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<i. 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.

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Create a free website or blog at WordPress.com.

%d bloggers like this: