Non-perturbative effects in the general Painlevé II equation

(1)

Non-perturbative effects in the

general Painlev´

e II equation

Olivier de Gaay Fortman

student number 10739521

Institute for Theoretical Physics Amsterdam

Faculty of Science

University of Amsterdam

Report Bachelor Project

Physics and Astronomy

Size 15 EC

Project conducted between 03/04/2017 and 30/06/2017

Submission date: 07/07/2017

Supervisor: dr. M.L. Vonk

Second assessor: dr. A. Castro Anich

(2)

(3)

Abstract

In this report the resurgent properties of the second Painlevé equation are presented as well as their interpretation in physics. It is a subject that has been studied in extensive detail for a specific form of the Painlevé II equation, but less is known for the general case. In particular, the aim of this research is to understand the effects of setting the constant term in the Painlevé II differential equation not equal to zero on the one-parameter transseries solution. It is shown that for a non-trivial constant term the perturbative series as a solution to the general Painlevé II equation is no longer expanded in the closed string coupling contrary to the trivial case.

(4)

Samenvatting

Figure 1: Groei van de coefficienten van een oplossing van PII voor de constante α = 0.001.

Paul Painlev´e (1863 - 1933) was vlak na de Eerste Wereldoorlog de Franse minis-ter van Oorlog. Deze intrigerende man was daarvoor ook wiskundig zeer actief geweest: rond 1900 bedacht Painlev´e een classificatie voor differentiaalvergeli-jkingen. Hij zocht naar vergelijkingen met polen als enige beweeglijke singular-iteiten. Later is men dit de Painlev´ e-eigenschap gaan noemen. Hoewel het er veel waren bleek het grootste deel met pen en papier op te lossen, dat wil zeggen: met een eindig aantal

rekenstap-pen en dus zonder computer. Slechts zes ervan zijn niet analytisch op te lossen of in versimpelde vorm weer te geven. Deze vergelijkingen worden de Painlev´ e-vergelijkingen genoemd; ze zijn wel op te lossen door oneindige reeksen te gebruiken. Op het moment dat Painlev´e stopte met wiskunde en politiek actief werd on-twikkelden Theodor Kaluza en Oskar Klein het idee dat elektromagnetisme afgeleid kan worden van de zwaartekracht. In hun costructie was een extra, onzichtbaar kleine ruimtedimensie nodig. Dit idee was oorspronkelijk een curiositeit, maar zou een halve eeuw later een centrale rol gaan spelen in de snaartheorie. Rond 1970 werd aange-toond dat de beschrijving van de sterke kernkracht ook voor de quantummechanica van zeer kleine trillende snaartjes moet gelden, met eigenschappen beschreven door trillingswijzen. Snaren kunnen in twee vormen voorkomen: open snaren met twee eindpunten, en gesloten snaren die een lusje vormen. Wat zou het wiskundige ver-schil zijn tussen deze twee snaren?

Hoewel voor deze ingewikkelde theorie te onderbouwen veel wiskunde nodig is, vindt niet alle wiskunde zijn toepassing in de natuurkunde. In dit onderzoek is vooral naar de wiskundige implicaties gekeken van het veranderen van de constante α in de tweede Painlev´e vergelijking (ook wel PII genoemd). Verder wordt er kort ingegaan op de snaar-theoretische interpretatie daarvan. Het is een onderwerp dat uitvoerig bestudeerd is voor een specifieke vorm van PII terwijl er minder bekend is voor de algemene vorm. In het bijzonder is gekeken naar wanneer een oneindige reeks een gesloten snaar dan wel een open snaar beschrijft. Reeds is bekend dat de meest

(5)

simpele oplossing van een specifieke vorm van PII een gesloten snaar beschrijft, terwijl een iets uitgebreidere oplossing dat voor een open snaar doet. In het algemene PII geval zal echter blijken dat alle oplossingen oneindige reeksen zijn in een combinatie van open en gesloten snaren. Ge¨ıntereseeerd in wat wiskunde achter snaartheorie? Lees dan mijn scriptie.

(6)

List of Figures

1 Groei van de coefficienten van een oplossing van PII voor de constante α = 0.001. . . 4 3.1 The growth of the coefficients of the perturbative solution of PI ( an

an−1) 12

3.2 Regular convergence towards A . . . 13 3.3 Accelerated convergence to A . . . 13 3.4 The growth of the first 50 coefficients of the perturbative solution of

PII in case α = 0. . . 18 3.5 PII perturbative solution coefficient growth for α = 0 showing the

convergence of (2n(2n − 1)an−1)/an towards A2 for values of n up to

n = 200. . . 19 3.6 PII perturbative solution coefficient growth for α = 0 showing the

convergence of (2n(2n − 1)an−1)/an towards A2 for values of n between

n = 180 and n = 200. . . 20 3.7 PII perturbative solution coefficient growth for α = 0 showing the first

Richardson convergence acceleration of (2n(2n − 1)an−1)/an towards

A2 _{for N = 100. . . .} ₂₁

3.8 First 11 coefficients for the perturbative solution of PII (u = P anz

1 2−

3n 2 )

for α = 0. . . 23 3.9 The growth of the coefficients of the perturbative solution to PII for

α = 1 and n up to 100. . . 24 3.10 The growth of the coefficients of the perturbative solution to PII for

α = −40 and n up to 200. . . 25 3.11 The growth of the coefficients of the perturbative solution to PII for

α = 100 and n up to 200. . . 25 3.12 Coefficient growth of the power series solution to 2u2 − 2zu − 1 = 0

with a0 := 1. . . 27

(9)

3.14 Coefficient growth of the power series solution to 2u2 − 2zu − 1 = 0 with a0 := 1. . . 28

3.15 Coefficient growth of the perturbative solution to PII for α = −1012. . 28 3.16 Perturbative coefficient growth an

an−1 for PII with α = 1/2 for steps 1n. 29

3.17 Perturbative coefficient growth for PII with α = 0.001 for steps 1n. . 30 3.18 Perturbative coefficient growth for PII with α = 0.001 for steps 2n. . 30 3.19 Perturbative coefficient growth an

an−2 for PII with α = 0.001 for steps 1n. 31

A.1 The first 10 coefficients of PI (exact fractions). . . 41 A.2 The first 10 coefficients of PI. You can see how fast they drop. . . 42 A.3 The first 15 coefficients of the PII perturbative solution in case of α =

0 (exact fractions). . . 42 A.4 The growth of the first 15 coefficients of the perturbative solution of

(10)

Chapter 1 Introduction

About a hundred years ago the French mathematician Paul Painlevé wrote down his six equations after an investigation of nonlinear second-order ordinary differential equations. He was searching for solutions in the form of new special functions, re-quiring the solution to be rational and movable singularities are limited to poles. He discovered that six equations in this classification remained irreducible and analyti-cally unsolvable: the Painlevé equations [8]. Although discovered by mathematical considerations, the equations have arisen in important physical applications such as quantum field theory, quantum gravity, general relativity, statistical mechanics, plasma physics, nonlinear optics, fibre optics and nonlinear waves [3]. In this light, perturbation theory is important as a method for finding an approximate solution to a problem. Being usually asymptotic, perturbation series occur in many fields of physics. Although perturbation theory and the Painlevé equations are not always directly applicable to a physical problem, they implicitly influence a wide variety of physical fields.

In this study the resurgent properties of the first two Painlevé equations (PI and PII) will be studied and their interpretation in physics will be considered. This sub-ject has been studied in extensive detail for the Painlevé I equation, but less is known for the general Painlevé II case. In particular, the aim is to understand the impli-cations of varying constant (denoted by α) in the second Painlevé equation. The perturbative solution will be regarded as well the one-parameter transseries solution, coefficient growth behaviour will be analyzed, physical conclusions deduced and all this will be compared with the particular case where the constant in PII equals zero. Ultimately we will see that in case of setting the constant α not equal to zero, solving the Painlevé II equation using a one-parameter transseries will not cause the pertur-bative series to be expansions in the closed string coupling constant as is the case for

(11)

α equals zero.

How does the coefficient growth behaviour of the perturbative solution depend on the constant term of the general PII differential equation? In addition, does this depen-dence effect the value of the constant in the exponent of the exponential term of the transseries solution, or of any other constant in that solution? Finally and most im-portantly, how does the one-parameter transseries solution vary with different values of the constant term in the general PII differential equation?

