On the Hardy constant of non-convex planar domains: the case of the quadrilateral

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1 On the Hardy constant of non-convex planar domains: the case of the quadrilateral Gerassimos Barbatis University of Athens joint work with Achilles Tertikas University of Crete Heraklion, Hardy constant September of quadrilateral2013 Heraklion, September

2 Hardy (1920 s) 0 u (t) 2 dt t 2 dt, for all u C c (0, ). Heraklion, September

3 Hardy (1920 s) 0 u (t) 2 dt t 2 dt, for all u C c (0, ). The inequality is optimal Heraklion, September

4 Hardy (1920 s) 0 u (t) 2 dt t 2 dt, for all u C c (0, ). The inequality is optimal There is no minimizer in H0 1 (0, + ) Heraklion, September

5 Higher dimensional analogue: R n bounded Heraklion, September

6 Higher dimensional analogue: R n bounded A. Hardy inequalities involving distance to a point: u 2 dx c x 2 dx, u C c (). ( ) 2 Best constant N 2 2. Heraklion, September

7 Higher dimensional analogue: R n bounded A. Hardy inequalities involving distance to a point: u 2 dx c x 2 dx, u C c (). ( ) 2 Best constant N 2 2. B. Hardy inequalities involving distance to the boundary: d(x) = dist(x, ). u 2 dx c d 2 dx, for all u C c () Valid if has a Lipschitz boundary. Heraklion, September

8 Higher dimensional analogue: R n bounded A. Hardy inequalities involving distance to a point: u 2 dx c x 2 dx, u C c (). ( ) 2 Best constant N 2 2. B. Hardy inequalities involving distance to the boundary: d(x) = dist(x, ). u 2 dx c d 2 dx, for all u C c () Valid if has a Lipschitz boundary. What about best constant? Heraklion, September

9 Higher dimensional analogue: R n bounded A. Hardy inequalities involving distance to a point: u 2 dx c x 2 dx, u C c (). ( ) 2 Best constant N 2 2. B. Hardy inequalities involving distance to the boundary: d(x) = dist(x, ). u 2 dx c d 2 dx, for all u C c () Valid if has a Lipschitz boundary. What about best constant? We always have c 1/4. Heraklion, September

10 What about u 2 dx 1 4 d 2 dx, for all u C c ()? Heraklion, September

11 What about u 2 dx 1 4 d 2 dx, for all u C c ()? Geometric assumptions are required. Heraklion, September

12 What about u 2 dx 1 4 d 2 dx, for all u C c ()? Geometric assumptions are required. Typical assumption: is convex Heraklion, September

13 What about u 2 dx 1 4 d 2 dx, for all u C c ()? Geometric assumptions are required. Typical assumption: is convex Various improvements Theorem (Brezis and Marcus, 97) u 2 dx 1 4 d 2 dx d 2 X 2 (d)dx, for all u C c () Heraklion, September

14 Theorem (B., Filippas and Tertikas 03) u 2 dx 1 4 d 2 dx d 2 X 1 2 dx d 2 X 1 2 X 2 2 dx +... Heraklion, September

15 Theorem (B., Filippas and Tertikas 03) u 2 dx 1 4 d 2 dx d 2 X 1 2 dx d 2 X 1 2 X 2 2 dx +... Convexity was replaced by weaker condition d(x) 0, in N = 2 convex N 3 strictly weaker (Armitage and Kuran 85) Heraklion, September

16 Theorem (B., Filippas and Tertikas 03) u 2 dx 1 4 d 2 dx d 2 X 1 2 dx d 2 X 1 2 X 2 2 dx +... Convexity was replaced by weaker condition d(x) 0, in N = 2 convex N 3 strictly weaker (Armitage and Kuran 85) Lewis, J. Li and Y.Y. Li (2012), Psaradakis (2012): If is C 2 and mean convex, then d(x) 0 in. Heraklion, September

17 Theorem (Avkhadiev 13) There exists a constant C = C(n) such that the following is true: if for every y there exists an open ball B y with y B y R n \ and with radius ρ satisffying ρ C n Inr(), then u 2 dx 1 4 d 2 dx, for all u C c (). Heraklion, September

18 Proof for convex domains. Heraklion, September

19 Proof for convex domains. For any φ > 0 we have 0 u φ φ u 2 dx = = u 2 φ 2 dx + dx φ φ dx φ 2 Heraklion, September

20 Proof for convex domains. For any φ > 0 we have 0 u φ φ u 2 dx = = u 2 φ 2 dx + dx φ φ dx = φ 2 u 2 φ dx φ u2 dx. Heraklion, September

21 Proof for convex domains. For any φ > 0 we have 0 u φ φ u 2 dx = = u 2 φ 2 dx + dx φ φ dx = φ 2 u 2 φ dx φ u2 dx. Hence: if there exists a positive function φ such that then u 2 dx 1 4 φ + 1 4d 2 φ 0 in, d 2 dx, for all u C c (). Heraklion, September

