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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

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

This leads to the solution

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

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 ?

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    $\begingroup$ The solution to the homogeneous problem usually involves. Easel Functiona. See "Heat Conduction" by M.N. Ozisik. $\endgroup$ Jun 3, 2017 at 2:15
  • $\begingroup$ 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 ? $\endgroup$
    – Zophikel
    Jun 3, 2017 at 3:39
  • $\begingroup$ That should have been Bessel Functions in my comment! $\endgroup$ Jun 3, 2017 at 3:49

1 Answer 1

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

\begin{equation} \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. \end{equation}

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

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  • $\begingroup$ Interesting approach, I didn't realize this was a valid approach, are there other ways of validating a general solution for a PDE. $\endgroup$
    – Zophikel
    Jun 4, 2017 at 17:47
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    $\begingroup$ @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. $\endgroup$ Jun 4, 2017 at 17:53
  • $\begingroup$ 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. $\endgroup$
    – Zophikel
    Jun 4, 2017 at 17:57

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