Potential Anal (2007) 26:189–212 DOI 10.1007/s11118-006-9034-0
Second Order Elliptic Equations in Rd with Piecewise Continuous Coefficients Doyoon Kim
Received: 25 July 2006 / Accepted: 20 October 2006 / Published online: 20 December 2006 © Springer Science + Business Media B.V. 2006
Abstract The existence and uniqueness of solutions of second order elliptic differential equations in Rd are proved. The coefficients of second order terms are allowed to have discontinuity at finitely many parallel hyper-planes in Rd and the first derivatives of solutions can have jumps at the hyper-planes. Key words elliptic equation · discontinuous coefficients · apriori estimate · multiplier Mathematics Subject Classifications (2000) 35J15 · 35R05
1 Introduction The main purpose of this paper is to study the L p -theory of the differential equation (in nondivergence form) a jk (x)ux j xk (x) + b j(x)ux j (x) + c(x)u(x) = f (x) in
Rd
when the leading coefficients a jk are discontinuous at finitely many parallel hyperplanes in Rd . It is well known that the above equation has a unique strong solution (functions in Sobolev spaces) if the coefficients a jk are uniformly continuous. More precisely, if the coefficients a jk , b j, and c are bounded, and a jk are uniformly elliptic as well as uniformly continuous in Rd , then for any f ∈ L p (Rd ), 1 < p < ∞, there exists a unique solution u ∈ W 2p (Rd ) of the above equation as long as there exists a positive number ε such that c(x) < −ε for all x ∈ Rd . It is also well known that if
D. Kim (B) Department of Mathematics and Statistics, University of Ottawa, 585 King Edward Avenue, Ottawa, ON K1N 6N5, Canada e-mail:
[email protected]
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the coefficients a jk are only measurable, then the equation may not have a unique solution (see [6, 7]). Between the two assumptions – uniformly continuous and measurable – on the leading coefficients, there are many possibilities where we can establish the unique solvability of the equation. Some possible cases were investigated before. For example, A. Lorenzi [4, 5] considered elliptic equations in Rd with piecewise constant coefficients a jk , b j, and c. In [1, 2], coefficients a jk in the space of VMO are considered. The parabolic case with piecewise constant coefficients was dealt with by S. Salsa [8]. In this paper we consider weaker assumptions on the coefficients than those in [4, 5]. Especially, we allow the coefficients a jk to be discontinuous at finitely many parallel hyperplanes in Rd . Specifically, we divide the whole Euclidean space into several subdomains using finitely many parallel hyper-planes. Then we assume that the leading coefficients a jk are uniformly continuous only on each subdomain. The simplest case is when we have a differential operator L of the form + L = a+jk (x)D jk + b+j (x)D j + c+ (x) in Rd+ L≡ , (1) L− = a−jk (x)D jk + b−j (x)D j + c− (x) in Rd− where a+jk and a−jk are uniformly continuous in Rd+ and Rd− , respectively. Here Rd+ = {(x1 , x ) : x1 > 0, x ∈ Rd−1 } and Rd− = {(x1 , x ) : x1 < 0, x ∈ Rd−1 }. Note that the leading coefficients of the elliptic operator L are not necessarily continuous at the hyper-plane {(0, x ) : x ∈ Rd−1 }. With assumptions on a jk as above, we find a unique solution u ∈ W 2p (Rd ) of the equation Lu − λu = f where f ∈ L p (Rd ) and λ is sufficiently large. In addition, we find a solution the derivative (with respect to x1 ) of which satisfies given jump conditions at the hyperplanes. For example, if the simplest case is considered, there exists a unique solution u satisfying the following: For given f ∈ L p (Rd ) and ϕ(x ) ∈ 1−1/ p (Rd−1 ), 1 < p < ∞, there exists a unique function Wp u+ in Rd+ u= u− in Rd− satisfying u+ ∈ W 2p Rd+ , u− (0, x ) = u+ (0, x ), and
u− ∈ W 2p Rd− ,
+ u− x1 (0, x ) = γ (x )ux1 (0, x ) + ϕ(x ),
L+ u+ − λu+ = f − −
−
L u − λu = f
Rd+ in Rd− in
for sufficiently large λ > 0. Note that u± (0, x ) and u± x1 (0, x ) are the traces of the ± 2 d d−1 functions u ∈ W p (R± ) to the hyper-plane {(0, x ) : x ∈ R }. As easy to see, if γ (x ) = 1 and ϕ(x ) = 0, then u ∈ W 2p (Rd ). If the coefficients of L are constant on each half space, γ (x ) = 1, and ϕ(x ) = 0, then the above result is the same as in [5]. The outline of this paper is the following. In Section 2 we prove the solvability of the equation when the number of hyper-planes at which a jk are discontinuous is one.
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In Section 3 we prove Lemma 2.4, which is a key estimate of the paper. In Section 4 we extend the result in Section 2 so that we can have two or more hyper-planes unless the number of parallel hyper-planes is infinite. One may think that by using a standard argument (see, for example, [3]) we can extend the results in this paper to the case where the domain is bounded and the second order coefficients are discontinuous at a d − 1 dimensional manifold in the domain. In fact, if the coefficients are discontinuous at a d − 1 dimensional closed smooth manifold (possibly disconnected) inside the domain, then this is true. Otherwise, especially when the manifold intersects the boundary of the domain, this is a nontrivial problem, and we do not know much about how to solve it.
2 Second Order Elliptic Equations in Rd with the Second Order Coefficients Being Discontinuous at a Hyper-plane We consider the differential operator (1) where the coefficients a±jk (x), b±j (x), and c± (x) satisfy the following assumption. Assumption 2.1 (a) a+jk (x), b+j (x), c+ (x), j, k = 1, · · · , d, are real valued measurable functions defined on Rd+ and a−jk (x), b−j (x), c− (x), j, k = 1, · · · , d, are real valued measurable functions defined on Rd− . a±jk = a± kj . (b) There are positive constants κ and K such that
κ|ϑ|2 ≤
d
a±jk (x)ϑ jϑk ≤ κ −1 |ϑ|2 ,
(2)
j,k=1
|a±jk (x)|, |b ±j (x)|, |c± (x)| ≤ K for all x ∈ Rd± , ϑ ∈ Rd . (c) The leading coefficients a±jk are uniformly continuous on each half space. More precisely, there is an increasing function ω such that ω(ε) → 0 as ε 0 and | a+jk (x) − a+jk (y)| ≤ ω(|x − y|) for x, y ∈ Rd+ , | a−jk (x) − a−jk (y)| ≤ ω(|x − y|) for x, y ∈ Rd− , where j, k = 1, · · · , d. Let p ∈ (1, ∞) and λ ∈ [0, ∞). We assume that γ (x ) is a function satisfying Assumption 2.2 γ (x ) is a function defined on Rd−1 . There exist positive constants δ, κ1 , and K1 such that γ C1−1/ p+δ (Rd−1 ) ≤ K1 and κ1 ≤ γ (x ) ≤ κ1−1 for all x ∈ Rd−1 .
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In this section we prove that, for given f ∈ L p (Rd ) and ϕ(x ) ∈ W p there is a unique function u+ in Rd+ u= u− in Rd−
1−1/ p
satisfying u+ ∈ W 2p (Rd+ ), u− ∈ W 2p (Rd− ), and L− u− − λu− = f in Rd− , −
+
u (0, x ) = u (0, x ),
u− x1 (0, x )
L+ u+ − λu+ = f in Rd+ ,
= γ (x )u+ x1 (0, x ) + ϕ(x ).
(Rd−1 ),
(3)
Remark 2.3 If γ (x ) ≡ 1 and ϕ(x ) ≡ 0, then the solution u+ in Rd+ u= u− in Rd− of the differential Eq. 3 is in W 2p (Rd ) because u− (0, x ) = u+ (0, x ) and u− x1 (0, x ) = + ux1 (0, x ).
