# Entire Solutions of an Integral Equation inR5

Authors: Xin Feng; Xingwang Xu
Journal: ISRN Mathematical Analysis
 (2013)

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## Abstract

We will study the entire positiveC0 solution of the geometrically and analytically interesting integral equation: u(x)=1/C5∫R5‍|x-y|u-q(y)dy with 0<q in R5. We will show that only when q=11, there are positive entire solutions which are given by the closed form u(x)=c(1+|x|2)1/2 up to dilation and translation. The paper consists of two parts. The first part is devoted to showing that q must be equal to 11 if there exists a positive entire solution to the integral equation. The tool to reach this conclusion is the well-known Pohozev identity. The amazing cancelation occurred in Pohozev’s identity helps us to conclude the claim. It is this exponent which makes the moving sphere method work. In the second part, as normal, we adopt the moving sphere method based on the integral form to solve the integral equation.


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## Body

## 1. Introduction

In this paper, we will study a very special type of the integral equation. From an analytical point of view, such equation is interesting to be studied. We should point out that even for radius case, the analysis of the equation has been already difficult.The method of moving planes has become a very powerful tool in the study of nonlinear elliptic equations; see Alexandrov [1], Serrin [2], Gidas et al. [3], and others. The moving plane method can be applied to prove the radial symmetry of solutions, and then one only needs to classify radial solutions. The method of moving planes in the integral form has been developed by Chen and Li [4]. This technique requires not only to prove that the solutions are radially symmetric, but also to take care of a possible singularity at origin. We note that in [5], Li and Zhang have given different proofs to some previous established Liouville type theorems based on the method of moving sphere. It is this method that we applied to the integral equation to capture the solutions directly. As usual, in order for such a method to work, Pohozev’s identity is a must. In our argument, we derive and use this powerful identity to conclude that the negative exponent q=11. Then it is the method of moving spheres in the integral form which helps us to deduce the exact solution to the integral equation in this paper, instead of only getting radius solutions or proving radial symmetry of solutions.The organization of the paper is as follows. In Section2, we prove that q=11. In this part, we use integration by parts to derive Pohozev’s identity in the form Δ3u=u-q. It is easy to observe that if u solves our integral equation, then u also solves the differential equation Δ3u=u-q. Thus we can put the integral form into Pohozev’s identity to calculate every term. It turns out that the boundary integral term tends to zero which forces q=11 unless u≡+∞ which is not allowed.In the third section, we adopt the moving sphere method in the integral form to solve the integral equation withq=11.


## 2.q=11

The main purpose in this section is to show the following theorem. It is this exact exponent that makes the moving sphere workable for the integral equation. Notice that it is not hard to see that everyC0 positive solution of the integral equation is C6. Thus in the following, we mainly assume that the solution is in C6(R5).Theorem 1.
Suppose thatu is a C6 entire positive solution of the integral equation
(1)u(x)=1C5∫R5‍|x-y|u-q(y)dyinR5,whereC5=24ω5
with q>0, then q=11.Lemma 2.
Ifu(x) satisfies (1), then Δ3u(x)=u-q(x).Proof.
By direct calculation, we have(2)Δxu(x)=1C5∫R5‍Δx(|x-y|)u-q(y)dyΔxu(x)=1C5∫R5‍4|x-y|u-q(y)dy;Δx2u(x)=1C5∫R5‍Δx(4|x-y|)u-q(y)dyΔx2u(x)=1C5∫R5‍-8|x-y|3u-q(y)dy.
SinceK(r)=(1/-3ω5)r-3 is the fundamental solution of Laplace equation in R5, it follows that
(3)Δx3u(x)=-8C5(-3ω5)∫R5‍Δx(1-3ω51|x-y|3)u-q(y)dy=24ω5C5u-q(x)=u-q(x).
Thus the lemma is true.Lemma 3.
Ifu>0 is the solution of the equation
(4)Δx3u(x)=Q(x)u-q(x)inR5,
then the following Pohozev identity holds:
(5)(12+51-q)∫Br(0)‍Qu1-qdx+P(u,r)=1q-1∫∂Br(0)‍rQu1-qdδ-11-q∫Br(0)‍(x·∇Q)u1-qdx,
where
(6)P(u,r)=∫∂Br(0)‍[r∂u∂r∂Δ2u∂r-Δ2u∂∂r(r∂u∂r)+12r(∂Δu∂r)2-12Δu∂∂r(r∂Δu∂r)+12rΔuΔ2u+12∂u∂rΔ2u-12u∂Δ2u∂r]dδ.Proof.
Multiplyingx·∇u on the both sides of (4) and integrating over the ball Br(0), one has
(7)0=∫Br(0)‍(x·∇u)[Δx3u(x)-Q(x)u-q(x)]dx=I2-I1,
where I1=∫Br(0)‍(x·∇u)Q(x)u-q(x)dx; I2=∫Br(0)‍(x·∇u)Δx3u(x).
First, rewriteI1 as
(8)I1=∫Br(0)‍[x·∇(u1-q(x)1-q)]Q(x)dx.
Using integration by parts, we have(9)I1=11-q∫∂Br(0)‍u1-q(x)Q(x)rdδ-11-q∫Br(0)‍(5Q+x·∇Q)u1-q(x)dx.
Now, the Second Green identity implies that(10)I2=∫∂Br(0)‍[(x·∇u)∂Δ2u∂r-(Δ2u)∂(x·∇u)∂r]dδ+∫Br(0)‍Δ(x·∇u)Δ2udx=∫∂Br(0)‍[r∂u∂r∂Δ2u∂r-(Δ2u)∂∂r(r∂u∂r)]dδ+2I3+I4,
where I3=∫Br(0)‍ΔuΔ2udx; I4=∫Br(0)‍x·∇(Δu)Δ2udx.
As forI4, since Δ(x·∇u)=2Δu+x·∇Δu and by the Second Green identity again, we obtain
(11)I4=∫Br(0)‍Δ[x·∇Δu]Δudx+∫∂Br(0)‍[x·∇Δu∂Δu∂r-∂(x·∇Δu)∂rΔu]dδ=∫Br(0)‍[2Δ2u+x·∇Δ2u]Δudx+∫∂Br(0)‍[r(∂Δu∂r)2-∂∂r(r∂Δu∂r)Δu]dδ.
Hence,(12)I4=2I3+∫Br(0)‍x·∇Δ2uΔudx+∫∂Br(0)‍[r(∂Δu∂r)2-∂∂r(r∂Δu∂r)Δu]dδ.
Observe that(13)I4+∫Br(0)‍x·∇Δ2uΔudx=∫Br(0)‍x·∇(ΔuΔ2u)dx=∫∂Br(0)‍rΔuΔ2udδ-5∫Br(0)‍ΔuΔ2udx=∫∂Br(0)‍rΔuΔ2udδ-5I3;
that is,
(14)I4=∫∂Br(0)‍rΔuΔ2udδ-5I3.
