The following lemma tells us that the Landesman–Lazer conditions force all components ${z}_{k}(t)=({x}_{k}(t),{y}_{k}(t))$ of the solutions to rotate around the origin, provided that they start sufficiently far away from the origin itself.

#### Proof.

It will be sufficient to analyze the behavior of each component ${z}_{k}(t)=({x}_{k}(t),{y}_{k}(t))$ of the solution $\mathbf{z}(t)$. Hence, we fix $k\in \{1,\mathrm{\dots},N\}$ and, to simplify the notation, we consider system (2.3) and denote by $z(t)=(x(t),y(t))$ its solutions.
Set

${D}_{k}:=\mathrm{max}\{{\parallel {\psi}_{k}^{+}\parallel}_{{L}^{1}},{\parallel {\psi}_{k}^{-}\parallel}_{{L}^{1}},{\parallel {\phi}_{k}^{+}\parallel}_{{L}^{1}},{\parallel {\phi}_{k}^{-}\parallel}_{{L}^{1}}\},$

where ${\psi}_{k}^{\pm}$ and ${\phi}_{k}^{\pm}$ are the functions introduced in condition (2.1).
The key information for the argument of the proof is contained in the following two propositions.

By the symmetry of our assumption (A2), we can write the analogous of Proposition 6 in the northern, western and southern regions, and the analogous of Proposition 7 in the north-western, south-western and south-eastern regions. For briefness, we leave this easy but tedious charge to the patient reader.

Let us now proceed with the proof of Lemma 5.
Let *M* be a positive integer and let ${R}_{1}>0$ be fixed: we can assume, without loss of generality, that ${R}_{1}\ge 2{d}_{1}$.
We will define two polygonal curves ${\mathrm{\Gamma}}_{1}^{k}$ and ${\mathrm{\Gamma}}_{2}^{k}$, represented in Figure 1, which will guide the components of the solutions: they are spiral-like curves, rotating counterclockwise around the origin infinitely many times as their distance from the origin goes to infinity. (For a similar approach, see also [10].)

Figure 1 The curves ${\mathrm{\Gamma}}_{1}^{k}$ and ${\mathrm{\Gamma}}_{2}^{k}$.

We start by fixing three constants ${\beta}_{1}\ge {R}_{1}$, ${\alpha}_{1}=-{\beta}_{1}$ and ${\gamma}_{1}\ge {R}_{1}$.

*First part of ${\mathrm{\Gamma}}_{\mathrm{1}}^{k}$.*
This is simply the segment $\{{\gamma}_{1}\}\times [-{\beta}_{1},{\beta}_{1}]$.

*First part of ${\mathrm{\Gamma}}_{\mathrm{2}}^{k}$.*
This is made up of three joined segments, which will now be defined. Using Proposition 6 (eastern region), with $\alpha ={\alpha}_{1}$, $\beta ={\beta}_{1}$ and $\gamma ={\gamma}_{1}$, we find a ${\gamma}_{1}^{*}>{\gamma}_{1}$, defined as in (2.11), i.e.,

${\gamma}_{1}^{*}={\gamma}_{1}+\left(\frac{2{\beta}_{1}+{\parallel {\psi}_{k}^{+}\parallel}_{{L}^{1}}}{\delta}+1\right)T{\parallel {{f}_{k}|}_{[0,T]\times [-{\beta}_{1},{\beta}_{1}+{\parallel {\psi}_{k}^{+}\parallel}_{{L}^{1}}]}\parallel}_{\mathrm{\infty}}.$

The first of the three segments is $\{{\gamma}_{1}^{*}\}\times [-{\beta}_{1},{\beta}_{1}]$. We now use Proposition 7 (north-eastern region), with $\mu ={\beta}_{1}$ and $\nu ={\gamma}_{1}^{*}$, and we find a ${\nu}_{1}^{*}>{\gamma}_{1}^{*}$, defined as in (2.12), i.e.,

${\nu}_{1}^{*}={\gamma}_{1}^{*}+{\parallel {\phi}_{k}^{+}\parallel}_{{L}^{1}}.$

The second of the three segments is $[{\gamma}_{1}^{*},{\nu}_{1}^{*}]\times \{{\beta}_{1}\}$. We now use the northern version of Proposition 6, with $\alpha =-{\nu}_{1}^{*}$, $\beta ={\nu}_{1}^{*}$ and $\gamma ={\beta}_{1}$, and we find a ${\gamma}_{2}^{*}>{\beta}_{1}$, defined similarly to (2.11), precisely

${\gamma}_{2}^{*}={\beta}_{1}+\left(\frac{2{\nu}_{1}^{*}+{\parallel {\phi}_{k}^{+}\parallel}_{{L}^{1}}}{\delta}+1\right)T{\parallel {{g}_{k}|}_{[0,T]\times [-{\nu}_{1}^{*}-{\parallel {\phi}_{k}^{+}\parallel}_{{L}^{1}},{\nu}_{1}^{*}]}\parallel}_{\mathrm{\infty}}.$

The third of the three segments is then $\{{\nu}_{1}^{*}\}\times [{\beta}_{1},{\gamma}_{2}^{*}]$.

