in , u = 0 on . (1) Minimal Solution of (1) Theorem. (Keener-Keller ’74, Crandall — Rabinowitz ’75) There exists so that (1) has > 0 the minimal solution for . only one solution when . no solution for . < = >
in , u = 0 on . (1) Second Solution of (1) Theorem 1. (Brezis-Nirenberg ’83) (1) has a second solution for . 0 < < The mountain pass theorem The Talenti function Keys:
at least 2 solutions for small . u = up + f in , u > 0 in , u = 0 on . The mountain pass theorem The Talenti function Keys: > 0 ( ) with : a = 0, b = 1
+ f in , u > 0 in , u = 0 on . There exists so that (3) has > 0 the minimal solution for . only one solution when . no solution for . < = > Theorem. (Naito-Sato ’12)
+ f in , u > 0 in , u = 0 on . Theorem 3. (Naito-Sato) 0 < < Let . Assume either: Then, (3) has a second solution. (i) and (ii) and 1 < 0 N 3 > 0 N = 3, 4, 5
0 the minimal solution for . only one solution when if . no solution for . < = > Theorem. (T) ( ) ( ) u + au = bup + f in , u > 0 in , u = 0 on . ( ) b > 0 in
bup + f in , u > 0 in , u = 0 on . ( ) Main Theorem. (T) Assume either: Then, (3) has a second solution. (i) and (ii) and (iii) , and (iv) , and N 3 m1 < 0 m1 > 0 N = 3, 4, 5 m1 = 0 m1 = 0 m2 > 0 m2 < 0 N 3 3 N < 6 + 2q
1 p + 1 (t+ + s)p+1 1 p + 1 sp+1 spt+ 1 2a(x)t2 + Here, is deﬁned by G: R R R We write as . G(v, u ) G(v(x), u (x), x) I (v) = 1 2 |Dv|2 dx G(v, u )dx v H1 0 ( ) is a solution of ( ) v is a critical point of
(t v ) The supremum is achieved: H (v, u ) I (v) = 1 2 |Dv|2 dx + 1 p + 1 bvp+1 dx b 1 p + 1 (v + u )p+1 1 p + 1vp+1 1 p + 1up+1 upv dx + 1 2 av2 dx. Divide into 3 lines: depending on q
+ C (N 2)/2 C (N 2)/4 + m1I1 + m2I2 Evaluation of . For small , > 0 Does there exist so that this ( )-part is negative? > 0 I av2 vp+1 and |Dv |2 H (t v , u )
> 0 I2 (N 2)/4 it suffice to have . as 0 3 N < 6 + 2q 1 + q 2 > N 2 4 , that is, Since I2 = O( 1+q/2 ) (N > q + 4), O( (N 2)/2|log |) (N = q + 4), O( (N 2)/2 ) (N < q + 4), this is equivalent to To satisfy C (N 2)/2 C (N 2)/4 + m2I2 < 0 ,