Solution $u(x,t)$ of Heat equation

Question

$$\begin{cases} u_t = u_{xx}, &x \in \mathbb{R},\quad t> 0\\ u(x,0) = e^{-x^2}\sin x, &x \in \mathbb{R} \end{cases}$$

I used the general formula of the heat equation in $\mathbb{R}$, but I can not come up with an explicit solution.

Answer 1

I thought the following:

$u(x,t)= \frac{1}{\sqrt{4\pi t}} \int_{-\infty}^\infty e^{-(x-y)^2/4t}\,e^{-y^2} \sin y \,dy$

The formula of the solution can be found in the book of Evans, chapter 2. But I do not know how to solve this integral

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

The solution was already given by @amathstudent. The integral evaluation requires the "complete the square" in the following evaluation:

\begin{align} \mrm{u}\pars{x,t} & \equiv {1 \over \root{4\pi t}}\,\Im\int_{-\infty}^{\infty} \exp\pars{-\,{\pars{x - y}^{2} \over 4t} - y^{2} + \ic y}\,\dd y \\[5mm] & = {1 \over \root{4\pi t}}\,\exp\pars{-\,{x^{2} + t \over 1 + 4t}}\, \Im\braces{\exp\pars{\ic x \over 1 + 4t}\int_{-\infty}^{\infty} \exp\pars{-\bracks{1 + {1 \over 4t}}\bracks{y - \tilde{y}}^{2}}\,\dd y} \end{align}

where $\ds{\tilde{y} \equiv {x + 2\ic t \over 1 + 4t}}$

\begin{align} \mrm{u}\pars{x,t} & = {1 \over \root{4\pi t}}\,\exp\pars{-\,{x^{2} + t \over 1 + 4t}}\, \Im\braces{\exp\pars{\ic x \over 1 + 4t} \int_{-\infty - \tilde{y}}^{\infty - \tilde{y}} \exp\pars{-\bracks{1 + {1 \over 4t}}y^{2}}\,\dd y} \\[5mm] & = {1 \over \root{4\pi t}}\,\exp\pars{-\,{x^{2} + t \over 1 + 4t}}\, \sin\pars{x \over 1 + 4t}\,{2\root{\pi} \over \root{4 + 1/t}} \\[5mm] & = \bbx{{1 \over \root{4t + 1}}\,\exp\pars{-\,{x^{2} + t \over 1 + 4t}}\, \sin\pars{x \over 1 + 4t}} \\ & \end{align}