Existence of infinitely many solutions for a nonlocal problem

1.   Introduction

In this paper, we consider the following nonlocal Hénon equation

with critical growth, where $\alpha > 0$ is a positive constant, $p = \frac{n+2s}{n-2s}, n\geq2+2s$, $\frac{1}{2} < s < 1$, $B_1(0)$ is the unit ball in ${\mathbb{R}}^n$ and $\mathcal{A}_s$ stands for the fractional Laplacian operator in $B_1(0)$ with zero Dirichlet boundary values on $\partial B_1(0)$.

Here, to define the fractional Laplacian operator $\mathcal{A}_s$ in $B_1(0)$, let $\{ \lambda_k, \varphi_k\}$ be the eigenvalues and corresponding eigenfunctions of the Laplacian operator $-\Delta$ in $B_1(0)$ with zero Dirichlet boundary values on $\partial B_1(0)$, namely, $\{ \lambda_k, \varphi_k\}$ satisfies

with $\| \varphi_k\|_{L^2(B_1(0))} = 1$. Then we can define the fractional Laplacian operator $\mathcal{A}_s$: $H_0^s(B_1(0))\rightarrow H_0^{-s}(B_1(0))$ as

where the fractional Sobolev space $H^s_0(B_1(0))(0 < s < 1)$ is given by

and equipped with the following inner product

Form the above definitions, it immediately follows that for any $u, v\in H^s_0(B_1(0))$,

It is well known that the nonlinear fractional equations appear in diverse areas including physics, biological modeling and mathematical finances and have attracted the considerable attention in the recent period. Also in recent years, there have been many investigations for the related fractional problem $\mathcal{A}_s u = f(u)$, where $f:{\mathbb{R}}^n\rightarrow{\mathbb{R}}$ is a certain function. But a complete review of the available results in this context goes beyond the aim of this paper. Here we just mention some very recent papers which study fractional equations involving the critical sobolev exponent (cf. ^{[3,7,19,20,21]}).

On the other hand, our main interest in the present paper is motivated by some works that have appeared in recent years related to the classical local Hénon equation of this kind,

Among pioneer works we mention Ni ^[13], where the author established a compactness result of $H_{0, rad}^1(B_1(0))\hookrightarrow L^{p+1}(B_1(0))$ and thus got the existence of one positive radial solution for (1.2) if $p\in(1, \frac{n+2+2\alpha}{n-2})$. Later, in ^[18], Smets, Su and Willem established some symmetry breaking phenomenon and obtained the non-radial property of the ground state solution of (1.2) if $1 < p < \frac{n+2}{n-2}$ and $\alpha$ is large enough. When $n\geq3$ and $p = \frac{n+2}{n-2}-\sigma$, Cao and Peng ^[1] verified that the ground state solutions of (1.2) are non-radial and blow up as $\sigma\rightarrow0$. Meanwhile, when $p = \frac{n+2}{n-2}$, Serra ^[17] showed that (1.2) has a non-radial solution if $n\geq4$ and $\alpha$ is large enough. More recently, Wei and Yan ^[23] proved the existence of infinitely many non-radial solutions for (1.2) for any $\alpha > 0$. For other results related to the Hénon problem (1.2), one can refer to ^[2,11,14,15] and the references therein.

Up to our knowledge, not much is obtained for the existence of multiple solutions of equation (1.2) with fractional operator. Motivated by ^[23] and ^[12], we want to exploit the finite dimensional reduction method to investigate the existence of infinitely many non-radial solutions for (1.1). To achieve our aim, we will study the following more general problem

where $\Phi(r)$ is a bounded function defined in $[0, 1]$. It is easy to check that a necessary condition for the existence of a solution of (1.3) is that $\Phi(r)$ is positive somewhere from Pohozaev identity (see^[16]). At this point we call attention to the recent work of ^[12], where we studied (1.3) in ${\mathbb{R}}^n$ and proved that if $n > 2+2s, 0 < s < 1$ and $\Phi(|x|)$ satisfies

with $\max\{2, n-2s-2\cdot\frac{(n-2s)^2}{n+2s}\} < m < n-2s, \Phi_0 > 0, r_0 > 0, \theta > 0, \ \delta > 0$, then (1.3) has infinitely many non-radial solutions. It is worth mentioning that from assumption (1.4), $r_0$ is a local maximum point of $\Phi(r)$ and then a critical point of $\Phi(r)$. Also the function $r^\alpha$ achieves its maximum on $[0, 1]$ at $r_0 = 1$ but $r_0 = 1$ is not a critical point of $r^\alpha$. However we will verify that if $\Phi(r)$ is increasing near $r_0 = 1$, through $r_0 = 1$ is not a critical point of $\Phi(r)$, the zero Dirichlet boundary condition makes it possible to construct infinitely many solutions of (1.3). Now we state our main result as follows:

