Geometric Asymptotics · Geometric Asymptotics Nalini Joshi for the ﬁrst Painlevé equation In...

Geometric Asymptotics

Nalini Joshi

for the first Painlevé equation

In collaboration with J.J. Duistermaat and H. Dullin

PI in Cy = 6 y2 + x

Is the solution space connected?

d2 y

dt2+ y3

dy

dt

2

= y2dy

dt

2 4dy

dt+ y4

y(t) = α tanα3 t+ β

y(t) =

43(t− γ)

1/3

E. Ince, ODEs, Dover (1956)

Water Waves• Dubrovin, Grava and Klein J. Nonlin. Sci (2009) analysed

critical behaviour of non-linear water waves under Hamiltonian perturbations

error proportional to a with a = 1.94, a correlation coefficient r = 0.9995 and standard errorσa = 0.03. In the non-symmetric case, we find a = 1.98, r = 0.999996 and σa = 0.003.

Close to the critical time the semiclassical solution only provides a satisfactory descrip-tion of the NLS solution for large values of |x − xc|. In the breakup region it fails to beaccurate since it develops a cusp at xc whereas the NLS solution stays smooth. This behaviorcan be well seen in Fig. 8 for the symmetric initial data. The largest difference between the

−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.51

2

3

4

5

6

7

8

x

u

Figure 8: The blue line is the function u of the solution to the focusing NLS equation forthe initial data u(x, 0) = 2 sech x and = 0.04 at the critical time, and the red line isthe corresponding semiclassical solution given by formulas (2.4). The green line gives themultiscales solution via the tritronquee solution of the Painleve I equation.

semiclassical and the NLS solution is always at the critical point. We find that the L∞ normof the difference scales roughly as 2/5 as suggested by the Main Conjecture. More preciselywe find a scaling proportional to a with a = 0.38 and r = 0.999997 and σa = 4.2 ∗ 10−4.For the non-symmetric initial data, we find a = 0.36, r = 0.9999 and σa = 0.002. Thecorresponding plot for u can be seen in Fig. 9.

The function v for the same situation as in Fig. 8 is shown in Fig. 10. It can be seen thatthe semiclassical solution is again a satisfactory description for |x − xc| large, but fails tobe accurate close to the breakup point. The phase for the non-symmetric initial data can beseen in Fig. 11. In the following we will always study the scaling for the function u withoutfurther notice.

28

Okamoto’s Space• Okamoto (1979) showed that the “space of initial

values” of the Painlevé equations can be compactified and regularized after exactly nine blow-ups.

• Sakai (2001) classified all equations (differential and difference) with this property, thereby providing a complete set of all Painlevé equations.

• We study Okamoto's space in an asymptotic limit |z|→ .∞

Boutroux’s Coordinates• Consider (Duistermaat & J: arXiv 1010:5563)

y’’=6y2+x in Boutroux’s coordinates

y(x) = x1/2 u(z), z =4x5/4

5

⇒ u = 6u2 + 1− u

z+

4u

25 z2

⇒

u1 = u2 −2u1

5 z

u2 = 6u21 + 1− 3u2

5 z

Projective Geometry

Affine coordinates [1 :

u011

u010:u012

u010] ⇔

Homogeneous coordinates [u010 : u011 : u012]

u010 = 0 ⇔ L0

CP1, the complex

projective space of dimension 1, is equivalent to the Riemann sphere.

In CP2 we have:

The Projective Plane ℂℙ2


First chart:u021 = −u021u022 + 2(5z)−1u021

u022 = u021 + 6u−1021 − u2

022 − (5z)−1u022

[u−11 : 1 : u−1

1 u2] = [u021 : 1 : u022]



u022 = u021 + 6u−1021 − u2

022 − (5z)−1u022

[u−11 : 1 : u−1

1 u2] = [u021 : 1 : u022]

Second chart: [u−12 : u1 u

−12 : 1] = [u031 : u032 : 1]

u031 = −u2031 − 6u2

032 + 3(5z)−1u031

u032 = −u031u032 − 6u−1031u

3032 + 1 + (5z)−1u032



u022 = u021 + 6u−1021 − u2

022 − (5z)−1u022

[u−11 : 1 : u−1

1 u2] = [u021 : 1 : u022]

Second chart: [u−12 : u1 u

−12 : 1] = [u031 : u032 : 1]

u031 = −u2031 − 6u2

032 + 3(5z)−1u031

u032 = −u031u032 − 6u−1031u

3032 + 1 + (5z)−1u032

base pt b0 : u031 = 0, u032 = 0

Blowing up at a base ptBlowing up at a base point

Figure 4.2.1: Real blowing up: a Mobius strip

106 Copyright Springer-Verlag 2009. No distribution is allowed. Any violation will be prosecuted.

From JJ Duistermaat QRT Maps and Elliptic Surfaces, Springer Verlag, 2010.

