Regularity of Solutions for a System of Integral Equations

Wenxiong Chen ^{∗} Congming Li

Abstract

In this paper, we study positive solutions of the following system of integral equations in R^{n}:

\{ \begin{matrix}u(x) = ∫_{R^{n}} |x − y|^{α−n}v(y)^{q}dy \\ v(x) = ∫_{R^{n}} |x − y|^{α−n}u(y)^{p}dy \end{matrix}  (0.1)

with \frac{1}{q+1} + \frac{1}{p+1} = \frac{n−α}{n}. In our previous paper, under the natural integrability conditions u ∈ L^{p+1}(R^{n}) and v ∈ L^{q+1}(R^{n}), we prove that all the solutions are radially symmetric and monotone decreasing about some point. In this paper, we go further to study the regularity of the solutions. We show that the solutions are bounded, and hence continuous and smooth. We also prove that if p = q, then u = v, and they both must assume the standard form

c(\frac{t}{t^{2} + |x − x_{o}|^{2}} )^{(n−α)/2}  (0.2)

with some constant c = c(n, α), and for some t > 0 and x_{o} ∈ R^{n}.

AMS Subject Classification 2000 35J99, 45E10, 45G05 Keywords Hardy-Littlewood-Sobolev inequalities, systems of integral

equations, boundedness, regularity, classification of solutions.

∗Partially supported by NSF Grant DMS-0072328

1 Introduction

Let 0 < α < n and let s, r > 1 such that \frac{1}{r} + \frac{1}{s} = \frac{n+α}{n}. Let ∥f∥_{p} be the L^{p}(R^{n}) norm of f. The well-known Hardy-Littlewood-Sobolev inequality states that:

∫_{R^{n}} ∫_{R^{n}} f(x)|x − y|^{α−n}g(y)dxdy ≤ C(n, s, α)||f||_{r}||g||_{s}  (1.3)

for any f ∈ L^{r}(R^{n}) and g ∈ L^{s}(R^{n}).

To find the best constant C = C(n, s, α) in the inequality, one can maximize the functional

J(f, g) = ∫_{R^{n}} ∫_{R^{n}} f(x)|x − y|^{α−n}g(y)dxdy  (1.4)

under the constraint

||f||_{r} = ||g||_{s} = 1.  (1.5)

Let (f, g) be a pair of positive maximizer, or more generally, a pair of critical points of (1.4) under the constraint (1.5). Letting u = λ_{1}f^{r−1}, v = λ_{2}g^{s−1}, p = \frac{1}{r−1}, q = \frac{1}{s−1}, and by a proper choice of constants λ_{1} and λ_{2}, then one can see that the pair (u, v) satisfies the following system of integral equations in R^{n}:

\{ \begin{matrix}u(x) = ∫_{R^{n}} |x − y|^{α−n}v^{q}(y)dy \\ v(x) = ∫_{R^{n}} |x − y|^{α−n}u^{p}(y)dy \end{matrix}  (1.6)

with

\frac{1}{q + 1} + \frac{1}{p + 1} = \frac{n − α}{n}.  (1.7)

The integral system is closely related to the system of partial differential equations

\{ \begin{matrix}(−Δ)^{α/2}u = v^{q}, u > 0, in R^{n}, \\ (−Δ)^{α/2}v = u^{p}, v > 0, in R^{n}. \end{matrix}  (1.8)

In the special case where p = q = \frac{n+α}{n−α}, and u(x) = v(x), the system becomes:

u(x) = ∫_{R^{n}} |x − y|^{α−n}u(y)^{\frac{n+α}{n−α}} dy, u > 0 in R^{n}.  (1.9)

The corresponding PDE is the well-known family of semi-linear equations

(−Δ)^{α/2}u = u^{(n+α)/(n−α)}, u > 0, in R^{n}  (1.10)

In particular, when n ≥ 3, and α = 2, (1.10) becomes

−Δu = u^{(n+2)/(n−2)}, u > 0, in R^{n}.  (1.11)

The classification of the solutions of (1.11) has provided an important ingredient in the study of the well-known Yamabe problem and the prescribing scalar curvature problem. It is also essential in deriving a priori estimates in many related nonlinear elliptic equations.