Combining (12) and (14), we have
(15)I4=-32I3+12∫∂Br(0)‍[r(∂Δu∂r)2-∂∂r(r∂Δu∂r)Δu+rΔuΔ2u]dδ.
Thus(16)I2=12I3+∫∂Br(0)‍[r∂u∂r∂Δ2u∂r-Δ2u∂∂r(r∂u∂r)+12r(∂Δu∂r)2-12Δu∂∂r(r∂Δu∂r)+12rΔuΔ2u]dδ.
To deal withI3, we observe that if multiplying u on both sides of (4) and integrating over the ball Br(0), one has
(17)0=∫Br(0)‍u[Δ3u-Q(x)u-q]dx.
Thus, with the help of the Second Green identity, one obtains
(18)∫Br(0)‍Q(x)u1-qdx=∫Br(0)‍uΔ3udx=∫Br(0)‍ΔuΔ2udx+∫∂Br(0)‍[u∂Δ2u∂r-∂u∂rΔ2u]dδ.
Therefore, we get(19)I3=∫Br(0)‍ΔuΔ2udx=∫Br(0)‍Q(x)u1-qdx+∫∂Br(0)‍[∂u∂rΔ2u-u∂Δ2u∂r]dδ.
Put it into (16) to conclude that
(20)I2=12∫Br(0)‍Qu1-qdx+∫∂Br(0)‍[r∂u∂r∂Δ2u∂r-Δ2u∂∂r(r∂u∂r)+12r(∂Δu∂r)2-12Δu∂∂r(r∂Δu∂r)+12rΔuΔ2u+12∂u∂rΔ2u-12u∂Δ2u∂r]dδ.
PutI2 and I1 into the equation I2-I1=0 and rearrange the terms; then we get the desired identity.
Thus the lemma holds.Lemma 4.
Ifu is an entire solution of the integral equation (1), then q=11.Proof.
Defined=(1/C5)∫R5‍u-q(y)dy, d1=(1/C5)∫R5‍|y|u-q(y)dy. Then it follows that
(21)u(x)=1C5∫R5‍|x-y|u-q(y)dy≤1C5∫R5‍|x|u-q(y)dy+1C5∫R5‍|y|u-q(y)dy=|x|d+d1.
If0<q≤6, then
(22)d1=u(0)=1C5∫R5‍|y|u-q(y)dy≥1C5∫R5‍|y|(|y|d+d1)-q(y)dy=ω5C5∫0+∞‍r5(rd+d1)-qdr=∞,
which contradicts the assumption that u is an entire solution. Therefore, we only consider the case q>6.
Rewrite the equation as(23)u(x)=d|x|+β1(x),
where
(24)β1(x)=1C5∫R5‍[|x-y|-|x|]u-q(y)dy.
Clearly, we have(25)|β1(x)|≤1C5∫R5‍|y|u-q(y)dy=d1.
And, by (1), one has
(26)∂u∂r=1C5∫R5‍x·(x-y)r|x-y|u-q(y)dy=1C5∫R5‍u-q(y)dy+1C5∫R5‍[(x-y)|x-y|·xr-1]u-q(y)dy:=d+β2(x)r,
where
(27)β2(x)=1C5∫R5‍(x-y)·x-r|x-y||x-y|u-q(y)dy.
Note thatβ2(x) can be controlled as follows:
(28)|β2(x)|=1C5|∫R5‍[|x-y|-|x|]|x-y|+y·(x-y)|x-y|u-q(y)dy|≤2C5∫R5‍|y|u-q(y)dy=2d1.
Clearly, with the definition ofβ2(x), the following holds true:
(29)r∂u∂r=dr+β2(x).
Then differentiate it to get(30)∂∂r(r∂u∂r)=d+1C5∫R5‍[(2x-y)|x-y|2-x·(x-y)(x-y)|x-y|3-x|x|]·xru-q(y)dy:=d+β3(x),
where
(31)β3(x)=1C5∫R5‍[(|x||x-y|-1)-x·(x-y)y·(x-y)r|x-y|3]u-q(y)dy.
One observes that(32)|x||β3(x)|=1C5|∫R5‍[|x|(|x||x-y|-1)-x·(x-y)y·(x-y)|x-y|3]×u-q(y)dy∫R5‍[|x|(|x||x-y|-1)-x·(x-y)y·(x-y)|x-y|3]∫R5‍|≤2C5∫R5‍|x||y||x-y|u-q(y)dy≤M3,
for a constant independent of x and for |x| sufficiently large. Here we have used the assumption that q>6 and u(x)=d|x|+β1(x), where β1(x) is bounded by a constant. The integral over R5 can be controlled as follows:
(33)∫R5‍|x||y||x-y|u-q(y)dy=(∫A1‍+∫A2‍)|x||y||x-y|u-q(y)dy∫R5‍|x||y||x-y|u-q(y)dy=S1+S2,
where A1={y∈R5:|x-y|≥|x|/2}, A2={y∈R5:|x-y|≤|x|/2}.
Then(34)S1=∫A1‍|x||y||x-y|u-q(y)dy≤2∫A1‍|y|u-q(y)dy≤2d1.
As forS2, y∈A2 implies that |y|≤|x|+|x-y|≤(3/2)|x| and |y|≥|x|-|x-y|≥(1/2)|x|.
Therefore(35)S2=∫A2‍|x||y||x-y|u-q(y)dy≤[min|x|/2≤|y|≤3|x|/2u(y)]-q3|x|22[∫A2‍1|x-y|dy]=[min|x|/2≤|y|≤3|x|/2u(y)]-q3|x|22[ω5∫0|x|/2‍r3dr]≤[min|x|/2≤|y|≤3|x|/2u(y)]-qω53|x|627.
Sinceu(x)=O(|x|) and q>6, so I2 is bounded for |x|≥R0 when R0 is large enough.
It follows that|x||β3(x)| is bounded for |x|≥R0 when R0 is large enough.
Now we deal with the higher order derivatives. First(36)Δu(x)=4C5∫R5‍1|x-y|u-q(y)dy=4|x|d+β4(x),
where
(37)β4(x)=4C5∫R5‍[1|x-y|-1|x|]u-q(y)dy.
For a constantM4 independent of x, when |x| is large enough, it is easy to see that
(38)|x|2|β4(x)|=4C5|∫R5‍|x|[|x|-|x-y||x||x-y|]u-q(y)dy||x|2|β4(x)|≤4C5∫R5‍|x||y||x-y|u-q(y)dy|x|2|β4(x)|≤M4,
by the estimation in β3.
Then, by computation, one gets(39)∂Δu∂r=4C5∫R5‍[-(x-y)·x|x-y|3|x|]u-q(y)dy=-4|x|2d+4C5∫R5‍[-(x-y)·x|x-y|3|x|+1|x|2]u-q(y)dy=-4|x|2d+β5(x)r,
where
(40)β5(x)=4C5∫R5‍[-(x-y)·x|x-y|3+1|x|]u-q(y)dy.
