I've been reading some papers about parabolic evolution problems between manifolds. We want to study the behaviour of maps from the domain $(0,T)\times \Omega$, where $\Omega\subset \Bbb R^m$, to some compact manifold $N$. In some of the papers, I've encounter the parabolic metric on $(0,T)\times \Omega$ $$ d((t_1,\mathbf x_1),(t_2,\mathbf x_2)):=\max\left\{\sqrt{|t_1-t_2|},|\mathbf x_1-\mathbf x_2| \right\}, $$ where $|\cdot|$ denotes the regular euclidean distance.

What is the motivation behind this definition?

From the look of it, it seems like we are considering the time dimension as if it's a $2-$dimensional space. This seems kind of weird to me.

If the answer to this turns out to be "because it gives the right estimate" I am OK with that. Nevertheless, I'd still want to know why it is a good idea to treat the time as if it has $2-$dimensions. If anyone can provide some insight I'd really appreciate that.

  • $\begingroup$ I don't know the answer, but the prototypical parabolic equation is the heat equation $(\partial _t - \Delta) u = 0$, where time carries an order-$1$ derivative, while space carries an order-$2$ one, so that one needs to "square" the time derivative to get the same order as the space one. A similar thing seems to happen in your formula. Of course, this is merely an intuition, not an answer. $\endgroup$ – Alex M. Jan 6 '18 at 20:15
  • $\begingroup$ @AlexM. That's an interesting observation, thanks! $\endgroup$ – BigbearZzz Jan 6 '18 at 20:17
  • $\begingroup$ Can you give a reference ? What makes it seem like time is 2D in the formula you wrote ? $\endgroup$ – Piyush Grover Jan 6 '18 at 20:25
  • $\begingroup$ @PiyushGrover It's just the square root there, maybe saying "$2-$dimensions" might not be the best wording. Perhaps I should just ask "why do we want the square root sign there"? $\endgroup$ – BigbearZzz Jan 6 '18 at 20:29
  • $\begingroup$ One of the books that use this metric is Roger Moser's "partial regularity for harmonic maps and related problems". $\endgroup$ – BigbearZzz Jan 6 '18 at 20:31

Here is one reason, the core of which can be found at page 17 of Moser's book mentioned in the comments. Many theorems about the heat equation are valid on "cylindrical" domains of the form $(0,T) \times \Omega$. If, in particular, $\Omega \subseteq \mathbb R^n$ is an open Euclidean ball $B(x_0,r)$ and $T = 2r^2$ (the same $r$), then

$$ \begin{align} (0,T) \times B(x_0, r) =& \{ (t, x) \in \mathbb R \times \mathbb R^n ; | t - r^2 | < r^2 \text{ and } \|x - x_0\| < r \} = \\ =& \{ (t, x) \in \mathbb R \times \mathbb R^n ; \sqrt {| t - r^2 |} < r \text{ and } \|x - x_0\| < r \} = \\ =& \{ (t, x) \in \mathbb R \times \mathbb R^n ; \max \Big (\sqrt {| t - r^2 |}, \|x - x_0\| \Big) < r \} = \\ =& \{ (t, x) \in \mathbb R \times \mathbb R^n ; d \big( (t,x), (r^2, x_0) \big) < r \} = B_d \big( (r^2, x_0), r) \big) \end{align} $$

where $B_d (p, R)$ is the ball of centerpoint $p$ and radius $R$ with respect to the distance $d$.

The use of this metric turns cylinders into balls, and balls are always nice to work with.

Another motivation (bottom of page 18) is that many parabolic problems find a natural framework in the setting of Hölder spaces. You might know that on a metric space with distance $d$, $f$ is $\alpha$-Hölder continuous (with $\alpha \in (0,1]$) if and only if

$$|f(x) - f(y)| \le C d(x,y)^\alpha \ \forall x, y$$

for some $C>0$. It turns out that when studying the heat equation, the correct spaces are a bit different; namely, they are called "parabolic Hölder spaces" and are made of those $u$ for which there exists $C>0$ such that

$$|u(s,x) - u(t,y)| \le C \Big( \sqrt {|s-t|} + \| x-y \| \Big)^\alpha \ ,$$

which does not look like the Hölder condition give above. In reality, since

$$\sqrt {|s-t|} + \| x-y \| \le 2 \max \Big( \sqrt {|t-s|} + \| x-y \| \Big) = 2 d \big( (s,x), (t,y) \big)$$

the "parabolic Hölder condition" becomes

$$|u(s,x) - u(t,y)| \le C' d\big( (s,x), (t,y) \big)$$

with $C' = 2^\alpha C$, which looks exactly like the "usual" Hölder condition on metric spaces - which allows you to use the full theory of Hölder spaces.

Of course, the deep reason of this all is the fact that the space variables enter parabolic equations with derivatives of order $2$, while time only contributes a derivative of oder $1$. In order to bring them on equal footing and restore the broken symmetry, it is useful to resort to this metric trick.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.