Entire Solutions of an Integral Equation in 𝑅⁵

Abstract

We will study the entire positive 𝐶⁰ 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.

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 Section 2, 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 with 𝑞=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 every 𝐶⁰ 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 that 𝑢 is a 𝐶⁶ entire positive solution of the integral equation 𝑢(𝑥)=1𝐶_5𝑅⁵𝑥−𝑦𝑢^−𝑞𝑦𝑑𝑦𝑖𝑛𝑅⁵,𝑤ℎ𝑒𝑟𝑒𝐶₅=24𝜔₅ (1) with 𝑞>0, then 𝑞=11.

Lemma 2. If 𝑢(𝑥) satisfies (1), then Δ³𝑢(𝑥)=𝑢^−𝑞(𝑥).

Proof. By direct calculation, we have Δ_𝑥𝑢(𝑥)=1𝐶_5𝑅⁵Δ_𝑥𝑥−𝑦𝑢^−𝑞𝑦=1𝑑𝑦𝐶_5𝑅⁵4𝑢𝑥−𝑦^−𝑞𝑦Δ𝑑𝑦;_2𝑥𝑢(𝑥)=1𝐶_5𝑅⁵Δ_𝑥4𝑢𝑥−𝑦^−𝑞𝑦=1𝑑𝑦𝐶_5𝑅⁵−8𝑥−𝑦³𝑢^−𝑞𝑦𝑑𝑦. (2)
Since 𝐾(𝑟)=(1/−3𝜔₅)𝑟⁻³ is the fundamental solution of Laplace equation in 𝑅⁵, it follows that Δ³_𝑥𝑢(𝑥)=−8𝐶₅−3𝜔_5𝑅5Δ_𝑥1−3𝜔₅1𝑥−𝑦³𝑢^−𝑞𝑦=𝑑𝑦24𝜔₅𝐶₅𝑢^−𝑞(𝑥)=𝑢^−𝑞(𝑥). (3)
Thus the lemma is true.

Lemma 3. If 𝑢>0 is the solution of the equation Δ³_𝑥𝑢(𝑥)=𝑄(𝑥)𝑢^−𝑞(𝑥)𝑖𝑛𝑅⁵, (4) then the following Pohozev identity holds: 12+51−𝑞_𝐵𝑟(0)𝑄𝑢^1−𝑞()=1𝑑𝑥+𝑃𝑢,𝑟𝑞−1_{𝜕𝐵𝑟(0)}𝑟𝑄𝑢^1−𝑞1𝑑𝛿−1−𝑞_𝐵𝑟(0)(𝑥⋅∇𝑄)𝑢^1−𝑞𝑑𝑥, (5) where 𝑃(𝑢,𝑟)=_{𝜕𝐵𝑟(0)}𝑟𝜕𝑢𝜕𝑟𝜕Δ²𝑢𝜕𝑟−Δ²𝑢𝜕𝑟𝜕𝑟𝜕𝑢+1𝜕𝑟2𝑟𝜕Δ𝑢𝜕𝑟²−12𝜕Δ𝑢𝑟𝜕𝑟𝜕Δ𝑢+1𝜕𝑟2𝑟Δ𝑢Δ²𝑢+12𝜕𝑢Δ𝜕𝑟²1𝑢−2𝑢𝜕Δ²𝑢𝜕𝑟𝑑𝛿. (6)

Proof. Multiplying 𝑥⋅∇𝑢 on the both sides of (4) and integrating over the ball 𝐵_𝑟(0), one has 0=_𝐵𝑟(0)(𝑥⋅∇𝑢)Δ³_𝑥𝑢(𝑥)−𝑄(𝑥)𝑢^−𝑞(𝑥)𝑑𝑥=𝐼₂−𝐼₁, (7) where 𝐼₁=∫_𝐵𝑟(0)(𝑥⋅∇𝑢)𝑄(𝑥)𝑢^−𝑞(𝑥)𝑑𝑥; 𝐼₂=∫_𝐵𝑟(0)(𝑥⋅∇𝑢)Δ³_𝑥𝑢(𝑥).
First, rewrite 𝐼₁ as 𝐼₁=_𝐵𝑟(0)𝑢𝑥⋅∇^1−𝑞(𝑥)𝑄(𝑥)1−𝑞𝑑𝑥. (8)
Using integration by parts, we have 𝐼₁=11−𝑞_{𝜕𝐵𝑟(0)}𝑢^1−𝑞(𝑥)𝑄(𝑥)−1𝑟𝑑𝛿1−𝑞_𝐵𝑟(0)(5𝑄+𝑥⋅∇𝑄)𝑢^1−𝑞(𝑥)𝑑𝑥. (9)
Now, the Second Green identity implies that 𝐼₂=_{𝜕𝐵𝑟(0)}(𝑥⋅∇𝑢)𝜕Δ²𝑢−Δ𝜕𝑟²𝑢𝜕(𝑥⋅∇𝑢)+𝜕𝑟𝑑𝛿_𝐵𝑟(0)Δ(𝑥⋅∇𝑢)Δ²=𝑢𝑑𝑥_{𝜕𝐵𝑟(0)}𝑟𝜕𝑢𝜕𝑟𝜕Δ²𝑢−Δ𝜕𝑟²𝑢𝜕𝑟𝜕𝑟𝜕𝑢𝜕𝑟𝑑𝛿+2𝐼₃+𝐼₄, (10) where 𝐼₃=∫_𝐵𝑟(0)Δ𝑢Δ²𝑢𝑑𝑥; 𝐼₄=∫_𝐵𝑟(0)𝑥⋅∇(Δ𝑢)Δ²𝑢𝑑𝑥.
As for 𝐼₄, since Δ(𝑥⋅∇𝑢)=2Δ𝑢+𝑥⋅∇Δ𝑢 and by the Second Green identity again, we obtain 𝐼₄=_𝐵𝑟(0)Δ[]+𝑥⋅∇Δ𝑢Δ𝑢𝑑𝑥_{𝜕𝐵𝑟(0)}𝑥⋅∇Δ𝑢𝜕Δ𝑢−𝜕()𝜕𝑟𝑥⋅∇Δ𝑢=𝜕𝑟Δ𝑢𝑑𝛿_𝐵𝑟(0)2Δ²𝑢+𝑥⋅∇Δ²𝑢Δ𝑢𝑑𝑥+_{𝜕𝐵𝑟(0)}𝑟𝜕Δ𝑢𝜕𝑟²−𝜕𝑟𝜕𝑟𝜕Δ𝑢𝜕𝑟Δ𝑢𝑑𝛿. (11)
Hence, 𝐼₄=2𝐼₃+_𝐵𝑟(0)𝑥⋅∇Δ²𝑢Δ𝑢𝑑𝑥+_{𝜕𝐵𝑟(0)}𝑟𝜕Δ𝑢𝜕𝑟²−𝜕𝑟𝜕𝑟𝜕Δ𝑢𝜕𝑟Δ𝑢𝑑𝛿. (12)
Observe that 𝐼₄+_𝐵𝑟(0)𝑥⋅∇Δ²=𝑢Δ𝑢𝑑𝑥_𝐵𝑟(0)𝑥⋅∇Δ𝑢Δ²𝑢=𝑑𝑥_{𝜕𝐵𝑟(0)}𝑟Δ𝑢Δ²𝑢𝑑𝛿−5_𝐵𝑟(0)Δ𝑢Δ²=𝑢𝑑𝑥_{𝜕𝐵𝑟(0)}𝑟Δ𝑢Δ²𝑢𝑑𝛿−5𝐼₃; (13) that is, 𝐼₄=_{𝜕𝐵𝑟(0)}𝑟Δ𝑢Δ²𝑢𝑑𝛿−5𝐼₃. (14)
Combining (12) and (14), we have 𝐼₄3=−2𝐼₃+12_{𝜕𝐵𝑟(0)}𝑟𝜕Δ𝑢𝜕𝑟²−𝜕𝑟𝜕𝑟𝜕Δ𝑢𝜕𝑟Δ𝑢+𝑟Δ𝑢Δ²𝑢𝑑𝛿. (15)
Thus 𝐼₂=12𝐼₃+_{𝜕𝐵𝑟(0)}𝑟𝜕𝑢𝜕𝑟𝜕Δ²𝑢𝜕𝑟−Δ²𝑢𝜕𝑟𝜕𝑟𝜕𝑢+1𝜕𝑟2𝑟𝜕Δ𝑢𝜕𝑟²−12𝜕Δ𝑢𝑟𝜕𝑟𝜕Δ𝑢+1𝜕𝑟2𝑟Δ𝑢Δ²𝑢𝑑𝛿. (16)
To deal with 𝐼₃, we observe that if multiplying 𝑢 on both sides of (4) and integrating over the ball 𝐵_𝑟(0), one has 0=_𝐵𝑟(0)𝑢Δ³𝑢−𝑄(𝑥)𝑢^−𝑞𝑑𝑥. (17) Thus, with the help of the Second Green identity, one obtains _𝐵𝑟(0)𝑄(𝑥)𝑢^1−𝑞=𝑑𝑥_𝐵𝑟(0)𝑢Δ³=𝑢𝑑𝑥_𝐵𝑟(0)Δ𝑢Δ²𝑢𝑑𝑥+_{𝜕𝐵𝑟(0)}𝑢𝜕Δ²𝑢−𝜕𝑟𝜕𝑢Δ𝜕𝑟²𝑢𝑑𝛿. (18)
Therefore, we get 𝐼₃=_𝐵𝑟(0)Δ𝑢Δ²=𝑢𝑑𝑥_𝐵𝑟(0)𝑄(𝑥)𝑢^1−𝑞𝑑𝑥+_{𝜕𝐵𝑟(0)}𝜕𝑢Δ𝜕𝑟²𝑢−𝑢𝜕Δ²𝑢𝜕𝑟𝑑𝛿. (19)
Put it into (16) to conclude that 𝐼₂=12_𝐵𝑟(0)𝑄𝑢^1−𝑞+𝑑𝑥_{𝜕𝐵𝑟(0)}𝑟𝜕𝑢𝜕𝑟𝜕Δ²𝑢𝜕𝑟−Δ²𝑢𝜕𝑟𝜕𝑟𝜕𝑢+1𝜕𝑟2𝑟𝜕Δ𝑢𝜕𝑟²−12𝜕Δ𝑢𝑟𝜕𝑟𝜕Δ𝑢+1𝜕𝑟2𝑟Δ𝑢Δ²𝑢+12𝜕𝑢Δ𝜕𝑟²1𝑢−2𝑢𝜕Δ²𝑢𝜕𝑟𝑑𝛿. (20)
Put 𝐼₂ and 𝐼₁ into the equation 𝐼₂−𝐼₁=0 and rearrange the terms; then we get the desired identity.
Thus the lemma holds.

