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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Additional info for Mathematical Methods in Electromagnetism: Linear Theory and Applications
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)
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.