Laplace's equation in polar coords

Question

Question: Suppose that the function u(r, $\phi$) satisfies Laplace’s equation for plane polar co-ordinates (r, $\phi$) i.e.

$$ ∇^2u = \frac{1}{r} \frac{∂}{∂r}(\frac{r∂u}{∂r}) + \frac{1}{r^2}\frac{∂^2u}{∂\phi^2} = 0 $$

Writing u(r,$\phi$) = P(r)Q($\phi$), show that the most general solution that is a bounded solution in this region and that satisfies u(r, 0) = u(r, π/2) = 0 for all r > a is $$ u(r,\phi) = \sum_{k=1}^∞ A_kr^{−2k} sin2k\phi. $$

Hence find the unique solution in this region that satisfies the above conditions, $u(a, \phi) = u_0$ for $0<\phi<π/4$ and $u(a,\phi)=−u_0$ for $π/4<\phi<π/2, $ where $u_0$ is a constant.

Now, I'm able to show the first part by going through the various steps and ending up with the general solution in the form

$$ u(r,\phi) = a_o + b_0lnr + \sum_{n=1}^∞ (A_nr^n + B_nr^{-n})(C_ncos(n\phi)+D_nsin(n\phi)) $$

After applying boundary conditions $u(r, 0) = 0$, I end up with $a_0 = b_0 = 0$ and $C_n = 0$ leaving

$$ u(r,\phi) = \sum_{n=1}^∞ (A_nr^n + B_nr^{-n})D_nsin(n\phi) $$

Applying $u(r, π) = 0$, I find that $r^n\to\infty$ in the expansion, so therefore $A_n = 0$. Further, $sin(\frac{nπ}{2}) = 0$ when $ n = 2k$. Letting $A_k = B_k D_k$, we have $$ u(r,\phi) = \sum_{k=1}^∞ A_kr^{-2k}sin(2k\phi) $$

However, with the second part, i.e. finding the unique solution, would I be correct in finding $A_k$ separately for $ 0 < \phi < π/4 $ and $ π/4 < \phi < π/2 $ by using Fourier sine series, or would I be better off (still using Fourier) integrating from 0 to π/2? I'm also not quite sure what to use as f(x) when integrating.

Answer 1

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ u\pars{a,\phi} = \sum_{k = 1}^{\infty}A_{k}a^{-2k}\sin\pars{2k\phi} $$ $$ u_{0}\int_{0}^{\pi/4}\sin\pars{2q\phi}\,\dd\phi - u_{0}\int_{\pi/4}^{\pi/2}\sin\pars{2q\phi}\,\dd\phi = \sum_{k = 1}^{\infty}A_{k}a^{-2k} \overbrace{\int_{0}^{\pi/2}\sin\pars{2q\phi}\sin\pars{2k\phi}\,\dd\phi} ^{\ds{{1 \over 4}\,\pi\,\delta_{qk}}} $$

\begin{align} \\[3mm] &{1 \over 4}\,\pi A_{q}a^{-2q} =u_{0}\int_{0}^{\pi/4}\sin\pars{2q\phi}\,\dd\phi - u_{0}\int_{\pi/4}^{\pi/2}\sin\pars{2q\phi}\,\dd\phi \\[3mm]&= u_{0}\int_{0}^{\pi/4}\sin\pars{2q\phi}\,\dd\phi - u_{0}\int_{-\pi/4}^{0}\sin\pars{2q\phi + q\pi}\,\dd\phi \\[3mm]&= u_{0}\int_{0}^{\pi/4}\sin\pars{2q\phi}\,\dd\phi - u_{0}\pars{-1}^{q}\int_{-\pi/4}^{0}\sin\pars{2q\phi}\,\dd\phi = u_{0}\bracks{1 - \pars{-1}^{q}}\int_{0}^{\pi/4}\sin\pars{2q\phi}\,\dd\phi \\[3mm]&= u_{0}\bracks{1 - \pars{-1}^{q}}{-\cos\pars{\pi q/2} + 1\over 2q} =u_{0}\,{1 - \pars{-1}^{q} \over q}\sin^{2}\pars{\pi q \over4} \end{align}

$A_{q}$ vanishes out for even $q$: \begin{align} {1 \over 4}\,\pi a^{-4k - 2}A_{2k + 1} &= {2u_{0} \over 2k + 1}\,\sin^{2}\pars{{\pi \over 4}\bracks{2k + 1}} \\[3mm]&= {2u_{0} \over 2k + 1}\,\bracks{% \sin\pars{\pi k \over 2}\cos\pars{\pi \over 4} + \cos\pars{\pi k \over 2}\sin\pars{\pi \over 4}}^{2} ={u_{0} \over 2k + 1} \\[3mm] A_{2k + 1} & = {4u_{0}/\pi \over 2k + 1}\,a^{4k + 2} \end{align}

$$\color{#0000ff}{\large% u\pars{r,\phi} = {4u_{0} \over \pi}\sum_{k = 0}^{\infty}{1 \over 2k + 1}\pars{a \over r}^{4k + 2} \sin\pars{2\bracks{2k + 1}\phi}} $$

