I believe that is correct and due to Matt Grayson. On the unit $S^2$, by the Gauss--Bonnet formula, $\int_c \kappa ds = 2\pi - A$, where $A$ is the enclosed area. So $dA/dt = - d/dt\int_c \kappa ds$. On the other hand, $d/dt\int_c \kappa ds = \int_c \kappa ds$ on the unit $S^2$. So solving the ODE $dA/dt = A -2\pi$ we get

(i) If $A(0)=2\pi$, then $A(t)=2\pi$ for all $t$.

(ii) If $A(0) \neq 2\pi$, then $A(t) =2\pi + e^t(A(0) -2\pi )$, so that the smaller area enclosed goes to zero at finite time $t_\max=\ln (2\pi /|A(0)-2\pi |)$.

The existence and convergence to a round point (or equator) is due to Grayson's extension of the Gage--Hamilton theorem on convex plane curves to both the embedded case and the case of an ambient curved surface.

I believe that is correct and due to Matt Grayson. Under the curve shortening flow on the unit $S^2$, the evolution of the area $A(t)$ enclosed by the curve $c(t)$ is given by $dA/dt = - \int_c \kappa ds = A -2\pi$, where the last equality is by the Gauss--Bonnet formula. Solving this ODE, we get $A(t) =2\pi + e^t(A(0) -2\pi )$. If $A(0)=2\pi$, then $A(t)=2\pi$ for all $t$ and the curve limits to a great circle as $t\rightarrow \infty$. On the other hand, if $A(0) \neq 2\pi$, then the smaller of the two enclosed areas goes to zero at finite time $t_\max=\ln (2\pi /|A(0)-2\pi |)$ and the curve limits to a round point as $t\rightarrow t_\max$.

This existence and convergence to a round point or equator is due to Grayson's extension of the Gage--Hamilton theorem on convex plane curves to both the embedded case and the case of an ambient curved surface.

I believe that is correct and due to Matt Grayson. On the unit $S^2$, by the Gauss--Bonnet formula, $\int_c \kappa ds = 2\pi - A$, where $A$ is the enclosed area. So $dA/dt = - d/dt\int_c \kappa ds$. On the other hand, $d/dt\int_c \kappa ds = \int_c \kappa ds$ on the unit $S^2$. So solving the ODE $dA/dt = A -2\pi$ we get

(i) If $A(0)=2\pi$, then $A(t)=2\pi$ for all $t$.

(ii) If $A(0) \neq 2\pi$, then $A(t) =2\pi e^t(A(0) -2\pi )$, so that the smaller area enclosed goes to zero at finite time $t_\max=\ln (2\pi /|A(0)-2\pi |)$.

The existence and convergence to a round point (or equator) is due to Grayson's extension of the Gage--Hamilton theorem on convex plane curves to the both the embedded case and the case of an ambient curved surface.

I believe that is correct and due to Matt Grayson. On the unit $S^2$, by the Gauss--Bonnet formula, $\int_c \kappa ds = 2\pi - A$, where $A$ is the enclosed area. So $dA/dt = - d/dt\int_c \kappa ds$. On the other hand, $d/dt\int_c \kappa ds = \int_c \kappa ds$ on the unit $S^2$. So solving the ODE $dA/dt = A -2\pi$ we get

(i) If $A(0)=2\pi$, then $A(t)=2\pi$ for all $t$.

(ii) If $A(0) \neq 2\pi$, then $A(t) =2\pi + e^t(A(0) -2\pi )$, so that the smaller area enclosed goes to zero at finite time $t_\max=\ln (2\pi /|A(0)-2\pi |)$.

The existence and convergence to a round point (or equator) is due to Grayson's extension of the Gage--Hamilton theorem on convex plane curves to both the embedded case and the case of an ambient curved surface.