Liouville-type theorem for a nonlocal operator on the half plane

Abstract

In this article we consider the following integral equation associated to the BO–ZK operator in the half plane. By combining the lifting regularity and the moving planes method for integral forms, we demonstrate that there is no positive solution for this integral equation.

1 Introduction

In [11] Dancer studied the nonexistence of positive solutions for the following nonlinear elliptic equation

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l@{\quad }l} -{\varDelta }u(x)=u^r(x),&{}x\in {\mathbb {R}}^2_+,\\ \\ u(x)=0,&{}x\in \partial {\mathbb {R}}^2_+\end{array}\right. \end{aligned}$$

(1)

proved a Liouville-type result by showing that problem (1) has no bounded positive solution. During the last years there has been an increasing interest in the study of linear and nonlinear integral operators, especially nonlocal and integral operators. Chen et al. [5] recently investigated the following integral equation on the upper half space

$$\begin{aligned} u(x)=c_n\int _{{{\mathbb {R}}}^n_+}\left( \frac{1}{|x-y|^{n-\alpha }}-\frac{1}{|\bar{x}-y|^{n-\alpha }}\right) u^r(y)\mathrm{d}y, \end{aligned}$$

where $r>1$ and $\alpha <n$, and established that the above integral equation is equivalent to the poly-harmonic semi-linear equation

$$\begin{aligned} (-{\varDelta })^\alpha u=u^r,\quad u>0, \end{aligned}$$

(2)

with Navier boundary conditions on the half-space. They applied the method of moving planes in integral forms and showed that there is no nonnegative solution $u\in L_{\mathrm{loc}}^{\frac{n(r-1)}{\alpha }}({{\mathbb {R}}}^n_+)$, if $\frac{n}{n-\alpha }<r<\frac{n+\alpha }{n-\alpha }$. One can see, for instance, [6, 21] for some good surveys on some Liouville-type theorems for (2).

In the present paper we study the the following integral equation on the upper half space

$$\begin{aligned} u(x)=\int _{{{\mathbb {R}}}^2_+}G_\alpha (x,y)u^r(y)\;\mathrm{d}y,\quad x=(x_1,x_2)\in {{\mathbb {R}}}^2_+={{\mathbb {R}}}\times {{\mathbb {R}}}_+, \end{aligned}$$

(3)

where $r>0$, $\alpha \in (0,1)$,

$$\begin{aligned} G_\alpha (x,y)=K_\alpha (x-y)-K_\alpha (\overline{x}-y), \end{aligned}$$

(4)

and $K_\alpha$ is the kernel of the Benjamin–Ono–Zakharov–Kuznetsov (BO–ZK) operator ${\mathcal {L}}_\alpha =I+D^{2\alpha }_{x_1}-\partial _{x_2}^2$, i.e.

$$\begin{aligned} K_\alpha (x)=C_\alpha \int _0^\infty \mathrm{e}^{-t}\mathrm{e}^{-\frac{x_2^2}{4t}}t^{-\frac{1}{2}}H_\alpha (x_1,t)\;\mathrm{d}t, \end{aligned}$$

(5)

with

$$\begin{aligned} H_\alpha ({x_1},t)=\int _{{{\mathbb {R}}}}\mathrm{e}^{-t|\xi |^{2\alpha }}\mathrm{e}^{\mathrm{i}x_1\xi }\mathrm{d}\xi , \end{aligned}$$

where $C_\alpha$ is a positive constant, depending on $\alpha$. Here $D_{x_1}$ is defined by $(-{\varDelta }_{x_1})^{1/2}$ and ${\bar{x}}$ is the reflection of x about $x_2 = 0$. It can be easily seen that under suitable decay assumptions on the solutions, (3) is equivalent to the equation

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l@{\quad }l} {\mathcal {L}}_\alpha u(x)=u^r(x),&{}x\in {\mathbb {R}}^2_+,\\ \\ u(x)=0,&{}x\in \partial {\mathbb {R}}^2_+. \end{array}\right. \end{aligned}$$

(6)

The operator ${\mathcal {L}}_\alpha$ appears in the study of toy models [1, 2, 10], parabolic equations for which local diffusions occur only in certain directions and nonlocal diffusions. See [16] for some results on regularity and rigidity properties of the operator ${\mathcal {L}}_\alpha$. Equation (6) appears in the study of solitary waves of the generalized BO–ZK equation

$$\begin{aligned} u_t+\partial _{x_1}\left( - D^{2\alpha }_{x_1}u+\partial _{x_2}^2u+u^{r}\right) =0. \end{aligned}$$

(7)

See also [22] for some local and global well-posedness results for (7). In the case $\alpha =1/2$, Eq. (7) turns into

$$\begin{aligned} u_t+ \partial _{{x_1}}\left( -{\mathscr {H}}\partial _{x_1}u+\partial _{x_2}^2u+u^{r}\right) =0, \end{aligned}$$

(8)

which was proposed as a model to describe the electromigration in thin nanoconductors on a dielectric substrate (see [13, 14]). Here ${\mathscr {H}}$ stands for the Hilbert transform in the ${x_1}$-variable such that ${\mathscr {H}}\partial _{x_1}=D_{x_1}$. In this case the kernel $K_{1/2}$ can be represented [14] by

$$\begin{aligned} K_{1/2}(x)=C_{1/2}\int _0^\infty \frac{\sqrt{t}}{t^2+x^2_1}\mathrm{e}^{-\frac{x_2^2}{4t}}\;\mathrm{d}t. \end{aligned}$$

It was proved in [14] that the regular solitary waves of (8) do exist in the fractional Sobolev–Liouville spaces (see [12]), if $2\le r<5$.

Before stating our main result, we recall that if $f\in L^q_{x_2} L^p_{x_1}({\mathbb {R}}^2_+)$, its norm is denoted by $\Vert f\Vert _{L^q_{x_2} L^p_{x_1}({\mathbb {R}}^2_+)}=\left\| \left\| f\right\| _{L^p_{x_1}({{\mathbb {R}}})}\right\| _{L^q_{x_2}({{\mathbb {R}}}^+)}$.

