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Direct Method of Scaling Spheres for the Laplacian and Fractional Laplacian Equations with Hardy-Hénon Type Nonlinearity

Meiqing Xu111School of Mathematical Sciences, CMA-Shanghai, Shanghai Jiao Tong University, China. The author is partially supported by NSFC-12031012, NSFC-11831003 and the Institute of Modern Analysis-A Frontier Research Center of Shanghai.
Abstract

In this paper, we focus on the partial differential equation

(Δ)α2u(x)=f(x,u(x)) in n,(-\Delta)^{\frac{\alpha}{2}}u(x)=f(x,u(x))\;\;\;\;\text{ in }\mathbb{R}^{n},

where 0<α20<\alpha\leq 2. By the direct method of scaling spheres investigated by Dai and Qin ([23], International Mathematics Research Notices, 2023), we derive a Liouville-type theorem. This mildly extends the previous researches on Liouville-type theorem for the semi-linear equation (Δ)α2u(x)=f(u(x))(-\Delta)^{\frac{\alpha}{2}}u(x)=f(u(x)) where the nonlinearity ff depends solely on the solution u(x)u(x), and covers the Liouville-type theorem for Hardy-Hénon equations (Δ)α2u(x)=|x|aup(x)(-\Delta)^{\frac{\alpha}{2}}u(x)=|x|^{a}u^{p}(x).

Keywords: Liouville-type theorem, equivalence, fractional Laplacians, method of scaling spheres.

MSC 2020: Primary: 35R11; Secondary: 35B50, 35C15.

1 Introduction

In this paper we consider the partial differential equation

(Δ)α2u(x)=f(x,u(x)) in n,(-\Delta)^{\frac{\alpha}{2}}u(x)=f(x,u(x))\;\;\;\;\text{ in }\mathbb{R}^{n}, (1.1)

where n>αn>\alpha, 0<α20<\alpha\leq 2, and the Hardy-Hénon type nonlinearity f:(n\{0})×[0,+)[0,+)f:(\mathbb{R}^{n}\backslash\{0\})\times[0,+\infty)\to[0,+\infty). For α=2\alpha=2, (Δ)α2(-\Delta)^{\frac{\alpha}{2}} is the classical Laplacians operator, that is, Δ=i=1n2xi2-\Delta=-\sum_{i=1}^{n}\frac{\partial^{2}}{\partial x_{i}^{2}}. For α\alpha taking any real number between 0 and 2, (Δ)α2(-\Delta)^{\frac{\alpha}{2}} is a nonlocal differential operator defined by

(Δ)α2u(x)=Cn,αP.V.nu(x)u(y)|xy|n+α𝑑y.(-\Delta)^{\frac{\alpha}{2}}u(x)=C_{n,\alpha}P.V.\int_{\mathbb{R}^{n}}\frac{u(x)-u(y)}{|x-y|^{n+\alpha}}dy. (1.2)

where P.V. is the Cauchy principal value. The integral on the right-hand side of (1.2) is well defined for uCloc[α],{α}+ϵ(n)α(n)u\in C_{loc}^{[\alpha],\{\alpha\}+\epsilon}(\mathbb{R}^{n})\cap\mathcal{L}_{\alpha}(\mathbb{R}^{n}), where ϵ>0\epsilon>0 is arbitrary small, [α][\alpha] denotes the integer part of α\alpha, {α}=α[α]\{\alpha\}=\alpha-[\alpha],

α(n)={u:n|n|u(x)|(1+|x|n+α)𝑑x<}.\mathcal{L}_{\alpha}(\mathbb{R}^{n})=\{u:\mathbb{R}^{n}\rightarrow\mathbb{R}|\int_{\mathbb{R}^{n}}\frac{|u(x)|}{(1+|x|^{n+\alpha})}dx<\infty\}. (1.3)

For more details, see [46, 13].

For α=2\alpha=2, we assume the solution uC2(n\{0})C(n)u\in C^{2}(\mathbb{R}^{n}\backslash\{0\})\cap C(\mathbb{R}^{n}). For 0<α<20<\alpha<2, we assume the solution uCloc[α],{α}+ϵ(n\{0})α(n)u\in C_{loc}^{[\alpha],\{\alpha\}+\epsilon}(\mathbb{R}^{n}\backslash\{0\})\cap\mathcal{L}_{\alpha}(\mathbb{R}^{n}).

Definition 1.1.

A function g(x,u)g(x,u) is called locally Lipschitz on uu in n×[0,+)\mathbb{R}^{n}\times[0,+\infty), provided that for any u0[0,+)u_{0}\in[0,+\infty) and Ωn\Omega\subset\mathbb{R}^{n} bounded, there exists a (relatively) open neighborhood U(u0)[0,+)U(u_{0})\subset[0,+\infty) such that gg is Lipschitz continuous on uu in Ω×U(u0)\Omega\times U(u_{0}).

We need the following assumptions about the Hardy-Hénon type nonlinearity f(x,u)f(x,u).

  1. (f0)(f_{0})

    The nonlinearity ff has subcritical growth, i.e.

    f(t2nαx,tu(x))/tn+αnα>f(x,u(x))for any t>1 and (x,u)(n\{0})×+n.f(t^{-\frac{2}{n-\alpha}}x,tu(x))/{t^{\frac{n+\alpha}{n-\alpha}}}>f(x,u(x))\;\;\;\;\text{for any $t>1$ and $(x,u)\in(\mathbb{R}^{n}\backslash\{0\})\times\mathbb{R}^{n}_{+}$}.
  2. (f1)(f_{1})

    The nonlinearity ff is nondecreasing about uu in (n\{0})×+n(\mathbb{R}^{n}\backslash\{0\})\times\mathbb{R}^{n}_{+}.

  3. (f2)(f_{2})

    There exists a σ<α\sigma<\alpha such that |x|σf(x,u)|x|^{\sigma}f(x,u) is locally Lipschitz on uu in n×[0,+)\mathbb{R}^{n}\times[0,+\infty).

