Functional decaying under the heat flow (?)

Let $(M, g)$ be a compact Riemannian manifold and let $a$, $p$ be two real numbers greater than $1$.

For any positive function $v$, I set

$$J(v) = \int_M \left|\nabla(v^a)\right|^p d\mu^g.$$

Assume now that $(v_t)_{t \geq 0}$ is a solution of the heat equation $$\frac{d}{dt} v_t = \Delta_g v_t.$$

Is it possible to prove the following statement?

There exists a constant $\lambda$ independent of the initial data $v_0$ such that $$J(v_t) \leq J(v_0) e^{\lambda t}$$

Computing the time derivative of $J(v_t)$ does not seem to be the solution since the expression we obtain is quite intricate (the initial problem is to prove that this derivative is "controllable" in some sense)

Any help would be invaluable!

As a reward: This question appeared in the course of proving a gradient estimate for the so called Lichnerowicz equation (see e.g. https://arxiv.org/abs/1403.5655). I will be happy to add the name of the first who finds the solution to the list of authors.

• I'm not quite sure I follow your argument for $\mathbb R^n$: The mapping $\mathbb R\times\mathbb R^n\ni (u,v)\mapsto [|u|^{a-1}|v|]^p$ is not convex. Commented Jun 1, 2018 at 6:05
• @fedja: this is right, the functional is not convex! I did a stupid mistake in my calculation... I removed the remark. Commented Jun 1, 2018 at 11:56
• Would you be interested in the $\mathbb R^n$ case then? (or in the flat torus if you prefer to keep everything compact)? Commented Jun 1, 2018 at 17:52
• Actually, I do know that the answer is bad on a flat torus: you can go up a bit (we figured it out with Ben Jaye once for another problem) but once you can go up, you can accelerate the time derivative as much as you want by considering $v(kx)$ instead of $v$, so no fixed $\lambda$ for you unless you are ready to allow a constant in front of the RHS (i.e., you care about large times only). Commented Jun 1, 2018 at 18:03
• Can you give me a reference to what you say? Commented Jun 1, 2018 at 18:05

OK, here is the story. Consider the case $a=2$. Then we want to figure out what happens with $\int |ff'|^p$. I want to create the situation when $ff'$ is the largest and positive at $0$ and goes up at that point. Then for large enough $p$ we are in trouble because once we went down from the maximum of $(f^2)'$, we can move $f^2$ around slowly and smoothly to close the period, so the maximum of $|ff'|$ after a short time will exceed the original one and thereby the same can be said about the integral of sufficiently large power. The local maximum conditions are $f,f'>0$, $f'^2+ff''=0$, $3f'f''+ff'''<0$. The going up condition is $f''f'+ff'''>0$. Now just take $f=1,f'=1,f''=-1,f'''=2$ (those are just the values at one point, so they are free).

• You are perfectly right. I did some numerics to see it explicitly. You can find the Python code here <lmpt.univ-tours.fr/~gicquaud/doc/test.py > Commented Jun 4, 2018 at 15:23
• @RomainGicquaud I'm curious what was the least value of $p$ that you could get. Commented Jun 5, 2018 at 18:33
• I did not spend time doing much numerics. With the particular choice of function you have in the Python source code, for p=6, I get decay but there is a slight increase when p=7. I will come back probably tomorrow to you with an argument that shows that you can choose some better range in 1d than what is found by @MarkusSprecher in higher dimension. Commented Jun 6, 2018 at 15:31

Here is a partial answer in 1d. I am assuming $M = \mathbb{S^1}$.

Let me set $u = v^a$ for simplicity. Then, since $v$ evolves according to the heat equation, we have $$\frac{du}{dt} = a v^{a-1} \frac{dv}{dt} = a v^{a-1} v'' = u'' - \frac{a-1}{a} \frac{(u')^2}{u}.$$ For simplicity, I set $$b = \frac{a-1}{a}.$$ Let me set $$J(v) = \frac{1}{p} \int_{\mathbb{S^1}} |(v^a)'|^p dx = \frac{1}{p} \int_{\mathbb{S^1}} |u'|^p dx$$ (I just add a factor $1/p$ for simplicity).

Note that $$\frac{d}{dt} |u'|^p = \frac{d}{dt} \left((u')^2\right)^{p/2} = \frac{p}{2} \left((u')^2\right)^{p/2-1} \left(2 u' \frac{du'}{dt}\right) = p |u'|^{p-2} u' \frac{du'}{dt} = p |u'|^{p-2} u' \left(u'' - b\frac{(u')^2}{u} \right)'$$ (I will use similar calculations later on).

Let $\lambda$ be a real number to be chosen later. We have \begin{eqnarray*} \frac{d}{dt} J(v) &=& \frac{1}{p} \frac{d}{dt} \int_{\mathbb{S^1}} |u'|^p dx\\ &=& \int_{\mathbb{S^1}} |u'|^{p-2} u' \left(u'' - b \frac{(u')^2}{u} \right)' dx\\ &=& \int_{\mathbb{S^1}} |u'|^{p-2} u' \left(u'' - \lambda b \frac{(u')^2}{u} \right)' dx + (1-\lambda) b \int_{\mathbb{S^1}} |u'|^{p-2} u' \left(\frac{(u')^2}{u} \right)' dx\\ &=& -(p-1)\int_{\mathbb{S^1}} \left(|u'|^{p-2} u''\right) \left(u'' - \lambda b \frac{(u')^2}{u} \right) dx\\ & & \qquad + (1-\lambda) b \int_{\mathbb{S^1}} |u'|^{p-2} u' \left(2\frac{u' u''}{u} - \frac{(u')^3}{u^2}\right)dx\\ &=& - (p-1) \int_{\mathbb{S^1}} |u'|^{p-2} (u'')^2 dx + b\left[\lambda (p-1) + 2(1-\lambda)\right] \int_{\mathbb{S^1}} |u'|^p \frac{u''}{u} dx\\ & & \qquad - (1-\lambda) b \int_{\mathbb{S}^1} \frac{|u'|^{p+2}}{u^2} dx. \end{eqnarray*} I now try to make the second term disappear by completing the square: \begin{eqnarray*} \frac{d}{dt} J(v) &=& - (p-1) \int_{\mathbb{S^1}} |u'|^{p-2} \left(u'' - \frac{b(\lambda (p-1) + 2(1-\lambda)}{2(p-1)} \frac{(u')^2}{u}\right)^2 dx\\ & & \qquad b\left[\frac{b\left(\lambda(p-1) + 2(1-\lambda)\right)^2}{4(p-1)}- (1-\lambda)\right] \int_{\mathbb{S}^1} \frac{|u'|^{p+2}}{u^2} dx. \end{eqnarray*}

Choosing $$\lambda = -\frac{2 (b p - 3 b + p - 1)}{b (p - 3)^2},$$ we get $$\frac{d}{dt} J(v) = - (p-1) \int_{\mathbb{S^1}} |u'|^{p-2} \left(u'' - \frac{b(\lambda (p-1) + 2(1-\lambda)}{2(p-1)} \frac{(u')^2}{u}\right)^2 dx\\ - \frac{4 (b (p - 3) + 1) (p - 1)^2}{b (p - 3)^2} \int_{\mathbb{S}^1} \frac{|u'|^{p+2}}{u^2} dx.$$ So, as long as $b(p-3)+1 \geq 0$, we have that $$\frac{d}{dt} J(v) \leq 0.$$