22 Let φ = d 1/2. Then φ + 1 4d 2 φ = 1 2 d 1/2 d 0, in. Heraklion, September

23 Let φ = d 1/2. Then φ + 1 4d 2 φ = 1 2 d 1/2 d 0, in. The general problem: Given a bounded domain R n determine (or find information about) the Hardy constant of. Heraklion, September

24 Theorem A. Ancona (1986) If R 2 is simply connected then u 2 dx 1 16 d 2 dx, for all u C c () Heraklion, September

25 Theorem A. Ancona (1986) If R 2 is simply connected then u 2 dx 1 16 d 2 dx, for all u C c () Proof. Uses Koebe 1/4 Thm g : D C, injective analytic ( ) B g(0), g (0) g(d) 4 Heraklion, September

26 Theorem A. Ancona (1986) If R 2 is simply connected then u 2 dx 1 16 d 2 dx, for all u C c () Proof. Uses Koebe 1/4 Thm g : D C, injective analytic This implies ( ) B g(0), g (0) g(d) 4 f : R 2 +, conformal mapping dist(f (z), ) y 2 f (z) (z = x + iy) = 1 y 2 1 4d 2 (f (z)) f (z) 2. Heraklion, September

27 Other results. Heraklion, September

28 Other results. Marcus, Mizel, Pinchover (1998) The Hardy constant of spherically symmetric annular domain is 1/4 when n 3. Heraklion, September

29 Other results. Marcus, Mizel, Pinchover (1998) The Hardy constant of spherically symmetric annular domain is 1/4 when n 3. Davies (1995) Computed the Hardy constant c β of a circular sector β of angle β. Heraklion, September

30 Proof. Using a positive supersolution: Heraklion, September

31 Proof. Using a positive supersolution: Euler-Lagrange u(x) + c d 2 u(x) = 0. (x) Heraklion, September

32 Proof. Using a positive supersolution: Euler-Lagrange u(x) + c d 2 u(x) = 0. (x) Seek a solution depending only on the angle θ: Heraklion, September

33 Proof. Using a positive supersolution: Euler-Lagrange u(x) + c d 2 u(x) = 0. (x) Seek a solution depending only on the angle θ: u = ψ(θ) where ψ solves { ψ (θ) + cv (θ)ψ(θ) = 0, 0 θ β, ψ(0) = ψ(β) = 0, where V (θ) = 1 sin 2 θ, 0 < θ < π 2, 1, π 2 < θ < β π 2, 1 sin 2 (β θ), β π 2 < θ < β. Heraklion, September

34 Natural questions Given a simply connected domain R 2 find (or obtain information about) the Hardy constant of. Heraklion, September

35 Natural questions Given a simply connected domain R 2 find (or obtain information about) the Hardy constant of. Find the best uniform Hardy constant valid for all simply connected domains R 2. Moreover, determine whether there are extremal domains, that is domains whose Hardy constant coincides with the best uniform Hardy constant. raised by A. Laptev (2005), R. Banuelos (2009). Heraklion, September

36 Natural questions Given a simply connected domain R 2 find (or obtain information about) the Hardy constant of. Find the best uniform Hardy constant valid for all simply connected domains R 2. Moreover, determine whether there are extremal domains, that is domains whose Hardy constant coincides with the best uniform Hardy constant. raised by A. Laptev (2005), R. Banuelos (2009). Not much progress! Heraklion, September

37 Theorem (Laptev and Sobolev 08) Let be simply connected and satisfy an external cone condition: each y is the vertex of an infinite cone C of angle θ with C. Then u 2 dx π2 4θ 2 d 2 dx, for all u C c (). Best case: convex θ = π Worst case: θ = 2π Heraklion, September

38 Theorem (Laptev and Sobolev 08) Let be simply connected and satisfy an external cone condition: each y is the vertex of an infinite cone C of angle θ with C. Then u 2 dx π2 4θ 2 d 2 dx, for all u C c (). Best case: convex θ = π Worst case: θ = 2π Proof. Uses a refined version of Koebe s theorem. Heraklion, September

39 Theorem Heraklion, September

40 Theorem Let β cr be the unique solution in (π, 2π) of the equation tan ( ( β cr π ) Γ( 3 = 4 4 ) ) 2 4 Γ( 1 4 ). Heraklion, September

41 Theorem Let β cr be the unique solution in (π, 2π) of the equation tan ( ( β cr π ) Γ( 3 = 4 4 ) ) 2 4 Γ( 1 4 ). Let be a non-convex quadrilateral with non-convex angle π < β < 2π. Then u 2 dx c β d 2 dx, where c β is the unique solution of the equation u C c (), cβ tan ( c β ( β π ( c Γ( β ) 2 )) 4 ) 2 = 2 Γ( 1+, 1 4c β 4 ) when β cr β < 2π and c β = 1/4 when π < β β cr. The constant c β is the best possible. Heraklion, September

42 About the proof. Heraklion, September

43 About the proof. One dimensional inequalities. Heraklion, September

44 About the proof. One dimensional inequalities. Eg. Let α denote the largest solution of α(1 α) = c β. Let 0 ω π/4. Then ψ (θ) ψ(θ) sin θ cos(θ + ω) + α cos ω 0, 0 θ π 2. Heraklion, September