# 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 $$(a * a)_k := \sum_{\substack{k_1 + k_2 = k \\ k_1, k_2 \geq 0}} a_{k_1} a_{k_2} ?$$

• could it be the negative coefficients are taken to be zero? Or, by odd,even properties of sine and cosine there is a necessary dependence between $k$ and $-k$? I'm not sure I know the culture from which your question stems to give the best answer here... May 28, 2014 at 14:39
• @JamesS.Cook Negative coefficients don't seem to be taken to be zero. However we do have the dependences $a_{-k} = a_k$ and $b_{-k} = - b_k$ (which implies $b_0 = 0$). May 28, 2014 at 14:47
• Indeed, that is what should be done, I suppose they do it because they can? This is not a very satisfying answer. Sorry. May 28, 2014 at 14:57
• You are told to expand $x(t)$ and $y(t)$ as Fourier series to find a periodic solution with period $2\pi$. How does this help?
– Did
May 31, 2014 at 8:15

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 $\pm m$.
• When you say we do this so that the coefficients of the square result from the "convolution" of the coefficient sequences, do you mean that $\sum_{k \geq 0} a_k \cos(kt) \cdot \sum_{k \geq 0} a_k \cos(kt) \neq \sum_{k \geq 0} (a * a)_k \cos(kt)$ or by "convolution" you imply we want the negative coefficients too ? May 28, 2014 at 17:59
• Yes, no. They are not equal since the product of $a_k\cos(kt)a_{n-k}\cos((n-k)t)$ contains both $\cos(nt)$, as expected of convolution, as well as $\cos(|n-2k|\,t)$, which is not compatible with the convolution product. May 28, 2014 at 18:03
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 $\lambda\leqslant1$.