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 ∂x2 + b(x, y) ∂2 ∂x∂y + c(x, y) ∂2 ∂y2 + +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 ∂x2 + b(x, y) ∂2 ∂x∂y + c(x, y) ∂2 ∂y2 + +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 ∂x2 + b(x, y) ∂2 ∂x∂y + c(x, y) ∂2 ∂y2 + +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 ∂

nu 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 ∂x2 + b(x, y) ∂2 ∂x∂y + c(x, y) ∂2 ∂y2 + +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 ∂

nu on the

boundary ∂Ω : we need to much informations.

◮ Is it possible to obtain a relation between u and ∂ nu 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 ∂ nψ are known on

∂D(a, r), ψ = C on the boundary ∂P of the plasma and div 1 x ∇ψ

• = 0

in D(a, r) \ P

Mathematical Formulation

◮ Poloidal field ψ and its normal derivative ∂ nψ are known on

∂D(a, r), ψ = C on the boundary ∂P of the plasma and div 1 x ∇ψ

• = 0

in D(a, r) \ P

◮ Generalized axisymetrical potential for α ∈ R :

div(xα∇u) = 0 ⇔ ∆u + α x ∂u ∂x = 0.

The Fokas method Fokas (1997) : a simple example.

◮

qt − qxx = 0 sur x ≥ 0, t ≥ 0 (H)

The Fokas method Fokas (1997) : a simple example.

◮

qt − qxx = 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) µx − ikµ = q µt − µxx = 0

The Fokas method Fokas (1997) : a simple example.

◮

qt − qxx = 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) µx − ikµ = q µt − µxx = 0

◮

• µx − ikµ = q

µt + k2µ = qx + ikq

The Fokas method Fokas (1997) : a simple example.

◮

qt − qxx = 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) µx − ikµ = q µt − µxx = 0

◮

• µx − ikµ = q

µt + k2µ = qx + ikq

◮

  

• µe−ikx+k2t

x

= qe−ikx+k2t

• µe−ikx+k2t

t

= (qx + ikq)e−ikx+k2t

The Fokas method Fokas (1997) : a simple example.

◮

qt − qxx = 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) µx − ikµ = q µt − µxx = 0

◮

• µx − ikµ = q

µt + k2µ = qx + ikq

◮

  

• µe−ikx+k2t

x

= qe−ikx+k2t

• µe−ikx+k2t

t

= (qx + ikq)e−ikx+k2t

◮

d

• µe−ikx+k2t

= e−ikx+k2t (qdx + (qx + ikq)dt)

◮ ν = e−ikx+k2t (qdx + (qx + ikq)dt) is closed

◮ ν = e−ikx+k2t (qdx + (qx + ikq)dt) is closed ◮ q(x, t) and qx(x, t) decay to 0 at ∞ sufficiently fast.

◮ ν = e−ikx+k2t (qdx + (qx + ikq)dt) is closed ◮ q(x, t) and qx(x, t) decay to 0 at ∞ sufficiently fast. ◮ arg k ∈

5π

4 , 7π 4

• ∞

x=0

e−ikxq(x, 0)dx = ∞

t=0

ek2t(qx(0, t) + ikq(0, t))dt (GR)

◮ ν = e−ikx+k2t (qdx + (qx + ikq)dt) is closed ◮ q(x, t) and qx(x, t) decay to 0 at ∞ sufficiently fast. ◮ arg k ∈

5π

4 , 7π 4

• ∞

x=0

e−ikxq(x, 0)dx = ∞

t=0

ek2t(qx(0, t) + ikq(0, t))dt (GR)

◮ Let us assume first that q(x, 0), qx(0, t), q(0, t) are known.

Reconstruction of q in R+ × R+ is equivalent to the reconstruction of µ since d(µe−ikx+k2t) = ν.

◮ ν = e−ikx+k2t (qdx + (qx + ikq)dt) is closed ◮ q(x, t) and qx(x, t) decay to 0 at ∞ sufficiently fast. ◮ arg k ∈

5π

4 , 7π 4

• ∞

x=0

e−ikxq(x, 0)dx = ∞

t=0

ek2t(qx(0, t) + ikq(0, t))dt (GR)

◮ Let us assume first that q(x, 0), qx(0, t), q(0, t) are known.