To deal with differential equations, as in Eq. 3, with conditions at a hyper-plane in Rd , we need trace theorems. In our case, we recall the well-known trace theorem (see [10]) which says that there is a bounded operator (the trace operator) from W 2p (Rd+ ) 2−1/ p 1−1/ p (Rd−1 ) × W p (Rd−1 ) such that to W p p 1−1/ p u(0, x ) W 2−1/ (Rd−1 ) + ux1 (0, x ) W p (Rd−1 ) ≤ N1 u W 2p (Rd+ ) p
(4)
for any u ∈ W 2p (Rd+ ), where N1 depends only on d and p. Moreover, there is 2−1/ p
an extension operator S from W p if v = S(ψ, ϕ) for ψ ∈ vx1 (0, x ) = ϕ(x ), and
2−1/ p (Rd−1 ) Wp
1−1/ p
(Rd−1 ) × W p
and ϕ ∈
(Rd−1 ) to W 2p (Rd+ ) such that
2−1/ p Wp (Rd−1 ),
then v(0, x ) = ψ(x ),
p 1−1/ p + ϕ v W 2p (Rd+ ) ≤ N2 ψ W 2−1/ d−1 d−1 (R ) Wp (R ) , p
(5)
where N2 depends only on d and p. The same results hold for W 2p (Rd− ). Assume that L in Eq. 1 is L0 of the form + in Rd+ L0 = a+jk Djk , L0 = − − L0 = a jk Djk in Rd− where a±jk , j, k = 1, · · · , d, are constant. With this L0 , we state the following lemma which is a key apriori estimate for solving Eq. 3. The proof is presented in Section 3.
Lemma 2.4 There is a constant N, depending only on d, K, p, and κ, such that, for any λ ≥ 0 and u ∈ C0∞ (Rd ), λ u L p (Rd ) + λ1/2 ux L p (Rd ) + uxx L p (Rd ) ≤ N L0 u − λu L p (Rd ) .
(6)
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Using the solvability of u − λu = f in Rd , Lemma 2.4, and the method of continuity, we can easily conclude Corollary 2.5 For any λ > 0 and f ∈ L p (Rd ), there is a unique function u ∈ W 2p (Rd ) satisfying in Rd+ L+ 0 u − λu = f . L− in Rd− 0 u − λu = f In addition, the estimate (6) holds true for the solution u ∈ W 2p (Rd ). Remark 2.6 The above result was proved in a slightly different way by A. Lorenzi in [5]. Now we obtain an apriori estimate for the solution u of Eq. 3 when the differential operator is L0 and γ (x ) is a constant γ . For shorter notations, set u± L p (Rd ) := u+ L p (Rd+ ) + u− L p (Rd− ) . We also set, for example, + − u± x L p (Rd ) = ux L p (Rd+ ) + ux L p (Rd− ) ,
(L± − λ)u± L p (Rd ) = (L+ − λ)u+ L p (Rd+ ) + (L− − λ)u− L p (Rd− ) . Let
s := 1 − 1/ p,
[ϕ]r :=
Rd−1
Rd−1
|ϕ(x ) − ϕ(y )| p dx dy |x − y |d−1+rp
1/ p ,
where 0 < r < 1. Lemma 2.7 Assume that γ is constant. Then there exist a constant N, depending only on d, K, p, κ, and κ1 , such that, for any λ ≥ 0 and u+ in Rd+ u= u− in Rd− satisfying u+ ∈ W 2p (Rd+ ), u− ∈ W 2p (Rd− ), and u− (0, x ) = u+ (0, x ), we have ± λ u± L p (Rd ) + λ1/2 u± x L p (Rd ) + uxx L p (Rd ) ± s/2 ≤ N (L± 0 − λ)u L p (Rd ) + λ ϕ L p (Rd−1 ) + [ϕ]s ,
(7)
+ where ϕ(x ) = u− x1 (0, x ) − γ ux1 (0, x ).
Proof Notice that by the multiplicative interpolation inequality and Young’s inequality, 1/2 λ1/2 u+ u+ L (Rd ) u+ x L p (Rd+ ) ≤ Nλ xx L p (Rd+ ) p + ≤ N λ u+ L p (Rd+ ) + u+ xx L p (Rd+ ) , 1/2
1/2
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where N is independent of u+ . The same inequality holds true with u− and Rd− in place of u+ and Rd+ , respectively. Thus we only need to prove λ u± L p (Rd ) + u± xx L p (Rd ) ± s/2 d) + λ d−1 ) + [ϕ]s . ≤ N (L± − λ)u ϕ L ( R L ( R 0 p p
(8)
If the above inequality holds true for λ > 0, then by letting λ → 0 we see that it holds true as well when λ = 0. Thus we prove the above inequality when λ > 0. Let us first assume that λ = 1. Set v = S(0, ϕ), where S is the extension operator in (5). The function v(x1 , x ) is 1−1/ p (Rd−1 ). Also set in W 2p (Rd+ ) because ϕ(x ) ∈ W p + + in Rd+ L+ 0u −u , f = − − L− in Rd− 0u −u fˆ(x1 , x ) = f (γ x1 , x ),
(x1 , x ) ∈ Rd+ ,
and ˆ +jk D jk , L+ 0,γ = a where + 2 aˆ + 11 = a11 /γ ,
+ aˆ + 1 j = a1 j /γ ,
j = 2, · · · , d,
aˆ +jk = a+jk ,
j, k = 2, · · · , d.
Consider a function uˆ defined by u+ (γ x1 , x ) + v(x1 , x ) in Rd+ ˆ 1, x ) = . u(x u− (x1 , x ) in Rd−
(9)
+ Since v(0, x ) = 0 and vx1 (0, x ) = ϕ(x ) = u− x1 (0, x ) − γ ux1 (0, x ), we have + u (γ x1 , x ) + v(x1 , x ) x1 =0 = u− (x1 , x ) x1 =0
and ∂ − ∂ (u+ (γ x1 , x ) + v(x1 , x )) x1 =0 = u (x1 , x ) x1 =0 . ∂ x1 ∂ x1 From this and the fact that u+ , v ∈ W 2p (Rd+ ) and u− ∈ W 2p (Rd− ), we see that uˆ ∈ W 2p (Rd ). By direct calculation, we also see that uˆ satisfies ˆ − uˆ = fˆ + L+ L+ in Rd+ 0,γ u 0,γ v − v . (10) ˆ − uˆ = f in Rd− L− 0u Thus, by the uniqueness of solution to the differential Eq. 10, the function uˆ is the solution of Eq. 10, and there is a constant N depending only on κ1 and N in (6) such that ˆ L p (Rd ) + uˆ xx L p (Rd ) ≤ N fˆ + L+ u (11) 0,γ v − v L p (Rd+ ) + f L p (Rd− ) .
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By combining this, (9), and (5), we have p u± L p (Rd ) + u± xx L p (Rd ) ≤ N f L p (Rd ) + ϕ W 1−1/ (Rd−1 ) , p
(12)
where N depends only K, κ1 , N2 in (5), and N in (11). Upon recalling that s = 1 − p 1/ p and ϕ W 1−1/ (Rd−1 ) = ϕ L p (Rd−1 ) + [ϕ]s , we see that (12) proves (8) when λ = 1. p To prove the inequality (8) for any λ > 0, consider
1 u¯ ± (x) = u± √ x . λ Then by the above argument, we have u¯ ± L p (Rd ) + u¯ ± xx L p (Rd ) ≤ N L0 u¯ ± − u¯ ± L p (Rd ) + ϕ ¯ L p (Rd−1 ) + [ϕ] ¯ s ,
(13)
¯+ where ϕ¯ = u¯ − x1 (0, x ) − γ u x1 (0, x ). Notice that
1 1 1 1 1 + 0, − γ u 0, . u− x x ϕ¯ = √ ϕ √ x = √ √ √ x1 x1 λ λ λ λ λ
Thus p
ϕ ¯ L p (Rd−1 ) =
√
d− p−1
λ
p
ϕ L p (Rd−1 )
and [ϕ] ¯ sp =
√ d−2 p p λ [ϕ]s .
