The Dirichlet boundary problems for second order parabolic operators satisfying a Carleson condition

Transcription

1 The Dirichlet boundary problems for second order parabolic operators satisfying a Martin Dindos Sukjung Hwang University of Edinburgh Satellite Conference in Harmonic Analysis Chosun University, Gwangju, Korea 06 August 2014

2 Main Question Solvability of the following 2nd order parabolic PDE { u t = div (A u) + B u in Ω, u = f on. where f L 2 (). What conditions on the coefficients? What is the time-varying domain Ω? The maximum principle provides L solvability

3 Parabolic distance and ball For the parabolic distance δ(x, t) = inf (Y,s) define a parabolic ball and surface ball that ( X Y + t s 1/2), B r (X, t) = {(Y, s) R n R : X Y + t s 1/2 < r}. Moreover, we define r (X, t) = B r (X, t), T ( r )(X, t) = Ω B r (X, t).

4 Carleson measure Definition Then a measure µ : Ω R + is said to be Carleson if there exists a constant C = C(r 0 ) such that for all r r 0 and all surface balls r µ (T ( r )) Cσ( r ). The best constant C(r 0 ) is called the Carleson norm. If lim r0 0 C(r 0 ) = 0, then we say that the measure µ satisfies the vanishing.

5 Conditions on the coefficients Ellipticity and Boundedness for A = [a ij ] λ ξ 2 a ij ξ i ξ j Λ ξ 2. Small s on the coefficients ( δ(x, t) A 2 + δ 3 (X, t) A t 2) dx dt, δ(x, t) B 2 (X, t) dx dt are Carleson measures with the norm ɛ. δ(x, t) A + δ 2 (X, t) A t ɛ 1/2, δ(x, t) B ɛ 1/2.

6 The time varying domain Ω For X = (x 0, x) R R n 1, consider the region given by Ω = {(x 0, x, t) : x 0 > ψ(x, t)} where the function ψ satisfies the followings ( ψ(x, t) ψ(y, s) L x y + t s 1/2), D t 1/2 ψ BMO L.

7 The time varying domain Ω For X = (x 0, x) R R n 1, consider the region given by Ω = {(x 0, x, t) : x 0 > ψ(x, t)} where the function ψ satisfies the followings ( ψ(x, t) ψ(y, s) L x y + t s 1/2), D t 1/2 ψ BMO L. For Elliptic equations, Ω = {(x 0, x) : x 0 > ψ(x)} where ψ(x) ψ(y) L x y. With only Lip(1, 1/2), parabolic measure and surface measure are failed to be mutually absolutely continuous. (Kaufmann & Wu 1988)

8 Goal to show L 2 solvability Goal inequality We wish to show that there exists a constant C such that N(u) L 2 () C f L 2 (). Definition For (y 0, y, s) U, define a parabolic cone Γ a (x 0, x, t) = {(y 0, y, s) Ω : x y + s t 1/2 a(y 0 x 0 ), y 0 > x 0 } and N a (u)(x 0, x, t) = sup u(y 0, y, s). (y 0,y,s) Γ a(x 0,x,t)

9 Pullback Transformation For a nonnegative function P(x, t) C 0 for (x, t) R n 1 R, set P λ (x, t) λ (n+1) P(x/λ, t/λ 2 ), P λ ψ(x, t) P λ (x y, t s)ψ(y, s) dy ds. R n 1 R

10 Non-tangential maximal function and Carleson measure We have integral inequalities that ( δ A 2 + δ 3 A t 2 + δ B 2) u 2 dx dt T ( r ) µ C r N 2 (u) dx dt where µ C is a Carleson norm. Symmetry of the coefficients is NOT assumed. Does NOT rely on layer potentials rather direct approach. Compare the non-tangential maximal function and the square function

11 Square and Area functions For u, a locally integrable distributional gradient, the square function is ( 1/2 S a (u)(x 0, x, t) = (y 0 x 0 ) n u 2 (y 0, y, s) dy 0 dy ds), Γ a(x 0,x,t) S a (u) 2 L 2 () = δ(y 0, y, s) u 2 (y 0, y, s) dy 0 dy ds Ω and the area function using the Stekelove averages to obtain u t by ( 1/2 A a (u)(x 0, x, t) = (y 0 x 0 ) n+2 u t 2 (y 0, y, s) dy 0 dy ds), Γ a(x 0,x,t) A a (u) 2 L 2 () = δ 3 (y 0, y, s) u t 2 (y 0, y, s) dy 0 dy ds. Ω