Reconstruction of q in R+ × R+ is equivalent to the reconstruction of µ since d(µe−ikx+k2t) = ν.

◮ Integration of ν between points of the boundary and (x, t).

◮

µ1(x, t, k) = eikx−k2t

• γ1

ν

◮

µ1(x, t, k) = eikx−k2t

• γ1

ν µ2(x, t, k) = eikx−k2t

• γ2

ν

◮

µ46(x, t, k) = eikx−k2t

• γ1

ν µ5(x, t, k) = eikx−k2t

• γ2

ν

◮

µ46(x, t, k) = eikx−k2t

• γ46

ν µ5(x, t, k) = eikx−k2t

• γ5

ν µ123(x, t, k) = eikx−k2t

• γ123

ν

Riemann-Hilbert Problems

◮

Riemann-Hilbert Problems

◮ ◮ µ(x, t, k) = O

1 k

Riemann-Hilbert Problems

◮ ◮ µ(x, t, k) = O

1 k

• ◮ µ+ − µ− = φ

Riemann-Hilbert Problems

◮ ◮ µ(x, t, k) = O

1 k

• ◮ µ+ − µ− = φ

◮ µ =

1 2πi

• L

φ(k′)dk′ k′ − k (Plemelj Formula)

Lax Pairs and closed differential form for the GASP equation.

◮ Lα(u) = ∆u + α

x ∂u ∂x = 0 in H = {z, Re z > 0}.

Lax Pairs and closed differential form for the GASP equation.

◮ Lα(u) = ∆u + α

x ∂u ∂x = 0 in H = {z, Re z > 0}.

◮ uz ¯ z +

α 2(z + ¯ z)(uz + u¯

z) = 0.

(GASP)

Lax Pairs and closed differential form for the GASP equation.

◮ Lα(u) = ∆u + α

x ∂u ∂x = 0 in H = {z, Re z > 0}.

◮ uz ¯ z +

α 2(z + ¯ z)(uz + u¯

z) = 0.

(GASP)

◮

   φz(z, k) = (k + ¯ z)α/2(k − z)α/2−1uz(z), φ¯

z(z, k) = (k + ¯

z)α/2−1(k − z)α/2u¯

z(z)

Lax Pairs and closed differential form for the GASP equation.

◮ Lα(u) = ∆u + α

x ∂u ∂x = 0 in H = {z, Re z > 0}.

◮ uz ¯ z +

α 2(z + ¯ z)(uz + u¯

z) = 0.

(GASP)

◮

   φz(z, k) = (k + ¯ z)α/2(k − z)α/2−1uz(z), φ¯

z(z, k) = (k + ¯

z)α/2−1(k − z)α/2u¯

z(z) ◮ W (z, k) =

• (k − z)(k + ¯

z) α/2−1 (k + ¯ z)uz(z)dz + (k − z)u¯

z(z)d ¯

z

• Note that, when α ∈ 2N∗, the differential form has no

singularity in Ω and k may be any complex number. Otherwise, for α ∈ R \ 2N∗, 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.

◮ Lα(u) = ∆u + α

x ∂u ∂x = 0 in H = {z, Re z > 0}.

◮ uz ¯ z +

α 2(z + ¯ z)(uz + u¯

z) = 0.

(GASP)

◮

   φz(z, k) = (k + ¯ z)α/2(k − z)α/2−1uz(z), φ¯

z(z, k) = (k + ¯

z)α/2−1(k − z)α/2u¯

z(z) ◮ W (z, k) =

• (k − z)(k + ¯

z) α/2−1 (k + ¯ z)uz(z)dz + (k − z)u¯

z(z)d ¯

z

• Note that, when α ∈ 2N∗, the differential form has no

singularity in Ω and k may be any complex number. Otherwise, for α ∈ R \ 2N∗, W (z, k) has a pole or a branching point in k or −¯ k if one of this point is in Ω.

◮ Lα(u) = 0 ⇔ L2−α(xα−1u) = 0.