Theorem 1

Suppose that $p_0,q_0\ge r$ satisfies

$$\begin{aligned} \frac{\alpha }{p_0}+\frac{1}{q_0}<\frac{2\alpha }{r-1}. \end{aligned}$$

(9)

Then there is no positive solution $u\in L_{x_2}^{p_0}L_{x_1}^{q_0}({\mathbb {R}}^2_+)$ of (3).

Corollary 1

Assume that $p_0,q_0\ge r$ satisfies (9). If $u\in L_{x_2}^{p_0}L_{x_1}^{q_0}({\mathbb {R}}^2_+)$ is a nonnegative solution of (6), then $u\equiv 0$.

To prove the non-existence of positive solutions for (3), we use regularity lifting by contracting operators appearing in the integral equations [7, 19] to boost the positive solutions for integral equation (3) to $L^1({\mathbb {R}}^2_+)\cap L^\infty ({\mathbb {R}}^2_+)$.

Theorem 2

Let u be a positive solution of (3). Suppose that $u\in L_{x_2}^{p_0}L_{x_1}^{q_0}({\mathbb {R}}^2_+)$, where $p_0,q_0\ge r$ satisfies (9). Then $u\in L^1({\mathbb {R}}^2_+)\cap L^\infty ({\mathbb {R}}^2_+)$.

Next step to prove Theorem 1 is to employ the method of moving planes in integral forms. We move the plane along $x_2$ direction to show that the solutions must be monotone increasing in $x_2$ and thus derive a contradiction.

Theorem 3

Under the assumption of Theorem 2, we have that u must be symmetric about the line $x_2=c$, for some constant c. Moreover u is strictly monotone increasing with respect to ${x_2}$.

For more related results regarding the method of moving planes and integral equations, refer the reader to [6, 8, 9, 15, 19, 20] and the references therein.

This paper is organized as follows. Section 2 is devoted to the preliminaries of the kernel $K_\alpha$ and also the proof of Theorem 2. The symmetry and the nonexistence result of the solutions are proved in Sect. 3.

For the simplicity and without loss of generality, we assume henceforth $C_\alpha =1$.

Throughout the paper, the notation $A\lesssim B$ means that there exists a constant $C > 0$ such that $A\le CB$. The notation $A\gtrsim B$ is similarly defined. We will also write $A\approx B$ to mean $A\lesssim B$ and $A\gtrsim B$.

2 Properties of $K_\alpha$

In this section we will give some key properties of the kernel $K_\alpha$.

Lemma 1

Let $\alpha >0$. The following properties hold:

• (i) $K_\alpha \in L_{x_2}^pL_{x_1}^q({{\mathbb {R}}}^2)\cap L_{x_1}^qL_{x_2}^p({{\mathbb {R}}}^2)$ for $p,q\ge 1$ with $\alpha (1+1/p)>1-1/q$.

• (ii) $K_\alpha (x)>0$, for $x\in {{\mathbb {R}}}^2$, and is an even function which is strictly decreasing in $|{x_1}|$ and $|{x_2}|$ and smooth for ${x_1}\ne 0$.

• (iii) For $0<\alpha <1$, we have the following bound

  $$\begin{aligned} K_\alpha (x)\lesssim |{x_1}|^{\alpha -1}\mathrm{e}^{-|{x_2}|},\quad x\in {{\mathbb {R}}}^2, \end{aligned}$$

  where C depends only on $\alpha$.

• (iv) There is a constant C, depending only on $\alpha$, such that

  $$\begin{aligned} K_\alpha (x)\lesssim |{x_1}|^{-1-2\alpha }\mathrm{e}^{-\frac{|{x_2}|}{4}},\quad \text{ if }\;\;|{x_1}|\ge 1. \end{aligned}$$

  (10)

  and

  $$\begin{aligned} K_\alpha (x)\gtrsim |{x_1}|^{-1-2\alpha }\mathrm{e}^{-\frac{x_2^2}{4}},\quad \text{ if }\;\;|{x_1}|\ge 1\ge |{x_2}|. \end{aligned}$$

  (11)

  In particular, $0<\lim _{|{x_1}|,|{x_2}|\rightarrow +\infty }|{x_1}|^{1+2\alpha }\mathrm{e}^{x_2^2/4}K_\alpha (x)<+\infty .$

Proof

The decay properties of $H_\alpha$ are obtained in [4] (see also [3]). In particular, it is proved that

$$\begin{aligned} \lim _{|{x_1}|\rightarrow \infty }|{x_1}|^{1+2\alpha }H_{\alpha }({x_1},1)<\infty . \end{aligned}$$

Using this estimate and the scaling property

$$\begin{aligned} H_\alpha ({x_1},t)=t^{-\frac{1}{2\alpha }}H_\alpha \left( t^{-\frac{1}{2\alpha }}{x_1},1\right) , \end{aligned}$$

it is easy to see that

$$\begin{aligned} H_\alpha ({x_1},t)\approx \min \left\{ t^{-1/2\alpha },t|{x_1}|^{-1-2\alpha }\right\} . \end{aligned}$$

(12)

So that

$$\begin{aligned} K_\alpha (x)\lesssim |{x_1}|^{-1-2\alpha }\int _{0}^{|{x_1}|^{2\alpha }} \mathrm{e}^{-t-\frac{x_2^2}{4t}}t^{\frac{1}{2}}\mathrm{d}t + \int _{|{x_1}|^{2\alpha }}^{\infty }\mathrm{e}^{-t-\frac{x_2^2}{4t}} t^{-\frac{1}{2}\left( 1+\frac{1}{\alpha }\right) }\mathrm{d}t. \end{aligned}$$

(13)

It follows then from a change of variable and the elementary inequality $s+y^2/s\ge 2|y|$, for all $s\ge 0$, that

$$\begin{aligned} K_\alpha (x)\lesssim |{x_1}|^{\alpha -1}\mathrm{e}^{-|{x_2}|}\left( \int _{0}^{1}t^{1/2}\mathrm{d}t+\int _1^\infty t^{-1/2(1+1/\alpha )}\mathrm{d}t\right) \lesssim |{x_1}|^{\alpha -1}\mathrm{e}^{-|{x_2}|}.\qquad \end{aligned}$$

