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

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.

′ 2. If {un(x)} is a representative Cauchy sequence for the class U, then {un(x)} is a Cauchy sequence in L2[0, l] and therefore corresponds to an element V ∈ L2[0, 1]. This element could be called the generalized derivative of U. More on this below.

3. The energy space ER is a particular example of a Sobolev space.

The displacement of the rod due to a distributed load f(x) It is instructive to outline the method of obtaining the fundamental equations that are the as opposed to simply writing these equations down. The so-called Principle of Virtual Work is used to find the displacement (i.e., stretching/compression of the rod – recall that it is fixed at one end). Very briefly, this principle states that the work done by internal forces in any virtual displacement is equal to the work done by external forces, i.e.,

δWi = δWe. (19)

A virtual displacement is one that is sufficiently smooth and which satisfies the geometric constraints imposed on the body.

Note: A virtual displacement is an assumed infinitesimal change of system coordinates occuring while time is held constant. It is called a virtual displacement as opposed to real displacement since that latter can take place only with the passage of time.

In general, as the particles comprising the continuum material are displaced, the internal forces, and possibly the external forces as well, may be affected. However, in computing δWi and δWe, such changes are ignored. In other words, one assumes that the internal and external forces are kept constant. For this reason, δWi and δWe are referred to as the virtual work done by internal and external forces, respectively. In the one-dimensional problem considered here, only longitudinal displacements are considered. A virtual displacement φ(x) is one that is sufficiently smooth, e.g., φ(x) ∈ C∞[0, l], but at least C1[0, l], and which satisfies the constraint,

φ(0) = 0, (20) imposed by the clamping condition. The virtual work then done by an external force f(x) will be given by l δWe = f(x)φ(x) dx. (21) Z0 The internal forces are longitudinal forces acting on the cross-section of the rod. At point x, the ′ strain through the cross-section A(x) is ǫxx = u (x) and the corresponding stress is σxx = Eǫxx = Eu′(x). A virtual displacement φ(x) of particles of the rod induces a virtual longitudinal strain

4 ′ ′ δǫxx = φ (x). The work done by the stress σxx = Eu (x) in the volume element A(x)δx under the virtual strain δǫxx is (force × distance × amount)

′ ′ (σxx)(δǫxx)(A(x)∆x)= Eu (x)φ (x)A(x) dx. (22)

Note: The reader may be wondering why we don’t simply let φ(x) = u(x) and φ′(x) = u′(x). The reason is that u(x) and u′(x) describe the physical “status” of the rod at some time. φ(x) represents an arbitrary virtual deviation produced from that “status.”

Therefore, with E = 1, the virtual work done by internal forces is

l ′ ′ δWi = A(x)u (x)φ (x) dx. (23) Z0 The Principle of Virtual Work in Eq. (19) then gives

l l A(x)u′(x)φ′(x) dx = f(x)φ(x) dx. (24) Z0 Z0 This is the fundamental equation from which the classical differential equation governing the rod, as well as the generalized solution of the problem, can be obtained.

Differential equation and classical solutions

In what follows, we assume that

A(x) ∈ C1[0, l], u(x) ∈ C2[0, l], f(x) ∈ C[0, l]. (25)

With these assumptions, integration by parts in Eq. (24) yields

l ′ l ′ ′ A(x)u (x)φ(x) 0 − [A(x)u (x)] + f(x) φ(x) dx = 0. (26) Z0    This relation must hold for φ(x) sufficiently smooth satisfying φ(0) = 0. It must therefore hold for the subset of functions φ(x) satisfying φ(l) = 0. This implies that Eq. (26) becomes

l [A(x)u′(x)]′ + f(x) φ(x) dx = 0. (27) Z0  Recall that the integrand is continuous on [0, l]. Because this equation is true for all φ(x) ∈ C[0, l] satisfying φ(0) = φ(l) = 0, it follows that the integrand is identically zero, i.e.,

[A(x)u′(x)]′ + f(x) = 0. (28)

This is the classical DE for the longitudinal extension of the rod. Now return to Eq. (26), recalling that it must hold for all φ(x) such that φ(0) = 0. If we now consider functions φ(x) for which φ(1) is nonzero, say, φ(1) = 1, then it follows that

A(l)u′(l) = 0. (29)

This is the classical “free end” condition.