α = −2m, m ∈ N.

◮ W (z, k) = (k + z)uzdz + (k − z)uzdz

[(k − z)(k + z)]m+1

α = −2m, m ∈ N.

◮ W (z, k) = (k + z)uzdz + (k − z)uzdz

[(k − z)(k + z)]m+1

◮

φ(z, k) =

• W (z′, k)

α = −2m, m ∈ N.

◮ W (z, k) = (k + z)uzdz + (k − z)uzdz

[(k − z)(k + z)]m+1

◮

φ(z, k) =

• W (z′, k)

◮ poles of order m in −zr, −z, z, zr if m > 0.

α = −2m, m ∈ N.

◮ W (z, k) = (k + z)uzdz + (k − z)uzdz

[(k − z)(k + z)]m+1

◮

φ(z, k) =

• W (z′, k)

◮ poles of order m in −zr, −z, z, zr if m > 0. ◮ φ(z, −k) = −φ(z, k)

α = −2m, m ∈ N.

◮ W (z, k) = (k + z)uzdz + (k − z)uzdz

[(k − z)(k + z)]m+1

◮

φ(z, k) =

• W (z′, k)

◮ poles of order m in −zr, −z, z, zr if m > 0. ◮ φ(z, −k) = −φ(z, k) ◮ Jump on (z, zr) ∪ (−zr, −z) equal to J(k) = −

• C

W (z′, k)

α = −2m, m ∈ N.

◮ W (z, k) = (k + z)uzdz + (k − z)uzdz

[(k − z)(k + z)]m+1

◮

φ(z, k) =

• W (z′, k)

◮ poles of order m in −zr, −z, z, zr if m > 0. ◮ φ(z, −k) = −φ(z, k) ◮ Jump on (z, zr) ∪ (−zr, −z) equal to J(k) = −

• C

W (z′, k)

◮ J has no singularity in z and zr.

α = −2m, m ∈ N.

◮ W (z, k) = (k + z)uzdz + (k − z)uzdz

[(k − z)(k + z)]m+1

◮

φ(z, k) =

• W (z′, k)

◮ poles of order m in −zr, −z, z, zr if m > 0. ◮ φ(z, −k) = −φ(z, k) ◮ Jump on (z, zr) ∪ (−zr, −z) equal to J(k) = −

• C

W (z′, k)

◮ J has no singularity in z and zr. ◮ φ ∼k→+∞ u(z)−u(zr) k2m−1

◮ Renormalization :

φ(z, k) = ((k − z)(k + z))mφ(z, k)

• J(z, k) = ((k − z)(k + z))mJ(z, k)

◮ Renormalization :

φ(z, k) = ((k − z)(k + z))mφ(z, k)

• J(z, k) = ((k − z)(k + z))mJ(z, k)

◮

φ(z, k) ∼k→+∞ u(z) − u(zr) k

◮ Renormalization :

φ(z, k) = ((k − z)(k + z))mφ(z, k)

• J(z, k) = ((k − z)(k + z))mJ(z, k)

◮

φ(z, k) ∼k→+∞ u(z) − u(zr) k

◮

φ regular in z and −z, polar singularities in zr et −zr.

◮ Renormalization :

φ(z, k) = ((k − z)(k + z))mφ(z, k)

• J(z, k) = ((k − z)(k + z))mJ(z, k)

◮

φ(z, k) ∼k→+∞ u(z) − u(zr) k

◮

φ regular in z and −z, polar singularities in zr et −zr.

◮

φzr,−zr (z, k) polar part in this point

◮ Renormalization :

φ(z, k) = ((k − z)(k + z))mφ(z, k)

• J(z, k) = ((k − z)(k + z))mJ(z, k)

◮

φ(z, k) ∼k→+∞ u(z) − u(zr) k

◮

φ regular in z and −z, polar singularities in zr et −zr.

◮

φzr,−zr (z, k) polar part in this point

◮

φ − φzr,−zr analytic outside (z, zr) ∪ (−zr, −z) and has jump equal to J(z, k)

◮ Renormalization :

φ(z, k) = ((k − z)(k + z))mφ(z, k)

• J(z, k) = ((k − z)(k + z))mJ(z, k)

◮

φ(z, k) ∼k→+∞ u(z) − u(zr) k

◮

φ regular in z and −z, polar singularities in zr et −zr.

