##### Cartesian Coordinates
ems.ca.sv.car

Example: infinite grounded metal plates

Consider two infinite grounded metal plates positioned parallel to the \(xz\) plane, one at \(y = 0\) and the other at \(y = a\).

At \(x = 0\), the setup is closed off with an infinite insulated strip maintained at potential \(\phi_0 (y)\).

Question: find the potential inside the slot.

Solution:

By translational symmetry along \(z\), the potential must be independent of \(z\). This thus falls back onto a 2d problem. We need to solve the 2d Laplace equation Lap_2d

\[ \frac{\partial^2 \phi}{\partial^2 x} + \frac{\partial^2 \phi}{\partial^2 y} = 0 \]

\begin{align} &(i) ~\phi(x, y=0) = 0, \hspace{5mm} &(ii) ~\phi(x, y=a) = 0, \nonumber \\ &(iii) ~\phi(x=0, y) = \phi_0 (y), &(iv) ~\phi (x \rightarrow \infty, y) \rightarrow 0. \end{align}

Let us look for solutions in the form

- Gr (3.23)

\[ \phi(x,y) = X(x) Y(y), \hspace{1cm} \frac{1}{X} \frac{d^2 X}{dx^2} + \frac{1}{Y} \frac{d^2 Y}{dy^2} = 0 \tag{Lap_sv_car}\label{Lap_sv_car} \] Since the \(x\) and \(y\) dependencies are separated, the only possibility is that each individual term in the Laplace equation equals a constant, and that these constants add up to zero. We can thus put (the sign choice anticipates the solution somewhat)

- Gr (3.26)

\[ \frac{d^2 X_n}{dx^2} = k_n^2 X_n, \hspace{1cm} \frac{d^2Y_n}{dy^2} = -k_n^2 Y_n \tag{Lap_sv_car_sep}\label{Lap_sv_car_sep} \] where \(k_n\) is some real number. Since Lap_2d is a linear equation for \(\phi\), we can linearly combine solutions of Lap_sv_car_sep for different \(k_n\) and still get a solution to Lap_sv_car.

Let's look first of all at the solutions of Lap_sv_car_sep for a given \(k_n\). Since this is a second-order linear differential equation, there are two linearly independent solutions. The most general solution for \(X\) and \(Y\) can be written

- Gr (3.27)

\begin{align} X_n(x) &= Ae^{k_nx} + Be^{-k_nx}, \nonumber \\ Y_n(y) &= C \sin k_ny + D \cos k_ny \tag{Lap_sv_car_solXY}\label{Lap_sv_car_solXY} \end{align}

The constants can be fixed by fitting the boundary conditions. From \((iv)\), we get \(A = 0\). From \((i)\), we get \(D = 0\). We are thus left with

\[ \phi(x,y) = C_n e^{-k_nx} \sin k_n y \] Going further, \((ii)\) requires the momenta \(k_n\) to be quantized according to

\[ k_n = \frac{n\pi}{a}, \hspace{1cm} n = 1, 2, 3, ... \] Since we can use as solution any linear combination of the functions defined by these momenta, we obtain the solution in the form of a Fourier series,

\[ \phi(x,y) = \sum_{n=1}^{\infty} C_n e^{-n\pi x/a} \sin (n\pi y/a) \] with yet-to-be-determined coefficients \(C_n\), which have to be chosen to fit boundary condition \((iii)\). Using the orthogonality relation

\[ \int_0^a dy ~\sin(n\pi y/a) \sin (n' \pi y/a) = \delta_{n n'} \frac{a}{2} \] we thus get that the \(C_n\) are Fourier coefficients of the boundary function \(\phi_0(y)\):

\[ C_n = \frac{2}{a} \int_0^a dy ~\phi_0(y) \sin(n\pi y/a) \]

Specific example: say that \(\phi_0(y) = \phi_0\), i.e. just a constant. Then,

- Gr (3.35)

\[ C_n = \frac{2\phi_0}{a} \int_0^a dy \sin(n\pi y/a) = \frac{2\phi_0}{n\pi} (1 - \cos n\pi) = \frac{4\phi_0}{n\pi} \delta_{n, odd} \tag{Cn}\label{Cn} \]

The logic of separation of variables exploited two characteristics of the basis functions in the Fourier decomposition, namely completeness and orthogonality.

Completeness simply means that any differentiable function can be expressed in terms of the basis functions:

\[ f(x) = \sum_{n=1}^{\infty} C_n f_n (y), \hspace{1cm} \forall ~f \in C^{\infty} \]

Orthogonality is a convenient choice made to keep things simple: each function, under the chosen integration domain and weighing, has no overlap with others, i.e.

\[ \int_0^a dx ~f_n (x) f_{n'} (x) = 0, \hspace{1cm} n \neq n' \]

The solution for the specific case \(\phi_0 (y) = \phi_0\) is thus \[ \phi(x,y) = \frac{4\phi_0}{\pi}\sum_{n=1, 3, 5, ...}^{\infty} \frac{1}{n} e^{-n\pi x/a} \sin (n\pi x/a) \]

Example: rectangular pipe

Take two infinitely long grounded plates at \(y = 0\) and \(y=a\), cut at \(x = \pm b\) and joined to metal strips maintained at constant \(\phi = \phi_0\).

Question: find the potential in the resulting rectangular pipe.

Solution: by translational invariance along \(z\), the potential must be independent of \(z\). We thus need to solve the 2d Laplace equation Lap_2d, \[ \partial^2 \phi/\partial x^2 + \partial^2 \phi/\partial y^2 = 0, \] with boundary conditions

\begin{align} &(i) ~\phi (x, y = 0) = 0, \hspace{5mm} &(ii) ~\phi (x, y = a) = 0, \nonumber \\ &(iii) ~\phi(x = b, y) = \phi_0, &(iv) ~\phi(x = -b, y) = \phi_0. \end{align}

The general solution is obtained from Lap_sv_car_XY, \[ \phi(x,y) = (Ae^{kx} + Be^{-kx}) (C\sin ky + D\cos ky). \] By symmetry, \(\phi(x,y) = \phi(-x,y)\) so \(A = B\). Since \(e^{kx} + e^{-kx} = 2\cosh kx\), the generic solution becomes (redefining \(C\) and \(D\)) \[ \phi(x,y) = \cosh kx ~(C\sin ky + D\cos ky). \] Boundary conditions \((i)\) and \((ii)\) require \(D = 0\), \(k = n\pi/a\) so

\[ \phi(x,y) = C \cosh(n\pi x/a) \sin(n\pi y/a) \] with \((iv)\) already satisfied if \((iii)\) is.

The full solution is then a linear combination of complete set of functions, \[ \phi(x,y) = \sum_{n=1}^{\infty} C_n \cosh(n\pi x/a) \sin(n\pi y/a). \] The coefficients must be chosen such that \((iii)\) is fulfilled, \(\phi(b,y) = \phi_0\). This simple case of a constant value \(\phi_0\) gives us the same relation as Cn, so

- Gr (3.42)

\[ \phi(x,y) = \frac{4\phi_0}{\pi} \sum_{n = 1, 3, 5, ...} \frac{1}{n} \frac{\cosh (n\pi x/a)}{\cosh(n\pi b/a)} \sin(n\pi y/a). \]

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42