1 Introduction
In recent years, there has been a great deal of interests in using the fractional Laplacian to model diverse physical phenomena, such as anomalous diffusion and quasi-geostrophic flows, turbulence and water waves, molecular dynamics, and relativistic quantum mechanics of stars (see [1‐6], and the references therein). The fractional Laplacian in \(R^{n}\) is a nonlocal operator, taking the form where \(0<\alpha<2\), P.V. stands for the Cauchy principal value. This operator is well defined in \(\mathcal {S}\), the Schwartz space of rapidly decreasing \(C^{\infty} \) functions in \(R^{n}\). One can extend this operator to a wider space \(\mathcal {L}_{\alpha} \) of distributions as follows.
$$ (-\Delta)^{\frac{\alpha}{2}}u(x)=C_{n,\alpha} \textit{ P.V. } \int _{R^{n}}\frac{u(x)-u(z)}{\vert x-z\vert ^{n+\alpha}}\,dz, $$
(1.1)
Let For \(u \in \mathcal {L}_{\alpha} \), we define \((-\Delta)^{\frac{\alpha }{2}}u(x)\) as a distribution:
$${\mathcal {L}}_{\alpha}= \biggl\{ u:R^{n}\rightarrow R \Bigm| \int_{R^{n}} \frac {\vert u(x)\vert }{1+\vert x\vert ^{n+\alpha}}\,dx < \infty \biggr\} . $$
$$\bigl\langle (-\Delta)^{\frac{\alpha}{2}}u(x), \phi\bigr\rangle = \bigl\langle u, (- \Delta)^{\frac {\alpha}{2}}\phi\bigr\rangle ,\quad \forall \phi\in \mathcal {S}. $$
Anzeige
Zhang and Cheng [7] considered the positive solutions of the following single equation in \(R^{n}\): They showed the following.
$$ (-\Delta)^{\frac{\alpha}{2}}u(x)=x^{\gamma}_{n}u^{p}(x),\quad u(x)>0, x\in R^{n}. $$
(1.2)
Proposition 1.1
[7]
Assume
\(p >\frac{n}{n-\alpha} \)
and
\(\gamma\geq0 \), if
\(u(x) \in L^{\frac{n(p-1)}{\alpha}}(R^{n}_{+})\)
is a non-negative solution of equation (1.2), then
\(u(x) \equiv0\).
Proposition 1.2
[7]
Anzeige
Assume
\(1< p\leq\frac{n+\alpha+2\gamma}{n-\alpha}\)
and
\(\gamma\geq 0 \), if
\(u(x)\)
is a locally bounded nonnegative solution of the equation (1.2), then
\(u(x)\equiv0\). In particular, when
\(\frac{n+\alpha}{n-\alpha}\leq p\leq \frac{n+\alpha+2\gamma}{n-\alpha}\), we only require
\(u(x) \in L^{\frac{n(p-1)}{\alpha}}(R^{n}_{+})\).
Motivated by [7], in this paper we consider the Dirichlet problem for the following pseudo differential system in \({R^{n}_{+}}\): where \(\gamma\geq0\), \(0<\alpha<2\), \(\alpha_{i}\), \(\beta_{i} >0\), \(i=1,2 \).
$$ \textstyle\begin{cases} (-\Delta)^{\frac{\alpha}{2}}u_{1}=x^{\gamma}_{n}u_{1}^{\alpha _{1}}(x)u_{2}^{\beta_{1}}(x), &\mbox{$x \in R^{n}_{+}$},\\ (-\Delta)^{\frac{\alpha}{2}}u_{2}=x^{\gamma}_{n}u_{1}^{\alpha _{2}}(x)u_{2}^{\beta_{2}}(x), &\mbox{$x \in R^{n}_{+}$},\\ u_{1}(x)=u_{2}(x)=0, &\mbox{$x \notin R^{n}_{+}$}, \end{cases} $$
(1.3)
First, we use the maximum principle and the Liouville theorem in \({R^{n}_{+}}\) in [6] to show that the positive solutions of problem (1.3) satisfy the following integral equations under some weak integrability condition: where is the Green’s function associated with \((-\Delta)^{\frac{\alpha}{2}}\) in \(R^{n}_{+}\) and \(A_{n,\alpha} \) is a constant depending on n and α.
$$ \textstyle\begin{cases} u_{1}(x)=c_{1}x^{\frac{\alpha}{2}}_{n}+\int _{R^{n}_{+}}G(x,y)y^{\gamma}_{n}u_{1}^{\alpha_{1}}(y)u_{2}^{\beta _{1}}(y)\,dy, &\mbox{$x \in R^{n}_{+}$},\\ u_{2}(x)=c_{2}x^{\frac{\alpha}{2}}_{n}+\int _{R^{n}_{+}}G(x,y)y^{\gamma}_{n}u_{1}^{\alpha_{2}}(y)u_{2}^{\beta _{2}}(y)\,dy, &\mbox{$x \in R^{n}_{+}$}, \end{cases} $$
(1.4)
$$ G(x,y)=\frac{A_{n,\alpha }}{\vert x-y\vert ^{n-\alpha}} \int_{0}^{\frac{4 x_{n} y_{n}}{\vert x-y\vert ^{2}}}\frac {b^{\frac{\alpha}{2}-1}}{(1+b)^{\frac{n}{2}}}\,db $$
(1.5)
Then we use Kelvin transform and the method of moving planes in integral forms to prove that \(c_{1}\) and \(c_{2}\) must be 0. We derive that (1.3) is equivalent to the following integral equations under some locally integrable conditions: In the subcritical case and critical case: \(1< \alpha_{i}+\beta _{i}\leq\frac{n+\alpha+2\gamma}{n-\alpha} \), we continue work on the integral systems (1.6) to show the nonexistence of positive solutions. That is, we have the following result.
$$ \textstyle\begin{cases} u_{1}(x)=\int_{R^{n}_{+}}G(x,y)y^{\gamma}_{n}u_{1}^{\alpha _{1}}(y)u_{2}^{\beta_{1}}(y)\,dy, &\mbox{$x \in R^{n}_{+}$},\\ u_{2}(x)=\int_{R^{n}_{+}}G(x,y)y^{\gamma}_{n}u_{1}^{\alpha _{2}}(y)u_{2}^{\beta_{2}}(y)\,dy, &\mbox{$x \in R^{n}_{+}$}. \end{cases} $$
(1.6)
Theorem 1.1
Assume that
\(u(x)=(u_{1}(x),u_{2}(x))\)
is a positive solution of equations (1.3). If
\(\vert u\vert \in L^{\frac{n(\alpha_{i}+\beta_{i}-1)}{\alpha }}_{\mathrm{loc}}(R^{n}_{+})\cap L^{\infty}_{\mathrm{loc}}(R^{n}_{+})\), then in the case
\(1<\alpha_{i}+\beta_{i}\leq\frac{n+\alpha+2\gamma }{n-\alpha} \), \(u(x)\)
is also a solution of integral equations (1.6), and vice versa.
Next, we establish the Liouville theorem for the integral equations as follows.
Theorem 1.2
Assume that
\(\alpha_{i}+\beta_{i} >\frac {n}{n-\alpha} \)
and
\(\gamma\geq0 \), if
\(\vert u\vert \in L^{\frac{n(\alpha _{i}+\beta_{i}-1)}{\alpha}}(R^{n}_{+})\)
is a nonnegative solution of the system of the integral equations (1.6), then
\(u(x)\equiv0\).
Theorem 1.3
Assume that
\(u(x)=(u_{1}(x),u_{2}(x))\)
is a nonnegative solution of equations (1.6). If
\(\vert u\vert \in L^{\frac{n(\alpha_{i}+\beta _{i}-1)}{\alpha}}_{\mathrm{loc}}(R^{n}_{+})\cap L^{\infty}_{\mathrm{loc}}(R^{n}_{+})\)
and
\(1<\alpha_{i}+\beta_{i}\leq\frac{n+\alpha+2\gamma}{n-\alpha} \), we have
\(u(x)\equiv0\).
2 Equivalence between the two systems
The proof of Theorem 1.1 is based on the following maximum principle and the Liouville theorem.
Proposition 2.1
[8]
Let Ω be a bounded open set, and let
\(f(x)\)
be a lower-semicontinuous function in Ω̄ such that
\((-\Delta )^{\frac{\alpha}{2}}f(x)\geq0 \)
in Ω and
\(f(x)\geq0 \)
in
\({R^{n}} \setminus\Omega \). Then
\(f(x)\geq0 \)
in
\({R^{n}} \).
Proposition 2.2
[8]
If
\(f (x)\in \mathcal {L}_{\alpha}\)
and
\((-\Delta)^{\frac{\alpha }{2}}f(x)\geq0\)
in an open set, then
\(f(x)\)
is lower semicontinuous in Ω.
Theorem 2.1
[9]
Let
\(0 < \alpha< 2\), \(u \in \mathcal {L}_{\alpha}\). Assume
u
is a nonnegative solution of
Then we have either
for some positive constant C.
$$ \textstyle\begin{cases} (-\Delta)^{\frac{\alpha}{2}}u(x)=0, &x \in{R^{n}_{+}}, \\ u(x)\equiv0, &x \notin{R^{n}_{+}}. \end{cases} $$
(2.1)
$$u(x)\equiv0 x\in R^{n}\quad\textit{or}\quad u(x)=\textstyle\begin{cases} Cx_{n}^{\frac{\alpha}{2}}, &x \in{R^{n}_{+}}, \\ 0, &x \notin{R^{n}_{+}}, \end{cases} $$
Proof
Assume \(u(x)\in \mathcal {L}_{\alpha}\) is a positive solution of the system of the fractional nonlinear PDEs system: where \(i=1,2\), \(\gamma\geq0\), \(0<\alpha<2\), \(\alpha_{i}\), \(\beta_{i} >0\).
$$ \textstyle\begin{cases} (-\Delta)^{\frac{\alpha}{2}}u_{i}(x)=x^{\gamma}_{n}u_{1}^{\alpha _{i}}(x)u_{2}^{\beta_{i}}(x), &\mbox{$x \in R^{n}_{+}$},\\ u_{i}(x)=0, &\mbox{$x \notin R^{n}_{+}$}, \end{cases} $$
(2.2)
We first show that Set \(P_{R}:=(0, \ldots, 0,R )\in R^{n}_{+} \), \(B^{+}_{R}(P_{R}):= \{x\in R^{n} : \vert x-P_{R} \vert < R \}\), the ball of radius R centered at \(P_{R}\). Let where \(G_{R}(x,y)\), the Green’s function on the ball \(B^{+}_{R}(P_{R})\), was given in [10], here \(s_{R}=\frac{\vert x-y\vert ^{2}}{R^{2}}\), \(t_{R}=(1-\frac {\vert x-P_{R}\vert ^{2}}{R^{2}})(1-\frac{\vert y-P_{R}\vert ^{2}}{R^{2}})\), \(A_{n,\alpha} \), and \(B_{n,\alpha}\) are constants depending on n and α.
