Nonlinear critical problems for the biharmonic operator with Hardy potential

Abstract

In this paper we study the problem

$$\begin{aligned} {\left\{ \begin{array}{ll}{\mathcal L}_\mu [u]:=\Delta ^2u -\mu \frac{u}{|x|^4}=\lambda u +|u|^{2^*-2}u\quad \hbox {in }\,\,\Omega \\ u=\frac{\partial u}{\partial n}=0\quad \hbox {on }\,\,\partial \Omega \end{array}\right. } \end{aligned}$$

where $\Omega \subset {\mathbb {R}^n}$ is a bounded open set containing the origin, $n\ge 5$ and $2^*=2n/(n-4)$. We find that this problem is critical (in the sense of Pucci–Serrin and Grunau) depending on the value of $\mu \in [0,\overline{\mu })$, $\overline{\mu }$ being the best constant in Rellich inequality. To achieve our existence results it is crucial to study the behavior of the radial solutions (whose analytic expression is not known) of the limit problem ${\mathcal L}_\mu u=u^{2^*-1}$ in the whole space ${\mathbb {R}^n}$. On the other hand, our non–existence results depend on a suitable Pohozaev-type identity, which in turn relies on some weighted Hardy–Rellich inequalities.

Appendices

Appendix A: Some results on the linear polyharmonic equation

In what follows we need two results which we quote here for the reader convenience. The first is a consequence of [4, Theorem 2.4]:

Proposition 5.1

Let $m\ge 1$ be an integer and $n> 2m$. Let $\mu$ be a positive Radon measure on ${\mathbb {R}^n}$ and $u\in L^1_{loc}({\mathbb {R}^n})$. The following statements are equivalent:

• (a) $u$ is a distributional solution to

  $$\begin{aligned} (-\Delta )^{m} u=\mu \qquad \mathrm {on\ \ } {\mathbb {R}^n} \end{aligned}$$

  (5.1)

  such that the following “ring condition” holds:

  $$\begin{aligned} \liminf _{R\rightarrow +\infty } \frac{1}{ R^n }\int _{B_{2R}(x)\setminus B_{R}(x)} \left| u\right| =0 \quad \text {a.e. in }{\mathbb {R}^n}; \end{aligned}$$

  (5.2)

• (b) $u$ is a distributional solution to (5.1), $\mathrm {essinf}\, u=0$ and $u$ is weakly polysuperharmonic, that is $(-\Delta )^i u$ are positive distributions for $i=0,\dots ,m-1$;

• (c) $u\in L^1_{loc}({\mathbb {R}^n})$ and

  $$\begin{aligned} u(x)=C_{2m} \int _{{\mathbb {R}^n}} \frac{d\mu (y)}{\left| x-y\right| ^{n-2m}}\quad \text {a.e. in }{\mathbb {R}^n}, \end{aligned}$$

  (5.3)

  where the constants $C_{2m}$ are defined in (1.5).

Moreover, if (a), (b) or (c) holds, then for any $i=1,\dots ,m$ the convolutions $\frac{1}{\left| \cdot \right| ^{n-2i}}* \mu$ are a.e. well posed and for $i=1,\dots ,m-1$ the distribution $(-\Delta )^i u$, which is a positive Radon measure, can be represented by

$$\begin{aligned} \langle (-\Delta )^i u,\varphi \rangle = \int _{{\mathbb {R}^n}} u(-\Delta )^i\varphi = C_{2(m-i)} \int _{{\mathbb {R}^n}}\varphi (x) \int _{{\mathbb {R}^n}}\frac{d\mu (y)}{\left| x-y\right| ^{n-2(m-i)}}dx. \end{aligned}$$

(5.4)

Remark 5.2

Applying Hölder inequality it is easy to see that if $u\in L^p({\mathbb {R}^n})$ with $1\le p< +\infty$, then $u$ satisfies (5.2).