We also have u¯ ± L p (Rd ) + u¯ ± xx L p (Rd ) = L0 u¯ ± − u¯ ± L p (Rd ) =
√ √
d λ p u± L p (Rd ) + λ−1 u± xx L p (Rd ) , d
λ p λ−1 L0 u± − λu± L p (Rd ) .
Therefore, the inequality (8) follows from (13). The lemma is proved.
We now prove the estimate (7) with L in place of L0 , where L is as in Eq. 1, but the coefficients a±jk , j, k = 1, · · · , d, are constant. This is done using the above lemma and the well-known interpolation inequality. Lemma 2.8 Assume that a±jk and γ are constant. Then there exist constants λ0 and N, depending only on d, K, p, κ, and κ1 , such that, for any λ > λ0 and u+ in Rd+ u= u− in Rd− satisfying u+ ∈ W 2p (Rd+ ), u− ∈ W 2p (Rd− ), and u− (0, x ) = u+ (0, x ), we have ± λ u± L p (Rd ) + λ1/2 u± x L p (Rd ) + uxx L p (Rd ) ≤ N (L± − λ)u± L p (Rd ) + λs/2 ϕ L p (Rd−1 ) + [ϕ]s , + where ϕ(x ) = u− x1 (0, x ) − γ ux1 (0, x ).
(14)
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Proof We only need to prove the estimate (8) with L in place of L0 . Notice that ± ± ± ± ± d d d d (L± − λ)u ≤ (L − λ)u + N u + u L p (R ) L p (R ) L p (R ) , x L p (R ) 0 where N depends only on d and K. Then it follows from Lemma 2.7 and the interpolation inequality that there exists a constant N1 , depending only on d, K, p, κ, and κ1 , such that ± ± s/2 d d d−1 ≤ N (L − λ)u + λ ϕ + [ϕ] λ u± L p (Rd ) + u± s L p (R ) L p (R ) xx L p (R ) 1 L (Rd ) + N1 u± L p (Rd ) . + u± 2 xx p This give us 2(λ − N1 ) u± L p (Rd ) + u± xx L p (Rd ) ≤ 2N (L± − λ)u± L p (Rd ) + λs/2 ϕ L p (Rd−1 ) + [ϕ]s . By observing that 2(λ − N1 ) ≥ λ, if λ ≥ 2N1 , we arrive at (8) with L and 2N in place
of L0 and N, respectively. This finishes the proof of this lemma. We extend the estimate (14) from the case where a±jk and γ are constants to the case where a±jk (x) and γ (x ) are as in Assumptions 2.1 and 2.2. To do this, we need the following observation. Lemma 2.9 If u ∈ W 1p (Rd+ ), then there exists a constant N depending only on d and p such that u(0, x ) L p (Rd−1 ) ≤ N u L
1−1/ p 1/ p . d u x L p (Rd+ ) p (R + )
(15)
From this and the multiplicative interpolation inequality, it follows that, for u ∈ W 2p (Rd+ ), 1
(1− 1p )
ux (0, x ) L p (Rd−1 ) ≤ N u L2
d p (R + )
1
(1+ 1p )
uxx L2
d p (R + )
,
where N depends only on d and p. Proof We know that (see [10])
u(0, ·) L p (Rd−1 ) ≤ N u L p (Rd+ ) + ux L p (Rd+ ) ,
where N depends only on d and p. Take a constant r ∈ (0, ∞) and put u(rx) in place of u, then we have r1/ p u(0, ·) L p (Rd−1 ) ≤ N u L p (Rd+ ) + r ux L p (Rd+ ) . By dividing both sides by r1/ p and minimizing with respect to r, we arrive at (15). The lemma is proved.
In the following lemma, we extend the estimate (14) to the case where L is as in Eq. 1 and γ (x ) is as in Assumption 2.2, but u has a support in a cylinder
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{(x1 , x0 ) : |x − x0 | < ε} for a sufficiently small ε. This is obtained by freezing γ (x ) 1−1/ p 1−1/ p as well as a±jk (x). Note that γ (x )ϕ(x ) ∈ W p (Rd−1 ) if ϕ(x ) ∈ W p (Rd−1 ) and γ (x ) ∈ C1−1/ p+δ (Rd−1 ). Lemma 2.10 There exists an ε = ε(d, K, K1 , p, δ, κ, κ1 , ω) > 0 such that if u+ in Rd+ u= u− in Rd− satisfies u+ ∈ W 2p (Rd+ ), u− ∈ W 2p (Rd− ), u− (0, x ) = u+ (0, x ), and supp u ⊂ {(x1 , x0 ) : |x − x0 | < ε} for some x0 ∈ Rd−1 , then there exist constant λ1 ≥ 1 and N, depending only on d, K, K1 , p, δ, κ, κ1 , and ω, such that, for any λ > λ1 , the estimate (14) holds true with γ (x ) + in place of γ so that ϕ(x ) = u− x1 (0, x ) − γ (x )ux1 (0, x ). Proof As in the proof of Lemma 2.8, it suffices to prove (8) with L in place of L0 . We can assume that x0 = 0 ∈ Rd−1 . Take an infinitely differentiable function η(x1 , x ) defined on Rd such that 1 if − ε/2 ≤ x1 ≤ ε/2 η ≥ 0, η(x1 , x ) = , 0 if x1 ≤ −ε or x1 ≥ ε where ε will be specified later. Set μ = 1 − η, then u± = ηu± + μu± . We prove this lemma by obtaining estimates for ηu± and μu± separately. For an estimate for μu± , we notice that since μu± ∈ W 2p (Rd± ) and μu± (0, x ) ≡ 0, there exist constants λ , N ∈ (0, ∞), depending only on d, K, p, κ, and ω, such that for any λ > λ , λ μu± L p (Rd ) + (μu± )xx L p (Rd ) ≤ N (L± − λ)(μu± ) L p (Rd ) .
(16)
Now we obtain an estimate for ηu± . Define γ0 = γ (0) and ± ± ± L± 0 = a jk (0)D jk + b (x)D j + c (x),
where a+jk (0) = lim a+jk (x1 , 0) x1 0
and a−jk (0) = lim a−jk (x1 , 0). x1 0
Then since ηu− (0, x ) = ηu+ (0, x ) and (ηu− )x1 (0, x ) − γ0 (ηu+ )x1 (0, x ) = ϕ0 (x ), + where ϕ0 (x ) = u− x1 (0, x ) − γ0 ux1 (0, x ), it follows by Lemma 2.8 that, for any λ > λ0 , λ ηu± L p (Rd ) + (ηu± )xx L p (Rd ) ± s/2 d) + λ d−1 ) + [ϕ0 ]s . ≤ N (L± − λ)(ηu ) ϕ 0 L ( R L ( R 0 p p
(17)
Now we have the following three observations for the last three terms of the above inequality. First, ± ± ± (L± 0 − λ)(ηu ) L p (Rd ) ≤ (L − λ)(ηu ) L p (Rd ) √ +Nω( 2ε) (ηu± )xx L p (Rd ) ,
(18)
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where N depends only on d. Second, λs/2 ϕ0 L p (Rd−1 ) ≤ λs/2 ϕ L p (Rd−1 ) +N
sup |γ (x ) − γ0 | λ ηu+ L p (Rd+ ) + (ηu+ )xx L p (Rd+ ) , (19)
x ∈B(0,ε)
where N depends only on d and p. Finally, [ϕ0 ]s ≤ [ϕ]s + N ε δ + sup |γ (x ) − γ0 | ηu+ L p (Rd+ ) + (ηu+ )xx L p (Rd+ ) , (20) x ∈B(0,ε)
where N depends only on d, K1 , p, and δ. The estimate (18) is straightforward. For the proof of (19), note that ϕ0 L p (Rd−1 ) ≤ ϕ L p (Rd−1 ) + (ϕ0 − ϕ) L p (Rd−1 ) ≤ ϕ L p (Rd−1 ) +
sup |γ (x ) − γ0 | u+ x1 (0, ·) L p (Rd−1 ) .