$$ \int_{R^{n}_{+}}G(x,y){y^{\gamma}_{n}} u^{\alpha_{i}}_{1}(y)u^{\beta _{i}}_{2}(y)\,dy< \infty. $$
(2.3)
$$ v_{i}^{R}(x)= \int_{B^{+}_{R}(P_{R})}G_{R}(x,y)y^{\gamma}_{n}u_{1}^{\alpha _{i}}(y)u_{2}^{\beta_{i}}(y) \,dy, $$
(2.4)
$$G_{R}(x,y)=\frac{A_{n,\alpha}}{ \vert x-y\vert ^{n-\alpha}} \biggl[1-\frac{B_{n,\alpha}}{(s_{R}+t_{R})^{\frac{(n-2)}{2}}} { \int_{0}^{\frac{s_{R}}{t_{R}}}\frac{(s_{R}-t_{R}b)^{\frac {(n-2)}{2}}}{b^{\frac{\alpha}{2}}(1+b)}}\,db \biggr],\quad x, y \in B^{+}_{ R }(P_{R}), $$
From the local bounded-ness assumption on u, one can see that, for each \(R > 0\), \(v_{i}^{R}(x)\) is well defined and is continuous. Moreover, Let \(w_{i}^{R}(x)=u_{i}(x)-v_{i}^{R}(x)\), by (2.2) and (2.5), we derive Applying the maximum principle (see Proposition 2.1), we derive that It is easy to prove that Obviously, Denote \(w_{i}(x)=u_{i}(x)-v_{i}(x)\). Using (2.2), (2.7), (2.8), and (2.9), we have Applying the Liouville theorem (see Theorem 2.1), we deduce that either for some positive constants \(c_{i}>0\), then we could write \(w_{i} = c_{i}x_{n}^{\frac{\alpha}{2}}\), \(c_{i}\geq0\). That is, the solutions of (2.2) satisfy where \(c_{i}\geq0\), \(G(x,y)\) is defined in (1.5).
$$ \textstyle\begin{cases} (-\Delta)^{\frac{\alpha}{2}}v_{i}^{R}(x)=x^{\gamma}_{n}u_{1}^{\alpha _{i}}(x)u_{2}^{\beta_{i}}(x), &\mbox{$x\in B^{+}_{R}(P_{R})$},\\ v_{i}^{R}(x)=0, &\mbox{$x\notin B^{+}_{R}(P_{R})$}. \end{cases} $$
(2.5)
$$ \textstyle\begin{cases} (-\Delta)^{\frac{\alpha}{2}}w_{i}^{R}(x)=0, &\mbox{$x\in B^{+}_{R}(P_{R})$},\\ w_{i}^{R}(x)\geq0, &\mbox{$ x\notin B^{+}_{R}(P_{R})$}. \end{cases} $$
(2.6)
$$ w_{i}^{R}(x) \geq0 ,\quad \forall x\in B^{+}_{R}(P_{R}). $$
(2.7)
$$ v_{i}^{R}(x)\rightarrow v_{i}(x)= \int_{R^{n}_{+}}G(x,y)y^{\gamma }_{n}u_{1}^{\alpha_{i}}(y)u_{2}^{\beta_{i}}(y) \,dy , \quad \mbox{as }R\rightarrow \infty. $$
(2.8)
$$ \textstyle\begin{cases} (-\Delta)^{\frac{\alpha}{2}}v_{i}(x)=x^{\gamma}_{n}u_{1}^{\alpha _{i}}(x)u_{2}^{\beta_{i}}(x),&\mbox{$x\in R^{n}_{+}$},\\ v_{i}(x)\equiv0,&\mbox{$x\notin R^{n}_{+}$}. \end{cases} $$
(2.9)
$$ \textstyle\begin{cases} (-\Delta)^{\frac{\alpha}{2}}w_{i}(x)=0,& w_{i}(x)\geq0,\mbox{$ x\in {R^{n}_{+}}$},\\ w_{i}(x)\equiv0, &\mbox{$ x\notin{R^{n}_{+}} $}. \end{cases} $$
$$w_{i}(x)\equiv0 ,\quad x \in{R^{n}} \quad\mbox{or}\quad w_{i} (x)\equiv c_{i}x_{n}^{\frac{\alpha}{2}},\quad \forall x \in{R^{n}_{+}}, i=1,2, $$
$$ u_{i}(x)=c_{i}x^{\frac{\alpha}{2}}_{n}+\int _{R^{n}_{+}}G(x,y)y^{\gamma}_{n} u_{1}^{\alpha_{i}}(y)u_{2}^{\beta_{i}}(y)\,dy, \quad x \in R^{n}_{+}, i=1,2, $$
(2.10)
Next we need to prove \(c_{i}\) must be zero for \(i=1,2\). To this end, we employ a certain type of Kelvin transform and the method of moving planes in integral forms.
For \(z^{0}=\{z^{0}_{1}, \ldots,z^{0}_{n-1} ,0 \}\in\partial {R^{n}_{+}} \), let \(\bar{u}_{i}^{z^{0}}(x)=\bar{u}_{i}(x) =\frac{1}{\vert x-z^{0}\vert ^{n-\alpha}}u_{i}(\frac{x-z^{0}}{\vert x-z^{0}\vert ^{2}}+z^{0})\), be the Kelvin transform of \(u_{i}(x)\) centered at \(z^{0}\).
Through a straightforward calculation by (2.10), we derive Let \(y=\frac{z-z^{0}}{\vert z-z^{0}\vert ^{2}}+z^{0} \), then \(dy=\frac {1}{\vert z-z^{0}\vert ^{2n}}\,dz\), where \(\epsilon>0\), \(\delta=n+\alpha+2\gamma-(n-\alpha)(\alpha _{i}+\beta_{i})\).
$$\bar{u}_{i}(x) =\frac{c_{i}x_{n}^{\frac{\alpha}{2}}}{\vert x-z^{0}\vert ^{n-\alpha}}+\frac {1}{\vert x-z^{0}\vert ^{n-\alpha}} \int_{R^{n}_{+}}G \biggl(\frac{x-z^{0}}{\vert x-z^{0}\vert ^{2}}+z^{0},y \biggr)y^{\gamma }_{n}u_{1}^{\alpha_{i}}(y)u_{2}^{\beta_{i}}(y) \,dy. $$
$$\begin{aligned} \bar{u}_{i}(x) ={}&\frac{c_{i}x_{n}^{\frac{\alpha}{2}}}{\vert x-z^{0}\vert ^{n-\alpha}}+ \frac {1}{\vert x-z^{0}\vert ^{n-\alpha}} \int_{R^{n}_{+}}G(x,z)\bigl\vert x-z^{0}\bigr\vert ^{n-\alpha} \bigl\vert z-z^{0}\bigr\vert ^{n-\alpha} \\ &{}\times\frac{\vert \frac{z_{n}}{\vert z-z^{0}\vert ^{2}}\vert ^{\gamma }u^{\alpha_{i}}_{1}(\frac{z-z^{0}}{\vert z-z^{0}\vert ^{2}}+z^{0})u^{\beta _{i}}_{2}(\frac{z-z^{0}}{\vert z-z^{0}\vert ^{2}}+z^{0})}{\vert z-z^{0}\vert ^{2n}}\,dz \\ ={}&\frac{c_{i}x_{n}^{\frac{\alpha}{2}}}{\vert x-z^{0}\vert ^{n-\alpha}}+ \int _{R^{n}_{+}}G(x,z)\frac{z_{n}^{\gamma}}{ \vert z-z^{0}\vert ^{n+\alpha+2\gamma}} \frac{u^{\alpha_{i}}_{1}(\frac {z-z^{0}}{\vert z-z^{0}\vert ^{2}}+z^{0})}{\vert z-z^{0}\vert ^{(n-\alpha)\alpha_{i}}} \\ &{}\times\frac{u^{\beta_{i}}_{2}(\frac {z-z^{0}}{\vert z-z^{0}\vert ^{2}}+z^{0})}{\vert z-z^{0}\vert ^{(n-\alpha)\beta_{i}}} \bigl\vert z-z^{0}\bigr\vert ^{(n-\alpha)\alpha_{i}} \bigl\vert z-z^{0}\bigr\vert ^{(n-\alpha)\beta_{i}}\,dz \\ ={}&\frac{c_{i}x_{n}^{\frac{\alpha}{2}}}{\vert x-z^{0}\vert ^{n-\alpha}}+ \int _{R^{n}_{+}}G(x,z) \frac{z_{n}^{\gamma}\bar{u}^{\alpha_{i}}_{1}(z)\bar{u}^{\beta _{i}}_{2}(z)}{\vert z-z^{0}\vert ^{n+2\gamma+\alpha-(n-\alpha)(\alpha _{i}+\beta_{i})}}\,dz \\ ={}&\frac{c_{i}x_{n}^{\frac{\alpha}{2}}}{\vert x-z^{0}\vert ^{n-\alpha}}+ \int _{R^{n}_{+}}G(x,y)\frac{y_{n}^{\gamma}\bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta_{i}}_{2}(y)}{\vert y-z^{0}\vert ^{\delta}}\,dy, \quad \forall x\in {R^{n}_{+}}\backslash B_{\epsilon}\bigl(z^{0} \bigr), \end{aligned}$$
(2.11)
Then we have \(\delta=0\)
i.e.
\(\alpha_{i}+\beta_{i}=\frac{n+\alpha+2\gamma}{n-\alpha}\), it is called critical case. When \(\delta>0\), we have \(1<\alpha_{i}+\beta_{i}<\frac{n+\alpha +2\gamma}{n-\alpha} \) and it is called the subcritical case. In this section, we consider these two cases \(1<\alpha_{i}+\beta_{i}\leq \frac{n+\alpha+2\gamma}{n-\alpha} \), then we have \(\delta\geq0\).
Now we introduce some basic notations in the method of moving planes. For a given real number λ, denote \({\Sigma_{\lambda}}=\{x=(x_{1},\ldots,x_{n})\in R^{n}_{+}\mid x_{1}< \lambda\}\), \({T_{\lambda}}=\{x=(x_{1},\ldots,x_{n})\in R^{n}_{+}\mid x_{1}= \lambda\}\). Let \(x^{\lambda}=(2\lambda-x_{1},x_{2},\ldots,x_{n})\) be the reflection of the point \(x=(x_{1},\ldots,x_{n}) \) about the plane \(T_{\lambda}\), and \(\bar{u}_{i}^{\lambda}(x)=\bar{u}_{i}(x^{\lambda})\), \(\bar{w}^{\lambda}_{i}(x)=\bar{u}^{\lambda}_{i}(x)-\bar{u}_{i}(x)\).
