On the global solvability of involutive systems

J. Math. Anal. Appl. 444 (2016) 527–549

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

On the global solvability of involutive systems Adalberto P. Bergamasco a,1 , Cleber de Medeira b , Alexandre Kirilov b,2 , Sérgio L. Zani a,∗,3 a b

Departamento de Matemática – ICMC-USP – Caixa Postal 668, 13560-970, São Carlos, SP, Brazil Departamento de Matemática – UFPR – Caixa Postal 19081, 81531-980, Curitiba, PR, Brazil

a r t i c l e

i n f o

Article history: Received 22 February 2016 Available online 29 June 2016 Submitted by H.R. Parks Keywords: Global solvability Complex vector fields Involutive systems Liouville number

a b s t r a c t We consider a class of involutive systems in Tn+1 associated with a closed 1-form defined on the torus Tn . We prove that, under a geometric condition, the global solvability of this class is equivalent to a diophantine condition involving Liouville forms and the connectedness of all sublevel and superlevel sets of a global primitive associated with the system. © 2016 Elsevier Inc. All rights reserved.

1. Introduction We consider an involutive structure on the torus Tn+1 that is associated to a closed 1-form c defined on Tn . More precisely, if we denote the coordinates of Tn+1  (R/2πZ)n+1 by (t, x), t ∈ Tnt , x ∈ T1x , then the 1-form dx − c(t) gives rise to a subbundle T  of C ⊗ T ∗ (Tn+1 ), and the orthogonal of which, L = (T  )⊥ ⊂ C ⊗ T (Tn+1 ), is an involutive structure of co-rank 1 that is spanned globally by the vector fields Lj =

∂ ∂ + cj (t) , ∂tj ∂x

j = 1, . . . , n,

where c(t) = c1 dt1 + · · · + cn dtn . Further explanations about these concepts can be found in the books [10] and [15]. n n We write c(t) = a(t) + ib(t), where a = j=1 aj dtj , and b = j=1 bj dtj , with aj , bj ∈ C ∞ (Tn ; R). * Corresponding author. E-mail addresses: [email protected] (A.P. Bergamasco), [email protected] (C. de Medeira), [email protected] (A. Kirilov), [email protected] (S.L. Zani). 1 Partially supported by CNPq (grants 305916/2011-4 and 307541/2015-0) and FAPESP (grant 2012/03168-7). 2 Partially supported by CAPES/IMI (grant 88881.068004/2014-01). 3 Partially supported by CNPq (grant 307473/2012-0) and FAPESP (grant 2012/03168-7). http://dx.doi.org/10.1016/j.jmaa.2016.06.045 0022-247X/© 2016 Elsevier Inc. All rights reserved.

(1.1)

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The system (1.1) gives rise to a complex of differential operators L which at the first level acts in the following way Lu = dt u + c(t) ∧

∂ u, ∂x

u ∈ C ∞ (Tn+1 ) (or D (Tn+1 )),

where dt denotes the exterior differential on the torus Tnt . The local solvability of this complex was completely solved by Treves in [14]. We are interested in studying the global solvability at the first level of this complex. We still denote by L the operator acting at this level. Some results are already known. The case when c is exact is contained in a result in [11] which implies that global solvability of L is equivalent to the connectedness of all superlevels and sublevels of the primitives of b in Tn . The case when the vector fields in (1.1) are real was treated in [9] and their results imply that L is globally solvable if and only if the 1-form a is either rational or non-Liouville (see Definition 2.1). In a more recent work, [4], the problem was solved when each bj depends only on the variable tj . The conditions involved both connectedness of superlevels and sublevels of the primitives of b in some covering space and the character of the 1-form a regarding whether it is an integral form, a rational form or a non-Liouville one. The articles [1,3–8,12,13] deal with similar questions. n In this work we are interested in global solvability in the case when the 1-form b = j=1 bj (t)dtj is n exact without imposing any further condition on the 1-form a = j=1 aj (t)dtj . Our main result is stated in Theorem 2.4. 2. Preliminaries and statement of the main result Let c be a smooth, complex, closed 1-form defined on Tn . Set a = Re c and b = Im c. Assume that b is exact. We study the global solvability of the equation Lu = dt u + c(t) ∧ where u ∈ D (Tn+1 ) and f ∈ C ∞ (Tnt × T1x ;

1,0

f=

∂ u = f, ∂x

), the space of all smooth 1-forms of the form n 

fj (t, x)dtj .

j=1

There are natural compatibility conditions on f for the existence of a solution u to the equation Lu = f . We now move on to describing them. Definition 2.1 (See [2]). Let a be a smooth, real, closed 1-form on Tn . We say that  1. a is integral if (2π)−1 σ a ∈ Z for any 1-cycle σ in Tn . 2. a is rational if qa is an integral 1-form for some q ∈ Z. 3. a is Liouville if a is not rational and there exists a sequence {pl } of integral 1-forms and a sequence 1 ∞ n {ql }, ql ≥ 2, of integers such that {(ql )l (a − (1/ql )pl )} is bounded in C (T ). n We may write a = a0 + dt A where A is a real valued smooth function of t ∈ Tn and a0 = j=1 aj0 dtj ∈  1 n 1 n R  Rn . We will identify the 1-form a0 ∈ R with the vector (a10 , . . . , an0 ) in Rn consisting of the periods of the real 1-form a given by

A.P. Bergamasco et al. / J. Math. Anal. Appl. 444 (2016) 527–549

aj0

1 = 2π

529

2π aj (0, . . . , τj , . . . , 0)dτj . 0

Proposition 2.2 (See [2]). Let a = a0 + dt A be a real closed 1-form on Tn . Then 1. a is integral if and only if a0 ∈ Zn . 2. a is rational if and only if a0 ∈ Qn . 3. a is Liouville if and only if a0 ∈ / Qn and there exist a constant C > 0 and a sequence {(pl , ql )} in Zn × Z (ql ≥ 2) such that (j)  p  C  max aj0 − l  ≤  l , ∀l ∈ N. j=1,...,n ql ql

Since b is exact, then there exists a real valued smooth function B on Tn such that dt B = b. Thus, we may write c = a0 + dt C where C(t) = A(t) + iB(t). 1,0 For each f ∈ C ∞ (Tnt × T1x ; ) we consider the x-Fourier series f (t, x) =



fˆ(t, ξ)eiξx ,

ξ∈Z

n where fˆ(t, ξ) = j=1 fˆj (t, ξ)dtj and fˆj (t, ξ) denotes the Fourier transform with respect to x. 1,0 ) and if there exists u ∈ D (Tn+1 ) such that Lu = f then Therefore, if f ∈ C ∞ (Tnt × T1x ; Lf = L(Lu) = 0, also fˆ(t, ξ)ei(ψξ (t)+ξC(t)) is exact when ξa0 is integral,

(2.1)

where ψξ ∈ C ∞ (Rn ; R) is such that dψξ = Π∗ (ξa0 ) and Π : Rn → Tn denotes the universal covering of Tn . Indeed, taking the x-Fourier series in the equation Lu = f , for each ξ we have dt u ˆ(t, ξ) + iξc(t)ˆ u(t, ξ) = fˆ(t, ξ) or equivalently dt u ˆ(t, ξ) + iξ(a0 + dt C(t))ˆ u(t, ξ) = fˆ(t, ξ), thus, if ξa0 is integral we obtain

 dt u ˆ(t, ξ)ei(ψξ (t)+ξC(t)) = fˆ(t, ξ)ei(ψξ (t)+ξC(t)) . We define the following set

1,0  E = f ∈ C ∞ (Tnt × T1x ; ); Lf = 0 and (2.1) holds . Definition 2.3. The operator L is said to be globally solvable on Tn+1 if for each f ∈ E there exists u ∈ D (Tn+1 ) such that Lu = f .

