Hammerstein integral equation

Remarks

It is the Urysohn equation with the product kernel $K(x, t, u) = k(x, t)\, g(t, u)$. Many authors write it as $\varphi + K N_g \varphi = f$, which only changes the sign of $g$.

Example: on $[0, 1]$, take $f = 0$, $k(x, t) = x t$ and $g(t, u) = u^2$. Try $\varphi(x) = c x$; the right side is $x \int_0^1 t \cdot c^2 t^2\, dt = \frac{c^2}{4} x$, so $c = c^2/4$, that is $c = 0$ or $c = 4$. So $\varphi = 0$ and $\varphi(x) = 4x$ are two solutions, and the equation has no uniqueness.

Non-example: $\varphi(x) = \int_0^1 \sin\bigl(x\, \varphi(t)\bigr)\, dt$ is a Urysohn equation but not a Hammerstein one with a single product kernel, because $\sin(xu)$ is not of the form $k(x, t)\, g(t, u)$. Indeed, the functions $u \mapsto \sin(x u)$ for different $x$ are not all multiples of one function.

Example of origin: on $[0, 1]$ the problem $-u'' = h(x, u)$, $u(0) = u(1) = 0$, with continuous $h$, is equivalent to $u(x) = \int_0^1 G(x, t)\, h(t, u(t))\, dt$, where $G(x, t) = \min(x, t)\bigl(1 - \max(x, t)\bigr)$ is the Green's function.