The research starts with determining the coefficients of the perturbative solution and discover its growth behaviour for each of the three cases PI, PII α = 0 and PII α 6= 0. From there we determine the value of the instanton action A numerically. Furthermore, by varying the value of α the influence this has on these steps will be observed. Is there be substantial difference in A between those cases? Then the one-parameter transseries will be determined. In addition, the recursion relation to the transseries solution can be obtained by solving the differential equation by plugging in an one-parameter Ansatz. In case α = 0, the perturbative coefficients u0,n 6= 0

for even n and 0 for odd n; but one might question whether u0,n = 0 for odd n in

case α 6= 0. Our expectation was that if α 6= 0 in PII, then the odd terms do not vanish. Indeed: it appears that solving the general Painlev´e II equation using a one-parameter transseries does not cause the perturbative series to be expansions in the closed string coupling constant, as is the case for α equals zero. However, although thus mathematically proved that the perturbative solution to α 6= 0 does not seem to describe closed strings, we have not been able to physically interpret this model as a string theory model; this could be the subject of future research.

(12)

Chapter 2 Preliminaries

2.1 Painlev´

e equations

Paul Painlevé searched for solutions of a suitable nonlinear differential equation in the form of new special functions. Difficulties with analyzing nonlinear differential equations could arise when the singularities of a solution are movable, i.e. their posi-tion depends on the soluposi-tion itself. If movable singularities are inevitable difficulties might still be solved by requiring that the singularities be as simple as possible: the only movable singularities are poles. An example is the differential equation of an elliptic function possessing this so-called Painlevé property. Accordingly, Painlevé proposed to classify all second order differential equation of the form

d2u

dz2 = f (z, u,

du

dz), (2.1)

with f a rational function and satisfying the Painlev´e property. He discovered that most equations in classification 2.1 were reducible to linear differential equations or solvable by elliptic functions. Only six equations remained irreducible and analytically unsolvable. These are now called the Painlev´e equations [8]. So generally speaking these equations are transcendental. The first two are defined as

Painlev´e I: 1 6 d2u dz2 = u 2_{− z} _(2.2) Painlev´e II: d2_u dz2 = 2u 3_{− 2zu − α} _(2.3)

Note that 2.3 in fact constitutes a family of equations, parameterized by a constant α. If the goal is to analyze particular differential equations that are not exactly solvable one could naturally look for other ways to solve the problem. One way to do this is by using perturbation theory.

(13)

2.2 Perturbation theory and asymptotic series

In short, perturbation theory reformulates the problem by adding a ’small’ term (usually denoted by ) to the mathematical equation of the problem. The formulation must be done in such way that in case this small term equals zero the equation is exactly solvable. Formal power series in this small parameter are called perturbation series. Quantitatively they resemble the deviation from the exactly solvable problem. The leading term in these series is the solution to the exactly solvable problem a0;

further terms describe the deviation from the initial problem [10]. As an example, consider the following equation [2].

x5+ x = 1. (2.4)

It is a quintic equation so it not have an exact answer. In order to find the real root 0.755... ones adds the small parameter: x5 _{+ x = 1.} _{The unperturbed}

problem = 0 gives x = 1 (real root). Then the solution is approximated by = P∞

n=0ann= 1 + a0 + a12+ .... The first three coefficients an are a0 = −1₅, a1 = −₂₅1

and a2 = −₁₂₅1 . Now let = 1 and add the series: 1 − .2 − .04 − .008 = 0.752 with an

error of 3 in 750 [2].

As we will need to use perturbation theory to analyze Painlev´e equations I and II, several notes should be made about asymptotic series and how they behave at zero and at infinity.

Consider a asymptotic series in w: let f (w) = ∞ X n=0 anwn. (2.5) So f (w) ≈ PN n=0anw

n _{for large N. There are two ways to proceed analyzing this}

series:

1. Keep w fixed and let N → ∞. This is called the Taylor sequence approach. The idea behind this approach is that the approximation to the function should be enhanced by choosing N larger; however, for the series we will encounter during this research - asymptotic series - this behaviour fails to occur.

2. Hence this method does not always provide a useful answer. Therefore, if you keep N fixed but large and let w → 0 you obtain so-called asymptotic sequences. This is the Poincar´e-Lindstedt method. For example, to calcu-late the fourteenth coefficient of 2.5 observe that f (w) −PN

(14)

w−N f (w) −PN n=0anwn ∼ w, so limw→0w−N f (w) − PN n=0anwn = 0 and limw→0 w−N f (w)−PN n=0anwn

w14 = a14. We conclude that whereas N → ∞ does not

always get you any further, the limit w → 0 in most cases does.

At this point we assume that although equation 2.2 is not exactly solvable, there might exist a perturbative solution of the form of some asymptotic series. The educated guess u(z) =P∞

n=0anzn unfortunately fails, which one can easily see by plugging it

in into 2.2. However, u(z) =P∞

n=0anzαn+β does work. Many authors [7] have shown

that α = −2₅ and β = 1₂ and consequently a0, a1, ... until a5 appear to be a0 = ±

1, (for which most authors choose a0 = 1), a1 = −₄₈1 and a2= −₄₆₀₈49 . The growth

behaviour of these coefficients is an ∼ _An!n, with the value of the constant A (called

the instanton action) A = ±8

√ 3

5 . Furthermore, if y is a solution, then y → ζ

3_{, t → ζt}

also, with ζ a complex number and fifth order root of 1.

2.3 Series acceleration

It appears to be possible to sum divergent series, even though they do not converge. Indeed, the summing of divergent series ”converges” faster than convergent series [2]. The fact that 1 +1₂ + 1₃ + ... = ζ(1)(= ∞) diverges does not necessarily lead to nowhere: it appears to be possible to assign a meaningful number to the sum of a series that does not converge. Instructive ways to handle these divergent series are linear and nonlinear sequence transformation. The first comprises Euler Summation, Borel Summation and Generic Summation; the second involves Shanks Transformation and Richardson Extrapolation. Especially this last method is useful when one wants to calculate a large amount of coefficients of a series [1].

2.3.1 Linear sequence transformation

To start with Euler, let P∞

n=0an be not convergent. Construct f (x) :=

P∞

n=0anxn,

for |x| < 1. Then define E, the Euler sum of the series, as E ≡ limx→1f (x). For

example, if 1 − 1 + 1 − 1 + −..., then f (x) = 1 − x + x2− x3_{+ −... =} 1

1+x(|x| < 1).

We conclude that the Euler sum equals [2] E = limx→1

1 1 + x =

1

2. (2.6)

For the Borel sum, again start withP∞

n=0anand meanwhile note that

R∞

0 dte

−t_tn _{= n!}

(to see why: R₀∞dte−ttn _{= (−t}n_e−t₎∞ 0 + R∞ 0 dt n e −t_tn−1 _{= n}R∞ 0 dt e −t_tn−1 _{= n(n −} 1(n − 2)...R₀∞dt e−ttn−(1+1+1...) = n! R₀∞dt e−t = n!). So 1 = R∞ 0 dte −t_tn n! . Therefore

(15)

P∞ n=0an ∗ 1 = P∞ n=0an R∞ 0 dte −t_tn

n! . Now the Borel sum B is defined as B(

P∞ n=0an) ≡ R∞ 0 dt e −t₍P∞ n=0 antn

n! ), so the sum and the integral are changed in order: first

perform the sum, which is in fact some function of t, then integrate the result over t. Borel Summation is more powerful than Euler Summation: the sum converges faster and the chance of convergence is better [2]. Let us now see what happens to the aforementioned sum: B = B(1 − 1 + 1 − 1 + −...) = Z ∞ 0 dt e−t( ∞ X n=0 −tn n! ) = Z ∞ 0 dt e−2t= 1 2, (2.7)

consistent with equation 2.6.

Finally, let S denote the Generic Summation ’transformation’, similar to what E and B meant above, and let the number that will be assigned to the summation be s. Then the properties of a general summation method must be:

property#1:

S(a0+ a1+ a2+ ...) = s = a0 + S(a1+ a2+ ...)

because if not the summation does not make sense. property#2:

S(P(αan+ βbn) = αS(P an) + βS(P bn) (linearity)

Now S(1−1+1−1+−...) = 1+S(−1+1−1+1−+...) = 1−S(1−1+1−1+−...), 2S(1− 1+1−1+−...) = 1, S(1−1+1−1+−...) = 12 = s which is consistent with equations 2.6

and 2.7. So Generic Summation results in the same value of 1₂, assuming linearity and summation, without knowing what the actual summation procedure was. Another example is the diverging series 1 + 0 − 1 + 1 + 0 − 1 + 1 + 0 − 1 + −..., Euler gives f (x) = 1−x2+x3−x5_+x6_−x8_{... = (1+x}3_+x6_+...)−(x2_+x5_+x7_{+..., (you can change}

positions because it is a Taylor series and inside the radius of convergence the order of the series does not matter, this holds only for rare cases). f (x) = _1−x1 3−

x2 1−x3 = 1−x2 1−x3, take limx→1f (x) = E = limx→1 1 − x2 1 − x3 = −2x −3x2(l’Hˆopital) → 2 3. (2.8)

Generic Summation shows that s = _{S(1 + 0 − 1 + 1 + 0 − 1 + 1 + 0 − 1 + ...), s =} 1 + S(0 − 1 + 1 + 0 − 1 + 1 + 0 − 1 + 1...), s = 1 + S(−1 + 10 − 1 + 1 + 0 − 1 + 0...). Adding these three gives 3s = 2 + S(0 + 0 + 0 + 0 + ...), s = 23 [2], consistent with

equation 2.8.