Lemma 4. If 𝑢 is an entire solution of the integral equation (1), then 𝑞=11.

Proof. Define 𝑑=(1/𝐶₅)∫_𝑅5𝑢^−𝑞(𝑦)𝑑𝑦, 𝑑₁=(1/𝐶₅)∫_𝑅5|𝑦|𝑢^−𝑞(𝑦)𝑑𝑦. Then it follows that 𝑢(𝑥)=1𝐶_5𝑅⁵𝑥−𝑦𝑢^−𝑞𝑦≤1𝑑𝑦𝐶_5𝑅⁵|𝑥|𝑢^−𝑞𝑦1𝑑𝑦+𝐶_5𝑅⁵𝑦𝑢^−𝑞𝑦=|𝑥|𝑑𝑦𝑑+𝑑₁. (21)
If 0<𝑞≤6, then 𝑑₁(0)=1=𝑢𝐶_5𝑅⁵𝑦𝑢^−𝑞𝑦≥1𝑑𝑦𝐶_5𝑅⁵𝑦𝑦𝑑+𝑑₁^−𝑞𝑦=𝜔𝑑𝑦₅𝐶₅₀^+∞𝑟⁵𝑟𝑑+𝑑₁^−𝑞𝑑𝑟=∞, (22) which contradicts the assumption that 𝑢 is an entire solution. Therefore, we only consider the case 𝑞>6.
Rewrite the equation as 𝑢(𝑥)=𝑑|𝑥|+𝛽₁(𝑥), (23) where 𝛽₁(𝑥)=1𝐶_5𝑅⁵𝑥−𝑦−|𝑥|𝑢^−𝑞𝑦𝑑𝑦. (24)
Clearly, we have 𝛽₁(𝑥≤1)𝐶_5𝑅⁵𝑦𝑢^−𝑞𝑦𝑑𝑦=𝑑₁. (25)
And, by (1), one has 𝜕𝑢=1𝜕𝑟𝐶_5𝑅⁵𝑥⋅𝑥−𝑦𝑟𝑢𝑥−𝑦^−𝑞𝑦=1𝑑𝑦𝐶_5𝑅⁵𝑢^−𝑞𝑦1𝑑𝑦+𝐶_5𝑅⁵𝑥−𝑦⋅𝑥𝑥−𝑦𝑟−1𝑢^−𝑞𝑦𝛽𝑑𝑦∶=𝑑+²(𝑥)𝑟, (26) where 𝛽₂(𝑥)=1𝐶_5𝑅⁵𝑥−𝑦⋅𝑥−𝑟𝑥−𝑦𝑢𝑥−𝑦^−𝑞𝑦𝑑𝑦. (27)
Note that 𝛽₂(𝑥) can be controlled as follows: 𝛽₂(𝑥=1)𝐶_5𝑅⁵𝑥−𝑦−|𝑥|𝑥−𝑦+𝑦⋅𝑥−𝑦𝑢𝑥−𝑦^−𝑞𝑦≤2𝑑𝑦𝐶_5𝑅⁵𝑦𝑢^−𝑞𝑦𝑑𝑦=2𝑑₁. (28)
Clearly, with the definition of 𝛽₂(𝑥), the following holds true: 𝑟𝜕𝑢𝜕𝑟=𝑑𝑟+𝛽₂(𝑥). (29)
Then differentiate it to get 𝜕𝑟𝜕𝑟𝜕𝑢1𝜕𝑟=𝑑+𝐶_5𝑅⁵2𝑥−𝑦𝑥−𝑦²−𝑥⋅𝑥−𝑦𝑥−𝑦𝑥−𝑦³−𝑥|𝑥|⋅𝑥𝑟𝑢^−𝑞𝑦𝑑𝑦∶=𝑑+𝛽₃(𝑥), (30) where 𝛽₃(𝑥)=1𝐶_5𝑅⁵|𝑥|−𝑥−𝑦−1𝑥⋅𝑥−𝑦𝑦⋅𝑥−𝑦𝑟𝑥−𝑦³𝑢^−𝑞𝑦𝑑𝑦. (31)
One observes that |𝑥|𝛽₃(𝑥=1)𝐶_5𝑅⁵𝑥||𝑥||−𝑥−𝑦−1𝑥⋅𝑥−𝑦𝑦⋅𝑥−𝑦𝑥−𝑦³×𝑢^−𝑞𝑦≤2𝑑𝑦𝐶_5𝑅⁵|𝑥|𝑦𝑢𝑥−𝑦^−𝑞𝑦𝑑𝑦≤𝑀₃, (32) 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: _𝑅5|𝑥|𝑦𝑢𝑥−𝑦^−𝑞𝑦𝑑𝑦=_𝐴1+_𝐴2|𝑥|𝑦𝑢𝑥−𝑦^−𝑞𝑦𝑑𝑦=𝑆₁+𝑆₂, (33) where 𝐴₁={𝑦∈𝑅⁵∶|𝑥−𝑦|≥|𝑥|/2}, 𝐴₂={𝑦∈𝑅⁵∶|𝑥−𝑦|≤|𝑥|/2}.
Then 𝑆₁=_𝐴1|𝑥|𝑦𝑢𝑥−𝑦^−𝑞𝑦𝑑𝑦≤2_𝐴1𝑦𝑢^−𝑞𝑦𝑑𝑦≤2𝑑₁. (34)
As for 𝑆₂, 𝑦∈𝐴₂ implies that |𝑦|≤|𝑥|+|𝑥−𝑦|≤(3/2)|𝑥| and |𝑦|≥|𝑥|−|𝑥−𝑦|≥(1/2)|𝑥|.
Therefore 𝑆₂=_𝐴2|𝑥|𝑦𝑢𝑥−𝑦^−𝑞𝑦≤𝑑𝑦min_{|𝑥|𝑦|𝑥|/2≤≤3/2}𝑢𝑦^−𝑞3|𝑥|²2_𝐴21=𝑥−𝑦𝑑𝑦min_{|𝑥|𝑦|𝑥|/2≤≤3/2}𝑢𝑦^−𝑞3|𝑥|²2𝜔₅₀^|𝑥|/2𝑟³≤𝑑𝑟min_{|𝑥|𝑦|𝑥|/2≤≤3/2}𝑢𝑦^−𝑞𝜔₅3|𝑥|⁶2⁷. (35)
Since 𝑢(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 (𝑥)=4Δ𝑢𝐶_5𝑅⁵1𝑢𝑥−𝑦^−𝑞𝑦4𝑑𝑦=|𝑥|𝑑+𝛽₄(𝑥), (36) where 𝛽₄(𝑥)=4𝐶_5𝑅⁵1−1𝑥−𝑦|𝑥|𝑢^−𝑞𝑦𝑑𝑦. (37)
For a constant 𝑀₄ independent of 𝑥, when |𝑥| is large enough, it is easy to see that |𝑥|²𝛽₄(𝑥=4)𝐶_5𝑅⁵|𝑥|𝑥|−|𝑥−𝑦|𝑥𝑢|𝑥−𝑦^−𝑞𝑦≤4𝑑𝑦𝐶_5𝑅⁵|𝑥|𝑦𝑢𝑥−𝑦^−𝑞𝑦𝑑𝑦≤𝑀₄, (38) by the estimation in 𝛽₃.
Then, by computation, one gets 𝜕Δ𝑢=4𝜕𝑟𝐶_5𝑅⁵−𝑥−𝑦⋅𝑥𝑥−𝑦³|𝑥|𝑢^−𝑞𝑦4𝑑𝑦=−|𝑥|²4𝑑+𝐶_5𝑅⁵−𝑥−𝑦⋅𝑥𝑥−𝑦³|𝑥|+1|𝑥|²𝑢^−𝑞𝑦4𝑑𝑦=−|𝑥|²𝛽𝑑+⁵(𝑥)𝑟, (39) where 𝛽₅(𝑥)=4𝐶_5𝑅⁵−𝑥−𝑦⋅𝑥𝑥−𝑦³+1|𝑥|𝑢^−𝑞𝑦𝑑𝑦. (40)
And as usual, we observe that |𝑥|²𝛽₅(𝑥=4)𝐶_5𝑅⁵−𝑥−𝑦⋅𝑥|𝑥|²𝑥−𝑦³+|𝑥|𝑢^−𝑞𝑦=4𝑑𝑦𝐶_5𝑅⁵|𝑥|𝑥−𝑦³−𝑥⋅𝑥−𝑦|𝑥|²𝑥−𝑦³𝑢^−𝑞𝑦=4𝑑𝑦𝐶_5𝑅⁵|𝑥|𝑥−𝑦𝑥−𝑦𝑥−𝑦−|𝑥|𝑥⋅𝑥−𝑦𝑥−𝑦³×𝑢^−𝑞𝑦=4𝑑𝑦𝐶_5𝑅⁵|𝑥|𝑥−𝑦−|𝑥|𝑥⋅𝑥−𝑦−𝑥−𝑦𝑦⋅𝑥−𝑦𝑥−𝑦³×𝑢^−𝑞𝑦≤4𝑑𝑦𝐶_5𝑅⁵|𝑥|²𝑦𝑥−𝑦²+|𝑥|𝑦𝑢𝑥−𝑦^−𝑞𝑦𝑑𝑦. (41)
Similarly, we have the following estimation: _𝑅5|𝑥|²𝑦𝑥−𝑦²𝑢^−𝑞𝑦=𝑑𝑦_𝐴1+_𝐴2|𝑥|²𝑦𝑥−𝑦²𝑢^−𝑞𝑦𝑑𝑦=𝑆₃+𝑆₄, (42) where 𝐴₁={𝑦∈𝑅⁵∶|𝑥−𝑦|≥|𝑥|/2}, 𝐴₂={𝑦∈𝑅⁵∶|𝑥−𝑦|≤|𝑥|/2}.
Then 𝑆₃=_𝐴1|𝑥|²𝑦𝑥−𝑦²𝑢^−𝑞𝑦𝑑𝑦≤4_𝐴1𝑦𝑢^−𝑞𝑦𝑑𝑦≤4𝑑₁. (43)
As for 𝑆₄, 𝑦∈𝐴₂ implies that |𝑦|≤|𝑥|+|𝑥−𝑦|≤(3/2)|𝑥| and |𝑦|≥|𝑥|−|𝑥−𝑦|≥(1/2)|𝑥|.
Therefore 𝑆₄=_𝐴2|𝑥|²𝑦𝑥−𝑦²𝑢^−𝑞𝑦≤𝑑𝑦min_{𝑦|𝑥|/2≤≤3|𝑥|/2}𝑢𝑦^−𝑞3|𝑥|³2_𝐴21𝑥−𝑦²=𝑑𝑦min_{𝑦|𝑥|/2≤≤3|𝑥|/2}𝑢𝑦^−𝑞3|𝑥|³2𝜔₅₀^|𝑥|/2𝑟²≤𝑑𝑟min_{𝑦|𝑥|/2≤≤3|𝑥|/2}𝑢𝑦^−𝑞𝜔₅|𝑥|⁶2⁴. (44)
Since 𝑢(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.
Since 𝑟(𝜕Δ𝑢/𝜕𝑟)=−(4/|𝑥|)𝑑+𝛽₅(𝑥), we obtain 𝜕𝑟𝜕𝑟𝜕Δ𝑢=4𝜕𝑟|𝑥|²𝑑+𝛽₆(𝑥), (45) where 𝛽₆(𝑥)=𝜕𝛽𝜕𝑟⁵(𝑥)=4𝐶_5𝑅⁵∇−𝑥−𝑦⋅𝑥𝑥−𝑦³+1|𝑥|⋅𝑥|𝑥|𝑢^−𝑞𝑦=4𝑑𝑦𝐶_5𝑅⁵2𝑥−𝑦𝑥−𝑦²−3𝑥−𝑦⋅𝑥𝑥−𝑦𝑥−𝑦⁵−𝑥|𝑥|³⋅𝑥|𝑥|𝑢^−𝑞𝑦=4𝑑𝑦𝐶_5𝑅⁵1|𝑥|𝑥−𝑦⁵3𝑥−𝑦²𝑥−𝑦⋅𝑥+3𝑦⋅𝑥−𝑦𝑥−𝑦⋅𝑥−𝑥⋅𝑥−𝑦𝑥−𝑦²−|𝑥|²𝑥−𝑦²−1|𝑥|²×𝑢^−𝑞𝑦=4𝑑𝑦𝐶_5𝑅⁵2𝑥−𝑦⋅𝑥|𝑥|𝑥−𝑦³+3𝑦⋅𝑥−𝑦𝑥−𝑦⋅𝑥|𝑥|𝑥−𝑦⁵−|𝑥|𝑥−𝑦³−1|𝑥|²𝑢^−𝑞𝑦𝑑𝑦. (46)
As before, we can estimate 𝛽₆(𝑥) as |𝑥|³𝛽₆(𝑥=4)𝐶_5𝑅⁵2𝑥−𝑦⋅𝑥|𝑥|²𝑥−𝑦³+3𝑦⋅𝑥−𝑦𝑥−𝑦⋅𝑥|𝑥|²𝑥−𝑦⁵−|𝑥|⁴𝑥−𝑦³−|𝑥|𝑢^−𝑞𝑦=4𝑑𝑦𝐶_5𝑅⁵𝑥−𝑦⋅𝑥|𝑥|²𝑥−𝑦³−|𝑥|⁴𝑥−𝑦³+𝑥−𝑦⋅𝑥|𝑥|²𝑥−𝑦³−|𝑥|+3𝑦⋅𝑥−𝑦𝑥−𝑦⋅𝑥|𝑥|²𝑥−𝑦⁵𝑢^−𝑞𝑦≤4𝑑𝑦𝐶_5𝑅⁵−|𝑥|²𝑥⋅𝑦𝑥−𝑦³+|𝑥|²−|𝑥+|𝑥𝑥⋅𝑥−𝑦|𝑥−𝑦𝑥⋅𝑥−𝑦|𝑥−𝑦𝑦⋅𝑥−𝑦𝑥−𝑦³⁻¹+3𝑦⋅𝑥−𝑦𝑥−𝑦⋅𝑥|𝑥|²𝑥−𝑦⁵𝑢^−𝑞𝑦≤4𝑑𝑦𝐶_5𝑅⁵|𝑥|³𝑦𝑥−𝑦³𝑢^−𝑞𝑦4𝑑𝑦+𝐶_5𝑅⁵|𝑥|²𝑦𝑥−𝑦²𝑢^−𝑞𝑦+4𝑑𝑦𝐶_5𝑅⁵|𝑥|𝑦𝑢𝑥−𝑦^−𝑞𝑦𝑑𝑦+12𝐶_5𝑅⁵|𝑥|³𝑦𝑥−𝑦³𝑢^−𝑞𝑦𝑑𝑦≤𝑀₆, (47) 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: _𝑅5|𝑥|³𝑦𝑥−𝑦³𝑢^−𝑞𝑦=𝑑𝑦_𝐴1+_𝐴2|𝑥|³𝑦𝑥−𝑦³𝑢^−𝑞𝑦𝑑𝑦=𝑆₅+𝑆₆, (48) where 𝐴₁={𝑦∈𝑅⁵∶|𝑥−𝑦|≥|𝑥|/2}, 𝐴₂={𝑦∈𝑅⁵∶|𝑥−𝑦|≤|𝑥|/2}.
