Sobolev spaces, Trace theorems and Green s functions.
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1 Sobolev spaces, Trace theorems and Green s functions. Boundary Element Methods for Waves Scattering Numerical Analysis Seminar. Orane Jecker October 21, 2010
2 Plan Introduction 1 Useful definitions 2 Distributions Main subjects 3 Sobolev spaces 4 Trace Theorems 5 Green s functions
3 Partial derivative Definition Let d N, α = (α 1,..., α d ) N d 0 be a multi index with absolute value α = α α d and x = (x 1,..., x d ) R d. For u a real valued function which is sufficiently smooth, the partial derivative is given by ( ) α1 ( ) αd D α u(x): =... u(x 1,..., x d ). x 1 x d
4 C k,κ (Ω) Let k N 0 and κ (0, 1). C k,κ (Ω) : = {u : Ω R D k u is Hölder continuous with exponent κ}. The associated norm is u C k,κ (Ω): = u C k (Ω)+ α =k D α u(x) D α u(y) sup x,y Ω,x y x y κ
5 Lipschitz domain Simplest case: There exists a function γ : R d 1 R such that Ω : = {x R d x d < γ( x) for all x = (x 1,..., x d 1 ) R d 1 }. Definition When γ is Lipschitz, then Ω is said to be a Lipschitz hypograph with boundary Ω = : Γ : = {x R d x d = γ( x) for all x R d 1 }.
6 Lipschitz domain Definition An open set Ω R d, d 2 is a Lipschitz domain if Γ is compact and if there exist finite families {W i } and {Ω i } such that: 1 {W i } is a finite open cover of Γ, that is W i R d is open for all i N and Γ i W i. 2 Each Ω i can be transformed into a Lipschitz hypograph by a rigid motion 3 For all i N the equality W i Ω = W i Ω i. The local representation of the boundary Γ is in general not unique.
7 Example of a non-lipschitz domain Figure: Example of a non-lipschitz domain in 2D.
8 L 1 loc (Ω) Definition L loc 1 : = {u : Ω R u is locally integrable }. That means, u is integrable with respect to any bounded closed subset K of Ω. Remark A function u : Ω R L loc 1 (Ω) is not, in general, in L 1 (Ω). On the other hand, u L 1 (Ω) implies that u L loc 1 (Ω). u(x)dx < u(x)dx <, K Ω. Ω K
9 L 1 loc (Ω) Example Let Ω = (0, 1) and u(x) = 1. We have x u(x)dx = lim ɛ 0 ɛ 1 dx = lim x ln ɛ 0 ( ) 1 =. ɛ That implies u / L 1 (Ω). Let K = [a, b] (0, 1) with 0 < a < b < 1. Then K u(x)dx = b a ( ) 1 b dx = ln <. x a That implies u L loc 1 (Ω).
10 General Partial Derivative Definition A function u L loc 1 (Ω) has a generalized partial derivative w.r.t. x i, if there exists v L loc 1 (Ω) such that Ω v(x)ϕ(x)dx = Ω u(x) x i ϕ(x)dx, for all ϕ C 0 (Ω). The GPD is denoted by x i u(x) : = v(x). We define the space of test functions by C 0 (Ω) : = D(Ω).
11 Convergence in D(Ω) Definition Let {ϕ n } n N Ω. {ϕ n } converges to ϕ in D(Ω) if 1 K Ω compact subset such that supp ϕ n K, n N 2 D α ϕ n C 0 D α ϕ, α N d. Definition A complex valued continuous linear map T : D(Ω) C is called a distribution. T is continuous if lim T (ϕ n) = T (ϕ), n for any {ϕ n } n N which converges to ϕ in D(Ω). The set of all distributions is denoted by D (Ω).
12 Distribution Definition Let T D (Ω). Its partial derivative w.r.t x i, 1 i d, in the sense of distribution is i T (ϕ) = T ( i ϕ), for all ϕ D(Ω) Definition For a function u L loc 1 (Ω) we define the distribution T u (ϕ) : = u(x)ϕ(x)dx, for ϕ D(Ω). Ω
13 Example of derivative in the sense of distribution Example Let v(x) = sign(x) L loc 1 ([ 1, 1]) and compute its derivative in the sense of distribution. 1 1 x sign(x)ϕ(x)dx = 1 for all ϕ D(Ω). We obtain x sign(x) = 2δ 0 D (Ω). 1 sign(x) ϕ(x)dx = 2ϕ(0), x
14 Example of derivative in the sense of distribution SOBOLEV SPACES
15 W k p (Ω) Definition Let k N 0, the Sobolev space is defined as W k p (Ω) : = C (Ω) W k p (Ω). The norm is given by ( ) 1 u W k p (Ω): = α k Dα u p p L p(ω), for 1 p <, max α k D α u L (Ω), for p =. W k p(ω) : = C 0 (Ω) W k p (Ω).
