# Verifying a Series Solution to Dirichlet's Problem via separation of variables

In Stein's Fourier Analysis i'm having trouble attempting to verify the series solution given in the problem in $(1.)$

$(1.)$

The Dirchlet problem is the annulus defined by ${{(r, \theta): p < r < 1}}$, where $0 \leq p \leq 1$ in the inner radius. The problem to solve:

$$\frac{\partial^{2}u}{\partial{r}^{2}} + \frac{1}{r} \frac{\partial{u}}{\partial{r}}+ \frac{1}{r^{2}}\frac{\partial^{2}u}{\partial{\theta}^{2}} = 0$$

subject to the boundary conditions:

{ \begin{align} u(1,\theta) = f(\theta) \\ u(p,\theta) = g(\theta). \end{align} } Stein's intial argument was to write solutions for the Dirchlet Problem as he did previously in the following form:

$$u(r,\theta)= \sum_{}^{}c_{n}(r)e^{in\theta}$$

with $c_{n}(r) = A_{n}r^{n} + B_{n}r^{-n}, n \neq 0$ Set: $$f(\theta) \sim \sum_{}^{} a_{n}e^{in\theta}$$ and

$$g(\theta) \sim \sum_{}^{}b_{n}e^{in\theta}$$

$$u(r,\theta) = \sum_{n \leq 0} (\frac{1}{p^{n}-p^{-n}})((p/r)^{n} - (r/p)^{n})a_{n} + (r^{n} - r^{-n})b_{n}]e^{in\theta} + a_{o} + (b_{o} - a_{o}\frac{logr}{logp}$$

From looking what was done in Chapter 1 as a prior example, and comparing the problem in $(1.)$the series the solution was obtained via Sepration of Variables. My initial attack can be followed in $(2.)$

$(2.)$

$$\frac{\partial^{2}u}{\partial{r}^{2}} + \frac{1}{r} \frac{\partial{u}}{\partial{r}}+ \frac{1}{r^{2}}\frac{\partial^{2}u}{\partial{\theta}^{2}} = \Delta{u}$$ $$r^{2}\frac{\partial^{2}u}{\partial{r}^{2}} + r\frac{\partial{u}}{\partial{r}} = -\frac{\partial^{2}u}{\partial{\theta}^{2}}$$

Plugging in our solution product: $(u(r,\theta))=F(r)G(\theta))$ $$r^{2}\frac{\partial^{2}u}{\partial{r}^{2}}F(r)G(\theta) + r\frac{\partial{u}}{\partial{r}}F(r)G(\theta)= -\frac{\partial^{2}u}{\partial{\theta}^{2}}F(r)G(\theta)$$

Now dividing by our Solution Product: $$\frac{r^{2}\frac{\partial^{2}u}{\partial{r}^{2}}F(r)G(\theta) + r\frac{\partial{u}}{\partial{r}}F(r)G(\theta) } {F{(r)}} = \frac{-\frac{\partial^{2}u}{\partial{\theta}^{2}}F(r)G(\theta)}{G{(\theta)}}$$

Finally one can observe in a more convenient form that we have the following:

$$\frac{r^{2}F(r)G''(\theta)+r(G(\theta))F'(r)}{F(r)} = \frac{F(r)-G''(\theta)}{G(\theta)}$$

\left\{ \begin{align} r^2G''(\theta) + r(G(\theta)F'(r)) = 0 \\ F(r) - \dfrac{F(r) - G''(\theta)}{G(\theta)} = 0 \end{align} \right.

\left\{ \begin{align} r^{2}G''(\theta)+r(G(\theta))F'(r)\lambda F(r)=0 \\ F(r) - \lambda \frac{G''(\theta)}{G''(\theta)} = 0 \end{align} \right. From the previous result above I'm stuck on working out a series solution to the above ODE's is their any fundamental observations I'm missing towards the problem ?

• The solution to the homogeneous problem usually involves. Easel Functiona. See "Heat Conduction" by M.N. Ozisik. Jun 3, 2017 at 2:15
• Intersting, In Stein's textbook he gives the solution written as the series: $u(r,\theta) = \sum_{n \leq 0} (\frac{1}{p^{n}-p^{-n}})((p/r)^{n} - (r/p)^{n})a_{n} + (r^{n} - r^{-n})b_{n}]e^{in\theta} + a_{o} + (b_{o} - a_{o}\frac{logr}{logp}$ i'm still trying to figure out how he initially derived it ? Jun 3, 2017 at 3:39
• That should have been Bessel Functions in my comment! Jun 3, 2017 at 3:49

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ The general solution of the $\ds{2D}$-Laplace Equation is given by

\begin{align} \mrm{u}\pars{r,\theta} & = \pars{A\,{\theta \over 2\pi} + B}\bracks{C\ln\pars{r \over p} + D} \\[5mm] & + \sum_{n = 1}^{\infty}\left\{\bracks{% A_{n}\pars{r \over p}^{n} + B_{n}\pars{p \over r}^{n}}\sin\pars{n\theta}\right. \\[3mm] & \left.\phantom{\sum_{n = 1}^{\infty}\!\!\braces{}}+ \bracks{% C_{n}\pars{r \over p}^{n} + D_{n}\pars{p \over r}^{n}}\cos\pars{n\theta}\right\}\label{1.a}\tag{1.a} \end{align}

where $\ds{A, B, C, D, \braces{A_{n}}, \braces{B_{n}}, \braces{C_{n}}\ \mbox{and}\ \braces{D_{n}}}$ are constants which are determined by the boundary conditions.

