Boundary value problems for elliptic partial differential equations Mourad Choulli
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Mourad Choulli. Boundary value problems for elliptic partial differential equations. Master. France. 2019. hal-02406106
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Boundary value problems for elliptic partial differential equations
Preface
This course is intended as an introduction to the analysis of elliptic partial differen- tial equations. The objective is to provide a large overview of the different aspects of elliptic partial differential equations and their modern treatment. Besides vari- ational and Schauder methods we study the unique continuation property and the stability for Cauchy problems. The derivation of the unique continuation property and the stability for Cauchy problems relies on a Carleman inequality. This inequal- ity is efficient to establish three-ball type inequalities which are the main tool in the continuation argument. We know that historically a central role in the analysis of partial differential equa- tions is played by their fundamental solutions. We added an appendix dealing with the construction of a fundamental solution by the so-called Levi parametrix method. We tried as much as possible to render this course self-contained. Moreover each chapter contains many exercices and problems. We have provided detailed solutions of these exercises and problems. The most parts of this course consist in an enhanced version of courses given by the author in both undergraduate and graduate levels during several years. Remarks and comments that can help to improve this course are welcome.
3
Contents
1 Sobolev spaces ...... 7 1.1 Lp spaces ...... 7 1.2 Approximation by smooth functions ...... 10 1.3 Weak derivatives ...... 15 1.4 W k,p spaces ...... 17 1.5 Extension and trace operators ...... 23 1.6 Imbedding Theorems ...... 31 1.7 Exercises and problems ...... 38 References ...... 44
2 Variational solutions ...... 45 2.1 Stampacchia’s theorem and Lax-Milgram lemma ...... 45 2.2 Elements of the spectral theory of compact operators ...... 48 2.3 Variational solutions for model problems ...... 60 2.3.1 Variational solutions ...... 60 2.3.2 H2-regularity of variational solutions ...... 64 2.3.3 Maximum principle ...... 72 2.3.4 Uniqueness of continuation across a non characteristic hypersurface ...... 74 2.4 General elliptic operators in divergence form ...... 81 2.4.1 Weak maximum principle ...... 82 2.4.2 The Dirichlet problem ...... 84 2.4.3 Harnack inequalities ...... 87 2.5 Exercises and problems ...... 98 References ...... 104
3 Classical solutions ...... 105 3.1 Holder¨ spaces ...... 105 3.2 Equivalent semi-norms on Holder¨ spaces ...... 111 3.3 Maximum principle ...... 114 3.4 Some estimates for Harmonic functions ...... 121
5 6 Contents
3.5 Interior Schauder estimates ...... 126 3.6 The Dirichlet problem in a ball ...... 130 3.7 Dirichlet problem on a bounded domain ...... 133 3.8 Exercises and problems ...... 136 References ...... 145
4 Classical inequalities, Cauchy problems and unique continuation . . . . 147 4.1 Classical inequalities for harmonic functions ...... 147 4.2 Two-dimensional Cauchy problems ...... 155 4.3 A Carleman inequality for a family of operators ...... 160 4.4 Three-ball inequalities ...... 163 4.5 Stability of the Cauchy problem ...... 168 4.6 Exercises and problems ...... 178 References ...... 184
Solutions of Exercises and Problems ...... 185
A Building a fundamental solution by the parametrix method ...... 237 A.1 Functions defined by singular integrals ...... 237 A.2 Weakly singular integral operators ...... 245 A.3 Canonical parametrix ...... 251 A.4 Fundamental solution ...... 258 References ...... 266
Index ...... 267 Chapter 1 Sobolev spaces
We define in this chapter the W k,p-spaces when k is non negative integer and 1 ≤ p < ∞ and we collect their main properties. We adopt the same approach as in the books by H. Brezis´ [2] and M. Willem [13]. That is we get around the theory of distributions. We precisely use the notion of weak derivative which in fact coincide with the derivative in the distributional sense. We refer the interested reader to the books by L. Hormander¨ [3] and C. Zuily [14] for a self-contained presentation of the theory of distributions. The most results and proofs of this chapter are borrowed from [2, 13].
1.1 Lp spaces
n In this chapter Ω denotes an open subset of R . As usual, we identify a measur- able function f on Ω by its equivalence class consisting of functions that are equal almost everywhere to f . For simplicity convenience the essential supremum and the essential minimum are simply denoted respectively by the supremum and the minimum. n Let dx be the Lebesgue measure on R . Define Z 1 L (Ω) = f : Ω → C measurable and | f (x)|dx < ∞ . Ω
Theorem 1.1. (Lebesgue’s dominated convergence theorem) Let ( fm) be a sequence in L1(Ω) satisfying (a) fm(x) → f (x) a.e. x ∈ Ω, 1 (b) there exists g ∈ L (Ω) so that for any m | fm(x)| ≤ g(x) a.e. x ∈ Ω. Then f ∈ L1(Ω) and Z | fm − f |dx → 0. Ω
7 8 1 Sobolev spaces
1 Theorem 1.2. (Fubini’s theorem) Let f ∈ L (Ω1 ×Ω2), where Ωi is an open subset n of R i , i = 1,2. Then Z 1 1 f (x,·) ∈ L (Ω2) a.e. x ∈ Ω1, f (·,y)dy ∈ L (Ω1) Ω2 and Z 1 1 f (·,y) ∈ L (Ω1) a.e. y ∈ Ω2, f (x,·)dx ∈ L (Ω2). Ω1 Moreover Z Z Z Z Z dx f (x,y)dy = dy f (x,y)dx = f (x,y)dxdy. Ω1 Ω2 Ω2 Ω1 Ω1×Ω2 Recall that Lp(Ω), p ≥ 1, denotes the Banach space of measurable functions f on Ω satisfying | f |p ∈ L1(Ω). We endow Lp(Ω) with its natural norm
Z 1/p p k f kp;Ω = k f kLp(Ω) = | f | dx . Ω
Let L∞(Ω) denotes the Banach space of bounded measurable functions f on Ω. This space equipped with the norm
k f k∞;Ω = k f kL∞(Ω) = sup| f |.
When there is no confusion k f kLp(Ω) is simply denoted by k f kp. The conjugate exponent of p, 1 < p < ∞, is given by the formula p p0 = . p − 1 Note that we have the identity 1 1 + = 1. p p0 If p = 1 (resp. p = ∞) we set p0 = ∞ (resp. p0 = 1). The following inequalities will be very useful in the rest of this text.
