Explicit variational forms for the inverses of integral operators - PowerPoint PPT Presentation

Explicit variational forms for the inverses of integral operators for the Laplace equation in the exterior of a flat disk in R 3 J.C. NEDELEC in collaboration with Pedro Ramaciotti Centre de Mathématiques Appliquées, Ecole Polytechnique, Palaiseau, France J.C. Nédélec (CMAP ) RICAM Workshop 2016 November 8, 2016 1 / 56

Contents Log-Kernel 1 The disc in R 3 2 Hilbert Spaces for a disc The unit sphere in R 3 and its equatorial disc Notations Traces Weighted Sobolev spaces Dirichlet Problems Average and jump decomposition Neumann problems The potential operators associated to the Laplace equations 3 Symmetric problem and weakly singular operator Antisymmetric problem and hypersingular operator Decomposition on basis functions 4 Spherical Harmonics and Associated Legendre functions Operators on the disc Images of the Spherical Harmonics Decomposition on basis functions Operators associated to the Laplace equation Expression of the kernels J.C. Nédélec (CMAP ) RICAM Workshop 2016 November 8, 2016 2 / 56

Abstract We introduce variational formulations for the weakly- and hyper-singular operators (as well as for their corresponding inverses) associated to the Laplace operator in the domain of R 3 exterior to a flat open disk in R 3 . Using adequate basis functions on the disk, we obtain an exact expression for the associated kernels. This work is an extension to R 3 of the article by Jerez-Hanckes and Nédélec (2012, Explicit variational forms for the inverses of integral logarithmic operators over an interval ([3])). J.C. Nédélec (CMAP ) RICAM Workshop 2016 November 8, 2016 3 / 56

Log-Kernel Log-Kernel Consider first the isotropic space R 2 divided into two half-planes: � � x ∈ R 2 : x 2 ≶ 0 π ± := (1) with interface Γ given by the line x 2 = 0. The interface is further divided into the open disjoint segments Γ m := ( − 1 , 1 ) × { 0 } and Γ f := Γ \ ¯ Γ m . Consequently, we have defined the domain Ω := R 2 \ ¯ Γ m . We seek u such that � − ∆ u = 0 x ∈ Ω for (2) g ∈ H 1 / 2 (Γ m ) . u = g for x ∈ Γ m ; with Then, the potential u can be represented as a single layer potential: � u ( x ) = L 1 ϕ = 1 1 | x − y | ϕ ( y ) d y , x ∈ Ω , log for (3) π Γ m Then ϕ is the solution of the logarithmic integral equation: � g ( x ) = 1 1 | x − y | ϕ ( y ) d y x ∈ Γ . log for (4) π Γ m J.C. Nédélec (CMAP ) RICAM Workshop 2016 November 8, 2016 4 / 56

Log-Kernel The equation (4) has a variational formulation in the space � H − 1 / 2 (Γ m ) which is: 0 � � � 1 1 g ( τ ) ϕ t ( τ ) d τ, ∀ ϕ t ∈ � H − 1 / 2 | τ − t | ϕ ( t ) ϕ t ( τ ) dtd τ = log (Γ m ) (5) 0 π Γ m Γ m Γ m This operator is a bijection between � H − 1 / 2 (Γ m ) and the space H 1 / 2 (Γ m ) of ∗ � 0 1 functions in H 1 / 2 (Γ m ) satisfying √ 1 − t 2 g ( t ) dt = 0 . and we have Γ m � � 1 1 | τ − t | ϕ ( t ) ϕ ( τ ) dtd τ ≥ C � ϕ � 2 (Γ m ) , ∀ ϕ ∈ � H − 1 / 2 log (Γ m ) . (6) H − 1 / 2 � 0 π 0 Γ m Γ m The inverse operator is a bijection of H 1 / 2 (Γ m ) onto � H − 1 / 2 (Γ m ) . This operator ∗ 0 N 1 is symmetric and coercive in the space H 1 / 2 (Γ m ) . It admits two variational ∗ formulations. Let M ( x , y ) be the function � � 2 � �� � M ( x , y ) = 1 ( y − x ) 2 + 1 − x 2 + 1 − y 2 (7) 2 � M ( x , y ) � � L 2 g = 1 log g ( y ) dy (8) π | x − y | Γ m J.C. Nédélec (CMAP ) RICAM Workshop 2016 November 8, 2016 5 / 56

