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.

  • 2
    $\begingroup$ Have you tried writing this out in index notation (or even just explicitly with $\partial_x, \partial_y, \partial_z$) ? That might give you a better understanding of the shorthand. $\endgroup$
    – J. Murray
    Jan 6, 2019 at 21:19
  • $\begingroup$ There should be an error there in the Lagrangian density. The laplaceian is squared and shouldn't be $\endgroup$
    – DanielC
    Jan 6, 2019 at 21:24
  • 1
    $\begingroup$ should the integral be over $d^3 x$? $\endgroup$ Jan 6, 2019 at 21:47
  • $\begingroup$ could you please tell us which is the reference you are using? book, lecture notes, website, etc. thanks $\endgroup$ Jan 7, 2019 at 4:51
  • $\begingroup$ Reference: tcm.phy.cam.ac.uk/~cc726/TP1/ExamFiles/exam18sol.pdf question 3c (page 6-7). You can do it by using the higher order EL equation derived in b, which I'm happy with and gives the results, but the solution also says it can be done using the usual E-L equations and partial integration, which I'm struggling with. Thanks! $\endgroup$ Jan 7, 2019 at 9:19

3 Answers 3


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.


Please read the edits/comments before continuing.

When in doubt, write out the indices.

So we start with

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


\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.

  • $\begingroup$ I don't think this is correct. When you do calculus of variations with higher derivatives, you have to treat all of the derivatives (first, second, third, etc.) as independent; so $\partial \mathcal{L}/\partial (\nabla h) = 0$ in this case. However, the form of the Euler-Lagrange equations changes, and it explains the additional minus sign. See here, for example. $\endgroup$ Jan 7, 2019 at 2:22
  • $\begingroup$ That’s interesting.. I didn’t know that I can fix my answer in a moment or I can delete it and let you write one if you want $\endgroup$ Jan 7, 2019 at 2:30
  • $\begingroup$ The modified Lagrange equations would be: $\frac{\delta L}{\delta h}+\frac{d^2}{dt^2}\frac{\delta L}{\delta \partial_t^2h}+\sum_i\partial_i^2\frac{\delta L}{\delta \partial_i^2h}=0$, and neither in that way we would recollect what OP states; I may be wrong but I think there's a mistake somewhere in what he wrote; $\endgroup$ Jan 7, 2019 at 5:06
  • $\begingroup$ Please find the comment/reference above, thanks! $\endgroup$ Jan 7, 2019 at 9:21

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:

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


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:

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

  • $\begingroup$ Thanks for your answer, but would you mind elaborating a bit more on the last sentence? What do you exactly mean by 'book dependence'? $\endgroup$ Jan 7, 2019 at 19:01
  • $\begingroup$ Oh, I just meant you can recover the solution you state in the OP (which I had assumed was from a book...) I'll clarify. $\endgroup$
    – jacob1729
    Jan 7, 2019 at 19:09

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