5 In the classical sense, given an f(x), one may now proceed to solve Eq. (28) using the standard techniques of ODEs. The result is a solution u(x) ∈ C2(0, l). However, it is often necessary to extend the class of solutions u(x) of this problem. For example, the force function f(x) may not necessarily be continuous – it may be well represented by a “point source” Dirac delta function. In this case, it is advantageous to consider generalized solutions to this problem and, in general, to PDEs. The Principle of Virtual Work will still be employed, however, in which case the assumption of smooth virtual displacements φ(x) is made. This is crucial.

Generalized solutions

We now seek to consider Eq. (24) (recall that it arose from the Principle of Virtual Work) for larger classes of functions A(x), u(x) and f(x). For example, consider the equation

l l A(x)U ′(x)φ′(x) dx = F (x)φ(x) dx, (30) Z0 Z0 1 where U ∈ ER and F ∈ L (0, l). (We can assume A(x) > 0 to be piecewise C[0, l].) It is normally supposed that φ ∈ C∞(0, l) with compact support on (0, l), i.e., φ(x) is zero outside intervals of the form [a, b + ǫ], ǫ > 0, where 0 < a < b ≤ l. Recall also that φ(0) = 0. For convenience this set will ∞ be denoted as Cc (0, l]. These functions are often called “test functions” – recall how the Dirac delta “function” is defined in terms of such functions. Test functions were mentioned in the first section of these notes. The integral on the left of (30) may be bounded (Schwartz inequality, using A(x) = [A(x)]1/2[A(x)]1/2):

l 2 l l A(x)U ′(x)φ′(x) dx ≤ A(x)[U ′(x)]2 dx A(x)[φ′(x)]2 dx . (31) Z0  Z0 Z0  The first integral is defined when U ∈ ER. From our earlier discussion,

l l A(x)[U ′(x)]2 dx = lim A(x)[u′ (x)]2 dx, (32) n→∞ n Z0 Z0 for a representative Cauchy sequence of functions {un} in the equivalence class U – convergence is in the norm k · kR defined in (12) so that

k U kR= lim k un kR . (33) n→∞ The integral on the right of (30) is defined for F ∈ L1(0, l) since F φ is also in L1(0, l).

∞ The solution U ∈ ER which satisfies Eq. (30) for all φ ∈ Cc (0, l] is called the generalized solution of the rod problem.

The generalized solution method outlined above is but one example of an entire arsenal of methods designed to obtain what are called weak solutions of partial differential equations. We shall elaborate further below.

Derivation of governing equation using Principle of Minimum Energy Classical case

Eq. (24) was derived above using the Principle of Virtual Work. However, the elastic rod is a conservative system (assuming that the external forces are conservative). In the classical case, i.e.,

6 the classical assumptions on A(x), u(x) and f(x) in Eq. (25), the solution u(x) of Eq. (24) is the minimizer of the total energy functional

1 l l J(u)= A(x)[u′(x)]2 dx − f(x)u(x) dx, (34) 2 Z0 Zo 2 2 on the subset of functions u ∈ C [0, l] satisfying u(0) = 0. To see this, let u0, v ∈ C [0, l] and u0(0) = 0 = v(0). Then

l l 1 ′ ′ 2 J(u0 + ǫv) = A(x)[u0(x)+ ǫv (x)] dx − f(x)[u0 + ǫv(x)] dx 2 0 0 Z l Z l ′ ′ = J(u0)+ ǫ A(x)u0(x)v (x) dx − f(x)v(x) dx Z0 Z0  ǫ2 l + A(x)[v′(x)]2 dx. (35) 2 Z0

The term in square brackets is the Fr´echet derivative DJ(u0) – more precisely, it is ǫDJ(u0)v. If u0 satisfies Eq. (24), then DJ(u0)=0 and J(u0) ≤ J(u0 + ǫv) for all ǫ and v(x).

Generalized case

The generalized problem, Eq. (30) can be derived by minimizing the functional

1 l l J(U)= A(x)[U ′(x)]2 dx − F (x)U(x) dx, (36) 2 Z0 Z0 1 for U ∈ ER and F ∈ L (0, l). The first term is simply

1 l 1 A(x)[U ′(x)]2 dx = k U k2 < ∞. (37) 2 2 R Z0 1 The second term defines a continuous linear functional in ER in the following sense. Let {un}∈ C (0, l) be a representative Cauchy sequence for the element U ∈ ER. Then

l l F (x)U(x) dx = lim F (x)un(x) dx. (38) n→∞ Z0 Z0 The limit exists because

l l F (x)un(x) dx ≤ |F (x)| dx · sup |un(x)|. (39) Z0 Z0  x∈[0,l]

