Existence and Regularity of Periodic Solutions to Certain First-Order Partial Differential Equations

Abstract

We present conditions on the coefficients of a class of vector fields on the torus which yield a characterization of global solvability as well as global hypoellipticity, in other words, the existence and regularity of periodic solutions. Diophantine conditions and connectedness of certain sublevel sets appear in a natural way in our results.

1 Introduction and Statement of the Results

For a positive integer m, let $\mathbb {T}^{m}\simeq \mathbb {R}^{m}/2\pi \mathbb {Z}^{m}$ be the m-dimensional torus. We will work on $\mathbb {T}^{N+1}$, where the coordinates are denoted by $(x,t)\in \mathbb {T}^{N}\times \mathbb {T}^1,$ with $x=(x_1,\ldots ,x_N)\in \mathbb {T}^N$.

We are interested in the existence of solutions in $\mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ to first-order partial differential equations given by $Lu=f,$ where $f\in \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ and L is a vector field on $\mathbb {T}^{N+1}$ of the type

$$\begin{aligned} L=\frac{\partial }{\partial t}+\sum _{j=1}^{N}\left( a_j+ib_j\right) (t)\frac{\partial }{\partial x_j}, \end{aligned}$$

(1.1)

where the coefficients $a_j$ and $b_j$ are real-valued smooth functions defined on $\mathbb {T}^1$, for $j=1,\ldots ,N$.

We recall that the existence of local solutions to $Lu=f$ is well-understood and is in fact characterized by the well-known Nirenberg-Treves condition ($\mathcal P$) (see [12, 14]).

The problem is still interesting if we consider the existence of solutions on the whole torus, a property which will be refered to as global solvability.

We say that L is globally solvable if the range of the operator $L:\mathcal {C}^{\infty }(\mathbb {T}^{N+1})\rightarrow \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ is closed. Standard arguments of Functional Analysis imply that $\overline{L\mathcal {C}^{\infty }(\mathbb {T}^{N+1})}=(\ker {}^tL)^\circ ,$ where $^tL$ denotes the transpose operator and $(\ker {}^tL)^{\circ }$ is the set of functions $\phi \in \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ such that $\mu (\phi )=0,$ for all $\mu \in \ker {}^tL\subset \mathcal {D}'(\mathbb {T}^{N+1}).$ It follows that L is globally solvable if and only if $L\mathcal {C}^{\infty }(\mathbb {T}^{N+1})=(\ker {}^tL)^{\circ }.$ We will use this characterization throughout this article.

It is well known that condition ($\mathcal P$) is not necessary for global solvability (see [11], and also [5]). On the other hand, it follows from [10] that the equation $Lu=f$ has a solution $u\in \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ for every $f\in \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ satisfying a finite number of compatibility conditions, provided L satisfies both condition ($\mathcal P$) and the following geometric condition (in which $\ell$ is the principal symbol of L and q is a complex-valued function):

(GC) every characteristic point of L lies on a compact interval of a bicharacteristic of $\mathfrak {R}(q\ell )$, on which $q\ne 0$, with no characteristic endpoint.

However, as shown in [5], our class of operators (in general) does not satisfy conditions ($\mathcal P$) and (GC) simultaneously. Hence, the global solvability of our class of operators cannot be obtained by directly applying results from [10].

When our operator L does not satisfy condition ($\mathcal {P}$), a necessary condition for solvability is given by the connectedness of certain sublevel sets [see Theorem 1.1-(II)]. This relation between solvability and connectedness of sublevel sets is due to Treves and it has appeared for instance in [13].

The approach and motivations for this paper are related to those in references such as [1–5, 8, 11, 13].

Our results extend those in [5], where global solvability and global hypoellipticity were characterized in the case where $N=2$ and L had the special form

$$\begin{aligned} \frac{\partial }{\partial t}+ib_1(t)\frac{\partial }{\partial x_1}+ib_2(t)\frac{\partial }{\partial x_2}. \end{aligned}$$

We introduce the following useful notations:

$$\begin{aligned} a_{j0}=\frac{1}{2\pi }\int _{0}^{2\pi }a_{j}(t)dt, \quad \text {for} \; j=1,\ldots ,N, \end{aligned}$$

and

$$\begin{aligned} b_{j0}=\frac{1}{2\pi }\int _{0}^{2\pi }b_{j}(t)dt, \quad \text {for} \; j=1,\ldots ,N. \end{aligned}$$

Comparing with [5], the presence of the coefficient $a_j$ makes the problem more difficult. For instance, it will be important to use some smooth cutoff functions in order to apply the Laplace method for integrals. Also, we must take into account the influence of the numbers $a_{j0}$ on the Diophantine conditions. As we know, for instance, from [1, 2, 5, 8], global solvability and global hypoellipticity are related to Diophantine conditions, which, in some cases, are linked with the notion of non-Liouville vector (or number). We say that $(\gamma _1,\gamma _2)\in \mathbb {R}^2\setminus \mathbb {Q}^2$ is a Liouville vector if there exist $C>0$ and a sequence $(p_n,q_n,j_n)\in \mathbb {Z}^2\times \mathbb {N}$ so that $j_n\ge 2,$ and

$$\begin{aligned} \left| \gamma _1-\frac{p_n}{j_n}\right| +\left| \gamma _2-\frac{q_n}{j_n}\right| <\frac{C}{(j_n)^{n}},\quad \text {for all}\,\,\,n\in \mathbb {N}. \end{aligned}$$

Motivated by [5], for a pair of vectors $(\alpha ,\beta )\in \mathbb {R}^N\times \mathbb {R}^N$ we define the following Diophantine conditions:

(DC1) there exist constants $C>0$ and $\gamma >0$ such that

$$\begin{aligned} |\tau +\langle \xi ,\alpha +i\beta \rangle |\ge C(|\xi |+|\tau |)^{-\gamma }, \end{aligned}$$

for all $(\xi ,\tau )=(\xi _1,\ldots ,\xi _N,\tau )\in \mathbb {Z}^{N+1}\setminus \{0\}.$

(DC2) there exist constants $C>0$ and $\gamma >0$ such that

$$\begin{aligned} |\tau +\langle \xi ,\alpha +i\beta \rangle |\ge C(|\xi |+|\tau |)^{-\gamma }, \end{aligned}$$

for all $(\xi ,\tau )\in \mathbb {Z}^{N+1}$ such that $\tau +\langle \xi ,\alpha +i\beta \rangle \ne 0.$

For instance, if $\lambda$ is a Liouville number then the pair $(\lambda ,0), (\lambda , 1)\in \mathbb {R}^2$ does not satisfy (DC2).

Note that (DC1) implies (DC2),  and for a pair $(\alpha ,\beta )$ such that $\tau +\langle \xi ,\alpha +i\beta \rangle \ne 0$ for all $(\xi ,\tau )\in \mathbb {Z}^{N+1}\setminus \{0\},$ (DC1) and (DC2) are equivalent.

The Diophantine condition (DC1) (respectively, (DC2)) is related to the notion of nonresonant (respectively, resonant) condition that appears in [7].

When $b_1\not \equiv 0$ and $b_1$ does not change sign, we will see (Example 3.5) that the operator

$$\begin{aligned} \frac{\partial }{\partial t}+(a_1(t)+i\lambda b_1(t))\frac{\partial }{\partial x_1}+ib_1(t)\frac{\partial }{\partial x_2}, \quad \lambda \in \mathbb {R}, \end{aligned}$$

(1.2)

is globally solvable if and only if $(a_{10},0,\lambda b_{10},b_{10})\in \mathbb {R}^2\times \mathbb {R}^2$ satisfies (DC2), which holds if and only if $(a_{10},\lambda )$ is not a Liouville vector. Comparing with [5], where $a_1$ vanishes identically, the operator (1.2) is globally solvable if and only if $\lambda$ is not a Liouville number. Recall that, in dimension 2, there is no influence of the real part of the coefficient on the global solvability for the operator

$$\begin{aligned} \frac{\partial }{\partial t}+\left( a_1(t)+i\lambda b_1(t)\right) \frac{\partial }{\partial x_1}\quad (\text {with}\; \lambda b_{10}\ne 0), \end{aligned}$$

which is globally solvable if and only if $b_1$ does not change sign (see [11, Theorem 3.2]).

Finally, we are ready to state our main results.

Define

$$\begin{aligned} \alpha _{0}=\left( a_{10},\ldots ,a_{N0}\right) , \quad \beta _{0}=\left( b_{10},\ldots ,b_{N0}\right) , \end{aligned}$$

$$\begin{aligned} \alpha (t)=(a_1(t),\ldots ,a_{N}(t)) \; \text {and} \; \beta (t)=(b_1(t),\ldots ,b_{N}(t)), \quad t\in \mathbb {T}^1. \end{aligned}$$

Theorem 1.1

Let L be given by (1.1).

• (I) If $b_j\equiv 0$ for every j,  then L is globally solvable if and only if the pair $(\alpha _0,0)$ satisfies (DC2).

• (II) If $b_{j0}=0,$ for every j,  and if $b_j\not \equiv 0$ for at least one j,  then L is globally solvable if and only if $(a_{10},\ldots ,a_{N0})\in \mathbb {Z}^N$ and all the sublevel sets

  $$\begin{aligned} \Omega _r^{\xi }=\left\{ t\in \mathbb {T}^1;\int _{0}^{t}\langle \xi ,\beta (\tau )\rangle d\tau <r\right\} , \quad r\in \mathbb {R} \; \text {and} \,\,\xi \in \mathbb {Z}^N, \end{aligned}$$

  are connected.

• (III) If $b_{j0}\ne 0$ for some j,  then L is globally solvable if and only if the following properties are satisfied:

  • (III.1) $\dim \text {span} \{b_1,\ldots ,b_N\}=1;$

  • (III.2) the functions $b_j$ do not change sign;

  • (III.3) the pair $(\alpha _{0},\beta _{0})$ satisfies (DC2).

Remark 1.2

In (III.1) the span is taken on $\mathbb {R};$ hence, condition (III.1) means that the functions $b_j$ are real multiples of a smooth real-valued function b defined on $\mathbb {T}^1.$

We now point out an important remark about Theorem 1.1-case (II). Consider

$$\begin{aligned} \frac{\partial }{\partial t}+i\sum _{j=1}^{N}b_j(t)\frac{\partial }{\partial x_j}, \end{aligned}$$

(1.3)

where $b_{j0}=0,$ for all j,  and at least one $b_j$ does not vanish identically. Of course, (1.3) does not satisfy condition ($\mathcal {P}$); however, a necessary condition for (1.3) to be globally solvable is that each function $\mathbb {T}^1\ni t\rightarrow \xi _1b_1(t)+\cdots +\xi _Nb_N(t)$ ($\xi \in \mathbb {Z}^N$) changes sign at most twice, which is weaker than condition ($\mathcal P$).