Then 𝑆₅=_𝐴1|𝑥|³𝑦𝑥−𝑦³𝑢^−𝑞𝑦𝑑𝑦≤8_𝐴1𝑦𝑢^−𝑞𝑦𝑑𝑦≤8𝑑₁. (49)
As for 𝑆₆, 𝑦∈𝐴₂ implies that |𝑦|≤|𝑥|+|𝑥−𝑦|≤(3/2)|𝑥| and |𝑦|≥|𝑥|−|𝑥−𝑦|≥(1/2)|𝑥|.
Therefore 𝑆₆=_𝐴2|𝑥|³𝑦𝑥−𝑦³𝑢^−𝑞𝑦≤𝑑𝑦min_{𝑦|𝑥|/2≤≤3|𝑥|/2}𝑢𝑦^−𝑞3|𝑥|⁴2_𝐴21𝑥−𝑦³=𝑑𝑦min_{𝑦|𝑥|/2≤≤3|𝑥|/2}𝑢𝑦^−𝑞3|𝑥|⁴2𝜔₅₀^|𝑥|/2≤𝑟𝑑𝑟min_{𝑦|𝑥|/2≤≤3|𝑥|/2}𝑢𝑦^−𝑞𝜔₅3|𝑥|⁶2⁴. (50)
Since 𝑢(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 Δ²𝑢(𝑥)=1𝐶_5𝑅⁵−8𝑥−𝑦³𝑢^−𝑞𝑦𝑑𝑦=−8|𝑥|³𝑑+𝛽₇(𝑥), (51) where 𝛽₇(𝑥)=8𝐶_5𝑅⁵1|𝑥|³−1𝑥−𝑦³𝑢^−𝑞𝑦𝑑𝑦. (52)
We still need to control 𝛽₇(𝑥) which can be done as follows: |𝑥|⁴𝛽₇(𝑥=8)𝐶_5𝑅⁵𝑥|−|𝑥||⁴𝑥−𝑦³𝑢^−𝑞𝑦=8𝑑𝑦𝐶_5𝑅⁵|𝑥|𝑥−𝑦|𝑥|²−2𝑥⋅𝑦+𝑦²−|𝑥|⁴𝑥−𝑦³𝑢^−𝑞𝑦=8𝑑𝑦𝐶_5𝑅⁵|𝑥|³𝑥−𝑦−|𝑥|+|𝑥|𝑥−𝑦−2𝑥⋅𝑦+𝑦²𝑥−𝑦³×𝑢^−𝑞𝑦≤8𝑑𝑦𝐶_5𝑅⁵|𝑥|³𝑦𝑥−𝑦³𝑢^−𝑞𝑦8𝑑𝑦+𝐶_5𝑅⁵2|𝑥|²𝑦𝑥−𝑦²𝑢^−𝑞𝑦+8𝑑𝑦𝐶_5𝑅⁵|𝑥|𝑦²𝑥−𝑦²𝑢^−𝑞𝑦𝑑𝑦≤𝑀₇, (53) for a constant independent of 𝑥 and for |𝑥| sufficiently large. Now, the third integral over 𝑅⁵ can be controlled as follows: _𝑅5|𝑥|𝑦²𝑥−𝑦²𝑢^−𝑞𝑦=𝑑𝑦_𝑅5|𝑥|𝑦𝑦𝑥−𝑦𝑢𝑥−𝑦^−𝑞𝑦≤𝑑𝑦_𝑅5|𝑥|𝑦+|𝑥|𝑥−𝑦𝑥−𝑦𝑢𝑥−𝑦^−𝑞𝑦≤𝑑𝑦_𝑅5|𝑥|𝑦𝑢𝑥−𝑦^−𝑞𝑦𝑑𝑦+_𝑅5|𝑥|²𝑦𝑥−𝑦²𝑢^−𝑞𝑦𝑑𝑦. (54)
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 𝜕Δ²𝑢=𝜕𝑟24𝐶_5𝑅⁵𝑥−𝑦⋅𝑥𝑥−𝑦⁵|𝑥|𝑢^−𝑞𝑦𝑑𝑦=24|𝑥|⁴𝑑+𝛽₈(𝑥), (55) where 𝛽₈(𝑥)=24𝐶_5𝑅⁵𝑥−𝑦⋅𝑥𝑥−𝑦⁵|𝑥|−1|𝑥|⁴𝑢^−𝑞𝑦𝑑𝑦. (56) The estimate on 𝛽₈(𝑥) can proceed as follows: |𝑥|⁵𝛽₈(𝑥=)24𝐶_5𝑅⁵𝑥−𝑦⋅𝑥|𝑥|⁴𝑥−𝑦⁵−|𝑥|𝑢^−𝑞𝑦=𝑑𝑦24𝐶_5𝑅⁵|𝑥|⁴𝑥−𝑦⋅𝑥−|𝑥|𝑥−𝑦⁵𝑥−𝑦⁵𝑢^−𝑞𝑦=𝑑𝑦24𝐶₅×_𝑅5|𝑥|³−𝑥−𝑦³|𝑥|𝑥⋅𝑥−𝑦+|𝑥|𝑥−𝑦³𝑦⋅𝑥−𝑦𝑥−𝑦³𝑦⋅𝑥−𝑦⁻¹×𝑢^−𝑞𝑦≤𝑑𝑦24𝐶_5𝑅⁵|𝑥|³−𝑥−𝑦|𝑥|²−2𝑥⋅𝑦+𝑦²|𝑥|²𝑥−𝑦⁴×𝑢^−𝑞𝑦+𝑑𝑦24𝐶_5𝑅⁵|𝑥|𝑦𝑢𝑥−𝑦^−𝑞𝑦≤𝑑𝑦24𝐶_5𝑅⁵|𝑥|⁴𝑦𝑥−𝑦⁴|𝑥|+2³𝑦𝑥−𝑦³+|𝑥|²𝑦²𝑥−𝑦³𝑢^−𝑞𝑦+𝑑𝑦24𝐶_5𝑅⁵|𝑥|𝑦𝑢𝑥−𝑦^−𝑞𝑦𝑑𝑦≤𝑀₈, (57) for a constant independent of 𝑥 and for |𝑥| sufficiently large. As before, the aforementioned first integral can be controlled as follows: _𝑅5|𝑥|⁴𝑦𝑥−𝑦⁴𝑢^−𝑞𝑦=𝑑𝑦_𝐴1+_𝐴2|𝑥|⁴𝑦𝑥−𝑦⁴𝑢^−𝑞𝑦𝑑𝑦=𝑆₇+𝑆₈, (58) where 𝐴₁={𝑦∈𝑅⁵∶|𝑥−𝑦|≥|𝑥|/2}, 𝐴₂={𝑦∈𝑅⁵∶|𝑥−𝑦|≤|𝑥|/2}.
Then 𝑆₇=_𝐴1|𝑥|⁴𝑦𝑥−𝑦⁴𝑢^−𝑞𝑦𝑑𝑦≤16_𝐴1𝑦𝑢^−𝑞𝑦𝑑𝑦≤16𝑑₁. (59)
As for 𝑆₈, 𝑦∈𝐴₂ implies that |𝑦|≤|𝑥|+|𝑥−𝑦|≤(3/2)|𝑥| and |𝑦|≥|𝑥|−|𝑥−𝑦|≥(1/2)|𝑥|.
Therefore 𝑆₈=_𝐴2|𝑥|⁴𝑦𝑥−𝑦⁴𝑢^−𝑞𝑦≤𝑑𝑦min_{𝑦|𝑥|/2≤≤3|𝑥|/2}𝑢𝑦^−𝑞3|𝑥|⁵2_𝐴2=1𝑑𝑦min_{𝑦|𝑥|/2≤≤3|𝑥|/2}𝑢𝑦^−𝑞3|𝑥|⁵2𝜔₅₀^|𝑥|/2≤1𝑑𝑟min_{𝑦|𝑥|/2≤≤3|𝑥|/2}𝑢𝑦^−𝑞𝜔₅3|𝑥|⁶2², (60) which is bounded, since 𝑢(𝑥)=𝑂(|𝑥|) and 𝑞>6, so 𝐼₈ is bounded for |𝑥|≥𝑅₀ when 𝑅₀ is large enough.
Besides, _𝑅5|𝑥|²𝑦²𝑥−𝑦³𝑢^−𝑞𝑦=𝑑𝑦_𝑅5|𝑥|²𝑦𝑥−𝑦²𝑦𝑢𝑥−𝑦^−𝑞𝑦≤𝑑𝑦_𝑅5|𝑥|²𝑦𝑥−𝑦²𝑢^−𝑞𝑦𝑑𝑦+_𝑅5|𝑥|³𝑦𝑥−𝑦³𝑢^−𝑞𝑦𝑑𝑦, (61) 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 𝑟𝜕𝑢𝜕𝑟𝜕Δ²𝑢𝜕𝑟−Δ²𝑢𝜕𝑟𝜕𝑟𝜕𝑢+1𝜕𝑟2𝑟𝜕Δ𝑢𝜕𝑟²−12𝜕Δ𝑢𝑟𝜕𝑟𝜕Δ𝑢+1𝜕𝑟2𝑟Δ𝑢Δ²1𝑢+2𝜕𝑢Δ𝜕𝑟²1𝑢−2𝑢𝜕Δ²𝑢𝛽𝜕𝑟=𝑟𝑑+²(𝑥)𝑟24|𝑥|⁴𝑑+𝛽₈(𝑥−)𝑑+𝛽₃(𝑥)−8|𝑥|³𝑑+𝛽₇(𝑥+1)2𝑟−4|𝑥|²𝛽𝑑+⁵(𝑥)𝑟²−124|𝑥|²𝑑+𝛽₆(𝑥4)|𝑥|𝑑+𝛽₄(𝑥+1)2𝑟4|𝑥|𝑑+𝛽₄(𝑥)−8|𝑥|³𝑑+𝛽₇(𝑥+1)2𝛽𝑑+²(𝑥)𝑟−8|𝑥|³𝑑+𝛽₇(𝑥−1)2𝑑|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+𝛽₂)(𝑥)𝛽₈(𝑥)−𝛽₃(𝑥)𝛽₇(𝑥)+1𝛽2𝑟⁵(𝑥)²−12𝛽₄(𝑥)𝛽₆(𝑥)+12𝑟𝛽₄(𝑥)𝛽₇(𝑥)+1𝑟𝛽₂(𝑥)𝛽₇(𝑥)−12𝛽₁(𝑥)𝛽₈(𝑥). (62)
Observe that |𝑥|⁴𝛽₂(𝑥)𝛽₈(𝑥)−𝛽₃(𝑥)𝛽₇(𝑥)+1𝛽2𝑟⁵(𝑥)²−12𝛽₄(𝑥)𝛽₆(𝑥)+12𝑟𝛽₄(𝑥)𝛽₇(𝑥)+1𝑟𝛽₂(𝑥)𝛽₇(𝑥)−12𝛽₁(𝑥)𝛽₈(𝑥)⟶0as|𝑥|⟶∞. (63) The coefficient of 𝑑 is given by −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|²3𝛽₄(𝑥)+2𝛽₅(𝑥)−2|𝑥|𝛽₆(𝑥)+32𝛽₇(𝑥)+|𝑥|2𝛽₈(𝑥=1)𝐶_5𝑅⁵−12𝑥−𝑦|𝑥|⁴+20𝑥−𝑦⋅𝑥|𝑥|⁴1𝑥−𝑦−8|𝑥|³+81|𝑥|²1𝑥−𝑦−8|𝑥|³−8𝑥⋅𝑥−𝑦𝑦⋅𝑥−𝑦|𝑥|⁴𝑥−𝑦³+1−24|𝑥|²1𝑥−𝑦+8|𝑥|³+16𝑥−𝑦⋅𝑥𝑥−𝑦³|𝑥|²+−16𝑥−𝑦⋅𝑥𝑥−𝑦³|𝑥|²−24𝑦⋅𝑥−𝑦𝑥−𝑦⋅𝑥|𝑥|²𝑥−𝑦⁵1+8𝑥−𝑦³1+8|𝑥|³+112|𝑥|³1−12𝑥−𝑦³+12𝑥−𝑦⋅𝑥𝑥−𝑦⁵1−12|𝑥|³𝑢^−𝑞𝑦=1𝑑𝑦𝐶_5𝑅⁵4|𝑥|⁴𝑥−𝑦⁵−3𝑥−𝑦⁶+5𝑥−𝑦⋅𝑥𝑥−𝑦⁴−4|𝑥|²𝑥−𝑦⁴𝑦⋅−2𝑥⋅𝑥−𝑦𝑥−𝑦𝑥−𝑦²|𝑥|−6𝑦⋅𝑥−𝑦𝑥−𝑦⋅𝑥²−|𝑥|⁴𝑥−𝑦²+3𝑥−𝑦⋅𝑥|𝑥|⁴×𝑢^−𝑞𝑦=1𝑑𝑦𝐶_5𝑅⁵4|𝑥|⁴𝑥−𝑦⁵𝑆𝑢^−𝑞𝑦𝑑𝑦, (64) where 𝑆=−3𝑥−𝑦⁶+5𝑥−𝑦⋅𝑥𝑥−𝑦⁴−4|𝑥|²𝑥−𝑦⁴−2𝑥⋅𝑥−𝑦𝑦⋅𝑥−𝑦𝑥−𝑦²|𝑥|−6𝑦⋅𝑥−𝑦𝑥−𝑦⋅𝑥²−|𝑥|⁴𝑥−𝑦²+3𝑥−𝑦⋅𝑥|𝑥|⁴=2𝑥⋅𝑥−𝑦𝑥−𝑦⁴+3𝑦⋅𝑥−𝑦𝑥−𝑦⁴−4|𝑥|²𝑥−𝑦²𝑥⋅𝑥−𝑦+4|𝑥|²𝑥−𝑦²𝑦⋅𝑥−𝑦−2𝑥⋅𝑥−𝑦𝑦⋅𝑥−𝑦𝑥−𝑦²|𝑥|−6𝑦⋅𝑥−𝑦𝑥−𝑦⋅𝑥²+2|𝑥|⁴𝑥⋅𝑥−𝑦+|𝑥|⁴2𝑦⋅𝑥−𝑦=𝑥⋅𝑥−𝑦𝑥−𝑦⁴−4|𝑥|²𝑥−𝑦²+2|𝑥|⁴+𝑦⋅𝑥−𝑦3𝑥−𝑦⁴+4|𝑥|²𝑥−𝑦²⋅−2𝑥𝑥−𝑦𝑥−𝑦²−6𝑥⋅𝑥−𝑦|𝑥|²+|𝑥|⁴. (65)
