Super Polyharmonic Property of Solutions for PDE Systems and Its Applications

In this paper, we prove that all the positive solutions for the PDE system (-\Delta)^{k}u_{i} = f_{i}(u_{1},..., u_{m}), x \in R^{n}, i = 1, 2,..., m are super polyharmonic, i.e. (-\Delta)^{j}u_{i}>0, j = 1, 2,..., k - 1; i = 1, 2,...,m. To prove this important super polyharmonic property, we introduced a few new ideas and derived some new estimates. As an interesting application, we establish the equivalence between the integral system u_{i}(x) = \int_{R^{n}} \frac{1}{|x - y|^{n-\alpha}}f_{i}(u_{1}(y),..., u_{m}(y))dy, x \in R^{n} and PDE system when \alpha? = 2k

To prove this important super polyharmonic property, we introduced a few new ideas and derived some new estimates.
In the last few years, a series of results on qualitative properties for solutions of integral systems (2) have been obtained, since the introduction of a powerful tool-the method of moving planes in integral forms. Now due to the equivalence established here, all these properties can be applied to the corresponding PDE systems.
We say that systems (1) and (2) are equivalent, if whenever u is a positive solution of (2), then u is also a solution of (−△) k u i = cf i (u 1 , · · · , u m ), x ∈ R n , i = 1, 2, · · · , m with some constant c; and vice versa.

Introduction
The method of moving planes in integral forms was introduced in [19] to solve an open problem posed by Lieb [38]: Can one classify all the positive solutions of the integral equation This is an Euler equation of the functional associated with the well-known Hardy-Littlewood-Sobolev inequality in a special case. The classification would provide the best constant in the corresponding inequality.
To show the radial symmetry of solutions for PDEs, usually a method of moving planes is applied, which relies heavily on maximum principles. While integral equations do not possess such local properties. To carry on the moving of planes, we explore global features of the integral equations and estimate certain integral norms. This is the essence of the method of moving planes in integral forms. It can be very conveniently applied to (3) to establish radial symmetry of the solutions under quite mild integrability assumptions, that is u ∈ L n+α n−α loc . As compare to the corresponding PDE to carry on the method of moving planes, one at least need to require the solution be α times differentiable and can only do this when α is an even number. While fractional powers of Laplacian in (4) have many applications in mathematical physics and related fields, such as anomalous diffusion and quasi-geostrophic flows, turbulence models and water waves, molecular dynamics and relativistic quantum mechanics of stars [5] [23]. Since its introduction, the method of moving planes in integral forms has become a powerful tool to establish symmetry, a priori estimates, and nonexistence for a variety of integral equations and systems, including (i) a system of m equations [11] ..m, 0 < α < n, and u(x) = (u 1 (x), u 2 (x), · · · , u m (x)); (5) (ii) the fully nonlinear integral systems involving Wolff potentials [14]: where (iii) equations and systems involving Bessel potentials [51] [52]: where * is the convolution and is the kernel of the Bessel potential, (iv) boundary values problems on half spaces [48] and in unit balls [25], (v) problems on complete Riemannian manifolds [57]. There are many more applications of this method to integral systems, and the interested readers may see [10]  In [35] [43] [44], asymptotic analysis for solutions of integral systems near the origin and at infinity were also obtained.
If one can show that an integral system is equivalent to a PDE system, then the results obtained for the integral system can be carried over to the PDE system.
For instance, in [19], we showed that (3) is equivalent to semi-linear elliptic PDE (4) and thus classified all the positive solutions for both integral equation (3) and PDE (4).
In [57], Qing and Raske considered a smooth family of equations on Riemannian manifolds. When α = 2, P α [g] is the conformal Laplacian, and when α = 4, it is the Paneitz operator. They showed that, under certain conditions, (7) is equivalent to a conformally invariant integral equation. Using the method of moving planes in integral forms, they obtained a priori estimate for the positive solutions for the integral equation as well as for PDE (7); and based on this estimate and a degree theory, they proved the existence of solutions.
In [54], Ma and Zhao were able to classified the positive solutions for the stationary Choquard equation and hence answered an open question posed by Lieb [39]. Their idea was to use Bessel potential to write equation (8) as an equivalent integral system Then they applied the method of moving planes in integral forms to derive the radial symmetry of the positive solutions of (9), and hence of (8) due to the equivalence. Since the method of moving planes in integral forms is applied directly to integral equations and systems, whether or not these results can be extended to the corresponding PDEs depends on whether one can establish the equivalence between the integral equations and PDEs. Is the integral system (5) equivalent to a PDE system? What can we say about (6)?
To establish the equivalence, a general approach is to multiply both sides of the PDEs by the Green's function on the ball of radius R, integrating by parts, then letting R→∞ and taking limits. In order to show that the boundary terms tend to zero, one important ingredient is to obtain the super polyharmonic property of the solutions for the PDEs. This is the main objective of our present paper.
Consider the following more general system of PDE inequalities: where u = (u 1 , · · · , u m ). We prove Theorem 1 Let u = (u 1 , · · · , u m ) be a positive solution of (10). Assume that Let w = u 1 + u 2 + · · · + u m . Suppose there exist p > 1, δ > 0, and C, C δ > 0, such that Then we have When f i (u) = u p i , the power of −△ need not to be the same, as illustrated in the following system of two inequalities: with positive integers t and s.
Theorem 2 Assume that p, q > 1. Let (u, v) be a pair of positive solutions for inequality system (14). Then Remark 1 One can verify that when (and only when) s = t, Theorem 2 is a special case of Theorem 1.
Theorem 1 contains the following result of Lin [41] (on a fourth order equation) and Wei and Xu [63] (on general even order equations) as a special case: To prove Theorem 1, we introduce some new ideas and obtained a few interesting estimates.
It is well-known that the super polyharmonic property (57) is very useful in analyzing the corresponding system. Now, as an important and immediate application, we use Theorem 1 and 2 to establish the equivalence between integral systems and the corresponding PDE systems.
We say that the integral system and the PDE system are equivalent, if whenever u = (u 1 , · · · , u m ) is a positive solution of (15), then u is also a solution of with some constant c; and vice versa.