And as usual, we observe that(41)|x|2|β5(x)|=4C5|∫R5‍[-(x-y)·x|x|2|x-y|3+|x|]u-q(y)dy|=4C5|∫R5‍|x||x-y|3-x·(x-y)|x|2|x-y|3u-q(y)dy|=4C5|∫R5‍|x|[|x-y|(x-y)(x-y)-|x|x·(x-y)]|x-y|3×u-q(y)dy∫R5‍|x|[|x-y|(x-y)(x-y)-|x|x·(x-y)]|x-y|3∫R5‍|=4C5|∫R5‍|x|[(|x-y|-|x|)x·(x-y)-|x-y|y·(x-y)]|x-y|3×u-q(y)dy∫R5‍|x|[(|x-y|-|x|)x·(x-y)-|x-y|y·(x-y)]|x-y|3|≤4C5∫R5‍[|x|2|y||x-y|2+|x||y||x-y|]u-q(y)dy.
Similarly, we have the following estimation:(42)∫R5‍|x|2|y||x-y|2u-q(y)dy=(∫A1‍+∫A2‍)|x|2|y||x-y|2u-q(y)dy=S3+S4,
where A1={y∈R5:|x-y|≥|x|/2}, A2={y∈R5:|x-y|≤|x|/2}.
Then(43)S3=∫A1‍|x|2|y||x-y|2u-q(y)dy≤4∫A1‍|y|u-q(y)dy≤4d1.
As forS4, y∈A2 implies that |y|≤|x|+|x-y|≤(3/2)|x| and |y|≥|x|-|x-y|≥(1/2)|x|.
Therefore(44)S4=∫A2‍|x|2|y||x-y|2u-q(y)dy≤[min|x|/2≤|y|≤3|x|/2u(y)]-q3|x|32[∫A2‍1|x-y|2dy]=[min|x|/2≤|y|≤3|x|/2u(y)]-q3|x|32[ω5∫0|x|/2‍r2dr]≤[min|x|/2≤|y|≤3|x|/2u(y)]-qω5|x|624.
Sinceu(x)=O(|x|) and q>6, so S4 is bounded for |x|≥R0 when R0 is large enough.
It follows that|x|2|β5(x)| is bounded for |x|≥R0 when R0 is large enough.
Sincer(∂Δu/∂r)=-(4/|x|)d+β5(x), we obtain
(45)∂∂r(r∂Δu∂r)=4|x|2d+β6(x),
where
(46)β6(x)=∂∂rβ5(x)=4C5∫R5‍∇[-(x-y)·x|x-y|3+1|x|]·x|x|u-q(y)dy=4C5∫R5‍[(2x-y)|x-y|2-3(x-y)·x(x-y)|x-y|5-x|x|3]·x|x|u-q(y)dy=4C5∫R5‍{1|x||x-y|5[3|x-y|2(x-y)·x+3y·(x-y)(x-y)·x-x·(x-y)|x-y|2-|x|2|x-y|2]-1|x|21|x||x-y|5[3|x-y|2(x-y)·x}×u-q(y)dy=4C5∫R5‍[2(x-y)·x|x||x-y|3+3y·(x-y)(x-y)·x|x||x-y|5-|x||x-y|3-1|x|2]u-q(y)dy.
As before, we can estimateβ6(x) as
(47)|x|3|β6(x)|=4C5|∫R5‍[2(x-y)·x|x|2|x-y|3+3y·(x-y)(x-y)·x|x|2|x-y|5-|x|4|x-y|3-|x|]u-q(y)dy|=4C5|∫R5‍[((x-y)·x|x|2|x-y|3-|x|4|x-y|3)+((x-y)·x|x|2|x-y|3-|x|)+3y·(x-y)(x-y)·x|x|2|x-y|5]u-q(y)dy|≤4C5|∫R5‍[-|x|2x·y|x-y|3+((|x-y|3)-1(|x||x-y|y|x|2x·(x-y)-|x||x-y|x·(x-y)+|x||x-y|y·(x-y))(|x-y|3)-1)+3y·(x-y)(x-y)·x|x|2|x-y|5]u-q(y)dy|≤4C5∫R5‍|x|3|y||x-y|3u-q(y)dy+4C5∫R5‍|x|2|y||x-y|2u-q(y)dy+4C5∫R5‍|x||y||x-y|u-q(y)dy+12C5∫R5‍|x|3|y||x-y|3u-q(y)dy≤M6,
for a constant independent of x and for |x| sufficiently large. We have proved that the first integral is bounded in the previous estimation. The second integral over R5 can be controlled as follows:
(48)∫R5‍|x|3|y||x-y|3u-q(y)dy=(∫A1‍+∫A2‍)|x|3|y||x-y|3u-q(y)dy=S5+S6,
where A1={y∈R5:|x-y|≥|x|/2}, A2={y∈R5:|x-y|≤|x|/2}.
Then(49)S5=∫A1‍|x|3|y||x-y|3u-q(y)dy≤8∫A1‍|y|u-q(y)dy≤8d1.
As forS6, y∈A2 implies that |y|≤|x|+|x-y|≤(3/2)|x| and |y|≥|x|-|x-y|≥(1/2)|x|.
Therefore(50)S6=∫A2‍|x|3|y||x-y|3u-q(y)dy≤[min|x|/2≤|y|≤3|x|/2u(y)]-q3|x|42[∫A2‍1|x-y|3dy]=[min|x|/2≤|y|≤3|x|/2u(y)]-q3|x|42[ω5∫0|x|/2‍rdr]≤[min|x|/2≤|y|≤3|x|/2u(y)]-qω53|x|624.
Sinceu(x)=O(|x|) and q>6, so S6 is bounded for |x|≥R0 when R0 is large enough.
It follows that|x|3|β6(x)| is bounded for |x|≥R0 when R0 is large enough.
Once again, calculation shows that(51)Δ2u(x)=1C5∫R5‍-8|x-y|3u-q(y)dy=-8|x|3d+β7(x),
where
(52)β7(x)=8C5∫R5‍[1|x|3-1|x-y|3]u-q(y)dy.
We still need to controlβ7(x) which can be done as follows:
(53)|x|4|β7(x)|=8C5|∫R5‍[|x|-|x|4|x-y|3]u-q(y)dy|=8C5|∫R5‍|x||x-y|(|x|2-2x·y+|y|2)-|x|4|x-y|3u-q(y)dy|=8C5|∫R5‍|x|3[|x-y|-|x|]+|x||x-y|(-2x·y+|y|2)|x-y|3×u-q(y)dy∫R5‍|x|3[|x-y|-|x|]+|x||x-y|(-2x·y+|y|2)|x-y|3|≤8C5∫R5‍|x|3|y||x-y|3u-q(y)dy+8C5∫R5‍2|x|2|y||x-y|2u-q(y)dy+8C5∫R5‍|x||y|2|x-y|2u-q(y)dy≤M7,
for a constant independent of x and for |x| sufficiently large. Now, the third integral over R5 can be controlled as follows:
(54)∫R5‍|x||y|2|x-y|2u-q(y)dy=∫R5‍|x||y||x-y||y||x-y|u-q(y)dy≤∫R5‍|x||y||x-y|[|x-y|+|x||x-y|]u-q(y)dy≤∫R5‍|x||y||x-y|u-q(y)dy+∫R5‍|x|2|y||x-y|2u-q(y)dy.
