I have a question about Euler-lagrange equation which you can check this file. http://www.bekirdizdaroglu.com/ceng/Downloads/ISCE10.pdf, specifically equation $8$.

There is a functional $F$ and we want to find a function $f$ which minimizes $F$. Then we attain the Euler-Lagrange function $E-L(f)$. And we iterate by $\frac{\delta f}{\delta t} = -(E - L)$ to get this $f$.

Why do we set up the problem as $\frac{\delta f}{\delta t} = -(E - L)$ and how we choose the $\delta t$?. The $f$ which meets the $E-L=0$ is a nextreme. Although the gradient descent method is also a iteration processing, the right side of its equation is $-\nabla f$ if one want the local minimum. But does $E-L$ is a type of $\nabla f$.

How we choose $\delta t$, if it is too big, the iteration process is not stable, if it is too small, it is time-consuming. There is a necessary condition - CFL condition, not sufficient condition .Also, this condition is decided by the right side of the equation. link In page three, part 3, $\delta$t is decided by the order spatial derivatives. But in my equation, I do not have any spatial derivatives, but I have a integral, $\int f$. So, how should I choose the $\delta t$, I ask a professor in my uni, the answer is 'It is not a analytical function, you need to try the number'.
PS: Could you understand my question?

  • $\begingroup$ possible duplicate of Euler-Lagrange, Gradient Descent, Heat Equation and Image Denoising $\endgroup$ – Hans Lundmark Jul 15 '11 at 9:36
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    $\begingroup$ @Hans: I'm not very familiar with this area of math (I've only taken an intro course on Lagrangian/Hamiltonian mechanics), but is it really appropriate to close as a duplicate of the other question? It seems that onlybluemoon is asking more of a fundamental question of why we should set things up as $\frac{df}{dt}=-(E-L)$, whereas the other question accepts this setup and is dealing with perhaps more advanced issues. Again, I'm not sure I'm able to judge this area, so please do tell me your thoughts. $\endgroup$ – Zev Chonoles Jul 16 '11 at 8:15
  • $\begingroup$ Thank you. The truth is that 'delta(f)/delta(t)=-(E-L function)' is difficult to me... I try to find the answer for all days. And then, I found math.stackexchange.com/questions/29703/… which have the same problem like me. Then I read the answers, but I can not understand fully. So I register on this web ,post a question and always check it. I try to earn some reputation but I can not answer these complex question... $\endgroup$ – Heather Jul 16 '11 at 8:41
  • $\begingroup$ To you, this is a fundamental question of why we should set things up as dfdt=−(E−L).So could you give me some information about it. I know the greatest gradient method, but my question is different from it. $\endgroup$ – Heather Jul 16 '11 at 8:47
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    $\begingroup$ Sorry, I am having a little bit of difficulty understanding your question. Is your question about "why do we set up the problem as $\delta f / \delta t = -(\mathrm{E-L})$ in the gradient descent method", or are you asking about something else? $\endgroup$ – Willie Wong Jul 16 '11 at 10:38

Thanks for clarifying your question. I'll assume that you're using $f$ to refer to the function denoted by $u$ in the paper you link to.

First, let me remark that I wouldn't put too much stock in that paper. It goes to a lot of effort just to derive the well-known linear Gaussian blurring convolution for noise reduction. The diffusion approach only adds value if you consider anisotropic diffusion; as long as you stick with the linear approach used in that paper, you might as well just do Gaussian blurring and forget about the whole diffusion aspect.

You seem to be thinking of an iterative gradient-descent scheme, in which some step $\delta t$ is chosen and the solution is modified by adding a corresponding multiple of the gradient to it. This is not what's happening in this paper -- there is no $\delta t$ in the paper; rather, the paper considers a continuous flow where at each point the flow is in the direction of the gradient. A discrete $\delta t$ that would have to be chosen would only enter into this if you were to try to calculate that flow numerically by discretizing it. This is not done in the paper; rather, the paper solves the heat flow equation analytically and derives the Gaussian blurring convolution from it, with the width of the Gaussian blurring kernel increasing with the time $t$.

I interpret the other part of your question to ask why the expression equated to zero in the Euler-Lagrange equations is like a gradient. The Euler-Lagrange equations are derived by varying the function $u$, calculating the resulting variation of the objective function $E$ and setting it to zero. The expression that is being set to zero is the variation of $E$ with respect to $u$ at each point. You can think of this in analogy to a function of several variables, where the gradient is obtained by considering the variation of the function under variation of each of the variables. Each function value of $u$ at a point is like a variable; the difference is that $u$ is defined at uncountably many points while a function of several variables only has a finite number of variables. The paper seems to equivocate somewhat on whether $u$ is defined on a continuum or on the pixels; if you consider it defined on the pixels, then each of its values at one of the pixels would be one of the variables, and the gradient of $u$ would be the collection of the variations of $u$ with respect to each of these variables, and this is what the expression in the Euler-Lagrange equations represents.

There might be more to explain, but I'll stop here and see whether this makes sense for you or whether I'm going off on a wrong track.

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    $\begingroup$ +1 for this answer! That's so much better than whatever I would have been able to come up with... $\endgroup$ – t.b. Jul 16 '11 at 15:05
  • $\begingroup$ @joriki,Thank you for your explanation. It is so helpful. I guess I understand all of what you was saying. In the fourth paragraph, you inspired me. $\delta$E is calculated according to f, the f which meet the equation $\delta$E=0 is a extrem - a stationary point, just like its gradient =0 . $\endgroup$ – Heather Jul 17 '11 at 2:04
  • $\begingroup$ @Theo Buehler, Thank you for your attention and comments. $\endgroup$ – Heather Jul 17 '11 at 2:08
  • $\begingroup$ @joriki, Hi, For the third paragraph,sorry I did not notice that. So I link the other paper, like link $\endgroup$ – Heather Jul 17 '11 at 2:19
  • $\begingroup$ @joriki, Hi, For the third paragraph,sorry I did not notice that. Let me specify in the above quesion. $\endgroup$ – Heather Jul 17 '11 at 2:43

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