# Ngô Quốc Anh

## May 20, 2013

### Super polyharmonic property of solutions: The case of single equations

Filed under: PDEs — Ngô Quốc Anh @ 9:39

Recently, I have read a paper by Chen and Li published in the journal CPAA [here or here] about the super polyharmonic property of solutions for some partial differential systems.

In this notes, we consider their result in a very particular case – the single equations. We shall prove the following.

Theorem 2.1. Let $p$ be a positive integer and $q>1$. For each positive solution $u$ of

$\displaystyle (-\Delta)^p u \geqslant u^q$

in $\mathbb R^n$, there holds

$\displaystyle (-\Delta)^i u >0 \quad i=\overline{1,p-1}.$

Proof. For simplicity, we write

$\displaystyle v_i = (-\Delta)^i u \quad i=\overline{1,p-1}.$

We must show that $v_i >0$ for all $i$. For simplicity, we divide the proof into two steps.

Step 1. Proving $v_{p-1}>0$.

Assume the contradiction, we then have two possible cases

Case 1. There is some $x^0 \in \mathbb R^n$ such that $v_{p-1}(x^0)<0$.

Case 2. $v_{p-1} \geqslant 0$ and there is a point $x_0$ such that $v_{p-1}(x_0)=0$.

As can be easily seen, in Case 2, $x_0$ is a minimum point for $v_{p-1}$, in particular, $-\Delta v_{p-1} |_{x_o} \leqslant 0$. However, this is a contradiction since

$\displaystyle 0 \geqslant -\Delta v_{p-1} |_{x_0}= (-\Delta )^p u |_{x_0}\geqslant u(x_0)^q>0.$

Hence, it suffices to consider Case 1. Without loss of generality, we may assume $x^0=0$.

Step 1.1.

Let us introduce the following notation

$\displaystyle \mathop {\overline f }\limits_x (r) = \frac{1}{{|\partial {B_r}(x)|}}\int_{\partial {B_r}(x)} {fd\sigma },$

called the average of $f$ with respect to a ball with radius $r$ centered at $x$. Sometimes, if $x \equiv 0$, we simply write $\overline f(r)$. Using this notation, we have the following well-known identity

$\displaystyle \mathop {\overline {\Delta f} }\limits_x (r) = \frac{1}{{|\partial {B_r}(x)|}}\int_{\partial {B_r}(x)} {{\Delta _z}f(z)d\sigma } = {\Delta _r}\left( {\frac{1}{{|\partial {B_r}(x)|}}\int_{\partial {B_r}(x)} {fd\sigma } } \right) = \Delta \mathop {\overline f }\limits_x (r).$

Using the Jensen inequality,

$\displaystyle {\left( {\frac{1}{{|\partial {B_r}(x)|}}\int_{\partial {B_r}(x)} {fd\sigma } } \right)^q} \leqslant \frac{1}{{|\partial {B_r}(x)|}}\int_{\partial {B_r}(x)} {{f^q}d\sigma }$

one immediately has

$\displaystyle\begin{array}{rcl} - \Delta {\overline v _{p - 1}} &\geqslant& {\overline u ^q}, \hfill \\ - \Delta {\overline v _{p - 2}} &=& {\overline v _{p - 1}}, \hfill \\ &\vdots & \hfill \\ - \Delta \overline u &=& {\overline v _1}. \hfill \\ \end{array}$

Using the first inequality, we obtain

$\begin{array}{rcl} \displaystyle\frac{d}{{dr}}\left( {{r^{n - 1}}\frac{d}{{dr}}{{\overline v }_{p - 1}}} \right) &=& \displaystyle (n - 1){r^{n - 2}}\frac{d}{{dr}}{\overline v _{p - 1}} + {r^{n - 1}}\frac{{{d^2}}}{{d{r^2}}}{\overline v _{p - 1}} \hfill \\ &=& \displaystyle {r^{n - 1}}\left( {\frac{{n - 1}}{r}\frac{d}{{dr}} + \frac{{{d^2}}}{{d{r^2}}}} \right){\overline v _{p - 1}} \hfill \\ &=& \displaystyle {r^{n - 1}}\Delta {\overline v _{p - 1}} < 0. \hfill \\ \end{array}$

By integrating over $[0,r]$, we get

$\displaystyle {r^{n - 1}}\frac{d}{{dr}}{\overline v _{p - 1}} = \int_0^r {\frac{d}{{ds}}\left( {{s^{n - 1}}\frac{d}{{ds}}{{\overline v }_{p - 1}}} \right)ds} \leqslant 0.$

