tailored to a numerical strategy. As examples of functions of those instruments and ideas, we resolve a number of basic difficulties of electromagentism, desk bound or time-dependent: scattering of an incident wave by means of a disadvantage, bounded or now not, by means of gratings; wave propogation in a waveguide, with junctions and cascades. it's the objective to offer mathematical notions that might let a greater figuring out of modelization in electromagnetism and emphasize the essentail beneficial properties on the topic of the geometery and nature of fabrics.

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**Sample text**

I / 2 ( c u r l D . This implies that the duality is continuous. Note also that from Corollary 2 and Proposition 3, we have the duality property (50) __, V u , v€ H"I/2(curl,lT), as a consequence of the Green formula (48). 0 The duality of Prop. 3 may seem fairly surprising, if we refer to the usual dualities for Sobolev spaces. We can verify this duality from two points of view. i) At first we use local charts to work on space R2, then we use a Fourier transform method as in Theorem 4. __

Summary of the main properties for a linear isotropic dielectric medium The electromagnetic behavior of a linear isotropic dielectric medium is given by the "constants" of the constitutive laws, the permittivity c or tc (linked by (42)) and the permeability p, with the following "natural" properties (see H4), HI 1)) (86) £C = E06 + OJY + VC, }is|io5 + x. Y the Heaviside function, oj constant > 0, KC, and x € L (R), real, with support in R+. Real and imaginary parts of the Fourier-Laplace transforms of vc and x have to satisfy the Kramers-Kronig relations (with H the Hilbert transformation, see (31)): (87) O^ffO;, 0 ^ - f f O ; .

Ii) We have to prove that y is onto. Let n be in H" 1/2 (0. Then there exists u in H (Q) solution of the Neumann problem: (18) -Au + u = 0 i n Q , | | l r = M on T. Then v=grad u is in H(div,Q) and satisfies ynv=p, which proves that yn is onto. Remark that the Green formula (14) is also true for v in H(div,&) and 0 in Hl(Cl). 1 SPACES FOR CURL AND DIV iii) Proof about the kernel of y. Let w be in ker yfl orthogonal to Z>(Q)n in H(div,Q). Then w satisfies: (19) ((v,w)) = (v,w) + (divv, divw)«0, Vv€Z)(Q)n.