0 Proposition 1.1. (Holder’s¨ inequality) Let f ∈ Lp(Ω) and g ∈ Lp (Ω), where 1 ≤ p ≤ ∞. Then f g ∈ L1(Ω) and Z | f g|dx ≤ k f kpkgkp0 . Ω Proposition 1.2. (Generalized Holder’s¨ inequality) Fix k ≥ 2 an integer. Let 1 < p j < ∞, 1 ≤ j ≤ k so that 1 1 + ... + = 1. p1 pk
p j k 1 If u j ∈ L (Ω), 1 ≤ j ≤ k, then ∏ j=1 u j ∈ L (Ω) and 1.1 Lp spaces 9
Z k k ∏ |u j|dx ≤ ∏ ku jkp j . Ω j=1 j=1
Proposition 1.3. (Interpolation inequality) Let 1 ≤ p < q < r < ∞ and 0 < λ < 1 given by 1 1 − λ λ = + . q p r If u ∈ Lp(Ω) ∩ Lr(Ω) then u ∈ Lq(Ω) and
1−λ λ kukq ≤ kukp kukr .
n As usual if E ⊂ R is a measurable set, its measure is denoted by |E|. Proposition 1.4. Let 1 ≤ p < q < ∞ and assume that |Ω| < ∞. If u ∈ Lq(Ω) then u ∈ Lp(Ω) and 1 − 1 kukp ≤ |Ω| p q kukq.
We collect in the following theorem the main properties of the Lp spaces.
Theorem 1.3. (i) Lp(Ω) is reflexive for any 1 < p < ∞. (ii) Lp(Ω) is separable for any 1 ≤ p < ∞. (iii) Let 1 ≤ p < ∞ and u ∈ [Lp(Ω)]0, the dual of Lp(Ω). Then there exists a unique p0 gu ∈ L (Ω) so Z p hu, f i = u( f ) = f gudx for any f ∈ L (Ω). Ω Furthermore 0 kuk[Lp(Ω)]0 = kgukLp (Ω).
Theorem 1.3 (iii) says that the mapping u 7→ gu is an isometric isomorphism 0 between [Lp(Ω)]0 and Lp (Ω). Therefore, we always identify in the sequel the dual 0 of Lp(Ω) by Lp (Ω). The following theorem gives sufficient conditions for a subset of Lp(Ω) to be relatively compact.
Theorem 1.4. (M. Riesz-Frechet-Kolmogorov)´ Let F ⊂ Lp(Ω), 1 ≤ p < ∞, admit- ting the following properties.
(i) sup f ∈F k f kLp(Ω) < ∞. (ii) For any ω b Ω, we have
lim kτh f − f kLp(ω) = 0, h→0 where τh f (x) = f (x − h), x ∈ ω, provided that |h| sufficiently small. (iii) For any ε > 0, there exists ω b Ω so that, for every f ∈ F , k f kLp(Ω\ω) < ε. Then F is relatively compact in Lp(Ω). 10 1 Sobolev spaces
We end the present section by a density result, where Cc(Ω) denotes the space of continuous functions with compact supported in Ω. p Theorem 1.5. Cc(Ω) is dense in L (Ω) for any 1 ≤ p < ∞.
1.2 Approximation by smooth functions
p Recall that Lloc(Ω), 1 ≤ p ≤ ∞, is the space of measurable functions f : Ω → R p satisfying f χK ∈ L (Ω), for every K b Ω. Here χK is the characteristic function of p K. We usually say that Lloc(Ω) is the space of locally p-integrable functions on Ω Define the space of test functions by
∞ D(Ω) = {ϕ ∈ C (Ω); supp(ϕ) b Ω}.
∂ n We use the notation ∂i = and, if α = (α1,...,αn) ∈ is a multi-index, we ∂xi N set α α1 αn ∂ = ∂1 ...∂n . Proposition 1.5. The function f defined by
e1/x, x < 0, f (x) = 0, x ≥ 0, belongs to C∞(R). Proof. We show, by using an induction with respect to the integer m, that
(m) (m) 1/x f (0) = 0, f (x) = Pm (1/x)e , x < 0, (1.1) where Pm is a polynomial. Clearly, the claim is true for m = 0. Assume then that (1.1) holds for some m. Then
f (m)(x) − f (m)(0) P (1/x)e1/x lim = lim m = 0, x<0, x→0 x x<0, x→0 x showing that f (m+1)(0) = 0. Finally, we have, for x < 0,
1 f (m+1)(x) = − P (1/x) + P0 (1/x)e1/x x2 m m 1/x = Pm+1 (1/x)e .
The proof is then complete. ut
We call a sequence of mollifiers any sequence of functions (ρm) satisfying Z n 0 ≤ ρm ∈ D(R ), supp(ρm) ⊂ B(0,1/m), ρmdx = 1. Rn 1.2 Approximation by smooth functions 11
Let 2 c−1e1/(|x| −1), |x| < 1, ρ(x) = 0, |x| ≥ 1, where Z 2 c = e1/(|x| −1)dx. B(0,1)
It is straightforward to check that (ρm) defined by
n n ρm = m ρ(mx), x ∈ R , is a sequence of mollifiers. 1 n Let f ∈ Lloc(Ω) and ϕ ∈ Cc(R ) satisfying supp(ϕ) ⊂ B(0,1/m) with m ≥ 1. Define the convolution ϕ ∗ f on
Ωm = {x ∈ Ω; dist(x,∂Ω) > 1/m} by Z Z ϕ ∗ f (x) = ϕ(x − y) f (y)dy = ϕ(y) f (x − y)dy. Ω B(0,1/m)
1 n Proposition 1.6. If f ∈ Lloc(Ω) and ϕ ∈ D(R ) is so that supp(ϕ) ⊂ B(0,1/m), ∞ n for m ≥ 1, then ϕ ∗ f ∈ C (Ωm) and, for every α ∈ N ,
∂ α (ϕ ∗ f ) = (∂ α ϕ) ∗ f .