16 W k p (Ω) Sobolev spaces can be define for all s R. For 0 < s, with s = k + κ, k N 0 and κ (0, 1), the norm is u W s p (Ω): = where u p Wp s (Ω) = α =k ( ) 1 u p Wp k (Ω) + u p p Wp s (Ω), Ω Ω D α u(x) D α (y) p x y d+pκ dxdy
17 W k p (Ω) For s < 0 and 1 < p <, W s p (Ω) : = 1 p + 1 q = 1. The norm is u W s p (Ω) := sup v W s u, v Ω v q (Ω),v 0 W s q ( W s q (Ω)) where (Ω)
18 W s 2 (Ω) The Sobolev space W2 s (Ω) admits an inner-product For s = k N 0 u, v W k 2 (Ω) : = D α u(x)d α v(x)dx α k Ω For s = k + κ with κ (0, 1) and k N 0 u, v W s 2 (Ω) : = u, v W k 2 (Ω)+ (D α u(x) D α u(y))(d α v(x) D α v(y)) dxdy x y d+2κ α =k Ω Ω
19 Tempered distribution Definition We define the space of rapidly decreasing functions where S(R d ) = {ϕ C (R d ) ϕ k,l < }, ϕ k,l = sup ( x k + 1) D α ϕ(x) <, for all k, l N 0. x R d α l The space of tempered distributions S (R d ) is S (R d ) : = {T : S(R d ) C T complex valued cont. lin. map}.
20 Tempered distribution For s R, the Bessel potential operator J s : S(R d ) R is given by J s u(x) : = 1 (1 + ξ 2 ) s (2π) d 2 û(ξ)e i x,ξ dξ, 2 R d for u S(R d ). (J s T )(ϕ) : = T (J s ϕ), for all ϕ S(R d ).
21 H s (R) Definition The Sobolev space over R d is defined as H s (R d ) : = {v S (R d ) J s v L 2 (R d )}, for all s R. The norm is v 2 H s (R d ) R : = (1 + ξ 2 ) s ˆv(ξ) 2 dξ. d
22 Relation between Wp k and Hk spaces Theorem For all s R, we have the following relation H s (R d ) = W s 2 (R d ). Let Ω be a bounded domain in R d, the norm is given by H s (Ω) : = {v = ṽ Ω ṽ H s (R d )}, v H s (Ω): = inf ṽ H s ṽ H s (R d (R ). d ),ṽ Ω =v
23 Relation between Wp k and Hk spaces Definition H s (Ω) : = C 0 (Ω) H s (R d ), H s 0(Ω) : = C 0 (Ω) H s (Ω)
24 Relation between Wp k and Hk spaces Theorem Let Ω R d be a Lipschitz domain. For s 0 we have H s (Ω) H0(Ω). s Moreover, H s (Ω) = H0(Ω) s for s / { 1 2, 3 2, 5 2,...}. H s (Ω) = [H s (Ω)], H s (Ω) = [ H s (Ω)]
25 Relation between Wp k and Hk spaces For s < 0, we define H s (Γ) : = (H s (Γ)) with the norm u H s (Γ): = u, v Γ sup 0 v H s (Γ) v H s (Γ)
26 Espace de Fréchet Definition The Sobolev spaces over a bounded domain Ω R d allow us to define the Fréchet space H 1 loc(ω) : = {v D (Ω) v H 1 (B)< + and the space for all bounded B Ω}, H 1 comp(ω) : = {v D (Ω) v H 1 (Ω), v has compact support} Remark We see that H 1 comp(ω) H 1 loc (Ω).
27 Sobolev Embedding Theorem Theorem (Sobolev Embedding Theorem.) Let Ω be an open subset of R d with a lipschitz continuous boundary. The following continuous embedding holds For all s R, H s+1 (Ω) H s (Ω). For k N 0, if 2(k m) > d, then H k (Ω) C m ( Ω). For all s R, if d s for p = 1, d p < s for p > 1 then W s p (Ω) C(Ω)
28 Sobolev Embedding Theorem Our goal is to study the continuity of function in Sobolev space. Take m = 0. 1 If d = 1, 2(k 0) > 1 is valid k 1. We have H 1 (Ω) C 0 ( Ω). Moreover, as H k+1 (Ω) H k (Ω), H k+1 (Ω) H k (Ω)... H 1 (Ω) C 0 ( Ω). 2 If d = 2, is H 1 (Ω) C 0 ( Ω)? No, H 1 (Ω) C 0 ( Ω)! We have H 2 (Ω) C 0 ( Ω).
