Oblique derivative problem for non-divergence parabolic equations with discontinuous in time coefficients 3 1 0 2 Vladimir Kozlov and Alexander Nazarov ∗ † n a J 8 1 Abstract ] Weconsideranobliquederivativeproblemfornon-divergenceparabolic P A equations with discontinuous in t coefficients in a half-space. We ob- . tain weighted coercive estimates of solutions in anisotropic Sobolev h spaces. We also give an application of this result to linear parabolic t a equations in a bounded domain. In particular, if the boundary is of m class 1,δ, δ (0,1], then we present a coercive estimate of solutions [ C ∈ in weighted anisotropic Sobolev spaces, where the weight is a power 1 of the distance to the boundary. v 5 1 4 1 Introduction 4 . 1 Consider the parabolic equation 0 3 1 ( u)(x,t) ∂ u(x,t) aij(t)D D u(x,t) = f(x,t) (1) 0 t i j : L ≡ − v i for x Rn and t R. Here and elsewhere D denotes the operator of X i ∈ ∈ differentiation with respect to x and ∂ u is the derivative of u with respect r i t a to t. ∗Department of Mathematics, University of Linköping, SE-581 83 Linköping, Sweden †St.-Petersburg Department of Steklov Mathematical Institute, Fontanka, 27, St.- Petersburg, 191023, Russia, and St.-Petersburg State University, Universitetskii pr. 28, St.-Petersburg, 198504,Russia 1 The only assumptions about the coefficients in (1) is that aij are measur- able real valued functions of t satisfying aij = aji and ν ξ 2 aijξ ξ ν−1 ξ 2, ξ Rn, ν = const > 0. (2) i j | | ≤ ≤ | | ∈ It was proved by Krylov [2, 3] that for f L (Rn R) with 1 < p,q < , p,q ∈ × ∞ equation (1) in Rn R has a unique solution such that ∂ u and D D u belong t i j × to L (Rn R) and p,q × ∂ u + D D u C f . (3) t p,q i j p,q p,q k k k k ≤ k k ij X Here L (Ω I) = L (I L (Ω)) is the space of functions on Ω I with p,q q p × → × finite norm q 1 f = f(x,t) pdx pdt q p,q k k | | (cid:16)ZI (cid:16)ZΩ (cid:17) (cid:17) (with natural change in the case p = or q = ). ∞ ∞ In the authors’ paper [4] estimate (3) was supplemented by a similar one in the space L (Rn R) p,q × e ∂tu p,q + DiDju p,q C f p,q. ||| ||| ||| ||| ≤ ||| ||| ij X Here L (Ω I) = L (Ω L (I)) is the space of functions on Ω I with p,q p q × → × finite norm p 1 e f = f(x,t) qdt qdx p. p,q ||| ||| | | (cid:16)ZΩ (cid:16)ZI (cid:17) (cid:17) (with natural change in the case p = or q = ). This space arises ∞ ∞ naturally in the theory of quasilinear non-divergence parabolic equations (see [9]). Note that for p = q we have L (Ω I) = L (Ω I) = L (Ω I); f = f = f . p,p p,p p p,p p,p p × × × ||| ||| k k k k The homogeneous Dirichlet problem for (1) in Rn R, where Rn the e + × + half-space x = (x′,x ) Rn : x > 0 , was considered in [2, 4]. It was n n { ∈ } proved that its solution satisfies the following weighted coercive estimate xµ∂ u + xµD D u C xµf , (4) k n t kp,q k n i j kp,q ≤ k n kp,q ij X 2 where 1 < p,q < and µ ( 1,2 1) (in [2] this estimate was proved only ∞ ∈ −p −p for µ (1 1,2 1)). An analog of estimate (4), where the norm is ∈ − p − p k·kp,q