I am asked to prove that given a surface $S$ and a point $p\in S$ non-umbilical, then there exists $U$ open in $\mathbb{R}^2$, there exists $Y:U\subset \mathbb{R}^2\longrightarrow \mathbb{R}^3$ a parametrization such that the coordinate lines are lines of curvature.
Let $X:(u,v)\in V\subset \mathbb{R}^2\longrightarrow\mathbb{R}^3$ a parametrization and suppose $X(0,0)=p$, and $p$ a non-umbilical point and $N=\dfrac{X_u \times X_v}{\lVert X_u \times X_v \rvert}$.
I take $\alpha:t\in I\longrightarrow \alpha(t)=X(u(t),v(t))\in\mathbb{R}^3$. Let $k_1(t),k_2(t)$ the principal curvatures.
We want $dN_{p}(\alpha'(t))=-k_1(t) \alpha'(t)$.
Suppose $dN_{p}=\left( \begin{array}{cc}
A & B \\
C & D \\
\end{array} \right)$ with respect to the basis $\{ {X}_u,{X}_v\}$.
Then we get $(Au'+Bv')X_u+(Cu'+Dv')X_v=-k_1(t)u'X_u-k_1(t)v'X_v$. Because $\{X_u,X_v\}$ is a basis, then we have the System of Differential Equations:
$\begin{array}{cccccc}
A+k_1(t) & u' & B & v' &=&0 \\
C & u' &D+k_1(t) &v'&=&0 \\\end{array}$
We know this system has an unique solution $u(t),v(t)$, so $\alpha(t)$ is a line of curvature.
Defining $\beta:t\in J\subset \mathbb{R}\longrightarrow \beta(t)=X(\hat{u}(t),\hat{v}(t))\in\mathbb{R}^3$ if we want that $dN_{p}(\beta'(t))=-k_2(t) \beta'(t)$ we get another system of Differential Equations that has solution.
So $\beta(t)$ is another line of curvature.
Setting $Y$ the parametrization such that $Y(u(t),0)=\alpha(t), Y(0,v(t))=\beta(t)$ is enough.
Is it correct?
Best Answer
Yes, for given $ u(t), v(t)$ the $\alpha(t), \beta (t) $ are both lines of principal curvature since derivatives vanish, and normal curvatures are maximum and minimum for these curvature lines.
For this to happen diagonal coefficients as also next mixed coefficients $ F, (f = M) $ of first and second fundamental form coefficients vanish.
Incidentally, other non-principal direction angular deviation on surface $ \psi $ can be referenced through $ u', v'$ with Euler relation for principal normal curvatures as $ k_n = k_1 \cos^2\psi + k_2 \sin^2 \psi. $
EDIT1:
Regarding Y(t,0)=α(t) and Y(0,t)=β(t) separation of functions on a single independent variable t.
While referring to a point on a surface we can never get rid of one of the parameters, but we can make it constant. It is like forgetting the road we came by after turning at an intersection.
We say Y(u,v) for any point referred on the surface, $Y(u,v1)$ for $v = v1$ = constant or $Y(u1,v)$ for $u = u1$ = another constant, Y(u,0) for the v = 0 line, Y(0,v) for the u = 0 line. The parametrization Y(u,v) implies we are referring to particular value of parameter u= u_1 = constant or v = v_1 = constant.
The following text-book material of surface theory is relevant to go through:
($E,F,G$ first fundamental form metric coefficients, $ e,f,g = L,M,N $ second fundamental form metric coefficients). The normal curvature
$$ k_n =\dfrac {L du^2 + 2 M du dv + N dv^2 }{E du^2 + 2 F du dv + G dv^2} $$
$$ = L (du/ds)^2 + 2 M (du/ds) (dv/ds) + N (dv/ds)^2 $$
Line of curvatures
$$ k_n = {(E M-F L) du^2 + (E N-G L) du dv + (F N-G M) dv^2 } $$
Necessary and sufficient condition for a line of curvature to be parametric $$ F =0, M =0 $$
Rodrigue's formula necessary and sufficient relation:
$$ k_n = - dN(u,v)/dY(u,v) $$
To derive Euler $k_n$ formula because $F=0$ ,
$$ \cos\psi = \sqrt {E} (du/ds) \;,\sin\psi =\sqrt {G} (dv/ds) $$
For a sphere lines of curvature ( meridians $\theta$ = const and parallels $\phi $ = constant ) are:
$$ x = a \cos\phi \cos \theta \;; y= a \cos\phi \sin\theta \; ; z = a \sin\phi ; $$
along a meridian $ \phi= \alpha $