Homework 5

#exercise-problems #math

Problem Statement

Show that if $F = 0$ then

K = \frac{- 1}{2 \sqrt{E G}} ({(\frac{E_{v}}{\sqrt{E G}})}_{v} + {(\frac{G_{u}}{\sqrt{E G}})}_{u})

and if $E = G$ we get the following equation:

{(\frac{G_{v}}{G})}_{v} + {(\frac{G_{u}}{G})}_{u} + 2 K G = 0

It is known that the Gaussian curvature takes the form:

K = \frac{1}{(E G - F^{2})^{2}} (| \begin{matrix} - \frac{1}{2} E_{u v} + F_{u v} - \frac{1}{2} G_{u u} & \frac{1}{2} E_{u} & F_{u} - \frac{1}{2} E_{v} \\ F_{v} - \frac{1}{2} G_{u} & E & F \\ \frac{1}{2} G_{v} & F & G \end{matrix} | - | \begin{matrix} 0 & \frac{1}{2} E_{v} & \frac{1}{2} G_{v} \\ \frac{1}{2} E_{v} & E & F \\ \frac{1}{2} G_{u} & F & G \end{matrix} |)

Let $F = 0$ , then:

\begin{aligned} K & = \frac{1}{(E G)^{2}} (| \begin{array}{c} - \frac{1}{2} E_{u v} - \frac{1}{2} G_{u u} & \frac{1}{2} E_{u} & - \frac{1}{2} E_{v} \\ - \frac{1}{2} G_{u} & E & 0 \\ \frac{1}{2} G_{v} & 0 & G \end{array} | - | \begin{array}{c} 0 & \frac{1}{2} E_{v} & \frac{1}{2} G_{v} \\ \frac{1}{2} E_{v} & E & 0 \\ \frac{1}{2} G_{u} & 0 & G \end{array} |) \\ = \frac{1}{(E G)^{2}} (\frac{1}{2} G_{v} (\frac{- 1}{2} E_{v} \cdot E) + G (E \cdot (- \frac{1}{2} E_{u v} - \frac{1}{2} G_{u u}) + \frac{1}{4} G_{u} \cdot E_{v}) - \frac{1}{2} G_{u} (- \frac{1}{2} E \cdot G_{v} + G (- \frac{1}{4} E_{v}^{2}))) \\ = \frac{- 1}{2 \sqrt{E G}} (\frac{E_{u v} \sqrt{E_{u v}} - \frac{1}{2} E_{v} {(E G)}^{\frac{- 1}{2}} (E_{v} G + G_{v} E)}{E G} + \frac{G_{u u} \sqrt{E G} - \frac{1}{2} G_{v} (E G)^{\frac{- 1}{2}} (E_{v} G + G_{v} E)}{E G}) \\ = \frac{- 1}{2 \sqrt{E G}} ({(\frac{E_{v}}{\sqrt{E G}})}_{v} + {(\frac{G_{u}}{\sqrt{E G}})}_{u}) \end{aligned}

If $E = G$ we get that:

K = \frac{- 1}{2 G} ({(\frac{G_{v}}{G})}_{v} + {(\frac{G_{u}}{G})}_{u})

thus rearranging we get that:

{(\frac{G_{v}}{G})}_{v} + {(\frac{G_{u}}{G})}_{u} + 2 K G = 0