(14)

To prove (10), we have for $|{x_1}|\ge 1$ from (13) and the inequality

$$\begin{aligned} t+\frac{x_2^2}{4t}\ge \frac{t}{2}+\frac{1}{2t}+\frac{|{x_2}|}{4} \end{aligned}$$

that

$$\begin{aligned} K_\alpha (x)\lesssim |{x_1}|^{-2\alpha -1}\mathrm{e}^{-\frac{|{x_2}|}{4}}\int _{0}^{\infty }\mathrm{e}^{-t}\mathrm{e}^{-\frac{1}{t}}t^{\frac{1}{2}}\mathrm{d}t \lesssim |{x_1}|^{-2\alpha -1}\mathrm{e}^{-\frac{|{x_2}|}{4}}. \end{aligned}$$

(15)

In order to prove (11), we have for $|{x_1}|\ge 1\ge |{x_2}|$ from (12) and the inequality $x_2^2/t\le x_2^2+1/t$ that

$$\begin{aligned} K_\alpha (x)\ge |{x_1}|^{-1-2\alpha }\int _0^1\mathrm{e}^{-t}\mathrm{e}^{-x_2^2/4t}t^{1/2}\gtrsim |{x_1}|^{-1-2\alpha }\mathrm{e}^{-x_2^2/4}. \end{aligned}$$

The properties of $K_\alpha$ in (ii) follow from the positivity and the monotonicity of $H_\alpha$ in [4, 17]. Finally the property (i) is deduced from (13) and the Minkowski inequality. $\square$

The following lemma gives a Hardy–Littlewood–Sobolev-type inequality; and is a direct consequence of Lemma 1 (see also [18]) and the Young inequality

$$\begin{aligned} \Vert f*g\Vert _{L_{x_2}^pL_{x_1}^q({\mathbb {R}}^2_+)}\le \Vert f\Vert _{L_{x_2}^{p_1}L_{x_1}^{q_1}({\mathbb {R}}^2_+)} \Vert g\Vert _{L_{x_2}^{p_2}L_{x_1}^{q_2}({\mathbb {R}}^2_+)}, \end{aligned}$$

(16)

with $1+\frac{1}{p}=\frac{1}{p_1}+\frac{1}{p_2}$ and $1+\frac{1}{q}=\frac{1}{q_1}+\frac{1}{q_2}$.

Lemma 2

Let $\alpha >0$ and $c>0$ and $f\in L_{x_2}^pL_{x_1}^q({{\mathbb {R}}}^2)$. Then $M_\alpha (f)=K_\alpha *f\in L_{x_2}^{p_1}L_{x_1}^{q_1}({{\mathbb {R}}}^2)$ and

$$\begin{aligned} \Vert M_\alpha (f)\Vert _{ L_{x_2}^{p_1}L_{x_1}^{q_1}({{\mathbb {R}}}^2)}\le C\Vert f\Vert _{ L_{x_2}^{p}L_{x_1}^{q}({{\mathbb {R}}}^2)}, \end{aligned}$$

provided $\alpha (2+1/p_1)+1/q_1>1/q+\alpha /p$, where $*$ is the convolution operator. The same result holds for $L_{x_1}^qL_{x_2}^p({{\mathbb {R}}}^2)$ and $L_{x_1}^{q_1} L_{x_2}^{p_1}({{\mathbb {R}}}^2)$.

To prove Theorem 2, we apply the regularity lifting by contracting operators.

Definition 1

Let V be a topological vector space with two extended norms $\Vert \cdot \Vert _X$ and $\Vert \cdot \Vert _Y$, where $X=\{v\in V;\;\Vert v\Vert _X<\infty \}$ and $Y=\{v\in V;\;\Vert v\Vert _Y<\infty \}$. The operator $T: X\rightarrow Y$ is said to be a contraction if

$$\begin{aligned} \Vert Tx-Ty\Vert _Y\le \theta \Vert x-y\Vert _X, \end{aligned}$$

for all $x,y\in X$ and some $0<\theta <1$.

We now recall the following regularity lifting theorem (see [7, 19]).

Theorem 4

[19, Lemma 2.2] Let T be a contracting operator from X to itself and from Y to itself, and assume that X, Y are both complete. If $f\in X$, and there exists $g \in Z = X\cap Y$ such that $f = Tf + g$ in X, then $g \in Z$.

The proof of Theorem 2 is a direct corollary of the following result and the Young inequality (16).

Theorem 5

Under the same conditions of Theorem 2, we have $u\in L_{x_2}^pL_{x_1}^q({\mathbb {R}}^2_+)$ for all $1< p,q<\infty$.

Proof

Define the linear operator

$$\begin{aligned} T_vw=\int _{{\mathbb {R}}^2_+} G_\alpha (x,y)(|v(y)|^{r-1}w(y))\;\mathrm{d}y. \end{aligned}$$

For a fixed real number $a > 0$, define

$$\begin{aligned} u_a(x)=\left\{ \begin{array}{ll} u(x),&{}|u(x)|>a\,\text{ or }\,|x|>a,\\ 0,&{}\text{ otherwise. } \end{array}\right. \end{aligned}$$

Write $u_b= u - u_a$, which is uniformly bounded by a in $B_a (0)$. It is evident that $u^q = (u_a + u_b )^q =u_a^q+u_b^q$ for all $q> 0$. Since $u=u_a+u_b$ satisfies (3), we have $u=T_{u_a}u_a+T_{u_b}u_b$. Let $g=T_{u_b}u_b-u_b$. Then we can see that $g\in L^\infty ({\mathbb {R}}^2_+)\cap L^1({\mathbb {R}}^2_+)$, so that $g\in L_{x_2}^pL_{x_1}^q({\mathbb {R}}^2_+)$ for all $1<p,q<\infty$. Thus $u_a=T_{u_a}u_a+g$. Now by using Lemma 2 and the Hölder inequality we have for $w\in L_{x_2}^pL_{x_1}^q({\mathbb {R}}^2_+)$ that

$$\begin{aligned} \Vert T_{u_a}w\Vert _{L_{x_2}^pL_{x_1}^q({\mathbb {R}}^2_+)}\le C\Vert u_a\Vert _{L_{x_2}^{p_0}L_{x_1}^{q_0}({\mathbb {R}}^2_+)}^{r-1}\Vert w\Vert _{L_{x_2}^pL_{x_1}^q({\mathbb {R}}^2_+)}. \end{aligned}$$

