Laplacian Problem

Question

I'm going through past exams for study and I've come across a question I can never seem to solve. Question: Given r = (x,y,z) and $r = \sqrt{x^2 + y^2 +z^2}$. Find $\nabla^2{\cosh(r)}$

I gave it a go and attempted to solve it directly, but the method was very tedious, and in an exam format would take too long. I know there is a way to solve using vector identities (nabla identities), but I'm not sure where to start.

Any help would be appreciated.

Thanks

Answer 1

$r = \sqrt{\mathbf{r} \cdot \mathbf{r}}$, so by the chain rule, $$ \nabla \cosh{r} = (\nabla r) \frac{\partial}{\partial r} \cosh{r} = (\nabla r) \sinh{r}, $$ and then $$ \nabla^2 \cosh{r} = \nabla \cdot \left( (\nabla r) \sinh{r} \right) = \lvert \nabla r \rvert^2 \cosh{r} + (\nabla^2 r)\sinh{r}. $$ Now, $$ \nabla r = \nabla \sqrt{\mathbf{r} \cdot \mathbf{r}} = \frac{\mathbf{r}}{(\mathbf{r} \cdot \mathbf{r})^{3/2}} $$ and then $$ \nabla \cdot \frac{\mathbf{r}}{(\mathbf{r} \cdot \mathbf{r})^{3/2}} = \frac{3}{r^{3/2}} - \frac{3 \mathbf{r} \cdot \mathbf{r} }{r^{5/2}}=0, $$ so we find $$ \nabla \cosh{r} = \frac{1}{r^2}\cosh{r}. $$ (Of course, the easiest way to do this is to use the Laplacian in spherical coordinates, $ \nabla^2 f = r^{-2}\partial_r r^2\partial_r f + \dotsb $.)

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\mbox{If a function}\ \mrm{f}\ \mbox{depends solely on}\ r = \verts{\vec{r}},\quad \nabla\mrm{f}\pars{r} = \mrm{f}'\pars{r}\,{\vec{r} \over r} = {\mrm{f}'\pars{r} \over r}\,\vec{r} \\[5mm] &\mbox{Note that}\ \partiald{\mrm{f}\pars{r}}{x} = \totald{\mrm{f}\pars{r}}{r}\,\partiald{r}{x} = \totald{\mrm{f}\pars{r}}{r}\,{1 \over 2r}\,\,\partiald{r^{2}}{x} = \totald{\mrm{f}\pars{r}}{r}\,{x \over r}\quad \pars{\substack{\mbox{Similarly for}\\[1mm] \ds{y}\ \mbox{and}\ z}} \\[1cm] &\mbox{In order to evaluate}\ \nabla^{2}\mrm{f}\pars{r} \stackrel{\mbox{def.}}{=} \nabla\cdot\bracks{\nabla\mrm{f}\pars{r}} = \nabla\cdot\bracks{{\mrm{f}'\pars{r} \over r}\,\vec{r}}\ \mbox{I'll use the identity} \\ & \nabla\cdot\pars{\varphi\vec{A}} = \pars{\nabla\varphi}\cdot\vec{A} + \varphi\nabla\cdot\vec{A}. \end{align}

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\begin{align} \nabla^{2}\mrm{f}\pars{r} & = \braces{\nabla\bracks{\mrm{f}'\pars{r} \over r}}\cdot\vec{r} + {\mrm{f}'\pars{r} \over r}\,\ \overbrace{\nabla\cdot\vec{r}}^{\ds{=\ 3}} = {\vec{r} \over r}\,\totald{}{r}\bracks{\mrm{f}'\pars{r} \over r}\cdot\vec{r} + {3\,\mrm{f}'\pars{r} \over r} \\[5mm] & = r\,{\mrm{f}''\pars{r}r - \mrm{f}'\pars{r} \over r^{2}} + {3\,\mrm{f}'\pars{r} \over r} = \bbx{\mrm{f}''\pars{r} + {2\,\mrm{f}'\pars{r} \over r}} \end{align}

With $\ds{\mrm{f}\pars{r} \equiv \cosh\pars{r}}$:

$$ \nabla^{2}\cosh\pars{r} = \bbx{\cosh\pars{r} + {2\sinh\pars{r} \over r}} $$