6
$\begingroup$

Solve $f(x)=\int_{x-1}^{x+1} f(t) \mathrm{d}t$.

When $f$ is a function, it looks like the only solution is $f(x)=0$. But what if we allow distributions, such as the Dirac delta?

$\endgroup$
2
  • 6
    $\begingroup$ $f(x) = A \cos(cx) + B \sin(cx)$ works where $c \approx 1.895494267$ is the positive solution of $c = 2 \sin c$. There should also be complex solutions $c$ of this equation, giving rise to further linearly independent solutions $f(x) = (A \cos sx + B \sin sx) \, e^{tx}$ where $c = s+it$. $\endgroup$ Jan 31, 2014 at 18:37
  • 1
    $\begingroup$ crossposted at math.stackexchange.com/questions/658615/… $\endgroup$ Jan 31, 2014 at 18:51

1 Answer 1

26
$\begingroup$

By differentiating we obtain $$f'(x)=f(x+1)-f(x-1)$$ This type of equations was addressed in the MO question On equation f(z+1)-f(z)=f'(z) Let $\lambda$ be any (complex) root of the equation $$\lambda=e^\lambda-e^{-\lambda},$$ which is equivalent to $z=2\sin z$, as Noam wrote. to this $\lambda$ a solution $f(x)=e^{\lambda x}$ is associated. So you have infinitely many exponential solutions. Any linear combination is also a solution. Then, depending on your assumptions of $f$ you can consider various limits of those linear combinations in the appropriate topology for your functions/distributions class.

Edit. For a complete theory of this kind of equations see "Fonctions moyenne-periodiques", a theory created by Delsart and Schwartz in 1940-s. This is generalized in the modern theory of "equations of convolution-type", see, for example Hormander, Linear Partial differential operators. In this theory one considers equations $u\star f=0$, where $u$ is a distribution. In your case, $u=\delta-\chi,$ where $\chi$ is the characteristic function of $[-1,1]$. The method of solution is an appropriate version of Fourier--Laplace transform, depending on your class of functions/distributions. Ordinary Fourier transform of $u$ is $U(\lambda)=1-2\sin\lambda/\lambda$, whose roots give you the exponential solutions.

The question whether we obtain all solutions in this way is addressed by the Malgrange approximation theorem (Theorem 16.4.1 in Hormander).

$\endgroup$
3

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.