Boundary Value Problem

Boundary Value Problem¶

2D Fourier Series

2D Fourier Series

2D Fourier Series¶

$p$ is a half of period

\begin{matrix} f (x, y) = \sum c_{m n} e^{j m x} e^{j n y} \\ c_{m n} = \frac{1}{2 p_{x}} \frac{1}{2 p_{y}} \int_{P y} \int_{P x} f (x, y) e^{- j m x} e^{- j n y} d x d y \end{matrix}

2D Fourier Sine Series¶

$p$ is a half of period

\begin{matrix} f (x, y) = \sum_{n, m \in ℕ} b_{m, n} \sin \frac{n π}{p_{x}} x \sin \frac{n π}{p_{y}} y \\ b_{m, n} = \frac{2}{p_{x}} \frac{2}{p_{y}} \int_{P_{y}} \int_{P_{x}} f (x, y) \sin \frac{n π}{p_{x}} x \sin \frac{n π}{p_{y}} y d x d y \end{matrix}

similar to 1D Fourier Sine Series

Possion's Equation¶

\begin{matrix} \nabla^{2} V = - \frac{ρ_{v}}{ε_{0}} \end{matrix}

we can solve Possion's equation by superposition which similar to how we solve non-homogeneous differential equation.

let $V = V_{L} + V_{P}$ , in which $V_{L}$ indicates the homogeneous differential equation case ( $ρ_{v} = 0$ ) i.e., Laplace's equation.

\begin{matrix} \nabla^{2} V_{L} = 0 \end{matrix}

as for the $V_{P}$ indicates the homogeneous boundary conditions, i.e., $\forall boundary V_{b} = 0$

\begin{matrix} \nabla^{2} V_{P} = - \frac{ρ_{v}}{ε_{0}} \end{matrix}

assume that

\begin{matrix} V_{p} = \sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} A_{m n} \sin \frac{m π x}{b} \sin \frac{n π y}{a} \\ ρ_{v} = \sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} C_{m n} \sin \frac{m π x}{p_{x}} \sin \frac{n π y}{p_{y}} \end{matrix}

notice that usually $C_{m n}$ is known, but the $A_{m n}$ is unknown

for $C_{m n}$

\begin{matrix} C_{m n} = \frac{2}{p_{x}} \frac{2}{p_{y}} \int_{X} \int_{Y} ρ_{v} \sin \frac{m π x}{p_{x}} \sin \frac{n π y}{p_{y}} d x d y \end{matrix}

\begin{array}{r} \nabla^{2} V_{P} = \sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} A_{m n} [{(\frac{m π}{p_{x}})}^{2} + {(\frac{n π}{p_{y}})}^{2}] \sin \frac{m π x}{b} \sin \frac{n π y}{a} \end{array}

\begin{matrix} ⟹ A_{m n} \end{matrix}

3D Laplace's Equation¶

Cartesian¶

\begin{matrix} V (x, y, z) = X (x) Y (y) Z (z) \end{matrix}

\begin{matrix} \nabla^{2} V = 0 \\ \frac{X^{″}}{X} + \frac{Y^{″}}{Y} + \frac{Z^{″}}{Z} = - k_{x}^{2} - k_{y}^{2} - k_{z}^{2} = 0 \end{matrix}

⟹ {\begin{matrix} X^{″} + k_{x}^{2} X & = 0 \\ Y^{″} + k_{y}^{2} Y & = 0 \\ Z^{″} + k_{z}^{2} Z & = 0 \\ k_{x}^{2} + k_{y}^{2} + k_{z}^{2} & = 0 \end{matrix}

Cylindrical¶

\begin{matrix} V (ρ, ϕ, z) = P (ρ) Φ (ϕ) Z (z) \\ \nabla^{2} V = \frac{1}{ρ} \frac{\partial}{\partial ρ} (ρ V_{ρ}) + \frac{1}{ρ^{2}} V_{ϕ ϕ} + V_{z z} = 0 \\ \frac{\nabla^{2} V}{P Φ Z} = \frac{1}{ρ} (ρ \frac{P^{″}}{P} + \frac{P^{'}}{P}) + \frac{1}{ρ^{2}} \frac{Φ^{″}}{Φ} + \frac{Z^{″}}{Z} = 0 \\ ⟹ \frac{1}{ρ} (ρ \frac{P^{″}}{P} + \frac{P^{'}}{P}) + \frac{1}{ρ^{2}} \frac{Φ^{″}}{Φ} = - \frac{Z^{″}}{Z} \equiv - λ^{2} \\ ρ^{2} \frac{P^{″}}{P} + ρ \frac{P^{'}}{P} + ρ^{2} λ^{2} = - \frac{Φ^{″}}{Φ} \equiv μ^{2} \end{matrix}

$$ \implies \left{ \begin{gather} Z''-\lambda^{2}Z=0 \\ \Phi'' + \mu^{2}\Phi = 0 \\ \rho^{2}P'' + \rho P' + \left(\rho^{2}\lambda - \mu^{2}\right)P = 0 & #Eq3 \end{gather}\right. $$

in which Eq. 3 is a Bessel Function

notice that if $λ = 0$ (e.g. the situation that $V$ is independent of $z$ ), Eq. 3 would be an Euler-Cauchy Equation, rather than Bessel.

for $λ > 0$ , let $x = ρ λ$ and $y = P$

note that

\begin{matrix} d x = d (ρ λ) = λ d ρ \\ y^{'} = \frac{d y}{d x} = \frac{d P}{d x} = \frac{1}{λ} \frac{d P}{d ρ} \end{matrix}

Eq.3 become

\begin{matrix} x^{2} y^{″} + x y + (x^{2} - μ^{2}) y = 0 \end{matrix}