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The question: determine the minimum and maximum values of the integral

$$\int_0^1 yy'dx$$

subject to the conditions $y(0)=0$ and $y(1)=1$.

There is no explicit y dependence, so our Euler-Lagrange Equation is

$$F-\frac{d}{dx}\frac{\partial F}{\partial y'}={\it constant}$$

However, when I do this, I get $y'y-y'y=0$. This tells me that

$$F=\frac{d}{dx}\frac{\partial F}{\partial y'}$$

Thus, the max/min correspond to those x-values where

$$\frac{dF}{dx}=\frac{d^2}{dx^2}\frac{\partial F}{\partial y'}=0$$

How do I take it from here? Did I do something wrong?

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1 Answer 1

Hint: $yy'=\frac12(y^2)'$. ${}{}{}{}$

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Okay, so I saw a mistake in my above derivation (I should have yy'-y'=c), but I don't really see how the above would help me. Where do the boundary conditions come in to play? –  Bronzeclocksofbenin Sep 17 '13 at 20:59
    
$\int_0^1 yy'dx = \frac12(y^2(1)-y^2(0))$. –  njguliyev Sep 17 '13 at 21:28

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