We now iterate such a procedure in the other regions, as briefly explained below.

*Second part of ${\mathrm{\Gamma}}_{\mathrm{1}}^{k}$.*
This is the segment $[-{\nu}_{1}^{*},{\gamma}_{1}]\times \{{\beta}_{1}\}$.

*Second part of ${\mathrm{\Gamma}}_{\mathrm{2}}^{k}$.*
As before, this is made up of three segments. The first one is $[-{\nu}_{1}^{*},{\nu}_{1}^{*}]\times \{{\gamma}_{2}^{*}\}$. We now use the north-western version of Proposition 7, with $\mu =-{\nu}_{1}^{*}$ and $\nu ={\gamma}_{2}^{*}$, and we find a ${\nu}_{2}^{*}>{\gamma}_{2}^{*}$, similarly to (2.12), precisely

${\nu}_{2}^{*}={\gamma}_{2}^{*}+{\parallel {\psi}_{k}^{+}\parallel}_{{L}^{1}}.$

The second of the three segments is $\{-{\nu}_{1}^{*}\}\times [{\gamma}_{2}^{*},{\nu}_{2}^{*}]$. We then use the western version of Proposition 6, with $\alpha =-{\nu}_{2}^{*}$, $\beta ={\nu}_{2}^{*}$ and $\gamma =-{\nu}_{1}^{*}$, and we find a ${\gamma}_{3}^{*}>{\nu}_{1}^{*}$ (we prefer writing ${\gamma}_{3}^{*}$ instead of $-{\gamma}_{3}^{*}$ in order to deal with a positive constant, even if $x(t)$ is negative in this region), similarly to (2.11), precisely

${\gamma}_{3}^{*}={\nu}_{1}^{*}+\left(\frac{2{\nu}_{2}^{*}+{\parallel {\psi}_{k}^{-}\parallel}_{{L}^{1}}}{\delta}+1\right)T{\parallel {{f}_{k}|}_{[0,T]\times [-{\nu}_{2}^{*}-{\parallel {\psi}_{k}^{-}\parallel}_{{L}^{1}},{\nu}_{2}^{*}]}\parallel}_{\mathrm{\infty}}.$

The third of the three segments is then $[-{\gamma}_{3}^{*},-{\nu}_{1}^{*}]\times \{{\nu}_{2}^{*}\}$.

*Third part of ${\mathrm{\Gamma}}_{\mathrm{1}}^{k}$.*
This is the segment $\{-{\nu}_{1}^{*}\}\times [-{\nu}_{2}^{*},{\beta}_{1}]$.

*Third part of ${\mathrm{\Gamma}}_{\mathrm{2}}^{k}$.*
This time, the first segment is $\{-{\gamma}_{3}^{*}\}\times [-{\nu}_{2}^{*},{\nu}_{2}^{*}]$. Using the south-western version of Proposition 7, with $\mu =-{\nu}_{2}^{*}$ and $\nu =-{\gamma}_{3}^{*}$, we find a ${\nu}_{3}^{*}>{\gamma}_{3}^{*}$ (again we prefer dealing with positive constants), precisely

${\nu}_{3}^{*}={\gamma}_{3}^{*}+{\parallel {\phi}_{k}^{-}\parallel}_{{L}^{1}}.$

The second segment is $[-{\nu}_{3}^{*},-{\gamma}_{3}^{*}]\times \{-{\nu}_{2}^{*}\}$. Using the southern version of Proposition 6, with $\alpha =-{\nu}_{3}^{*}$, $\beta ={\nu}_{3}^{*}$ and $\gamma =-{\nu}_{2}^{*}$, and we find a ${\gamma}_{4}^{*}>{\nu}_{2}^{*}$ (again a positive constant), precisely

${\gamma}_{4}^{*}={\nu}_{2}^{*}+\left(\frac{2{\nu}_{3}^{*}+{\parallel {\phi}_{k}^{-}\parallel}_{{L}^{1}}}{\delta}+1\right)T{\parallel {{g}_{k}|}_{[0,T]\times [-{\nu}_{3}^{*},{\nu}_{3}^{*}+{\parallel {\phi}_{k}^{-}\parallel}_{{L}^{1}}]}\parallel}_{\mathrm{\infty}}.$

The third segment is then $\{{\nu}_{3}^{*}\}\times [-{\gamma}_{4}^{*},-{\nu}_{2}^{*}]$.