  4. (f3)(f_{3})

    There exists a cone 𝒞n\mathcal{C}\subset\mathbb{R}^{n} with vertex at 0, constants C>0C>0, α<a<+-\alpha<a<+\infty, 0<p<pc(a)0<p<p_{c}(a) such that

    f(x,u)C|x|aup(x)in 𝒞×[0,+).f(x,u)\geq C|x|^{a}u^{p}(x)\;\;\;\;\text{in $\mathcal{C}\times[0,+\infty)$.}

The Liouville type theorems of the Lane-Emden equations, Hardy-Hénon equations and equations with general nonlinearities have been extensively studied (see [35, 52, 33, 34, 31, 49, 5, 6, 7, 53, 17, 29]). In the study of Liouville-type theorems concerning the nonexistence of non-trivial solutions, the most commonly employed approach is the method of moving planes. This method has been extensively explored in prior works such as [3, 1, 8, 9, 32] and further developed in recent research contributions, including [39, 10, 15, 11, 12, 9, 20]. However, the method of moving planes has limitations. For example, it requires the solution to be bounded or integrable. In [23], Dai and Qin introduced the method of scaling spheres, which offers significant advantages. Notably, in the fractional case, it does not necessitate any boundedness or integrability of the solution, in contrast to the method of moving planes. This variation can potentially provide novel perspectives for exploring Liouville-type theorems. Another distinct feature is that the potential term in nonlinearity does not need to be strictly decreasing after the Kelvin transform, as required by the method of moving planes. This extension greatly broadens the range of the exponent pp, from 1<p<p+a+αnα1<p<\frac{p+a+\alpha}{n-\alpha} to 0<p<p+2a+αnα0<p<\frac{p+2a+\alpha}{n-\alpha}. To be precise, by the direct method of scaling spheres, Dai and Qin expanded the range of pp in the following ways (only relevant parts of the results are listed here):

  1. 1.

    For nonnegative solution uu of the equation

    (Δ)α2u(x)=|x|aup(x),xn,(-\Delta)^{\frac{\alpha}{2}}u(x)=|x|^{a}u^{p}(x),\;\;\;\;x\in\mathbb{R}^{n},

    where 0<α20<\alpha\leq 2, n>αn>\alpha, α<a<+-\alpha<a<+\infty, 0<p<p+2a+αnα0<p<\frac{p+2a+\alpha}{n-\alpha}, uu can only be trivial.

  2. 2.

    For a non-negative solution uu of the equation

    {(Δ)α2u(x)=f(x,u),x+n,u(x)=0,x(+n)c,\begin{cases}(-\Delta)^{\frac{\alpha}{2}}u(x)=f(x,u),\;\;\;\;&x\in\mathbb{R}^{n}_{+},\\ u(x)=0,\;\;\;\;&x\in(\mathbb{R}^{n}_{+})^{c},\end{cases}

    where 0<α20<\alpha\leq 2, n>αn>\alpha, and Hardy-Hénon type nonlinearity ff is subcritical and satisfies assumptions (f1)(f_{1}), (f2)(f_{2}) and (f3)(f_{3}), uu can only be trivial.

Dai and Qin also indicated that the problems addressable with the direct method of scaling spheres are not limited to the ones mentioned above. In fact, they explicitly point out that method of scaling spheres can also be employed to equation

(Δ)α2u(x)=f(x,u),xn.(-\Delta)^{\frac{\alpha}{2}}u(x)=f(x,u),\;\;\;\;x\in\mathbb{R}^{n}.

with Hardy-Hénon type nonlinearity ff. By further developing their idea and applying the method of scaling spheres, we derive the following Liouville-type theorem.

Theorem 1.2.

Assume n>αn>\alpha, 0<α20<\alpha\leq 2, 0<pc(a):=n+α+2anα0<p_{c}(a):=\frac{n+\alpha+2a}{n-\alpha}. Suppose ff is subcritical and satisfies the assumption (f1)(f_{1}), (f2)(f_{2}) and (f3)(f_{3}), and uu is a nonnegative solution of (1.1). Then u0u\equiv 0, i.e. any nonnegative solution of (1.1) is trivial.

To establish Theorem 1.2, we begin by giving the equivalence (Theorem 2.1) between (1.1) and the corresponding integral equation

u(x)=Cn,αnf(y,u(y))|xy|nα𝑑y.u(x)=C_{n,\alpha}\int_{\mathbb{R}^{n}}\frac{f(y,u(y))}{|x-y|^{n-\alpha}}dy. (1.4)

Then we establish the narrow region principle (Theorem 3.2), and use the (direct) method of scaling spheres.

In particular, f(x,u)=|x|aupf(x,u)=|x|^{a}u^{p} with a>αa>-\alpha satisfies all the assumptions in Theorem 1.2. Therefore, this result covers the Liouville-type theorem presented in [23] for the equation (Δ)α2u(x)=|x|aup(-\Delta)^{\frac{\alpha}{2}}u(x)=|x|^{a}u^{p}. In [47], Yu applied the method of moving planes in the integral form to get the Liouville-type theorem of the integral equation

u(x)=nf(u(y))|xy|nα𝑑y,xn,u(x)=\int_{\mathbb{R}^{n}}\frac{f(u(y))}{|x-y|^{n-\alpha}}dy,\;\;\;\;x\in\mathbb{R}^{n},

where f(u)f(u) is continuous and non-decreasing, and f(t)/tn+αnαf(t)/t^{\frac{n+\alpha}{n-\alpha}} is non-decreasing. Also in [14], Chen, Li and Zhang applied the direct method of moving spheres to get the Liouville-type theorem of the partial differential equation

(Δ)α2u(x)=f(u(x)),xn,(-\Delta)^{\frac{\alpha}{2}}u(x)=f(u(x)),\;\;\;\;x\in\mathbb{R}^{n},

where f(u)f(u) is locally bounded and f(t)/tn+αnαf(t)/t^{\frac{n+\alpha}{n-\alpha}} is non-decreasing. Distinguishing our research from previous works where the nonlinearity f(u)f(u) depends solely on the solution uu, we midly extend the scope by considering the spatial dependence in the nonlinearity f(x,u(x))f(x,u(x)). For additional research on fractional Laplacians with non-linearity, see [22, 23, 44, 28, 27, 43, 26, 45, 38, 37, 36, 48]. For further studies using the method of scaling spheres, see [41, 40, 42, 25, 21, 19, 30, 24, 18, 4]

As mentioned in [23], when dealing with fractional Laplacian, it is often more convenient to apply the Hardy-Littlewood-Sobolev inequality and the method of scaling spheres in integral form. We hope to apply the latter approach to the higher-order fractional Laplacian equations with general nonlinearity and get the Liouville-type theorem.

