I'm trying understand the article "The Heat Equation Shrinking Convex Planes" by Gage and Hamilton and I'm get stuck in the proof of Theorem $3.2.1$, specifically, in the proof of this inequality:
$\frac{\partial^2 g}{\partial s_1^2} \frac{\partial^2 g}{\partial s_2^2} - \left( \frac{\partial^2 g}{\partial s_1s_2} \right)^2 \geq 0,$
$\textbf{EDIT:}$
where the left expression on this inequality is evaluated in a minimum point of $g$, $g$ is defined by $g(u_1,u_2,t) = f(u_1,u_2,t) + \varepsilon t$, $f$ is defined by $f: S^1 \times S^1 \times [0,T) \longrightarrow \mathbb{R}$ ($S^1$ is interpreted by the authors like $\mathbb{R}/\mod 2 \pi$), $f(u_1,u_2,t) := |F(u_1, t) - F(u_2,t)|^2$ and $F$ is a parametrization of a closed convex regular curve which trace is contained in $\mathbb{R}^2$.
The parameters $s_1$ and $s_2$ are the parameters by arc length with respect the curve $F$ in the instante $t$ in the points $F(u_1,t)$ and $F(u_2,t)$ respectively.
It is stated that an easy calculation verifies this inequality, but I can't see why this inequality it's true. I know that $\frac{\partial }{\partial s} = \frac{1}{v} \frac{\partial }{\partial u}$ $\left(v = ||\frac{\partial F}{\partial u}|| \right)$. Adopting the following notation to simplify the computations, $v_1 = ||\frac{\partial F}{\partial u_1}||$ and $v_2 = ||\frac{\partial F}{\partial u_2}||$, I found
$\frac{\partial^2 g}{\partial s_1^2} = \frac{\partial^2 f}{\partial s_1^2} = \frac{1}{v_1} \frac{\partial }{\partial u_1} \left( \frac{1}{v_1} \frac{\partial f }{\partial u_1} \right) = \frac{1}{v_1} \left[ \left( - \frac{1}{v_1^2} \right) \frac{\partial v_1}{\partial u_1} \frac{\partial f }{\partial u_1} + \frac{1}{v_1} \frac{\partial^2 f}{\partial u_1^2} \right] = \frac{1}{v_1} \left[ \left( - \frac{1}{v_1^2} \right) \langle \frac{\partial^2 f}{\partial u_1^2},\frac{\partial f }{\partial u_1} \rangle \frac{\partial f }{\partial u_1} + \frac{1}{v_1} \frac{\partial^2 f}{\partial u_1^2} \right]$
$\frac{\partial^2 g}{\partial s_2^2} = \frac{\partial^2 f}{\partial s_2^2} = \frac{1}{v_2} \frac{\partial }{\partial u_2} \left( \frac{1}{v_2} \frac{\partial f }{\partial u_2} \right) = \frac{1}{v_2} \left[ \left( - \frac{1}{v_2^2} \right) \frac{\partial v_2}{\partial u_2} \frac{\partial f }{\partial u_2} + \frac{1}{v_2} \frac{\partial^2 f}{\partial u_2^2} \right] = \frac{1}{v_2} \left[ \left( - \frac{1}{v_2^2} \right) \langle \frac{\partial^2 f}{\partial u_2^2}, \frac{\partial f }{\partial u_2} \rangle \frac{\partial f }{\partial u_2} + \frac{1}{v_2} \frac{\partial^2 f}{\partial u_2^2} \right]$
$\frac{\partial^2 g}{\partial s_1s_2} = \frac{\partial^2 f}{\partial s_1s_2} = \frac{1}{v_1} \frac{\partial }{\partial u_1} \left( \frac{\partial f}{\partial s_2} \right) = \frac{1}{v_1} \frac{\partial }{\partial u_1} \left( \frac{1}{v_2} \frac{\partial f}{\partial u_2} \right) = \frac{1}{v_1} \left[ \left( - \frac{1}{v_2^2} \right) \frac{\partial v_2}{\partial u_1} \frac{\partial f }{\partial u_2} + \frac{1}{v_2} \frac{\partial^2 f}{\partial u_1 u_2} \right] = \frac{1}{v_1} \left[ \left( - \frac{1}{v_2^2} \right) \langle \frac{\partial^2 f}{\partial u_1 u_2}, \frac{\partial f}{\partial u_2} \rangle \frac{\partial f }{\partial u_2} + \frac{1}{v_2} \frac{\partial^2 f}{\partial u_1 u_2} \right]$
Computing $\frac{\partial^2 g}{\partial s_1^2} \frac{\partial^2 g}{\partial s_2^2} - \left( \frac{\partial^2 g}{\partial s_1s_2} \right)^2$, I obtained something plus the Hessian of $f$, I tried compute $\langle \frac{\partial^2 f}{\partial u_1^2},\frac{\partial f }{\partial u_1} \rangle$, $\langle \frac{\partial^2 f}{\partial u_2^2},\frac{\partial f }{\partial u_2} \rangle$ and $\langle \frac{\partial^2 f}{\partial u_1 u_2},\frac{\partial f }{\partial u_2} \rangle$, I tried study the signs of $\frac{\partial v_1}{\partial u_1}$, $\frac{\partial v_2}{\partial u_2}$ and $\frac{\partial v_2}{\partial u_1}$ in order to express $\frac{\partial^2 g}{\partial s_1^2} \frac{\partial^2 g}{\partial s_2^2} - \left( \frac{\partial^2 g}{\partial s_1s_2} \right)^2$ in terms of the Hessian of $f$, because I know that in a minimum point $\det Hess \ f \geq 0$ and I would obtain the inequality desired, but I didnt have success with my attempts. Can anyone help me why this inequality its true?
Thanks in advance!