And directly calculation gives 3𝑥−𝑦⁴+4|𝑥|²𝑥−𝑦²−2𝑥⋅𝑥−𝑦𝑥−𝑦²−6𝑥⋅𝑥−𝑦|𝑥|²+|𝑥|⁴=𝑥⋅𝑥−𝑦𝑥−𝑦²+4|𝑥|²𝑥−𝑦²⋅𝑥|−6𝑥𝑥−𝑦|²+|𝑥|⁴−3𝑦⋅𝑥−𝑦𝑥−𝑦²=𝑥⋅𝑦²+𝑦⋅𝑥−𝑦4𝑥⋅𝑥−𝑦+4|𝑥|²−3𝑦⋅𝑥−𝑦. (66)
By substitution, we have 𝑦⋅+𝑆=2𝑥⋅𝑥−𝑦𝑦⋅𝑥−𝑦𝑦⋅𝑥−𝑦+2𝑥⋅𝑦𝑥−𝑦𝑥⋅𝑦²+𝑦⋅𝑥−𝑦𝑥⋅𝑦²−𝑦⋅𝑥−𝑦²×4𝑥⋅𝑥−𝑦+4|𝑥|²−3𝑦⋅𝑥−𝑦=−2𝑥⋅𝑥−𝑦𝑦⋅𝑥−𝑦²+4𝑥⋅𝑦𝑥⋅𝑥−𝑦𝑦⋅𝑥−𝑦−4𝑥⋅𝑥𝑦⋅𝑥−𝑦²+2𝑥⋅𝑦²𝑥⋅𝑥−𝑦+𝑦⋅𝑥−𝑦𝑥⋅𝑦²+3𝑦⋅𝑥−𝑦³×−|𝑥|=2𝑦⋅𝑥−𝑦₂𝑥⋅𝑦+3|𝑥|²𝑦²−𝑥⋅𝑦²−𝑦²𝑥⋅𝑦+2𝑥⋅𝑦²𝑥⋅𝑥−𝑦+𝑦⋅𝑥−𝑦𝑥⋅𝑦²+3𝑦⋅𝑥−𝑦³=−2|𝑥|²𝑥⋅𝑦𝑦⋅𝑥−𝑦+2𝑥⋅𝑦²6|𝑥|𝑥⋅𝑥−𝑦+𝑦⋅𝑥−𝑦²𝑦²−𝑥⋅𝑦²+3𝑦⋅𝑥−𝑦²−2𝑦²|𝑥|𝑥⋅𝑦𝑦⋅𝑥−𝑦=8²𝑦²𝑥⋅𝑦−6|𝑥|²𝑦⁴−10𝑦²𝑥⋅𝑦²+11𝑦⁴𝑥⋅𝑦−3𝑦⁶. (67)
Hence 4|𝑥|⁴−3𝛽₁(𝑥)+5𝛽₂(𝑥)+8|𝑥|³𝛽₃(𝑥)−2|𝑥|²3𝛽₄(𝑥)+2𝛽₅(𝑥)−2|𝑥|𝛽₆(𝑥)+32𝛽₇(𝑥)+|𝑥|2𝛽₈(𝑥=1)𝐶_5𝑅⁵4|𝑥|⁴𝑥−𝑦⁵8|𝑥|²𝑦²𝑥⋅𝑦−6|𝑥|²𝑦⁴−10𝑦²𝑥⋅𝑦²+11𝑦⁴𝑦𝑥⋅𝑦−3⁶𝑢^−𝑞𝑦𝑑𝑦. (68)
Now, integrate it over 𝜕𝐵_𝑟(0) with 𝑟=|𝑥|, and we obtain that 𝐴∶=_{𝜕𝐵𝑟(0)}1𝐶_5𝑅⁵4|𝑥|⁴𝑥−𝑦⁵×8|𝑥|²𝑦²𝑥⋅𝑦−6|𝑥|²𝑦⁴−10𝑦²𝑥⋅𝑦²+11𝑦⁴𝑥⋅𝑦−3𝑦⁶×𝑢^−𝑞𝑦=1𝑑𝑦𝑑𝛿𝐶_{5|𝑦|<|𝑥|}+_{|𝑦|>|𝑥|}×_{𝜕𝐵𝑟(0)}4|𝑥|⁴𝑥−𝑦⁵×8|𝑥|²𝑦²𝑥⋅𝑦−8|𝑥|²𝑦⁴+2|𝑥|²𝑦⁴−10𝑦²𝑥⋅𝑦²+11𝑦⁴𝑥⋅𝑦−3𝑦⁶×𝑢^−𝑞𝑦𝑑𝛿𝑑𝑦∶=𝐼₁+𝐼₂+𝐼₃+𝐼𝐼₁+𝐼𝐼₂+𝐼𝐼₃, (69) where 𝐼₁=_{|𝑦|<|𝑥|𝜕𝐵𝑟(0)}4|𝑥|⁴𝑥−𝑦⁵8|𝑥|²𝑦²𝑥⋅𝑦−8|𝑥|²𝑦⁴×𝑢^−𝑞𝑦=𝑑𝑦𝑑𝛿_{|𝑦|<|𝑥|}32𝑦²|𝑥|²𝑢^−𝑞𝑦_{𝜕𝐵𝑟(0)}𝑥⋅𝑦−𝑦²𝑥−𝑦⁵𝑑𝛿𝑑𝑦. (70) It is clear that Δ_𝑥[(𝑥⋅𝑦)−|𝑦|²]=0, by Poisson representation formula for harmonic function, the previous integral equals 𝐼₁=_{|𝑦|<|𝑥|}32𝑦²|𝑥|²𝑢^−𝑞𝑦|𝑥|𝜔₅|𝑥|²−𝑦²𝑦⋅𝑦−𝑦²𝑑𝑦=0. (71)
Similarly, calculation gives Δ_𝑥[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 𝐼₂=_{|𝑦|<|𝑥|𝜕𝐵𝑟(0)}4|𝑥|⁴𝑥−𝑦⁵×2|𝑥|²𝑦⁴−10𝑦²𝑥⋅𝑦²𝑢^−𝑞𝑦=𝑑𝑦𝑑𝛿_{|𝑦|<|𝑥|}4|𝑥|⁴𝑢^−𝑞𝑦×_{𝜕𝐵𝑟(0)}2|𝑥|²𝑦⁴−10𝑦²𝑥⋅𝑦²𝑥−𝑦⁵=𝑑𝛿𝑑𝑦_{|𝑦|<|𝑥|}4|𝑥|⁴𝑢^−𝑞𝑦|𝑥|𝜔₅|𝑥|²−𝑦²2𝑦⁶−10𝑦⁶=𝑑𝑦_{|𝑦|<|𝑥|}4|𝑥|⁴𝑢^−𝑞𝑦|𝑥|𝜔₅|𝑥|²−𝑦²−8𝑦⁶𝑑𝑦. (72)
Once again, Δ_𝑥[11|𝑦|⁴(𝑥⋅𝑦)−3|𝑦|⁶]=0, and Poisson formula leads to 𝐼₃=_{|𝑦|<|𝑥|𝜕𝐵𝑟(0)}4|𝑥|⁴𝑥−𝑦⁵×11𝑦⁴𝑥⋅𝑦−3𝑦⁶𝑢^−𝑞𝑦=𝑑𝛿𝑑𝑦_{|𝑦|<|𝑥|}4|𝑥|⁴𝑢^−𝑞𝑦_{𝜕𝐵𝑟(0)}11𝑦⁴𝑥⋅𝑦−3𝑦⁶𝑥−𝑦⁵=𝑑𝛿𝑑𝑦_{|𝑦|<|𝑥|}4|𝑥|⁴𝑢^−𝑞𝑦|𝑥|𝜔₅|𝑥|²−𝑦²8𝑦⁶𝑑𝑦. (73)
Notice that 𝐼₂+𝐼₃=0.
Next, consider the following integral: 𝐼𝐼₁=_{|𝑦|>|𝑥|𝜕𝐵𝑟(0)}4|𝑥|⁴𝑥−𝑦⁵×8|𝑥|²𝑦²𝑥⋅𝑦+11𝑦⁴𝑥⋅𝑦×𝑢^−𝑞𝑦=𝑑𝛿𝑑𝑦_{|𝑦|>|𝑥|}4|𝑥|⁴𝑢^−𝑞𝑦8|𝑥|²𝑦²+11𝑦⁴×_{𝜕𝐵𝑟(0)}𝑥⋅𝑦𝑥−𝑦⁵𝑑𝛿𝑑𝑦. (74) Setting 𝑧=(|𝑥|²𝑦)/|𝑦|², the previous integral is equal to 𝐼𝐼₁=_{|𝑦|>|𝑥|}4|𝑥|⁴𝑢^−𝑞𝑦8|𝑥|²𝑦²+11𝑦⁴×_{𝜕𝐵𝑟(0)}𝑥⋅|𝑥|²𝑧/|𝑧|²𝑥−|𝑥|²𝑧/|𝑧|²⁵=𝑑𝛿𝑑𝑦_{|𝑦|>|𝑥|}4|𝑥|⁴𝑢^−𝑞𝑦8|𝑥|²𝑦²+11𝑦⁴×_{𝜕𝐵𝑟(0)}|𝑥|²/|𝑧|²𝑥⋅𝑧(|𝑥|/|𝑧|)⁵|𝑥−𝑧|⁵=𝑑𝛿𝑑𝑦_{|𝑦|>|𝑥|}4|𝑥|⁴𝑢^−𝑞𝑦8|𝑥|²𝑦²+11𝑦⁴|𝑥||𝑧|⁻³×|𝑥|𝜔₅|𝑥|²−|𝑧|²𝑧⋅𝑧𝑑𝑦=_{|𝑦|>|𝑥|}𝑢^−𝑞𝑦×32𝜔₅|𝑥|⁴𝑦𝑦²−|𝑥|²+44𝜔₅|𝑥|²𝑦𝑦²−|𝑥|²𝑑𝑦. (75)
Also, we have 𝐼𝐼₂=_{|𝑦|>|𝑥|𝜕𝐵𝑟(0)}4|𝑥|⁴𝑥−𝑦⁵−8|𝑥|²𝑦⁴−3𝑦⁶×𝑢^−𝑞𝑦=𝑑𝛿𝑑𝑦_{|𝑦|>|𝑥|}4|𝑥|⁴𝑢^−𝑞𝑦−8|𝑥|²𝑦⁴−3𝑦⁶×_{𝜕𝐵𝑟(0)}1𝑥−𝑦⁵=𝑑𝛿𝑑𝑦_{|𝑦|>|𝑥|}4|𝑥|⁴𝑢^−𝑞𝑦−8|𝑥|²𝑦⁴−3𝑦⁶×_{𝜕𝐵𝑟(0)}1𝑥−|𝑥|²𝑧/|𝑧|²⁵=𝑑𝛿𝑑𝑦_{|𝑦|>|𝑥|}4|𝑥|⁴𝑢^−𝑞𝑦−8|𝑥|²𝑦⁴−3𝑦⁶|𝑥||𝑧|⁻⁵×_{𝜕𝐵𝑟(0)}1|𝑥−𝑧|⁵=𝑑𝛿𝑑𝑦_{|𝑦|>|𝑥|}4|𝑥|⁴𝑢^−𝑞𝑦−8|𝑥|²𝑦⁴−3𝑦⁶×|𝑥||𝑧|⁻⁵|𝑥|𝜔₅|𝑥|²−|𝑧|²𝑑𝑦=_{|𝑦|>|𝑥|}𝑢^−𝑞𝑦−32𝜔₅𝑦|𝑥|²𝑦²−|𝑥|²−12𝜔₅𝑦³𝑦²−|𝑥|²𝑑𝑦. (76)
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 𝐼𝐼₃=_{|𝑦|>|𝑥|𝜕𝐵𝑟(0)}4|𝑥|⁴𝑥−𝑦⁵×−10𝑦²𝑥⋅𝑦²+2|𝑥|²𝑦⁴×𝑢^−𝑞𝑦=𝑑𝛿𝑑𝑦_{|𝑦|>|𝑥|}4𝑦²|𝑥|⁴𝑢^−𝑞𝑦×_{𝜕𝐵𝑟(0)}1𝑥−|𝑥|²𝑧/|𝑧|²⁵×|𝑥|−10𝑥⋅²𝑧|𝑧|²²|𝑥|+2⁶|𝑧|²=𝑑𝛿𝑑𝑦_{|𝑦|>|𝑥|}4𝑦²|𝑥|⁴𝑢^−𝑞𝑦×_{𝜕𝐵𝑟(0)}1(|𝑥|/|𝑧|)⁵|𝑥−𝑧|⁵×|𝑥|−10²|𝑧|²²(𝑥⋅𝑧)²|𝑥|+2⁶|𝑧|²=𝑑𝛿𝑑𝑦_{|𝑦|>|𝑥|}4𝑦²|𝑥|⁴|𝑥|𝑦𝑢^−𝑞𝑦×_{𝜕𝐵𝑟(0)}−10(𝑥⋅𝑧)²+2|𝑥|²|𝑧|²|𝑥−𝑧|⁵𝑑𝛿𝑑𝑦=_{|𝑦|>|𝑥|}4𝑦|𝑥|³𝑢^−𝑞𝑦|𝑥|𝜔₅|𝑥|²−|𝑧|²−8|𝑧|⁴𝑑𝑦=−32𝜔_{5|𝑦|>|𝑥|}𝑢^−𝑞𝑦|𝑥|⁴𝑦²−|𝑥|²𝑦𝑑𝑦. (77)
Combining those previous calculations, we can conclude that 𝐴=𝐼₁+𝐼₂+𝐼₃+𝐼𝐼₁+𝐼𝐼₂+𝐼𝐼₃=_{|𝑦|>|𝑥|}𝑢^−𝑞𝑦×32𝜔₅|𝑥|⁴𝑦𝑦²−|𝑥|²+44𝜔₅|𝑥|²𝑦𝑦²−|𝑥|²+𝑑𝑦_{|𝑦|>|𝑥|}𝑢^−𝑞𝑦−32𝜔₅𝑦|𝑥|²𝑦²−|𝑥|²−12𝜔₅𝑦³𝑦²−|𝑥|²𝑑𝑦−32𝜔_{5|𝑦|>|𝑥|}𝑢^−𝑞𝑦|𝑥|⁴𝑦²−|𝑥|²𝑦𝑑𝑦=−12𝜔_{5|𝑦|>|𝑥|}𝑢^−𝑞𝑦𝑦𝑑𝑦⟶0, (78) as |𝑥|→∞, since 𝑢(0)=∫_𝑅5𝑢^−𝑞(𝑦)|𝑦|𝑑𝑦<+∞.
What we have shown so far is that, if 𝑢 is regular, that is, 𝑞>6, then 𝑃(𝑢,𝑟)⟶0as𝑟⟶∞. (79)
Setting 𝑄=1, when 𝑞>6, we have _{𝜕𝐵𝑟(0)}𝑟𝑢^1−𝑞𝑑𝛿⟶0, (80) 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 exponent 𝑞=11.