replaced by , is also proved in [4]. p,q |||·||| In the paper [5] the homogeneous Dirichlet problem for (1) in cones and wedges was considered, and coercive estimates for solutions were obtained in the scales of weighted L and L spaces, where the weight is a power of p,q p,q the distance to the vertex (edge). Let us turn to the oblique deerivative problem in the half-space Rn. Now + equation (1) is satisfied for x > 0 and ∂u = 0 for x = 0. Here γ is a n ∂γ n constant vector field with γ > 0. n By changing the spatial variables one can reduce the boundary condition to the case D u = 0 for x = 0. (5) n n One of the main results of this paper is the proof of estimate (4) and its analog for the norm , for solutions of the oblique derivative problem p,q ||| · ||| (1), (5) with arbitrary p,q (1, ) and for µ satisfying ∈ ∞ 1 1 < µ < 1 . −p − p In the case of time independent coefficients such estimates for the Neumann problem were proved in [9]. Weuseanapproachbased onthestudyoftheGreenfunctions. InSection 2 we collect (partiallyknown) results ontheestimate of theGreenfunction of the Dirichlet problem for equation (1). Section 3 is devoted to the estimates of the Green function of problem (1), (5). In Section 4 we apply the obtained estimates to the oblique derivative problem for linear non-divergence parabolic equations with discontinuous in time coefficients in cylinders Ω (0,T), where Ω is a bounded domain in × Rn. We prove solvability results in weighted L and L spaces, where the p,q p,q weight is a power of the distance to the boundary of Ω. The smoothness of the boundary is characterized by smoothness of leocal isomorphisms in neighborhoodsofboundarypoints, whichflattentheboundary. Inparticular, if the boundary is of the class 1,δ with δ (0,1], then for solutions to the C ∈ equation (1)1 in Ω (0,T) with zero initial and boundary conditions the × 1Herethecoefficientsaij maydependonx(namely,weassumeaij (Ω L (0,T))). ∞ ∈C → 3 following coercive estimate is proved in Theorem 4 (see Remark 1): (d(x))µ∂ u + (d(x))µD D u C (d(x))µf , t p,q i j p,q p,q k k k k ≤ k k ij X (db(x))µ∂ u + (db(x))µD D u C (db(x))µf , t p,q i j p,q p,q ||| ||| ||| ||| ≤ ||| ||| ij X b b b where µ, p, q and δ satisfy 1 < p,q < , 1 δ 1 < µ < 1 1. ∞ − − p − p Let us recall some notation: x = (x ,...,x ) = (x′,x ) is a point in Rn; 1 n n Du = (D u,...,D u) is the gradient of u. 1 n We denote Q (x0,t0) = (x,t) : x x0 < R, 0 < t0 t < R2 ; R { | − | − } Q+(x0,t0) = (x,t) : x x0 < R, x > 0, 0 < t0 t < R2 . R { | − | n − } The last notation will be used only for x0 Rn. ∈ + Set x y = n , = n . x y R x +√t s R y +√t s n n − − In what follows we denote by the same letter the kernel and the corresponding integral operator, i.e. t ( h)(x,t) = (x,y;t,s)h(y,s)dyds. K K −Z∞RZn Here we expand functions and h by zero to whole space-time if necessary. K We adopt the convention regarding summation from 1 to n with respect to repeated indices. We use the letter C to denote various positive constants. To indicate