For \(x,y \in{\Sigma_{\lambda}}\), \(x\neq y\), by [5], we have Obviously, we have By an elementary calculation, we derive The proof consists of two steps. In step 1, we will show that, for λ sufficiently negative, In step 2, we deduce that \(T_{\lambda}\) can be moved to the right all the way to \(z_{1}^{0}\). Furthermore, we obtain \(\bar{w}_{z^{0}_{1}}\equiv0\), \(\forall x \in\Sigma_{z_{1}^{0}}\).
$$ G(x,y) =G \bigl(x^{\lambda},y^{\lambda} \bigr) > G \bigl(x,y^{\lambda} \bigr)=G \bigl(x^{\lambda},y \bigr). $$
(2.12)
$$\begin{aligned} \bar{u}_{i}(x)={}&\frac{c_{i}x_{n}^{\frac{\alpha }{2}}}{\vert x-z^{0}\vert ^{n-\alpha}}+ \int_{R^{n}_{+}}G(x,y) \frac{y_{n}^{\gamma}\bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta _{i}}_{2}(y)}{\vert y-z^{0}\vert ^{\delta}}\,dy \\ ={}&\frac{c_{i}x_{n}^{\frac{\alpha}{2}}}{\vert x-z^{0}\vert ^{n-\alpha}}+ \int _{\Sigma_{\lambda}}G(x,y) \frac{y_{n}^{\gamma}\bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta _{i}}_{2}(y)}{\vert y-z^{0}\vert ^{\delta}}\,dy \\ &{}+ \int_{\Sigma_{\lambda}}G \bigl(x^{\lambda},y \bigr)\frac{y_{n}^{\gamma}\bar{u}^{\alpha_{i}}_{1}(y^{\lambda}) \bar{u}^{\beta_{i}}_{2}(y^{\lambda})}{\vert y^{\lambda}-z^{0}\vert ^{\delta}} \,dy , \\ \bar{u}^{\lambda}_{i}(x) ={}& \frac{c_{i}x_{n}^{\frac{\alpha}{2}}}{\vert x^{\lambda}-z^{0}\vert ^{n-\alpha }}+ \int_{\Sigma_{\lambda}}G \bigl(x^{\lambda},y \bigr) \frac{y_{n}^{\gamma}\bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta _{i}}_{2}(y)}{\vert y-z^{0}\vert ^{\delta}} \,dy \\ &{}+ \int_{\Sigma_{\lambda }}G(x,y)\frac{y_{n}^{\gamma}\bar{u}^{\alpha_{i}}_{1}(y^{\lambda })\bar{u}^{\beta_{i}}_{2}(y^{\lambda})}{\vert y^{\lambda}-z^{0} \vert ^{\delta}}\,dy . \end{aligned}$$
$$\begin{aligned} \bar{u}_{i}(x)- \bar{u}^{\lambda}_{i}(x) ={}& \frac{c_{i}x_{n}^{\frac{\alpha}{2}}}{\vert x-z^{0}\vert ^{n-\alpha}}- \frac {c_{i}x_{n}^{\frac{\alpha}{2}}}{\vert x^{\lambda}-z^{0}\vert ^{n-\alpha}} \\ &{}+ \int_{\Sigma_{\lambda}} \bigl[G(x,y)-G \bigl(x^{\lambda},y \bigr) \bigr]y_{n}^{\gamma } \biggl[\frac{\bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta _{i}}_{2}(y)}{\vert y-z^{0}\vert ^{\delta}}-\frac{\bar{u}^{\alpha _{i}}_{1}(y^{\lambda})\bar{u}^{\beta_{i}}_{2}(y^{\lambda })}{\vert y^{\lambda}-z^{0}\vert ^{\delta}} \biggr]\,dy \\ \leq{}& \int_{\Sigma_{\lambda}} \bigl[G(x,y)-G \bigl(x^{\lambda},y \bigr) \bigr]y_{n}^{\gamma } \biggl[\frac{\bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta _{i}}_{2}(y)}{\vert y-z^{0}\vert ^{\delta}}-\frac{\bar{u}^{\alpha _{i}}_{1}(y^{\lambda})\bar{u}^{\beta_{i}}_{2}(y^{\lambda })}{\vert y^{\lambda}-z^{0}\vert ^{\delta}} \biggr]\,dy \\ \leq{}& \int_{\Sigma_{\lambda}} \bigl[G(x,y)-G \bigl(x^{\lambda},y \bigr) \bigr] \frac {y_{n}^{\gamma}[\bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta _{i}}_{2}(y)-\bar{u}^{\alpha_{i}}_{1}(y^{\lambda})\bar{u}^{\beta _{i}}_{2}(y^{\lambda})]}{\vert y-z^{0}\vert ^{\delta}}\,dy. \end{aligned}$$
(2.13)
$$\bar{w}_{i}^{\lambda}(x)= \bar{u}_{i}^{\lambda}(x)- \bar{u}_{i}(x)\geq0 ,\quad \mbox{a.e. } \forall x\in \Sigma_{\lambda} . $$
Step 1. (Prepare to move the plane from near \(x_{1}=-\infty\).) In this step, we will show that, for λ sufficiently negative, \(\epsilon>0\) sufficiently small where \(({z^{0}})^{\lambda}\) is the reflection of \(z^{0}\) about the plane \(T_{\lambda}\). Define \(\Gamma^{\lambda}_{i}=\{x\in\Sigma_{\lambda} \backslash B_{\epsilon}(({z^{0}})^{\lambda}) \mid\bar{u}^{\lambda}_{i}(x)<\bar{u}_{i}(x) \} \), the sets where the inequalities (2.14) are violated. We will prove that \(\Gamma^{\lambda}_{i}\) are empty, where \(i=1,2\).
$$ \bar{u}_{i}^{\lambda}(x)\geq\bar{u}_{i}(x) ,\quad \mbox{a.e. } \forall x\in{\Sigma_{\lambda} \backslash{B_{\epsilon } \bigl( \bigl({z^{0}} \bigr)^{\lambda} \bigr)}}, $$
(2.14)
Without loss of generality, we consider \(\bar{u}_{1} \). Denote \(\Sigma ^{\lambda}_{i}=\{x\in\Sigma_{\lambda} \backslash B_{\epsilon }(({z^{0}})^{\lambda}) \mid\bar{u}_{1}^{\alpha_{i}}(x^{\lambda})\times \bar{u}_{2}^{\beta_{i}}(x^{\lambda})<\bar{u}_{1}^{\alpha_{i}}(x)\bar{u}_{2}^{\beta_{i}}(x) \} \), for \(y\in\Sigma^{\lambda}_{1}\), we may assume that \(\bar{u}_{1}(y)>\bar{u}^{\lambda}_{1}(y) \) and \(\bar{u}_{2}(y)\leq\bar{u}^{\lambda}_{2}(y)\). Define and \(\bar{w}^{\lambda}(y)=(\bar{w}^{\lambda}_{1}(y),\bar{w}^{\lambda}_{2}(y) )\). By the expression of \(G(x, y)\), it is easy to see Applying the mean value theorem, combining (2.13) and (2.15), we have, for \(x\in\Gamma^{\lambda}_{i}\), Noticing \(\Sigma_{i}^{\lambda} \subseteq\Gamma^{\lambda}_{j} \) for some j, applying the Hardy-Littlewood-Sobolev inequality and the Hölder inequality we obtain for any \(q>\frac{n}{n-\alpha}\), where \(\Gamma_{\lambda} =\Gamma^{\lambda}_{1}\cup\Gamma ^{\lambda}_{2}\). Since \(\gamma\geq0\), we can easily see that \(y^{\gamma}_{n}\) is bounded in each bounded domain \(\Omega\subset R^{n}_{+}\). Therefore, by our assumption \(\vert u\vert \in L^{\frac{n(\alpha_{i}+\beta _{i}-1)}{\alpha}}_{\mathrm{loc}}(R^{n}_{+})\), i.e.
\(\vert u\vert ^{\alpha _{i}+\beta_{i}-1}\in L^{\frac{n}{\alpha}}_{\mathrm{loc}}(R^{n}_{+})\), we derive Hence, we obtain for any domain Ω that is a positive distance away from the \(z^{0}\) and Ω̂ is the image of Ω about the Kelvin transform. By (2.19), for λ sufficiently negative and for \(\epsilon > 0\) sufficiently small, \(c\Vert \frac{y^{\gamma}_{n}\vert \bar{u}\vert ^{\alpha_{i}+\beta_{i}-1}}{\vert y-z^{0}\vert ^{\delta}} \Vert _{L^{\frac{n}{\alpha }}({\Gamma_{\lambda}}) } \) can be made very small. Combining this with (2.17), we arrive at \(\Vert \bar{w}_{\lambda} \Vert _{L^{q}({\Gamma_{\lambda}})}=0 \), and \(\Gamma^{\lambda}_{i}\) must be of measure zero. Hence (2.14) holds.
$$\bar{w}^{\lambda}_{i}(y)=\textstyle\begin{cases} 0,&\mbox{for } \bar{u}_{i}(y)< \bar{u}^{\lambda}_{i}(y) , \\ \bar{u}_{i}(y)-\bar{u}^{\lambda}_{i}(y), &\mbox{for } \bar{u}_{i}(y)>\bar{u}^{\lambda}_{i}(y), \end{cases} $$
$$ G(x, y)\leq\frac{A_{n,\alpha}}{ \vert x-y\vert ^{n-\alpha}}. $$
(2.15)
$$\begin{aligned} \bar{u}_{i}(x)-\bar{u}^{\lambda}_{i}(x) \leq{}& \int_{\Sigma_{\lambda}}G(x,y)\frac{y_{n}^{\gamma}[\bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta_{i}}_{2}(y)-\bar{u}^{\alpha _{i}}_{1}(y^{\lambda})\bar{u}^{\beta_{i}}_{2}(y^{\lambda })]}{\vert y-z^{0}\vert ^{\delta}}\,dy \\ \leq{}& \int_{\Sigma^{\lambda}_{i}}G(x,y)\frac{y_{n}^{\gamma}[\bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta_{i}}_{2}(y)-\bar{u}^{\alpha _{i}}_{1}(y^{\lambda})\bar{u}^{\beta_{i}}_{2}(y^{\lambda })]}{\vert y-z^{0}\vert ^{\delta}}\,dy \\ ={}& \int_{\Sigma^{\lambda}_{i}}G(x,y)\frac{y_{n}^{\gamma}\{[\bar{u}^{\alpha_{i}}_{1}(y)-\bar{u}^{\alpha_{i}}_{1}(y^{\lambda})]\bar{u}^{\beta_{i}}_{2}(y)+ \bar{u}^{\alpha_{i}}_{1}(y)[\bar{u}^{\beta_{i}}_{2}(y)-u^{\beta _{i}}_{2}(y^{\lambda})]\} }{\vert y-z^{0}\vert ^{\delta}} \,dy \\ \leq{}&c \int_{\Sigma^{\lambda}_{i}}G(x,y)\frac{y_{n}^{\gamma}[\bar{u}^{\alpha_{i}}_{1}(y)-\bar{u}^{\alpha_{i}}_{1}(y^{\lambda})]\bar{u}^{\beta_{i}}_{2}(y)}{\vert y-z^{0}\vert ^{\delta}} \,dy \\ \leq{}&c \int_{\Sigma^{\lambda}_{i}}G(x,y)\frac{y_{n}^{\gamma}\psi ^{\alpha_{i}-1}_{1}(y)[\bar{u}_{1}(y)-\bar{u}_{1}(y^{\lambda})]\bar{u}^{\beta_{i}}_{2}(y)}{\vert y-z^{0}\vert ^{\delta}} \,dy \\ \leq{}&c \int_{\Sigma^{\lambda}_{i}}\frac{1}{\vert x-y\vert ^{n-\alpha}}\frac {y_{n}^{\gamma}\bar{u}^{\alpha_{i}-1}_{1}(y)[\bar{u}_{1}(y)-\bar{u}_{1}(y^{\lambda})]\bar{u}^{\beta_{i}}_{2}(y)}{\vert y-z^{0}\vert ^{\delta}} \,dy. \end{aligned}$$
(2.16)
$$ \Vert \bar{w}_{\lambda} \Vert _{L^{q}({\Gamma_{\lambda}})}\leq c\biggl\Vert \frac{y_{n}^{\gamma} \vert \bar{u}\vert ^{\alpha_{i}+\beta_{i}-1}}{\vert y-z^{0}\vert ^{\delta}} \bar{w}_{\lambda}\biggr\Vert _{L^{\frac{nq}{n+\alpha q}}({\Gamma_{\lambda}}) } \leq c\biggl\Vert \frac{y_{n}^{\gamma} \vert \bar{u}\vert ^{\alpha_{i}+\beta_{i}-1}}{\vert y-z^{0}\vert ^{\delta}}\biggr\Vert _{L^{\frac {n}{\alpha}}({\Gamma_{\lambda}}) }\Vert \bar{w}_{\lambda} \Vert _{L^{q}({ \Gamma_{\lambda}})}, $$
(2.17)
$$ y^{\gamma}_{n} \vert u\vert ^{\alpha_{i}+\beta_{i}-1} \in L^{\frac{n}{\alpha }}_{\mathrm{loc}} \bigl(R^{n}_{+} \bigr). $$
(2.18)
$$ \int_{ \hat{\Omega}} \biggl[\frac{y^{\gamma}_{n}\vert \bar{u}\vert ^{\alpha_{i}+\beta _{i}-1}(y)}{{\vert y-z^{0}\vert ^{\delta}}} \biggr]^{\frac{n}{\alpha}} \,dy= \int_{ \Omega} \bigl(y^{\gamma}_{n} \vert u \vert ^{\alpha_{i}+\beta_{i}-1}(y) \bigr)^{\frac{n}{\alpha}}\,dy < \infty $$
(2.19)
Step 2. (Move the plane to the limiting position to derive symmetry.)