By virtue of $u\in L_{x_2}^{p_0}L_{x_1}^{q_0}({\mathbb {R}}^2_+)$, choose a large enough such that

$$\begin{aligned} C\Vert u_a\Vert _{L_{x_2}^{p_0}L_{x_1}^{q_0}({\mathbb {R}}^2_+)}^{r-1}<1/2. \end{aligned}$$

This combining with $T_{u_a}$ being a linear operator, implies that $T_{u_a}$ is a contraction map from $L_{x_2}^{p}L_{x_1}^{q}({\mathbb {R}}^2_+)$ into itself, for all $p,q>r$. Applying Theorem 4, the solution $w=T_{u_a}w+g$ belongs to $L_{x_2}^{p_0}L_{x_1}^{q_0}({\mathbb {R}}^2_+)\cap L_{x_2}^{p}L_{x_1}^{q}({\mathbb {R}}^2_+)$; and by uniqueness of solution, $u_a\in L_{x_2}^{p_0}L_{x_1}^{q_0}({\mathbb {R}}^2_+)\cap L_{x_2}^{p}L_{x_1}^{q}({\mathbb {R}}^2_+)$, since $u_a$ is a solution of $w=T_{u_a}w+g$. Therefore $u\in L_{x_2}^{p}L_{x_1}^{q}({\mathbb {R}}^2_+)$ for all $r<p,q<\infty$.

Finally an application of Lemma 1 shows that $u\in L_{x_2}^{p}L_{x_1}^{q}({\mathbb {R}}^2_+)$ for all $1<p,q<\infty$. $\square$

Remark 1

Lemmas 1 and 2 show that Theorem 5 is also true if we consider $L_{x_1}^{q}L_{x_2}^{p}({\mathbb {R}}^2_+)$ instead of $L_{x_2}^{p}L_{x_1}^{q}({\mathbb {R}}^2_+)$.

We show that the positive solution u of (3) are continuous. We believe that u is also Lipschitz continuous, but we are not able to show it.

Proposition 1

Under the same conditions of Theorem 2, the positive solution u of (3) is continuous.

Proof

It follows from (3) that

$$\begin{aligned} \begin{aligned} u(x)-u(y)&=\int _{{\mathbb {R}}^2_+}\left( G_\alpha (x,z)-G_\alpha (y,z)\right) u^r(z)\mathrm{d}z\\&=\int _{B_\delta (x)}\left( G_\alpha (x,z)-G_\alpha (y,z)\right) u^r(z)\mathrm{d}z\\&\quad + \int _{{\mathbb {R}}^2_+\setminus B_\delta (x)}\left( G_\alpha (x,z)-G_\alpha (y,z) \right) u^r(z)\mathrm{d}z. \end{aligned} \end{aligned}$$

(17)

By Theorem 2, $\int _{{\mathbb {R}}^2_+}G_\alpha (x,z)u^r(z)\mathrm{d}z<+\infty$, and thus the second term of the right hand side of (17) is small enough if we choose $\delta$ sufficiently large. On the other hand, we have from Lemma 1 that $K_\alpha (x-z)-K_\alpha (y-z)\rightarrow 0$, as $|x-y|\rightarrow 0$, for any $z\in B_\delta (x)$. Hence

$$\begin{aligned} \lim _{x\rightarrow y}\int _{B_\delta (x)}\left( K_\alpha (x-z)-K_\alpha (y-z)\right) u^r(z)\mathrm{d}z=0. \end{aligned}$$

Since $|{\bar{x}}-{\bar{y}}|\rightarrow 0$ as $| x- y|\rightarrow 0$, we have

$$\begin{aligned} \lim _{x\rightarrow y}\int _{B_\delta (x)}\left( K_\alpha ({\bar{x}}-z)-K_\alpha (\bar{y}-z)\right) u^r(z)\mathrm{d}z=0. \end{aligned}$$

Therefore we deduce

$$\begin{aligned} \lim _{x\rightarrow y}\int _{B_\delta (x)}\left( G_\alpha (x,z)-G_\alpha (y,z)\right) u^r(z)\mathrm{d}z=0. \end{aligned}$$

Thus the first term of the right hand side of (17) is finite, and consequently the solution u of (3) is continuous. $\square$

3 Symmetry

For a given real number ${\lambda }>0$, we may define a family of moving planes

$$\begin{aligned} {\varOmega }_{\lambda }=\{x\in {\mathbb {R}}^2_+,\;{x_2}={\lambda }\}. \end{aligned}$$

and moving regions ${\varSigma }_{\lambda }=\{x\in {\mathbb {R}}^2_+,\;0<{x_2}<\lambda \}$. Let us list some properties of $G_\alpha$. For any $x\in {\mathbb {R}}^2_+$, we denote its reflection through the plane ${\varOmega }_{\lambda }$ by $x_{\lambda }=({x_1},2{\lambda }-{x_2})$.

Lemma 3

• (i) For any $x,y\in {\varSigma }_{\lambda }$ with $x\ne y$, we have

  $$\begin{aligned} \max \{G_\alpha (x_{\lambda },y),G_\alpha (x,y_{\lambda })\}<G_\alpha (x_{\lambda },y_{\lambda }) \end{aligned}$$

  (18)

  and

  $$\begin{aligned} |G_\alpha (x_{\lambda },y)-G_\alpha (x,y_{\lambda })|<G_\alpha (x_{\lambda },y_{\lambda })-G_\alpha (x,y). \end{aligned}$$

  (19)

• (ii) For any $x\in {\varSigma }_{\lambda }$ and $y\in {\varSigma }_{\lambda }^c={\mathbb {R}}^2_+\setminus {\varSigma }_\lambda$, it holds that

  $$\begin{aligned} G_\alpha (x,y)<G_\alpha (x_{\lambda },y). \end{aligned}$$

  (20)