Another question of our interest is the regularity of solutions $\mu \in \mathcal {D}'(\mathbb {T}^{N+1})$ to $L\mu =f,$ where L is given by (1.1) and f belongs to $\mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$ This regularity will be described by the expression global hypoellipticity. We say that L is globally hypoelliptic if $\mu \in \mathcal {D}'(\mathbb {T}^{N+1})$ and $L\mu \in \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ imply that $\mu \in \mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$

As in Theorem 1.1, Diophantine properties of the coefficients of L are linked to hypoellipticity.

Theorem 1.3

Let L be given by (1.1). Then, L is globally hypoelliptic if and only if the functions $a_j$ and $b_j$ satisfy the following properties:

• (1) the functions $b_j$ do not change sign;

• (2) $\dim \text {span} \{b_1,\ldots ,b_N\}\le 1$;

• (3) the pair $(\alpha _{0},\beta _{0})$ satisfies (DC1).

Note that, in contrast to Theorem 1.1, condition ($\mathcal {P}$) is a necessary condition for global hypoellipticity.

As a consequence of Theorem 1.3, the operator L given by (1.2) is globally hypoelliptic if and only if $(a_{10},\lambda )\in \mathbb {R}^2\setminus \mathbb {Q}^2$ is not a Liouville vector.

When the coefficients of L are constant, Theorems 1.1 and 1.3 give us results which are in agreement with those of Greenfield and Wallach in [8]. In fact, when

$$\begin{aligned} L=\frac{\partial }{\partial t}+\sum _{j=1}^{N}\left( \alpha _j+i\beta _j\right) \frac{\partial }{\partial x_j}, \end{aligned}$$

where $\alpha _j$ and $\beta _j$ are real numbers, for $j=1,\ldots ,N,$ then it follows from simple adaptations of the techniques used in [8] that L is globally solvable if and only if the pair $(\alpha ,\beta )$ satisfies (DC2),  where $\alpha =(\alpha _1,\ldots ,\alpha _j)$ and $\beta =(\beta _1,\ldots ,\beta _N).$ Likewise, this constant coefficient vector field is globally hypoelliptic if and only if the pair $(\alpha ,\beta )$ satisfies (DC1).

We dedicate Sects. 2 and 3 to the proof of Theorem 1.1, and Sect. 4 to the proof of Theorem 1.3.

2 Beginning of the Proof of Theorem 1.1: Cases (I) and (II)

First, recall that

$$\begin{aligned} L=\frac{\partial }{\partial t}+\sum _{j=1}^{N}\left( a_j+ib_j\right) (t)\frac{\partial }{\partial x_j} \end{aligned}$$

is globally solvable if and only if $L\mathcal {C}^{\infty }(\mathbb {T}^{N+1})=(\ker {}^tL)^\circ ,$ where $^tL$ denotes the transpose operator of L,  and $(\ker {}^tL)^\circ$ is the set of the functions $\phi \in \mathcal {C}^\infty (\mathbb {T}^{N+1})$ such that $\mu (\phi )=0$ for all $\mu \in \ker {}^tL.$

As a first step in our proofs, we claim that there is no loss of generality in assuming that the real part of our operator L has constant coefficients. Indeed, by using partial Fourier series in the variables $(x_1,\ldots ,x_N),$ we may define the automorphism $T:\mathcal {C}^{\infty }(\mathbb {T}^{N+1})\rightarrow \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ given by

$$\begin{aligned} \widehat{Tu}(\xi ,t)=\hat{u}(\xi ,t)\exp \left\{ i\int _{0}^{t}\langle \xi ,\alpha (\tau )-\alpha _0\rangle d\tau \right\} , \quad \text {for all} \quad \xi \in \mathbb {Z}^N, \end{aligned}$$

(2.1)

with inverse $T^{-1}:\mathcal {C}^{\infty }(\mathbb {T}^{N+1})\rightarrow \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ given by

$$\begin{aligned} \widehat{T^{-1}v}(\xi ,t)=\hat{v}(\xi ,t)\exp \left\{ -i\int _{0}^{t}\langle \xi ,\alpha (\tau )-\alpha _0\rangle d\tau \right\} , \quad \text {for all} \quad \xi \in \mathbb {Z}^N. \end{aligned}$$

(2.2)

The operator L is globally solvable if and only if the new operator $TLT^{-1}$ is globally solvable; the latter has the form

$$\begin{aligned} TLT^{-1}=\frac{\partial }{\partial t}+\sum _{j=1}^{N}\left( a_{j0}+ib_j(t)\right) \frac{\partial }{\partial x_j}, \end{aligned}$$

(2.3)

where now the real part of the coefficients is constant; moreover, it is easy to see that $a_{j0}$ can be replaced by zero if $a_{j0}$ belongs to $\mathbb {Z}.$

As a consequence, when $b_j\equiv 0$ for all j,  we may assume that L is the constant coefficient operator

$$\begin{aligned}\frac{\partial }{\partial t}+\sum _{j=1}^{N}a_{j0}\frac{\partial }{\partial x_j};\end{aligned}$$

hence, as in [8], we can prove that L is globally solvable if and only if $(\alpha _0,0)\in \mathbb {R}^N\times \mathbb {R}^N$ satisfies (DC2), where $\alpha _0=(a_{10},\ldots ,a_{N0}).$ This concludes the proof of Theorem 1.1 in case (I).

For the remainder of this section, we will proceed with the proof of Theorem 1.1 by analyzing the situation (II), where $b_{j0}=0,$ for all $j=1,\ldots ,N,$ and at least one function $b_j$ does not vanish identically. Propositions 2.1 and 2.2 show the necessity of the conditions in (II), while Proposition 2.5 shows the sufficiency.

Proposition 2.1

Let L be given by the right-hand side of (2.3). Suppose that $b_{j0}=0$, for all j,  and that at least one $b_j$ does not vanish identically. If $(a_{10},\ldots ,a_{N0})\not \in \mathbb {Z}^N,$ then L is not globally solvable.

Proof

The hypotheses imply the existence of m and $\ell$ in $\{1,\ldots ,N\}$ such that $b_m$ does not vanish identically, $b_{m0}=0$ and $a_{\ell 0}\not \in \mathbb {Z}.$ Assume first that $m=\ell .$ Without loss of generality we may assume that $m=\ell =1.$ In this case, it follows from Theorem 3.2 of [11] that the vector field

$$\begin{aligned} L_1=\frac{\partial }{\partial t}+\left( a_{10}+ib_1(t)\right) \frac{\partial }{\partial x_1}, \end{aligned}$$

viewed as a vector field on $\mathbb {T}^2,$ is not globally solvable. By using partial Fourier series in the variables $(x_2,\ldots ,x_N),$ we can verify that the fact that $L_1$ is not globally solvable easily implies that L itself is not globally solvable.

Now, assume that $m\ne \ell .$ Again, there is no loss of generality in assuming that $m=1$ and $\ell =2;$ hence $b_1$ does not vanish identically, $b_{10}=0,$ $a_{10}\in \mathbb {Z},$ $b_2$ vanishes identically and $a_{20}\not \in \mathbb {Z}.$ As above, in order to prove that L is not globally solvable it is enough to prove that

$$\begin{aligned} \frac{\partial }{\partial t}+\left( a_{10}+ib_1(t)\right) \frac{\partial }{\partial x_1}+a_{20}\frac{\partial }{\partial x_2} \end{aligned}$$

is not globally solvable. Recall that, since $a_{10}\in \mathbb {Z},$ it is enough to prove that

$$\begin{aligned} L_{12}\doteq \frac{\partial }{\partial t}+ib_1(t)\frac{\partial }{\partial x_1}+a_{20}\frac{\partial }{\partial x_2} \end{aligned}$$

is not globally solvable.

We will construct a function f in $(\ker {}^tL_{12})^{\circ }\setminus L_{12} \mathcal {C}^{\infty }(\mathbb {T}^3)$ by using partial Fourier series in the variables $(x_1,x_2).$ Note that, since $a_{20}\not \in \mathbb {Z},$ there is an increasing sequence $(k_n)_{n\in \mathbb {N}}\subset \mathbb {Z}$ such that $k_n\ge n>0$ and $a_{20}k_n\not \in \mathbb {Z},$ for all $n\in \mathbb {N}.$ We will construct the function f of the form

$$\begin{aligned} f(x_1,x_2,t)=\sum _{n=1}^{\infty }\hat{f}\left( k_n,k_n,t\right) e^{ik_n(x_1+x_2)}. \end{aligned}$$

(2.4)

The next step is to construct the sequence $\hat{f}(k_n,k_n,\cdot )$ (the Fourier coefficients of f).

Define

$$\begin{aligned} H(s,t)=\int _{t-s}^{t}b_1(\tau )d\tau ,\quad 0\le s,t\le 2\pi , \end{aligned}$$

and set

$$\begin{aligned} A=H(s_0,t_0)=\max _{0\le s,t\le 2\pi }H(s,t). \end{aligned}$$

Since $b_1$ changes sign we have $A>0;$ also, we may assume that $0<s_0,\,t_0,\,t_0-s_0<2\pi .$ Set $\sigma _0=t_0-s_0$ and take a function $\phi \in \mathcal {C}^{\infty }_c((\sigma _0-\delta ,\sigma _0+\delta ))$ which satisfies $0\le \phi (t)\le 1$ and $\phi (t)\equiv 1$ in a neighborhood of $[\sigma _0-\delta /2,\sigma _0+\delta /2],$ where $\delta >0$ is small enough so that $(\sigma _0-\delta ,\sigma _0+\delta )\subset (0,t_0).$

We finally define $\hat{f}(k_n,k_n,\cdot )\in \mathcal {C}^{\infty }(\mathbb {T}^1)$ to be the $2\pi$- periodic extension of

$$\begin{aligned} \phi (t)e^{ik_na_{20}(t_0-t)}e^{-k_nA} , \quad t\in [0,2\pi ], \end{aligned}$$

for all $n\in \mathbb {N}.$

Since $A>0,$ the sequence $(\hat{f}(k_n,k_n,\cdot ))_{n\in \mathbb {N}},$ which consists of smooth functions on $\mathbb {T}^1,$ decays rapidly; hence, the function f given by (2.4) belongs to $\mathcal {C}^{\infty }(\mathbb {T}^3).$ Moreover, our assumption $k_na_{20}\not \in \mathbb {Z}$ implies that $\hat{\mu }(-k_n,-k_n,\cdot )\equiv 0,$ for all $\mu \in \ker {}^tL_{12}$ and $n\in \mathbb {N}.$ Hence $f\in (\ker {}^tL_{12})^{\circ }.$