Inequality (2.14) provides a starting point to move the plane \(T_{\lambda}\). Now we start to move the plane \(T_{\lambda}\) along the \(x_{1}\) direction as long as (2.14) holds. Define We prove that \(\lambda_{0}= z^{0}_{1} \). On the contrary, suppose that \(\lambda_{0}< z^{0}_{1}\). We will show that \(\bar{u}(x)\) is symmetric about the plane \(T_{\lambda _{0}}\), that is, Suppose (2.20) is not true, then, for such \(\lambda_{0}< z^{0}_{1} \), for all \(i=1,2\), we have In fact, by (2.14), we have two cases for all \(i=1,2\); one is For the other case, without loss of generality, we may assume that \(\bar{u}_{1}^{\lambda_{0}}(z_{1})>\bar{u}_{1}(z_{1})\) and For the first case, (2.21) is proved. For the other case, we have Combining (2.23) with (2.24), we obtain This is impossible. Hence (2.21) holds. Next based on (2.21), we will verify that the plane can be moved further to the right. More precisely, there exists a \(\zeta> 0\) such that, for all \(\lambda\in[\lambda_{0}, \lambda_{0} + \zeta)\)
\(\bar{u}_{i}^{\lambda}(x)\geq\bar{u}_{i} (x)\), a.e. \(\forall x\in\Sigma _{\lambda}\backslash B_{\epsilon}(z^{0})^{\lambda_{0}}\). By inequality (2.17), we have Equation (2.19) ensures that one can choose η sufficiently small so that, for all λ in \([\lambda_{0}, \lambda _{0}+\eta) \), We postpone the proof of this inequality (2.26) for a moment.
$$\lambda_{0} =\operatorname{sup} \bigl\{ \lambda\leq z^{0}_{1} \mid \bar{u}_{i}^{\mu}(x)\geq \bar{u}_{i}(x) , \mbox{ a.e. } \forall x\in{ \Sigma_{\mu}} , \mu\leq\lambda \bigr\} . $$
$$ \bar{u}(x)\equiv\bar{u}^{\lambda_{0}}(x) , \quad\mbox{a.e. } \forall x\in{ \Sigma _{\lambda_{0}}\backslash{B_{\epsilon} \bigl( \bigl(z^{0} \bigr)^{\lambda_{0}} \bigr)}}. $$
(2.20)
$$ \bar{u}_{i}^{\lambda_{0}}(x) >\bar{u}_{i}(x)\quad \mbox{a.e. } x \in{\Sigma_{\lambda _{0}}\backslash{B_{\epsilon} \bigl( \bigl(z^{0} \bigr)^{\lambda_{0}} \bigr)}}. $$
(2.21)
$$ \bar{u}_{i}^{\lambda_{0}}(y)>\bar{u}_{i}(y) \quad\mbox{on a set of positive measure, $i=1$ and 2.} $$
(2.22)
$$ \bar{u}_{2}^{\lambda_{0}}(z_{2})=\bar{u}_{2}(z_{2}). $$
(2.23)
$$ \bar{u}_{1}^{\alpha_{i}}(y)\bar{u}_{2}^{\beta_{i}}(y)- \bar{u}_{1}^{\alpha_{i}} \bigl(y^{\lambda_{0}} \bigr)\bar{u}_{2}^{\beta_{i}} \bigl(y^{\lambda _{0}} \bigr)< 0. $$
(2.24)
$$0=\bar{u}_{i}(z_{i})-\bar{u}_{i}^{\lambda_{0}}(z_{i}) < \int_{\Sigma_{\lambda_{0}} } \bigl[G(z_{i},y)-G \bigl({z_{i}}^{\lambda},y \bigr) \bigr]\frac {y^{\gamma}_{n} [\bar{u}_{1}^{\alpha_{i}}(y)\bar{u}_{2}^{\beta_{i}}(y)-\bar{u}_{1}^{\alpha_{i}}(y^{\lambda_{0}})\bar{u}_{2}^{\beta_{i}}(y^{\lambda _{0}})]}{\vert y-z^{0}\vert ^{\delta}}\,dy< 0. $$
$$ \Vert \bar{w}_{\lambda} \Vert _{L^{q}(\Gamma_{\lambda})}\leq \biggl\{ \int_{\Gamma_{\lambda}} \biggl(\frac{y^{\gamma}_{n}\vert \bar{u}\vert ^{\alpha_{i}+\beta_{i}-1}(y)}{\vert y-z^{0}\vert ^{\delta}} \biggr)^{\frac{n}{\alpha}}\,dy \biggr\} ^{\frac{\alpha}{n}} \Vert \bar{w}_{\lambda} \Vert _{L^{q}(\Gamma_{\lambda})}. $$
(2.25)
$$ c \biggl\{ \int_{\Gamma_{\lambda}} \biggl(\frac{y^{\gamma}_{n}\vert \bar{u}\vert ^{\alpha_{i}+\beta_{i}-1}(y)}{\vert y-z^{0}\vert ^{\delta}} \biggr)^{\frac{n}{\alpha}}\,dy \biggr\} ^{\frac{\alpha}{n}} \leq\frac{1}{2}. $$
(2.26)
Now combining (2.25) and (2.26), we have \(\Vert \bar{w}_{\lambda} \Vert _{L^{q}(\Gamma_{\lambda})}=0 \), and \(\Gamma_{\lambda} \) must be of measure zero. Hence, for these values of \(\lambda> \lambda_{0}\), we have \(\bar{w}_{\lambda}(x)\geq0\), a.e. \(\forall x\in\Sigma_{\lambda _{0}}\backslash B_{\epsilon}((z^{0})^{\lambda})\), \(\epsilon>0 \). This contradicts the definition of \(\lambda_{0}\). Therefore (2.20) must hold. That is, if \(\lambda_{0}< z^{0}_{1} \), then we must have
$$ \bar{u}_{i}(x)\equiv\bar{u}_{i}^{\lambda}(x),\quad \mbox{a.e. } \forall x \in\Sigma _{\lambda_{0}}\backslash B_{\epsilon}\bigl( \bigl(z^{0} \bigr)^{\lambda_{0}} \bigr). $$
(2.27)
Recall that, by our assumption, \(c_{i_{0}}>0\) and It follows that \(\bar{u}_{i_{0}}\) is singular at \(z^{0}\), hence by (2.27), \(\bar{u}_{i_{0}}\) must also be singular at \((z^{0})^{\lambda}\). This is impossible, because \(z^{0}\) is the only singularity of ū. Hence, we must have \(\lambda_{0}=z^{0}_{1} \). Since ϵ is an arbitrary positive number, we have actually derived that Entirely similarly, we can move the plane from near \(x_{1} = +\infty\) to the left and obtain \(\bar{u}_{i}^{\lambda_{0}}(x)\leq\bar{u}_{i}(x)\), a.e. \(\forall x\in\Sigma_{\lambda_{0}}\), \(\lambda_{0}= z^{0}_{1} \). Therefore we have \(\bar{w}_{\lambda_{0}} (x)\equiv0\), a.e. \(\forall x\in\Sigma_{\lambda _{0}}\), \(\lambda_{0}=z_{1}^{0}\).
$$ \bar{u}_{i_{0}}(x)=\frac{c_{i_{0}}x^{\frac{\alpha }{2}}_{n}}{\vert x-z^{0}\vert ^{n-\alpha}}+ \int_{R^{n}_{+}}G(x,y)\frac{y^{\gamma}_{n}\bar{u}_{1}^{\alpha _{1}}(y)\bar{u}_{2}^{\beta_{1}}(y)}{\vert y-z^{0}\vert ^{\delta}}\,dy. $$
(2.28)
$$\bar{u}_{i}^{\lambda_{0}}(x)\geq\bar{u}_{i}(x), \quad\mbox{a.e. } \forall x\in\Sigma _{\lambda_{0}}, \lambda_{0}=z^{0}_{1} . $$
Now we prove inequality (2.26). For any small \(\eta>0\), \(\forall\varepsilon>0\), we can choose R sufficiently large so that For any \(\tau>0\), define \(E_{i}^{\tau}=\{x \in({\Sigma_{\lambda_{0}}}\backslash {B_{\varepsilon}((z^{0})^{\lambda_{0}})}) \cap{B_{R}(0)} \mid\bar{u}_{i}^{\lambda_{0}}(x)- \bar{u}_{i}(x)>\tau \} \), and \(F_{i}^{\tau}=\{ ({\Sigma_{\lambda_{0}}}\backslash {B_{\varepsilon}((z^{0})^{\lambda_{0}})}) \cap{B_{R}(0)} \} \backslash E_{i}^{\tau}\). Obviously, \({\lim_{\tau\rightarrow0}}\mu(F_{i}^{\tau})=0 \).
$$ \biggl( \int_{(R^{n}_{+} \backslash B_{\varepsilon}(z^{0}))\backslash B_{R}} \biggl[\frac{y_{n}^{\gamma} \vert \bar{u}\vert ^{\alpha _{i}+\beta_{i}-1}(y)}{\vert y-z^{0}\vert ^{\delta}}\,dy \biggr]^{\frac{n}{\alpha}} \biggr)^{\frac {\alpha}{n}} \leq\eta. $$
(2.29)
For \(\lambda> {\lambda_{0}}\), let \(D_{\lambda}=\{ ({\Sigma_{\lambda}}\backslash{B_{\varepsilon }((z^{0})^{\lambda})}) \backslash({\Sigma_{\lambda_{0}}}\backslash {B_{\varepsilon}((z^{0})^{\lambda_{0}})})\cap{B_{R}(0)} \} \).