(16)

to assign a meaningful number to a divergent series. And because perturbation series are typically divergent, we can use these techniques on perturbation series. However, especially nonlinear sequence transformations appear to provide powerful numerical methods for the summation of divergent series that among others arise in perturbation theory. Indeed, nonlinear sequence transformation is regarded as a highly effective extrapolation method. It is very useful in the acceleration of convergence - as our research encompasses the determination of coefficient growth up to high orders we will at this point clarify its most important aspects.

2.3.2 Nonlinear sequence transformation

Nonlinear sequence transformation is used to improve the rate of convergence of a converging sequence; most part of the absolute error is eliminated by it [4].

By using Shanks Transformation [1] on a sequence {am}m∈N the series A =P ∞ m=0am

can be determined. The partial sum An is defined as An =

Pn

m=0am and forms a

new sequence {An}n∈N. If the series converge, An approaches the limit as n → ∞.

The Shanks Transformation S(An) of the sequence Anis the new sequence defined by

S(An) := An+1An−1− A2n An+1− 2An+ An−1 = An+1− (An+1− An)2 (An+1− An) − (An− An−1) . (2.9) This new sequence S(An) converges often more rapidly than the sequence An. It is

possible to accelerate more: S3(An) = S(S(S(An))) for example.

Richardson Extrapolation is another sequence acceleration method. It is similarly used to improve the rate of convergence of a sequence. The method comprises the following steps [1]:

1. Start by approximating the sequence SN by SN = S +_Na +_Nb2 +

c

N3..., where S

denotes the limit of the sequence, and choose up to which order of N you want to calculate the approximation: the higher the order the better the approximation of the limit. Choosing to take only one order, 1/N , for example, gives

2. SN ∼ S + _Na for very large N. Then N SN ∼ N S + a, SN +1 ∼ S + _{N +1}a and

(N + 1)SN +1 ∼ (N + 1)S + a.

3. Extracting then gives (N + 1)SN +1− N SN ≡ R1 ∼ S (as N→ ∞). This is the

(17)

4. Analogously one finds (N +2)2SN +2−2(N +1)2SN +1+N2SN 2 ≡ R2 ∼ S (as N→ ∞) and (N +3)3_S N +3−3(N +2)3SN +2+3(N +1)3SN +1−N3SN 3! ≡ R3 ∼ S (as N→ ∞) and so forth.

The higher the Richardson correction (so the larger the n in Rn), the better the

ap-proximation of the limit. With the Richardson extrapolation one can rapidly find the number where a sequence converges to. Consider for exampleP∞

n=1 1

n2. The limit of

this sequence should be 1 + 1₄ + 1₉ + ... ≈ 1.644934066848 = P1012

n=1 1

n2. Calculating

without accelerated convergence provides P1000

n=1 1

n2 ≈ 1.643934566681 by letting the

computer calculate 1000 terms. For a first Richardson extrapolation with N = 50 the value of the asymptotic series as calculated by the computer, is 1.644740576759. The same extrapolation with N = 100 provides the number 1.644884890284. A second Richardson extrapolation is much more accurate in spite of using a lower value of N, as we will see in the following chapter.

Now that we have all the tools to analyze the first two Painlev´e equations, let us proceed by determining the most straightforward solution of the first: PI.

(18)

Chapter 3 Results

3.1 Painlev´

e I

3.1.1 Perturbative solution to PI

In order to see how the coefficients of the perturbative solution to PI behave we shall need to calculate the coefficient recursion relation. To do so we need an expression for the solution; recall that the differential equation reads

u00(z) −1 6u

2_{(z) − z = 0,} _(3.1)

and we plug u(z) = P∞

n=0anz

1 2−

5n

2 into this expression. Then

u00(z) = ∞ X n=0 25n2_{− 1} 4 anz −3 2− 5n 2 u2(z) = ∞ X N =0 N X n=0 anaN −nz1− 5 2(N ) (3.2) which implies ∞ X N =0 N X n=0 anaN −nz1− 5 2(N )− ∞ X N =1 25(N − 1)2− 1 24 aN −1z 1−5N₂ _{− z = 0.} _(3.3)

N = 0 gives the coefficient before z: a2

0 = 1. Proceeding: ∞ X N =1 ( N X n=0 anaN −n− 25(N − 1)2_{− 1} 24 aN −1)z 1−5₂(N ) _{= 0,} _(3.4) resulting in ( N X n=0 anaN −n) − 25(N − 1)2_{− 1} 24 aN −1= 0. (3.5)

(19)

By taking some factors out and renaming the coefficients, one obtains a0 = ±1 an = 25(n − 1)2 _{− 1} 48 an−1− 1 2 n−1 X l=1 alan−l. (3.6)

The two possibilities for a0 (and thus the two possibilities for the series of solutions)

coincide with the two roots of 1 in the complex plane: P∞

n=0an(±

√

z)1−5n_.

By letting the computer calculate the coefficients we conclude the 0th instanton so-lution equals u(0)(z) ≈ √z(1 − 1 48z −5 2 − 49 4608z −10 2 − 1225 55296z −15 2 − ). (3.7)

Note that this solution is a power series in z−52. This parameter is known to be the

coupling constant of the minimal string theory. Nonetheless this string theory is a closed string theory so we expect its perturbative free energy to be a function of the closed string coupling constant. On the other hand, the non-perturbative effects in string theory are associated with open strings because they are related with D -branes. The closed string coupling constant is the square of the open string coupling constant. Concluding this section one expects the non-perturbative contributions to the free energy to be expansions in z−54 [7].

3.1.2 Transseries solution to PI

To determine the full transseries solution we start with the perturbation series and with 3.1 of course. A general solution would be of the form u(z) =P∞

n=0anzβ

0

n+γ_{; we}

need find the values of β0 and γ. The leading term in the series is the first, a0zγ, so

u00→ zγ−2 _{and u}2 _{→ z}2γ_{. Then the possibilities are γ − 2 = 1, γ − 2 = 2γ and 2γ = 1,}

implying γ = 3, γ = −2 or γ = 1₂, respectively. However, z = ∞ is the point where the string coupling equals zero, so where we expect a finite solution. If γ = −2, then u ∼ _z12 and the entire solution vanishes at z = ∞ which is a contradiction: plugging

∞ into 2.2 then would give 0 − 0 − ∞ 6= 0 so it would not be a solution to PI at all. We conclude that

γ = 1

2 (3.8)

(if the expansion is around z = 0 in stead of z = ∞, γ = 3 would be the other possibility). Then follows that β0 should equal

β0 = −5

(20)

because if you take another term (u ∼ z1/2_{+ z}β0+1/2_{), then u}00 _{gives −}3

2 and β

0

− 3 2,

u2 _{gives 1, β}0_{+ 1 and 2β}0_{+ 1 and z gives 1; 2β}0_{+ 1 = β}0₋3

2 and β

0

+ 1 = −3₂ leading to the answer. To proceed determining the unknown constants in the the transseries solution we again start with equation 3.1 but now we plug in the Ansatz

u(z) = ∞ X l=0 ∞ X n=0 Clzβl+β 0 n+γ_e−lAzδ . (3.10)

The first 100 coefficients of the perturbative solution are determined (see the list with coefficients in Appendix 1) and because β0 and γ are now known we have:

u(z) ∼ ((z1/2+ 1 48z −2 + ...) + e−Azδ Czβ+γ+ Dzβ+γ+ (3.11) Collecting terms that contain e−Azδ we see that u2 _{provides 2Cz}β+1_{, u}00 _{gives (1)}

zβ+2δ−32, (2) zβ+δ− 3 2, (3) zβ+δ− 3 2 and (4) zβ− 3 2. So β + 1 = β + 2δ − 3 2, δ = 5 4. Then −1 6(A 2_δ2_{C) + 2C = 0 implying} A = ±8 √ 3 5 (3.12)

for which we choose the positive value so that e−Azδ will suppress the solution instead of blow it up for large z. Proceed with the terms (2) and (3) from the list above. These terms should cancel out (both have same z exponent), so −Aδ(β+δ−1

2−Aδ( 1

2+β) = 0,

(β + δ −1₂) + (1₂ + β) = 0, from which we ultimately obtain β = −5

8. (3.13)

Concluding, we now know that 3.10 takes the form u(z) = ∞ X l=0 ∞ X n=0 Clz−58l− 5 2n+ 1 2e−lAz 5 4 . (3.14)

When one calculates the coefficients for the one-instanton correction l = 1 one obtains the following solution [7]:

u(1)(z) ≈ Cz−18e−Az 5 4 (1 − 5 64√3z −5 4 + 75 8192z −5 2 + .. (3.15)

Indeed this is a series in the open string coupling z−54: the closed string coupling of

the perturbative expansion z−52 is the square of the open string coupling of the first

instanton correction z−54.

(21)

Figure 3.1: The growth of the coefficients of the perturbative solution of PI ( an

an−1)

exponent e in 3.14 is correct. Assume that angrows with (2n)!

A2n [7]: it increases as a

fac-ulty. But how fast do the coefficients of PI exactly grow? And one might also wonder: why 2n instead of n? In case of PI, the series√zP∞

n=0a n_z−5n

2 are an approximation

of the solution for large n. However, the transseries √zP∞

l=0 P∞ n=0a (l) n z −5n 2 e−Alz 5 4

form a better approximation. Here the perturbative solution l = 0 coincides with the regular series and closed strings, g(0)s = z−

5

2. Furthermore, l > 0 coincides with open

strings: e.g. gs(1) = z−

5

4. In this scenario, we find

√ zP∞

m=0lmz−

5m

4 . So b_2n = a_n and

b2n+1 = 0. Therefore b2n = an ∼ (2n)!_A2n , which explains the 2n. With this knowledge,

consider an an−1 = (2n)! (2n − 2)! A2n−2 A2n = 1 A22n(2n − 1). (3.16)

So the coefficient growth should be quadratic. In figure 3.1 it can be seen that this quadratic behaviour in fact agrees with what you see if you plot an

an−1 over n. Now

be-cause an an−1 = 1 A22n(2n−1), A ≈ q 2n(2n−1)an−1

an for large n and the positive sign chosen

for A. First this is done without series acceleration. Taking n up to 1000 coefficients it takes considerably much time and the result is not too accurate (see figure 3.2). The value of A for n up to 100 equals A ≈ 2.79227, which is not bad, considering the literature value [6] is equal to A = 8

√ 3

5 ≈ 2.771281292110204. However, afterwards

the Richardson Extrapolation method is used to accelerate convergence to A and to improve the accuracy of the answer. By looking at graphs 3.2 and 3.3 one observes the difference in convergence speed graphically.

The first Richardson Transform for N = 50 resulted in A ≈ 2.770468668577 (13 digits). Obviously this is a lot more accurate; besides, the computer calculation last only seconds. However, in case of the first Richardson approximation you take powers of 1/n, whereas at the second up to 1/n2 _{and at the third up to 1/n}3_{. Thus you can}

(22)

Figure 3.2: Regular convergence towards A

Figure 3.3: Accelerated convergence to A

value of N. Indeed, the second Richardson Transform gave A ≈ 2.771297025280546 for N = 50.

A final remark about the slope of the curve: when you consider equation 3.14 the A resides in the exponent of e, so you might wonder what the relation could be between the coefficient growth and the some constant term in the exponent of e. Start with

(23)

an educated guess by assuming that an ∼ (kn)! and see what happens. Let Bn:= an an−1 = (n)! (k(n − 1))! = (kn)! (kn − k))! = (kn)(kn − 1)(kn − 2)... (kn − k)(kn − k − 1)... = (kn)(kn − 1)(kn − 2)...(kn − k)(kn − k − 1)... (kn − k)(kn − k − 1)... = kn(kn − 1)(kn − 2)...(kn − (k − 1)) ∼ nk (3.17)

If you plot this, realize that the result could be a straight line, a quadratic form or even something else. It will tell you the behaviour of the coefficient growth. Because as Bn ∼ nk, log(Bn) ∼ k log(n), so plotting log(Bn) over log(n) will give you the

slope of the curve (this even appears to be true if k is not a integer).

3.1.3 Physical interpretation of the expansion behaviour of

the solution to PI

We derived the fact that the perturbative solution to PI expands in z−52 which

led us to equation 3.14. The two-parameter transseries solution to PI looks like P∞ n=0σ1nσ2me −A(n−m)xP∞ g=0a (n|m) g xg, with P∞_g=0a(n|m)g xg = Φ(n|m)(x). Furthermore,

gclosed is the coupling constant. In case of closed strings (l = 0), g (0) s = z−

5

2. These

imply regular series and closed strings. l > 0 coincides with open strings: e.g. gs(1) = z−

5

4. Therefore, and g2

open = gclosed: for the expansion in closed strings the

coupling constant is g2_open.

Moreover, Φ(n|0) resembles open strings on D-branes. For Φ(n|m) the general conjec-ture is that open strings are described on n D-branes and m ’generalized D-branes’, that have a lot in common with anti-branes but are not precisely the same. What they truly are is in fact not yet known. In practice we find that Φ(n|n) _{with m = n are}

again expansions in z−5/2, the closed string coupling. The D-branes and the general-ized D-branes cancel each other out so that open strings do not seem to end anywhere and we find an expansion in the closed string coupling constant.

Finally, the physical interpretation of A, appearing in e−lAzδ, is the ’instanton action’. If the action is very large, the tunneling requires more effort.

(24)

3.2 Painlev´

e II

3.2.1 Particular PII equation

3.2.1.1 Perturbative solution to PII with α = 0 Recall that the second Painlev´e equation reads

u00− 2u3 _{+ 2zu + α = 0.} _(3.18)

However, later on when several findings are done for this general form of the PII equation it will be convenient to check these findings with the particular case where α = 0. Therefore we need to investigate the solution to this equation for which α is set to equal zero. For that reason we will start looking into

u00− 2u3_{+ 2zu = 0} _(3.19)

and its solution.

In order to determine the transseries solution to 3.19 we need to determine some coefficients of the perturbative solution. In addition, in order to determine the co-efficient recursion relation of the perturbative solution u(0)_{(z) =} P∞

n=0anzβ

0

n+γ _to

3.19 we need to know the precise form of this solution. Accordingly, similar steps as performed in section 3.1.2 are performed; they will not be provided in much detail again. Starting with γ, we have u ∼ zγ as the leading term, u00 provides zγ−2, u3 gives z3γ _{and zu coincides with z}γ+1_{. Then γ + 1 = 3γ so γ =} 1

2. Now take two

leading terms: u ∼ zγ_{+ z}β0+γ _{= z}1₂ _{+ z}β0+1₂_{. Then u}00 _{gives z}−3 2 + zβ 0 −3 2, u3 gives z32+ 3zβ 0

+3₂_{+ 3z}2β0+3₂_{+ z}3β0+₂3 _{and zu gives z}3₂_{+ z}β0+3₂_{. It follows that 2β}0₊3 2 = β

0

−3 2

or 2β0 +3₂ = −3₂ so β0 = −3 or β0 = −3₂. Therefore the perturbative solution is upert(z) = ∞ X n=0 anz−3n+ 1 2, (3.20)

as β= -3 for α = 0 and (β = −3₂ for α 6= 0 - in fact, β = −3₂ is always good, but then in the 0th instanton case all the coefficients for odd n will be zero, and all the even terms will make the exponents also multiples of three).

Knowing the perturbative solution to 3.19 one is now able to derive the recursion relation for the coefficients of this solution, in a way similar to what we did in section 3.1: an= an−1 36(n − 1)2_{− 1} 16 − n−1 X l=1 an−lal 2 − n−1 X k=1 k X l=0 an−kak−lal 2 , (3.21)

(25)

yielding u(0)(z) =√z 1 − 1 16z −3₋ 73 512z −6₋ 10657 8192 z −9₋ 13912277 542888 z −12_{− ...} . (3.22) See figure A.1 for the first 15 coefficients in fractions presented in a tabular form and their growth presented graphically.

