# The Poincaré theorem on the nonexistence of first integrals

Nel documento HAMILTONIAN MECHANICS —————- (pagine 73-87)

### 4.3The Poincaré theorem on the nonexistence of first in-tegrals

Another theorem due to Poincaré states that, under the same hypotheses of non-degeneracy of the integrable part, genericness of the perturbation (to leading order), and under strong regularity assumptions, the perturbed system admits only trivial first integrals, i.e. the func-tions of the Hamiltonian itself. In what follows, for the sake of convenience, the notation is slightly changed and the integrable part of the Hamiltonian is denoted by H0(J), whereas the perturbation terms are denoted by Hj(θ, J), j ≥ 1.

Theorem 4.3. Let H(θ, J;ε) = H0(J) + εH1(θ, J) + ε2. . . be analytic in all its arguments, with H0non degenerate and H1 generic in B. Then, no first integral F(θ, J;ε) exists, analytic in all its arguments and independent ofH.

C PROOF. The scheme of the proof is the following:

1. one proves that F0:= F(θ, J;0) is independent of θ;

2. one proves that F0=F0(H0);

3. one proves that if F is independent of H then there exists another first integral whose zero order term is independent of H0, contrary to what proven in the previous point.

The first integral F =P

j≥0εjFj can be formally constructed by imposing that its Poisson bracket with H =P

j≥0εjHj be zero. The condition 0 = {F, H} = X

j,`≥0εj+`{Fj, H`} =X

s≥0

εs Ã

X

j+`=s

{Fj, H`}

!

is equivalent to the hierarchy of conditions X

j,`≥0 j+`=s

{Fj, H`} = 0 , s = 0,1,2... , (4.25)

to be solved one after the other.

1. The first of conditions (4.25) is

{F0, H0} =∂F0

∂θ ·∂H0

∂J = ω(J) ·∂F0

∂θ = 0 , or, passing to the Fourier coefficients

ik · ω(J) bF0,k(J) = 0 , ∀k ∈ Zn , ∀J ∈ B .

The above equation is obviously satisfied for k = 0, for any J ∈ B. For non resonant actions, i.e. J ∈ B \Sn−1

r=1Br, k · ω(J) 6= 0, so that bF0,k(J) = 0 for any k ∈ Zn\ {0} on a dense, full measure set of actions. By requiring the continuity of the Fourier coefficients with

respect to the actions one immediately getsFb0,k(J) = 0 for any k ∈ Zn\ {0} and any J ∈ B.

One is thus left with F0(θ, J) =Fb0,0(J) := F0(J), i.e. F0is independent of the angles.

An alternative argument is the following. The condition {F0, H0} = 0 means that one is looking for a first integral of H0, i.e. a function F0 which is invariant with respect to the flow of H0, so that F0(θ, J) = F0(θ + ω(J)t, J). For non resonant frequencies, i.e. for non resonant actions, the flowθ 7→ θ +ω(J)t is ergodic on Tn. As a consequence F0= 〈F0〉, but F0= F0 by invariance of F0, so that F0(θ, J) = 〈F0〉 (J) for anyθ ∈ Tn, any J in the dense set of non resonant actions and, by continuity, any J ∈ B.

2. The second of conditions (4.25) reads {F0, H1} + {F1, H0} = −∂F0

∂J ·∂H1

∂θ + ω(J) ·∂F1

∂θ = 0 , or, passing to the Fourier coefficients

k · ω(J) bF1,k(J) = k ·∂F0(J)

∂J Hb1,k(J), (4.26)

which must hold for any integer vector k and any action J in B. In particular, for any given J in Bn−1= ω−1(Ωn−1∩ ω(B)) there exist n − 1 linearly independent integer vectors k(1), . . . , k(n−1) such that k( j)· ω(J) = 0 ( j = 1, . . . , n − 1). On the other hand, by the generic-ness of H1there exist other n − 1 integer vectors q(1), . . . , q(n−1), with each q( j) parallel to k( j), such thatHb1,q( j)(J) 6= 0. Since obviously q( j)·ω(J) = 0 for any j = 1, . . . , n −1, equation (4.26) implies q( j)· ∂F0/∂J = 0. Thus, for any completely resonant J there exist n − 1 lin-early independent vectors that are orthogonal toω(J) = ∂H0/∂J and, as a consequence, to∂F0/∂J. Thus, for any such completely resonant J the gradients of F0 and of H0 are parallel to each other, i.e. there exists a functionλ(J) such that

∂F0

∂J = λ(J)∂H0

∂J , ∀J ∈ Bn−1.

