Variations on the p Laplacian

Transcription

1 kawohl

2 Toics -harmonic functions u = 1 overdetermined roblems oen roblems

3 Lalace oerator u = u x1 x u xnx n = u νν + u ν div(ν) where ν(x) = u(x) u(x) is direction of steeest descent. In fact, div(ν) = u u + u x i u xj u xi x j u 3 = u u + u νν u so that u = u νν u div(ν) = u νν + u ν div(ν) or u = u νν + u ν (n 1)H with H denoting mean curvature of a level set of u. For radial u recall u = u rr + n 1 r u r.

4 For (1, ) one can write the -Lalace oerator as u = div ( u 2 u ) = u 2 [ u + ( 2)u νν ] = u 2 [( 1)u νν + (n 1)Hu ν ] and the normalized or game-theoretic -Lalace oerator as N u = 1 u 2 div ( u 2 u ) = 1 u νν + 1 (n 1)Hu ν = 1 N u + 1 N 1 u. Observe N u = u νν, N 2 u = 1 2 u and N 1 u = u div( u u ).

5 -harmonic functions Given Ω R n bounded, Ω of class C 2,α and g(x) W 1, (Ω) u = 0 in Ω, (1) u(x) = g(x) on Ω. (2) u can be charactzerized as the unique (weak) solution of the strictly convex variational roblem Minimize I (v) = v L (Ω) on g(x) + W 1, 0 (Ω), (3) so that Ω u 2 u φ dx = 0 for every φ W 1, 0 (Ω). (4) It is well known, that weak solutions are locally of class C 1,α. They are even of class C wherever their gradient does not vanish.

6 One can show (Juutinen, Lindqvist, Manfredi 2001) that weak solutions are also viscosity solutions of the associated Euler equation F (Du, D 2 u) = Du 4 ( Du 2 traced 2 u + D 2 udu, Du ) = 0 Incidentally, only for (1, 2) does this imly that they are also viscosity solutions of the normalized equation F N (Du, D 2 u) = 1 traced2 u 2 D 2 udu, Du Du 2 = 0

7 What haens as? For g W 1, (Ω) the family u is uniformly bounded in W 1, because I (u ) I (g) g Ω. Wolog Ω := 1. For q > n fixed and > q one finds u q u Ω ( q)/q g Ω 1+1/q +1 as, so u u in some C α. By the stability theorem for viscosity solutions u should be viscosity solution to a limit equation F (Du, D 2 u) = 0. What is this equation? Let us check the condition for subsolutions.

8 Let ϕ be a C 2 testfunction s.th. ϕ u has a min at x and ϕ(x ) 0. Then wolog ϕ u has a min at x near x and x x as. Since u is viscosity subsolution or Dϕ 4 [ Dϕ 2 ϕ + ( 2) D 2 ϕdϕ, Dϕ ] (x ) 0, 2 D2 ϕdϕ, Dϕ (x ) 1 Dϕ 2 ϕ(x ). gives D 2 ϕdϕ, Dϕ (x ) := ϕ Thus u is (unique) viscosity solution of u = 0 in Ω, u = g on Ω.

9 It is worth noting that the variational roblem Minimize I (v) = v L (Ω) on g(x) + W 1, 0 (Ω), (5) can have many solutions,

10 It is worth noting that the variational roblem Minimize I (v) = v L (Ω) on g(x) + W 1, 0 (Ω), (5) can have many solutions, e.g. the minimum of two cones (not C 1 ) or u C 1,α (Savin). Kawohl, Shagholian 2005

11 What haens to -harmonic functions as 1? I. g. no uniform convergence, but Juutinen (2005) found sufficient conditions: If g C(Ω) and Ω convex, then u u 1 uniformly as 1. Moreover, u 1 is unique minimizer of { } E 1 (v) = su u divσdx; σ C0 (Ω, R n ), σ(x) 1 in Ω Ω on {v BV (Ω) C(Ω), v = g on Ω}. Here the limiting variational roblem has a unique solution, while the limiting Euler equation can have many viscosity solutions.

12 Heuristic reason for uniqueness of minimizer of the TV-functional: If there are two minimizers u and v (for simlicity in W 1,1 (Ω)) of E 1, then any convex combination w = tu + (1 t)v would also be minimizer, hence level lines of u are also level lines of v, u v, v = f (u). Dirichlet cond. imlies f (g) = g, so that f = Id on range Ω (g). But since min Ω g u, v max Ω g in Ω we find f (u) = u in Ω.