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n Let B be a global primitive of the real 1-form b = j=1 bj (t)dtj . Since b is exact, the primitive B may be thought as a 2π-periodic smooth function defined on Rn, that is, we may consider the function B as a global primitive of the pull-back Π∗ b via the universal covering Π : Rn → Tn . We will denote by M the maximum of B over the cube [0, 2π]n and by Mj the maximum of B over the face [0, 2π]n ∩ {t; tj = 0}. Also, we will consider the set . J = {j ∈ {1, . . . , n}; bj ≡ 0}. Note that, J = ∅ means that the system (1.1) has real vector fields. n . Let a = j=1 aj (t)dtj be the real part of the 1-form c. If J = ∅ we consider the 1-form aJ =  j∈J aj (t)dtj . When the 1-form a is rational and J = ∅ we denote by qJ the smallest positive integer such that qJ aJ is integral and by q∗ the smallest positive integer such that q∗ a is integral; thus qJ ≤ q∗ and qJ divides q∗ . Under the previous notation the main result of this work is the following theorem: n Theorem 2.4. Let c = a + ib be a smooth closed 1-form defined on Tn where b = j=1 bj dtj is exact, and let B be a global primitive of b. If Mj < M for each j such that bj ≡ 0, then the operator L = dt + c(t) ∧ ∂/∂x is globally solvable on Tn+1 if and only if one of the following conditions holds: (I) J = ∅, the 1-form a is integral and the sublevels Ωs = {t ∈ Tn ; B(t) < s} and superlevels Ωs = {t ∈ Tn ; B(t) > s} are connected for all s ∈ R. (II ) J = ∅, the 1-form a is rational, qJ = q∗ and the sublevels Ωs = {t ∈ Tn ; B(t) < s} and superlevels Ωs = {t ∈ Tn ; B(t) > s} are connected for all s ∈ R. (III ) J = ∅ and the 1-form aJ is non-Liouville. When each function bj is non-identically zero (J = ∅) the condition Mj < M implies that the maximum of primitive B over the cube [0, 2π]n is not attained at the boundary. When J = ∅, a simple change of variables shows that we can consider J = {1, . . ., m} where 1 ≤ m ≤ n. If 1 ≤ m ≤ n − 1 then the system (1.1) has two blocks of vector fields, the first one of them consisting of all the real vector fields in the original system. Note that when J = {1, . . . , n} we have qJ = q∗ and b = 0, thus B is constant, hence B has only connected sublevels and superlevels on Tn . In this case Theorem 2.4 says that L is globally solvable if and only if the 1-form a is either rational or non-Liouville, which was proved by Bergamasco and Petronilho in [9]. The differential operator L is globally solvable if and only if the differential operator dt + (a0 + ib(t)) ∧

∂ ∂x

(2.2)

is globally solvable. Indeed, we write a in the form a = a0 + dt A where A ∈ C ∞ (Tn ; R) and define u=

 ξ∈Z

u ˆ(t, ξ)eiξx → Su =



u ˆ(t, ξ)eiξA(t) eiξx .

ξ∈Z

Observe that S is an automorphism of D (Tn+1 ) and of C ∞ (Tn+1 ). Moreover, we have SLS −1 = dt + (a0 + ib(t)) ∧

∂ , ∂x

which ensures our statement. Therefore, it is enough to prove Theorem 2.4 for the operator (2.2). From now on we will denote by L the operator (2.2), that is,

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L = dt + (a0 + ib(t)) ∧

∂ ∂x

531

(2.3)

and by E the corresponding space of compatibility conditions. The new operator L is associated with the following vector fields Lj =

∂ ∂ + (aj0 + ibj (t)) , j = 1, . . . , n. ∂tj ∂x

(2.4)

3. Sufficiency in Theorem 2.4 Suppose first that condition (I) holds. Since the real 1-form a is integral, then a0 ∈ Zn and u=



u ˆ(t, ξ)eiξx → T u =



ξ∈Z

u ˆ(t, ξ)eiξa0 eiξx

ξ∈Z

is an automorphism of D (Tn+1 ) and of C ∞ (Tn+1 ). Also, T LT −1 = dt + ib(t) ∧

∂ . ∂x

(3.1)

Therefore, when a is integral, by (3.1) the operator L given by (2.3) is globally solvable if and only if the ∂ . is globally solvable. operator L0 = dt + ib(t) ∧ ∂x On the other hand, if B is a global primitive of b defined on Tn , it follows from [11] that the operator L0 is globally solvable if and only if B has only connected sublevels and superlevels sets in Tn . Therefore, condition (I) together with (3.1) implies that L is globally solvable. Assume now that condition (III ) holds. Since aJ is non-Liouville then (a10 , . . . , am0 ) ∈ / Qm and it follows that there exist a constant C > 0 and an integer d > 1 such that max |qaj0 − pj | ≥ C|q|−(d−1) ,

1≤j≤m

(p, q) ∈ Zm × N.

(3.2)

We will denote by (tJ , tJ  , x) the coordinates on Tnt ×T1x , where tJ = (t1 , . . . , tm ) and tJ  = (tm+1 , . . . , tn ). n Let f (t, x) = j=1 fj (t, x)dtj ∈ E. Consider the (tJ , x)-Fourier series as follows u(t, x) =



u ˆ(tJ  , p, q)ei(p·tJ +qx) ,

(3.3)

fˆj (tJ  , p, q)ei(p·tJ +qx) ,

(3.4)

(p,q)∈Zm ×Z

and for each j = 1, . . . , n fj (t, x) =

 (p,q)∈Zm ×Z

where u ˆ(tJ  , p, q) and fˆj (tJ  , p, q) denote the coefficients of the partial Fourier series with respect to variables (t1 , . . . , tm , x). Replacing the formal series (3.3) and (3.4) in the equations Lj u = fj , j = 1, . . . , m, for each (p, q) = (0, 0) we obtain i(pj + qaj0 )ˆ u(tJ  , p, q) = fˆj (tJ  , p, q).

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On the other hand, from the compatibility condition Lf = 0 or equivalently Lj fk = Lk fj , j, k = 1, . . . , n we obtain the following equations (pj + qaj0 )fˆk (tJ  , p, q) = (pk + qak0 )fˆj (tJ  , p, q). Therefore, by the preceding equations we have u ˆ(tJ  , p, q) =

1 fˆN (tJ  , p, q), i(pN + qaN 0 )

(p, q) = (0, 0),

(3.5)

where N = N (q), 1 ≤ N ≤ m, is such that |pN + qaN 0 | = max |pj + qaj0 | = 0. 1≤j≤m

ˆ If (p, q) = (0, 0), since fˆ(tJ  , 0, 0) is exact, there exists v ∈ C ∞ (Tn−m tJ  ) such that dv = f (·, 0, 0). Thus, we choose u ˆ(tJ  , 0, 0) = v(tJ  ). we obtain from (3.2) and (3.5) the following estimative, Given α ∈ Zn−m + |∂ α u ˆ(tJ  , p, q)| ≤

1 d−1 α ˆ |q| |∂ fN (tJ  , p, q)|, C

for all tJ  and (p, q) ∈ Zm × Z. Since each fj is a smooth function we conclude that 

u(t, x) =

u ˆ(tJ  , p, q)ei(p·tJ +qx) ∈ C ∞ (Tn+1 ).