It is easy to see that For λ sufficiently close to \(\lambda_{0}\), \(\mu (D_{\lambda}) \) is very small. We will show that \(\mu(\Gamma^{\lambda}_{i} \cap E_{i}^{\tau} ) \) is sufficiently small as λ close to \(\lambda_{0}\).
$$ \bigl\{ {\Gamma^{\lambda}_{i}} \cap{B_{R}(0)} \bigr\} \subset \bigl({\Gamma ^{\lambda}_{i}} \cap E_{i}^{\tau} \bigr)\cup F_{i}^{\tau} \cup D_{\lambda}. $$
(2.30)
In fact, \(\bar{w}_{i}^{\lambda}(x)=\bar{u}_{i}^{\lambda}(x)-\bar{u}_{i}(x) = \bar{u}_{i}^{\lambda}(x)-\bar{u}_{i}^{\lambda_{0}}(x)+\bar{u}_{i}^{\lambda _{0}}(x) -\bar{u}_{i}(x) < 0\), \(\forall x \in(\Gamma^{\lambda}_{i} \cap E_{i}^{\tau} )\). Therefore, \(\bar{u}_{i}^{\lambda_{0}}(x)-\bar{u}_{i}^{\lambda}(x)>\bar{u}_{i}^{\lambda }(x)-\bar{u}_{i}^{\lambda}(x) >\tau\), \(\forall x \in(\Gamma^{\lambda }_{i} \cap E_{i}^{\tau} )\). It follows that By the well-known Chebyshev inequality, for fixed τ, as λ is close to \(\lambda_{0} \), \(\mu (E_{i}^{\tau}) \) can be sufficiently small. By (2.30) and (2.31), we derive that \(\mu (\Gamma^{\lambda}_{i} \cap B_{R}(0) ) \) can be made as small as we wish. Combining this with (2.29), we deduce that (2.26) holds.
$$ \bigl(\Gamma^{\lambda}_{i} \cap E_{i}^{\tau} \bigr) \subset H_{i}^{\tau} = \bigl\{ x \in B_{R}(0) \mid\bar{u}_{i}^{\lambda_{0}}(x)- \bar{u}_{i}^{\lambda}(x) >\tau \bigr\} . $$
(2.31)
Since we can choose any direction that is perpendicular to the \(x_{n}\)-axis as the \(x_{1}\) direction, we have actually shown that the Kelvin transform of the solution \(\bar{u}(x)\) is rotationally symmetric about the line parallel to the \(x_{n}\)-axis and passing through \(z^{0}\). Now we take any two points \(X^{1}\) and \(X^{2}\), with \(X^{l}=(x^{\prime l},x_{n})\in R^{n-1}\times[0,\infty)\), \(l=1,2\). Let \(z^{0}\) be the projection of \(\bar{X}= \frac{X^{1}+X^{2}}{2}\) on \(\partial R^{n}_{+}\). Set \(Y^{l}= \frac{X^{l}-z^{0}}{\vert X^{l}-z^{0}\vert ^{2}}+z^{0}\), \(l=1,2\). From the above arguments, it is easy to see \(\bar{u}(Y ^{1}) = \bar{u}(Y^{2})\), hence \(u(X^{1}) = u(X^{2})\). This implies that \(u_{i}\) is independent of \(x'=(x_{1},\ldots,x_{n-1}) \). That is, \(u_{i}=u_{i}(x_{n})\), and we will show that this will contradict the finiteness of the integral \(\int_{R^{n}_{+}}G(x,y)y^{\gamma}_{n} u^{\alpha_{i}}_{1}(y)u^{\beta _{i}}_{2}(y)\,dy\). To continue, we need the following lemma.
Lemma 2.1
[5]
If
\(\frac{t}{s}\)
is sufficiently small, then
\(\forall x=(x',x_{n})\), \(y=(y',y_{n})\in R^{n}_{+}\), one can derive that
Here
\(s=\vert x-y\vert ^{2} \), \(t=4x_{n}y_{n}\), \(c_{n,\alpha}\), and
\(C_{n,\alpha}\)
stand for different positive constants that only depend on
n
and
α.
$$\frac{c_{n,\alpha}}{s^{\frac{n-\alpha}{2}}}\cdot\frac{t^{\frac {\alpha}{2}}}{s^{\frac{\alpha}{2}}}\leq G(x , y)\leq \frac{C_{n,\alpha}}{s^{\frac{n-\alpha}{2}}} \cdot\frac{t^{\frac {\alpha}{2}}}{s^{\frac{\alpha}{2}}}, \quad\textit{i.e. } G(x , y) \sim\frac{t^{\frac{\alpha}{2}}}{s^{\frac{n}{2}}}. $$
Set \(x=(x',x_{n})\), \(y=(y',y_{n})\in R^{n-1}\times(0,+\infty)\), \(r^{2}=\vert x'-y'\vert ^{2} \) and \(a^{2}=\vert x_{n}-y_{n}\vert ^{2} \). If \(u_{i}=u_{i}(x_{n}) \) is a solution of then, for each fixed \(x\in R^{n}_{+} \), letting R be large enough, by elementary calculations, we have Equation (2.33) implies that there exists a sequence \(\{ y^{k}_{n}\}\rightarrow\infty \) as \(k\rightarrow\infty \), such that Similarly to (2.33), for any \(x=(0,x_{n})\in R^{n}_{+} \), we derive that Let \(x_{n} = 2R\) be sufficiently large. By (2.35), we deduce that Then by (2.35) and (2.36), for \(x_{n} = 2R\) sufficiently large, we also obtain Continuing this way m times, for \(x_{n} = 2R\), we have For any fixed α and γ in their respective domain, we choose m to be an integer greater than \(\frac{-\alpha^{2}-\alpha \gamma+\gamma+3}{\alpha+\gamma}\) and 1. That is, where \(\lceil a\rfloor\) is the integer part of a.
$$ u_{i}(x)= \int_{R^{n}_{+}}G(x,y)y^{\gamma}_{n} u^{\alpha_{i}}_{1}(y)u^{\beta _{i}}_{2} (y)\,dy, $$
(2.32)
$$\begin{aligned} +\infty>u_{i}(x_{n})={}& \int^{\infty}_{0}y^{\gamma}_{n} u^{\alpha _{i}}_{1}(y_{n})u^{\beta_{i}}_{2}(y_{n}) \int_{R^{n-1}} G(x,y)\,dy'\,dy_{n} \\ \geq{}&C \int^{\infty}_{R}y^{\gamma}_{n} u^{\alpha_{i}}_{1}(y_{n})u^{\beta _{i}}_{2}(y_{n})y_{n}^{\frac{\alpha}{2}} \int_{R^{n-1}\backslash B_{R}(0)} \frac{1}{\vert x-y\vert ^{n}}\,dy' \,dy_{n} \\ \geq{}&C \int^{\infty}_{R}y^{\gamma}_{n} u^{\alpha_{i}}_{1}(y_{n})u^{\beta _{i}}_{2}(y_{n})y_{n}^{\frac{\alpha}{n-2}}{ \int^{\infty}_{R}} \frac {r^{2}}{(r^{2}+a^{2})^{\frac{n}{2}}}\,dr \,dy_{n} \\ \geq{}&C \int^{\infty}_{R}y^{\gamma}_{n} u^{\alpha_{i}}_{1}(y_{n})u^{\beta _{i}}_{2}(y_{n})y_{n}^{\frac{\alpha}{2}} \frac{1}{\vert x_{n}-y_{n}\vert }{ \int ^{\infty}_{\frac{R}{a}}} \frac{\tau^{n-2}}{(\tau^{2}+1)^{\frac {n}{2}}}\,d\tau \,dy_{n} \\ \geq{}&C \int^{\infty}_{R} u^{\alpha_{i}}_{1}(y_{n})u^{\beta _{i}}_{2}(y_{n})y_{n}^{\gamma+\frac{\alpha}{2}-1} \,dy_{n}. \end{aligned}$$
(2.33)
$$ u^{\alpha_{i}}_{1} \bigl(y^{k}_{n} \bigr)u^{\beta_{i}}_{2} \bigl(y^{k}_{n} \bigr) \bigl(y^{k}_{n} \bigr)^{\gamma +\frac{\alpha}{2}}\rightarrow0. $$
(2.34)
$$ +\infty>u_{i}(x_{n})\geq C_{0}{ \int_{0}^{\infty}} y_{n}^{\gamma}u^{\alpha _{i}}_{1}(y_{n})u^{\beta_{i}}_{2}(y_{n})y_{n}^{\frac{ \alpha}{2}}\frac{1}{\vert x_{n}-y_{n}\vert } \,dy_{n} x_{n}^{\frac{ \alpha}{2}}. $$
(2.35)
$$\begin{aligned} +\infty>u_{i}(x_{n})\geq{}& C_{0}{ \int_{0}^{1}} y_{n}^{\gamma}u^{\alpha _{i}}_{1}(y_{n})u^{\beta_{i}}_{2}(y_{n})y_{n}^{\frac{ \alpha}{2}}\frac{1}{\vert x_{n}-y_{n}\vert } \,dy_{n} x_{n}^{\frac{ \alpha}{2}} \\ \geq{}&\frac{C_{0}}{2R}(2R)^{\frac{\alpha}{2}} { \int_{0}^{1}} y_{n}^{\gamma }u^{\alpha_{i}}_{1}(y_{n})u^{\beta_{i}}_{2}(y_{n})y_{n}^{\frac{ \alpha }{2}}\,dy_{n} \geq C_{1}(2R)^{\frac{\alpha}{2}-1} =C_{1}x_{n}^{\frac{\alpha}{2}-1}. \end{aligned}$$
(2.36)
$$\begin{aligned} u_{i}(x_{n})\geq{}&C_{0}{ \int_{\frac{R}{2}^{R}}}y_{n}^{\gamma}u^{\alpha _{i}}_{1}(y_{n})u^{\beta_{i}}_{2}(y_{n})y_{n}^{\frac{ \alpha}{2}} \frac{1}{\vert x_{n}-y_{n}\vert }\,dy_{n} x_{n}^{\frac{ \alpha}{2}} \\ \geq{}&C_{0}{ \int_{\frac{R}{2}^{R}}}y_{n}^{\gamma}C_{1}^{\alpha_{i}+\beta _{i}}y_{n}^{(\alpha_{i}+\beta_{i})(\frac{\alpha}{2}-1)}y_{n}^{\frac{ \alpha}{2}} \frac{1}{\vert x_{n}-y_{n}\vert }\,dy_{n} x_{n}^{\frac{ \alpha}{2}} \\ \geq{}&C_{0} C_{1}^{\alpha_{i}+\beta_{i}}R^{(\alpha_{i}+\beta_{i})(\frac{\alpha }{2}-1)+\gamma}(2R)^{\frac{ \alpha}{2}}\frac{2}{3R} { \int_{\frac{R}{2}^{R}}}y_{n}^{\frac{ \alpha}{2}}\,dy_{n} \\ :={}&AR^{(\alpha_{i}+\beta_{i})(\frac{\alpha}{2}-1)+\gamma+\alpha} :=A_{1}x_{n}^{(\alpha_{i}+\beta_{i})(\frac{\alpha}{2}-1)+\gamma+\alpha}. \end{aligned}$$
$$ u_{i}(x_{n})\geq A(m,\alpha_{i}+ \beta_{i},\alpha,\gamma)x_{n}^{(\alpha _{i}+\beta_{i})^{m}(\frac{\alpha}{2}-1)+\frac{(\alpha_{i}+\beta _{i})^{m}-1}{\alpha_{i}+\beta_{i}-1}(\gamma+\alpha)}. $$
(2.37)
$$ m\geq \operatorname{max} \biggl\{ \biggl\lceil \frac{-\alpha^{2}- \alpha\gamma+\gamma+3}{\alpha +\gamma} \biggr\rfloor +1,1 \biggr\} , $$
(2.38)
We claim that, for such a choice of m, we have We postpone the proof of (2.39) for a moment. Now by (2.37) and (2.39), we derive that for all \(x_{n}\) sufficiently large. This contradicts (2.34). So there is no positive solution of (2.32). This implies that \(u(x)\) must be constant. By our positive assumption on u, we have \(u_{i}(x) = b_{i} > 0\), \(i = 1, 2\). Taking \(u_{i}\) into (2.2), we have \(0=(-\Delta)^{\frac{\alpha }{2}}u_{i}(x)=x^{\gamma}_{n}u_{1}^{\alpha_{i}}(x)u_{2}^{\beta_{i}}(x)>0\). This is impossible. Hence, in (2.10), \(c_{i}\) must be zero, \(i = 1, 2\).
