# Analyse numérique et équations aux dérivées partielles : by Niçaise

By Niçaise

Nicaise S. examine numerique et equations aux derivees partielles (Dunod, 2000)(fr)(ISBN 2100049410)

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Additional info for Analyse numérique et équations aux dérivées partielles : cours et problèmes résolus

Example text

We thus obtain A* =^^-h°. Since now (11) h^h°, V-A = fr + h° ^ 2h° ^ 2^-^ (12) follows from (11) and (9). Hence d^2. Since, on the other hand, d>\, we find that έ/=2. (13) Moreover, it can be also inferred that only = is possible in (12), instead of ^ , and consequently that h° = ^ and therefore that %-l. e. (16) We shall now very soon achieve our aim. From (14) and (16), we can write (17) Substituting this into (10) and taking (13) into account we obtain Simple comparison of coefficients gives the system of equations As 2 ^ 0 , the second equation implies and thus the first and the third equations give Since, accordingly, 2μ = (1 +μ) 2 , we also have: for suitable v.

Then the validity of both divisibility relations F(x)\f(x), x*-x, as well as yet the further one F(x) / ' ( * ) can be inferred trivially (but they hold for very different reasons). Consequently both divisibilities F(x)\g(x), M |g'(*)_i follow from (3). Finally, by multiplication of the latter, the divisibility f(x)\g(x)(g'(x)-l) (5) follows. The left-hand side is of degree q, while the degree of the right-hand side is not greater than q, as a consequence of (4). As it also follows from (2) and (3) that the right-hand side does not vanish, it can be concluded that both sides are associated with one another and that in (4) = is valid, instead of ^ .

As σ 6 F3, (36j) remains invariant, whereas (37) becomes This equation is actually the case p = 3 for (35). ) We now state THEOREM 6. The λ-differential equation (g(xf)' \λ € F ; g(x) e F[x] ; g° = ^ ± l ) = x«-x + kg(x) (38) is solvable if and only if λ = 3σ (σ = ± 1 ) ; (39) moreover, in this case all its solutions are obtained by substituting the solutions a(x), b(x) of the a, b-polynomial equation xa(x)p + b(x)p = χ*α(χ)ρ~2 + a(x)p-3b(x) into r = l; P a(x)£t[x]; l b(x)£t[x]; b° ^ a° = (40) 2 ^ retaining only those of the resulting expressions g(x) that fall in F[x].