The paper is organized as follows. In Section 2 we prove the equivalence between PDE (1.1) and IE (1.4). In Section 3 we leverage contradiction arguments, the narrow region principle (Theorem 3.2) and the direct method of scaling spheres to prove Theorem 1.2.

2 Equivalence between PDE and IE

We first show the equivalence between (1.1) and the integral equation i.e. the solution of the PDE also solves the IE (1.4). The idea comes from [51, 23].

Theorem 2.1.

Assume n>αn>\alpha, 0<α20<\alpha\leq 2 and 0<p<+0<p<+\infty. Suppose ff is subcritical and ff satisfies the assumption (f1)(f_{1}), (f2)(f_{2}) and (f3)(f_{3}). Suppose uu is a non-trivial solution of (1.1). Then it also solves the IE (1.4), and vice versa.

One can check that if uu solves the IE then it solves the PDE. Thus we only check that any solution of the PDE is also a solution of IE. The idea of proof basically comes from [23], Theorem 2.1.

Proof.

For any R>0R>0, let

vR(x)=BRGRα(x,y)f(y,u(y))𝑑y,v_{R}(x)=\int_{B_{R}}G_{R}^{\alpha}(x,y)f(y,u(y))dy,

where BR={xn|x|<R}B_{R}=\{x\in\mathbb{R}^{n}\mid|x|<R\}, and GRα(x,y)G_{R}^{\alpha}(x,y) is the Green function for Laplacian (Δ)α2(-\Delta)^{\frac{\alpha}{2}} with 0<α20<\alpha\leq 2 in BRB_{R}, i.e.

GRα(x,y)={Cn,α|xy|nα0(R2|x|2)(R2|y|2)R2|xy|2bα21(1+b)n2𝑑b,x,yBR,0,x or yBRc.G_{R}^{\alpha}(x,y)=\begin{cases}\frac{C_{n,\alpha}}{|x-y|^{n-\alpha}}\int_{0}^{\frac{(R^{2}-|x|^{2})(R^{2}-|y|^{2})}{R^{2}|x-y|^{2}}}\frac{b^{\frac{\alpha}{2}-1}}{(1+b)^{\frac{n}{2}}}db,\;\;\;\;&x,y\in B_{R},\\ 0,&x\text{ or }y\in B_{R}^{c}.\end{cases}

By assumption (f3)(f_{3}), we have vRCloc1,1(BR\{0})C(n)α(n)v_{R}\in C_{loc}^{1,1}(B_{R}\backslash\{0\})\cap C(\mathbb{R}^{n})\cap\mathcal{L}_{\alpha}(\mathbb{R}^{n}) (vRC2(BR\{0})C(n)v_{R}\in C^{2}(B_{R}\backslash\{0\})\cap C(\mathbb{R}^{n}) if α=2\alpha=2), and

{(Δ)α2vR(x)=f(x,u(x)),xBR\{0},vR(x)=0,xBRc.\begin{cases}(-\Delta)^{\frac{\alpha}{2}}v_{R}(x)=f(x,u(x)),\;\;\;\;&x\in B_{R}\backslash\{0\},\\ v_{R}(x)=0,&x\in B_{R}^{c}.\end{cases}

Let wR(x)=u(x)vR(x)w_{R}(x)=u(x)-v_{R}(x).According to Lemma 2.2 in [23], we can establish that

{(Δ)α2wR(x)=0,xBR,wR(x)0,xBRc.\begin{cases}(-\Delta)^{\frac{\alpha}{2}}w_{R}(x)=0,\;\;\;\;&x\in B_{R},\\ w_{R}(x)\geq 0,&x\in B_{R}^{c}.\end{cases}

According to the maximum principle for Laplacians and fractional Laplacians (as discussed in [46] and also in Theorem 2.1 of [11]), we can conclude that

wR(x)=u(x)vR(x)0,xn.w_{R}(x)=u(x)-v_{R}(x)\geq 0,\;\;\;\;\forall x\in\mathbb{R}^{n}.

Fixing xx and letting RR\to\infty, we have

u(x)Cn,αnf(y,u(y))|xy|nα𝑑y0.u(x)\geq C_{n,\alpha}\int_{\mathbb{R}^{n}}\frac{f(y,u(y))}{|x-y|^{n-\alpha}}dy\geq 0.

Let x=0x=0. By assumption (f3)(f_{3}),

+>u(0)Cn,αnf(y,u(y))|y|nα𝑑yCn,α𝒞up(y)|y|nαa𝑑y.+\infty>u(0)\geq C_{n,\alpha}\int_{\mathbb{R}^{n}}\frac{f(y,u(y))}{|y|^{n-\alpha}}dy\geq C_{n,\alpha}\int_{\mathcal{C}}\frac{u^{p}(y)}{|y|^{n-\alpha-a}}dy. (2.1)

Let v(x)=Cn,αnf(y,u(y))|xy|nα𝑑y+CC0.v(x)=C_{n,\alpha}\int_{\mathbb{R}^{n}}\frac{f(y,u(y))}{|x-y|^{n-\alpha}}dy+C\geq C\geq 0.. One can see v(x)v(x) also solves

(Δ)α2v(x)=f(x,u(x))x0.(-\Delta)^{\frac{\alpha}{2}}v(x)=f(x,u(x))\;\;\;\;\forall x\neq 0.

Let w(x)=u(x)v(x)w(x)=u(x)-v(x). By  Lemma 2.2 in [23],

{(Δ)α2w(x)=0,xn,w(x)0,xn.\begin{cases}(-\Delta)^{\frac{\alpha}{2}}w(x)=0,\;\;\;\;&x\in\mathbb{R}^{n},\\ w(x)\geq 0,&x\in\mathbb{R}^{n}.\end{cases}

By Liouville Theorem for Laplacians and fractional Laplacians (see Theorem 1 in [50]), we derive that

u(x)=Cn,αnf(y,u(y))|xy|nα𝑑y+CC0.u(x)=C_{n,\alpha}\int_{\mathbb{R}^{n}}\frac{f(y,u(y))}{|x-y|^{n-\alpha}}dy+C\geq C\geq 0. (2.2)

Combining (2.2) and (2.1), it follows that

Cn,α𝒞Cp|y|nαa𝑑y<+.C_{n,\alpha}\int_{\mathcal{C}}\frac{C^{p}}{|y|^{n-\alpha-a}}dy<+\infty.

