2
$\begingroup$

(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.

Improve this question
$\endgroup$
5
  • $\begingroup$ Is $p\in R$ really negative, or is it just a typo? In which sense do you understand $\varphi '$ (I guess $\partial \varphi/\partial t$) if you only have $\varphi\in \mathcal{C}^1(0,T;L^2(\Omega))$ as a time regularity? Of course with your assumptions you also have $\varphi\in L^{2}(0,T;H^1(\Omega))\hookrightarrow L^{2}(0,T;H^{1/2}(\partial\Omega))\hookrightarrow L^{2}(0,T;L^2(\partial\Omega))$, but this still doesn't make sense of the time derivative on the boundary. I'm a little surprised by your $\partial\Omega$ boundary terms, can you tell us what is your (degenerate) parabolic PDE? $\endgroup$
    – leo monsaingeon
    Commented Apr 16, 2014 at 18:21
  • $\begingroup$ @leomonsaingeon $p$ is negative. Hmm, I guess I should think more carefully about the space $\varphi$ belongs to.. perhaps also $\varphi \in L^2(0,T;H^1(\partial\Omega)) \cap H^{-1}(0,T;L^{2}(\partial\Omega))$. Maybe $\varphi$ should be smooth on the closure. I am just starting with this formulation, i just called it degenerate due to the negative power. $\endgroup$
    – riem
    Commented Apr 16, 2014 at 19:39
  • $\begingroup$ What is the PDE, then? it seems you have time derivatives on the boundary, which I've never seen before... $\endgroup$
    – leo monsaingeon
    Commented Apr 16, 2014 at 20:36
  • $\begingroup$ @leomonsaingeon Sorry for late response. There is no PDE so to speak, it's jsut something I thought about. I have heavily changed the question (it's much easier now). $\endgroup$
    – riem
    Commented Apr 21, 2014 at 14:31
  • $\begingroup$ So there is a PDE in the end, and you were basically trying to compare a subsolution and a supersolution of your PDE. Your computation must have gone wrong at some point: the weak formulation shouldn't involve time derivatives on the boundary, but only in the interior $\Omega\times(0,T)$ of the parabolic domain. $\endgroup$
    – leo monsaingeon
    Commented Apr 21, 2014 at 15:52

1 Answer 1

Reset to default
2
$\begingroup$

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).

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thank you Leo. I in fact was reading the very book. The author claims on page 95 the topic of my OP (near equation 5.27) but does not say why it is true. The theorem you cited I believe is more general (i.e. the weak formulation is different) so I believe whatever the comparison principle the author refers to on page 94 probably is simpler than the one on page 132 hence my question. $\endgroup$
    – riem
    Commented Apr 21, 2014 at 16:06
  • $\begingroup$ I updated my previous answer $\endgroup$
    – leo monsaingeon
    Commented Apr 21, 2014 at 16:31
  • $\begingroup$ One last comment: the "$a=\frac{F(u)-F(v)}{u-v}$" trick is preciesly what makes the duality method work for very weak solutions, see the proof of theorem 6.5 (with the extra difficulty that one first has to truncate to avoid the degenerate levelset $u=0$). $\endgroup$
    – leo monsaingeon
    Commented Apr 21, 2014 at 16:37
  • $\begingroup$ No problem, you're very welcome! $\endgroup$
    – leo monsaingeon
    Commented Apr 22, 2014 at 22:22
  • $\begingroup$ Dear Leo, as the linear maximum principle (applied to $z$) that you mentioned at the end, could you please provide a reference? Further, the $a$ appearing here is the function $a=a(x,t)$ or the constant $a=a(x_0,t_0)$? Many thanks for the clarification $\endgroup$
    – Fawen90
    Commented Jun 7, 2023 at 13:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .