Flaw in solving a PDE and Fourier Series

Question

• Consider the Laplacian equation in cylindrical coordinates with no theta $\left(~\theta~\right)$ dependence: \begin{align} &\mbox{So,}\quad\frac{1}{r} (r(u)_r)_r + (u)_{zz} = 0 \\[2mm] &\mbox{subject to}\quad \left\{\begin{array}{rcl} u\left(r=R,z\right) & = & 1 \\ u\left(r=0,z\right) & = & \text{finite} \\ u\left(r,z=\pm H\right) & = & 0 \end{array}\right. \end{align}
• Using separation of variables, $u(r,z) = A(r)B(z)$, we get $$B''=-\lambda ^2 B \\ rA'' + A' - \lambda^2 r A = 0$$
• Solving the $B(z)$ equation and imposing the boundary conditions, I get, $$c_1 \cos(\lambda H) + c_2\sin(\lambda H) = 0 \\ c_1 \cos(\lambda H) - c_2\sin(\lambda H) = 0$$ I was told that this was incorrect, and I should be using $\sinh$ and $\cosh$ functions to this DE. My question is, why is that allowed? Aren't they non-periodic?

What am I doing wrong here, and how am I supposed to know when to use $\sin$ and $\cos$, and when to use $\sinh$ and $\cosh$ when solving PDEs?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $ {\Large ?}:\quad\left\{\begin{array}{l} \ds{{1 \over r}\partiald{}{r} \bracks{r\,\partiald{\on{u}\pars{r,z}}{r}} + \partiald[2]{\on{u}\pars{r,z}}{z} = 0} \\[3mm] \ds{\on{u}\pars{r,\pm H} = 0} \\[3mm] \left.\begin{array}{rcl} \ds{\on{u}\pars{0,z}} & \mbox{is} & \ds{finite}. \\[1mm] \ds{\on{u}\pars{R,z}} & \ds{=} & \ds{1} \end{array}\right\} \end{array}\right. $

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• Lets $\ds{\on{u}\pars{r,z} \equiv \sum_{n = 0}^{\infty}a_{n}\pars{r} \cos\pars{k_{n}z}}$ where $\ds{k_{n} \equiv \pars{2n + 1}{\pi \over 2H}}$.

Note that $\ds{\on{u}\pars{r,z}}$ already satisfies the boundary conditions at $\ds{z = \pm H}$ and, in addition, it exhibits the $\ds{z \mapsto -z}$ symmetry. Also, \begin{equation} \int_{-H}^{H}\cos\pars{k_{m}z} \cos\pars{k_{n}z}\,\dd z = H\,\delta_{mn} \label{1}\tag{1} \end{equation}

• $\ds{\on{u}\pars{r,z}}$ must satisfy the above differential equation: \begin{align} &\sum_{n = 1}^{\infty}\braces{% {1 \over r}\totald{}{r} \bracks{r\,\totald{a_{n}\pars{r}}{r}} - k_{n}^{2}\,a_{n}\pars{r}}\cos\pars{k_{n}z} = 0 \end{align} Use (\ref{1}) to obtain $\ds{{1 \over r}\totald{}{r}\bracks{r\,\totald{a_{n}\pars{r}}{r}} - k_{n}^{2}\,a_{n}\pars{r} = 0}$
• The last equation solution is a linear combination of $\ds{\on{J}_{0}\pars{\ic k_{n}r}\ \mbox{and}\ \on{Y}_{0}\pars{\ic k_{n}r}}$ which are Bessel Functions. Note that, as $\ds{r \to 0^{+}}$, $$ \on{J}_{0}\pars{\ic k_{n}r} \to 1\quad \mbox{and}\quad \on{Y}_{0}\pars{\ic k_{n}r} \sim {2 \over \pi}\ln\pars{\ic k_{n}r}. $$ See this link. The finite solution, when $\ds{r \to 0^{+}}$, is given by $\ds{b_{n}\on{J}_{0}\pars{\ic k_{n}r}}$ where $\ds{b_{n}}$ is a constant.
• Then, $$ \on{u}\pars{r,z} = \sum_{n = 0}^{\infty} b_{n}\on{J}_{0}\pars{\ic k_{n}r} \cos\pars{k_{n}z} $$
• Moreover, $$ 1 = \sum_{n = 0}^{\infty} b_{n}\on{J}_{0}\pars{\ic k_{n}\color{red}{R}} \cos\pars{k_{n}z} $$ Multiply both members by $\ds{\cos\pars{k_{n}z}/H}$ and use (\ref{1}) to integrate over $\ds{z \in \pars{-H,H}}$: \begin{align} &\overbrace{{1 \over H} \int_{-H}^{H}\cos\pars{k_{n}z}\dd z} ^{\ds{{4 \over \pi}\,{\pars{-1}^{n} \over 2n + 1}}} \\[5mm] = &\ \sum_{m = 0}^{\infty} b_{m}\on{J}_{0}\pars{\ic k_{m}R} \underbrace{\bracks{{1 \over H}\int_{-H}^{H} \cos\pars{k_{n}z}\cos\pars{k_{m}z}\dd z}} _{\ds{\delta_{nm}}} \\[5mm] & \implies b_{m} = {4 \over \pi}\,{\pars{-1}^{n} \over \pars{2n + 1}\on{J}_{0}\pars{\ic k_{n}R}} \end{align}
• $\ds{\large\underline{Finally},}$ \begin{align} &\on{u}\pars{r,z} \\[2mm] = &\ {4 \over \pi}\sum_{n = 0}^{\infty} {\pars{-1}^{n} \over \pars{2n + 1} \on{J}_{0}\pars{\ic k_{n}R}} \on{J}_{0}\pars{\ic k_{n}r} \cos\pars{k_{n}z} \end{align}

