Geometric Asymptotics · Geometric Asymptotics Nalini Joshi for the first Painlevé equation In...
Transcript of Geometric Asymptotics · Geometric Asymptotics Nalini Joshi for the first Painlevé equation In...
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Geometric Asymptotics
Nalini Joshi
for the first Painlevé equation
In collaboration with J.J. Duistermaat and H. Dullin
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PI in Cy = 6 y2 + x
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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)
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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.
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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.
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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|→ .∞
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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
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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:
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The Projective Plane ℂℙ2
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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]
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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]
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
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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]
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
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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]
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
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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
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First Blow-up
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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):
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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):
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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):
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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):
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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.
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Second Blow-Up
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Second Blow-Up•Chart (2,1): [1 : u211 : u212] = [1 : u111/u112 : u112]
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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
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Second Blow-Up•Chart (2,1): [1 : u211 : u212] = [1 : u111/u112 : u112]
•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
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Second Blow-Up•Chart (2,1): [1 : u211 : u212] = [1 : u111/u112 : u112]
•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
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Second Blow-Up•Chart (2,1): [1 : u211 : u212] = [1 : u111/u112 : u112]
•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
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Second Blow-Up•Chart (2,1): [1 : u211 : u212] = [1 : u111/u112 : u112]
•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
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Exceptional LinesExceptional Curves
L2 L1
L0
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Third Blow-Up
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Third Blow-Up•Chart (3,1): [1 : u311 : u312] = [1 : u211/u212 : u212]
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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
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Third Blow-Up•Chart (3,1): [1 : u311 : u312] = [1 : u211/u212 : u212]
•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
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Third Blow-Up•Chart (3,1): [1 : u311 : u312] = [1 : u211/u212 : u212]
•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
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Third Blow-Up•Chart (3,1): [1 : u311 : u312] = [1 : u211/u212 : u212]
•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
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Third Blow-Up•Chart (3,1): [1 : u311 : u312] = [1 : u211/u212 : u212]
•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
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Exceptional LinesExceptional Curves
L3 L2 L1
L0
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Fourth Blow-Up
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Fourth Blow-Up•Chart (4,1): [1 : u411 : u412] = [1 :
u311 − 4
/u312 : u312]
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Fourth Blow-Up•Chart (4,1): [1 : u411 : u412] = [1 :
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
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Fourth Blow-Up•Chart (4,1): [1 : u411 : u412] = [1 :
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
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Fourth Blow-Up•Chart (4,1): [1 : u411 : u412] = [1 :
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
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Fourth Blow-Up
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Fourth Blow-Up• Chart (4,2):
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Fourth Blow-Up• Chart (4,2):
[1 : u421 : u422] = [1 : (u311 − 4) : u312(u311−4) ]
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Fourth Blow-Up• Chart (4,2):
[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.
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Exceptional LinesExceptional Curves
L4
L3 L2 L1
L0
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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.
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Exceptional Curves
L9
L8
L7L6
L5L4
L3 L2 L1L0
Exceptional Lines
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The Poles
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The Poles•The vector field is regular along and transversal to u912=0 which is L9.
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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
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Pole Dancing
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Pole Dancing
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Pole Dancing
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Pole Dancing
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Pole Dancing
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Pole Dancing
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Pole Dancing
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Pole Dancing
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Pole Dancing
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Pole Dancing
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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.
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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
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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.
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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
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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.