Elliptic PDE - max principle The maximum principle for elliptic PDEs is established for the nondivergence form as in http://www.ann.jussieu.fr/~frey/cours/UdC/ma691/ma691_ch3.pdf.
But what if we are dealing with the divergence form? Is there a corresponding maximum principle?
Say we have $-\text{div}(A(x)\nabla u(x))=0$. I can surely split this up into two components, one of which involves the non-divergence form. Can I sort of omit the other term to have a subsolution or supersolution? 
 A: No need in omitting any terms. The divergence form is much better,
reducing the proof of the elliptic PDE maximum principle to a rather trivial one,
while making the principle valid even for weak solutions, and sometimes called the essential maximum principle. Thus consider a weak solution $\,u\in H^1(\Omega)\,$ 
of the equation $\,\text{div}(A(x)\nabla u(x))=0\,$ in a bounded Lipschitz domain
$\,\Omega\subset\mathbb{R}^n$, $n\geqslant 2$, i.e., 
$$
\int\limits_{\Omega}\bigl(A(x)\nabla u,\nabla v\bigr)\,dx=0\quad \forall\, 
v\in C^{\infty}_0(\Omega)\tag{$\ast$}
$$
for some positive definite matrix $A=A(x)$ with elements $a_{ij}\in L^{\infty}(\Omega)$. The latter assumption is just meant to imply an equivalence of the integral identity $(\ast)$ to that with the test functions $\,v\in H^1_0(\Omega)$ instead.  Note that in the identity $(\ast)$, to designate the inner product in $\mathbb{R}^n$, an alternative notation $(\cdot\,,\cdot)$ for the dot product is used.  Denote by $\,{\rm ess\,sup}_{\Omega}\,$ the essential supremum over the domain $\,\Omega\,$ w.r.t. the $\,n$-dimensional Lebesgue measure, and let $\,{\rm ess\,sup}_{\partial\Omega}\,$ denote the essential supremum over the Lipschitz boundary $\,\partial\Omega\,$ w.r.t. the $\,(n-1)$-dimensional Lebesgue measure. Proof of the essential maximum principle is based on employing a one parameter function $\,\eta_{m}\,$ of the form
$$
\eta_{m}(\xi)=
\begin{cases}
0,\quad \xi\leqslant m,\\
\xi-m,\quad m<\xi\leqslant m+1,\\
1,\quad \xi\geqslant m+1.
\end{cases}
$$
Most important is the fact of $\,\eta_m\,$ being Lipschitz on $\mathbb{R}$.
Essential maximum principle. Let $\,u\in H^1(\Omega)\,$ be a weak solution
of the equation $\,\text{div}(A(x)\nabla u(x))=0\,$ in a bounded Lipschitz domain
$\Omega$. Then 
$$
{\rm ess\,sup}_{\Omega}u(x)\leqslant{\rm ess\,sup}_{\partial\Omega}u(x)
\tag{$\ast\ast$}
$$
Proof. In case the RHS of $(\ast\ast)$ is finite, denote it by $m$, and
notice that for any $\,u\in H^1(\Omega)$, a function $\,\eta_m(u)\in H^1(\Omega)\,$ since it is subject to the chain rule $\,\nabla \eta_m(u)=\eta'_m(u)\nabla u$. 
The choice of $\,m={\rm ess\,sup}_{\partial\Omega}u(x)\,$ implies that the trace 
$\,\eta_m(u)|_{\partial\Omega}=0$, and hence $\,\eta_m(u)\in H^1_0(\Omega)$. 
Choosing the test function $\,v=\eta_m(u)\,$ in $(\ast)$ yields
$$
\int\limits_{\Omega}\bigl(A(x)\nabla u(x),
\nabla u(x)\bigr)\eta'_m\bigl(u(x)\bigr)\,dx=0
$$
whence follows $\,\bigl(A(x)\nabla u(x),\nabla u(x)\bigr)\eta'_m\bigl(u(x)\bigr)=0\,$ a.e. in $\,\Omega$. Multiply the latter by $\,\eta'_m(u)\,$ to get $\,\bigl(A\nabla \eta_m(u),\nabla \eta_m(u)\bigr)=0\,$ which implies $\,|\nabla\eta_m(u)|=0$, and hence $\,\eta_m\bigl(u(x)\bigr)=C\,$ a.e. in $\,\Omega\,$. But it is clear that $\,C=0\,$ since the trace $\,\eta_m(u)|_{\partial\Omega}=0\,$. Therefore,  $\,\eta_m\bigl(u(x)\bigr)=0\,$ a.e. in $\,\Omega\,$ which implies $(\ast\ast)$. Q.E.D.
