gentile g., mastropietro v., procesi m. - periodic solutions for completely resonant nonlinear wave...

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Periodic solutions for completely resonantnonlinear wave equations

Guido Gentile , Vieri Mastropietro , Michela Procesi Dipartimento di Matematica, Universit` a di Roma Tre, Roma, I-00146

Dipartimento di Matematica, Universit` a di Roma Tor Vergata, Roma, I-00133 SISSA, Trieste, I-34014

Abstract. We consider the nonlinear string equation with Dirichlet boundary

conditions uxx u tt = (u), with (u) = u3

+ O(u5) odd and analytic, = 0 , and we construct small amplitude periodic solutions with frequency for a large Lebesgue

measure set of close to 1. This extends previous results where only a zero-measure set of frequencies could be treated (the ones for which no small divisors appear).The proof is based on combining the Lyapunov-Schmidt decomposition, which leads to two separate sets of equations dealing with the resonant and nonresonant Fourier components, respectively the Q and the P equations, with resummation techniques of divergent powers series, allowing us to control the small divisors problem. The main difficulty with respect the nonlinear wave equations uxx u tt + Mu = (u), M = 0 ,is that not only the P equation but also the Q equation is innite-dimensional.

1. Introduction

We consider the nonlinear wave equation in d = 1 given by

u tt uxx = (u),u(0, t ) = u(, t ) = 0 , (1.1)where Dirichlet boundary conditions allow us to use as a basis in L2([0, ]) the set of functions

{sin mx,m N }, and (u) is any odd analytic function (u) = u3 + O(u5) with = 0. We shallconsider the problem of existence of periodic solutions for (1.1), which represents a completelyresonant case for the nonlinear wave equation as in the absence of nonlinearities all the frequencies

are resonant.In the nite dimensional case the problem has its analogous in the study of periodic orbits close

to elliptic equilibrium points: results of existence have been obtained in such a case by Lyapunov[27] in the nonresonant case, by Birkhoff and Lewis [6] in case of resonances of order greater thanfour, and by Weinstein [33] in case of any kind of resonances. Systems with innitely many degreesof freedom (as the nonlinear wave equation, the nonlinear Schr odinger equation and other PDEsystems) have been studied much more recently; the problem is much more difficult because of the presence of a small divisors problem, which is absent in the nite dimensional case. For thenonlinear wave equations utt uxx + M u = (u), with mass M strictly positive, existence of periodic solutions has been proved by Craig and Wayne [13], by P oschel [29] (by adapting theanalogous result found by Kuksin and P oschel [25] for the nonlinear Schrodinger equation) and

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by Bourgain [8] (see also the review [12]). In order to solve the small divisors problem one has

to require that the amplitude and frequency of the solution must belong to a Cantor set, and themain difficulty is to prove that such a set can be chosen with non-zero Lebesgue measure. Werecall that for such systems also quasi-periodic solutions have been proved to exist in [25], [29],[9] (in many other papers the case in which the coefficient M of the linear term is replaced bya function depending on parameters is considered; see for instance [32], [7] and the reviews [23],[24]).

In all the quoted papers only non-resonant cases are considered. Some cases with some low-orderresonances between the frequencies have been studied by Craig and Wayne [14]. The completelyresonant case (1.1) has been studied with variational methods starting from Rabinowitz [30], [31],[11], [10], [15], where periodic solutions with period which is a rational multiple of have beenobtained; such solutions correspond to a zero-measure set of values of the amplitudes. The caseof irrational periods, which in principle could provide a large measure of values, has been studiedso far only under strong Diophantine conditions (as the ones introduced in [2]) which essentiallyremove the small divisors problem leaving in fact again a zero-measure set of values [26], [3], [4].It is however conjectured that also for M = 0 periodic solutions should exist for a large measureset of values of the amplitudes, see for instance [24], and indeed we prove in this paper that this isactually the case: the unperturbed periodic solutions with periods j = 2/j can be continued intoperiodic solutions with period ,j close to j . We rely on the Renormalization Group approachproposed in [19], which consists in a Lyapunov-Schmidt decomposition followed by a tree expansionof the solution which allows us to control the small divisors problem.

Note that a naive implementation of the Craig and Wayne approach to the resonant case en-counters some obvious difficulties. Also in such an approach the rst step consists in a Lyapunov-Schmidt decomposition, which leads to two separate sets of equations dealing with the resonant

and nonresonant Fourier components, respectively the Q and the P equations. In the nonresonantcase, by calling v the solution of the Q equation and w the solution of the P equation, if onesuppose to x v in the P equation then one gets a solution w = w(v), which inserted into the Qequation gives a closed nite-dimensional equation for v. In the resonant case the Q equation isinnite-dimensional. We can again nd the solution w(v) of the P equation by assuming that theFourier coefficients of v are exponentially decreasing. However one consumes part of the expo-nential decay to bound the small divisors, hence the function w(v) has Fourier coefficients decayingwith a smaller rate, and when inserted into the Q equation the v solution has not the propertiesassumed at the beginning. Such a problem can be avoided if there are no small divisors, as in sucha case even a power law decay is sufficient to nd w(v) and there is no loss of smoothness; this iswhat is done in the quoted papers [26], [3] and [4]: the drawback is that only a zero-measure setis found. The advantage of our approach is that we treat the P and Q equation simultaneously,solving them together. As in [3] and [5] we also consider the problem of nding how many solutionscan be obtained with given period, and we study their minimal period.

If = 0 every real solution of (1.1) can be written as

u(x, t ) =

n =1U n sin nx cos(n t + n ), (1.2)

where n = n and U n R for all n N .For > 0 we set = F , with = sgn and F > 0, and rescale u

/Fu in (1.1), so

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obtaining

u tt uxx = u3 + O(2),u(0, t ) = u(, t ) = 0 , (1.3)where O(2) denotes an analytic function of u and of order at least 2 in , and we dene = 1 , with R , so that = 1 for = 0.

As the nonlinearity is odd the solution of (1.3) can be extended in the x variable to an odd 2 periodic function (even in the variable t). We shall consider small and we shall show that thereexists a solution of (1.3), which is 2 / -periodic in t and -close to the function

u0(x, t) = a0 ( t + x) a0( t x), (1.4)provided that is in an appropriate Cantor set and a0( ) is the odd 2 -periodic solution of the

integro-differential equation a0 = 3 a20 a0 a30 , (1.5)where the dot denotes derivative with respect to , and, given any periodic function F ( ) withperiod T , we denote by

F = 1T T 0 d F ( ) (1.6)

its average. Then a 2 / -periodic solution of (1.1) is simply obtained by scaling back the solutionof (1.3).

The equation (1.5) has odd 2 -periodic solutions, provided that one sets > 0; we shall choose = 1 in the following. An explicit computation gives [3]

a0( ) = V m sn(m , m ) (1.7)

for m a suitable negative constant ( m 0.2554), with m = 2K (m )/ and V m = 2m m ,where sn( m , m ) is the sine-amplitude function and K (m ) is the elliptic integral of the rst kind,with modulus m [21]; see Appendix A1 for further details. Call 2 the width of the analyticitystrip of the function a0( ) and the maximum value it can assume in such a strip; then one has

|a0,n | e 2k | n | . (1.8)Our result (including also the cases of frequencies which are multiples of ) can be more preciselystated as follows.

Theorem. Consider the equation (1.1), where (u) = u3 + O(u5) is an odd analytic function,with F = || = 0 . Dene u0(x, t ) = a0(t + x) a0(t x), with a0( ) the odd 2-periodic solution of (1.5). There are a positive constant 0 and for all j N a set E [0, 0 /j 2] satisfying

lim 0

meas(E [0, ])

= 1 , (1.9)

such that for all E , by setting = 1 and f (x, t ) r =

(n,m )Z 2f n,m er ( | n | + | m | ) , (1.10)

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for analytic 2-periodic functions, there exist 2/j -periodic solutions u ,j (x, t ) of (1.1), analytic

in (t, x ), with u,j (x, t ) j /Fu 0( jx, j t) C j , (1.11)

for some constants C > 0 and 0 < < .

Note that such a result provides a solution of the open problem 7.4 in [24], as far as periodicsolutions are concerned.

As we shall see for (u) = F u 3 for all j N one can take the set E = [0, 0], independently of j ,so that for xed E no restriction on j have to be imposed.

We look for a solution of (1.3) of the form

u(x, t ) =(n,m )Z 2

einjt + ijmx un,m = v(x, t ) + w(x, t ),

v(x, t ) = a( ) a( ), = t + x, = t x,a( ) =

nZein an ,

w(x, t ) =(n,m )Z 2

| n | = | m |

einjt + ijmx wn,m ,

(1.12)

with = , such that one has w(x, t ) = 0 and a( ) = a0( ) for = 0. Of course by the symmetryof (1.1), hence of (1.4), we can look for solutions (if any) which verify

un,m = un, m = u n,m (1.13)for all n, m Z .

Inserting (1.12) into (1.3) gives two sets of equations, called the Q and P equations [13], whichare given, respectively, by

Qn2an = [(v + w)]n,n ,

n2an = [(v + w)]n, n ,P 2n2 + m2 wn,m = [(v + w)]n,m , |m| = |n|,

(1.14)

where we denote by [ F ]n,m the Fourier component of the function F (x, t ) with labels ( n, m ), sothatF (x, t ) =

(n,m )Z 2eint + mx [F ]n,m . (1.15)

In the same way we shall call [ F ]n the Fourier component of the function F ( ) with label n; inparticular one has [ F ]0 = F . Note also that the two equations Q are in fact the same, by thesymmetry property [ (v + w)]n,m = [(v + w)]n, m , which follows from (1.13).

We start by considering the case (u) = u3 and j = 1, for simplicity. We shall discuss at theend how the other cases can be dealt with; see Section 8.

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2. Lindstedt series expansion

One could try to write a power series expansion in for u(x, t ), using (1.14) to get recursiveequations for the coefficients. However by proceeding in this way one nds that the coefficient of order k is given by a sum of terms some of which of order O(k! ), for some constant . This is thesame phenomenon occurring in the Lindstedt series for invariant KAM tori in the case of quasi-integrable Hamiltonian systems; in such a case however one can show that there are cancellations between the terms contributing to the coefficient of order k, which at the end admits a bound C k ,for a suitable constant C . On the contrary such cancellations are absent in the present case andwe have to proceed in a different way, equivalent to a resummation (see [19] where such procedurewas applied to the same nonlinear wave equation with a mass term, uxx u tt + Mu = (u)).Denition 1. Given a sequence { m ()}| m | 1 , such that m = m , we dene the renormalizedfrequencies as 2m 2m + m , m = |m|, (2.1)and the quantities m will be called the counterterms .

