Carleson measures and elliptic boundary value problems Abstract. Mathematics Subject Classiﬁcation (2010). Keywords. 1.

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Brandon Kennedy
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1 A BMO BMO R n R n+ + BMO Q R n l(q) T Q = {(x, t) R n+ + : x I, 0 < t < l(q)} Q T Q dµ R n+ + C Q R n µ(t (Q)) < C Q Q

2 Q L p f L p (R n ) f < p < u(x, t) L p u x 0 R n x 0 Γ a (x 0 ) = {(x, t) : x x 0 < at} u(x, t) f L p (R n ) x 0 u(x, t) f(x 0 ) (x, t) Γ a (x 0 ) x 0 L p u(x, t) f BMO dµ = t u 2 dxdt R n+ + L p R n L := A(x), 60 6 R n+ A (n+) (n+) λ ξ 2 A(x)ξ, ξ := n+ i,j= A ij (x)ξ j ξ i, A L (R n ) λ, λ > 0 ξ R n+ x R n Lu = 0 Ω u W,2 loc (Ω) A u Ψ = 0, Ψ C0 (Ω) Ω R n+ + := {(x, t) R n (0, )}

3 L L p (dx) ω L L L ω R n A C θ Q F Q ( ) θ F ω(f ) C ω(q). Q ω A (dx) q > k := dω/dx ( Q /q k(x) dx) q C k(x) dx, Q Q q ω L q > L f L p p q L p ( ) /2 S α (u)(x) := u(y, t) 2 dydt x y <αt t n, 82 N α (u)(x) := u(y, t) (y,t): x y <αt q L p p q Lu = 0 R n+ + t 0 u(, t) = f L p (R n ).. D p N (u) Lp (R n ) < C f p. u f.. (y,t) (x,0) u(y, t) = f(x), a.e. x R n (y, t) Γ(x) := {(y, t) R n+ + : y x < t} C α L p L Lu = 0 R n+ +

4 B 2R (X) R n+ + X R n+ + R > 0 ( ) µ Y Z u(y ) u(z) C R B 2R (X) u 2 B 2 R(X) 2, Y, Z B R (X), µ > 0 C > 0 p > 0 p u(y ) C u p, Y, Z B R (X). B 2 R(X) B 2R (X) t L p (dx) < p < u(x, t) f N(u) p C f p L := A(x), A C L v u L A L 2 D 2 D p p < 2 t = ϕ(x) ϕ(x)

5 (x, t) (x, t ϕ(x)) R n+ + L t D 2 R n+ + t L 2 L t Lu = 0 e R n+ + (A u. u e) = 2 (D n+ (u)a u). L 2 L 2 L t R n+ + ω t L := A(x, t), R n+ + L 0 := A(x, 0) A(x, t) A(x, 0) ω L L L 0 t L 0 = A 0 L = A a(x, t) a(x, t) = { A 0 (y, s) A (y, s) : y x < t, t/2 < s < 2t}. a 2 (x, t)t dxdt ω L0 A ω L A

6 A L L 2 t L := A(x), A ϵ t u L (R n+ + ) u ϵ > 0 u ϵ Q 0 R n φ = φ Q0 W, (T Q0 ) u φ L (T Q0 ) < ϵ, φ(x, t) dxdt C ϵ, Q Q 0 Q T Q C ϵ Q ϵ R n+ + L 2 L ϵ L p L := A(x), R n+ + ϵ L Lu = 0 R n+ + u ϵ ω A A η > 0 δ > 0 Q R n E Q E / Q < η ω(e)/ω(q) < δ

7 Q r E ω(e) ϕ ϵ u E L ϕ r A r (ϕ)(x) := ( ) /2 dydt ϕ(y, t) x y <t<r t k = k(ϵ) n E k 2 < A 2 r(ϕ)(x)dx < A 2 r(ϕ)(x)dx < φ(x, t) dxdt E Q T Q C ϵ Q E / Q < η η /k 2 A(ϕ) u Lu = 0 u x E u ϕ ϕ ϕ x Q R n ω {I j,l } Q R n j 0 R n = l I j,l I j,l I j,l2 = l l 2 I j,l B(2 j, x l ) M B(M2 j, x l ) B(2 j, x l ) 2 j x l R n I j,l I j,l I j,l I j,l I j,l I j,l C M < ω(i j,l ) < C M ω(i j,l) I j,l I j,l O O = I j,l I j,l I j,l p j,l p j,l I j,l I j,l 228 ϵ E Q ϵ E k {O i } k i= E O 229 k O k... O 0 Q O i = Sl i 23 Sl i < i < k ω(o i S i l ) < ϵω(s i l ) ϵ Sj i 233 Si l k > i > m > 0 ω(sj m O i ) < ϵ i m ω(sj m) 234 ϵ > 0 δ > 0 ω(e) < δ E ϵ k k ω(e) 0 ϵ k f u L k f = ( ) i X Oi. i=0

8 u Lu = 0 u(x, 0) = f f 0 u x E X m = (x m, t m ) Γ(x) 0 < m < k u(x m ) > c 0 < m < k u(x m ) < c 2 c c 2 > c(ϵ) X m Sl m O m x l(s) S X m m Sl m m X m = (x m, t m ) t m t ηl(sl m ) X m t m < ρt m ρ < m u(x, t) = K(x, t; y, 0)f(y)dω(y) m u(x m ) = u (X m ) + u 2 (X m ) u (x, 0) = f (x) := m i=0 X O i. u > 0 c u(x m ) > K(x m, t m ; y, 0)f(y)dω(y) c ω(sl m f(y)dω(y). ) m f = Sl m u > c k ω(sl m f 2 (y)dω(y) < ) Sl m ω(sl m ω(o i Sl m ) < 2ϵ, ) i=m+ u(x m ) > c 2ϵ > c m f f Sl m (x m, t m ) t m l(sl m ) u(x m, t m ) < c 2 < c ϵ ϵ η {x m, t m } k m=0 Γ(x) u(x m, t m ) u(x m, t m ) > ϵ t m < ρt m L A(u) A(ϕ) ϕ u L L A BMO L := A(x), R n+ + ω A u Lu = 0 f BMO t u(x, t) 2 dxdt C f 2 Q Q BMO, T Q S m l

9 A η > 0 δ > 0 Q R n E Q ω(e)/ω(q) < η E / Q < δ E BMO f X E BMO ω(e)/ω(q) f A BMO u f A(u) E Q ω(e) S(u) Q Γ j (x) = {(y, t) Γ(x) : 2 j < t < 2 j+ } S(u)(x) = t n u 2 dydt j Γ j (x) Γ j (x) t n u 2 dydt u u u {x m, t m } k m=0 Γ(x) O m A L f O m Sl m Sl m Sl m S l m m f m l (Sm l \ S l m ) 0 m f m f m f m = f m = ) f = k m=0 f m u Lu = 0 f m 0 C, c > 0 x E {x m, t m } k m=0 ct m < t m < Ct m u(x m, t m ) u(x m, t m ) > ϵ L L A L p p > BMO L L := A(x), R n+ + ω A u Lu = 0 f t u(x, t) 2 dxdt C. Q Q T Q

10 ϵ L p p D p A ϵ t A L 2 R 2 L := A(x) R 2 e A(x, t) = A((x;, t) e) ω L A R 2 L 2 L p L p L p L t p < D p Q R n L ω L A (Q) Q A ϵ

11 u u(x, t) 2 tdtdx C u L Q Q (Ω), T Q C A A A ϵ L 2 L R n

12 L p p A C

13 L p L H (R n )

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