The two integrals are bounded as we have shown before. It follows that|x|4|β7(x)| is bounded for |x|≥R0 when R0 is large enough.
Finally, by computation, one obtains(55)∂Δ2u∂r=24C5∫R5‍(x-y)·x|x-y|5|x|u-q(y)dy=24|x|4d+β8(x),
where
(56)β8(x)=24C5∫R5‍[(x-y)·x|x-y|5|x|-1|x|4]u-q(y)dy.
The estimate on β8(x) can proceed as follows:
(57)|x|5|β8(x)|=24C5|∫R5‍[(x-y)·x|x|4|x-y|5-|x|]u-q(y)dy|=24C5|∫R5‍|x|4(x-y)·x-|x||x-y|5|x-y|5u-q(y)dy|=24C5×|∫R5((|x-y|3y·(x-y))-1([|x|3-|x-y|3]|x|x·(x-y)+|x||x-y|3y·(x-y)[|x|3-|x-y|3]|x|x)(|x-y|3y·(x-y))-1)×u-q(y)dy∫R5‍|≤24C5|∫R5‍[|x|3-|x-y|(|x|2-2x·y+|y|2)]|x|2|x-y|4×u-q(y)dy∫R5‍[|x|3-|x-y|(|x|2-2x·y+|y|2)]|x|2|x-y|4|+24C5∫R5‍|x||y||x-y|u-q(y)dy≤24C5∫R5‍[|x|4|y||x-y|4+2|x|3|y||x-y|3+|x|2|y|2|x-y|3]u-q(y)dy+24C5∫R5‍|x||y||x-y|u-q(y)dy≤M8,
for a constant independent of x and for |x| sufficiently large. As before, the aforementioned first integral can be controlled as follows:
(58)∫R5‍|x|4|y||x-y|4u-q(y)dy=(∫A1‍+∫A2‍)|x|4|y||x-y|4u-q(y)dy=S7+S8,
where A1={y∈R5:|x-y|≥|x|/2}, A2={y∈R5:|x-y|≤|x|/2}.
Then(59)S7=∫A1‍|x|4|y||x-y|4u-q(y)dy≤16∫A1‍|y|u-q(y)dy≤16d1.
As forS8, y∈A2 implies that |y|≤|x|+|x-y|≤(3/2)|x| and |y|≥|x|-|x-y|≥(1/2)|x|.
Therefore(60)S8=∫A2‍|x|4|y||x-y|4u-q(y)dy≤[min|x|/2≤|y|≤3|x|/2u(y)]-q3|x|52[∫A2‍1dy]=[min|x|/2≤|y|≤3|x|/2u(y)]-q3|x|52[ω5∫0|x|/2‍1dr]≤[min|x|/2≤|y|≤3|x|/2u(y)]-qω53|x|622,
which is bounded, since u(x)=O(|x|) and q>6, so I8 is bounded for |x|≥R0 when R0 is large enough.
Besides,(61)∫R5‍|x|2|y|2|x-y|3u-q(y)dy=∫R5‍|x|2|y||x-y|2|y||x-y|u-q(y)dy≤∫R5‍|x|2|y||x-y|2u-q(y)dy+∫R5‍|x|3|y||x-y|3u-q(y)dy,
and the two integrals have been shown to be bounded before.
It follows that|x||β8(x)| is bounded for |x|≥R0 when R0 is large enough.
Combining these estimates, we can see that(1)
|x|4β2(x)β8(x)→0 as |x|→∞;(2)
|x|4β3(x)β7(x)→0 as |x|→∞;(3)
|x|3β5(x)2→0 as |x|→∞;(4)
|x|4β4(x)β6(x)→0 as |x|→∞;(5)
|x|5β4(x)β7(x)→0 as |x|→∞;(6)
|x|3β2(x)β7(x)→0 as |x|→∞;(7)
|x|4β1(x)β8(x)→0 as |x|→∞.
Also we have(62)[r∂u∂r∂Δ2u∂r-Δ2u∂∂r(r∂u∂r)+12r(∂Δu∂r)2-12Δu∂∂r(r∂Δu∂r)+12rΔuΔ2u+12∂u∂rΔ2u-12u∂Δ2u∂r]=r[d+β2(x)r][24|x|4d+β8(x)]-[d+β3(x)][-8|x|3d+β7(x)]+12r[-4|x|2d+β5(x)r]2-12[4|x|2d+β6(x)][4|x|d+β4(x)]+12r[4|x|d+β4(x)][-8|x|3d+β7(x)]+12[d+β2(x)r][-8|x|3d+β7(x)]-12[d|x|+β1(x)][24|x|4d+β8(x)]=d[-12|x|4β1(x)+20|x|4β2(x)+8|x|3β3(x)+-6|x|2β4(x)+-4|x|2β5(x)+-2|x|β6(x)+32β7(x)+|x|2β8(x)]+[12β2(x)β8(x)-β3(x)β7(x)+12rβ5(x)2-12β4(x)β6(x)+12rβ4(x)β7(x)+1rβ2(x)β7(x)-12β1(x)β8(x)].
Observe that(63)|x|4[β2(x)β8(x)-β3(x)β7(x)+12rβ5(x)2-12β4(x)β6(x)+12rβ4(x)β7(x)+1rβ2(x)β7(x)-12β1(x)β8(x)]⟶0as|x|⟶∞.