Since the solution must be invariant under $\ds{\theta \mapsto \theta + 2\pi}$, I'll conclude that $\ds{A = 0}$. In such a case, $\ds{B}$ is 'redundant' such that I can set $\ds{B = 1}$. \eqref{1.a} is reduced to: \begin{align} \mrm{u}\pars{r,\theta} & = C\ln\pars{r \over p} + D + \sum_{n = 1}^{\infty}\left\{\bracks{% A_{n}\pars{r \over p}^{n} + B_{n}\pars{p \over r}^{n}}\sin\pars{n\theta}\right. \\[3mm] & \left.\phantom{\sum_{n = 1}^{\infty}\!\!\braces{}AAAAAAAAAAA\,}+ \bracks{% C_{n}\pars{r \over p}^{n} + D_{n}\pars{p \over r}^{n}}\cos\pars{n\theta}\right\} \label{1.b}\tag{1.b} \end{align}
Then, \begin{align} \mrm{f}\pars{\theta} & = -C\ln\pars{p} + D + \sum_{n = 1}^{\infty}\bracks{% \pars{A_{n}p^{-n} + B_{n}p^{n}}\sin\pars{n\theta} + \pars{C_{n}p^{-n} + D_{n}p^{n}}\sin\pars{n\theta}}\label{2}\tag{2} \\[2mm] \mrm{g}\pars{\theta} & = D + \sum_{n = 1}^{\infty}\bracks{% \pars{A_{n} + B_{n}}\sin\pars{n\theta} + \pars{C_{n} + D_{n}}\sin\pars{n\theta}} \label{3}\tag{3} \end{align}

Integrating both members of \eqref{2} and \eqref{3} over $\ds{\pars{0,2\pi}}$:

$$\left\{\begin{array}{rcrcl} \ds{-\ln\pars{p}C} & \ds{+} & \ds{D} & \ds{=} & \ds{{1 \over 2\pi}\int_{0}^{2\pi}\mrm{f}\pars{\theta}\,\dd\theta} \\[2mm] && \ds{D} & \ds{=} & \ds{{1 \over 2\pi}\int_{0}^{2\pi}\mrm{g}\pars{\theta}\,\dd\theta} \end{array}\right.$$

Those relations determines $\ds{C\ \mbox{and}\ D}$.

Similarly, \begin{align} \int_{0}^{2\pi}\mrm{f}\pars{\theta}\sin\pars{n\theta}\,{\dd\theta \over \pi} & = A_{n}p^{-n} + B_{n}p_{n} \\[2mm] \int_{0}^{2\pi}\mrm{f}\pars{\theta}\cos\pars{n\theta}\,{\dd\theta \over \pi} & = C_{n}p^{-n} + D_{n}p_{n} \\[2mm] \int_{0}^{2\pi}\mrm{g}\pars{\theta}\sin\pars{n\theta}\,{\dd\theta \over \pi} & = A_{n} + B_{n} \\[2mm] \int_{0}^{2\pi}\mrm{g}\pars{\theta}\cos\pars{n\theta}\,{\dd\theta \over \pi} & = C_{n} + D_{n} \end{align}

Those equations determine $\ds{\braces{A_{n}},\braces{B_{n}},\braces{C_{n}}\ \mbox{and}\ \braces{D_{n}}}$.

• Interesting approach, I didn't realize this was a valid approach, are there other ways of validating a general solution for a PDE. Jun 4, 2017 at 17:47
• @Zophikel I guess this is the straightforward way because we know, at the very beginning, what is the general solution. If you make 'variable separation' you arrive to that expression $\left(1.a\right)$. So, we never do that because we already know we'll arrive to it anyway. Usually, we know something else for $\,\mathrm{f}\ \mbox{and}\ \,\mathrm{g}$ $\left(~symmetries,\ etc,\ \ldots~\right)$ which enables further simplifications of the solutions $\mathsf{before}$ we impose the boundary conditions. Jun 4, 2017 at 17:53
• Ahhh ok now I understand this is my first time dealing with PDE in Stein's Analysis text, so when dealing with verifying general solutionsPDE one should simplify our general solution before imposing boundary conditions. Jun 4, 2017 at 17:57