So I have the integral equation :

$$\int_{-\pi}^{\pi} f(t)f(x-t) dt = -\cos (x).$$

I know that I should use Fourier transform or Laplace transform and to use the convolution theorem, but I'm not sure exactly how.

$f(x)$ should also be $2\pi$ periodic.

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    $\begingroup$ What does the convolution theorem say ? $\endgroup$ – Yves Daoust Sep 18 '15 at 20:10
  • $\begingroup$ that the Fourier transform of the convolution of f and g equal to the Fourier transform of f times the Fourier transform of g. see: en.wikipedia.org/wiki/Convolution_theorem $\endgroup$ – YaG32 Sep 18 '15 at 20:15
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    $\begingroup$ Can you identify a convolution in the given problem statement ? If yes, apply the theorem. $\endgroup$ – Yves Daoust Sep 18 '15 at 20:18

If the function $f$ is periodic, then it is clear we can write $f$ in terms of its Fourier series:

$$f(t) = \frac{a_0}{2} + \sum_{k=1}^{\infty} \left (a_k \cos{k t} + b_k \sin{k t} \right ) $$

Now, the RHS is simply proportional to $\cos{x}$. By the orthogonality of the sine and cosine, all we need worry about then are the $k=1$ terms.

Plugging away, we find that the integral is equal to

$$\int_{-\pi}^{\pi} dt \left (a_1 \cos{t} + b_1 \sin{t} \right ) \left (a_1 \cos{(x-t)} + b_1 \sin{(x-t)} \right )$$

which the reader can verify is, again using orthogonality to make life easy,

$$(a_1^2-b_1^2) \pi \cos{x} + 2 a_1 b_1 \pi \sin{x} $$

This expression must be identically equal to $-\cos{x}$. This means that $a_1 b_1=0$, so either $a_1$ or $b_1$ is zero. But if $b_1$ were zero, the expression on the LHS must be positive, which is false. Thus, $a_1=0$ and hence,

$$f(t) = \pm \frac1{\sqrt{\pi}} \sin{t}$$


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