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In the book of Filho Lopes, Weak solutions for the equations of incompressible and inviscid Fuid dynamics. Page 59 They want to prove the following:

Take $w^{\epsilon}_0$ a $C^{\infty}_c(\mathbb{R})^2$ initial vorticity, then defining $u^{\epsilon}_0$ the initial velocity field by Biot Savart law, we have it in the space $C^{\infty}(\mathbb{R}^2)\cap L^2_{loc}$, so that we can apply to it a theorem of existence of classical solutions and get a velocity flow in $C^1([0;\infty) \times \mathbb{R}^2)$ that solves the 2-D Euler equations.

Now they want to prove that $ ||u^{\epsilon}(.,t)||_{L^{\infty}} \leq C(||w^{\epsilon}_0||_{L^{\infty}} + ||w^{\epsilon}_0||_{L^1})$ For this they argue with the following, they consider the representation of $u^{\epsilon}(.,t)$ by $w^{\epsilon}(.,t)$ through the biot savart law and are able to show that $||u^{\epsilon}(.,t)||_{L^{\infty}} \leq C(||w^{\epsilon}(.,t)||_{L^{\infty}} + ||w^{\epsilon}(.,t)||_{L^1})$ And then they directly claim that this is controlled by the same quantities but at time 0.

A little above the theorem they quickly say that since the vorticity satisfy the transport equation $\partial_tw + u\cdot \nabla w =0$ its $L^{\infty}$ and other $L^p$ norms are preserved in time. They say that this comes from the fact that $w(.,t)$ is given from $w(.,0)$ by means of particle trajectories.

My problem is that I don't see why would it be that at each point the particles trajectories exists globally and are smooth so that we could use them to show theses conservations. Especially that a little later they claim that the estimates that they prove in this theorem allow them to have smooth and global particles trajectories. The estimates they show are the following:

$||u^{\epsilon}(.,t)||_{L^{\infty}} \leq C(||w^{\epsilon}_0||_{L^{\infty}} + ||w^{\epsilon}_0||_{L^1})$

$|u^{\epsilon}(x,t)-u^{\epsilon}(y,t)| \leq C|x-y|(||w_0||_{L^{\infty}} + ||w_0||_{L^1} - ||w_0||_{L^{\infty}}log^-(|x-y|)$

Am I wrong to think that global existence of the particles trajectories and their smoothness come only from the fact that $u^{\epsilon}$ is smooth and from the first estimate, or is really the second estimate necessary ? And also how could we justify rigourously preservation of the $L^{\infty}$norm of the vorticity without having the global existence of particle trajectories.

incas
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1 Answers1

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Well, you can do that by standard energy estimates. Consider the equation $$ \partial_t w+u\cdot\nabla w=0, $$ where $u$ is divergence free. Then you have the following estimates $$ \frac{d}{dt}\frac{1}{1+p}\int w^{p+1}=\int w^p\partial_tw=-\int u\cdot \nabla w w^p=-\frac{1}{1+p}\int u\cdot \nabla(w^{p+1})=\frac{1}{1+p}\int \nabla\cdot u w^{p+1}. $$ The last integral vanish due to the divergence free condition. Consequently, integrating in time, we have $$ \|w(t)\|_{L^{p+1}}=\|w(0)\|_{L^{p+1}} $$ To recover the $L^\infty$ case, just take the appropriate limit in $p$. Is this helping?

guacho
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  • Thanks,

    you mean that $||w(0)||{L^{p}} \to ||w(0)||{L^{\infty}} $ as $p \to \infty$ ?

     – incas Apr 14 '15 at 10:25
  • Ok, If you were meaning this reasoning then I have I problem with it then, because you need to already now that $w(.,t)$ is in $L^{\infty}$ in order to be able to say that $||w(.,t)||{L^p} \to ||w(.,t)||{L^{\infty}}$ as $p\to\infty$ – incas Apr 14 '15 at 14:14
  • 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 http://math.stackexchange.com/questions/242779/limit-of-lp-norm) – guacho Apr 14 '15 at 15:24
  • So what I mean precisely is the folowing:

    I do understand that for any $t$ : $||w(t)||{L^2}=||w(0)||{L^2}$ And since also $w(0) \in L^{\infty}$ then $\lim\limits_{p\to\infty}||w(t)||{L^p}=\lim\limits{p\to\infty}||w(0)||{L^p}= ||w(0)||{L^{\infty}}$ but how do we get on the other hand that $\lim\limits_{p\to\infty}||w(t)||{L^p}=||w(t)||{L^{\infty}}$

     – incas Apr 14 '15 at 15:48
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    Ok my apologies, $\lim\limits_{p \to \infty} ||w(t)||{L^p} = ||w(0)||{L^{\infty}}$ means that the function $p \to ||w(t)||{L^p}$ is bounded which implies that $w(t)$ is in $L^{\infty}$ and therefore we can apply the proposition you mentionned to have the equality $||w(t)||{L^{\infty}} = ||w(0)||_{L^{\infty}}$ – incas Apr 14 '15 at 15:54
  • Yes :-). Everything is clear now, right? – guacho Apr 14 '15 at 18:40
  • yes thank you ! – incas Apr 14 '15 at 21:54