= Solution
Let $P=\{w\in L^2(\Omega):w\geq0\text{ almost everywhere}\}$. For $g\in L^2(\Omega)$, define $Gg=z\in H_0^1(\Omega)$ by
$$
\int_\Omega\nabla z\cdot\nabla v\,dx=\int_\Omega gv\,dx
\qquad(v\in H_0^1(\Omega)).
$$
The <Lax-Milgram theorem> guarantees a unique $z$: the <bilinear form> on the left is bounded and <coercive> in the <norm> $\|\nabla\cdot\|_2$, and the right side is a bounded <linear functional> by the <Poincaré inequality>. Testing with $z$ gives
$$
\|\nabla Gg\|_2\leq C_P\|g\|_2,
\qquad\|Gg\|_2\leq C_P^2\|g\|_2.
$$
Define $T:P\to L^2(\Omega)$ by $Tw=G(1+\sqrt w)$. This actually maps into $P$. Indeed the <weak maximum principle> says that a <weak solution> $z\in H_0^1(\Omega)$ of $-\Delta z=g\geq0$ is nonnegative <almost everywhere>. Here $1+\sqrt w\geq0$ and belongs to $L^2(\Omega)$. Directly, taking the <negative part> $z^-=\max\{-z,0\}$ as a <test function> gives $-\|\nabla z^-\|_2^2=\int_\Omega gz^-\,dx\geq0$, so $z^-=0$ by the <Poincaré inequality>.
The required <fixed point> formulation is
$$
\boxed{u\in P,\qquad u=Tu=G(1+\sqrt u).}
$$
Every such <fixed point> belongs automatically to $H_0^1(\Omega)$ and satisfies part (b)(i). Conversely, uniqueness in the <Lax-Milgram theorem> makes every nonnegative <weak solution> a <fixed point> of $T$.
Back to article page