First, define 𝑣(𝑥)=|𝑥|𝑢(𝑥/|𝑥|²) and 𝑣_𝜆(𝑥)=𝜆𝑢(𝑥/|𝜆|²); it is rather easy to see that 𝑣_𝜆(𝑥)=(|𝑥|/𝜆)𝜐((𝜆²𝑥)/|𝑥|²).

Lemma 5. If one sets 𝜔_𝜆(𝑥)=−𝜐(𝑥)+𝜐_𝜆(𝑥), then 𝜔_𝜆(𝑥)=∫_𝐵𝜆[𝐺(𝑥,𝑧)𝜓(𝑧)𝜔_𝜆(𝑧)]𝑑𝑧, where 𝜓=(1/(𝜐¹¹𝜐_𝜆¹¹))[∑¹⁰_𝑖=0𝜐^𝑖𝜐_𝜆^10−𝑖]≥0; 𝐺(𝑥,𝑧)=[(|𝑧|/𝜆)|𝑥−((𝜆²𝑧)/|𝑧|²)|−|𝑥−𝑧|]≥0 when |𝑥|≤𝜆 and |𝑧|≤𝜆. In particular, 𝜔_𝜆(0)=∫_𝐵𝜆[𝜆−|𝑧|]𝜓(𝑧)𝜔_𝜆(𝑧)𝑑𝑧, which belongs to (0,+∞).

Proof. Since 𝑢(𝑥)=(1/𝐶₅)∫_𝑅5|𝑥−𝑦|𝑢⁻¹¹(𝑦)𝑑𝑦, 𝑣(𝑥)=|𝑥|𝑢𝑥|𝑥|²=1𝐶_5𝑅⁵|𝑥𝑥||𝑥|²−𝑦𝑢⁻¹¹𝑦=1𝑑𝑦𝐶_5𝑅⁵𝑦𝑦𝑥−𝑦²𝑢⁻¹¹𝑦=1𝑑𝑦𝐶₅𝐼₁+𝐼₂, (81) 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 for 𝐼₁, set 𝑦=𝑧/𝜆² to obtain 𝐼₁=_𝐵𝜆|𝑧|𝜆²𝑥−𝑧/𝜆²𝑧/𝜆²²𝑢⁻¹¹|𝑧|𝜆²𝑑𝑧𝜆²=_𝐵𝜆𝜆⁻¹²|𝑧𝜆|𝑥−²𝑧|𝑧|²𝑢⁻¹¹𝑧𝜆²𝑑𝑧. (82)
As for 𝐼₂, if we set 𝑦=𝑧/|𝑧|², then we obtain 𝐼₂=_𝐵𝜆|𝑧||𝑧|²|𝑧|𝑥−𝑧/²𝑧/|𝑧|²²𝑢⁻¹¹|𝑧||𝑧|²𝑑𝑧|𝑧|²=_𝐵𝜆|𝑧|⁻¹¹|𝑥−𝑧|𝑢⁻¹¹𝑧|𝑧|²𝑑𝑧. (83)
Similarly, one can write 𝑣_𝜆(𝑥) as 𝑣_𝜆(𝑥)𝑥=𝜆𝑢𝜆²=1𝐶_5𝐵1/𝜆+_{𝑅5/𝐵1/𝜆}𝜆𝑥𝜆²−𝑦𝑢⁻¹¹𝑦=1𝑑𝑦𝐶₅𝐼₃+𝐼₄. (84)
As for 𝐼₃, by setting 𝑦=𝑧/𝜆², one concludes that 𝐼₃=_𝐵𝜆𝜆⁻¹¹|𝑥−𝑧|𝑢⁻¹¹𝑧𝜆²𝑑𝑧. (85) For 𝐼₄, as before, set 𝑦=𝑧/|𝑧|² to get 𝐼₄=_𝐵𝜆𝜆⁻¹|𝑧|⁻¹⁰𝜆𝑥−²𝑧|𝑧|²𝑢⁻¹¹𝑧|𝑧|²𝑑𝑧. (86) Therefore, combining these previous calculations, we get 𝜔_𝜆(𝑥)=1𝐶₅𝐼₁−𝐼₄+𝐼₂−𝐼₃=_𝐵𝜆|𝑧|𝜆𝜆𝑥−²𝑧|𝑧|²−|𝑥−𝑧|𝑣⁻¹¹(𝑧)−𝑣_𝜆⁻¹¹(𝑧)𝑑𝑧=_𝐵𝜆|𝑧|𝜆𝜆𝑥−²𝑧|𝑧|²−|×𝜐𝑥−𝑧|₋₁₁(𝑧)−𝜐_𝜆⁻¹¹(𝑧)𝜔_𝜆(𝑧)𝜔_𝜆(𝑧)𝑑𝑧=_𝐵𝜆𝐺(𝑥,𝑧)𝜓(𝑧)𝜔_𝜆(𝑧)𝑑𝑧, (87) where 𝜓=[(𝜐⁻¹¹(𝑧)−𝜐_𝜆⁻¹¹(𝑧))/𝜔_𝜆(𝑧)]=(1/𝜐¹¹𝜐_𝜆¹¹)[∑¹⁰_𝑖=0𝜐^𝑖𝜐_𝜆^10−𝑖]≥0 and 𝐺=[(|𝑧|/𝜆)|𝑥−(𝜆²𝑧/|𝑧|²)|−|𝑥−𝑧|].
Now, since |𝑧|²𝜆²𝜆𝑥−²𝑧|𝑧|²²−|𝑥−𝑧|²=|𝑧|²𝜆|𝑥|²𝜆−2²|𝑧|²⟨⟩+𝜆𝑥,𝑧⁴|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=1𝜆²𝜆²−|𝑥|²𝜆²−|𝑧|², (88)𝐺(𝑥,𝑧)=[(|𝑧|/𝜆)|𝑥−(𝜆²𝑧/|𝑧|²)|−|𝑥−𝑧|]≥0 when |𝑥|≤𝜆 and |𝑧|≤𝜆.
Put 𝑥=0 to obtain 𝜔_𝜆(0)=_𝐵𝜆𝐺(0,𝑧)𝜓(𝑧)𝜔_𝜆(𝑧=)𝑑𝑧_𝐵𝜆[𝜆−|𝑧|]𝜓(𝑧)𝜔_𝜆(𝑧)𝑑𝑧, (89) which belongs to (0,+∞).
Thus we complete the proof.

Proposition 6. If 𝜆>0 is large enough, then 𝜔_𝜆(𝑥)≥0 for all 𝑥 such that |𝑥|≤𝜆.

Proof. We prove that 𝜔_𝜆(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 𝜔_𝜆≥0.
It is clear to see that |𝑥/|𝑥|²|=1/|𝑥|≤1/𝑅₀; |𝑥/𝜆²|≤1/2𝜆≤1/4𝑅₀.
When 𝑅₀ is large enough, 𝜔_𝜆(𝑥)|𝑥|𝑢𝑥=−|𝑥|²𝑥+𝜆𝑢𝜆²|𝑥𝑢(0)1=−|+𝑂|𝑥|𝑢(0)|𝑥|+𝜆+𝑂𝜆²≥|𝑥|2𝑢(0)>0. (90)
Step 2. There exists an 𝑅₁>𝑅₀ such that for 𝑅₁≤𝜆/2≤|𝑥|≤𝜆, we have Δ𝜔_𝜆≤0 and 𝜔_𝜆≥0.
First, we calculate −Δ𝜔_𝜆. In fact −Δ𝜔_𝜆=−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, for 𝑛=5 one observes that Δ|𝑥|=4|𝑥|,∇|𝑥|𝑥⋅∇𝑢|𝑥|²=−|𝑥|⁵⁻³_𝑖=1𝑢_𝑖𝑥|𝑥|²⋅𝑥_𝑖,𝑥Δ𝑢𝑥²=|𝑥|⁻⁴()𝑥Δ𝑢|𝑥|²+(−6)|𝑥|⁵⁻⁴_𝑖=1𝑢_𝑖𝑥|𝑥|²⋅𝑥_𝑖. (91) Thus −Δ𝜔_𝜆=−|𝑥|⁻²|𝑥|²𝜆³()𝑥Δ𝑢𝜆−1|𝑥|()𝑥Δ𝑢|𝑥|²+8|𝑥|⁵_𝑖=1𝑢_𝑖𝑥|𝑥|²⋅𝑥_𝑖|𝑥|²−4𝑣(𝑥). (92)
If 𝑅₁>1, then |𝑥/|𝑥|²|=1/|𝑥|≤1/𝑅₁≤1 if |𝑥|≥𝑅₁. Let 𝐶=max_𝐵1|∇𝑢|. Then |8∑⁵_𝑖=1𝑢_𝑖(𝑥/|𝑥|²)⋅(𝑥_𝑖/|𝑥|²)|≤8𝐶/|𝑥| when 𝑅₁ is large enough. Now let 𝐵=max_𝐵1|Δ𝑢|. Hence we have |𝑥|²𝜆³()𝑥Δ𝑢𝜆²−1|𝑥|()𝑥Δ𝑢|𝑥|²≤𝜆²𝜆³1𝐵+|𝑥|3𝐵≤𝜆𝐵. (93)
Notice that if |𝑥| is sufficiently large, then 𝑣(𝑥)≥(𝑢(0)/2)|𝑥|. Therefore if 𝜆>0 is large enough, then −Δ𝜔_𝜆(𝑥)≥0 for 𝑅₁≤𝜆/2≤|𝑥|≤𝜆.
Besides, 𝜔_𝜆(𝑥)|_|𝑥|=𝜆=0 by definition, and 𝜔_𝜆(𝑥)|_{|𝑥|=𝜆/2}≥0 by Step 1. Thus the maximal principle implies that 𝜔_𝜆(𝑥)≥0 on the domain 𝑅₁≤𝜆/2≤|𝑥|≤𝜆.
Step 3. There exists an 𝑅₂≥𝑅₁ such that 𝜔_𝜆(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 𝑣_𝜆(𝑥)=𝜆𝑢(𝑥/𝜆²) implies that 𝑣_𝜆(𝑥)≥(𝑢(0)/2)𝜆 when 𝜆 is large enough. Therefore, 𝜔_𝜆(𝑥)>0 for 0<|𝑥|≤𝑅₀ when 𝜆 is large enough.
Combining Steps 1 to 3, we have 𝜔_𝜆(𝑥)≥0.
Thus we complete the proof of this proposition.

Now, for any 𝑏∈𝑅⁵, we denote 𝑢(𝑥+𝑏) by 𝑢_𝑏(𝑥), 𝑣_𝑏(𝑥) by |𝑥|𝑢_𝑏(𝑥/|𝑥|²), and set 𝜔_𝜆,𝑏(𝑥)=−𝑣_𝑏(𝑥)+(𝑣_𝑏)_𝜆(𝑥).

By Proposition 6, if 𝜆 is large enough, then 𝜔_𝜆,𝑏≥0. Thus, we define 𝜆_𝑏=inf{𝜆>0∶𝜔_𝜇,𝑏(𝑥)≥0 in 𝐵_𝜇(0) for 𝜆<𝜇<+∞}.

Proposition 7. There exists 𝑏∈𝑅⁵ such that 𝜆_𝑏>0.

Proof. Suppose that the proposition is not true. Hence 𝜆_𝑏=0 for all 𝑏; that is, for all 𝑏∈𝑅⁵ and all 𝜆>0, we always have 𝜔_𝜆,𝑏(𝑥)≥0 for all 𝑥∈𝐵_𝜆(0).
Note that 𝜔_|𝑥|,𝑏(𝑥)=−𝑣_𝑏|𝑥|(𝑥)+𝜆𝑣_𝑏𝜆²𝑥|𝑥|²_𝜆=|𝑥|𝜔=0,_|𝑥|,𝑏(𝑟𝑥)=−𝑣_𝑏(𝑟𝑥)+𝑟𝑣_𝑏𝑥𝑟≥0when0<𝑟<1. (94)
Therefore, it follows that (𝑑/(𝑑𝑟))𝜔_|𝑥|,𝑏(𝑟𝑥)|_𝑟=1≤0; that is, −∇𝑣_𝑏(𝑟𝑥)⋅𝑥+𝑣_𝑏𝑥𝑟+𝑟∇𝑣_𝑏𝑥𝑟−𝑥𝑟²_𝑟=1≤0. (95)
Simplifying this, we get −2∇𝑣_𝑏⋅𝑥+𝑣_𝑏(𝑥)≤0, (96) where 𝑣_𝑏(𝑥)=|𝑥|𝑢((𝑥/|𝑥|²)+𝑏).
Setting 𝑦=𝑥/|𝑥|², by calculation we get ∇𝑣_𝑏|𝑥|𝑢𝑥⋅𝑥=|𝑥|²+|𝑥)𝑥+𝑏|(∇𝑢|𝑥|²⋅𝑥+𝑏|𝑥|²=|𝑥|𝑢𝑦+𝑏−|𝑥|(∇𝑢)𝑦+𝑏⋅𝑦. (97)
Substituting (97) into (96) and doing variable change, we obtain −12𝑢𝑦+(∇𝑢)𝑦⋅𝑦−𝑏≤0. (98)
Then both sides being divided by |𝑏| and sending |𝑏| to +∞, let 𝑏=|𝑏|𝜔_𝑏, we get (∇𝑢)𝑦⋅𝜔_𝑏≤0. (99)
Since the previous inequality holds for all 𝑏∈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 point 𝑏∈𝑅⁵ such that 𝜆_𝑏>0; then 𝜔_{𝜆𝑏,𝑏}≡0 for all 𝑥∈𝑅⁵.

Proof. Without loss of generality, we assume that 𝑏=0. Suppose that the proposition is not true; then 𝜔_𝜆0,0(𝑥)≢0.
In order to complete the argument, we need to do some preparations.
First of all, setting 𝐶₀=min_{𝑥∈𝐵𝜆0/2}𝜔_𝜆0,0(𝑥)>0, then it is clear that 𝜔_𝜆0(𝑥)≥𝐶₀>0∀𝑥∈𝐵_𝜆0/2. (100)
Next, calculate (𝜕𝐺(𝑥,𝑧)/𝜕𝜂_𝑥)|_|𝑥|=𝜆0 as follows.
Since we have the formula |𝑧|𝜆𝜆𝑥−²𝑧|𝑧|²=|𝑥|𝜆𝜆𝑧−²𝑥|𝑥|², (101) which is because |𝑧|𝜆𝜆𝑥−²𝑧|𝑧|²²=|𝑧|²𝜆²𝜆𝑥−²𝑧|𝑧|²𝜆,𝑥−²𝑧|𝑧|²=|𝑧|²𝜆²⟩𝜆⟨𝑥,𝑥−2²|𝑧|²⟨⟩+𝜆𝑥,𝑧⁴|𝑧|²=|𝑧|²|𝑥|²𝜆²−2⟨𝑥,𝑧⟩+𝜆²=|𝑥|²𝜆²⟩𝜆⟨𝑧,𝑧−2²|𝑥|²⟨⟩+𝜆𝑥,𝑧⁴|𝑥|²=|𝑥|𝜆𝜆𝑧−²𝑥|𝑥|²², (102) thus, 𝐺()=|𝑧|𝑥,𝑧𝜆𝜆𝑥−²𝑧|𝑧|²−||=|𝑥|𝑥−𝑧𝜆𝜆𝑧−²𝑥|𝑥|²−|𝑥−𝑧|. (103)
Hence, by direct calculation, one obtains 𝜕𝐺(𝑥,𝑧)𝜕𝜂_{𝑥|𝑥|=𝜆0}=_𝑖𝜕𝜕𝑥_𝑖𝐺()⋅𝑥𝑥,𝑧^𝑖|𝑥|_|𝑥|=𝜆0=⁵_𝑖=1|𝑧|𝜆𝜕𝜕𝑥_𝑖𝜆𝑥−²𝑧|𝑧|²⋅𝑥_𝑖|𝑥|−⁵_𝑖=1𝜕𝜕𝑥_𝑖||⋅𝑥𝑥−𝑧^𝑖|𝑥|=⁵_𝑖=1|𝑧|𝜆𝑥_𝑖−𝜆²𝑧_𝑖/|𝑧|²𝑥−𝜆²𝑧/|𝑧|²−𝑥_𝑖−𝑧_𝑖||⋅𝑥𝑥−𝑧|𝑥^𝑖|_|𝑥|=𝜆0=⁵_𝑖=1|𝑧|𝜆𝑥_𝑖−𝜆²𝑧_𝑖/|𝑧|²(|𝑥|/|𝑧|)𝑧−𝜆²𝑥/|𝑥|²−𝑥_𝑖−𝑧_𝑖||⋅𝑥𝑥−𝑧_𝑖|𝑥|_|𝑥|=𝜆0=|𝑧|²−𝜆₀²𝜆₀||.𝑥−𝑧 (104)
It clearly implies that 𝜕𝐺(𝑥,𝑧)𝜕𝜂_{𝑥𝜕𝐵𝜆0(0)}≤0 (105) whenever |𝑧|≤𝜆.
Therefore, by Lemma 5, 𝜕𝜔_𝜆0(𝑥)𝜕𝜂_{𝑥𝜕𝐵𝜆0(0)}=1𝐶_{5𝐵𝜆0(0)}|𝑧|²−𝜆²₀𝜆₀||𝜓(𝑧)𝜔𝑥−𝑧^𝜆0(𝑧)𝑑𝑧. (106)
By the definition of 𝜆₀ and our assumption that 𝜔_𝜆0,0(𝑥)≢0, from Lemma 5, we see that 𝜔_𝜆0(𝑥)>0 inside the ball 𝐵_𝜆0(0). Then, at every boundary point 𝑥, we have 𝜕𝜔_𝜆0(𝑥)𝜕𝜂_{𝑥𝜕𝐵𝜆0(0)}<0. (107)
Besides, the definition of 𝜆₀ implies that there is a sequence 𝜆_𝑘 such that 𝜆_𝑘⟶𝜆₀,𝜆_𝑘<𝜆₀,inf_{𝐵𝜆𝑘(0)}𝜔_𝜆𝑘(𝑥)<0. (108)
It follows from (108) that there exists 𝑥_𝑘∈𝐵_𝜆𝑘(0) such that 𝜔_𝜆𝑘𝑥_𝑘=min_{𝐵𝜆𝑘(0)}(𝑥)<0. (109) Since 𝜔_𝜆𝑘(𝑥)≡0 for |𝑥|=𝜆_𝑘, we have |𝑥_𝑘|<𝜆_𝑘<𝜆₀. Clearly, 𝑥_𝑘 has a convergent subsequence, still denoted by 𝑥_𝑘, with a limit point 𝑥₀. Also, ∇𝜔_𝜆𝑘(𝑥_𝑘)=0. Then 𝑥₀ satisfies ∇𝜔_𝜆0𝑥₀=0,𝜔_𝜆0𝑥₀≤0. (110)
Now consider the following two cases.
Case 1 (|𝑥₀|<𝜆₀). The conclusion 𝜔_𝜆0(𝑥₀)≤0 contradicts the fact that 𝜔_𝜆0(𝑥₀) is strictly positive inside the ball 𝐵_𝜆0(0).
Case 2 (|𝑥₀|=𝜆₀). Then the fact that ∇𝜔_𝜆0(𝑥₀)=0 contradicts (107).
Hence, the proposition follows.

Proposition 9. For all 𝑏∈𝑅⁵, one has 𝜆_𝑏>0.

Proof. It follows from Propositions 7 and 8 that there exists some 𝑏∈𝑅⁵ such that 𝜆_𝑏>0 and 𝜔_𝜆𝑏(𝑥)=0; that is, 𝑣_𝑏(𝑥)=(𝑣_𝑏(𝑥))_𝜆; that is, |𝑥|𝑢_𝑏𝑥|𝑥|²=𝜆_𝑏𝑢_𝑏𝑥𝜆²_𝑏. (111)
Setting 𝑦=𝑥/|𝑥|², we get 𝑦⁻¹𝑢_𝑏𝑦=𝜆_𝑏𝑢_𝑏𝑦𝜆²_𝑏𝑦². (112)
It follows that lim_𝑦→∞𝑦⁻¹𝑢_𝑏𝑦=𝜆_𝑏𝑢_𝑏(0). (113)
Suppose that there exists a point 𝑏∈𝑅⁵ such that 𝜆_𝑏=0. Then 𝜔_𝜆,𝑏(𝑥)=−𝑣_𝑏(𝑥)+(𝑣_𝑏(𝑥))_𝜆≥0 for all 𝜆>0 and 𝑥∈𝐵_𝜆(0)/{0}.
That is, 𝜆𝑢_𝑏(|𝑥|/𝜆²)≥|𝑥|𝑢_𝑏(𝑥/|𝑥|²) for all 𝜆>0 and 𝑥∈𝐵_𝜆(0)⧵{0}.
Fixing 𝜆>0 and letting |𝑥|→0, we have 𝜆𝑢_𝑏(0)≥𝜆_𝑏𝑢_𝑏(0).
Therefore, 𝜆𝜆_𝑏𝑢(𝑏)≥𝑢𝑏. (114)
Letting 𝜆 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. If 𝑢>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 Propositions 8 and 9, we know that for all 𝑏∈𝑅⁵, we have 𝜔_{𝜆𝑏,𝑏}≡0; that is, 𝑣_𝑏(𝑥)=(|𝑥|/𝜆_𝑏)𝑣_𝑏(𝜆²_𝑏𝑥/|𝑥|²).
Using 𝑣(𝑥)=|𝑥|𝑢(𝑥/|𝑥|²), setting 𝑦=𝑥/𝜆²_𝑏 and 𝜇_𝑏=𝜆_𝑏⁻¹, we can rewrite it as follows: 𝑢𝑦+𝑏=𝑦𝜇_𝑏⁻¹𝑢𝜇²_𝑏𝑦𝑦²+𝑏. (115)
Now setting 𝑥=𝑦+𝑏, then we have 𝑢(𝑥)=𝜇_𝑏⁻¹||𝑢𝜇𝑥−𝑏^2𝑏(𝑥−𝑏)|𝑥−𝑏|²+𝑏. (116)
It follows that the limit lim_|𝑥|→∞|𝑥|⁻¹𝑢(𝑥) exists and is equal to 𝜇_𝑏⁻¹𝑢(𝑏) for all 𝑏∈𝑅⁵.
Then set 𝐴=lim_|𝑥|→∞|𝑥|⁻¹𝑢(𝑥)=𝜇_𝑏⁻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, 𝑢(𝑥)=𝜇₀⁻¹|𝑥|⁻¹𝑢(0)+𝜕𝑢𝜕𝑥_𝑖(0)⋅𝜇²₀𝑥_𝑖|𝑥|²1+𝑜|𝑥|, (117) so the coefficient of 𝑥_𝑖 is (𝜇₀/|𝑥|)(𝜕𝑢/𝜕𝑥_𝑖)(0), and 𝑢(𝑥)=𝜇_𝑏⁻¹|𝑥−𝑏|⁻¹𝑢(𝑏)+𝜕𝑢𝜕𝑥_𝑖(𝑏)⋅𝜇²_𝑏𝑥_𝑖−𝑏_𝑖|𝑥−𝑏|²1+𝑜|𝑥|=𝜇_𝑏⁻¹𝑥|+𝑥|^𝑖|𝑥|−𝑏_𝑖1+𝑜|𝑥|×𝑢(𝑏)+𝜕𝑢𝜕𝑥_𝑖(𝑏)𝜇²_𝑏|𝑥|𝑥_𝑖1+𝑜|𝑥|, (118) so the coefficient of 𝑥_𝑖 is (1/|𝑥|)+(𝜕𝑢/𝜕𝑥_𝑖)(𝑏)(𝜇_𝑏/|𝑥|).
Comparing these two coefficients and 𝐴=1, we obtain 1|𝑥|+𝜕𝑢𝜕𝑥_𝑖(𝑏)𝜇_𝑏|𝑥|=𝜇₀|𝑥|𝜕𝑢𝜕𝑥_𝑖(0). (119)
Then 𝑢_𝑏𝜕𝑢𝜕𝑥_𝑖(𝑏)=𝑢₀𝜕𝑢𝜕𝑥_𝑖(0)+𝑏_𝑖. (120)
Thus, 12𝜕𝑢²𝜕𝑥_𝑖(𝑥)_𝑥=𝑏=12𝜕𝑢²𝜕𝑥_𝑖(𝑥)_𝑥=0+12𝜕|𝑥|²𝜕𝑥_𝑖(𝑥)_𝑥=𝑏. (121)
It gives 𝜕𝑢²(𝑥)−|𝑥|²𝜕𝑥_{𝑖𝑥=𝑏}=𝐶_𝑖. (122)
By solving the previous equation, we obtain 𝑢²(𝑥)=|𝑥|²𝐶+𝐶𝑥+𝐷=𝑥−2²+𝐸=𝑥−𝑥₀²+𝑑, (123) for some 𝑥₀∈𝑅⁵, equivalent to 𝑢(𝑥)=𝑎𝑥−𝑥₀²+𝑑^1/2. (124)

Acknowledgments

The paper is mainly taken from first author’s master thesis at National University of Singapore. She is sincerely grateful to her family, friends, and all other people who helped her during the completion of the project. The research of the second author is supported through his NUS research Grant R146-000-112-127.