I have a set of equations involving two anti periodic functions $f(t+T)=-f(t)$ and $g(t+T)=-g(t)$

\begin{cases} f(t)&=J^2g(t)^3 \\ \mathcal{F}[g](\omega_n)&=\frac{1}{-i\omega_n-\mathcal{F}[f](\omega_n)} \end{cases}

Where $\mathcal{F}[f](\omega_n)$ is the Fourier transform of the function $f$ and $\omega_n =\frac{\pi(2n-1)}{T}$. I would like to numerically solve these equations to obtain $f(t)$ and $g(t)$ but I'm not sure what function in Mathematica to use. The Fourier series can be recast as an integral, which is why I thought DSolve or NDSolve would work. However, none of these worked.

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    $\begingroup$ Publish equations and code. $\endgroup$ Commented May 20, 2019 at 19:50
  • $\begingroup$ What do you mean by "I'm not sure what function to use for this type of equation"? What are you actually trying to do? You've given expressions for the function and its Fourier transform, but no differential equations, so I'm not sure why NDSolve or DSolve would be relevant. Please provide more details by editing your post and including all necessary relevant information. $\endgroup$
    – march
    Commented May 20, 2019 at 21:12
  • $\begingroup$ Try to set up the equations yourself, by this I mean 1. Try an attempt at the code to see what you can do with it and 2. Use your mathematics knowledge and apply it to your budding Mathematica skills! And I say all of this because 3. I do this in my research and it’s just a couple lines here, a dot product there, and you’re off to the races! You list an expansion basis, no? If so, expand across the basis! Then see what your minimum number is needed for an appropriate approximation, by this I mean your minimum number of summations needed. I might look more into how Fourier series actually 1/2 $\endgroup$ Commented May 21, 2019 at 13:07
  • $\begingroup$ 2/2 work and then try to replicate this directly in Mathematica. Make it rough and ugly looking, your code that is! Then clean it up and see how it goes :) be sure to share it here so we can help if you need! Good luck, I’m excited to see what you get to :)) $\endgroup$ Commented May 21, 2019 at 13:08


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