Solutions to (1.11) were studied by Gidas, Ni, and Nirenberg [GNN]. They proved that all the positive solutions of (1.11) with reasonable behavior at infinity

u(x) = O(\frac{1}{|x|^{n−2}} )  (1.12)

are radially symmetric and therefore assume the form of

c(\frac{t}{t^{2} + |x − x_{o}|^{2}} )^{(n−2)/2}  (1.13)

with some positive constants c and t.

Later, in [CGS], Caffarelli, Gidas, and Spruck removed the growth condition (1.12) and obtained the same result. Then Chen and Li [CL1], and Li [Li] simplified their proof. Recently, Wei and Xu [WX] generalized this result to the solutions of more general equation (1.10) with α being any even numbers between 0 and n.

Apparently, for other real values of α between 0 and n, equation (1.10) is also of practical interest and importance. For instance, it arises as the Euler-Lagrange equation of the functional

I(u) = ∫_{R^{n}} |(−Δ)^{\frac{α}{4}} u|^{2}dx/(∫_{R^{n}} |u|^{\frac{2n}{n−α}} dx)^{\frac{n−α}{n}} .

The classification of the solutions would provide the best constant in the inequality of the critical Sobolev imbedding from H^{\frac{α}{2}} (R^{n}) to L^{\frac{2n}{n−α}} (R^{n}):

(∫_{R^{n}} |u|^{\frac{2n}{n−α}} dx)^{\frac{n−α}{n}} ≤ C ∫_{R^{n}} |(−Δ)^{\frac{α}{4}} u|^{2}dx.

In [L], Lieb classified all the maximizers of the functional (1.3) under the constraint (1.5) in the special case where p = q = \frac{n+α}{n−α}, and thus obtained the best constant in the H-L-S inequalities in that case. He then posed the classification of all the critical points of the functional – the solutions of the integral equation (1.9) as an open problem.

We solved this open problem in one of our previous paper [CLO]. We proved

Proposition 1 All solutions of partial differential equation (1.10) satisfy the integral equation (1.9), and vise versa. Every positive solution u(x) ∈

Lloc^{\frac{2n}{n−α}} (R^{n}) of (1.9) or (1.10) is radially symmetric and decreasing about some point x_{o} and therefore assumes the form of (0.2).

Later, Y. Li [LiY] obtained the same classification result by using a different method–the method of moving spheres.

Then in [CLO3], we considered more general system (1.6) and showed that

Proposition 2 Let (u, v) be a pair of solutions of (1.6) and p, q ≥ 1 . Assume that u ∈ L^{p+1}(R^{n}) and v ∈ L^{q+1}(R^{n}). Then u and v are radially symmetric and decreasing about some point x_{o}.

To prove the radial symmetry and monotonicity of the solutions, we use an integral form of the method of moving planes. This was a new idea we introduced in [CLO]. Then in [CLO3], we modified this idea so that it can be applied to systems of integral equations. It is entirely different from the traditional methods of moving planes used for partial differential equations. For PDEs, the local properties of the differential operators are exploited extensively. This lack of knowledge of the local properties prevents us from using many known results, such as maximum principles. However, by exploring various special features possessed by the integral equation in its global form, and by estimating certain integral norms, we are still able to establish the symmetry results.

In this paper, we go further to study the regularity of the solutions. We prove

Theorem 1 Assume that (u(x), v(x)) is a pair of positive solutions of (1.6), and u ∈ L^{p+1}(R^{n}) and v ∈ L^{q+1}(R^{n}). Then u(x) and v(x) are uniformly bounded in R^{n}.

It follows from Theorem 1, the standard integration theory, and regularity argument that u(x) and v(x) are continuous and hence smooth everywhere.

We also prove

Theorem 2 In the system (1.6), if p = q, then u = v, and they both must assume the standard form

c(\frac{t}{t^{2} + |x − x_{o}|^{2}} )^{(n−α)/2}  (1.14)

with some constant c = c(n, α), and for some t > 0 and x_{o} ∈ R^{n}.