By setting um = 0 in Eq. (17), we have

1/2 |un(x)|≤ (l/m1) k un kR, (40) so that l 1/2 F (x)U(x) dx ≤ (l/m1) k F k1k U kR . (41) Z0

Thus, J(U) may be written as follows

1 J(U)= k U k2 +Φ(U), (42) 2 R

7 where Φ(U) is a continuous linear functional in ER. We now consider the minimization of this functional in the same way as was done for the classical case. For U0,V ∈ ER, 1 J(U + ǫV ) = k U + ǫV k2 +Φ(U + ǫV ) 0 2 0 R 0 1 1 = k U k2 +ǫhU ,V i + ǫ2 k V k2 +Φ(U )+ ǫΦ(V ) 2 0 R 0 R 2 R 0 1 = J(U )+ ǫ [hU ,V i + φ(V )] + ǫ2 k V k2 . 0 0 R 2 R

For U0 to be a minimizer, since ǫ is arbitrary, it must satisfy the condition,

hU0,V iR + Φ(V ) = 0 for all V ∈ ER . (43)

As in the classical case, this is the condition that the Fr´echet derivative of J vanish U0, i.e.,

DJ(U0) = 0 . (44)

In integral form, this condition is given by

l l ′ ′ A(x)U0(x)V (x) dx = F (x)V (x) dx, (45) Z0 Z0 ∞ which has the same form as (30), except that φ(x) ∈ Cc (0, l) has been replaced with V ∈ ER. These two equations are equivalent: If Eq. (45) holds for every V ∈ ER, then it will hold for every ∞ ∞ φ ∈ Cc (0, l). On the other hand, if is holds for every φ ∈ Cc (0, l), then it will hold for every V ∈ ER ∞ since the Cc (0, l) is dense in ER.

The two-dimensional membrane

We discuss very briefly the extension of the energy method described above to two-dimensional prob- lems. For example, consider a taut membrane stretched with uniform tension T across a domain D ⊂ R2. The strain energy of the membrane is T U = |∇u|2 dA, (46) 2 ZD where ∇u denotes the gradient of u. For simplicity, set T = 1 and consider the clamped membrane with boundary conditions u(x,y) = 0, (x,y) ∈ ∂D, (47) where ∂D denotes the boundary of D. In the same way as for the one-dimensional rod, we can use U to define a metric, norm and inner product on the subset S ∈ C1(D) satisfying the above boundary condition and |∇u|2 dA < ∞. (48) ZD Thus

1/2 2 dM (u, v) = |∇u −∇v| dA , ZD  1/2 2 k u kM = |∇u| dA , (49) ZD  hu, viM = ∇u ·∇v dA. ZD 8 (Here, the subscript M refers to membrane.) The completion of the space S in the norm k · kM will be called the energy space EM for the clamped membrane. First, one requires the Poincar´einequality, Theorem 4.9 of the Course Notes, p. 68-69: For a bounded domain D, there exists a constant C, depending on D, such that

u2 dA ≤ C |∇u|2 dA. (50) ZD ZD ∞ ∞ This result is first proved for u ∈ Cc (D) and then extended to S – the latter is possible because Cc is dense in S. From this result we have the following: If {un} is a representative Cauchy sequence for 2 U ∈ EM then {un} is a representative Cauchy sequence for U ∈ L (D), which implies that

k U k2 ≤ m k U kM (51) for some constant m. This is a kind of analogue to Eq. (18) for the one-dimensional rod, although it stops at L2(D). The membrane problem may now be analysed in terms of either the Principle of Minimum Energy or Principle of Virtual Work. For the former, suppose that there is a “distributed load”, or force f(x,y), acting throughout the membrane so that the total energy is 1 J(u)= |∇u|2 dA − f dA. (52) 2 ZD ZD 2 For the classical analysis, we assume that u ∈ C (D) and f ∈ C(D). Then setting u = u0 + ǫv as 2 before, where u0, v ∈ C (D) and v =0 on ∂D, one finds that

[∇uo ·∇v − fv] dA = 0. (53) ZD This is also the equation that one would obtain from the Principle of Virtual Work. Now use the identity 2 ∇u0 ·∇v = div(v∇u0) − v∇ u0 (54) and the divergence theorem to give

2 ∇u0 ·∇v dA = v∇u0 · n ds − v∇ u0 dA. (55) ZD Z∂D ZD The integral over the boundary is zero (since v =0 on ∂D) so that Eqs. (53) and (55) yield

2 [∇ u0 + f] dA = 0. (56) ZD Once again because of the continuity of the integrand, we have the classical differential equation for the loaded membrane 2 ∇ u0 + f = 0, (x,y) ∈ D. (57) Equations (52) and (57) represent special case of the so-called Dirichlet problem in which the boundary value, i.e., the value of u on ∂D, is zero. We shall return to the more general form of the Dirichlet problem in a later section.

