TITLE: The equation $\,\,\Delta u+\cos u=0\,\,$ possesses a weak solution in $\,W^{1,2}_0(D)$
QUESTION [5 upvotes]: Let $D$ be an open bounded subset in $\mathbb{R}^{n}$, with sufficiently smooth boundary. Prove that there is a weak solution in $W^{1,2}_0$$(D)$ to following equation
$$\Delta u+\cos u=0.$$
Help me some hints to start. 
Thanks in advanced.

REPLY [6 votes]: Existence of a weak solution. An alternative approach using a fixed point method:
Let $v\in L^2(D)$ and $\varphi(v)\in W^{1,2}_0(D)$ be a weak solution of
$$
\Delta u+\cos v=0.
$$
Such a weak solution exists as this means for $\varphi(v)$ that
$$
\int_D \nabla\varphi(v)\cdot\nabla w\,dx= 
\int_{D}w\cos v\,dx \quad\text{for all $w\in W^{1,2}_0(D)$}.
$$
And since $\ell(w)=\int_{D}w\cos v\,dx$ is a bounded linear 
functional on $W^{1,2}_0(D)$, then there exists a $\varphi(v)\in W^{1,2}_0(D)$, 
such that 
$$
\ell(w)=\langle w,\varphi(v)\rangle_{W^{1,2}_0(\Omega)}=\int_\Omega \nabla\varphi(v)\cdot\nabla w\,dx.
$$ 
Note that $W^{1,2}_0(D)$ is a Hilbert space with inner product $\langle w,w'\rangle_{W^{1,2}_0(D)}=\int_D \nabla w\cdot\nabla w'\,dx$.
Also, for every $v\in L^2(D)$:
$$
\|\varphi(v)\|_{W^{1,2}_0(D)}=\|\ell\|\le \left(\int_D \cos^2 v\,dx|\right)^{1/2}
\le |D|^{1/2}=:M.
$$ 
Hence the
nonlinear functional $\varphi$ maps $L^2(D)$ into 
$$
B=\{w\in {W^{1,2}_0(D)}: \|w\|_{{W^{1,2}_0(D)}}\le M\}.
$$
Now $B\subset \{u\in L^2(D) : \|u\|_{L^2}\le N\}$, for some $N>0$, due to Poincaré inequality. In particular, $\varphi$ maps $B$ into $B$, and $\varphi[B]\subset B$. But $B$
is compact subset of $L^2(D)$, due to Rellich compactness theorem. Hence, 
Schauder fixed point theorem guarantees a fixed point $u$ for $\varphi$. Clearly $u\in L^2(D)$, but $u=\varphi(u)\in B\subset W^{1,2}_0(D)$.
The weak solution is also a classical solution. We have obtained a function $u\in W_0^{1,2}(D)$ satisfying $$\Delta u=-\cos u,$$ is the sense of distributions. But if $u\in W_0^{1,2}(D)$, then $\cos u\in W^{1,2}(D)$ and hence $\Delta u\in W^{1,2}(D)$, which implies that $u\in W^{3,2}(D)$. This in turn implies that $\cos u\in W^{3,2}(D)$ and recursively we obtain that
$$u\in W^{k,2}(D), \quad \text{for all $k\in\mathbb N$},$$ and thus $u$ is a classical solution.