Note that this solution is a power series in z−3. Being a closed string theory we expect its perturbative free energy to be a function of the closed string coupling constant and the non-perturbative effects in string theory to be associated with open strings, just as in the PI case. The closed string coupling constant is the square of the open string coupling constant; accordingly we expect the non-perturbative contributions to the free energy to be expansions in z−32. In other words: in case of setting the constant α

in equation 3.41 equal to zero, solving 3.41 using a one-parameter transseries would cause the perturbative series to be expansions in the closed string coupling constant g_open2 .

3.2.1.2 Transseries solution to PII with α = 0

Now that we have calculated the perturbative solution and some of its coefficients we can proceed determining the transseries solution to equation 3.19. As an Ansatz, using the notation, (l)(z) =P∞

n=0ul,nz

β0n_{, use the solution}

u(z) = ∞ X l=0 Clu(l)(z) = ∞ X l=0 Clzβl+γe−lAzδ(l)(z), z → ∞ (3.23) just like the PI case. The unknown values will have to be determined by plugging the transseries Ansatz back into equation 3.19. It should be checked that the analytically determined value of A coincides with its numerical counterpart. The numerical way to find A is assuming that an ∼ (2n)!_A2n which implies

an

an−1 = A

−2_{(2n)(2n − 1) so it is}

quadratic which agrees with figure A.4. With this relation one is able to determine A2 ₌ (2n)(2n−1)an−1

an numerically; by calculating A analytically, we can check whether

the two values coincide. For the latter, rewrite the Ansatz as u(z) = ∞ X l=0 ∞ X n=0 Clzβl+γe−lAzδul,nzβ 0 n_, _{z → ∞.} (3.24)

Starting with only a couple of terms: u ∼ (z1/2− 1 16z

−5

2) + e−Az δ

(Czβ+12). Collecting

(26)

u3 ∼ zβ+3₂_{, z}β−3₂_{, z}β+3₂ _{and z}β−3₂_{. Finally, z u ∼: z}β+3₂_{. We can see that β +}3

2 should

equal β + 2δ −3₂ and hence

δ = 3

2. (3.25)

To determine A, we have the terms A2δ2Ce−Azδzβ+2δ−32 and 3Ce−Az δ zβ+32, Ce−Az δ zβ+32, so A2δ2C − 6C + 2C = 0, A2δ2 = 4, A2 = _δ42 = 16

9 so we conclude that the instanton

action

A = ±4

3. (3.26)

There is a choice of sign. In the one-parameter transseries it is common to choose the positive sign, as by doing so the instanton factor exp(−Azδ_{) is exponentially}

suppressed for large values of z instead of making the solution blow up for positive values of l as z → ∞. To determine β, again observe that u ∼ (z1/2 − 1

16z −5

2) +

e−Azδ(Czβ+12), so u00has the following terms that contain Ce−Az δ : (1) A2δ2Czβ+2δ−32, (2) −AδC(β+δ−1₂)zβ+δ−32, (3) (β+1 2)(−Aδ)Cz β+δ−3₂ _{and (4) (β+}1 2)(β− 1 2)Cz β−3₂_{. (2)}

and (3) should cancel out: −AδC(β +δ −1₂)+(β +1₂)(−AδC) = 0, β +δ −1₂+β +1₂ = 0 implying

β = −3

4. (3.27)

Thus the transseries solution takes the following form: u(z) =√zP∞ l=0C l_z−3l 4+ 1 2e−lAz 3 2 (l)(z), with z → ∞, or u(z) = ∞ X l=0 ∞ X n=0 Clz12− 3l 4− 3n 2 e−lAz 3 2 ul,n, (3.28) where A equals ±4₃.

Now let us calculate A numerically to see whether the two values agree with each other. Recall that for α = 0, an

an−1 = A

−2_{(2n)(2n − 1), which is quadratic in n and}

thus agrees with figure 3.4. Plotting q

(2n)(2n−1)an−1

an and determining the number to

which the graph converges is one numerical way to find the (positive) value of A (A takes a positive and a negative value as we have also seen above).

The analytic value of A for α = 0 is |A| = 4 3,so A

2 ₌ 16

9 ≈ 1.77778. The value of A 2

obtained by using the first Richardson extrapolation method in Mathematica is A2 _≈

1.77636 for N = 50 and A2 _{≈ 1.77743 for N = 100. Below the graphs of convergence}

towards A2 _{are plotted.}

To conclude, the only thing that is left to do is calculate the actual coefficients of all the instanton solutions to PII (l = 1, l = 2, ...). Only in this manner will we be able to plug the transseries solution in into equation 3.41 to check its correctness.

(27)

Figure 3.4: The growth of the first 50 coefficients of the perturbative solution of PII in case α = 0.

Start with equations 3.41 and 3.28. Firstly, the following coefficients ul,0 were

ob-tained by the following relation for PII with α 6= 0. This will result in the recursion relation for ul,0 ul,0= 1 2l2_{− 2}( l−1 X k=1 u0,0ul−k,0uk,0+ l−1 X N =1 N X k=0 ul−N,0uN −k,0uk,0). (3.29)

The check with l = 0 gives u0,0 = u30,0. This is correct. Furthermore, setting u1,0 = 1,

then for l = 2, u2,0 = 1₂; the first 10 coefficients are

ul,0= 1, 1,1 2, 1 4, 1 8, 1 16, 1 32, 1 64, 1 128, 1 256, 1 512... (3.30) which does coincide with literature [9].

However, we are interested in all ul,n, so we need to adopt a different approach. Let

u(z) = ∞ X l=0 ∞ X n=0 Clz−3l4− 3n 2 + 1 2e−lAz 3 2 ul,n. (3.31) Then u00(z) = ∞ X l=0 ∞ X n=0 (14 − 6n − 3l 4 )( 10 − 6n − 3l 4 )ul,n−2 − 3l( 9 − 6n − 3l 4 )Aul,n−1+ 9 4l 2_A2_u l,n Clz6−6n−3l4 e−lAz 3 2 (3.32)

(28)

Figure 3.5: PII perturbative solution coefficient growth for α = 0 showing the con-vergence of (2n(2n − 1)an−1)/an towards A2 for values of n up to n = 200.

provided that ul,m ≡ 0 for m < 0; writing −2u3 and 2zu in a similar manner one

obtains the implicit recursion relation

0 = 14 − 6n − 3l 4 10 − 6n − 3l 4 ul,n−2 − 3l 9 − 6n − 3l 4 Aul,n−1 + 9 4l 2_A2_{+ 2} ul,n− 2 l X m=0 l−m X k=0 n X p=0 n−p X q=0 ul−m−k,n−p−qum,puk,q (3.33)

This leads to explicit relations u0,n= 1 4( 7 − 3n 2 )( 5 − 3n 2 )u0,n−2− 1 2 n−1 X p=1 n−p X q=1 u0,n−p−qu0,pu0,q− n−1 X q=1 u0,n−qu0,0u0,q (3.34) for l = 0, u1,n = (− 1 3A)( 2 3n)( 5 − 6n 4 )( 1 − 6n 4 )u1,n−1 +(2 A)( 2 3n) n X p=2 n+1−p X q=2 u1,n+1−p−qu0,pu0,q +(4 A)( 2 3n) n+1 X q=2 u1,n+1−qu0,q (3.35)

(29)

Figure 3.6: PII perturbative solution coefficient growth for α = 0 showing the con-vergence of (2n(2n − 1)an−1)/an towards A2 for values of n between n = 180 and

n = 200.

for n ≥ 1 (for n = 0 we can choose u0,1 = 1 by absorbing its value in C) and

ul,n= − 1 (9₄l2_A2 _{− 4)}( 14 − 6n − 3l 4 )( 10 − 6n − 3l 4 )ul,n−2+ 3l (9₄l2_A2_{− 4)}( 9 − 6n − 3l 4 )Aul,n−1 + 2 (9₄l2_A2_{− 4)} l−1 X m=1 l−m X k=0 n X p=0 n−p X q=0 ul−m−k,n−p−qum,puk,q + − 2 (9₄l2_A2_{− 4)} l−1 X k=1 n X p=0 n−p X q=0 ul−k,n−p−qu0,puk,q + 6 (9₄l2_A2_{− 4)} n X p=1 n−p X q=1 ul,n−p−qu0,pu0,q+ 12 (9₄l2_A2 _{− 4)} n X p=1 ul,n−pu0,pu0,0 (3.36) for l > 1.