By the above denition and the parity properties (1.13) the P equation in (1.14) can be rewrittenas

wn,m 2n2 + 2m = m wn,m + [(v + w)]n,m= (a )m wn,m +

(b)m wn, m + [(v + w)]n,m ,

(2.2)

where (a )m (b)m = m . (2.3)

With the notations of (1.15), and recalling that we are considering (u) = u3 , we can write

(v + w)3 n,n = [v3]n,n + [w3]n,n + 3[v2w]n,n + 3[w2v]n,n

[v3]n,n + [g(v, w)]n,n ,(2.4)

where, again by using the parity properties (1.13),

[v3]n,n = [a3]n + 3 a2 an . (2.5)

Then the rst Q equation in (1.13) can be rewritten as

n2an = [a3]n + 3 a2 an + [g(v, w)]n,n , (2.6)

so that an is the Fourier coefficient of the 2 -periodic solution of the equation

a = a3 + 3 a2 a + G(v, w) , (2.7)where we have introduced the function

G(v, w) =nZ

ein [g(v, w)]n,n . (2.8)

To study the equations (2.2) and (2.6) we introduce an auxiliary parameter , which at the endwill be set equal to 1, by writing (2.2) as

wn,m 2n2 + 2m = (a )m wn,m + (b)m wn, m + [(v + w)]n,m , (2.9)5


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and we shall look for un,m in the form of a power series expansion in ,

un,m =

k=0

k u (k )n,m , (2.10)

with u (k )n,m depending on and on the parameters (c)m , with c = a, b and |m| 1.In (2.10) k = 0 requires un, n = a0,n and un,m = 0 for |n| = |m|, while for k 1, as we shall

see later on, the dependence on the parameters (c )

m will be of the form

m =2 c = a,b

(c )

m k ( c

)m , (2.11)

with |k| = k(a )1 + k(b)1 + k(a )2 + k(b)2 + . . . k 1. Of course we are using the symmetry property torestrict the dependence only on the positive labels m .

We derive recursive equations for the coefficients u(k )n,m of the expansion. We start from thecoefficients with |n | = |m|.

By (1.12) and (2.10) we can write

a = a0 +

k=1

k A(k ) , (2.12)

and inserting this expression into (2.7) we obtain for A (k ) the equation

A(k ) = 3 a20A(k ) + a20 A(k ) + 2 a0A(k ) a0 + f (k ) , (2.13)with

f (k ) = k1 + k2 + k3 = k

k i = k| n i | = | m i |n 1 + n 2 + n 3 = n

m 1 + m 2 + m 3 = m

u (k1 )n 1 ,m 1 u(k2 )n 2 ,m 2 u

(k3 )n 3 ,m 3 , (2.14)

where we have used the notations

u (k )n,m =v(k )n,m , if |n | = |m| ,w(k )n,m , if |n | = |m| ,

(2.15)

withv(k )n,n =

A(k )n , if k = 0,a0,n , if k = 0, v(k )n, n = A

(k )n , if k = 0,

a0,n , if k = 0 . (2.16)

Before studying how to nd the solution of this equation we introduce some preliminary deni-tions. To shorten notations we write

c( ) cn(m , m ), s( ) sn(m , m ) d( ) dn(m , m ), (2.17)and set cd( ) = cn( m , m ) dn( m , m ). Moreover given an analytic periodic function F ( ) wedene

P [F ]( ) = F ( ) F . (2.18)6


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and we introduce a linear operator I acting on 2 -periodic zero-mean functions and dened by its

action on the basis en ( ) = ein , n Z \ {0},I [en ]( ) =

en ( )in

. (2.19)

Note that if P [F ] = 0 then P [I [F ]] = 0 (I [F ] is simply the zero-mean primitive of F ); moreover Iswitches parities.

In order to nd an odd solution of (2.13) we replace rst a0A(k ) with a parameter C (k ) , andwe study the modied equation

A(k ) = 3 a20A(k ) + a20 A(k ) + 2 C (k ) a0 + f (k ) . (2.20)Then we have the following result (proved in Appendix A1).

Lemma 1. Given an odd analytic 2-periodic function h( ), the equation

y = 3 a20 + a20 y + h, (2.21)admits one and only one odd analytic 2-periodic solution y( ), given by

y = L [h] Bm 2m D 2m s s h + 1m D m (s I [cd h]cd I [P [s h ]]) + cd I [I [cd h]] , (2.22)with Bm = m / (1 m ) and Dm = 1/ m .

As a0 is analytic and odd, we nd immediately, by induction on k and using lemma 1, that f (k )

is analytic and odd, and that the solution of the equation (2.20) is odd and given by

A(k ) = L [6C (k ) a0 + f (k ) ]. (2.23)The function A(k ) so found depends of course on the parameter C (k ) ; in order to obtain A(k ) =

A(k ) , we have to impose the constraint

C (k ) = a0A(k ) , (2.24)

and by (2.23) this givesC (k ) = 6C (k ) a0L[a0] + a0L [f (k ) ] , (2.25)

which can be rewritten as

(1 + 6 a0L [a0] ) C (k ) = a0L [f (k ) ] . (2.26)

An explicit computation (see Appendix A3) gives

a0 L [a0] = 12

V 2m 2m B m 2D m 12

s4 + 2D m (D m 1) + 12

s2 2 , (2.27)

which yields r 0 = (1 + 6 a0L [a0] ) = 0. At the end we obtain the recursive denitionA(k ) = L [f (k ) 6C (k ) a0],C (k ) = r 10 a0L [f (k ) ] .

(2.28)

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In Fourier space the rst of (2.28) becomes

A(k )n = Bm 2m D 2m sn

n 1 + n 2 =0sn 1 f (

k )n 2 6C (k ) a0,n 2

+ Bmn 1 + n 2 + n 3 = n

1i2(n2 + n3)2

cdn 1 cdn 2 f (k )n 3 6C (k ) a0,n 3

+ Bm 1m D mn 1 + n 2 + n 3 = n

1i(n2 + n3)

sn 1 cdn 2 f (k )

n 3 6C (k ) a0,n 3

Bm 1m D mn 1 + n 2 + n 3 = n

1i(n2 + n3)

cdn 1 sn 2 f (k )

n 3 6C (k ) a0,n 3

n L nn f

(k )n

6C (k ) a0,n ,

(2.29)

where the constants Bm and Dm are dened after (2.22), and the in the sums means that onehas the constraint n2 + n3 = 0, while the second of (2.28) can be written as

C (k ) = r 10n,n Z

a0, n L n,n f (k )n . (2.30)

Now we consider the coefficients u(k )n,m with |n | = |m|. The coefficients w(k )n,m verify the recursive

equationsw(k )n,m 2n2 + 2m = (a )m w(k 1)n,m + (b)m w

(k 1)n, m + [(v + w)

3 ](k 1)n,m , (2.31)

where[(v + w)3 ](k )n,m =

k1 + k2 + k3 = k n 1 + n 2 + n 3 = nm 1 + m 2 + m 3 = m

u (k1 )n 1 ,m 1 u(k2 )n 2 ,m 2 u

(k3 )n 3 ,m 3 , (2.32)

if we use the same notations (2.15) and (2.16) as in (2.14).The equations (2.29) and (2.31), together with (2.32), (2.14), (2.30) and (2.32), dene recursively

the coefficients u (k )n,m .To prove the theorem we shall proceed in two steps. The rst step consists in looking for

the solution of the equations (2.29) and (2.31) by considering = {m }| m | 1 as a given setof parameters satisfying the Diophantine conditions (called respectively the rst and the secondMel nikov conditions)

|n m | C 0|n| n Z \ {0} and m Z \ {0} such that |m| = |n|,|n (m m )| C 0|n| (2.33)

n Z \ {0} and m, m Z \ {0} such that |n| = |m m |,with positive constants C 0 , . We shall prove in Sections 3 to 5 the following result.

Proposition 1. Let be given a sequence = {m }| m | 1 verifying (2.33), with = = 1 and such that |m |m|| C/ |m| for some constant C . For all 0 > 0 there exists 0 > 0 such that for || 0 and 0 < < 0 there is a sequence (, ; ) = { m (, ; )}| m | 1, where each m (, ; ) is analytic in , such that the coefficients u

(k )n,m which solve (2.29) and (2.31) dene via

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(2.10) a function u(x, t ; , ; ) which is analytic in , analytic in (x, t ) and 2-periodic in t and

solves n2an = a3 n,n + 3 a

2 an + [g(v, w)]n,n ,

n2an = a3 n, n + 3 a2 a n + [g(v, w)]n, n ,2n2 + 2m wn,m = m (, ; ) wn,m + [(v + w)]n,m , |m| = |n |,

(2.34)

with the same notations as in (1.14).

If 2 then one can require only the rst Melnikov conditions in (2.33), as we shall show inSection 7.

Then in Proposition 1 one can x 0 = 1, so that one can choose = 1 and set u(x, t ; , ) =u(x, t ; , ;1) and m (, ) = m (, ;1).

The second step, to be proved in Section 6, consists in inverting (2.1), with m = m (, )and verifying (2.33). This requires some preliminary conditions on , given by the Diophantineconditions

|n m| C 1|n| 0 n Z \ {0} and m Z \ {0} such that |m| = |n|, (2.35)with positive constants C 1 and 0 > 1.

This allows to solve iteratively (2.1), by imposing further non-resonance conditions besides (2.35),provided that one takes C 1 = 2C 0 and 0 < 1, which requires > 2. At each iterative stepone has to exclude some further values of , and at the end the left values ll a Cantor set E withlarge relative measure in [0 , 0] and verify (2.35).

If 1 < 2 the rst Melnikov conditions, which, as we said above, become sufficient to proveProposition 1, can be obtained by requiring (2.35) with 0 = ; again this leaves a large measureset of allowed values of . This is discussed in Section 7.

The result of this second step can be summarized as follows.

