# Analytic Semigroups and Semilinear Initial Boundary Value by Kazuaki Taira By Kazuaki Taira

This cautious and obtainable textual content makes a speciality of the connection among interrelated matters in research: analytic semigroups and preliminary boundary price difficulties. This semigroup method will be traced again to the pioneering paintings of Fujita and Kato at the Navier-Stokes equation. the writer reports nonhomogeneous boundary price difficulties for second-order elliptic differential operators, within the framework of Sobolev areas of Lp sort, which come with as specific circumstances the Dirichlet and Neumann difficulties, and proves that those boundary worth difficulties offer an instance of analytic semigroups in Lp.

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Extra resources for Analytic Semigroups and Semilinear Initial Boundary Value Problems

Example text

4) Now we consider the following non-homogeneous Cauchy problem: (NP) de = Ax(t) + f (t, (-A)-"y(t)), 0 < t < T, x(0) = x0. But we know that f (t, (-A)-"y(t)) belongs to C([to, tl]; E) fl CQ'((to, tl]; E) where 0 < Q' < min(y,1 - a). 16, we obtain that the function u(t) = U(t - to)xo + f U(t - s) f (s, (-A)-"y(s)) ds tot belongs to C([to, ti]; E) fl C1((to, ti ); E), and is a unique solution of problem (NP). 60) which proves that the function u is a solution of problem (SLP). 60). (5) Finally, we prove the uniqueness of solutions of problem (SLP).

Proof. 3) 1U(X) u(y)I (lu(s) - u(y)I Ix ylIx-yl (lu(s) - u(y)I -_ lx -yl Hence, if 0 < a < b < c < 1, then we have lulb. )(b-a)/(c-a) (hula < (I ul c oo)(c-b)/(c-a) Thus we have only to consider the following two cases: a=0 and c=1. v/nlx - yl. But we remark that 8u 01xi (x)I - lim h-0 u(x + hei) - u(x) I h < sup XOY lu(x) - u(y)l lx - yl (c-b)/(c-a) II. SOBOLEV IMBEDDING THEOREMS 50 so that < sup Iu(x) - u(y)I lull,,,. = max sup I

3 lul-n = lull,. )(nA+l)/n(A-0) (lull/µ)(nA+l)/n(A-µ) _ `Y Hence it follows that lull/µ <_ 7 (lull/v)(nµ+l)/(nv+l) ((lull! a)(nµ+l)/n(µ-A) (lull/")(na+l)/n(a-µ) X = (lull/v) (nµ+l)/(nv+l) "ate =,, -+1 A-4 np+l -4 (lull/a) -+1 4-X so that lull/µ _< y (lull/A)(v-µ)/(v-A) (lull/v)(µ-\)/(v-a) .