Log-Kernel The first one is: � � � M ( x , y ) � � � � ′ dydx = ( N 1 g , g t ) = 1 g ′ ( x ) g t ( y ) ϕ ( x ) g t ( x ) dx log (9) π | x − y | Γ m Γ m Γ m for all g t ∈ H 1 / 2 (Γ m ) , which gives a first norm on the space H 1 / 2 (Γ m ) : ∗ ∗ � M ( x , y ) � � � 1 g ′ ( x ) g ′ ( y ) dy dx ≥ C � g � 2 (Γ m ) ; ∀ g ∈ H 1 / 2 log (Γ m ) (10) H 1 / 2 ∗ | x − y | π ∗ Γ m Γ m The second one is � � � M ( x , y ) � � d 2 � � 1 g t ( x ) − g t ( y ) ϕ ( x ) g t ( x ) dx ( g ( x ) − g ( y )) dydx = dxdy log (11) 2 π | x − y | Γ m Γ m Γ m for all g t ∈ H 1 / 2 (Γ m ) , ∗ So we have a second norm on the space H 1 / 2 (Γ m ) which is: ∗ � 1 − xy � ( g ( x ) − g ( y )) 2 � � 1 dydx ≥ C � g � 2 Γ m ) , ∀ g ∈ H 1 / 2 ∗ ( Γ m ) (12) H 1 / 2 ( x − y ) 2 ∗ ( 2 π w ( x ) w ( y ) Γ m Γ m where the weight function w is given by � w ( x ) := 1 − x 2 for x ∈ ( − 1 , 1 ) . (13) J.C. Nédélec (CMAP ) RICAM Workshop 2016 November 8, 2016 6 / 56

Log-Kernel We can also consider the Neumann problem � − ∆ u = 0 x ∈ Ω for (14) m ∂ n u = γ − γ + ϕ ∈ H − 1 / 2 (Γ m ) m ∂ n u = ϕ for x ∈ Γ m , which can be represent as a double layer potential of harmonic solution in the domain Ω of the form . � u ( x ) = 1 x 2 | x − y | 2 α ( y ) d y , for x ∈ Ω , (15) π Γ m Then the unknown α is the solution of the hyper singular integral equation: � ϕ ( x ) = N 2 α = 1 1 | x − y | 2 α ( y ) dy for x ∈ Γ . (16) π Γ m where α is also the jump of the Dirichlet trace of the solution of problem (14). J.C. Nédélec (CMAP ) RICAM Workshop 2016 November 8, 2016 7 / 56

Log-Kernel A variational formulation of the integral equation (16) in the space � H 1 / 2 (Γ m ) is � � � 1 1 ϕ ( τ ) α t ( τ ) d τ, ∀ α t ∈ � ′ dtd τ = | τ − t | α ′ ( t )( α t ( τ )) H 1 / 2 (Γ m ) log (17) π Γ m Γ m Γ m The associated operator � D is a bijection from � H 1 / 2 (Γ m ) to H − 1 / 2 (Γ m ) . Moreover, this bilinear form is coercive, i.e., � � 1 1 | τ − t | α ′ ( t ) α ( τ ) ′ dtd τ ≥ C � α � 2 H 1 / 2 (Γ m ) , ∀ α ∈ � H 1 / 2 (Γ m ) . log (18) � π Γ m Γ m This operator admits a second variational formulation which is � � � � � � α t ( x ) − α t ( y ) ( α ( x ) − α ( y )) α ( x ) α t ( x ) 1 dxdy + 1 ϕ ( x ) α t ( x ) dx (19) dx = | x − y | 2 1 − x 2 2 π π Γ m Γ m Γ m Γ m for all α t ∈ � H 1 / 2 (Γ m ) , and the next expression is a norm on � H 1 / 2 (Γ m ) � � � ( α ( x ) − α ( y )) 2 α ( x ) 2 1 dxdy + 1 1 − x 2 dx ≥ C � α � 2 H 1 / 2 (Γ m ) , ∀ α ∈ � H 1 / 2 (Γ m ) (20) � 2 π | x − y | 2 π Γ m Γ m Γ m J.C. Nédélec (CMAP ) RICAM Workshop 2016 November 8, 2016 8 / 56

