Fokas Methods applied to a Boundary Valued Problem for Conjugate - PowerPoint PPT Presentation

Fokas Methods applied to a Boundary Valued Problem for Conjugate Conductivity Equations Conference in memory of Gennadi Henkin Joint work with Slah Chaabi and Franck Wielonsky (Universit´ e Aix-Marseille) Moscow, the 15th of september, 2016

Different motivations. ◮ Ω is a smooth domain in C .

Different motivations. ◮ Ω is a smooth domain in C . ◮ L an elliptic differential operator in Ω L = a ( x , y ) ∂ 2 ∂ x 2 + b ( x , y ) ∂ 2 ∂ x ∂ y + c ( x , y ) ∂ 2 ∂ y 2 + + d ( x , y ) ∂ ∂ x + e ( x , y ) ∂ ∂ y + f ( x , y )

Different motivations. ◮ Ω is a smooth domain in C . ◮ L an elliptic differential operator in Ω L = a ( x , y ) ∂ 2 ∂ x 2 + b ( x , y ) ∂ 2 ∂ x ∂ y + c ( x , y ) ∂ 2 ∂ y 2 + + d ( x , y ) ∂ ∂ x + e ( x , y ) ∂ ∂ y + f ( x , y ) ◮ Lax-Milgram theorem gives the existence and unicity of the solution to Lu = 0 if u is given on ∂ Ω (and continuous).

Different motivations. ◮ Ω is a smooth domain in C . ◮ L an elliptic differential operator in Ω L = a ( x , y ) ∂ 2 ∂ x 2 + b ( x , y ) ∂ 2 ∂ x ∂ y + c ( x , y ) ∂ 2 ∂ y 2 + + d ( x , y ) ∂ ∂ x + e ( x , y ) ∂ ∂ y + f ( x , y ) ◮ Lax-Milgram theorem gives the existence and unicity of the solution to Lu = 0 if u is given on ∂ Ω (and continuous). ◮ Construction of fundamental solutions gives a representation formula in terms of u and the normal derivative ∂ � n u on the boundary ∂ Ω : we need to much informations.

Different motivations. ◮ Ω is a smooth domain in C . ◮ L an elliptic differential operator in Ω L = a ( x , y ) ∂ 2 ∂ x 2 + b ( x , y ) ∂ 2 ∂ x ∂ y + c ( x , y ) ∂ 2 ∂ y 2 + + d ( x , y ) ∂ ∂ x + e ( x , y ) ∂ ∂ y + f ( x , y ) ◮ Lax-Milgram theorem gives the existence and unicity of the solution to Lu = 0 if u is given on ∂ Ω (and continuous). ◮ Construction of fundamental solutions gives a representation formula in terms of u and the normal derivative ∂ � n u on the boundary ∂ Ω : we need to much informations. ◮ Is it possible to obtain a relation between u and ∂ � n u on ∂ Ω without computing u in Ω ?

One other motivation : What is a Tokamak ?

Mathematical Formulation

Mathematical Formulation ◮ Poloidal field ψ and its normal derivative ∂ � n ψ are known on ∂ D ( a , r ), ψ = C on the boundary ∂ P of the plasma and � 1 � div x ∇ ψ = 0 in D ( a , r ) \ P

Mathematical Formulation ◮ Poloidal field ψ and its normal derivative ∂ � n ψ are known on ∂ D ( a , r ), ψ = C on the boundary ∂ P of the plasma and � 1 � div x ∇ ψ = 0 in D ( a , r ) \ P ◮ Generalized axisymetrical potential for α ∈ R : ∆ u + α ∂ u div( x α ∇ u ) = 0 ⇔ ∂ x = 0 . x

The Fokas method Fokas (1997) : a simple example. ◮ q t − q xx = 0 sur x ≥ 0 , t ≥ 0 ( H )

The Fokas method Fokas (1997) : a simple example. ◮ q t − q xx = 0 sur x ≥ 0 , t ≥ 0 ( H ) ◮ Formulation in terms of Lax pair : splitting of this equation into two ordinary differential equations, compatible if and only if q is a solution to ( H ) � µ x − ik µ = q µ t − µ xx = 0

