The Dirichlet problem for non-divergence parabolic equations with discontinuous in time coefficients in a wedge

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1 The Dirichlet problem for non-divergence parabolic equations with discontinuous in time coefficients in a wedge Vladimir Kozlov (Linköping University, Sweden) 2010 joint work with A.Nazarov

2 Lu t u a ij (t)d i D j u = f, (1) a ij are measurable real valued functions of t satisfying a ij = a ji and ν ξ 2 a ij ξ i ξ j ν 1 ξ 2, ξ R n ν = const > 0. We use the space L p,q (Ω R) with the norm f p,q = ( R ( Ω f(x, t) p dx ) q/pdt ) 1/q. N.V. Krylov (2001): for f L p,q (R n R), 1 < p, q <, equation (1) has a unique solution s.t. t u p,q + ij D i D j u p,q C f p,q. He proved also coercive estimates for u in spaces L q (R; C 2+α ), α (0, 1).

3 The Dirichlet BVP in the half-space R n + = {x = (x, x n ) R n : x n > 0}. Now equation (1) is satisfied for x n > 0 and u = 0 for x n = 0. The weighted coercive estimate x µ n tu p,q + ij x µ n D id j u p,q C x µ n f p,q, (2) was proved by Krylov (2001), with 1 < p, q < and µ (1 1/p, 2 1/p). In Vladimir Kozlov and Alexander Nazarov, The Dirichlet problem for non-divergence parabolic equations with discontinuous in time coefficients, Math. Nachr. 282 (2009), No. 9, estimate (2) is proved for solutions of the Dirichlet problem to (1) for the same p, q and 1/p < µ < 2 1/p. (3)

4 Remarks In the paper [2007] Krylov and Kim proved, in particular, estimate (2) in the halfspace for µ = 0 and p = q. In D. Kim, Parabolic Equations with Partially BMO Coefficients and Boundary Value Problems in Sobolev Spaces with Mixed Norms, Potential Anal., published on line in estimate (2) is proved for µ = 0 and arbitrary 1 < p, q <.

5 Dirichlet problem in bounded domain Ω Let Q = Ω R. We introduce the spaces L p,q,(µ) (Q) with the norm f p,q,(µ) = ( d(x)) µ f p,q, where d(x) is the distance from x Ω to Ω. We consider the boundary value problem t u a ij (x, t)d i D j u + b i (x, t)d i u = f(x, t) in Q; u Q = 0; the matrix (a ij ) C(Ω L (0, T )) is symmetric and uniformly elliptic. Here Q is the boundary of Q. K.-N.,2009:Let Ω C 1,δ with δ [0, 1], 1 < p, q <, and let 1 δ 1 p < µ < 2 1 p. Then, for bi in a suitable class and for any f L p,q,(µ) (Q), the above problem has a unique solution in L p,q,(µ) (Q). Moreover, this solution satisfies t u p,q,(µ) + ij D i D j u p,q,(µ) C f p,q,(µ), Remarks. For p = q and δ = 0 this theorem was proved by Kim and Krylov (2004).

6 Coercive estimates for the heat equation with constant coefficients in a wedge. Conical points 1. V. A. Kozlov and V. G.Maz ya, On singularities of a solution to the first boundary-value problem for the heat equation in domains with conical points, Izv. Vyssh. Uchebn. Zaved., Ser. Mat., No. 2, (1987) and No.3, (1987). 2. V. A. Kozlov, On asymptotic of the Green function and the Poisson kernels of the mixed parabolic problem in a cone, Zeitschr. Anal. Anw., 8 (1989), No. 2, and 10 (1991), No. 1,

7 Dihedral angles and wedges 5. Solonnikov, V. A., L p -estimates for solutions of the heat equation in a dihedral angle, Rend. Mat. Appl. (7) 21 (2001), N1-4, Nazarov, A. I., L p -estimates for a solution to the Dirichlet problem and to the Neumann problem for the heat equation in a wedge with edge of arbitrary codimension, Probl. Mat. Anal., 22 (2001), (Russian); English transl.: J. Math. Sci., 106 (2001), N3,

8 We use the notation (x, x ) R n, where x R and x R n m. Let K be a cone in R m such that the boundary K \ O is of class C 2. We put K = K R n m. For µ R and 1 < p, q < we introduce spaces L p,q,µ = L p,q,µ (K R) with the norm u p,q,µ = ( R ( K x µp u(x, t) p dx ) q/pdt ) 1/q Let also Q K R (t 0) = ( B R (0) K ) (t 0 R 2, t 0 ) where B R (x 0 ) is the ball x x 0 < R.

9 By V (Q K R (t 0)) we denote the set of functions u with finite norm u V (Q K R (t 0 )) = + D x u L 2 (Q K R (t 0))) + t0 sup u(τ, ) L τ (t 0 R 2 2 (B R (0) K),t 0 ) t 0 R 2 D tu(t, ) W 1 (B R (0) K) dt. We define the critical exponent for the operator L and the wedge K as the supremum of all λ such that u(x, t) C λ ( x R for (x, t) Q K R/2 (t 0). λ sup u(y, τ) (4) (y,τ) Q K R (t 0) This inequality must be satisfied for all t 0, R > 0 and u V (Q K R (t 0)) subject to and Lu = 0 in Q K R (t 0) (5) u = 0 on Q K R (t 0) K R. We shall denote this critical exponent by λ c. Since λ = 0 satisfies (4) we conclude that λ c 0. Below we give some estimates for λ c for various geometries of K.