Proposition 2. There are > 0 and a set E [0, 0] with complement of relative Lebesgue measure of order 0 such that for all E there exists = () which solves (2.1) and satisfy the Diophantine conditions (2.33) with |m |m|| C/ |m| for some constant C .

As we said, our approach is based on constructing the periodic solution of the string equationby a perturbative expansion which is the analogue of the Lindstedt series for (maximal) KAMinvariant tori in nite-dimensional Hamiltonian systems. Such an approach immediately encountersa difficulty; while the invariant KAM tori are analytic in the perturbative parameter , the periodicsolutions we are looking for are not analytic; hence a power series construction seems at rst sight

hopeless. Nevertheless it turns out that the Fourier coefficients of the periodic solution have theform un,m ((, ), ; ); while such functions are not analytic in , it turns out to be analyticin , provided that satises the condition (2.33) and is small enough; this is the content of Proposition 1. As far as we know, in the case M = 0, besides [19], the smoothness in at xed was only discussed in a paper by Bourgain [8]. This smoothness is what allow us to write as aseries expansion un,m (, ; ); this strategy was already applied in [19] in the massive case.

3. Tree expansion: the diagrammatic rules

A (connected) graph G is a collection of points (vertices) and lines connecting all of them. Thepoints of a graph are most commonly known as graph vertices, but may also be called nodes or

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points. Similarly, the lines connecting the vertices of a graph are most commonly known as graph

edges, but may also be called branches or simply lines, as we shall do. We denote with P (G ) andL(G ) the set of vertices and the set of lines, respectively. A path between two vertices is a subsetof L(G ) connecting the two vertices. A graph is planar if it can be drawn in a plane without graphlines crossing (i.e. it has graph crossing number 0).

Denition 2. A tree is a planar graph G containing no closed loops (cycles); in other words,it is a connected acyclic graph. One can consider a tree G with a single special vertex V 0 : this introduces a natural partial ordering on the set of lines and vertices, and one can imagine that each line carries an arrow pointing toward the vertex V 0 . We can add an extra (oriented) line 0connecting the special vertex V 0 to another point which will be called the root of the tree; the added line will be called the root line . In this way we obtain a rooted tree dened by P () = P (G ) and L() = L(G )

0 . A labeled tree is a rooted tree together with a label function dened on the sets L() and P ().

Note that the denition of rooted tree given above is sligthly different from the one which isusually adopted in literature [20], [22] according to which a rooted tree is just a tree with aprivileged vertex, without any extra line. However the modied denition that we gave will bemore convenient for our purposes. In the following we shall denote with the symbol both rootedtrees and labels rooted trees, when no confusion arises

We shall call equivalent two rooted trees which can be transformed into each other by continuously

deforming the lines in the plane in such a way that the latter do not cross each other (i.e. withoutdestroying the graph structure). We can extend the notion of equivalence also to labeled trees,simply by considering equivalent two labeled trees if they can be transformed into each other insuch a way that also the labels match.

Given two points V , W P (), we say that V W if V is on the path connecting W to the root line.We can identify a line with the points it connects; given a line = ( V , W ) we say that enters V andcomes out of W .

In the following we shall deal mostly with labeled trees: for simplicity, where no confusion canarise, we shall call them just trees. We consider the following diagramamtic rules to construct thetrees we have to deal with; this will implicitly dene also the label function.

(1) We call nodes the vertices such that there is at least one line entering them. We call end-points the vertices which have no entering line. We denote with L(), V () and E () the set of lines,nodes and end-points, respectively. Of course P () = V ()E ().

(2) There can be two types of lines, w-lines and v-lines, so we can associate to each line L() abadge label {v, w} and a momentum (n , m ) Z 2 , to be dened in item (8) below. If = vone has |n| = |m|, while if = w one has |n | = |m|. One can have ( n , m ) = (0 , 0) only if is a v-line. To the v-lines with n = 0 we also associate a label {1, 2}. All the lines coming

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out from the end-points are v-lines.

(3) To each line coming out from a node we associate a propagator

g g(n , m ) =

1

2n2 + 2m , if = w,

1(in )

, if = v, n = 0 ,1, if = v, n = 0 ,

(3.1)

with momentum ( n , m ). We can associate also a propagator to the lines coming out fromend-points, simply by setting g = 1.

(4) Given any node V V () denote with s the number of entering lines ( branching number ): onecan have only either s

= 1 or s

= 3. Also the nodes V can be of w-type and v-type: we saythat a node is of v-type if the line coming out from it has label = v; analogously the nodes of w-type are dened. We can write V () = V v ()V w (), with obvious meaning of the symbols; wealso call V sw (), s = 1 , 3, the set of nodes in V w () with s entering lines, and analogously we deneV sv (), s = 1 , 3. If V V 3v and two entering lines come out of end points then the remaining lineentering V has to be a w-line. If V V 1w then the line entering V has to be a w-line. If V V 1v thenits entering line comes out of an end-node.

(5) To the nodes V of v-type we associate a label jV {1, 2, 3, 4} and, if s = 1, an order labelk

, with k 1. Moreover we associate to each node V of v-type two mode labels (n , m ), with

m

= n , and ( n , m ), with m = n , and such that one has

m n

= m n

=

s

i=1

mi

s

i=1

ni

, (3.2)

where i are the lines entering V . We shall refer to them as the rst mode label and the second mode label, respectively. To a node V of v-type we associate also a node factor

dened as

=

Bm 2m D 2m sn sn , if j = 1 and s = 3 ,6Bm 2m D 2m sn sn C (k ) , if j = 1 and s = 1 ,Bm cdn cdn , if j = 2 and s = 3 ,6Bm cdn cdn C (k ) , if j = 2 and s = 1 ,Bm 1m D m sn cdn , if j = 3 and s = 3 ,

6Bm 1m D m sn

cdn

C (k ) , if j

= 3 and s

= 1 ,Bm 1m Dm cdn

sn

, if j

= 4 and s

= 3 ,6Bm 1m D m cdn

sn

C (k ) , if j

= 4 and s

= 1 .

(3.3)

Note that the factors C (k ) = r 10 a0L [f (k )] depend on the coefficients u(k )n,m , with k < k , so that

they have to been dened iteratively. The label of the line coming out from a node V of v-typeis related to the label j

of v: if j

= 1 then n = 0, while if j > 1 then n = 0 and = 1 + j ,2 ,where j

,1 denotes the Kronecker delta (so that = 2 if j + 2 and = 1 otherwise).

(6) To the nodes V of w-type we simply associate a node factor

given by

=, if s

= 3, (c)m , if s = 1.

(3.4)

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In the latter case ( n, m ) is the momentum of the line coming out from V , and one has c = a if

the momentum of the entering line is ( n , m ) and c = b if the momentum of the entering line is(n, m); we shall call -vertices the nodes V of w-type with s = 1. In order to unify notationswe can associate also to the nodes V of w-type two mode labels, by setting ( n

, m

) = (0 , 0) and(n

, m

) = (0 , 0).

(7) To the end-points V we associate only a rst mode label ( n

, m

), with |m | = |n |, and anend-point factor

V

= ( 1)1+ n

,m a0,n

= a0,m

. (3.5)

The line coming out from an end-point has to be a v-line.

(8) The momentum ( n, m ) of a line is related to the mode labels of the nodes preceding ; if aline comes out from a node v one has

n = n + V ( )

(n

+ n

) + E ( )

n

,

m = m + V ( )

(m

+ m

) + E ( )

m

,(3.6)

where some of the mode labels can be vanishing according to the notations introduced above. If comes out from an end-point we set ( n , m ) = (0 , 0).

We dene (k )n,m as the set of unequivalent labeled trees , formed by following the rules (1) to (8)given above, and with the further following constraints:(i) if (n 0 , m 0 ) denotes the momentum owing through the root line 0 and (n 0 , m 0 ) is the rstmode label associated to the node V 0 which 0 comes out from (special vertex), then one hasn = n0 + n 0 and m = m0 + m

0 ;

(ii) one hask = |V w ()|+

V 1v ( )

k

, (3.7)

with k called the order of the tree.An example of tree is given in Figure 3.1, where only the labels v, w of the nodes have been

explicitly written.

Denition 3. For all (k )n,m , we call

Val( ) = L ( ) g V ( ) E ( ) V , (3.8)the value of the tree .

Then the main result about the formal expansion of the solution is provided by the followingresult.

Lemma 2. We can write u (k )n,m =

( k )n,m

Val( ), (3.9)

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v

w

w

w

w

v

w

v

v

w

w

v

w

Figure 3.1

and if the root line 0 is a v-line the tree value is a contribution to v(k )n, n , while if 0 is a w-line

the tree value is a contribution to w(k )n,m . The factors C (k ) are dened as

C (k ) = r 10( k )n,n

a0, n Val( ), (3.10)

where the in the sum means the extra constraint s 0 = 3 for the node immediately preceding the root (which is the special vertex of the rooted tree).

Proof. The proof is done by induction in k. Imagine to represent graphically a0,n as a (small)white bullet with a line coming out from it, as in Figure 3.2a, and u(k )n,m , k 1, as a (big) blackbullet with a line coming out from it, as in Figure 3.2b.

a bFigure 3.2

One should imagine that labels k,n, m are associated to the black bullet representing u(k )n,m , whilea white bullet representing a0,n carries the labels n, m = n.For k = 1 the proof of (3.9) and (3.10) is just a check from the diagrammatic rules and therecursive denitions (2.27) and (2.29), and it can be performed as follows.

Consider rst the case |n| = |m|, so that u(1)n,m = w(1)n,m . By taking into account only the badge

labels of the lines, by item (4) there is only one tree whose root line is a w-line, and it has one nodeV 0 (the special vertex of the tree) with s 0 = 3, hence three end-points V 1 , V 2 and V 3 . By applyingthe rules listed above one obtains, for |n| = |m|,

w(1)n,m = 1

2n2 + 2m n 1 + n 2 + n 3 = nm 1 + m 2 + m 3 = m

v(0)n 1 ,m 1 v(0)n 2 ,m 2 v

(0)n 3 ,m 3 =

(1)n,m

Val( ), (3.11)

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w

Figure 3.3

where the sum is over all trees which can be obtained from the tree appearing in Figure 3.3 bysumming over all labels which are not explicitly written.

