My Math Forum Variational problem: quadratic functional

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 April 9th, 2015, 05:05 AM #1 Newbie   Joined: Apr 2015 From: Moscow, Russia Posts: 1 Thanks: 0 Variational problem: quadratic functional Let me bring to your attention the following problem. Suppose we have the functional $\displaystyle F = \int\limits_{a}^{b} f(y(x))\cdot(\frac{dy}{dx})^2 dx$ with differential constraint $\displaystyle \frac{dy}{dx} / n(y(x)) - 1 = 0$ - DE. We write the Lagrangian for this problem $\displaystyle L = f(y(x))\cdot(\frac{dy}{dx})^2 + \lambda(x) \cdot ((\frac{dy}{dx} / n(y(x))) - 1 )$ Value at the end of the interval y(b) is not fixed (free end point) EL equation is: $\displaystyle \frac{d( f(y(x))\cdot (\dot{y})^2)}{dx} + \dot{\lambda} =0$ or $\displaystyle f(y(x))\cdot (\dot{y})^2 + {\lambda(x)} =const$. Very strange EL equation. From my point of view, it means the following. Lagrange multiplier is a function only of the argument x, then it (and therefore the Lagrangian, as can be seen from equation) can be represented as a derivative with respect to x of a function p. And this is in accordance with the well-known theorem sufficient criterion of "zero Lagrangian" (EL equation is an identity - every function which satisfies the boundary conditions is the extremum). Boundary condition for the non fixed end point: $\displaystyle (\frac{dL}{d\dot{y}})_b = 2(f(y)\cdot (\dot{y}))_b + ({\lambda(x)}/ n(y(x))_b = 0$ The question is how to determine whether the extremum is minimum or maximum? Thank you for your help.

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