The coefficient of d is given by
(64)[-12|x|4β1(x)+20|x|4β2(x)+8|x|3β3(x)+-6|x|2β4(x)+-4|x|2β5(x)+-2|x|β6(x)+32β7(x)+|x|2β8(x)]=[4|x|4(-3β1(x)+5β2(x))+8|x|3β3(x)-2|x|2(3β4(x)+2β5(x))-2|x|β6(x)+32β7(x)+|x|2β8(x)]=1C5∫R5‍[(-12|x-y||x|4+20(x-y)·x|x|4|x-y|-81|x|3)+(81|x|2|x-y|-81|x|3-8x·(x-y)y·(x-y)|x|4|x-y|3)+(-241|x|2|x-y|+81|x|3+16(x-y)·x|x-y|3|x|2)+(-16(x-y)·x|x-y|3|x|2-24y·(x-y)(x-y)·x|x|2|x-y|5+81|x-y|3+81|x|3(x-y)·x|x-y|3|x|2)+(121|x|3-121|x-y|3)+(12(x-y)·x|x-y|5-121|x|3)]u-q(y)dy=1C5∫R5‍{4|x|4|x-y|5[-3|x-y|6+5(x-y)·x|x-y|4-4|x|2|x-y|4-2x·(x-y)y·(x-y)|x-y|2-6y·(x-y)(x-y)·x|x|2-|x|4|x-y|2+3(x-y)·x|x|4]4|x|4|x-y|5}×u-q(y)dy=1C5∫R5‍{4|x|4|x-y|5S}u-q(y)dy,
where
(65)S=[-3|x-y|6+5(x-y)·x|x-y|4-4|x|2|x-y|4-2x·(x-y)y·(x-y)|x-y|2-6y·(x-y)(x-y)·x|x|2-|x|4|x-y|2+3(x-y)·x|x|4]=[2x·(x-y)|x-y|4+3y·(x-y)|x-y|4-4|x|2|x-y|2x·(x-y)+4|x|2|x-y|2y·(x-y)-2x·(x-y)y·(x-y)|x-y|2-6y·(x-y)(x-y)·x|x|2+2|x|4x·(x-y)+|x|4y·(x-y)]=x·(x-y)[2|x-y|4-4|x|2|x-y|2+2|x|4]+y·(x-y)[3|x-y|4+4|x|2|x-y|2-2x·(x-y)|x-y|2-6x·(x-y)|x|2+|x|4].
And directly calculation gives(66)[3|x-y|4+4|x|2|x-y|2-2x·(x-y)|x-y|2-6x·(x-y)|x|2+|x|4]=x·(x-y)|x-y|2+4|x|2|x-y|2-6x·(x-y)|x|2+|x|4-3y·(x-y)|x-y|2=(x·y)2+y·(x-y)[4x·(x-y)+4|x|2-3y·(x-y)].
By substitution, we have(67)S=2x·(x-y)[(x·y)2y·(x-y)y·(x-y)+2(x·y)y·(x-y)+(x·y)2]+y·(x-y)(x·y)2-[y·(x-y)]2×[4x·(x-y)+4|x|2-3y·(x-y)]=-2x·(x-y)[y·(x-y)]2+4(x·y)x·(x-y)y·(x-y)-4x·x[y·(x-y)]2+2(x·y)2x·(x-y)+y·(x-y)(x·y)2+3[y·(x-y)]3=2y·(x-y)×[-|x|2(x·y)+3|x|2|y|2-(x·y)2-|y|2(x·y)]+2(x·y)2x·(x-y)+y·(x-y)(x·y)2+3[y·(x-y)]3=-2|x|2(x·y)y·(x-y)+2(x·y)2x·(x-y)+y·(x-y)[6|x|2|y|2-(x·y)2+3[y·(x-y)]2]-2|y|2(x·y)y·(x-y)=8|x|2|y|2(x·y)-6|x|2|y|4-10|y|2(x·y)2+11|y|4(x·y)-3|y|6.
Hence(68)[4|x|4(-3β1(x)+5β2(x))+8|x|3β3(x)-2|x|2(3β4(x)+2β5(x))-2|x|β6(x)+32β7(x)+|x|2β8(x)]=1C5∫R5‍{4|x|4|x-y|5[8|x|2|y|2(x·y)-6|x|2|y|4-10|y|2(x·y)2+11|y|4(x·y)-3|y|6]u-q(y)4|x|4|x-y|5}dy.
Now, integrate it over∂Br(0) with r=|x|, and we obtain that
(69)A≔∫∂Br(0)‍1C5∫R5‍{4|x|4|x-y|5×[8|x|2|y|2(x·y)-6|x|2|y|4-10|y|2(x·y)2+11|y|4(x·y)-3|y|6]×u-q(y)4|x|4|x-y|5}dydδ=1C5(∫|y|<|x|‍+∫|y|>|x|‍)×{∫∂Br(0)‍4|x|4|x-y|5×[8|x|2|y|2(x·y)-8|x|2|y|4+2|x|2|y|4-10|y|2(x·y)2+11|y|4(x·y)-3|y|6]×u-q(y)dδ∫∂Br(0)‍4|x|4|x-y|5}dy:=I1+I2+I3+II1+II2+II3,
where
(70)I1=∫|y|<|x|‍∫∂Br(0)‍4|x|4|x-y|5[8|x|2|y|2(x·y)-8|x|2|y|4]×u-q(y)dydδ=∫|y|<|x|‍32|y|2|x|2u-q(y)[∫∂Br(0)‍(x·y)-|y|2|x-y|5dδ]dy.
It is clear that Δx[(x·y)-|y|2]=0, by Poisson representation formula for harmonic function, the previous integral equals
(71)I1=∫|y|<|x|‍32|y|2|x|2u-q(y)[|x|ω5|x|2-|y|2(y·y-|y|2)]dy=0.
Similarly, calculation givesΔx[2|x|2|y|4-10|y|2(x·y)2]=20|y|2-20|y|2=0, and then Poisson representation formula implies that
(72)I2=∫|y|<|x|‍∫∂Br(0)‍4|x|4|x-y|5×[2|x|2|y|4-10|y|2(x·y)2]u-q(y)dydδ=∫|y|<|x|‍4|x|4u-q(y)×∫∂Br(0)‍2|x|2|y|4-10|y|2(x·y)2|x-y|5dδdy=∫|y|<|x|‍4|x|4u-q(y)[|x|ω5|x|2-|y|2(2|y|6-10|y|6)]dy=∫|y|<|x|‍4|x|4u-q(y)[|x|ω5|x|2-|y|2(-8|y|6)]dy.
Once again,Δx[11|y|4(x·y)-3|y|6]=0, and Poisson formula leads to
(73)I3=∫|y|<|x|‍∫∂Br(0)‍4|x|4|x-y|5×[11|y|4(x·y)-3|y|6]u-q(y)dδdy=∫|y|<|x|‍4|x|4u-q(y)∫∂Br(0)‍11|y|4(x·y)-3|y|6|x-y|5dδdy=∫|y|<|x|‍4|x|4u-q(y)[|x|ω5|x|2-|y|2(8|y|6)]dy.
Notice thatI2+I3=0.
Next, consider the following integral:(74)II1=∫|y|>|x|‍∫∂Br(0)‍4|x|4|x-y|5×[8|x|2|y|2(x·y)+11|y|4(x·y)]×u-q(y)dδdy=∫|y|>|x|‍4|x|4u-q(y)[8|x|2|y|2+11|y|4]×∫∂Br(0)‍x·y|x-y|5dδdy.
Setting z=(|x|2y)/|y|2, the previous integral is equal to
(75)II1=∫|y|>|x|‍4|x|4u-q(y)[8|x|2|y|2+11|y|4]×∫∂Br(0)‍x·(|x|2z/|z|2)|x-(|x|2z/|z|2)|5dδdy=∫|y|>|x|‍4|x|4u-q(y)[8|x|2|y|2+11|y|4]×∫∂Br(0)‍(|x|2/|z|2)x·z(|x|/|z|)5|x-z|5dδdy=∫|y|>|x|‍4|x|4u-q(y)[8|x|2|y|2+11|y|4](|x||z|)-3×|x|ω5|x|2-|z|2z·zdy=∫|y|>|x|‍u-q(y)×[32ω5|x|4|y|(|y|2-|x|2)+44ω5|x|2|y||y|2-|x|2]dy.