The Fokas method Fokas (1997) : a simple example. ◮ q t − q xx = 0 sur x ≥ 0 , t ≥ 0 ( H ) ◮ Formulation in terms of Lax pair : splitting of this equation into two ordinary differential equations, compatible if and only if q is a solution to ( H ) � µ x − ik µ = q µ t − µ xx = 0 ◮ � µ x − ik µ = q µ t + k 2 µ = q x + ikq

The Fokas method Fokas (1997) : a simple example. ◮ q t − q xx = 0 sur x ≥ 0 , t ≥ 0 ( H ) ◮ Formulation in terms of Lax pair : splitting of this equation into two ordinary differential equations, compatible if and only if q is a solution to ( H ) � µ x − ik µ = q µ t − µ xx = 0 ◮ � µ x − ik µ = q µ t + k 2 µ = q x + ikq  � µ e − ikx + k 2 t � ◮ qe − ikx + k 2 t  = � µ e − ikx + k 2 t � x  ( q x + ikq ) e − ikx + k 2 t = t

The Fokas method Fokas (1997) : a simple example. ◮ q t − q xx = 0 sur x ≥ 0 , t ≥ 0 ( H ) ◮ Formulation in terms of Lax pair : splitting of this equation into two ordinary differential equations, compatible if and only if q is a solution to ( H ) � µ x − ik µ = q µ t − µ xx = 0 ◮ � µ x − ik µ = q µ t + k 2 µ = q x + ikq  � µ e − ikx + k 2 t � ◮ qe − ikx + k 2 t  = � µ e − ikx + k 2 t � x  ( q x + ikq ) e − ikx + k 2 t = t � µ e − ikx + k 2 t � ◮ = e − ikx + k 2 t ( qdx + ( q x + ikq ) dt ) d

◮ ν = e − ikx + k 2 t ( qdx + ( q x + ikq ) dt ) is closed

◮ ν = e − ikx + k 2 t ( qdx + ( q x + ikq ) dt ) is closed ◮ q ( x , t ) and q x ( x , t ) decay to 0 at ∞ sufficiently fast.

◮ ν = e − ikx + k 2 t ( qdx + ( q x + ikq ) dt ) is closed ◮ q ( x , t ) and q x ( x , t ) decay to 0 at ∞ sufficiently fast. � 5 π � 4 , 7 π ◮ arg k ∈ 4 � ∞ � ∞ e k 2 t ( q x (0 , t ) + ikq (0 , t )) dt ( GR ) e − ikx q ( x , 0) dx = x =0 t =0

◮ ν = e − ikx + k 2 t ( qdx + ( q x + ikq ) dt ) is closed ◮ q ( x , t ) and q x ( x , t ) decay to 0 at ∞ sufficiently fast. � 5 π � 4 , 7 π ◮ arg k ∈ 4 � ∞ � ∞ e k 2 t ( q x (0 , t ) + ikq (0 , t )) dt ( GR ) e − ikx q ( x , 0) dx = x =0 t =0 ◮ Let us assume first that q ( x , 0) , q x (0 , t ) , q (0 , t ) are known. Reconstruction of q in R + × R + is equivalent to the reconstruction of µ since d ( µ e − ikx + k 2 t ) = ν.

◮ ν = e − ikx + k 2 t ( qdx + ( q x + ikq ) dt ) is closed ◮ q ( x , t ) and q x ( x , t ) decay to 0 at ∞ sufficiently fast. � 5 π � 4 , 7 π ◮ arg k ∈ 4 � ∞ � ∞ e k 2 t ( q x (0 , t ) + ikq (0 , t )) dt ( GR ) e − ikx q ( x , 0) dx = x =0 t =0 ◮ Let us assume first that q ( x , 0) , q x (0 , t ) , q (0 , t ) are known. Reconstruction of q in R + × R + is equivalent to the reconstruction of µ since d ( µ e − ikx + k 2 t ) = ν. ◮ Integration of ν between points of the boundary and ( x , t ).