Proof. Assume first that |α| = ∑αi = 1. Let x ∈ Ωm and r > 0 such that B(x,r) ⊂ Ωm. Whence, ω = B(x,r + 1/m) b Ω and, for 0 < |ε| < r, ϕ ∗ f (x + εα) − ϕ ∗ f (x) Z ϕ(x + εα − y) − ϕ(x − y) = f (y)dy. ε ω ε But ϕ(x + εα − y) − ϕ(x − y) lim = ∂ α ϕ(x − y) ε→0 ε and ϕ(x + εα − y) − ϕ(x − y) sup < ∞. y∈Ω, 0<|ε| Define, for |y| < 1/m, the translation operator τy as follows 1 τy : f ∈ Lloc(Ω) → τy f : τy f (x) = f (x − y), x ∈ Ωm. 12 1 Sobolev spaces Lemma 1.1. Let ω b Ω. (i) If f ∈ C(Ω) then we have, for sufficiently large m, sup|ρm ∗ f (x) − f (x)| ≤ sup sup τy f (x) − f (x) . x∈ω |y|<1/m x∈ω p (ii) If f ∈ Lloc(Ω), 1 ≤ p < ∞ then we have, for sufficiently large m, kρm ∗ f − f kLp(ω) ≤ sup kτy f − f kLp(ω). |y|<1/m Proof. Note first that ω ⊂ Ωm provided that m is sufficiently large. Let f ∈ C(Ω). As Z ρm(y)dy = 1, B(0,1/m) we obtain Z |ρm ∗ f (x) − f (x)| = ρm(y)( f (x − y) − f (x))dy 1 B(0, m ) ≤ sup sup|τy f (x) − f (x)|. 1 x∈ω |y|< m That is (i) holds. p Next, let f ∈ Lloc(Ω), 1 ≤ p < ∞. We have by Holder’s¨ inequality, for any x ∈ ω, Z |ρm ∗ f (x) − f (x)| = ρm(y)[ f (x − y) − f (x)]dy B(0,1/m) Z 1 1 0 ≤ ρm(y) p ρm(y) p [ f (x − y) − f (x)]dy B(0,1/m) Z 1/p p ≤ ρm(y)| f (x − y) − f (x)| dy . B(0,1/m) We get by applying Fubini’s theorem Z p Z Z p ρm ∗ f (x) − f (x) dx ≤ dx ρm(y) f (x − y) − f (x) dy ω ω B(0,1/m) Z Z p = dy ρm(y) f (x − y) − f (x) dx B(0,1/m) ω Z Z p ≤ ρm(y)dy sup | f (x − z) − f (x)| dx B(0,1/m) |z|<1/m ω Z ≤ sup | f (x − z) − f (x)|pdx. |z|<1/m ω Assertion (ii) then follows. ut 1.2 Approximation by smooth functions 13 Lemma 1.2. (Continuity of translation operator) Let ω b Ω. (i) If f ∈ C(Ω) then lim sup τy f (x) − f (x) = 0. y→0 x∈ω p (ii) If f ∈ Lloc(Ω), 1 ≤ p < ∞, then lim kτy f − f kLp(ω) = 0. y→0 Proof. Let f ∈ C(Ω) and ω˜ be so that ω b ω˜ b Ω. Then (i) follows from the fact that f is uniformly continuous on ω˜ . p Next, let f ∈ Lloc(Ω), 1 ≤ p < ∞ and ε > 0. According to Theorem 1.5, we find g ∈ Cc(ω˜ ) so that k f − gkLp(ω˜ ) ≤ ε. By (i), there exists 0 < δ < dist(ω,ω˜ ) so that sup|τyg(x) − g(x)| ≤ ε for any |y| < δ. x∈ω Thus we have, for |y| < δ, kτy f − f kLp(ω) ≤ kτy f − τygkLp(ω) + kτyg − gkLp(ω) + kg − f kLp(ω) 1/p ≤ 2kg − f kLp(ω) + |ω| sup τyg(x) − g(x) e x∈ω ≤ (2 + |ω|1/p)ε. Hence (ii) follows because ε > 0 is arbitrary. ut A combination of Lemmas 1.1 and 1.2 yields the following regularization theo- rem. Theorem 1.6. (i) If f ∈ C(Ω) then ρm ∗ f converges to f uniformly in any compact subset of Ω. p p (ii) If u ∈ Lloc(Ω), 1 ≤ p < ∞, then ρm ∗ f converges to f in Lloc(Ω). 1 Theorem 1.7. (Cancellation theorem) Let f ∈ Lloc(Ω) satisfying Z f ϕdx = 0 for all ϕ ∈ D(Ω). Ω Then f = 0 a.e. in Ω. Proof. We have in particular ρm ∗ f = 0, for every m. The proof is completed by applying Theorem 1.6 (ii). ut Theorem 1.8. D(Ω) is dense Lp(Ω) for any 1 ≤ p < ∞. p p Proof. Let f ∈ L (Ω) and ε > 0. Since Cc(Ω) is dense L (Ω) by Theorem 1.5, we find g ∈ Cc(Ω) so that 14 1 Sobolev spaces kg − f kp ≤ ε/2. As supp(ρm ∗g) ⊂ B(0,1/m)+supp(g) (see Exercise 1.5), there exists a compact set K ⊂ Ω such that supp(ρm ∗ g) ⊂ K if m ≥ m0, for some integer m0. By Proposition p 1.6, ρm ∗ g ∈ D(Ω), and from Theorem 1.6, ρm ∗ g converges to g in L (Ω) (note that g has compact support). Hence, we find k ≥ m0 so that kρk ∗ g − gkp ≤ ε/2, from which we deduce that k(ρk ∗ g) − f kp ≤ ε. ut Here is another application of the regularization theorem Lemma 1.3. Let K ⊂ Ω be compact. There exists ϕ ∈ D(Ω) so that 0 ≤ ϕ ≤ 1 and ϕ = 1 in K. Proof. Set, for r > 0, n Kr = {x ∈ R ; dist(x,K) ≤ r}. Let ε > 0 so that K2ε ⊂ Ω. Let g be the function which is equal to 1 in Kε and n equal to 0 in R \ Kε . Fix m > 1/ε. Then the function ϕ = ρm ∗ g takes its values n between 0 and 1, it is equal to 1 in K and equal to 0 in R \ K2ε . Finally, ϕ ∈ D(Ω) by Proposition 1.6. ut n Theorem 1.9. (Partition of unity) Let K be a compact subset of R and Ω1,...,Ωk n k n be open subsets of R so that K ⊂ ∪ j=1Ω j. Then there exist ψ0,ψ1,...