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Generalized Solutions, Sobolev Spaces (2017) AMATH 731: Applied Functional Analysis Fall 2017 Generalized solutions, Sobolev spaces (To accompany Section 4.6 of the AMATH 731 Course Notes) Introduction As a simple motivating example, consider the following inhomogeneous DE of the form Lu = f, (Lu)(x)= u′′(x)+ g(x)u(x)= f(x). (1) In the usual introductory courses of ODEs, f(x) is assumed sufficiently “nice,” i.e., at least continuous, so that u′′(x) is continuous, implying that u(x) is twice differentiable. At worst, f(x) could be piecewise continuous, which still does not represent a problem. One may then resort to a number of “classical” techniques to solve for u(x), or at least approximations to it (e.g., Fourier series, power series). But what if f(x) is not so “nice”? What if f(x) is discontinuous or, at best, integrable? Classical methods will break down here. One could, however, still try to use Fourier series (assuming that the Fourier expansion of f(x) exists). But the question remains: In what space does the solution u(x) live? What does it mean to say that u′(x) or u′′(x) is a function in L2? This is a fundamental problem in the study of partial differential equations (PDEs). Only a small class of partial differential equations, e.g., Laplace’s equation, can be solved in the classical sense. Many others, if not most, cannot. One must resort to other methods to construct “less smooth” solutions. An example of such a method is that of generalized or weak solutions. Sobolev spaces represent the completion of the appropriate classical function spaces in order to accomodate such solutions. The basic “trick” behind generalized solutions is the well-known integration by parts procedure. For example, let us return to Eq. (1), and now assume that f(x) ∈ L1[a, b]. Now consider a set of infinitely smooth “test functions” φ(x) with compact support on (a, b), with the additional condition ∞ that φ(a) = φ(b) = 0. We’ll call this space Cc (a, b). (This space includes all functions that are ∞ supported on smaller subintervals of [a, b].) Now multiply both sides of Eq. (1) by a φ ∈ Cc (a, b), i.e., b b b u′′(x)φ(x) dx + g(x)u(x)φ(x) dx = f(x)φ(x) dx. (2) Za Za Za Now integrate the first term by parts to give b b b ′′ ′ b ′ ′ ′ ′ u (x)φ(x) dx = u (x)φ(x) a − u (x)φ (x) dx, = − u (x)φ (x) dx, (3) Za Za Za where the boundary terms have vanished because φ(a)= φ(b) = 0. Eq. (2) now becomes b b b − u′(x)φ′(x) dx + g(x)u(x)φ(x) dx = f(x)φ(x) dx. (4) Za Za Za This equation has the general form B(u, φ)= hf, φi, (5) where h, i denotes the usual inner product on L2[a, b] and B(·, ·) denotes a bilinear form, i.e., a (bounded) functional that is linear in each of its two arguments. We shall see that this bilinear form will make sense only if both of its arguments come from the same set of functions. This set will be 1 an appropriate Hilbert space, the Sobolev space of functions – call it H(a, b) for now – that satisfy the boundary conditions. (It is not a problem to move the second argument from the test function space ∞ Cc (a, b) to the space H(a, b) since the former is dense in the latter. And, as we shall discuss below, it is not a problem to move the first argument from the space u′ ∈ C(a, b) to u ∈ H(a, b).) In this case, we shall write that the function u ∈ H is a weak solution of the Eq. (1) if B(u, v)= hf, vi (6) for all v ∈ H. The so-called Lax-Milgram Theorem for bounded bilinear functionals will guarantee the existence of a unique weak solution to Eq. (1). A physical motivation: Longitudinal extension of a one-dimensional elastic rod We begin with a physical application as a motivation for the study of Sobolev spaces: the longitudinal extension of an elastic rod. The discussion below follows Section 3.1 of the book, Functional Analysis, Applications in Mechanics and Inverse Problems, by Lebedev et al.. Suppose that the rod may be modelled as a one-dimensional elastic object of length l and cross- sectional area A(x), 