The resulting instanton solutions are, for l = 0, 1, 2, 3: (0)(z) = 1 − 1 16z −6/2₋ 73 512z −12/2₋ 10657 8192 z −18/2₋13912277 524288 z −24/2_{− ...} _(3.37) (1)(z) = 1 − 17 96z −3/2 + 1513 18432z −6/2₋ 850193 5308416z −9/2 + 407117521 2038431744z −12/2_{− ...} (3.38) (2)(z) = 1 2 − 41 96z −3/2 + 5461 9216z −6/2₋ 1734407 1327104z −9/2 +925779217 254803968z −12/2_{− ... (3.39)} (3)(z) = 1 4− 47 128z −3/2₊5285 8192z −6/2₋1193755 786432 z −9/2₊ 440251337 100663296z −12/2_{− ... (3.40)}

As expected, the first instanton correction is a series in the open string coupling z−32:

the closed string coupling of the perturbative expansion z−3 is the square of the open string coupling of the first instanton correction z−32. We are now in the position to

(30)

Figure 3.7: PII perturbative solution coefficient growth for α = 0 showing the first Richardson convergence acceleration of (2n(2n − 1)an−1)/antowards A2 for N = 100.

analogously analyze the general PII equation to see if a one-parameter transseries solution equivalently causes all the perturbative series to be expansions in the closed string coupling constant g_open2 .

3.2.2 General PII equation

3.2.2.1 Perturbative solution to PII with α 6= 0

Let us start with writing down 3.18 again, the general form of the second Painlev´e differential equation:

u00− 2u3 _{+ 2zu + α = 0.} _(3.41)

In order to be able to derive information about the expansion behaviour of the solution to this equation and hopefully thereof deduce any physical interpretations of that behaviour, possibly varying with different values of α, the same procedures as those performed in the PI and PII with α = 0 cases will at this point need to be undertaken. However, to not repeat oneself only the most necessary steps will be shown.

First of all, to obtain the perturbative solution u0_{(z) =} P∞

n=0anz

γ+β0n_{, plugging}

this equation into 3.41, the terms u00, u3_{, zu and α give z}γ−2_{, z}3γ_{, z}γ+1 _{and z}0_:

γ = 1₂ or γ = −1.

Taken into account the value of β0 as well, one has the following combination: γ = −1 β0 = 3₂

(31)

γ = 1₂ β0 = −3₂. Then either u0(z) = P∞ n=0anz−1+ 3n 2 or u0(z) =P∞ n=0anz 1 2− 3n 2 .

The Painlev´e I analog is γ = 1₂, β0 = −5₂ or γ = 3, β0 = −5. Because the formula must hold in all cases, also for α = 0, we know that γ must equal 1

2 so the

second option is the correct one:

u0(z) = ∞ X n=0 anz 1 2− 3n 2 . (3.42)

Another reason for this fact is that if u ∼ 1_z, the entire solution vanishes at z = ∞, whereas just as we saw in the PI case, z = ∞ is the point where the string coupling equals zero, so where we expect a finite solution.

Let us proceed with finding the coefficients of this infinite series. Consider equa-tion 3.41. Plug in 3.42; after several calculaequa-tions one finds a0 = 0 or a20 = 1, and

a1 = α₄. There the α appears for the first time; afterwards it will just appear

indi-rectly, because of its presence in a1. Deciding to normalize the solution by setting

a0 = 1, one has a0 = 1, a1 = α 4 (3.43) and an= an−2 9(n − 2)2_{− 1} 16 − n−1 X l=1 alan−l(a0) 2 − n−1 X k=1 k X l=0 alak−lan−k 2 . (3.44) Notice the similarity with equation 3.21. The first five coefficients are

1,α₄, −3α₃₂2 − 1 16, − α3 128 − 3 4 −3α2 32 − 1 16 α +α₈ and −3 32 −3α2 32 − 1 16 α2−3 2 −3α2 32 − 1 16 2 +35₁₆−3α2 32 − 1 16 −3 4 −α3 128 − 3 4 −3α2 32 − 1 16 α +α₈α. By observing these coefficients, that follow from the value of α in equation 3.43,

note that a1 only equals zero for α = 0. This seemingly trivial statement does

indi-cate that whereas beforehand we could have expected that also for other values of α the odd coefficients would vanish, now we are sure that this cannot be true. α = 0 is the only ’closed string theory’ in the PII family of equations.

Before proceeding with calculating A and the unknown constants of the transseries solution we should make sure the above mentioned relation is correct. Firstly this could be done by checking whether α = 0 returns equation 3.22, in other words: check

(32)

whether the first six coefficients are 1, −₁₆1 , −₅₁₂73, −10657₈₁₉₂, −13912277₅₄₂₈₈₈ and−8045883943_8388608 . It should have odd terms vanishing, as u(z) for α = 0 is u = P anz

1 2−3n instead of u =P anz 1 2− 3n

2 ; indeed, the first 11 coefficients are:

Figure 3.8: First 11 coefficients for the perturbative solution of PII (u =P anz

1 2−

3n 2 )

for α = 0.

But if the odd terms vanish anyway we might as well replace n by 2n in equation 3.44 and let the sum go until 2n − 2 as a2n−1 vanishes:

a2n = a2n−2 9(2n − 2)2_{− 1} 16 − 2n−1 X l=1 ala2n−l(a0) 2 − 2n−1 X k=1 k X l=0 alak−la2n−k 2 , = a2(n−1) 36(n − 1)2 _{− 1} 16 − 2(n−1) X l=1 ala2n−l 2 − 2(n−1) X k=1 k X l=0 alak−la2n−k 2 . (3.45)

Observe the resemblance with 3.21.

Now as a second check write u(z) ≈ a1z1/2 + a2z(1/2)−(3/2) + ... (the first couple

of coefficients, start with α = 0) and plug it into the PII differential equation, as u00(z) − 3u(z) + 3zu(z) + α should be smaller the larger the highest order of the so-lution you plug in. In other words: the minimum exponent of z−1 should be larger as terms of high order are closer to zero than terms of smaller order as z approaches infinity. By computer the minimum and maximum exponent appearing in that result can be calculated. For u(z) itself, the maximum exponent is (1/2) for any number of terms and minimum exponent is −(11/2) if you take five terms (remember the odd coefficients are zero). After plugging it into the differential equation, the minimum exponent equals −(15/2), and the maximum as well: everything has vanished except for that term. If you take u(z) up to ten terms its maximum exponent is (1/2); how-ever, the maximum and minimum exponent of the outcome (they are the same) of plugging it into the differential equation is −(33/2). Everything has vanished except for the term with _z33/21 . As the outcome gets smaller the larger the largest order of

the solution you plug in (33/2 is larger than 15/2), the coefficient recursion relation must be correct.

For α 6= 0 the maximum and minimum order of the result after plugging solutions up to 10 and 11 terms are: 15 and 16.5 for the solution of up to ten terms and 16.5 and 18 for the solution of up to 11 terms (the orders mentioned are negative, of course).

(33)

Figure 3.9: The growth of the coefficients of the perturbative solution to PII for α = 1 and n up to 100.

Therefore we conclude that as we take more terms of the solution, the result after plugging the solution into PII with α 6= 0 becomes smaller; this validates the form and coefficients of the solution for PII with α 6= 0.

As a final check the analytic value of A should be compared to its numerical coun-terpart. In addition, plotting the growth of the coefficients gives us a graphical representation of the expansion behaviour of the perturbative solution to the general PII equation; by changing the value of α we expect different expansion behaviour. Recall that for α = 0, an ∼ (2n)!_A2n so

an

an−1 = A

−2_{(2n)(2n − 1), which is quadratic in}

n. Analogously, by plotting an

an−1 for α = 1 it is possible form an expectation about

how an

an−1 should look like, and consequently how an is related to n and A.

an

an−1 is

plotted as a function of n in figure 3.9. The relation between an

an−1 and n is clearly

linear; this implies an ∼ n! instead of ∼ (2n!); furthermore, the slope is negative so

for α = 1, we could use an ∼ _(−A)n!n as an educated guess. Indeed:

an

an−1 = −

n A so

if this is true then A = −nan

an−1. Richardson Transforms 1 and 2 with N=100 indeed

provide 1.333240063133154 and 1.333332101080588 consistent with the analytically obtained A = 4₃.

Because α = −1 gives the same curve and the same value of A in spite of a minus sign - now we get +A instead of -A - the relation of A with n and α in case of these two values of α actually depends on the sign of α: we guess that A = −sign(α) n an

an−1 , with

sign defined as the function mapping its argument on its sign. Indeed, this holds for other values of α with |α| > 1 as well, integer as well as real. The reason why

(34)

this sign of the curve depends on the sign of α has not been investigated during this research, but it nicely coincides with the fact that for α = 0 the odd terms vanish: naturally these cannot show both types of growth behaviour, whereas the even terms, containing even powers of ±A, can show positive and as well as negative slopes in their growth behaviour.