Log-Kernel The inverse operator is a bijection of H − 1 / 2 (Γ m ) onto � H 1 / 2 (Γ m ) . The associated operator is symmetric and coercive in the space H − 1 / 2 (Γ m ) . It admits the following variational formulation: � � � M ( x , y ) � � 1 ϕ ( x ) ϕ t ( y ) dydx = α ( x ) ϕ t ( x ) dx , ∀ ϕ ∈ H − 1 / 2 (Γ m ) log (21) π | x − y | Γ m Γ m Γ m and thus the following expression is a norm on the space H − 1 / 2 (Γ m ) � � � M ( x , y ) � 1 ϕ ( x ) ϕ ( y ) dydx ≥ C � ϕ � 2 1 / 2 (Γ m ) , ∀ ϕ ∈ H − 1 / 2 (Γ m ) log (22) H − π | x − y | Γ m Γ m J.C. Nédélec (CMAP ) RICAM Workshop 2016 November 8, 2016 9 / 56

Log-Kernel The operators L 1 , L 2 , N 1 , N 2 , D , D ∗ are linked by the identities L 2 ◦ N 2 = − L 2 ◦ D ∗ ◦ L 1 ◦ D = I , I ∈ � H 1 / 2 (Γ m ) L 1 ◦ N 1 = − L 1 ◦ D ◦ L 2 ◦ D ∗ = I , I ∈ H 1 / 2 (Γ m ) ∗ N 1 ◦ L 1 = − D ◦ L 2 ◦ D ∗ ◦ L 1 = I , H − 1 / 2 I ∈ � (Γ m ) 0 N 2 ◦ L 2 = − D ∗ ◦ L 2 ◦ D ◦ L 1 = I , I ∈ H − 1 / 2 (Γ m ) H 1 / 2 (Γ m ) into H 1 / 2 L 1 ◦ D is continuous and invertible from � (Γ m ) . ∗ L 2 ◦ D ∗ is continuous and invertible from H 1 / 2 (Γ m ) into � H 1 / 2 (Γ m ) . ∗ D ∗ ◦ L 1 is continuous and invertible from � H − 1 / 2 (Γ m ) into H − 1 / 2 (Γ m ) . 0 D ◦ L 2 is continuous and invertible from H − 1 / 2 (Γ m ) into � H − 1 / 2 (Γ m ) . 0 The Dirichlet and Neumann Laplacian ∆ D , ∆ N are linked to L 1 , L 2 and N 1 , N 2 : L 1 = ( − ∆ D ) − 1 1 2 ; 2 ; − N 1 = ( − ∆ D ) L 2 = ( − ∆ N ) − 1 1 2 ; 2 . − N 2 = ( − ∆ N ) J.C. Nédélec (CMAP ) RICAM Workshop 2016 November 8, 2016 10 / 56

The disc in R 3 Hilbert Spaces for a disc The disc in R 3 We try now to extend these results to the unit disc in R 3 . We Introduce the splitting of the space R 3 into two half-spaces � � x ∈ R 3 : x 3 ≷ 0 π ± := , by the plane x 3 = 0 that will be denote as Γ . Let c be the circle of center at the origin and of radius 1 in the plane Γ . Let D be the plane disc delimitated by the circle c and D the associated flat domain in R 3 . Now its complement in R 2 , is Γ f := Γ \ ¯ D . Henceforth, the problem domain is denoted by Ω := R 3 \ ¯ D . We also consider the sphere S of radius 1 and center at the origin in R 3 . The disc D divide this sphere into two half-sphere that we denote respectively S + and S − . J.C. Nédélec (CMAP ) RICAM Workshop 2016 November 8, 2016 11 / 56

The disc in R 3 The unit sphere in R 3 and its equatorial disc The unit sphere in R 3 and its equatorial disc We consider the unit sphere S in R 3 (Fig. 1) and the spherical coordinates: ( r , θ, ϕ ) , where r is the radius and θ, ϕ the two Euler angles.  x 1 = r sin θ cos ϕ,  x 2 = r sin θ sin ϕ, (25) x 3  x 3 = r cos θ. ! � e ' M ! � e � � O x 2 ' m x 1 Fig. 1: Spherical coordinates The vectors e θ and e ϕ are unitary. The vector e ρ directed along Om is unitary. J.C. Nédélec (CMAP ) RICAM Workshop 2016 November 8, 2016 12 / 56