We prove
Theorem 3 Let t and s be positive integers less than n 2 , and assume that p, q > 1. Then the PDE system is equivalent to the integral system Theorem 4 Let α be an even integer less than n. Then under the assumption of Theorem 1, PDE system (16) is equivalent to integral system (15).
For other real values of α, we define the positive solutions of in the distribution sense, i.e. u ∈ H α 2 (R n ) and satisfies for any φ ∈ C ∞ 0 (R n ) and φ(x) ≥ 0. Here, as usual, whereû i andφ are the Fourier transform of u i and φ respectively. Theorem 5 PDE system (18) as defined above is equivalent to integral system (15).
Based on these equivalences, a series of previous results obtained by many authors for integral equations and systems can now be applied to the corresponding PDE systems.
In order to better illustrate our new idea, in Section 2, we will apply it to a single inequality. In Section 3, we apply our new idea to a system of two inequalities and prove Theorem 2. In Section 4, we consider a more general system of m inequalities and obtain Theorem 1. Finally, in Section 5, we establish the equivalences between integral and PDE systems and thus prove Theorem 3, 4, and 5.
We would like to mention that, in Section 4 and 5, when considering a general system, we introduce an interesting idea, so that we can reduce the treatment of a system into that of a single equation or inequality and hence simplify the proofs remarkably.
Throughout this paper, as usual, when we say "a solution" to a PDE or inequality system, we mean "a classical solution" if not otherwise indicate.

Super Polyharmonic Properties for Single Inequalities
Let p be a positive integer and q > 1. Consider Theorem 2.1 For each positive solution u of (20), it holds Proof of Theorem 2.1.
Suppose in contrary, there are two possible cases: In this case,x is a local minimum of v p−1 , and we must have Therefore we only need to consider Case i). Without loss of generality, we may assume that x o = 0.
In this step, we will show that, after a few times of re-centers, if we denote the resulting functions byṽ k , theñ The positiveness of u implies that p must even. Hence by this alternating sign nature, we have − △ũ ≡ṽ 1 < 0, and thereforeũ (r) ≥ũ(0) > 0.
Here, we assume that after re-centers,ũ is radially symmetric about the origin.
To derive (23), letū be the average of u. Then by Jensen's inequality and the well-known property that △u = △ū, From the first inequality, we havē Then from the second equation, we have Integrating yields v ′ p−2 (r) > cr andv p−2 (r) ≥v p−2 (0) + cr 2 , ∀ r ≥ 0. Obviously,ū andv i , i = 1, · · · , p − 1, still satisfy (26). Applying (27) From (28) and repeat the same argument as forv p−2 , we obtain Continuing this way, after a few steps of re-centers (denote the resulting functions byṽ k andũ), we arrive at This implies immediately that p must be even, since if p is odd, then (31) implies that u should be negative somewhere, a contradiction with our assumption that u > 0. Hence in the following, we assume that p is even. Let be the re-scaling of u. Then equation (20) is invariant under this re-scaling, and for any λ > 0, u λ is still a positive solution of (20). By (31), −△ũ < 0, and henceũ ′ (r) > 0.
Hence one can choose λ sufficiently large, such that u λ be as large as we wish. Without loss of generality, we may assume that for any given a o > 0, we already haveũ Here we choose σ o , such that The value of a o will be determined later.
From (32), Integrating both sides twice and taking into account of (31) yield This implies Continuing this way, we arrive at Here we have denoted 2p + n as m. Denote σ k+1 = 2σ k q ≥ σ k q + m and a k+1 = a q k (2σ k q) m .
Suppose we haveũ (r) ≥ a k r σ k .
We want to use induction to show that Obviously, we can make (38) true for k = 0 by choosing a 0 sufficiently large. Assume it is true for k. Then we have Therefore, (38) is true for all integer k. Now by (35) and (38), This is obviously impossible. Therefore (22) must hold.