*Fourth part of ${\mathrm{\Gamma}}_{\mathrm{1}}^{k}$.*
This is simply the segment $[-{\nu}_{1}^{*},{\nu}_{3}^{*}]\times \{-{\nu}_{2}^{*}\}$.

*Fourth part of ${\mathrm{\Gamma}}_{\mathrm{2}}^{k}$.*
As usual, this is made up of three segments. The first one is $[-{\nu}_{3}^{*},{\nu}_{3}^{*}]\times \{-{\gamma}_{4}^{*}\}$. We now use the south-eastern version of Proposition 7, with $\mu ={\nu}_{3}^{*}$ and $\nu =-{\gamma}_{4}^{*}$, and we find a ${\nu}_{4}^{*}>{\gamma}_{4}^{*}$ (again positive), precisely

${\nu}_{4}^{*}={\gamma}_{4}^{*}+{\parallel {\psi}_{k}^{-}\parallel}_{{L}^{1}}.$

The second segment is $\{{\nu}_{3}^{*}\}\times [-{\nu}_{4}^{*},-{\gamma}_{4}^{*}]$. We now use the eastern version of Proposition 6, with $\alpha =-{\nu}_{4}^{*}$, $\beta ={\nu}_{4}^{*}$ and $\gamma ={\nu}_{3}^{*}$, and we find a ${\gamma}_{5}^{*}>{\nu}_{3}^{*}$, precisely

${\gamma}_{5}^{*}={\nu}_{3}^{*}+\left(\frac{2{\nu}_{4}^{*}+{\parallel {\psi}_{k}^{+}\parallel}_{{L}^{1}}}{\delta}+1\right)T{\parallel {{f}_{k}|}_{[0,T]\times [-{\nu}_{4}^{*},{\nu}_{4}^{*}+{\parallel {\psi}_{k}^{+}\parallel}_{{L}^{1}}]}\parallel}_{\mathrm{\infty}}.$

The third segment is then $[{\nu}_{3}^{*},{\gamma}_{5}^{*}]\times \{-{\nu}_{4}^{*}\}$.

*Fifth part of ${\mathrm{\Gamma}}_{\mathrm{1}}^{k}$.*
This is simply the segment $\{{\nu}_{3}^{*}\}\times [-{\nu}_{2}^{*},{\nu}_{4}^{*}]$.

*Fifth part of ${\mathrm{\Gamma}}_{\mathrm{2}}^{k}$.*
This is constructed exactly as the first part, starting with the segment $\{{\gamma}_{5}^{*}\}\times [-{\nu}_{4}^{*},{\nu}_{4}^{*}]$, and then continuing analogously.

After having completed the first lap, we can now proceed recursively, until the curves ${\mathrm{\Gamma}}_{1}^{k}$ and ${\mathrm{\Gamma}}_{2}^{k}$ have completed $M+1$ rotations around the origin.

Fix ${R}_{2}>0$ so that the curves ${\mathrm{\Gamma}}_{1}^{k}$ and ${\mathrm{\Gamma}}_{2}^{k}$ are contained in the ball centered at the origin, with radius ${R}_{2}$.
Choose $R\ge {R}_{2}$, and let $z(t)=(x(t),y(t))$ be a solution of (2.3) such that, for some
${t}_{0}\in \mathbb{R}$, one has that $|z({t}_{0})|=R$. We will analyze the behavior of $z(t)$ showing that its
orbit is controlled, and in some sense guided, by the curves ${\mathrm{\Gamma}}_{1}^{k}$ and ${\mathrm{\Gamma}}_{2}^{k}$. Indeed, the curve ${\mathrm{\Gamma}}_{1}^{k}$ keeps $z(t)$ from getting too close to the origin, while ${\mathrm{\Gamma}}_{2}^{k}$ provides some reference lines which must be crossed by the orbit of $z(t)$, forcing it to rotate around the origin. Moreover, the estimates given in Propositions 6 and 7, and their analogues in the other regions of the plane, show that the amplitudes of the orbit and the times needed by the orbit to cross the different regions of the plane are all controlled by some constants which can be chosen to depend only on *R*.

More precisely, let $z(t)$ be a solution with $|z({t}_{0})|=R\ge {R}_{2}$.
It is possible to determine the region where $z({t}_{0})$ is located with respect to the last lap of ${\mathrm{\Gamma}}_{2}^{k}$. Assume, for instance, that it is in the “northern region”, by which we mean that $\alpha \le x({t}_{0})\le \beta $ and $y({t}_{0})\ge \gamma $, where $\alpha =-\beta $ and γ are as shown in Figure 2.

