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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?

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  • $\begingroup$ 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? $\endgroup$ – Quickbeam2k1 Jan 15 '14 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 $L^1(\Omega)$. $\endgroup$ – gerw Jan 16 '14 at 8:32
  • $\begingroup$ Imho, the weak solution should satisfy $\int -s(t) \, \partial_t \xi -s_0 \, \xi(0) +\nabla [\varphi(s)]\nabla \xi =0$? $\endgroup$ – gerw Jan 16 '14 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. $\endgroup$ – Quickbeam2k1 Jan 16 '14 at 9:06
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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 $L^2(\Omega)$.

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  • $\begingroup$ 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 $\endgroup$ – Quickbeam2k1 Jan 16 '14 at 9:17
  • $\begingroup$ Sorry: does strong convergence in $H^{-1}$ imply weak convergence in $L^2$? $\endgroup$ – Quickbeam2k1 Jan 16 '14 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 $L^2(\Omega)$. $\endgroup$ – gerw Jan 16 '14 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 :) $\endgroup$ – Quickbeam2k1 Jan 16 '14 at 19:41

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