Solving heat equation using radial solution 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 $u(x,t)= v(r)$.
 A: $\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}\,}\,}
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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.
