RENDICONTI DEL SEMINARIO MATEMATICO

80 M. Francawith slow decay.Moreover, exploiting the existence of the G.S. with slow decay in the case q 2 =2(q 1 − 1) > 2 ∗ , Flores was able to prove the following result in [1].THEOREM 16. Assume that 2 ∗ < q 1 < 2 ∗ < q 2 and q 1 > 2 N+2√ N−1−2N+2 √ N−1−4 . Then,given any integer k ≥ 1, there is a number ǫ k > 0 such that, if |q 2 − 2(q 1 − 1)| < ǫ k ,then there are at least k radial G.S. with fast decay for (2). In particular if q 2 =2(q 1 − 1) there are infinitely many G.S. with fast decay.The condition q 1 > 2 N+2√ N−1−2N+2 √ guarantees that P(+∞) is a focus and thisN−1−4point is crucial for the proof.We think that all the Theorems of this section could be generalized to the casep ̸= 2 using the new change of coordinates (4). Moreover we think that these techniquescould be adapted to generalize Theorems 13, 14, 15 also to the spatial dependentcase, that is when f (u, r) = k 1 (r)u|u| q1−2 + k k (r)u|u| q2−2 , where k 1 and k 2 are actuallyfunctions. The last Theorem 16 crucially depends on the existence of the G.S. withslow decay, that seems to be structurally unstable. However we have been able to computethis solution also for the corresponding equation (1). So perhaps also Theorem 16can be extended to the case p ̸= 2.(q 1 −1)pp−1REMARK 8. Consider (1) where f (u, r) = u|u| q1−1 + u|u| q2−1 , where q 2 =and p ∗ < q 1 < p ∗ < q 2 . Then there is a radial G.S. with slow decaywhere A = [∣ ∣ pq 1 −pu(r) = A(∣ p−1 (n −p(q 1 −1))] 1q 1 −p1B + r pp−16. f negative for u small and positive for u large) p−1q 1 −pq 1 −pand B = ( n − p(q 1−1)q 1 −p)Aq 1 −1p−1 .In this section we will consider (3), assuming that f (u, r) is negative for u small andpositive for u large and r small. The prototypical non-linearity we are interested in isthe following(22) f (u, r) = −k 1 (r)u|u| q 1−2 + k 2 (r)u|u| q 2−2where the functions k i (r) are nonnegative and continuous. When p = q 1 = 2 (2)describes a Bose-Einstein condensate, and the G.S., if it exists, is the least energysolution. In order to have G.S. we need to have a balance between the gain of energydue to the negative terms and the loss of energy due to the positive terms. The strengthof the contribution is proportional to the corresponding value of |Ji+ (r)|, so it dependsstrongly on the exponent q i . When f is as in (22) and q 1 < p ∗ ≤ q 2 the contribution