Simultaneous Reconstruction of Coefficients and Source Parameters in Elliptic Systems Modelled with Many Boundary Values Problems Nilson C. Roberty Nuclear Engineering Program Federal University of Rio de Janeiro-Brazil
The practical expression of linear elliptic partial differential equations found in most of the engineering application is represented by the following system, in which the fields may be a vector and coefficients can be represented by matrices and vectors according: To find u ( x ) such that ∇ . ( − c ∇ u − α u + γ ) + β. ∇ u + au = f if x ∈ Ω; hu = g if x ∈ ∂ Ω D ; (1) ν. ( c ∇ u + α u − γ ) + qu = g ν − h ∗ µ if x ∈ ∂ Ω N ; where ν is the outward unit normal vector on ∂ Ω := ∂ Ω D ∪ Π ∪ ∂ Ω N ,
◮ Boundary integral formulation for the inverse problem: Let G ξ be the fundamental solution for the strongly elliptic system (1). ◮ Then the Calder´ on projector gap is: � γ u ( ξ ) � � � � Ω γ ξ [ G ξ ]( y ) f ( y , u ) dy = + � B ν u ( ξ ) Ω B ν ξ [ G ξ ]( y ) f ( y ) dy � � γ u ( ξ ) � 1 � 2 ( I x → ξ − T x → ξ ) S x → ξ , ( x , ξ ) ∈ Γ × Γ 1 R x → ξ 2 ( I x → ξ + T x → ξ ) B ν u ( ξ )
◮ Boundary integral formulation for the inverse problem: Let v be the regular fundamental solution for the strongly elliptic system (1). ◮ Variational : for all v ∈ ( H L ∗ (Ω) m ) ∗ . d d � � � � v ( x ) f ( x ) dx = − ( ν j A jk ∂ k v ( x )+ A j v ( x )) u ( x ) d σ x Ω Γ j =1 k =1 d d � � � + v ( x ) A jk ∂ k u ( x ) d σ x ν j Γ j =1 k =1
◮ Direct problem (Closed operators): Variational formulation and stabilization via Babuska-Brezzi-Necas-Banach condition. ◮ Inverse problem (Compact operators): Variational or strong formulation and stabilization thorough Picard-Tikhonov-Landweber-Morozov-Banach regularization. ◮ Functional Analysis Framework: Closed Range Banach Theorem and Fredholm Operator Theory. ◮ Solution of Direct problems with minimization of Least square discrepancy between calculated and measured Neumann data. ◮ Many solutions of two equivalent direct problems and minimization of the discrepancy between these two solutions.(This work !)
◮ Given The Dirichlet to Newmann map 1 2 ( ∂ Ω) → H − 1 2 ( ∂ Ω) Λ c : H ◮ To find u ∈ H 1 (Ω) , c ( x ) ∈ L ∞ (Ω) such that � ∇ . ( − c ∇ u ) = 0 in Ω; (2) g ν = Λ c [ g ] on ∂ Ω; ◮ Q c = Ω c ( x ) ||∇ u ( x ) || 2 dx = � � ∂ Ω g ( x )Λ c [ g ]( x ) d σ ( x )
◮ In 1980, (Seminar on Numerical Analysis and its Applications to Continuum Physics, SBM, Rio de Janeiro), Calder´ on posed the following problem: ◮ Decide whether c is uniquely determined by Q c , and, if so, calculate c in terms of Q c . ◮ the uniqueness problem, for conductivity or any other coefficients, is an open problem that has been only partially solved. ◮ ”Intrinsic non uniqueness in the inverse source problem”; ◮ The Calderon Projector Gap is the same for all Cauchy datum used to estimate it.
◮ Some uniqueness class ◮ The regular affine class ”If the sources are restricted to the affine class of functions C ( D ; F ) = { f ∈ H 2 (Ω) : Df = F } then we have uniqueness of the associated inverse source problem (Alves, Martins, Roberty, Cola¸ co, Olander, 2007 ); ◮ The characteristic class f = F χ ω ( P. Novikov, 1938, Isakov, 1990); ◮ The mono and dipolar source class A = { f := � m 1 j =1 λ j δ x j + � m 1 j =1 p j · ∇ δ x j (El Badia e Ha Duong, 2000).
