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A Conservation law form of Navier Stokes Equation

  1. Nov 19, 2016 #1
    I am pretty confused about how to write Navier-Stokes Equation into conservation form, it seems that from my notes,
    first, the density term with the pressure gradient dropped out.
    and second, du^2/dx seems to be equal to udu/dx.
    Why is it so? I attached my notes here for your reference.
    upload_2016-11-19_22-38-9.png
    upload_2016-11-19_22-37-37.png
     
  2. jcsd
  3. Nov 24, 2016 #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.
     
  4. Nov 25, 2016 #3
    This is pretty straightforward to do. Integrating by parts, $$u\frac{\partial u}{\partial x}+v\frac{\partial v}{\partial y}=u\frac{\partial u}{\partial x}+\frac{\partial (uv)}{\partial y}-u\frac{\partial v}{\partial y}$$Next, adding and subtracting ##u(\partial u/\partial x)## gives:
    $$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=2u\frac{\partial u}{\partial x}+\frac{\partial (uv)}{\partial y}-u\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)$$
    But, from the continuity equation, the last term in parenthesis in this equation is zero. So$$u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=2u\frac{\partial u}{\partial x}+\frac{\partial (uv)}{\partial y}=\frac{\partial u^2}{\partial x}+\frac{\partial (uv)}{\partial y}$$
    The rest of the derivation is straightforward.
     
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