Next, it will be shown that there is no $u\in \mathcal {C}^{\infty }(\mathbb {T}^3)$ such that $L_{12}u=f.$ Indeed, suppose that $u\in \mathcal {C}^{\infty }(\mathbb {T}^3)$ is such that $L_{12}u=f.$ Then the partial Fourier series in the variables $(x_1,x_2)$ gives

$$\begin{aligned} \frac{d}{dt}\hat{u}(j,k,t)+\left( -jb_1(t)+kia_{20}\right) \hat{u}(j,k,t)=\hat{f}(j,k,t), \end{aligned}$$

for all $(j,k)\in \mathbb {Z}^2$ and $t\in \mathbb {T}^1.$ In particular, for each $n\in \mathbb {N}$ we have

$$\begin{aligned} \frac{d}{dt}\hat{u}\left( k_n,k_n,t\right) +k_n\left( -b_1(t)+ia_{20}\right) \hat{u}\left( k_n,k_n,t\right) =\hat{f}\left( k_n,k_n,t\right) , \end{aligned}$$

which has a unique solution, since $k_na_{20}\not \in \mathbb {Z}.$ This solution can be written as

$$\begin{aligned} \hat{u}\left( k_n,k_n,t\right) =\left( 1-e^{-2\pi ik_na_{20}}\right) ^{-1}\int _{0}^{2\pi }\hat{f}\left( k_n,k_n,t-s\right) e^{-isk_na_{20}}e^{k_n\int _{t-s}^{t}b_1(\tau )d\tau }ds.\quad \quad \quad \quad \end{aligned}$$

For $t=t_0,$ the definition of $\hat{f}(k_n,k_n,\cdot )$ implies that

$$\begin{aligned} |\hat{u}(k_n,k_n,t_0)|= & {} |1-e^{-2\pi ik_na_{20}}|^{-1}\int _{|s-s_0|<\delta }\phi (t_0-s)e^{-k_n(A-H(s,t_0))}ds\\\ge & {} \frac{1}{2}\int _{|s-s_0|<\delta /2}e^{-k_n(A-H(s,t_0))}ds. \end{aligned}$$

Since $A-H(s_0,t_0)=0,$ $s_0$ is a point of minimum of $A-H(s,t_0)\ge 0.$ In particular, $s_0$ is either a zero of infinite order or a zero of even order. By proceeding as in the proof of Proposition 3.3 in [5], we may apply the Laplace method for integrals (see [6]) so that we obtain a constant $C>0,$ which does not depend on n,  such that

$$\begin{aligned} |\hat{u}(k_n,k_n,t_0)|\ge C(k_n)^{-1/2}, \end{aligned}$$

contradicting the fact that $\hat{u}(k_n,k_n,t_0)$ decays rapidly.

Therefore, there is no $u\in \mathcal {C}^{\infty }(\mathbb {T}^3)$ such that $L_{12}u=f.$

This completes the proof of Proposition 2.1. $\square$

Proposition 2.2

Under the assumptions of Proposition 2.1, but now supposing that $(a_{10},\ldots ,a_{N0})\in \mathbb {Z}^N,$ the existence of a disconnected sublevel set $\Omega _r^{\xi }$ implies that L is not globally solvable.

Proof

Recall that, since $a_{j0}\in \mathbb {Z},$ for $j=1,\ldots ,N,$ we may assume that

$$\begin{aligned} L=\frac{\partial }{\partial t}+i\sum _{j=1}^{N}b_j(t)\frac{\partial }{\partial x_j}. \end{aligned}$$

(2.5)

The proof will be by contradiction. Suppose that L is globally solvable. Due to Lemma 6.1.2 of [9], it is well known that the solvability of L implies an inequality involving the transpose operator. As in the proof of necessity in Proposition 2.1 of [5], a convenient version of this lemma yields the existence of constants $m\in \mathbb {N}$ and $C>0$ such that

$$\begin{aligned} \left| \int _{\mathbb {T}^{N+1}}fv\right| \le C\cdot \left( \sum _{|\alpha |\le m}\sup _{\mathbb {T}^{N+1}}|\partial ^{\alpha }f|\right) \cdot \left( \sum _{|\alpha |\le m}\sup _{\mathbb {T}^{N+1}}|\partial ^{\alpha }({}^tLv)|\right) , \end{aligned}$$

(2.6)

for every $f\in (\ker {}^tL)^{\circ }$ and every $v\in \mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$

The proof is then reduced to the study of inequality (2.6). By making use of a disconnected sublevel set $\Omega _r^{\xi },$ we will construct sequences of functions, $\{f_n\}$ and $\{v_n\}$, which will produce a contradiction with (2.6).

Take a disconnected sublevel

$$\begin{aligned} \Omega _r^{\xi '}=\left\{ t\in \mathbb {T}^1;\int _{0}^{t}\langle \xi ',\beta (\tau )\rangle d\tau <r\right\} ,\quad \text {where}\,\,\,\beta (\tau )=(b_1(\tau ),\ldots ,b_N(\tau )). \end{aligned}$$

As in [5, 11], we can find a real number $r_0<r$ such that

$$\begin{aligned} \Omega _{r_0}^{\xi '}=\left\{ t\in \mathbb {T}^1;\int _{0}^{t}\langle \xi ',\beta (\tau )\rangle d\tau <r_0\right\} \end{aligned}$$

has two connected components with disjoint closure. Also, we can construct functions $f_0,v_0\in \mathcal {C}^{\infty }(\mathbb {T}^1),$ satisfying the following conditions:

$$\begin{aligned}&\int _{0}^{2\pi }f_0(t)dt=0,\ supp(f_0)\cap \Omega _{r_0}^{\xi '}=\emptyset ,\\&supp(v_0')\subset \Omega _{r_0}^{\xi '}, \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{2\pi }f_0(t)v_0(t)dt>0. \end{aligned}$$

For $n\in \mathbb {N}$, define the functions $f_n,v_n:\mathbb {T}^{N+1}\rightarrow \mathbb {C}$ by

$$\begin{aligned} f_n(x,t)=\exp \left\{ -n\int _{0}^{t}\langle \xi ',\beta (\tau )\rangle d\tau \right\} f_0(t)e^{-in\langle \xi ',x\rangle }, \end{aligned}$$

and

$$\begin{aligned} v_n(x,t)=\exp \left\{ n\int _{0}^{t}\langle \xi ',\beta (\tau )\rangle d\tau \right\} v_0(t)e^{in\langle \xi ',x\rangle }. \end{aligned}$$

Since $b_{j0}=0,$ for all j,  it follows that $f_n$ and $v_n$ belong to $\mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$ Moreover, we claim that $f_n\in (\ker {}^tL)^{\circ }.$ In fact, for each $\mu \in \ker {}^tL\subset \mathcal {D}'(\mathbb {T}^{N+1}),$ by using partial Fourier series in the variables $(x_1,\ldots ,x_N)$ we can write

$$\begin{aligned}\mu =\sum _{\xi \in \mathbb {Z}^N}c_{\xi }\exp \left\{ \int _{0}^{t}\langle \xi ,\beta (\tau )\rangle d\tau \right\} e^{i\langle \xi ,x\rangle }, \,\,\text {with } c_{\xi }\in \mathbb {C},\end{aligned}$$

from which we obtain

$$\begin{aligned} \langle \mu ,f_n\rangle =(2\pi )^Nc_n\int _{0}^{2\pi }f_0(t)dt=0, \quad \text {for every} \quad n\in \mathbb {N}, \end{aligned}$$

where $c_n$ is a complex number. Thus $f_n\in (\ker {}^tL)^{\circ },$ for all $n\in \mathbb {N}.$

The inequality (2.6) for $f_n$ and $v_n$ yields

$$\begin{aligned} \left| \int _{\mathbb {T}^{N+1}}f_nv_n\right| \le C\cdot \left( \sum _{|\alpha |\le m}\sup _{\mathbb {T}^{N+1}}|\partial ^{\alpha }f_n|\right) \cdot \left( \sum _{|\alpha |\le m}\sup _{\mathbb {T}^{N+1}}|\partial ^{\alpha }({}^tLv_n)|\right) , \end{aligned}$$

(2.7)

for every $n\in \mathbb {N}.$

Notice that the left-hand side of the inequality (2.7) satisfies

$$\begin{aligned} \int _{\mathbb {T}^{N+1}}f_n(t)v_n(t)dt=(2\pi )^N\int _{0}^{2\pi }f_0(t)v_0(t)dt>0. \end{aligned}$$

On the other hand, since $(supp(f_0))\cap \Omega _{r_0}^{\xi '}=\emptyset ,$ there exists $C_1>0,$ which does not depend on n,  such that

$$\begin{aligned} \sum _{|\alpha |\le m}\sup _{\mathbb {T}^{N+1}}|\partial ^{\alpha }f_n|\le C_1n^{m}\sup _{t\in \mathbb {T}^1\setminus \Omega _{r_0}^{\xi '}}e^{-n\int _{0}^{t}\langle \xi ',\beta (\tau )\rangle d\tau }\le C_1n^{m}e^{-nr_0}; \end{aligned}$$

(2.8)

also, since $(supp(v_0'))\subset \Omega _{r_0}^{\xi '},$ there exists $C_2>0,$ which does not depend on n,  such that

$$\begin{aligned} \sum _{|\alpha |\le m}\sup _{\mathbb {T}^{N+1}}|\partial ^{\alpha }({}^tLv_n)|&=\sum _{|\alpha |\le m}\sup _{\mathbb {T}^{N+1}}|\partial ^{\alpha }[e^{n\int _{0}^{t}\langle \xi ',\beta (\tau )\rangle d\tau }v_0'(t)e^{in\langle \xi ',x\rangle }]|\nonumber \\&\le C_2n^{m}\sup _{t\in supp(v_0')}e^{n\int _{0}^{t}\langle \xi ',\beta (\tau )\rangle d\tau }\nonumber \\&\le C_2n^{m}\left( \sup _{t\in supp(v_0')}e^{\int _{0}^{t}\langle \xi ',\beta (\tau )\rangle d\tau }\right) ^{n}\nonumber \\&=C_2n^{m}e^{n\int _{0}^{t_1}\langle \xi ',\beta (\tau )\rangle d\tau }, \end{aligned}$$

(2.9)

where $t_1$ is a maximum point of the function $t\mapsto e^{\int _{0}^{t}\langle \xi ',\beta (\tau )\rangle d\tau }$ restricted to $supp(v_0')\subset \Omega _{r_0}^{\xi '}.$

We now set $\displaystyle c\doteq \int _{0}^{t_1}\langle \xi ',\beta (\tau )\rangle d\tau -r_0.$ Notice that $c<0$ and that c does not depend on $n\in \mathbb {N}$. Moreover, from (2.8) and (2.9) we have

$$\begin{aligned} \left( \sum _{|\alpha |\le m}\sup _{\mathbb {T}^{N+1}}|\partial ^{\alpha }f_n|\right) \cdot \left( \sum _{|\alpha |\le m}\sup _{\mathbb {T}^{N+1}}|\partial ^{\alpha }({}^tLv_n)|\right) \le C_1C_2n^{2m}e^{nc}, \end{aligned}$$

(2.10)

for all $n\in \mathbb {N}.$

Finally, from (2.7) and (2.10) we have

$$\begin{aligned} 0<(2\pi )^N\int _{0}^{2\pi }f_0(s)v_0(s)ds\le CC_1C_2n^{2m}e^{nc},\,\,\text{ for }\,\,\text{ all }\,\,n\in \mathbb {N}, \end{aligned}$$

which is a contradiction, since $\displaystyle \lim _{n\rightarrow \infty } n^{2m}e^{nc}=0.$

The proof of Proposition 2.2 is complete. $\square$

Example 2.3

The operator

$$\begin{aligned} \frac{\partial }{\partial t}+\frac{\partial }{\partial x_1} +i\cos (2t)\frac{\partial }{\partial x_2} \end{aligned}$$

defined on $\mathbb {T}^3$, is not globally solvable. Indeed, the sublevel set

$$\begin{aligned} \left\{ t\in \mathbb {T}^1;\int _{0}^{t}\cos (2\tau )d\tau <0\right\} \end{aligned}$$

is disconnected.

