I am currently reading 'Lectures on Ricci Flow' by Peter Topping and I have got to Chapter 3 where he states the 'weak maximum principle for scalars'. Suppose for $t \in [0,T]$ for finite $T$ that $g(t)$ is a smooth family of metrics and that $X(t)$ is a smooth family of vector fields on a closed manifold $M$. We let $F\: : \mathbb{R} \: \times \: [0,T] \rightarrow \mathbb{R}$. Suppose that $u \in C^{\infty}(M \times [0,T], \mathbb{R})$ solves


$\frac{\partial u}{\partial t} \leq \:\Delta\_{g(t)}u \: + \: <X(t),\nabla u> \: + \: F(u,t)$.


Also, suppose that $\phi \:: [0,T]\rightarrow \mathbb{R}$ solves the system


$\frac{d \phi}{d t} = F(\phi(t),t), $


$\phi(0)= \alpha \in \mathbb{R}$.


If $u(.,0) \leq \alpha$, then $u(.,t) \leq \phi(t)$ for all $t \in [0,t]$.


I follow that $u(.,t)< \phi(t)$ for finite $t$ unless $u(x,t)=\phi(t)$ for all $x$ in the manifold $M$ but I don't really understand why the principle is defined in this way or how it is proved. I see how the principle can be used to get some preliminary estimates on evolution of curvatures but I am not sure what the relation is with what is called the strong/weak maximum principle in texts on differential geometry and PDEs.



What is the Weak Maximum Principle for Scalars and how is it Derived?
This is probably better suited for <https://math.stackexchange.com/> but it probably doesn't hurt to say a few things here given that it's a fundamental part of studying parabolic equations and geometric evolution equations such as the Ricci flow. It's also a prototype for other maximum principal arguments and the comparison principle for various weak forms of solutions to fully non-linear equations such as viscosity solutions.


The function $\varphi$ in fact solves the equation
$$
\partial\_t \varphi = \Delta\_{g(t)} \varphi + \langle X(t), \nabla \varphi\rangle + F(\varphi(t), t)
$$
since $\varphi$ depends on $t$ only and hence $\Delta\_{g(t)} \varphi = 0$ and $\nabla \varphi = 0$ and has initial data satisfying $\sup\_M u(\cdot, 0) \leq \varphi(0)$. Then the weak maximum principal says that this inequality is preserved for all $t>0$: $\sup\_M u(\cdot, t) \leq \varphi(t)$.


To prove that the inequality is preserved, one typically proceeds by contradiction. First suppose that $\sup\_M u(\cdot, 0) < \varphi(0)$. Consider the function
$$
v(x, t) = \varphi(t) - u(x, t).
$$
We have that $\inf\_M v(x, 0) > 0$ and need to prove that $v(x, t) \geq 0$ for all $x \in M$ and $t \geq 0$. In fact, we can prove that $v(x, t) > 0$.


Now the contradiction: suppose not. There there will be first time $t\_0$ and a point $x\_0 \in M$ such that $v(x\_0, t\_0) = 0$. That is, $v(x, t) > 0$ for all $x \in M$ and all $t < t\_0$ and $v(x, t\_0) \geq 0$ for all $x \in M$. Thus $\partial\_t|\_{t=t\_0} v(x\_0, t) \leq 0$ and $x\_0$ is a spatial minimum of the function $x \mapsto v(x, t\_0)$. Then we also have $\Delta\_{g(t\_0)} v|\_{x=x\_0} \geq 0$ and $\nabla v|\_{x=x\_0} = 0$.


Therefore at $(x\_0, t\_0)$, we have
$$
\tag{1}\label{eq:min}
\partial\_t v - \Delta\_{g(t)} v - \langle X(t), \nabla v \rangle \leq 0.
$$
On the hand, since $u$ is a sub-solution, $\varphi$ is a solution and everything (except perhaps $F$) is linear:
$$
\begin{split}
\partial\_t v - \Delta\_{g(t)} v + \langle X(t), \nabla v \rangle &= \partial\_t \varphi - \left(\partial\_t u - \Delta\_{g(t)} u - \langle X(t, \nabla u \rangle\right) \\
&\geq F(\varphi(t\_0), t\_0) - F(u(x\_0, t\_0), t\_0) = 0
\end{split}
\tag{2}\label{eq:solns}
$$
since $\varphi(t\_0) = u(x\_0, t\_0)$.


Equation \eqref{eq:solns} does not quite contradict equation $\eqref{eq:min}$ because both could in fact be $0$. To obtain the contradiction, consider something like $v\_{\epsilon} (x, t) = \varphi(t) - u(x, t) + \epsilon e^t$ for $\epsilon > 0$ and then send $\epsilon \to 0$.


This new $v\_{\epsilon}$ will also allow you to get the result with only weak inequality: $\sup\_M u(\cdot, 0) \leq \varphi(0)$. Essentially that is a continuity argument.


Another way to approach the contradiction argument, which works already with weak inequality (but you'll still need to add $\epsilon e^t$ to get a contradiction), is to suppose the result is false, so that there exists a $T > 0$ such that
$$
\inf\_{x \in M, t \in [0, T]} \varphi(t) - u(x, t) < 0.
$$
Then choose $x\_0 \in M$ and $t\_0 \in [0, T]$ to realise the $\inf$ and the same argument works. Some messing around is also necessary here to deal with the $F$ term when you do it this way. The way around is to add $\epsilon e^{\lambda t}$ for suitably chosen $\lambda$.


This should be in many books dealing with PDE such as Evans. Another source for maximum principles in particular is Protter and Weinberger.