The latter condition can be equivalently written as dF0= λdH0, so thatλ(J) = dF0/dH0, which implies F0(J) =F0(H0(J)) (λ =F00) for any J ∈ Bn−1 and, by regularity, for any J ∈ B.

3. Going back to equation (4.26), for non resonant actions and k 6= 0 one gets bF1,k = F00(H0)Hb1,k, which extends by regularity to the whole set B. This implies

F1(θ, J) =F00(H0(J))H1(θ, J) +F1(J),

where 〈F〉1 is the undetermined mean value of F1 on Tn. The latter expression of F1 yields

F0+ εF1+ ε2. . . = F0(H0) + εF00(H0)H1+ ε 〈F〉1+ ε2· · · =

= F0(H0+ εH1+ ε2. . . ) + ε〈F〉1+ ε2. . . . (4.27) On the other hand, by studying the third equation of the hierarchy (4.25), namely

{F0, H2} + {F1, H1} + {F2, H0} = 0 (4.28)

4.3. THE POINCARÉ THEOREM ON THE NONEXISTENCE OF FIRST INTEGRALS 75

with the same reasonings made above, one finds that

〈F〉1=F1(H0), (4.29)

F2=F00(H0)H2+1

2F000(H0)H12+F10(H0)H1+ 〈F〉2 , (4.30) where 〈F〉2is undetermined. Equations (4.29) and (4.30) imply

F0+ εF1+ εF2+ ε3. . . = F0(H0+ εH1+ ε2H2+ ε3. . . ) +

+ εF1(H0+ εH1+ ε2. . . ) + ε2〈F〉2+ ε3. . . ,

which suggests that one is actually building up a function of H. In order to prove this, suppose that

F(0):=F0(H0) + εF1+ ε2F2+ ε3. . .

is a first integral of the perturbed system independent of H. Let us prove that starting from F(0) one can construct another first integral such that its term of order zero (inε), denoted as usual by the subscript 0, is independent of H0. Obviously

F(1):=F(0)−F0(H)

ε ,

as a linear combination of first integrals, is a first integral too. Now, if F0(1)is independent of H0one stops, otherwise one defines F0(1):=F1(H0) and goes on defining

F(2):=F(1)−F1(H)

ε ,

which is still a first integral. Again, either F0(2) is independent of H0 or one goes on this way by defining F0(2) :=F2(H0). Such a procedure must stop after a finite number m of steps, yielding a first integral F(m) whose zero order term F0(m) is independent of H0. Indeed, if this were not the case, one would have

F(0) = F0(H) + εF(1)=F0(H) + εF1(H) + ε2F(2)+

= F0(H) + εF1(H) + ε2F2(H) + ε3F(3)+ . . . =X

j≥0

εjFj(H),

i.e. F(0) would be a function of H, against the hypothesis. Thus a first integral F(m)such that F0(m)is independent of H0exists, but this is absurd, since in the previous point 2. it was proven that the zero order term of any first integral must necessarily be a function of H0. The conclusion is that any first integral F must be a function of H.B

Exercise 4.2. Study equation (4.28) and get equations (4.29) and (4.30).

## Hierarchy of structures in topology andtheir utility

Definition A.1. A metric space is a pair (X, d) where X is a set and d : X ×X → R+is a distance, i.e. a nonnegative function which satisfies

1. d(x, y) = 0 iff x = y (nondegeneracy);

2. d(x, y) = d(y, x) ∀x, y ∈ X (symmetry);

3. d(x, z) ≤ d(x, y) + d(y, z) ∀x, y, z ∈ X (triangle inequality).

In metric spaces one can define limits of functions and thus continuity. Indeed, given two metric spaces (X, d) and (X0, d0), one says that a function f : X → X0has (finite) limit` ∈ X0 as x tends to x0, and writes limx→x0f (x) = `, if for any ² > 0 there exists a δ²> 0 such that for all x ∈ X \ {x0} satisfying d(x, x0) < δ²one has d0( f (x),`) < ². The definition of continuous function is obtained by setting` = f (x0).