◮ � µ 1 ( x , t , k ) = e ikx − k 2 t ν γ 1

◮ � µ 1 ( x , t , k ) = e ikx − k 2 t ν γ 1 � µ 2 ( x , t , k ) = e ikx − k 2 t ν γ 2

◮ � µ 46 ( x , t , k ) = e ikx − k 2 t ν γ 1 � µ 5 ( x , t , k ) = e ikx − k 2 t ν γ 2

◮ � µ 46 ( x , t , k ) = e ikx − k 2 t ν γ 46 � µ 5 ( x , t , k ) = e ikx − k 2 t ν γ 5 � µ 123 ( x , t , k ) = e ikx − k 2 t ν γ 123

Riemann-Hilbert Problems ◮

Riemann-Hilbert Problems � 1 � ◮ µ ( x , t , k ) = O ◮ k

Riemann-Hilbert Problems � 1 � ◮ µ ( x , t , k ) = O ◮ k ◮ µ + − µ − = φ

Riemann-Hilbert Problems � 1 � ◮ µ ( x , t , k ) = O ◮ k ◮ µ + − µ − = φ � φ ( k ′ ) dk ′ 1 ◮ µ = (Plemelj Formula) k ′ − k 2 π i L

Lax Pairs and closed differential form for the GASP equation. ◮ L α ( u ) = ∆ u + α ∂ u ∂ x = 0 in H = { z , Re z > 0 } . x

Lax Pairs and closed differential form for the GASP equation. ◮ L α ( u ) = ∆ u + α ∂ u ∂ x = 0 in H = { z , Re z > 0 } . x α ◮ u z ¯ z + z )( u z + u ¯ z ) = 0 . (GASP) 2( z + ¯

Lax Pairs and closed differential form for the GASP equation. ◮ L α ( u ) = ∆ u + α ∂ u ∂ x = 0 in H = { z , Re z > 0 } . x α ◮ u z ¯ z + z )( u z + u ¯ z ) = 0 . (GASP) 2( z + ¯  z ) α/ 2 ( k − z ) α/ 2 − 1 u z ( z ) , φ z ( z , k ) = ( k + ¯  ◮  z ) α/ 2 − 1 ( k − z ) α/ 2 u ¯ φ ¯ z ( z , k ) = ( k + ¯ z ( z )

Lax Pairs and closed differential form for the GASP equation. ◮ L α ( u ) = ∆ u + α ∂ u ∂ x = 0 in H = { z , Re z > 0 } . x α ◮ u z ¯ z + z )( u z + u ¯ z ) = 0 . (GASP) 2( z + ¯  z ) α/ 2 ( k − z ) α/ 2 − 1 u z ( z ) , φ z ( z , k ) = ( k + ¯  ◮  z ) α/ 2 − 1 ( k − z ) α/ 2 u ¯ φ ¯ z ( z , k ) = ( k + ¯ z ( z ) ◮ W ( z , k ) = � � α/ 2 − 1 � � ( k − z )( k + ¯ z ) ( k + ¯ z ) u z ( z ) dz + ( k − z ) u ¯ z ( z ) d ¯ z Note that, when α ∈ 2 N ∗ , the differential form has no singularity in Ω and k may be any complex number. Otherwise, for α ∈ R \ 2 N ∗ , W ( z , k ) has a pole or a branching point in k or − ¯ k if one of this point is in Ω.

Lax Pairs and closed differential form for the GASP equation. ◮ L α ( u ) = ∆ u + α ∂ u ∂ x = 0 in H = { z , Re z > 0 } . x α ◮ u z ¯ z + z )( u z + u ¯ z ) = 0 . (GASP) 2( z + ¯  z ) α/ 2 ( k − z ) α/ 2 − 1 u z ( z ) , φ z ( z , k ) = ( k + ¯  ◮  z ) α/ 2 − 1 ( k − z ) α/ 2 u ¯ φ ¯ z ( z , k ) = ( k + ¯ z ( z ) ◮ W ( z , k ) = � � α/ 2 − 1 � � ( k − z )( k + ¯ z ) ( k + ¯ z ) u z ( z ) dz + ( k − z ) u ¯ z ( z ) d ¯ z Note that, when α ∈ 2 N ∗ , the differential form has no singularity in Ω and k may be any complex number. Otherwise, for α ∈ R \ 2 N ∗ , W ( z , k ) has a pole or a branching point in k or − ¯ k if one of this point is in Ω. ◮ L α ( u ) = 0 ⇔ L 2 − α ( x α − 1 u ) = 0.

α = − 2 m , m ∈ N . ◮ W ( z , k ) = ( k + z ) u z dz + ( k − z ) u z dz [( k − z )( k + z )] m +1

α = − 2 m , m ∈ N . ◮ W ( z , k ) = ( k + z ) u z dz + ( k − z ) u z dz [( k − z )( k + z )] m +1 � ◮ W ( z ′ , k ) φ ( z , k ) =