Part II.
Base on the positiveness of v p−1 , we can now show that all other v k must also be positive: v p−i (x) > 0, for i = 2, 3, · · · , p − 1.
Suppose for some i, v p−i is negative somewhere, then through the same arguments as in Step 1 of Part I, after a few steps of re-centers, if we denote the resulting functions byṽ k , then the signs ofṽ k are alternating, and by the positiveness ofũ, we must have − △ũ < 0 andũ > 0.
Coming back to the original equation By a direct integration as in Part I, we would arrive at v p−1 < 0, somewhere .
This contradicts with the proven fact that v p−1 > 0, and thus completes the proof.

A System of Two Inequalities
Consider a system of two inequalities in R n with positive integers t and s. An entirely similar approach will be applied to systems of more inequalities and more general right hand sides functions in the next section. Nevertheless, the system in the next section does not include (40), because there we require the powers of all −△ be the same, and here in (40), they are allowed to be different. Let Then one can easily verified that system (40) is invariant under this re-scaling. Let u k = (−△) k u and v k = (−△) k v.
We want to show that u k , v k > 0 for k = 1, 2, · · · up to t − 1 or s − 1 .
Part I. We first show that Suppose in contrary, there is some point x o , such that Then use the same argument as we did in deriving (31) for the single equation, we obtain whereũ k is obtained from u k through several re-centers. (43) implies that t must be even. From (43), −△ũ(r) < 0, and henceũ ′ (r) > 0 andũ(r) >ũ(0) > 0.
From this and the inequality −△ṽ s−1 ≥ũ p (r) we derive thatṽ s−1 must be negative somewhere. Now repeat the above argument and after a few steps of re-centers, we arrive at Here we still denote the resulting functions after re-centers byṽ k . From here, we also obtainṽ (r) >ṽ(0) > 0.
After re-scaling, we can makeṽ λ as large as we wish. Hence, without loss of generality, we may assume that, for any fixed Using this and (−△) tũ ≥ṽ q and through the same argument as in the previous section, we arrive at Here we have let m = max{n + 2t, n + 2s}. Choose σ o large so that σ o q ≥ m. Then (46) becomes Using the second equation and repeat the same process as in deriving (46), Then similar as in the single equation case, we can show by induction that a k →∞ as k→∞.
Instead of carrying out the detail, we rather do it heuristically, which will better illustrate the idea. First notice that σ k →∞. To derive (49), it suffice to show that for some l to be determined later. We again use induction. We can choose a o large, so (50) is true for k = 0. Now assume (50) holds for k, and we want to show that it is true for k + 1. In fact, we have Now choose l, such that the power l(h − 1) − m(p + 1) of σ k is positive, say greater that 1, and also choose σ o large (note σ k ≥ σ o ) to ensure that (50) holds for k + 1. This verifies (42). Similarly, one can derive that v s−1 (r) ≡ (−△) s−1 v > 0.

Part II.
Base on the positiveness of v s−1 , we can now show that all other u k must also be positive: Suppose for some i, u t−i is negative somewhere, then through the same arguments as in Part I, after a few steps of re-centers, if we denote the resulting functions byũ k , then the signs ofũ k are alternating, and by the positiveness ofũ, we must have − △ũ < 0 andũ > 0.