In Section 2, we study the regularity of the solutions and prove Theorem 1. In Section 3, we prove Theorem 2.

2 Regularity of the Solutions

In this section, we consider the regularity of the solutions and prove Theorem 1. To this end, we first need the following lemma.

Lemma 2.1 Let

h(x) = ∫_{R^{n}} |x − y|^{α−n}f(y) dy

Then for any t > \frac{n}{n−α}, we have

∥h∥_{t} ≤ C(n, s, α)∥f∥_{\frac{nt}{n+αt}} .  (2.15)

Proof. From the definition of h(x) and the Hardy-Littlewood-Sobolev Inequality (1.3), we have

∫_{R^{n}} h(x)g(x) dx ≤ C(n, s, α)∥f∥_{r}∥g∥_{s}.

with \frac{1}{r} + \frac{1}{s} = \frac{n+α}{n}.

Consequently,

sup_{∥g∥_{s}=1} ∫_{R^{n}} h(x)g(x) dx ≤ C(n, s, α)∥f∥_{r}.  (2.16)

It follows from (2.16) and the duality that,

∥h∥_{s/(s−1)} ≤ C(n, s, α)∥f∥_{r}.  (2.17)

Let t = \frac{s}{s−1}, then r = \frac{nt}{n+αt}. Now (2.15) follows immediately from (2.17). This completes the proof of the Lemma.

The Proof of Theorem 1.

Let (u, v) be a pair of solutions of the system (1.6).

Given any r > \frac{p+1}{2}, choose s, such that

\frac{1}{s} − \frac{1}{r} = \frac{1}{q + 1} − \frac{1}{p + 1}.  (2.18)

Then one can verify that

s > \frac{nr}{n + αr} and s > \frac{n}{n − α}.  (2.19)

This guarantees that there exists an l > 1, such that

\frac{l − 1}{l}s = \frac{nr}{n + αr}.  (2.20)

(1.7), (2.18), and (2.20) imply that

l(q − 1)\frac{nr}{n + αr} = q + 1.  (2.21)

Let f ∈ L^{r}(R^{n}) and g ∈ L^{s}(R^{n}) respectively. Define the operators T_{1} and T_{2} by

(T_{1}g)(x) = ∫_{R^{n}} |x−y|^{α−n}v^{q−1}(y)g(y) dy, (T_{2}f)(x) = ∫_{R^{n}} |x−y|^{α−n}u^{p−1}(y)f(y) dy.

Applying first (2.15), then Holder inequality, and by virtue of (2.21), we obtain

∥T_{1}g∥_{r} ≤ C_{1}∥v^{q−1}g∥\frac{nr}{n+αr} ≤ C_{1}∥v∥_{q+1}^{q−1}∥g∥_{s}.  (2.22)

Similarly, we have

∥T_{2}f∥_{s} ≤ C_{2}∥u∥_{p+1}^{p−1}∥f∥_{r}.  (2.23)

Hence T_{1} maps L^{s}(R^{n}) into L^{r}(R^{n}) and T_{2} maps L^{r}(R^{n}) into L^{s}(R^{n}).

Define the norm in the cross product L^{r}(R^{n}) × L^{s}(R^{n}) by

∥(f, g)∥_{r×s} = ∥f∥_{r} + ∥g∥_{s},

and define T : L^{r}(R^{n}) × L^{s}(R^{n})→L^{r}(R^{n}) × L^{s}(R^{n}) by

T (f, g) = (T_{1}g, T_{2}f).

Let

v_{A}(x) = \{ \begin{matrix}v(x) & if v(x) > A \\ 0 & if v(x) ≤ A. \end{matrix}

And similarly for u_{A}(x).

Define

T_{1}^{A}g = ∫_{R^{n}} |x − y|^{α−n}v_{A}^{q−1}(y)g(y) dy.

And define T_{2}^{A} similarly.