x ∈B(0,ε)
+ Since u+ x1 (0, x ) = (ηu )x1 (0, x ) and s = 1 − 1/ p, Lemma 2.9 and Young’s inequality show that + + λs/2 u+ x1 (0, ·) L p (Rd−1 ) ≤ N λ ηu L p (Rd+ ) + (ηu )xx L p (Rd+ ) ,
where N depends only on d and p. Combining the above two inequalities gives us (19). For the proof of (20), note that [ϕ0 ]s ≤ [ϕ]s + [ϕ0 − ϕ]s .
(21)
If we set σ (x ) = γ (x ) − γ0 , then last term of the above inequality is no greater than 1/ p
σ (x )u+ (0, x ) − σ (x )u+ (0, y )I{|x |<ε} p x1 x1 dx dy |x − y |d+ p−2 Rd−1 Rd−1
+
Rd−1
σ (x )u+ (0, y )I{|x |<ε} − σ (y )u+ (0, y ) p x1
x1
|x − y |d+ p−2
Rd−1
1/ p
dx dy
. (22)
Since u+ x1 (0, y ) = 0 for |y | ≥ ε, the pth power of the second term in (22) equals σ (x )u+ (0, y ) − σ (y )u+ (0, y ) p x1 x1 dx dy |x − y |d+ p−2 |y |<ε |x |<ε σ (y )u+ (0, x ) − σ (y )u+ (0, y ) p x1 x1 + dx dy . (23) |x − y |d+ p−2 |y |<ε |x |≥ε Using again the fact that u+ x1 (0, x ) = 0 for |x | ≥ ε, we see that the first term in (22) and the pth root of the second term in (23) are no greater than
sup |γ (x ) − γ0 | [u+ x1 (0, ·)]s .
x ∈B(0,ε)
Hence we have [ϕ0 − ϕ]s ≤ I 1/ p + 2 sup |γ (x ) − γ0 | [u+ x1 (0, ·)]s , x ∈B(0,ε)
(24)
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where I is the first term in (23). Recall that γ (x ) C1−1/ p+δ (Rd−1 ) ≤ K1 , where we can assume without loss of generality that 0 < 1 − 1/ p + δ < 1, thus |γ (x ) − γ (y )| ≤ K1 |x − y |1−1/ p+δ for any x , y ∈ Rd−1 . Then we have the following estimate for I. + u (0, y ) p |x − y |1−d+ pδ dx dy I ≤ K1 x1 |y |<ε
≤ K1
|y |<ε
|x |<ε
+ u (0, y ) p x1
p ≤ ε pδ N u+ x1 (0, ·) L
p (R
d−1 )
|x |<2ε
|x |1−d+ pδ dx dy
,
where N depends only on d, K1 , p, and δ. This and (24) imply that p [ϕ0 − ϕ]s ≤ N ε δ + sup |γ (x ) − γ0 | u+ x1 (0, ·) W 1−1/ (Rd−1 ) , p x ∈B(0,ε)
(25)
where N depends only on d, K1 , p, and δ. Moreover, using the fact that u+ x1 (0, x ) = (ηu+ )x1 (0, x ), the trace theorem, and the interpolation inequality, we have + + 1−1/ p d + (ηu ) xx d (0, ·) ≤ N ηu u+ d−1 L p (R + ) L p (R + ) , x1 Wp (R )
where N depends only on d and p. From this, (21), and (25), we obtain the inequality (20). Now the inequalities (18), (19), and (20) indicate that if we choose a sufficiently small ε, which depends on d, K, K1 , p, δ, κ, κ1 , and ω, then, for λ > max{λ0 , 1}, the inequality (17) can be replaced by λ ηu± L p (Rd ) + (ηu± )xx L p (Rd ) ≤ N (L± − λ)(ηu± ) L p (Rd ) + λs/2 ϕ L p (Rd−1 ) + [ϕ]s +
1 1 λ ηu+ L p (Rd+ ) + (ηu± )xx L p (Rd ) . 2 2
This proves that λ ηu± L p (Rd ) + (ηu± )xx L p (Rd ) ≤ 2N (L± − λ)(ηu± ) L p (Rd ) + λs/2 η(0, ·)ϕ L p (Rd−1 ) + [ϕ]s ,
(26)
where N is as in (17). Hence it follows from (16) and (26) that, for any λ > max{λ0 , λ , 1}, d ≤ N (L± − λ)(ηu± ) L p (Rd ) λ u± L p (Rd ) + u± xx L p (R ) + (L± − λ)(μu± ) L p (Rd ) + λs/2 ϕ L p (Rd−1 ) + [ϕ]s , where N depends only on d, K, p, κ, κ1 , and ω. To finish the proof, we use the argument in the proof of Lemma 2.8. Especially we need the sup norms of η, ηx , and ηxx . In the above argument actually we choose ε first, then choose η, thus we
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have to say that those sup norms depend on ε. This is why λ1 and N in the statement of this lemma depend on d, K, K1 , p, δ, κ, κ1 , and ω. The lemma is proved.
We finally prove the apriori estimate of solution u of Eq. 3. Lemma 2.11 There exist λ2 ≥ 1 and N, depending only on d, K, K1 , p, δ, κ, κ1 , and ω, such that, for any λ > λ2 and u+ in Rd+ u= u− in Rd− satisfying
u+ ∈ W 2p Rd+ ,
u− ∈ W 2p Rd− ,
u− (0, x ) = u+ (0, x ),
the estimate (14) holds true with γ (x ) in place of γ so that ϕ(x ) = u− x1 (0, x ) − + γ (x )ux1 (0, x ).
Proof Again it is enough to prove (8) with L in place of L0 . Take an infinitely differentiable function ζ defined on Rd−1 with ζ L p (Rd−1 ) = 1 and supp ζ ⊂ B(0, ε) ⊂ Rd−1 , where ε is as in Lemma 2.10. Then p p + + ux j xk (x1 , x )ζ (x − y ) dy , x1 > 0, ux j xk (x1 , x ) = Rd−1
p − (x , x ) ux j xk 1 =
Rd−1
p − ux j xk (x1 , x )ζ (x − y ) dy ,
x1 < 0.
Notice that p ± ux j xk (x1 , x )ζ (x − y ) = (u± (x1 , x )ζ (x − y ))x j xk − u± x j (x1 , x )ζxk (x − y ) p ± − u± (x , x )ζ (x − y ) − u (x , x )ζ (x − y ) 1 x 1 x x j j k xk p ≤ 2 p (u± (x1 , x )ζ (x − y ))x j xk p p +N u± (x1 , x ) + u± η(x − y ), x (x1 , x ) where η = |ζx | p + |ζx x | p ∈ L1 (Rd−1 ). Thus + p + u (u (·)ζ (· − y ))xx p d dy ≤ N xx L p (Rd ) L p (R ) +
+
Rd−1
p p +u+ L p (Rd ) + u+ x L p (R d ) . +
+
The same inequality holds with u− and Rd− in place of u+ and Rd+ , respectively. Since u± (x1 , x )ζ (x − y ) satisfy the assumptions in Lemma 2.10, we have ± ± (u (·)ζ (· − y ))xx (L − λ)(u± (·)ζ (· − y )) ≤ N d L p (R ) L p (R d ) ϕ(·)ζ (· − y ) L p (Rd−1 ) + [ϕ(·)ζ (· − y )]s
s/2
+λ
Second order elliptic equations in Rd with piecewise continuous coefficients
201
for λ > λ1 , where λ1 and N are as in Lemma 2.10. Hence ± p p u (L± − λ)(u± (·)ζ (· − y )) L p (Rd ) dy xx L p (Rd ) ≤ N Rd−1
+
Rd−1
p λ ps/2 ϕ(·)ζ (· − y ) L p (Rd−1 ) + [ϕ(·)ζ (· − y )]sp dy
p p + u± L p (Rd ) + u± x L p (R d ) .