13 Nonuniqueness of viscosity solutions to the Dirichlet roblem 1 u = 0 in Ω, u = g on Ω. Sternberg, Ziemer (1994) gave counterexamle: Ω = B(0, 1) R 2, g(x 1, x 2 ) = cos(2ϕ) has a whole family u λ of viscosity solutions, λ [ 1, 1], but only u 0 minimizes E 1. In fact, 2x1 2 1 left and right of rectangle u λ (x 1, x 2 ) = λ in the rectangle generated by cos(2ϕ) = λ 1 2x2 2 on to and bottom is viscosity sol. of both 1 u = 0 and N 1 u = κ u = 0 in Ω.

14 wobei hier u λ (r, ϕ) die Darstellung der Funktion u λ in Polarkoordinaten ist. (b) Die Niveaulinien von u λ sehen entsrechend der linken Abbildung aus. 1 λ u λ 1 ϕ λ 1 1 Diese Funktion löst formell das Randwertroblem (level) lot of u λ 1 u = 0 in B, u = ψ auf B. Geben Sie u λ exlizit an und berechnen Sie

15 definition of viscosity solutions for discontinuous F u is a viscosity solution of F (Du, D 2 u) = 0, iff sub- and suersol. u is subsol. if for every x Ω and ϕ C 2 s.th. ϕ u has min at x the ineq. F (Dϕ, D 2 ϕ) 0 holds. Here F = lsc hull of F. u is suersol. if for every x Ω and ϕ C 2 s.th. ϕ u has max at x the ineq. F (Dϕ, D 2 ϕ) 0 holds. Here F = usc hull of F. { ( ) F N 1 δ ij + ( 1) q i q j X q (q, X ) = 2 ij if q 0? if q = 0

16 Symm. matrix X has eigenvalues λ 1 (X ) λ 2 (X )... λ n (X ) { F N 1 n 1 i=1 (0, X ) = λ i 1 λ n if [2, ] 1 n i=2 λ i 1 λ 1 if [1, 2] { F N 1 n (0, X ) = i=2 λ i 1 λ 1 if [2, ] 1 n 1 i=1 λ i 1 λ n if [1, 2]

17 Symm. matrix X has eigenvalues λ 1 (X ) λ 2 (X )... λ n (X ) { F N 1 n 1 i=1 (0, X ) = λ i 1 λ n if [2, ] 1 n i=2 λ i 1 λ 1 if [1, 2] { F N 1 n (0, X ) = i=2 λ i 1 λ 1 if [2, ] 1 n 1 i=1 λ i 1 λ n if [1, 2] In articular, for n = 2, F1 N (0, X ) = λ 2 and F1 N (0, X ) = λ1,

18 Symm. matrix X has eigenvalues λ 1 (X ) λ 2 (X )... λ n (X ) { F N 1 n 1 i=1 (0, X ) = λ i 1 λ n if [2, ] 1 n i=2 λ i 1 λ 1 if [1, 2] { F N 1 n (0, X ) = i=2 λ i 1 λ 1 if [2, ] 1 n 1 i=1 λ i 1 λ n if [1, 2] In articular, for n = 2, F1 N (0, X ) = λ 2 and F1 N (0, X ) = λ1, so that we require λ 2 (D 2 ϕ) 0 for subsols. if ϕ(x) = 0 and λ 1 (D 2 ϕ) 0 for suersols. if ϕ(x) = 0

19 u = 1 The Dirichlet roblem u = 1 in Ω, u = 0 on Ω can be treated in a similar way. Again surrises as or 1. lim u (x) = d(x, Ω) (Kawohl 1990) and the limiting deq and bv is Du = 1 in Ω, u = 0 on Ω (Bhattacharya, DiBenedetto, Manfredi 1991) It has a unique viscosity solution, but many distributional solutions.

20 u = 1 The Dirichlet roblem u = 1 in Ω, u = 0 on Ω can be treated in a similar way. Again surrises as or 1. lim u (x) = d(x, Ω) (Kawohl 1990) and the limiting deq and bv is Du = 1 in Ω, u = 0 on Ω (Bhattacharya, DiBenedetto, Manfredi 1991) It has a unique viscosity solution, but many distributional solutions. 0 if Ω is small lim u (x) = discontinuous if Ω is inbetween 1 + if Ω is large

21 Why this strange behaviour as 1? The limiting equation 1 u = 1 reads (n 1)H = 1 or H = 1 n 1 in intrinsic coordinates.

