Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In signal processing theory, I found this integral equation which I suppose to be of Hammerstein type: $$u(t)-\int_0^1\frac{\cos(\omega t+\phi)}{u(\phi)}d\phi=0$$ I didn't find anything in literature apart this:

Could someone give me some hints how to solve this equation? Thanks.

share|cite|improve this question
up vote 1 down vote accepted

Note that we have $$u''(t) + \omega^2 \int_0^1 \dfrac{\cos(\omega t+ \phi)}{u(\phi)} d\phi =0\implies u''(t)+\omega^2 u(t) = 0 \implies u(t) = ae^{i\omega t}+b e^{-i \omega t}$$ Obtain $a$ and $b$ by plugging it back into the equations.

share|cite|improve this answer

$u(t)-\int_0^1\dfrac{\cos(\omega t+\phi)}{u(\phi)}d\phi=0$

$u(t)=\int_0^1\dfrac{\cos\omega t\cos\phi-\sin\omega t\sin\phi}{u(\phi)}d\phi$

$u(t)=\cos\omega t\int_0^1\dfrac{\cos\phi}{u(\phi)}d\phi-\sin\omega t\int_0^1\dfrac{\sin\phi}{u(\phi)}d\phi$

$u(t)=C_1\cos\omega t+C_2\sin\omega t$

$\therefore C_1\cos\omega t+C_2\sin\omega t\equiv\cos\omega t\int_0^1\dfrac{\cos\phi}{C_1\cos\omega\phi+C_2\sin\omega\phi}d\phi-\sin\omega t\int_0^1\dfrac{\sin\phi}{C_1\cos\omega\phi+C_2\sin\omega\phi}d\phi$

The only problem is to determine $C_1$ and $C_2$ .

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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