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Mohamed's question via email about solving an integral equation using Fourier transforms

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Well-known member
MHB Math Helper
Jan 26, 2012
1,403
$\displaystyle \begin{align*} f(t) \end{align*}$ satisfies the integral equation

$\displaystyle \begin{align*} f(t) + 3\int_{-\infty}^{\infty}{f(t-u)\,\mathrm{e}^{-2u}\,\mathrm{H}(u)\,\mathrm{d}u} = 15\mathrm{e}^{-2t^2} - 12t\,\mathrm{e}^{-2t^2} \end{align*}$

Find the solution to the integral equation using Fourier Transforms.
The first thing you need to do is take the Fourier Transform of both sides of the equation.

$\displaystyle \begin{align*} \mathcal{F} \left\{ f(t) + 3\int_{-\infty}^{\infty}{f(t-u)\,\mathrm{e}^{-2u} \, \mathrm{H}(u)\,\mathrm{d}u} \right\} &= \mathcal{F} \left\{ 15\mathrm{e}^{-2t^2} - 12t\,\mathrm{e}^{-2t^2} \right\} \end{align*}$

The integral you should recognise as a convolution integral:

$\displaystyle \begin{align*} \mathcal{F}\left\{ f(t) * g(t) \right\}= \mathcal{ F} \left\{ \int_{-\infty}^{\infty}{f(t-u)\,g(u)\,\mathrm{d}u} \right\} = F( \omega ) \, G( \omega ) \end{align*}$

with $\displaystyle \begin{align*} g(u) = \mathrm{e}^{-2u}\,\mathrm{H}(u) \end{align*}$.

You will also need the following transforms from your Fourier Transform table:

$\displaystyle \begin{align*} \mathcal{F} \left\{ \mathrm{e}^{-a\,t^2} \right\} &= \sqrt{ \frac{\pi}{a} } \, \mathrm{e}^{-\frac{\omega ^2}{4a}} \\ \\ \mathcal{F} \left\{ t\,\mathrm{e}^{-a\,t^2} \right\} &= -\frac{\mathrm{i}\,\sqrt{\pi}}{2a^{\frac{3}{2}}}\,\omega\,\mathrm{e}^{-\frac{\omega ^2}{4a} } \\ \\ \mathcal{F} \left\{ \mathrm{e}^{-a\,t}\,\mathrm{H}(t) \right\} &= \frac{1}{a+ \mathrm{i}\,\omega } \end{align*}$

and so now when we have taken the Fourier Transform of both sides we should get:

$\displaystyle \begin{align*} \mathcal{F} \left\{ f(t) + \int_{-\infty}^{\infty}{ f(t-u)\,\mathrm{e}^{-2u}\,\mathrm{H}(u)\,\mathrm{d}u } \right\} &= \mathcal{F} \left\{ 15\mathrm{e}^{-2t^2} - 12t\,\mathrm{e}^{-2t^2} \right\} \\ F(\omega ) + \mathcal{F} \left\{ f(t) \right\} \cdot \mathcal{ F} \left\{ \mathrm{e}^{-2t}\,\mathrm{H}(t) \right\} &= 15\,\sqrt{ \frac{\pi}{2} } \,\mathrm{e}^{ -\frac{ \omega ^2}{4 \cdot 2} } + \frac{12\, \mathrm{ i }\,\sqrt{\pi}}{2 \cdot 2^{\frac{3}{2}} } \,\omega \, \mathrm{e}^{-\frac{\omega ^2}{4\cdot 2} } \\ F( \omega ) + F( \omega ) \, \frac{1}{2 + \mathrm{i }\,\omega} &= \frac{15\,\sqrt{2\pi} }{2} \,\mathrm{e}^{-\frac{\omega ^2}{8}} + \frac{ 3\,\mathrm{i } \,\sqrt{2\pi} }{2}\,\omega \, \mathrm{e}^{-\frac{\omega ^2}{8} } \\ \left( 1 + \frac{1}{2 + \mathrm{i }\,\omega} \right)\,F(\omega ) &= \frac{ 15\,\sqrt{2\pi}}{2}\,\mathrm{e}^{-\frac{\omega ^2}{8} } + \frac{3\,\mathrm{i }\,\sqrt{2\pi}}{2}\,\omega\,\mathrm{e}^{-\frac{\omega ^2}{8}} \end{align*}$

And now it is simply a case of solving for $\displaystyle \begin{align*} F(\omega ) \end{align*}$ and doing the inverse Fourier Transform to get $\displaystyle \begin{align*} f(t) \end{align*}$, which is the solution to your equation.