29 Sobolev Embedding Theorem TRACE THEOREMS
30 Trace Theorem We recall the interior trace operator γ D : = γ int 0 : C (Ω) C ( Ω). Theorem (Trace Theorem.) Let Ω be a C k 1,1 -domain. For 1 < s k the interior trace 2 operator γ D : H s (Ω) H s 1 2 (Γ), where γ D v : = v Γ, is bounded. There exists C T > 0 such that γ D v H s 1 2 (Γ) C T v H s (Ω) for all v H s (Ω). For a lipschitz domain Ω, put k = 1. γ D : H s (Ω) H s 1 2 (Ω) for s ( 1, 1]. That is true for 2 s ( 1, 3) 2 2
31 Inverse Trace Theorem Theorem (Inverse Trace Theorem.) The trace operator γ D : H s (Ω) H s 1 2 (Γ) has a continuous right inverse operator E : H s 1 2 (Γ) H s (Ω) satisfying (γ D E)(w) = w for all w H s 1 2 (Γ). There exists a constant C I > 0 such that Ew H s (Ω) C I w H s 1 2 (Γ), for all w H s 1 2 (Γ).
32 Inverse Trace Theorem γ D : H s (Ω) H s 1 2 (Γ) is surjective and its continuous right inverse E : H s 1 2 (Γ) H s (Ω) is injective. With the two last Theorems, we can redefine the Sobolev space H s (Γ). For s > 0. H s (Γ) can be seen as the space of traces of H s+ 1 2 (Ω). The interest of fractional Sobolev spaces comes from the use of the Green s theorems in BEM.
33 Inverse Trace Theorem GREEN S FORMULA AND FUNCTIONS - Fundamental solutions
34 Green s Theorem Theorem 1 For u H 2 (Ω) and v H 1 (Ω), the first Green s formula is u, v Ω = u, v Ω + γ D u n, γ Dv Ω. 2 For u, v H 2 (Ω), the second Green s formula is u, v Ω v, u Ω = γ D u n, γ Dv Ω γ D v n, γ Du Ω.
35 Fundamental solution Let us consider a scalar partial differential equation (Lu)(x) = f (x), x Ω R d. Definition A fundamental solution of the PDE is the solution of (L y G(x, y))(x, y) = δ 0 (y x), x, y R d, in the distributional sense. Green s function of a PDE is a fundamental solution satisfying the boundary conditions. Green s functions are distributions.
36 Example of Green s function Example Compute the Green s function G(x, y) such that u(x) = 1 0 G(x, y)f (y)dy, for x (0, 1), is the unique solution of the Dirichelet BVP { u (x) = f (x), for x (0, 1) u(0) = u(1) = 0.
37 Example of Green s function Solve G (x, y) = δ(x, y) and split (0, 1) into 0 < y < x and x < y < 1. G = 0 on each side { a 1 (x)y + b 1 (x), y (0, x) G(x, y) = a 2 (x)y + b 2 (x), y (x, 1) Boundary conditions b 1 (x) = 0 and b 2 (x) = a 2 (x). ϕ D([0, 1]) solve G (x, ), ϕ = ϕ(x) G(x, ), ϕ = ϕ(x). G(x, y) = { (1 x)y, y (0, x) x(1 y), y (x, 1)
38 Fundamental solution of Laplace operator Let us consider the Laplace operator (Lu)(x) : = u(x) for x R d, d = 2, 3. The fundamental solution G(x, y) is the distributional solution of the PDE y G(x, y) = δ 0 (y x) for x, y R d. First put G(x, y) = v(z), where z = y x. Hence solve v(z) = δ 0 (z), z R d.
39 Fundamental solution of Laplace operator We apply the Fourier transformation and obtain ˆv(ξ) = 1 (2π) d 2 1 ξ 2 S (R d ) Definition For a distribution T S (R d ), the Fourier transform is given by ˆT (ϕ) = T ( ˆϕ), for all ϕ S(R d ). Hence, we have to solve ˆv, ϕ L 2 (R d ) = v, ˆϕ L 2 (R d ), for ϕ S(R d ).
40 3D case For d = 3. The fundamental solution is G(x, y) = 1 1 4π x y, x, y R3. G(x, y) is continuous (C ) except when x = y. The fundamental solution is bounded at the infinity. When x or y G(x, y) 0. G(x, y) / L 2 (R 3 ). Consider v(z) = 1 z v(z) L 2 (R 3 )= 1 z 2 dz = 0 1 z dz }{{ 2 } + 1 z 2 dz 0 } {{ }
41 2D case For d = 2. The fundamental solution is G(x, y) = 1 2π log( x y ), x, y R2. G(x, y) C except on x = y. The fundamental solution is unbounded at the infinity. When x or y U(x, y). The problem of at the infinity is solved with the boundary conditions in the Green function.
42 2D case The fundamental solution for the Helmolzt equation u(x) 2ku(x) = 0 for x R d, k R, is for d = 3 G k (x, y) = 1 e ik x y 4π x y, x, y R3. for d = 2 G k (x, y) = 1 2π Y 0(k x y ), x, y R 3, where Y 0 = second Bessel function of order zero.
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