# When the Euler Lagrange equation simplifies to zero

My question is rather simple, and I'm sure I'm missing something simple, and yet...

I'm trying to calculate the Euler Lagrange Equations for the example function here:

http://en.wikipedia.org/wiki/First_variation

$E(y(x)) = \int yy'dx$

From what I can see it is:

$y'-y'=0$

which is clearly not very usefull. If I carry forward the first variation from (http://en.wikipedia.org/wiki/First_variation) I beleive it becomes the same thing. To me this implies that the first variation is always zero, but that can't be since some functions, $y$, clearly have more energy than others...

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Your equation is absolutely correct, and the first variation is indeed always zero in this case. Note that the Euler-Lagrange equation is derived using integration by parts on the assumption that all candidate functions take the same values at the endpoints. Thus

$$E=\int_a^byy'\mathrm dx=\int_a^b\frac12(y^2)'\mathrm dx=\frac12\left(y(b)^2-y(a)^2\right)$$

is the same for all candidate functions. The calculation in the Wikipedia article you link to doesn't yield zero because it doesn't make that assumption, doesn't integrate by parts and just lets the result stand for arbitrary $h$ without $h=0$ at the endpoints. You can rewrite the result they obtain as

$$\int_a^b(yh'+y'h)\mathrm dx=\int_a^b(yh)'\mathrm dx=y(b)h(b)-y(a)h(a)\;,$$

which is again zero if we require $h(a)=h(b)=0$, as the derivation of the Euler-Lagrange equation does.

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Thank you! The Euler-Lagrange is often used to solve for the minimum of a functional using gradient descent. Does this mean that once initialized it will only explore pertubations of y for which y(a),y(b) remain fixed? –  Tim Feb 24 '12 at 17:52
@Tim: I don't know that method. I might be able to say more if you describe how it's done. –  joriki Feb 25 '12 at 11:03
Sure, I'll post a new question as its hard to format things in these comments. –  Tim Feb 27 '12 at 17:29