a computational study of a class of multivaluedsanum.github.io/2016/slides/fasondini.pdf · a...

A computational study of a class of multivaluedtronquée solutions of the third Painlevé equation

Marco Fasondini University of the Free State

Supervisors:

André Weideman Stellenbosch University

Bengt Fornberg University of Colorado at Boulder

Introducing PIII

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

Phase portrait

u(z) = r(z)eiθ(z)

Modulus plot

• Poles: u ∼ cz − z0

, z −→ z0

c = ±1, (red/yellow)• Zeros: u ∼ c(z − z0), z −→ z0c = ±1, (purple/green)

Introducing PIII

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

uPhase portrait

u(z) = r(z)eiθ(z)

Modulus plot

• Poles: u ∼ cz − z0

, z −→ z0

c = ±1, (red/yellow)• Zeros: u ∼ c(z − z0), z −→ z0c = ±1, (purple/green)

Tronquée PIII solutions

Tronquée solutions, Boutroux [1913]

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

Lin, Dai and Tibboel [2014] provedthe existence of tronquée PIIIsolutions whose pole-free sectorshave angular widths of

• π and 2π if γ = 1 and δ = −1, and

•3π

2and 3π if α = 1,γ = 0 and δ = −1.

McCoy, Tracy and Wu [1977] derived asymptotic formulaefor tronquée solutions with parameters α = −β = 2ν, γ = 1and δ = −1 and applied their results to the Ising model.

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

The MTW solutions: α = −β = 2ν, γ = 1, δ = −1

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+

σ(λ) =2

πarcsin(πλ) ,

B(σ, ν) = 2−2σΓ2(12(1− σ)

)Γ(12(1 + σ) + ν

)Γ2(12(1 + σ)

)Γ(12(1− σ) + ν

)−1π< λ <

1

π

u(x; ν,−λ) = 1u(x; ν, λ)

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

Tronquée PIII solutionsThe MTW solutions: α = −β = 2ν, γ = 1, δ = −1. For the case λ > 1

π,

let λ =cosh(πµ)

π, µ > 0, then [u(x; ν,−λ) = 1/u(x; ν, λ)]

Green u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

Redu(x)

x∼ 1

2µ

{ν

µ(1− cos [φ(x, ν, µ)]) + sin [φ(x, ν, µ)]

}, x→ 0+

φ(x, ν, µ) = 2µ ln(x/4)− 4 arg [Γ(µi)] + 2 arg [Γ(ν + µi)]

π,

let λ =cosh(πµ)

π, µ > 0, then [u(x; ν,−λ) = 1/u(x; ν, λ)]

)2−2νx−ν−(1/2)e−2x, x→∞

Redu(x)

x∼ 1

2µ

{ν

}, x→ 0+

π,

let λ =cosh(πµ)

π, µ > 0, then [u(x; ν,−λ) = 1/u(x; ν, λ)]

)2−2νx−ν−(1/2)e−2x, x→∞

Redu(x)

x∼ 1

2µ

{ν

}, x→ 0+

π,

let λ =cosh(πµ)

π, µ > 0, then [u(x; ν,−λ) = 1/u(x; ν, λ)]

)2−2νx−ν−(1/2)e−2x, x→∞

Redu(x)

x∼ 1

2µ

{ν

}, x→ 0+

π,

let λ =cosh(πµ)

π, µ > 0, then [u(x; ν,−λ) = 1/u(x; ν, λ)]

)2−2νx−ν−(1/2)e−2x, x→∞

Redu(x)

x∼ 1

2µ

{ν

}, x→ 0+

π,

let λ =cosh(πµ)

π, µ > 0, then [u(x; ν,−λ) = 1/u(x; ν, λ)]

)2−2νx−ν−(1/2)e−2x, x→∞

Redu(x)

x∼ 1

2µ

{ν

}, x→ 0+

π,

let λ =cosh(πµ)

π, µ > 0, then [u(x; ν,−λ) = 1/u(x; ν, λ)]

)2−2νx−ν−(1/2)e−2x, x→∞

Redu(x)

x∼ 1

2µ

{ν

}, x→ 0+

π,

let λ =cosh(πµ)