Proof

For any x, y, let $d(x,y)=|x_2-y_2|^2$. Recalling (4), one has

$$\begin{aligned} \begin{aligned} G_\alpha (x,y)&=K_\alpha (x-y)-K_\alpha ({\bar{x}}-y)\\&=\int _0^\infty \mathrm{e}^{-t}t^{-\frac{1}{2}}H_\alpha (x_1-y_1,t) \left( \mathrm{e}^{-\frac{(x_2-y_2)^2}{4t}}-\mathrm{e}^{-\frac{(-x_2-y_2)^2}{4t}}\right) \mathrm{d}t\\&=\int _0^\infty \mathrm{e}^{-t}t^{-\frac{1}{2}}H_\alpha (x_1-y_1,t) \left( \mathrm{e}^{-\frac{d(x,y)}{4t}}-\mathrm{e}^{-\frac{d(x,y)+\psi (x,y)}{4t}}\right) \mathrm{d}t, \end{aligned} \end{aligned}$$

where $\psi (x,y)=4x_2y_2$. It is clear that $G_\alpha (x,y)>0$. By direct computations, one obtains that

$$\begin{aligned} \frac{\partial G_\alpha }{\partial d}<0,\quad \frac{\partial G_\alpha }{\partial d}>0,\quad \frac{\partial ^2 G_\alpha }{\partial \psi \partial d}<0. \end{aligned}$$

(21)

On the other hand, it is obvious to see for any $x,y\in {\varSigma }_{\lambda }$ that

$$\begin{aligned} d(x_{\lambda },y_{\lambda })=d(x,y)<d(x_{\lambda },y)=d(x,y_{\lambda }), \end{aligned}$$

(22)

and

$$\begin{aligned} \psi (x,y)\le \max \{\psi (x_{\lambda },y),\psi (x,y_{\lambda })\}\le \psi (x_{\lambda },y_{\lambda }). \end{aligned}$$

(23)

The proof of lemma follows from (21)–(23). $\square$

Lemma 4

For any positive solution u of (3), we have for any $x\in {\varSigma }_{\lambda }$ that

$$\begin{aligned} \begin{aligned} u(x)-u_{\lambda }(x)&\le \int _{{\varSigma }_{\lambda }} \left( G_\alpha (x_{\lambda },y_{\lambda })-G_\alpha (x,y_{\lambda })\right) \left( u^r(y)-u^r_{\lambda }(y)\right) \;\mathrm{d}y, \end{aligned} \end{aligned}$$

where $u_{\lambda }(x)=u(x_{\lambda })$.

Proof

Let ${{\tilde{{\varSigma }}}}_{\lambda }=\{x_{\lambda },\;x\in {\varSigma }_{\lambda }\}$. It is easy to see that

$$\begin{aligned} \begin{aligned} u(x)&=\int _{{\varSigma }_{\lambda }}G_\alpha (x,y)u^r(y)\mathrm{d}y+ \int _{{{\tilde{{\varSigma }}}}_{\lambda }}G_\alpha (x,y)u^r(y)\mathrm{d}y \\&\quad + \int _{{\varSigma }_{\lambda }^c\setminus {{\tilde{{\varSigma }}}}_{\lambda }} G_\alpha (x,y)u^r(y)\mathrm{d}y\\&=\int _{{\varSigma }_{\lambda }}G_\alpha (x,y)u^r(y)\mathrm{d}y+ \int _{{\varSigma }_{\lambda }}G_\alpha (x,y_{\lambda })u^r_{\lambda }(y)\mathrm{d}y \\&\quad + \int _{{\varSigma }_{\lambda }^c\setminus {\tilde{{\varSigma }}}_{\lambda }} G_\alpha (x,y)u^r(y)\mathrm{d}y. \end{aligned} \end{aligned}$$

(24)

Substituting x by $x_{\lambda }$, we get

$$\begin{aligned} \begin{aligned} u(x)-u(x_{\lambda })&=\int _{{\varSigma }_{\lambda }}\left( G_\alpha (x,y)-G_\alpha (x_{\lambda },y)\right) u^r(y)\mathrm{d}y\\&\quad +\int _{{\varSigma }_{\lambda }}\left( G_\alpha (x,y_{\lambda })-G_\alpha (x_{\lambda },y_{\lambda })\right) u^r_{\lambda }(y)\mathrm{d}y\\&\quad +\int _{{\varSigma }_{\lambda }^c\setminus {\tilde{{\varSigma }}}_{\lambda }}\left( G_\alpha (x,y)-G_\alpha (x_{\lambda },y)\right) u^r(y)\mathrm{d}y. \end{aligned} \end{aligned}$$

(25)

It is deduced from Lemma 3 that

$$\begin{aligned} \begin{aligned} u(x)-u(x_{\lambda })&\le \int _{{\varSigma }_{\lambda }}\left( G_\alpha (x,y)-G_\alpha (x_{\lambda },y)\right) u^r(y)\mathrm{d}y \\&\quad - \int _{{\varSigma }_{\lambda }}\left( G_\alpha (x_{\lambda },y_{\lambda })- G_\alpha (x,y_{\lambda })\right) u^r_{\lambda }(y)\mathrm{d}y\\&\le \int _{{\varSigma }_{\lambda }}\left( G_\alpha (x_{\lambda },y_{\lambda })- G_\alpha (x,y_{\lambda })\right) u^r(y)\mathrm{d}y \\&\quad - \int _{{\varSigma }_{\lambda }}\left( G_\alpha (x_{\lambda },y_{\lambda })- G_\alpha (x,y_{\lambda })\right) u^r_{\lambda }(y)\mathrm{d}y\\&= \int _{{\varSigma }_{\lambda }}\left( G_\alpha (x_{\lambda },y_{\lambda })- G_\alpha (x,y_{\lambda })\right) \left( u^r(y)-u^r_{\lambda }(y)\right) \mathrm{d}y; \end{aligned} \end{aligned}$$

and the proof is completed. $\square$

Lemma 5

For $0<{\lambda }\ll 1$, ${\varSigma }_{\lambda }^-=\{x\in {\varSigma }_{\lambda },\; u(x,y)>u_{\lambda }(x,y)\}$ has measure zero.

