On the Existence of a Component-wise Positive Radially Symmetric Solution for a Superlinear System

The system under consideration is −∆u + a u u = u 3 − βuv 2 , u = u(x), −∆v + a v v = v 3 − βu 2 v, v = v(x), x ∈ R 3 , u| |x|→∞ = v| |x|→∞ = 0, where a u , a v and β are positive constants. We prove the existence of a component-wise positive smooth radially symmetric solution of this system. This result is a part of the results presented in the recent paper [1]; in our opinion, our method allows one to treat the problem simpler and shorter.


INTRODUCTION. RESULT
We consider the problem in the spatial dimension 3, where all the quantities are real, a u , a v and β are positive constants and Δ = is the Laplace operator.Our aim is to prove the existence of a pair of functions (u, v) ∈ C 2 (R 3 )×C 2 (R 3 ) satisfying (1)Ä(3), radially symmetric and component-wise positive in R 3 .Of course, (1)Ä(3) is a model problem which naturally arises when one considers standing waves for a coupled system of nonlinear Schré odinger equations and which has various applications in different areas of physics, for instance, in the heat and diffusion theory, in the theory of nonlinear waves, for example, in plasma or in water, etc.The author's interest to this problem was mainly stimulated by the quite recent article [1], on the one hand, and by his publication [2], on the other hand.In fact, with the present work we improve the results in [2], where it is assumed that β ∈ (0, 1], and obtain a result similar to one of those in [1] by another method in a simpler and shorter way; in fact, we proceed as in [2].Here, our main result is the following. Theorem.Let a u , a v and β be positive constants.Then, problem (1)Ä(3) has a C 2 -solution radially symmetric and component-wise positive in R 3 .
Remark 1.Of course, if β ∈ (0, 1) and a u = a v , then the problem has a solution (u, v) satisfying u ≡ v (see, for example, [3]).However, it seems to be surprising that the solution in the theorem above exists if β 1.A similar statement was already presented in [1].Now, we introduce some notation.Let H 1 = H 1 (0, ∞) be the standard Sobolev space of functions deˇned in (0, ∞) and equal to 0 at the point 0, with the ]dr be the equivalent norms in this space.

PROOF OF THE THEOREM
In the class of radially symmetric solutions, system (1)Ä(3) reduces to the following: where the prime denotes the differentiation in r.By the substitution y(r) = ru(r), z(r) = rv(r) we also reduce problem (4)Ä(6) to the following: System (7)Ä( 9) is variational, and X-extremals of the functional are formally its solutions.In view of estimate (15) and the proof of Lemma 2 (see below), it is determined in X.In the following, we exploit a variant of the method of S. I. Pokhozhaev described, for example, in [3].Let S = {(y, z) ∈ X : y 2 u = 1 and z 2 v = 1}.Consider an arbitrary (y 0 , z 0 ) ∈ S, a, b > 0 and a point (y, z) = (ay 0 , bz 0 ).For this point (y, z) to be an X-extremal of H, it is necessary that ∂H(y, z) ∂a = ∂H(y, z) ∂b = 0.This easily yields the following two conditions for X-extremals of H: Lemma 1.Let (y 0 , z 0 ) ∈ S.Then, a point (y, z) = (ay 0 , bz 0 ), where a, b > 0, satisfying (10) and ( 11) exists if and only if p 0 q 0 − s 2 0 > 0, where p 0 = p(y 0 ), q 0 = q(z 0 ) and s 0 = s(y 0 , z 0 ).
Proof.Substitute (y, z) = (ay 0 , bz 0 ) in ( 10) and (11).Then, we obtain the system with the unknown quantities a and b.It is easily seen that ( 12) has a real solution (a, b), where a, b > 0, if and only if p 0 q 0 − s 2 0 > 0, and this solution is given by Lemma 1 is proved.2 Consider the set and denote T = {(ay 0 , bz 0 ) : a, b > 0 are given by ( 13) and (y 0 , z 0 ) ∈ S 0 }.By (10) and ( 11) so that the functional H is bounded from below on T .
Lemma 2. The functionals p, q and s are weakly continuous in X.
Proof.Let a sequence {(y n , z n )} n=1,2,3,... be weakly converging in X.Then, it is bounded in X and, consequently, in C(0, ∞) × C(0, ∞).We have the estimate (and by analogy for z) which shows that for any > 0 there exists δ > 0 such that dr < and In addition, and by analogy for s and q.These estimates complete our proof of Lemma 2.2 Remark 2. Using estimates of the type of estimate (15), one can easily prove that the functional H = H(y, z) in X is continuously differentiable in each of its arguments when the other one is ˇxed.
Let {(y n , z n )} n=1,2,3,... be an arbitrary minimizing sequence for the functional H on T .By (14), it is bounded in X and therefore, weakly compact.Without the loss of generality we accept that it is weakly converging in X to a point (y, z), and that there exist limits of y n u and z n v as n → ∞.Lemma 3.There exists c > 0 such that y n c and z n c for all n.
Proof.Consider an arbitrary (y 0 , z 0 ) ∈ S 0 , the corresponding (y, z) = (ay 0 , bz 0 ) ∈ T , where a, b > 0, and the condition where y 1 = a 1 y, a 1 > 0. By (15), for example, theorem II.2.2 in [3]) y is a critical point of the functional H(y, z) taken with the ˇxed second argument, and z is a critical point of the functional H(y, z) taken with the ˇxed ˇrst argument.Therefore, by standard arguments, the pair (y, z) belongs to , and it is a solution of problem ( 7)-(9).