Functional decaying under the heat flow (?)

Question

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.

Answer 1

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).

Answer 2

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 $$ \frac{d}{dt} J(v) \leq 0. $$