This result implies that C=0C=0 since 𝒞\mathcal{C} is an infinite cone. Therefore, (2.2) implies that

u(x)=Cn,αnf(y,u(y))|xy|nα𝑑y.u(x)=C_{n,\alpha}\int_{\mathbb{R}^{n}}\frac{f(y,u(y))}{|x-y|^{n-\alpha}}dy.

Therefore, uu also solves the IE (1.4). ∎

Moreover, we can derive a lower bound estimate of uu from the equivalent IE. To this end, we can see that for any |x|1|x|\geq 1,

u(x)\displaystyle u(x) Cn,α|x|nαB1/2¯𝒞f(y,u(y))𝑑y\displaystyle\geq\frac{C_{n,\alpha}}{|x|^{n-\alpha}}\int_{\overline{B_{1/2}}\cap\mathcal{C}}f(y,u(y))dy (2.3)
(f3)Cn,α|x|nαB1/2¯𝒞|y|aup(y)𝑑y\displaystyle\overset{(f_{3})}{\geq}\frac{C_{n,\alpha}}{|x|^{n-\alpha}}\int_{\overline{B_{1/2}}\cap\mathcal{C}}|y|^{a}u^{p}(y)dy
Cn,α|x|nα.\displaystyle\geq\frac{C_{n,\alpha}}{|x|^{n-\alpha}}.

3 Proof of Theorem 1.2

Next we will apply the method of scaling spheres to give the following lower bounds for positive uu, which will lead a contraction to IE (1.4).

Theorem 3.1.

Assume n>αn>\alpha, 0<α20<\alpha\leq 2, 0<p<pc(a):=n+α+2anα0<p<p_{c}(a):=\frac{n+\alpha+2a}{n-\alpha}. Suppose ff is subcritical and ff satisfies the assumption (f1)(f_{1}), (f2)(f_{2}) and (f3)(f_{3}). Suppose uu is a non-trivial solution of (1.1). Then we have the following lower bound estimates.

  1. 1.

    If 0<p<10<p<1, then u(x)Ck|x|ku(x)\geq C_{k}|x|^{k} for any |x|1|x|\geq 1 and any k<α+a1pk<\frac{\alpha+a}{1-p};

  2. 2.

    If 1p<pc(a)=n+α+anα1\leq p<p_{c}(a)=\frac{n+\alpha+a}{n-\alpha}, then u(x)Ck|x|ku(x)\geq C_{k}|x|^{k} for any |x|1|x|\geq 1 and any k<+k<+\infty.

Proof.

For any λ>0\lambda>0, define the Kelvin transform of uu by

uλ(x)=(λ|x|)nαu(λ2x|x|2)x0.u_{\lambda}(x)=\left(\frac{\lambda}{|x|}\right)^{n-\alpha}u\left(\frac{\lambda^{2}x}{|x|^{2}}\right)\;\;\;\;\forall x\neq 0. (3.1)

One may verify that uλα(n)Cloc1,1(+n)C(+n¯\{0})u_{\lambda}\in\mathcal{L}_{\alpha}(\mathbb{R}^{n})\cap C_{loc}^{1,1}(\mathbb{R}^{n}_{+})\cap C(\overline{\mathbb{R}^{n}_{+}}\backslash\{0\}) for 0<α<20<\alpha<2 and uλC2(+n)C(+n¯\{0})u_{\lambda}\in C^{2}(\mathbb{R}^{n}_{+})\cap C(\overline{\mathbb{R}^{n}_{+}}\backslash\{0\}) for α=2\alpha=2. It is well-known that

(Δ)α2uλ(x)=(λ|x|)n+α((Δ)α2u)(λ2x|x|2).(-\Delta)^{\frac{\alpha}{2}}u_{\lambda}(x)=\left(\frac{\lambda}{|x|}\right)^{n+\alpha}((-\Delta)^{\frac{\alpha}{2}}u)\left(\frac{\lambda^{2}x}{|x|^{2}}\right). (3.2)

Combine (3.2) and the assumption that f is subcritical, we deduce that for any |x|<λ|x|<\lambda,

(Δ)α2uλ(x)=(λ|x|)n+αf(λ2x|x|2,u(λ2x|x|2))\displaystyle(-\Delta)^{\frac{\alpha}{2}}u_{\lambda}(x)=\left(\frac{\lambda}{|x|}\right)^{n+\alpha}f\left(\frac{\lambda^{2}x}{|x|^{2}},u\left(\frac{\lambda^{2}x}{|x|^{2}}\right)\right) =(λ|x|)n+αf(λ2|x|2x,(λ|x|)(nα)uλ(x))\displaystyle=\left(\frac{\lambda}{|x|}\right)^{n+\alpha}f\left(\frac{\lambda^{2}}{|x|^{2}}x,\left(\frac{\lambda}{|x|}\right)^{-(n-\alpha)}u_{\lambda}(x)\right) (3.3)
>f(x,uλ(x)).\displaystyle>f(x,u_{\lambda}(x)).

Before applying the method of scaling sphere, we give some definitions. Define Bλ={x:|x|<λ}B_{\lambda}=\{x:|x|<\lambda\} and

wλ(x)=uλ(x)u(x)xBλ\{0}.w^{\lambda}(x)=u_{\lambda}(x)-u(x)\;\;\;\;\forall x\in B_{\lambda}\backslash\{0\}.

By direct calculation, one can check that wλw^{\lambda} is spherically anti-symmetric, i.e.

wλ(x)=(wλ)λ(x).w^{\lambda}(x)=-(w^{\lambda})_{\lambda}(x).

The process of scaling a sphere can be broken down into two steps.

Step 1. We will show that for λ>0\lambda>0 sufficiently small, it holds that wλ0w_{\lambda}\geq 0 for any xBλ\{0}x\in B_{\lambda}\backslash\{0\}.

Define

Bλ={xBλ\{0}wλ(x)<0}.B_{\lambda}^{-}=\{x\in B_{\lambda}\backslash\{0\}\mid w^{\lambda}(x)<0\}.

We will show through contradiction arguments that in fact Bλ=B_{\lambda}^{-}=\emptyset for λ>0\lambda>0 sufficiently small, which finishes the proof of Step 1.