From JJ Duistermaat, QRT Maps and Elliptic Surfaces, Springer Verlag, 2010

First Blow-up

First Blow-up[1 : u111 : u112] = [1 : u031/u032 : u032]

u111 = −u111u−1112 + 2(5z)−1u111

u112 = 1− u111u2112 − 6u−1

111u2112 + (5z)−1u112

• Chart (1,1):

First Blow-up

• Chart (1,2): [1 : u121 : u122] = [1 : u031 : u032/u031]

u121 = u2121

−6u2

122 − 1+ 3 (5z)−1 u121

u122 = u−1121 − 2 (5z)−1 u122

[1 : u111 : u112] = [1 : u031/u032 : u032]

u111 = −u111u−1112 + 2(5z)−1u111

u112 = 1− u111u2112 − 6u−1

111u2112 + (5z)−1u112

• Chart (1,1):

First Blow-up

• Chart (1,2): [1 : u121 : u122] = [1 : u031 : u032/u031]

u121 = u2121

−6u2

122 − 1+ 3 (5z)−1 u121

u122 = u−1121 − 2 (5z)−1 u122

base pt b1 : u111 = 0, u112 = 0

[1 : u111 : u112] = [1 : u031/u032 : u032]

u111 = −u111u−1112 + 2(5z)−1u111

u112 = 1− u111u2112 − 6u−1

111u2112 + (5z)−1u112

• Chart (1,1):

Exceptional Curves

L1

L0

Exceptional Lines

• In Chart (1, 1), u111=0 defines the proper transform L0

(1), while u112=0 is L1.

• In Chart (1,2), L0 is not visible.

Second Blow-Up

Second Blow-Up•Chart (2,1): [1 : u211 : u212] = [1 : u111/u112 : u112]


u211 = u2211u

2212 − 2u211u

−1212 + 6 + (5z)−1u211

u212 = −u211u3212 − 6u−1

211u212 + 1 + (5z)−1u212


•Chart (2,2): [1 : u221 : u222] = [1 : u111 : u112/u111]

u211 = u2211u

2212 − 2u211u

−1212 + 6 + (5z)−1u211

u212 = −u211u3212 − 6u−1

211u212 + 1 + (5z)−1u212


•Chart (2,2): [1 : u221 : u222] = [1 : u111 : u112/u111]

u211 = u2211u

2212 − 2u211u

−1212 + 6 + (5z)−1u211

u212 = −u211u3212 − 6u−1

211u212 + 1 + (5z)−1u212

u221 = −u−1222 + 2(5z)−1u221

u222 = −u2221u

2222 + 2u−1

221 − 6u2222 − (5z)−1u222


•Chart (2,2): [1 : u221 : u222] = [1 : u111 : u112/u111]

u211 = u2211u

2212 − 2u211u

−1212 + 6 + (5z)−1u211

u212 = −u211u3212 − 6u−1

211u212 + 1 + (5z)−1u212

u221 = −u−1222 + 2(5z)−1u221

u222 = −u2221u

2222 + 2u−1

221 − 6u2222 − (5z)−1u222

b2 : u211 = 0, u212 = 0

Exceptional LinesExceptional Curves

L2 L1

L0

Third Blow-Up

Third Blow-Up•Chart (3,1): [1 : u311 : u312] = [1 : u211/u212 : u212]


u311 = u−1312

−3

u311 − 4

+ 2u2

311u4312

u312 = −u311u4312 − 6u−1

311 + 1 + (5z)−1u312


•Chart (3,2): [1 : u321 : u322] = [1 : u211 : u212/u211]

u311 = u−1312

−3

u311 − 4

+ 2u2

311u4312

u312 = −u311u4312 − 6u−1

311 + 1 + (5z)−1u312


•Chart (3,2): [1 : u321 : u322] = [1 : u211 : u212/u211]

u311 = u−1312

−3

u311 − 4

+ 2u2

311u4312

u312 = −u311u4312 − 6u−1

311 + 1 + (5z)−1u312

u321 = u4321u

2322 − 2u−1

322 + 6 + (5z)−1u321

u322 = −u−1321

2u4

321u3322 + 12u322 − 3


•Chart (3,2): [1 : u321 : u322] = [1 : u211 : u212/u211]

u311 = u−1312

−3

u311 − 4

+ 2u2

311u4312

u312 = −u311u4312 − 6u−1

311 + 1 + (5z)−1u312

u321 = u4321u

2322 − 2u−1

322 + 6 + (5z)−1u321

u322 = −u−1321

2u4

321u3322 + 12u322 − 3

b3 : u311 = 4, u312 = 0


L3 L2 L1

L0

Fourth Blow-Up

Fourth Blow-Up•Chart (4,1): [1 : u411 : u412] = [1 :