Lemma Let V , W Banach spaces and V ′ , W ′ its respective dual. Let A ∈ L ( V , W ) and A T ∈ L ( W ′ , V ′ ) its transposed. Let ker ( A ) , ker ( A T ) , im ( A ) and im ( A T ) denotes its respective kernel and range. For M ⊂ V and N ⊂ W ′ , let M ⊥ := { v ′ ∈ V ′ |∀ m ∈ M , � v ′ , m � V ′ × V = 0 } and N ⊥ := { v ∈ V |∀ n ∈ N , � n , v � V ′ × V = 0 } . Then, the following properties hold: ◮ ker ( A ) = ( im ( A T )) ⊥ and ker ( A T ) = ( im ( A )) ⊥ ; ◮ im ( A ) = ( ker ( A T )) ⊥ and im ( A T ) = ( ker ( A )) ⊥
We first apply this Fundamental Lemma to the operator 0 (Ω)) m → ( L 2 (Ω)) m L 0 : ( H 2 0 (Ω)) m := { v ∈ ( H 2 (Ω)) m | v | ∂ Ω = 0; B ν v | ∂ Ω = 0 } has where ( H 2 transpose 0 : ( L 2 (Ω)) m → ( H − 2 (Ω)) m L ∗ Note that in this case, by (ii), 0 (Ω) ) m = ker ( L ∗ 0 ) = ( im ( L 0 )) ⊥ = ( L 0 (( H 2 0 (Ω)) m )) ⊥ and, by ( H L ∗ (iii) L 2 (Ω) ( L 2 (Ω)) m = ( H L ∗ 0 (Ω)) m ⊕ L 0 (( H 2 0 (Ω) m )) (3)
Now we define the following space D , N (Ω)) m := { v ∈ ( H 2 (Ω)) m | v | Γ D = 0; B| Γ N v = 0 } ( H 2 where Γ D ⊂ ∂ Ω and Γ N ⊂ ∂ Ω are arbitrary. Note that 0 (Ω)) m ⊂ ( H 2 H 2 D , N (Ω)) m D , N (Ω)) m → ( L 2 (Ω)) m has transpose The operator L 0 , D , N : ( H 2 0 , D c , N c : ( L 2 (Ω)) m → (( H 2 D , N (Ω)) m ) ∗ = ( H − 2 D c , N c (Ω)) m ⊂ L ∗ ( H − 2 (Ω)) m where ( H − 2 D c , N c (Ω)) m is a set of distribution with trace support in ∂ Ω \ Γ D and conormal trace with support in ∂ Ω \ Γ N and whose kernel is 0 , Dc , Nc (Ω)) m = { v ∈ ( L 2 (Ω)) m |L ∗ ( H L ∗ 0 , D c , N c v = 0 }
0 , Dc , Nc (Ω) ) m = ker ( L ∗ Note that in this case, by (ii), ( H L ∗ 0 , D c , N c ) = ( im ( L 0 , D , N )) ⊥ = ( L 0 (( H 2 D , N (Ω))) m ) ⊥ and, by (iii) L 2 (Ω) ( L 2 (Ω)) m = ( H L ∗ 0 , Dc , Nc (Ω)) m ⊕ L 0 , D , N (( H 2 D , N (Ω)) m ) (4) Note that for Γ D = Γ N = Γ, decomposition (4) reduces to (3). Also that when Γ D and Γ N are a Lipschitz dissection of ∂ Ω, the fact that the unique solution v χ of L ∗ 0 , D c , N c v = χ if x ∈ Ω; P c ∗ γ [ v ] = 0 if x ∈ ∂ Ω N = ∂ Ω \ Γ D ; (5) χ, 0 , 0 ˜ B ν v = 0 if x ∈ ∂ Ω D = ∂ Ω \ Γ D ; 0 , Dc , Nc (Ω)) m = ( { 0 } ) m is the trivial when χ ( x ) = 0,and ( H L ∗
◮ Characterizes materials parameters and source is a central question in the engineering project; ◮ it is important adequate existing engineering and multiphysics software to handle uncertainties in these parameters; ◮ be used as a tool for process experimental data; ◮ but respecting the actual engineering project project status of art. ◮ Applications when we have incomplete information about these coefficient and sources.