22 Why this strange behaviour as 1? The limiting equation 1 u = 1 reads (n 1)H = 1 or H = 1 n 1 in intrinsic coordinates. Level surfaces satisfying this curvature condition in Ω are boundaries of so-called Cheeger sets. A set C Ω is a Cheeger set of Ω if it infimizes D / D among all smooth subsets of Ω,... this would be an extra talk.

23 N u = 1 What about N u = 1 in Ω, u = 0 on Ω? For (1, ] there exists a unique viscosity solution (Lu Wang 2008), details in the ublished version of this talk. For = the equation reads u νν = 1 in Ω, and for = 1 it is u (n 1)H = 1 in Ω. They are degenerate ellitic in the sense of viscosity solutions.

24 Serrin and Weinberger roved in 1971 that the following overdet. bv cannot have a solution in a smooth simly connected domain unless Ω is a ball. u = 1 in Ω, u = 0 and u ν = a = const. on Ω. Physical interretation: Laminar flow in a noncircular ie cannot have constant shear stress on the wall of the ie.

25 Serrin s roof uses the moving lane method and alies to ositive classical solutions of autonomous strongly ellitic equations n a ij (u, u )u xi x j i,j=1 = f (u, u ), while Weinberger s roof is given only for u = 1 and uses both variational methods and (other) maximum rinciles. Does the roof at least extend to u = 1?

26 Serrin s roof uses the moving lane method and alies to ositive classical solutions of autonomous strongly ellitic equations n a ij (u, u )u xi x j i,j=1 = f (u, u ), while Weinberger s roof is given only for u = 1 and uses both variational methods and (other) maximum rinciles. Does the roof at least extend to u = 1? Yes (Farina Kawohl 2008)

27 1) P(x) := 2( 1) u(x) + 2 nu(x) attains max. over Ω on Ω, and thus P(x) 2( 1) a =: c in Ω.

28 1) P(x) := 2( 1) u(x) + 2 nu(x) attains max. over Ω on Ω, and thus P(x) 2( 1) a =: c in Ω. 2) Show that Ω P(x)dx = c Ω, then P(x) c on Ω.

29 1) P(x) := 2( 1) u(x) + 2 nu(x) attains max. over Ω on Ω, and thus P(x) 2( 1) a =: c in Ω. 2) Show that Ω P(x)dx = c Ω, then P(x) c on Ω. 3) Show that P c in Ω imlies radial symmetry.

30 1) P(x) := 2( 1) u(x) + 2 nu(x) attains max. over Ω on Ω, and thus P(x) 2( 1) a =: c in Ω. 2) Show that Ω P(x)dx = c Ω, then P(x) c on Ω. 3) Show that P c in Ω imlies radial symmetry. Caution with 1) and 2): 1) u C 3, so P is roblematic. Regularize 2) u C 2, so classical Pohožaev identities need adjustments to C 1 -functions by Degiovanni, Musesti, Squassina (2003).

31 Proof that P c in Ω: Testing u = 1 with u gives Ω u dx = Ω udx (1)

32 Proof that P c in Ω: Testing u = 1 with u gives Ω u dx = Ω udx (1) test with (x, u): Ω u(x, u) = Ω (x, u) = n Ω u (2)

33 Proof that P c in Ω: Testing u = 1 with u gives Ω u dx = Ω udx (1) test with (x, u): Ω u(x, u) = Ω (x, u) = n Ω u (2) lhs of (2)= Ω u 2 u (x, u) Ω a 2 u ν (x, u)

34 Proof that P c in Ω: Testing u = 1 with u gives Ω u dx = Ω udx (1) test with (x, u): Ω u(x, u) = Ω (x, u) = n Ω u (2) lhs of (2)= Ω u 2 u (x, u) Ω a 2 u ν (x, u) = [ ] Ω u 2 u 2 + (x, ( u 2 2 )) Ω a (x, ν)

35 Proof that P c in Ω: Testing u = 1 with u gives Ω u dx = Ω udx (1) test with (x, u): Ω u(x, u) = Ω (x, u) = n Ω u (2) lhs of (2)= Ω u 2 u (x, u) Ω a 2 u ν (x, u) = [ ] Ω u 2 u 2 + (x, ( u 2 2 )) Ω a (x, ν) = Ω u + (x, ( u ))dx a n Ω

