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.


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.

  • $\begingroup$ So is it the case that every initial or boundary-value problem essentially has its own maximum principle, such that the principle we are looking at here is the maximum principle for the Ricci flow equation, or it just for geometric flows in general? $\endgroup$ Jan 24 '19 at 21:04
  • $\begingroup$ For scalar, parabolic equations: yes, more or less. One requires something like $u$ is a sub solution initially below a supersolution $v$ and below $v$ on the boundary of the domain. Then this will be true on the entire domain and for all $t$ such that both solutions are defined. For fully non-linear equations, the differential operator just needs to be monotone in the second derivative. At first glance it appears one also needs monotonicity in the degree 0 variable $u$, but in fact it just needs to be bounded and the $\lambda$ trick may be applied. See Protter-Weinberger. $\endgroup$
    – Paul Bryan
    Jan 24 '19 at 21:54
  • $\begingroup$ Could I double-check how the $v_{\epsilon}$ part obtains the contradiction? So basically it is no longer the case that both could be 0 because the derivative of $v_\epsilon$ is never 0 due to the exponential term, but then $v_\epsilon$ tends to $v$ uniformly as $\epsilon$ tends to 0. $\endgroup$ Jan 27 '19 at 14:59
  • 2
    $\begingroup$ Regarding "being slow": On this Reddit (reddit.com/r/math/comments/2bv740/doing_math_slowly) you will find the pearl of wisdom: "Math is not a speed game, it's not a carnival trick; Ultimately, you are being taught how to approach and solve problems. This is why it's one of the most employable degrees you can get." $\endgroup$
    – Paul Bryan
    Jan 29 '19 at 22:10
  • 1
    $\begingroup$ If $\varphi$ solves the ODE, then $\bar{\varphi}$ solves the PDE. It's a function of two variables that is independent of one of them. For example, $\varphi(t) = 1/t$ satisfies $\partial_t \varphi = - \varphi^2$ while $\bar{\varphi} (x, t) = 1/t$ satisfies $\partial_x \bar{\varphi} = 0$ and $\partial_t \bar{\varphi} = - \bar{\varphi}^2$. Thus $\bar{\varphi}$ solves the PDE $\partial_t \bar{\varphi} = \Delta \bar{\varphi} - \bar{\varphi}^2$. $\endgroup$
    – Paul Bryan
    Jan 30 '19 at 12:30

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