Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Shallow water Lagrangian

  1. Dec 2, 2010 #1
    1. The problem statement, all variables and given/known data

    In a shallow layer of water, the velocity of water in the z direction may be ignored and is therefore [tex](\dot{x},\dot{y})[/tex]. We can define the Lagrangian coordinates such that the depth of water h is satisfied by the relations

    Given that [tex]h = \frac{1}{\alpha}[/tex] and [tex]\alpha = \frac{\partial(x,y)}{\partial(a,b)}[/tex]

    and the Lagrangian density is given as

    [tex]L = \frac{1}{2}\dot{x}^2 + \frac{1}{2}\dot{y}^2 - \frac{1}{2}gh(x_a,x_b,y_a,y_b)[/tex]

    where [tex]p_q = \frac{\partial p}{\partial q}[/tex].

    Given further that Lagrange's equations for a 2D continuous system are known to be

    \frac{D}{Dt}\left( \frac{\partial L}{\partial\dot{x}} \right) + \frac{\partial}{\partial a} \left( \frac{\partial L}{\partial x_a} \right) + \frac{\partial}{\partial b} \left( \frac{\partial L}{\partial x_b} \right) - \frac{\partial L}{\partial x} = 0

    with a similar equation for the y variable, prove that

    [tex] \frac{D\dot{x}}{Dt} + g \frac{\partial h}{\partial x} = 0 [/tex]

    2. Relevant equations

    I know the general approach of this problem, but my main problem comes in substituting

    [tex] \frac{\partial}{\partial a} \frac{\partial L}{\partial x_a} [/tex].

    If I apply chain rule on the Lagrangian here,
    [tex] \frac{\partial}{\partial a} \frac{\partial L}{\partial h} \frac{\partial h}{\partial x_a} = -\frac{g}{2} \frac{\partial}{\partial a} \frac{\partial h}{\partial x_a} [/tex]
    How do I proceed after this point?
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted
Similar Discussions: Shallow water Lagrangian
  1. QFT lagrangian (Replies: 1)