Notice that a metric space is not necessarily a linear (or vector) space. On the other hand, if one wants to build up differential calculus, one needs the linear structure. Indeed, the following definition of (weak, or Gateaux) differential of a function f : X → X0 in x0∈ X with increment h ∈ X is quite natural:

d f (x0; h) := lim

t→0

1

t[ f (x0+ th) − f (x0)] .

Observe that one needs the linear combination x0+ th ∈ X and the differential d f (x0; h) ∈ X0, the convergence of the above limit being meant in the metric d0 of X0. Actually, one needs to measure the size of the objects in X and X0, in order to have some control on the above formula (for example, one would like to have that d f is “small” when h is “small”). This is achieved when the distances d on the linear spaces X is induced by a norm k · k, such that d(x, y) = kx − yk.

Definition A.2. A normed space is a pair (X, k · k) where X is a linear space and k · k : X → R+ is a norm, i.e. a nonnegative function satisfying

1. kxk = 0 iff x = 0 (nondegeneracy);

77

2. kλxk = |λ|kxk ∀λ ∈ R and ∀x ∈ X (homogeneity of degree one);

3. kx − zk ≤ kx − yk + ky − zk (triangle inequality).

One easily checks that a normed linear space is a metric space with the distance

d(x, y) := kx − yk . (A.1)

Differential calculus is performed in normed spaces, but in order to set up all the machinery of the differential equations one needs a stronger property. More precisely, in order to have the uniqueness of the solution to a given differential equation one needs the completeness of the space, namely that every Cauchy sequence converges (recall that a sequence {xn}n∈N⊂ X is a Cauchy sequence if ∀² > 0 there exist N²∈ N such that for any pair n, m ∈ N satisfying n ≥ m > N²one has d(xn, xm) < ²).

Definition A.3. A complete normed space is called Banach space.

Almost all the classical results in the theory of differential equations is (existence, unique-ness and regularity) hold in Banach spaces. Completeunique-ness enters the game in the proof of the Picard existence and uniqueness theorem, as follows. We recall that a differential equation

˙x(t) = u(x(t)), with initial condition x(0) = x0, is naturally solved by iterating the map P : X → X defined by

x0(t) = P(x) := x0+ Z t

0

u(x(s)) ds ,

with initial point x(s) ≡ x0. Now, P acts on the space B(I, X ) of bounded functions t 7→ x(t), defined on a suitable real interval I, with value in the Banach space X . Such a space, endowed with the metric d(x, y) := supt∈Ikx(t)− y(t)k, k·k being the norm on X , turns out to be complete.

A local Lipschitz condition on u guarantees (and is optimal) that P : B → B is a contraction, i.e. that there existsρ ∈]0,1[ such that d(P(x), P(y)) ≤ ρd(x, y) for any x, y close enough to x0. Since a contraction in a complete metric space admits a unique fixed point, then there exists a unique ˆx(t) ∈ B(I, X ) such that ˆx(t) = P( ˆx(t)), i.e. ˆx(t) is the local solution of the given differential equation.

It may happen that the linear space where the problem at hand is set up possesses a Eu-clidean structure, i.e. a scalar (or internal) product between its elements. When this happens it is very useful, since then concepts like angle, direction and projection become meaningful.

Definition A.4. A Euclidean space is a pair (X, 〈,〉) where X is a linear space and 〈,〉 : X ×X → R is a scalar product, i.e. a real function satisfying

1. 〈x, x〉 > 0 ∀x ∈ X \ {0}, 〈0,0〉 = 0 (nondegeneracy);

2. 〈x, y〉 = 〈y, x〉 ∀x, y ∈ X (symmetry);

3. ­

λx + µy, z® = λ〈x, z〉 + µ〈y, z〉 ∀λ,µ ∈ R and ∀x, y, z ∈ X (linearity in the first entry).

79 One easily checks that a Euclidean space is a normed (and thus a metric, see (A.1)) space with the norm

kxk :=p

〈x, x〉 . (A.2)

Two vectors x, y of (X , 〈,〉) are said to be mutually orthogonal if 〈x, y〉 = 0. In a Euclidean space, for example, the Pythagoras theorem holds, namely kx + yk2= kxk2+ kyk2 for all pairs x, y of mutually orthogonal vectors.

