ELLIPTIC AND PARABOLIC EQUATIONS WITH MEASURABLE COEFFICIENTS IN L P -SPACES WITH MIXED NORMS

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1 METHODS AND APPLICATIONS OF ANALYSIS. c 2008 International Press Vol. 15, No. 4, pp , December ELLIPTIC AND PARABOLIC EQUATIONS WITH MEASURABLE COEFFICIENTS IN L P -SPACES WITH MIXED NORMS DOYOON KIM Abstract. The unique solvability results for second order parabolic and elliptic equations in Sobolev spaces with mixed norms are presented. The second order coefficients are measurable in one spatial variable and VMO vanishing mean oscillation) in the other spatial variables. In the parabolic case, the coefficients except a 11 ) are further allowed to be only measurable in time. We first prove the solvability results for equations in the whole Euclidean space. Then, using these results as well as some extension techniques, we prove the solvability results for equations on a half space without any boundary estimates. The mixed norms we present here are more general than the usual mixed norm L t ql x p. Key words. Second order elliptic and parabolic equations, vanishing mean oscillation, Sobolev spaces with mixed norms. AMS subject classifications. 35J15, 35K10, 35R05, 35A05 1. Introduction. In this paper we study elliptic and parabolic equations of nondivergence type in Sobolev spaces with mixed norms. The differential equations we consider are elliptic and parabolic, respectively) a ij x)u x i x j + bi x)u x i + cx)u = f, 1) u t + a ij t, x)u x i x j + bi t, x)u x i + ct, x)u = f. 2) The equations are assumed to be uniformly non-degenerate with bounded coefficients. The regularity assumptions on the coefficients a ij no regularity assumptions are needed for coefficients b i and c) are, roughly speaking, that they are merely measurable i.e., no regularity assumptions) in one spatial direction, and belong to the space of VMO mean vanishing oscillation) as functions of the other variables. In the parabolic case, they except a 11 ) are measurable in two variables including the time variable and in the space of VMO as functions of the remaining variables. The coefficient a 11 is measurable in one spatial variable and VMO in the other variables. There has been, in fact, considerable study of parabolic equations in mixed norm spaces in the literature see, e.g., [4, 18, 11, 10, 12, 21, 3, 6, 20, 14] and references therein). The usual mixed norms are of the form L q 0, T), L p ), that is, q summability in the time variable and p summability in the spatial variables. For example, in [12, 21, 20, 14] parabolic equations as in 2) quasi-linear equations in [20]) are investigated in Sobolev spaces with the mixed norm L q 0, T), L p Ω)), where Ω R d. In [3] one sees parabolic systems in L q 0, T), X), where X is an L p space with a Muckenhoupt weight. In this paper, however, by mixed norms we mean, not the usual mixed norms, but more general ones defined as follows. First we explain some notations. By R d we mean a d-dimensional Euclidean space and a point in R d is denoted by x. In the parabolic case we consider R d+1, where a point in R d+1 is denoted by t, x). Throughout the paper, we fix two nonnegative Received October 29, 2008; accepted for publication November 7, Department of Mathematics, University of Southern California, 3620 Vermont Avenue, KAP108 Los Angeles, CA 90089, USA doyoonki@usc.edu). 437

2 438 D. KIM integers d 1 and d 2 such that d 1 +d 2 = d. We also fix all different integers i 1, i 2,, i d1 and j 1, j 2,, j d2 from {1, 2,, d}. Then for x = x 1, x 2,, x d ) R d, we denote x 1 = x i1, x i2,, x i d 1) R d 1, x 2 = x j1, x j2,, x j d 2) R d 2. 3) Note that, as one can see, for example, x 1 consists of d 1 coordinates, but x 1 is not necessarily the first d 1 coordinates of x R d. Likewise, R d1 above is a subspace of R d, but not necessarily consists of the first d 1 coordinates of points in R d. If D is a subset in R d+1, we define u L t,x 1 q L x 2 p D) = R R d 1 R d 2 where I D is the indicator function. Similarly, u L x 1 q L t,x 2 p D) = R d 1 For elliptic equations, we define u L x 1 q L x 2 p Ω) = R d 1 R R d 2 R d 2 where Ω is a subset in R d. Finally we set ut, x) p I D t, x)dx 2 ) q/p dx 1 dt) 1/q, ut, x) p I D t, x)dx 2 dt) q/p dx 1 ) 1/q. ux) p I Ω x)dx 2 ) q/p x 1 ) 1/q, L q,p D) := L t,x1 q L x2 p D) or Lx1 q Lt,x2 p D), 4) L q,p Ω) := L x1 q L x2 p Ω). Note that, in case p = q, L p D) = L p,p D) = L t,x1 p L x2 p D) = Lx1 p Lt,x2 p D) and L p Ω) = L p,p Ω). Apparently our mixed norms are more general than the usual mixed norms such as L q 0, T), L p R d )). According to the notation above, L q 0, T), L p R d )) is denoted by L t ql x p0, T) R d ). In our case we have q summability in any given variables, which may or may not include the time variable, and p summability in the remaining variables. With the above mixed norms in hand, we prove that, for a given f L q,p 0, T) R d ), q > p 2, there exists a unique solution u Wq,p 1,20, T) Rd ) to the parabolic equation 2) with an appropriate initial condition, where u Wq,p 1,2 0, T) R d ) means that u, u t, u x i, u x i x j L q,p0, T) R d ). We also prove the unique solvability of the elliptic equation 1) in Sobolev spaces with mixed norms, which are somewhat more general than the usual homogenous Sobolev spaces no mixed norms) for elliptic equations as in [5, 13, 8]. Considering elliptic equations in mixed norm spaces as ours must be another expansion of L p -theory for elliptic equations. The proof for the elliptic case is rather simple and uses the solvability result for parabolic equations. As explained in [14], the necessity of mixed norms for differential equations arises when one has to improve the regularity of a solution in some directions. For example, if the derivatives of a solution are in L t q Lx p 0, T) Rd ) and q is large enough, then using embedding theorems we might be able to increase the regularity of the solution in the

3 EQUATIONS IN L p-spaces WITH MIXED NORMS 439 time direction. Indeed, in [7] the solvability of parabolic equations in Sobolev space with mixed norms is used when proving results for equations in Sobolev spaces without mixed norms for instance, the Hölder continuity of solutions to parabolic equations in Sobolev spaces without mixed norms). Moreover, by having mixed norms as ours we are able to have better regularity of solutions as functions of some given variables not as functions of only time or only the whole spatial variables). An example of this argument is presented in the proof of Lemma 5.5 of this paper. We also prove the existence and uniqueness of solutions to equations defined on a half space R d + = {x = x1,, x d ) R d, x 1 > 0}. Traditionally, in order to solve equations on a half space or on a bounded domain, one has to have boundary estimates. However, we do not need any boundary estimates but the results for equations in the whole space. As is seen in the proof of Theorem 2.5, this is made possible by our assumption that a ij are measurable in one spatial direction. This shows that the class of coefficients a ij in this paper is very useful. Certainly, our results for equations on a half space can be used, via a partition of unity and the usual flattening argument, when dealing with equations on a bounded domain if the leading coefficients are, for instance, in the class of VMO as functions of t, x). The same or similar) class of coefficients is presented in [8, 9, 7], where homogeneous Sobolev spaces are considered. Before the current class of coefficients a ij was introduced, one considerably general assumption on a ij had been that they are in the space of VMO, which were first investigated in [2, 1]. Then recently a new class of coefficients called V MO x coefficients) was suggested by N. Krylov in [13]. The V MO x coefficients are characterized as being measurable in the time variable and VMO in the spatial variables. For more information about general elliptic and parabolic equations in Sobolev spaces, see [5, 17, 19, 15] and references therein. On the other hand, for example, in [6, 12, 14] one can see VMO or measurable coefficients for parabolic equations in mixed norm spaces. The coefficients in [6] are VMO in x R d, but independent of t R, whereas the coefficients a ij in [12] are measurable functions of only t, and the coefficients a ij in [14] are V MO x coefficients, the same class as in [13]. Since in our case in the parabolic case) the coefficients a ij t, x) except a 11 ) are measurable as functions of t, x 1 ) and VMO as functions of x, where x = x 2,, x d ) R d 1 a 11 is measurable in x 1 and VMO in t, x )), as far as coefficients a ij, i 1 or j 1, are concerned, the class of coefficients in this paper is bigger than those in [6, 12, 14]. Our approach to dealing with mixed norms is based on the method presented in [14]. In order to have, for example, an L t,x1 q L x2 p Rd+1 )-estimate, we first prove pointwise estimates of the sharp functions of u x i x jt,x 1, ) LpR d 2) as functions of t,x 1 ) R R d1. Then we obtain the desired mixed norm estimate using the Hardy- Littlewood theorem and Fefferman-Stein theorem. This approach is quite different from ones used in many papers about equations in Sobolev spaces with or without mixed norms. Especially, semigroup or singular integral based methods seem not to be applicable in our situation. This paper consist of two parts. In the first part we solve the equation 2) in Sobolev spaces with mixed norms when the coefficients a ij are measurable in x 1 R and VMO in t, x ) R R d 1. This result serves as one of main steps in [7]. Then using the results in [7] as well as in the first part of this paper, we prove the main results of this paper. The first part consists of section 3 and 4; the second part consists of section 5 and 6. In section 2 we states the assumptions and the main results. A few more words about notation: We denote x the last d 1 coordinates of x,

