Find a solution to the heat equation with the initial values
$\begin{cases} u_t - 4u_{xx} = 0 & -\infty < x < \infty, \; t>0 \\ u(x,0)= 2e^{-x^2} & -\infty < x < \infty \end{cases}$
I know how to do this in separable method. But I want the solution in terms of radial method.
I would start by assuming $u(x,t)= v(r)$.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ There is not any cylindrical or spherical symmetry which justify your initial claim !!!.
Hereafter, you'll see a straightforward approach:
Multiply both member of the Heat Equation by $\ds{\expo{-st}}$ and integrate over $\ds{t > 0}$. Namely, \begin{align} 0 & = \int_{0}^{\infty}\expo{-st}\partiald{\on{u}\pars{x,t}}{t}\,\dd t - 4\,\partiald[2]{}{x}\ \overbrace{\int_{0}^{\infty}\on{u}\pars{x,t}\expo{-st}\,\dd t} ^{\ds{\hat{\on{u}}\pars{x,s}}} \\[5mm] & \stackrel{\substack{{\rm IBP} \\[0.5mm] {\rm first\ integral}}}{=}\,\,\,\,\,\, -\on{u}\pars{x,0} + s\,\hat{\on{u}}\pars{x,s} -4\,\partiald[2]{\,\hat{\on{u}}\pars{x,s}}{x} \\[5mm] \implies & \pars{\partiald[2]{}{x} - {1 \over 4}\,s}\hat{\on{u}}\pars{x,s} = -\,{1 \over 4}\,\on{u}\pars{x,0} = -\,{1 \over 2}\expo{-x^{2}} \\[5mm] & \mbox{which has the solution} \\[2mm] & \bbx{\hat{\on{u}}\pars{x,s} = \hat{\on{u}}_{p}\pars{x,s} -\,{1 \over 2}\int_{-\infty}^{\infty}\on{G}\pars{x,x'} \expo{-x\,'^{2}}\,\,\dd x\,'} \\ & \end{align} $\ds{\hat{\on{u}}_{p}\pars{x,s}}$ is a particular solution -a homogeneous equation solution- which is usually chosen to satisfies the boundary conditions. In this way, $\ds{\on{G}\pars{x,x'}}$ satisfies homogeneous boundary conditions. Namely, \begin{align} & \left\{\begin{array}{rcl} \ds{\pars{\partiald[2]{}{x} - {1 \over 4}\,s}\on{G}\pars{x,x'}} & \ds{=} & \ds{0\,,\quad x \not= x\,'} \\[1mm] \ds{\on{G}\pars{0,x'}} & \ds{=} & \ds{0} \\[1mm] \ds{\left.\partiald{\on{G}\pars{x,x'}}{x} \,\right\vert_{\,x\ = x\ '^{\,-}}^{\,x\ = x\ '^{\,+}}} & \ds{=} & \ds{1} \\[3mm] \substack{\ds{\on{G}\pars{0,x'}}\ \mbox{continuous} \\ \mbox{at}\ \ds{x = x'}} && \end{array}\right. \end{align} Then, \begin{align} \on{G}\pars{x,x\ '} & = \left\{\begin{array}{lcl} \ds{0} & \mbox{if} & \ds{x < x\ '} \\ \ds{A\sinh\pars{{1 \over 2}\root{s}\pars{x - x\ '}}} & \mbox{if} & \ds{x > x\ '} \end{array}\right. \\[5mm] \left.\partiald{\on{G}\pars{x,x'}}{x} \,\right\vert_{\,x\ = x\ '^{\,-}}^{\,x\ = x\ '^{\,+}} & = 1\ \implies A = {2 \over \root{s}} \end{align} Then, \begin{align} \hat{\on{u}}\pars{x,s} & = \hat{\on{u}}_{p}\pars{x,s} -\int_{-\infty}^{x}{\sinh\pars{\root{s}\bracks{x - x'}/2} \over \root{s}}\expo{-x\,'^{2}}\,\,\dd x\,' \\[5mm] & = \hat{\on{u}}_{p}\pars{x,s} - {\expo{\root{s}x/2} \over 2\root{s}} \int_{-\infty}^{x}\exp\pars{-x\ '^{2} - \root{s}x\ '/2}\,\dd x\ ' \\[2mm] & + {\expo{-\root{s}x/2} \over 2\root{s}} \int_{-\infty}^{x}\exp\pars{-x\ '^{2} + \root{s}x\ '/2}\,\dd x\ ' \end{align} I'll leave the OP the evaluation of the remaining integrals.