$$ \tau(\alpha_{i}+\beta_{i}):= \biggl[({ \alpha_{i}+\beta_{i}})^{m} \biggl( \frac{\alpha }{2}-1 \biggr)+\frac{({\alpha_{i}+\beta_{i}})^{m}-1}{{\alpha_{i}+\beta _{i}}-1}(\alpha+\gamma) \biggr]({ \alpha_{i}+\beta_{i}})+\frac{\alpha }{2}+\gamma\geq0. $$
(2.39)
$$u^{\alpha_{i}+\beta_{i}}_{i}(x_{n})x^{\frac{\alpha}{2}+\gamma}_{n} \geq A(m,{\alpha_{i}+\beta_{i}},\alpha, \gamma)x^{\tau({\alpha_{i}+\beta _{i}})}_{n} \geq A(m,{\alpha_{i}+ \beta_{i}},\alpha,\gamma)>0, $$
Now it is left to verify (2.39). In fact, if we let then We show that \(f'({\alpha_{i}+\beta_{i}})>0\), for \(1<{\alpha_{i}+\beta_{i}}\leq \frac{n+\alpha+2\gamma}{n-\alpha} \). Since \({\alpha_{i}+\beta_{i}}>1\), it suffices to show \((m+2)(\frac{\alpha}{2}-1)({\alpha_{i}+\beta_{i}})+(m+1)(\frac{\alpha }{2}+\gamma+1) \geq\frac{\alpha}{2}\). Due to the fact \(\frac{\alpha}{2}-1 < 0\), \(n \geq3\), and \({\alpha _{i}+\beta_{i}} \leq\frac{n+\alpha+2\gamma}{n-\alpha} \), we only need to verify that which can be derived directly from (2.38).
$$\begin{aligned} f({\alpha_{i}+\beta_{i}}):={}&\tau({ \alpha_{i}+ \beta_{i}}) ({\alpha_{i}+\beta _{i}}-1) \\ ={}& ({\alpha_{i}+\beta_{i}})^{m+2} \biggl( \frac{\alpha}{2}-1 \biggr)+({\alpha_{i}+\beta _{i}})^{m+1} \biggl(\frac{\alpha}{2}+\gamma+1 \biggr) -\frac{\alpha}{2}({ \alpha_{i}+\beta_{i}})-\frac{\alpha}{2}-\gamma, \end{aligned}$$
$$f'({\alpha_{i}+\beta_{i}})=({ \alpha_{i}+\beta_{i}})^{m} \biggl[(m+2) \biggl( \frac{\alpha }{2}-1 \biggr) ({\alpha_{i}+\beta_{i}})+ ({m+1}) \biggl(\frac{\alpha}{2}+\gamma+1 \biggr) \biggr]-\frac{\alpha}{2}. $$
$$(m+2) \biggl(\frac{\alpha}{2}-1 \biggr)\frac{3+\alpha+2\gamma}{3-\alpha}+(m+1) \biggl( \frac{\alpha}{2}+\gamma+1 \biggr) \geq\frac{\alpha}{2}, $$
On the other hand, assume that \(u(x)\) is a solution of the integral equation (1.6). Then, for any \(\phi\in C^{\infty }_{0}({R^{n}_{+}}) \), we have Hence u also satisfies equations (1.3).
$$\begin{aligned} \bigl\langle (-\Delta)^{\frac{\alpha}{2}}u_{i}, \phi\bigr\rangle ={}& \biggl\langle \int_{R^{n}_{+}}G_{\infty}(x,y)y^{\gamma}_{n}u_{1}^{\alpha _{i}}(y)u_{2}^{\beta_{i}}(y) \,dy, (-\Delta)^{\frac{\alpha}{2}}\phi (x)\biggr\rangle \\ ={}& \int_{R^{n}_{+}} \biggl\{ \int_{R^{n}_{+}}G_{\infty}(x,y)y^{\gamma }_{n}u_{1}^{\alpha_{i}}(y)u_{2}^{\beta_{i}}(y) \,dy \biggr\} (-\Delta)^{\frac {\alpha}{2}}\phi(x)\,dx \\ ={}& \int_{R^{n}_{+}} \biggl\{ \int_{R^{n}_{+}}G_{\infty}(x,y) (-\Delta)^{\frac {\alpha}{2}} \phi(x)\,dx \biggr\} y^{\gamma}_{n}u_{1}^{\alpha _{i}}(y)u_{2}^{\beta_{i}}(y) \,dy \\ ={}& \int_{R^{n}_{+}} \biggl\{ \int_{R^{n}_{+}}\delta(x-y)\phi(x)\,dx \biggr\} y^{\gamma}_{n}u_{1}^{\alpha_{i}}(y)u_{2}^{\beta_{i}}(y) \,dy \\ ={}& \int_{R^{n}_{+}}y^{\gamma}_{n}u_{1}^{\alpha_{i}}(y)u_{2}^{\beta _{i}}(y) \phi(y)\,dy =\bigl\langle y^{\gamma}_{n}u_{1}^{\alpha_{i}}u_{2}^{\beta_{i}}, \phi\bigr\rangle . \end{aligned}$$
This completes the proof of Theorem 1.1. □
3 Liouville theorems
3.1 The proof of Theorem 1.2
In this section, we will establish the nonexistence of the solutions to (1.6) by using the method of moving planes base up in the positive \(x_{n}\) direction.
For a given positive real number λ, define \(\hat{\Sigma}_{\lambda}=\{x=(x_{1},\ldots,x_{n-1},x_{n})\in R^{n}_{+}\mid0< x_{n}< \lambda\}\), \(\hat{T}_{\lambda}=\{x=(x_{1},\ldots,x_{n-1},x_{n})\in R^{n}_{+}\mid x_{n}= \lambda\}\).
Let \(x^{\lambda}=(x_{1},x_{2},\ldots,x_{n-1},2\lambda-x_{n})\) be the reflection of the point \(x=(x_{1},\ldots,x_{n}) \) about the plane \(\hat{T}_{\lambda}\), set \(\Sigma^{c}_{\lambda }=R^{n}_{+}\backslash\hat{\Sigma}_{\lambda} \) the complement of \(\hat{\Sigma}_{\lambda} \), and write \(u_{i}^{\lambda }(x)=u_{i}(x^{\lambda}) \) and \(w_{i}^{\lambda}(x)=u_{i}^{\lambda}(x)-u_{i}(x)\).
The following two lemmas are the key ingredient in our integral estimate.
Lemma 3.1
[5]
(i) For any
\(x, y \in\hat{\Sigma}_{\lambda}\), \(x\neq y\), we have
$$\begin{aligned} &G \bigl(x^{\lambda},y^{\lambda}\bigr)>\operatorname{max} \bigl\{ G \bigl(x^{\lambda},y \bigr),G \bigl(x,y^{\lambda}\bigr) \bigr\} , \\ & G \bigl(x^{\lambda},y^{\lambda}\bigr)-G(x,y)>\bigl\vert G \bigl(x^{\lambda},y \bigr)-G \bigl(x,y^{\lambda}\bigr)\bigr\vert . \end{aligned}$$
(ii) For any
\(x \in\hat{\Sigma}_{\lambda}\), \(y \in\Sigma_{\lambda}^{C}\), we have
\(G(x^{\lambda},y)>G(x,y)\).
