= Solution
Use the real <zero-trace Sobolev space> $H_0^1(\Omega)$ with <norm> $\|\nabla u\|_2$. The <Poincaré inequality> on the bounded smooth <domain> gives $\|u\|_2\leq C_P\|\nabla u\|_2$, making this an equivalent <Sobolev norm>. The <weak formulation> is to find $u\in H_0^1(\Omega)$ with $u\geq0$ <almost everywhere> and
$$
\boxed{\int_\Omega\nabla u\cdot\nabla v\,dx
=\int_\Omega(1+\sqrt u)v\,dx\qquad\text{for every }v\in H_0^1(\Omega).}
$$
The zero <Dirichlet boundary condition> is encoded in the <Sobolev trace>. Formally, the identity follows by multiplying the <partial differential equation> by a <test function> and using <integration by parts>; it then extends to all $v\in H_0^1(\Omega)$ by density.
To check all terms, boundedness of $\Omega$ gives finite <Lebesgue measure> $m=|\Omega|$. The <Cauchy-Schwarz inequality> gives $u\in L^1(\Omega)$ and
$$
\|\sqrt u\|_2^2=\int_\Omega u\,dx\leq m^{1/2}\|u\|_2,
\qquad
\|1+\sqrt u\|_2\leq m^{1/2}+m^{1/4}\|u\|_2^{1/2}.
$$
Thus the right side is a bounded <linear functional> of $v$, since its absolute value is at most $\|1+\sqrt u\|_2 C_P\|\nabla v\|_2$. The left side is finite by the <Cauchy-Schwarz inequality>. These estimates work for every stated $n\geq2$.
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