9 Sobolev spaces

Sobolev spaces can be considered as mathematical generalizations of the energy spaces ER and EM introduced above. They may also be regarded as generalizations of the Lebesgue spaces, involving not only the distance between two functions but also the distance(s) between selected derivatives. Let D ⊂ RN be a nonempty open set, the domain of concern. A convenient notation for the mixed partial derivative of a function of N variables is

|α| α ∂ f(x1,x2, · · · ,xN ) D f = α1 α2 αN , |α| = α1 + α2 + · · · + αN . (58) ∂x1 ∂x2 · · · ∂xN

m Let m be a non-negative integer, and let CB (D) denote the set of functions f(x), x ∈ D, which have bounded continuous derivatives Dαf for |α|≤ m. Recall that a semi-norm, k · k, on a linear space X is a real-valued function that satisfies all properties of a norm except “k x k= 0 iff x = 0.” This property is replaced by: “k x k=0 if x = 0. Now introduce the semi-norm

1/p |f| = (k Dαf k )p , (59) m,p  p  |αX|=m   where k · kp denotes the Lebesgue p-norm.

Particular examples:

1. m = 0: 1/p p |f|0,p =k f kp= |f| dV , (60) ZD  2. m = 1,N = 1: 1/p ′ p |f|1,p = |f | dx , (61) ZD  3. m = 2,N = 2: 1/p p p p |f|2,p = (|f,11| + 2|f,12| + |f,22| ) dA . (62) ZD 

The energy norm ER discussed earlier was the semi-norm |f|1,2, with N = 1, and the membrane norm EM was |f|1,2, with N = 2. These two semi-norms turn out to be norms on their appropriate spaces – see note below. We now introduce the norm m 1/p p k f km,p= (|f|n,p) . (63) ! nX=0 For example, in the case N = 2, m = 2, p = 2, then

1/2 2 2 2 2 2 2 k f k2,2= [f + (f,1 + f,2) + (f,11 + 2f,12 + f,22)] dV . (64) ZD  m,p m Definition 1 We define W (D) to be the completion of CB (D) in the norm k · km,p.

m,p α If {fn} is a representative Cauchy sequence for F ∈ W (D), then {D fn} is a Cauchy sequence s in Lp(D) for any α such that |α|

10 ∞ Recall that Cc (D) was defined to be the set of functions having continuous derivatives of all m,p ∞ orders in D, and having compact support in D. We define Wc (D) to be the completion of Cc (D) m,p m,p in the norm k · km,p. Wc (D) is a subspace of W (D). m,p The spaces Wc (D) form the generalization of the energy spaces for the clamped membrane EM . Furthermore, it can be shown (see Functional Analysis, by Lebedev et al., pp. 89-90), using Poincar´e’s inequality, that |f|1,2 is a norm for EM .

1,2 Important special case: The Sobolev space Wc (D) 1,2 The space Wc (D) is a Hilbert space and of great importance in the study of linear second-order partial differential equations. In one dimension, N = 1, the norm of this space assumes the form,

1,2 2 ′ 2 kfk1,2 = [f(x)) + (f (x)) ] dx . (65) ZD 

Generalized, or “weak”, derivatives in Sobolev spaces

We now return to the idea, sketched in the Introduction, of using integration by parts to “transfer” the derivative process onto another function. For simplicity, we first consider functions of a single real ∞ ∞ variable. Let D = (a, b) and C0 (D) denote the space of all C functions with compact support in ∞ D. This implies that if u ∈ C0 , then u(a)= u(b)=0. The integration-by-parts formula is as follows: For a u ∈ C1(D), ′ ′ ∞ uv dx = − u v dx, for all v ∈ C0 (D). (66) ZD ZD The “trick,” i.e., the first part of the procedure, is to set

w(x)= u′(x), x ∈ D. (67)

We thus obtain the formula,

′ ∞ uv dx = − wv dx, for all v ∈ C0 (D). (68) ZD ZD Definition 2 Let u, w ∈ L2(D) and suppose that Eq. (68) holds. Then the function w is called a generalized derivative of the function u on the set D. As in the usual, classical case, we shall write w = u′.

