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In the calculus of variations (unconstrained), one applies a first-order variation on a general functional of the form

$$\int_{a}^{b}F(x,y,y')\,dx$$

to obtain the first-order necessary condition for optimality as expressed by the Euler-Lagrange equality

$$\frac{d}{dx}F_{y'} -F_y = 0.$$

What is the necessary condition for a minima? How about a sufficient condition?

Josh Allen
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2 Answers2

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Coming from the world of optimal control, I tend to view the calculus of variations from a Pontryagin point of view. The conditions stated by Pontryagin are necessary, and sufficient under certain hypotheses (mainly related to the convexity of the function $F$). I used to refer to this article during my PhD:

O. L. Mangasarian, Sufficient Conditions for the Optimal Control of Nonlinear Systems

fpacaud
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It is possible to apply the notions of stationary points and the second derivative of a function to functionals.

For $|\varepsilon|\ll1$ and and a differentiable function $h$, we can write, using Taylor series,$$F(x,y+\varepsilon h,y'+\varepsilon h')=F(x,y,y')+\varepsilon\mathcal{I}(\Delta[y,h])+\frac{\varepsilon^2}2\mathcal I(\Delta_2[y,h])+\mathcal O(\varepsilon^3)$$ where $\mathcal I$ denotes the integrand, $\Delta$ the Gâteaux differential and $$\Delta_2[y,h]=\int_a^b\left(F_{y'y'}h'^2+\left(F_{yy}-\frac {dF_{yy'}}{dx}\right)h^2\right)\,dx$$ over the domain $[a,b]$. Analogous to functions, at a stationary point path, the Gâteaux differential is zero. Note that $\Delta_2$ is also analogous to the second derivative.

Therefore, it is sufficient that $\Delta_2>0$ for all non-zero $h$ for a minimum to occur, whereas the necessary condition is weaker with $\Delta_2\ge0$.