Lemma 3.2
For any
\(x \in\hat{\Sigma}_{\lambda}\), \(u_{i}\)
are the positive solution of (1.6), we have
$$u_{i}(x)- u_{i}^{\lambda}(x)\leq \int_{\hat{\Sigma}_{\lambda }} \bigl[G \bigl(x^{\lambda},y^{\lambda}\bigr)-G \bigl(x,y^{\lambda}\bigr) \bigr]y_{n}^{\gamma}\bigl[u^{\alpha_{i}}_{1}(y) u^{\beta_{i}}_{2}(y)- u^{\alpha _{i}}_{1} \bigl(y^{\lambda} \bigr) u^{\beta_{i}}_{2} \bigl(y^{\lambda} \bigr) \bigr] \,dy. $$
Proof
Let \(\tilde{\Sigma}_{\lambda}\) be the reflection of \(\hat{\Sigma}_{\lambda}\) about the plane \(\hat{T}_{\lambda}\). we have By Lemma 3.1, we arrive at This completes the proof of Lemma 3.2. □
$$\begin{aligned} &u_{i}(x)= \int_{R^{n}_{+}}G(x,y) y_{n}^{\gamma} u^{\alpha_{i}}_{1}(y) u^{\beta_{i}}_{2}(y)\,dy \\ &\phantom{u_{i}(x)}= \int _{\hat{\Sigma}_{\lambda}}G(x,y) y_{n}^{\gamma} u^{\alpha_{i}}_{1}(y) u^{\beta_{i}}_{2}(y)\,dy \\ &\phantom{\phantom{u_{i}(x)}={}}+ \int_{\hat{\Sigma}_{\lambda}}G \bigl(x,y^{\lambda} \bigr) \bigl(y_{n}^{\lambda } \bigr)^{\gamma} u^{\alpha_{i}}_{1} \bigl(y^{\lambda} \bigr) u^{\beta _{i}}_{2} \bigl(y^{\lambda} \bigr)\,dy+{ \int_{\Sigma^{c}_{\lambda}\backslash\tilde {\Sigma_{\lambda}}}}G(x,y)y_{n}^{\gamma} u_{\lambda}^{\alpha _{i}}(y) u_{\lambda}^{\beta_{i}}(y)\,dy , \\ &u_{i} \bigl(x^{\lambda} \bigr) = \int_{\hat{\Sigma}_{\lambda}}G \bigl(x^{\lambda},y \bigr) y_{n}^{\gamma} u^{\alpha_{i}}_{1}(y) u^{\beta_{i}}_{2}(y)\,dy \\ &\phantom{u_{i} \bigl(x^{\lambda} \bigr) ={}}+ \int_{\hat{\Sigma}_{\lambda}}G \bigl(x^{\lambda},y^{\lambda } \bigr) \bigl(y_{n}^{\lambda} \bigr)^{\gamma} u^{\alpha_{i}}_{1} \bigl(y^{\lambda} \bigr) u^{\beta_{i}}_{2} \bigl(y^{\lambda} \bigr)\,dy + \int_{\Sigma^{c}_{\lambda}\backslash\tilde{\Sigma}_{\lambda }}G \bigl(x^{\lambda},y \bigr)y_{n}^{\gamma} u^{\alpha_{i}}_{1}(y) u^{\beta _{i}}_{2}(y)\,dy . \end{aligned}$$
$$\begin{aligned} u_{i}(x)- u_{i} \bigl(x^{\lambda} \bigr) \leq& \int_{\hat{\Sigma}_{\lambda}} \bigl[G(x,y)-G \bigl(x^{\lambda},y \bigr) \bigr] y_{n}^{\gamma} u^{\alpha_{i}}_{1}(y) u^{\beta_{i}}_{2}(y)\,dy \\ &{}- \int_{\hat{\Sigma}_{\lambda}} \bigl[G \bigl(x^{\lambda},y^{\lambda } \bigr)-G \bigl(x,y^{\lambda} \bigr) \bigr] \bigl(y_{n}^{\lambda} \bigr)^{\gamma} u^{\alpha _{i}}_{1} \bigl(y^{\lambda} \bigr) u^{\beta_{i}}_{2} \bigl(y^{\lambda} \bigr)\,dy \\ =& \int_{\hat{\Sigma}_{\lambda}} \bigl[G \bigl(x^{\lambda},y^{\lambda } \bigr)-G \bigl(x,y^{\lambda} \bigr) \bigr] \bigl[y_{n}^{\gamma} u^{\alpha_{i}}_{1}(y) u^{\beta _{i}}_{2}(y)- \bigl(y_{n}^{\lambda} \bigr)^{\gamma} u^{\alpha _{i}}_{1} \bigl(y^{\lambda} \bigr) u^{\beta_{i}}_{2} \bigl(y^{\lambda} \bigr) \bigr]\,dy \\ \leq& \int_{\hat{\Sigma}_{\lambda}} \bigl[G \bigl(x^{\lambda},y^{\lambda } \bigr)-G \bigl(x,y^{\lambda} \bigr) \bigr] \bigl[y_{n}^{\gamma} u^{\alpha_{i}}_{1}(y) u^{\beta _{i}}_{2}(y)- \bigl(y_{n}^{\lambda} \bigr)^{\gamma} u^{\alpha _{i}}_{1} \bigl(y^{\lambda} \bigr) u^{\beta_{i}}_{2} \bigl(y^{\lambda} \bigr) \bigr]\,dy \\ \leq& \int_{\hat{\Sigma}_{\lambda}} \bigl[G \bigl(x^{\lambda},y^{\lambda } \bigr)-G \bigl(x,y^{\lambda} \bigr) \bigr] y_{n}^{\gamma} \bigl[ u^{\alpha_{i}}_{1}(y) u^{\beta_{i}}_{2}(y)- u^{\alpha_{i}}_{1} \bigl(y^{\lambda} \bigr) u^{\beta_{i}}_{2} \bigl(y^{\lambda} \bigr) \bigr]\,dy . \end{aligned}$$
In order to prove the Liouville theorem, Theorem 1.2, we carry out the method of moving planes in integral form in the positive \(x_{n}\) direction.
The proof consists of two steps. In the first step, we start from the very low end of our region \(R^{n}_{+}\), i.e. near \(x_{n} = 0\) and show that, for λ sufficiently small, In the second step, we will move our plane \(\hat{T}_{\lambda}\) up in the positive \(x_{n}\) direction as long as the inequality (3.1) holds and show that \(u(x)\) is monotone increasing in \(x_{n}\) and thus derive a contradiction.
$$ w_{i}^{\lambda}(x)=u_{i}^{\lambda}(x)- u_{i}(x)\geq0,\quad \mbox{a.e. } \forall x\in\hat{\Sigma}_{\lambda}. $$
(3.1)
Step 1. Define \(\Gamma^{\lambda}_{i}=\{x\in\hat{\Sigma}_{\lambda} \mid u^{\lambda }_{i}(x)< u_{i}(x) \}\), \(i=1,2\) and \(\Sigma^{\lambda}_{i}=\{x\in\hat{\Sigma}_{\lambda} \mid u_{1}^{\alpha_{i}}(x^{\lambda}) u_{2}^{\beta_{i}}(x^{\lambda})< u_{1}^{\alpha_{i}}(x) u_{2}^{\beta_{i}}(x) \} \). We show that, for λ sufficiently small, \(\Gamma^{\lambda }_{i} \) must be measure zero. In fact, for any \(x \in\Gamma^{\lambda}_{i} \), by the mean value theorem similar to (2.16) and Lemma 3.2, we obtain By the expression of \(G(x, y)\), it is easy to see \(G(x, y)\leq\frac{A_{n,\alpha}}{ \vert x-y\vert ^{n-\alpha}}\). From (3.2), we have
Notice that now γ is only a little larger than 0, so \(y_{n}\) is bounded within \(\Sigma^{\lambda}_{i} \), and since \(\gamma\geq0\), we get \(\vert y^{\gamma}_{n}\vert \leq C\), hence we derive (3.4) from (3.3).
$$\begin{aligned} u_{i}(x)-u_{i}^{\lambda}(x) \leq{}& \int_{\Sigma^{\lambda }_{i}}G \bigl(x^{\lambda},y^{\lambda} \bigr) y_{n}^{\gamma} \bigl[ u^{\alpha_{i}}_{1}(y) u^{\beta_{i}}_{2}(y)- u^{\alpha_{i}}_{1} \bigl(y^{\lambda} \bigr) u^{\beta_{i}}_{2} \bigl(y^{\lambda} \bigr) \bigr]\,dy \\ = {}& \int_{\Sigma^{\lambda}_{i}}G \bigl(x^{\lambda},y^{\lambda} \bigr) y_{n}^{\gamma} \bigl[ \bigl(u^{\alpha_{i}}_{1}(y)- u^{\alpha _{i}}_{1} \bigl(y^{\lambda} \bigr) \bigr) u^{\beta_{i}}_{2}(y)+ u^{\alpha_{i}}_{1}(y) \bigl( u^{\beta_{i}}_{2}(y)-u^{\beta _{i}}_{2} \bigl(y^{\lambda} \bigr) \bigr) \bigr]\,dy \\ \leq{}& \int_{\Sigma^{\lambda}_{i}}G \bigl(x^{\lambda},y^{\lambda} \bigr) y_{n}^{\gamma} \bigl[ u^{\alpha_{i}}_{1}(y)- u^{\alpha_{i}}_{1} \bigl(y^{\lambda } \bigr) \bigr] u^{\beta_{i}}_{2}(y)\,dy \\ \leq{}& c \int_{\Sigma^{\lambda}_{i}}G \bigl(x^{\lambda},y^{\lambda} \bigr) y_{n}^{\gamma} u^{\alpha_{i}-1}_{1}(y) \bigl[ u_{1}(y)- u_{1} \bigl(y^{\lambda } \bigr) \bigr] u^{\beta_{i}}_{2}(y)\,dy. \end{aligned}$$
(3.2)
$$\begin{aligned} {1} u_{i}(x)-u_{i}^{\lambda}(x)\leq{}&c \int_{\Sigma^{\lambda}_{i}}\frac {1}{\vert x-y\vert ^{n-\alpha}}\bigl\vert y^{\gamma}_{n} u^{\alpha_{i}+\beta _{i}-1}(y)\bigr|\bigl| w^{\lambda}(y)\bigr\vert \end{aligned}$$
(3.3)
$$\begin{aligned} \leq{}&c \int_{\Sigma^{\lambda}_{i}}\frac{1}{\vert x-y\vert ^{n-\alpha}}\bigl\vert u^{\alpha_{i}+\beta_{i}-1}(y) \bigr|\bigl| w^{\lambda}(y)\bigr\vert . \end{aligned}$$
(3.4)
Noticing \(\Sigma_{i}^{\lambda} \subseteq\Gamma^{\lambda}_{j} \) for some j, applying Hardy-Littlewood-Sobolev inequality and the Hölder inequality we obtain, for any \(q>\frac{n}{n-\alpha}\), where \(\Gamma_{\lambda} =\Gamma^{\lambda}_{1}\cup\Gamma ^{\lambda}_{2}\).
$$ \Vert w_{\lambda} \Vert _{L^{q}({\Gamma_{\lambda}})}\leq c\bigl\Vert \vert u\vert ^{\alpha _{i}+\beta_{i}-1}w_{\lambda} \bigr\Vert _{L^{\frac{nq}{n+\alpha q}}({\Gamma _{\lambda}}) } \leq c\bigl\Vert \vert u\vert ^{\alpha_{i}+\beta_{i}-1}\bigr\Vert _{L^{\frac{n}{\alpha }}({\Gamma_{\lambda}}) }\Vert w_{\lambda} \Vert _{L^{q}({\Gamma_{\lambda }})}, $$
(3.5)
Since \(\vert u\vert \in L^{\frac{n(\alpha_{i}+\beta_{i}-1)}{\alpha}}(R^{n}_{+}) \), we can choose sufficiently small positive λ such that By (3.5) and (3.6), we derive \(\Vert w_{\lambda} \Vert _{L^{q}({\Gamma_{\lambda}})}=0 \), and \(\Gamma^{\lambda}_{i} \) must be of measure zero, hence (3.1) holds. This provides us a starting point for moving the plane.
$$ c\bigl\Vert u^{\alpha_{i}+\beta_{i}-1} \bigr\Vert _{L^{\frac{n}{\alpha}}(\Gamma _{\lambda})}= c \biggl\{ { \int_{{\Gamma_{\lambda}}}} \vert u\vert ^{\frac{n(\alpha _{i}+\beta_{i}-1)}{\alpha}} (y) \,dy \biggr\} ^{\frac{\alpha}{n}} \leq\frac{1}{2}. $$
(3.6)
Step 2. Now we start from such small λ and move the plane \(\hat{T}_{\lambda}\) up as long as (3.1) holds. Define We will prove Suppose to the contrary that \(\lambda_{0}<+\infty\), we will show that \(u_{i}(x)\) is symmetric about the plane \(\hat{T}_{\lambda_{0}}\), i.e.
