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A Linearised N.S.E - advection operator help

  1. Mar 16, 2017 #1
    Hi PF,

    So I am trying to separate the adjoint advection operator taken from a linearised Navier-Stokes equation. It has been a while since I have delved into these equations, so need some assistance. Would someone be so kind as to check my working? The equation I am trying to solve:
    $$A = -\left(\mathbf{U} \cdot \nabla \right)\mathbf{u}^* + \left(\nabla \mathbf{U}\right)^{\rm{T}} \cdot \mathbf{u}^*,$$
    where
    $$\mathbf{U} = U\mathbf{i} + V\mathbf{j} + W\mathbf{k},$$
    $$\mathbf{u}^*=u^*\mathbf{i} + v^*\mathbf{j} + w^*\mathbf{k}.$$
    For the current system we also have the conditions, ##\frac{\partial \mathbf{U}}{\partial z} = U = V = 0##and ##\frac{\partial \mathbf{u}^*}{\partial z} = ik\mathbf{u}^* ## , where ## k ## is a streamwise wavenumber (i.e. ##z##-direction, ##W##-velocity direction). ## i## represents a complex number.

    Applying this to the first equation we have (I think),
    $$A = -\left[W\frac{\partial\left(u^*\mathbf{i} + v^*\mathbf{j} + w^*\mathbf{k}\right)}{\partial z}\right] + \left[\mathbf{k}\frac{\partial W}{\partial x} + \mathbf{k}\frac{\partial W}{\partial y}\right] \cdot \left(u^*\mathbf{i} + v^*\mathbf{j} + w^*\mathbf{k}\right) $$
    $$A = -\left[i k W \mathbf{u}^*\right] + w^*\left(\frac{\partial W}{\partial x}\mathbf{i} + \frac{\partial W}{\partial y}\mathbf{j}\right)$$

    I get a bit confused with the## \nabla \mathbf{U}## as I don't really remember covariant derivatives that well. In fact, I didn't even do the tensor calculus, I just remembered some assumptions based on the conditions I stated. Not that sure it's even close to correct.

    Thanks for your help in advance!!!!!
     
  2. jcsd
  3. Mar 21, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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