(p,q)∈Zm ×Z

By construction u is a solution of the equations Lj u = fj , 1 ≤ j ≤ m. Also, since the system is involutive it is not difficult to verify that u is a solution of the equations Lj u = fj , m + 1 ≤ j ≤ n. Finally, assume that condition (II ) holds. Consider q∗ the smallest positive integer such that q∗ a is integral. Thus, by hypothesis if qJ is the smallest positive integer such that qJ aJ is integral then qJ = q∗ . We define  .  DA (Tn+1 ) = u ∈ D (Tn+1 );

u(t, x) =



 u ˆ(t, ξ)eiξx ,

ξ∈A

where A = qJ Z.  (Tn+1 ) and LB the operator L acting on DB (Tn+1 ), where Let LA be the operator L acting on DA B = Z \ A. The operator L is globally solvable if and only if the operators LA and LB are globally solvable on Tn+1 (see [3]). Proposition 3.1. The operator LA is globally solvable. Proof. Since qJ a0 ∈ Zn , we define u=

 ξ∈A

u ˆ(t, ξ)eiξx → TA u =

 ξ∈A

u ˆ(t, ξ)eiξa0 eiξx .

A.P. Bergamasco et al. / J. Math. Anal. Appl. 444 (2016) 527–549

533

 ∞ Note that TA is an automorphism of DA (Tn+1 ) and of CA (Tn+1 ), and the following relation holds: −1 TA LA TA = L0,A ,

(3.6)

∂ . where L0 = dt + ib(t) ∧ ∂x Let B be a global primitive of b defined on Tn . Since B has only connected sublevels and superlevels sets in Tn , then by results in [11] we conclude that L0 is globally solvable, hence L0,A is globally solvable, which implies from (3.6) that LA is globally solvable. 2 If qJ = 1 then the proof is complete, otherwise: Proposition 3.2. The operator LB is globally solvable. Proof. Let (p, q) = (p1 , . . . , pm , q) ∈ Zm × B. Since q∗ is the smallest positive integer such that q∗ (a10 , . . . , am0 ) ∈ Zm , then there exists k ∈ {1, . . . , m} such that  C pk   , ak0 −  ≥ q |q| where C = 1/q∗ , therefore  C pj   pk   , max aj0 −  ≥ ak0 −  ≥ 1≤j≤m q q |q|

(p, q) ∈ Zm × B.

Note that, if the denominators q ∈ B then aJ behaves as non-Liouville (see inequality (3.2)). Thus, the rest of the proof is analogous to the case where the 1-form aJ is non-Liouville. 2 4. Necessity in Theorem 2.4 We are going to prove that if each one of the conditions (I)–(III ) fails then L is not globally solvable. We separate the proof in two parts. In the first part we suppose that the system (2.4) has no real vector fields (J = ∅) and in the second one we consider the remaining case (J = ∅). Part 1: J = ∅ Assume first that the 1-form a is integral and a global primitive B of b has a disconnected sublevel or superlevel in Tn . Then, by [11], the operator L0 is not globally solvable, hence by (3.1) the operator L is not globally solvable. / Zn . Since a0 = (a10 , . . . , an0 ) ∈ / Zn , reordering Assume now that the 1-form a is not integral, hence a0 ∈ n the coordinates in T we may assume that a10 , . . . , am0 ∈ / Z and a(m+1)0 , . . ., an0 ∈ Z where 1 ≤ m ≤ n. From now on, we will denote by (t , t , x) the coordinates on Tn+1 where t = (t1 , . . . , tm ) and t = (tm+1 , . . . , tn ). Also, we will use the notation a0 = (a0 , a0 ) where a0 = (a10 , . . . , am0 ) and a0 = (a(m+1)0 , . . . , an0 ). If m = n we have only t = t = (t1 , . . . , tn ). Let B be a global primitive of the real 1-form b. We may consider the primitive B as a 2π-periodic smooth function defined on Rn , that is, we will consider B as a global primitive of the pull-back Π∗ b via the universal covering Π : Rn → Tn . Also, by hypothesis, the maximum of B over the cube [0, 2π]n is not attained at the boundary. Furthermore, we may assume that B(0) = 0, otherwise we consider B − B(0).

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In order to construct an f ∈ E such that the equation Lu = f has no solution u ∈ D (Tn+1 ) we need to n define an f = j=1 fj (t, x)dtj satisfying the compatibility condition Lf = 0 or equivalently Lj fk = Lk fj ,

j, k = 1, . . . , n.

Thus, if there exists a solution u of the equations Lj u = fj we have Lj Lk u = Lk Lj u,

j, k = 1, . . . , n.

In particular, this condition implies the following equality Lm Lm−1 . . . L1 u = Lσ(m) Lσ(m−1) . . . Lσ(1) u, σ ∈ Sm ,

(4.1)

where Sm is the group of all permutations of the natural numbers 1, . . . , m. The idea is to use (4.1) to find the x-Fourier coefficients of the functions f1 , . . . , fm . Also, if m < n we will choose the functions fm+1 , . . . , fn identically zero. Consider now the set F = {a10 , . . . , am0 } ∩ Q. If F = ∅ we write aj0 = rj /sj ∈ F where rj ∈ Z and sj ∈ N are coprime, and we define the set G=



{sj Z; aj0 = rj /sj ∈ F }.

(4.2)

Since a10 , . . . , am0 ∈ / Z, we have that sj > 1 and that both Z+ \ G and Z− \ G have infinitely many elements. Motivated by (4.1), we substitute the formal series u(t, x) =



u ˆ(t, ξ)eiξx and h(t, x) =

ξ∈Z



ˆ ξ)eiξx h(t,

ξ∈Z

in the equation . Lm Lm−1 . . . L1 u = h, where the function h will be chosen later on. We obtain ˆ ξ) = h(t,



∂ 

∂ + iξcm (t) · · · + iξc1 (t) u ˆ(t, ξ), ∂tm ∂t1

ξ ∈ Z,

where cj (t) = aj0 + ibj (t). For each ξ ∈ Z+ \ G the preceding equation has exactly one solution given by  u ˆ(t, ξ) = dξ





eiξ(C(t −s ,t



h(t − s , t , ξ)ds ,

)−C(t)) ˆ

(4.3)

[0,2π]m

n where dξ = d1ξ d2ξ . . . dmξ with djξ = (1 − e−i2πξaj0 )−1 and C(t) = A(t) + iB(t) with A(t) = j=1 aj0 tj . If m = 1 we will choose f1 = h. If m > 1 we consider v(t, ξ) as in the right-hand side of (4.3), and for each ξ ∈ Z+ \ G we define . fˆj (t, ξ) =

∂  + iξcj (t) v(t, ξ), ∂tj

j = 1, . . . , m.

A.P. Bergamasco et al. / J. Math. Anal. Appl. 444 (2016) 527–549

535

Hence 

dξ fˆj (t, ξ) = djξ





eiξ(C(t −sj ,t



h(t − sj , t , ξ)dsj ,

)−C(t)) ˆ

(4.4)

[0,2π]m−1

where sj = (s1 , . . . , sj−1 , 0, sj+1 , . . . , sm ) and dsj = ds1 . . . dsj−1 dsj+1 . . . dsn . We define . fj (t, x) =



fˆj (t, ξ)eiξx , j = 1, . . . , m.