◮

φzr,−zr (z, k) polar part in this point

◮

φ − φzr,−zr analytic outside (z, zr) ∪ (−zr, −z) and has jump equal to J(z, k)

◮ Plemelj formula

• φ(z, k) −

φzr,−zr (z, k) = 1 2πi

• (−zr,−z)∪(z,zr)
• J(z, k′)dk′

k′ − k

◮ Renormalization :

φ(z, k) = ((k − z)(k + z))mφ(z, k)

• J(z, k) = ((k − z)(k + z))mJ(z, k)

◮

φ(z, k) ∼k→+∞ u(z) − u(zr) k

◮

φ regular in z and −z, polar singularities in zr et −zr.

◮

φzr,−zr (z, k) polar part in this point

◮

φ − φzr,−zr analytic outside (z, zr) ∪ (−zr, −z) and has jump equal to J(z, k)

◮ Plemelj formula

• φ(z, k) −

φzr,−zr (z, k) = 1 2πi

• (−zr,−z)∪(z,zr)
• J(z, k′)dk′

k′ − k

◮

u(z) − u(zr) = 2 Re ar − 1 πIm

• (z,zr)
• J(z, k′)dk′

Computation of the residue ar of φ in zr

◮ uzdz = 1 2(ut + iun)ds,

uzdz = 1

2(ut − iun)ds.

Computation of the residue ar of φ in zr

◮ uzdz = 1 2(ut + iun)ds,

uzdz = 1

2(ut − iun)ds. ◮ φ(z, k) =

• W (z′, k) =

[(k − iy′)ut + ix′un)]ds [(k − z′)(k + z′)]m+1

Computation of the residue ar of φ in zr

◮ uzdz = 1 2(ut + iun)ds,

uzdz = 1

2(ut − iun)ds. ◮ φ(z, k) =

• W (z′, k) =

[(k − iy′)ut + ix′un)]ds [(k − z′)(k + z′)]m+1

◮ φ(z, k) =

• (k − z′)−m−1w(z′, k)dz′ where

w(z′, k) = (k + z′)−m−1((k − iy′)ut + ix′un)τ −1(z′)dz′ τ(z′) is the unit tangent vector to C in z′.

Computation of the residue ar of φ in zr

◮ uzdz = 1 2(ut + iun)ds,

uzdz = 1

2(ut − iun)ds. ◮ φ(z, k) =

• W (z′, k) =

[(k − iy′)ut + ix′un)]ds [(k − z′)(k + z′)]m+1

◮ φ(z, k) =

• (k − z′)−m−1w(z′, k)dz′ where

w(z′, k) = (k + z′)−m−1((k − iy′)ut + ix′un)τ −1(z′)dz′ τ(z′) is the unit tangent vector to C in z′.

◮ m integrations by parts give

φzr,−zr , then ar.

Let u be a solution to the equation ∆u + αx−1∂xu = 0, α = −2m, m ∈ N, in the domain D with smooth tangential derivatives ut and normal derivatives un on the boundary C. u(z) = − 1 πIm

• (z,zr)
• (k − z)(k + ¯

z) mJ(z, k)dk + 2Re ar + u(zr), (1) where ar can be explicitly computed in terms of the tangential derivative along C of ut and un, up to the order m − 1, in zr. Function J(z, k) is given by J(z, k) = −

• C

W (z′, k), where W (z, k) is the differential form W (z, k) =

• (k − z)(k + ¯

z) −m−1 ((k + ¯ z)uz(z)dz + (k − z)u¯

z(z)d ¯

z) (2) =

• (k − z)(k + ¯

z) −m−1 ((k − iy)ut(z) + ixun(z)) ds, (3) with z = x + iy and ds the unit length element on C.

The case α = −1.

◮ W (z, k) = (k + z)uzdz + (k − z)uzdz

• (k − z)(k + z)

The case α = −1.

