# struggling with this advection problem

• Sep 4th 2009, 03:12 PM
nicedream
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?

• Jan 18th 2010, 03:36 PM
Rebesques
(...Ah, the Navier-Stokes equations, always a treat (Clapping)
Not!
I hate 'em (Punch))

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 :p), 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.