The second result we need is standard in potential theory, see for instance [13]:

Proposition 5.3

Let $1<p,q,r<+\infty$ with $1+\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$. Let $f\in L^p(\mathbb {R}^n)$ and $g\in L^q_w(\mathbb {R}^n)$. Then,

$$\begin{aligned} \left\| f* g\right\| _r\le C(n,p,q) \left\| g\right\| _{q,w}\left\| f\right\| _p. \end{aligned}$$

(5.5)

Now we may state the following result regarding the polyharmonic operator:

Proposition 5.4

Let $n>2m\ge 2$ and $n/2m>p>1$. Let $f\in L^p({\mathbb {R}^n})$. Then the problem

$$\begin{aligned} (-\Delta )^{m} u=f \qquad \mathrm {on\ \ } {\mathbb {R}^n}, \end{aligned}$$

(5.6)

has a unique distributional solution $u\in L^1_{loc}({\mathbb {R}^n})$ satisfying (5.2). Moreover, a.e. in $\mathbb {R}^n$:

$$\begin{aligned} u(x)&= C_{2m}\int \frac{f(y)}{\left| x-y\right| ^{n-2m}}dy,\end{aligned}$$

(5.7)

$$\begin{aligned} (-\Delta u)^i(x)&= C_{2(m-i)}\int _{{\mathbb {R}^n}}\frac{f(y)}{\left| x-y\right| ^{n-2(m-i)}}dy, \qquad i=1,\dots ,m-1, \end{aligned}$$

(5.8)

and

$$\begin{aligned} u\in L^{\frac{np}{n-2mp}}({\mathbb {R}^n}),\quad \left| \nabla u\right| \in L^{\frac{np}{n-(2m-1)p}}({\mathbb {R}^n}),\quad \left| \Delta ^i u\right| \in L^{\frac{np}{n-2(m-i)p}}({\mathbb {R}^n}) \end{aligned}$$

(5.9)

where the constants $C_{2k}$ are defined in (1.5).

In particular:

• (1) if $f\not \equiv 0$ is nonnegative, then $u>0$ and $u$ is superharmonic;

• (2) if $f\not \equiv 0$ is nonnegative and radial, then $u$ is radial, nonincreasing (decreasing for $m\ge 2$) and vanishes at infinity;

• (3) if $f$ is radial, then $u$ is radial and vanishes at infinity;

• (4) if $f$ is radial, and $v$ is any distributional radial solution to (5.6), then there exist $c_0, c_1,\dots , c_{m-1}$ real numbers such that

  $$\begin{aligned} v(x) = u(x)+c_0+c_1 \left| x\right| ^2 + \dots +c_{m-1} \left| x\right| ^{2(m-1)}. \end{aligned}$$

  In particular, if $v$ vanishes at infinity then $v\equiv u$;

• (5) if $m=1$, then $u\in \mathcal {D}^{2,p}(\mathbb {R}^n)$;

• (6) let $m=1$ and let $v$ be the solution to $-\Delta v = f^*$ on ${\mathbb {R}^n}$ satisfying (5.2). Then

  $$\begin{aligned} v\in L^{\frac{np}{n-2p}}({\mathbb {R}^n})\cap \mathcal {D}^{2,p}({\mathbb {R}^n}),\quad \left| \nabla v\right| \in L^{\frac{np}{n-p}}({\mathbb {R}^n}),\quad v\ge u^*\,\,\text {a.e.} \end{aligned}$$

  (5.10)

Proof

Uniqueness. Let $v,w$ be two solutions to (5.6) satisfying (5.2). Then also $u:=v-w$ satisfies (5.2) and solves $(-\Delta )^{m} u =0$. Applying Proposition () with $\mu =0$, from (5.3) we obtain $v\equiv w$.

Existence. Let ${\mathcal N}_{2m}:=C_{2m}\left| x\right| ^{n-2m}$; then ${\mathcal N}_{2m}$ is the fundamental solution of $(-\Delta )^m$ with the pole at the origin. Let $f\in L^p(\mathbb {R}^n)$ with $1<p<n/2m$. We first assume that $f\ge 0$. Since ${\mathcal N}_{2m}\in L_w^{n/(n-2m)}(\mathbb {R}^n)$, from (5.5) we have that

$$\begin{aligned} u:=f* {\mathcal N}_{2m}\in L^r(\mathbb {R}^n)\quad \text {with}\quad r=\frac{np}{n-2mp}, \end{aligned}$$

(5.11)

i.e. the first part of (5.9); moreover, (5.7) holds.

