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I would like to find a continuous function $y : [0,4] \to \mathbb{R}$ that minimizes the following functional

$$I (y) := \displaystyle\int_{0}^4\sqrt{y\left(1+(y^{\prime})^2\right)} dx$$

subject to the boundary conditions $y (0) = 5/4$ and $y (4) = 13/4$. How do I solve this minimization problem? I tried and tried, but I can't get rid of the $y^{\prime}$. Whatever I do, I still have a big ugly equation with $y$ and $y^{\prime}$, and even if I change it to $\frac{dy}{dx}$, it doesn't get any better. Anyone has an idea?

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    Equation? What equation? I see none. Moreover, what are the limits of integration? You should integrate from $a$ to $b$ so that you obtain a scalar instead of a function, which is what a functional does.2012-09-22
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    Sorry for being so not-precise. So, again, for an integral from 0 to 4, and y(0)=5/4, y(4)=13/4, the answer should be y = ((x-1)/2)^2 + 1, but like I said, I can't get rid of y' or y in order to solve the equation.2012-09-22
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    @mimi: Have you tried using the Euler-Lagrange equations? http://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation2012-09-22
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    Of course I did. I just got stuck at one point. Also, if the function is f=f(y,y'), the Euler-Lagrange equation changes into f - y'* df/dy = c (c=const.) right?2012-09-22
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    What about $y=0$? Note that in order to take a square root, $y(\sqrt{(y')^2+1})$ must be non-negative (otherwise you have a complex valued function, which upon integration may or may not give a real number).2012-09-22
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    @BabyDragon That's not the problem.2012-09-22
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    are u sure, there is y' instead of y?2012-09-22
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    I checked it, that's how it's written in my book. Also, for an integral from 0 to 4, and y(0)=5/4, y(4)=13/4, the answer should be y = ((x-1)/2)^2 + 1, but like I said, I can't get rid of y' or y in order to solve the equation.2012-09-22
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    @Rut welcome to math.se! Posts of this nature should usually be comments on the main question rather than answers. The latter are generally reserved for complete solutions to the problem at hand.2012-09-22
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    I see, you have some endpoints.2013-02-11

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