It is easy to realize that (3.11) corresponds to (2.31) for k = 1. Each end-point V i is graphicallya white bullet with rst mode labels ( n i , m i ) and second mode labels (0 , 0), and has associated anend-point factor ( 1)1+ n i ,m i a0,n i (see (3.5) in item (7)). The node V 0 is represented as a (small)gray bullet, with mode labels (0 , 0) and (0 , 0), and the factor associated to it is

0 = (see (3.4)in item (6)). We associate to the line coming out from the node V 0 a momentum ( n, n ), withn = n, and a propagator g = 1 / (2n2 + 2m ) (see (3.1) in item (3)).

Now we consider the case |n | = |m|, so that u(1)n,m = A

(1)n (see (2.16)). By taking into account

only the badge labels of the lines, there are four trees contributing to A(1)n : they are representedby the four trees in Figure 3.4 (the tree b and c are simply obtained from the tree a by a differentchoice of the w-line entering the last node).

v

w

v w v

w

v

a b c d

Figure 3.4

In the trees of Figures 3.4a, 3.4b and 3.4c the root line comes out from a node V 0 (the specialvertex of the tree) with s

0 = 3, and two of the entering lines come out from end-points: then theother line has to be a w-line (by item (4)), and (3.7) requires that the subtree which has such aline as root line is exactly the tree represented in Figure 3.2. In the tree of Figure 4.4d the rootline comes out from a node V 0 with s 0 = 1, hence the line entering V 0 is a v-line coming out froman end-point (again see item (4)).

By dening (1)n,n as the set of all labeled trees which can be obtained by assigning to the treesin Figure 3.4 the labels which are not explicitly written, one nds

A(1)n =(1)n,n

Val( ), (3.12)

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which corresponds to the sum of two contributions. The rst one arises from the trees of Figures

3.4a, 3.4b and 3.4c, and it is given by

3n Z

L n,n n 1 + n

2 + n

3 = n

m 1 + m2 + m

3 = n

v(0)n 1 ,m 1 v(0)n 2 ,m

2w(1)n 3 ,m 3 , (3.13)

where one has

L n,n = Bm 2m D 2mn 1 + n 2 = n n

n 2 = n

sn 1 sn 2 + Bmn 1 + n 2 = n n

1i2(n2 + n )2

cdn 1 cdn 2

+ B m 1m D m

n 1 + n 2 = n n

1i(n2 + n )

sn 1 cdn 2

B m 1m D mn 1 + n 2 = n n

1i(n2 + n )

cdn 1 sn 2 ,

(3.14)

with the denoting the constraint n2 + n = 0. The rst and second mode labels associated tothe node V 0 are, respectively, ( m

0 , n

0 ) = ( n1 , n 1) and ( m 0 , n 0 ) = ( n2 , n 2), while the momentumowing through the root line is given by ( n, m ), with |m | = |n | expressed according to thedenition (3.6) in item (8): the corresponding propagator is ( n) for n = 0 and 1 for n = 0, asin (3.1) in item (3).

The second contribution corresponds to the tree of Figure 3.4d, and it is given by

n Z

L n,n C (1) a0,n , (3.15)

with the same expression (3.14) for Ln,n and C (1) still undetermined. The mode labels of thenode V 0 and the momentum of the root line are as before.

Then one immediately realize that the sum of (3.13) and (3.15) corresponds to (2.27) for k = 1.Finally that C (1) is given by (3.9) follows from (2.12). This completes the check of the case

k = 1.In general from (2.31) one gets, for

(k )n,m contributing to w(

k )n,m , that the tree value Val( ) is

obtained by summing all contribution either of the form

1

2n2 + 2m

n 1 + n 2 + n 3 = n

m 1 + m 2 + m 3 = m

k1 + k2 + k3 = k 1

1( k 1 )n 1 ,m 1 2

( k 2 )n 2 ,m 2 3

( k 3 )n 3 ,m 3

Val( 1) Val( 2 ) Val( 3),(3.16)

or of the form1

2n2 + 2m c= a,b (c)m

1( k, 1)n,m ( c )

Val( 1), (3.17)

with m (a ) = m and m (b) = m; the corresponding graphical representations are as in Figure 3.5.Therefore, by simply applying the diagrammatic rules given above, we see that by summing

together the contribution (3.16) and (3.17) we obtain (3.9) for |n | = |m|.15


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w

w

w

w

w

w

w

w

a b c d

e f g h

Figure 3.5

A similar discussion applies to A(k )n , and one nds that A(k )n can be written as sum of contributioneither of the form

n Z

L n,n n 1 + n

2 + n

3 = n

m 1 + m2 + m

3 = n

k1 + k2 + k3 = k

1( k 1 )n 1 ,m 1 2

( k 2 )n 2 ,m 2 3

( k 3 )n 3 ,m 3

Val( 1) Val( 2 ) Val( 3),

(3.18)

or of the form

n Z

L n,n C (k ) a0,n , (3.19)

with C (k ) still undetermined. Both (3.18) and (3.19) are of the form Val( ), for (k )n,m . A

graphical representation is in Figure 3.6.

v v v v v

a b c d e

Figure 3.6

Analogously to the case k = 1 the coefficients C (k ) are found to be expressed by (3.10). Thenthe lemma is proved.

Lemma 3. For any rooted tree one has |V 3v ()| 2|V 3w ()|+ 2 |V 1v ()| and |E ()| 2(|V 3v ()|+16


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|V 3w ()|) + 1 .Proof. First of all note that |V 3w ()| = 0 requires |V 1v ()| 1, so that one has |V 3w ()|+ |V 1v ()| 1for all trees .

We prove by induction on the number N of nodes the bound

V 3v () 2|V 3w ()|+ 2 |V 1v ()| 1, if the root line is a v-line ,2|V 3w ()|+ 2 |V 1v ()| 2 if the root line is a w-line ,

(3.20)

which will immediately imply the rst assertion.For N = 1 the bound is trivially satised, as Figures 3.3 and 3.4 show.Then assume that (3.20) holds for the trees with N nodes, for all N < N , and consider a tree

with V () = N .If the special vertex V 0 of is not in V 3v () (hence it is in V w ()) the bound (3.20) follows trivially

by the inductive hypothesis.If V 0 V 3v () then we can write

|V 3v ()| = 1 +s

i=1|V 3v (i )|, (3.21)

where 1 , . . . , s are the subtrees whose root line is one of the lines entering V 0 . One must haves 1, as s = 0 would correspond to have all the entering lines of V 0 coming out from end-points,hence to have N = 1.

If s 2 one has from (3.21) and from the inductive hypothesis

|V 3v ()

| 1 +

s

i=1

2

|V 3w (i )

|+ 2

|V 1v (i )

| 1

1 + 2

|V 3w ()

|+ 2

|V 1v ()

| 2, (3.22)

and the bound (3.20) follows.If s = 1 then the root line of 1 has to be a w-line by item (4), so that one has

|V 3v ()| 1 + 2|V 3w (1 )|+ 2 |V 1v ()| 2 (3.23)which again yields (3.20).

Finally the second assertion follows from the standard (trivial) property of trees

V ( )

(s 1) = |E ()| 1, (3.24)

and the observation that in our case one has s

3.4. Tree expansion: the multiscale decomposition

We assume the Diophantine conditions (2.33). We introduce a multiscale decomposition of thepropagators of the w-lines. Let (x) be a C non-increasing function such that (x) = 0 i f

|x| C 0 and (x) = 1 i f |x| 2C 0 (C 0 is the same constant appearing in (2.33)), and let h (x) = (2h x) (2h +1 x) for h 0, and 1(x) = 1 (x); such functions realize a smoothpartition of the unity as

1 = 1(x) +

h =0

h (x) =

h = 1

h (x). (4.1)

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If h (x) = 0 for h 0 one has 2 h 1C 0 |x| 2 h +1 C 0 , while if 1(x) = 0 one has |x| C 0 .We write the propagator of any w-line as sum of propagators on single scales in the following

way:

g(n,m ) =

h = 1

h (|n| m )2n2 + 2m

=

h = 1

g(h ) (n,m ). (4.2)

Note that we can bound |g(h ) (n,m )| 2 h +1 C 0 .This means that we can attach to each w-line in L() a scale label h 1, which is the scale

of the propagator which is associated to . We can denote with (k )n,m the set of trees which differfrom the previous ones simply because the lines carry also the scale labels. The set (k )n,m is denedaccording to the rules (1) to (8) of Section 3, by changing item (3) into the following one.

(3 ) To each line coming out from nodes of w-type we associate a scale label h 1. Fornotational convenience we associate a scale label h = 1 to the lines coming out from the nodes of v-type and to the lines coming out from the end-points. To each line we associate a propagator

g(h ) g(h ) (n , m ) = h (|n| m )2n2 + 2m

, if = w,

1(in )

, if = v, n = 0 ,1, if = v, n = 0 ,

(4.3)

with momentum ( n , m ).

Denition 4. For all (k )n,m , we dene

Val( ) = L ( ) g(h )

V ( )

E () V

, (4.4)

the value of the tree .

Then (3.9) and (3.10) are replaced, respectively, with

u (k )n,m = ( k )n,m

Val( ), (4.5)

andC (k ) = r 10

( k )n,n

a0, n Val( ), (4.6)

with the new denition for the tree value Val( ) and with meaning the same constraint as in(3.10).Denition 5. A cluster T is a connected set of nodes which are linked by a continuous path of lines with the same scale label hT or a lower one and which are maximal; we shall say that the cluster has scale hT . We shall denote with V (T ) and E (T ) the set of nodes and the set of end-points,respectively, which are contained inside the cluster T , and with L(T ) the set of lines connecting them. As for trees we call V v (T ) and V w (T ) the sets of nodes V V (T ) which are of v-type and of w-type respectively.

We dene the order kT of a cluster T as the order of a tree (see item (ii) before Denition 3),with the sums restricted to the nodes internal to the cluster.

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An inclusion relation is established between clusters, in such a way that the innermost clusters

are the clusters with lowest scale, and so on. Each cluster T can have an arbitrary number of linesentering it (incoming lines), but only one or zero line coming from it (outcoming line); we shalldenote the latter (when it exists) with 1T . We shall call external lines of the cluster T the lineswhich either enter or come out from T , and we shall denote by h(e)T the minimum among the scalesof the external lines of T . Dene also

K () = V ( )

(|n |+ |n |) + E ( )

|n |,

K (T ) = V (T )

(|n |+ |n |) + E ( )

|n |,(4.7)

where we recall that one has ( n

, m

) = ( n

, m

) = 0 if V

V () is of w-type.

