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I have to solve that equation with Fourier method: $y'-iy=1+\delta'(x)$

Fourier transform is defined like this:

$F[\varphi](k)=\int\limits_{-\infty}^{\infty}\varphi(x)e^{ikx}dx$

$F^{-1}[\varphi](x)=\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}\varphi(k)e^{-ikx}dk$, where $\varphi \in \mathcal{S}$.

Applying Fourier method:

$F[y]\cdot(k-1) = k - 2\pi i\delta(k)$

The solution is:

$F[y]=A\delta(k-1)+\frac{k}{k-1+i0}+2\pi i\delta(k)$

Now I want to get $y(x)$ applying inverse transform:

$y(x)=\frac{A}{2\pi}e^{-ix}+i+\delta(x)-ie^{-ix}\theta(x)$

So, I know the correct answer

$y(x)=\frac{A}{2\pi}e^{ix}+i+\delta(x)+ie^{ix}\theta(x)$

but I do not see where I've made mistake.

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  • $\begingroup$ It is about the inverse FT of $\frac{k}{k-1} = 1+\frac{1}{k-1}$ $\endgroup$ – reuns Mar 24 '17 at 2:20
  • $\begingroup$ How it can be that the first summand become $\e^{+ix}$ is the inverse FT defined as I wrote above and it should be $\e^{-ix}$? $\endgroup$ – danielleontiev Mar 24 '17 at 2:53
  • $\begingroup$ The FT of $\frac{A}{2\pi} e^{ix}$ is $A\, \delta(k-1)$ and the FT of $i$ is $2i \pi \delta(k)$. So we are left with the two other terms $\endgroup$ – reuns Mar 24 '17 at 3:14
  • $\begingroup$ One problem I'm worried about it is the integration by parts: It requires something like $\,\mathrm{y}\left(\pm\infty\right) \to 0$ and it isn't consistent with the expected solution. $\endgroup$ – Felix Marin Mar 24 '17 at 4:16
  • $\begingroup$ @FelixMarin but in fact $y(\pm\infty) < \infty$. Maybe it's enough $\endgroup$ – danielleontiev Mar 24 '17 at 15:08
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Notation: $\ds{\mrm{f}\pars{x} = \int_{-\infty}^{\infty}\hat{\mrm{f}}\pars{k}\expo{\ic kx} \,{\dd k \over 2\pi}\iff \,\hat{\mrm{f}}\pars{k} = \int_{-\infty}^{\infty}\mrm{f}\pars{x}\expo{-\ic kx}\,\dd k}$.

\begin{align} &\mrm{y}'\pars{x} - \ic\,\mrm{y}\pars{x} = 1 +\delta\,'\pars{x} \implies \ic k\,\hat{\mrm{y}}\pars{k} - \ic\,\hat{\mrm{y}}\pars{k} = 2\pi\,\delta\pars{k} + \ic k \\[5mm] \implies & \hat{\mrm{y}}\pars{k} = {k - 2\pi\,\delta\pars{k}\ic \over k - 1} = 1 + {1 \over k - 1} + 2\pi\,\delta\pars{k}\ic \end{align}


\begin{align} \mrm{y}_{\pm}\pars{x} & = \int_{-\infty}^{\infty}\bracks{1 + {1 \over k - 1 \pm \ic 0^{+}} + 2\pi\,\delta\pars{k}\ic}\expo{\ic kx}\,{\dd k \over 2\pi} = \delta\pars{x} + \ic + \expo{\ic x}\int_{-\infty}^{\infty} {\expo{\ic kx} \over k \pm \ic 0^{+}}\,{\dd k \over 2\pi} \\[5mm] & = \delta\pars{x} + \ic + \expo{\ic x}\bracks{% \mrm{P.V.}\int_{-\infty}^{\infty}{\expo{\ic kx} \over k}\,{\dd k \over 2\pi} + \int_{-\infty}^{\infty}\expo{\ic kx}\bracks{\mp\pi\ic\,\delta\pars{k}} \,{\dd k \over 2\pi}} \\[5mm] & = \delta\pars{x} + \ic + \expo{\ic x}\bracks{% \int_{0}^{\infty}{2\ic\sin\pars{kx} \over k}\,{\dd k \over 2\pi} \mp {1 \over 2}\,\ic} = \delta\pars{x} + \ic + \expo{\ic x}\bracks{% {\ic \over \pi}\,\mrm{sgn}\pars{x}\,{\pi \over 2} \mp {1 \over 2}\,\ic} \\[5mm] & = \delta\pars{x} + \ic + \expo{\ic x} \bracks{{2\Theta\pars{x} - 1 \mp 1}}{\ic \over 2} \end{align}
$$\bbox[15px,#ffe,border:1px dotted navy]{\ds{% \left\{\begin{array}{rcl} \ds{\quad\mrm{y}_{-}\pars{x}} & \ds{=} & \ds{\delta\pars{x} + \ic + \Theta\pars{x}\expo{\ic x}\ic} \\[3mm] \ds{\quad\mrm{y}_{+}\pars{x}} & \ds{=} & \ds{\delta\pars{x} + \ic - \Theta\pars{-x}\expo{\ic x}\ic} \end{array}\right.}} $$

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