Also, we have(76)II2=∫|y|>|x|‍∫∂Br(0)‍4|x|4|x-y|5[-8|x|2|y|4-3|y|6]×u-q(y)dδdy=∫|y|>|x|‍4|x|4u-q(y)[-8|x|2|y|4-3|y|6]×∫∂Br(0)‍1|x-y|5dδdy=∫|y|>|x|‍4|x|4u-q(y)[-8|x|2|y|4-3|y|6]×∫∂Br(0)‍1|x-(|x|2z/|z|2)|5dδdy=∫|y|>|x|‍4|x|4u-q(y)[-8|x|2|y|4-3|y|6](|x||z|)-5×∫∂Br(0)‍1|x-z|5dδdy=∫|y|>|x|‍4|x|4u-q(y)[-8|x|2|y|4-3|y|6]×(|x||z|)-5|x|ω5|x|2-|z|2dy=∫|y|>|x|‍u-q(y)[-32ω5|y||x|2|y|2-|x|2-12ω5|y|3|y|2-|x|2]dy.
Finally, it follows from variable change and Poisson formula for the harmonic function(-10|y|2(x·y)2+2|x|2|y|4) that
(77)II3=∫|y|>|x|‍∫∂Br(0)‍4|x|4|x-y|5×[-10|y|2(x·y)2+2|x|2|y|4]×u-q(y)dδdy=∫|y|>|x|‍4|y|2|x|4u-q(y)×∫∂Br(0)‍1|x-(|x|2z/|z|2)|5×[-10(x·|x|2z|z|2)2+2|x|6|z|2]dδdy=∫|y|>|x|‍4|y|2|x|4u-q(y)×∫∂Br(0)‍1(|x|/|z|)5|x-z|5×[-10(|x|2|z|2)2(x·z)2+2|x|6|z|2]dδdy=∫|y|>|x|‍4|y|2|x|4|x||y|u-q(y)×∫∂Br(0)‍-10(x·z)2+2|x|2|z|2|x-z|5dδdy=∫|y|>|x|‍4|y||x|3u-q(y)|x|ω5|x|2-|z|2-[8|z|4]dy=-32ω5∫|y|>|x|‍u-q(y)|x|4(|y|2-|x|2)|y|dy.
Combining those previous calculations, we can conclude that(78)A=I1+I2+I3+II1+II2+II3=∫|y|>|x|‍u-q(y)×[32ω5|x|4|y|(|y|2-|x|2)+44ω5|x|2|y||y|2-|x|2]dy+∫|y|>|x|‍u-q(y)[-32ω5|y||x|2|y|2-|x|2-12ω5|y|3|y|2-|x|2]dy-32ω5∫|y|>|x|‍u-q(y)|x|4(|y|2-|x|2)|y|dy=-12ω5∫|y|>|x|‍u-q(y)|y|dy⟶0,
as |x|→∞, since u(0)=∫R5‍u-q(y)|y|dy<+∞.
What we have shown so far is that, ifu is regular, that is, q>6, then
(79)P(u,r)⟶0asr⟶∞.
SettingQ=1, when q>6, we have
(80)∫∂Br(0)‍ru1-qdδ⟶0,
as |x|→∞, since u(x)=O(|x|) when |x| is large enough.
Hence, let|x|→∞, and we conclude that q=11 from (4).


## 3. Moving Sphere Method

This section is to run the method of moving sphere for the integral equation with the exact exponentq=11.First, definev(x)=|x|u(x/|x|2) and vλ(x)=λu(x/|λ|2); it is rather easy to see that vλ(x)=(|x|/λ)υ((λ2x)/|x|2).Lemma 5.
If one setsωλ(x)=-υ(x)+υλ(x), then ωλ(x)=∫Bλ‍[G(x,z)ψ(z)ωλ(z)]dz, where ψ=(1/(υ11υλ11))[∑i=010‍υiυλ10-i]≥0; G(x,z)=[(|z|/λ)|x-((λ2z)/|z|2)|-|x-z|]≥0 when |x|≤λ and |z|≤λ. In particular, ωλ(0)=∫Bλ‍[λ-|z|]ψ(z)ωλ(z)dz, which belongs to (0,+∞).Proof.
Sinceu(x)=(1/C5)∫R5‍|x-y|u-11(y)dy,
(81)v(x)=|x|u(x|x|2)=1C5∫R5‍|x||x|x|2-y|u-11(y)dy=1C5∫R5‍|y||x-y|y|2|u-11(y)dy=1C5(I1+I2),
where I1=∫B1/λ‍[|y||x-(y/|y|2)|u-11(y)]dy, I2=∫R5/B1/λ‍[|y||x-(y/|y|2)|u-11(y)]dy.
As forI1, set y=z/λ2 to obtain
(82)I1=∫Bλ‍[|z|λ2|x-z/λ2|z/λ2|2|u-11(|z|λ2)]dzλ2=∫Bλ‍[λ-12|z||x-(λ2z|z|2)|u-11(zλ2)]dz.
As forI2, if we set y=z/|z|2, then we obtain
(83)I2=∫Bλ‍[|z||z|2|x-z/|z|2|z/|z|2|2|u-11(|z||z|2)]dz|z|2=∫Bλ‍[|z|-11|x-z|u-11(z|z|2)]dz.
Similarly, one can writevλ(x) as
(84)vλ(x)=λu(xλ2)=1C5(∫B1/λ‍+∫R5/B1/λ‍)[λ|xλ2-y|u-11(y)]dy=1C5(I3+I4).
As forI3, by setting y=z/λ2, one concludes that
(85)I3=∫Bλ‍[λ-11|x-z|u-11(zλ2)]dz.
For I4, as before, set y=z/|z|2 to get
(86)I4=∫Bλ‍[λ-1|z|-10|x-λ2z|z|2|u-11(z|z|2)]dz.
Therefore, combining these previous calculations, we get
(87)ωλ(x)=1C5[(I1-I4)+(I2-I3)]=∫Bλ‍[|z|λ|x-λ2z|z|2|-|x-z|][v-11(z)-vλ-11(z)]dz=∫Bλ‍[|z|λ|x-λ2z|z|2|-|x-z|]×[υ-11(z)-υλ-11(z)ωλ(z)]ωλ(z)dz=∫Bλ‍[G(x,z)ψ(z)ωλ(z)]dz,
where ψ=[(υ-11(z)-υλ-11(z))/ωλ(z)]=(1/υ11υλ11)[∑i=010‍υiυλ10-i]≥0 and G=[(|z|/λ)|x-(λ2z/|z|2)|-|x-z|].