π, µ > 0, then [u(x; ν,−λ) = 1/u(x; ν, λ)]

)2−2νx−ν−(1/2)e−2x, x→∞

Redu(x)

x∼ 1

2µ

{ν

}, x→ 0+

π,

let λ =cosh(πµ)

π, µ > 0, then [u(x; ν,−λ) = 1/u(x; ν, λ)]

)2−2νx−ν−(1/2)e−2x, x→∞

Redu(x)

x∼ 1

2µ

{ν

}, x→ 0+

π,

let λ =cosh(πµ)

π, µ > 0, then [u(x; ν,−λ) = 1/u(x; ν, λ)]

)2−2νx−ν−(1/2)e−2x, x→∞

Redu(x)

x∼ 1

2µ

{ν

}, x→ 0+

π,

let λ =cosh(πµ)

π, µ > 0, then [u(x; ν,−λ) = 1/u(x; ν, λ)]

)2−2νx−ν−(1/2)e−2x, x→∞

Redu(x)

x∼ 1

2µ

{ν

}, x→ 0+

π,

let λ =cosh(πµ)

π, µ > 0, then [u(x; ν,−λ) = 1/u(x; ν, λ)]

)2−2νx−ν−(1/2)e−2x, x→∞

Redu(x)

x∼ 1

2µ

{ν

}, x→ 0+

π,

let λ =cosh(πµ)

π, µ > 0, then [u(x; ν,−λ) = 1/u(x; ν, λ)]

)2−2νx−ν−(1/2)e−2x, x→∞

Redu(x)

x∼ 1

2µ

{ν

}, x→ 0+

As λ is varied, multiple tronquée solutions occur on sheetsother than the main sheet of the MTW solutions

Lin, Dai and Tibboel [2014] proved the existence of tronquéePIII solutions whose pole-free sectors have angular widthsof

• π and 2π if γ = 1 and δ = −1, and

•3π

2and 3π if α = 1,γ = 0 and δ = −1.

Tronquée PIII solutionsAs λ is varied, multiple tronquée solutions occur on sheets other thanthe main sheet of the MTW solutions, u(z)

On the region {z ∈ C | π ≤ arg z ≤ 9π} we have found

• 16 tronquée solutions if ν 6∈ Z. They have the properties

u(z) ∼ −1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 8, 6, 4, 2

k = 8, 6, 4

k = 8, 6

k = 8

u(z) ∼ 1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 7, 5, 3

k = 7, 5

k = 7

Conjecture: The sequences continue indefinitely

• 16 tronquée solutions if ν 6∈ Z.

u(z) ∼ −1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 8, 6, 4, 2

k = 8, 6, 4

k = 8, 6

k = 8

u(z) ∼ 1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 7, 5, 3

k = 7, 5

k = 7

u(z) ∼ −1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 8, 6, 4, 2

k = 8, 6, 4

k = 8, 6

k = 8

u(z) ∼ 1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 7, 5, 3

k = 7, 5

k = 7

u(z) ∼ −1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 8, 6, 4, 2

k = 8, 6, 4

k = 8, 6

k = 8

u(z) ∼ 1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 7, 5, 3

k = 7, 5

k = 7

u(z) ∼ −1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 8, 6, 4, 2

k = 8, 6, 4

k = 8, 6

k = 8

u(z) ∼ 1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 7, 5, 3

k = 7, 5

k = 7

u(z) ∼ −1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 8, 6, 4, 2

k = 8, 6, 4

k = 8, 6

k = 8

u(z) ∼ 1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 7, 5, 3

k = 7, 5

k = 7

u(z) ∼ −1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 8, 6, 4, 2

k = 8, 6, 4

k = 8, 6

k = 8

u(z) ∼ 1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 7, 5, 3

k = 7, 5

k = 7

• 72 tronquée solutions if ν ∈ Z. They have the properties

u(z) ∼ i, |z| → ∞, kπ < arg z < (k + 1) π

k = 8, 7, 6, 5, 4, 3, 2, 1

k = 8, 7, 6, 5, 4, 3

k = 8, 7, 6, 5

k = 8, 7

u(z) ∼ −i, |z| → ∞, kπ < arg z < (k + 1) π

k = 8, 7, 6, 5, 4, 3, 2

k = 8, 7, 6, 5, 4

k = 8, 7, 6

k = 8

• 72 tronquée solutions if ν ∈ Z. They have the properties

u(z) ∼ −1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 8, 7, 6, 5, 4, 3, 2, 1

k = 8, 7, 6, 5, 4, 3

k = 8, 7, 6, 5

k = 8, 7

u(z) ∼ 1, |z| → ∞,(k +

1

2

)π < arg z <

(k +

3

2

)π

k = 8, 7, 6, 5, 4, 3, 2

k = 8, 7, 6, 5, 4

k = 8, 7, 6

k = 8

Future plans

• Survey tronquée PIII solutions with α = 1,γ = 0 andδ = −1• Survey single-valued PIII solutions• Survey triply branched PIII solutions• Extend computational method for PIII to

PV :d2u

dz2=

(1

2u+

1

u− 1

)(du

dz

)2− 1z

du

dz+

(u− 1)2

z2

(αu +

β

u

)+ γ

u

z+ δ

u(u + 1)

u− 1

and

PV I :d2u

dz2=

1

2

(1

u+

1

u− 1+

1

u− z

)(du

dz

)2−(

1

z+

1

z − 1+

1

u− z

)(du

dz

)+u(u− 1)(u− z)z2(z − 1)2

(α + β

z

u2+ γ

z − 1(u− 1)2

+ δz(z − 1)(u− z)2

)

How to compute PIII?Challenges:

• Moveable poles: the ‘Pole Field Solver’ (PFS), Fornberg & Weideman [2011]

u(z), u′(z) −→ Taylor coefficients −→ Padé approximation at z + heiθ

Stage 1: pole avoidance on a coarse grid

Stage 2: compute the solution on a fine grid

Single-valued Painlevé transcendents: PI, F & W [2011] ;PII, F & W [2014, 2015] ; PIV , Reeger & F [2013, 2014]• Multivaluedness

How to compute PIII?Challenges:

• Moveable poles: the ‘Pole Field Solver’ (PFS), Fornberg & Weideman [2011]

u(z), u′(z) −→ Taylor coefficients −→ Padé approximation at z + heiθ

Stage 1: pole avoidance on a coarse grid

Stage 2: compute the solution on a fine grid

Single-valued Painlevé transcendents: PI, F & W [2011] ;PII, F & W [2014, 2015] ; PIV , Reeger & F [2013, 2014]• Multivaluedness

How to compute PIII?Second challenge: MultivaluednessMake the paths run in the right direction

−10 −5 0 5 10−4

−2

0

2

4

1st sheet

How to compute PIII?Second challenge: MultivaluednessMake the paths run in the right direction

−10 −5 0 5 10−4

−2

0

2

4

1st sheet

−10 −8 −6 −4 −2 0 2−4

−2

0

2

4

counterclockwise sheet

How to compute PIII?

Second challenge: Multivaluedness

PIII :d2u

dz2=

1

u

(du

dz

)2− 1z

du

dz+αu2 + β

z+ γu3 +

δ

u

Let z = eζ/2, u(z) = e−ζ/2w(ζ) in PIII , then

P̃III :d2w

dζ2=

1

w

(dw

dζ

)2+

1

4

(αw2 + γw3 + βeζ +

δe2ζ

w

),

and all solutions of P̃III are single-valued

Hinkkanen & Laine [2001]. (z = 0⇒ ζ = −∞)

How to compute PIII?

z = eζ/2, u(z) = e−ζ/2w(ζ)

u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

u(x) ∼ Bxσ, x→ 0+, −1π< λ <

1

π

MTW solutions: u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

Limiting solution: u(x) ∼ 1− kx−ν−(1/2)e−2x, x→∞

MTW solutions: u(x) ∼ 1− λΓ(ν + 12

)2−2νx−ν−(1/2)e−2x, x→∞

Limiting solution: u(x) ∼ 1− kx−ν−(1/2)e−2x, x→∞

a computational study of a class of multivaluedsanum.github.io/2016/slides/fasondini.pdf · a...

Documents