Example 2.4

The operator

$$\begin{aligned} \frac{\partial }{\partial t}+\left( \sin ^3t+i\sin ^3(2t)\right) \left( \frac{\partial }{\partial x_1}+\frac{\partial }{\partial x_2}\right) \end{aligned}$$

defined on $\mathbb {T}^3$, is not globally solvable. Indeed, the sublevel set

$$\begin{aligned} \left\{ t\in \mathbb {T}^1;\int _{0}^{t}\sin ^3(2\tau )d\tau <\frac{1}{2}\right\} \end{aligned}$$

is disconnected.

The next result shows that L is globally solvable under the conditions given in (II)-Theorem 1.1. In fact, the correct choice of the point $t_\xi$ in (2.14) will imply that we can construct a continuous right inverse for L,  namely, $L^{-1}:Im(L)\rightarrow \mathcal {C}^{\infty }(\mathbb {T}^{N+1}),$ given by means of formula (2.16).

Proposition 2.5

Let L be the operator given by (2.3). Suppose that $b_{j0}=0$, for all j,  and that at least one $b_j$ does not vanish identically. If $(a_{10},\ldots ,a_{N0})\in \mathbb {Z}^N$ and, moreover, all the sublevel sets $\Omega _r^{\xi }$ are connected, then L is globally solvable.

Proof

Given a function $f\in (\ker {}^tL)^{\circ },$ we will find $u\in \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ such that $Lu=f.$ As in the proof of Proposition 2.2, we may assume that L is given by (2.5). By using partial Fourier series in the variables $(x_1,\ldots ,x_N),$ we are led to solve

$$\begin{aligned} \frac{d}{dt}\hat{u}(\xi ,t)-\langle \xi ,\beta (t)\rangle \hat{u}(\xi ,t)=\hat{f}(\xi ,t), \quad t\in \mathbb {T}^1, \; \xi \in \mathbb {Z}^N, \end{aligned}$$

(2.11)

where we recall that $\beta (t)=(b_1(t),\ldots ,b_N(t)).$

Our assumption $b_{j0}=0$ (for all $j=1,\ldots ,N$) implies that, for each $\xi \in \mathbb {Z}^N,$ the function

$$\begin{aligned} \phi _\xi (x,t)=\exp \left\{ -\int _{0}^{t}\langle \xi ,\beta (\tau )\rangle d\tau \right\} e^{-i\langle \xi ,x\rangle } \end{aligned}$$

belongs to $\mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$ Moreover, ${}^tL(\phi _\xi )=-L(\phi _\xi )=0;$ that is, $\phi _\xi \in \ker {}^tL.$ Since $f\in (\ker {}^tL)^{\circ }$ we have

$$\begin{aligned} \int _{0}^{2\pi }e^{-\int _{0}^{t} \langle \xi ,\beta (\tau )\rangle d\tau }\hat{f}(\xi ,t)dt=0, \quad \text {for all} \quad \xi \in \mathbb {Z}^{N}. \end{aligned}$$

(2.12)

It follows from (2.12) that, for an arbitrary $t_\xi \in \mathbb {T}^1,$ the function $\hat{u}(\xi ,\cdot )$ defined by

$$\begin{aligned} \hat{u}(\xi ,t)=e^{\int _{0}^{t}\langle \xi ,\beta (\tau )\rangle d\tau }\int _{t_\xi }^{t}e^{-\int _{0}^{s}\langle \xi ,\beta (\tau )\rangle d\tau }\hat{f}(\xi ,s)ds, \end{aligned}$$

(2.13)

with $t_\xi \in \mathbb {T}^1$ to be chosen later, is well defined and belongs to $\mathcal {C}^{\infty }(\mathbb {T}^1).$ Simple calculations show that $\hat{u}(\xi ,\cdot )$ is a solution to (2.11) [in particular, there are infinitely many solutions to (2.11)].

In order to obtain a sequence of solutions $\hat{u}(\xi ,\cdot )$ which decays rapidly, we will choose a convenient point $t_\xi$ in (2.13). For each $\xi \in \mathbb {Z}^N,$ choose $t_{\xi }\in \mathbb {T}^1$ such that

$$\begin{aligned} \int _{0}^{t_{\xi }}\langle \xi ,\beta (\tau )\rangle d\tau =\sup _{t\in \mathbb {T}^1}\left\{ \int _{0}^{t}\langle \xi ,\beta (\tau )\rangle d\tau \right\} , \end{aligned}$$

or equivalently

$$\begin{aligned} -\int _{0}^{t_{\xi }}\langle \xi ,\beta (\tau )\rangle d\tau =\inf _{t\in \mathbb {T}^1}\left\{ -\int _{0}^{t}\langle \xi ,\beta (\tau )\rangle d\tau \right\} . \end{aligned}$$

(2.14)

For $t_\xi$ satisfying (2.14), we claim that the sequence $\hat{u}(\xi ,\cdot ),$ given by (2.13), decays rapidly. In fact, for fixed $\xi \in \mathbb {Z}^N$ and for fixed $t\in \mathbb {T}^1,$ define

$$\begin{aligned} r_t\doteq -\int _{0}^{t}\langle \xi ,\beta (\tau )\rangle d\tau \end{aligned}$$

and

$$\begin{aligned} \Omega _t\doteq \left\{ s\in \mathbb {T}^1;-\int _{0}^{s}\langle \xi ,\beta (\tau )\rangle d\tau \le r_t\right\} . \end{aligned}$$

It follows from our assumptions that $\Omega _t$ is connected. Since t and $t_\xi$ belong to $\Omega _{t},$ there is an arc $\Gamma _t$ entirely contained in $\Omega _t$ and joining $t_{\xi }$ to t. Hence

$$\begin{aligned} -\int _{0}^{s}\langle \xi ,\beta (\tau )\rangle d\tau \le -\int _{0}^{t}\langle \xi ,\beta (\tau )\rangle d\tau , \quad \text {for all} \quad s\in \Gamma _t. \end{aligned}$$

(2.15)

Again, by using (2.12) we can write

$$\begin{aligned} \hat{u}(\xi ,t)=\int _{\Gamma _t}\exp \left\{ \int _{0}^{t}\langle \xi ,\beta (\tau )\rangle d\tau -\int _{0}^{s}\langle \xi ,\beta (\tau )\rangle d\tau \right\} \hat{f}(\xi ,s)ds. \end{aligned}$$

(2.16)

Finally, the conjunction of (2.16) with the estimate (2.15) implies that

$$\begin{aligned} |\hat{u}(\xi ,t)|\le 2\pi \sup _{t\in \mathbb {T}^1}|\hat{f}(\xi ,t)|; \end{aligned}$$

(2.17)

hence, from the rapid decay of $\hat{f}(\xi ,\cdot ),$ for every $n\in \mathbb {Z}_{+},$ we obtain a constant $C=C(n)>0$ such that

$$\begin{aligned} |\hat{u}(\xi ,t)|\le \frac{C}{|\xi |^n}, \,\,\,\text {for all}\,\,\,t\in \mathbb {T}^1. \end{aligned}$$

By proceeding as above we can prove analogous estimates for all the derivatives $\hat{u}^{(m)}(\xi ,\cdot ),$ $m\in \mathbb {N}.$

Therefore, $\hat{u}(\xi ,\cdot )$ decays rapidly and

$$\begin{aligned} u(x,t)=\sum _{\xi \in \mathbb {Z}^N}\hat{u}(\xi ,t)e^{i\langle \xi ,x\rangle } \end{aligned}$$

is a solution to $Lu=f$ in $\mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$

This concludes the proof of Proposition 2.5. $\square$

Example 2.6

As mentioned in [5], the operator

$$\begin{aligned} \frac{\partial }{\partial t}+i\cos (t)\frac{\partial }{\partial x_1}+i\sin (t)\frac{\partial }{\partial x_2} \end{aligned}$$

is globally solvable. According to Propositions 2.1 and 2.5, by adding a real part in the coefficient we may either maintain or lose solvability; for instance,

$$\begin{aligned} \frac{\partial }{\partial t}+\left( \sin (t)+i\cos (t)\right) \frac{\partial }{\partial x_1}+\left( \cos (t)+i\sin (t)\right) \frac{\partial }{\partial x_2} \end{aligned}$$

is still globally solvable; on the other hand,

$$\begin{aligned} \frac{\partial }{\partial t}+\left( \sin ^2(t)+i\cos (t)\right) \frac{\partial }{\partial x_1}+i\sin (t)\frac{\partial }{\partial x_2} \end{aligned}$$

is not globally solvable, since $(1/2\pi )\int _{0}^{2\pi }\sin ^2(t)dt=1/2$ (is not an integer number).