36 Proof that P c in Ω: Testing u = 1 with u gives Ω u dx = Ω udx (1) test with (x, u): Ω u(x, u) = Ω (x, u) = n Ω u (2) lhs of (2)= Ω u 2 u (x, u) Ω a 2 u ν (x, u) = [ ] Ω u 2 u 2 + (x, ( u 2 2 )) Ω a (x, ν) = Ω u + (x, ( u = Ω u n u dx + Ω ))dx a n Ω a (x, ν)ds a n Ω

37 Proof that P c in Ω: Testing u = 1 with u gives Ω u dx = Ω udx (1) test with (x, u): Ω u(x, u) = Ω (x, u) = n Ω u (2) lhs of (2)= Ω u 2 u (x, u) Ω a 2 u ν (x, u) = [ ] Ω u 2 u 2 + (x, ( u 2 2 )) Ω a (x, ν) = Ω u + (x, ( u = Ω u n u dx + Ω = Ω n [ 1 n u u ] ))dx a n Ω dx 1 a (x, ν)ds a n Ω a n Ω

38 Proof that P c in Ω: Testing u = 1 with u gives Ω u dx = Ω udx (1) test with (x, u): Ω u(x, u) = Ω (x, u) = n Ω u (2) lhs of (2)= Ω u 2 u (x, u) Ω a 2 u ν (x, u) = [ ] Ω u 2 u 2 + (x, ( u 2 2 )) Ω a (x, ν) now 2 n (2) = Ω = Ω u + (x, ( u = Ω u n u dx + Ω = Ω n [ 1 n u u ] ))dx a n Ω dx 1 a (x, ν)ds a n Ω a n Ω 2 n u 2 u dx c Ω = 2 Ω u

39 Proof that P c in Ω: Testing u = 1 with u gives Ω u dx = Ω udx (1) test with (x, u): Ω u(x, u) = Ω (x, u) = n Ω u (2) lhs of (2)= Ω u 2 u (x, u) Ω a 2 u ν (x, u) = [ ] Ω u 2 u 2 + (x, ( u 2 2 )) Ω a (x, ν) now 2 n (2) = Ω so by (1): = Ω u + (x, ( u = Ω u n u dx + Ω = Ω n [ 1 n u u Ω ] ))dx a n Ω dx 1 a (x, ν)ds a n Ω a n Ω 2 n u 2 u dx c Ω = 2 Ω u 2 n u + 2( 1) u dx = c Ω (= Ω P(x)dx)

40 P c in Ω imlies symmetry: a) If Ω C 2,α, thenp ν = 0 on Ω imlies H = 1 n a1, because P ν = 2( 1) u ν 2 u ν u νν + 2 n u ν = [ ( 1) u ν 2 u νν + 1 ] 2u ν = 0 n

41 P c in Ω imlies symmetry: a) If Ω C 2,α, thenp ν = 0 on Ω imlies H = 1 n a1, because P ν = 2( 1) u ν 2 u ν u νν + 2 n u ν = [ ( 1) u ν 2 u νν + 1 ] 2u ν = 0 n and u = 1 = ( 1) u ν 2 u νν + (n 1)H u ν 2 u ν imly H = 1 n a1 on Ω. Done.

42 P c in Ω imlies symmetry: a) If Ω C 2,α, thenp ν = 0 on Ω imlies H = 1 n a1, because P ν = 2( 1) u ν 2 u ν u νν + 2 n u ν = [ ( 1) u ν 2 u νν + 1 ] 2u ν = 0 n and u = 1 = ( 1) u ν 2 u νν + (n 1)H u ν 2 u ν imly H = 1 n a1 on Ω. Done. b) If Ω is not smooth, consider Γ = {x u(x) = ε}.

43 P c in Ω imlies symmetry: a) If Ω C 2,α, thenp ν = 0 on Ω imlies H = 1 n a1, because P ν = 2( 1) u ν 2 u ν u νν + 2 n u ν = [ ( 1) u ν 2 u νν + 1 ] 2u ν = 0 n and u = 1 = ( 1) u ν 2 u νν + (n 1)H u ν 2 u ν imly H = 1 n a1 on Ω. Done. b) If Ω is not smooth, consider Γ = {x u(x) = ε}. u C 1,β (Ω) & u ν = a on Ω imly u 0 and u C 2,β near Γ.