4 440 D. KIM that is, x = x 2,, x d ) R d 1, so that x = x 1, x ). By u x we mean, depending on the context, one of u x j, i = 2,, d, or the whole collection {u x 2,, u x d}. As usual, u x represents one of u x i, i = 1,, d, or the whole collection of {u x 1,, u x d}. Thus u xx is one of u x i xj, where i {1,, d} and j {2,, d}, or the collection of them. For a function g defined on R m or on a subset in R m ), m d + 1, the average of g over an open set Ω R m is denoted by u) Ω, i.e., g) Ω = 1 gx)dx = gx) dx, Ω Ω where Ω is the m-dimensional volume of Ω. Finally, various constants are denoted by N, their values may vary from one occurrence to another. We write Nd, δ,... ) if N depends only on d, δ, Main results. We consider the elliptic 1) and parabolic equation 2) with coefficients a ij, b i, and c satisfying the following assumptions. If the elliptic equation is considered, we assume that all coefficients are independent of t R. Assumption 2.1. The coefficients a ij, b i, and c are measurable functions defined on R d+1, a ij = a ji. There exist constants δ 0, 1) and K > 0 such that b i t, x) K, ct, x) K, Ω δ ϑ 2 for any t, x) R d+1 and ϑ R d. d a ij t, x)ϑ i ϑ j δ 1 ϑ 2 i,j=1 Another assumption on the coefficients a ij t, x) is that they are, in case p 2, ), measurable in t, x 1 ) R 2 and VMO in x = x 2,, x d ) R d 1 the coefficient a 11 t, x) is measurable in x 1 R and VMO in t, x ) R d ). In case p = 2, the coefficients a ij t, x) are measurable functions of only t, x 1 ) R 2, but a 11 t, x 1 ) is VMO in t R. If a ij are independent of t i.e., the elliptic case), they are measurable in x 1 and VMO in x R d 1. To state this assumption precisely, we introduce the following notations. Let B r x) = {y R d : x y < r}, B r x ) = {y R d 1 : x y < r}, Γ r x) = x 1 r, x 1 + r) B rx ), Q r t, x) = t, t + r 2 ) B r x), Λ r t, x) = t, t + r 2 ) Γ r x). Set B r = B r 0), B r = B r 0), Q r = Q r 0) and so on. By B r we mean the d 1- dimensional volume of B r 0). Denote osc x a ij, Λ r t, x) ) t+r 2 = r 3 B r 2 t x 1 +r x 1 r A ij x s, τ)dτ ds, osc t,x ) a ij, Λ r t, x) ) x 1 +r = r 5 B r 2 x 1 r A ij t,x ) τ)dτ,

5 EQUATIONS IN L p-spaces WITH MIXED NORMS 441 where A ij x s, τ) = y,z B r x ) a ij s, τ, y ) a ij s, τ, z ) dy dz, A ij t,x ) τ) = Also denote σ, t,t+r 2 ) y,z B r t,x ) a ij σ, τ, y ) a ij, τ, z ) dy dz dσ d. OR x aij ) = sup sup t,x) R d+1 r R osc x a ij, Λ r t, x) ), O t,x ) R a ij ) = sup t,x) R d+1 sup r R osc t,x ) a ij, Λ r t, x) ). Finally set a # R = Ot,x ) R a 11 ) + i 1 or j 1 O x R a ij ). If elliptic equations are considered, we set and osc x a ij, Γ r x) ) x 1 = r 1 B r +r 2 x 1 r a # R = sup sup x R d r R y,z B r x ) osc x a ij, Γ r x) ). a ij τ, y ) a ij τ, z ) dy dz dτ Assumption 2.2. There is a continuous function ωr) defined on [0, ) such that ω0) = 0 and a # R ωr) for all R [0, ). By W 1,2 q,p D), where D is an open set in R d+1, we mean the collection of all functions defined on D such that u W 1,2 q,pd) := u L q,pd) + u x Lq,pD) + u xx Lq,pD) + u t Lq,pD) <, where L q,p D) is defined in 4). For elliptic equations, we say u Wq,p 2 Ω) if u, u x, u xx L q,p Ω), where Ω is an open set in R d. Again note that Wp,pD) 1,2 = Wp 1,2 D) and Wp,p 2 Ω) = W p 2 Ω). We denote the parabolic and elliptic differential operators by L and E, respectively, that is, Lu = u t + a ij u xi x j + bi u x i + cu, Eu = a ij u x i x j + bi u x i + cu. The following are our main results. First we state our result about parabolic equations in the whole space. Theorem 2.3. Let q > p 2, 0 < T <, and the coefficients of L satisfy Assumption 2.1 and 2.2. In addition, we assume that, in case p = 2, a ij are independent of x R d 1. Then for any f L q,p 0, T) R d ), there exists a unique

6 442 D. KIM u W 1,2 q,p 0, T) Rd ) such that ut, x) = 0 and Lu = f in 0, T) R d. Furthermore, there is a constant N, depending only on d 1, d 2, δ, K, p, q, T, and the function ω, such that u W 1,2 q,p0,t) R d ) N Lu L q,p0,t) R d ) for any u W 1,2 q,p 0, T) R d ) satisfying ut, x) = 0. We have the following result for elliptic equations in the whole space. Theorem 2.4. Let q > p 2 and the coefficients of E satisfy Assumption 2.1 and 2.2. In addition, we assume that, in case p = 2, a ij are independent of x R d 1. Then there exist constants λ 0 0 and N, depending only on d 1, d 2, δ, K, p, q, and the function ω, such that λ u Lq,pR d ) + λ u x Lq,pR d ) + u xx Lq,pR d ) N Eu λu Lq,pR d ) 5) for any u W 2 q,pr d ) and λ λ 0. Moreover, for any λ > λ 0 and f L q,p R d ), there exists a unique u W 2 q,p Rd ) such that Eu λu = f in R d. We present results about equations on a half space. Recall the definition of R d + given in the introduction. The proofs of the following two theorems show how useful is the assumption that a ij are allowed to be only measurable in one spatial direction. Since their proofs are almost the same, we here give only a proof of Theorem 2.5. Also see Theorem 2.7 in [9] and Theorem 2.7 in [8]. Theorem 2.5. Let q > p 2, 0 < T <, and the coefficients of L satisfy Assumption 2.1 and 2.2. In addition, we assume that, in case p = 2, a ij are independent of x R d 1. Then for any f L q,p 0, T) R d +), there exists a unique u Wq,p 1,20, T) Rd + ) such that ut, x) = ut, 0, x ) = 0 and Lu = f in 0, T) R d +. Proof. Define a new operator ˆL by ˆL = t + âij 2 x i x j + ˆb i x i + ĉ, where â ij, ˆb i, and ĉ are either even or odd extensions of a ij, b i, and c. Specifically, for i = j = 1 or i, j {2,, d}, the coefficients â ij are even extensions of a ij : â ij t, x 1, x ) = a ij t, x 1, x )I x1 0 + a ij t, x 1, x )I x1 <0. For j = 2,, d, the coefficients â 1j are odd extensions of a 1j : â 1j t, x 1, x ) = a 1j t, x 1, x )I x 1 0 a 1j t, x 1, x )I x 1 <0. Similarly, ˆb 1 is the odd extension of b 1, and ˆb i, j = 2, d, as well as ĉ are the even extensions of b j and c. We see that the coefficients â ij, ˆb i, and ĉ satisfy Assumption 2.1. In addition, the coefficients â ij satisfy Assumption 2.2 with 2ω. Let ˆf be the odd extension of f. By Theorem 2.3 there exists a unique u Wq,p 1,20, T) Rd ) such that ˆLu = ˆf and ut, x) = 0. It is easy to check that ut, x 1, x ) Wq,p 1,20, T) Rd ) also satisfies the same equation, so by uniqueness we have ut, x 1, x ) = ut, x 1, x ). From this it follows that ut, 0, x ) = 0 for t, x ) 0, T) R d 1. This together with the fact that ut, 0) = 0 shows that that u, as a function defined on 0, T) R+ d,

