Eigenvalues in Riemannian Geometry by Isaac Chavel

By Isaac Chavel

The fundamental targets of the ebook are: (i) to introduce the topic to these drawn to researching it, (ii) to coherently current a few uncomplicated concepts and effects, presently utilized in the topic, to these operating in it, and (iii) to give a few of the effects which are appealing of their personal correct, and which lend themselves to a presentation no longer overburdened with technical equipment.

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All the basic facts in Section 3 about solutions to the eigenvalue problems remain unchanged except, of course, the differentiability of the solutions at the singularity of the boundary. We also note, in this case, that the functions, which are C" on M and the smooth part of aM, are dense in X ( M ) . To distinguish between the cases when aM is C", and when aM is possibly only piecewise C", we refer to connected M with compact closure and nonempty 16 1. The Laplacian C“ boundary as a regular domain, and to connected M with compact closure and nonempty piecewise C“ boundary as a normal domain.

Thus (78) will be valid for 4 E C Z ( M )f, E &@(M). For the Neumann eigenvalue problem we start with the validity of (78) for 4 E C'(M), satisfying v4 = 0 on d M , and f E C"(M)-the appropriate Green formula is (39). Again one extends the validity of (78) to allow f E &@(MI. For the Dirichlet eigenvalue problem we proceed as follows: (78) is valid for 4 E CZ(M), satisfying 4 = 0 on a M , and f E C m ( M ) with compact support-again by (39). The validity of (78) will now be extended to let f be in the completion of C" functions, with compact support in M , in X ( M ) .

M, add Neumann data to aR, n M and leave original data on aR, n aM unchanged. Again arrange all the eigenvalues of R,, . -. 19 5. Rayleigh and Max-Min Methods Then for each k = 1,2,. . we have (86) pk 5 PROOF:Let y : M + Iw be the eigenfunction of pi when Y; is restricted to the appropriate subdomain, and let Vl be identically zero otherwise. Now if f is any function in $(M), then f E for every r = 1, . , m. We can therefore argue that iff is orthogonal to Yl, . , Yk- in L2(M)then s(Qr) , But there exists a nontrivial k f = 2 ajbj j= 1 orthogonal to Y l ,.

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