3 End of the Proof of Theorem 1.1: Case (III)

We proceed with the proof of Theorem 1.1 and we now deal with case (III), in which $b_{j0}\ne 0$ for at least one $j\in \{1,\ldots ,N\}.$ Proposition 3.2 shows the necessity of conditions (III.1)–(III.3), while Proposition 3.4 shows the sufficiency. First, we will prove the following equivalence:

Lemma 3.1

For a pair of vectors $\alpha =(\alpha _1,\ldots ,\alpha _N)$ and $\beta =(\beta _1,\ldots ,\beta _N)$ in $\mathbb {R}^N,$ the condition (DC2) is equivalent to the following Diophantine condition:

(DC3) there exist constants $C>0$ and $\gamma >0$ such that

$$\begin{aligned} |1-e^{2\pi \langle \xi ,\beta -i\alpha \rangle }|\ge C|\xi |^{-\gamma }, \end{aligned}$$

for all $\xi \in \mathbb {Z}^N$ such that $\langle \xi ,\beta -i\alpha \rangle \not \in i\mathbb {Z}.$

Proof

Suppose that $(\alpha ,\beta )$ does not satisfy (DC3). Thus we can find a sequence $(\xi (n))_{n\in \mathbb {N}}\subset \mathbb {Z}^N$ such that $|\xi (n)|\ge n,$ $\langle \xi (n),\beta -i\alpha \rangle \not \in i\mathbb {Z},$ and

$$\begin{aligned} |1-e^{2\pi \langle \xi (n),\beta -i\alpha \rangle }|<|\xi (n)|^{-n}, \end{aligned}$$

(3.1)

for all $n\in \mathbb {N}.$ In particular, $|\langle \xi (n),\beta \rangle |\rightarrow 0,$ as $n\rightarrow \infty ,$ which implies that

$$\begin{aligned} |1-e^{2\pi \langle \xi (n),\beta \rangle }|\ge e^{-1}2\pi |\langle \xi (n),\beta \rangle |, \end{aligned}$$

(3.2)

for $n\in \mathbb {N}$ sufficiently large.

It follows from (3.1) that there is a sequence of integers $(\tau _n)$ such that $\tau _n+\langle \xi (n),\alpha \rangle \rightarrow 0,$ as $n\rightarrow \infty$ (in fact, it is enough, for each $n\in \mathbb {N},$ to choose $\tau _n$ to be either the greatest integer not exceeding $-\langle \xi (n),\alpha \rangle$ or the least integer greater than $-\langle \xi (n),\alpha \rangle$). Hence, if $n\in \mathbb {N}$ is sufficiently large, we have

$$\begin{aligned} |\sin (2\pi [\tau _n+\langle \xi (n),\alpha \rangle ])|\ge \pi |\tau _n+\langle \xi (n),\alpha \rangle |. \end{aligned}$$

(3.3)

Inequalities (3.1), (3.2) and (3.3) imply that, for $n\in \mathbb {N}$ sufficiently large,

$$\begin{aligned}e^{-1}2\pi |\langle \xi (n),\beta \rangle |\le |1-e^{2\pi \langle \xi (n),\beta \rangle }|\le |1-e^{2\pi \langle \xi (n),\beta -i\alpha \rangle }|<|\xi (n)|^{-n}\end{aligned}$$

and

$$\begin{aligned} \pi \left| \tau _n+\langle \xi (n),\alpha \rangle \right|\le & {} \left| \sin (2\pi [\tau _n+\langle \xi (n),\alpha \rangle ])\right| \\\le & {} 2e^{2\pi \langle \xi (n),\beta \rangle }\left| \sin (2\pi [\tau _n+\langle \xi (n),\alpha \rangle ])\right| \\\le & {} 2|1-e^{2\pi \langle \xi (n),\beta -i\alpha \rangle }|<2|\xi (n)|^{-n}. \end{aligned}$$

Finally, from these last inequalities we obtain

$$\begin{aligned} |\tau _n+\langle \xi (n),\alpha +i\beta \rangle |\le |\tau _n+\langle \xi (n),\alpha \rangle |+|\langle \xi (n),\beta \rangle |<|\xi (n)|^{-n}\frac{2}{\pi }<\frac{2n^{-n}}{\pi }, \end{aligned}$$

for all $n\in \mathbb {N}$ sufficiently large, which implies that $(\alpha ,\beta )$ does not satisfy (DC2).

Conversely, suppose that $(\alpha ,\beta )$ does not satisfy (DC2). In particular, we can find a sequence $(\xi (n),\tau _n)\in \mathbb {Z}^N\times \mathbb {Z}$ such that $|\xi (n)|\ge n,$ $\tau _n+\langle \xi (n),\alpha +i\beta \rangle \ne 0,$ $\langle \xi (n),\beta \rangle \le 0$ and

$$\begin{aligned} \left| \tau _n+\langle \xi (n),\alpha +i\beta \rangle \right| <\left( |\xi (n)|+|\tau _n|\right) ^{-n}<|\xi (n)|^{-n}, \end{aligned}$$

for every $n\in \mathbb {N}.$ For $n\in \mathbb {N}$ sufficiently large we have $\cos (2\pi \langle \xi (n),\alpha \rangle )\ne 0,$ and since $\langle \xi (n),\beta \rangle \le 0,$ it follows that

$$\begin{aligned} |1-e^{2\pi \langle \xi (n),\beta -i\alpha \rangle }|&\le |1-e^{2\pi \langle \xi (n),\beta \rangle }\cos (2\pi \langle \xi (n),\alpha \rangle )|+ |\sin (2\pi \langle \xi (n),\alpha \rangle )|\\&\le |1-\cos (2\pi \langle \xi (n),\alpha \rangle )|+|1-e^{2\pi \langle \xi (n),\beta \rangle }|\\&\quad +\,|\sin (2\pi \langle \xi (n),\alpha \rangle )|\\&\le |1-\cos (2\pi [\tau _n+\langle \xi (n),\alpha \rangle ])|+|1-e^{2\pi \langle \xi (n),\beta \rangle }|\\&\quad +\,|\sin (2\pi [\tau _n+\langle \xi (n),\alpha \rangle ])|\\&\le 4\pi |\tau _n+\langle \xi (n),\alpha \rangle |+2\pi |\langle \xi (n),\beta \rangle |\\&\le 6\pi |\tau _n+\langle \xi (n),\alpha +i\beta \rangle |<6\pi |\xi (n)|^{-n}<6\pi n^{-n}, \end{aligned}$$

which implies that $(\alpha ,\beta )$ does not satisfy (DC3). $\square$

We now prove the necessity of conditions (III.1)–(III.3) in Theorem 1.1.

Proposition 3.2

The operator L given by (2.3) is not globally solvable in the case where $b_{j0}\ne 0,$ for some $j=1,\ldots ,N,$ and at least one of the following conditions fails:

• (III.1) $\dim \text {span} \{b_1,\ldots ,b_N\}=1;$

• (III.2) the functions $b_j$ do not change sign;

• (III.3) the pair $(\alpha _{0},\beta _{0})$ satisfies (DC2).

Proof

Under the general assumption that $b_{j0}\ne 0$ for some j,  the proof can be divided into three parts.

First, suppose that (III.1) fails; that is, there exist m and $\ell$ in $\{1,\ldots ,N\}$ such that $b_{m0}\ne 0,$ and $b_m(t)$ and $b_\ell (t)$ are $\mathbb {R}$-linearly independent functions. There is no loss of generality in assuming that $m=1$ and $\ell =2.$ We claim that

$$\begin{aligned} L_{12}=\frac{\partial }{\partial t}+\left( a_{10}+ib_1(t)\right) \frac{\partial }{\partial x_1}+\left( a_{20}+ib_2(t)\right) \frac{\partial }{\partial x_2} \end{aligned}$$

is not globally solvable. The arguments to prove our claim are similar to that of Proposition 2.1. We will sketch the proof.

Recall that Lemma 3.1 of [5] implies the existence of integers p and q such that

$$\begin{aligned}\mathbb {T}^1\ni t\mapsto \psi (t)=pb_1(t)+qb_2(t)\end{aligned}$$

changes sign and $\psi _0\doteq (2\pi )^{-1}\int _{0}^{2\pi }\psi (t)dt<0.$ We now define

$$\begin{aligned} H(s,t)=\int _{t-s}^{t}\psi (\tau )d\tau ,\quad 0\le s,t\le 2\pi , \end{aligned}$$

and

$$\begin{aligned} A=H(s_0,t_0)=\max _{0\le s,t\le 2\pi }H(s,t), \end{aligned}$$

where, without loss of generality, we can assume that $0<s_0,\,t_0,\,t_0-s_0<2\pi .$ Set $\sigma _0\doteq t_0-s_0$ and let $\delta >0$ small enough so that $(\sigma _0-\delta ,\sigma _0+\delta )\subset (0,t_0).$ Next, let $\phi \in \mathcal {C}^{\infty }_{c}((\sigma _0-\delta ,\sigma _0+\delta ))$ such that $\phi (t)\equiv 1$ in a neighborhood of $[\sigma _0-\delta /2,\sigma _0+\delta /2]$ and $0\le \phi (t)\le 1,$ for all $t\in \mathbb {T}^1.$

As in the proof of Proposition 2.1, by using partial Fourier series we may consider a function $f\in (\ker {}^tL_{12})^{\circ }\setminus L_{12} \mathcal {C}^{\infty }(\mathbb {T}^3),$ given by

$$\begin{aligned} f(x_1,x_2,t)=\sum _{n=1}^{\infty }\hat{f}(np,nq,t)e^{in(px_1+qx_2)}, \end{aligned}$$

where $\hat{f}(np,nq,\cdot )$ is the function in $\mathcal {C}^{\infty }(\mathbb {T}^1)$ which is the $2\pi$-periodic extension of

$$\begin{aligned}\phi (t)e^{in(pa_{10}+qa_{20})(t_0-t)}e^{-nA}, \quad t\in [0,2\pi ],\end{aligned}$$

for all $n\in \mathbb {N}.$

By proceeding as in the proof of Proposition 2.1, we can verify that f belongs to $(\ker {}^tL_{12})^{\circ }\setminus L_{12} \mathcal {C}^{\infty }(\mathbb {T}^3),$ which implies that L is not globally solvable.

Now, assume that (III.1) satisfied and that (III.2) fails. In this case, there exists $m\in \{1,\ldots ,N\}$ such that $b_{m0}\ne 0$ and $b_m(t)$ changes sign. Thus, it follows from [11]-Theorem 3.2 that

$$\begin{aligned} L_m=\frac{\partial }{\partial t}+\left( a_{m0}+ib_m(t)\right) \frac{\partial }{\partial x_m} \end{aligned}$$

is not globally solvable, which implies that L is not globally solvable.