7 EQUATIONS IN L p-spaces WITH MIXED NORMS 443 is a solution to Lu = f satisfying ut, x) = ut, 0, x) = 0. Uniqueness follows from the fact that the odd extension of a solution belongs to Wq,p 1,2 0, T) R d ) and the uniqueness of solutions to equations on 0, T) R d. Theorem 2.6. Let q > p 2 and the coefficients of E satisfy Assumption 2.1 and 2.2. In addition, we assume that, in case p = 2, a ij are independent of x R d 1. Then there exist constants λ 0 0 and N, depending only on d 1, d 2, δ, K, p, q, and the function ω, such that, for any λ > λ 0 and f L q,p R d + ), there exists a unique u Wq,pR 2 d +) satisfying u0, x ) = 0 and Eu λu = f in R d +. Remark 2.7. The above four theorems for the case q = p 2 follow from the results in [8, 9, 7]. If q = p = 2, all of the coefficients a ij t, x) for the parabolic operator L are further allowed to be measurable in t, x 1 ) see [9]) as long as they are functions of only t, x 1 ). In the above theorems, the condition that a ij are independent of x R d 1 can be obviously replaced by the uniform continuity of a ij as functions of x R d 1 uniformly in t, x 1 ). Regarding Theorem 2.5 and 2.6, appropriate L q,p - estimates can be stated. Moreover, it is possible to replace the Dirichlet boundary condition by the Neumann boundary condition. That is, in Theorem 2.5 the condition ut, 0, x) = 0 can be replaced by u x 1t, 0, x) = 0. In Theorem 2.6 u x 10, x) = 0 in place of u0, x) = 0. The proofs are the same, but instead of the odd extension of f, one has to use its even extension. For details, see Theorem 2.8 in [8] or Theorem 2.8 in [9]. 3. Parabolic equations with a ij measurable in x 1 R and VMO in t, x ) R d. To move toward the proofs of the main results above, we first need to deal with the parabolic operator L when the coefficients a ij t, x 1, x ) are measurable in x 1 R and VMO in t, x ) R d. Note that in Theorem 2.3 and 2.5 all coefficients a ij t, x 1, x ) except a 11 are assumed to be measurable in t, x 1 ) and VMO in x, so the class of coefficients we consider in this section is less general than those in Theorem 2.3 and 2.5. To state this assumption on the coefficients a ij precisely, set = a #t,x ) R i,j=1 O t,x ) R a ij ). Assumption 3.1. There is a continuous function ωr) defined on [0, ) such that ω0) = 0 and a #t,x ) R ωr) for all R [0, ). The following is the main result of this section. Theorem 3.2. Let q > p 2, 0 < T <, and the coefficients of L satisfy Assumption 2.1 and 3.1. In addition, we assume that, in case p = 2, a ij are independent of x R d 1. Then for any f L q,p 0, T) R d ), there exists a unique u Wq,p 1,20, T) Rd ) such that ut, x) = 0 and Lu = f in 0, T) R d. Furthermore, there is a constant N, depending only on d 1, d 2, δ, K, p, q, T, and the function ω, such that u W 1,2 q,p0,t) R d ) N Lu L q,p0,t) R d ) for any u W 1,2 q,p 0, T) R d ) satisfying ut, x) = 0. This theorem is proved in the next section after presenting some preliminary results. Throughout this section, we set L λ u = u t + a ij x 1 )u x i x j λu,

8 444 D. KIM where λ 0 and a ij are measurable functions of only x 1 R satisfying Assumption 2.1. We start with a theorem which can be derived from results in [9]. Theorem 3.3. Let p 2 and T [, ). For any λ > 0 and f L p T, ) R d ), there exists a unique solution u Wp 1,2 T, ) R d ) to the equation L λ u = f. Furthermore, there is a constant N = Nd, p, δ) such that, for any λ 0 and u T, ) Rd ), we have W 1,2 p u t LpT, ) R d ) + u xx LpT, ) R d ) + λ u x LpT, ) R d ) +λ u LpT, ) R d ) N L λ u LpT, ) R d ). If T =, this theorem is obtained from Theorem 3.2 in [9] for p = 2 and Lemma 5.3 in [9] for p > 2. For the case T, ), we use the case T = and the argument following Corollary 5.14 in [14]. The following three lemmas are L p -versions of Lemma 4.2, 4.3, and 4.4 in [9]. Since the estimate in Theorem 3.3 is available, their proofs can be done by repeating the proofs of Lemma 4.2, 4.3, and 4.4 in [9] with p in place of 2. Lemma 3.4. Let p [2, ). For any u W 1,2 p,loc Rd+1 ), we have u t LpQ r) + u xx LpQ r) + u x LpQ r) N L 0 u LpQ R) + u LpQ R)), where 0 < r < R < and N = Nd, p, δ, r, R). Lemma 3.5. Let p [2, ), 0 < r < R <, and γ = γ 1,, γ d ) be a multiindex such that γ 1 = 0, 1, 2. If v Cloc Rd+1 ) is a function such that L 0 v = 0 in Q R, then Dt m Dxv γ p dxdt N v p dxdt, Q r Q R where m is a nonnegative integer and N = Nd, p, δ, γ, m, r, R). Lemma 3.6. Let p 2 and v C loc Rd+1 ) be a function such that L 0 v = 0 in Q 4. Then where N = Nd, p, δ). sup Q 1 v tt + sup Q 1 v tx + sup Q 1 v txx + sup Q 1 v xxx N v LpQ 4), The proofs of the lemmas and theorem below follow the ideas in [14], specifically, those in the proofs of Lemma 5.9, Theorem 5.10, and Theorem 5.1 in [14]. Lemma 3.7. Let p 2 and λ 0. For every v C loc Rd+1 ) such that L λ v = 0 in Q 4, we have sup v tt + sup v tx + sup v txx + sup v xxx Q 1 Q 1 Q 1 Q 1 Nd, p, δ) v xx LpQ 4) + v t LpQ 4) + ) λ v x LpQ 4). In fact, Lemma 5.3 in [9] says that the estimate in Theorem 3.3 holds for all λ λ 0 0, where λ 0 is not necessarily 0. However, since the coefficients a ij of L λ are measurable functions of only x 1 R and b i = c = 0, it can be proved, using a dilation argument, that λ 0 = 0 in our case.

9 EQUATIONS IN L p-spaces WITH MIXED NORMS 445 Proof. We first note that, in case λ = 0, by Lemma 3.6 I := sup Q 1 v tt + sup Q 1 v tx + sup Q 1 v txx + sup Q 1 v xxx N v LpQ 4), The function u := v v) Q4 x i v x i) Q4 can replace v in the above inequality since L 0 u = 0 in Q 4. This together with the fact that D m t D γ xv = D m t D γ xu for m 1 or γ 2 gives I N v v) Q4 x i v x i) Q4 LpQ 4). This and Lemma 5.4 in [14] prove the inequality in the lemma for λ = 0. In case λ > 0, we extend vt, x) to a function defined on R d+2. Specifically, set vt, x, ξ) = vt, x)cos λξ), where ξ R and t, x, ξ) R d+2. Notice that L 0 v + v ξξ = 0 in Q 4, Dt m Dγ x vt, x) = Dm t Dγ xvt, x, 0), sup Dt m Dγ xv sup Dt m Dγ x v, Q 1 Q 1 where Q r = 0, r 2 ) {x, ξ) R d+1 : x 2 + ξ 2 < r 2 }. Thus by the argument above for the case λ = 0 we have I N v xx LpQ 4) + v xξ LpQ 4) + v ξξ LpQ 4) + v t LpQ 4)). 6) Note that, for example, v xx = v xx cos λξ), v xξ = λv x sin λξ), v ξξ = λv cos λξ). Therefore, the right-hand side of the inequality 6) is not greater than a constant times v xx LpQ 4) + v t LpQ 4) + λ v x LpQ 4) + λ v LpQ 4). This is bounded by the right-hand side of the inequality in the lemma note that λv = L 0 v in Q 4 ). The lemma is proved. Lemma 3.8. Let p 2, λ 0, κ 4, and r 0, ). Let v Cloc Rd+1 ) be such that L λ v = 0 in Q κr. Then there is a constant N, depending only on d, p, and δ, such that v t t, x) v t ) Qr p dxdt + v xx t, x) v xx ) Qr p dxdt Q r Q r Nκ p v xx p + v t p + λ p/2 v x p) Q κr. 7) Proof. Due to a dilation argument see the proof of Theorem 5.10 in [14]), it is enough to prove the inequality 7) when r = 1. For v Cloc Rd+1 ) such that L λ v = 0 in Q κ, κ 4, set κ ) ) 2t, κ ˇvt, x) = v 4 4 x, ǎ ij x 1 ) = a ij κx 1 /4).

