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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 $\displaystyle \Omega \subset \mathbb{R}^3$ be open and bounded. Suppose that $\displaystyle \overrightarrow{u} = \overrightarrow{u}(x,t)$ and $\displaystyle \theta = \theta(x,t)$ are regular for $\displaystyle t \geq 0$, 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 $\displaystyle | \overrightarrow{a} (\theta) | \leq \alpha | \theta |$ and $\displaystyle | b(\overrightarrow{u})| \leq \beta |\overrightarrow{u}|$

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

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

    I though about multiply the first equation by $\displaystyle u$, second by $\displaystyle \theta$, integrate over $\displaystyle \Omega$ 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 $\displaystyle
    \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

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

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

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

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

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

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

    Now, the left hand side can be made less than $\displaystyle \phi(t,\nu)\Phi(u)(t)$
    for some function $\displaystyle \phi$ (calculations up to you ), while the right hand side will be less that $\displaystyle \alpha \rho(t)\Phi(u)(t)\Psi(\theta)(t)$ for some non singular function $\displaystyle \rho$.

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

    In this way, the half of $\displaystyle (R)$ 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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