ψk ∈ D(R ) so that k (a) 0 ≤ ψ j ≤ 1, 0 ≤ j ≤ k, ∑ j=0 ψ j = 1, n (b) supp(ψ0) ⊂ R \ K, ψ j ∈ D(Ω j), 1 ≤ j ≤ k. Proof. By Borel-Lebesgue’s theorem K can be covered by finite number of balls B(xm,rm), 1 ≤ m ≤ `, such that, for any m, B(xm,2rm) ⊂ Ω j(m) for some 1 ≤ j(m) ≤ k. If 1 ≤ j ≤ k then set ` ` [ [ [ Kj = B(xm,2rm), K0 = B(xm,rm), Ω = B(xm,2rm). j(m)= j m=1 m=1 According to Lemma 1.3, there exists, where 1 ≤ j ≤ k, ϕ j ∈ D(Ω j) satisfying 0 ≤ ϕ j ≤ 1 and ϕ j = 1 sur Kj. There exists also ϕ ∈ D(Ω) satisfying 0 ≤ ϕ ≤ 1 and ϕ = 1 in K0. Take then ϕ0 = 1 − ϕ and, for 0 ≤ j ≤ k, ϕ j ψ j = k . ∑i=0 ϕi Then it is straightforward to check that ψ0,ψ1,...ψk have the expected properties. ut 1.3 Weak derivatives 15 1.3 Weak derivatives The notion of weak derivative consists in generalizing the derivative in classical 1 sense by means of an integration by parts formula. Precisely if f ,g ∈ Lloc(Ω) and n α ∈ N , we say that g = ∂ α f in the weak sense provided that Z Z f ∂ α ϕdx = (−1)|α| gϕdx for any ϕ ∈ D(Ω). Ω Ω Note that by the cancellation theorem ∂ α f is uniquely determined. In light of to this definition we can state the following result. 1 Lemma 1.4. (Closing lemma) If ( fm) converges to f in Lloc(Ω) and (gm) converge 1 α α vers g in Lloc(Ω) and if gm = ∂ fm in the weak sense, for any m, then g = ∂ f in the weak sense. 1 n α Proposition 1.7. Let f ,g ∈ Lloc(Ω) and α ∈ N . If g = ∂ f in the weak sense then in Ωm = {x ∈ Ω; dist(x,∂Ω) > 1/m} we have α ∂ (ρm ∗ f ) = ρm ∗ g. ∞ Proof. As ρm ∗ f ∈ C (Ωm) by Proposition 1.6, we have, for x ∈ Ωm, Z α α ∂ (ρm ∗ f )(x) = ∂x ρm(x − y) f (y)dy Ω Z |α| α = (−1) ∂y ρm(x − y) f (y)dy. Ω Whence Z α 2|α| ∂ (ρm ∗ f )(x) = (−1) ρm(x − y)g(y)dy = ρm ∗ g(x) Ω as expected. ut Lemma 1.5. Let f ,g ∈ C(Ω) and |α| = 1. If f has a compact support and g = ∂ α f in the classical sense then Z gdx = 0. Ω Proof. We may assume, without loss of generality, that α = (0,...,0,1). As supp( f ) n− is compact, it is contained in R 1×]a,b[ for some finite interval ]a,b[. We naturally extend f and g by 0 outside Ω. We still denote by f and g these extensions. Using 0 n−1 the notation x = (x ,xn) ∈ R × R, we get Z Z b 0 0 0 0 g(x ,xn)dxn = g(x ,xn)dxn = f (x ,a) − f (x ,b) = 0. R a Whence, we obtain by applying Fubini’s theorem 16 1 Sobolev spaces Z Z Z Z 0 gdx = gdx = dx gdxn = 0. Ω Rn Rn−1 R This completes the proof. ut Theorem 1.10. (Du Bois-Reymond lemma) Let f ,g ∈ C(Ω) and |α| = 1. Then g = ∂ α f in the weak sense if and only if g = ∂ α f in the classical sense. Proof. If g = ∂ α f in the weak sense then, according to Proposition 1.7, we have in the classical sense α ∂ (ρm ∗ f ) = ρm ∗ g. Hence Z ε Z ε α ρm ∗ g(x +tα)dt = ∂ (ρm ∗ f )(x +tα)dt = ρm ∗ f (x + εα) − ρm ∗ f (x). 0 0 That is we have Z ε ρm ∗ f (x + εα) = ρm ∗ f (x) + ρm ∗ g(x +tα)dt. (1.2) 0 From the regularization theorem ρm ∗ f (resp. ρm ∗ g) tends to f (resp. g) uniformly in any compact subset of Ω, as m → ∞. We get by passing to the limit in (1.2) Z ε f (x + εα) = f (x) + g(x +tα)dt. 0 Whence, g = ∂ α f in the classical sense. Conversely, if g = ∂ α f in the classical sense then by virtue of Lemma 1.5 we have Z Z 0 = ∂ α ( f ϕ)dx = ( f ∂ α ϕ + gϕ)dx for every ϕ ∈ D(Ω). Ω Ω Thus, g = ∂ α f in the weak sense. ut We close this section by an example. Let, for −n < θ ≤ 1, θ 2 θ/2 f (x) = |x| and fε (x) = (|x| + ε) , ε > 0. n n The function fε is continuous in R and therefore it is measurable in R . As ( fε ) converges a.e. (in fact everywhere except at 0), we deduce that f is measurable. On the other hand, for any R > 0, we have by passing to polar coordinates Z Z Z R |x|θ dx = dσ rn+θ−1dr < ∞, B(0,R) Sn−1 0 n− n where S 1 is the unit sphere of R . We have, when θ ≤ 0, 0 ≤ fε ≤ f , and since fε converges a.e. to f , we may apply Lebesgue’s dominated convergence theorem. Therefore, ( fε ) converges to f 1 n in Lloc(R ). We proceed similarly in the case θ ≥ 0. We note that fε ≤ f1, 0 < ε ≤ 1, 1 n and we conclude that we still have the convergence of ( fε ) to f in Lloc(R ). 1.4 W k,p spaces 17 ∞ n We get by simple computations that fε ∈ C (R ) and 2 θ−2 ∂ j fε = θx j(|x| + ε) 2 . θ−1 1 θ−2 Hence |∂ j fε | ≤ θ|x| and then ∂ j fε converges in Lloc(R) to g = θx j|x| if θ > 1 − n. In light of the closing lemma, we obtain that g = ∂ j f in the weak sense. 