Gradient represented through fundamental form

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February 24th 2011, 01:26 PM
lllll

Gradient represented through fundamental form

If $E, F, G$ are the coefficients of the first fundamental form in a parametrization $\mathbf{x}:U \subset \mathbb{R}^2 \mapsto S$ , then $grad \ f$ on $\mathbf{x}(U)$ is given by:

$grad \ f =\frac{f_uG-f_vF}{EG-F^2}\mathbf{x}_u +\frac{f_vE-f_uF}{EG-F^2}\mathbf{x}_v$

Now what I have come to realize is that we can represent $grad \ f$ as $(\mathbf{x}_u, \mathbf{x}_v) \begin{pmatrix} E & F\\ F & G \end{pmatrix}^{-1} \begin{pmatrix} f_u\\ f_v \end{pmatrix}$

Now we get $\begin{pmatrix} E & F\\ F & G \end{pmatrix}^{-1}$ from the second fundamental form.

where we have
$N_u = a_{11}\mathbf{x}_u+a_{21}\mathbf{x}_v$

$N_v = a_{12}\mathbf{x}_u+a_{22}\mathbf{x}_v$

Now is there a way that I can incorporate these elements to get it to work out? Can I for instance carry on with the following $II_p(a) = \langle D (grad \ f(v)),v\rangle$ where $v$ is a tangent vector on a surface?
February 24th 2011, 07:51 PM
xxp9

What is your question? From my understanding you're trying to express grad(f) using E, F, G and you did it. The result is correct so what are you really asking?
February 24th 2011, 08:35 PM
lllll

I forget to write down the first part of the question, which defines $df_p= \langle grad \ f(p),v \rangle_p$ where $grad \ f(p) \in T_p(S) \subset \mathbb{R}^3$
Then from the given definition I'm trying to derive the formula $grad \ f =\frac{f_uG-f_vF}{EG-F^2}\mathbf{x}_u +\frac{f_vE-f_uF}{EG-F^2}\mathbf{x}_v$
February 24th 2011, 08:56 PM
xxp9

since grad(f) is a tangent vector it is expressible by the base vectors:
grad(f) = a Xu + b Xv, then we need to compute a and b by computing the inner products with Xu and Xv:
Xu . grad(f) = a Xu . Xu + b Xu . Xv
df( Xu) = a E + b F
fu = a E + b F.
Similarly:
fv = a F + b G
So we get a system of linear equations. Solve it we get
(a, b)' = ( E F; F G)^{-1} (fu, fv)'