# Fourier series problem

• Apr 29th 2010, 04:14 AM
sitho
Fourier series problem
Hi! I wish to find Fourier coefficients $\displaystyle a_k$ satisfying

$\displaystyle \sum_{k}{a_k e^{i k x}} = 0~\forall x \notin \mathbb{P}$ (1)

$\displaystyle k = \frac{2 \pi m}{X},~m \in \mathbb{Z}$ (2)

$\displaystyle a_k = b_k c_k + d_k$ (3)

$\displaystyle \sum_{k}{b_k e^{i k x}} = 0~\forall x \in \mathbb{P}$ (4)

$\displaystyle \mathbb{P} \equiv \lbrace x : \exists n \in \mathbb{Z} : 0 \leq x + n X \leq L_x \rbrace,~L_x < X$ (5)

Here, $\displaystyle a_k$ and $\displaystyle b_k$ are unknown coefficients, $\displaystyle c_k$ and $\displaystyle d_k$ are known coefficients, and $\displaystyle \mathbb{P}$ is a periodic set in $\displaystyle x$ with periodicity $\displaystyle X$, and is defined by eq. (5). The sum in eq. (1) and (4) runs over integers $\displaystyle m$ according to eq. (2). My actual problem is two dimensional. Hence, the sums are over $\displaystyle k_x$ and $\displaystyle k_y$, and the set $\displaystyle \mathbb{P}$ is defined in two dimensions. I don't know if this changes the problem very much. I'm thankful for any help at all. (Blush)
• Apr 30th 2010, 01:34 PM
CaptainBlack
Quote:

Originally Posted by sitho
Hi! I wish to find Fourier coefficients $\displaystyle a_k$ satisfying

$\displaystyle \sum_{k}{a_k e^{i k x}} = 0~\forall x \notin \mathbb{P}$ (1)

$\displaystyle k = \frac{2 \pi m}{X},~m \in \mathbb{Z}$ (2)

$\displaystyle a_k = b_k c_k + d_k$ (3)

$\displaystyle \sum_{k}{b_k e^{i k x}} = 0~\forall x \in \mathbb{P}$ (4)

$\displaystyle \mathbb{P} \equiv \lbrace x : \exists n \in \mathbb{Z} : 0 \leq x + n X \leq L_x \rbrace,~L_x < X$ (5)

Here, $\displaystyle a_k$ and $\displaystyle b_k$ are unknown coefficients, $\displaystyle c_k$ and $\displaystyle d_k$ are known coefficients, and $\displaystyle \mathbb{P}$ is a periodic set in $\displaystyle x$ with periodicity $\displaystyle X$, and is defined by eq. (5). The sum in eq. (1) and (4) runs over integers $\displaystyle m$ according to eq. (2). My actual problem is two dimensional. Hence, the sums are over $\displaystyle k_x$ and $\displaystyle k_y$, and the set $\displaystyle \mathbb{P}$ is defined in two dimensions. I don't know if this changes the problem very much. I'm thankful for any help at all. (Blush)

This might be easier to understand if you posted the original problem.

CB
• May 1st 2010, 03:43 PM
sitho
So the original two dimensional problem is:

$\displaystyle \sum_{\boldsymbol{k}}{a_{\boldsymbol{k}} e^{i \boldsymbol{k}\cdot \boldsymbol{x}}} = 0~\forall \boldsymbol{x} \notin \mathbb{P}$ (1)

$\displaystyle \boldsymbol{k} = \left(\frac{2 \pi m_x}{X}, \frac{2 \pi m_y}{Y}\right),~m_x, m_y \in \mathbb{Z}$ (2)

$\displaystyle a_{\boldsymbol{k}} = b_{\boldsymbol{k}} c_{\boldsymbol{k}} + d_{\boldsymbol{k}}$ (3)

$\displaystyle \sum_{\boldsymbol{k}}{b_{\boldsymbol{k}} e^{i \boldsymbol{k}\cdot \boldsymbol{x}}} = 0~\forall \boldsymbol{x} \in \mathbb{P}$ (4)

$\displaystyle \mathbb{P} \equiv \lbrace \boldsymbol{x} : \exists n_x, n_y \in \mathbb{Z} : 0 \leq x + n_x X \leq L_x, 0 \leq y + n_y Y \leq L_y \rbrace,~L_x < X, L_y < Y$ (5)