In particular,

$\displaystyle\frac{d}{{dr}}{\overline v _{p - 1}} <0$

and therefore

$\displaystyle {\overline v _{p - 1}}(r) < {\overline v _{p - 1}}(0) < 0 \qquad \forall r > 0.$

From the second equation, $- \Delta {\overline v _{p - 2}} = {\overline v _{p - 1}}$, we have

$\displaystyle - \frac{1}{{{r^{n - 1}}}}\frac{d}{{dr}}\left( {{r^{n - 1}}\frac{d}{{dr}}{{\overline v }_{p - 2}}} \right) = - \Delta {\overline v _{p - 2}} = {\overline v _{p - 1}} < \underbrace {{{\overline v }_{p - 1}}(0)}_{ - C}.$

Integrating yields

$\displaystyle {\frac{d}{dr}\overline v _{p - 2}} > Cr$

and

$\displaystyle {\overline v _{p - 2}}(r) \geqslant {{\bar v}_{p - 2}}(0) + \frac{C}{2}{r^2} \qquad \forall r \geqslant 0.$

Hence, there is some $r_0$ such that $\overline v_{p-1}(r_0)>0$. Take a point $x^1$ with $|x^1|=r_0$ as the new center to make average, then from

$\displaystyle {\mathop {\overline v }\limits_{x^0} {}_{p - 1}}(r) < 0$

we now have

$\displaystyle {\mathop {\overline v }\limits_{x^1} {}_{p - 2}}(0) > 0.$

Since the system

$\displaystyle\begin{array}{rcl} - \Delta {\overline v _{p - 1}} &\geqslant& {\overline u ^q}, \hfill \\ - \Delta {\overline v _{p - 2}} &=& {\overline v _{p - 1}}, \hfill \\ &\vdots & \hfill \\ - \Delta \overline u &=& {\overline v _1}\hfill \\ \end{array}$

still holds for the new center $x^1$, we also have

$\displaystyle {\mathop {\overline v }\limits_{x^1} {}_{p - 1}}(r) < 0 \qquad \forall r>0.$

As before, this helps us to write

$\displaystyle \mathop{\overline v}\limits_{x^1}{} _{p - 2}(r) \geqslant \mathop{\overline v}\limits_{x^1}{}_{p - 2}(0) + \frac{C}{2}{r^2} \qquad \forall r \geqslant 0.$

But now we can conclude that

$\displaystyle \mathop{\overline v}\limits_{x^1}{} _{p - 2}(r) > 0 \qquad \forall r \geqslant 0.$

Using $\overline v_{p-2}$ we can consider $\overline v_{p-3}$. However, in order to make sure that our inequalities hold for all $r \geqslant 0$, we need to find a new center. Hence, there is a sequence of points $\{x^i\}_{i=0}$ such that

$\displaystyle (-1)^i\mathop{\overline v}\limits_{x^{i-1}}{} _{p - i}(r) > 0 \qquad \forall r \geqslant 0.$

As a consequence of the preceding inequality, $p$ must be even.

Step 1.2.

For the sake of simplicity, let us denote

$\displaystyle \widetilde v_{p-i}=\mathop{\overline v}\limits_{x^{i-1}}{} _{p - i}$

and

$\displaystyle \widetilde u=\mathop{\overline u}\limits_{x^{p-1}}.$

First, we observe that we can make $\widetilde u$ to be as large as we want in $r \in [0,1]$. Indeed, if we let

$\displaystyle {u_\lambda }(x) = {\lambda ^{\frac{{2p}}{{q - 1}}}}u(\lambda x)$

then $u_\lambda$ still solves

$\displaystyle (-\Delta)^p u_\lambda \geqslant (u_\lambda)^q.$

Since $-\Delta \widetilde u<0$, we know that $\widetilde u'(r)>0$ for all $r \geqslant 0$. It follows that $\widetilde u(r) > \widetilde u(0)=c_0 >0$. Hence, we can choose $\lambda$ sufficiently large such that $u_\lambda$ is as large as we want. Therefore, we can use $\widetilde u_\lambda$ instead of $\widetilde u$. Having this observation, for any given $a_0>0$ to be determined later, we can assume