(27)
Observe that (L± − λ)(u± (x1 , x )ζ (x − y )) = ζ (x − y )(L± − λ)u± (x1 , x ) + 2a±jk (x)u± x j (x1 , x )ζxk (x − y ) + u± (x1 , x ) a±jk (x)ζx j xk (x − y ) + b ±j (x)ζx j (x − y ) .
It follows then that |(L± − λ)(u± (x1 , x )ζ (x − y ))| p p ± p ≤ 2 p |ζ (x − y )(L± − λ)u± (x1 , x )| p + N(|u± x (x1 , x)| + |u (x1 , x)| )η(x − y ).
This gives us the following estimate. p (L+ − λ)(u+ (·)ζ (· − y )) L Rd−1
p ≤ N (L+ − λ)u+ (x) L
d p (R + )
dy
+ u+ L p
d p (R + )
+ u+ x L p
d p (R + )
d p (R + )
.
(28)
The same inequality holds with u− and Rd− in place of u+ and Rd+ , respectively. Also observe that p p ϕ(·)ζ (· − y ) L p (Rd−1 ) dy = ϕ L p (Rd−1 ) (29) Rd−1
and Rd−1
=
[ϕ(·)ζ (· − y )] p dy
Rd−1
Rd−1
Rd−1
ϕ(x )ζ (x − y ) − ϕ(z )ζ (z − y ) p |x − z |d+ p−2
dx dz dy
ϕ(x )ζ (x − y ) − ϕ(x )ζ (z − y ) p ≤2 dx dz dy |x − z |d+ p−2 Rd−1 Rd−1 Rd−1 ϕ(x )ζ (z − y ) − ϕ(z )ζ (z − y ) p p +2 dx dz dy |x − z |d+ p−2 Rd−1 Rd−1 Rd−1 p ≤ N ϕ L p (Rd−1 ) + [ϕ]sp . p
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From this, (27), (28), and (29), we have p p p ± ± ps/2 u± ϕ L p (Rd−1 ) + [ϕ]sp xx L p (Rd ) ≤ N (L − λ)u (x) L p (Rd ) + λ p p + u± L p (Rd ) + u± x L p (R d ) .
(30)
Similarly,
λ
p
p u± L p (Rd )
≤2 λ
u± (·)ζ (· − y ) L p (Rd ) dy p
p p
Rd−1
p p ≤ N (L± − λ)u± (x) L p (Rd ) + λ ps/2 ϕ L p (Rd−1 ) + [ϕ]sp p p + u± L p (Rd ) + u± x L p (R d ) .
(31)
Therefore, from (30) and (31) we have λ u± L p (Rd ) + u± xx L p (Rd ) ≤ N1 (L± − λ)u± L p (Rd ) + λs/2 ϕ L p (Rd−1 ) + [ϕ]s ± +N2 u± x L p (Rd ) + u L p (Rd ) , where N1 and N2 depend only on d, K, K1 , p, δ, κ, κ1 , and ω. Using again the argument in the proof of Lemma 2.8, we finish the proof.
For t ∈ [0, 1], let Lt = (1 − t) + tL and γt (x ) = (1 − t)γ0 + tγ (x ), where L and γ (x ) are as in Eq. 3 and γ0 is a constant between κ1 and κ1−1 . The proof of Lemma 2.7 actually shows that there is a unique solution to Eq. 3 when we have and γ0 in place of L and γ (x ). Hence, by making use of the above estimate and the method of continuity on Lt and γt (x ), we obtain 1−1/ p
Theorem 2.12 For any λ > λ2 , f ∈ L p (Rd ), and ϕ ∈ W p from Lemma 2.11, there exists a unique function u=
u+ u−
Rd+ in Rd− in
satisfying u+ ∈ W 2p (Rd+ ), u− ∈ W 2p (Rd− ), and Eq. 3.
(Rd−1 ), where λ2 is taken
Second order elliptic equations in Rd with piecewise continuous coefficients
203
3 Proof of Lemma 2.4 Lemma 2.4 can be proved using dilation and the results in [5], but we present here a slightly different and short proof. First we introduce some notations, which are from [4]. Set α ± (ξ ) = i
d
a± 1 jξ j,
ξ = (ξ2 , · · · , ξd ) ∈ Rd−1 ,
i=
√ −1,
j=2 d
β ± (ξ ) =
a±jk ξ jξk + 1,
j,k=2
and 2 ± H ± (ξ ) = α ± (ξ ) + a± 11 β (ξ ) ⎛ ⎞ ⎞2 ⎛ d d ⎠ + a± ⎝ = −⎝ a± a±jk ξ jξk + 1⎠ . 11 1 jξ j j=2
Also let z+ (ξ ) =
j,k=2
√ −α + (ξ ) − H + (ξ ) , a+ 11
z− (ξ ) =
√ −α − (ξ ) + H − (ξ ) . a− 11
Then we have the following estimate, the proof of which can be found in [4]. For completeness, we present its proof below. Lemma 3.1 H ± (ξ ) ≥ N(1 + |ξ |2 ),
ξ ∈ Rd−1 ,
where N depends only on κ. Hence [z+ (ξ )] ≤ −N(1 + |ξ |2 ) 2 1
and [z− (ξ )] ≥ N(1 + |ξ |2 ) 2 , 1
where N also depends only on κ. Proof First we consider H + (ξ ). Let ϑ = (t, ξ2 , · · · , ξd ), then by the first inequality in (2) we have 2 a+ 11 t + 2t
d j=2
a+ 1 jξ j +
d
a+jk ξ jξk ≥ κ(t2 + |ξ |2 ).
j,k=2
Since for ξ ∈ Rd−1 this inequality holds for every t, we must have ⎞2 ⎞ ⎛ ⎛ d d ⎠ − (a+ ⎝ ⎝ a+ a+jk ξ jξk − κ|ξ |2 ⎠ ≤ 0. 11 − κ) 1 jξ j j=2
j,k=2
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Hence ⎞2 ⎛ ⎞ ⎛ d d d + ⎝ + + 2 + ⎠ ⎠ a+ ξ + a a ξ ξ + 1 ≥ κa |ξ | + a + κ a+jk ξ jξk − κ 2 |ξ |2 −⎝ 11 11 11 1j j jk j k j=2
j,k=2
j,k=2
≥ min{κ, κ 2 }(1 + |ξ |2 ).
The remaining is proved similarly. Lemma 3.2 Let T be an operator such that = Th
z+ (ξ )
1 h, − z− (ξ )
h ∈ C0∞ (Rd−1 ),
and where Th h are Fourier transforms of Th and h in Rd−1 , respectively. Then, for each nonnegative integer k, there is a constant N, depending only on d, K, p, k, and κ, such that Th W k+1 d−1 ) ≤ N h W k (Rd−1 ) . p p (R d−1 ). Hence T can be extended to a bounded operator from W kp (Rd−1 ) into W k+1 p (R
Proof Let
K(ξ ) =
1 . z+ (ξ ) − z− (ξ )
First notice that by Lemma 3.1 this is well-defined. By multiplier theorems (in particular, in [9]) we only need to show that, for any multi-index α, there exists a constant N, depending only on K, α, and κ, such that |Dαξ K(ξ )| ≤ N(1 + |ξ |2 )−
|α|+1 2
for all ξ ∈ Rd−1 . This can be shown by using Lemma 3.1, direct calculation, and induction. The lemma is proved.
Proof of Lemma 2.4 Since we can use dilation, we only prove (6) when λ = 1. That is, we prove u W 2p (Rd ) ≤ N f L p (Rd ) for u ∈ C0∞ (Rd ), where f (x) = L0 u − u. Let g(x ) = u(0, x ). Then g(x ) ∈ C0∞ (Rd−1 ) and there exist unique solutions v ± ∈ W 2p (Rd± ) and w ± ∈ W 2p (Rd± ) of differential equations + + L0 v − v + = 0 in Rd+ , v + (0, x ) = g(x )
+ + L+ 0w −w = f w + (0, x ) = 0
in Rd+
,
(32)
Second order elliptic equations in Rd with piecewise continuous coefficients
and
− − d L− 0 v − v = 0 in R− , − v (0, x ) = g(x )
− − L− 0w −w = f − w (0, x ) = 0
205
in Rd−
.
(33)
Notice that w + W 2p (Rd+ ) ≤ N f L p (Rd ) ,
w − W 2p (Rd− ) ≤ N f L p (Rd ) ,
and v + W 2p (Rd+ ) ≤ N g
2− 1p
Wp
(Rd−1 )
,
v − W 2p (Rd− ) ≤ N g
2− 1p
Wp
(Rd−1 )
,
(34)
where N depends only on d, K, p, and κ. The inequalities in (34) can be proved by using the extension S(g, 0) ∈ W 2p (Rd ) of g, where S is in (5). Also notice that v + + w + in Rd+ (35) u= v − + w − in Rd− and u W 2p (Rd ) ≤ v + W 2p (Rd+ ) + w + W 2p (Rd+ ) + v − W 2p (Rd− ) + w − W 2p (Rd− ) . Therefore, to finish the proof we only need to show that g
2− 1p
Wp
(Rd−1 )
≤ N f L p (Rd ) ,
(36)
where N depends only on d, K, p, and κ. From (35) we have ux1 (0, x ) = vx+1 (0, x ) + wx+1 (0, x ) = vx−1 (0, x ) + wx−1 (0, x ). This shows that vx+1 (0, x ) − vx−1 (0, x ) = wx−1 (0, x ) − wx+1 (0, x ). 1−
Let h(x ) = wx−1 (0, x ) − wx+1 (0, x ). Then h ∈ W p h
1− 1p
Wp
(Rd−1 )
1 p
(37)
(Rd−1 ) and
≤ N f L p (Rd ) ,
where N depends only on d, K, p, κ, and N1 in (4). Hence the inequality (36) holds true if we prove g
2− 1p
Wp
(Rd−1 )
≤ N h
1− 1p
Wp
(Rd−1 )
,
(38)
where N depends only on d, K, p, and κ. To prove the above inequality we first observe that, as shown in Lemma 3.3 below, + + (x , ξ ) = g(ξ )ex1 z (ξ ) v 1
− − (x , ξ ) = and v g(ξ )ex1 z (ξ ) , 1
± (x , ξ ) and where v g(ξ ) are the Fourier transforms of v ± (x1 , x ) and g(x ) with 1 respect to x , respectively. Then we take Fourier transforms of both sides of (37) so that we have
(z+ (ξ ) − z− (ξ )) g(ξ ) = h(ξ ).
(39)
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D. Kim +
Indeed, since g(ξ )ex1 z (ξ ) ∈ S(Rd−1 ), where S(Rd−1 ) is the Schwartz space of all complex valued rapidly decreasing infinitely differentiable functions on Rd−1 , we have + g(ξ )ex1 z (ξ ) dξ v + (x1 , x ) = cd−1 eix ·ξ Rd−1
and vx+1 (0, x )
= lim cd−1 h→0
Rd−1
= cd−1
+
Rd−1
e
ix ·ξ
g(ξ )
ehz
(ξ )
−1
h
dξ
g(ξ )z+ (ξ ) dξ , eix ·ξ
where cd−1 is an appropriate constant. The same argument holds for vx−1 (0, x ). Hence Eq. 39 follows. Now we let T be an operator such that 1 h z+ (ξ ) − z− (ξ )
= Th
for h ∈ C0∞ (Rd−1 ). Then, as proved in Lemma 3.2, T is extended to a bounded d−1 operator from W kp (Rd−1 ) into W k+1 ), k = 0, 1, satisfying p (R Th W k+1 d−1 ) ≤ N h W k (Rd−1 ) , p p (R
k = 0, 1,
where N depends only on d, K, p, and κ. Moreover, by interpolation arguments, T 1− 1p
is extended to a bounded operator from W p Th
2− 1p
Wp
(Rd−1 )
2− 1p
(Rd−1 ) into W p
≤ N h
1− 1p
Wp
(Rd−1 )
(Rd−1 ) such that
,
(40)
where N depends only on d, K, p, and κ. Notice that, for h and g in (38), the Eq. 39 gives us Th = g.
This and (40) prove (38). This finishes the proof. Lemma 3.3 For v ± in Eqs. 32 and 33, + + (x , ξ ) = v g(ξ )ex1 z (ξ ) 1
− − (x , ξ ) = and v g(ξ )ex1 z (ξ ) , 1
(41)
± (x , ξ ) and g(ξ ) are the Fourier transforms of v ± (x1 , x ) and g(x ) with where v 1 respect to x , respectively.
Proof If L0 = , then we have an exact expression for v + (x1 , x ), which is ∞ x21 ei t x1 + v (x1 , x ) = cd g(y ) dt dy , d+1 Rd−1 −∞ (|x − y |2 + x21 (t2 + 1)) 2 where c−1 d
=
Rd
1 (|y|2
+ 1)
d+1 2
dy.
Second order elliptic equations in Rd with piecewise continuous coefficients
207
Calculating the Fourier transform of this v + (x1 , x ) with respect to x , we have √ 2 + (x , ξ ) = v g(ξ )e−x1 1+|ξ | . 1 For the given L0 , using a change of coordinates, we prove Eq. 41. The lemma is proved.
