Solving PDE in 3D

Question

Could anyone solve below pde in 3d? r is dposition vector in spherical coordinate.

$$ \nabla^{2}\mathrm{f} = \frac{a r^4}{1}\,\left[-1 + 3\left(\hat{e} \cdot \hat{r}\right)^{2}\right] $$

$\hat{e}$ is a defined vector in the x_y plane.

Answer 1

WLOG choose our coordinates such that $\hat{e} = \hat{z}$. Using the fact that $$ 3 (\hat{e} \cdot \hat{r})^2 - 1 = 3 \cos^2 \theta - 1 = 2 P_2(\cos \theta), $$ where $P_2(x)$ is the second-order Legendre polynomial, we guess a solution of the form $$ f(r, \theta) = R(r) P_2(\cos \theta). $$ Plugging this solution into the Laplacian, we get $$ \nabla^2 f = \frac{P_2(\cos \theta)}{r^2} \frac{d}{dr} \left( r^2 R'(r) \right) + \frac{R(r)}{r^2 \sin \theta} \frac{d}{d\theta} \left( \sin \theta \frac{d P_2(\cos \theta)}{d\theta} \right) \\ = \frac{P_2(u)}{r^2} \frac{d}{dr} \left( r^2 R'(r) \right) + \frac{R(r)}{r^2 } \frac{d}{du} \left( (1- u^2) \frac{d P_2(u)}{du} \right) $$ where $u \equiv \cos \theta$. But by the definition of Legendre's equation, $$ \frac{d}{du} \left( (1- u^2) \frac{d P_n(u)}{du} \right) = - n(n+1) P_n(u), $$ and so the original equation $\nabla^2 f= \frac{a}{r^4} (-1+ 3 (\hat{e} \cdot \hat{r})^2)$ becomes $$ \frac{P_2(u)}{r^2} \frac{d}{dr} \left( r^2 R'(r) \right) + \frac{R(r)}{r^2} \left(-6 P_2(u) \right) = \frac{2a}{r^4} P_2(u), $$ or $$ \frac{d}{dr} \left( r^2 R'(r) \right) - 6 R(r) = \frac{2a}{r^2}. $$ This is a non-homogenous Euler equation and can be solved for $R(r)$ using standard techniques from ODEs. One of the arbitrary coefficients in the resulting solution will vanish due to the demand that $f \to 0$ as $r \to \infty$. Note, however, that no solutions exist such that $f \to 0$ as $r \to 0$ (regardless of the boundary conditions at infinity.)

Note that throughout this answer, I am using the "physics" convention for spherical coordinates where $\theta$ is the angle between the position vector and the $z$-axis, hence the need to pick $\hat{e} = \hat{z}$ above. Once the overall solution for $f$ has been found, the coordinates in the solution can then be rotated such that $\hat{e}$ coincides with its original definition in the $xy$-plane (if desired.)

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\mathsf{I'll\ slightly\ modify\ the\ original\ post,\ as\ follows,}\ \mathsf{in\ order\ to\ discuss\ the\ convergence\ of\ radial}\ \\ \mathsf{integrals}}$:

$\ds{\nabla^{2}\mathrm{f}\pars{\vec{r}} = {1 \over a^{3}}\pars{r \over a}^{\nu}\bracks{-1 + 3\pars{\hat{\mrm{e}}\cdot\hat{r}}^{2}} = -4\pi\bracks{{1 \over 2\pi a^{3}}\pars{r \over a}^{\nu}\,\, {3\cos^{2}\pars{\angle\pars{\hat{e},\hat{r}}} - 1 \over 2}}:\ {\large ?}}$.

Note that \begin{align} &{3\cos^{2}\pars{\angle\pars{\hat{e},\hat{r}}} - 1 \over 2} = \,\mrm{P}_{2}\pars{\cos\pars{\angle\pars{\hat{e},\hat{r}}}} = {4\pi \over 5} \sum_{m = -2}^{2}\mrm{Y}_{2m}\pars{\hat{r}} \overline{\mrm{Y}_{2m}\pars{\hat{\mrm{e}}}} \\[5mm] &\mbox{such that}\ \nabla^{2}\mathrm{f}\pars{\vec{r}} = -4\pi\bracks{% {1 \over 2\pi a^{3}}\pars{r \over a}^{\nu}\,{4 \pi \over 5} \sum_{m = -2}^{2}\mrm{Y}_{2m}\pars{\hat{r}} \overline{\mrm{Y}_{2m}\pars{\hat{\mrm{e}}}}} \end{align} where $\ds{\,\mrm{P}_{\ell}}$ is a Legendre Polynomial and $\ds{\,\mrm{Y}_{\ell m}}$ is a Harmonic Spherical. $\ds{\,\mrm{f}\pars{\vec{r}}}$ is given by \begin{align} \mrm{f}\pars{\vec{r}} & = \iiint_{\mathbb{R}^{3}} \bracks{{1 \over 2\pi a^{3}}\pars{r' \over a}^{\nu}\,{4 \pi \over 5} \sum_{m = -2}^{2}\mrm{Y}_{2m}\pars{\hat{r}'} \overline{\mrm{Y}_{2m}\pars{\hat{\mrm{e}}}}} \,{\dd^{3}\vec{r}\,' \over \verts{\vec{r} - \vec{r}\,'}} \end{align}

Hereafter, I'll use some properties which are explained in the above mentioned links.

