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Consider

$$F(\theta)=\sin \theta \int_{-L}^{+L}h(z)e^{-ikz\cos \theta} \,dz$$

$$|z|\le L$$ $$0 \le \theta \le \pi$$


By having knowledge of $F(\theta)$, how can one approximate $h(z)$? In addition, I know that $F$ is differentiable with respect to $\theta$.

How can I model and solve such problem with unknown function $h(z)$ in Mathematica?

There is an idea in which we write Fourier series for both $F$ and $h$ then by solving or manipulation of the integral we approximate the coefficients.

** $k$ is the wave number, don't confuse it with anything else.
** $z$ might be a complex number, but in this case you can consider it as real.
**Any idea on this problem would be extremely appreciated.

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    $\begingroup$ its an inverse pb. call x= k cos(theta); F(theta)/sin(theta)= G(x), Z=2Pi z/L and you get something related to G(x) = fourier transform of h(Z) ? $\endgroup$ – chris Jan 3 '15 at 9:47
  • $\begingroup$ Might be able to approximate it as an integral equation. There are several posts on this sort of thing scattered in MSE and the Mathematica section of StackOverflow. $\endgroup$ – Daniel Lichtblau Jan 3 '15 at 21:00
  • $\begingroup$ Could you please give me sources or at least give some hints to approximate it using fourier series? $\endgroup$ – FreeMind Jan 3 '15 at 22:02
  • $\begingroup$ @chris there is a complex contour integral for it, but i wanna approximate fourier coefficients. $\endgroup$ – FreeMind Jan 3 '15 at 22:03
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    $\begingroup$ To me, when you say "By having knowledge of $F(θ)$...", you imply you do know $F$. If both $h$ and $F$ are unknown, my thought is that the problem is underspecified and cannot be solved. You can write $F$ in terms of $h$ or $h$ in terms $F$, but I doubt one can solve one integral equation for both functions. $\endgroup$ – Michael E2 Jan 10 '15 at 3:50
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Well, I'll take a crack at it, although I think chris identified the nub of the problem in the first comment. We need to tighten up the language from the comments. First, write $$F(\theta)=F(\theta,k)=\sin \theta \int_{-L}^{+L}h(z)\,e^{-ikz\cos \theta} \,dz=\sin \theta \int_{-\infty}^{+\infty}\tilde{h}(z)\,e^{-ikz\cos \theta} \,dz$$ where $$\tilde{h}(z)=\cases{h(z) & $-L\le z \le L$ \cr 0 & otherwise \cr}$$ Technically, then, substituting $w = k\cos\theta$, $F(\theta,k)\,/\sin \theta$ is given by the inverse Fourier transform of $\tilde{h}$, $${F(\theta,\,w\,/\cos\theta) \over \sin\theta} = \int_{-\infty}^{+\infty}\tilde{h}(z)\,e^{-iwz} \,dz$$

The OP indicates that $k$ is the wave number, which suggests a discrete spectrum. But given that the length $2L$ is arbitrary and not a multiple of the period of the exponential factor, I think we have to fall back on the continuous integral transform. They're related anyway, so it's should be no great concern.

Consequently we have $$\tilde{h}(z) = {1 \over 2\pi} \int_{-\infty}^{+\infty}{F(\theta,\,w\,/\cos\theta)\over\sin\theta}\,e^{iwz} \,dz$$ or, in Mathematica,

FourierTransform[1/Sqrt[2 Pi] F[t, w/Cos[t]] / Sin[t], w, z]

where the factor 1/Sqrt[2 Pi] is needed to balance the default coefficients of the transforms in Mathematica.

Example: Take h[z_] = z, L = 3.

F[t_, k_] = Sin[t] Integrate[z Exp[-I k z Cos[t]], {z, -3, 3}]
(*  (2 I (3 k Cos[3 k Cos[t]] - Sec[t] Sin[3 k Cos[t]]) Tan[t]) / k^2  *)

Recover h:

Assuming[0 < t < Pi && z ∈ Reals,
 FourierTransform[1/Sqrt[2 Pi] F[t, w/Cos[t]]/Sin[t], w, z] // 
   Simplify // PiecewiseExpand
 ]

Mathematica graphics

Of course, we get the midpoints at the discontinuities, which agrees with the theory of Fourier integrals.

Of course, one can do the integral with NIntegrate (or Integrate) when appropriate.

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  • $\begingroup$ Bravo for your attempt and especially for teasing out the information that OP actually does not have $F(\theta)$ given, although just to note that this is basically the same approach shown in the textbook that all of OP's questions (image links: 1 2 3) are paraphrasing. It's still very unclear to me exactly what OP expects here. $\endgroup$ – Oleksandr R. Jan 10 '15 at 15:27
  • $\begingroup$ @OleksandrR. Thanks. It's definitely a textbook approach. After looking at image 1, I doubt the answer will be helpful, except maybe to convince OP to read the textbook again. Perhaps the concrete example will help. I've certainly gotten confused learning new stuff before and had to reread the book several times to sort it out. It occurs to me that some feel the reputed power of M allows it to solve an arbitrary generic equation and tell you the answer is 42 or something. $\endgroup$ – Michael E2 Jan 10 '15 at 16:08
  • $\begingroup$ @MichaelE2 Thanks for you attempt, but I have pointed out this approach as classical one in another topic, what I am looking for is stated in the question, writing fourier series for both sides, for $F$ and $h$ then calculating the coefficients.If I had to take that classical approach, I would never ask this question here, because it is stated in a better way in the text book! Anyway, Plus one. $\endgroup$ – FreeMind Jan 10 '15 at 18:58
  • $\begingroup$ In addition, in the inverse fourier transform, you should write $e^{iwz}$ $\endgroup$ – FreeMind Jan 10 '15 at 19:01
  • $\begingroup$ @FreeMind Thanks for pointing out the typo. The way it's written, the primary, or at least the first, question is how to get h, which turns out to be in the book. The "There is an idea" part is obscure, and only now am I thinking you might be thinking of Fourier coefficients w.r.t θ. Since the Fourier integral defining F is a transform between the z and k domains, this seemed a stretch, probably to everybody -- even more so, since θ is restricted from 0 to π, instead of to 2π, which is what is needed for the Fourier coefficients w.r.t θ, even if the restriction is mathematically unnecessary. $\endgroup$ – Michael E2 Jan 10 '15 at 20:07

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