If a cluster has only one entering line 2T and (n, m ) is the momentum of such a line, for any line L(T ) one can write ( n, m ) = ( n0 , m 0 ) + (n, m ), where = 1 if the line is along the pathconnecting the external lines of T and = 0 otherwise.

Denition 6. A cluster T with only one incoming line 2T such that one has

n1T = n2T and m1T = m2T (4.8)will be called a self-energy graph or a resonance . In such a case we shall call a resonant line the line 1T , and we shall refer to its momentum as the momentum of the self-energy graph.

Examples of self-energy graphs T with kT = 1 are represented in Figure 4.1. The lines crossingthe encircling bubbles are the external lines, and in the gure are assumed to be on scales higherthan the lines internal to the bubbles. There are 9 self-energy graphs with kT = 1: they are allobtained by the three drawn in Figure 4.1, simply by considering all possible unequivalent trees.

w w

v

a b

Figure 4.1

Denition 7. The value of the self-energy graph T with momentum (n, m ) associated to the line 2T is dened as

V hT (n,m ) =T

g(h ) VV (T ) VE (T ) V . (4.9)19


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where h = h(e)T is the minimum between the scales of the two external lines of T (they can differ

at most by a unit), and one has

n(T ) V (T )

(n

+ n

) + E (T )

n

= 0 ,

m(T ) V (T )

(m

+ m

) + E (T )

m {0, 2m},

(4.10)

by denition of self-energy graph; one says that T is a resonance of type c = a when m(T ) = 0 and a resonance of type c = b when m(T ) = 2 m.

Denition 8. Given a tree , we shall denote by N h () the number of lines with scale h, and by C h () the number of clusters with scale h.

Then the product of propagators appearing in (4.4) can be bounded as

L ( )

g(h )

h =0

2hN h ( ) L ( ) = w

h (|n| m )|n |+ m L ( )

= v, n =0

1

|n|, (4.11)

and this will be used later.

Lemma 4. Assume 0 < C 0 < 1/ 2 and that there is a constant C 1 such that one has |m |m|| C 1/ |m|. If is small enough for any tree

(k )n,m and for any line on a scale h 0 one has

min{m , n } 1/ 2.Proof. If a line with momentum ( n, m ) is on scale h 0 then one has

12

> C 0 ||n | m | 1 1 |n|+ ( |n| |m|) C 1/ |m|

|n |1 + 1

(|n | |m|) C 1/ |m| ,(4.12)

with |n | = |m|, hence |n m| 1, so that |n | 1/ 2. Moreover one has ||n| m | 1/ 2 andm |m| = O(), and one obtains also |m| > 1/ 2.Lemma 5. Dene h0 such that 2h 0 16C 0/ < 2h 0 +1 , and assume that there is a constant C 1such that one has |m |m|| C 1/ |m|. If is small enough for any tree

(k )n,m and for all

h h0 one has N h () 4K ()2(2 h ) / C h () + S h () + M h (), (4.13)where K () is dened in (4.7), while S h () is the number of self-energy graphs T in with h

(e)T = h

and M h () is the number of -vertices in such that the maximum scale of the two external lines is h.

Proof. We prove inductively the bound

N h () max{0, 2K ()2(2 h ) / 1}, (4.14)where N h () is the number of non-resonant lines in L() on scale h

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First of all note that for a tree to have a line on scale h the condition K () > 2(h 1) / is

necessary, by the rst Diophantine conditions in (2.33). This means that one can have N h () 1only if K = K () is such that K > k 0 2(h 1) / : therefore for values K k0 the bound (4.14) issatised.

If K = K () > k 0 , we assume that the bound holds for all trees with K ( ) < K . DeneE h = 2 1(2(2 h ) / ) 1 : so we have to prove that N h () max{0, K ()E 1h 1}.

Call the root line of and 1 , . . . , m the m 0 lines on scale h which are the closest to (i.e. such that no other line along the paths connecting the lines 1 , . . . , m to the root line is onscale h).

If the root line of is either on scale < h or on scale h and resonant, then

N h () =m

i=1

N h (i ), (4.15)

where i is the subtree with i as root line, hence the bound follows by the inductive hypothesis.If the root line has scale h and is non-resonant, then 1 , . . . , m are the entering line of a

cluster T .By denoting again with i the subtree having i as root line, one has

N h () = 1 +m

i=1

N h (i ), (4.16)

so that the bound becomes trivial if either m = 0 or m 2.If m = 1 then one has a cluster T with two external lines and 1 , which are both with scales

h; then

||n | m | 2 h +1 C 0 , |n 1 | m 1 2 h +1 C 0 , (4.17)and recall that T is not a self-energy graph.

Note that the validity of both inequalities in (4.17) for h h0 imply that one has |n n1 | =|m m1 |, as we are going to show.

By Lemma 4 we know that one has min {m , n } 1/ 2. Then from (4.17) we have, for some, 1 {1},

2 h +2 C 0 (n n 1 ) + m + 1 m 1 , (4.18)so that if one had |n n 1 | = |m m1 | we would obtain for small enough

2 h +2 C 0

1 + 1

|n n1 | C 1

|m

|

C 1

|m 1

|

2 4C 1

2 > 4

, (4.19)

which is contradictory with h h0 ; hence one has |n n1 | = |m m1 |.Then, by (4.17) and for |n n1 | = |m m 1 |, one has, for suitable , 1 {+ , },

2 h +2 C 0 (n n 1 ) + m + 1 m 1 C 0|n n1 | , (4.20)where the second Diophantine conditions in (2.33) have been used. Hence K () K (1) > E h ,which, inserted into (4.16) with m = 1, gives, by using the inductive hypothesis,

N h () = 1 + N

h (1) 1 + K (1)E 1h 1 1 + K () E h E 1h 1 K ()E 1h 1,

(4.21)

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hence the bound is proved also if the root line is on scale h.In the same way one proves that, if we denote with C h () the number of clusters on scale h, one

hasC h () max{0, 2K ()2(2 h ) / 1}; (4.22)

see [19] for details.

Note that the argument above is very close to [19]: this is due to the fact that the external linesof any self-energy graph T are both w-lines, so that the only effect of the presence of the v-linesand of the nodes of v-type is in the contribution to K (T ).

The following lemma deals with the lines on scale h < h 0 .

Lemma 6. Let h0 be dened as in Lemma 2 and C 0 < 1/ 2, and assume that there is a constant C 1 such that one has |m |m|| C 1. If is small enough for h < h 0 one has |g(h ) | 32.Proof. Either if h = h of h = h = 1 the bound is trivial. If h = h 0 one has

g(h ) = h (|n| m )

|n|+ m1

|n |+ m, (4.23)

where |n|+ m 1/ 2 by Lemma 4. Then one has1

|n |+ m 2, (4.24)

which, inserted in (4.23), gives |g(h ) | 2h +2 /C 0 32, so that the lemma is proved.

5. The renormalized expansion

It is an immediate consequence of Lemma 5 and Lemma 6 that all the trees with no self-energygraphs or -vertices admit a bound O(C k k ), where C is a constant. However the generic tree with S h () = 0 admits a much worse bound, namely O(C k k k! ), for some constant , andthe presence of factorials prevent us to prove the convergence of the series; in KAM theory thisis called accumulation of small divisors . It is convenient then to consider another expansion forun,m , which is essentially a resummation of the one introduced in Sections 3 and 4.

We dene the set (k )Rn,m of renormalized trees , which are dened as (k )n,m except that the followingrules are added.

(9) To each self-energy graph (with |m| 1) the R = 11 L operation is applied, where L acts onthe self-energy graphs in the following way, for h 0 and |m| 1,

LV hT (n,m ) = V hT (sgn( n) m , m), (5.1)

R is called regularization operator ; its action simply means that each self-energy graph V hT (n,m )must be replaced by RV hT (n,m ).(10) To the nodes V of w-type with s

= 1 (which we still call -vertices) and with h 0 theminimal scale among the lines entering or exiting V , we associate a factor 2 h (c)h,m , c = a, b, where

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(n, m ) and ( n, m), with |m| 1, are the momenta of the lines and a corresponds to the sign +and b to the sign in m.(11) The set {h} of the scales associated to the lines L() must satisfy the following constraint(which we call compatibility ): xed (n, m ) for any L() and replaced R with 11 at eachself-energy graph, one must have h (|n | m ) = 0.(12) The factors C (k ) in (3.3) are replaced with , to be considered a parameter.

The set (k )Rn,m is dened as (k )n,m with the new rules and with the constrain that the order k isgiven by k = |V w ()|+ |V 1v ()|.

We consider the following expansion

un,m =

k=1

k

( k ) Rn,m

Val( ), (5.2)

where, for |m| 1 and h 0, (c)h,m is given by

2 h (c)h,m = (c)m +

12 = T T ( c )


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First (Lemma 7) we will show that the expansion (5.2) is well dened, for h,m , = O(); then

(Lemma 8 and 9) we show that under the same conditions also the r.h.s. of (5.3) is well dened;moreover (Lemma 10) we prove by using (5.3) that it is indeed possible to choose (c)m such that h,m = O() for any h; then (Lemma 11 and 12) we show that (5.2) admit a solution = O();nally (Lemma 13) we show that indeed (5.2) solves the last of (1.14) and (2.2); this completesthe proof of Proposition 1. In the next section we will solve the implicit function problem of (2.1)so completing the proof of Theorem 1.

We start from the following lemma stating that, if the (c)h,m and functions are bounded, thenthe expansion (5.2) is well dened.