Coming back to the original inequality
By a direct integration as in Part I, we would arrive at v s−1 < 0, somewhere .
This contradicts with the proven fact that v s−1 > 0 and thus (52) must hold. Similarly, based on the positiveness of u t−1 we can derive v s−i (x) > 0, for i = 2, 3, · · · , s − 1.
This completes the proof.
Here we introduce an interesting idea, so that we can reduce the problem for system into one for single equation.
Suppose (57) is not true. Without loss of generality, we may assume for some j o < k, Let for some ǫ > 0. Then by (56), we have Now for each ǫ > 0, applying the same arguments as for single equation in Section 2, we can derive This would contradict with (58) for sufficiently small ǫ. Hence (58) must be false and this completes the proof of the theorem.

The Equivalence between Integral and PDE Systems
In this section, we establish the equivalence between the integral system and the PDE system where u = (u 1 , · · · , u m ). We say that systems (60) and (61) are equivalent, if whenever u is a positive solution of (60), then u is also a solution of with some constant c; and vice versa.
We first prove Theorem 5.1 Let α = 2k be an even number less than n. Then every positive solution of PDE (61) satisfies integral equation (60) with f i (u) replaced by cf i (u).
Remark 5.1 i) The converse of Theorem 5.1 is much easier to obtain by elementary argument, and we here skip its proof.
ii) The proof of Theorem 3 is entirely similar to that of Theorem 5.1, we also skip it.
Notice that here we do not assume any asymptotic behavior of the solution near infinity for PDE system (61), hence to prove Theorem 5.1, we first need to show that the integral on the right hand side of (60) is finite. We will multiply both sides of (61) by proper functions and integrate by parts on B r (0), then let r→∞ and take limits. To this end, we need some estimates of the solutions as stated in the following lemma. One will see that Theorem 4.1 (the super polyharmonic property) is a key ingredient in the proof of the following lemma.
As an immediate consequence of Lemma 5.1, we derive Lemma 5.2 There exists a sequence r h →∞, such that Proof of Lemma 5.1 Let δ(x) be the Dirac Delta function. Let φ be the solution of the following boundary value problem By the maximum principle, one can easily verify that Sum over the m equations in (61), we obtain Multiply both side of the equation (68) by φ and integrate on B r (0). After integrating by parts several times and applying Theorem 4.1 and (67), we arrive at Now letting r→∞, one can see that (63) is just a direct consequence of the following fact φ(x)→ c |x| n−2k with some constant c.
The Proof of Lemma 5.2.
To verify (65) for each fixed i, we sum over (63) and (64) to obtain This implies that there exists a sequence r h →∞, such that Since each term in the above summation is nonnegative, we have and For any ǫ > 0, it is obvious that where χ ǫ is the characteristic function on the set {x ∈ ∂B r h | w(x) ≥ ǫ}.
Since each u i is nonnegative, we have Combining this with (74), we arrive at (65). This completes the proof of Lemma 5.2.
Proof of Theorem 5.1.
For each r > 0, let φ r (x) be the solution of (66). Then as in the previous lemma, one verifies that and It follows that, Also one can verify that, on ∂B r (0), Multiply both side of the equation by φ r (x) and integrate on B r (0). After integrating by parts several times and applying Theorem 4.1 and (67), we arrive at By virtue of (79) and (65), there exist a sequence r h →∞, such that the boundary integral on ∂B r h (0) in (80) approaches 0 as r h →∞.
By a translation, that is, for each fixed x, integrating on B r (x) instead of on B r (0), we can derive that u i (x) is a solution of u i (x) = c R n 1 |x − y| n−2k f i (u(y))dy, i = 1, · · · , m.
This completes the proof of the theorem.
So far, we have proved that if α = 2k is an even number, then every solution of the PDE system (61) is a solution of our integral system (81). Now for other real values of α, we define the positive solution of (61) in the distribution sense, i.e. u ∈ H α 2 (R n ) and satisfies Theorem 5.2 PDE system (61) as defined above is equivalent to integral system (60).
Choose an appropriate constant c, so that (−△) α/2 ψ = φ, consequently ψ ∈ H α (R n ) ⊂ H α 2 (R n ), and hence (82) holds for ψ: Integration by parts of the left hand side and exchange the order of integration of the right hand side yields Since φ is any nonnegative C ∞ 0 function, we conclude that u i satisfies the integral equation u i (x) = c R n 1 |x − y| n−α f i (u(y))dy.

It follows that
That is, u i is a solution of (−△) α/2 u i = cf i (u(x)), x ∈ R n in the sense of distributions.
This completes the proof of the theorem.