12 Property the parabolic PDE u t 2 3 A 2 2 u ( A 2 + B 2) u 2. A Cacciopoli inequality for the second gradient We are getting that A a (u)(x 0, x, t) CS 2a (u)(x 0, x, t) and A a (u) L 2 () C S a (u) L 2 ().

13 The strategy to show our goal inequality The first inequality is that S 2 (u) dx dt C f 2 dx dt + ɛ N 2 (u) dx dt where ɛ comes from the smallness of the Carleson measure of the coefficients.

14 The strategy to show our goal inequality The first inequality is that S 2 (u) dx dt C f 2 dx dt + ɛ N 2 (u) dx dt where ɛ comes from the smallness of the Carleson measure of the coefficients. The second inequality is given by [ N 2 (u) dx dt C S 2 (u) dx dt + ] f 2 dx dt.

15 The strategy to show our goal inequality The first inequality is that S 2 (u) dx dt C f 2 dx dt + ɛ N 2 (u) dx dt where ɛ comes from the smallness of the Carleson measure of the coefficients. The second inequality is given by [ N 2 (u) dx dt C S 2 (u) dx dt + Combining two inequalities above provides our goal that N(u) L 2 () C f L 2 (). ] f 2 dx dt.

16 S N We begin with the integral quantity r a ij u xi u xj x 0 ζ 2 dx dt dx 0 a 00 i,j 0 B 2r

17 S N We begin with the integral quantity r a ij u xi u xj x 0 ζ 2 dx dt dx 0 a 00 i,j 0 B 2r We obtain C r u 2 x 0 dx dt dx r u 2 (x 0, x, t) dx dt dx r 0 u 2 (r, x, t) dx dt + u 2 (0, x, t) dx dt + ɛ Nr 2 (u) dx dt.

18 S N We begin with the integral quantity r a ij u xi u xj x 0 ζ 2 dx dt dx 0 a 00 i,j 0 B 2r We obtain C r u 2 x 0 dx dt dx r u 2 (x 0, x, t) dx dt dx r 0 u 2 (r, x, t) dx dt + u 2 (0, x, t) dx dt + ɛ Nr 2 (u) dx dt. Interior estimates, Maximum principle, Growth condition u 2 (r, x, t) dx dt 2 r u 2 (r, x, t) dx dt+ɛ N 2 (u) dx dt r 0

19 N S Good λ inequality 1. {N(u) λ}

20 N S Good λ inequality 1. {N(u) λ} 2. {N(u) > λ} {S(u) + A(u) ρλ}

21 N S Good λ inequality 1. {N(u) λ} 2. {N(u) > λ} {S(u) + A(u) ρλ} 3. {N(u) > λ} {S(u) + A(u) ρλ}

22 N S Good λ inequality 1. {N(u) λ} 2. {N(u) > λ} {S(u) + A(u) ρλ} 3. Search {N(u) > λ} {S(u) + A(u) ρλ} h λ,a (u)(x, t) = sup{x 0 0 : sup u > λ} Lip(1, 1/2). Γ a(x,t) There is a point on Γ a (x, t) where u = λ.

23 There exists a parabolic cube where u > λ/2 so M(u)(x, t) > cλ because u oscillates small (arbitrary control of the square and area functions).

24 There exists a parabolic cube where u > λ/2 so M(u)(x, t) > cλ because u oscillates small (arbitrary control of the square and area functions). On the curve h, we obtain 2 u(h α (x, t), x, t) 2 dx dt dα 1 B ( r [ ] C N(u) L 2 (B 2r ) S(u) L 2 (B 2r ) + A(u) L 2 (B 2r ) ) + S(u) 2 L 2 (B 2r ) + A(u) 2 L 2 (B 2r ) + r n+1 u(a r ) 2 where A r corkscrew point. By averaging interior points, it absorbed to L 2 norm of the boundary data.