Vector calculus in classical fields

Question

The action is defined as: $$S = \int d^2\textbf{x}\,dt \left[\left(\frac{\partial h}{\partial t}\right)^2 + (\nu \,\nabla^2h)^2\right]$$

The equation of motion is asked for, so use Euler-Lagrange: $$\frac{\partial\mathcal{L}}{\partial h} = \frac{\partial}{\partial t} \frac{\partial\mathcal{L}}{\partial(\partial h/\partial t)} + \nabla \cdot \frac{\partial \mathcal{L}}{\partial(\nabla h)}$$

1. Now I'm having trouble evaluating $\frac{\partial \mathcal{L}}{\partial(\nabla h)}$. How do you evaluate: $$\frac{\partial}{\partial\,\nabla h}\left( (\nabla^2h)^2 \right)$$

2. The solution simply states $$\frac{\partial \mathcal{L}}{\partial(\nabla h)}=-2\nu\,\nabla\,\nabla^2h$$ with a comment saying 'where we have freely integrated by parts'. Can someone write out this approach explicitely? I'm not sure how to integrate by parts with gradient/laplacian operators.

Thanks in advance.

Answer 1

The solution plays fast and loose with the calculus of variations, and uses a trick (or a rule of thumb) that you can usually get away with, but is not obvious to the beginner. Here's how it works:

The action is $$ S = \int d^2\textbf{x}\,dt \left[\left(\frac{\partial h}{\partial t}\right)^2 + (\nu \,\nabla^2h)^2\right]. $$ If we integrate the last term by parts (this is what is meant by "freely integrating by parts"), the action becomes to $$ S = \int d^2\textbf{x}\,dt \left[\left(\frac{\partial h}{\partial t}\right)^2 + \nu^2 [\nabla^2h] [\nabla \cdot (\nabla h)] \right] \\= \int d^2\textbf{x}\,dt \left[\left(\frac{\partial h}{\partial t}\right)^2 - \nu^2 [\nabla (\nabla^2h)] \cdot (\nabla h) \right] $$ If we take the derivative of this quantity with respect to $\nabla h$, we should obtain $$ \frac{\partial \mathcal{L}}{\partial (\nabla h)} = - \nu^2 \nabla (\nabla^2h), $$ which is pretty close to what's provided.

But where does the factor of two come from? The answer is that there are two "copies" of $h$ in the term in question; and when we take the functional derivative of $- \nu^2 [\nabla (\nabla^2h)] \cdot (\nabla h)$ with respect to $\nabla h$, we should really be differentiating with respect to both of them. This is much more obvious if you actually write out the variation of this term in terms of a variation $h + \delta h$: $$ \delta \left[\nu^2 (\nabla^2h)^2 \right] = 2 \nu^2 (\nabla^2 h) (\nabla^2 \delta h). $$ You can view the factor of two as coming from the fact that there are two "copies" of $\nabla^2 h$ in this expression. When you vary this term, you have to vary both the first "copy" and the second "copy", thereby picking up a factor of 2.

Answer 2

Please read the edits/comments before continuing.

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When in doubt, write out the indices.

So we start with

$$ {\cal L} = \left(\frac{\partial h}{\partial t}\right)^2 + (\nu \partial_i \partial^i h)^2 = \left(\frac{\partial h}{\partial t}\right)^2 + (\nu \partial_i \partial^i h)(\nu \partial_k \partial^k h) $$

and

\begin{align*} &\frac{\partial \cal L}{\partial( \partial_j h)} = \frac{\partial }{\partial( \partial_j h)} \nu^2 ( \partial_i \partial^i h)( \partial_k \partial^k h)\\ &= \nu^2 (\partial_i \delta_{ij})( \partial_k \partial^k h) + \nu^2(\partial_i \delta_{kj})( \partial_i \partial^i h)\\ &=\nu^2 \partial_j (\partial_i\partial^i h)\\ &=2 \nu^2 \partial_j \nabla^2h \end{align*}

This is true for each component $j$ so we may adopt the short hand

$$\frac{\partial \cal L}{\partial( \nabla h)} = 2 \nu^2 \nabla \nabla^2h $$

As for the negative sign, I have no idea where that comes from.

Edit: As pointed out by @Michael Seifert, the Euler Lagrange equations change with higher order derivatives! Therefore this answer is incorrect and I will type out the right solution as soon as possible.

Answer 3

I'll add an alternative perspective. The Lagrangian $\mathcal{L}$ is a function of $h$ and its derivatives, $\mathcal{L}=\mathcal{L}(h,\nabla h , \nabla^2 h, ... )$. In this case clearly $\mathcal{L}$ does not depend on $\nabla h$. As another answer notes, you can integrate the action by parts (which changes the Lagrangian by a total derivative) and get a new Lagrangian which depends on $\nabla h$. Or, you note the full Euler-Lagrange equations are actually:

$$ \frac{\partial \mathcal{L}}{\partial h} - \nabla\frac{\partial \mathcal{L}}{\partial (\nabla h)} + \nabla^2 \frac{\partial \mathcal{L}}{\partial (\nabla^2 h)} - ...= 0 $$

Where each term has alternating signs, and higher derivatives. This gives you your E-L equation:

$$\frac{\partial \mathcal{L}}{\partial h} + \nabla^2(2\nu^2 \nabla^2 h) = 0 $$

Edit

The solution in the OP can be found by requiring the equation of motion to be in the form:

$$\frac{\partial \mathcal{L}}{\partial h} - \nabla\frac{\partial \mathcal{L}} {\partial (\nabla h)} $$

and equating with the equation of motion above, which gives that:

$$\nabla\frac{\partial \mathcal{L}} {\partial (\nabla h)} = -\nabla^2(2\nu^2 \nabla^2 h) = -\nabla \cdot (\nabla 2\nu^2 \nabla^2 h) $$ $$\therefore \frac{\partial \mathcal{L}} {\partial (\nabla h)} = -2\nu^2\nabla \cdot\nabla^2 h $$