Proof

It is easy to see from Lemma 3, for any $x\in {\varSigma }_{\lambda }^-$, that

$$\begin{aligned} 0<u(x)-u_{\lambda }(x)\le & {} \int _{{\varSigma }_{\lambda }}\left( G_\alpha (x_{\lambda },y_{\lambda })-G_\alpha (x,y_{\lambda })\right) \left( u^r(y)-u^r_{\lambda }(y)\right) \mathrm{d}y\\= & {} \int _{{\varSigma }_{\lambda }^-}\left( G_\alpha (x_{\lambda },y_{\lambda })-G_\alpha (x,y_{\lambda })\right) \left( u^r(y)-u^r_{\lambda }(y)\right) \mathrm{d}y\\&+\,\int _{{\varSigma }_{\lambda }\setminus {\varSigma }_{\lambda }^-}\left( G_\alpha (x_{\lambda },y_{\lambda })-G_\alpha (x,y_{\lambda })\right) \left( u^r(y)-u^r_{\lambda }(y)\right) \mathrm{d}y. \end{aligned}$$

By Lemma 3, $u^r(x)\le u^r_{\lambda }(x)$ on ${\varSigma }_{\lambda }\setminus {\varSigma }_{\lambda }^-$ and $G_\alpha (x_{\lambda },y_{\lambda })\ge G_\alpha (x,y_{\lambda })$ for $y\in {\varSigma }_{\lambda }\setminus {\varSigma }_{\lambda }^-$, then the last integral in the above inequality is negative. Hence we get

$$\begin{aligned} \begin{aligned} u(x)-u_{\lambda }(x)&\le \int _{{\varSigma }_{\lambda }^-} \left( G_\alpha (x_{\lambda },y_{\lambda })-G_\alpha (x,y_{\lambda })\right) \left( u^r(y)-u_{\lambda }^r(y)\right) \mathrm{d}y\\&\le \int _{{\varSigma }_{\lambda }^-} G_\alpha (x_{\lambda },y_{\lambda })\left( u^r(y)-u_{\lambda }^r(y)\right) \mathrm{d}y\\&\le \int _{{\varSigma }_{\lambda }^-} K_\alpha (x_{\lambda }-y_{\lambda })\left( u^r(y)-u_{\lambda }^r(y)\right) \mathrm{d}y\\&\lesssim \int _{{\varSigma }_{\lambda }^-} G_\alpha (x_{\lambda },y_{\lambda })\varphi ^{r-1}(y)\left( u(y)-u_{\lambda }(y)\right) \mathrm{d}y\\&\lesssim \int _{{\varSigma }_{\lambda }^-} G_\alpha (x_{\lambda },y_{\lambda })u^{r-1}(y)\left( u(y)-u_{\lambda }(y)\right) \mathrm{d}y, \end{aligned} \end{aligned}$$

(26)

where we have used the mean value theorem with $\varphi (y)$ valued between u(y) and $u_{\lambda }(y)$, and the fact that $0\le u_{\lambda }(y)\le \varphi (y)\le u(y)$ on ${\varSigma }_{\lambda }^-$.

It follows first from Lemma 2 and then the Hölder inequality that

$$\begin{aligned} \begin{aligned} \Vert u-u_{\lambda }\Vert _{L^{p_0}_{x_2}L^{q_0}_{x_1}({\varSigma }_{\lambda }^-)}&\lesssim \Vert u\Vert _{L^{p_0}_{x_2}L^{q_0}_{x_1}({\varSigma }_{\lambda })}^{r-1}\Vert u-u_{\lambda }\Vert _{L^{p_0}_{x_2}L^{q_0}_{x_1}({\varSigma }_{\lambda }^-)}. \end{aligned} \end{aligned}$$

Since $u\in L^{p_0}_{x_2}L^{q_0}_{x_1}({{\mathbb {R}}}^2)$, by choosing ${\lambda }\ll 1$ we deduced that $\Vert u-u_{\lambda }\Vert _{L^{p_0}_{x_2}L^{q_0}_{x_1}({\varSigma }_{\lambda }^-)}=0$, and therefore ${\varSigma }_{\lambda }^-$ has measure zero. $\square$

Proof of Theorem 3

Define

$$\begin{aligned} {\lambda }_0=\sup \{{\lambda },\;u_\mu (x)\ge u(x), \forall \mu \le {\lambda },\,x\in {\varSigma }_\mu \}. \end{aligned}$$

We assume that ${\lambda }_0<+\infty$, because the case ${\lambda }_0=+\infty$ gives the proof by the definition of ${\lambda }_0$ and the assumption $u\in L_{x_2}^{p_0}L_{x_1}^{q_0}({\mathbb {R}}^2_+)$. We show that the solution u(x) is monotone increasing with respect to the $x_2$-variable and symmetric about ${\varOmega }_{{\lambda }_0}$, that is, $u(x)=u_{{\lambda }_0}(x)$ on ${\varSigma }_{{\lambda }_0}$. Suppose by the contradiction argument that $u\le u_{{\lambda }_0}$ and $u\not \equiv u_{\lambda }$ on ${\varSigma }_{{\lambda }_0}$. We prove that there exists an $\epsilon >0$ such that, for any ${\lambda }_0\le {\lambda }<{\lambda }_0+\epsilon$, it holds on ${\varSigma }_{\lambda }$ that

$$\begin{aligned} u(x)\le u_{\lambda }(x). \end{aligned}$$

By using an argument analogous to the proof of Lemma 5, we can obtain that

$$\begin{aligned} \begin{aligned} \Vert u-u_{\lambda }\Vert _{L^{p_0}_{x_2}L^{q_0}_{x_1}({\varSigma }_{\lambda })}&\le C \Vert u\Vert _{L^{p_0}_{x_2}L^{q_0}_{x_1}({\varSigma }_{\lambda }^-)}^{r-1}\Vert u-u_{\lambda }\Vert _{L^{p_0}_{x_2}L^{q_0}_{x_1}({\varSigma }_{\lambda })}. \end{aligned} \end{aligned}$$

(27)

Now if we establish for $\epsilon \ll 1$ that

$$\begin{aligned} C\Vert u\Vert _{L^{p_0}_{x_2}L^{q_0}_{x_1}({\varSigma }_{\lambda }^-)}^{r-1}\le \frac{1}{4}, \end{aligned}$$

(28)

then it follows from (27) that ${\varSigma }_{\lambda }^-$ is a set of zero measure; and consequently $u_{\lambda }(x)\ge u(x)$ for any $x\in {\varSigma }_{\lambda }$ and ${\lambda }\in [{\lambda }_0,{\lambda }_0+\epsilon )$. This contradicts with the definition of ${\lambda }_0$ and the result follows.