This will contradict the strict positivity of \(u_{i}(x)\). Suppose (3.8) does not hold. Then, for such a \(\lambda_{0}\), we have \(u^{\lambda_{0} }_{i}(x)\geq u_{i}(x)\), but \(u^{\lambda_{0} }_{i}(x)\not \equiv u_{i}(x)\) a.e. on \(\hat{\Sigma}_{\lambda_{0}} \). We show that the plane can be moved further up. More precisely, there exists an \(\epsilon> 0\) such that for all \(\lambda\in[\lambda_{0}, \lambda _{0}+\epsilon) \)
To verify this, we will again resort to inequality (3.5). If one can show that, for ϵ sufficiently small so that, for all λ in \([\lambda_{0}, \lambda_{0}+\epsilon)\), we have then by (3.3) and (3.10), we have \(\Vert w_{\lambda_{0} }\Vert _{L^{q}(\Gamma_{\lambda})}=0\), and therefore \(\Gamma^{\lambda}\) must be of measure zero. Hence, for these values of \(\lambda> \lambda_{0}\), we have (3.9). This contradicts the definition of \(\lambda_{0}\). Therefore (3.8) must hold.
$$\lambda_{0}=\operatorname{sup} \bigl\{ \lambda\mid w_{\rho}(x)\geq0, \rho\leq \lambda, \forall x\in\hat{\Sigma}_{\rho}\bigr\} . $$
$$ \lambda_{0}=+\infty. $$
(3.7)
$$ u^{\lambda_{0} }_{i}(x)\equiv u_{i}(x),\quad\mbox{a.e. } \forall x\in\hat{\Sigma}_{\lambda_{0}}. $$
(3.8)
$$ u^{\lambda}_{i}(x)\geq u_{i}(x),\quad \mbox{a.e. on } \hat{\Sigma}_{\lambda}. $$
(3.9)
$$ c \biggl\{ { \int_{{\Gamma_{\lambda}}}} \vert u\vert ^{\frac{n(\alpha_{i}+\beta _{i}-1)}{\alpha}}(y) \,dy \biggr\} ^{\frac{\alpha}{n}} \leq\frac{1}{2}, $$
(3.10)
The proof of the inequality (3.10) is similar to the argument of the inequality (2.26) in Section 2 and the proof is standard.
By (3.8), we derive that \(u_{i}(x) = 0\) on the plane \(x_{n} = 2\lambda_{0}\), the symmetric image of the boundary \(\partial R^{n}_{+}\) with respect to the plane \(\hat{T}_{\lambda_{0}}\). This contradicts our assumption \(u_{i}(x) > 0\) in \(R^{n}_{+}\). Therefore, (3.7) must be valid. Now we have proved that the positive solution of (1.6) is monotone increasing with respect to \(x_{n}\), and this contradicts \(u^{\alpha_{i}+\beta _{i}-1}\in L^{\frac{n}{\alpha}}(R^{n}_{+}) \). Therefore the positive solutions of (1.6) do not exist.
This completes the proof of Theorem 1.2.
3.2 The proof of Theorem 1.3
In this section, we will use a proper Kelvin type transforms and derive the nonexistence of positive solutions for (1.6) in \(R^{n}_{+}\) under much weaker conditions, i.e. the solution u of (1.6) is only locally integrable and locally bounded.
With no explicit global integrability assumptions on the solution u, we cannot directly carry out the method of moving planes on u. To overcome this difficulty, we employ Kelvin type transforms.
For \(z^{0} \in R^{n}_{+}\), let be the Kelvin transform of \(u_{i}(x) \) centered at \(z^{0}\).
$$ \bar{u}_{i}(x)=\frac{1}{\vert x-z^{0}\vert ^{n-\alpha}}u_{i} \biggl(\frac {x-z^{0}}{\vert x-z^{0}\vert ^{2}}+z^{0} \biggr) $$
(3.11)
Through a straightforward calculation, we have \(\bar{u}_{i}(x) =\int_{R^{n}_{+}}G(x,y)\frac{y_{n}^{\gamma}\bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta_{i}}_{2}(y)}{\vert y-z^{0}\vert ^{\delta }}\,dy\), \(\forall x\in{R^{n}_{+}}\backslash{B_{\epsilon}(z^{0})}\), where \(\delta=n+\alpha+2\gamma-(n-\alpha)(\alpha_{i}+\beta_{i})\), \(\epsilon>0\), \(i=1,2\).
Proof of Theorem 1.3 in the subcritical case
\(1<\alpha_{i}+\beta_{i}<\frac{n+\alpha+2\gamma}{n-\alpha} \): where \(\delta=n+\alpha+2\gamma-(n-\alpha)(\alpha_{i}+\beta _{i})>0\), \(\epsilon>0\).
$$ \bar{u}_{i}(x) = \int _{R^{n}_{+}}G(x,y)\frac{y_{n}^{\gamma}\bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta_{i}}_{2}(y)}{\vert y-z^{0}\vert ^{\delta}}\,dy,\quad \forall x\in {R^{n}_{+}}\backslash{B_{\epsilon}\bigl(z^{0} \bigr)}, $$
(3.12)
This specific proof is the same as the proof of Theorem 1.1 and we omit here. □
Remark
Proof of Theorem 1.3 in the critical case
\(1<\alpha_{i}+\beta_{i}=\frac{n+\alpha+2\gamma}{n-\alpha} \): By the Kelvin transform of \(u_{i}(x)\) we derive If \(u(x)\) is a solution of (3.13), then \(\bar{u} (x)\) is also a solution of (3.14). Therefore, by our assumption \(\vert u\vert \in L^{\frac{n(\alpha_{i}+\beta_{i}-1)}{\alpha}}_{\mathrm{loc}}(R^{n}_{+})\), we derive \(y^{\gamma}_{n} \vert u\vert ^{\alpha_{i}+\beta_{i}-1}\in L^{\frac{n}{\alpha }}_{\mathrm{loc}}(R^{n}_{+})\). then where Ω̂ is the image of Ω about the Kelvin transform. Now we consider two possibilities.
$$ u_{i}(x)= \int_{R^{n}_{+}}G(x,y) y_{n}^{\gamma} u^{\alpha_{i}}_{1}(y) u^{\beta_{i}}_{2}(y)\,dy. $$
(3.13)
$$ \bar{u}_{i}(x)= \int_{R^{n}_{+}}G(x,y) y_{n}^{\gamma}\bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta_{i}}_{2}(y)\,dy. $$
(3.14)
$$ \int_{\hat{\Omega}} \bigl[y^{\gamma}_{n} \bar{u}^{\alpha_{i}+\beta_{i}-1}(y) \bigr]^{\frac{n}{\alpha}}\,dy= \int_{ \Omega} \bigl[y^{\gamma}_{n} u^{\alpha _{i}+\beta_{i}-1}(y) \bigr]^{\frac{n}{\alpha}}\,dy< \infty, $$
(3.15)
Possibility 1. If there is a \(z^{0}=(z^{0}_{1}, \ldots ,z^{0}_{n-1}, 0 )\in\partial R^{n}_{+} \) such that \(\bar{u}_{i}(x)\) is bounded near \(z^{0}\), then by (3.11), we obtain And we further deduce Since \(\alpha_{i}+\beta_{i}=\frac{n+\alpha+2\gamma}{n-\alpha }>\frac{n}{n-\alpha}\) and \(\vert u\vert \in L^{\frac{n(\alpha_{i}+\beta _{i}-1)}{\alpha}}_{\mathrm{loc}}(R^{n}_{+}) \), together with (3.16), we have In this situation, we still carry on the moving planes on u. Going through exactly the same arguments as in the proof of Theorem 1.2, we obtain the nonexistence of positive solutions for (3.13).
$$u_{i}(y)=\frac{1}{\vert y-z^{0}\vert ^{n-\alpha}}\bar{u}_{i} \biggl( \frac {y-z^{0}}{\vert y-z^{0}\vert ^{2}}+z^{0} \biggr) . $$
$$ u_{i}(y)=O \biggl(\frac{1}{\vert y\vert ^{n-\alpha}} \biggr), \quad \mbox{as } \vert y\vert \rightarrow\infty. $$
(3.16)
$$ \int_{ R^{n}_{+}} u^{\frac{n(\alpha_{i}+\beta_{i}-1)}{\alpha}}(y) \,dy\leq c \int_{ R^{n}_{+}}\frac{1}{\vert y\vert ^{\frac{n(\alpha_{i}+\beta _{i}-1)(n-\alpha)}{\alpha}}}\,dy< \infty. $$
(3.17)
Possibility 2. For all \(z^{0}=(z^{0}_{1}, \ldots ,z^{0}_{n-1}, 0 )\in\partial R^{n}_{+} \), \(\bar{u}_{i}(x)\) is unbounded near \(z^{0}\), we will carry out the method of moving planes on \(\bar{u}(x)\) in \(R^{n-1}\) to prove that it is rotationally symmetric about the line passing through \(z^{0}\) and parallel to the \(x_{n}\)-axis. From this, we will deduce that u is independent of the first \(n-1\) variables \(x_{1}, \ldots,x_{n-1}\). That is \(u = u(x_{n})\), and we will derive a contradiction with the finiteness of \(\int_{R^{n}_{+}}G(x,y)y^{\gamma}_{n} u^{\alpha_{i}}_{1}(y)u^{\beta _{i}}_{2}(y)\,dy\).
For a given real number λ, the notations such as \(\Sigma_{\lambda}\), \(T_{\lambda}\) are the same as the ones in Section 2. By (2.12), obviously we have By elementary calculation we derive We will move the plane \({T _{\lambda}}\) along the direction of the \(x_{1}\)-axis to show that the solution is rotationally symmetric about the line passing through \(z^{0}\) and parallel to the \(x_{n}\)-axis. The proof is the same as the proof of \(c_{i}=0\) in Section 2, in fact we only need to apply arguments of the inequality (2.13) to equation (3.18). Similarly, we derive that \(u_{i}=u_{i}(x_{n})\) and any positive solution u of (3.13) must be \(u(x)\equiv0\). This implies that there is no positive solution of (3.13) in the critical case.
$$\begin{aligned} &\bar{u}_{i}(x) = \int_{\Sigma_{\lambda}}G(x,y) y_{n}^{\gamma}\bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta_{i}}_{2}(y)\,dy + \int_{\Sigma_{\lambda}}G \bigl(x^{\lambda},y \bigr)y_{n}^{\gamma} \bar{u}^{\alpha_{i}}_{1} \bigl(y^{\lambda} \bigr) \bar{u}^{\beta_{i}}_{2} \bigl(y^{\lambda} \bigr)\,dy, \\ &\bar{u}^{\lambda}_{i}(x) = \int_{\Sigma_{\lambda}}G \bigl(x^{\lambda},y \bigr) y_{n}^{\gamma} \bar{u}^{\alpha_{i}}_{1}(y)\bar{u}^{\beta_{i}}_{2}(y)\,dy + \int_{\Sigma_{\lambda}}G(x,y)y_{n}^{\gamma}\bar{u}^{\alpha _{i}}_{1} \bigl(y^{\lambda} \bigr)\bar{u}^{\beta_{i}}_{2} \bigl(y^{\lambda} \bigr)\,dy . \end{aligned}$$
$$\begin{aligned} \bar{u}_{i}(x)- \bar{u}^{\lambda}_{i}(x) = \int_{\Sigma_{\lambda}} \bigl[G(x,y)-G \bigl(x^{\lambda},y \bigr) \bigr]y_{n}^{\gamma } \bigl[\bar{u}^{\alpha_{i}}_{1}(y) \bar{u}^{\beta_{i}}_{2}(y)-\bar{u}^{\alpha _{i}}_{1} \bigl(y^{\lambda} \bigr)\bar{u}^{\beta_{i}}_{2} \bigl(y^{\lambda} \bigr) \bigr]\,dy. \end{aligned}$$
(3.18)
This completes the proof of Theorem 1.3. □
Acknowledgements
This work is partially supported by NSFC (No. U1304101 and No. 11671121) and HASTIT (No. 15HASTIT012).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
ZD participated in the method of moving plane studies in the paper and drafted the manuscript; LC carried out the Liouville type theorem, and PW carried out the evaluation of inequalities. All authors read and approved the final manuscript.