10 Estimates for the critical exponent 1. Using weighted energy estimates one can show that λ c 2 m + ν Λ D + (m 2) 2 /4, 2 where Λ D is the first positive eigenvalue of the Dirichlet-Laplacian on K S m Using barrier technique one can show that a) the critical exponent is positive provided the complement of K is non-empty; b) if K is contained in a half-space then λ c > 1. 3). If L = t then λ c = 2 m 2 + Λ D + (m 2) 2 /4

11 Theorem Let λ c be the critical exponent. Then for µ + n p m < λ c + m 2 2 the following estimate holds: u t p,q,µ + u p,q,µ C f p,q,µ

12 For δ > 0 we define K δ = {x K : dist(x, K) > δ x } and K δ = K δ R. The next statement can be found (up to scaling) in [LSU]. Proposition 1. (i)let u W 2,1 (Q R (x 0, t 0 )) solve the equation L u = 0 in Q R (x 0, t 0 ). Then Du C R sup u in Q R/2 (x 0, t 0 ). Q R (x 0,t 0 ) (ii) For sufficiently small δ > 0, x 0 K \ K δ and x 0 = 1 the following assertion is valid. Let u W 2,1 2 (Q+ R (x 0, t 0 )) solve the equation L u = 0 in Q + R (x 0, t 0 ), where R 1/2, and let u(x, t) = 0 for x K. Then Du C R sup u in Q + Q + R (x R/2 (x 0, t 0 ). 0,t 0 ) Here C depends only on ν and K and δ. We used the notations Q R (x 0, t 0 ) = B R (x 0 ) (t 0 R 2, t 0 ) and Q + R (x 0, t 0 ) = ( B R (x 0 ) K ) (t 0 R 2, t 0 ).

13 Iterating the inequality from Proposition (i) we arrive at Lemma 1. Let u W 2,1 2 (Q R(x 0, t 0 )) solve the equation L u = 0 in Q R (x 0, t 0 ). Then D α u C R α sup u in Q R/2 (x 0, t 0 ). Q R (x 0,t 0 ) Next Lemma is actually proved in [KN]. Lemma 2. For sufficiently small δ > 0, x 0 K \ K δ, x 0 Rn m and x 0 = 1 the following assertion is valid. Let u W 2,1 2 (Q+ R (x 0, t 0 )) solve the equation L u = 0 in Q + R (x 0, t 0 ), where R < 1/2, and let u(x, t) = 0 for x K. For α 2 and arbitrary small ε > 0 d(x) α 2+ε D α x u C R 2 ε sup u, (6) Q + R (x 0,t 0 ) in Q + R/8 α (x 0, t 0 ), where C is a positive constant depending on ν, α, K, δ and ε.

14 Green s function in K R Let us consider (1) in the whole space. Using the Fourier transform with respect to x we obtain: t u(x, t) = Γ(x, y; t, s)f(y, s) ds, (7) R n where Γ is the Green function of the operator L 0 given by Γ(x, y; t, s) = det ( ts A(τ)dτ ) 1 2 (4π) n 2 ( ( ts A(τ)dτ ) ) 1 (x y), (x y) exp ( for t > s and 0 otherwise. Here by A(t) is denoted the matrix {a ij (t)}. The above representation implies: k t D α xd β y Γ(x, y; t, s) exp ( σ x y 2 t s ) 4 ) C k,α,β (t s) (n+2k+ α + β )/2, (8) where k 1 and α and β are arbitrary indexes. Here σ is a positive constant depending on ν.

15 We denote by Γ K = Γ K (x, y; t, s) Green s function to the homogeneous Dirichlet problem of (1), in the half-space. Clearly, Γ K (x, y; t, s) Γ(x, y; t, s) and therefore Γ K (x, y; t, s) C (t s) n/2 exp We shall use the notations ( σ x y 2 t s ) in K R. (9) R x = x x + t s, R y = y y + t s and r x = d(x)( x + t s) x t s, r y = d(y)( y + t s) y, t s where d(x) is the distance from x to the boundary K.

16 Proposition 2. The following inequality Γ K (x, y; t, s) C R λ xr λ y(t s) n/2 exp σ 1 x y 2 t s (10) holds for x, y K and s < t. Here σ 1 is a positive constant depending only on the ellipticity constant ν and C may depend on ν and λ. ( ) The proof of this proposition and the next theorem essentially uses the above local estimates and the definition of the critical exponent.

17 Theorem 1. Let α, β 2. For x, y K, 0 s < t the following estimate is valid D α x D β y Γ K (x, y; t; s) C R λ α (t s) n+ α + β 2 exp ( x σ 1 x y 2 t s Ry λ β ) rx ε ry ε, (11) where σ 1 is a positive constant depending on ν, ε is an arbitrary small positive number and C may depend on ν, α, β and ε. If α 1 (or β 1) then the factor rx ε the right-hand side respectively. (ry ε ) must be removed from

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