The relation is only linear for α = ±1, however. The larger the absolute value of α, the larger the coefficient growth deviates from linear behaviour. For values α 6 40

an

an−1 is still fairly approximated by a straight line, see figures 3.10 and 3.11.

Figure 3.10: The growth of the coefficients of the perturbative solution to PII for α = −40 and n up to 200.

Figure 3.11: The growth of the coefficients of the perturbative solution to PII for α = 100 and n up to 200.

(35)

Clearly the larger the absolute value of α, the larger the n one must take in order to approximate the straight line again. Consequently, to approximate A one must evaluate −sign(α) n an

an−1 also for large values of n. With very large values of α the solution

to the differential equation 2.1 seems initially, for lower values of n, to approximate a solution to another equation. In this light, note that as one substitutes z with α2/3_z

and u with α1/3_{u, the terms with u}3 _{and zu will both obtain a factor α, but the term}

with u00 obtains a factor α−1. Accordingly, for very large α, u_α00 becomes negligible and 2.1 approximates the (algebraic) equation

2u3− 2zu − 1 = 0. (3.46) One easily derives the asymptotic series solution to this equation, which are of the form ∞ X n=0 a0z 1 2− 3n 2 _(3.47)

with coefficient recursion relation

a2₀ = 0 or a2₀ = 1, a1 = 1 3a2 0− 2 , an= −1 (3a2 0− 1) ( n−1 X k=1 akan−ka0+ n−1 X p=1 p X k=0 akap−kan−p) for n ≥ 2. (3.48)

One can see there are three choices for the first coefficient a0. Choosing a0 = 0

implies the coefficients for even n to vanish and shows growth behaviour similar to 3.12; recall that the odd terms vanish for the perturbative solution to PII if α = 0. Plotting the coefficient growth an

an−1 for the other two choices of a0 shows that they

coincide with the choice of sign for α in PII, see figures 3.12, 3.13, 3.14 and 3.15. Notice the similarity: this proves the fact that 3.41 becomes 3.46 for α very large. It forms another proof our perturbative solution to 3.41 is correct.

(36)

Figure 3.12: Coefficient growth of the power series solution to 2u2_{− 2zu − 1 = 0 with}

a0 := 1.

(37)

Figure 3.14: Coefficient growth of the power series solution to 2u2− 2zu − 1 = 0 with a0 := 1.

Figure 3.15: Coefficient growth of the perturbative solution to PII for α = −1012_.

The fact that an

an−1 is only linear for very large n does not pose a problem, as from

the Borel-methods we know that we can only be certain of growth behaviour of an

for n very large; even though it will complicate the process of finding A numerically. Typical values are shown in table 3.1.

Finally, we should mention what happens for α < 1 as that behaviour might seem unexpected. From [3] and [5] we know that for 3.41 for each integer alpha there exists one rational solution and that whenever α is half-integer the solution can be expressed in terms of Airy functions. Can we observe this difference graphically in case of the growth behaviour of the coefficients? Generally, the answer is no, an

an−1

does not behave differently in case of half integers than in case of integers α; in fact, all real values seem to imply comparable behaviour.

(38)

α N A

±10 100 1.335377272434656 ±50 100 1.339791169153759 ±100 100 1.209293552469041 ±100 200 1.333733545883948

Table 3.1: Several values of A for PII constant α and first Richardson Transforms taken at N .

However, although for α < 1 half-integer or integer does not seem to matter ei-ther - see figures 3.16 and 3.17 for the growth behaviour that starts to occur when α gets further away from 1 and closer to 0 - one clearly sees that this growth behaviour is much different than for α ≥ 1.

Figure 3.16: Perturbative coefficient growth an

an−1 for PII with α = 1/2 for steps 1n.

Indeed, when one analyzes an

an−1 for α approaching zero, the coefficients are divided

into two sections: one set of values goes towards zero for each step 2n (starting at n = 1), and one set of coefficients grows linearly (steps of 2n starting at n = 0). Below these results are plotted for α = 0.001.

The result from figure 3.19 is expected: as α goes to zero, general PII 3.41 approx-imates its particular form 3.19 and the growth behaviour of the coefficients of the solution becomes quadratic in n again.

(39)

Figure 3.17: Perturbative coefficient growth for PII with α = 0.001 for steps 1n.

Figure 3.18: Perturbative coefficient growth for PII with α = 0.001 for steps 2n. 3.2.2.2 Transseries solution to PII with α 6= 0

To proceed obtaining the one-parameter transseries solution to the general form of PII, recall equations 3.43 and 3.44.

a0 = 1, a1 = α 4, (3.49) an= an−2 9(n − 2)2− 1 16 − n−1 X l=1 alan−l(a0) 2 − n−1 X k=1 k X l=0 alak−lan−k 2 . (3.50) Then for α = {1, -1, 1₂ and −1₂}, the perturbative solutions

u0(z) = ∞ X n=0 anz 1 2− 3n 2 (3.51)

(40)

Figure 3.19: Perturbative coefficient growth an

an−2 for PII with α = 0.001 for steps 1n.

look like √ z +_4z1 − 5 32z5/2 + 15 64z4 − 1105 2048z11/2 + 1695 1024z7 + ... α = 1 √ z − 1 4z − 5 32z5/2 − 1695 1024z7 − _2048z110511/2 − 15 64z4 − ... α = −1 √ z +_8z1 − 11 128z5/2 + 2361 4096z7 − 7609 32768z11/2 + 3 32z4 + ... α = 1 2 √ z −_8z1 − 11 128z5/2 − 2361 4096z7 − 7609 32768z11/2 − 3 32z4 − ... α = 1 2.

Alternatively, we could write down the first terms of 3.51 in the most general way: √ z + α 4z + −3α2 32 − 1 16 z5/2 + −α3 128− 3 4 −3α2 32 − 1 16 α + α₈ z4 + −3 32 −3α2 32 − 1 16 α2 ₋3 2 −3α2 32 − 1 16 2 + 35₁₆−3α2 32 − 1 16 − 3 4 −α3 128− 3 4 −3α2 32 − 1 16 α + α₈α z11/2 +... (3.52)

Note that as β0 = −3₂ so the open string coupling is z−32, equal to the α = 0 case,

z−3 should be the closed string coupling constant to make the to be expansions in the open string coupling constant g2_open = gclosed. Put differently: for α 6= 0, coefficients

for odd n need to vanish in case of the one-instanton correction l = 1.

To determine the constants in the transseries solution, we concentrate on the first two terms and add the first term of the first order transseries part:

u(z) ∼ (√z +_4zα + ...) + e−Azδ(Czβ+1₂ _{+ ...). Collecting terms that contain C e}−Azδ

gives u00 ∼ zβ+2δ−3₂_{, z}β+δ−3₂ _{and z}β−₂3_.u3 _{∼ z}β+3/2_{, z}β _{and z}β−3/2_{. zu ∼: z}β+3₂_{. Then}

β +3₂ should equal β + 2δ −3₂ and hence δ = 3

(41)

which is consistent with zβ+δ−3/2 = zβ. Furthermore, collecting terms that contain C e−Azδzβ+3/2: u00 provides A2δ2Ce−Azδzβ+2δ−32, u3 gives 3Ce−Az

δ

zβ+32 and zu

con-tributes Ce−Azδzβ+32, (α does not contribute), hence A2δ2C − 6C + 2C = 0 and

A = ±4

3, (3.54)

just as in the α = 0 case. Again there is a choice of sign; we choose the positive sign as by doing so the instanton factor exp(−Azδ_{) is exponentially suppressed for large}

values of z. To determine β, again observe that u ∼ (z1/2₋ 1 16z

−5

2) + e−Azδ(Czβ+ 1 2),

so u00 has the following terms that contain Ce−Azδ: (1) A2_δ2_Czβ+2δ−3

2, (2) −AδC(β + δ − 1₂)zβ+δ−32, (3) (β + 1 2)(−Aδ)Cz β+δ−3 2 and (4) (β + 1 2)(β − 1 2)Cz β−3 2. (2) and (3)

should cancel out: −AδC(β + δ − 1₂) + (β + 1₂)(−AδC) = 0, β + δ − 1₂ + β + 1₂ = 0 implying

β = −3

4. (3.55)

Thus the transseries solution to PII with α 6= 0 takes the form of 3.28 again: u(z) = √z ∞ X l=0 Clz−34le−lAz 3 2 (l)(z), z → ∞ (3.56) or u(z) = ∞ X l=0 ∞ X n=0 Clz12− 3l 4− 3n 2 e−lz 3 2 ul,n A = 4 3. (3.57)

