In search of periodic solutions of a system of ODEs by means of Fourier series Consider the following non-linear system of ODEs :
\begin{cases}
x' = y \\
y' = x^2-\lambda x.
\end{cases}
In search of a solution such that $y(0) = y(2 \pi) = 0$, I am being told to seek $x$ and $y$ as
$$
x(t) = \sum_{k \in \mathbb{Z}} a_k \cos(kt) \\
y(t) = \sum_{k \in \mathbb{Z}} b_k \sin(kt).
$$
Question : Why do we let $k \in \mathbb{Z}$ ? As far as I know, for real periodic functions, Fourier series have coefficients $(a_k)_{k \geq 0}$ and $(b_k)_{k \geq 1}$.
Is it because, for example,
$$
\sum_{k \in \mathbb{Z}} a_k \cos(kt) \cdot \sum_{k \in \mathbb{Z}} a_k \cos(kt) = \sum_{k \in \mathbb{Z}} (a \star a)_k \cos(kt)
$$
where
$$
(a \star a)_k := \sum_{k_1 + k_2 = k} a_{k_1} a_{k_2}
$$
but
$$
\sum_{k \geq 0} a_k \cos(kt) \cdot \sum_{k \geq 0} a_k \cos(kt) \neq \sum_{k \geq 0} (a * a)_k \cos(kt)
$$
where
$$
(a * a)_k := \sum_{\substack{k_1 + k_2 = k \\ k_1, k_2 \geq 0}} a_{k_1} a_{k_2} ?
$$
 A: This is done so that the coefficients of the square result from the convolution of the coefficient sequences. An easier procedure would be to use the complex Fourier series with reality conditions on the coefficients.
The basis for these games is
$$
2\cos kx\cos mx
=\cos(k+m)x+\cos(k-m)x
$$
where you combine and separate all the 4 cross terms involving $\pm k$ and $\pm m$.
A: To find periodic solutions, consider
$$
H_\lambda (x)=\lambda x^2-\frac23x^3,
$$
and note that 
$$
V_\lambda (x,y)=y^2+H_\lambda (x),
$$ is constant on every solution $t\mapsto(x(t),y(t))$ since, for every solution,
$$
\frac{\mathrm d}{\mathrm dt}V_\lambda (x(t),y(t))=2y(t)y'(t)+H'_\lambda (x(t))x'(t)=0.
$$ 
Let $v_\lambda =\frac13\lambda^3$. For every $v$ in $(0,v_\lambda )$, the function 
$$v-H_\lambda :x\mapsto v-H_\lambda (x)$$ is increasing on $(-\infty,0)$ from $-\infty$ to $v$, decreasing on $(0,\lambda)$ from $v$ to $v-v_\lambda $ and increasing on $(\lambda,+\infty)$ from $v-v_\lambda $ to $+\infty$, hence the function $v-H_\lambda $ has three roots $(-a_\lambda (v),b_\lambda (v),c_\lambda (v))$ with $-a_\lambda (v)$ in $(-\infty,0)$, $b_\lambda (v)$ in $(0,\lambda)$ and $c_\lambda (v)$ in $(\lambda,+\infty)$ and 
$$
v-H_\lambda (x)=\frac23(x+a_\lambda (v))(x-b_\lambda (v))(x-c_\lambda (v)).
$$ 
If a solution is in the upper halfplane $y\geqslant0$ for every time in $(0,t)$, one has
$$
t=\int_{x(0)}^{x(t)}\frac{\mathrm dx}{\sqrt{v-H_\lambda (x)}}.
$$
In particular, solutions starting from $(x,y)=(-a_\lambda (v),0)$ reach $(x,y)=(b_\lambda (v),0)$ through the upper halfplane, passing by the point $(0,\sqrt{v})$, then go back to $(-a_\lambda (v),0)$ using the symmetrical path in the lower halfplane $y\leqslant0$, passing by the point $(0,-\sqrt{v})$, hence their period $T_\lambda (v)$ is
$$
T_\lambda (v)=2\int_{-a_\lambda (v)}^{b_\lambda (v)}\frac{\mathrm dx}{\sqrt{v-H_\lambda (x)}}.
$$
The RHS is a function of $v$ hence, to get a solution with period $2\pi$, one should solve for $v$ the implicit equation 
$$
T_\lambda (v)=2\pi.
$$
Note that
$$
T_\lambda (v)=\int_0^1\frac{\sqrt6\,\mathrm dx}{\sqrt{x(1-x)(c_\lambda (v)+a_\lambda (v)-(b_\lambda (v)+a_\lambda (v))x)}}.
$$
Back-of-the-envelope computations seem to indicate that $T_\lambda :v\mapsto T_\lambda (v)$ is increasing on $(0,v_\lambda )$ with 
$$
\lim\limits_{v\to0}T_\lambda (v)=2\pi/\sqrt{\lambda},\qquad\lim\limits_{v\to v_\lambda }T_\lambda (v)=+\infty.$$
If this is true, a solution with period $2\pi$ would exist for every $\lambda\gt1$ but not for $\lambda\leqslant1$.