Of course, the Euclidean structure is very useful in dealing with differential equations, whose appropriate environment, as explained above, is the Banach space. This justifies the introduction of a stronger topological structure, namely that of a Euclidean space which is complete with respect to the norm (A.2) naturally induced by the scalar product, which one can shortly refer as to a Euclidean-Banach space.

Definition A.5. A Euclidean-Banach space is called a Hilbert space.

Remark A.1. Wherever needed, the linear structure of the space X can be considered on the complex fieldC, which only requires the symmetry property 2. of the scalar product to change as follows: 〈x, y〉 = 〈y, x〉, a star denoting complex conjugation.

Example A.1. Quantum mechanics is the theory of certain unitary evolutions in Hilbert spaces, described by equations of the form

u = −i ˆ˙ Hu

where ˆH: H → H is a real self-adjoint operator on an infinite-dimensional Hilbert space H. The formal solution of the above equation reads u(t) = e−i ˆHtu(0). The meaning of such a solution is completely specified in terms of the basis of H formed by the eigenvectors of ˆH, satisfying Hˆϕω= Eωϕω, where the set of the eigenvalues {Eω} contains, in general, both a discrete and a continuous component. If one writesu(t) =P

ωcω(t)ϕω, thenu(t) =P

ωcω(0)e−iEωtϕω. Example A.2. The hydrodynamic inviscid Burgers-Hopf equation

vt+ vvx= 0 ,

where v(t, ·) is defined on X = R or X = R/Z, is a highly nontrivial example of nonlinear Hamil-tonian PDE. A HamilHamil-tonian is H(v) = −16R v3dx with Poisson tensor x, so that vt= ∂xδH/δv.

One easily checks that the flow of the above equation preserves all the Lp(X) norms k · kp :=

¡R

X| · |pdx¢1/p

, for any p ≥ 1.

## Fourier series expansion of functions onthe torus

Let us consider a smooth function f :Tn→ R and its Fourier series expansion f (θ) = X

k∈Zn

keik·θ , (B.1)

with Fourier coefficients ˆfkgiven by fˆk= 1

(2π)n Z

Tne−ik·θf (θ) dnθ . (B.2)

The larger function space where the above formal relations become meaningful isL2(Tn,µT) . Here we limit ourselves to prove the following statements linking the decay of the Fourier coefficients of the function f to its regularity. In the sequel we make use of theRn-norms

|ξ|s:=

Cs,n being a numerical constant independent of f . Conversely, suppose that three positive con-stants As, s and R exist, such that the bounded coefficients ˆfk satisfy | ˆfk| ≤ As/|k|s1 for any

|k|1> R; then the Fourier seriesP

kkeik·θ is absolutely convergent, uniformly onTn, if s > n.

C PROOF. From (4.14) it follows that (−ikj)sk = 1

Taking the modulus of both sides in the above identity yields

Summing the latter inequalities over j one gets

|k|ss| ˆfk| ≤ max

This differs from (B.3) in the presence of the norm |k|s in place of |k|1. The inequality (B.3) is obtained by recalling that all norms are equivalent in Rn, so that a constant Cs,n> 0 exists, such that C1/ss,n|ξ|1≤ |ξ|s∀ξ ∈ Rn.

On the other hand, suppose that three positive constants As, s and R exist, such that the bounded sequence { ˆfk}k∈Zn satisfies | ˆfk| ≤ As/|k|s1∀|k|1> R. Then

where C is a positive constant bounding the finite sum over |k|1≤ R, and the (optimal) estimate

has been used in the last step. Obviously, the generalized harmonic series appearing in the last row of (B.4) converges if s − n > 0.B

Remark B.1. Under suitable hypotheses, the above result can be improved: actually the Fourier coefficients of absolutely continuous (a bit more than uniformly continuous) functions decay as 1/|k|1.

Remark B.2. A large torus dimension n does not allow to preserve the same degree of regularity in passing from the function to the coefficient and back. For example, if | ˆfk| ≤ As/|k|n+11 for any

|k|1> R, then the Fourier series of f is absolutely and uniformly convergent on Tn, but the series of ∇f is not.