Theorem 1.1. Suppose that $n\geq2+2s, \, \, \frac{1}{2} < s < 1$. If $\Phi(1) > 0$ and $\Phi'(1) > 0$, then problem (1.3) has infinitely many non-radial solutions. Particularly, the Hénon equation (1.1) has infinitely many non-radial solutions.

In the end of this part, let us outline the main idea in the proof of Theorem 1.1.

Given any $\varepsilon > 0$ and $y^0\in{\mathbb{R}}^n$, let

for $x\in{\mathbb{R}}^n$ and $\sigma_{n, s} = 2^{\frac{n-2s}{2}}\Big(\frac{\Gamma(\frac{n+2s}{2})}{\Gamma(\frac{n-2s}{2})}\Big)^{\frac{n-2s}{4s}}.$ In 1983, Lieb ^[10] (also see ^[6,7,8,9]) proved that $U_{ \varepsilon, y^0}(x)$ solves the following critical fractional equation

Also, very recently, J. DÁvila, M. del Pino and Y. Sire ^[5] obtained the non-degeneracy of $U_{ \varepsilon, y^0}(x)$. More precisely, if we define the corresponding functional of (1.5) as

then $I_0$ possesses a finite-dimensional manifold $Z$ of least energy critical points, given by

and

Now let us fix a positive integer $k\geq k_0$, where $k_0$ is large, which is to be determined later and set

to be the scaling parameter. Using the transformation $u(x)\mapsto\nu^{-\frac{n-2s}{2}}u(\frac {x}{\nu})$, (1.3) becomes

Since $U_{ \varepsilon, y^0}$ is not zero on $\partial B_\nu(0)$, we define $PU_{ \varepsilon, y^0}$ as the solution of the following problem

and we will use the solution $PU_{ \varepsilon, y^0}$ to build up the approximate solutions for (1.6).

For $x = (x', x'')\in {\mathbb{R}}^2\times{\mathbb{R}}^{n-2},$ we define

Also, we denote

where

with $0$ is the zero vector in ${\mathbb{R}}^{n-2}$. And throughout this paper, we always assume that $r\in\big[\nu(1-\frac{r_0}{k}), \nu(1-\frac{r_1}{k})\big]$, $\varepsilon_0\leq \varepsilon\leq \varepsilon_1$ for some constants $r_1 > r_0 > 0$ and $\varepsilon_1 > \varepsilon_0 > 0$.

To prove Theorem 1.1, it suffices to verify the following result:

Theorem 1.2. Under the assumption of Theorem 1.1, there is an integer $k_0 > 0$, such that for any integer $k\geq k_0$, (1.6) has a solution $u_k$ of the form

where $\omega_k\in H_k$, and as $k\rightarrow +\infty, \|\omega_k\|_{L^\infty(B_\nu(0))}\rightarrow0$, $r_k\in\Big[\nu(1-\frac{r_0}{k}), \nu(1-\frac{r_1}{k})\Big]$, $\varepsilon_0\leq \varepsilon\leq \varepsilon_1$.

We want to point out that compared with ^[23], due to the fact that the fractional Laplacian operator is nonlocal and very few things on this topic are known about the fractional Laplacian, we have to face much difficulties in the reduction process and need some more delicate estimates in the proof of our results.

The rest of the paper is organized as follows. In Section 2, we will carry out a reduction procedure and we prove our main result in Section 3. Finally, in Appendix, some basic estimates and an energy expansion for the functional corresponding to problem (1.6) will be established.

2.   Finite-dimension reduction

In this section, we perform a finite-dimensional reduction. Let

and

where $\tau = \frac{n-2s}{n-2s+1}.$ For this choice of $\tau$, we find that

Let

Now we consider

for some numbers $c_l$, where $\langle u, v\rangle = \int_{B_\nu(0)} uv.$ Then we have

Lemma 2.1. Suppose that $\varphi_{k}$ solves (2.3) for $g = g_{k}$. If $\|g_k\|_{**}$ goes to zero as $k$ goes to infinity, so does $\|\varphi_{k}\|_{*}$.

