Monotone operator method for Hammerstein equations

Remarks

Let $S$ be the positive square root of $K$, so $K = S^2$ with $S$ self-adjoint. Look for $u$ of the form $u = f - S v$, and put $G(v) = v - S F(f - S v)$. For $a = f - S v$ and $b = f - S w$ one finds $\ip{G(v) - G(w)}{v - w} = \norm{v - w}^2 + \ip{F(a) - F(b)}{a - b} \ge \norm{v - w}^2$. So $G$ is strongly monotone, hemicontinuous and coercive, and Minty's theorem in Hilbert space, the case of the Browder–Minty theorem, gives a unique $v$ with $G(v) = 0$. Then $u = f - S v$ solves the equation, and each solution $u$ gives a zero $v = S F(u)$ of $G$, which yields uniqueness.

Application: let $\Omega \subseteq \R^m$ be measurable with finite measure and $H = L^2(\Omega)$. Let $k$ be a symmetric square-integrable kernel with $\int\!\!\int k(x, t)\, v(x) v(t)\, dx\, dt \ge 0$ for all $v$, and let $g(x, u)$ be a Carathéodory function that is nondecreasing in $u$ with $\abs{g(x, u)} \le a(x) + b \abs{u}$, $a \in L^2(\Omega)$. Then the Nemytskii operator $F = N_g$ is continuous and monotone on $L^2(\Omega)$, and $\varphi(x) + \int_\Omega k(x, t)\, g(t, \varphi(t))\, dt = f(x)$ has exactly one solution in $L^2(\Omega)$ for each $f \in L^2(\Omega)$.

The signs matter. In $H = \R$ with $K = 1$ and $F(u) = -u$, which is not monotone, the equation reads $0 = f$ and has no solution for $f \ne 0$.

Going deeper

For a student

The theorem is one of the main uses of monotone operator theory outside partial differential equations. The trick with the square root $S$ moves the problem to one where the operator is strongly monotone, which Browder–Minty theory handles directly. Positive kernels arise as Green's functions of self-adjoint boundary value problems, so the theorem gives existence and uniqueness for $-u'' + g(x, u) = f$ with Dirichlet conditions when $g$ is nondecreasing in $u$. Versions without symmetry of $K$ and in Banach spaces exist, where $K$ maps $X^*$ to $X$ and $F$ maps $X$ to $X^*$.