◮ W (z, k) = (k + z)uzdz + (k − z)uzdz

• (k − z)(k + z)

◮ Riemann surface Sz : two copies of C, Sz,1 et Sz,2 glued

together along the branching cut [−z, z].

The case α = −1.

◮ W (z, k) = (k + z)uzdz + (k − z)uzdz

• (k − z)(k + z)

◮ Riemann surface Sz : two copies of C, Sz,1 et Sz,2 glued

together along the branching cut [−z, z].

◮

The case α = −1.

◮ W (z, k) = (k + z)uzdz + (k − z)uzdz

• (k − z)(k + z)

◮ Riemann surface Sz : two copies of C, Sz,1 et Sz,2 glued

together along the branching cut [−z, z].

◮ ◮ λ1(z, k) ∼ k when k → ∞1 on the sheet above Sz,1

λ2(z, k) ∼ −k when k → ∞2 on the sheet below Sz,2

Integration of the form.

◮ φ(z, k) =

(k + z′)uzdz′ + (k − z′)uzdz′ λ(z′, k)

Integration of the form.

◮ φ(z, k) =

(k + z′)uzdz′ + (k − z′)uzdz′ λ(z′, k)

◮ Sheet 1

Integration of the form.

◮ φ(z, k) =

(k + z′)uzdz′ + (k − z′)uzdz′ λ(z′, k)

◮ Sheet 1 ◮ Sheet 2

Integration of the form.

◮ φ(z, k) =

(k + z′)uzdz′ + (k − z′)uzdz′ λ(z′, k)

◮ Sheet 1 ◮ Sheet 2 ◮ φ(z, ∞1) = −φ(z, ∞2).

◮

◮ ◮

φ(z, k) = 1 4iπ

• Cup∪−Cup

J(z, k′) λ(z, k) λ1(z, k′) + 1

• dk′

k′ − k + 1 4iπ

• Clow∪−Clow

J(z, k′) λ(z, k) λ2(z, k′) + 1

• dk′

k′ − k , (4)

The notion of good Lax pair.

◮

• ∂Ω

W (ut, un, k)ds = 0 for all k ∈ K ⊂ C.

The notion of good Lax pair.

◮

• ∂Ω

W (ut, un, k)ds = 0 for all k ∈ K ⊂ C.

◮ Is this relation useful ? In other words, if u = 0 ⇒ un = 0 on

∂Ω ? If u is known on ∂Ω, is it possible to obtain un on ∂Ω ?

The notion of good Lax pair.

◮

• ∂Ω

W (ut, un, k)ds = 0 for all k ∈ K ⊂ C.

◮ Is this relation useful ? In other words, if u = 0 ⇒ un = 0 on

∂Ω ? If u is known on ∂Ω, is it possible to obtain un on ∂Ω ?

◮ In full generality, no

The notion of good Lax pair.

◮

• ∂Ω

W (ut, un, k)ds = 0 for all k ∈ K ⊂ C.

◮ Is this relation useful ? In other words, if u = 0 ⇒ un = 0 on

∂Ω ? If u is known on ∂Ω, is it possible to obtain un on ∂Ω ?

◮ In full generality, no ◮ Example : Ω = D, W = uz k−z dz + u¯

z

k+z dz. ∆u = 0.

The notion of good Lax pair.

◮

• ∂Ω

W (ut, un, k)ds = 0 for all k ∈ K ⊂ C.

◮ Is this relation useful ? In other words, if u = 0 ⇒ un = 0 on

∂Ω ? If u is known on ∂Ω, is it possible to obtain un on ∂Ω ?

◮ In full generality, no ◮ Example : Ω = D, W = uz k−z dz + u¯

z

k+z dz. ∆u = 0. ◮ uzdz = 1 2(ut + iun)ds, u¯ z = 1 2(ut − iun)ds.

The notion of good Lax pair.

◮

• ∂Ω

W (ut, un, k)ds = 0 for all k ∈ K ⊂ C.

◮ Is this relation useful ? In other words, if u = 0 ⇒ un = 0 on

∂Ω ? If u is known on ∂Ω, is it possible to obtain un on ∂Ω ?