To complete the proof, we now assume that (III.1) and (III.2) are satisfied, but (III.3) fails. In this last case, we can write the operator L as

$$\begin{aligned} L=\frac{\partial }{\partial t}+\sum _{j=0}^{N}\left( a_{j0}+i\lambda _jb(t)\right) \frac{\partial }{\partial x_j}, \end{aligned}$$

(3.4)

where $b\in \mathcal {C}^{\infty }(\mathbb {T}^1,\mathbb {R}),$ b does not change sign, $b_0\doteq (2\pi )^{-1}\int _{0}^{2\pi }b(t)dt\ne 0,$ and $\lambda \doteq (\lambda _1,\ldots ,\lambda _N)\in \mathbb {R}^N\setminus \{0\}.$

Recall that $\alpha _0=(a_{10},\ldots ,a_{N0}),$ $\beta _0=(b_{10},\ldots ,b_{N0})=b_0(\lambda _1,\ldots ,\lambda _N)=b_0\lambda ,$ and $\beta (t)=(b_1(t),\ldots ,b_N(t))=b(t)(\lambda _1,\ldots ,\lambda _N)=b(t)\lambda .$ Since we are supposing that $(\alpha _0,\beta _0)$ does not satisfy (DC2), Lemma 3.1 implies that (DC3) does not hold. Thus, one of the following situations must occur:

(i) there exists a sequence $(\xi (n))_{n\in \mathbb {N}}\subset \mathbb {Z}^N$ such that $|\xi (n+1)|>|\xi (n)|\ge n,$ $\langle \xi (n),\beta _0\rangle >0,$ and

$$\begin{aligned} |1-\exp \left\{ 2\pi \langle \xi (n),\beta _0-i\alpha _0\rangle \right\} |< |\xi (n)|^{-n}, \quad \text {for all} \quad n\in \mathbb {N}, \end{aligned}$$

(3.5)

or

(ii) there exists a sequence $(\xi (n))_{n\in \mathbb {N}}\subset \mathbb {Z}^N$ such that $|\xi (n+1)|>|\xi (n)|\ge n,$ $\langle \xi (n),\beta _0\rangle =0,$ $\langle \xi (n),\alpha _0\rangle \not \in \mathbb {Z},$ and

$$\begin{aligned} |1-\exp \{-2\pi i\langle \xi (n),\alpha _0\rangle \}|< |\xi (n)|^{-n}, \quad \text {for all} \quad n\in \mathbb {N}. \end{aligned}$$

(3.6)

In order to finish this proof we must show that L given by (3.4) is not globally solvable when either (i) or (ii) occurs.

Assume first that (i) occurs. For L given by (3.4) we will construct a function f belonging to $(\ker {}^tL)^{\circ }\setminus L \mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$ By using partial Fourier series in the variables $(x_1,\ldots ,x_N),$ we consider the function

$$\begin{aligned} f(x,t)=\sum _{n=1}^{\infty }\hat{f}\left( \xi (n),t\right) e^{i\langle \xi (n),x\rangle }, \end{aligned}$$

(3.7)

with partial Fourier coefficients, $\hat{f}(\xi (n),\cdot ),$ given by the $2\pi$-periodic extension of

$$\begin{aligned} |\xi (n)|^{-n/2}\varphi (t)e^{i(\pi -t)\langle \xi (n),\alpha _0\rangle }, \quad t\in [0,2\pi ], \end{aligned}$$

where $\varphi (t)\in \mathcal {C}^{\infty }_c((\frac{\pi }{2}-\delta ,\frac{\pi }{2}+\delta )),$ $\varphi (t)\equiv 1$ in a neighborhood of $[\frac{\pi -\delta }{2},\frac{\pi +\delta }{2}],$ $0\le \varphi (t)\le 1,$ and $\delta >0$ is small enough so that $(\frac{\pi }{2}-\delta ,\frac{\pi }{2}+\delta )\subset (0,\pi ).$ The factor $|\xi (n)|^{-n/2}$ implies that $\hat{f}(\xi (n),\cdot )$ decays rapidly. Hence $f\in \mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$ Moreover, for all $\mu \in \ker {}^tL$ and for $\xi \in \mathbb {Z}^N$ satisfying $\langle \xi ,\beta _0\rangle \ne 0,$ we have $\hat{\mu }(\xi ,\cdot )=0.$ Thus $f\in (\ker {}^tL)^{\circ }.$

We claim that $f\not \in L \mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$ Indeed, suppose by contradiction that there exists $u\in \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ such that $Lu=f.$ The partial Fourier series in the variables $(x_1,\ldots ,x_N)$ gives

$$\begin{aligned} \hat{f}(\xi (n),t)= & {} \frac{d}{dt}\hat{u}(\xi (n),t)+i\langle \xi (n),\alpha _0+i\beta (t)\rangle \hat{u}(\xi (n),t)\nonumber \\= & {} \frac{d}{dt}\hat{u}(\xi (n),t)+i\langle \xi (n),\alpha _0+ib(t)\lambda \rangle \hat{u}(\xi (n),t), \end{aligned}$$

(3.8)

for all $n\in \mathbb {N}$ and $t\in \mathbb {T}^1.$ Since $\langle \xi (n),\beta _0\rangle >0,$ Eq. (3.8) has a unique solution, which can be written as

$$\begin{aligned}&\hat{u}(\xi (n),t)\\&\quad = \left( 1-e^{2\pi c_n}\right) ^{-1}\int _{0}^{2\pi }\hat{f}(\xi (n),t-s)\exp \left\{ \langle \xi (n),\lambda \rangle \int _{t-s}^{t}b(\tau )d\tau \right\} e^{-is\langle \xi (n),\alpha _0\rangle }ds, \end{aligned}$$

where, to simplify notation, we have defined

$$\begin{aligned} c_n\doteq \langle \xi (n),\beta _0-i\alpha _0\rangle =\langle \xi (n),b_0\lambda -i\alpha _0\rangle \quad (n\in \mathbb {N}). \end{aligned}$$

The definition of the function $\hat{f}(\xi (n),\cdot )$ implies that

$$\begin{aligned} |\hat{u}(\xi (n),\pi )|\ge |1-e^{2\pi c_n}|^{-1}|\xi (n)|^{-n/2}\int _{|s-\pi /2|\le \delta /2}e^{\langle \xi (n),\lambda \rangle \int _{\pi -s}^{\pi }b(\tau )d\tau }ds. \end{aligned}$$

Recall that b does not change sign and that $\langle \xi (n),\lambda b_0\rangle >0.$ Hence

$$\begin{aligned} \langle \xi (n),\lambda \rangle \int _{\pi -s}^{\pi }b(\tau )d\tau \ge 0, \quad \text {for all} \quad |s-\pi /2|\le \delta /2. \end{aligned}$$

Moreover, (3.5) implies that

$$\begin{aligned}|1-e^{2\pi c_n}|^{-1}=|1-\exp \{2\pi \langle \xi (n),\beta _0-i\alpha _0\rangle \}|^{-1}>|\xi (n)|^{n}.\end{aligned}$$

Summarizing, we obtain

$$\begin{aligned}|\hat{u}(\xi (n),\pi )|\ge |1-e^{2\pi c_n}|^{-1}|\xi (n)|^{-n/2}\delta >|\xi (n)|^{n}|\xi (n)|^{-n/2}\delta \ge n^{n/2}\delta ,\end{aligned}$$

which is a contradiction, since $|\hat{u}(\xi (n),\pi )|$ should decay rapidly.

Therefore, f given by (3.7) belongs to $(\ker {}^tL)^{\circ }\setminus L \mathcal {C}^{\infty }(\mathbb {T}^{N+1}),$ which implies that L is not globally solvable when (i) occurs.

To finish the proof we just mention that similar computations show that L is not globally solvable when (ii) occurs. Indeed, under conditions (ii), the same f,  given by (3.7), belongs to $(\ker {}^tL)^{\circ }\setminus L \mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$

The proof of Proposition 3.2 is complete. $\square$

Example 3.3

By Proposition 3.2, for any functions $a_1,a_2\in \mathcal {C}^\infty (\mathbb {T}^1,\mathbb {R}),$ the operator

$$\begin{aligned}\frac{\partial }{\partial t}+\left( a_1(t)+i\sin ^2(t)\right) \frac{\partial }{\partial x_1}+\left( a_2(t)+i\cos ^2(t)\right) \frac{\partial }{\partial x_2}\end{aligned}$$

is not globally solvable, since $\sin ^2(t)$ and $\cos ^2(t)$ are $\mathbb {R}$-linearly independent functions.

We finish the proof of Theorem 1.1 by showing that conditions (III.1)–(III.3) are sufficient for global solvability.

Proposition 3.4

Let L be given by (2.3) and suppose that $b_{j0}\ne 0,$ for some $j=1,\ldots ,N.$ Then L is globally solvable if the following conditions are verified:

• (III.1) $\dim \text {span} \{b_1,\ldots ,b_N\}=1;$

• (III.2) the functions $b_j$ do not change sign;

• (III.3) the pair $(\alpha _{0},\beta _{0})$ satisfies (DC2).

Proof

Given $f\in (\ker {}^tL)^{\circ },$ our goal is to seek $u\in \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ solution of $Lu=f.$

Notice that our assumptions imply that L can be written as

$$\begin{aligned} \frac{\partial }{\partial t}+\sum _{j=0}^{N}\left( a_{j0}+i\lambda _jb(t)\right) \frac{\partial }{\partial x_j}, \end{aligned}$$

where $b\in \mathcal {C}^{\infty }(\mathbb {T}^1,\mathbb {R}),$ b does not change sign, $b_0\doteq (2\pi )^{-1}\int _{0}^{2\pi }b(t)dt\ne 0,$ and $\lambda \doteq (\lambda _1,\ldots ,\lambda _N)\in \mathbb {R}^N\setminus \{0\}.$

Partial Fourier series in the variables $(x_1,\ldots ,x_N)$ leads us to solve the equations

$$\begin{aligned} \frac{d}{dt}\hat{u}(\xi ,t)+i\langle \xi ,\alpha _0+ib(t)\lambda \rangle \hat{u}(\xi ,t)=\hat{f}(\xi ,t), \quad t\in \mathbb {T}^1, \quad \xi \in \mathbb {Z}^N, \end{aligned}$$

(3.9)

where we recall that $\alpha _0=(a_{10},\ldots ,a_{N0}).$

We now divide the task of finding solutions to (3.9) into three cases.

First: $\xi \in \mathbb {Z}^N$ is such that $\langle \xi ,b_0\lambda \rangle =0$ and $\langle \xi ,\alpha _0\rangle \in \mathbb {Z}.$

In this case, the function $\phi _{\xi }(x,t)\doteq e^{it\langle \xi ,\alpha _0\rangle }e^{-i\langle \xi ,x\rangle }$ belongs to $\ker {}^tL,$ which implies that

$$\begin{aligned} 0=\langle \phi _{\xi },f\rangle =(2\pi )^N\int _{0}^{2\pi }\hat{f}(\xi ,t)e^{it\langle \xi ,\alpha _0\rangle }dt. \end{aligned}$$

Hence, Eq. (3.9) has infinitely many solutions and we fix

$$\begin{aligned} \hat{u}(\xi ,t)=\int _{0}^{t}\hat{f}(\xi ,s)e^{i(s-t)\langle \xi ,\alpha _0\rangle }ds, \quad t\in \mathbb {T}^1, \end{aligned}$$

(3.10)

as a solution of (3.9). Notice that

$$\begin{aligned} |\hat{u}(\xi ,t)|\le \int _{0}^{t}|\hat{f}(\xi ,s)|ds. \end{aligned}$$

(3.11)

Second: $\xi \in \mathbb {Z}^N$ is such that $\langle \xi ,b_0\lambda \rangle \ne 0.$

Now Eq. (3.9) has a unique solution in $\mathcal {C}^{\infty }(\mathbb {T}^1).$ It is worth pointing out that this solution can be written in two different ways.