Suppose BλB_{\lambda}^{-}\neq\emptyset. For any xBλx\in B_{\lambda}, by (3.3), we deduce that

(Δ)α2wλ(x)=(Δ)α2uλ(x)(Δ)α2u(x)>f(x,uλ(x))f(x,u(x))=cλ(x)wλ(x),(-\Delta)^{\frac{\alpha}{2}}w^{\lambda}(x)=(-\Delta)^{\frac{\alpha}{2}}u_{\lambda}(x)-(-\Delta)^{\frac{\alpha}{2}}u(x)>f(x,u_{\lambda}(x))-f(x,u(x))=c_{\lambda}(x)w^{\lambda}(x),

where

cλ(x):=f(x,uλ(x))f(x,u(x))uλ(x)u(x).c_{\lambda}(x):=\frac{f(x,u_{\lambda}(x))-f(x,u(x))}{u_{\lambda}(x)-u(x)}.

By (f1)(f_{1}), it follows that cλ(x)0c_{\lambda}(x)\geq 0 for any xBλx\in B_{\lambda}^{-}. Since 0<uλ(x)<u(x)supBλu<+0<u_{\lambda}(x)<u(x)\leq\sup_{B_{\lambda}}u<+\infty, we use (f2)(f_{2}) to get that

cλ(x)=|x|σ|x|σf(x,uλ)|x|σf(x,u)uλuLλ|x|σxBλ,c_{\lambda}(x)=|x|^{-\sigma}\frac{|x|^{\sigma}f(x,u_{\lambda})-|x|^{\sigma}f(x,u)}{u_{\lambda}-u}\leq L_{\lambda}|x|^{-\sigma}\;\;\;\;\forall x\in B_{\lambda}^{-}, (3.4)

where Lλ>0L_{\lambda}>0 is dependent of λ\lambda. Thus

(Δ)α2wλ(x)Lλ|x|σwλ(x)xBλ.(-\Delta)^{\frac{\alpha}{2}}w^{\lambda}(x)\geq L_{\lambda}|x|^{-\sigma}w^{\lambda}(x)\;\;\;\;\forall x\in B_{\lambda}^{-}. (3.5)

To show that Bλ=B_{\lambda}^{-}=\emptyset, we need the following narrow region principle, and the idea comes from [14, 23, 16, 2, 1].

Theorem 3.2 (Narrow region principle).

Assume n>αn>\alpha, 0<α20<\alpha\leq 2 and 1p<+1\leq p<+\infty. Let λ>0\lambda>0 and Aλ,l={xnλl<|x|<λ}A_{\lambda,l}=\{x\in\mathbb{R}^{n}\mid\lambda-l<|x|<\lambda\} be an annulus with thickness l(0,λ)l\in(0,\lambda). Let BλB_{\lambda}^{-} be as above. Suppose wλα(n)Cloc1,1(Aλ,l)w^{\lambda}\in\mathcal{L}_{\alpha}(\mathbb{R}^{n})\cap C_{loc}^{1,1}(A_{\lambda,l}) for 0<α<20<\alpha<2 and wλC2(Aλ,l)w^{\lambda}\in C^{2}(A_{\lambda,l}) for α=2\alpha=2 and satisfies

  1. 1.

    (Δ)α2wλ(x)cλ(x)wλ(x)0(-\Delta)^{\frac{\alpha}{2}}w^{\lambda}(x)-c_{\lambda}(x)w^{\lambda}(x)\geq 0 in Aλ,lBλA_{\lambda,l}\cap B_{\lambda}^{-}, where 0cλ(x)Lλ|x|σ0\leq c_{\lambda}(x)\leq L_{\lambda}|x|^{-\sigma} with σ<α\sigma<\alpha;

  2. 2.

    negative minimum of wλw^{\lambda} is attained in Aλ,lA_{\lambda,l}.

Then we have

  1. (i)

    there exists a sufficiently small δ0>0\delta_{0}>0 such that, if 0<λδ00<\lambda\leq\delta_{0}, then

    wλ(x)0xAλ,l;w^{\lambda}(x)\geq 0\;\;\;\;\forall x\in A_{\lambda,l}; (3.6)
  2. (ii)

    for arbitrarily fixed θ(0,1)\theta\in(0,1), there exists a sufficiently small l0>0l_{0}>0 depending on λ\lambda continuously such that, if 0<ll00<l\leq l_{0} and 0<lθλ0<l\leq\theta\lambda, then

    wλ(x)0xAλ,l.w^{\lambda}(x)\geq 0\;\;\;\;\forall x\in A_{\lambda,l}. (3.7)
Proof.

Suppose on the contrary that (3.6) and (3.7) do not hold. In other words, there exists a small δ0\delta_{0}, λ(0,δ0)\lambda\in(0,\delta_{0}), and x1Aλ,lx_{1}\in A_{\lambda,l} such that wλ(x1)<0w^{\lambda}(x_{1})<0. Also, for sufficiently small l0(λ)l_{0}(\lambda), which depends continuously on λ\lambda, there exists l(0,l0(λ))l\in(0,l_{0}(\lambda)) and x2Aλ,lx_{2}\in A_{\lambda,l} such that wλ(x2)<0w^{\lambda}(x_{2})<0.

By the same process in [23], Theorem 2.8, it follows that there exists x~Aλ,lBλ\tilde{x}\in A_{\lambda,l}\cap B_{\lambda}^{-} such that

(Δ)α2wλ(x~)Clαwλ(x~)<0(-\Delta)^{\frac{\alpha}{2}}w^{\lambda}(\tilde{x})\leq\frac{C}{l^{\alpha}}w^{\lambda}(\tilde{x})<0 (3.8)

for 0<α20<\alpha\leq 2. (3.8) and assumption 1 yield that

Lλ|x~|σcλ(x~)Clα.L_{\lambda}|\tilde{x}|^{-\sigma}\geq c_{\lambda}(\tilde{x})\geq\frac{C}{l^{\alpha}}. (3.9)

We can derive from (3.9) that for σ(0,α)\sigma\in(0,\alpha),

CLλlα|x~|σLλλα|λl|σ,C\leq L_{\lambda}\frac{l^{\alpha}}{|\tilde{x}|^{\sigma}}\leq L_{\lambda}\frac{\lambda^{\alpha}}{|\lambda-l|^{\sigma}},

and for σ(,0]\sigma\in(-\infty,0],

CLλlα|x~|σLλλασ.C\leq L_{\lambda}\frac{l^{\alpha}}{|\tilde{x}|^{\sigma}}\leq L_{\lambda}\lambda^{\alpha-\sigma}.