10 446 D. KIM Then Ľ κ 4 λˇvt, x) := )2 t + ǎij x 1 ) 2 x i x j κ 4 ) 2 λ ) ˇvt, x) κ ) 2 κ ) ) 2t, κ = Lλ v) x = 0 in Q 4. Thus by Lemma 3.7, it follows that Ǐ N ˇv xx LpQ 4) + ˇv t LpQ 4) + κ ) λ ˇvx LpQ 4 4), 8) where Note that 4/κ) 3 Ǐ = κ/4) Ǐ := sup Q 1 ˇv tt + sup Q 1 ˇv tx + sup Q 1 ˇv txx + sup Q 1 ˇv xxx. sup v tt + sup v txx Q κ/4 Q κ/4 ) + sup Q κ/4 v tx + sup Q κ/4 v xxx. Using this, the inequality 8), and the fact that κ 4, we have v t t, x) v t ) Q1 p dxdt + v xx t, x) v xx ) Q1 p dxdt Q 1 Q 1 N sup v tt + sup v tx + sup v txx + sup v xxx Q κ/4 Q κ/4 Q κ/4 Q κ/4 ) p Nκ 3p Ǐ p Nκ 3p ˇv xx p L pq 4) + ˇv t p L pq 4) + κp λ p/2 ˇv x p L pq 4) ) This finishes the proof. = Nκ p v xx p + v t p + λ p/2 v x p) L pq κ). Theorem 3.9. Let p 2. Then there is a constant N, depending only on d, p, and δ, such that, for any u Wp 1,2Rd+1 ), r 0, ), and κ 8, u t t, x) u t ) Qr p dxdt + u xx t, x) u xx ) Qr p dxdt Q r Q r Nκ d+2 L 0 u p ) Qκr + Nκ p u xx p ) Qκr. Proof. Since C0 Rd+1 ) is dense in Wp 1,2Rd+1 ), it is enough to have u C0 Rd+1 ). In addition, we can assume that a ij x 1 ) are infinitely differentiable. Take a λ > 0 and, for u C0 R d+1 ), let f := f λ = L λ u.

11 EQUATIONS IN L p-spaces WITH MIXED NORMS 447 We see f C 0 Rd+1 ). For given r > 0 and κ 8, let η C 0 Rd+1 ) be a function such that η = 1 on Q κr/2 and η = 0 outside κr) 2, κr) 2) B κr. Also let g := fη, h := f1 η). Then by Theorem 3.3 there exists a unique solution v Wp 1,2 R d+1 ) note that λ > 0) to the equation L λ v = h. From the classical theory we see that the function v is infinitely differentiable. Moreover, since L λ v = h = 0 in Q κr/2 and κ/2 4, by Lemma 3.8, we have v t t, x) v t ) Qr p dxdt + v xx t, x) v xx ) Qr p dxdt Q r Q r Nκ p v xx p + v t p + λ p/2 v x p) Q κr/2 Nκ p v xx p + v t p + λ p/2 v x p) Q κr. Set w := u v Wp 1,2Rd+1 ). Then from the above inequality it follows that u xx t, x) u xx ) Qr p dxdt Q r 2 p w xx t, x) w xx ) Qr p dxdt + 2 p v xx t, x) v xx ) Qr p dxdt Q r Q r N w xx p ) Qr + Nκ p v xx p + v t p + λ p/2 v x p) Q κr. Similar inequalities are possible with u t in place of u xx. Thus we have u t t, x) u t ) Qr p dxdt + u xx t, x) u xx ) Qr p dxdt Q r Q r Now we observe that N w t p + w xx p ) Qr + Nκ p v xx p + v t p + λ p/2 v x p) Q κr. 9) L λ w = L λ u v) = f h = g and, by Theorem 3.3, w t p dxdt + w xx p dxdt Q r Q r w t p L p0, ) R d ) + w xx p L p0, ) R d ) + λp/2 w x p L p0, ) R d ) N g p L p0, ) R d ) Q = N g p dxdt N f p dxdt. κr Q κr From this we see that w t p ) Qr + w xx p ) Qr Nκ d+2 f p ) Qκr,

12 448 D. KIM w xx p + w t p + λ p/2 w x p) Q κr N f p ) Qκr. Now we use the these inequalities as well as the inequality 9). We also use the fact u = w + v and κ 8. Then we obtain u t t, x) u t ) Qr p dxdt + u xx t, x) u xx ) Qr p dxdt Q r Q r Nκ d+2 f p ) Qκr + Nκ p w xx p + w t p + λ p/2 w x p) Q κr +Nκ p u xx p + u t p + λ p/2 u x p) Q κr Nκ d+2 f p ) Qκr + Nκ p u xx p + u t p + λ p/2 u x p) Q κr. To complete the proof, we use the fact that u t = f + λu a ij u x i xj, and then let λ ց Proof of Theorem 3.2. Set L 0 u = u t + a ij t, x)u xi x j, where coefficients a ij satisfies Assumption 2.1 and 3.1. Lemma 4.1. Let p > q 2, and r 0, 1]. Assume that v W 1,2 p,loc Rd+1 ) satisfies L 0 v = 0 in Q 2r. Then v xx p ) 1/p Q r N v xx 2) 1/2 Q 2r N v xx q ) 1/q Q 2r, where N depends only on d, p, δ, and the function ω. Proof. This lemma is almost the same as Corollary 6.4 in [14] if L 0 is replaced by the operator used there. In our case, we can repeat the argument in Corollary 6.4 of [14] if we have the estimate u xx LpQ r) N L 0 u LpQ κr) + r 1 u x LpQ κr) + r 2 ) u LpQ κr) for p 2, ) and u W 1,2 p,loc Rd+1 ), where r 0, 1], κ 1, ), and N depends only on d, p, δ, κ, and the function ω. This is obtained using Theorem 2.5 in [9] and the argument in the proof of Lemma 6.3 in [14]. The following theorem is proved in the same way as Lemma 3.1 in [14]. However, because of the different conditions on a ij between our operator L 0 and the operator defined in [14], we give a complete proof here. Theorem 4.2. Let p 2. In case p = 2, the coefficients a ij of L 0 are assumed to be independent of x R d 1. Then there exists a constant N, depending on d, p, δ, and the function ω, such that, for any u C0 Rd+1 ), κ 16, and r 0, 1/κ], we have u t t, x) u t ) Qr p dxdt + u xx t, x) u xx ) Qr p dxdt Q r Q r

13 EQUATIONS IN L p-spaces WITH MIXED NORMS 449 Nκ d+2 L 0 u p ) Qκr + N κ p + κ d+2 â 1/2) u xx p ) Qκr, where â = a #t,x ) κr. Proof. For given u C0 R d+1 ), κ 16, and r 0, 1/κ], find a unique function w Wp 1,2 3, 4) Rd ) satisfying w4, x) = 0 and L 0 w = fi Qκr, where f := L 0 u. This is possible by Theorem 2.2 and 2.5 in [9]. Moreover, w Wq 1,2 3, 4) R d ) for all q 2, ) because fi Qκr L q 3, 4) R d ) for all q > 2. Let wt, x) = ηt) wt, x), where ηt) is an infinitely differentiable function defined on R such that ηt) = 1, 1 t 2, ηt) = 0, t 2 or t 3. We see that w Wp 1,2Rd+1 ) and, in addition, w Wq 1,2 R d+1 ) for all q 2, ). From the estimates from Theorem 2.2 and 2.5 in [9] we have w t p + w xx p ) dxdt Q κr w t p + w xx p ) dxdt N 3,4) R d f p dxdt, Q κr where N depends only on d, δ, p, and the function ω it also depends on the time interval, but the time interval here is fixed as 3, 4)). Thus w t p + w xx p ) Qκr N f p ) Qκr, 10) w t p + w xx p ) Qr Nκ d+2 f p ) Qκr. 11) where N = Nd, δ, p, ω). Now we set v = u w. Then v Wp 1,2Rd+1 ), v Wq 1,2 R d+1 ) for all q 2, ), and L 0 v = 0 in Q κr. Let where ā ij x 1 ) = L 0 = t + āij x 1 ) 2 i j, κr/2) 2 0 B κr/2 a ij s, x 1, y )dy ds. Since v Wp 1,2 R d+1 ) and κ/2 8, by Theorem 3.9 applied to the operator L 0, we have v t t, x) v t ) Qr p dxdt + v xx t, x) v xx ) Qr p dxdt Q r Q r