When $\langle \xi ,\lambda \rangle b_0<0,$ we write

$$\begin{aligned} \hat{u}(\xi ,t)=(1-e^{2\pi c_{\xi }})^{-1}\int _{0}^{2\pi }\hat{f}(\xi ,t-s)e^{\langle \xi ,\lambda \rangle \int _{t-s}^{t}b(\tau )d\tau }e^{-is\langle \xi ,\alpha _0\rangle }ds, \end{aligned}$$

(3.12)

where $c_{\xi }=\langle \xi ,b_0\lambda -i\alpha _0\rangle .$ Since b does not change sign and since $\langle \xi ,\lambda \rangle b_0<0$ we have

$$\begin{aligned} \langle \xi ,\lambda \rangle \int _{t-s}^{t}b(\tau )d\tau \le 0,\quad \text {for all} \quad t,s\in [0,2\pi ]. \end{aligned}$$

(3.13)

Recall now that the validity of (III.3) and Lemma 3.1 imply that $(\alpha _0,b_0\lambda )$ satisfies (DC3). It follows from (3.13) and (DC3) that the solution given by (3.12) satisfies

$$\begin{aligned} |\hat{u}(\xi ,t)|\le \frac{1}{C}|\xi |^{\gamma }\int _{0}^{2\pi }|\hat{f}(\xi ,t-s)|ds. \end{aligned}$$

(3.14)

On the other hand, when $\langle \xi ,\lambda \rangle b_0>0$ we write

$$\begin{aligned} \hat{u}(\xi ,t)=(e^{-2\pi c_{\xi }}-1)^{-1}\int _{0}^{2\pi }\hat{f}(\xi ,t+s)e^{-\langle \xi ,\lambda \rangle \int _{t}^{t+s}b(\tau )d\tau }e^{is\langle \xi ,\alpha _0\rangle }ds.\end{aligned}$$

(3.15)

Since b does not change sign and, also, $\langle \xi ,\lambda \rangle b_0>0,$ we have

$$\begin{aligned} \langle \xi ,\lambda \rangle \int _{t}^{t+s}b(\tau )d\tau \ge 0,\quad \text {for all} \quad t,s\in [0,2\pi ]. \end{aligned}$$

(3.16)

As before, (3.16) and (DC3) imply that the solution given by (3.15) satisfies

$$\begin{aligned} |\hat{u}(\xi ,t)|\le \frac{1}{C}|\xi |^{\gamma }\int _{0}^{2\pi }|\hat{f}(\xi ,t+s)|ds. \end{aligned}$$

(3.17)

Third: $\xi \in \mathbb {Z}^N$ is such that $\langle \xi ,b_0\lambda \rangle =0$ and $\langle \xi ,\alpha _0\rangle \not \in \mathbb {Z}.$ Again Eq. (3.9) has a unique solution in $\mathcal {C}^{\infty }(\mathbb {T}^1),$ which can be written as

$$\begin{aligned} \hat{u}(\xi ,t)=(1-e^{-2\pi i\langle \xi ,\alpha _0\rangle })^{-1}\int _{0}^{2\pi }\hat{f}(\xi ,t-s)e^{-is\langle \xi ,\alpha _0\rangle }ds. \end{aligned}$$

(3.18)

By using estimate (DC3) we obtain

$$\begin{aligned} |\hat{u}(\xi ,t)|\le \frac{1}{C}|\xi |^{\gamma }\int _{0}^{2\pi }|\hat{f}(\xi ,t-s)|ds. \end{aligned}$$

(3.19)

Up to this point we have defined a sequence $(\hat{u}(\xi ,\cdot ))_{\xi \in \mathbb {Z}^N}$ of solutions of the Eq. (3.9). The proof will be concluded if we show that this sequence decays rapidly. Since $\hat{f}(\xi ,\cdot )$ decays rapidly, it follows from (3.11), (3.14), (3.17) and (3.19) that, for each $n\in \mathbb {Z}_+$ there exists $C(n)>0$ such that

$$\begin{aligned}|\hat{u}(\xi ,t)|\le \frac{C(n)}{|\xi |^{n}}, \quad \text {for all} \quad t\in \mathbb {T} \; \text {and} \; \xi \in \mathbb {Z}^N.\end{aligned}$$

Similar estimates can be found for each derivative $\hat{u}^{(m)}(\xi ,\cdot ),$ $m\in \mathbb {Z}_+.$ Therefore, $(\hat{u}(\xi ,\cdot ))_{\xi \in \mathbb {Z}^N}$ decays rapidly, which completes the proof of Proposition 3.4. $\square$

Example 3.5

Consider

$$\begin{aligned} L=\frac{\partial }{\partial t}+\left( a_1(t)+i\lambda b_1(t)\right) \frac{\partial }{\partial x_1} +ib_1(t)\frac{\partial }{\partial x_2},\quad b_1\not \equiv 0, \end{aligned}$$

defined on $\mathbb {T}^3$, where $\lambda \in \mathbb {R}$, $a_1,b_1\in \mathcal {C}^\infty (\mathbb {T}^1,\mathbb {R}),$ and $b_1$ does not change sign. It follows from Propositions 3.2 and 3.4 that L is globally solvable if and only if the pair $(a_{10},0),(\lambda b_{10},b_{10})\in \mathbb {R}^2$ satisfies (DC2),  which holds if and only if either $(a_{10},\lambda )\in \mathbb {Q}^2$ or $(a_{10},\lambda )\in \mathbb {R}^2\setminus \mathbb {Q}^2$ is not a Liouville vector. In particular, if $\alpha$ is a Liouville number ($\alpha \in \mathbb {R}\setminus \mathbb {Q}$), then

$$\begin{aligned} L_1=\frac{\partial }{\partial t}+\left( \alpha +i \cos ^2t\right) \frac{\partial }{\partial x_1} +i\cos ^2t\frac{\partial }{\partial x_2} \end{aligned}$$

is not globally solvable. On the other hand, there are $\alpha$ and $\lambda$ Liouville numbers such that the pair $(\alpha ,\lambda )$ is not a Liouville vector (see [2, Example 4.9]), hence

$$\begin{aligned} L_2=\frac{\partial }{\partial t}+(\alpha +i \lambda \cos ^2t)\frac{\partial }{\partial x_1} +i\cos ^2t\frac{\partial }{\partial x_2} \end{aligned}$$

is globally solvable.

4 Proof of Theorem 1.3

In this section we will deal with global hypoellipticity by proving Theorem 1.3. As in the preceding section (in which we studied global solvability), by using an automorphism we may assume that L is given by (2.3), that is,

$$\begin{aligned} L=\frac{\partial }{\partial t}+\sum _{j=1}^{N}(a_{j0}+ib_j(t))\frac{\partial }{\partial x_j}. \end{aligned}$$

(4.1)

We begin by proving the necessity of the conditions (1)-(3).

Suppose that $b_j$ changes sign, for some $j\in \{1,\ldots ,N\}.$ Without loss of generality we may assume that $j=1.$ It follows from Theorem 2.2 of [11] that there exists $\nu \in \mathcal {D}'(\mathbb {T}^2_{(x_1,t)})\setminus \mathcal {C}^{\infty }(\mathbb {T}^2_{(x_1,t)})$ for which

$$\begin{aligned} f\dot{=}\left( \frac{\partial }{\partial t}+(a_{10}+ib_1(t))\frac{\partial }{\partial x_1}\right) \nu \end{aligned}$$

belongs to $\mathcal {C}^{\infty }(\mathbb {T}^2_{(x_1,t)})$. Hence, by setting $x'=(x_2,\ldots ,x_N)$ it follows that $\omega \doteq \nu \otimes 1_{x'}\in \mathcal {D}'(\mathbb {T}^{N+1})\setminus \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ and, also, $L\omega \in \mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$ Therefore, L is not globally hypoelliptic.

Hence we have the right to restrict ourselves to the case where $b_j$ does not change sign, for $j=1,\ldots ,N.$

Suppose now that $\dim \text {span}\{b_1,\ldots ,b_N\}>1.$ Then there exist functions $b_m$ and $b_\ell$ which are $\mathbb {R}$-linearly independent in $\mathcal {C}^{\infty }(\mathbb {T}^1,\mathbb {R}).$ Since $b_m\not \equiv 0$ and $b_m$ does not change sign, we have $b_{m0}\ne 0.$ Similarly, $b_{\ell 0}\ne 0.$ Again, we may assume that $m=1$ and $\ell =2.$ Under these assumptions, recall that in the proof of Proposition 3.2 we exhibited a smooth function $f\in (\ker {}^tL)^{\circ },$ for which the equation

$$\begin{aligned} L_{12}u=\left( \frac{\partial }{\partial t}+(a_{10}+ib_1(t))\frac{\partial }{\partial x_1}+(a_{20}+ib_2(t))\frac{\partial }{\partial x_2}\right) u=f \end{aligned}$$

does not have smooth solutions u in $\mathbb {T}^3.$ However, we claim that for such f we can find a distribution solution, $\mu ,$ for $L_{12}\mu =f.$ Indeed, with the same notation used in the proof of Proposition 3.2, we define

$$\begin{aligned} \mu =\sum _{n=1}^{\infty }\hat{\mu }(np,nq,t)e^{in(px_1+qx_2)}, \end{aligned}$$

where

$$\begin{aligned} \hat{\mu }(np,nq,t)=(1-e^{2\pi d_n})^{-1}\int _{0}^{2\pi }\hat{f}(np,nq,t-s)e^{nH(s,t)}e^{-ins(pa_{10}+qa_{20})}ds \end{aligned}$$

and $d_n\doteq n(\psi _0-i(pa_{10}+qa_{20}))=n(pb_{10}+qb_{20})-in(pa_{10}+qa_{20}).$ Recall that $\hat{f}(np,nq,\cdot )$ is the function belonging to $\mathcal {C}^{\infty }(\mathbb {T}^1)$ which is the $2\pi$-periodic extension of

$$\begin{aligned} \phi (t)e^{in(pa_{10}+qa_{20})(t_0-t)}e^{-nA}, \quad t\in [0,2\pi ], \end{aligned}$$

for all $n\in \mathbb {N}.$ Since $\psi _0<0$ and $0\le \phi (t)\le 1,$ for each $n\in \mathbb {N}$ we have

$$\begin{aligned} |<\hat{\mu }(np,nq,t),\theta (t)>|\le (2\pi )^2\Vert \theta \Vert _{\infty }|1-e^{2\pi d_n}|^{-1}\le (2\pi )^2\Vert \theta \Vert _{\infty }(1-e^{2\pi \psi _0})^{-1}, \end{aligned}$$

for every $\theta \in \mathcal {C}^{\infty }(\mathbb {T}^1).$ Hence $\mu \in \mathcal {D}'(\mathbb {T}^3)\setminus \mathcal {C}^{\infty }(\mathbb {T}^3)$ and $L_{12}\mu =f;$ consequently, $L_{12}$ is not globally hypoelliptic, which implies that L is not globally hypoelliptic.