ξ∈Z+ \G

In Proposition 4.2 we prove that f1 , . . . , fm are smooth functions for a convenient choice of the function h. n It will then follow from the definition of the functions fj and the choice of h that f (t, x) = j=1 fj (t, x)dtj ∈ E. Also, if there existed u such that Lu = f then the x-Fourier coefficients of u would have to verify u ˆ(t, ξ) = v(t, ξ). Construction of the function h: Given 0 < δ < π/2, consider a function χδ ∈ Cc∞ (Rm ) such that  χδ (t ) =

1 if t ∈ (−δ, δ)m 0 if t ∈ / (−2δ, 2δ)m

and 0 ≤ χδ (t ) ≤ 1, for all t ∈ Rm .  the maximum of B over the We will denote by M the maximum of B over the cube [0, 2π]n and by M n   boundary of [0, 2π] . Let t∗ = (t∗ , t∗ ) be an interior point of the cube [0, 2π]n such that B(t∗ ) = M . We define the coefficients ˆ ξ) = h(t,

      dξ −1 eξ[B(t)−M −Kq(t )+λ+i(p(t )+a0 ·(t −t∗ ))] 0

if ξ ∈ Z+ \ G otherwise

(4.5)

)/2 and K > 0 are constants. The constant K will be chosen later on. Furthermore, where 0 < λ < (M − M p and q are 2π-periodic smooth functions as follows p(t ) =



a0 · (t∗ − t + 2πN )χδ (t + 2πN ),

t ∈ Rm ,

N ∈Zm

and q(t ) = 1 +



(|t + 2πN |2 − 1)χδ (t + 2πN ),

t ∈ Rm .

N ∈Zm

Note that ˆ ξ) is a 2π-periodic function; • Since a0 ∈ Zn−m then for each ξ the coefficient h(t,       • If |t | < δ then p(t ) = a0 · (t∗ − t ) and q(t ) = |t |2 ; • q(t ) ≥ 0 for all t ∈ Rm and q(t ) = 0 if and only if t = 2πN for some N ∈ Zm .

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A.P. Bergamasco et al. / J. Math. Anal. Appl. 444 (2016) 527–549

Finally, we define 

h(t, x) =

ˆ ξ)eiξx . h(t,

ξ∈Z+ \G

Proposition 4.1. h ∈ C ∞ (Tn+1 ). Proof. The function ϕ(t) = B(t) − M − Kq(t ) + λ satisfies the inequality  − M )/2, ϕ(t) ≤ (M

t ∈ [0, 2π]n .

Indeed, consider the set Q = {t ∈ [0, 2π]m ; q(t ) = 0}. Thus, by definition of the function q we have that t ∈ Q if and only if t = 2πN for some N ∈ Zm . Note that, for each t = (t , t ) ∈ Q × [0, 2π]n−m  − M + λ. ϕ(t) = B(2πN, t ) − M + λ = B(0, t ) − M + λ ≤ M  − M )/2 for all t = (t , t ) ∈ Q × [0, 2π]n−m . Moreover, there is a neighborhood U Therefore ϕ(t) < (M  − M )/2, for all t ∈ U . of the compact set Q × [0, 2π]n−m such that ϕ(t) ≤ (M  Let μ be the minimum of the function t → q(t ) over [0, 2π]n \ U . Since μ > 0, we choose the constant K > 0 sufficiently large such that  − M )/2, ϕ(t) ≤ B(t) − M − Kμ + λ ≤ (M

t ∈ [0, 2π]n \ U,

which ensures our above statement. m Since |d−1 ξ | ≤ 2 , we have    ˆ h(t, ξ) ≤ 2m e−ξ(M −M )/2 ,

t ∈ [0, 2π]n ,

ξ ∈ Z+ \ G.

ˆ ξ) where Pξ (t) is a polynomial involving ˆ ξ) = Pξ (t)h(t, Given α = (α1 , . . . , αn ) ∈ Zn+ we have ∂ α h(t, . only powers of ξ (powers less than |α| = α1 + · · · + αn ) and derivatives of p, q and B, which are bounded on [0, 2π]n . Thus, there exists a positive constant Cα such that |Pξ (t)| ≤ Cα ξ |α| for all ξ ∈ Z+ \ G and t ∈ [0, 2π]n . Therefore, 

ˆ ξ)| ≤ 2m Cα ξ |α| e−ξ(M −M )/2 , |∂ α h(t, whence we conclude that h ∈ C ∞ (Tn+1 ).

t ∈ [0, 2π]n ,

ξ ∈ Z+ \ G,

2

Replacing the coefficients (4.5) into the equations (4.3) and (4.4), for each ξ ∈ Z+ \ G, we have 





eξϕ1 (t,s ) eiξϕ2 (t,s ) ds ,

u ˆ(t, ξ) =

(4.6)

[0,2π]m

and if m > 1 then for each j = 1, . . . , m we have fˆj (t, ξ) = d−1 jξ







eξϕ1 (t,sj ) eiξϕ2 (t,sj ) dsj ,

[0,2π]m−1

where ϕ1 and ϕ2 are real-valued smooth functions given by

(4.7)

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. ϕ1 (t, s ) = B(t) − M − Kq(t − s ) + λ and . ϕ2 (t, s ) = p(t − s ) − a0 · s − a0 · (t − t∗ ). Proposition 4.2. f1 , . . . , fm ∈ C ∞ (Tn+1 ). Proof. If m = 1 Proposition 4.2 reduces to Proposition 4.1. Suppose that m > 1 and for each fixed j = 1, . . . , m, we identify the points sj = (s1 , . . . , sj−1 , 0, sj+1 , . . . , sm ) with (s1 , . . . , sj−1 , sj+1 , . . . , sn ) and consider the set Qj = {(t , sj ) ∈ [0, 2π]m × [0, 2π]m−1 ; q(t − sj ) = 0}. Since q(t − sj ) = 0 if and only if t − sj = 2πN for some N ∈ Zn , for each (t , sj ) ∈ Qj and t ∈ [0, 2π]n−m we obtain ϕ1 (t, sj ) = B(sj + 2πN, t ) − M + λ = B(s1 , . . . , 0, . . . , sm , t ) − M + λ  − M + λ < (M  − M )/2. ≤M Now, we consider a neighborhood U of the compact set Qj × [0, 2π]n−m such that t  − M )/2, ϕ1 (t, sj ) ≤ (M

(t, sj ) ∈ U.

Let μ be the minimum of the function (t, sj ) → q(t − sj ) over the set [0, 2π]n \ U , then μ > 0. Therefore, if necessary, we take a larger K > 0 to obtain  − M )/2, ϕ1 (t, sj ) ≤ −Kμ + λ ≤ (M for all (t, sj ) ∈ [0, 2π]n \ U .  − M )/2 for all (t, s ) ∈ [0, 2π]n × [0, 2π]m−1 , then there exists a positive constant Since ϕ1 (t, sj ) ≤ (M j C that does not depend on t and ξ, such that from (4.7) we obtain 

|fˆj (t, ξ)| ≤ Ce−ξ(M −M )/2 ,

t ∈ [0, 2π]n ,

ξ ∈ Z+ \ G.

Given α ∈ Zn+ , it follows from (4.7) that 

∂ fˆj (t, ξ) = d−1 jξ α



Pξ (t, sj )eξϕ(t,sj ) dsj ,

[0,2π]m−1

where Pξ (t, s) is a polynomial involving only powers of ξ less than |α|, and derivatives of ϕ(t, sj ) = ϕ1 (t, sj ) + iϕ2 (t, sj ), which are bounded on [0, 2π]n × [0, 2π]m−1 . Thus, there exists a positive constant Cα such that |Pξ (t, sj )| ≤ Cα ξ |α| which implies the inequality |∂ α fˆj (t, ξ)| ≤ |d−1 jξ |





|Pξ (t, sj )||eξϕ(t,sj ) |dsj

[0,2π]m−1 

≤ CCα ξ |α| e−ξ(M −M )/2 , for all t ∈ [0, 2π]n and ξ ∈ Z+ \ G. We conclude that fj ∈ C ∞ (Tn+1 ).

