If $\textbf{x}\in \Omega \subset\mathbb{R}^n,$ where $\Omega$ is a bounded open set, $u:\Omega\rightarrow\mathbb{R}, \;\eta:\Omega\rightarrow\mathbb{R},\;u'=\nabla u = \displaystyle\left(\frac{\partial u}{\partial x_1}\cdots\frac{\partial u}{\partial x_n}\right), \;\;\eta'=\nabla \eta = \displaystyle\left(\frac{\partial \eta}{\partial x_1}\cdots\frac{\partial \eta}{\partial x_n}\right)$ and I have the functions:

$$J(u)=\int_\Omega F(\textbf{x}, u, u')\;d\textbf{x}=\int_\Omega F(\textbf{x},u,u')\;dx_1\cdots dx_n$$ $$\Phi(\varepsilon)=J(u+\varepsilon\eta)=J(z)=\int_\Omega F(\textbf{x}, z,z')\;d\textbf{x},$$

where $\varepsilon$ is some small number and $F$ is a function of $\textbf{x}, z$ and $z'$. My problem is to calculate the derivative of $\Phi$ at the point $\varepsilon = 0$ and equating it to $0$. I get the following:

$$0=\Phi'(\varepsilon)=\int_\Omega F_z\eta+\sum_{i=1}^n\left(\frac{\partial F}{\partial p_i}\frac{\partial \eta}{\partial x_i}\right)\;d\textbf{x},\;\;\;(1)$$

where $p_i=\displaystyle\frac{\partial u}{\partial x_i}+\varepsilon\frac{\partial \eta}{\partial x_i}$. Now by integrating by parts I should get $(1)$ to be equal to: (this is given to me in reference material)

$$\int_\Omega F_z\eta+\sum_{i=1}^n\left(\frac{\partial F}{\partial p_i}\frac{\partial \eta}{\partial x_i}\right)\;d\textbf{x}=F_z-\sum_{i=1}^n\frac{d}{dx_i}\left(\frac{\partial F}{\partial p_i}\right)\;\;\;(2)$$

Can someone show me the steps how does one get from $(1)$ to $(2)$. My teacher tells me that simple integration by parts is applied, but I can't get it. Any advices? thank you =)

  • $\begingroup$ there still should be an integral on the right side of equation 2. Regarding your question: it's defined in the usual way for every single term in the sum. $\endgroup$
    – tired
    Jan 23, 2015 at 11:17
  • $\begingroup$ Thank you for your help @tired there is no integral on the right side of eqution 2 in my reference material? Could this be wrong then? In my lecture material I mean...here you can see my reference material in case you want to check: math.stackexchange.com/questions/1115621/… $\endgroup$
    – jjepsuomi
    Jan 23, 2015 at 11:18
  • $\begingroup$ I think it should be $$\int_\Omega F_z\eta+\sum_{i=1}^n\left(\frac{\partial F}{\partial p_i}\frac{\partial \eta}{\partial x_i}\right)\;d\textbf{x}=\int_{\Omega}\eta F_z-\sum_{i=1}^n\frac{d}{dx_i}\left( \frac{\partial F}{\partial p_i} \right) \eta$$ $\endgroup$
    – tired
    Jan 23, 2015 at 11:20
  • $\begingroup$ Thank you :) @tired so do you think there is a mistake in the reference? P.S. I used my own notation in tmy question, because I think the notation in the reference is bad. $\endgroup$
    – jjepsuomi
    Jan 23, 2015 at 11:22

1 Answer 1


Maybe it's better to put it in an answer:

$$\int_\Omega F_z\eta+\sum_{i=1}^n\left(\frac{\partial F}{\partial p_i}\frac{\partial \eta}{\partial x_i}\right)\;d\textbf{x}=\int_{\Omega}\eta \left( F_z-\sum_{i=1}^n\frac{d}{dx_i}\left( \frac{\partial F}{\partial p_i} \right) \right)\text{d}\textbf{x} \rightarrow \\ F_z-\sum_{i=1}^n\frac{d}{dx_i}\left( \frac{\partial F}{\partial p_i} \right)=0 $$

As mentioned in the comment, we used integration by parts for every term in the sum to get the first line.

Rememeber that our goal was to find the minimum of our functional derivative $ \Phi'(0)$ therefore we introduced the auxillary function $\eta$. It should be clear that our final equation should be independent of this quantity. To assure this the term in the second line has to be zero.

For further informations please have a look at this excellent book Mathematics for Physics - M. Stone & P. Goldbart

  • $\begingroup$ Thank you very much! =) Can you emphasize a little bit more your answer, why is the first equation true? Where did $d\textbf{x}$ disappear and why does the last equation follow? =) I'm sorry for being so demanding, the multidimensionality troubles me here. Could you maybe write out few more details into your answer that explicitly show why it is true? Thank you very much if you could do this! =) For example when you stated: "The second follows from the fact, that our integral should be zero regardless of the value of the parameter η". My question is why? $\endgroup$
    – jjepsuomi
    Jan 23, 2015 at 11:30
  • $\begingroup$ hey, the missing "dx" was my fault... $\endgroup$
    – tired
    Jan 23, 2015 at 11:34
  • $\begingroup$ Okay, no problem at all :) Sorry if I'm bugging you ;D I've just tried to figure this out for some time and I want everything to be crystal clear in my mind why the above is true with no question marks left unsolved :) $\endgroup$
    – jjepsuomi
    Jan 23, 2015 at 11:37
  • $\begingroup$ i added a few comments... $\endgroup$
    – tired
    Jan 23, 2015 at 11:42
  • $\begingroup$ Hi @tired one last question and then everything is clear =) Why is $\displaystyle \frac{\partial F}{\partial p_i}\eta = 0$? $\endgroup$
    – jjepsuomi
    Jan 23, 2015 at 12:12

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