Expanding $\ds{\verts{\vec{r} - \vec{r}\,'}^{\,-1}}$ in Harmonic Spherical: \begin{align} \mrm{f}\pars{\vec{r}} & = {1 \over 2\pi a^{3}}\iiint_{\mathbb{R}^{3}} \bracks{\pars{r' \over a}^{\nu} {4\pi \over 5}\sum_{m = -2}^{2}\mrm{Y}_{2m}\pars{\hat{r}'} \overline{\mrm{Y}_{2m}\pars{\hat{\mrm{e}}}}}\,\times \\[3mm] & \phantom{=\,\,\,}\ \underbrace{% \bracks{\sum_{\ell' = 0}^{\infty}\sum_{m' = -\ell'}^{\ell'}{4\pi \over 2\ell' + 1}{r_{<}^{\ell'} \over r_{>}^{\ell' + 1}}\,\mrm{Y}_{\ell'm'}\pars{\hat{r}} \overline{\,\mrm{Y}_{\ell'm'}\pars{\vec{r}\,'}}}} _{\ds{=\ {1 \over \verts{\vec{r} - \vec{r}\,'}}}}\ \,\dd^{3}\vec{r}\,'\,,\qquad \left\{\substack{\ds{r_{<} = \min\braces{r,r'}}\\[2mm] \ds{r_{>} = \max\braces{r,r'}}}\right. \\[1cm] & = {1 \over 2\pi a^{3}}\,{4\pi\over 5}\sum_{m = -2}^{2} \sum_{\ell' = 0}^{\infty}\sum_{m' = -\ell'}^{\ell'}{4\pi \over 2\ell' + 1} \int_{0}^{\infty}\pars{r' \over a}^{\nu}{r_{<}^{\ell'} \over r_{>}^{\ell' + 1}} \overline{\mrm{Y}_{2m}\pars{\hat{\mrm{e}}}} \,\mrm{Y}_{\ell'm'}\pars{\hat{r}}\,\times \\[3mm] & \phantom{=\,\,\,}\ \underbrace{% \int_{\Omega_{\vec{r}\,'}}\mrm{Y}_{2m}\pars{\hat{r}\,'} \overline{\,\mrm{Y}_{\ell'm'}\pars{\hat{r}\,'}}\,\dd\Omega_{\vec{r}\,'}} _{\ds{\delta_{\ell' 2}\delta_{mm'}}}\ \,r'^{2}\,\dd r' \\[1cm] & = {2 \over a^{3}}\ \overbrace{\bracks{{4\pi \over 5}\sum_{m = -2}^{2} \mrm{Y}_{2m}\pars{\hat{r}}\overline{\,\mrm{Y}_{2m}\pars{\hat{e}}}}} ^{\ds{3\cos^{2}\pars{\angle\pars{\hat{r},\hat{e}}} - 1 \over 2}} {1 \over a^{\nu}} \int_{0}^{\infty}r'^{\nu + 2}\,{r_{<}^{2} \over r_{>}^{3}}\,\dd r' \\[5mm] & = {3\cos^{2}\pars{\angle\pars{\hat{r},\hat{e}}} - 1 \over a} \bracks{{1 \over a^{\nu + 2}}\int_{0}^{r}{r'^{\nu + 4} \over r^{3}}\,\dd r' + {1 \over a^{\nu + 2}}\int_{r}^{\infty}r'^{\nu - 1}r^{2}\,\dd r'} \end{align}

Both integrals converge whenever $\ds{\nu + 4 > -1\ \mbox{and}\ \nu < 0 \implies -5 < \nu < 0}$. Then,

\begin{align} \mrm{f}\pars{\vec{r}} & = {3\cos^{2}\pars{\angle\pars{\hat{r},\hat{e}}} - 1 \over a} \bracks{{1 \over a^{\nu + 2}}\,{r^{\nu + 2} \over \nu + 5} - {1 \over a^{\nu + 2}}\,{r^{\nu + 2} \over \nu}} \\[5mm] & =\ \bbox[#ffe,15px,border:1px dotted navy]{\ds{5\,{3\cos^{2}\pars{\angle\pars{\hat{r},\hat{e}}} - 1 \over a} {\pars{r/a}^{\nu + 2} \over -\nu\pars{\nu + 5}}}}\,,\qquad \nu \in \pars{-5,0} \end{align}