◮ In full generality, no ◮ Example : Ω = D, W = uz k−z dz + u¯

z

k+z dz. ∆u = 0. ◮ uzdz = 1 2(ut + iun)ds, u¯ z = 1 2(ut − iun)ds. ◮

• ∂D

un(z)

• 1

k − z − 1 k + z

• ds(z) =

i

• ∂D

ut(z)

• 1

k − z + 1 k + z

• ds(z)

The notion of good Lax pair.

◮

• ∂Ω

W (ut, un, k)ds = 0 for all k ∈ K ⊂ C.

◮ Is this relation useful ? In other words, if u = 0 ⇒ un = 0 on

∂Ω ? If u is known on ∂Ω, is it possible to obtain un on ∂Ω ?

◮ In full generality, no ◮ Example : Ω = D, W = uz k−z dz + u¯

z

k+z dz. ∆u = 0. ◮ uzdz = 1 2(ut + iun)ds, u¯ z = 1 2(ut − iun)ds. ◮

• ∂D

un(z)

• 1

k − z − 1 k + z

• ds(z) =

i

• ∂D

ut(z)

• 1

k − z + 1 k + z

• ds(z)

◮ ut ≡ 0 et

1 k − z =

∞

• n=1

zn−1 kn , 1 ¯ z + k =

∞

• n=1

(−z)n−1 kn

The notion of good Lax pair.

◮

• ∂Ω

W (ut, un, k)ds = 0 for all k ∈ K ⊂ C.

◮ Is this relation useful ? In other words, if u = 0 ⇒ un = 0 on

∂Ω ? If u is known on ∂Ω, is it possible to obtain un on ∂Ω ?

◮ In full generality, no ◮ Example : Ω = D, W = uz k−z dz + u¯

z

k+z dz. ∆u = 0. ◮ uzdz = 1 2(ut + iun)ds, u¯ z = 1 2(ut − iun)ds. ◮

• ∂D

un(z)

• 1

k − z − 1 k + z

• ds(z) =

i

• ∂D

ut(z)

• 1

k − z + 1 k + z

• ds(z)

◮ ut ≡ 0 et

1 k − z =

∞

• n=1

zn−1 kn , 1 ¯ z + k =

∞

• n=1

(−z)n−1 kn

◮

un(z) =

∞

• n=−∞

bnzn,

+∞

• n=1

bn−1 + (−1)nbn−1 kn = 0.

Good Lax pair for α = −2(m − 1).

◮ ut ≡ 0.

Good Lax pair for α = −2(m − 1).

◮ ut ≡ 0. ◮

• ∂D(a,1)

xun(z)ds(z) [(k − z)(k + z)]m = 0, ∀k ∈ C \ (Da ∪ D−a)

Good Lax pair for α = −2(m − 1).

◮ ut ≡ 0. ◮

• ∂D(a,1)

xun(z)ds(z) [(k − z)(k + z)]m = 0, ∀k ∈ C \ (Da ∪ D−a)

◮ Da → D. z − a = 1 z−a.

Good Lax pair for α = −2(m − 1).

◮ ut ≡ 0. ◮

• ∂D(a,1)

xun(z)ds(z) [(k − z)(k + z)]m = 0, ∀k ∈ C \ (Da ∪ D−a)

◮ Da → D. z − a = 1 z−a. ◮

• T

zm−1f (z)dz (z − (k − a))m z +

1 k+a

m = 0, f (z) = (x+a)un(z+a).

Good Lax pair for α = −2(m − 1).

◮ ut ≡ 0. ◮

• ∂D(a,1)

xun(z)ds(z) [(k − z)(k + z)]m = 0, ∀k ∈ C \ (Da ∪ D−a)

◮ Da → D. z − a = 1 z−a. ◮

• T

zm−1f (z)dz (z − (k − a))m z +

1 k+a

m = 0, f (z) = (x+a)un(z+a).

◮ f is real valued, f (z) = g(z) + g(1/z), g ∈ H(D),

g(1/z) ∈ H(C \ D).

◮

• T

zm−1(g(z) + g(1/z))dz (z − (k − a))m z +

1 k+a

m = 0.