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n Let f = j=1 fj (t, x)dtj be the 1-form where f1 , . . . , fm are the smooth functions defined previously and fm+1 , . . . , fn are identically zero. By construction Lj fk = Lk fj for all j, k = 1, . . . , m and Lj fk = Lk fj = 0 for all j, k = m + 1, . . . , n. Also, from definition of the function h and by (4.7) it follows that Lk fj = 0 if j = 1, . . . , m and k = m + 1, . . . , n. Thus, Lf = 0. On the other hand, from definition of the coefficients fˆj (t, ξ), if ξa0 is integral for some ξ ∈ Zn (or equivalently ξ(a10 , . . . , an0 ) ∈ Zn ) then ξ = 0 or ξ ∈ G when G = ∅. In this case fˆ(t, ξ) = 0 and the second compatibility condition holds. Therefore f ∈ E and by construction if there existed u ∈ D (Tn+1 ) such that Lu = f then the x-Fourier coefficients of u would be given by (4.6). Proposition 4.3. There is no distribution u ∈ D (Tn+1 ) whose sequence of x-Fourier coefficients is the sequence {ˆ u(t, ξ)} given by (4.6). Proof. Let t∗ = (t∗ , t∗ ) be the interior point of the cube [0, 2π]n , which was considered in the construction of the function h, where B(t∗ ) = M . Consider the equation u ˆ(t∗ , ξ) = Iξ + Jξ , where 

























eiξ(p(t∗ −s )−a0 ·s ) e−ξ(Kq(t∗ −s )−λ) ds ,

Iξ = |t∗ −s |<δ



eiξ(p(t∗ −s )−a0 ·s ) e−ξ(Kq(t∗ −s )−λ) ds .

Jξ = |t∗ −s |≥δ

If |t∗ − s | < δ then p(t∗ − s ) = a0 · s and q(t∗ − s ) = |t∗ − s |2 . Hence, if we make the change of variables √ σ = Kξ(t∗ − s ) we obtain  Iξ =

e−ξ(K|t

|t∗ −s |<δ

eξλ = (Kξ)m/2 = eξλ  where γξ =

∗

−s|2 −λ)



ds

e−|σ| dσ 2

√ |σ|<δ Kξ

γξ γ1 ≥ eξλ , (Kξ)m/2 (Kξ)m/2

e−|σ| dσ, ξ ∈ Z+ \ G, is an increasing sequence of real numbers. 2

√ |σ|<δ Kξ

On the other hand, if μ is the minimum of the function t → q(t ) over the cube Q∗ = t∗ − [0, 2π]m , then μ > 0. Now, if necessary, we take a larger K > 0 to obtain  |Jξ | ≤

e

−ξ(Kμ−λ)





ds ≤

|t∗ −s |≥δ

where C > 0 is a constant that does not depend on ξ.

|t∗ −s |≥δ

e−ξλ ds = Ce−ξλ ,

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 m/2 for all ξ ≥ ξ0 , where C  = (K m/2 /γ1 )(1 + C) is constant. Consider now ξ0 ∈ Z+ such that eξλ/2 ≥ Cξ Therefore 3 γ1 eξλ/2 ≥ 1 + C ≥ 1 + Ce− 2 λξ . (ξK)m/2

Multiplying the above inequality by eξλ/2 we obtain |ˆ u(t∗ , ξ)| ≥ |Iξ | − |Jξ | ≥

γ1 eξλ − Ce−ξλ ≥ eξλ/2 , (Kξ)m/2

for all ξ ∈ Z+ \ G such that ξ ≥ ξ0 . Therefore, the sequence {ˆ u(t, ξ)} given by (4.6) does not correspond to any sequence of x-Fourier coefficients of a distribution u ∈ D (Tn+1 ). 2 Part 2: J = ∅ Recall that when J = {1, . . . , n} (b = 0) the problem was solved in [9]. Thus, in order to prove the non-solvability in part 2 we consider J = {1, . . . , m} where 1 ≤ m ≤ n − 1 and J  = {m + 1, . . . , n}. We will denote by (tJ , tJ  , x) the coordinates on Tn+1 where tJ = (t1 , . . . , tm ) and tJ  = (tm+1 , . . . , tn ). We will prove the non-solvability of L in two cases. Namely, aJ rational and aJ Liouville. Case 1: aJ is rational Assume first that the 1-form aJ  is not rational. Since dt b = 0 or equivalently ∂ ∂ bj = bk , ∂tk ∂tj

j, k = 1, . . . , n,

and bj = 0 for all 1 ≤ j ≤ m, it follows that bk = bk (tJ  ) for each k = m + 1, . . . , n, that is, each non-identically zero function bk depends only on the variables tm+1 , . . . , tn . Let B be a global primitive of the 1-form b such that Mj < M for each m + 1 ≤ j ≤ n, where M is the maximum of B over [0, 2π]n and Mj is the maximum of B over [0, 2π]n ∩ {t; tj = 0}. Since aJ  is not rational, then by part 1 the involutive system LJ  generated by the vector fields Lj =

∂ ∂ + (aj0 + ibj (tJ  )) , ∂tj ∂x

j ∈ J  = {m + 1, . . . , n},

is not globally solvable on TptJ  × T1x , where p = n − m. Let A = qJ Z where qJ is the smallest positive integer such that qJ aJ is integral. Since qJ aJ  is not rational, as in part 1, it is possible to conclude that the involutive system LJ  ,A is not solvable on Tp+1 ,  where LJ  ,A is the system LJ  acting on DA (Tp+1 ). Therefore, the system LA is not globally solvable on n+1 T by the following proposition. Proposition 4.4. If the 1-form aJ is rational and the involutive system LJ  ,A is not globally solvable on Tp+1 then LA is not globally solvable on Tn+1 . Proof. Consider the space of compatibility conditions EJ  ,A associated to LJ  ,A . Since LJ  ,A is not globally solvable on Tp+1 then there exists g(tJ  , x) =



gk (tJ  , x)dtk ∈ EJ  ,A

k∈J   such that LJ  ,A v = g, has no solution v in DA (Tp+1 ).

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Now, we will define smooth functions f1 , . . . , fn on the torus Tn+1 such that f =  and LA u = f has no solution u ∈ DA (Tn+1 ). For each j = 1, . . . , m we choose fj ≡ 0 and for each k = m + 1, . . . , n we define fk (t, x) =



n j=1

fj (t, x)dtj ∈ EA

fˆk (t, ξ)eiξx

ξ∈A

where fˆk (t, ξ) = gˆk (tJ  , ξ)e−iξψ(tJ ) ,

ξ ∈ A,

m and ψ(tJ ) = j=1 aj0 tj . Note that, for each k ∈ J  and each ξ ∈ A the function fˆk (t, ξ) is defined on Tn since ξaj0 ∈ Z, j ∈ J. n It is easy to check that each fj is a smooth function and f = j=1 fj (t, x)dtj ∈ EA .  (Tn+1 ) such that LA u = f . Then, for each j ∈ J we have Suppose that there exists u ∈ DA ∂ u ˆ(t, ξ) + iξaj0 u ˆ(t, ξ) = 0, ∂tj

ξ ∈ A,

or equivalently  ∂

u ˆ(t, ξ)eiξψ(tJ ) = 0, ∂tj

ξ ∈ A.

Therefore, for each ξ ∈ A u ˆ(t, ξ)eiξψ(tJ ) = vξ (tJ  ).

(4.8)

.   Since u is a distribution then v(tJ  , ξ) = ξ∈A vξ (tJ  )eiξx ∈ DA (Tp+1 ). On the other hand, if k ∈ J  we have ∂ u ˆ(t, ξ) + iξ(ak0 + ibk (t))ˆ u(t, ξ) = fˆk (t, ξ), ∂tk

ξ ∈ A.

Replacing (4.8) in the above equality it follows that ∂ vξ (tJ  ) + iξ(ak0 + ibk (t))vξ (tJ  ) = gˆk (t, ξ), ∂tk

ξ ∈ A.

