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Math Help - Fourier series problem

  1. #1
    Junior Member
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    Fourier series problem

    Hi! I wish to find Fourier coefficients a_k satisfying

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

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

    a_k = b_k c_k + d_k (3)

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

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

    Here, a_k and b_k are unknown coefficients, c_k and d_k are known coefficients, and \mathbb{P} is a periodic set in x with periodicity X, and is defined by eq. (5). The sum in eq. (1) and (4) runs over integers m according to eq. (2). My actual problem is two dimensional. Hence, the sums are over k_x and k_y, and the set \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.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by sitho View Post
    Hi! I wish to find Fourier coefficients a_k satisfying

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

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

    a_k = b_k c_k + d_k (3)

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

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

    Here, a_k and b_k are unknown coefficients, c_k and d_k are known coefficients, and \mathbb{P} is a periodic set in x with periodicity X, and is defined by eq. (5). The sum in eq. (1) and (4) runs over integers m according to eq. (2). My actual problem is two dimensional. Hence, the sums are over k_x and k_y, and the set \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.
    This might be easier to understand if you posted the original problem.

    CB
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  3. #3
    Junior Member
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    So the original two dimensional problem is:

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

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

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

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

    \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 c_{\boldsymbol{k}} and d_{\boldsymbol{k}} are given by:

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

    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])} - {}
     {} - \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]

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

    Every single quantity in the expressions for c_{\boldsymbol{k}} and d_{\boldsymbol{k}} are given. \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 a(\boldsymbol{x}) and 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 \boldsymbol{x} and transform \boldsymbol{x} and \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 x and y. An analytical solution for a_{\boldsymbol{k}} as a function of c_{\boldsymbol{k}} and d_{\boldsymbol{k}} would have been a bit nicer. =P
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