0
$\begingroup$

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 $\hat{f}$?

$\endgroup$
1
$\begingroup$

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

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

\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} $$\color{#0000ff}{\large% \uu\pars{x,t} = \int_{-\infty}^{\infty}\fermi\pars{x'} {\expo{-\pars{x\ -\ x'\ +\ at}^{2}/4t} \over \root{2\pi}\root{2t}}\,\dd x'} $$

$\endgroup$
  • $\begingroup$ 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 $\hat{f}$) $\endgroup$ – Kunal Nov 17 '13 at 18:03
  • $\begingroup$ @Kunal I rewrote somehow the solution such that you can see your required solution is already here $\left(~\mbox{see the}\ \color{#ff0000}{\large\bf\mbox{RED}}\ \mbox{formula}~\right)$. $\endgroup$ – Felix Marin Nov 17 '13 at 20:35
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.