Let M be a complete,n dimensional Riemannian manifold without boundary. Suppose $g_1,g_2$ are two metrics on M and $g_1\leqslant g_2$. Suppose that there exists $T>0$ such that for $i=1,2$, the Ricci flow $\frac{\partial g}{\partial t}=-2 Ric$ with initial condition $g(0)=g_i$ exist for $t\in [0,T]$. Do we have $g_1(t)\leqslant g_2(t)$ for $t\in (0,T]$?

When n=2, the metric can be written in the conformal coordinates $$ g_i(t)=e^{2u_i(t)}(dr^2+r^2 d\theta^2). $$ $u_i(t)$ satisfies the parabolic equation $$ \frac{\partial u_i}{\partial t}=e^{-2u_i}\Delta u_i=-K_i $$ where $K_i$ are the Gauss curvature. We have $u_1(0)\leqslant u_2(0)$, we need to prove $u_1(t)\leqslant u_2(t)$.

Let $v(t)=\min_{x\in \mathbb{R}^2} u_2(t,x)-u_1(t,x)$, suppose the minimum are attained at $x_t$, then $\Delta(u_2-u_1)|_{x_t} \leqslant 0$ and thus $$ \frac{\partial v}{\partial t} \leqslant (e^{-2u_2}-e^{-2u_1})\Delta u_1|_{x_t}. $$ It seems that we can't expect that $\frac{\partial v}{\partial t}\leqslant 0$. So how can we prove $u_1(t)\leqslant u_2(t)$ ?