Answer 2

There is more than one representation of the solution in term of series and Bessel functions. I'll try to derive the one with $\sinh$ and $\cosh$.

First, let's make the $r = R$ conditions homogenous. Let $u = 1 - v$. The new unknown $v$ satisfies the same equation $$ \frac{1}{r}\frac{\partial}{\partial r}(r v_r) + v_{zz} = 0, $$ but the boundary conditions are $$ v(r=R, z) = 0\\ |v(r=0, z)| < \infty\\ v(r, z=\pm H) = 1 $$ By separation of variables $v = Q(r) Z(z)$ we have $$ Z'' = \lambda Z\\ (rQ')' + \lambda r Q = 0. $$ Taking $\lambda < 0$ leads us to $\sin$ and $\cos$ for $Z$ (the path you've originally taken). There's noting wrong with it, but it leads to Bessel functions with imaginary arguments.

So let's assume $\lambda = \kappa^2 > 0$. $$ Z'' = \kappa^2 Z\\ (rQ')' + \kappa^2 r Q = 0. $$ The solutions are $Z(z) = A \cosh \kappa z + B \sinh \kappa z$ and $Q(r) = C J_0(\kappa r) + D Y_0(\kappa r)$. And since $Y_0(x)$ is infinite at $x = 0$ we need to state that $D = 0$.

The equation $Q(R) = 0$ is $$ J_0(\kappa R) = 0. $$ If $\mu_n$ are the roots of $J_0(\mu) = 0$ then $$ \kappa_n = \frac{\mu_n}{R}. $$ The roots of Bessel functions are quite common in such problems. Moreover (see Fourier-Bessel series, A&S 11.4.5), $$ \int_0^R J_0(\kappa_n r) J_0(\kappa_m r) r dr = \delta_{nm} R^2 \frac{J_1(\mu_n)^2}{2}. $$

So the solution should be sought as $$ v(r,z) = \sum_{n=1}^\infty (A_n \cosh \kappa_n z + B_n \sinh \kappa_n z) J_0(\kappa_n r). $$ Let's put $z = \pm H$: $$ 1 = v(r,\pm H) = \sum_{n=1}^\infty (A_n \cosh \kappa_n H \pm B_n \sinh \kappa_n H) J_0(\kappa_n r). $$ Multiplying both sides with $J_0(\kappa_m r)$ and applying $\int_0^R \bullet rdr$ on the both sides gives: $$ \int_0^R J_0(\kappa_m r) rdr = (A_m \cosh \kappa_m H \pm B_m \sinh \kappa_m H) R^2 \frac{J_1(\mu_m)^2}{2}. $$ The integral on the left is $R^2 \frac{J_1(\mu_m)}{\mu_m}$ (see A&S 11.3.20), so $$ A_m \cosh \kappa_m H \pm B_m \sinh \kappa_m H = \frac{2}{\mu_m J_1(\mu_m)}. $$ Clearly, $B_m = 0$ (we might guess this from $z$-symmetry) and $$ A_m = \frac{2}{\mu_m J_1(\mu_m)}\frac{1}{\cosh \kappa_m H}. $$

Finally, the answer is $$ u = 1 - \sum_{n=1}^\infty \frac{2}{\mu_n J_1(\mu_n)} \frac{\cosh \kappa_n z}{\cosh \kappa_n H} J_0\left(\kappa_n r\right) $$

I made a plot with 20 terms of Felix Marin's solution (red) and mine (blue): The $\cos$-series has oscillations near $r = R$ while $\cosh$-series has oscillations near $z = \pm H$. These oscillations are in fact, Gibb's phenomenon and are due no approximation of constant $1$ with either Fourier series $\cos k_n z$ or by Fourier-Bessel series $J(\kappa_n r)$. Far from the oscillations the solutions are almost equal up to small difference (due to truncated infinite series).