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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.

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2 Answers 2

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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.

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

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

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

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

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

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

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