1.4 W k,p spaces If k ≥ 1 is an integer and 1 ≤ p < ∞, we define the Sobolev W k,p(Ω) by W k,p(Ω) = {u ∈ Lp(Ω); ∂ α u ∈ Lp(Ω) for any |α| ≤ k}. Here ∂ α u is understood as the derivative in the weak sense. We endow this space with the norm 1/p Z ! α p kukW k,p(Ω) = kukk,p = ∑ |D u| dx . |α|≤k Ω The space Hk(Ω) = W k,2(Ω) is equipped with the scalar product α α (u|v)Hk(Ω) = ∑ (∂ u|∂ v)L2(Ω). |α|≤k k,p Define also the local Sobolev space Wloc (Ω) as follows k,p p k,p Wloc (Ω) = {u ∈ Lloc(Ω); u|ω ∈ W (ω, for each ω b Ω}. k,p We say that the sequence (um) converges to u in Wloc (Ω) if, for every ω b Ω, we have kum − ukW k,p(ω) → 0, as m → ∞. p d Consider L (Ω,R ) that we equip with its natural norm 1/p d Z ! p k(vs)kp = ∑ |vs| dx . s=1 Ω Let Γ = {1,...,d} × Ω and define on L = {u : Γ → R; u(s,·) ∈ Cc(Ω), 1 ≤ s ≤ d} the elementary integral 18 1 Sobolev spaces Z d Z udµ = ∑ u(s,x)dx. Γ s=1 Ω p d Then it is not difficult to check that L (Ω,R ) is isometrically isomorphic to Lp(Γ ,dµ). An extension of Theorem 1.3 to a measure space with positive σ-finite measure yields the following result. p d Lemma 1.6. Any closed subspace of L (Ω,R ) is a separable Banach space. We get by applying Riesz’s representation theorem1 to Lp(Γ , µ) the following lemma. p d 0 Lemma 1.7. Let 1 < p < ∞ and Φ ∈ [L (Ω,R )] . Then there exists a unique ( fs) ∈ p0 d L (Ω,R ) so that d Z hΦ,(vs)i = ∑ fsvsdx. s=1 Ω Moreover k( fs)kp0 = kΦk. Let 1 ≤ p < ∞, k ≥ 1 an integer and d = ∑|α|≤k 1. Then the mapping k,p p d α A : W (Ω) → L (Ω,R ) : u → Au = (∂ u)|α|≤k is isometric, i.e. kAukp = kukk,p. Theorem 1.11. For 1 ≤ p < ∞ and k ≥ 1 an integer, the Sobolev space W k,p(Ω) is separable Banach space. k,p p d Proof. According to the closing lemma, F = A(W (Ω)) is closed in L (Ω,R ). The theorem follows then from Lemma 1.6. ut 0 Theorem 1.12. Let k ≥ 1 be an integer and 1 < p < ∞. For any Φ ∈ W k,p(Ω) , p0 d there exists a unique ( fα ) ∈ L (Ω,R ) so that k( fα )kp0 = kΦk and α k,p hΦ,ui = Φ(u) = ∑ fα D u, for all u ∈ W (Ω). |α|≤k 1 Riesz’s representation theorem Let (X,dµ) be a measure space, where dµ is positive σ-finite p 0 p0 measure. For each Φ ∈ [L (X, µ)] , there exists a unique gΦ ∈ L (X, µ) so that Z p Φ( f ) = hΦ, f i = f gΦ dµ for any f ∈ L (X,dµ). X Furthermore kΦk = kgΦ kp0 . 1.4 W k,p spaces 19 Proof. By Hahn-Banach’s extension theorem2, there exists a continuous linear form p d extending Φ to L (Ω,R ) without increasing its norm. The theorem follows then from Lemma 1.7. ut Let k ≥ 1 be an integer and 1 ≤ p < ∞. Denote the closure of D(Ω) in W k,p(Ω) k,p k,2 k by W0 (Ω). The space W0 (Ω) is usually denoted by H0 (Ω). Let W −k,p(Ω) be the space of continuous linear forms given as follows Z α Φ : D(Ω) → R : u → ∑ gα ∂ udx, |α|≤k Ω k,p p0 d when D(Ω) is seen as a subspace of W (Ω) and where (gα ) ∈ L (Ω,R ). We endow W −k,p(Ω) with its natural quotient norm kΦkW −k,p(Ω) = kΦk−k,p ( Z ) α =inf k(gα )kp0 ; hΦ,ui = ∑ gα ∂ udx for any u ∈ D(Ω) . |α|≤k Ω The space W −k,2(Ω) is usually denoted by H−k(Ω). Theorem 1.13. For any integer k ≥ 1 and 1 ≤ p < ∞, the space W −k,p(Ω) is iso- h k,p i0 metrically isomorphic to W0 (Ω) . −k,p p0 d Proof. If Φ ∈ W (Ω) then there exists (gα ) ∈ L (Ω,R ) so that Z α hΦ,ui = ∑ gα ∂ udx, for any u ∈ D(Ω). |α|≤k Ω Hence |hΦ,ui| ≤ k(gα )kp0 kukk,p for any u ∈ D(Ω). k,p We can extend uniquely Φ to W0 (Ω) by density. We still denote this exten- sion by Φ. Moreover kΦkh k,p i0 ≤ k(gα )kp0 in such a way that kΦkh k,p i0 ≤ W0 (Ω) W0 (Ω) kΦk −k,p . W0 (Ω) 2 Hahn-Banach’s extension theorem. Let V be a real normed vector space with norm k·k. Let V0 be a subspace of V and let Φ0 : V0 → R be a continuous linear form with norm kΦ0k 0 = sup Φ0(x). V0 x∈V0, kxk≤1 0 Then there exists Φ ∈ V extending Φ0 so that kΦk 0 = kΦ0k 0 . V V0 20 1 Sobolev spaces h k,p i0 Conversely, let Φ ∈ W0 (Ω) . Then by Hahn-Banach’s extension theorem Φ has an extension, still denoted by Φ, to W k,p(Ω) that does not increase its norm. But p d from Theorem 1.12 there exists (gα ) ∈ L (Ω,R ) so that k(gα )kp0 = kΦkh k,p i0 W0 (Ω) and Z α hΦ,ui = ∑ gα D udx for all u ∈ D(Ω). |α|≤k Ω Whence, Φ ∈ W −k,p(Ω) and kΦkW −k,p(Ω) ≤ k(gα )kp0 = kΦkh k,p i0 . W0 (Ω) This completes the proof. ut Next, we extend some classical rules of differential calculus to weak derivatives. 