44 P c in Ω imlies symmetry: a) If Ω C 2,α, thenp ν = 0 on Ω imlies H = 1 n a1, because P ν = 2( 1) u ν 2 u ν u νν + 2 n u ν = [ ( 1) u ν 2 u νν + 1 ] 2u ν = 0 n and u = 1 = ( 1) u ν 2 u νν + (n 1)H u ν 2 u ν imly H = 1 n a1 on Ω. Done. b) If Ω is not smooth, consider Γ = {x u(x) = ε}. u C 1,β (Ω) & u ν = a on Ω imly u 0 and u C 2,β near Γ. Thus Γ C 2,α and P ν = 0 on Γ, i.e. [ ( 1) u ν 2 u νν + 1 n] = 0

45 P c in Ω imlies symmetry: a) If Ω C 2,α, thenp ν = 0 on Ω imlies H = 1 n a1, because P ν = 2( 1) u ν 2 u ν u νν + 2 n u ν = [ ( 1) u ν 2 u νν + 1 ] 2u ν = 0 n and u = 1 = ( 1) u ν 2 u νν + (n 1)H u ν 2 u ν imly H = 1 n a1 on Ω. Done. b) If Ω is not smooth, consider Γ = {x u(x) = ε}. u C 1,β (Ω) & u ν = a on Ω imly u 0 and u C 2,β near Γ. Thus Γ C 2,α and P ν = 0 on Γ, i.e. [ ( 1) u ν 2 u νν + 1 n] = 0 Now we get 1 (n 1)H u ν n = 0 or H = h( u ν ) on Γ.

46 P c in Ω imlies symmetry: a) If Ω C 2,α, thenp ν = 0 on Ω imlies H = 1 n a1, because P ν = 2( 1) u ν 2 u ν u νν + 2 n u ν = [ ( 1) u ν 2 u νν + 1 ] 2u ν = 0 n and u = 1 = ( 1) u ν 2 u νν + (n 1)H u ν 2 u ν imly H = 1 n a1 on Ω. Done. b) If Ω is not smooth, consider Γ = {x u(x) = ε}. u C 1,β (Ω) & u ν = a on Ω imly u 0 and u C 2,β near Γ. Thus Γ C 2,α and P ν = 0 on Γ, i.e. [ ( 1) u ν 2 u νν + 1 n] = 0 Now we get 1 (n 1)H u ν n = 0 or H = h( u ν ) on Γ. But since P c, u = g(u) and H = h(g(ε)) = const. on Γ.

47 There is also an anisotroic version of the Serrin/Weinberger result Theorem (A. Cianchi & P. Salani, Dec 2008) Suose that H is a norm with a strictly convex unit ball and that u is a minimizer of ( 1 2 H( v)2 v ) dx in W 1,2 0 (Ω), and H( u) = a on Ω. Ω Then Ω is a ball in the dual norm H 0 to H of suitable radius r and u(x) = r 2 H 0 (x) 2 2n. The roof of Cianchi and Salani uses entirely different methods. Indeendently, in May 2009 Guofang Wang and Chao Xia gave another roof that follows our method.

48 What about N u = 1 in Ω, u = 0 AND u = a on Ω? For = 1 we look at u (n 1)H = 1 in Ω, u = a and u = 0 on Ω. So a C 2 solution on a smooth domain satisfies H 1/(a(n 1)) on Ω. Aly Alexandrov to see that Ω = ball of radius (n 1)a. For = the overdetermined bv. u νν = 1 in Ω, u = a and u = 0 on Ω can have C 1 viscosity solutions on secial (non-ball) domains, e.g. stadium domains. Buttazzo Kawohl 2011

49 oen roblems For fixed (1, ) consider the second eigenfunction u 2 + λ 2 u 2 2 u 2 = 0 in Ω, u = 0 on Ω. It changes sign, it has two nodal domains, it can be characterized as a mountain ass going from u 1 to u 1. (Cuesta, de Figuereido, Gossez 1999)

50 For (1, ) and Ω R 2 the eigenfunction u 2 has a nodal line. Conjectures: a) For Ω a disk, the nodal line is a diameter. b) For Ω a square the nodal line is diagonal if (2, ) and horizontal or vertical if (1, 2). Conjectures a) and b) hold for = 1 (Enea Parini 2009), = 2, and = (Juutinen & Lindqvist 2005). Moreover, they are suorted for general by numerical evidence of Jiří Horák (2009).

51 Ω a disk, = 1.1, courtesy of J. Horák

52 Ω a square, = 5, courtesy of J. Horák

53 Ω a square, = 1.1, courtesy of J. Horák