Define

T_{A}(f, g) = (T_{1}^{A}g, T_{2}^{A}f).

It follows from (2.21) and (2.22) that

∥T_{1}^{A}g∥_{r} ≤ C_{1}∥v_{A}∥_{q+1}^{q−1}∥g∥_{s}, and ∥T_{2}^{A}f∥_{s} ≤ C_{2}∥u_{A}∥_{p+1}^{p−1}∥f∥_{r}.  (2.24)

Since v ∈ L^{q+1}(R^{n}) and u ∈ L^{p+1}(R^{n}), one can choose A sufficiently large, so that both ∥v_{A}∥_{q+q} and ∥u_{A}∥_{p+1} are sufficiently small. Hence, by (2.24), there exists a large A, such that

∥T_{A}(f, g)∥_{r×s} ≤ \frac{1}{2}∥g∥_{s} + \frac{1}{2}∥f∥_{r} = \frac{1}{2}∥(f, g)∥_{r×s}.  (2.25)

Therefore, T_{A} is a contract mapping from L^{r}(R^{n}) × L^{s}(R^{n}) to itself.

Since (u, v) is a pair of solutions of the integral system (1.6), it is easy to verify that

(u_{A}, v_{A}) = T_{A}(u_{A}, v_{A}) + (F, G)  (2.26)

where both functions F (x) and G(x) are uniformly bounded in R^{n} and

∥F ∥_{p+1}, ∥G∥_{q+1} < ∞.  (2.27)

By the Contracting Mapping Theorem, we conclude that the system

(f, g) = T_{A}(f, g) + (F, G)  (2.28)

has a unique pair of solutions in both L^{p+1}×L^{q+1} and (L^{p+1}×L^{q+1})∩(L^{r}×L^{s}). From (2.26) we see that (u_{A}, v_{A}) is a pair of solutions of (2.28), hence by the uniqueness, (u_{A}, v_{A}) ∈ (L^{p+1} × L^{q+1}) ∩ (L^{r} × L^{s}). By virtue of (2.18), one can choose r to be arbitrarily large, while s is still greater than q + 1. One can also choose s to be arbitrarily large, and while r remains greater than p + 1. It follows that, for any pair (r, s), we have

(u_{A}, v_{A}) ∈ (L^{p+1} × L^{q+1}) ∩ (L^{r} × L^{s}).

Finally, through a standard argument, one can conclude that the pair (u_{A}, v_{A}) is bounded in R^{n}, so does (u, v). This completes the proof of the Theorem.

3 Classification of Solutions

In this section, we prove Theorem 2, i.e. we show that when p = q, then we must have u(x) = v(x). Let τ = \frac{n+α}{n−α}, then from (1.7) and (1.6), we have

u(x) = ∫_{R^{n}} |x − y|^{α−n}v^{τ}(y)dy,  (3.29)

and

v(x) = ∫_{R^{n}} |x − y|^{α−n}u^{τ}(y)dy.  (3.30)

Multiplying (3.29) by u^{τ}(x) and (3.30) by v^{τ}(x), and then integrating on R^{n}, we obtain

∫_{R^{n}} u^{τ+1}(x)dx = ∫_{R^{n}} v^{τ+1}(x)dx.  (3.31)

The Proof of Theorem 2.

From Theorem 1, one can see that the limits lim_{|x|→∞} |x|^{n−α}u(x) and lim_{|x|→∞} |x|^{n−α}v(x) exist, and we denote them by u(∞) and v(∞) respectively.

Let r^{n−α} = \frac{u(∞)}{u(0)}, we claim that

u(rx) = \frac{1}{|x|^{n−α}} u(\frac{rx}{|x|^{2}} ).  (3.32)

More generally, for any point a ∈ R^{n}, if

\frac{u(∞)}{u(a)} = r^{n−α},  (3.33)

then we have

u(rx + a) = \frac{1}{|x|^{n−α}} u(\frac{rx}{|x|^{2}} + a).  (3.34)

To verify the (3.32), we consider

u_{t}(x) = t^{(n−α)/2}u(tx), and v_{t}(x) = t^{(n−α)/2}v(tx).

