(Crossposted from <https://math.stackexchange.com/questions/757672/how-to-prove-comparison-principle-for-parabolic-pde-nonlinear>)


Suppose $F:\mathbb{R} \to \mathbb{R}$ is smooth with $F(x) > 0$ for $x > 0$ and $F:(0,\infty) \to (0,\infty)$ continuous and increasing with $F(0) = 0$.


Consider the PDE 
$$u\_t = \Delta F(u) \quad \text{on $\Omega$}$$
$$u(x,0) = u\_0$$
$$u(x,t) = C\_1\quad \text{on $\partial \Omega$}$$


Suppose we have two weak solutions $u$ and $v$ in $L^2(0,T;H^1)\cap H^{-1}(0,T;H^{-1})$ with initial data $u\_0$ and $v\_0$ and boundary data $C\_u$ and $C\_v$ respectively, where
$u\_0 \leq v\_0$ and $C\_u \leq C\_v$.


How do I show that the comparison principle holds: that $u \leq v$?


Assume more smoothness of the solutions if necessary. I can't do it. I know we need to test with $(u(t)-v(t))^+$ but I don't know what to do with the nonlinear term.



A comparison principle for parabolic equation
With your assumptions on $F$ your PDE $\partial\_t u=\Delta F(u)$ is usually referred to as the "generalized Porous Medium Equation" (the "real" PME would be for the specific choice of the nonlinearity $F(u)=u^m$ for fixed $m>1$ and non-negative solutions, or $F(u)=|u|^{m-1}u$ if you're interested in signed solutions as well). This is a widely studied topics and you should definitely give a look at Vazquez's book ["The Porous Medium Equation", Oxford science publications, 2007].


Regarding your comparison principle: the result is true both for weak and very weak solutions, but it requires $u,F(u)\in L^{2}(\Omega\times(0,T))$ and there are various issues/subtlelties depending on the weak formulation you want to use. The exact statement in Vazquez's book is theorem 6.5 page 132 (at least in my version). The proof is not trivial and relies on the so-called duality method.


**Update** on the last OP's comment (too long for one more comment box): at this stage of the proof (his equation 5.27) Vazquez is arguing for classical positive solutions $u,v\geq \varepsilon>0$ (actually $u\_n$ and $u\_{n+1}$). For these solution the trick is the following: the function $a(x,t)=\frac{F(u)-F(v)}{u-v}(x,t)$ can be defined as a positive function everywhere, for if a contact point $u=v$ occurs at some $(x\_0,t\_0)$ then you can extend by continuity $a(x\_0,t\_0)=F'(u(x\_0,t\_0))>0$. This really exploits the fact that $u,v$ are bounded away from zero, so that $F'>0$ there. Then substracting the two differential inequalities you see that $z=u-v$ satisfies
$$\partial\_t z\leq\Delta(a z)=a\Delta z+\nabla a\cdot\nabla z.$$
Because $a>0$ you can now use the linear maximum principle to deduce that $z=u-v\leq 0$ (of course the initial and boundary data come into play, but with the good sign since this is part of your asusmption).