In both cases, we encounter a contradiction when ll is fixed and λ\lambda is sufficiently small. Thus, (i) is verified.

Similarly,

CLλlα((1θ)λ)σfor σ(0,α),\displaystyle C\leq L_{\lambda}\frac{l^{\alpha}}{((1-\theta)\lambda)^{\sigma}}\;\;\;\;\text{for }\sigma\in(0,\alpha),
CLλlαλσfor σ(,0].\displaystyle C\leq L_{\lambda}\frac{l^{\alpha}}{\lambda^{\sigma}}\;\;\;\;\text{for }\sigma\in(-\infty,0].

And we get a contraction when l0l_{0} is sufficiently small and l(0,min{θλ,l0})l\in(0,\min\{\theta\lambda,l_{0}\}). This verifies (ii). ∎

We claim that there exists a sufficiently small η0>0\eta_{0}>0 such that the following argument holds: if 0<λ<η00<\lambda<\eta_{0}, then there exists a sufficiently small θ>0\theta>0 such that

wλ(x)1,xBθλ¯\{0}.w^{\lambda}(x)\geq 1,\;\;\;\;\forall x\in\overline{B_{\theta\lambda}}\backslash\{0\}. (3.10)

In fact, we hold λ\lambda fixed and choose θ>0\theta>0 small enough such that λ2x|x|2\frac{\lambda^{2}x}{|x|^{2}} attains large values. Thus by the estimate (2.3), for any xBθλ¯\{0}x\in\overline{B_{\theta\lambda}}\backslash\{0\}, we have

uλ(x)=(λ|x|)nαu(λ2x|x|2)C(λ|x|)nα(|x|λ2)nα=Cλnα.u_{\lambda}(x)=(\frac{\lambda}{|x|})^{n-\alpha}u(\frac{\lambda^{2}x}{|x|^{2}})\geq C(\frac{\lambda}{|x|})^{n-\alpha}(\frac{|x|}{\lambda^{2}})^{n-\alpha}=\frac{C}{\lambda^{n-\alpha}}.

Then for λ\lambda sufficiently small, it holds that

wλ(x)CλnαC1.w^{\lambda}(x)\geq\frac{C}{\lambda^{n-\alpha}}-C\geq 1.

As a result, (3.10) holds.

Now let ϵ0=min{δ0,η0}\epsilon_{0}=\min\{\delta_{0},\eta_{0}\}, l=(1θ)λl=(1-\theta)\lambda. For 0<λϵ00<\lambda\leq\epsilon_{0}, (3.5) and (3.10) imply that the assumptions in Theorem 3.2 hold. Consequently, by Theorem 3.2 (i) we conclude that

wλ0in Aλ,l.w^{\lambda}\geq 0\;\;\;\;\text{in }A_{\lambda,l}.

Hence,

Bλ=for λ>0 sufficiently small.B_{\lambda}^{-}=\emptyset\;\;\;\;\text{for $\lambda>0$ sufficiently small}.

This completes Step 1.

Step 2. We will dilate the sphere {xn|x|=λ}\{x\in\mathbb{R}^{n}\mid|x|=\lambda\} outward until λ=+\lambda=+\infty to derive lower bound estimates on uu.

In fact, Step 1 provides a starting point to carry out the method of scaling spheres. Then we will continuously increase λ\lambda as long as

wλ0in Bλ\{0}.w_{\lambda}\geq 0\;\;\;\;\text{in }B_{\lambda}\backslash\{0\}.

Define

λ0=sup{λ>0wμ0 in Bμ\{0},0<μλ}.\lambda_{0}=\sup\{\lambda>0\mid w^{\mu}\geq 0\text{ in }B_{\mu}\backslash\{0\},\;\forall 0<\mu\leq\lambda\}.

It implies that

wλ00xBλ0\{0}.w^{\lambda_{0}}\geq 0\;\;\;\;\forall x\in B_{\lambda_{0}}\backslash\{0\}.

We will show that λ0=+\lambda_{0}=+\infty.

Suppose, to the contrary, that λ0<+\lambda_{0}<+\infty. To establish a contradiction, we will prove the following claim:

wλ00xBλ0\{0}.w^{\lambda_{0}}\equiv 0\;\;\;\;\forall x\in B_{\lambda_{0}}\backslash\{0\}. (3.11)

We use a contradiction argument to show the claim. Suppose, to the contrary, that (3.11) does not hold. Then wλ0w^{\lambda_{0}} is nonnegative in Bλ0\{0}B_{\lambda_{0}}\backslash\{0\}, and there exists x0Bλ0\{0}x_{0}\in B_{\lambda_{0}}\backslash\{0\} such that wλ0(x0)>0w^{\lambda_{0}}(x_{0})>0. We will first prove that

wλ0>0in Bλ0\{0},w^{\lambda_{0}}>0\;\;\;\;\text{in }B_{\lambda_{0}}\backslash\{0\}, (3.12)

and then apply Theorem 3.2 to derive a contradiction.

Since the solution uu of PDE (1.1) also satisfies IE (1.4), we can get through direct calculations that

u(x)=Bλ0f(y,u(y))|xy|nα+f(λ02|y|2y,u(λ02|y|2y))||y|λ0xλ0|y|y|nαdy.u(x)=\int_{B_{\lambda_{0}}}\frac{f(y,u(y))}{|x-y|^{n-\alpha}}+\frac{f\left(\frac{\lambda_{0}^{2}}{|y|^{2}}y,u(\frac{\lambda_{0}^{2}}{|y|^{2}}y)\right)}{|\frac{|y|}{\lambda_{0}}x-\frac{\lambda_{0}}{|y|}y|^{n-\alpha}}dy. (3.13)

Apply (1.4) and the equality |x|x|2z|z|2|=|xz||x||z||\frac{x}{|x|^{2}}-\frac{z}{|z|^{2}}|=\frac{|x-z|}{|x||z|}, we can derive that

uλ0(x)=nf(λ02|y|2y,u(λ02|y|2y))|xy|nα(λ0|y|)n+α𝑑y.u_{\lambda_{0}}(x)=\int_{\mathbb{R}^{n}}\frac{f\left(\frac{\lambda_{0}^{2}}{|y|^{2}}y,u(\frac{\lambda_{0}^{2}}{|y|^{2}}y)\right)}{|x-y|^{n-\alpha}}\left(\frac{\lambda_{0}}{|y|}\right)^{n+\alpha}dy.