14 450 D. KIM Nκ d+2 L 0 v p) Q κr/2 + Nκ p v xx p ) Qκr/2. Using the fact that L 0 v = 0 in Q κr, we have L0 v p) = ā ij x 1 ) a ij t, x) ) v Q xi κr/2 x j p dxdt Q κr/2 ) 1/2 ) 1/2 ā ij x 1 ) a ij t, x) 2p dxdt v xi x j 2p dxdt, Q κr/2 Q κr/2 where we see ā ij x 1 ) a ij t, x) 2p dxdt N ā ij x 1 ) a ij t, x) dxdt Na #t,x ) κr/2. Q κr/2 Q κr/2 From Lemma 4.1 we also see ) 1/2 ) v x i x j 2p dxdt Nd, p, δ, ω) v xx p dxdt. Q κr/2 Q κr Hence v t t, x) v t ) Qr p dxdt + v xx t, x) v xx ) Qr p dxdt Q r Q r N κ p + κ d+2 â 1/2) v xx p ) Qκr. Note that v xx p ) Qκr N u xx p ) Qκr + N w xx p ) Qκr N u xx p ) Qκr + N f p ) Qκr, where the second inequality is due to 10). Also note that, using the inequality 11), w xx t, x) w xx ) Qr p dxdt N w xx p ) Qr Nκ d+2 f p ) Qκr, Q r w t t, x) w t ) Qr p dxdt N w t p ) Qr Nκ d+2 f p ) Qκr. Q r Therefore, u xx t, x) u xx ) Qr p dxdt Q r N v xx t, x) v xx ) Qr p dxdt + N w xx t, x) w xx ) Qr p dxdt Q r Q r N κ p + κ d+2 â 1/2) u xx p + f p ) Qκr + Nκ d+2 f p ) Qκr.

15 EQUATIONS IN L p-spaces WITH MIXED NORMS 451 Similarly, we have u t t, x) u t ) Qr p dxdt Q r N κ p + κ d+2 â 1/2) u xx p + f p ) Qκr + Nκ d+2 f p ) Qκr. The theorem is proved. Here we introduce some notations we use below. Let B d1 r x 1) = { x 1 y 1 < r : y 1 R d1 }, Q d1 r t,x 1) = t, t + r 2 ) B d1 r x 1), B d2 r x 2) = { x 2 y 2 < r : y 2 R d2 }, Q d2 r t,x 2) = t, t + r 2 ) B d2 r x 2). Recall that x 1 and x 2 are those defined in 3). As before, we set, for example, Br d1 = Br d1 0) and Qd1 r = Q d1 r 0, 0). For a function f defined on Rd+1, denote ft,x 1, ) pd2) = R d 2 ft, x) p dx 2 ) 1/p. Note that ft,x 1, ) pd2) is a function of t,x 1 ). To define the above norm more precisely as well as explain some notations below, we add the following remark. Remark 4.3. As indicated in the introduction, we have fixed two nonnegative integers d 1 and d 2 d 1 +d 2 = d) and all different integers i 1, i 2,, i d1, j 1, j 2,, j d2 from {1, 2,, d}. For x = x 1, x 2,, x d ) R d, we have denoted x 1 = x i1, x i2,, x i d 1) R d 1, x 2 = x j1, x j2,, x j d 2) R d 2. Let Π 1 be a projection from R d onto R d1 such that Π 1 x) = x 1. Likewise, let Π 2 be a projection from R d onto R d2 such that Π 2 x) = x 2. Define Π to be a function from R d onto R d1 R d2 such that Πx) = Π 1 x), Π 2 x)) = x 1,x 2 ). We see that the inverse Π 1 of Π exists. Then by ft,x 1, ) pd2) we mean R d 2 f t, Π 1 x 1,x 2 ) ) p dx 2 ) 1/p. In the following we use, for example, ut,x 1,x 2 ), the precise notation for which is therefore ut, Π 1 x 1,x 2 )). Lemma 4.4. Let p 2. In case p = 2, we assume that the coefficients a ij of L 0 are independent of x R d 1. Then there exists a constant N = Nd 1, d 2, p, δ, ω) such that r r ut t,x 1, ) pd2) u t s,y 1, ) pd2) p dx 1 dt dy 1 ds + uxx t,x 1, ) pd2) u xx s,y 1, ) pd2) p dx 1 dt dy 1 ds r r

16 452 D. KIM N κ p + κ d+2 â 1/2) κr u xx t,x 1, ) p pd 2) dx 1 dt +Nκ d+2 L 0 ut,x 1, ) p Q d pd dx 1 2) 1 dt, κr for any u C 0 Rd+1 ), r > 0, and κ 16 2 satisfying κr 1, where â = a #t,x ) κr. Proof. Let us denote by I the left-hand side of the inequality in the lemma. Note that ut t,x 1, ) pd2) u t s,y 1, ) pd2) p u t t,x 1, ) u t s,y 1, ) p pd. 2) A similar inequality holds for the integrand of the second integral of I. Thus I u t t,x 1, ) u t s,y 1, ) p pd dx 2) 1 dt dy 1 ds + r Note that Hence I 1 = R d 2 r r u xx t,x 1, ) u xx s,y 1, ) p Q d pd dx 1 2) 1 dt dy 1 ds := I 1 + I 2. r u t t,x 1, ) u t s,y 1, ) p pd 2) = u t t,x 1,z 2 + w 2 ) u t s,y 1,z 2 + w 2 ) p dz 2 dw 2 B d 2 r R d 2 Also note that and = r R d 2 r u t t,x 1,w 2 ) u t s,y 1,w 2 ) p dw 2 dz 2. B d 2 r z 2) B d 2 r z 2) u t t,x 1,w 2 ) u t s,y 1,w 2 ) p dw 2 dx 1 dt dy 1 ds dz 2. u t t,x 1,w 2 ) u t s,y 1,w 2 ) p 2 p ut t,x 1,w 2 ) u t ) Q 2r 0,z 2) p + 2 p ut s,y 1,w 2 ) u t ) Q 2r 0,z 2) Q d1 r Bd2 r z 2) Q 2r 0,z 2), where, and throughout the proof, 0,z 2 ) = 0, Π 1 0,z 2 )) see Remark 4.3). Thus I 1 Nd, p) u t t,x 1,w 2 ) u t ) Q 2r 0,z 2) p dw 2 dx 1 dt dz 2. R d 2 Q 2r 0,z2) p

17 EQUATIONS IN L p-spaces WITH MIXED NORMS 453 Then by Theorem 4.2 with an appropriate translation) we have I 1 Nκ d+2 f p ) dz Qκr0,z 2) 2 R d 2 ) ) 1/2 +N κ p + κ d+2 a #t,x ) κr R d 2 u xx p ) Qκr0,z 2) dz 2, where f = L 0 u, κ 16 2, and κr 1. Observe that f p ) dz Qκr0,z 2) 2 = R d 2 R d 2 ft,x 1,z 2 + w 2 ) p dw 2 dx 1 dt dz 2 Q κr Nd) R d 2 = Nd) κr R d 2 Similarly, Therefore, R d 2 κr ft,x 1,z 2 + w 2 ) p dw 2 dx 1 dt dz 2 B d 2 κr ft,x 1,z 2 ) p dz 2 dx 1 dt = Nd) ft,x 1, ) p Q d pd dx 1 2) 1 dt. κr u xx p ) dz Qκr0,z 2) 2 Nd) u xx t,x 1, ) p Q d pd dx 1 2) 1 dt. κr I 1 Nκ d+2 ft,x 1, ) p Q d pd dx 1 2) 1 dt κr ) ) 1/2 +N κ p + κ d+2 a #t,x ) κr u xx t,x 1, ) p Q d pd dx 1 2) 1 dt. κr By following the same argument above, we arrive at the above inequality with I 2 in place of I 1. This finishes the proof. Set φt,x 1 ) = u t t,x 1, ) pd2), ϕt,x 1 ) = u xx t,x 1, ) pd2), ζt,x 1 ) = u xx t,x 1, ) pd2), ψt,x 1 ) = L 0 ut,x 1, ) pd2). Then they are functions of t,x 1 ) R R d1. Let Q d1 be the collection of all Q d1 r t,x 1), t,x 1 ) R R d1, r 0, ). The maximal and sharp function of gt,x 1 ), t,x 1 ) R R d1, are defined by Mgt,x 1 ) = g # t,x 1 ) = sup t,x 1) sup t,x 1) gs,y 1 ) dy 1 ds, gs,y 1 ) g) dy 1 ds,

