Boundary Value Problem
Boundary Value Problem
2D Fourier Series
2D Fourier Series
- \(p\) is a half of period
\[
\begin{gather}
f(x,y) = \sum{c_{mn}e^{jmx}e^{jny}}
\\\\
c_{mn} = \frac{1}{2p_x}\frac{1}{2p_y}\int_{Py}\int_{Px}{f(x, y)e^{-jmx}e^{-jny}\ dxdy}
\end{gather}
\]
2D Fourier Sine Series
- \(p\) is a half of period
\[
\begin{gather}
f(x,y) = \sum_{n, m \in ℕ}{b_{m, n}\sin{\frac{n\pi}{p_x} x}\,\sin{\frac{n\pi}{p_y} y}}
\\\\
b_{m, n} = \frac{2}{p_x}\frac{2}{p_y}\int_{P_y}\int_{P_x}{f(x,y)\sin{\frac{n\pi}{p_x}x}\,\sin{\frac{n\pi}{p_y}y}\,dxdy}
\end{gather}
\]
Possion's Equation
\[
\begin{gather}
\nabla^{2}V=-\frac{\rho_v}{\varepsilon_0}
\end{gather}
\]
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 (\(\rho_v = 0\)) i.e., Laplace's equation.
\[
\begin{gather}
\nabla^{2} V_L =0
\end{gather}
\]
as for the \(V_P\) indicates the homogeneous boundary conditions, i.e., \(\forall \,\text{boundary }V_b = 0\)
\[
\begin{gather}
\nabla^{2}V_P=-\frac{\rho_v}{\varepsilon_0}
\end{gather}
\]
assume that
\[
\begin{gather}
V_p = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}{A_{mn}\sin{\frac{m\pi x}{b}}\sin{\frac{n\pi y}{a}}}
\\\\
\rho_v = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}{C_{mn}\sin{\frac{m\pi x}{p_x}}\sin{\frac{n\pi y}{p_y}}}
\end{gather}
\]
notice that usually \(C_{mn}\) is known, but the \(A_{mn}\) is unknown
for \(C_{mn}\)
\[
\begin{gather}
C_{mn} = \frac{2}{p_x}\frac{2}{p_y}\int_X\int_Y {\rho_v\sin\frac{m\pi x}{p_x}\sin\frac{n\pi y}{p_y}dxdy}
\end{gather}
\]
\[
\begin{align}
\nabla^{2}V_P=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}{A_{mn}\left[\left(\frac{m\pi}{p_x}\right)^{2}+\left(\frac{n\pi}{p_y}\right)^{2}\right]\sin{\frac{m\pi x}{b}}\sin{\frac{n\pi y}{a}}}
\end{align}
\]
\[
\begin{gather}
\implies A_{mn}
\end{gather}
\]
3D Laplace's Equation
Cartesian
\[
\begin{gather}
V(x, y, z) = X(x) Y(y) Z(z)
\\\\
\end{gather}
\]
\[
\begin{gather}
\nabla^{2}V = 0
\\\\
\frac{X''}{X}+\frac{Y''}{Y}+\frac{Z''}{Z} =-k_x^{2}-k_y^{2}-k_z^{2} = 0
\end{gather}
\]
\[
\implies \left\{
\begin{gather}
X''+k_x^{2}X &= 0
\\\\
Y''+k_y^{2}Y &= 0
\\\\
Z''+k_z^{2}Z &= 0
\\\\
k_{x}^{2}+k_y^{2}+k_z^{2} &= 0
\end{gather}\right.
\]
Cylindrical
\[
\begin{gather}
V(\rho, \phi, z) = P(\rho)\Phi(\phi)Z(z)
\\\\
\nabla^{2}V=\frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho V_\rho\right)+\frac{1}{\rho^{2}}V_{\phi\phi}+V_{zz} = 0
\\\\
\frac{\nabla^{2}V}{P\Phi Z}=\frac{1}{\rho}\left(\rho \frac{P''}{P}+\frac{P'}{P}\right)+\frac{1}{\rho^{2}}\frac{\Phi''}{\Phi}+\frac{Z''}{Z} =0
\\\\
\implies
\frac{1}{\rho}\left(\rho \frac{P''}{P}+\frac{P'}{P}\right)+\frac{1}{\rho^{2}}\frac{\Phi''}{\Phi}=-\frac{Z''}{Z} \equiv \mathbf{-\lambda^{2}}
\\\\
\rho^{2} \frac{P''}{P}+\rho\frac{P'}{P}+\rho^{2}\lambda^{2}=-\frac{\Phi''}{\Phi} \equiv \mu^{2}
\\\\
\end{gather}
\]
$$
\implies \left{
\begin{gather}
Z''-\lambda^{2}Z=0
\\
\Phi'' + \mu^{2}\Phi = 0
\\
\rho^{2}P'' + \rho P' + \left(\rho{2}\lambda - \mu^{2}\right)P = 0 & #Eq3
\end{gather}\right.
$$
in which Eq. 3 is a Bessel Function
notice that if \(\lambda = 0\) (e.g. the situation that \(V\) is independent of \(z\)), Eq. 3 would be an Euler-Cauchy Equation, rather than Bessel.
for \(\lambda > 0\), let \(x=\rho\lambda\)
and \(y=P\)
note that
\[
\begin{gather}
dx=d(\rho\lambda)=\lambda d\rho
\\\\
y'=\frac{dy}{dx}=\frac{dP}{dx}=\frac{1}{\lambda }\frac{dP}{d\rho}
\end{gather}
\]
Eq.3 become
\[
\begin{gather}
x^{2}y''+xy+(x^{2}-\mu^{2})y=0
\end{gather}
\]