Thus, for each k ∈ J  we have Lk v = gk which implies LJ  ,A v = g, that is a contradiction. 2 Assume now the 1-form aJ  is rational and qJ < q∗ . Thus, the 1-form a is rational, but qJ a is not integral. Consider the operators LA and LB where A = qJ Z and B = Z \ A. By Proposition 3.2 the operator LB is globally solvable on Tn+1 . However, LA is not globally solvable. Indeed, since qJ a is not integral, then qJ aJ  is not integral. Therefore, as in part 1, the operator LJ  ,A is not globally solvable on Tp+1 . By Proposition 4.4 we conclude the non-solvability of LA which implies the non-solvability of L. Finally assume a rational, qJ = q∗ and if B is a primitive of b then B has a disconnected sublevel or superlevel on Tn . Therefore, by [11] the operator L0,A is not globally solvable, hence from (3.6) we have LA not globally solvable.

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Case 2: aJ is Liouville We will prove this case in two subcases, namely, when aJ  is rational and when aJ  is not rational. Case 2.1: aJ is Liouville and aJ  is rational Let LJ be the operator associate with the vector fields Lj , j ∈ J. Consider the space of compatibility conditions EJ associated to LJ and a positive integer η such that ηaJ  is integral. We claim that LJ,A is not globally solvable on Tm+1 , where A = ηZ. Indeed, since aJ is Liouville then (a10 , . . . , am0 ) ∈ / Qm and there are a constant C > 0 and a sequence {(pl , ql )} in Zm × Z, ql ≥ 2, such that (j)  p   max aj0 − l  ≤ C(ql )−l , ∀l ∈ N, j∈J ql (j)

where for each l ∈ N it is possible to choose pl , ql ∈ A. Indeed, if η = 1 then the conclusion is obvious. Suppose that η > 1, since we may consider ql < ql+1 , therefore η ≤ qη ≤ qηl ≤ (qηl )η−1 , l ∈ N. Consider now the subsequence {(pηl , qηl )}, then (j)  ηpηl   −(η−1)l max aj0 − )(ηqηl )−l ≤ C(ηqηl )−l ,  ≤ C(qηl )−ηl = C (η l qηl j∈J ηqηl    ≤1

for all l ∈ N. Now, it suffices to choose ( pl , ql ) = (ηpηl , ηqηl ), l ∈ N which ensures our above statement. By [9] and by the previous explanation we conclude that LJ,A is not globally solvable on Tm+1 . In order to prove that L is not globally solvable we will prove that LA is not globally solvable on Tn . Let EJ,A be the space of compatibility conditions associate to LJ,A . Since LJ,A is not globally solvable  on Tm+1 , there exists g(tJ , x) = j∈J gj (tJ , x)dtj ∈ EJ such that LJ,A v = g  has no solution v ∈ DA (Tm+1 ). n The natural idea is to define smooth functions f1 , . . . , fn on Tn+1 such that f = j=1 fj (t, x)dtj ∈ EA  and LA u = f has no solution u ∈ DA (Tn+1 ). Let B be a global primitive of the 1-form b. Since ∂t∂j B = bj and for each j ∈ J the function bj ≡ 0, then B depends only on the variables tJ  = (tm+1 , . . . , tn ), that is, B = B(tJ  ). Furthermore, consider the  function ψ(tJ  ) = k∈J  ak0 tk . For each k ∈ J  we choose fk ≡ 0 and for j ∈ J we define

.  ˆ fj (t, x) = fj (t, ξ)eiξx , ξ∈A

where  . fˆj (t, ξ) =

gˆj (tJ , ξ)eξ(B(t)−M −iψ(tJ  ))

if ξ ∈ A ∩ Z+

gˆj (tJ , ξ)eξ(B(t)−μ−iψ(tJ  ))

if ξ ∈ A ∩ Z−

M and μ are, respectively, the maximum and minimum of B over Tn .

,

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Note that, fˆj (t, ξ) is a function of t ∈ Tn and |fˆj (t, ξ)| ≤ |ˆ gj (tJ , ξ)| for all (t, ξ) ∈ Tn × A. Similarly, α α ˆ |∂ fj (t, ξ)| ≤ Cα |ξ ||∂ gˆj (tJ , ξ)|. Therefore, since gj is smooth then so is fj . Also, it is easy to verify that n f = j=1 fj (t, x)dtj ∈ EA .   Suppose now that there exists u ∈ DA (Tn+1 ) such that LA u = f . Then, if u(t, x) = ξ∈A u ˆ(t, ξ)eiξx we have for each ξ ∈ A the following equations α

∂ u ˆ(t, ξ) + iξaj0 u ˆ(t, ξ) = fˆj (t, ξ), ∂tj

j ∈ J,

(4.9)

and ∂ u ˆ(t, ξ) + iξ(ak0 + ibk (t))ˆ u(t, ξ) = 0, ∂tk

k ∈ J .

(4.10)

Thus, for each k ∈ J  we may write (4.10) as follows ∂  u ˆ(t, ξ)e−ξ(B(t)−M −iψ(tJ  )) = 0, ∂tk ∂  u ˆ(t, ξ)e−ξ(B(t)−μ−iψ(tJ  )) = 0, ∂tk

if ξ ∈ A ∩ Z+ , if ξ ∈ A ∩ Z− .

Therefore, . u ˆ(t, ξ)e−ξ(B(t)−M −iψ(tJ  )) = ϕξ (tJ ), . u ˆ(t, ξ)e−ξ(B(t)−μ−iψ(tJ  )) = ϕξ (tJ ),

ξ ∈ A ∩ Z+ ,

(4.11)

ξ ∈ A ∩ Z− .

∗ Let t∗ and t∗ in Tm u(tJ , t∗ , ξ)| if ξ ∈ A ∩ Z+ J  such that B(t ) = M and B(t∗ ) = μ. Thus, |ϕξ (tJ )| = |ˆ −  n+1 and |ϕξ (tJ )| = |ˆ u(tJ , t∗ , ξ)| if ξ ∈ A ∩ Z for all tJ . Since u ∈ DA (T ) then

.   v(tJ , x) = ϕξ (tJ )eiξx ∈ DA (Tm+1 ).

(4.12)

ξ∈A

On the other hand, replacing (4.11) in (4.9), for each j ∈ J and each ξ ∈ A we have ∂ ϕξ (tJ ) + iξaj0 ϕξ (tJ ) = gˆj (tJ , ξ), ∂tj which implies that v given by (4.12) is a solution of LJ v = g, that is a contradiction. Case 2.2: aJ is Liouville and aJ  is not rational Since the 1-form aJ is Liouville and the 1-form aJ  is not rational, there exist j ∈ J and k ∈ J  such that the periods aj0 and ak0 are not rational. Reordering the variables t1 , . . . , tn we may assume that for some nonnegative integers , r, 1 ≤  < r ≤ n, we have {1, . . . , } ⊂ J, { + 1, . . . , r} ⊂ J  , with a10 , . . . , ar0 not rational, while a(r+1)0 , . . . , an0 are rational. Note that {r + 1, . . . , n} ⊂ J ∪ J  . It is easy to see that (a10 , . . . , a 0 ) is still Liouville. In the news coordinate, the original system can be rewritten as follows ⎧ ∂ ∂ ⎪ + aj0 , Lj = ⎪ ⎪ ⎪ ∂t ∂x j ⎪ ⎨ ∂ ∂ + (aj0 + ibj (t)) , Lj = ∂tj ∂x ⎪ ⎪ ⎪ ⎪ ∂ ∂ ⎪ ⎩Lj = + (aj0 + ibj (t)) , ∂tj ∂x

j = 1, . . . , ,

aj0 ∈ / Q,

j =  + 1, . . . , r,

aj0 ∈ / Q,

bj = 0,

j = r + 1, . . . , n,

aj0 ∈ Q (if r < n).

