Fourier Transforms for ODE

Question

QUESTION

Use the Fourier Transform to solve the ODE

$\frac {df}{dx}-bf(x)=g(x)$

Subject to the boundary condition at infinity $f(x)\to0 $ as $\lvert x \rvert \to \infty$

where $ g(x) = \begin{cases} e^{-bx}, & \text{if $x \ge 0$} \\[2ex] 0, & \text{if $x \lt 0$ } \end{cases} $

Attempt
I have applied the Fourier Transform to the equation given, and applied the inverse Fourier Transform after that. But I got stuck at this step:

$f(x) = \frac{-1}{2 \pi} \int_{\xi=-\infty}^{\infty} \frac{e^{-i \xi x}}{i \xi +b} \; \int_{y=0}^\infty \ e^{(i \xi - b)y} \;\; dy \ d\xi$

Could someone advise how I can proceed further? I think I am missing out on something here..

Answer 1

If you assume $f'$ and $f$ are square-integrable, then $\mathcal{F}\{f'\}|_{s} = is\mathcal{F}\{f\}|_{s}$ holds without further assumptions. And, if $\Re b > 0$, then $g$ is also square integrable, with \begin{align} \mathcal{F}\{g\} & =\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-bx}e^{-isx}dx \\ & = \left.\frac{1}{\sqrt{2\pi}}\frac{e^{x(-b-is)}}{-b-is}\right|_{x=0}^{\infty} =\frac{1}{\sqrt{2\pi}}\frac{1}{b+is} \end{align} Therefore, \begin{align} \mathcal{F}\{f\}& =\frac{1}{\sqrt{2\pi}}\frac{1}{(b+is)(is-b)} \\ & = -\frac{1}{\sqrt{2\pi}}\frac{1}{(s-ib)(s+ib)} \end{align} The inversion integral gives $$ f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{e^{isx}}{(s-ib)(s+ib)}ds. $$ The line integral can be converted to a contour integral if you choose a circular contour where the integrand decays. To do this, note that $|e^{isx}|=e^{\Re(isx)}=e^{-x\Im s}$ decays in the upper half plane if $x > 0$ and decays in the lower half plane if $x < 0$. Because $\Re b > 0$, $ib$ is in the upper half plane and $-ib$ is in the lower half plane. Therefore, if $x > 0$, you close the contour in the upper half plane and $1/(s-ib)$ has a pole at $s=ib$ in the upper half plane. The contour is counterclockwise. So you get $$ f(x) = i\frac{e^{i(ib)x}}{ib+ib}=\frac{1}{2b}e^{-bx},\;\; x > 0. $$ The first factor of $i$ is because you need $\frac{1}{2\pi i}$ outside the integral and you only have $\frac{1}{2\pi}$. For $x < 0$, you close the contour in the lower half plane, and you end up with a clockwise contour, which introduces a negative times a residue. There is a pole at $-ib$ and a missing $i$ multiplying $2\pi$ again, which gives $$ f(x) = -i\frac{e^{i(-ib)x}}{-ib-ib}=\frac{1}{2b}e^{bx},\;\;\; x < 0. $$ You can put these two pieces together: $$ f(x)=\frac{1}{2b}e^{-b|x|},\;\;\; x\in\mathbb{R}. $$ Note that $f$, $f'$ are square integrable. So the assumptions are justified for $\Re b > 0$.