Laplace Equation -> Boundary Conditions

Question

Given $\nabla^2 \phi=0$,

With B.C.

1. $\Gamma_1 =-V$ (Left side of rectangle)

2. $\Gamma_2 = V$ (right side of rectangle)

3. $\Gamma_0 = 0$ (top and bottom sides of rectangle)

Can Separation of Variables be applied to this problem? Will the superposition principle have to be applied to the second ODE because there are two non-zero BC?

Answer 1

Absolutely. First write $\phi(x,y) = \phi_1(x,y)+\phi_2(x,y)$, where:

$$\nabla^2 \phi_1 = 0$$ $$\phi_1(x,0) = 0$$ $$\phi_1(x,d)=0$$ $$\phi_1(0,y) = -V$$ $$\phi_1(w,y) = 0$$

and

$$\nabla^2 \phi_2 = 0$$ $$\phi_2(x,0) = 0$$ $$\phi_2(x,d)=0$$ $$\phi_2(0,y) = 0$$ $$\phi_2(w,y) = V$$

We may write $\phi_1(x,y) = X(x) Y(y)$ to find that

$$\frac{X''}{X} = -\frac{Y''}{Y} = k^2$$

Then

$$Y(y) = A \cos{k y} + B \sin{k y}$$

The boundary conditions in $y$ imply that $A=0$ and $k = n \pi/d$. For the $x$ equation, we may write

$$X(x) = C_n \cosh{\frac{ n \pi (w-x)}{d}} + D_n \sinh{\frac{ n \pi (w-x)}{d}}$$

I wrote the solution this way so that we can have $C=0$ from the boundary condition at $x=w$. The general solution is then

$$\phi_1(x,y) = \sum_{n=1}^{\infty} D_n \sinh{\frac{ n \pi (w-x)}{d}} \sin{\frac{ n \pi y}{d}}$$

We find the $D_n$ from the B.C. at $x=0$. By orthogonality, we have

$$D_n = \frac{2}{d \sinh{\frac{ n \pi w}{d}}} \int_0^d dy (-V) \sin{\frac{ n \pi y}{d}} = -\frac{2 V}{n \pi \sinh{\frac{ n \pi w}{d}}} \left ( 1-\cos{n \pi}\right )$$

We find $\phi_2$ similarly, except the expansion is now given by

$$\phi_2(x,y) = \sum_{n=1}^{\infty} C_n \sinh{\frac{ n \pi x}{d}} \sin{\frac{ n \pi y}{d}}$$

We find the $C_n$ from the B.C. at $x=w$:

$$D_n = \frac{2}{d \sinh{\frac{ n \pi w}{d}}} \int_0^d dy (V) \sin{\frac{ n \pi y}{d}} = \frac{2 V}{n \pi \sinh{\frac{ n \pi w}{d}}} \left ( 1-\cos{n \pi}\right )$$

Putting this altogether, I get as the sought-after solution:

$$\phi(x,y) = \frac{4 V}{\pi} \sum_{k=0}^{\infty} \frac{\sinh{\frac{(2 k+1) \pi x}{d}}-\sinh{\frac{(2 k+1) \pi (w-x)}{d}}}{(2 k+1) \sinh{\frac{(2 k+1) \pi w}{d}}} \sin{\frac{(2 k+1) \pi y}{d}} $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\underline{\tt Domain}:\ {\large\pars{0,b}\times\pars{0,h}}}$

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It's always better, at the very beginning, to take advantage of:

• the homogeneous boundary conditions : It's $\ds{\underline{\mbox{equivalent to}\ variable\ separation}}$ and it's less "painful".
• the problem symmetries: The solution must changes its sign when $\ds{x \mapsto b - x}$ and it's invariant when $\ds{y \mapsto h - y}$.
• The above statements lead to $\ds{\phi\pars{x,y} = \sum_{n = 0}^{\infty}\on{a}_{n}\pars{x}\sin\pars{k_{n}y}}$ where $\ds{k_{n} \equiv \pars{2n + 1}\,{\pi \over h}}$ and $\ds{\on{a}_{n}\pars{x} = -\on{a}_{n}\pars{b -x}}$. Note that, with $\ds{m, n \in \mathbb{N}_{\,\geq\ 1}}$, $$ \left\{\begin{array}{rcl} \ds{\int_{0}^{h}\sin\pars{k_{m}y}\sin\pars{k_{n}y}\,\dd y} & \ds{=} & \ds{{1 \over 2}\,h\,\delta_{mn}} \\ \ds{\int_{0}^{h}\sin\pars{k_{n}y}\,\dd y} & \ds{=} & \ds{{2 \over \pars{2n + 1}\pi}\,h\,\delta_{mn}} \end{array}\right. \label{1}\tag{1} $$
• $\ds{\phi\pars{x,y}}$ must satify the differential equation: $$ \sum_{n = 0}^{\infty} \bracks{\pars{\totald[2]{}{x} - k_{n}^{2}}\on{a}_{n}\pars{x}}\sin\pars{k_{n}y} = 0 \label{2}\tag{2} $$
• Use (\ref{1}) and (\ref{2}). I get \begin{align} &\pars{\totald[2]{}{x} - k_{n}^{2}}\on{a}_{n}\pars{x} = 0 \implies \on{a}_{n}\pars{x} = c_{n}\sinh\pars{k_{n}\bracks{x - {b \over 2}}} \end{align} The general solution is reduced to: $$ \phi\pars{x,y} = \sum_{n = 0}^{\infty} c_{n}\sinh\pars{k_{n}\bracks{x - {b \over 2}}}\sin\pars{k_{n}y} $$
• The boundary conditions yield \begin{align} &V = \sum_{n = 0}^{\infty} c_{n}\sinh\pars{k_{n}\bracks{b \over 2}}\sin\pars{k_{n}y} \\[5mm] \implies &\ \overbrace{V\,{2 \over h}\int_{0}^{h}\sin\pars{k_{n}y} \dd y}^{\ds{4V \over \pars{2n + 1}\pi}} \\[2mm] = &\ \sum_{m = 0}^{\infty} c_{m}\sinh\pars{k_{m}\,{b \over 2}}\ \underbrace{\bracks{{2 \over h} \int_{0}^{h}\sin\pars{k_{n}y}\sin\pars{k_{m}y}\dd y}} _{\ds{\delta_{nm}}} \\[5mm] &\ \implies c_{n} = {4V \over \pars{2n + 1}\pi\sinh\pars{k_{n}b/2}} \end{align}
• Finally, $$ \bbx{\phi\pars{x,y} = {4V \over \pi}\sum_{n = 0}^{\infty} {\sinh\pars{k_{n}\bracks{x - b/2}}\sin\pars{k_{n}y} \over \pars{2n + 1}\sinh\pars{k_{n}b/2}}} \\ $$