Now, since(88)|z|2λ2|x-λ2z|z|2|2-|x-z|2=|z|2λ[|x|2-2λ2|z|2〈x,z〉+λ4|z|2]-[|x|2-2〈x,z〉+|z|2]=|x|2|z|2λ2+λ2-|z|2-|x|2=|x|2(|z|2λ2-1)+(λ2-|z|2)=(|x|2λ2-1)(|z|2-λ2)=1λ2(λ2-|x|2)(λ2-|z|2),G(x,z)=[(|z|/λ)|x-(λ2z/|z|2)|-|x-z|]≥0 when |x|≤λ and |z|≤λ.
Putx=0 to obtain
(89)ωλ(0)=∫Bλ‍[G(0,z)ψ(z)ωλ(z)]dzωλ(0)=∫Bλ‍[λ-|z|]ψ(z)ωλ(z)dz,
which belongs to (0,+∞).
Thus we complete the proof.Proposition 6.
Ifλ>0 is large enough, then ωλ(x)≥0 for all x such that |x|≤λ.Proof.
We prove thatωλ(x)≥0 when λ>0 is large enough by three steps.
Step
1. There exists R0>0 such that for R0≤|x|≤λ/2, we have ωλ≥0.
It is clear to see that|x/|x|2|=1/|x|≤1/R0; |x/λ2|≤1/2λ≤1/4R0.
WhenR0 is large enough,
(90)ωλ(x)=-|x|u(x|x|2)+λu(xλ2)ωλ(x)=-|x|[u(0)+O(1|x|)]+λ[u(0)+O(|x|λ2)]ωλ(x)≥|x|2u(0)ωλ(x)>0.
Step
2. There exists an R1>R0 such that for R1≤λ/2≤|x|≤λ, we have Δωλ≤0 and ωλ≥0.
First, we calculate-Δωλ. In fact -Δωλ=-λΔ(u(x/λ2))+Δ(|x|u(x/|x|2)), where Δ(u(x/λ2))=  (1/λ4)(Δu)(x/λ2) and Δ|x|u(x/|x|2)=(Δ|x|)u(x/|x|2)+(Δu(x/|x|2))|x|+2∇|x|∇u(x/|x|2).
To carry out the detail, forn=5 one observes that
(91)Δ|x|=4|x|,∇|x|·∇u(x|x|2)=-|x|-3∑i=15‍ui(x|x|2)·xi,Δu(x|x2|)=|x|-4(Δu)(x|x|2)+(-6)|x|-4∑i=15‍ui(x|x|2)·xi.
Thus
(92)-Δωλ=-|x|-2[|x|2λ3(Δu)(xλ)-1|x|(Δu)(x|x|2)+8|x|∑i=15‍ui(x|x|2)·xi|x|2-4v(x)].
IfR1>1, then |x/|x|2|=1/|x|≤1/R1≤1 if |x|≥R1. Let C=maxB¯1|∇u|. Then |8∑i=15‍ui(x/|x|2)·(xi/|x|2)|≤8C/|x| when R1 is large enough. Now let B=maxB¯1|Δu|. Hence we have
(93)||x|2λ3(Δu)(xλ2)-1|x|(Δu)(x|x|2)|≤λ2λ3B+1|x|B≤3λB.
Notice that if|x| is sufficiently large, then v(x)≥(u(0)/2)|x|. Therefore if λ>0 is large enough, then -Δωλ(x)≥0 for R1≤λ/2≤|x|≤λ.
Besides,ωλ(x)||x|=λ=0 by definition, and ωλ(x)||x|=λ/2≥0 by Step 1. Thus the maximal principle implies that ωλ(x)≥0 on the domain R1≤λ/2≤|x|≤λ.
Step
3. There exists an R2≥R1 such that ωλ(x)≥0 for x≤R0 for λ large enough.
On the one hand, when|x| is small, v(x)=|x|u(x/|x|2)≤|x|·c|x/|x|2|=c, and v is continuous on the rest of ball, so v is bounded for x≤R0. On the other hand, the definition vλ(x)=λu(x/λ2) implies that vλ(x)≥(u(0)/2)λ when λ is large enough. Therefore, ωλ(x)>0 for 0<|x|≤R0 when λ is large enough.
Combining Steps 1 to 3, we haveωλ(x)≥0.
Thus we complete the proof of this proposition.Now, for anyb∈R5, we denote u(x+b) by ub(x), vb(x) by |x|ub(x/|x|2), and set ωλ,b(x)=-vb(x)+(vb)λ(x).By Proposition6, if λ is large enough, then ωλ,b≥0. Thus, we define λb=inf{λ>0:ωμ,b(x)≥0 in B¯μ(0) for λ<μ<+∞}.Proposition 7.
There existsb¯∈R5 such that λb¯>0.Proof.
Suppose that the proposition is not true. Henceλb=0 for all b; that is, for all b∈R5 and all λ>0, we always have ωλ,b(x)≥0 for all x∈B¯λ(0).
Note that(94)ω|x|,b(x)=-vb(x)+|x|λvb(λ2x|x|2)|λ=|x|=0,ω|x|,b(rx)=-vb(rx)+rvb(xr)≥0when0<r<1.
Therefore, it follows that(d/(dr))ω|x|,b(rx)|r=1≤0; that is,
(95)-∇vb(rx)·x+vb(xr)+r∇vb(xr)(-xr2)|r=1≤0.
Simplifying this, we get(96)-2(∇vb)·x+vb(x)≤0,
where vb(x)=|x|u((x/|x|2)+b).
Settingy=x/|x|2, by calculation we get
(97)∇vb·x=|x|u(x|x|2+b)+|x|(∇u)(x|x|2+b)·x|x|2=|x|u(y+b)-|x|(∇u)(y+b)·y.
Substituting (97) into (96) and doing variable change, we obtain
(98)-12u(y)+(∇u)(y)·(y-b)≤0.
Then both sides being divided by|b| and sending |b| to +∞, let b=|b|ωb, we get
(99)(∇u)(y)·(ωb)≤0.
Since the previous inequality holds for allb∈R5, so (∇u)=0. It implies that u is a constant which contradicts u(x)=(1/C5)∫R5‍|x-y|u-11(y)dy.
Thus this proposition holds.Proposition 8.
Suppose that there exists a pointb∈R5 such that λb>0; then ωλb,b≡0 for all x∈R5.Proof.
Without loss of generality, we assume thatb=0. Suppose that the proposition is not true; then ωλ0,0(x)≢0.
In order to complete the argument, we need to do some preparations.
First of all, settingC0=minx∈Bλ0/2¯ωλ0,0(x)>0, then it is clear that
(100)ωλ0(x)≥C0>0∀x∈Bλ0/2¯.
Next, calculate(∂G(x,z)/∂ηx)||x|=λ0 as follows.