Proof. We will argue by an indirect method. Suppose by contradiction that there exist $k\rightarrow +\infty, g = g_k, \varepsilon_{k}\in[\varepsilon_{0}, \varepsilon_{1}], r_k\in\Big[\nu(1-\frac{r_0}{k}), \nu(1-\frac{r_1}{k})\Big]$ and $\varphi_{k}$ solving (2.3) for $g = g_k, \varepsilon = \varepsilon_{k}, r = r_{k}$ with $\|g_k\|_{**}\rightarrow0$ and $\|\varphi_{k}\|_{*}\geq c > 0.$ We may assume that $\|\varphi_{k}\|_{*} = 1$. Also for simplicity, we drop the subscript $k$.

Note that we can rewrite (2.3) as

Now we estimate each terms in (2.4). Analogous to Lemma A.3, we have

Meanwhile, by Lemma A.2, we get

and

Next, we estimate $c_{l}, l = 1, 2.$ Multiplying (2.3) by $Z_{1, t}(t = 1, 2)$ and integrating, we see that $c_{l}$ satisfies

First, it follows from Lemma A.1 that

On the other hand,

Define

Observe that for $y\in \Omega_1$, $|y-x^i|\geq |y-x^1|$ and then

which implies

and

As a result, from (2.9), we have

Using the same argument used for proving (A.7), it follows from (2.9) and (2.10) that

Finally, we have

Hence using (2.8), we get

So, combining (2.4)-(2.7) and (2.12), one has

Being $\|\varphi\|_{*} = 1$, we obtain from (2.13) that there is $R > 0$ such that

for some $i$. But, by using (2.3), $\tilde{\varphi}(x) = \varphi(x-x^i)$ converges, uniformly in any compact set, to a solution $\phi$ of the following equation

for some $\varepsilon\in[\varepsilon_{0}, \varepsilon_{1}]$. Due to the non-degeneracy of $U_{ \varepsilon, 0}$, we can infer that $\phi = 0$, which yields a contradiction with (2.14) and then this proof has been proved.

From Lemma 2.1, arguing as proving Proposition 4.1 in ^[6] or Proposition 2.2 in ^[12], we can show the following result.

Proposition 2.2. There exists $k_0 > 0$ and a constant $C > 0,$ independent of $k,$ such that for all $k\geq k_0,$ and $g\in L^\infty({\mathbb{R}}^n)$, problem (2.3) has a unique solution $\varphi = L_k(g)$. Also

and

To prove our results, we consider

In order to use the contraction mapping theorem to prove that (2.15) is uniquely solvable in the set that $\|\varphi\|_*$ is small, we rewrite (2.15) as

where

and

Lemma 2.3. There holds

Proof. Firstly, we deal with the case $p\leq2.$ By Hölder inequality, we find

Using the same argument as above, if $p > 2$, we also have

which completes our proof.

Lemma 2.4. Assume that $r\in\Big[\nu(1-\frac{r_0}{k}), \nu(1-\frac{r_1}{k})\Big]$. Then there is a small $\epsilon > 0,$ such that

Proof. Recall that

and

By symmetry, we can suppose that $x\in\Omega_{1}$ and for any $x\in\Omega_1$,

Thus

By Lemma A.1, taking any $1 < \theta\leq \frac{n+2s}{2}$, we obtain that for any $x\in\Omega_{1},$

Choosing $\theta > \frac{n-2s+1}{2}$ with $n+2s-\theta\geq \frac{n+2s}{2}+\tau$, we have

On the other hand, for $x\in \Omega_{1}$, by Lemma A.1 again, we get

So

which gives

Combining (2.17)-(2.19), we have

Now, we estimate $J_2$. Let $H(x, y)$ be the regular part of the Green function for $\mathcal{A}_s$ in $B_1(0)$ with the Dirichlet boundary condition (see the definition in Appendix) and let $\bar{x}^i_*$ be the reflection point of $\bar{x}^i$ with respect to $\partial B_1(0)$, where $\bar{x}^i = \frac{x^i}{\nu}$. Then

Take $\kappa = 1-\theta$ with $\theta > 0$ small. By (A.1), we have

since $\theta\frac{n-2s}{n-2s+1} > \frac{1}{2}$ for $n\geq 2+2s$.