Lemma 7. Assume that there exist a constant C such that one has || C and | (c)h,m | C,

with c = a, b, for all |m| 1 and all h 0. Then for all 0 > 0 there exists 0 > 0 such that for all

|

| 0 and for all 0 < < 0 and for all (n, m )

Z 2 one has

|un,m | D0 e ( | n | + | m | ) / 4, (5.8)where D0 is a positive constant. Moreover un,m is analytic in and in the parameters

(c)m ,h , with

c = a, b and |m | 1.Proof. In order to take into account the R operation we write (5.6) as

RV h eT T (nT , m T ) = nT m T 10 dt V h eT T (nT + t(nT m T ), m T ), (5.9)

where denotes the derivative with respect to the argument nT + t(nT m T ).By (5.7) we see that the derivatives can be applied either on the propagators in L0(T ), or on the

RV heT

T . In the rst case there is an extra factor 2 h( e )T + h T with respect to the bound (4.11): 2 h ( e )T

is obtained from nT m T while g(h T ) is bounded proportionally to 2 2h T ; in the second casenote that tRV

h eT T = tV

h eT T as LV

h ( e )T

T is independent of t; if the derivative acts on the propagatorof a line L(T ), we get a gain factor

2 h( e )T + h T 2 h

( e )T + h T 2 h

( e )T

+ h T , (5.10)

as h (e)T hT . We can iterate this procedure until all the R operations are applied on propagators;at the end (i) the propagators are derived at most one time; (ii) the number of terms so generatedis k; (iii) to each self-energy graph T a factor 2 h

( e )T + h T is associated.

Assuming that

| (c)h,m

| C and

|

| C, for any one obtains, for suitable constants D

|Val( )| | V w ( ) | + | V (1)

v ( ) | D| V ( ) |

h = h 0

exp h log2 4K ()2 (h 2) / C h () + S h () + M h ()

T S ( )h ( e )T h 0

2 h( e )T + h T

h = h 0

2 hM h ( )

VV ( )E ( ) e ( | n |+ | n | ) VV ( )E ( ) e ( | m |+ | m | ) ,(5.11)

24


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where the second line is a bound for

h h 0 2

hN h ( ) and we have used that by item (12) N h ()

can be bounded through Lemma 5), and Lemma 4 has been used for the lines on scales h < h 0;moreover h = h 0 2 hM h ( ) takes into account the factors 2 h arising from the running couplingconstants (c)h,m and the action of R produces, as discussed above, the factor T S ( ) 2 h ( e )T + h T .Then one has

h = h 0

2hS h ( ) T S ( ) 2 h ( e )T = 1 ,

h = h 0

2 hC h ( ) T S ( ) 2h T 1.(5.12)

We have to sum the values of all trees, so we have to worry about the sum of the labels. Recallthat a labeled tree is obtained from an unlabeled tree by assigning all the labels to the points andthe lines: so the sum over all possible labeled trees can be written as sum over all unlabeled treesand of labels. For a xed unlabeled tree with a given number of nodes, say N , we can assign rstthe mode labels {(n , m ), (n , m )}vV ( )E ( ) , and we sum over all the other labels, which gives4| V v ( ) | (for the labels j

) times 2 | L ( ) | (for the scale labels): then all the other labels are uniquelyxed. Then we can perform the sum over the mode labels by using the exponential decay arisingfrom the node factors (3.3) and end-point factors (3.5). Finally we have to sum over the unlabeledtrees, and this gives a factor 4 N [22]. By Lemma 3, one has |V ()| = |V w ()|+ |V

(1)v ()|+ |V

(3)v ()|

3(|V (3)

w ()|+ |V (1)v ()|), hence N 3k, so that ( k ) Rn,m |Val( )| D k k , for some positive constant

D .Therefore, for xed ( n, m ) one has

k=1 ( k )n,m

k |Val( )| D0 e ( | n | + | m | ) / 4, (5.13)

for some positive constant D 0 , so that (5.8) is proved.

From (5.3) we know that the quantities (c)h,m , for h 0 and |m| 1, verify the recursive relations

(c)h +1 ,m = 2 (c)h,m +

(c)h,m (,, {

(c )h ,m }), (5.14)

where, by dening T (c)

h as the set of self-energy graphs in T (c)


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and, in the latter case, to the end-point V E 0 () we associate a rst mode label ( m , n ), wherem = m / , a second mode label (0 , 0) and an end-point factor V = 1.

Then we have the following generalization of Lemma 4.

Lemma 8. If is small enough for any tree (k )Rn,m one has

N h () 4K ()2(2 h ) / C h () + S h () + M h (), (5.16)where the notations are as in lemma 4.

Proof. Lemma 4 holds for E 0() = 0; we mimic the proof of Lemma 4 proving that

N h () max{0, 2K ()2(2 h ) / }, (5.17)for all trees with E 0() = , again by induction on K ().For any line L() set = 1 if the line is along the path connecting 0 to the root and = 0otherwise, and write

n = n0 + m , m = m0 + m, (5.18)

which implicitly denes n0 and m0 .Dene k0 = 2 (h 1) / . One has N h () = 0 for K () < k 0 , because if a line L() is indeed on

scale h then |n m | < C 021 h , so that (5.18) and the Diophantine conditions implyK () n0 > 2(h 1) / k0 . (5.19)

Then, for K k0 , we assume that the bound (5.17) holds for all K () = K < K , and we showthat it follows also for K () = K .

If the root line of is either on scale < h or on scale h and resonant, the bound (5.17) followsimmediately from the bound (4.13) and from the inductive hypothesis.

The same occurs if the root line is on scale h and non-resonant, and, by calling 1 , . . . , m thelines on scale h which are the closest to , one has m 2: in fact in such a case at least m 1among the subtrees 1 , . . . , m having 1 , . . . , m , respectively, as root lines have E (i ) = , so thatwe can write, by (4.13) and by the inductive hypothesis,

N h () = 1 +m

i=1

N h (i ) 1 + E 1hm

i=1

K (i ) (m 1) E h K (), (5.20)

so that (5.17) follows.If m = 0 then N

h() = 1 and K ()2(2 h ) /

1 because one must have K ()

k

0.

So the only non-trivial case is when one has m = 1. If this happens 1 is, by construction, theroot line of a tree 1 such that K () = K (T ) + K (1), where T is the cluster which has and 1as external lines and K (T ), dened in (4.7), satises the bound K (T ) |n 1 n|.

Moreover, if E 0(1) = , one has|n0 + m | m 2 h +1 C 0 ,|n01 + m | m 1 2 h +1 C 0 ,

(5.21)

so that, for suitable , 1 {, + }, we obtain2 h +2 C 0 (n0 n01 ) + m + 1 m 1 C 0|n0 n01 | C 0|n n1 | , (5.22)

26


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by the second Diophantine conditions in (2.33), as the quantities m appearing in (5.21) cancel

out. Therefore one obtains by the inductive hypothesis

N h () 1 + K (1)E 1h 1 + K ()E 1h K (T )E 1h K ()E 1h , (5.23)hence the rst bound in (5.17) is proved.

If E 0 (1) = , one hasN h () 1 + K (1)E 1h 1 1 + K ()E 1h 1 K ()E 1h , (5.24)

and (5.17) follows also in such a case.

The following bound for V h +1T (m , m ), h h0 , can then be obtained.

Lemma 9. Assume that there exist a constant C such that one has || C and | (c)

h,m | C,with c = a, b, for all |m| 1 and all h 0. Then if is small enough for all h 0 and for all T T

(c)h one has

|V h +1T (m , m)| B | V (T ) | e 2( h 1) / / 4e K (T ) / 4| V

(1)v (T ) | + | V w (T ) | , (5.25)

where B is a constant and K (T ) is dened in (4.7).

Proof. By using lemma 7 we obtain for all T T (c)

h and assuming h h0 we get the bound

V h +1T (m , m) B| V (T ) |

| V (1)

v (T ) | + | V w (T ) |

h

h = h 0

exp 4K (T )log2h 2(2 h ) /

C h (T ) + S h (T ) + M h (T )

T T h ( )

T h 0

2 h( e )T + h T

h

h = h 0

2 h M h (T ) e | K (T ) | / 2 , (5.26)

where B is a suitable constant. If h < h 0 the bound trivializes as the r.h.s. reduces simply toC | V (T ) | ||| V

(1)v (T ) | + | V w (T ) | e

2 | K (T ) | . The main difference with respect to Lemma 6 is that, given a

self-energy graph T T (c)

h , there is at least a line L(T ) on scale h = h and with propagator1

2(n0 + m )2 + 2m 0 + m, (5.27)

where = 1 if the line belongs to the path of lines connecting the entering line (carrying amomentum ( n, m )) of T with the line coming out of T , and = 0 otherwise. Then one has by theMelnikov conditions

C 0 |n0 | n0 + m m 0 + m C 02 h +1 , (5.28)so that |n0 | 2(h 1) / . On the other hand one has |n0 | K (T ), hence K (T ) 2(h 1) / ; so weget the bound (5.25).

It is an immediate consequence of the above Lemma that for all 0 > 0 there exists 0 > 0 suchthat for all || 0 and 0 < < 0 one has |

(c)h,m | B1||, with B1 a suitable constant.

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We have then proved convergence assuming that the parameters h,m and are bounded; we

have to show that this is actually the case, if the (c)m in (2.34) are chosen in a proper way.We start proving that it is possible to choose (c) = {

(c)m }| m | 1 such that, for a suitable positive

constant C , one has | (c)h,m | C for all h 0 and for all |m| 1.

For any sequence a {am }| m | 1 we introduce the norma = sup

| m | 1 |am |. (5.29)Then we have the following result.

Lemma 10. Assume that there exists a constant C such that || C. Then for all 0 > 0 there exists 0 > such that, for all || 0 and for all 0 < < 0 there is a family of intervals I

( h )c,m ,

h

0,

|m

| 1, c = a, b, such that I ( h +1)c,m

I ( h )c,m ,

|I h

c,m | 2( 2) ( h +1) and, if (c)m

I ( h )c,m , then

(c)h D, h h 0, (5.30) for some positive constant D . Finally one has (c)h, m =

(c)h,m , c = a, b, for all h h 0 and for all

|m| 1. Therefore one has h C for all h 0, for some positive constant D ; in particular | m | D for all m 1.Proof. The proof is done by induction on h. Let us dene J (h )c,m = [, ] and call J (h ) =| m | 1,c = a,b J

(h )c,m and I (h ) = | m | 1,c = a,b I

(h )c,m .

