I have a surface $f : \Omega \rightarrow \mathbb{R}^3$ that is represented by $$f(t, \phi) = (ae^t \cos(\phi),ae^t \sin(\phi), \int_0^t \sqrt{1-a^2 e^{2x}} dx)$$
I also calculated the matrix representation of the Shape operator (or Weingarten map):
$$L(t,\phi):=\left(\begin{matrix} - \frac{ae^{t}}{\sqrt{1-a^2e^{2t}}} & 0 \\ 0 & \sqrt{1-a^2 e^{2t}} e^t a \\ \end{matrix} \right)$$
Now, I want to calculate the asymptotic lines, which are the ones satisfying $II(c'(s),c'(s))=0$.
To do this, I calculated $g_p ( L c'(s),c'(s))=0$ for $c'(s)= (t'(s),\phi'(s))$ and got $-(t'(s))^2+(1-a^2 e^{2t})(\phi'(s))^2=0.$ Thus $t'(s) = \pm |\phi'(s)|\sqrt{1-a^2e^{2t}}.$
Does anybody know how to continue in order to actually get the curves on the surface now?
I mean the problem is that this is an ODE containing two functions, so I think this problem is somehow underdetermined or am I completely wrong?
You've already realized that the lines of curvature are the $t$- and $\phi$-curves. (This is the case for any surface of revolution.)
For the asymptotic curves, you want to solve the differential equation $$\frac{d\phi}{dt} = \pm\frac1{\sqrt{1-a^2e^{2t}}}.$$ This comes from your equation relating $v_1$ and $v_2$. (The slope of an asymptotic curve, according to your calculations, is given by the ratio $v_2/v_1$.)