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

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  • $\begingroup$ 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... $\endgroup$ May 28, 2014 at 14:39
  • $\begingroup$ @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 $b_0 = 0$). $\endgroup$
    – Amateur
    May 28, 2014 at 14:47
  • $\begingroup$ Indeed, that is what should be done, I suppose they do it because they can? This is not a very satisfying answer. Sorry. $\endgroup$ May 28, 2014 at 14:57
  • $\begingroup$ 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? $\endgroup$
    – Did
    May 31, 2014 at 8:15

2 Answers 2

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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 $\pm m$.

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  • $\begingroup$ 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 ? $\endgroup$
    – Amateur
    May 28, 2014 at 17:59
  • $\begingroup$ 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. $\endgroup$ May 28, 2014 at 18:03
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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 $\lambda\leqslant1$.

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