u311 − 4

/u312 : u312]


u311 − 4

/u312 : u312]

u411 = u−1412 (u411u412 + 4)−1

×−10u411 − 4u2

411u412 + 128u3412

+ 112u411u4412 + 32u2

411u4412

+ 3u3411u

6412

− (5z)−1u411

u412 = − (u411u412 + 4)−1

×2− u411u412 + 16u4

412 + u2411u

6412 + 8u411u

5412

+ (5z)−1u412


u311 − 4

/u312 : u312]

u411 = u−1412 (u411u412 + 4)−1

×−10u411 − 4u2

411u412 + 128u3412

+ 112u411u4412 + 32u2

411u4412

+ 3u3411u

6412

− (5z)−1u411

u412 = − (u411u412 + 4)−1

×2− u411u412 + 16u4

412 + u2411u

6412 + 8u411u

5412

+ (5z)−1u412

b4 : u411 = 4, u412 = 0

Fourth Blow-Up

Fourth Blow-Up• Chart (4,2):


[1 : u421 : u422] = [1 : (u311 − 4) : u312(u311−4) ]


[1 : u421 : u422] = [1 : (u311 − 4) : u312(u311−4) ]

u421 = u−1422

−3 + 32u3

421u4422 + 16u4

421u4422 + 2u5

421u4422

u422 = u−1421 (u421 + 4)−1

×10 + 4u421 − 128u3

421u4422

− 112u4421u

4422 − 3u6

421u4422 − 32u5

421u4422

+ (5z)−1u422

No base point in this chart.


L4

L3 L2 L1

L0

From Fifth to Ninth

• There are four more blow-ups:b5 : u511 = 0, u512 = 0

b6 : u611 = 0, u612 = 0

b7 : u711 = 32, u712 = 0

b8 : u811 = − 28

(5 z), u812 = 0

• Only the last one differs from the elliptic case.

Exceptional Curves

L9

L8

L7L6

L5L4

L3 L2 L1L0

Exceptional Lines

The Poles

The Poles•The vector field is regular along and transversal to u912=0 which is L9.

The Poles•The vector field is regular along and transversal to u912=0 which is L9. u911(z) = a+O

z − ζ

, u912(z) = −1

2(z − ζ) +O

z − ζ

2

u(z) = (z − ζ)−2 − 1

5ζ(z − ζ)−1 +

3

22 · 5 · ζ2− 31 · (z − ζ)

2 · 53 · ζ3

+

19 · 283

26 · 56 · ζ4− 1

2 · 5

(z − ζ)2

−3 · 11 · 727

24 · 56 · ζ5+

11

2 · 4 · 52 · ζ

(z − ζ)3

+

197 · 443

26 · 56 · ζ6+

29

23 · 3 · 52 · ζ2− a

28 · 7

(z − ζ)4

+ O(z − ζ)5

Pole Dancing

The Repellor Set• Definition: For z ∈ ℂ\0, let S denote the

fibre bundle of the Okamoto surfaces S9(z) and

I(z) := ∪8i=0L

(9−i)i (z)

This is the infinity set.

• Lemma: I(z) is a repellor for the flow.

Elements of Proof• The “energy” function E := u2

22 − 2u3

1 − u1

and the Jacobian of the coordinate change to each chart

wij =∂uij1

∂u1

∂uij2

∂u2− ∂uij1

∂u2

∂uij2

∂u1

provide a “distance” function to I which allow us to bound the flow near I.

• Near we use while near we use . In the overlap,

I\L(1)8 L(1)

8

w92 2E w92 → 11/E

A Fragment•E.g., near where u922 → 0L(1)

8 \L(2)7

u921 ∼ −2−1 u922−1

w92 ∼ 26 u922

w92/w92 = 6 (5 z)−1 +O(u9222) = 6 (5 z)−1 +O(w92

2)

2E w92 ∼ 1− 28 (5 z)−1 u921−1.

The Limit Set• Definition: For every solution U(z) ∈ S9(z)\I(z),

let

This is the limit set.

ΩU =s ∈ S9(∞)\I(∞)

∃ zj s.t. zj → ∞,

U(zj) → s as j → ∞

• Lemma: is a non-empty, connected and compact subset of Okamoto’s space.

ΩU

Summary• Okamoto’s space of initial values provides

complete information about the Painlevé transcendents.

• The ninth coordinate charts provide detailed information about the poles.

• We proved that the space of asymptotic behaviours is connected. As a corollary, we show that the solutions have an infinite number of poles in C.

• We also described solutions near equilibria.

Geometric Asymptotics · Geometric Asymptotics Nalini Joshi for the ﬁrst Painlevé equation In...

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