$\displaystyle \widetilde u(r) \geqslant a_0 \geqslant a_0 r^{\sigma_0} \qquad \forall r \in [0,1]$

where $\sigma_0 > 0$ is chosen in such a way that $\sigma_0 q \geqslant n+2p$. Clearly,

$\displaystyle - \frac{1}{{{r^{n - 1}}}}\frac{d}{{dr}}\left( {{r^{n - 1}}\frac{d}{{dr}}{{\widetilde v}_{p - 1}}} \right) = - \Delta {\widetilde v_{p - 1}} \geqslant {\widetilde u^q} \geqslant a_0^q{r^{{\sigma _0}q}}.$

Integrating twice gives

$\displaystyle {\widetilde v_{p - 1}}(r) \leqslant - \frac{{a_0^q}}{{({\sigma _0}q + n)({\sigma _0}q + 2)}}{r^{{\sigma _0}q + 2}}$

which implies

$\displaystyle - \frac{1}{{{r^{n - 1}}}}\frac{d}{{dr}}\left( {{r^{n - 1}}\frac{d}{{dr}}{{\widetilde v}_{p - 2}}} \right) \leqslant - \frac{{a_0^q}}{{({\sigma _0}q + n)({\sigma _0}q + 2)}}{r^{{\sigma _0}q + 2}}.$

Again by integrating twice, we arrive at

$\displaystyle {\widetilde v_{p - 2}}(r) \geqslant \frac{{a_0^q}}{{({\sigma _0}q + n)({\sigma _0}q + 2)({\sigma _0}q + n + 2)({\sigma _0}q + 4)}}{r^{{\sigma _0}q + 4}} \geqslant \frac{{a_0^q}}{{{{({\sigma _0}q + n + 4)}^4}}}{r^{{\sigma _0}q + 4}}.$

By repeating this procedure, we eventually obtain

$\displaystyle\widetilde u(r) \geqslant \frac{{a_0^q}}{{{{({\sigma _0}q + n + 2p)}^{2p}}}}{r^{{\sigma _0}q + 2p}} \geqslant \frac{{a_0^q}}{{{{({\sigma _0}q + n + 2p)}^{2p + n}}}}{r^{{\sigma _0}q + 2p + n}} = \frac{{a_0^q}}{{{{({\sigma _0}q + m)}^m}}}{r^{{\sigma _0}q + m}},$

with $m=2p+n$. We now denote

$\displaystyle {\sigma _{k + 1}} = 2{\sigma _k}q \geqslant {\sigma _0}q + m, \qquad {a_{k + 1}} = \frac{{a_k^q}}{{{{(2{\sigma _k}q)}^m}}}$

we then have $\widetilde u(r) \geqslant {a_1}{r^{{\sigma _1}}}$. By mathematical induction, we easily prove that

$\displaystyle\widetilde u(r) \geqslant {a_k}{r^{{\sigma _k}}}$

for any $k$. We now choose $l>\frac{2}{q-1}$ and then $\sigma_0$ such that

$\displaystyle \sigma_0 \geqslant 2^{l+q+1}q^{2(l+1)+q}.$

By choosing $a_0$ sufficiently large and by mathematical induction, we can show that

$\displaystyle a_k^q \geqslant (\sigma_k q)^{m(l+1)} \quad \forall k \geqslant 0, k \in \mathbb Z.$

Having all these preparation,

$\displaystyle \widetilde u(1) \geqslant a_k \geqslant (\sigma_k q)^\frac{m(l+1)}{q} \to +\infty$

as $k \to \infty$. This is a contradiction.

Step 2. Proving $v_{p-i}>0$ for $i \geqslant 2$.

Suppose for some $i>1$, $v_{p-i}<0$ at some point and $v_{p-j}>0$ for all $j. Going through Step 1.1 above, the signs of $\widetilde v_k$ are alternating. Since $\widetilde u>0$, we must have $-\Delta \widetilde u<0$. In particular, $\widetilde u \geqslant c_0>0$. Using the equation $-\Delta v_{p-1}=u^q$, we obtain $-\Delta \widetilde v_{p-1} \geqslant \widetilde u^q \geqslant c_1>0$. Integrating

$\displaystyle - \frac{1}{{{r^{n - 1}}}}\frac{d}{{dr}}\left( {{r^{n - 1}}\frac{d}{{dr}}{{\widetilde v}_{p - 1}}} \right) = - \Delta {\widetilde v_{p - 1}}$

shows that $\widetilde v_{p-1}<0$ somewhere. This is also a contradiction.