that C depends on some parameter a, we sometimes write C . a 2 Preliminary results 2.1 The estimates in the whole space and in the half- space under the Dirichlet boundary condition Let us consider equation (1) in the whole space Rn. Using the Fourier trans- form with respect to x one can obtain the following representation of solution 4 through the right-hand side: t u(x,t) = Γ(x,y;t,s)f(y,s) dyds, (6) −Z∞RZn where Γ is the Green function of the operator given by 0 L t −1 tA(τ)dτ −1(x y),(x y) det A(τ)dτ 2 Γ(x,y;t,s) = s exp s − − (cid:0)R(4π)n2 (cid:1) (cid:18)−(cid:16)(cid:0)R (cid:1) 4 (cid:17)(cid:19) for t > s and 0 otherwise. Here A(t) is the matrix aij(t) n . The above { }i,j=1 representation implies, in particular, the following evident estimates. Proposition 1. Let α and β be two arbitrary multi-indices. Then σ x y 2 DαDβΓ(x,y;t,s) C(t s)−n+|α2|+|β| exp | − | | x y | ≤ − − t s (cid:18) − (cid:19) and σ x y 2 |∂sDxαDyβΓ(x,y;t,s)| ≤ C(t−s)−n+|α2|+|β|−1 exp − |t −s| (cid:18) − (cid:19) for x,y Rn and s < t. Here σ depends only on the ellipticity constant ν ∈ and C may depend on ν, α and β. In the next proposition we present solvability results for equation (1) in the whole space. Proposition 2. Let p,q (1, ). ∈ ∞ (i) If f L (Rn R), then the solution of equation (1) given by (6) p,q ∈ × satisfies ∂ u + D D u C f , t p,q i j p,q p,q k k k k ≤ k k ij X where C depends only on ν, p, q. (ii) If f L (Rn R), then the solution of equation (1) given by (6) p,q ∈ × satisfies e ∂ u + D D u C f , (7) t p,q i j p,q p,q ||| ||| ||| ||| ≤ ||| ||| ij X where C depends only on ν, p, q. 5 The first assertion is proved in [2] and the second one in [4]. We denote by ΓD(x,y;t,s) the Green function of the operator in the 0 L half-space Rn subject to the homogeneous Dirichlet boundary condition on + the boundary x = 0. n The next statement is proved in [4, Theorem 3.6]. Proposition 3. For x,y Rn and t > s the following estimate is valid: ∈ + 2−αn−ε 2−βn−ε σ x y 2 DαDβΓD(x,y;t,s) C Rx Ry exp | − | , (8) | x y | ≤ (t s)n+|α2|+|β| (cid:18)− t−s (cid:19) − where σ is a positive number dependent on ν and n, ε is an arbitrary small positive number and C may depend on ν, α, β and ε. If α 1 (or β 1) n n ≤ ≤ then 2 α ε (2 β ε) must be replaced by 1 α (1 β ) respectively n n n n − − − − − − in the corresponding exponents. Since (∂ +a ( s)D D )ΓD(x,y;t,s) = 0 for s < t, we obtain s ij − yi yj Corollary 1. For x,y Rn and t > s ∈ + 2−αn−ε −βn−ε σ x y 2 DαDβ∂ ΓD(x,y;t,s) C Rx Ry exp | − | . (9) | x y s | ≤ (t s)n+2+|2α|+|β| (cid:18)− t−s (cid:19) − If α 1 then 2 α ε must be replaced by 1 α . n n n ≤ − − − 2.2 Coercive estimates for weak solutions to the Dirich- let problem in the half-space We formulate two auxiliary results on estimates of integral operators. The first statement is a particular case m = 1 of [4, Lemmas A.1 and A.3 and Remark A.2], see also [9, Lemmas 2.1 and 2.2]. Proposition 4. Let 1 p , σ > 0, 0 < r 2, λ +λ > 1, and let 1 2 ≤ ≤ ∞ ≤ − 1 1 λ < µ < 1 +λ . (10) − p − 1 − p 2 Suppose also that the kernel (x,y;t,s) satisfies the inequality T λ1+r λ2 xµ−r σ x y 2 (x,y;t,s) C Rx Ry n exp | − | , |T | ≤ (t−s)n+22−r ynµ (cid:18)− t−s (cid:19) 6 for t > s. Then the integral operator is bounded in L (Rn R) and in p T × L (Rn R). p,∞ × The next proposition is a particular case m = 1 of [4, Lemma A.4], see e also [9, Lemma 3.2]. Proposition 5. Let 1 < p < , σ > 0, κ > 0, 0 r 2, λ +λ > 1 and 1 2 ∞ ≤ ≤ − let µ be subject to (10). Also let the kernel (x,y;t,s) satisfy the inequality T λ1+r λ2 xµ−r δ κ σ x y 2 (x,y;t,s) C Rx Ry n exp | − | , |T | ≤ (t−s)n+22−r ynµ (cid:18)t−s(cid:19) (cid:18)− t−s (cid:19) for t > s+δ. Then for any s0 > 0 the norm of the operator : L (Rn (s0 δ,s0 +δ)) L (Rn (s0 +2δ, )) p,1 p,1 T × − → × ∞ does not exceed a constant C independent of δ and s0. Now we consider the problem Lu = f +div(f) in Rn R (11) 0 + × (here f = (f ,...,f )) with the boundary condition 1 n u = 0 for x = 0. (12) n Theorem 1. Let 1 < p,q < and µ ( 1,1 1). ∞ ∈ −p − p (i) Suppose that xµ+1f , xµf L (Rn R). Then the function n 0 n ∈ p,q + × t e u(x,t) = ΓD(x,y;t,s)f (y) D ΓD(x,y;t,s) f(y) dyds (13) 0 y − · −Z∞RZn (cid:16) (cid:17) + gives a weak solution of problem (11), (12) and satisfies the estimate xµDu + xµ−1u C( xµ+1f + xµf ). (14) ||| n |||p,q ||| n |||p,q ≤ ||| n 0|||p,q ||| n |||p,q (ii) Suppose that xµ+1f , xµf L (Rn R). Then the function (13) gives n 0 n ∈ p,q + × a weak solution of problem (11), (12) and satisfies the estimate xµDu + xµ−1u C( xµ+1f + xµf ). (15) k n kp,q k n kp,q ≤ k n 0kp,q k n kp,q 7 Proof. First, function (13) obviously solves problem (11), (12) in the sence of distributions. Thus, it is sufficient to prove estimates (14), (15). Put xµ−1 xµ−1 (x,y;t,s) = n ΓD(x,y;t,s); (x,y;t,s) = n D ΓD(x,y;t,s); K0 yµ+1 K1 yµ y n n xµ xµ (x,y;t,s) = n D Γ (x,y;t,s); (x,y;t,s) = nD D Γ (x,y;t,s). K2 yµ+1 x D K3 yµ x y D n n (i)ByProposition3thekernels and satisfytheconditionsofPropo- 0 1 K K sition 4 with r = 1 and with λ = 1, λ = 1 and µ replaced by µ+1 for the kernel ; 1 2 0 − K with λ = λ = 0 for the kernel , respectively. 1 2 1 K This implies that for µ ( 1,1 1) ∈ −p − p xµ−1u C( xµ+1f + xµf ) (16) k n kp ≤ k n 0kp k n kp and xµ−1u C( xµ+1f + xµf ). (17) ||| n |||p,∞ ≤ ||| n 0|||p,∞ ||| n |||p,∞ Interpolating (16) and (17) we arrive at xµ−1u C( xµ+1f + xµf ), (18) ||| n |||p,q ≤ ||| n 0|||p,q ||| n |||p,q for 1 < p q < and µ ( 1,1 1). Now duality argument gives (18) ≤ ∞ ∈ −p − p for all 1 < p,q < and for the same interval for µ. ∞ To estimate the first term in the left-hand side of (14) we use local estimates. We put ρ B (ξ) = x Rn : x′ ξ′ < ρ, < x < ρ . ρ,ϑ { ∈ | − | ϑ n } Localization of estimate (7) using an appropriate cut-off function, which is equal to 1 on B and 0 outside B , gives ρ,2 2ρ,8 p p Du qdt qdx C ( u qρ−q +ρq f q + f qdt qdx. 0 | | ≤ | | | | | | Bρ,Z2(ξ) (cid:16)ZR (cid:17) B2ρZ,8(ξ) (cid:16)ZR (cid:17) 8 Using a proper partition of unity in Rn, we arrive at + p/q p/q Du qdt xµpdx C u qdt xµp−pdx | | n ≤ | | n RZn (cid:18)ZR (cid:19) (cid:18) RZn (cid:18)ZR (cid:19) + + p/q p/q + f qdt xµpdx+ f qdt xµp+pdx . | | n | 0| n RZn (cid:18)ZR (cid:19) RZn (cid:18)ZR (cid:19) (cid:19) + + This immediately implies (14) with regard of (18). (ii) To deal with the scale L , we need the following lemma. p,q Lemma 1. Let a function h be supported in the layer s s0 δ and satisfy | − | ≤ h(y;s) ds 0. Also let p (1, ) and µ ( 1,1 1). Then the operators ≡ ∈ ∞ ∈ −p −p , j = 0,1,2,3, satisfy j RK ( h)( ;t) dt C h , j p p,1 k K · k ≤ k k Z |t−s0|>2δ where C does not depend on δ and s0. Proof. By h(y;s)ds 0, we have ≡ R t ( h)(x;t) = (x,y;t,s) (x,y;t,s0) h(y;s)dyds (19) j j j K K −K −Z∞RZn (cid:16) (cid:17) (we recall that all functions are assumed to be extended by zero). We choose ε > 0 such that 1 1 < µ < 1 ε. (20) − p − p − For s s0 < δ and t s0 > 2δ, estimate (9) implies | − | − s (x,y;t,s) (x,y;t,s0) ∂ (x,y;t,τ) dτ j j τ j K −K ≤ | K | Z (cid:12) (cid:12) s0 (cid:12) ℓ1 ℓ2−ε xℓ3 δ (cid:12) σ x y 2 C Rx Ry n exp | − | , ≤ (t−s)n+22−r ynℓ4 t−s (cid:18)− t−s (cid:19) 9 with r = 2, ℓ = 1, ℓ = 0, ℓ = µ 1, ℓ = µ+1 for the kernel ; 1 2 3 4 0 − K with r = 1, ℓ = 1, ℓ = 1, ℓ = µ 1, ℓ = µ for the kernel ; 1 2 3 4 1 − − K with r = 1, ℓ = 0, ℓ = 0, ℓ = µ, ℓ = µ+1 for the kernel ; 1 2 3 4 2 K with r = 0, ℓ = 0, ℓ = 1, ℓ = µ, ℓ = µ for the kernel . 1 2 3 4 3 − K On the other hand, estimate (8) implies ℓ1 ℓ2+1 xℓ3 σ x y 2 (x,y;t,s) (x,y;t,s0) C Rx Ry n exp | − | . Kj −Kj ≤ (t−s)n+22−r ynℓ4 (cid:18)− t−s (cid:19) (cid:12) (cid:12) (cid:12) (cid:12) Combination of these estimates gives Cδκ ℓ1 ℓ2+1−ε xℓ3 σ x y 2 (x,y;t,s) (x,y;t,s0) Rx Ry n exp | − | , Kj −Kj ≤ (t−s)n+22−r+κ ynℓ4 (cid:18)− t−s (cid:19) (cid:12) (cid:12) (cid:12) (cid:12) where κ = ε . Thus, the kernels in (19) satisfy the assumptions of Propo- 1+ε sition 5 with λ = 1, λ = 1 ε and µ replaced by µ+1 for kernels and ; 1 2 0 2 − − K K with λ = 0, λ = ε for kernels and , respectively. 1 2 1 3 − K K Inequality (20) becomes (10), and the Lemma follows. We continue the proof of the second statement of Theorem 1. Estimate (14) for q = p provides boundedness of the operators , j = 0,1,2,3, in j K L (Rn R), which gives the first condition in [1, Theorem 3.8]. Lemma 1 p × is equivalent to the second condition in this theorem. Therefore, Theorem 3.8 [1] ensures that these operators are bounded in L (Rn R) for any p,q × q (1,p). For q (p, ) this statement follows by duality arguments. This ∈ ∈ ∞ implies estimate (15). 3 Oblique derivative problem 3.1 The Green function Theorem 2. There exists a Green function ΓN = ΓN(x,y;t,s) of problem (1), (5) and for arbitrary x,y Rn and t > s it satisfies the estimate ∈ + αbn βbn σ x y 2 DαDβΓN(x,y;t,s) C Rx Ry exp | − | , (21) | x y | ≤ (t s)n+|α2|+|β| (cid:18)− t−s (cid:19) − 10

Oblique derivative problem for non-divergence parabolic equations with discontinuous in time coefficients

Similar Oblique derivative problem for non-divergence parabolic equations with discontinuous in time coefficients