◮

• T

zm−1(g(z) + g(1/z))dz (z − (k − a))m z +

1 k+a

m = 0.

◮

−1 k + a m zm−1g(z) (z − (k − a))m (m−1) −1 k + a

• +
• 1

k − a m zm−1¯ g(z) (z + (k + a))m (m−1) 1 k − a

• = 0.

◮

• T

zm−1(g(z) + g(1/z))dz (z − (k − a))m z +

1 k+a

m = 0.

◮

−1 k + a m zm−1g(z) (z − (k − a))m (m−1) −1 k + a

• +
• 1

k − a m zm−1¯ g(z) (z + (k + a))m (m−1) 1 k − a

• = 0.

◮

Φ(z) =

m−1

• p=0

(−1)m−1+pzm−1−p (2m − p − 2)! (m − p − 1)!p!× ×(zm−1g(z))(p)(z2 + 2az + 1)p,

◮

ϕ(z) = − z 1 + 2az ϕ

• −

1 k + a

• =

1 k − a.

◮

ϕ(z) = − z 1 + 2az ϕ

• −

1 k + a

• =

1 k − a.

◮

Φ(z) + ¯ Φ(ϕ(z)) = 0, z ∈ A.

◮

ϕ(z) = − z 1 + 2az ϕ

• −

1 k + a

• =

1 k − a.

◮

Φ(z) + ¯ Φ(ϕ(z)) = 0, z ∈ A.

◮

Φ(z) =

m−1

• p=0

αpzm−1−p(z2 + 2az + 1)ph(p)(z) αp = (−1)m−1+p (2m − p − 2)! (m − p − 1)!p! h(z) = zm−1g(z).

◮

ϕ(z) = − z 1 + 2az ϕ

• −

1 k + a

• =

1 k − a.

◮

Φ(z) + ¯ Φ(ϕ(z)) = 0, z ∈ A.

◮

Φ(z) =

m−1

• p=0

αpzm−1−p(z2 + 2az + 1)ph(p)(z) αp = (−1)m−1+p (2m − p − 2)! (m − p − 1)!p! h(z) = zm−1g(z).

◮

◮

Φ(z) =

m−1

• p=0

αpzm−1−p(z2 + 2az + 1)ph(p)(z) = P2m−2

◮

Φ(z) =

m−1

• p=0

αpzm−1−p(z2 + 2az + 1)ph(p)(z) = P2m−2

◮ h(z) = +∞ n=0 an(z − z1)n.

◮

Φ(z) =

m−1

• p=0

αpzm−1−p(z2 + 2az + 1)ph(p)(z) = P2m−2

◮ h(z) = +∞ n=0 an(z − z1)n. ◮ m−1 ℓ=0 an−ℓβn ℓ = 0, n ≥ 2m − 1.

◮

Φ(z) =

m−1

• p=0

αpzm−1−p(z2 + 2az + 1)ph(p)(z) = P2m−2

◮ h(z) = +∞ n=0 an(z − z1)n. ◮ m−1 ℓ=0 an−ℓβn ℓ = 0, n ≥ 2m − 1. ◮ For N ∈ {0, . . . , m − 2} and n = 2m − 1 + N,

βn

N+1 = · · · = βn m−1 = 0

and βn

0 = 0, . . . , βn N = 0.

◮

Φ(z) =

m−1

• p=0

αpzm−1−p(z2 + 2az + 1)ph(p)(z) = P2m−2

◮ h(z) = +∞ n=0 an(z − z1)n. ◮ m−1 ℓ=0 an−ℓβn ℓ = 0, n ≥ 2m − 1. ◮ For N ∈ {0, . . . , m − 2} and n = 2m − 1 + N,

βn

N+1 = · · · = βn m−1 = 0

and βn

0 = 0, . . . , βn N = 0. ◮ h is a polynomial of degree less than 2m − 2.

