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I'm currently working on maxwell equations and in order to lower the fields dimension, we perform a Fourier decomposition (according to $\theta$) due to the system symmetry. For any vector field $\mathbf{U}(r, \theta, z)$, we have, $$ \mathbf{U}(r,\theta,z) = \sum_\alpha \tilde{\mathbf{U}}^\alpha(t,z)e^{i\alpha \theta}.$$ Together with the usual Maxwell equations (in cylindrical coordinate system) we decide to work with metallic boundary conditions. Let $\Omega$ be our domain and $\Gamma$ its boundary, then, for the electric field, $$ \mathbf{E} \times \mathbf{n} = 0 \quad \text{on $\Gamma$}$$ Where $\mathbf{n}$ is the normal vector oriented outside $\Gamma$.

Then, my question is how to get the boundary conditions verified by $\mathbf{\tilde{E}}$ instead of $\mathbf{E}$ ? Does it lead to $\mathbf{\tilde{E}} \times \mathbf{n} = 0$ ? An additional question would be : what about a non homegenous conditions such as $\mathbf{E} \times \mathbf{n} = f$ ?

I know that's pretty dumb but I can't convice myself. Thanks in advance.

Amzocks
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    Presumably you mean $\tilde{U}_\alpha (t,z)$? Otherwise it just pulls out of the sum. You just plug in whatever boundary conditions you need and do the Fourier inversion to get the boundary conditions on the Fourier components. – Michael Jul 16 '13 at 07:17
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    Exact Duplicate by the OP on Math.SE: http://math.stackexchange.com/questions/444268/maxwell-equations-and-fourier-decomposition. – Abhimanyu Pallavi Sudhir Jul 16 '13 at 07:28
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    This question appears to be off-topic because it is about mathematics and has already been posted on math.SE – Michael Jul 16 '13 at 07:42
  • Michael Brown you're right about the indice $\alpha$ but concerning math.SE someone told me to post here, then ... – Amzocks Jul 16 '13 at 07:47
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    Yeah well, it's still a math question. But since you got an acceptable answer here no harm no foul... – Michael Jul 16 '13 at 11:30
  • I'm inclined to let this stay - it is not purely math. @Amzocks For the record (and I know no one told you this) we discourage cross-posting. A better alternative if you learn the question is better at another site is to flag it and request that the moderators migrate it to that site. –  Jul 16 '13 at 16:06

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Your boundary condition (I assume you mean that $\Gamma$ is a cylinder centred on $r=0$ in cylindrical co-ordinates and with radius $r_\Gamma$ as you imply that the problem has circular symmetry) is:

$\mathbf{U}(r_\Gamma,\theta,z) \wedge \mathbf{n} = \sum\limits_{\alpha = -\infty}^\infty \tilde{\mathbf{U}}_\alpha(r_\Gamma, z) \wedge \mathbf{n}\, e^{I\,\alpha\,\theta} = \mathbf{0};\;\forall \theta, z \in \mathbb{R}$

The next step comes from the $\mathbf{L}^2$-completeness of the set of functions $\left\{f_\alpha : [0,\,2\pi)\rightarrow [0,\,2\pi),\, \alpha\in\mathbb{Z}: f_\alpha(\theta)=e^{I\,\alpha\,\theta}\right\}$ on the interval $[0, 2\pi)$ and their known linear independence. You simply take this as a proven mathematical fact: i.e. the only way a Fourier series can be identically nought is if all of its co-efficients are nought. Hence you conclude $\tilde{\mathbf{U}}_\alpha(r_\Gamma, z) \wedge \mathbf{n} = \mathbf{0}$ for each and every $\alpha \in \mathbb{Z}$.

Selene Routley
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  • Thanks for your input ! But I was wondering, Cantor proved the uniqueness of trigonometric representation Cantor's proof, but did he use the $\mathbb{L}^2$-completeness to do it ? I refer to the atatched link – Amzocks Jul 16 '13 at 09:31
  • @Amzocks You are quite right - there are many proofs of this uniqueness, many making slightly different assumptions. The $\mathbf{L}^2$ completeness framework is simply an "intuitive" one, because it frames the problem in a Hilbert space wherein thinks of series as vectors and of inner products between them. I'm just saying that this is the result that will get you where you need to go. – Selene Routley Jul 16 '13 at 10:15