99

# Semilinear elliptic equations with critical Sobolev exponent and non-homogeneous term

RIMS Workshop: Shapes and other properties of solutions of PDEs
11 Nov 2015
Research Institute for Mathematical Sciences, Kyoto University

## Kazune Takahashi

November 11, 2015

## Transcript

1. ### Semilinear elliptic equations with critical Sobolev exponent and non-homogeneous term

Kazune Takahashi (The University of Tokyo) RIMS Workshop — Shapes and other properties of solutions of PDEs November, 11, 2015

3. ### Constraints RN (N 3) Bounded smooth domain a L (

) is the ﬁrst eigenvalue of . 1 D a in is a constant such that . > 1 b L ( ) b 0 in , b 0 * * *
4. ### Main Problem u + au = bup + f in

, u > 0 in , u = 0 on . ( ) f H 1 ( ), f 0, f 0 > 0 Parameter Critical Sobolev Exponent p = N+2 N 2
5. ### u + au = bup + f in , u

> 0 in , u = 0 on . ( ) Existence of solutions depends on Involving critical Sobolev exponent p = N+2 N 2 Dimension Shape of N
6. ### H1 0 ( ) Lq ( ) Compact 1 q

< p + 1 Subspace, but not compact Background: Sobolev Embedding Not subspace generally H1 0 ( ) Lq ( ) * * * (q > p + 1) H1 0 ( ) Lp+1 ( )
7. ### Why consider Sobolev Embedding? Deﬁnition. u H1 0 ( )

is a weak solution of ( ) (Du · D + au ) dx = bup dx + f dx H1 0 ( ) th power of function H1 0 ( ) (p + 1) Integrable, but no convergence property

9. ### Known Result (1) u = (1 + u)p in ,

u > 0 in , u = 0 on . (1) Chase two solutions : 1. Minimal Solution 2. Second Solution u u
10. ### Minimal/Second Solution Deﬁnition. 1. is the minimal solution u for

any solution . u 2. is a second solution u is other than . u u u u in
11. ### u = (1 + u)p in , u > 0

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 . < = >
12. ### u = (1 + u)p in , u > 0

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:
13. ### Mountain Pass Theorem The Highest point of the “lowest pass”

is a critical point = solution. Can we get this point as the limit of a sequence? The Talenti function

15. ### Known Result (2) (2) Theorem 2. (Tarantello ’92) (2) has

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
16. ### Known Result (3) (3) ( ) with : a =

, b = 1 u + u = up + f in , u > 0 in , u = 0 on .
17. ### Minimal Solution of (3) (3) u + u = up

+ 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)
18. ### Second Solution of (3) (3) u + u = up

+ 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
19. ### Solutions of (3) with . Minimal Solution 0 · L

Second Solution N = 3, 4, 5 N 6 > 0
20. ### Minimal Solution of . There exists so that has >

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
21. ### Second Solution of . ( ) * 0 < <

* b achieves its maximum at . x0 M is continuous near . x0 * b * near . x0 Assumptions. a(x) = m1 + m2|x x0|q + o(|x x0|q)
22. ### Second Solution of . ( ) u + au =

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
23. ### Solutions of .- (iv) Minimal Solution 0 · L Second

Solution ( ) 3 N < 6 + 2q a(x) = m2|x x0|q
24. ### Comparison between (3) and . ( ) N < 6

+ 2q a = m2|x x0|q -(iv) ( ) a(x) x0 (3)-(ii) (3)-(i) a = 0 N < N < 6 a = > 0 0 > 0 0 (3)-(i) (3)-(ii) N < 6 N <

26. ### — Equation of . v Instead of a second solution

, we chase . v = u u u ( : the minimal solution ) u v + av = b (v + u )p up in , v > 0 in , v = 0 on . ( ) ( )
27. ### Catch by variational method v G(t, s, x) = b(x)

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
28. ### I (v) = 1 2 |Dv|2 dx b 1 p

+ 1 (v + u )p+1 1 p + 1up+1 upv dx + 1 2 av2 dx.
29. ### Sufficient conditions The mountain pass theorem A critical point of

I (v) condition (PS)c details omitted The Talenti function sup t>0 I (tv0) < 1 NM(N 2)/2 SN/2 v0 H1 0 ( ) : nonnegative ( )
30. ### Best Sobolev constant Deﬁnition. S = inf u H1 0

(V),u 0 Du 2 L2(V) u 2 Lp+1(V) V RN For , is the best Sobolev constant. Proposition. The inﬁmum is achieved by the Talenti function when . V = RN
31. ### Talenti function Deﬁnition. The Talenti function is U(x) = 1

(1 + |x|2)(N 2)/2 (x RN ) We mend this: u (x) = (x) ( + |x|2)(N 2)/2 , v (x) = u (x) b1/(p+1)u Lp+1( ) ( > 0) cut-off function
32. ### Calculation of Integrals sup t>0 I (tv ) = I

(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
33. ### Calculation of ﬁrst line 1 2t2 Dv 2 L2( )

1 p + 1tp+1 sup t>0 1 2t2 Dv 2 L2( ) 1 p + 1tp+1 = 1 N Dv 2 L2( ) N/2 = 1 NM(N 2)/2 SN/2 + O( (N 2)/2 ). ( ) Left side of
34. ### Integral of . H (t, s, x) = b(x) 1

p + 1 (t + s)p+1 1 p + 1tp+1 1 p + 1 sp+1 spt Cb(x)tp for . x , t t0, s s0 H H (t v , u )dx C {|x| } b(x)(t v )pdx = C (N 2)/4 .
35. ### Integral of . au2 dx = O(1) + m1I1 +

m2I2 I1 = {|x |<r0 } 1 ( + |x|2)N 2 dx, I2 = {|x |<r0 } |x|q ( + |x|2)N 2 dx Here, We set We evaluate x0 = 0 av2
36. ### Calculation of . I1, I2 I1 = O( (N 4)/2

) (N 5), O(|log |) (N = 4), O(1) (N = 3). : famous (referring Brezis-Nirenberg ’83) I1 I2 : calculated by ourselves I2 = O( (N q 4)/2 ) (N > q + 4), O(|log |) (N = q + 4), O(1) (N < q + 4).
37. ### b1/(p+1)u 2 Lp+1( ) = O( (N 2)/2 ) We

also have Then, we conclude av2 dx = O(1) + m1I1 + m2I2, I1 = O( ) (N 5), O( |log |) (N = 4), O( 1/2 ) (N = 3), I2 = O( 1+q/2 ) (N > q + 4), O( (N 2)/2|log |) (N = q + 4), O( (N 2)/2 ) (N < q + 4)
38. ### sup t>0 I (tv ) 1 N M(N 2)/2 SN/2

+ 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 )
39. ### Proof of (iv) : . . m1 = 0, m2

> 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 ,
40. ### Solutions of .- (iv) Minimal Solution 0 · L Second

Solution ( ) 3 N < 6 + 2q a(x) = m2|x x0|q