◮

Φ(z) =

m−1

• p=0

αpzm−1−p(z2 + 2az + 1)ph(p)(z) = P2m−2

◮ h(z) = +∞ n=0 an(z − z1)n. ◮ m−1 ℓ=0 an−ℓβn ℓ = 0, n ≥ 2m − 1. ◮ For N ∈ {0, . . . , m − 2} and n = 2m − 1 + N,

βn

N+1 = · · · = βn m−1 = 0

and βn

0 = 0, . . . , βn N = 0. ◮ h is a polynomial of degree less than 2m − 2. ◮ g is a polynomial de degree less than m − 1.

◮

• T

zm−1f (z) (z − (k − a))m(z + (k + a)−1)m = 0, ∀k ∈ C\{Da∪−Da}, where f (z) = g(z) + g 1 z

• .

◮

• T

zm−1f (z) (z − (k − a))m(z + (k + a)−1)m = 0, ∀k ∈ C\{Da∪−Da}, where f (z) = g(z) + g 1 z

• .

◮ We put ξ = −(k + a)−1, we get

• T

zm−1f (z) (ξz + 2aξ + 1)m (z − ξ)m = 0 ∀ξ ∈ D \ D

• −

2a 4a2 − 1, 1 4a2 − 1

• .

◮

∂m−1 ∂zm−1

• zm−1f (z)

(ξz + 2aξ + 1)m

• |z=ξ

= 0.

◮

∂m−1 ∂zm−1

• zm−1f (z)

(ξz + 2aξ + 1)m

• |z=ξ

= 0.

◮ Let us denote F(z) = zm−1f (z) ∈ C2m−2[z]. m−1

• k=0

(−1)k(m + k − 1)! k!(m − 1 − k)! ξk (ξ2 + 2aξ + 1)m+k F (m−1−k)(ξ) = 0.

◮

∂m−1 ∂zm−1

• zm−1f (z)

(ξz + 2aξ + 1)m

• |z=ξ

= 0.

◮ Let us denote F(z) = zm−1f (z) ∈ C2m−2[z]. m−1

• k=0

(−1)k(m + k − 1)! k!(m − 1 − k)! ξk (ξ2 + 2aξ + 1)m+k F (m−1−k)(ξ) = 0.

◮ We can extend F to C.

◮

∂m−1 ∂zm−1

• zm−1f (z)

(ξz + 2aξ + 1)m

• |z=ξ

= 0.

◮ Let us denote F(z) = zm−1f (z) ∈ C2m−2[z]. m−1

• k=0

(−1)k(m + k − 1)! k!(m − 1 − k)! ξk (ξ2 + 2aξ + 1)m+k F (m−1−k)(ξ) = 0.

◮ We can extend F to C. ◮ Denotes z1 = −a +

√ a2 − 1 and z2 = −a − √ a2 − 1 ∀k ∈ {0, 1, . . . , m − 1}, F (k)(z1) = 0, F (k)(z2) = 0.

◮

∂m−1 ∂zm−1

• zm−1f (z)

(ξz + 2aξ + 1)m

• |z=ξ

= 0.

◮ Let us denote F(z) = zm−1f (z) ∈ C2m−2[z]. m−1

• k=0

(−1)k(m + k − 1)! k!(m − 1 − k)! ξk (ξ2 + 2aξ + 1)m+k F (m−1−k)(ξ) = 0.

◮ We can extend F to C. ◮ Denotes z1 = −a +

√ a2 − 1 and z2 = −a − √ a2 − 1 ∀k ∈ {0, 1, . . . , m − 1}, F (k)(z1) = 0, F (k)(z2) = 0.

◮ F ≡ 0.

◮

∂m−1 ∂zm−1

• zm−1f (z)

(ξz + 2aξ + 1)m

• |z=ξ

= 0.

◮ Let us denote F(z) = zm−1f (z) ∈ C2m−2[z]. m−1

• k=0

(−1)k(m + k − 1)! k!(m − 1 − k)! ξk (ξ2 + 2aξ + 1)m+k F (m−1−k)(ξ) = 0.

◮ We can extend F to C. ◮ Denotes z1 = −a +

√ a2 − 1 and z2 = −a − √ a2 − 1 ∀k ∈ {0, 1, . . . , m − 1}, F (k)(z1) = 0, F (k)(z2) = 0.

◮ F ≡ 0. ◮ Thank you very much for your attention !