Calculus of variations and geometric evolution problems: by F. Bethuel

By F. Bethuel

The overseas summer season college on Calculus of diversifications and Geometric Evolution difficulties used to be held at Cetraro, Italy, 1996. The contributions to this quantity replicate fairly heavily the lectures given at Cetraro that have supplied a picture of a pretty wide box in research the place in recent times we've seen many very important contributions. one of the subject matters taken care of within the classes have been variational equipment for Ginzburg-Landau equations, variational types for microstructure and part transitions, a variational remedy of the Plateau challenge for surfaces of prescribed suggest curvature in Riemannian manifolds - either from the classical standpoint and within the atmosphere of geometric degree thought.

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Additional info for Calculus of variations and geometric evolution problems: lectures given at the 2nd session of the Centro Internazionale Matematico Estivo

Example text

The first term (u - ax2)2 of the Lagrangian forces the minimizer u and its derivative u' to increase with x, until u' at some point reaches the interval of nonconvexity of Fn(u'). The derivative u' must vary outside of the forbidden interval of nonconvexity of the function Fn at all times .. 5); therefore, it needs relaxation. 20)) is independent of u' and it does not change after the convexification). The convex envelope CFp is equal to 1. Relaxation of One-Dimensional Variational Problems 26 Let us compute CFn(v) (again we use the notation v = u /).

The relaxed form of the problem with zero boundary conditions min (I CG(u, u'), u 10 u(O) = u(l) = 0, has an obvious solution, u(x) = u'(x) = 0, that yields the minimal (zero) value of the functional. It corresponds to the constant optimal value mopt of m(x): mopt(x) = ~ "Ix E [0,1]. The relaxed Lagrangian is minimized over four functions u, ml, VI, V2 bounded by one equality, u' = mlvl + (1 - ml)v2 and the inequalities o ~ m ~ 1, while the original Lagrangian is minimized over one function u. In contrast to the initial problem, the relaxed one has a differentiable solution in tenns of these four controls.

To estimate, we use the smallness of the interval of variation. Replace uE(x) with uo(xo) UE(x) = uo(xo) + O(e) and compute F(x, uE(x), Vi + u~(xo)) = F(xo, uo(xo), u~(xo) + u~(x)) + O(e) for any x E [xo, Xo + e]. The Lagrangian (rounded up to O(e)) is piecewise constant in the interval [xo, Xo + c]. The impact IE(u E) becomes L miF(xO, uE(xo), uo(xo) + Vi) + O(e). 14) with respect to the arguments VI, ... , vp and mlo ... 13)) p p Lmi = 1, i=1 Lmivi =0. 7)): L miF(x, U, Vi) = CFv(x, Uo, V). 17) parts so that V = u' takes k + 1 values.

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