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-   -   Shallow water Lagrangian (http://www.physicsforums.com/showthread.php?t=453072)

 bigevil Dec2-10 10:01 AM

Shallow water Lagrangian

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 $$(\dot{x},\dot{y})$$. We can define the Lagrangian coordinates such that the depth of water h is satisfied by the relations

Given that $$h = \frac{1}{\alpha}$$ and $$\alpha = \frac{\partial(x,y)}{\partial(a,b)}$$

and the Lagrangian density is given as

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

where $$p_q = \frac{\partial p}{\partial q}$$.

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

$$\frac{D\dot{x}}{Dt} + g \frac{\partial h}{\partial x} = 0$$

2. Relevant equations

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

$$\frac{\partial}{\partial a} \frac{\partial L}{\partial x_a}$$.

If I apply chain rule on the Lagrangian here,
$$\frac{\partial}{\partial a} \frac{\partial L}{\partial h} \frac{\partial h}{\partial x_a} = -\frac{g}{2} \frac{\partial}{\partial a} \frac{\partial h}{\partial x_a}$$
How do I proceed after this point?

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