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April 9th, 2015, 04:05 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 wellknown 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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functional, problem, quadratic, variational 
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