0 ≤ x ≤ l. (A point x ∈ [0, l] identifies a particular section of the rod at rest.) The longitudinal displacement of the rod at a point x will be denoted as u(x). We assume that the Young’s modulus of the rod is E(x). In this one-dimensional idealization, the strain tensor has only ′ one component, ǫxx = u (x), so that the stress tensor is given by σxx = E(x)ǫxx. (This is essentially Hooke’s “Law”: force – stress – is proportional to displacement – strain.) The reason for the derivative u′(x) is briefly as follows: If u(x) and u(x +∆x) are the displacements of the rod at, respectively, x and x +∆x, then the length of the rod is now u(x +∆x) − u(x) ≈ u′(x)∆x.) The potential energy stored in an infinitesimal element of the rod of length dx centered at x associated with the strain u′(x) is 1 dU = E(x)A(x)[u′(x)]2dx. (7) 2 The total strain energy of the rod is then given by 1 l U = E(x)A(x)[u′(x)]2 dx. (8) 2 Z0 For additional simplicity, we shall assume that E is constant and units chosen so that E = 1. Fur- thermore, A(x) is assumed to be bounded as follows, 0 <m1 ≤ A(x) ≤ m2. (9) Furthermore, we assume that the rod is clamped at one end, so that u(0) = 0. (10) We can use U to define a metric, norm and inner product in the subset of functions in C1(0, l) satisfying (10) for which l A(x)[u′(x)] dx < ∞. (11) Z0 The metric, norm and inner product are given by, respectively, l 1/2 ′ ′ 2 dR(u, v) = A(x)[u (x) − v (x)] dx , Z0 l 1/2 ′ 2 k u kR = A(x)[u (x)] dx , (12) Z0 l ′ ′ hu, viR = A(x)u (x)v (x) dx. Z0 2 (The subscript R refers to rod.) One can confirm that the above do, in fact, satisfy the requirements for a metric, norm and inner product – in particular, that k u kA= 0 iff u(x) = 0. We shall call the set S of all u ∈ C1(0, l) satisfying (10) and (11), the subspace of functions with finite energy. S is an incomplete (metric, normed, inner product) space. (This is the subject of Question No. 5 in Problem Set 4 of the Course Notes.) For example, consider the sequence {un} defined as follows, 0, 0 ≤ x ≤ l/2, un(x)= 1+1/n (13) ( (x − l/2) , l/2 ≤ x ≤ l. In the limit n →∞, the un converge to the function 0, 0 ≤ x ≤ l/2, u(x)= (14) ( (x − l/2), l/2 ≤ x ≤ l, Note that u′(l/2) does not exist. It is now desired to complete this space – one may resort to the Completion Theorem for metric spaces (and associated ones for normed and inner product spaces). We let the completion of S in the metric (12) be denoted as ER. Recall that an element in ER is a class U of Cauchy sequences in S that are equivalent in the metric (12). First of all, it is quite straightforward to show that if u(x) ∈ S, then it is uniformly continuous in (0, l), therefore in C[0, l]: If u ∈ S, then x u(x)= u′(t) dt, (15) Z0 so that x x |u(x) − u(y)| = u′(t) dt ≤ |u′(t)| dt Zy Zy x 1/2 l 1/2 ≤ 12 dt |u′(t)|2 dt Zy Z0 l 1/2 1/2 1/2 ′ 2 ≤ |x − y| (l/m1 ) A(t)[u (t)] dt Z0 ≤ M|x − y|1/2 (16) Thus |u(x) − u(y)| <ǫ is satisfied when |x − y| < δ, where Mδ1/2 <ǫ. This implies that u is uniformly continuous on (0,l), implying that it is in C[0, l]. Now suppose that {un} is a representative Cauchy sequence in a class U ∈ ER. Then l ′ ′ |um(x) − un(x)| ≤ |un(t) − um(t)| Z0 l 1/2 l 1/2 2 ′ ′ 2 ≤ 1 dt |un(t) − um(t)| dt Z0 Z0 l 1/2 1/2 ′ ′ 2 ≤ (l/m1 ) A(t)[un(t) − um(t)] dt → 0 (17) Z0 as m,n →∞. This implies that {un} converges in the uniform norm, implying uniform convergence. From the completeness of C[0, l], it follows that the limit function u(x) is uniformly continuous. It is left as an exercise to show that the limit u is independent of the choice of representative Cauchy sequences from the class U. 3 Some important remarks: 1. To each class U ∈ ER corresponds a unique limit function u(x) ∈ C[0, l]. This limit function does not necessarily belong to C1[0, l]. A little extra work show that k u k∞≤ m k U kR, (18) for some constant m. This implies that the mapping of an element U ∈ ER to its corresponding element u ∈ C[0, l] is bounded, therefore continuous.
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