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Thread: struggling with this advection problem

  1. #1
    Sep 2009

    struggling with this advection problem

    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

    <br />
\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 \\<br />
    <br />
\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<br />

    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.
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  2. #2
    Super Member Rebesques's Avatar
    Jul 2005
    My house.
    (...Ah, the Navier-Stokes equations, always a treat
    I hate 'em )

    Well... the relation <br />
\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<br />

    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)<br />

    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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