ISRN Mathematical Analysis

Volume 2013 (2013), Article ID 384394, 17 pages

http://dx.doi.org/10.1155/2013/384394

Research Article

Entire Solutions of an Integral Equation in

Xin Feng and Xingwang Xu

Department of Mathematics, National University of Singapore, Block S17 (SOC1), 10 Lower Kent Ridge Road, Singapore, 119076

Received 29 May 2013; Accepted 17 June 2013

Academic Editors: G. Gripenberg, M. McKibben, and M. Winter

Copyright © 2013 Xin Feng and Xingwang Xu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

We will study the entire positive solution of the geometrically and analytically interesting integral equation: with in . We will show that only when , there are positive entire solutions which are given by the closed form up to dilation and translation. The paper consists of two parts. The first part is devoted to showing that 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 . 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 . In this part, we use integration by parts to derive Pohozev’s identity in the form . It is easy to observe that if solves our integral equation, then also solves the differential equation . 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 unless which is not allowed.

In the third section, we adopt the moving sphere method in the integral form to solve the integral equation with .

2.

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 positive solution of the integral equation is . Thus in the following, we mainly assume that the solution is in .

Theorem 1. Suppose that is a entire positive solution of the integral equation

with , then .

Lemma 2. If satisfies (1), then .

Proof. By direct calculation, we have

Since is the fundamental solution of Laplace equation in , it follows that

Thus the lemma is true.

Lemma 3. If is the solution of the equation

then the following Pohozev identity holds:

where

Proof. Multiplying on the both sides of (4) and integrating over the ball , one has

where ; .

First, rewrite as

Using integration by parts, we have

Now, the Second Green identity implies that

where ; .

As for , since and by the Second Green identity again, we obtain

Hence,

Observe that

that is,

Combining (12) and (14), we have

Thus

To deal with , we observe that if multiplying on both sides of (4) and integrating over the ball , one has

Thus, with the help of the Second Green identity, one obtains

Therefore, we get

Put it into (16) to conclude that

Put and into the equation and rearrange the terms; then we get the desired identity.

Thus the lemma holds.

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

Proof. Define , . Then it follows that

If , then

which contradicts the assumption that is an entire solution. Therefore, we only consider the case .

Rewrite the equation as

where

Clearly, we have

And, by (1), one has

where

Note that can be controlled as follows:

Clearly, with the definition of , the following holds true:

Then differentiate it to get

where

One observes that

for a constant independent of and for sufficiently large. Here we have used the assumption that and , where is bounded by a constant. The integral over can be controlled as follows:

where , .

Then

As for , implies that and .

Therefore

Since and , so is bounded for when is large enough.

It follows that is bounded for when is large enough.

Now we deal with the higher order derivatives. First

where

For a constant independent of , when is large enough, it is easy to see that

by the estimation in .

Then, by computation, one gets

where

And as usual, we observe that

Similarly, we have the following estimation:

where , .

Then

As for , implies that and .

Therefore

Since and , so is bounded for when is large enough.

It follows that is bounded for when is large enough.

Since , we obtain

where

As before, we can estimate as

for a constant independent of and for sufficiently large. We have proved that the first integral is bounded in the previous estimation. The second integral over can be controlled as follows:

where , .

Then

As for , implies that and .

Therefore

Since and , so is bounded for when is large enough.

It follows that is bounded for when is large enough.

Once again, calculation shows that

where

We still need to control which can be done as follows:

for a constant independent of and for sufficiently large. Now, the third integral over can be controlled as follows:

The two integrals are bounded as we have shown before. It follows that is bounded for when is large enough.

Finally, by computation, one obtains

where

The estimate on can proceed as follows:

for a constant independent of and for sufficiently large. As before, the aforementioned first integral can be controlled as follows:

where , .

Then

As for , implies that and .

Therefore

which is bounded, since and , so is bounded for when is large enough.

Besides,

and the two integrals have been shown to be bounded before.

It follows that is bounded for when is large enough.

Combining these estimates, we can see that

(1) as ;(2) as ;(3) as ;(4) as ;(5) as ;(6) as ;(7) as .

Also we have

Observe that

The coefficient of is given by

where

And directly calculation gives

By substitution, we have

Hence

Now, integrate it over with , and we obtain that

where

It is clear that , by Poisson representation formula for harmonic function, the previous integral equals

Similarly, calculation gives , and then Poisson representation formula implies that

Once again, , and Poisson formula leads to

Notice that .