Finally, we estimate $J_3.$ For any $x\in\Omega_{1}$ and $i = 2, \cdots, k$, applying Lemma A.1, we get

which gives that

On the other hand, if $x\in \Omega_1$ and $||x|-\nu|\geq\delta\nu$, where $\delta > 0$ is a fixed constant, then

and thus

If $x\in \Omega_1$ and $||x|-\nu|\leq\delta\nu$, we find

and $||x|-|x^1||\leq||x|-\nu|+|\nu-|x^1||\leq2\delta\nu.$ Thus,

which, together with (2.20)-(2.22), yields that

This finished this proof.

Proposition 2.5. There is an integer $k_0 > 0,$ such that for each $k\geq k_0, \ \varepsilon_0\leq \varepsilon\leq \varepsilon_1, \ r\in\Big[\nu(1-\frac{r_0}{k}), \nu(1-\frac{r_1}{k})\Big],$ (2.16) has a unique solution $\varphi = \varphi(r, \varepsilon)$ satisfying

where $\epsilon > 0$ is a small constant.

Proof. Recall that $\nu = k^{\frac{n-2s+1}{n-2s}}$ and set

where $0 < \alpha_0 < s$ and $i = 1, 2, ..., k, l = 1, 2.$ From Proposition 2.2, solving (2.16) is equivalent to solving

where $L_{k}$ is defined in Proposition 2.2.

Firstly, we find

which tells that $\mathcal {B}$ maps $\mathcal {N}$ to $\mathcal {N}.$

On the other hand, since

we get

Taking into account that

we have

Thus $\mathcal{B}$ is a contraction map.

Therefore, Applying the contraction mapping theorem, we can find a unique $\varphi = \varphi(r, \varepsilon)\in \mathcal {N}$ such that

and

Moreover, we get the estimate of $c_{l}$ from (2.12).

3.   Proof of Theorem 1.2

Let $F(\ell, \varepsilon) = I(U_{ \varepsilon, r}+\varphi),$ where $r = |x^1|, \ell = 1-\frac{r}{\nu}$, $\varphi$ is the function obtained in Proposition 2.5, and

Proposition 3.1. We have

where $A, A_1, A_2$ are some positive constants, $\epsilon > 0$ is a small constant.

Proof. Since

there is $t\in(0, 1)$ such that

Firstly,

From $\sum\limits_{i = 2}^k\frac{1}{|x^i-x^{1}|^{\tau}}\leq C$ and Lemma A.1, we obtain

Therefore, we see

On the other hand, by Hölder inequality, we have

Combining the estimates above and applying Proposition A.5, we have proved

Proposition 3.2. We have

and

where $A_1, A_2$ are the same constants as in Proposition 3.1, and $\epsilon > 0$ is a small constant.

Proof. Note that

Considering

and from Proposition 2.5, we can get

On the other hand, being $\varphi\in\mathcal{N}$, we have

and

Thus, using the estimates above,

from which and Proposition A.6, the conclusion follows.

Finally, noting that $\frac{\partial}{\partial \ell} = -\nu\frac{\partial}{\partial r}$ and arguing as before, we can get the estimate $\frac{\partial F(\ell, \varepsilon)}{\partial\ell}$.

To prove Theorem 1.2, now we estimate $H(\bar{x}^1, \bar{x}^1)$ and $G(\bar{x}^i, \bar{x}^1), i\geq2$. Let $\bar{x}^1_* = \big(\frac{1}{|\bar{x}^1|}, 0, \cdots, 0\big)$ be the reflection of $\bar{x}^1$ with respect to the unit sphere. Then

for some constant $C > 0$.

On the other hand,

where $\theta_i$ is the angle between $\bar{x}^i-\bar{x}^1$ and $(1, 0, \cdots, 0)$ and then $\theta_i = \frac{\pi}{2}+\frac{(i-1)\pi}{k}$. Thus

As

$\ell k\rightarrow c > 0$ and

we find

for some constants $c_2\geq c_1 > 0$.

So, there exists a constant $A_3 > 0$ such that

Therefore, it follows from Propositions 3.1 and 3.2 that there are constants $B_1, B_2, B_3$ such that

and

Proof of Theorem 1.2. Recall that $\ell = 1-\frac{r}{\nu}$ and $\nu = k^{\frac{n-2s+1}{n-2s}}$. Denote $L = \ell k$. Then from (3.2) and (3.3), $\frac{\partial F(\ell, \varepsilon)}{\partial \varepsilon} = 0$ and $\frac{\partial F(\ell, \varepsilon)}{\partial \ell} = 0$ are equivalent to

and

respectively.