We suppose that there exists I ( h ) such that, if spans I ( h ) then h spans J ( h ) and | (c)h,m | D for

h h 0; we want to show that the same holds for h + 1. Let us call J ( h +1) the interval spannedby {

(c)

h +1 ,m }| m | 1,c = a,b when { (c)

m }| m | 1,c = a,b span I ( h )

. For any { (c)

m }| m | 1,c = a,b I ( h )

one has{ h +1 ,m }| m | 1,c = a,b [2 D 2 , 2 + D 2], where the bound (5.25) has been used. This meansthat J ( h +1) is strictly contained in J ( h +1) . On the other hand it is obvious that there is a one-to-onecorrespondence between {

(c)m }| m | > 1,c = a,b and the sequence {

(c)h,m }| m | 1,c = a,b , h +1 h 0. Hence

there is a set I ( h +1) I ( h ) such that, if {

(c)m }| m | 1,c = a,b spans I ( h +1) , then {

(c)h +1 ,m }| m | 1,c = a,b

spans the interval J ( h ) and, for small enough, | h | C for h + 1 h 0.The previous computations also show that the inductive hypothesis is veried also for h = 0 so

that we have proved that there exists a decreasing sets of intervals I ( h ) such that if

{ (c)m }| m | > 1,c = a,b I ( h ) then the sequence {

(c)h,m }| m | 1,c = a,b is well dened for h h and it veries

| (c)h,m | C ||. In order to prove the bound on the size of I

( h )c,m let us denote by {

(c)h,m }| m | 1,c = a,b

and

{ (c)h,m

}| m | 1,c = a,b , 0

h

h , the sequences corresponding to

{ (c)m

}| m | 1,c = a,b and

{ (c)m }| m | 1,c = a,b in I ( h ) . We have (c)h +1 ,m

(c)h +1 ,m = 2

(c)h,m

(c)h,m +

(c)h,m

(c)h,m , (5.31)

where (c)h,m and (c)h,m are shorthands for the beta functions. Then, as | k k | | h h | for

all k h, we have| h h |

12 | h +1

h +1 | + D 2| h h | . (5.32)

Hence if is small enough then one has

( 2) ( h +1) h h . (5.33)28


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Since, by denition, if spans I ( h ) , then h spans the interval J ( h ) , of size 2||, the size of I ( h ) isbounded by 2 ||( 2)(

h 1) .

Finally note that one can choose (c)m = (c) m and then (c)h,m =

(c)h, m for any |m| 1 and any

h h 0; this follows from the fact that the function (c)k,m in (5.15) is even under the exchange

m m; it depends on m through m (which is an even function of m), through the end-pointsv E () (which are odd under the exchange m m; but their number must be even) and nallythrough (q 1)k,m which are assumed inductively to be even.

It will be useful to explicitly construct the (c)h,m by a contraction method. By iterating (5.14) wend, for |m| 1

(c)h,m = 2h +1 (c)m +

h 1

k= 1

2 k 2 (c)k,m (,,

{ (c

)k ,m

}) , (5.34)

where (c)k,m (,, { (c )k ,m }) depends on k ,m with k k 1. If we put h = h in (5.34) we get

(c)m = h 1

k= 1

2 k 2 (c)k,m (,, { (c )k ,m }) + 2 h 1

(c)h,m (5.35)

and, combining (5.34) with (5.35), we nd, for h > h 0,

(c)h,m = 2h +1h 1

k= h

2 k 2 (c)k,m (,, { (c )k ,m }) + 2 h h

(c)h,m . (5.36)

The sequences { (c)h,m }| m | > 1 , h > h h0 , parametrized by {

(c)h,m }| m | 2 such that

(c)h C,

can be obtained as the limit as q of the sequences { (c)( q)h,m }, q 0, dened recursively as

(c)(0)h,m = 0 ,

(c)( q)h,m = 2h +1h 1

k= h

2 k 2 (c)k,m (,, { (c )( q 1)k ,m }) + 2 h h

(c)h,m .

(5.37)

In fact, it is easy to show inductively that, if is small enough, (q)h C, so that (5.37) is

meaningful, and

max0 h h (q)h

(q 1)h (C )q . (5.38)

For q = 1 this is true as (c)(0)h = 0; for q > 1 it follows by the fact that (c)k (,, {

(c )( q 1)k ,m })

(c)k (,, { (c )( q 2)k ,m }) can be written as a sum of terms in which there are at least one -vertex,

with a difference (c )( q 1)

h (c )( q 2)h , with h

k, in place of the corresponding (c )h , and one

node carrying an . Then (q)h converges as q , for h < h 1, to a limit h , satisfying thebound h C. Since the solution is unique, it must coincide with one in Lemma 10.

We have then constructed a sequence of (c)h,m solving (5.36) for any h > 1 and any (c)h,m ; we shall

call (c)h,m () the solution of (5.36) with h = and (c) ,m = 0, to stress the dependence on .

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We will prove the following Lemma.

Lemma 11. Under the the same conditions of Lemma 10 it holds that for any h 0 h (1 ) h (2) D |1 2|, (5.39)

for a suitable constant D.

Proof. Calling (q)h () the l.h.s. of (5.37) with h = and ,m = 0, we can show by induction onq that

(q)h (1 )

(q)h (

2 ) D |1 2|, (5.40)We nd convenient to write explicitly the dependence of the function (c)h,m from the parameter

, so that we rewrite (c)k,m (,,

{ (c

)( q 1)k ,m

} in the r.h.s. of (5.37) as (c)k,m (,, ,

{ (c

)( q 1)k ,m ()

}.

Then from (5.37) we get

(c)( q)m (1) (c)( q)m (2) =

k= h

2h k 1[ (c)k,m (,, 1 , {

(c )( q 1)k ,m (

1)}) (c)k,m (,,

2 , { (c )( q 1)k ,m (

2)})](5.41)

When q = 1 we have that (c)k,m (,, 1 , {

(c )( q 1)k ,m (

1)}) (c)k,m (,,

2 , { (c )( q 1)k ,m (

2)}) is givenby a sum of self energy graphs with one node V with a factor

with replaced by 1 2 ; asthere is at least a vertex V of w-type by the denition of the self energy graphs we obtain

(1)h (1)

(1)h (

2 )

D 1 + D 12

|1

2

|, (5.42)

for positive constants D1 < 2D and D 2 , where D1|1 2| is a bound for the self-energy rstorder contribution.

For q > 1 we can write the difference in (5.41) as

(c)k,m (,, 1, {

(c )( q 1)k ,m (

1)}) (c)k,m (,,

2 , { (c)( q 1)k ,m (

1 )})+ (c)k,m (,,

2 , { (c)( q 1)k ,m (

1)}) (c)k,m (,,

2 , { (c)( q 1)k ,m (

2)}) .(5.43)

The rst factor is given by a sum over self-energy graphs with one node V with a factor

with replaced by 1 2 ; the other difference is given by a sum over self energy graphs with a -vertexto which is associated a factor (c)( q 1)k ,m (

1 )

(c)( q 1)k ,m (

2); hence we nd

(q)h (1)

(q)h (

2) D 1 + D 32 |1 2|+ D 2 suph 0 (q 1)h (

1 ) (q 1)h (

2 ) , (5.44)

where D 1|1 2| is a bound for the rst order contribution coming from the rst line in (5.43),while the last summand in (5.44) is a bound from the terms from the last line of (5.43). Then(5.40) follows with D = 2D 1 , for small enough.

By using Lemma 11 we can show that the self consistence equation for (5.4) has a uniquesolution = O().

Lemma 12. For all 0 > 0 there exists 0 > 0 such that, for all || 0 and for all 0 < < 0 ,30


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given (c)m () chosen as in Lemma 9 (with h =

and (c) ,m = 0) it holds that (5.4) has a solution

|| C where C is a suitable constant.Proof. The solution of (5.4) can be obtained as the limit as q of the sequence (q) , q 0,dened recursively as

(0) = 0 ,

(q) = r 10

k=1 ( k )( q 1)) Rn,n

a0, n Val( ), (5.45)

where we dene (k )( q)Rn,m as the set of trees identical to (k )Rn,m except that in is replaced by(q) and each h,m is replaced by h,m ( (q) ), for all h

0,

|m

| 1. (5.45) is a contraction dened

on the set || C , for small. In fact if |(q 1) | C , then by (5.45) |(q) | C 1 + C 2C 2 ,where we have used that the rst order contribution to the r.h.s. of (5.45) is -indipendent (seeSection 3), and C 1 < C/ is a bound for it; hence for small enough (5.45) send the interval

|| C to itself.Moreover we can show inductively that

(q) (q 1) (C )q . (5.46)

For q = 1 this is true; for q > 1 (q) (q 1) can be written as sum of trees in which a)or toa node V is associated a factor proportional to (q 1) (q 2) ; or b) to a vertex is associated h ,m ( (q 1) ) h ,m ( (q 2) ) for some h , m . In the rst case we note that the constraint inthe sum in the r.h.s. of (5.45) implies that sv0 = 3 for the special vertex of ; hence, item (4)in Section 3, says that V = v0 so that such terms are bounded by D1|(1) (2) | (a term orderO(1 2) should have three v lines entering v0 and two of them coming from end points, whichis impossible). In the second case we use (5.39), and we bound such terms by D2|(1) (2) | .Hence by induction (5.46) is found, if C D1 / 4, D 2/ 4, C 1 / 4).

We have nally to prove that un,m solves the last of (1.14) and (2.2).

Lemma 13. For all 0 > 0 there exists 0 > such that, for all || 0 and for all 0 < < 0 ,given (c)m () chosen as in Lemma 9 and chosen as in lemma 12 then un,m solves the last of (1.14) and (2.2).

