# Differential equation for heat equation

Consider the equation $$\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2} + a\frac{\partial u}{\partial x}$$ for a function $u(x,t)$ with initial value $$u(x,0)=f(x).$$ Let $\hat{u}(y,t)$ and $\hat{f}(y)$ denote the Fourier transform in the $x$ variable of $u$ and $f$. For each fixed $y$, find the ordinary differential equation for $\hat{u}(y,t)$ formally (assuming the derivatives all make sense). Then solve the equation for $\hat{u}$ in terms of $\hat{f}$.

Taking the Fourier transform, I get $$\frac{\partial}{\partial t}\hat{u}(y,t)=(iy)^2\hat{u}(y,t)+aiy\hat{u}(y,t)=-y^2\hat{u}(y,t)+aiy\hat{u}(y,t)$$

But how can I solve the equation for $\hat{u}$ in terms of $\hat{f}$?

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}% \newcommand{\uu}{\,{\rm u}}$ $\ds{\partiald{\uu}{t} = \partiald[2]{\uu}{x} + a\partiald{\uu}{x}\,,\quad \uu\pars{x,0} = \fermi\pars{x}}$.

$$\uu\pars{x,t} \equiv \int_{-\infty}^{\infty} \tilde{\uu}\pars{k}\exp\pars{\ic kx - {t \over \tau_{k}}}\,{\dd k \over 2\pi} \quad\imp\quad -\,{1 \over \tau_{k}} = -k^{2} + \ic ka$$

\begin{align} &\uu\pars{x,t} \equiv \int_{-\infty}^{\infty} \tilde{\uu}\pars{k}\expo{\ic kx + \pars{-k^{2} + \ic ka}t}\,{\dd k \over 2\pi}\ \imp\ \uu\pars{x,0} = \fermi\pars{x} = \int_{-\infty}^{\infty} \tilde{\uu}\pars{k}\expo{\ic kx}\,{\dd k \over 2\pi}\tag{1} \end{align}

Notice that $\tilde{\uu}\pars{k}$ is the $\fermi\pars{x}$ Fourier transform $\hat{\fermi}\pars{k}$ ( in the OP notation ): $\hat{\fermi}\pars{k} \equiv \tilde{\uu}\pars{k}$. In terms of it ( $\underline{\mbox{as required by the OP}}$ ) the solution is ( see expressions $\pars{1}$ ): $$\color{#ff0000}{\large\bf% \uu\pars{x,t} = \int_{-\infty}^{\infty} \hat{\fermi}\pars{k}\expo{\ic kx + \pars{-k^{2} + \ic ka}t}\,{\dd k \over 2\pi}} \tag{2}$$

\begin{align} &\tilde{\uu}\pars{k} = \int_{-\infty}^{\infty}\fermi\pars{x}\expo{-\ic k x}\,\dd x \quad\imp\quad \uu\pars{x,t} = \int_{-\infty}^{\infty}{\rm K}\pars{x - x',t}\fermi\pars{x'}\,\dd x' \\[3mm]& \mbox{where}\quad {\rm K}\pars{x,t} \equiv \int_{-\infty}^{\infty}\expo{\ic kx - \pars{k^{2} - \ic ka}t}\,{\dd k \over 2\pi} \end{align} Let's evaluate ${\rm K}\pars{x,t}$: \begin{align} {\rm K}\pars{x,t} &\equiv \int_{-\infty}^{\infty}\expo{-t\bracks{k^{2} - \ic\pars{x/t + a}k}}\,{\dd k \over 2\pi} = \int_{-\infty}^{\infty} \expo{-t\bracks{k - \ic\pars{x/t + a}/2}^{2} - t\pars{x/t +a}^{2}/4} \,{\dd k \over 2\pi} \\[3mm]&= \expo{-\pars{x + at}^{2}/4t}\int_{-\infty}^{\infty}\expo{-tk^{2}} \,{\dd k \over 2\pi} \quad\imp\quad {\rm K}\pars{x,t} = {\expo{-\pars{x + at}^{2}/4t} \over \root{2\pi}\root{2t}} \end{align} $$\color{#0000ff}{\large% \uu\pars{x,t} = \int_{-\infty}^{\infty}\fermi\pars{x'} {\expo{-\pars{x\ -\ x'\ +\ at}^{2}/4t} \over \root{2\pi}\root{2t}}\,\dd x'}$$

• Hi Felix. Thanks for your answer, but sorry I'm really lost how this helps with my question (solving for $\hat{u}$ in terms of $\hat{f}$) – Kunal Nov 17 '13 at 18:03
• @Kunal I rewrote somehow the solution such that you can see your required solution is already here $\left(~\mbox{see the}\ \color{#ff0000}{\large\bf\mbox{RED}}\ \mbox{formula}~\right)$. – Felix Marin Nov 17 '13 at 20:35

Taking the Fourier transform in $x$ converts the $x$ derivatives to multiplication, so the only derivative left will be with respect to $t$ - hence giving an ODE.