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It is well known that the problem of minimizing

$$ J[y] = \int_{0}^{1} \sqrt{y(x)^2 + \dot{y}(x)^2} dx $$

with $y \in C^2[0,1]$ and $y(0) = 1$ and $y(1) = 0$ has no solutions. However, if we remove the condition $y(1) = 0$ and instead let the value of $y$ at $x = 1$ be free, then an optimal solution does exist.

An easy way to see this is to observe that $J[y]$ is really just the arc length of the plane curve with polar equation $r(\theta) = y(\theta)$. Clearly then, the function $y$ which traces out in that way the shortest line segment joining the point $(0,1)$ (given in polar coordinates) and the ray $\theta = 1$ is the (unique) solution to this new problem.

Inspired by this little example, I wonder: are there results regarding the existence of solutions to variational problems with freedom at one or both endpoints and similar integrands?

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1 Answer 1

Sure. You can take any smooth $f(x)$ with $f(0) = 0,$ then minimize $$ \int_0^1 \sqrt{1 + \left( f(\dot{y}(x)) \right)^2} \; dx $$ with $y(0) = 73.$ The minimizer is constant $y.$

More interesting is the free boundary problem for surface area. Given a wire frame that describes a nice curve $\gamma$ in $\mathbb R^3,$ once $\gamma$ is close enough to the $xy$ plane, there is a surface (topologically an annulus) with one boundary component being $\gamma$ and the other being a curve in the $xy$ plane, that minimizes the surface area among all such surfaces. The optimal surface meets the $xy$ plane orthogonally.

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I was hoping for moreso general result which makes use of the freedom at the end-points. –  user18063 Mar 10 '12 at 6:50

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