(BQ) Part 2 book "Functional analysis, sobolev spaces and partial differential equations" has contents: Sobolev spaces and the variational formulation of boundary value problems in one dimension, miscellaneous complements, evolution problems-the heat equation and the wave equation,.and other contents. | Chapter 8 Sobolev Spaces and the Variational Formulation of Boundary Value Problems in One Dimension Motivation Consider the following problem. Given f ∈ C([a, b]), find a function u satisfying −u + u = f on [a, b], u(a) = u(b) = 0. (1) A classical—or strong—solution of (1) is a C 2 function on [a, b] satisfying (1) in the usual sense. It is well known that (1) can be solved explicitly by a very simple calculation, but we ignore this feature so as to illustrate the method on this elementary example. Multiply (1) by ϕ ∈ C 1 ([a, b]) and integrate by parts; we obtain b (2) a b uϕ + a b uϕ = fϕ ∀ϕ ∈ C 1 ([a, b]), ϕ(a) = ϕ(b) = 0. a Note that (2) makes sense as soon as u ∈ C 1 ([a, b]) (whereas (1) requires two derivatives on u); in fact, it suffices to know that u, u ∈ L1 (a, b), where u has a meaning yet to be made precise. Let us say (provisionally) that a C 1 function u that satisfies (2) is a weak solution of (1). The following program outlines the main steps of the variational approach in the theory of partial differential equations: Step A. The notion of weak solution is made precise. This involves Sobolev spaces, which are our basic tools. Step B. Existence and uniqueness of a weak solution is established by a variational method via the Lax–Milgram theorem. Step C. The weak solution is proved to be of class C 2 (for example): this is a regularity result. H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, DOI , © Springer Science+Business Media, LLC 2011 201 202 8 Sobolev Spaces and the Variational Formulation of Boundary Value Problems in 1D Step D. A classical solution is recovered by showing that any weak solution that is C 2 is a classical solution. To carry out Step D is very simple. In fact, suppose that u ∈ C 2 ([a, b]), u(a) = u(b) = 0, and that u satisfies (2). Integrating (2) by parts we obtain b (−u + u − f )ϕ = 0 ∀ϕ ∈ C 1 ([a, b]), ϕ(a) = ϕ(b) = 0 a and .