Up to this point we have shown that the global hypoellipticity of L implies that each $b_j$ does not change sign and that $\dim \text {span}\{b_1,\ldots ,b_N\}\le 1.$ Notice that, under these assumptions the operator L can be written in the form

$$\begin{aligned} L=\frac{\partial }{\partial t}+\sum _{j=1}^{N}(a_{j0}+i\lambda _jb(t))\frac{\partial }{\partial x_j}, \end{aligned}$$

where $b\in \mathcal {C}^{\infty }(\mathbb {T}^1,\mathbb {R}),$ b does not change sign and $\lambda =(\lambda _1,\ldots ,\lambda _N)\in \mathbb {R}^N\setminus \{0\}.$

In order to complete the proof of necessity in Theorem 1.3 we must verify that (DC1) is also a necessary condition for the global hypoellipticity of L.

When $b\equiv 0,$ the necessity of the condition (DC1) follows by applying the techniques of [8].

Suppose now that $b\not \equiv 0.$ In particular, $b_0\ne 0.$ If now (DC1) does not hold, then one of the following situations must occur:

• (i) there exists a sequence $(\xi (n),\tau _n)_{n\in \mathbb {N}}\subset \mathbb {Z}^{N+1}$ such that $|\xi (n)|+|\tau _n|\ge n,$ $\langle \xi (n),b_0\lambda \rangle =0,$ and $\tau _n=-\langle \xi (n),\alpha _0\rangle \in \mathbb {Z}.$

• (ii) the pair $(\alpha _0,b_0\lambda )$ does not satisfy (DC2).

Suppose first that (i) occurs. Notice that $|\xi (n)|$ is unbounded. By taking a subsequence if necessary we can assume that $|\xi (n)|$ is increasing. We now set

$$\begin{aligned} \mu \dot{=}\sum _{n=1}^{\infty }e^{-it\langle \xi (n),\alpha _0\rangle }e^{i\langle \xi (n),x\rangle }, \end{aligned}$$

which belongs to $\mathcal {D}'(\mathbb {T}^{N+1})\setminus \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ and satisfies $L\mu =0.$ Therefore, L is not globally hypoelliptic.

Now, assume that (ii) occurs. We will proceed as in the proof of Proposition 3.2 (the part in which (III.1) and (III.2) are true, but (III.3) fails) in order to construct a function f in $\mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ so that $L\mu =f$ has a solution in $\mathcal {D}'(\mathbb {T}^{N+1})\setminus \mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$ Since $(\alpha _0,\beta _0)=(\alpha _0,b_0\lambda )$ does not satisfy (DC2), Lemma 3.1 implies that (DC3) does not hold; hence, either (3.5) or (3.6) occurs. When (3.5) occurs, we have, with the same notations as in the proof of Proposition 3.2, $|1-e^{2\pi c_{n}}|\le |\xi (n)|^{-n}$ (where $c_n\doteq \langle \xi (n),b_0\lambda -i\alpha _0\rangle$) and $\langle \xi (n),b_0\lambda \rangle >0.$ It follows that the sequence $\hat{f}(\xi (n),\cdot )$ given by the $2\pi$-periodic extension of

$$\begin{aligned} (1-e^{2\pi c_n})e^{-\langle \xi (n),2\pi b_0\lambda \rangle }\varphi (t)e^{i(\pi -t)\langle \xi (n),\alpha _0\rangle }, \quad t\in [0,2\pi ], \end{aligned}$$

is rapidly decreasing (recall that $\varphi (t)\in \mathcal {C}^{\infty }_c((\frac{\pi }{2}-\delta ,\frac{\pi }{2}+\delta )),$ $\varphi (t)\equiv 1$ in a neighborhood of $[\frac{\pi -\delta }{2},\frac{\pi +\delta }{2}],$ $0\le \varphi (t)\le 1,$ and $\delta >0$ is small enough so that $(\frac{\pi }{2}-\delta ,\frac{\pi }{2}+\delta )\subset (0,\pi )$). Thus we may define a function $f\in \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ by

$$\begin{aligned} f(x,t)=\sum _{n=1}^{\infty }\hat{f}(\xi (n),t)e^{i\langle \xi (n),x\rangle }. \end{aligned}$$

For this f,  we define the distribution

$$\begin{aligned} \mu =\sum _{n=1}^{\infty }\hat{\mu }(\xi (n),t)e^{i\langle \xi (n),x\rangle }, \end{aligned}$$

where

$$\begin{aligned}&\hat{\mu }(\xi (n),t)\nonumber \\&\quad =(1-e^{2\pi c_n})^{-1}\int _{0}^{2\pi }\hat{f}(\xi (n),t-s)e^{-is\langle \xi (n),\alpha _0\rangle }e^{\langle \xi (n),\lambda \rangle \int _{t-s}^{t}b(\tau )d\tau }ds\in \mathcal {C}^{\infty }(\mathbb {T}^{1}). \end{aligned}$$

Notice that, since $\langle \xi (n),b_0\lambda \rangle >0$ and since $\langle \xi (n),b_0\lambda \rangle \rightarrow 0$ as $n\rightarrow \infty$ (see (3.5)), the definition of $\hat{f}(\xi (n),\cdot )$ implies

$$\begin{aligned} |\hat{\mu }(\xi (n),\pi )|\ge \delta e^{-\langle \xi (n),2\pi b_0\lambda \rangle }\ge \delta /2; \end{aligned}$$

hence $\mu \not \in \mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$ On the other hand, for every $\theta \in \mathcal {C}^{\infty }(\mathbb {T}^1)$ we have

$$\begin{aligned} |\langle \hat{\mu }(\xi (n),t),\theta (t)\rangle |\le (2\pi )^2\Vert \theta \Vert _{\infty }, \end{aligned}$$

which implies that $\mu \in \mathcal {D}'(\mathbb {T}^{N+1})\setminus \mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$ A simple computation shows that $L\mu =f.$ Therefore, L is not globally hypoelliptic when (3.5) occurs.

Finally, suppose that (3.6) occurs. In this case, we consider the function $f\in \mathcal {C}^\infty (\mathbb {T}^{N+1})$ given by

$$\begin{aligned} f(x,t)=\sum _{n=1}^{\infty }\hat{f}(\xi (n),t)e^{i\langle \xi (n),x\rangle }, \end{aligned}$$

where $\hat{f}(\xi (n),\cdot )$ is the $2\pi$-periodic extension of

$$\begin{aligned} \left( 1-e^{-2\pi i\langle \xi (n),\alpha _0\rangle }\right) \varphi (t)e^{i(\pi -t)\langle \xi (n),\alpha _0\rangle }, \quad t\in [0,2\pi ]. \end{aligned}$$

Notice that f is well defined since the estimate $|1-e^{-2\pi i\langle \xi (n),\alpha _0\rangle }|\le |\xi (n)|^{-n}$ [see (3.6)] implies that $\hat{f}(\xi (n),\cdot )$ decays rapidly.

For such f,  consider the distribution

$$\begin{aligned} \mu =\sum _{n=1}^{\infty }\hat{\mu }(\xi (n),t)e^{i\langle \xi (n),x\rangle }, \end{aligned}$$

where

$$\begin{aligned} \hat{\mu }(\xi (n),t)=\left( 1-e^{-2\pi \langle \xi (n),\alpha _0\rangle }\right) ^{-1}\int _{0}^{2\pi }\hat{f}(\xi (n),t-s)e^{-is\langle \xi (n),\alpha _0\rangle }ds\in \mathcal {C}^{\infty }(\mathbb {T}^{1}). \end{aligned}$$

As before, simple computations show that $\mu \in \mathcal {D}'(\mathbb {T}^{N+1})\setminus \mathcal {C}^{\infty }(\mathbb {T}^{N+1}),$ and that $L\mu =f.$ Therefore, L is not globally hypoelliptic when (3.6) occurs.

The proof of necessity in Theorem 1.3 is complete.

We now move on to the proof of sufficiency. Assuming that (1)-(3) hold, let $\mu$ be a distribution in $\mathcal {D}'(\mathbb {T}^{N+1})$ and suppose that $L\mu =f\in \mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$ Our goal is to prove that $\mu \in \mathcal {C}^{\infty }(\mathbb {T}^{N+1}).$

By using partial Fourier series in the variables $x=(x_1,\ldots ,x_N)$ we can write

$$\begin{aligned} f(x,t)=\sum _{\xi \in \mathbb {Z}^N}\hat{f}(\xi ,t)e^{i\langle \xi ,x\rangle } \end{aligned}$$

and

$$\begin{aligned} \mu =\sum _{\xi \in \mathbb {Z}^N}\hat{\mu }(\xi ,t)e^{i\langle \xi ,x\rangle }, \end{aligned}$$

where a priori $\hat{\mu }(\xi ,\cdot )$ belongs to $\mathcal {D}'(\mathbb {T}^1).$ The equation $L\mu =f$ imply that, for each $\xi \in \mathbb {Z}^N,$ $\hat{\mu }(\xi ,\cdot )$ is a solution of

$$\begin{aligned} \frac{d}{dt}\hat{\mu }(\xi ,t)+\langle \xi ,i\alpha _0-b(t)\lambda \rangle \hat{\mu }(\xi ,t)=\hat{f}(\xi ,t). \end{aligned}$$

(4.2)

Now, recall that (DC1) implies (DC2). Hence, Lemma 3.1 implies that (DC3) is satisfied. Also, (DC1) implies that we simultaneously have $\langle \xi ,\alpha _0\rangle \in \mathbb {Z}$ and $\langle \xi ,b_0\lambda \rangle =0$ only for $\xi =0.$ Thus, by proceeding as in the proof of Proposition 3.4 we see that $\hat{\mu }(\xi ,\cdot )$ is a smooth function on $\mathbb {T}^1,$ which is uniquely determined for all indices $\xi \in \mathbb {Z}^N\setminus \{0\}.$ Moreover, since (DC3) holds we see that $\hat{\mu }(\xi ,\cdot )$ is rapidly decreasing. Therefore, $\mu \in \mathcal {C}^{\infty }(\mathbb {T}^{N+1})$ and the proof of sufficiency in Theorem 1.3 is complete.