2
$\begingroup$

Let $\varphi:[0,1]\to\mathbb{R}$ be nondecreasing and continuous, and let $s\in L^\infty(\Omega\times[0,T],[0,1])$ be such that $\varphi(s)\in L^1(0,T;H^1(\Omega))$ be a solution of $$ \partial_t s =\Delta[\varphi(s)]$$ $$ s(x,0)=s_0(x)$$ for a measurable function $s_0(x)$. ($\Omega$ is a bounded domain)

Now a weak definition in the homogeneous Dirichlet case should look like this $$\int (s_0(x)-s(t))\partial_t \xi +\nabla [\varphi(s)]\nabla \xi =0$$ for every $\xi\in C_c^\infty([0,T)\times \Omega)$.

From this, it shall be possible to show, that $s(t)\rightharpoonup s_0(x)$ as $t\to 0$ in $L^1(\Omega)$ (and $t$ does not hit a set of zero measure $E$). A prior step could be to show that $s(t)\rightharpoonup* s_0$ in $L^1(\Omega)$(and the probably the dunford-pettis theorem is also used)

I don't see what a good starting point is. At least I can imagine where the set $E$ is coming from. Choosing in the weak definition $\xi$ to be a cut off function of an interval $[0,\tau ]$, i get pointwise (in $t$) convergence of the integrals. But I don't see how this should help with weak(*) $L^1$-convergence since $\xi$ has to be differentiable in space.

Ar there any suggestions or ideas? Am I just missing something?

Edit: What one receives quite directly is the distributional convergence of $s(t)\to s_0$ for $t\to 0$.

Let $\xi=\alpha(t)\beta(x)$ and $\alpha$ be a cut-off function of some $[0,t_0]$ where $t_0$ is a Lebesgue point of $\int_\Omega s(x,t) dx$. Then we obtain (using the properties of Lebesgue points $$\int_\Omega [s_0(x)-s(x,t_0)]\beta(x) dx = -\int_0^{t_0} \int_\Omega \nabla(\varphi(s))\nabla \beta. $$

Now clearly, the right hand side goes to zero as $E\not\ni t_0\to 0$ for $\beta\in C_c^\infty(\Omega)$ and we obtain the distributional convergence.

Note that $s(x,t)\in [0,1]$. Is it then possible to deduce weak convergence in some $L^p$ space (preferably for $p=1$)? Due to the bound on $s$ there should be something possible, right?

$\endgroup$
4
  • $\begingroup$ Shouldn't the distrbiutional convergence suffice? since $\{s(t)\}$ is a bounded set in $L^p(\Omega)$ for $t\in (0,T)\setminus E$, we know that it is weakly compact (fpr $p=1$ we use that it is also equiintegrable). Because of the distributional convergence the weak limit is characterized and hence we infer that for any sequence $t_k\not\in E\downarrow 0$ there is a subsequence such that $s(t_{k_l})\to s_0(x)$ in $L^p(\Omega)$. With a standard argument we conclude that even $s(t_k)\to s_0(x)$ in $L^1(\Omega)$. Any comments? $\endgroup$ Commented Jan 15, 2014 at 14:04
  • $\begingroup$ $L^1(\Omega)$ is not a dual space (indeed not isometric isomorph to any dual space) - hence you cannot speak about weak-* convergence in $L^1(\Omega)$. $\endgroup$ Commented Jan 16, 2014 at 8:32
  • $\begingroup$ Imho, the weak solution should satisfy $\int -s(t) \, \partial_t \xi -s_0 \, \xi(0) +\nabla [\varphi(s)]\nabla \xi =0$? $\endgroup$ Commented Jan 16, 2014 at 8:34
  • $\begingroup$ @ your first comment: I think in that case one considers $L^1(\Omega)$ to be a subspace of the signed Radon measures. @ your second comment: you just evaluated the integral. $\endgroup$ Commented Jan 16, 2014 at 9:06

1 Answer 1

Reset to default
1
$\begingroup$

From your PDE, you get $\partial_t s \in L^2(0,T; H^{-1}(\Omega))$ in the weak sense. Hence, $s \in H^1(0,T; H^{-1}(\Omega)) \hookrightarrow C(0,T; H^{-1})$. This gives the convergence of $s(t)$ towards $s(0) = s_0$ in $H^{-1}(\Omega)$.

If you even have $s \in L^2(0,T; H^1_0(\Omega))$, then you have $s \in C(0,T; L^2(\Omega))$, hence convergence in $L^2(\Omega)$.

$\endgroup$
4
  • $\begingroup$ You probably mean convergence in $H^{-1}(\Omega)$, right? Unfortunately, I won't have $s\in L^2(0,T;H^1_0(\Omega))$. Maybe I need to think more about my model problem, somehow your answer seems to easy though I definitely see your point. Thanks very much $\endgroup$ Commented Jan 16, 2014 at 9:17
  • $\begingroup$ Sorry: does strong convergence in $H^{-1}$ imply weak convergence in $L^2$? $\endgroup$ Commented Jan 16, 2014 at 9:24
  • $\begingroup$ Yes, I mean strong convergence in $H^{-1}$. In general, this does not imply weak convergence in $L^2$. However, one can employ more information: Since $s(t)$ is bounded in $L^2$, you can extract a weakly convergent subsequence for $t \to 0$. Since this subsequence converges also in $H^{-1}$, the limit is unique. Hence, $s(t)$ converges weakly to $s_0$ in $L^2(\Omega)$. $\endgroup$ Commented Jan 16, 2014 at 11:28
  • $\begingroup$ yeah, somehow this is what i explained in my last edit but considering the weaker distributional convergence. But thanks for agreeing to my idea :) $\endgroup$ Commented Jan 16, 2014 at 19:41

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.