18 454 D. KIM where the supremums are taken over all Q d1 Q d1 containing t,x 1 ). Lemma 4.5. Let p 2. In case p = 2, we assume that the coefficients a ij of L 0 are independent of x R d 1. Let R 0, 1] and u be a function in C 0 Rd+1 ) such that Then ut, x) = ut,x 1,x 2 ) = 0 if t,x 1 ) / 0, R 4 ) B d1 R 2. φ # t,x 1 ) + ϕ # t,x 1 ) Nκ d+2)/p Mψ p t,x 1 )) 1/p + NκR) 2 2/p Mφ p t,x 1 )) 1/p +N κr) d1+2)1 1/p) + κ 1 + κ d+2)/p ωr)) 1/2p)) Mζ p t,x 1 )) 1/p for all κ 16 2 and t,x 1 ) R R d1, where N = Nd 1, d 2, p, δ, ω). Proof. Take a κ such that κ If r R/κ, then Thus by Lemma 4.4, r κr R 1, a #t,x ) κr a #t,x ) R ωr). φs,y 1 ) φ) Q d 1 r p dy 1 ds + r ϕs,y 1 ) ϕ) Q d 1 r p dy 1 ds Nκ d+2 ψ p ) Q d 1 κr + N κ p + κ d+2 ωr)) 1/2) φ p ) Q d 1 κr. From this and an appropriate translation we obtain, for t, x 1 ) R R d1, φs,y 1 ) φ) d Q 1 p dy 1 ds + ϕs,y 1 ) ϕ) d Q 1 r t, x 1) r t, x 1) r t, x 1) r t, x 1) p dy 1 ds Nκ d+2 ψ p ) Q d 1 κr t, x 1) + N κ p + κ d+2 ωr)) 1/2) ζ p ) Q d 1 κr t, x 1) if r R/κ. Then using the Hölder s inequality it follows that, for r R/κ, φs,y 1 ) φ) d Q 1 dy 1 ds + ϕs,y 1 ) ϕ) d Q 1 dy 1 ds r t, x 1) r t, x 1) r t, x 1) r t, x 1) Nκ d+2)/p ψ p ) 1/p + N κ 1 + κ d+2)/p ωr)) 1/2p)) ζ p ) 1/p κr t, x 1) On the other hand, if r > R/κ, φs,y 1 ) φ) d Q 1 2 r t, x 1) r t, x 1) I Q d 1 R 2 dy 1 ds ) 1 1/p r t, x 1) r t, x 1) dy 1 ds φs,y 1 ) p dy 1 ds κr t, x 1). ) 1/p

19 EQUATIONS IN L p-spaces WITH MIXED NORMS 455 Nd 1 ) R 2 /r ) d 1+2)1 1/p) φ p ) 1/p r t, x 1) Nd 1 )κr) d1+2)1 1/p) φ p ) 1/p r t, x 1). By the same reasoning, if r > R/κ, we have ϕs,y 1 ) ϕ) d Q 1 dy 1 ds N κr) d1+2)1 1/p) ϕ p ) 1/p r t, x 1) Therefore, r t, x 1) φs,y 1 ) φ) d Q 1 r t, x 1) r t, x 1) Nκ d+2)/p ψ p ) 1/p κr t, x 1) + N dy 1 ds+ r t, x 1) ϕs,y 1 ) ϕ) d Q 1 κ 1 + κ d+2)/p ωr)) 1/2p)) ζ p ) 1/p + N κr) d1+2)1 1/p) φ p ) 1/p + r t, x ϕp ) 1/p 1) for all r > 0. Now we observe that, for any t,x 1 ) Q r t, x 1 ), ψ p ) Q d 1 κr t, x 1) Mψp t,x 1 ), ζ p ) Q d 1 κr t, x 1) Mζp t,x 1 ),. r t, x 1) r t, x 1) κr t, x 1) r t, x 1) dy 1 ds ) 12) φ p ) Q d 1 κr t, x 1) Mφp t,x 1 ), ϕ p ) Q d 1 κr t, x 1) Mζp t,x 1 ). The last two inequalities also hold true if κr is replaced by r. From these inequalities as well as 12) it follows that, for any Q d1 Q d1 such that Q d1 t,x 1 ), φs,y1 ) φ) dy1 ds + ϕs,y1 ) ϕ) dy1 ds +N Nκ d+2)/p Mψ p t,x 1 )) 1/p + N κr) d1+2)1 1/p) Mφ p t,x 1 )) 1/p κr) d1+2)1 1/p) + κ 1 + κ d+2)/p ωr)) 1/2p)) Mζ p t,x 1 )) 1/p. Taking the supremum of the left-hand side of the above inequality over all Q d1 Q d1 such that Q d1 t,x 1 ), we obtain the inequality in the lemma. The lemma is proved. Now we set u Lq,p = u L t,x 1 q L x 2 p R R d ). Corollary 4.6. Let q > p 2. In case p = 2, we assume that the coefficients a ij of L 0 are independent of x R d 1. Then there exist constants R and N, depending only on d 1, d 2, p, q, δ, and the function ω, such that u t Lq,p + u xx Lq,p N L 0 u Lq,p 13)

20 456 D. KIM for any u C 0 Rd+1 ) satisfying ut, x) = ut,x 1,x 2 ) = 0 if t,x 1 ) / 0, R 4 ) B d1 R 2. Proof. Let u C 0 Rd+1 ) be a function such that ut, x) = ut,x 1,x 2 ) = 0 if t, x ) / 0, R 4 ) B d1 R 2, where R 0, 1] will be specified below. Using the inequality in Lemma 4.5 as well as the Hardy-Littlewood theorem and Fefferman-Stein theorem note that q/p > 1), we have u t Lq,p + u xx Lq,p Nκ d+2)/p L 0 u Lq,p + NκR) 2 2/p u t Lq,p +N κr) d1+2)1 1/p) + κ 1 + κ d+2)/p ωr)) 1/2p)) u xx Lq,p. for all κ The left-hand side of the above inequality can be replaced by u t Lq,p + u xx Lq,p because u x 1 x 1 = 1 L a 11 0 u u t a ij u x 1 x j. i 1 or j 1 Now we choose a large κ and then a small R so that N κr) d1+2)1 1/p) + κ 1 + κ d+2)/p ωr)) 1/2p)) < 1/2 and It then follows that This finishes the proof. NκR) 2 2/p < 1/2. u t Lq,p + u xx Lq,p 2Nκ d+2)/p L 0 u Lq,p. Remark 4.7. In the above corollary the estimate is prove when L q,p = L t,x1 q L x2 p R R d ). However, by making appropriates changes in Lemma 4.4 and Lemma 4.5, we prove the same estimate when L q,p = L x1 q L t,x2 p R R d ). Theorem 4.8. Under the assumptions on L given in Theorem 3.2, there exist λ 0 0 and N, depending only on d 1, d 2, δ, K, p, q, and the function ω, such that λ u Lq,pT, ) R d ) + λ u x Lq,pT, ) R d ) + u xx Lq,pT, ) R d ) + u t Lq,pT, ) R d ) N L λ)u Lq,pT, ) R d ) 14) for all λ λ 0 and u W 1,2 q,p T, ) Rd ), where T <. Proof. By the argument following Corollary 5.14 in [14] it is enough to prove 14) for T =. Moreover, it is enough to prove 14) for u C 0 Rd+1 ). Then the inequality 14) follows from Corollary 4.6 also see Remark 4.7) and a partition of unity. Details can be obtained by following the proofs in section 3 [14], more precisely, those of Lemma 3.4 and Theorem 3.5 in [14]. Proof of Theorem 3.2. The estimate in Theorem 4.8 along with the method of continuity implies Theorem 3.2. For details, see the proof of Theorem 2.1 in [13].

21 EQUATIONS IN L p-spaces WITH MIXED NORMS Equations with a ij measurable in t, x 1 ) R 2. Throughout this section, we set L λ u = u t + a ij t, x 1 )u x i x j λu, where λ 0 and a ij are functions of only t, x 1 ) R 2, a 11 x 1 ) is a function of x 1 R, satisfying Assumption 2.1. As is seen in [7], one of key steps there is based on Theorem 3.2 in this paper when d 1 = 0 and d 2 = d. Now that we have proved Theorem 3.2, it is legitimate to use the results in [7]. Thus we can state the following theorem, which is due to the result in [7] as well as in [9]. Theorem 5.1. Let p 2 and T [, ). For any λ > 0 and f L p T, ) R d ), there exists a unique solution u Wp 1,2 T, ) R d ) to the equation L λ u = f. Furthermore, there is a constant N = Nd, p, δ) such that, for any λ 0 and u T, ) R d ), we have W 1,2 p u t LpT, ) R d ) + u xx LpT, ) R d ) + λ u x LpT, ) R d ) +λ u LpT, ) R d ) N L λ u LpT, ) R d ). More precisely, this theorem follows, in case p = 2, from Theorem 3.2 in [9] and, in case p > 2, from Corollary 4.2 in [7] as well as the argument in the proof of Theorem 4.1 in [13] also see the discussion following Theorem 3.3). Based on the estimate in the above theorem, we have the following lemma, the proof of which is almost identical to those of Lemma 3.4 and Lemma 5.3 a complete proof of Lemma 5.3 is given). Recall that Λ r = 0, r 2 ) r, r) B r. Lemma 5.2. Let p [2, ). For any u W 1,2 p,loc Rd+1 ), we have u t LpΛ r) + u xx LpΛ r) + u x LpΛ r) N L 0 u LpΛ R) + u LpΛ R)), where 0 < r < R < and N = Nd, p, δ, r, R). In the following we set u Lq,pΛ r) = u L t,x 1 u Lq,pΛ r) = r 2 r 0 r B r q L x p Λr), i.e., ) q/p ut, x 1, x ) p dx dx 1 dt 1/q. Lemma 5.3. Let L 0 u = u t + a 11 x 1 )u x 1 x 1 + d 1u, where d 1 u = d j=2 u x j xj. Then for p, q [2, ) such that q p, we have u t Lq,pΛ r) + u xx Lq,pΛ r) + u x Lq,pΛ r) N ) L 0 u Lq,pΛ R) + u Lq,pΛ R) for u C loc Rd+1 ), where 0 < r < R < and N = Nd, p, q, δ, r, R).

22 458 D. KIM Proof. The case p = q is proved in Lemma 5.2, so we assume that q > p. Set r 0 = r, r m = r + R r) m k=1 1, m = 1, 2,, 2k Λ m = 0, r m ) 2 r m, r m ) B r m, m = 0, 1,. Then take η m C0 R d+1 ) such that { 1 if t, x) Λm, η m t, x) = 0 if t, x) / rm+1, 2 rm+1) 2 r m+1, r m+1 ) B r m+1, η m ) x N 2m+1 R r, η m) xx N 22m+2 R r) 2, η m) t N 22m+2 R r) 2, where N depends only on d. To construct them take an infinitely differentiable function gs), s, ), such that gs) = 1 for s 0, gs) = 0 for s 1, and 0 g 1. After this define where η m t, x 1, x ) = η 1m t)η 2m x 1 )η 3m x ), η 1m t) = g 2 m+1 R r) 1 t r m ) ), η 2m x 1 ) = g 2 m+1 R r) 1 x 1 r m ) ), η 3 m x ) = g 2 m+1 R r) 1 x r m ) ). Now we observe that the coefficients L 0 satisfy the assumptions in Theorem 3.2, so the estimate 14) is available. Let d 1 = 1 especially, we set x 1 = x 1 ) and d 2 = d 1. For u C loc Rd+1 ), we take an appropriate λ > 0 and apply the estimate 14) with T = 0 to η m u. Then A m := η m u W 1,2 q,p0, ) R d ) N L 0 λ)η m u) Lq,p0, ) R d ) where N = Nd, p, q, δ) and 22m NB + N η mx u x Lq,p0, ) R d ) + N C, 15) R r) 2 B := L 0 λ)u Lq,pΛ R), C := u Lq,pΛ R). Observe that, for arbitrary ε > 0, η mx u x Lq,p0, ) R d ) = η mx η m+1 u) x Lq,p0, ) R d ) 2 2m N 2m R r η m+1u) x Lq,p0, ) R d ) εa m+1 + Nε 1 R r) 2 C, where the last inequality is due to the interpolation inequality. Thus, 15) yields 2 2m A m NB + εa m+1 + Nε 1 R r) 2 C.

23 EQUATIONS IN L p-spaces WITH MIXED NORMS 459 Set ε = 1/8. Then by multiplying both sides by ε and summing up, we get A 0 + ε m A m NB + m=1 ε m 1 A m + N R r) 2 C. m=1 Indeed, the series of ε m A m = 8 m A m converges since A m N2 2m R r) 2 u W 1,2 q,pλ R). Therefore, after taking care of similar terms we see that A 0 is less than or equal to the right-hand side of the inequality in the lemma. The left-hand side is obviously less than A 0, so the lemma is proved. Denote q r t, x 1 ) := t, t + r 2 ) x 1 r, x 1 + r) R R. Especially, q r = q r 0, 0) = 0, r 2 ) r, r). Lemma 5.4. Let p, q [2, ) be such that q p and 1/q 1/p 1/3. Also let 0 < r < R < and γ = γ 1,, γ d ) be a multi-index such that γ 1 = 0. If v C loc Rd+1 ) is a function such that L 0 v = 0 in Λ R, then D γ x v t Lq,pΛ r) + 2 D γ x Dm x 1v L q,pλ r) N v LpΛ R), m=0 where N = Nd 1, d 2, p, q, δ, γ, r, R). Proof. It is enough to prove v Lq,pΛ r) + v x 1 Lq,pΛ r) + v x 1 x 1 L q,pλ r) + v t Lq,pΛ r) N v LpΛ r1 ), 16) where r 1 is a number such that r < r 1 < R. To see this, using the fact that D γ x v satisfies L 0 D γ x v = 0 in Λ R, we have the above inequality with D γ x v in place of v. Then using Lemma 5.2 as many times as needed, we arrive at the desired inequality in the lemma. If p = q, the inequality 16) follows directly from Lemma 5.2, so we consider the case q > p. Since L 0 v = 0 in Λ R, we have v t + a 11 x 1 )v x1 x 1 + d 1v = d 1 v a ij v x i x j in Λ R. i 1 or j 1 If τ is a number between r and r 1, by Lemma 5.3, we have v Lq,pΛ r) + v x 1 Lq,pΛ r) + v x 1 x 1 L q,pλ r) + v t Lq,pΛ r) N v xx Lq,pΛ τ) + v Lq,pΛ τ)), where N = Nd, p, q, r, τ, δ). Thus we obtain the inequality 16) once we prove However, we prove v xx Lq,pΛ τ) + v Lq,pΛ τ) N v LpΛ r1 ). v xx L x p Lt,x1 q Λ τ) + v L x p Lt,x1 q Λ τ) N v L pλ r1 ) 17)