Next, consider the following integral:

Setting , the previous integral is equal to

Also, we have

Finally, it follows from variable change and Poisson formula for the harmonic function that

Combining those previous calculations, we can conclude that

as , since .

What we have shown so far is that, if is regular, that is, , then

Setting , when , we have

as , since when is large enough.

Hence, let , and we conclude that from (4).

3. Moving Sphere Method

This section is to run the method of moving sphere for the integral equation with the exact exponent .

First, define and ; it is rather easy to see that .

Lemma 5. If one sets , then , where ; when and . In particular, , which belongs to .

Proof. Since ,

where , .

As for , set to obtain

As for , if we set , then we obtain

Similarly, one can write as

As for , by setting , one concludes that

For , as before, set to get

Therefore, combining these previous calculations, we get

where and .

Now, since

when and .

Put to obtain

which belongs to .

Thus we complete the proof.

Proposition 6. If is large enough, then for all such that .

Proof. We prove that when is large enough by three steps.

Step 1. There exists such that for , we have .

It is clear to see that ; .

When is large enough,

Step 2. There exists an such that for , we have and .

First, we calculate . In fact , where    and .

To carry out the detail, for one observes that

Thus

If , then if . Let . Then when is large enough. Now let . Hence we have

Notice that if is sufficiently large, then . Therefore if is large enough, then for .

Besides, by definition, and by Step 1. Thus the maximal principle implies that on the domain .

Step 3. There exists an such that for for large enough.

On the one hand, when is small, , and is continuous on the rest of ball, so is bounded for . On the other hand, the definition implies that when is large enough. Therefore, for when is large enough.

Combining Steps 1 to 3, we have .

Thus we complete the proof of this proposition.

Now, for any , we denote by , by , and set .

By Proposition 6, if is large enough, then . Thus, we define in for .

Proposition 7. There exists such that .

Proof. Suppose that the proposition is not true. Hence for all ; that is, for all and all , we always have for all .

Note that

Therefore, it follows that ; that is,

Simplifying this, we get

where .

Setting , by calculation we get

Substituting (97) into (96) and doing variable change, we obtain

Then both sides being divided by and sending to , let , we get

Since the previous inequality holds for all , so . It implies that is a constant which contradicts .

Thus this proposition holds.

Proposition 8. Suppose that there exists a point such that ; then for all .

Proof. Without loss of generality, we assume that . Suppose that the proposition is not true; then .

In order to complete the argument, we need to do some preparations.

First of all, setting , then it is clear that

Next, calculate as follows.

Since we have the formula

which is because

thus,

Hence, by direct calculation, one obtains

It clearly implies that

whenever .

Therefore, by Lemma 5,

By the definition of and our assumption that , from Lemma 5, we see that inside the ball . Then, at every boundary point , we have

Besides, the definition of implies that there is a sequence such that

It follows from (108) that there exists such that

Since for , we have . Clearly, has a convergent subsequence, still denoted by , with a limit point . Also, . Then satisfies

Now consider the following two cases.

Case 1 . The conclusion contradicts the fact that is strictly positive inside the ball .

Case 2 . Then the fact that contradicts (107).

Hence, the proposition follows.

Proposition 9. For all , one has .

Proof. It follows from Propositions 7 and 8 that there exists some such that and ; that is, ; that is,

Setting , we get

It follows that

Suppose that there exists a point such that . Then for all and .

That is, for all and .

Fixing and letting , we have .

Therefore,

Letting go to , we have , which contradicts the fact that is a positive entire solution.

Hence, this proposition holds.

Proposition 10. If satisfies , then there exist some and constants , such that .

Proof. From Propositions 8 and 9, we know that for all , we have ; that is, .

Using , setting and , we can rewrite it as follows:

Now setting , then we have

It follows that the limit exists and is equal to for all .

Then set .

Case 1. If , then which is impossible to be a positive solution.

Case 2. If , the definition of implies that . Without loss of generality, we can assume that . Now for large,

so the coefficient of is , and

so the coefficient of is .

Comparing these two coefficients and , we obtain

Then

Thus,

It gives

By solving the previous equation, we obtain

for some , equivalent to

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.

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