Using a similar calculation, we obtain

uλ0(x)=Bλ0f(y,u(y))||y|λ0xλ0|y|y|nα+(λ0|y|)n+αf(λ02|y|2y,u(λ02|y|2y))|xy|nαdy.u_{\lambda_{0}}(x)=\int_{B_{\lambda_{0}}}\frac{f(y,u(y))}{|\frac{|y|}{\lambda_{0}}x-\frac{\lambda_{0}}{|y|}y|^{n-\alpha}}+\left(\frac{\lambda_{0}}{|y|}\right)^{n+\alpha}\frac{f\left(\frac{\lambda_{0}^{2}}{|y|^{2}}y,u(\frac{\lambda_{0}^{2}}{|y|^{2}}y)\right)}{|x-y|^{n-\alpha}}dy. (3.14)

Combine (3.13) and (3.14), we obtain that for any xBλ0\{0}x\in B_{\lambda_{0}}\backslash\{0\},

wλ0(x)=Bλ0(1|xy|nα1||y|λ0xλ0|y|y|nα)((λ0|y|)n+αf(λ02|y|2y,u(λ02|y|2y))f(y,u(y)))𝑑y.w^{\lambda_{0}}(x)=\int_{B_{\lambda_{0}}}\left(\frac{1}{|x-y|^{n-\alpha}}-\frac{1}{|\frac{|y|}{\lambda_{0}}x-\frac{\lambda_{0}}{|y|}y|^{n-\alpha}}\right)\left(\left(\frac{\lambda_{0}}{|y|}\right)^{n+\alpha}f\left(\frac{\lambda_{0}^{2}}{|y|^{2}}y,u(\frac{\lambda_{0}^{2}}{|y|^{2}}y)\right)-f(y,u(y))\right)dy.

One may check that

1|xy|nα1||y|λ0xλ0|y|y|nα>0x,yBλ0\{0}.\frac{1}{|x-y|^{n-\alpha}}-\frac{1}{|\frac{|y|}{\lambda_{0}}x-\frac{\lambda_{0}}{|y|}y|^{n-\alpha}}>0\;\;\;\;\;\forall x,y\in B_{\lambda_{0}}\backslash\{0\}.

Since f(x,u(x))f(x,u(x)) is subcritical and satisfies assumption (f1)(f_{1}), it follows that

wλ0(x)>(1|xy|nα1||y|λ0xλ0|y|y|nα)(f(y,uλ0(y))f(y,u(y)))dy0xBλ0\{0}.w^{\lambda_{0}}(x)>\left(\frac{1}{|x-y|^{n-\alpha}}-\frac{1}{|\frac{|y|}{\lambda_{0}}x-\frac{\lambda_{0}}{|y|}y|^{n-\alpha}}\right)\big{(}f(y,u_{\lambda_{0}}(y))-f(y,u(y))\big{)}dy\geq 0\;\;\;\;\forall x\in B_{\lambda_{0}}\backslash\{0\}. (3.15)

Thus (3.12) is verified.

Next, we claim that there exists ϵ>0\epsilon>0 such that wλ(x)0w^{\lambda}(x)\geq 0 in Bλ\{0}B_{\lambda}\backslash\{0\} for all λ[λ0,λ0+ϵ)\lambda\in[\lambda_{0},\lambda_{0}+\epsilon). Consequently, this implies that the sphere can be dilated outward slightly more than λ0\lambda_{0}, which contradicts the definition of λ0\lambda_{0}.

Define

l0~=min{θλ,minλ[λ0,2λ0]l0(λ)}>0,\tilde{l_{0}}=\min\{\theta\lambda,\min_{\lambda\in[\lambda_{0},2\lambda_{0}]}l_{0}(\lambda)\}>0,

where l0(λ)l_{0}(\lambda) and θ\theta are given in Theorem 3.2 (ii). For a fixed small 0<r0<12min{l0~,λ0}0<r_{0}<\frac{1}{2}\min\{\tilde{l_{0}},\lambda_{0}\}, define

m0=infxBλ0r0¯\{0}wλ0(x).m_{0}=\inf_{x\in\overline{B_{\lambda_{0}-r_{0}}}\backslash\{0\}}w^{\lambda_{0}}(x).

By (3.12), m0>0m_{0}>0. Using the same uniform-continuity arguments as in the proof of Theorem 1.3, Step 2 in [23], we can establish the existence of a sufficiently small ϵ1\epsilon_{1} in the interval (0,12minl0~,λ0)(0,\frac{1}{2}\min{\tilde{l_{0}},\lambda_{0}}). This small value of ϵ1\epsilon_{1} ensures that for all λ\lambda in the range [λ0,λ0+ϵ1][\lambda_{0},\lambda_{0}+\epsilon_{1}], the following inequality holds:

wλ(x)m02>00<|x|λ0r0.w^{\lambda}(x)\geq\frac{m_{0}}{2}>0\;\;\;\;\forall 0<|x|\leq\lambda_{0}-r_{0}. (3.16)

That means

wλ attains negative minimum in Aλ,l,where l=λλ0+r0.w^{\lambda}\text{ attains negative minimum in }A_{\lambda,l},\;\text{where }l=\lambda-\lambda_{0}+r_{0}. (3.17)

For any λ[λ0,λ0+ϵ1]\lambda\in[\lambda_{0},\lambda_{0}+\epsilon_{1}], let l=λλ0+r0l=\lambda-\lambda_{0}+r_{0}. Then l(0,l0~)l\in(0,\tilde{l_{0}}). By combining (3.17) and (3.5), one can see that the assumptions in Theorem 3.2 are satisfied. Thus,

wλ0,xAλ,l={xnλ0r0<|x|<λ}.w^{\lambda}\geq 0,\;\;\;\;\forall x\in A_{\lambda,l}=\{x\in\mathbb{R}^{n}\mid\lambda_{0}-r_{0}<|x|<\lambda\}. (3.18)

(3.16) and (3.18) imply that for any λ[λ0,λ0+ϵ1]\lambda\in[\lambda_{0},\lambda_{0}+\epsilon_{1}],

wλ0,,xBλ\{0},w^{\lambda}\geq 0,,\;\;\;\;\forall x\in B_{\lambda}\backslash\{0\},

which contradicts with the definition of λ0\lambda_{0}. As a result, (3.11) is satisfied, implying that

wλ00xBλ0\{0}.w^{\lambda_{0}}\equiv 0\;\;\;\;\forall x\in B_{\lambda_{0}}\backslash\{0\}.