Since we have the formula(101)|z|λ|x-λ2z|z|2|=|x|λ|z-λ2x|x|2|,
which is because
(102)(|z|λ|x-λ2z|z|2|)2=|z|2λ2〈x-λ2z|z|2,x-λ2z|z|2〉=|z|2λ2[〈x,x〉-2λ2|z|2〈x,z〉+λ4|z|2]=|z|2|x|2λ2-2〈x,z〉+λ2=|x|2λ2[〈z,z〉-2λ2|x|2〈x,z〉+λ4|x|2]=(|x|λ|z-λ2x|x|2|)2,
thus,
(103)G(x,z)=|z|λ|x-λ2z|z|2|-|x-z|=|x|λ|z-λ2x|x|2|-|x-z|.
Hence, by direct calculation, one obtains(104)∂G(x,z)∂ηx||x|=λ0=∑i‍(∂∂xiG(x,z)·xi|x|)||x|=λ0=∑i=15‍[|z|λ∂∂xi(|x-λ2z|z|2|)·xi|x|]-∑i=15‍[(∂∂xi|x-z|)·xi|x|]=∑i=15‍[(|z|λxi-(λ2zi/|z|2)|x-(λ2z/|z|2)|-xi-zi|x-z|)·xi|x|]||x|=λ0=∑i=15‍[(|z|λxi-(λ2zi/|z|2)(|x|/|z|)|z-(λ2x/|x|2)|-xi-zi|x-z|)·xi|x|(|z|λxi-((λ2zi)/|z|2)(|x|/|z|)|z-((λ2x)/|x|2)|-xi-zi|x-z|)]||x|=λ0=|z|2-λ02λ0|x-z|.
It clearly implies that(105)∂G(x,z)∂ηx|∂Bλ0(0)≤0
whenever |z|≤λ.
Therefore, by Lemma5,
(106)∂ωλ0(x)∂ηx|∂Bλ0(0)=1C5∫Bλ0(0)‍|z|2-λ02λ0|x-z|ψ(z)ωλ0(z)dz.
By the definition ofλ0 and our assumption that ωλ0,0(x)≢0, from Lemma 5, we see that ωλ0(x)>0 inside the ball Bλ0(0). Then, at every boundary point x, we have
(107)∂ωλ0(x)∂ηx|∂Bλ0(0)<0.
Besides, the definition ofλ0 implies that there is a sequence λk such that
(108)λk⟶λ0,λk<λ0,infBλk¯(0)ωλk(x)<0.
It follows from (108) that there exists xk∈Bλk¯(0) such that
(109)ωλk(xk)=minBλk¯(0)(x)<0.
Since ωλk(x)≡0 for |x|=λk, we have |xk|<λk<λ0. Clearly, xk has a convergent subsequence, still denoted by xk, with a limit point x0. Also, ∇ωλk(xk)=0. Then x0 satisfies
(110)∇ωλ0(x0)=0,ωλ0(x0)≤0.
Now consider the following two cases.
Case
1
(
|
x
0
|
<
λ
0
). The conclusion ωλ0(x0)≤0 contradicts the fact that ωλ0(x0) is strictly positive inside the ball Bλ0(0).
Case
2
(
|
x
0
|
=
λ
0
). Then the fact that ∇ωλ0(x0)=0 contradicts (107).
Hence, the proposition follows.Proposition 9.
For allb∈R5, one has λb>0.Proof.
It follows from Propositions7 and 8 that there exists some b¯∈R5 such that λb¯>0 and ωλb¯(x)=0; that is, vb¯(x)=(vb¯(x))λ; that is,
(111)|x|ub¯(x|x|2)=λb¯ub¯(xλb-2).
Settingy=x/|x|2, we get
(112)|y|-1ub¯(y)=λb¯ub¯(yλb-2|y|2).
It follows that(113)lim|y|→∞[|y|-1ub¯(y)]=λb¯ub¯(0).
Suppose that there exists a pointb∈R5 such that λb=0. Then ωλ,b(x)=-vb(x)+(vb(x))λ≥0 for all λ>0 and x∈Bλ(0)/{0}.
That is,λub(|x|/λ2)≥|x|ub(x/|x|2) for all λ>0 and x∈Bλ(0)∖{0}.
Fixingλ>0 and letting |x|→0, we have λub(0)≥λb¯ub¯(0).
Therefore,(114)λλb¯u(b)≥u(b¯).
Lettingλ go to 0, we have 0≥u(b¯), which contradicts the fact that u is a positive entire solution.
Hence, this proposition holds.Proposition 10.
Ifu>0 satisfies u(x)=(1/C5)∫R5‍|x-y|u-11(y)dy, then there exist some x0 and constants a, d such that u(x)=(a|x-x0|2+d)1/2.Proof.
From Propositions8 and 9, we know that for all b∈R5, we have ωλb,b≡0; that is, vb(x)=(|x|/λb)vb(λb2x/|x|2).
Usingv(x)=|x|u(x/|x|2), setting y=x/λb2 and μb=λb-1, we can rewrite it as follows:
(115)u(y+b)=|y|μb-1u(μb2y|y|2+b).
Now settingx=y+b, then we have
(116)u(x)=μb-1|x-b|u(μb2(x-b)|x-b|2+b).
It follows that the limitlim|x|→∞|x|-1u(x) exists and is equal to μb-1u(b) for all b∈R5.
Then setA=lim|x|→∞|x|-1u(x)=μb-1u(b).
Case
1. If A=0, then u≡0 which is impossible to be a positive solution.
Case
2. If A≠0, the definition of A implies that A>0. Without loss of generality, we can assume that A=1. Now for |x| large,
(117)u(x)=μ0-1|x|-1{u(0)+∂u∂xi(0)·μ02xi|x|2+o(1|x|)},
so the coefficient of xi is (μ0/|x|)(∂u/∂xi)(0), and
(118)u(x)=μb-1|x-b|-1{u(b)+∂u∂xi(b)·μb2(xi-bi)|x-b|2+o(1|x|)}=μb-1[|x|+xi|x|(-bi)+o(1|x|)]×[u(b)+∂u∂xi(b)μb2|x|xi+o(1|x|)],
so the coefficient of xi is (1/|x|)+(∂u/∂xi)(b)(μb/|x|).
Comparing these two coefficients andA=1, we obtain
(119)1|x|+∂u∂xi(b)μb|x|=μ0|x|∂u∂xi(0).
Then(120)ub∂u∂xi(b)=u0∂u∂xi(0)+bi.
Thus,(121)12∂u2∂xi(x)|x=b=12∂u2∂xi(x)|x=0+12∂|x|2∂xi(x)|x=b.
It gives(122)∂u2(x)-|x|2∂xi|x=b=Ci.
By solving the previous equation, we obtain(123)u2(x)=|x|2+Cx+D=|x-C2|2+E=|x-x0|2+d,
for some x0∈R5, equivalent to
(124)u(x)=a(|x-x0|2+d)1/2.



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*Source: 384394-2013-07-24.xml*