4 Second Order Elliptic Equations in Rd with the Second Order Coefficients Being Discontinuous at Two Different Hyper-planes We can extend the results in Section 2 to the case where the coefficients of the second order terms of the differential operator are discontinuous at finitely many parallel hyper-planes. In this section we consider especially the case where the number of such hyper-planes is two. Take two real numbers s and t such that −∞ < s < t < ∞ and set
Rds = {(x1 , x ) : x1 < s, x ∈ Rd−1 }, Rdm = {(x1 , x ) : s < x1 < t, x ∈ Rd−1 }, Rdt = {(x1 , x ) : x1 > t, x ∈ Rd−1 }. Let λ be a positive real number and γi (x ), i = 1, 2, be functions on Rd−1 satisfying Assumption 2.2. Also let L be a differential operator such that ⎧ ˆ = aˆ jk (x)D jk + bˆ j(x)D j + c(x) ⎪ ˆ L in Rds ⎪ ⎨ ¯ = a¯ jk (x)D jk + b¯ j(x)D j + c(x) L≡ L ¯ in Rdm , ⎪ ⎪ ⎩ ˇ ˇ L = aˇ jk (x)D jk + bˇ j(x)D j + c(x) in Rdt ˆ L, ¯ and where the coefficients are as in Assumption 2.1, that is, the coefficients of L, d d d ˇ L are bounded measurable functions defined on Rs , Rm , and Rt , respectively, and the coefficients aˆ jk , a¯ jk , aˇ jk satisfy the ellipticity condition (2). In addition, aˆ jk , a¯ jk , and aˇ jk are uniformly continuous in the domains where they are defined. (In this section we use the same κ, K, and ω as in Assumption 2.1.) With these assumptions we investigate the solvability of the differential equation ⎧ d d d ˆ ¯ ˇ ⎪ ⎨ Luˆ − λuˆ = f in Rs , Lu¯ − λu¯ = f in Rm , Luˇ − λuˇ = f in Rt , (42) ˆ x ) = u(s, ¯ x ), u(s, uˆ x1 (s, x ) = γ1 (x )u¯ x1 (s, x ) + ϕ1 (x ), ⎪ ⎩ ¯ x ) = u(t, ˇ x ), u(t, u¯ x1 (t, x ) = γ2 (x )uˇ x1 (t, x ) + ϕ2 (x ), (Rd−1 ), i = 1, 2, and f ∈ L p (Rd ) are arbitrarily given. where ϕi (x ) ∈ W p Let μ be an infinitely differentiable function on Rd such that 1 if x1 ≤ s + 13 (t − s), 0 ≤ μ ≤ 1, μ(x1 , x ) = 0 if x1 ≥ s + 23 (t − s). √ √ Set η1 = μ and η2 = 1 − μ, then η12 + η22 = 1. We also take infinitely differentiable functions νi , i = 1, 2 on Rd such that 0 ≤ νi ≤ 1, νi = 1 on supp ηi , and 1−1/ p
ν1 = 0
if
5 x1 ≥ s + (t − s), 6
ν2 = 0
if
1 x1 ≤ s + (t − s). 6
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Set Rdsc = {(x1 , x ) : x1 > s, x ∈ Rd−1 }, Rdtc = {(x1 , x ) : x1 < t, x ∈ Rd−1 } and define Li , i = 1, 2 to be differential operators defined on Rdsc and Rdtc , respectively, such that ¯ + (1 − ν1 ) , L1 = ν1 L
¯ + (1 − ν2 ) . L2 = ν2 L
Then by Theorem 2.12, for any λ > λ2 , g− ∈ L p (Rds ), g+ ∈ L p (Rdsc ), h− ∈ L p (Rdtc ), 1−1/ p (Rd−1 ), i = 1, 2, there exist unique solutions of the h+ ∈ L p (Rdt ), and ϕi ∈ W p differential equations − ˆ Lv − λv − = g− in Rds , L1 v + − λv + = g+ in Rdsc , (43) − + − v (s, x ) = v (s, x ), vx1 (s, x ) = γ1 (x )vx+1 (s, x ) + ϕ1 (x ), and
ˇ + − λw + = h+ in Rdt , Lw L2 w − − λw − = h− in Rdtc , − + − w (t, x ) = w (t, x ), wx1 (t, x ) = γ2 (x )wx+1 (t, x ) + ϕ2 (x ).
(44)
Let R(i) λ , i = 1, 2 be the inverses of the differential operators in Eqs. 43 and 44, respectively. More precisely, (1,−) Rλ (g− , g+ ; ϕ1 ) in W 2p (Rds ) (1) − + Rλ (g , g ; ϕ1 ) = (g− , g+ ; ϕ1 ) in W 2p (Rdsc ) R(1,+) λ is the unique solution of Eq. 43 and − R(2) λ (h ,
+
h ; ϕ2 ) =
(h− , h+ ; ϕ2 ) R(2,−) λ
in
W 2p (Rdtc )
(h− , h+ ; ϕ2 ) R(2,+) λ
in
W 2p (Rdt )
is the unique solution of Eq. 44. It can be easily checked that, for gi− ∈ L p (Rds ), gi+ ∈ L p (Rdsc ), hi− ∈ L p (Rdtc ), hi+ ∈ L p (Rdt ), i = 1, 2, − + − + − g1 , g1 ; ϕ1 − R(1) g2 , g2 ; ϕ1 = R(1) g1 − g2− , g1+ − g2+ ; 0 , R (1) λ λ λ − + − + − + + h1 , h1 ; ϕ1 − R(2) h2 , h2 ; ϕ1 = R(2) h1 − h− (45) R (2) λ λ λ 2 , h 1 − h2 ; 0 . ¯ (i) u¯ = u(¯ ¯ a jk ηi x j xk + b¯ jηi x j ) + 2¯a jk ηi x j u¯ xk , i = 1, 2. That is, For u¯ ∈ W 1p (Rdm ), let L (i) 2 ¯ ¯ ¯ ¯ − ηi Lu¯ if u¯ ∈ W p (Rdm ). L u¯ = L(ηi u) Lemma 4.1 Let λ > λ2 , where λ2 is as in Lemma 2.11, and ⎧ d ⎪ ⎨ uˆ in Rs u = u¯ in Rdm ⎪ ⎩ uˇ in Rdt satisfy uˆ ∈ W 2p (Rds ), u¯ ∈ W 2p (Rdm ), uˇ ∈ W 2p (Rdt ), and Eq. 42. Then it holds true that ⎧ (1,−) ¯ (1) ⎪ ⎨ uˆ = Rλ ( f, η1 f + L u¯ ; ϕ1 ) (1,+) ¯ (1) u¯ ; ϕ1 ) + η2 R(2,−) (η2 f + L ¯ (2) u, ¯ f ; ϕ2 ) u¯ = η1 Rλ ( f, η1 f + L λ ⎪ ⎩ (2,+) ¯ (2) u, ¯ f ; ϕ2 ) uˇ = Rλ (η2 f + L
in Rds in Rdm , (46) in Rdt
Second order elliptic equations in Rd with piecewise continuous coefficients
209
¯ (i) u, ¯ i = 1, 2, are defined to be zero outside supp ηi , thus functions on where ηi f + L
Rdsc and Rdtc , respectively. Proof Extend η1 u¯ by zero outside the support of η1 so that it is a function on Rdsc . Then η1 u¯ ∈ W 2p (Rdsc ) and ¯ u¯ − λu) ¯ (1) u¯ = η1 f + L ¯ (1) u¯ in ¯ − λη1 u¯ = η1 ( L ¯ +L L1 (η1 u)
Rdsc .
ˆ uˆ − λuˆ = f in Rds . In addition, uˆ and η1 u¯ satisfy We also have that uˆ ∈ W 2p (Rds ) and L ˆ x ) = (η1 u)(s, ¯ u(s, x ) and
¯ x1 (s, x ) + ϕ1 (x ). uˆ x1 (s, x ) = γ1 (x )(η1 u)
Hence, by uniqueness argument on the differential equation as in (43), we have ¯ (1) u; ¯ ϕ1 ), ( f, η1 f + L uˆ = R(1,−) λ
¯ (1) u; ¯ ϕ1 ). η1 u¯ = R(1,+) ( f, η1 f + L λ
By the same reasoning ¯ (2) u, ¯ f ; ϕ2 ), (η2 f + L η2 u¯ = R(2,−) λ
¯ (2) u, ¯ f ; ϕ2 ). uˇ = R(2,+) (η2 f + L λ
¯ + η2 (η2 u). ¯ The lemma is proved. It only remains to notice that u¯ = η1 (η1 u)
In the following, we prove the inverse of the above lemma. Lemma 4.2 There exists λ3 ≥ 1, depending only on t − s, d, K, K1 , p, δ, κ, κ1 , and ω, 1−1/ p (Rd−1 ), i = 1, 2, if such that, for any λ > λ3 , f ∈ L p (Rd ), and ϕi ∈ W p ⎧ d ⎪ ⎨ uˆ in Rs u = u¯ in Rdm ⎪ ⎩ uˇ in Rdt satisfies uˆ ∈ W 2p (Rds ), u¯ ∈ W 2p (Rdm ), uˇ ∈ W 2p (Rdt ), and Eq. 46, then it is a solution of Eq. 42. ˆ and ηi , i = 1, 2, the functions u, Proof First notice that by the definitions of R(i,±) λ ¯ and uˇ satisfy the conditions at hyper-planes {(s, x ) : x ∈ Rd−1 } and {(t, x ) : x ∈ u, Rd−1 }. Let g := L¯ u¯ − λu¯ in Rdm , then we also notice that Lˆ uˆ − λuˆ = f in Rds , L¯ u¯ − ˇ uˇ − λuˇ = f in Rdt . Hence u, ¯ u¯ − λu¯ = ˆ u, ¯ and uˇ satisfy Eq. 42 with L λu¯ = g in Rdm , and L ¯ g in place of Lu¯ − λu¯ = f . Therefore, to prove this lemma we only need to show that f = g in Rdm . ¯ u¯ − λu¯ = g in ˆ u, ¯ and uˇ satisfy Eq. 42 with L Due to Lemma 4.1 and the fact that u, ¯ place of Lu¯ − λu¯ = f , we have ¯ (1) u¯ ; ϕ1 ) + η2 R(2,−) (η2 g + L ¯ (2) u, ¯ f ; ϕ2 ) in Rdm . u¯ = η1 R(1,+) ( f, η1 g + L λ λ ˆ u, ¯ and uˇ satisfy Eq. 46, we have On the other hand, since u, ¯ (1) u¯ ; ϕ1 ) + η2 R(2,−) (η2 f + L ¯ (2) u, ¯ f ; ϕ2 ) u¯ = η1 R(1,+) ( f, η1 f + L λ λ
in Rdm .