Now we prove inequality (28). Choose, for any small $\varsigma >0$, a large enough number $\delta >0$ such that

$$\begin{aligned} \Vert u\Vert _{L^{p_0}_{x_2}L^{q_0}_{x_1}({\mathbb {R}}^2_+\setminus B_\delta (0) )}^{r-1}<\varsigma , \end{aligned}$$

(29)

where $B_\delta (0)$ is the ball of radius $\delta >0$ centered at zero in ${\mathbb {R}}^2_+$. It is straightforward to see that $u< u_{\lambda }$ in ${\varSigma }_{{\lambda }_0}$. Indeed by contrary suppose that $u_{\lambda }(x_0)=u(x_0)$, for some $x_0\in {\varSigma }_{{\lambda }_0}$. It follows then from Lemma 3 and the proof of Lemma 4 that

$$\begin{aligned} \begin{aligned} 0= u_{\lambda }(x_0)-u(x_0)&\ge \int _{{\varSigma }_{{\lambda }_0}}\left( K_\alpha (x_{{\lambda }_0}, y_{{\lambda }_0})-K_\alpha (x,y_{{\lambda }_0})\right) \left( u_{{\lambda }_0}^r(y)-u^r(y)\right) \mathrm{d}y\\&\quad + \int _{{\varSigma }_{{\lambda }_0}^c\setminus {\tilde{{\varSigma }}}_{{\lambda }_0}} \left( K_\alpha (x_{{\lambda }_0},y)-K_\alpha (x,y)\right) u_{{\lambda }_0}^r(y)\mathrm{d}y\\&\ge \int _{{\varSigma }_{{\lambda }_0}^c\setminus {\tilde{{\varSigma }}}_{{\lambda }_0}}\left( K_\alpha (x_{{\lambda }_0},y) -K_\alpha (x,y)\right) u_{{\lambda }_0}^r(y)\mathrm{d}y. \end{aligned} \end{aligned}$$

By applying again Lemma 3 it yields that $u(y)=0$ for all $y\in {\varSigma }_{{\lambda }_0}^c\setminus {\tilde{{\varSigma }}}_{{\lambda }_0}$ which contradicts with the positivity of u. Now for any $\kappa >0$ define

$$\begin{aligned} B_\kappa =\{x\in {\varSigma }_{{\lambda }_0}\cap B_\delta (0),\;u_{{\lambda }_0}(x)-u(x)>\kappa \},\quad B_\kappa ^c={\varSigma }_{{\lambda }_0}\cap B_\delta (0)\setminus B_\kappa ; \end{aligned}$$

and denote, for ${\lambda }>{\lambda }_0$,

$$\begin{aligned} {\tilde{B}}_{\lambda }=({\varSigma }_{\lambda }\setminus {\varSigma }_{{\lambda }_0})\cap B_\delta (0). \end{aligned}$$

Note that the measure of $B_\kappa ^c$ tends to zero as $\kappa \rightarrow 0$. Moreover

$$\begin{aligned} {\varSigma }^-_{\lambda }\cap B_\delta (0)\subset ({\varSigma }_{\lambda }^-\cap B_\kappa )\cup B_\kappa ^c\cup {\tilde{B}}_{\lambda }\end{aligned}$$

(30)

and the measure of ${\tilde{B}}_{\lambda }$ is small as ${\lambda }$ is close to ${\lambda }_0$. We show that the measure of ${\varSigma }_{\lambda }^-\cap B_\kappa$ is sufficiently small as ${\lambda }$ is close to ${\lambda }_0$. Actually, since for any $x\in {\varSigma }_{\lambda }^-\cap B_\kappa$ we have

$$\begin{aligned} u_{\lambda }(x)-u(x)=u_{\lambda }(x)-u_{{\lambda }_0}(x)+u_{{\lambda }_0}(x)-u(x)<0, \end{aligned}$$

then $u_{{\lambda }_0}(x)-u_{\lambda }(x)>\kappa$. And hence

$$\begin{aligned} {\varTheta }_\kappa :=\{x\in B_\delta (0),\; u_{{\lambda }_0}(x)-u_{\lambda }(x)>\kappa \}\supset {\varSigma }^-_{\lambda }\cap B_\kappa . \end{aligned}$$

(31)

Therefore it is deduced from the Chebyshev inequality that

$$\begin{aligned} |{\varTheta }_\kappa |\le \frac{1}{\kappa ^{r+1}}\int _{{\varTheta }_\kappa }|u_{{\lambda }_0}(x)-u_{\lambda }(x)|^{r+1}\mathrm{d}x \le \frac{1}{\kappa ^{r+1}}\int _{B_\delta (0)}|u_{{\lambda }_0}(x)-u_{\lambda }(x)|^{r+1}\mathrm{d}x. \end{aligned}$$

The above integral and consequently ${\varSigma }_{\lambda }^-\cap B_\kappa$ is small enough as ${\lambda }$ is close to ${\lambda }_0$. Finally by combining (30) and (31) we obtain that the measure of ${\varSigma }_{\lambda }^-\cap B_\delta (0)$ is sufficiently small for ${\lambda }$ close to ${\lambda }_0$. This completes the proof. $\square$

Proof of Theorem 1

Suppose that u is a nontrivial nonnegative solution of (3). Then there exists $y_0\in {\mathbb {R}}^2_+$ such that $u(y_0 ) > 0$. By the continuity of u from Proposition 1, there exists a neighborhood $N_{y_0}$ of $y_0$ in ${\mathbb {R}}^2_+$ such that $u(y) > 0$ for any $y \in N.$ Since $G_\alpha >0$ in ${\mathbb {R}}^2_+$, then

$$\begin{aligned} u(x)=\int _{{\mathbb {R}}^2_+} G_\alpha (x,y)u^r(y)\mathrm{d}y\ge \int _{N_{y_0}} G_\alpha (x,y)u^r(y)\mathrm{d}y>0. \end{aligned}$$