83 Exercise B.1. Prove the estimate (B.5). Hint: first, check thatNn(0) = 1, N1(s) = 2,N2(s) = 4s;

secondly, prove that

Nn(s) =Nn−1(s) + 2[Nn−1(s − 1) + ··· +Nn−1(1) +Nn−1(0)] ; finally, use induction forn ≥ 3.

Lemma B.2. If a real number ρ > 0 exists such that f is analytic in the complex extension of the torusTρn:= {z ∈ Cn: Re(zj) ∈ T ; |Im(zj)| ≤ ρ ; j = 1,..., n}, with Mρ:= maxTρn| f (z)|, then

| ˆfk| ≤ Mρe−ρ|k|1 , ∀k ∈ Zn. (B.6) Conversely, suppose that two positive constantsB andρ exist, such that the coefficients ˆfksatisfy

| ˆfk| ≤ Be−ρ|k|1 ∀k ∈ Zn. Then the function f (z) :=P

kkeik·z is analytic in Tnρ−δ, with the upper boundMρ−δ≤ B/(tanh(δ/2))n, for any 0 < δ < ρ.

C PROOF. Consider first the case n = 1, and define the piecewise linear, counter-clockwise oriented closed path in the complex plane

C+:= {0 ≤ x ≤ 2π; y = 0} ∪ {x = 2π;0 ≤ y ≤ ρ} ∪ {0 ≤ x ≤ 2π; y = ρ} ∪ {x = 0;0 ≤ y ≤ ρ} . Let us consider also the clockwise oriented, closed pathC, reflected ofC+ with respect to the x-axis. Since e−ikzf (z) is analytic in an open set containingC+, one has

0 = I

C+e−ikzf (z) dz = Z 2π

0

e−ikxf (x) dx + Z 0

2πe−ik(x+iρ)f (x + iρ) dx + +

Z ρ

0

e−ik(2π+i y)f (2π + i y) idy + Z 0

ρ e−ik(0+i y)f (0 + i y) idy .

The two integrals in the second row above cancel out each other by periodicity of f along the x direction, so that

k= 1 2π

Z 2π

0

e−ikxf (x) dx = ekρ 2π

Z 2π

0

e−ikxf (x + iρ) dx .

Now, if k = −|k| ≤ 0, the latter relation yields the searched estimate

| ˆfk| ≤ Mρe−|k|ρ .

In the case k ≥ 0 the same result is reached by choosing C as the starting contour of integration. The generalization to the case n > 1 is almost trivial. One defines the contourCj± relative to the variable zj= xj+ i yj as above ( j = 1,..., n). The analyticity of e−ik·zf (z) with respect to zjimplies

I

Cj±e−i(k1z1+···knzn)f (z1, . . . , zn) dzj= 0 ∀ j = 1, . . . , n .

By integrating e−ik·zf (z) with respect to each variable zj on the contour Cjσj, where σj= −sign(kj), implies

k=e−ρ(|k1|+···+|kn|) (2π)n

Z

Tne−i(k1x1+···knxn)f (x1+ iσ1y1, . . . , xn+ iσnyn) dx1· · · dxn. The latter relation provides then the estimate (B.6).

Suppose now, conversely, that two positive constants B and ρ exist, such that the se-quence { ˆfk}k∈Zn satisfies | ˆfk| ≤ Be−ρ|k|1 ∀k ∈ Zn. One has

¯

¯

¯

¯

¯ X

k∈Zn

keik·z

¯

¯

¯

¯

¯

≤ X

k∈Zn

Be|k1|(|y1|−ρ)+···+|kn|(|yn|−ρ).

Notice that the series on the right hand side requires |yj| < ρ for any j = 1, . . . , n in order to converge. Setting |yj| = ρ − δ one gets the estimate

¯

¯

¯

¯

¯ X

k∈Zn

keik·z

¯

¯

¯

¯

¯

≤ B X

k∈Zn

e−δ|k|1= B Ã

X

j∈Z

e−δ j

!n

= B

(tanh(δ/2))n , which proves the upper bound on the maximum Mρ−δ of f (z) :=P

kkeik·z in Tρn. The analyticity of f (z) in the same domain is an obvious consequence of the exponential decay of its Fourier coefficients, which in turn implies that any partial derivative of f , of any order, admits a Fourier series expansion absolutely and uniformly convergent onTnρ.B

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Nel documento HAMILTONIAN MECHANICS —————- (pagine 73-87)