For the value of r defined above, we have

u_{r}(0) = u_{r}(∞).  (3.35)

Let e = (1, 0, ⋯, 0). Make Kelvin transforms:

\{ \begin{matrix}ū(x) = \frac{1}{|x|^{n−α}} u_{r}(\frac{x}{|x|^{2}} − e) \\ v̄(x) = \frac{1}{|x|^{n−α}} v_{r}(\frac{x}{|x|^{2}} − e). \end{matrix}  (3.36)

Then (ū, v̄) is still a pair of solutions of (1.6), and

ū(0) = u_{r}(∞) = u_{r}(0) = ū(e).

From Theorem 1, we see that ū must be radially symmetric about the point \frac{1}{2}e, and so must v̄ about the same point. It follows that, for any h,

\frac{r^{(n−α)/2}}{|1/2 − h|^{n−α}} u(r\frac{1/2 + h}{1/2 − h}e) = \frac{r^{(n−α)/2}}{|1/2 + h|^{n−α}} u(r\frac{1/2 − h}{1/2 + h}e).  (3.37)

Let λ = \frac{1/2−h}{1/2+h}, we arrive at

u(rλe) = \frac{1}{λ^{n−α}} u(\frac{r}{λ}e).  (3.38)

This proves (3.32). Then (3.34) follows by a translation. Similarly, for the same value r, we also have

v(rx + a) = \frac{1}{|x|^{n−α}} v(\frac{rx}{|x|^{2}} + a).  (3.39)

Using an entirely similar argument, we can derive that if

s^{n−α} = \frac{v(∞)}{v(0)},  (3.40)

then

u(sx + a) = \frac{1}{|x|^{n−α}} u(\frac{sx}{|x|^{2}} + a).  (3.41)

From (3.34) and (3.41), we will be able to conclude that

r = s.  (3.42)

We now postpone the proof of (3.42) for a moment, and continue to complete the proof of the Theorem. (3.33), (3.40), and (3.42) implies that for any point a ∈ R^{n}, we have

\frac{u(a)}{u(∞)} = \frac{v(a)}{v(∞)}.

It follows that

u(a) = cv(a) ∀a ∈ R^{n}.

Then by (3.31), we must have

u(x) ≡ v(x).

Now what left is to prove (3.42). To better illustrate the idea, we only present the proof for the case n = 1 and a = (0, ⋯, 0). The arguments in other cases are entirely similar.

Let h be any positive real number. Then (3.32) and (3.41) imply that

\{ \begin{matrix}h^{n−α}u(rh) = u(\frac{r}{h}) \\ h^{n−α}u(sh) = u(\frac{s}{h}). \end{matrix}  (3.43)

It follows that

h^{n−α}u(sh) = u(r(\frac{s}{rh})) = (\frac{rh}{s})^{n−α}u(\frac{r^{2}h}{s}).

Consequently

u(sh) = (\frac{r}{s})^{n−α}u((\frac{r}{s})^{2}sh).  (3.44)

Suppose r ≠ s. Without loss of generality, we may assume that r > s. Let t = \frac{r}{s} and rewrite sh in (3.44) as h, we have

u(th) = t^{(α−n)/2}u(h).  (3.45)

In (3.45), we first let h = 1, then let h = t, and then continue this way, we arrive at

(t^{k})^{(n−α)/2}u(t^{k}) = u(1), ∀k = 1, 2, · · · .  (3.46)

Notice that

|x|^{n−α}u(x)→u(∞), as |x|→∞,

we have

(t^{k})^{(n−α)/2}u(t^{k})→0, as k→∞.

This contradicts with (3.46). Therefore, we must have r = s. This completes the proof of the Theorem.

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Addresses and E-mails

Wenxiong Chen Department of Mathematics Yeshiva University 500 W. 185th St. New York NY 10033 wchen@yu.edu

Congming Li Department of Applied Mathematics Campus Box 526 University of Colorado at Boulder Boulder CO 80309 cli@colorado.edu