Proposition 1 The generalized derivative w = u′ is uniquely determined up to the values of w on a set of Lebesgue measure zero.

2 Proof: Suppose that Eq. (68) holds for w1, w2 ∈ L (D). Then

∞ (w1 − w2)v dx = 0, for all v ∈ C0 (D). (69) ZD 2 This implies that w1(x)= w2(x) in the L -sense, i.e., for almost all x ∈ D.

Example 1: Consider the function u : (−1, 1) → R given by

u(x)= |x|, x ∈ (−1, 1). (70)

11 Let −1, −1

∞ We now verify the above statement. For all v ∈ C0 (−1, 1), integration by parts yields

1 0 1 uv′ dx = (−x)v′ dx + xv′ dx Z−1 Z−1 Z0 −ǫ 1 = lim (−x)v′ dx + xv′ dx ǫ→0+ Z−1 Zǫ  −ǫ −ǫ 1 1 = lim −xv(x) + v dx + xv(x) − vdx ǫ→0+ −1 −1 ǫ ǫ  Z Z  0 1 = (0)v(0) − (−1) v(−1) + v dx + (1) v(1) − (0)v(0) − v dx Z−1 Z0 0 1 = v dx − v dx Z−1 Z0 0 1 = − (−1)v dx − (1)v dx Z−1 Z0 1 = − wv dx, (72) Z−1

Example 2: Consider the function u : (0, 2) → R given by

x, 0

The graph of u(x) is sketched below.

y

2

u(x) 1

x 0 1 2

12 The weak derivative u′ does not exist, as we now show.

2 1−ǫ 2 uv′ dx = lim uv′ dx + uv′ dx ǫ→0 Z0 Z0 Z1+ǫ  1−ǫ 2 = lim u(1 − ǫ)v(1 − ǫ) − u(0)v(0) − u′v dx + u(2)v(2) − u(1 + ǫ)v(1 + ǫ) − u′v dx ǫ→0  Z0 Z1+ǫ  2 = u(1−)v(1−) − u(1+v(1+) − u′v dx (v(0) = v(2) = 0) 0 2 Z = [u(1−) − u(1+)]v(1) − u′v dx 0 2 Z = −v(1) − u′v dx 0 2 Z 6= − u′v dx. (74) Z0 The discontinuity of u at x = 1 prevents the existence of the weak derivative.

Note: This result does not contradict the original definition of weak derivative. The function u is certainly an L2 function, but Eq. (68) does not hold. Therefore it does not have a weak derivative. (Of course, since u(x) is differentiable on any subinterval (a, b) ⊂ (0, 1) or (a, b) ⊂ (1, 2), the weak derivative u′ exists over any domain D ⊂ (0, 1) or D ⊂ (1, 2).)

That being said, we can assign another type of derivative to u at x = 1, namely, a derivative involving the “Dirac delta function,” but that is another story, and another section.

The idea of generalized derivatives extends to partial derivatives. There is a variety of notations for such derivatives, as there are for partial derivatives themselves. Two common notations are illustrated below for the single partial derivative analogue to the integration-by-parts formula (66):

∞ uvxi dx = − uxi v dx, for all v ∈ C0 (D). (75) ZD ZD and ∞ u(∂iv) dx = − (∂iu)v dx, for all v ∈ C0 (D). (76) ZD ZD We shall use the second notation below. The notations for mixed partial derivatives can get compli- cated. We omit them here since they will not be used in any discussion. It will suffice to define the single partial generalized derivative:

Definition 3 Let D ⊂ RN be a nonempty, open subset. Let u, w ∈ L2(D) and suppose that Eq. (76) holds. Then the function w is called a generalized derivative of the function u on the set D. As in the usual, classical case, we shall write w = ∂iu.

The connection between generalized derivatives and Sobolev spaces will now be stated (see Partial Differential Equations, L.C. Adams, AMS, 1998).

Definition 4 The Sobolev space W k,p(D) consists of all locally summable functions u : U → R (i.e., integrals are finite) such that for each multi-index α, with |α|≤ k, Dαu exists in the weak sense and belongs to Lp(D).

13