Proof. Let us consider rst the case in which |n | = |m| and we call Rn,m = k (k )Rn,m ; assume also

(what of course is not restrictive) that n, m is such that h 0 (|n| m ) + h 0 +1 (|n | m ) = 1.We call Rn,m, the set of trees Rn,m with root line at scale h0 , so that

un,m = Rn,m

Val( ) = Rn,m,h 0

Val( ) + Rn,m,h 0 +1

Val( ), (5.47)

and we write Rn,m,h 0 = , Rn,m,h 0

, Rn,m,h 0 , where

, Rn,m,h 0 are the trees with s 0 = 1, while

, Rn,m, h

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are the trees with s 0 = 3 and V 0 is the special vertex (see Denition 2). Then

, Rn,m,h 0

Val( ) =c= a,b

g( h ) (n, m )2 h 0 (c)h 0 ,m Rn,m c ,h 0

Val( )

+ g( h ) (n, m )2 h (c)h 0 ,m Rn,m c ,h 0 +1

Val( )

+ g(h 0 +1) (n, m )2 h 0 (c)h 0 ,m Rn,m,h 0

Val( )

+ g(h 0 +1) (n, m )2 h 0 1 (c)h 0 +1 ,m Rn,m c ,h 0 +1

Val( ) ,

(5.48)

where mc is such that ma = m and mb = m. On the other hand we can write , Rn,m,h 0 = 1,Rn,m,h 0

2,Rn,m,h 0 , where

1,Rn,m,h 0 are the trees such that the root line 0 is the external line of a

self-energy graph, and 2,Rn,m,h 0 is the complement. Then we can write

1 , Rn,m,h 0

Val( ) =c= a,b

g(h 0 ) (n, m )T T ( c )


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The last line is equal to

12n2 + 2m

[ (a )m un,m + (b)m un, m ], (5.51)

while adding the rst three lines in (5.49) to 1 , Rn,m,h 0 Val( ) we get

n 1 + n 2 + n 3 = nm 1 + m 2 + m 3 = m

un 1 ,m 1 un 2 ,m 2 un 3 ,m 3 , (5.52)

from which we get that Rn,m Val( ), for = 1, is a formal solution of (2.2). A similar resultholds for |n | = |m|.6. Construction of the perturbed frequencies

In the following it will be convenient to set = {m }| m | 2 . By the analysis of the previous sectionswe have found the counterterms { m (, )}| m | 2 as functions of and . We have now to invertthe relations

2m m (, ) = m2 , (6.1)in order to prove Proposition 2.

We shall show that there exists a sequence of sets {E ( p)} p=0 in [0, 0], such that E ( p+1) E ( p) ,and a sequence of functions {( p) ()} p=0 , with each ( p) ( p) () dened for E ( p) , such thatfor all E , with

E =

p=0E ( p) = lim p E ( p) , (6.2)

there exists the limit( ) () = lim p

( p) (), (6.3)

and it solves (6.1).To fulll the program above we shall dene since the beginning = = 1 , and we shall

follow an iterative scheme by setting, for |m| 1,(0)2m = m

2 ,

( p)2m ( p)2m () = m2 + m (( p 1) , ), p 1,(6.4)

and reducing recursively the set of admissible values of .We start by imposing on the Diophantine conditions

|n m| 2C 0|n | 0 n Z \ {0} and m Z \ {0} such that |m| = |n|, (6.5)where C 0 and 0 are two positive constants.

This will imply some restrictions on the admissible values of , as the following result shows.

Lemma 14. For all 0 < C 0 1/ 2 there exist 0 > 0 and 0 , 0 > 0 such that the set E (0) of values [0, 0] for which (6.5) are satised has Lebesgue measure |E (0) | 0(1 000 ) provided that one has 0 > 1.

Proof. For (n, m ) such that |n m| 2C 0 the Diophantine conditions in (6.5) are triviallysatised. We consider then ( n, m ) such that |n m| < 2C 0 and we write, if 0 < C 0 1/ 2

1 > 2C 0 > |n m| n

1 + 1 n m , (6.6)

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and as |n m| 1 one gets |n | 1/ 2 1/ 20 . Moreover, for xed n, the set M of ms such that|n m| < 1 contains at most 2 + 0|n | values. By writing

f ((t)) = n 1 (t) m = t 2C 0|n | 0 , t [0, 1], (6.7)and, calling I (0) the set of such that |n m| < 2C 0|n | 0 is veried for some (n, m ), one ndsfor the Lebesgue measure of I (0)

meas( I (0) ) = I (0) d = | n | 1/ 2 0 | m |M 1

1dt

d(t)dt

. (6.8)

We have from (6.7)df

dt =

df

d

d

dt =

2C 0

|n| 0, (6.9)

so that, noting that one has |f/ | |n |/ 4, we have to exclude a set of measure

| n | N

8C 0

|n | 0 +1 (2 + 0|n|) const . 00 , (6.10)

and one has to impose 0 = 1 + 0 , with 0 > 0.

For p 1 the sets E ( p) will be dened recursively as

E ( p) = E ( p 1) : |n ( p)m | > C 0 |n| |m| = |n|,

|n

(( p)m

( p)m )

| > C 0

|n

|

|n

| =

|m

m

|, p

1,

(6.11)

for > 0 to be xed.In Appendix A4 we prove the following result.

Lemma 15. For all p 1 one has ( p) () ( p 1) () C

p0 E ( p) , (6.12)

for some constant C .

Therefore we can conclude that there exist a sequence {( p) ()} p=0 converging to ( ) () for E . We have now to show that the set E has positive (large) measure.

It is convenient to introduce a set of variables (, ) such that

m + m (, ) = m |m|; (6.13)the variables (, ) and the counterterms are trivially related by

m (, ) = 2m (, ) + 2 m m (, ). (6.14)One can write ( p)m = m m (( p 1) , ), according to (6.4). We shall impose the Diophantine

conditions

|n (m m (( p 1) , )) | > C 0 |n| , (6.15)|n ((m m ) (m (( p 1) , ) m (( p 1) , ))) | > C 0 |n| .

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Suppose that for E ( p 1) the functions m (( p 1) , ) are well dened; then dene I ( p) = I ( p)1 I ( p)2 I ( p)3 , where I ( p)1 is the set of values E ( p 1) verifying the conditions

n m m (( p 1) , ) C 0 |n| , (6.16)

I ( p)2 is the set of values verifying the conditions

n (m m ) (m (( p 1) , ) m (( p 1) , )) C 0 |n| , (6.17)

and I ( p)3 is the set of values verifying the conditions

n

(m + m )

(m (( p 1) , ) + m (( p 1) , ))

C 0

|n

| . (6.18)

For future convenience we shall call, for i = 1 , 2, 3, I ( p)i (n) the subsets of I

( p)i which verify the

Diophantine conditions (6.16), (6.17) and (6.18), respectively, for xed n .We want to bound the measure of the set I ( p) . First we need to know a little better the

dependence on and of the counterterms: this is provided by the following result.

Lemma 16. For all p 1 and for all E ( p) there exists a positive constant C such that m (( p) , ) C, ( p )m m (

( p) , ) C, m 1,m (( p) , )

Cm

, ( p )m

m (( p) , ) Cm

, m 1, ( p)m (, ) C, m 1,

(6.19)

where the derivatives are in the sense of Whitney [34].

Proof. The bound for m (( p) , ) is obvious by construction.In order to prove the bound for ( p )

m m (( p) , ) note that one has

g(h j )j (

) g(h j )j () ( ) g

(h j )j () C n2j 2 3h j

2 , (6.20)

from the compact support properties of the propagator.Let us consider the quantity ( , ) (, ) ( ) (, ), where (, ) denotes the

derivative in the sense of Whitney, and note that it can be expressed as a sum over trees each onecontaining a line with propagator g

(h j )j (

) g(h j )j () ( ) g

(h j )j (), by proceeding as in

the proof of lemma 9 of [19]. Then we nd

( , ) (, ) ( ) (, ) C 2 , (6.21)

and the second bound in (6.19) follows.The bounds for m (( p) , ) and ( p )

m m (( p) , ) simply follow from (6.14) which gives

m (( p) , ) = m (( p) , )

|m|1

1 +

1 m (( p) , )/m

. (6.22)

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In order to prove the last bound in (6.19) we prove by induction the bound

( p)m (, ) C, m 1, (6.23)

by assuming that it holds for ( p 1) ; then from (6.4) we have

2( p) ( p)m () = m (( p 1) (), ) (6.24)=

m Z ( p 1)

m m (( p 1) (), )

( p 1)m () m (( p 1)m (), ) = ,

as it is easy to realize by noting that m can depend on m when |m | > |m| only if the sum of the absolute values of the mode labels mv is greater than |m m|, while we can bound the sum of the contributions with |m

| < |m| by a constant times , simply by using the second line in (6.19).Hence from the inductive hypothesis and the proved bounds in (6.19), we obtain( p) ( ) ( p) () ( ) ( p) () C | | , (6.25)

so that also the bound (6.23) and hence the last bound in (6.16) follow.

Now we can bound the measure of the set we have to exclude: this will conclude the proof of Proposition 2.

We start with the estimate of the measure of the set I ( p)1 . When (6.16) is satised one must have

C 2

|m

| |m

m (( p 1) , )

|

|n

|+ C 0

|n

|

C 2

|n

|,

C 1|m| |m m (( p 1) , )| |n| C 0|n | C 1|n|, (6.26)which implies

M 1|n| |m| M 2|n|, M 1 = C 1C 1

, M 2 = C 2C 2

, (6.27)

and it is easy to see that, for xed n , the set M0(n) of ms such that (6.16) are satised containsat most 2 + 0|n | values.

Furthermore (by using also (6.5)) from (6.16) one obtains also

2C 0|n | 0 ||n| m | ||n | m + m (( p 1) , )|+ |m (( p 1) , )|

C 0

|n

| + C 0 /

|m

|,

(6.28)

which implies, together with (6.36), for 0 ,

|n | N 0 C 0M 1C 0

1/ ( 0 1)

. (6.29)

Let us write () = = 1 and consider the function (( p 1) , ): we can dene a mapt (t) such that

f ((t)) ((t)) |n| m + m (( p 1) , (t)) = t C 0

|n | , t [1, 1], (6.30)

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describes the interval dened by (6.16); then the Lebesgue measure of

I

( p)1 is

meas( I ( p)1 ) = I ( p )1 d = | n | N 0 mM 0 (n )

1

1dt

d(t)dt

. (6.31)

We have from (6.30)df dt

= df d

ddt

= C 0

|n | , (6.32)

hence

meas( I ( p)1 ) =

| n | N 0 mM 0 (n )

C 0

|n| 1

1dt

df ((t))d(t)

1

. (6.33)

In order to perform the derivative in (6.30) we write

dmd

= m +| m | 1

( p 1)m

m ( p 1)m , (6.34)

where m and ( p 1)m are bounded through lemma 11. Moreover one has

( p 1)m

m C e | m m | / 2 , |m | > |m|, (6.35)

and the sum over m can be dealt with as in (6.24).At the end we get that the sum in (6.34) is O(), and from (6.29) and (6.30) we obtain, if 0 is

small enough,f ((t))

(t) |n|4

, (6.36)

so that one has I ( p )1 d const. | n | N 0 C 0|n| +1 (2 + 0|n|) const . 0

( 1) / ( 0 1)0 +

( 0 +1) / ( 0 1)0 .

(6.37)

Therefore for max{1, 0 1} the Lebesgue measure of the

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