However, this contradicts (3.15). Therefore, we can conclude that

λ0=+.\lambda_{0}=+\infty.

That is, for all λ(0,+)\lambda\in(0,+\infty),

u(x)(λ|x|)nαu(λ2x|x|2)u(x),0<|x|λ.u(x)\leq\left(\frac{\lambda}{|x|}\right)^{n-\alpha}u(\frac{\lambda^{2}x}{|x|^{2}})\geq u(x),\;\;\;\;\forall 0<|x|\leq\lambda.

And it is equivalent to that for all λ(0,+)\lambda\in(0,+\infty)

u(x)(λ|x|)nαu(λ2x|x|2),|x|λ.u(x)\geq\left(\frac{\lambda}{|x|}\right)^{n-\alpha}u(\frac{\lambda^{2}x}{|x|^{2}}),\;\;\;\;\forall|x|\geq\lambda.

For |x|1|x|\geq 1, let λ=x\lambda=\sqrt{x}. Then

u(x)1|x|nα2u(x|x|),|x|1.u(x)\geq\frac{1}{|x|^{\frac{n-\alpha}{2}}}u(\frac{x}{|x|}),\;\;\;\;\forall|x|\geq 1.

So,

u(x)(min|x|=1u(x))1|x|nα2=C0|x|nα2,|x|1.u(x)\geq(\min_{|x|=1}u(x))\frac{1}{|x|^{\frac{n-\alpha}{2}}}=\frac{C_{0}}{|x|^{\frac{n-\alpha}{2}}},\;\;\;\;\forall|x|\geq 1. (3.19)

Let μ0=nα2\mu_{0}=\frac{n-\alpha}{2}. The assumption (f3)(f_{3}) and (3.19) yield that for any |x|1|x|\geq 1,

u(x)\displaystyle u(x) =nf(y,u(y))|xy|nα𝑑y\displaystyle=\int_{\mathbb{R}^{n}}\frac{f(y,u(y))}{|x-y|^{n-\alpha}}dy (3.20)
C{2|x||y|4|x|}𝒞|y|aup(y)|xy|nα𝑑y\displaystyle\geq C\int_{\{2|x|\leq|y|\leq 4|x|\}\cap\mathcal{C}}\frac{|y|^{a}u^{p}(y)}{|x-y|^{n-\alpha}}dy
C|x|nα{2|x||y|4|x|}𝒞1|y|pμ0α𝑑y\displaystyle\geq\frac{C}{|x|^{n-\alpha}}\int_{\{2|x|\leq|y|\leq 4|x|\}\cap\mathcal{C}}\frac{1}{|y|^{p\mu_{0}-\alpha}}dy
C|x|nα{2|x||y|4|x|}dy|y|pμ0α\displaystyle\geq\frac{C}{|x|^{n-\alpha}}\int_{\{2|x|\leq|y|\leq 4|x|\}}\frac{dy}{|y|^{p\mu_{0}-\alpha}}
C1|x|pμ0(a+α).\displaystyle\geq\frac{C_{1}}{|x|^{p\mu_{0}-(a+\alpha)}}.

Let μ1=pμ0(a+α)\mu_{1}=p\mu_{0}-(a+\alpha). Since 0<p<pc(a):=n+α+2anα0<p<p_{c}(a):=\frac{n+\alpha+2a}{n-\alpha}, it follows that μ1<μ0\mu_{1}<\mu_{0}. Thus, we have obtained a better lower bound estimate than (3.19) after one iteration. For k=0,1,2,k=0,1,2,..., define

μk+1=pμk(a+α).\mu_{k+1}=p\mu_{k}-(a+\alpha).

Since p<n+α+2anαp<\frac{n+\alpha+2a}{n-\alpha}, one can see that {μk}\{\mu_{k}\} is a decreasing sequence, and

μka+α1pif 0<p<1;\displaystyle\mu_{k}\to-\frac{a+\alpha}{1-p}\;\;\;\;\text{if }0<p<1;
μkif 1p<n+α+2anα.\displaystyle\mu_{k}\to-\infty\;\;\;\;\text{if }1\leq p<\frac{n+\alpha+2a}{n-\alpha}.

Thus, continuing the iteration process as (3.20), we have the following lower bound estimates:

  1. 1.

    If 0<p<10<p<1, then u(x)Ck|x|ku(x)\geq C_{k}|x|^{k} for any |x|1|x|\geq 1 and any k<α+a1pk<\frac{\alpha+a}{1-p};

  2. 2.

    If 1p<pc(a)=n+α+anα1\leq p<p_{c}(a)=\frac{n+\alpha+a}{n-\alpha}, then u(x)Ck|x|ku(x)\geq C_{k}|x|^{k} for any |x|1|x|\geq 1 and any k<+k<+\infty.

This finishes Theorem 3.1. ∎

Now we choose x~𝒞,|x~|=1\tilde{x}\in\mathcal{C},\;|\tilde{x}|=1, where 𝒞\mathcal{C} is give in assumption (f3)(f_{3}). One can see that the lower bound estimates in Theorem 3.1 actually contradicts the following intergrability indicated by IE (1.4), that is,

+>u(x~)\displaystyle+\infty>u(\tilde{x}) C{|y|2}𝒞|y|aup(y)|x~y|nα𝑑y\displaystyle\geq C\int_{\{|y|\geq 2\}\cap\mathcal{C}}\frac{|y|^{a}u^{p}(y)}{|\tilde{x}-y|^{n-\alpha}}dy
C{|y|2}𝒞up(y)|y|nαa𝑑y.\displaystyle\geq C\int_{\{|y|\geq 2\}\cap\mathcal{C}}\frac{u^{p}(y)}{|y|^{n-\alpha-a}}dy.

Therefore, uu must be trivial, that is, u0u\equiv 0 in n\mathbb{R}^{n}.

COI: The authors declared that they had no conflict of interest.

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