0
$\begingroup$

I'm trying to solve the following integral equation using Fourier transforms: $$u(t)+ \int_{-\infty}^{t} e^{\tau-t} u(\tau)\,d\tau=e^{-2|\tau|}$$

I tried to transform both sides of the equation using $\mathfrak{F}\left[\int_{\infty}^{t} f(\tau)\,d\tau\right]=\frac{F(\omega)}{i\omega}+\pi F(0)\delta(\omega)$ which I found in a table of Fourier transforms but this did not seem to help me.

Is using Fourier transform a good approach for this equation? If it is, how do I proceed?

$\endgroup$
0
$\begingroup$

Let $$ \phi(t)=\begin{cases}e^t & \text{if }t\le0,\\0 & \text{if }t>0.\end{cases} $$ Then $$ \int_{-\infty}^{t} e^{\tau-t} u(\tau)\,d\tau=\int_{-\infty}^{\infty} \phi(\tau-t)\, u(\tau)\,d\tau=\phi\ast u(t). $$ Since the Fourier transform of a convolution is the product of the Fourier transforms, taking the Fourier transform on the equation we get $$ \hat u+\hat\phi\,\hat u=\widehat{\bigl(e^{-2|t|}\bigr)}. $$

$\endgroup$

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.