91Ó°ÊÓ

Let \(S \subset R^{3}\) be a regular surface with Gaussian curvature \(K>0\) and without umbilical points. Prove that there exists no point on \(S\) where \(H\) is a maximum and \(K\) is a minimum.

(Calculus of Variations.) Geodesics are particular cases of solutions to variational problems. In this exercise, we shall discuss some points of a simple, although quite representative, variational problem. In the next exercise we shall make some applications of the ideas presented here. Let \(y=y(x), x \in\left[x_{1}, x_{2}\right]\) be a differentiable curve in the \(x y\) plane and let a variation of \(y\) be given by a differentiable map \(y=y(x, t)\), \(t \in(-\epsilon, \epsilon)\). Here \(y(x, 0)=y(x)\) for all \(x \in\left[x_{1}, x_{2}\right]\), and \(y\left(x_{1}, t\right)=\) \(y\left(x_{1}\right), y\left(x_{2}, t\right)=y\left(x_{2}\right)\) for all \(t \in(-\epsilon, \epsilon)\) (i.e., the end points of the variation are fixed). Consider the integral 4\. First and Second Variations of Are Length, Bonnet's Theorem $$ I(t)=\int_{x_{1}}^{x_{2}} F\left(x, y(x, t), y^{\prime}(x, t)\right) d x, \quad t \in(-\epsilon, \epsilon) $$ where \(F\left(x, y, y^{\prime}\right)\) is a differentiable function of three variables and \(y^{\prime}=\partial y / \partial x\). The problem of finding the critical points of \(I(t)\) is called a variational problem with integrand \(F\). a. Assume that the curve \(y=y(x)\) is a critical point of \(I(t)\) (i.e., \(d I / d t=0\) for \(t=0\) ). Use integration by parts to conclude that \((I=d I / d t)\) $$ \begin{aligned} I(t) &=\int_{x_{1}}^{x_{2}}\left(F_{y} \frac{\partial y}{\partial t}+F_{y} \frac{\partial y^{\prime}}{\partial t}\right) d x \\ &=\left[\frac{\partial y}{\partial t} F_{y}\right]_{x_{1}}^{x_{2}}+\int_{x_{1}}^{x_{2}} \frac{\partial y}{\partial t}\left(F_{y}-\frac{d}{d x} F_{y^{\prime}}\right) d x \end{aligned} $$