Figure 2 The northern region.

Then, by the analogue of Proposition 6, there is a first time ${t}_{1}\ge {t}_{0}$ at which the orbit reaches a point $z({t}_{1})=(x({t}_{1}),y({t}_{1}))$, with $x({t}_{1})=\beta $, and

$\alpha -{D}_{k}\le x(t)\le \beta ,\mu \le y(t)\le {\kappa}_{1}(R)\mathit{\hspace{1em}\hspace{1em}}\text{for every}t\in [{t}_{0},{t}_{1}],$

where $\mu >0$ is determined by the inner curve ${\mathrm{\Gamma}}_{1}^{k}$, and

${\kappa}_{1}(R)=R+\left(\frac{2R+{D}_{k}}{\delta}+1\right)T{\parallel {{g}_{k}|}_{[0,T]\times [-R-{D}_{k},R]}\parallel}_{\mathrm{\infty}}.$

Moreover, by the analogue of Proposition 6, the time interval ${t}_{1}-{t}_{0}$ is controlled from above by a constant which may be chosen to depend only on *R*, since the starting point lies on a compact set.

Therefore, we have that $z({t}_{1})\in \{\beta \}\times [\mu ,{\kappa}_{1}(R)]$. The solution now enters the “north-eastern region” depicted in Figure 3 and, by Proposition 7, there is a first time ${t}_{2}\ge {t}_{1}$ at which the orbit reaches a point $z({t}_{2})=(x({t}_{2}),y({t}_{2}))$ with $y({t}_{2})=\mu $, and

$\nu \le x(t)\le {\kappa}_{2}(R),\mu \le y(t)\le {\kappa}_{1}(R)+{D}_{k}\mathit{\hspace{1em}\hspace{1em}}\text{for every}t\in [{t}_{1},{t}_{2}],$

where $\nu =\beta -{D}_{k}$ and

${\kappa}_{2}(R)={\kappa}_{1}(R)+\left(\frac{{\kappa}_{1}(R)+{D}_{k}}{\delta}+1\right)T{\parallel {{f}_{k}|}_{[0,T]\times [0,{\kappa}_{1}(R)+{D}_{k}]}\parallel}_{\mathrm{\infty}}.$

Figure 3 The north-eastern region.

By Proposition 7, the time interval ${t}_{2}-{t}_{1}$ is controlled from above by a constant which may be chosen to depend only on *R* since we started from a compact set.

Now the solution has arrived at $z({t}_{2})\in [\beta -{D}_{k},{\kappa}_{2}(R)]\times \{\mu \}$, and it enters the “eastern region”, where it behaves similarly to the northern region: we will find a first time ${t}_{3}\ge {t}_{2}$ at which the orbit reaches a point $z({t}_{3})=(x({t}_{3}),y({t}_{3}))$, with $y({t}_{3})=-\mu $, and

$\rho \le x(t)\le {\kappa}_{3}(R),-\mu \le y(t)\le \mu +{D}_{k}\mathit{\hspace{1em}\hspace{1em}}\text{for every}t\in [{t}_{2},{t}_{3}],$

where $\rho >0$ is determined by ${\mathrm{\Gamma}}_{1}^{k}$, and ${\kappa}_{3}(R)$ is a constant depending only on *R* (see Figure 4).
Again, the time interval ${t}_{3}-{t}_{2}$ is controlled from above by a constant which only depends on *R*.

Figure 4 The eastern region.

And this can be repeated on and on, until the solution has completed one rotation around the origin. Observe that, while crossing the different regions, the orbit of $z(t)$ is always “controlled from below” by the inner curve ${\mathrm{\Gamma}}_{1}^{k}$, which will guarantee that, during all the time needed to perform a complete rotation, the distance from the origin will remain greater than ${R}_{1}$.

Clearly enough, the same type of reasoning applies when $z({t}_{0})$, instead of being in the “northern region”, belongs to the “north-eastern region”. The estimates will still depend only on *R* by continuity and compactness. When $z({t}_{0})$ belongs to any of the other regions, the situation is perfectly symmetrical with the above, as can be seen by rotating Figure 1 by a multiple of 90 degrees.

After the solution has completed one rotation around the origin, it could have approached the origin, but not too much, due to the fact that it cannot intersect the curve ${\mathrm{\Gamma}}_{1}^{k}$. Hence, we can repeat the same argument, taking this time as reference regions those determined by the inner lap of ${\mathrm{\Gamma}}_{2}^{k}$, until the solution has completed the second rotation around the origin. And all this can be repeated until the solution has performed $M+1$ rotations around the origin, thus completing the proof.
∎

We have thus proved global existence in the future. Concerning the past, this can be obtained by reversing the time and arguing similarly.

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