The calculations to obtain the explicit instanton solutions for l = 0, 1, 2, ... are done in a similar way as what we did in section 3.2.1 and can be found in appendix B.1. The results are as following. First of all,

u0,0 = 0 or u20,0 = 1 (3.58)

for which we choose the latter option: u2

0,0 = 1. Furthermore:

u0,1 =

α

4 (3.59)

which agrees equation 3.43. The explicit relation we found for l = 0 is

u0,0 = 1, u0,1 = α 4 u0,n= 1 4( 7 − 3n 2 )( 5 − 3n 2 )u0,n−2− 1 2 n−1 X p=1 n−p X q=1 u0,n−p−qu0,pu0,q− n−1 X q=1 u0,n−qu0,0u0,q (3.60)

(42)

The perturbative relation we found earlier is a0 = 1, a1 = α 4 an= an−2 9(n − 2)2− 1 16 − n−1 X l=1 alan−l(a0) 2 − n−1 X k=1 k X l=0 alak−lan−k 2 . (3.61)

For α = 0 these two equations 3.60 and 3.61 give the same coefficients. For α = 1 3.60 gives 1,1 4, − 5 32, 15 64, − 1105 2048, 1695 1024, − 414125 65536 , 59025 2048 , − 1282031525 8388608 , 242183775 262144 , ... (3.62) and 3.61 gives 1,1 4, − 5 32, 15 64, − 1105 2048, 1695 1024, − 414125 65536 , 59025 2048 , − 1282031525 8388608 , 242183775 262144 , ... (3.63) so we assume that the obtained recursion relations are the same for all values of α. The explicit relation obtained for l = 1 is

u1,n= 4 (9 − 18n − 9l) A + 48u0,1 8 − 6n − 3l 4 4 − 6n − 3l 4 u1,n−1 −6 4 (9 − 18n − 9l) A + 48u0,1 n X p=1 n+1−p X q=1 u1,n+1−p−qu0,pu0,q −12 4 (9 − 18n − 9l) A + 48u0,1 n+1 X q=2 u1,n+1−qu0,q (3.64) for n ≥ 1.

Finally, the explicit relation for l > 1 is ul,n = −1 9 4l2A2− 4 14 − 6n − 3l 4 10 − 6n − 3l 4 ul,n−2 +3l ₉ 1 4l2A2− 4 9 − 6n − 3l 4 Aul,n−1+ 2 1 9 4l2A2− 4 l−1 X m=1 l−m X k=0 n X p=0 n−p X q=0 ul−m−k,n−p−qum,puk,q +2 ₉ 1 4l 2_A2_{− 4} l−1 X k=1 n X p=0 n−p X q=0 ul−k,n−p−qu0,puk,q + 6 1 9 4l 2_A2_{− 4} n X p=1 n−p X q=1 ul,n−p−qu0,pu0,q +12 ₉ 1 4l 2_A2_{− 4} n X p=1 ul,n−pu0,pu0,0. (3.65)

(43)

The resulting instanton corrections are, for l = 0 until l = 5 for general α: (0)(z) = √z(1 + α 4z −3/2_{− (}3α2 32 + 1 16)z −6/2 + −... (3.66) (1)(z) =√z(1 + 2(−3α 2 32 − 1 16) − 5 96) z−3/2+ −... (3.67) (2)(z) =√z(1 2+ 1 6 4 −3α 2 32 − 1 16 +α 4 − 5 48 + ... z−3/2+ −... (3.68) (3)(z) =√z(1 4 + 1 16 2 −3α 2 32 − 1 16 + 1 3 4 −3α 2 32 − 1 16 + ... + ... z−3/2+ −... + −... (3.69) (4)(z) =√z(1 8 + 1 30( 1 2(2(− 3α2 32 + −...)z −3/2 + −... (3.70) (5)(z) =√z( 1 16+ α 512 + 1 48( 103α 640 + 1 4(2(− 3α2 32 − 1 16)z −3/2_{+ −...} _(3.71)

Indeed, the zeroth instanton correction is a series in the open string coupling z−32.

Hence the coupling of the perturbative expansion z−3/2 is not the square of the open string coupling of the first instanton correction z−32.

3.2.3 Conclusion

We conclude that the one-parameter transseries solution to the general form of PII 3.41 does not cause the perturbative series to be expansions in the closed string coupling constant g2

(44)

Chapter 4 Discussion

In order to fully understand the behaviour of solutions - perturbative as well as and one-parameter transseries - to the general form PII, we had to investigate PI, PII (α = 0) and PII (α 6= 0). This investigation led us to the 0th instanton solution to PI, equation 3.22. We observed its perturbative free energy to be a function of the closed string coupling constant, as expected. According to literature [7], u(1)_{(z) is an}

expansion in z−54 so the non-perturbative contributions to the free energy are indeed

expansions in the square of the open string coupling. Comparing equation 3.22 with literature has been sufficient reassurance to us that the obtained perturbative coeffi-cient recursion relation is correct.

The literature value of the instanton action A in case of PI [6] is equal to A =

8√3

5 = 2.771281292110204. Numerically we obtained A = 2.79227, without using

se-ries acceleration, taking n up to 1000 coefficients. The first Richardson Transform for N = 50 resulted in A ≈ 2.770468668577 and the second Richardson Transform gave A ≈ 2.771297025280546 for N = 50. However, a more accurate answer with higher order approximations could have been obtained by using higher order Richardson Transforms - e.g. the third correction taken up to terms of order 1/n3_{. This holds}

for later PII calculations as well.

For the particular case of PII 3.19 we obtained the perturbative solution u(0)(z) = √

z (0)(z) in equation 3.37. The closed string coupling constant is the square of the open string coupling constant; accordingly we expected the non-perturbative con-tributions to the free energy to be expansions in z−32. We obtained the transseries

solution 3.28 for which A equals A = ±4₃. The first coefficients ul,0 were calculated

first, although this was not necessary and this process could have been skipped. They were shown in 3.30 and agreed with literature [9]. Then the recursion relations for

(45)

l = 0, n ≥ 1, for l = 1 and for l > 1 resulted in instanton solutions 3.37 and 3.38, coinciding with literature. As predicted [9] the first instanton correction is a series in the open string coupling z−32 and the closed string coupling of the perturbative

expansion z−3 is the square of the open string coupling of the first instanton correc-tion z−32. Hence the perturbative series are expansions in the closed string coupling

constant g2

open in case α = 0.

The analytic value of A for α = 0 is |A| = 4₃, so A2 ₌ 16

9 ≈ 1.77778. The value

of A2 _{obtained by using the first Richardson extrapolation method via the computer}

is A2 _{≈ 1.77636 for N = 50 and A}2 _{≈ 1.77743 for N = 100. Again, a much more}

accurate value could have been obtained by taking higher order Richardson Trans-forms. However, we can reasonable consider the analytically obtained value of A to be correct, mostly because literature shows that it equals 4₃ [9].

For the general case α 6= 0 we obtained the perturbative recursion relations 3.43 and 3.44 starting with coefficients 1,α₄, and −3α₃₂2 − 1

16. Checking the correctness of

this recursion relation was done by checking whether α = 0 returns equation 3.22, in other words: check whether odd terms vanished and the other coefficients were 1, −₁₆1 , −₅₁₂73, −10657₈₁₉₂ and − 13912277₅₄₂₈₈₈ . This was indeed the case. The recursion relation without the odd coefficients (being zero anyway) is 3.21; this was not equal to 3.45 obtained by substituting n 7→ 2n in 3.44. Where this difference comes from could be possibility fur further research.

As a second check it was observed that the minimum (negative) exponent of the solution after plugging it into the general PII equation became larger by taking more terms of that solution. As the outcome gets smaller the more terms of the solution you plug in, it gave an extra reassurance that the coefficient recursion relation should be correct.

This was also done for α 6= 0: the maximum and minimum order of the result after plugging solutions up to 10 and 11 terms were 15 and 16.5 for the solution of up to ten terms and 16.5 and 18 for the solution of up to 11 terms (the orders mentioned being negative). Taking more terms of the solution caused the result after plugging the solution into PII with α 6= 0 to become more and more negligible, which validated the form and coefficients of the solution for PII with α 6= 0. As a final check the analytic value of A for the general PII equation was 4

3 as well. To numerically check

this for α = 1 we guessed the relation between an

an−1 and n to be linear by looking at

Non-perturbative effects in the general Painlevé II equation