1,1 1 Proposition 1.8. (Derivative of a product) if u ∈ Wloc (Ω) and f ∈ C (Ω) then f u ∈ 1,1 Wloc (Ω) and ∂ j( f u) = f ∂ ju + ∂ j f u. Proof. If um = ρm ∗ u then we have in the classical sense ∂ j( f um) = f ∂ jum + ∂ j f um. 1 By the regularization theorem, um → u and ∂ jum = ρm ∗∂ ju → ∂ ju in Lloc(Ω). Thus f um → f u, ∂ j( f um) = f ∂ jum + ∂ j f um → f ∂ ju + ∂ j f u 1 in Lloc(Ω). The expected result follows then from the closing lemma. ut This proposition will be used to prove the following result. 1,p n 1,p n Theorem 1.14. If 1 ≤ p < ∞ then W0 (R ) = W (R ). n ,p n Proof. It is sufficient to prove that D(R ) is dense in W 1 (R ). We use truncation and regularization procedures. We fix θ ∈ C∞(R) satisfying 0 ≤ θ ≤ 1 and θ(t) = 1, t ≤ 1, θ(t) = 0, t ≥ 2. We define a truncation sequence by setting n θm(x) = θ (|x|/m), x ∈ R . ,p n Let u ∈ W 1 (R ). The formula giving the weak derivative of a product shows that 1,p n um = θmu ∈W (R ). With the help of Lebesgue’s dominated convergence theorem 1,p n one can check that um converges to u in W (R ) θm converges a.e. to u because α θm tends to 1 and we have |θmu| ≤ |u|. On the other hand, supp(∂ θm) ⊂ {m ≤ |x| ≤ α α α 2m} and ∂ um = θmD u + D θmu . This construction guarantees that the support of um is contained in B(0,2m). 1.4 W k,p spaces 21 We now proceed to regularization. From the previous step we need only to con- ,p n n sider u ∈ W 1 (R ) with compact support. Let K be a compact subset of de R so ∞ n that, for any m, the support of um = ρm ∗ u is contained in K. As um ∈ C (R ) by n Proposition 1.6, we deduce that um belongs to D(R ). We have from the regulariza- tion theorem p n um → u, ∂ jum = ρm ∗ ∂ ju → ∂ ju in L (R ). The proof is then complete. ut n Let Ω and ω be two open subsets of R . Recall that f : ω → Ω is a diffeo- morphism if f is bijective, it is continuously differentiable and satisfies Jf (x) = det(∂ j fi(x)) 6= 0, for every x ∈ ω. 1 The following result follows from the density of Cc(Ω) in L (Ω) and the classical formula of change of variable for smooth functions. 1 1 Theorem 1.15. If u ∈ L (Ω) then (u ◦ f )|Jf | ∈ L (ω) and Z Z u( f (x))|Jf (x)|dx = u(y)dy. ω Ω Proposition 1.9. (Change of variable formula) Let ω and Ω be two open subsets of n 1,1 1,1 R and let f : ω → Ω be a diffeomorphism. If u ∈ Wloc (Ω) then u ◦ f ∈ Wloc (ω) and n ∂ j(u ◦ f ) = ∑ (∂ku ◦ f )∂ j fk. k=1 Proof. Let um = ρm ∗ u and v ∈ D(ω). We have according to the definition of weak derivatives Z Z n (um ◦ f )(y)∂ jv(y)dy = − ∑ (∂kum ◦ f )(y)∂ j fk(y)v(y)dy. ω ω k=1 If g = f −1 then Theorem 1.15 yields Z Z (um ◦ f )(y)∂ jv(y)dy = um(x)(∂ jv ◦ g)(x)|det(Jg(x))|dx ω Ω Z n = − ∑ ∂kum(x)(∂ j fk ◦ g)(x)(v ◦ g)(x)|det(Jg(x))|dx Ω k=1 Z n = ∑ (∂kum ◦ f )(y)∂ j fk(y)v(y)dy. (1.3) ω k=1 Now, we have by the regularization theorem 1 um → u, ∂ jum → ∂ ju in Lloc(ω). We pass to the limit, when m → ∞, in the second and the third members of (1.3). We get 22 1 Sobolev spaces Z u(x)(∂ jv ◦ g)(x)|det(Jg(x))|dx Ω Z n = − ∑ ∂ku(x)(∂ j fk ◦ g)(x)(v ◦ g)(x)|det(Jg(x))|dx Ω k=1 and hence Z Z n (u ◦ f )(y)∂ jv(y)dy = − ∑ (∂ku ◦ f )(y)∂ j fk(y)v(y)dy ω ω k=1 by Theorem 1.15. The proposition is then proved. ut 1 1,1 Proposition 1.10. (Derivative of a composition) Let f ∈ C (R) and u ∈ Wloc (Ω). If 0 1,1 M = sup| f | < ∞ then f ◦ u ∈ Wloc (Ω) and 0 ∂ j( f ◦ u) = f ◦ u ∂ ju. Proof. If um = ρm ∗ u then we have in the classical sense 0 ∂ j( f ◦ um) = f ◦ um∂ jum. So, by the regularization theorem, we obtain 1 um → u, ∂ jum → ∂ ju in Lloc(Ω). If ω b Ω we get subtracting if necessary a subsequence we may assume that the convergence holds also almost everywhere in ω again from the regularization the- orem Z Z |( f ◦ um) − ( f ◦ u)|dx ≤ M |um − u|dx → 0, ω ω Z 0 0 | f ◦ um ∂ jum − f ◦ u ∂ ju|dx ω Z Z 0 0 ≤ M |∂ jum − ∂ ju|dx + | f ◦ um − f ◦ u ||∂ ju|dx → 0. ω ω Hence 0 0 1 f ◦ um → f ◦ u, f ◦ um ∂ jum → f ◦ u ∂ ju in Lloc(Ω) from Lebesgue’s dominated convergence theorem. The proof is then completed by using the closing lemma. ut 1,1 + − 1,1 Corollary 1.1. If u ∈ Wloc (Ω) then u ,u ,|u| ∈ Wloc (Ω) and ∂ ju in {u > 0}, ∂ j|u| = −∂ ju in {u < 0}, 0, in {u = 0}. 