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If r = n we have just the first two blocks of vector fields and J = {1, . . . , }. If r < n then the third block may contain real vector fields. Let B be a global primitive of pull-back Π∗ b, where Π : Rn → Tn is the universal covering. In the new coordinates, by hypothesis, the maximum of B over the cube [0, 2π]n is not attained over the faces [0, 2π]n ∩ {t; tj = 0}, j =  + 1, . . . , r. Also, since ∂t∂j B = bj = 0 for all j = 1, . . . ,  then B = B(t +1 . . . , tn ). We will denote by t = (t , t , t ) the coordinates in Tn , where t = (t1 , . . . , t ), t = (t +1 , . . . , tr ) and    t = (tr+1 , . . . , tn ). Also, we use the following notation a0 = (a0 , a0 , a 0 ), where a0 = (a10 , . . . , a 0 ), a0 =    (a( +1)0 , . . . , ar0 ) and a0 = (a(r+1)0 , . . . , an0 ). If r = n we have only t = (t1 , . . . , t ) and t = (t +1 , . . . , tn ). Since (a10 , . . . , a 0 ) ∈ (R \ Q) is Liouville then it follows that there are a constant C > 0 and a sequence {(pl , ql )}, ql ≥ 2, in Z × Z such that  (j)  max pl + ql aj0  ≤

1≤j≤

C , |(pl , ql )|l

l ∈ N.

(4.13)

. (j) As in case 2.1 we may consider pl , ql ∈ A = ηZ where η is an integer number such that ηaj0 ∈ Z for each j = r + 1, . . . , n. Analogously to (4.1), we obtain the equality Lr Lr−1 . . . L1 u = Lσ(r) Lσ(r−1) . . . Lσ(1) u, σ ∈ Sr , where Sr is the group of all permutations of the natural numbers 1, . . . , r. As in part 1, we choose fr+1 = · · · = fn = 0 (if r < n) and consider the equation h = Lr Lr−1 . . . L1 u, to define smooth functions f1 , . . . , fr such that f = u ∈ D (Tn+1 ). The function h will be chosen later on. Substituting the formal series 

u(t, x) =

n j=1

(4.14)

fj (t, x)dtj ∈ E and Lu = f has no solution



u ˆ(t , t , ξ, κ)ei(κ·t +ξx) ,

(ξ,κ)∈Z×Z

and 

h(t, x) =

ˆ  , t , ξ, κ)ei(κ·t +ξx) , h(t

(ξ,κ)∈Z×Z

in the equation (4.14), for each (κ, ξ) ∈ Z × Z, we obtain ˆ  , t , κ, ξ) = h(t

r     ∂  + iξcj (t) i(κj + ξaj0 ) u ˆ(t , t , κ, ξ). ∂tj j=1

(4.15)

j= +1

Therefore, for each l ∈ N the equation (4.15) with (κ, ξ) = (pl , ql ) has exactly one solution given by 





u ˆ(t , t , pl , ql ) = dl [0,2π]r−

where

eiql (C(t



−s ,t )−C(t)) ˆ

h(t − s , t , pl , ql )ds ,

(4.16)

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544

dl =



(j)

i(pl

r 

+ ql aj0 )−1

j=1

(1 − e−i2πql aj0 )−1 ,

j= +1

s = (s +1 , . . . , sr ), ds = ds +1 . . . , dsr , and C(t) = A(t) + iB(t) with A(t) = a0 · t. Let v(t , t , pl , ql ) denote the right-hand side of (4.16). We define fj (t, x) =

∞ 

 fˆj (t , t , pl , ql )ei(pl ·t +ql x) ,

l=1

where ⎧ (j)   ⎪ ⎨i(pl + ql aj0 )v(t , t , pl , ql ),  fˆj (t , t , pl , ql ) = ∂ ⎪ ⎩ + iql cj (t) v(t , t , pl , ql ), ∂tj

if j = 1, . . . ,  if j =  + 1, . . . , r.

Therefore, if j = 1, . . . ,  we obtain fˆj (t , t , pl , ql ) = dl ejl

 eiql (C(t



−s ,t )−C(t)) ˆ

h(t − s , t , pl , ql )ds ,

(4.17)

[0,2π]r− (j)

where ejl = i(pl + ql aj0 ). If j =  + 1, . . . , r and r >  + 1 dl fˆj (t , t , pl , ql ) = djl

 eiql (C(t



 −s j ,t )−C(t)) ˆ

h(t − sj , t , pl , ql )dsj ,

(4.18)

[0,2π]r−−1

where djl = (1 − e−iql 2πaj0 ), sj = (s +1 , . . . , 0, . . . , sr ), and dsj = ds +1 . . . dsj−1 dsj+1 . . . dsr . If r =  + 1 we have dl ˆ   fˆr (t , t , pl , ql ) = h(t , t , pl , ql ). drl

(4.19)

We are going to prove that for a convenient choice of the function h the functions f1 , . . . , fr are smooth r and f = j=1 fj (t, x)dtj ∈ E. Also, from the definition of the functions fj , if there is a solution u to the equation Lu = f , then the (t , x)-Fourier coefficients of u must satisfy u ˆ(t , t , pl , ql ) = v(t , t , pl , ql ). Construction of the function h: Let {(pl , ql )} be the sequence given by (4.13). Consider the sets 

N1 = {l ∈ N|

eql (M −M )/2 pl + ql a0  ≤ 1},

N2 = {l ∈ N|

eql (M −M )/2 pl + ql a0  > 1},

and 

where  (j)  pl + ql a0  = max pl + ql aj0 . 1≤j≤

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545

We define the sequence {El } as follows  El =

if l ∈ N1

eλql , pl +

ql a0 −1/2

if l ∈ N2 ,

)/2 is a constant. where 0 < λ < (M − M Let 0 < δ < π/2. Consider a function χδ ∈ Cc∞ (Rr− ) such that 

χδ (t ) =

 1 0

if t ∈ (−δ, δ)r− if t ∈ / (−2δ, 2δ)r− ,

and 0 ≤ χδ (t ) ≤ 1 for all t . We define p(t ) =



a0 · (t∗ − t − 2πN )χδ (t + 2πN ),

N ∈Zr−

q(t ) = 1 +



(|t + 2πN |2 − 1)χδ (t + 2πN ),

N ∈Zr− n− where (t∗ , t such that B(t∗ , t ∗ ) is a point of the cube [0, 2π] ∗ ) = M . Therefore p and q are 2π-periodic smooth functions satisfying the following conditions:

• q(t ) ≥ 0, for all t ∈ Rr− ; q(t ) = 0 if and only if t = 2πN , N ∈ Zr− . • If |t | < δ then p(t ) = a0 · (t∗ − t ) and q(t ) = |t |2 . Finally, we define

h(t, x) =

∞ 

ˆ  , t , pl , ql )ei(pl ·t +ql x) , h(t

l=1

where   ˆ  , t , pl , ql ) = El eql [B(t)−M −Kq(t )+i(p(t )+a 0 ·(t −t∗ ))] . h(t dl

(4.20)

The constant K > 0 will be chosen later on. n−r ˆ  , t , pl , ql ) is a 2π-periodic function. Observe that, since ql a , for each l the coefficient h(t 0 ∈Z Proposition 4.5. h ∈ C ∞ (Tn+1 ). Proof. We claim that the function φ(t , t ) = B(t) − M − Kq(t ) satisfies the following estimative  − M )/2, |φ(t , t )| ≤ (M

(t , t ) ∈ [0, 2π]n− .

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Indeed, let Q = {t ∈ [0, 2π]r− ; q(t ) = 0}. Thus t ∈ Q if and only if t = 2πN , N ∈ Zr− . Note that Q is the set of the vertices of the cube [0, 2π]r− . Also,  − M )/2, |φ(t , t )| < (M

(t , t ) ∈ Q × [0, 2π]n−r .

