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 Apr14 comment Conservation of norms by the 2-d euler vorticity equation Yes :-). Everything is clear now, right? Apr14 comment Conservation of norms by the 2-d euler vorticity equation Mmm, I don't know if I'm understanding you correctly. What I'm saying is, that if you assume $w(0)\in L^1\cap L^\infty$, then definitely, you will have a global bound $w(t)\in L^2$. Then you have that $$\|f\|_{L^p}\rightarrow \|f\|_{L^\infty},$$ as $p\rightarrow\infty$ for $f\in L^q$ (see math.stackexchange.com/questions/242779/limit-of-lp-norm) Apr13 answered Conservation of norms by the 2-d euler vorticity equation Apr9 comment $L^{\infty}$ estimate on the boundary by means of Sobolev norms in the interior jajajaja ouch, my bad. No, no, that was what I said. It's just that for some reason I though that you were using the Evans version of the Theorem with a full derivative. I'm sorry for the confusion. Apr7 comment $L^{\infty}$ estimate on the boundary by means of Sobolev norms in the interior Well, maybe it's better if you, instead of applying the trace theorem with a full derivative, apply the trace theorem with half derivative: $$\|u\|_{H^s(\partial \Omega)}\leq C\|u\|_{H^{s+0.5}(\Omega)}.$$ Does this help? Mar17 comment Regularity of a parabolic equation I already did :-). Check chapter 7 in the Evans's book. By parabolic smoothing I mean that the solution will gain some space derivatives in some integral sense in time. Something like $$\mu\in L^2_t H^1_x$$ Mar17 comment Regularity of a parabolic equation Well, you can read about this kind of parabolic pde in the book by Evans "Partial differential equations". Actually,if you assume $\nu$ to be smooth you can apply the standard energy methods straightforwardly to check that the solution is smooth. If the initial data is merely $L^1$, then you have to argue with parabolic smoothing. Mar10 comment $\|u\|_{L^{3}(\mathbb R)} \leq C \|Du\|_{L^{2}(\mathbb R)}^{\alpha} \|u\|_{L^{2}(\mathbb R)}^{1-\alpha}$? In one dimension you have $$\|u\|_{L^\infty}^2\leq C\|u\|_{L^2}\|D u\|_{L^2}$$ Mar10 answered $\|u\|_{L^{3}(\mathbb R)} \leq C \|Du\|_{L^{2}(\mathbb R)}^{\alpha} \|u\|_{L^{2}(\mathbb R)}^{1-\alpha}$? Mar3 answered Gronwall's inequality and polynomial Mar3 answered Show that $2\nabla \sqrt f\,+\,x \sqrt f=0$ (a.e.). $\implies$ $\sqrt f\in \mathcal C_0$. (Derivatives are in weak sense) Jun12 comment Convergence weak* in $L^{\infty}([0,T];L^2(\mathbb{T}^2))$ implies convergence weak* a.e in $L^2(\mathbb{T}^2)$? Why does $x_n$ tend weakly to zero? I see that this should be true, but I don't see the proof now. Please, can you elaborate? Jun4 accepted A bound in Sobolev spaces of negative order Jun3 asked A bound in Sobolev spaces of negative order May15 awarded Yearling May6 comment Bounded data means bounded solution to parabolic PDE Concerning you comment on the sign of the laplacian: the approximate solution will be as smooth as you want, isn't it? However, the problem with the boundary is pertinent. Even if you mollify the initial $L^\infty$ data (which gives you $C^1$ function with the appropriate $L^\infty$ bound) there is a problem close to the boundary. May6 comment Bounded data means bounded solution to parabolic PDE Well, I don't know it it's seems unlikely or not. You can NOT establish differentiability. What you do is to prove that M(t) is Lipschitz, ergo, it is differentiable a.e.. Once you have this, you prove that $M'(t)=\partial_t u(x_t,t)$. Of course that $M(t)$ may be attained in more than one point, however, this is not relevant since at every point will decay. This "overcomplicated" approach is widely spread (check this, for instance, uam.es/personal_pdi/ciencias/acordoba/documentos/articulos/…) Apr30 comment Bounded data means bounded solution to parabolic PDE Actually you need $C^1([0,T]\times\Omega)$ for the approximated solutions. Apr30 answered Bounded data means bounded solution to parabolic PDE Apr29 answered PDE Evans Chapter 7 problem 16