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I'm considering Dido's problem:

Consider 2 differentiable arcs $C$ and $C_0$ in $\mathbb{R}^2$ from the point $P$ to $Q$ and back. We keep $C_0,P,Q$ fixed, and want to choose the arc $C$ such that under all arcs of a specified length larther then $|PQ|$ the area $A$ enclosed by the 2 curves is maximized.

$$A = \frac{1}{2}\int_{C\cup C_0}xdy-ydx $$

Solutions to this problem using variational calculus are sketched in (1), (2)

(1) http://galileo.phys.virginia.edu/classes/321.jvn.fall02/var_meth.pdf

(2) http://mathematicalgarden.wordpress.com/2008/12/21/the-problem-of-dido/

I want to show that a necessary condition is that the curvature $\kappa$ is constant

$$\kappa = \frac{\dot{x}\ddot{y}-\dot{y}\ddot{x}}{(\dot{x}^2+\dot{y}^2)^{3/2}}$$

Ofcourse knowing that the solution to this problem is a circular arc, we know that it is. But I want to derive this...

It appears that the Lagrangian of this problem (see (2)) is

$$\frac{1}{2}(x\dot{y}-y\dot{x})+\lambda\sqrt{\dot{x}^2+\dot{y}^2} $$

And in (1) we see that using Eulers equations

\begin{align*} \dot{y}\kappa+\lambda x =0\\ -\dot{x}\kappa + \lambda y =0 \end{align*}

These can be combined to see that $\lambda(x\dot{x}+y\dot{y}) =0$ with solution $x^2+y^2 = C$. But I simply want to show that $\kappa$ is constant is a necessary condition, but I cant see how. How can we derive this?

Thanks for any enlighting remark.

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1 Answer 1

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I follows when you take the Euler-Langrange equation of

$ L: =\frac{1}{2}(xy'-yx')+\lambda \sqrt{x'^2+y'^2}. $

So begin with

$ \frac{d}{dt}\frac{d}{dx'}L - \frac{d}{dx} L =0 \\ \frac{d}{dt}\frac{d}{dy'}L - \frac{d}{dy} L =0. $

After some manipulation you will end up with

$ y' \left(1+\lambda \frac{x'y''-y'x''}{(x'^2+y'^2)^{\frac{3}{2}}} \right)=0 \\ x' \left(1+\lambda \frac{x'y''-y'x''}{(x'^2+y'^2)^{\frac{3}{2}}} \right)=0 $

Since $x'$ and $y'$ cannot be zero everywhere, unless $|PQ|=0$, it follows that

$ \frac{x'y''-y'x''}{(x'^2+y'^2)^{\frac{3}{2}}}=-\frac{1}{\lambda} $

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