 − M )/2 in a neighborhood U of the compact set Q × Since φ is continuous we have φ(t , t ) ≤ (M n−r . [0, 2π] If μ is the minimum of (t , t ) → q(t ) over the compact set [0, 2π]n− \ U , then μ > 0.  − M )/2 on the set [0, 2π]n− \ U , which ensure Now we take a larger K > 0 to obtain φ(t , t ) ≤ (M our statement. Furthermore, r− |d−1 pl + ql a0  , l |≤2

∀l ∈ N,

hence, for each l ∈ N, we obtain from (4.20) the following estimate 

ˆ  , t , pl , ql )| ≤ 2r− El pl + ql a  e−ql (M −M )/2 |h(t 0 ⎧ )/2−λ] r−  ⎨2 pl + ql a0  e−ql [(M −M = )/2 ⎩2r− p + q a (2 −1)/2 e−ql (M −M l l 0

if l ∈ N1 if l ∈ N2 .

ˆ  , t , ·, ·). Hence h ∈ C ∞ (Tn+1 ). Similar estimates may be obtained for the derivatives of h(t

2

As consequence of Proposition 4.5 and (4.19), if r =  + 1 then the function fr ∈ C ∞ (Tn+1 ). In this case, the functions fr+1 , . . . , fn were chosen to be identically zero. Replacing the coefficients (4.20) in the equations (4.16) – (4.18) we obtain, for each l ∈ N, the following 





eql ϕ1 (t

u ˆ(t , t , pl , ql ) = El



,t ,s ) iql ϕ2 (t ,t ,s )

e

ds ,

(4.21)

[0,2π]r−

and for j = 1, . . . ,  fˆj (t , t , pl , ql ) = ejl El

 eql ϕ1 (t



,t ,s ) iql ϕ2 (t ,t ,s )

e

ds ,

(4.22)

[0,2π]r−

and if r >  + 1 for j =  + 1, . . . , r El fˆj (t , t , pl , ql ) = djl

 eql ϕ1 (t



   ,t ,s j ) iξl ϕ2 (t ,t ,sj )

e

[0,2π]r−−1

where ϕ1 (t , t , s ) = B(t , t ) − M − Kq(t − s ) and   ϕ2 (t , t , s ) = p(t − s ) − a0 · s − a 0 · (t − t∗ )

are real valued smooth functions.

dsj ,

(4.23)

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547

Proposition 4.6. f1 , . . . , f ∈ C ∞ (Tn+1 ). Proof. Note that ϕ1 (t , t , s ) ≤ 0, for all (t , t , s ) ∈ [0, 2π]n− × [0, 2π]r− . Thus, for each fixed j = 1, . . . ,  we obtain from (4.22) the following estimative |fˆj (t , t , pl , ql )| ≤ (2π)r− El pl + ql a0   )/2−λ] −ql [(M −M r− e ≤ (2π) pl + ql a0 1/2

if l ∈ N1 if l ∈ N2 .

As in Proposition 4.5, we conclude that f1 , . . . , f ∈ C ∞ (Tn+1 ). 2 Proposition 4.7. f +1 , . . . , fr ∈ C ∞ (Tn+1 ). Proof. First we will show that  − M )/2, ϕ1 (t , t , sj ) ≤ (M for all (t , t , sj ) ∈ [0, 2π]n− × [0, 2π]r− −1 . Indeed, for each j = 1, . . . ,  we define the set   Qj = (t , t , sj ) ∈ [0, 2π]n− × [0, 2π]r− −1 | q(t − sj ) = 0 . Therefore, (t , t , sj ) ∈ Qj if and only if t − sj = 2πN for some N ∈ Zr− and  − M, ϕ1 (t , t , sj ) = B(sj + 2πN, t ) − M = B(sj , t ) − M ≤ M for all (t , t , sj ) ∈ Qj . Thus, ϕ1 (t , t , sj ) <

−M M , 2

(t , t , sj ) ∈ Qj .

Also, there exists a neighborhood U of the compact set Qj such that ϕ1 (t , t , sj ) ≤

−M M , 2

(t , t , sj ) ∈ U.

If μ is the minimum of the function (t , t , sj ) → q(t − sj ) over the compact set ([0, 2π]n− × [0, 2π]r− −1 ) \ U , then μ > 0. Finally, if necessary, we take a larger K > 0 such that ϕ1 (t , t , sj ) ≤ −Kμ ≤

−M M , 2

for all (t , t , sj ) ∈ ([0, 2π]n− × [0, 2π]r− −1 ) \ U , which ensures our statement. It follows from (4.23) that there exists a constant C > 0 that does not depends on (t , t , pl , ql ) such that   Ce−ql [(M −M )/2−λ] if l ∈ N1  |fˆj (t , t , pl , ql )| ≤ CEl e−ql (M −M )/2 ≤ Cpl + ql a0 1/2 if l ∈ N2 . As in Proposition 4.5, we conclude that f +1 , . . . , fr ∈ C ∞ (Tn+1 ).

2

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Proposition 4.8. There is no distribution u ∈ D (Tn+1 ) whose sequence of (t , x)-Fourier coefficients is the sequence {ˆ u(t , t , pl , ql )} given by (4.21).   Proof. Let (t∗ , t ∗ ) be the point considered in the construction of the function h satisfying B(t∗ , t∗ ) = M . Consider the following equation

u ˆ(t∗ , t ∗ , pl , ql ) = El (Il + Jl ), where 





eql ϕ1 (t∗ ,t∗

Il =

  ,s ) iql ϕ2 (t ∗ ,t∗ ,s )

e

ds ,

 |t ∗ −s |<δ







eql ϕ1 (t∗ ,t∗

Jl =

  ,s ) iql ϕ2 (t ∗ ,t∗ ,s )

e

ds .

 |t ∗ −s |≥δ

If |t∗ − s | < δ then p(t∗ − s ) = a0 · s and q(t∗ − s ) = |t∗ − s |2 , it follows that    2 Il = e−ql K|t∗ −s | ds .  |t ∗ −s |<δ

√

Kql (t∗ − s ) we obtain  2 1 Il = e−|σ| dσ (r− )/2 (Kql ) √

Now, if we make the change of variables σ =

|σ|<δ Kql

2 .  where γl = |σ|<δ√Kql e−|σ| dσ is an increasing sequence of positive real numbers. On the other hand, consider the cube Q∗ = t∗ −[0, 2π]r− . If μ is the minimum of the function τ  → q(τ  ) over the set Q∗ ∩ {τ  ; |τ  | ≥ δ} then μ > 0. Thus, if necessary, we take a larger K > 0 such that

    r− ϕ1 (t∗ , t , ∗ , s ) = Kq(t∗ − s ) ≤ −Kμ < −λ, s ∈ [0, 2π]

hence, there exists a constant C > 0 such that     |Jl | ≤ eql ϕ1 (t∗ ,t∗ ,s ) ds ≤ Ce−ql λ .  |t ∗ −s |≥δ

Now, we take a natural number l0 such that

eq l λ γ 1 ≥ C for all ql > l0 . Therefore, (Kql )(r− )/2

|ˆ u(t∗ , t ∗ , pl , ql )| ≥ El (|Il | − |Jl |)

 γl −ql λ ≥ El − Ce (Kql )(r− )/2

 γ1 γl ≥ El − (Kql )(r− )/2 (Kql )(r− )/2 =

(γl − γ1 ) El , (Kql )(r− )/2

∀ql > l0 .

Since the sequence {El } increases faster than any polynomial, it follows that the sequence {ˆ u(t , t , pl , ql )}  n+1 ). 2 does not correspond to any u in D (T

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549

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