To prove that $u$ is solution to (5.6) in distributional sense, it is sufficient to apply the distribution $u$ to the test function $(-\Delta )^m \varphi$ and to use (5.11) (see [4] for further details).

The relation (5.8) follows from Proposition 5.1.

As for the summability of $\nabla u$ we compute the distributional derivative of $u$. Let $\varphi \in C_c^\infty (\mathbb {R}^n)$.

$$\begin{aligned} (\partial _i u)(\varphi )&= -\int u(x) \varphi _{x_i}(x)\,dx= -C_{2m}\int \int f(y)\frac{1}{\left| x-y\right| ^{n-2m}}\varphi _{x_i}(x) \\&= C_{2m}(2m-n)\int \int f(y)\frac{x_i-y_i}{\left| x-y\right| ^{n-2m+2}} \varphi (x) =\int (R_i* f)(x) \varphi (x)\, dx \end{aligned}$$

where $R_i(x):=C_{2m}(2m-n)\frac{x_i}{\left| x\right| ^{n-2m+2}}\in L_w^{n/(n-2m+1)}(\mathbb {R}^n)$. Now, from (5.5) we have that $R_i* f$ is well defined a.e. on ${\mathbb {R}^n}$ and $R_i* f\in L^{r_1}(\mathbb {R}^n)$ with $r_1=\frac{np}{n-(2m-1)p}$. The remaining integrability properties of $(-\Delta )^i u$ follow from (5.8) and (5.5).

The general case, without assumption on the sign of $f$, follows by writing $f$ as $f=f^+-f^-$, where $f^+$ and $f^-$ are the positive and negative part of $f$. Indeed, setting $u_1:=f^+* {\mathcal N}_{2m}$ and $u_2:=f^-* {\mathcal N}_{2m}$ we have the thesis taking $u:=u_1-u_2$.

Now we pass to the other statements.

Proof of (1) Statement (1) is an immediate consequence of (5.7) and (5.8).

Proof of (2) Let $m\ge 2$. From (5.7) it follows that $u$ is radial and positive, while (5.8) with $i=1$ implies that $u$ is strictly superharmonic; hence

$$\begin{aligned} -(\rho ^{n-1} u'(\rho ))'>0, \end{aligned}$$

(5.12)

i.e. the function $\rho ^{n-1} u'(\rho )$ is decreasing, so that there exists $l:=\lim \limits _{\rho \rightarrow 0} \rho ^{n-1} u'(\rho )$. We claim that $l=0$, hence $u'(\rho )<0$ for $\rho >0$, i.e. $u$ is decreasing.

To see that $l=0$, let us observe that, being $\nabla u\in L^{r_1}({\mathbb {R}^n})$ with $r_1=\frac{np}{n-(2m-1)p}>\frac{n}{n-1}$, we have

$$\begin{aligned} \int _0^\infty |u'(\rho )|^{r_1}\rho ^{n-1}\,d\rho <\infty \Longrightarrow \liminf _{\rho \rightarrow 0}|u'(\rho )|\rho ^{n/r_1}=0\Longrightarrow \liminf _{\rho \rightarrow 0}|u'(\rho )|\rho ^{n-1}=0, \end{aligned}$$

hence $l=0$.

The proof in the case $m=1$ is quite analogous, but now (5.12) must be replaced by

$$\begin{aligned} -(\rho ^{n-1} u'(\rho ))'\ge 0, \end{aligned}$$

that is $\rho ^{n-1} u'(\rho )$ is only nonincreasing. Arguing as above, we conclude in this case that $u$ is nonincreasing (but, in general, decreasing for $\rho$ sufficiently large).

Proof of (3) Writing $u=u_1-u_2$ as above, namely

$$\begin{aligned} u=u_1-u_2={\mathcal N}_{2m}* f^+ -{\mathcal N}_{2m}* f^-, \end{aligned}$$

we see that statement (3) is an immediate consequence of statement (2).

Proof of (4) Let $w$ be a distribution on ${\mathbb {R}^n}$. We claim that

$$\begin{aligned} (-\Delta )^m w=0,\quad w\text { radial }\Longleftrightarrow w=c_0+c_1 \left| x\right| ^2 + \dots +c_{m-1} \left| x\right| ^{2(m-1)}. \end{aligned}$$

(5.13)

Indeed, part $\Leftarrow$ of (5.13) is an obvious consequence of

$$\begin{aligned} -\Delta |x|^{2k}=-2k(2k+n-2)|x|^{2k-2}. \end{aligned}$$

(5.14)

Now we prove part $\Rightarrow$ of (5.13) by induction on $m$. When $m=1$, then $-\Delta w=0$, i.e. $w$ is a radial harmonic function; hence, by the mean property, $w$ is constant. Now let us suppose that part $\Rightarrow$ of (5.13) holds for a generic $m$, and let $w$ be a radial distribution such that $(-\Delta )^{m+1} w=-\Delta (-\Delta )^m w=0$; then, arguing as before, $(-\Delta )^m w=c$. By (5.14) we get that there exists a suitable constant $c_m$ such that $(-\Delta )^m c_m |x|^{2m}=c$; therefore $(-\Delta )^m (w-c_m|x|^{2m})=0$, and we may apply induction.

Finally, statement (4) follows by applying (5.13) to $v-u$.

Proof of (5) We begin by proving (5) for $f\in C_c^\infty ({\mathbb {R}^n})$. We have to show that $u:={\mathcal N}_2* f$ belongs to $\mathcal {D}^{2,p}$, i.e. $u$ is the limit of a sequence of functions belonging to $C_c^\infty ({\mathbb {R}^n})$ in the norm $\Vert \cdot \Vert _{2,p}$.

First we notice that, being $f\in L^r(\mathbb {R}^n)$ for any $r$, from (5.7) and (5.9) we deduce

$$\begin{aligned} u\in C^\infty ({\mathbb {R}^n})\cap L^{p}(\mathbb {R}^n)\,;\quad \left| \nabla u\right| \in L^{p}(\mathbb {R}^n). \end{aligned}$$

(5.15)

Let $\varphi _k(x):=\varphi _1(\left| x\right| /k)$ where $\varphi _1$ is a standard cut off function, that is

$$\begin{aligned} \varphi _1\in C_c^\infty (\mathbb {R}),\quad 0\le \varphi _1\le 1,\quad \varphi _1(t)={\left\{ \begin{array}{ll} 1&{} \left| t\right| \le 1;\\ 0&{} \left| t\right| \ge 2. \end{array}\right. } \end{aligned}$$

(5.16)

We claim that $\varphi _ku\rightarrow u$ in $\mathcal {D}^{2,p}({\mathbb {R}^n})$ as $k\rightarrow +\infty$. Indeed, taking into account that $-\Delta u= f$, for $k$ sufficiently large (so that $(1-\varphi _k)f\equiv 0$) we have

$$\begin{aligned} \Vert \Delta (\varphi _ku)- \Delta u\Vert _p&= \Vert (\varphi _k-1) \Delta u+u \Delta \varphi _k+ 2\nabla u\cdot \nabla \varphi _k\Vert _p\nonumber \\&\le \Vert u \Delta \varphi _k\Vert _p+2\Vert \nabla u\cdot \nabla \varphi _k\Vert _p \le C_{\varphi _1}(k^{-2}\Vert u\Vert _p+k^{-1}\Vert \nabla u\Vert _p). \end{aligned}$$

(5.17)

To conclude the proof, let $f\in L^p(\mathbb {R}^n)$ and let $(f_n)\subset C_c^\infty (\mathbb {R}^n)$ be a sequence of functions such that $f_n\rightarrow f$ in $L^p$. Set $u_n:={\mathcal N}_2* f_n$, $u:={\mathcal N}_2* f$. We have already proved that $u_n\in \mathcal {D}^{2,p}(\mathbb {R}^n)$, and since $\left\| u_n-u\right\| _{2,p} =\left\| \Delta u_n-\Delta u\right\| _p=\left\| f_n-f\right\| _p$, the claim follows.

Proof of (6) First of all let us recall that, when $\textstyle {p=\frac{2n}{n+2}}$, then statement (6) follows from [17].

Now, let $f\in L^p(\mathbb {R}^n)$ and let $u$, $v$ be as in the claim. Since the symmetrization operator

$$\begin{aligned} g\in L^q({\mathbb {R}^n})\rightarrow g^*\in L^q({\mathbb {R}^n}) \end{aligned}$$

(5.18)

preserves the $L^q$ norm for any $q>1$, we have that $f^*\in L^p$, so that (5.9) and statement (5) hold true.

In order to show that $v\ge u^*$, we argue as follows. Let $(f_n)\subset C_c^\infty (\mathbb {R}^n)$ be a sequence of functions such that $f_n\rightarrow f$ in $L^p$. Let $u_n$ and $v_n$ be the solutions to

$$\begin{aligned} -\Delta u_n=f_n,\qquad -\Delta v_n=f_n^* \end{aligned}$$

satisfying (5.2); as $f_n\in L^{\frac{2n}{n+2}}(\mathbb {R}^n)$, we may apply Talenti comparison (see [17]), obtaining

$$\begin{aligned} v_n-u_n^*\ge 0\quad \text {a.e.} \end{aligned}$$

(5.19)

Since the convolution operator

$$\begin{aligned} g\in L^p({\mathbb {R}^n})\rightarrow {\mathcal N}_2* g\in L^{\frac{np}{n-2p}}({\mathbb {R}^n}) \end{aligned}$$

is continuous, we have that

$$\begin{aligned} u_n= {\mathcal N}_2* f_n \rightarrow u\,,\quad v_n= {\mathcal N}_2* f_n^* \rightarrow v \quad \text {in }L^{\frac{np}{n-2p}}({\mathbb {R}^n}). \end{aligned}$$

Next, being the symmetrization operator (5.18) non expansive, $u_n^*\rightarrow u^*$ in $L^{\frac{np}{n-2p}}({\mathbb {R}^n})$. Hence $v_n-u_n^*\rightarrow v-u^*$ in $L^{\frac{np}{n-2p}}({\mathbb {R}^n})$, and the claim follows taking (5.19) into account. $\square$

Appendix B: Hardy inequalities

In this section we present some suitable generalizations of the classical Hardy inequality (see [9]) in the radial case for unbounded functions (see also [16]).

Proposition 5.5

Let $h>0$. Then, for any $f\in C^1((0,1])$ with $f(1)=0$, the inequality

$$\begin{aligned} \int _0^1 f'(\rho )^2\rho ^{h+1}\,d\rho \,\ge \frac{h^2}{4}\int _0^1 f(\rho )^2\rho ^{h-1}\,d\rho \end{aligned}$$

(5.20)

holds.

Proof

Let $f\not \equiv 0$ be as in the hypothesis (if $f\equiv 0$ the claim is trivial). Integrating by parts on $[1/n,1]$ we have

$$\begin{aligned} \int _{1/n}^1 f(\rho )f'(\rho )\rho ^h\,d\rho = -\frac{f(1/n)^2}{2n^h} -\frac{h}{2}\int _{1/n}^1 f(\rho )^2\rho ^{h-1}\,d\rho ; \end{aligned}$$

hence

$$\begin{aligned} \left( \int _{1/n}^1 f(\rho )f'(\rho )\rho ^h\,d\rho \right) ^2=\left( \frac{h}{2}\int _{1/n}^1 f(\rho )^2\rho ^{h-1}\,d\rho + \frac{f(1/n)^2}{2n^h}\right) ^2. \end{aligned}$$

(5.21)

On the other hand, by Cauchy–Schwarz inequality, we obtain

$$\begin{aligned} \left( \int _{1/n}^1 f(\rho )f'(\rho )\rho ^h\,d\rho \right) ^2&= \left( \int _{1/n}^1 f(\rho )\rho ^\frac{h-1}{2} f'(\rho )\rho ^\frac{h+1}{2}\,d\rho \right) ^2\\&\le \left( \int _{1/n}^1 f(\rho )^2\rho ^{h-1}\,d\rho \right) \left( \int _{1/n}^1 f'(\rho )^2\rho ^{h+1}\,d\rho \right) \end{aligned}$$

which, by means of (5.21), implies

$$\begin{aligned} \int _{1/n}^1 f'(\rho )^2\rho ^{h+1}\,d\rho \ge \frac{\displaystyle {\left( \frac{h}{2}\int _{1/n}^1 f(\rho )^2\rho ^{h-1}\,d\rho \, +\frac{f(1/n)^2}{2n^h}\right) ^2}}{\displaystyle {\int _{1/n}^1 f(\rho )^2\rho ^{h-1}\,d\rho }}\ge \frac{h^2}{4}\int _{1/n}^1 f(\rho )^2\rho ^{h-1}\,d\rho . \end{aligned}$$

Letting $n\rightarrow \infty$ we achieve the claim. $\square$

Now we state a second order Hardy inequality for unbounded functions.

Proposition 5.6

Let $h>0$ and $c\ge 0$. Then for any $u\in C^2((0,1])$ such that

$$\begin{aligned} u(1)=u'(1)=0,\quad \int _0^1 u''(\rho )^2\rho ^{h+3}\,d\rho \,<\infty , \end{aligned}$$

the inequality

$$\begin{aligned} \int _0^1 u''(\rho )^2\rho ^{h+3}\,d\rho - \left( \frac{h^2}{2}+h+1-c\right) \int _0^1 u'(\rho )^2\rho ^{h+1}\,d\rho \nonumber \\ +\frac{h^2}{4}\left( \frac{h^2}{4}-c\right) \int _0^1 u(\rho )^2\rho ^{h-1}\,d\rho \ge 0 \end{aligned}$$

(5.22)

holds.

Proof

First of all, hypothesis $\int _0^1 u''(\rho )^2\rho ^{h+3}<\infty$ implies, by means of (5.20), that $\int _0^1 u'(\rho )^2\rho ^{h+1}$ and $\int _0^1 u(\rho )^2\rho ^{h-1}$ are finite, hence (5.22) makes sense.

Now fix $u\in C^2((0,1])$ such that $u(1)=u'(1)=0$, and set $v(\rho ):=\rho ^{h/2}u(\rho )$. Taking into account Lemma 4.3 and integrating by parts we obtain

$$\begin{aligned} \int _0^1 v'(\rho )^2\rho \,d\rho =&\int _0^1 u'(\rho )^2\rho ^{h+1}\,d\rho -\frac{h^2}{4} \int _0^1 u(\rho )^2\rho ^{h-1}\,d\rho ;\nonumber \\ \int _0^1 v''(\rho )^2\rho ^{3}\,d\rho =&\int _0^1 u''(\rho )^2\rho ^{h+3}\,d\rho -\left( \frac{h^2}{2}+h\right) \int _0^1 u'(\rho )^2\rho ^{h+1}\,d\rho \nonumber \\&+\frac{h^2}{4}\left( \frac{h^2}{4}-1\right) \int _0^1 u(\rho )^2\rho ^{h-1}\,d\rho . \end{aligned}$$

(5.23)

Next, by Hardy inequality (5.20) we have

$$\begin{aligned} \int _0^1 v''(\rho )^2\rho ^{3}\,d\rho -\int _0^1 v'(\rho )^2\rho \,d\rho \ge 0. \end{aligned}$$

(5.24)

Combining (5.23) e (5.24), we deduce

$$\begin{aligned} 0&\le \int _0^1 v''(\rho )^2\rho ^{3}\,d\rho -\int _0^1 v'(\rho )^2\rho \,d\rho \nonumber \\&=\int _0^1 u''(\rho )^2\rho ^{h+3}\,d\rho - \left( \frac{h^2}{2}+h+1\right) \int _0^1 u'(\rho )^2\rho ^{h+1}\,d\rho +\frac{h^4}{16} \int _0^1 u(\rho )^2\rho ^{h-1}\,d\rho \end{aligned}$$

(5.25)

which concludes the proof in the case $c=0$.

The case $c>0$ immediately follows by adding the positive (thanks again to 5.20) quantity

$$\begin{aligned} c \left( \int _0^1 u'(\rho )^2\rho ^{h+1}\,d\rho - \frac{h^2}{4}\int _0^1 u(\rho )^2\rho ^{h-1}\,d\rho \right) \end{aligned}$$

to (5.25). $\square$