2 2 1/2 Proof. Let, for ε > 0, fε (t) = (t + ε ) and 1.5 Extension and trace operators 23 ∂ ju in {u > 0}, v = −∂ ju in {u < 0}, 0 in {u = 0}. We get by using Proposition 1.10 u ∂ j( fε ◦ u) = ∂ ju. (u2 + ε2)1/2 Hence 1 fε ◦ u → |u|, ∂ j( fε ◦ u) → v in Lloc(Ω) and ∂ j|u| = v according to the closing lemma. Finally, for completing the proof we observe that 2u+ = |u| + u and 2u− = |u| − u. ut 1.5 Extension and trace operators We start by extension operators using the simple idea of reflexion. If ω is an open n− subset of R 1 and 0 < δ ≤ +∞, we let Q = ω×] − δ,+δ[, Q+ = ω×]0,δ[. For an arbitrary u : Q+ → R, we define σu and τu on Q by 0 0 u(x ,xn) if xn > 0, σu(x ,xn) = 0 u(x ,−xn) if xn < 0, and 0 0 u(x ,xn) if xn > 0, τu(x ,xn) = 0 −u(x ,−xn) if xn < 0. 1,p Lemma 1.8. (Extension by reflexion) Let 1 ≤ p < ∞. If u ∈ W (Q+) then σu ∈ W 1,p(Q) and 1/p 1/p k uk p ≤ kuk p , k uk 1,p ≤ kuk 1,p . σ L (Q) 2 L (Q+) σ W (Q) 2 W (Q+) Proof. We first prove that ∂ jσu = σ∂ ju, 1 ≤ j ≤ n − 1. If v ∈ D(Q), we have Z Z 0 0 σu∂ jvdx = uD jwdx, with w(x ,xn) = v(x,xn) + v(x ,−xn). (1.4) Q Q+ Fix η ∈ C ∞(R) satisfying 24 1 Sobolev spaces 0 if t < 1 , η(t) = 2 1 if t > 1, and set ηm = η(mt). 0 As ηm(xn)w(x ,xn) ∈ D(Q+), we obtain Z Z Z uηmD jw = u∂ j(ηmw) = − ∂ juηmw. Q+ Q+ Q+ Il light of Lebesgue’s dominated convergence theorem, we can pass to the limit, when m tends to infinity, in the first and third terms. We obtain Z Z Z u∂ jwdx = − ∂ juwdx = − σ∂ juvdx Q+ Q+ Q which, combined with (1.4), entails Z Z σu∂ jvdx = − σ∂ juvdx. Q Q That is we have ∂ j(σu) = σ∂ ju, 1 ≤ j ≤ n − 1. (1.5) Next, we prove that ∂n(σu) = τ∂nu. For this purpose, we note that Z Z 0 0 σu∂nvdx = u∂nwdx˜ , withw ˜(x ,xn) = v(x,xn) − v(x ,−xn). (1.6) Q Q+ 0 0 Asw ˜(x ,0) = 0, there exists C0 > 0 so that |w˜(x ,xn)| ≤ C0|xn| in Q+. Using the fact 0 that ηm(xn)w˜(x ,xn) ∈ D(Q+), we find Z Z u∂n(ηmw˜)dx = − ∂nuηmwdx˜ . Q+ Q+ But 0 ∂n(ηmw˜) = ηmDnw˜ + mη (mxn)w˜. 0 Let C1 = kη k∞. We get, observing thatw ˜ has a compact support, that there exists a compact subset K of ω so that Z Z 0 mη (mxn)uwdx˜ ≤ C0C1m |u|xndx Q+ K×]0,1/m[ Z ≤ C0C1 |u|dx → 0. K×]0,1/m[ We get by applying again Lebesgue’s dominated convergence theorem Z Z Z u∂nwdx˜ = − ∂nuwdx˜ = − τ∂nuvdx. Q+ Q+ Q 1.5 Extension and trace operators 25 This identity together with (1.6) imply Z Z Z σu∂nvdx = ∂nuwdx˜ = − τ∂nuvdx. Q Q+ Q In other words, we demonstrated that ∂n(σu) = τ∂nu. (1.7) Finally, identities (1.5) and (1.7) yield the expected result. ut We use in the sequel the following notations n D(Ω) = {u|Ω ; u ∈ D(R )}, n 0 n 0 n−1 R+ = {(x ,xn) ∈ R ; x ∈ R , xn > 0}. n Lemma 1.9. (Trace inequality) Let 1 ≤ p < ∞. We have, for u ∈ D(R+), Z 0 p 0 p−1 |u(x ,0)| dx ≤ pkuk p n k∂nuk p n . L (R ) L (R+) Rn−1 + n Proof. Consider first the case 1 < p < ∞. Let u ∈ D(R+). As u has compact support, 0 n−1 0 0 for each x ∈ R , we find yn = yn(x ) so that u(x ,yn) = 0, yn ≥ yn. Whence Z y 0 p n 0 p−1 0 |u(x ,0)| = − p|u(x ,xn)| ∂nu(x ,xn)dxn 0 and hence Z ∞ 0 p 0 p−1 0 |u(x ,0)| ≤ p|u(x ,xn)| |∂nu(x ,xn)|dxn. 0 We get by applying Fubini’s theorem and then Holder’s¨ inequality Z Z 0 p 0 p−1 |u(x ,0)| dx ≤ p |u| |∂nu|dx n−1 n R R+ 1 1 Z 0 Z p (p−1)p0 p p ≤ p |u| dx |∂nu| dx n n R+ R+ 1 1 Z 1− p Z p p p ≤ p |u| dx |∂nu| dx , n n R+ R+ which yields the expected inequality. The proof in the case p = 1 is quite similar to that of the case 1 < p < ∞. ut Proposition 1.11. Let 1 ≤ p < ∞. There exists a unique linear bounded operator 1,p n p n−1 γ0 : W (R+) → L (R ) n satisfying γ0u = u(·,0), for each u ∈ D(R+). 26 1 Sobolev spaces n Proof. If u ∈ D(R+), we set γ0u = u(·,0). From Lemma 1.9, we have 1 p kγ uk p n−1 ≤ p kuk 1,p n . 0 L (R ) W (R+) n We deduce, using the theorem of extension by reflexion and the density of D(R ) 1,p n n 1,p n in W (R ) (a consequence of Theorem 1.14), that D(R+) is dense W (R+). We complete the proof by extending γ0 by density. ut 1,p n Proposition 1.12. (Integration by parts) Let 1 ≤ p < ∞. If u ∈ W (R+) and v ∈ n D(R+) then Z Z Z 0 v∂nudx = − ∂nvudx − γ0vγ0udx n n n−1 R+ R+ R and Z Z v∂ judx = − ∂ jvudx, 1 ≤ j ≤ n − 1. n n R+ R+ n Proof. Assume first that u ∈ D(R+). The classical integration by parts formula 0 n− yields, for each x ∈ R 1, Z +∞ Z +∞ 0 0 0 0 v(x ,xn)∂nu(x ,xn)dxn = − u(x ,xn)∂nv(x ,xn)dxn 0 0 − u(x0,0)v(x0,0). We then obtain by applying Fubini’s theorem Z Z Z 0 0 0 v∂nudx = − ∂nvudx − v(x ,0)u(x ,0)dx , n n n−1 R+ R+ R that we