Also $$ u\pars{r,\phi} = -\,{4u_{0} \over \pi}\Im\sum_{k = 0}^{\infty}{1 \over 2k + 1} \pars{{a \over r}\,\expo{-\ic\phi}}^{4k + 2} = -\,{4u_{0} \over \pi}\Im\fermi\pars{{a \over r\expo{\ic\phi}}} $$ where $\fermi\pars{z} = \sum_{k = 0}^{\infty}z^{4k + 2}/\pars{2k + 1}$. Then, \begin{align} \fermi'\pars{z} &= 2\sum_{k = 0}^{\infty}z^{4k + 1} = 2\,{z \over 1 - z^{4}} = {z \over z^{2} + 1} - {z \over z^{2} - 1} \\ \Im\fermi\pars{z}&=\half\,\ln\pars{z^{2} + 1 \over z^{2} - 1} =\half\,\Im\ln\pars{1 + z^{-2} \over 1 - z^{-2}} =\half\,\Im\ln\pars{% 1 + \bracks{x + \ic y}^{2}/a^{2} \over 1 - \bracks{x + \ic y}^{2}/a^{2}} \\[3mm]&=\half\,\Im\ln\pars{% x^{2} - y^{2} + a^{2} + 2xy\ic \over -x^{2} + y^{2} + a^{2} - 2xy\ic} \\[3mm]&=\half\,\Im\ln\pars{% \bracks{x^{2} - y^{2} + a^{2} + 2xy\ic}\bracks{-x^{2} + y^{2} + a^{2} + 2xy\ic}} \\[3mm]&= \half\,\Im\ln\pars{% -\bracks{x^{2} - y^{2}}^{2} + a^{4} - 4x^{2}y^{2} + 4a^{2}xy\ic} \\[3mm]&= \half\,\arctan\pars{% 4a^{2}xy \over -\bracks{x^{2} - y^{2}}^{2} + a^{4} - 4x^{2}y^{2}} \end{align}

In Cartesian coordinates: $$\color{#0000ff}{\large% u\pars{x,y} =-\,{2u_{0} \over \pi}\,\arctan\pars{% 4a^{2}xy \over -\bracks{x^{2} - y^{2}}^{2} + a^{4} - 4x^{2}y^{2}}} $$

$\large\tt\mbox{Please, check for some typo !!!}$.

Answer 2

The functions

$$ \sin 2\phi,\; \sin 4\phi,\; \sin 8\phi,\; \ldots,\; \sin 2n\phi,\;\ldots $$ are the unique eigenfunctions of the $L^{2}[0,\pi/2]$ Sturm-Liouville problem $$ -\frac{d^{2}}{d\phi^{2}}f = \lambda f, \;\;\; f(0)=f(\pi/2) = 0. $$ As such, these functions automatically form a complete orthogonal set of functions in $L^{2}[0,\pi/2]$. Because of this, you'll be able to satisfy an arbitrary condition at $r=a$: $$ \sum_{n=1}^{\infty} A_{n}a^{-2n}\sin 2n\phi=\left\{\begin{array}{cc}u_{0} & 0 < \phi < \phi/4, \\ -u_{0} & \pi/4 < \phi < \pi/2\end{array}.\right. $$ Furthermore, by orthogonality of these eigenfunctions on $[0,\pi/2]$, $$ A_{n}a^{-2n}\frac{\pi}{4} = A_{n}a^{-2n}\int_{0}^{\pi/2}\sin^{2}2n\phi\,d\phi = u_{0}\int_{0}^{\pi/4}\sin 2n\phi\,d\phi-u_{0}\int_{\pi/4}^{\pi/2}\sin 2n\phi\,d\phi. $$ In this case, the eigenvalues $\lambda=2^{2},4^{2},6^{2},\ldots$ of the Sturm-Liouville problem on $[0,\pi/2]$ dictates the powers $1/r^{2},1/r^{4},1/r^{6},\ldots$, as well as the orthogonal eigenfunction expansion. More explicitly, if $$ u(r,\phi) = \sum_{k=1}^{\infty}A_{k}r^{2k}\sin 2k\phi, $$ and $u(a, \phi)$ is known (as it is here) for $0\le \phi \le \pi/2$, then the automatic orthogonality of the eigenfunctions of the polar equation leads to $$ \int_{0}^{\pi/2}u(a,\phi)\sin 2n\phi d\phi = \sum_{k=1}^{\infty}A_{k}a^{2k}\int_{0}^{\pi/2}\sin 2k\phi \sin 2n\phi d\phi = A_{n}a^{2n}\int_{0}^{\pi/2}\sin^{2}2n\phi d\phi $$ for all $n \ge 1$. The series at $r=a$ converges to the desired function except at $0,\pi/4,\pi/2$, where it converges to $0$.