The explicit expressions for $\displaystyle c_{\boldsymbol{k}}$ and $\displaystyle d_{\boldsymbol{k}}$ are given by:

$\displaystyle c_{\boldsymbol{k}} \equiv {-\frac{\cosh(\kappa[x_{pr} - x_{sr}])}{\kappa\cosh(\kappa[x_{pr} - x_c])\sinh(\kappa[x_c - x_{sr}])}}$

$\displaystyle d_{\boldsymbol{k}} \equiv b_{z 0}(\boldsymbol{k})\frac{e^{-i k_x x_{pr}} + R e^{i k_x x_{pr}}}{\cosh(\kappa[x_{pr} - x_c])} - {}$
$\displaystyle {} - \mu_0 j_{a, y}(\boldsymbol{k})\bigg[\theta(x_c - x_a)\frac{\cosh(\kappa[x_{pr} - x_a])}{\cosh(\kappa[x_{pr} - x_c])} - \theta(x_a - x_c)\frac{\sinh(\kappa[x_a - x_{sr}])}{\sinh(\kappa[x_c - x_{sr}])}\bigg]$

$\displaystyle \kappa = \sqrt{k_x^2 + k_y^2 - \left(\frac{\omega}{c}\right)^2}$
$\displaystyle b_{z 0}(\boldsymbol{k}) = \frac{s(\boldsymbol{k})}{(b_1(\boldsymbol{k}) + i k_x)e^{-i k_x x_{pr}} + (b_1(\boldsymbol{k}) - i k_x)e^{i k_x x_{pr}}}$
$\displaystyle s(\boldsymbol{k}) = \mu_0 \kappa \frac{j_{a, y}(\boldsymbol{k})\sinh(\kappa[x_a - x_{sr}])}{\cosh(\kappa[x_{pr} - x_{sr}])}$
$\displaystyle b_1(\boldsymbol{k}) = \kappa \tanh(\kappa[x_{pr} - x_{sr}])$
$\displaystyle j_{a, y}(\boldsymbol{k}) = \left\lbrace \begin{array}{lcc} -I_a\frac{e^{-i k_x L_x} - 1}{2 \pi R k_x}\frac{e^{-i k_y L_y} - 1}{2 \pi r k_y} & : & k_x \neq 0,~k_y \neq 0 \\ I_a \frac{i L_x}{2 \pi R}\frac{e^{-i k_y L_y} - 1}{2 \pi r k_y} & : & k_x = 0,~k_y \neq 0 \end{array}\right.$
$\displaystyle j_{a, y}(\boldsymbol{k}) = \left\lbrace \begin{array}{lcc} I_a \frac{e^{-i k_x L_x} - 1}{2 \pi R k_x}\frac{i L_y}{2 \pi r} & : & k_x \neq 0,~k_y = 0 \\ I_a \frac{L_x}{2 \pi R}\frac{L_y}{2 \pi r} & : & k_x = 0,~k_y = 0 \end{array}\right.$

Every single quantity in the expressions for $\displaystyle c_{\boldsymbol{k}}$ and $\displaystyle d_{\boldsymbol{k}}$ are given. $\displaystyle \theta(x)$ is the Heaviside step function. I don't know if these expressions make life much easier. Just to clearify, eqs (1) and (4) just tell that the functions that the Fourier coefficients describe (we can call them $\displaystyle a(\boldsymbol{x})$ and $\displaystyle b(\boldsymbol{x})$) are determined to be zero outside and inside the region defined by eq. (5) respectively.

Actually I've found at least one way to solve the problem, and that is to discretize $\displaystyle \boldsymbol{x}$ and transform $\displaystyle \boldsymbol{x}$ and $\displaystyle \boldsymbol{k}$ into one dimensional vectors. This turns all the equations into linear matrix equations that can be solved. However, this only gives an approximate numerical solution that depends on the choice of the grid size in $\displaystyle x$ and $\displaystyle y$. An analytical solution for $\displaystyle a_{\boldsymbol{k}}$ as a function of $\displaystyle c_{\boldsymbol{k}}$ and $\displaystyle d_{\boldsymbol{k}}$ would have been a bit nicer. =P