Solution of the PDE

Question

I am trying to solve the following PDE:

$\begin{cases} r^2u_{rr}+ru_r+u_{\varphi\varphi}=0, \\ u(1,\varphi)=cos^3\varphi, & u(0,\varphi)<\infty \\ u(r,\varphi+2\pi)=u(r,\varphi) \end{cases}$

My idea is to use the Fourier method. By using the substitution $r\frac{d}{dr}=\frac{d}{dz}\Rightarrow \frac{d}{dz}=\frac{dr}{dz}\frac{d}{dr}$, from which we can deduce:

$\frac{dr}{dz}=r\Rightarrow r=e^z$

My question is, is it now enough to use the Fourier method with the given boundary conditions, i.e. $\sum_{k=0}^{\infty}\alpha_k(r)U(\varphi)$?

Answer 1

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Following your suggestion, you can write the solution as $\ds{\mrm{a}\pars{r}\cos\pars{\varphi} + \mrm{b}\pars{r}\cos\pars{3\varphi}}$ because $\ds{\cos^{3}\pars{\varphi}}$ is a linear combination of $\ds{\cos\pars{\varphi}}$ and $\ds{\cos\pars{3\varphi}}$. That yields simple differential equations for $\ds{\mrm{a}\pars{r}}$ and $\ds{\mrm{b}\pars{r}}$.

Hereafter, I'll try the general method ( it starts from the general solution ).

The $\ds{2D}$-Laplace equation general solution, in polar coordinates, is given by: \begin{align} \mrm{u}\pars{r,\varphi} & = \pars{A\varphi + B}\bracks{C\ln\pars{r} + D} + \sum_{n = 1}^{\infty}\bracks{a_{n}\sin\pars{n\varphi + \alpha_{n}}r^{n} + b_{n}\sin\pars{n\varphi + \beta_{n}}{1 \over r^{n}}}\label{1}\tag{1} \end{align} where $\ds{A, B, C, D, \braces{a_{n}},\braces{b_{n}},\braces{\alpha_{n}},\braces{\beta_{n}}}$ are constants. Since $\ds{\mrm{u}\pars{r,\varphi + 2\pi} = \mrm{u}\pars{r,\varphi}}$, $\ds{A = 0}$ and, in such case, I can perform the scales $\ds{BC \mapsto C}$ and $\ds{BD \mapsto D}$. Then, \eqref{1} is reduced to \begin{align} \mrm{u}\pars{r,\varphi} & = C\ln\pars{r} + D + \sum_{n = 1}^{\infty}\bracks{a_{n}\sin\pars{n\varphi + \alpha_{n}}r^{n} + b_{n}\sin\pars{n\varphi + \beta_{n}}{1 \over r^{n}}}\label{2}\tag{2} \end{align} $\ds{\mrm{u}\pars{0,\varphi} < \infty \implies C = 0\ \mbox{and}\ b_{n} = 0\ \forall\ n \in \mathbb{N}_{\ \geq\ 1}}$. Also, $\ds{\mrm{u}\pars{1,\varphi} = \cos^{3}\pars{\varphi} = {3 \over 4}\,\cos\pars{\varphi} + {1 \over 4}\,\cos\pars{3\varphi}}$ and

\begin{align} \mrm{u}\pars{1,\varphi} & = D + \sum_{n = 1}^{\infty}a_{n}\sin\pars{n\varphi + \alpha_{n}} = D + a_{1}\sin\pars{\varphi + \alpha_{1}} + a_{3}\sin\pars{3\varphi + \alpha_{3}} \\[2mm] & + \sum_{{\large n = 2} \atop {\large n \not= 3}}^{\infty}a_{n}\sin\pars{n\varphi + \alpha_{n}}r^{n} \label{3}\tag{3} \end{align}

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\begin{align} &\mbox{Set}\qquad D = 0\,;\quad \alpha_{1} = \alpha_{3} = -\,{\pi \over 2}\,;\quad -a_{1} = {3 \over 4}\,,\quad -a_{3} = {1 \over 4} \\ & \mbox{and}\qquad a_{n} = b_{n} = 0\,,\quad n = 2,4,5,6,\ldots \end{align}

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The final solution becomes: \begin{align} \mrm{u}\pars{r,\varphi} & = {3 \over 4}r\cos\pars{\varphi} + {1 \over 4}\,r^{3}\cos\pars{3\varphi} \end{align}