struggling with this advection problem

**nicedream** · September 4th 2009, 02:12 PM

This is the exercise I'm trying to solve:

Let $\Omega \subset \mathbb{R}^3$ be open and bounded. Suppose that $\overrightarrow{u} = \overrightarrow{u}(x,t)$ and $\theta = \theta(x,t)$ are regular for $t \geq 0$ , and satisfy

$ \frac{\partial \overrightarrow{u}}{\partial t} + ( \overrightarrow{u} . \nabla ) \overrightarrow{u} - \nu \Delta \overrightarrow{u} + \nabla p = \overrightarrow{a}(\theta), \hspace{15pt} \nabla . \overrightarrow{u} = 0 \\ $
$ \frac{\partial \theta}{\partial t} + ( \overrightarrow{u} . \nabla ) \theta - k \Delta \theta = b(\overrightarrow{u}) \hspace{25pt} \overrightarrow{u}_{|\partial \Omega} = \overrightarrow{0} \text{ and } \theta_{| \partial \Omega} = 0 $

with $| \overrightarrow{a} (\theta) | \leq \alpha | \theta |$ and $| b(\overrightarrow{u})| \leq \beta |\overrightarrow{u}|$

The problem is: give sufficient conditions over the constants $\alpha, \beta, \nu, \kappa > 0$ so that there exists $\delta > 0$ and the following estimation holds, $\forall t > 0$ :

$\int_{\Omega} \Big( |\overrightarrow{u}|^2 + \theta^2 \Big)(x,t) dx \leq e^{-\delta t} \int_{\Omega}\Big( |\overrightarrow{u}|^2 + \theta^2 \Big)(x,0) dx$

I though about multiply the first equation by $u$ , second by $\theta$ , integrate over $\Omega$ and try get somewhere, but I'm still lost. Any suggestions?

Thanks in advance.

**Rebesques** · January 18th 2010, 02:36 PM

(...Ah, the Navier-Stokes equations, always a treat

Not!
I hate 'em

)

Well... the relation $ \int_{\Omega} \Big( |u|^2 + \theta^2 \Big)(x,t) dx \leq e^{-\delta t} \int_{\Omega}\Big( |u|^2 + \theta^2 \Big)(x,0) dx $

is equivalent to

$\frac{1}{t}\log\left(\frac{\int_{\Omega}\Big( |u|^2 + \theta^2 \Big)(x,0) dx}{\int_{\Omega}\Big( |u|^2 + \theta^2 \Big)(x,t) dx}\right)\geq-\delta$ ,

which will ofcourse hold true if-f $\int_{\Omega}\Big( |u|^2 + \theta^2 \Big)(x,t) dx=: \Phi(u)(t)+\Psi(\theta)(t)$ does not have zero limit as $t\rightarrow 0+$ . Call this requirement $(R) $ .

So, the whole idea is to get $\Phi,\Psi$ (the Bochner norms, I think) bounded as $t\rightarrow+0$ .

Remember that spatially $u_{|\partial \Omega}=0, \theta_{|\partial \Omega}=0$ so by the Poincare inequality, these norms are bounded above and below by the $L^2$ norms of the respective gradient times a non-singular function of $t$ . So, we need just bound the $L^2$ norms of the gradients.

Back to the equations. Multiply the first by $u$ and integrate. Using some calculus and the fact that $u$ is divergence-free, we will obtain

$\int_{\Omega}\frac{1}{2}\left(|u^2|_t+\nu|\nabla u|^2\right)+(\nabla p)u dx= \int_{\Omega}a(\theta) udx$ .

Now, the left hand side can be made less than $\phi(t,\nu)\Phi(u)(t)$
for some function $\phi$ (calculations up to you

), while the right hand side will be less that $\alpha \rho(t)\Phi(u)(t)\Psi(\theta)(t)$ for some non singular function $\rho$ .

Let us demand $\phi(t,\nu)\leq \alpha \rho(t)\Psi(\theta)$ . This gives $\Psi(\theta)(t)\geq\frac{\phi(t,\nu)}{\alpha \rho(t)}$ , and by making an even grander demand we take the right hand side to have a nonzero limes inferior as $t\rightarrow 0$ .

In this way, the half of $(R)$ is met.
Proceed in the same manner with the other equation.

Ps. Ofcourse, this proof is a bit rough around the edges,
but I bet you' re a smart kid and will straighten things out.

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