Denote

and

Thus $h_1 = 0$ and $h_2 = 0$ have a unique solution

Moreover, it is easy to verify that

and

which means that $h_1 = 0$ and $h_2 = 0$ at $(L_0, \varepsilon_0)$ is invertible. So, (3.4) and (3.5) have a solution near $(L_0, \varepsilon_0)$.

Acknowledgments

This work was partially supported by NSFC (No.11601194) and the author thanks the referee's thoughtful reading of details of the paper and nice suggestions to improve the results.

Conflict of interest

The author declares that there are no conflicts of interest.

A.   Energy expansion

In this section, we will give some basic estimates and the energy expansion for the approximate solutions. First, recall that

where $0\in {\mathbb{R}}^{n-2}$, $r\in\Big[\nu(1-\frac{r_0}{k}), \nu(1-\frac{r_1}{k})\Big]$, and

Now, we introduce the following two lemmas which are important in this sequel and have been proved in ^[22] and ^[12] respectively.

Lemma A.1. For any constant $0 < \sigma\leq < /italic > < italic > \min\{ \alpha, \beta\}$, there exists a constant $C > 0$, such that

Lemma A.2. For any constant $0 < \kappa < < /italic > < italic > n-2s$, there is a constant $C > 0$ such that

Note that

and

for some suitable $C_{n, s}$. We have

Lemma A.3. There is a small $\epsilon > 0$ and some constant $C > 0$ such that

Proof. We can find this proof in ^[12]. Here we just need to use

Let $G(x, y)$ be the Green function of $\mathcal{A}_s$ in $B_1(0)$ with the Dirichlet boundary condition (see ^[4]), namely, $G(x, y)$ solves

and the regular part of $G$ is given by

where $\alpha_{n, s} = \frac{1}{|S^{n-1}|}\frac{2^{1-2s}\Gamma(\frac{n-2s}{2})}{\Gamma(\frac{n}{2})\Gamma(s)}$.

Let $\bar{x}^i = \frac{x^i}{\nu}$. Then

Proposition A.4. We have

where $\ell = 1-|\bar{x}^i| = 1-\frac{r}{\nu}$.

Proof. First, letting $\varphi_{x^i, \varepsilon} = U_{ \varepsilon, x^i}-PU_{ \varepsilon, x^i}$, from the equations satisfied by $U_{ \varepsilon, x^i}$ and $PU_{ \varepsilon, x^i}$, we can get that

Denote by $\tilde{G}(x, y)$ the Green function of $\mathcal{A}_s$ in $B_\nu(0)$ with the Dirichlet boundary condition and by $\tilde{H}(x, y)$ the regular part of $\tilde{G}(x, y)$. So we can find

which, together with (A.2), yields that

As a result,

for some constant $C > 0$.

Using (A.3), we have proved

Proposition A.5. There holds

where $A, A_1, A_2$ are some positive constants, and $\epsilon$ is a small constant.

Proof. Note that

Using the symmetry and (A.1), we have

where $B_0 = \int_{{\mathbb{R}}^n}U_{1, 0}^{p}$.

Recalling that

we can obtain that

Next, we estimate each terms in (A.5). For $x\in \Omega_1$, $|x-x^i|\geq |x-x^1|,$ we obtain for any $\kappa\in(1, n-2s)$,

which implies that if $\kappa$ close to $n-2s$,

On the other hand, from (A.1) again, we have

where $\tilde{\Omega}_1 = \{\frac{y}{ \varepsilon}+x^1\in \Omega_1\}$.

Now we estimate $\int_{\Omega_1}\big|\Phi(\frac{|x|}{\nu})-1\big|U_{ \varepsilon, x^1}^{p}\sum_{i = 2}^kU_{ \varepsilon, x^i}$. For $x\in \Omega_1$ and $||x|-\nu|\geq\delta\nu$, where $\delta > 0$ is a fixed constant, we have

and from Lemma A.1,

If $x\in \Omega_1$ and $||x|-\nu|\leq\delta\nu$, we have

and $||x|-|x^1||\leq||x|-\nu|+|\nu-|x^1||\leq2\delta\nu.$ Choosing $\kappa\in(1, n-2s)$, we obtain that

This, together with (A.6), tells us that

Finally,

But

Thus, from the estimates above, we have proved

which, together with (A.4), implies that our desired result holds.

Similar to Proposition A.5, we also have

Proposition A.6.

and

where $A_1, A_2$ are the same constants as in Proposition A.5.