write in the form Z Z Z 0 v∂nudx = − ∂nvudx − γ0vγ0udx . (1.8) n n n−1 R+ R+ R n 1,p n We know from the proof of Proposition 1.11 that D(R+) is dense in W (R+). So 1,p n n if u ∈ W (R+) then we may find a sequence (um) in D(R+) that converges to u 1,p n 1,p n in W (R+). This and the fact that γ0 is a bounded operator from W (R+) into p n−1 p n−1 L (R ) entail that γ0um converges to γ0u in L (R ). We obtain from (1.8) Z Z Z 0 v∂numdx = − ∂nvumdx − γ0vγ0umdx . n n n−1 R+ R+ R We then pass to the limit, when m → ∞, to get the first formula. The second formula can be established analogously. ut n n Hereafter, if u is a function defined on R+ then its extension to R by 0 is denoted by u. 1.5 Extension and trace operators 27 1,p n Proposition 1.13. Let 1 ≤ p < ∞ and u ∈ W (R+). The following assertions are equivalent. 1,p n (i) u ∈ W0 (R+). (ii) γ0u = 0. 1,p n (iii) u ∈ W (R ) and ∂ ju = ∂ ju, 1 ≤ j ≤ n. 1,p n n Proof. If u ∈ W0 (R+) then there exits a sequence (um) in D(R+) converging to 1,p n p n−1 u in W (R+). Therefore γ0um → γ0u dans L (R ). But γ0um = 0, for each m. Whence, γ0u = 0. That is (i) implies (ii). n If γ0u = 0 then by Proposition 1.12 we have, for every v ∈ D(R ), Z Z v∂ judx = ∂ jvudx, 1 ≤ j ≤ n. Rn Rn In other words, we proved that (ii) implies (iii). Assume finally that (iii) holds. If (θm) is the truncation sequence introduced in 1,p n the proof of Theorem 1.14 then the sequence um = θmu converges to u in W (R ) n and um has its support contained in B(0,2m) ∩ R+. We are then reduced to consider n the case where additionally u has a compact support in R+. Let ym = (0,...,0,1/m) and vm = τym u. Since ∂ jvm = τym ∂ ju, the continuity of 1,p n translation operators guarantees that vm → u in W (R+). That is we are lead to n the case where u has a compact support in R+. Therefore, we may find a compact n ∞ n subset K of R+ so that, for each m, supp(ρm ∗ u) ⊂ K. As wm = ρm ∗ u ∈ C (R+), n we have wm ∈ D(R+). According to the regularization theorem, wn tends to u in 1,p n 1,p n W (R+). In consequence, u ∈ W0 (R+) and then (iii) entails (i). This completes the proof. ut n Prior to considering extension and trace theorems for an arbitrary domain of R , we introduce the definition of an open subset of class Ck. We say that an open subset n k Ω of R is of class C if, for each x ∈ Γ = ∂Ω, we can find a neighborhood U of n n− k x in R , an open subset ω of R 1, ψ ∈ C (ω) 3 and δ > 0 so that, modulo a rigid transform, U = {(y0,ψ(y0) +t); y0 ∈ ω, |t| < δ}, Ω ∩U = {(y0,ψ(y0) +t); y0 ∈ ω, 0 < t < δ}, Γ ∩U = {(y0,ψ(y0)); y0 ∈ ω}. In other words, an open subset Ω is of class Ck if any point of its boundary admits a neighborhood U so that U ∩ Ω coincide with the epigraph of a function of class Ck. We leave to the reader to check that, with the aid of the implicit function theorem, n k the above definition of an open subset of R of class C is equivalent to the following one: let 3 k k n Recall that C (ω) = {u|ω ; u ∈ C (R )}. 28 1 Sobolev spaces 0 n 0 Q = {x = (x ,xn) ∈ R ; |x | < 1 and |xn| < 1}, n Q+ = Q ∩ R+, 0 n 0 Q0 = {x = (x ,xn) ∈ R ; |x | < 1 and xn = 0}. Ω is said of class Ck, k ≥ 1 is an integer if, for each x ∈ Γ , there exists a neighbor- n hood U of x in R and a bijective mapping Φ : Q → U satisfying k −1 k Φ ∈ C (Q), Φ ∈ C (U), Φ(Q+) = U ∩ Ω, Φ(Q0) = U ∩Γ . If the open subset Ω is of class Ck and has bounded boundary then there exist n (think to the compactness of Γ ) a finite number of open subsets of R , U1,...,U`, n−1 open subsets of R , ω1,...,ω`, functions ψ1,...ψ` and positive real numbers δ1,...,δ` satisfying all the conditions of the preceding definition and are so that ` [ Γ ⊂ Uj. j=1 ∞ n By the theorem of partition of unity, there exist φ0,...,φ` ∈ C (R ) satisfying ` (i) 0 ≤ φ j ≤ 1, 0 ≤ j ≤ `, ∑ j=0 φ j = 1, n (ii) supp(φ0) ⊂ R \Γ , φ j ∈ D(Uj), j = 1,...,`. Theorem 1.16. (Extension theorem) Let 1 ≤ p < ∞ and let Ω be an open subset of n R of class C1 with bounded boundary. There exists a bounded operator 1,p 1,p n P : W (Ω) → W (R ) so that Pu|Ω = u. Proof. We use the notations that we introduced above in the definition of Ω of class Ck with bounded boundary. Fix u ∈ W 1,p(Ω) and 1 ≤ j ≤ `. From the change of variable formula, we have