24 460 D. KIM because by Minkowski s inequality note that q/p > 1), for example, v xx Lq,pΛ τ) = v xx L t,x 1 v q L x p Λτ) xx. L x p Lt,x1 q Λ τ) To prove 17), for each x B R, we view vt, x1, x ) as a function of t, x 1 ). By the Sobolev embedding theorem Lemma 3.3 in Chapter 2 [16]) v, x ) Lqq τ) + v x 1, x ) Lqq τ) N v, x ) W 1,2 p q τ), where N = Np, q, τ). In the above inequality we can replace vt, x 1, x ) by v x t, x 1, x ) or v x x t, x1, x ) as functions of t, x 1 ). Then, for each x B R, we have v, x ) Lqq τ) + v xx, x ) Lqq τ) N 2 m=0 D m x v, x ) W 1,2 p q τ), where Dx m v, m = 0, 1, 2, is v, v x, and v x x, respectively. By integrating pth power of both sides of the above inequality over B τ with respect to x R d 1, we get v L x p Lt,x1 q Λ τ) + v xx L x N p Lt,x1 q 2 Dx m v t p L + pλ τ) m=0 2 N Λ τ) 2 l,m=0 m=0 B τ Dx m v, x ) p Wp 1,2 q dx τ) Dx l 1Dm x v p L pλ τ) N v p L pλ r1 ), where the last inequality is due to the fact L 0 v = 0 in Λ R and repeated use of Lemma 5.2. Hence the inequality 17) is proved, and so is the lemma. Below we use the following notations 0 < µ < 1). f 0,Qr := ft, x) fs, y) sup ft, x), [f] µ,qr := sup t,x) Q r t,x),s,y) Q r t s µ/2 + x y µ. t,x) s,y) Lemma 5.5. Let p 2. Assume that v C loc Rd+1 ) is a function such that L 0 v = 0 in Q 4. Then where µ = µp) 0, 1). Proof. We prove [v xx ] µ,q1 Nd 1, d 2, p, δ) v LpQ 4), [v] µ,q1 + [v x 1] µ,q1 N v LpΛ τ), 18) where 2 < τ < 8. If this is done, we can finish the proof using Lemma 5.4 when p = q) and the fact that Lv x = Lv x x = 0 in Q 4 also note that Λ 8 Q 4). Fix a number q as follows. If p > 3, then q = p. If 2 p 3, then q is a number such that 3 < q 6. We see that q p and 1/q 1/p 1/3.

25 EQUATIONS IN L p-spaces WITH MIXED NORMS 461 We first prove max x B 1 v, x ) W 1,2 q q + max 1) v x, x ) x B 1 W 1,2 q q N v 1) L pq τ). 19) Consider vt, x 1, x ) as a function of x B 1. By the Sobolev embedding theorem there exist N and k such that, for each t, x 1 ) q 1, max vt, x 1, x ) N vt, x 1, ) W x B 1 k p B 1 ), where vt, x 1, ) W k p B 1 ) is the W k p B 1 ) norm of v as a function of x. Similarly, by considering v t t, x 1, x ) as well as D m x 1 vt, x 1, x ), m = 1, 2, as functions of x B 1, we have the same inequalities as above with v t or D m x 1 v in place of v. Thus, for each t, x 1 ) q 1, N This implies that max x B 1 v t t, x 1, x ) + v t t, x 1, ) W k p B 1 ) + 2 max Dx m 1vt, x1, x ) x m=0 B 1 ) 2 Dx m 1vt, x1, ) W k p B 1 ). m=0 max v, x ) x B 1 W 1,2 q q 1) N q 1 v t t, x 1, ) q W k p B 1 ) + N D γ x v t Lq,pΛ 1) + γ k 2 Dx m 1vt, x1, ) q Wp kb 1 ) dx1 dt m=0 ) 1/q ) 2 D γ x Dm x 1v L q,pλ 1) N v LpΛ τ), m=0 where the last inequality is from Lemma 5.4. Thus the first term in 19) is proved to be less than or equal to the right hand side of the inequality. To complete the inequality 19), we repeat the same argument as above with v x in place of v. Now we obtain the following inequalities. For each x B 1, view vt, x1, x ) and v x t, x 1, x ) as functions of t, x 1 ) defined on q 1. Then by the embedding theorem Lemma 3.3 in Chapter 2 [16]) v x, x ) 0,q1 + v x 1 x, x ) 0,q1 N v x, x ) W 1,2 q q 1), 20) [v, x )] µ,q1 + [v x 1, x )] µ,q1 N v, x ) W 1,2 q q 1) 21) for each x B 1, where µ = 1 3/q and N = Nq). Finally, we prove the inequality 18). Note that v x 1t, x) v x 1s, y) I 1 + I 2,

26 462 D. KIM where I 1 = v x 1t, x 1, x ) v x 1s, y 1, x ), I 2 = v x 1s, y 1, x ) v x 1s, y 1, y ). By 21) and 19) I 1 N t s µ/2 + x 1 y 1 µ) v LpΛ τ). To take care of I 2, we use 20) and 19). I 2 x y max v x z B 1 1 x s, y1, z ) x y max v x, z ) z B 1 W 1,2 q q 1) This shows that x y v LpΛ τ). [v x 1] µ,ν,q1 N v LpQ τ). By repeating the above argument with v in place of v x 1, we complete the proof of the inequality 18). The lemma is proved. Lemmas similar to Lemma 3.7 and 3.8 in section 3 are repeated below, but since the operator L λ is being dealt with, the lemmas have to be modified as follows. Lemma 5.6. Let p 2 and λ 0. For every v Cloc Rd+1 ) such that L λ v = 0 in Q 4, we have [v xx ] µ,q1 N v xx LpQ 4) + v t LpQ 4) + ) λ v x LpQ 4), where µ = µp) 0, 1) and N = Nd, p, δ). Proof. We follow the steps in the proof of Lemma 3.7, but the sup-norms of the derivatives of v on Q 1 have to be replaced by [v xx ] µ,q1. Lemma 5.7. Let p 2, λ 0, κ 4, and r 0, ). Let v Cloc Rd+1 ) be such that L λ v = 0 in Q κr. Then there is a constant N, depending only on d, p, and δ, such that v xx t, x) v xx ) Qr p dxdt Nκ µp Q r where µ = µp) 0, 1). and v xx p + v t p + λ p/2 v x p) Q κr, Proof. We use Lemma 5.6. Note that v xx t, x) v xx ) Q1 p dxdt N[v xx ] p µ,q 1 Q 1 κ ) µ+2 κ ) ) 2t, κ [ˇv xx ] µ,q1 = [vxx ] µ,qκ/4 if ˇvt, x) = v x. Using these as well as the argument in the proof of Lemma 3.8, one can complete the proof.

27 EQUATIONS IN L p-spaces WITH MIXED NORMS 463 Now we arrive at the following theorem, the proof of which is almost identical to that of Theorem 3.9. Theorem 5.8. Let p 2. Then there is a constant N, depending only on d, p, and δ, such that, for any u Wp 1,2Rd+1 ), r 0, ), and κ 8, u xx t, x) u xx ) Qr p dxdt Nκ d+2 L 0 u p ) Qκr + Nκ µp u xx p ) Qκr, Q r where µ = µp) 0, 1). 6. Proof of Theorem 2.3 and 2.4. In this section, as stated in Theorem 2.3, we deal with coefficients of L satisfying Assumption 2.1 and 2.2. Also note that the coefficients a ij t, x) are independent of x R d 1 if p = 2. Throughout the section, we denote L 0 u = u t + a ij t, x)u x i x j. As noted earlier, due to Theorem 3.2 in this paper, the results in [7] are now available. This implies that, by the same reasoning as in the proof of Lemma 4.1, the inequalities in Lemma 4.1 are possible with L 0 defined above. Then using the results in section 5 and repeating the proof of Theorem 4.2 with necessary changes), we obtain Theorem 6.1. Let p 2. In case p = 2, we assume that the coefficients a ij t, x) of L 0 are independent of x R d 1. Then there exists a constant N, depending on d, p, δ, and the function ω, such that, for any u C0 Rd+1 ), κ 16, and r 0, 1/κ], we have u xx t, x) u xx ) Qr p dxdt Q r Nκ d+2 L 0 u p ) Qκr + N where µ = µp) 0, 1). κ µp + κ d+2 a # κr )1/2) u xx p ) Qκr, In the following we state two lemmas corresponding to Lemma 4.4 and Lemma 4.5, respectively. Since we are dealing with a ij different from those in section 4, we have different statements, but the proofs are almost the same as those for Lemma 4.4 and Lemma 4.5. Lemma 6.2. Let p 2. In case p = 2, we assume that the coefficients a ij t, x) of L 0 are independent of x R d 1. Then there exists a constant N, depending on d 1, d 2, p, δ, and the function ω, such that, for any u C0 R d+1 ), κ 16 2, and r 0, 1/κ], we have r ϕt,x 1 ) ϕs,y 1 ) p dx 1 dt dy 1 ds r Nκ d+2 ψ p ) Q d 1 κr + N κ µp + κ d+2 a # κr) 1/2) ζ p ) Q d 1 κr,