(32)

Due to Theorem 3, we know that the plane ${\varOmega }_{\lambda }$ can be moved to the limiting position ${\varOmega }_{{\lambda }_0}$. We show that ${\lambda }_0=+\infty$, which gives a simple contradiction argument. Assume ${\lambda }_0<+\infty$, then the symmetric image of the boundary of ${\mathbb {R}}^2_+$ through the line ${\varOmega }_{{\lambda }_0}$ is the plane ${x_2}=2{\lambda }_0$. And therefore $u(x) = 0$ for any $x\in {\varOmega }_{2{\lambda }_0}$, which is a contradiction to (32). Now making use again of Theorem 3, it can be derived that u(x) is monotone increasing with respect to ${x_2}$. This leads to a contradiction with the assumption $u\in L_{x_2}^{p_0}L_{x_1}^{q_0}({\mathbb {R}}^2_+)$. As a result, the positive solution of (3) does not exist. $\square$

Remark 2

Under the assumptions of Theorem 1, one can easily repeat, with some modifications, the proof of Theorem 1 and demonstrate the non-existence of positive solution for the integral equation (3) in ${{\mathbb {R}}}^+\times {{\mathbb {R}}}$. The key point is that by Bernstein’s theorem [17], one can write $H_\alpha$ in terms of subordination formula

$$\begin{aligned} H_\alpha (x_1,t)=\int _0^\infty \frac{1}{\sqrt{2s}}\mathrm{e}^{-\frac{t^2x^2_1}{4s}}\mathrm{d}\mu _\alpha (s), \end{aligned}$$

with some nonnegative finite measure $\mu _\alpha \ge 0$ with $\mu _\alpha \not \equiv 0$.

Theorem 1 can be extended to a general nonlinearity.

Theorem 6

Let the assumptions of Theorem 1 hold. Suppose that $u\in L_{x_2}^{p_0}L_{x_1}^{q_0}({\mathbb {R}}^2_+)$ is the nonnegative solution of

$$\begin{aligned} u(x)=\int _{{{\mathbb {R}}}^2_+}G_\alpha (x,y)f(y,u(y))\;\mathrm{d}y,\quad x=(x_1,x_2)\in {{\mathbb {R}}}^2_+={{\mathbb {R}}}\times {{\mathbb {R}}}_+, \end{aligned}$$

(33)

Suppose also that f(x, u) is nondecreasing in the variable $x_2$ and nondecreasing with respect to u, and $\frac{\partial f}{\partial u}\in L_{x_2}^{p_1}L_{x_1}^{q_1}({\mathbb {R}}^2_+)$ is non-decreasing with respect to u, where $1\le p_1,q_1\le \infty$ with

$$\begin{aligned} \frac{1}{p_1}+\frac{1}{p_0}\le 1\quad \text{ and }\quad \frac{1}{q_1}+\frac{1}{q_0}\le 1. \end{aligned}$$

Then u is identically equal to zero.

The proof is basically the same as that of Theorem 1. In order to avoid repetition, we explain the key modifications of the proof of Theorem 6.

Applying the same arguments as in Lemma 4, we can observe for any $x\in {\varSigma }_{\lambda }$ from the monotonicity of f(x, u) with respect to $x_2$ that

$$\begin{aligned} u(x)-u_{\lambda }(x)\le & {} \int _{{\varSigma }_{\lambda }}\left( G_\alpha (x_{\lambda },y_{\lambda })-G_\alpha (x,y_{\lambda })\right) \left( f(y,u(y))-f(y,u_{\lambda }(y))\right) \mathrm{d}y\nonumber \\&+ \int _{{\varSigma }_{\lambda }}\left( G_\alpha (x_{\lambda },y_{\lambda })-G_\alpha (x,y_{\lambda })\right) \left( f(y,u_{\lambda }(y))-f(y_{\lambda },u_{\lambda }(y))\right) \mathrm{d}y\nonumber \\\le & {} \int _{{\varSigma }_{\lambda }}\left( G_\alpha (x_{\lambda },y_{\lambda })-G_\alpha (x,y_{\lambda })\right) \left( f(y,u(y))-f(y,u_{\lambda }(y))\right) \mathrm{d}y.\nonumber \\ \end{aligned}$$

(34)

Recall the definition of ${\varSigma }_{\lambda }^-$ in Lemma 5. Now since f is nondecreasing in u, we have

$$\begin{aligned} u(x)-u_{\lambda }(x)\le & {} \int _{{\varSigma }_{\lambda }}\left( G_\alpha (x_{\lambda },y_{\lambda })-G_\alpha (x,y_{\lambda })\right) \left( f(y,u(y))-f(y,u_{\lambda }(y))\right) \mathrm{d}y\nonumber \\\le & {} \int _{{\varSigma }^-_{\lambda }}\left( G_\alpha (x_{\lambda },y_{\lambda })-G_\alpha (x,y_{\lambda })\right) \left( f(y,u(y))-f(y,u_{\lambda }(y))\right) \mathrm{d}y\nonumber \\\le & {} \int _{{\varSigma }^-_{\lambda }}G_\alpha (x,y)\left( f(y,u(y))-f(y,u_{\lambda }(y))\right) \mathrm{d}y\nonumber \\\le & {} \int _{{\varSigma }^-_{\lambda }}K_\alpha (x-y)\left( f(y,u(y))-f(y,u_{\lambda }(y))\right) \mathrm{d}y\nonumber \\\le & {} \int _{{\varSigma }^-_{\lambda }}G_\alpha (x,y)\frac{\partial f}{\partial u}(y,u(y))(u(y)-u_{\lambda }(y)) \mathrm{d}y, \end{aligned}$$

(35)

where in the last inequality we have used the mean value theorem and the monotonicity of $\frac{\partial f}{\partial u}$ in u. The rest of proof is the same as that of Theorem 1 by using Lemma 2.