◮ This work is addressed to investigate the class of problems in which we want determine unknown parameters in the functions that characterize these coefficients and sources. ◮ To compensate this incomplete information that ill-posed the problem, we suppose that both, Neumann and Dirichlet data, are prescribed for many boundary value problems. ◮ These problems are formulated for the same physical coefficients and source which depend on the same set of unknown parameters.
◮ L u = − � d j =1 ( � d k =1 ∂ j ( A jk ∂ k ) u + A j ∂ j u ) + Au ◮ ( A jk , A j , A ) : Ω → R m × m . ◮ u is a column vector with m scalar fields and L u : Ω → R m ◮ strongly elliptic system. d d � � L 0 u = − ∂ j B j u where B j = A jk ∂ k (6) j =1 k =1 ◮ Ω is a Lipschitz domain and γ is the trace operator ◮ the conormal derivative is d � B ν u = ν j γ [ B j u ] (7) j =1
Let Ω a domain with Lipschitz dissection boundary ∂ Ω = ∂ Ω N ∪ Π ∪ ∂ Ω N . The mixed boundary value problem for the physical model given by (1) is given by the well posed problem P f , g D , g N : To find u ∈ H 1 (Ω) m such that L u = f if x ∈ Ω; γ [ u ] = g D if x ∈ ∂ Ω D ; P f , g D , g N (8) B ν u = g N if x ∈ ∂ Ω N ; we can show that (8) has the following weak formulation W f , g D , g N ( L u , v ) Ω + ( B ν u , γ [ v ]) ∂ Ω = Φ( u , v ) = if v ∈ H 1 D (Ω) m ; = ( f , v ) Ω + ( g N , γ [ v ]) ∂ Ω N (9) γ [ u ] = g D if x ∈ ∂ Ω D ;
Definition When u = u + + u − ∈ L 2 ( R d ) m , with u ± ∈ H 1 (Ω ± ) m , has compact support in R d and f = f + + f − ∈ H − 1 ( R d ) m , we can enunciate the Third Green Identity u = G f + DL [ u ] Γ − SL [ B ν u ] Γ on R d . (10)
Definition When the mixed boundary value problem is posed with a non null source, P f , g D , g N ν , we have a gap in the Calder´ on projector: � γ u ( ξ ) � � � � � = Ω γ ξ [ G ξ ]( y ) f ( y ) dy Ω B ν ξ [ G ξ ]( y ) f ( y ) dy + B ν u ( ξ ) � � γ u ( ξ ) � 1 2 ( I x → ξ − T x → ξ ) S x → ξ � , ( x , ξ ) ∈ Γ × Γ 1 R x → ξ 2 ( I x → ξ + T x → ξ ) B ν u ( ξ )
Matrix equation for Calder´ on Projector Gap Lipschtiz Boundary Dissection: � γ u ( ξ ) | Γ D Ω γ ξ G ξ | Γ D ( y ) f ( y ) dy � γ u ( ξ ) | Γ N Ω γ x iG ξ | Γ N ( y ) f ( y ) dy = + � B ν u ( ξ ) | Γ D Ω B ν ξ G ξ | Γ D ( y ) f ( y ) dy � B ν u ( ξ ) | Γ N Ω B ν ξ G ξ | Γ N ( y ) f ( y ) dy 1 2 ( I DD x → ξ − T DD − T ND S DD S ND x → ξ ) x → ξ x → ξ x → ξ 1 − T DN 2 ( I NN x → ξ − T NxN S DN S NN x → ξ ) x → ξ x → ξ x → ξ 1 x → ξ + ˜ ˜ R DD R ND 2 ( I DD T ∗ DD T ∗ ND x → ξ ) x → ξ x → ξ x → ξ ˜ 1 x → ξ + ˜ R DN R NN T ∗ DN 2 ( I NN T ∗ NN x → ξ x → ξ x → ξ x → ξ
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