The above two equalities and Eq. 45 imply that (0, η1 ( f − g); 0) + η2 Rλ(2,−) (η2 ( f − g), 0; 0) in 0 = η1 R(1,+) λ
Rdm .
(47)
210
D. Kim
Let v0+ := R(1,+) (0, η1 ( f − g); 0) and w0− := R(2,−) (η2 ( f − g), 0; 0). Then by the defλ λ ¯ 1 v + ) = L1 (η1 v + ) and L(η ¯ 2 w − ) = L2 (η2 w − ). initions of Li , i = 1, 2, we see that L(η 0 0 0 0 d Thus it follows from Eq. 47 that in Rm ¯ − λ)(η1 v + + η2 w − ) 0 = (L 0 0 ¯ (1) v + + η2 (L2 − λ)w − + L ¯ (2) w − = η1 (L1 − λ)v0+ + L 0 0 0 ¯ (1) v + + L ¯ (2) w − . = f −g+L 0 0 ¯ (i) , i = 1, 2, do not include second order derivatives imply This and the fact that L ¯ (1) v + + L ¯ (2) w − L (Rd ) g − f L p (Rdm ) = L 0 0 p m + ≤ N v0 W 1p (Rdc ) + w0− W 1p (Rdc ) , s
s
(48)
where N only depends on K and t − s. By applying the estimate in Lemma 2.11, we have v0+ W 1p (Rdc ) + w0− W 1p (Rdc ) ≤ λ−1/2 N f − g L p (Rdm ) , s
s
(49)
where N depends only on d, K, K1 , p, δ, κ, κ1 , and ω. Now we combine (48) and (49), then it follows that f = g for all sufficiently large λ. This finishes the proof.
ˆ u, ¯ and uˇ satisfying the Eq. 46, then we This lemma indicates that if we can find u, can prove the solvability of the Eq. 42. Thus we need the existence and uniqueness of the solution of Eq. 46. Lemma 4.3 There exists λ3 ≥ 1, depending only on t − s, d, K, K1 , p, δ, κ, κ1 , and ω, 1−1/ p (Rd−1 ), i = 1, 2, there exists a such that, for any λ > λ3 , f ∈ L p (Rd ), and ϕi ∈ W p unique function ⎧ ⎪ ⎨ uˆ u = u¯ ⎪ ⎩ uˇ
in Rds in Rdm in Rdt
satisfying uˆ ∈ W 2p (Rds ), u¯ ∈ W 2p (Rdm ), uˇ ∈ W 2p (Rdt ), and Eq. 46. In addition, ¯ L p (Rdm ) + u ˇ L p (Rdt ) ˆ L p (Rds ) + u λ u ˆ W 1p (Rds ) + u ¯ W 1p (Rdm ) + u ˇ W 1p (Rdt ) +λ1/2 u p ϕi W 1−1/ ≤ N f L p (Rd ) + λs/2 d−1 ) , ( R p i=1,2
where s = 1 − 1/ p and N depends only on t − s, d, K, K1 , p, δ, κ, κ1 , and ω.
(50)
Second order elliptic equations in Rd with piecewise continuous coefficients
211
Proof Let λ > λ2 , where λ2 is the one in Lemma 2.11, and u˜ ∈ W 2p (Rdm ). Then for 1−1/ p
given f ∈ L p (Rd ), ϕi ∈ W p (Rd−1 ), i = 1, 2, we consider ⎧ (1,−) ¯ (1) in Rds ⎪ ⎨ uˆ = Rλ ( f, η1 f + L u˜ ; ϕ1 ) (1,+) (2,−) ¯ (1) u˜ ; ϕ1 ) + η2 R ¯ (2) u, ˜ f ; ϕ2 ) in Rdm . u¯ = η1 Rλ ( f, η1 f + L (η2 f + L λ ⎪ ⎩ (2,+) ¯ (2) u, ˜ f ; ϕ2 ) uˇ = Rλ (η2 f + L in Rdt ¯ (i) , i = 1, 2, it follows easily By the estimate in Lemma 2.11 and the definitions of L that ¯ L p (Rdm ) + u ˇ L p (Rdt ) ˆ L p (Rds ) + u λ u ¯ W 1p (Rdm ) + u ˇ W 1p (Rdt ) ˆ W 1p (Rds ) + u + λ1/2 u
s/2 p ˜ W 1p (Rdm ) + λ ≤ N f L p (Rd ) + u ϕi W 1−1/ (51) (Rd−1 ) , p i=1,2
where s = 1 − 1/ p and N depends only on t − s, d, K, K1 , p, δ, κ, κ1 , and ω. This estimate and the contraction argument prove the lemma. Indeed, to finish the proof we can proceed as follows. The uniqueness is proved easily. To prove the existence, take u¯ 0 = 0 in Rdm and, for λ > λ2 , set (1,−) ¯ (1) u¯ n ; ϕ1 ), v + = R(1,+) ( f, η1 f + L ¯ (1) u¯ n ; ϕ1 ), ( f, η1 f + L uˆ − λ n+1 = Rλ n+1 − ¯ (2) u¯ n , f ; ϕ2 ), uˇ n+1 = R(2,+) (η2 f + L ¯ (2) u¯ n , f ; ϕ2 ), (52) = R(2,−) (η2 f + L wn+1 λ λ
and + − u¯ n+1 = η1 vn+1 + η2 wn+1
in Rdm ,
(53)
where n = 0, 1, · · · . Then by Eq. 45 we have ¯ (1) (u¯ n − u¯ n−1 ) ; 0) uˆ n+1 − uˆ n = Rλ(1,−) (0, L ¯ (1) (u¯ n − u¯ n−1 ) ; 0) + η2 R(2,−) ( L ¯ (2) (u¯ n − u¯ n−1 ), 0 ; 0) (0, L u¯ n+1 − u¯ n = η1 R(1,+) λ λ ¯ (2) (u¯ n − u¯ n−1 ), 0 ; 0) uˇ n+1 − uˇ n = Rλ(2,+) ( L for n ≥ 1. This and the estimate (51) allow us to take a λ3 (≥ λ2 ) such that, for all λ > λ3 , {uˆ n }, {u¯ n }, and {uˇ n } defined in Eqs. 52 and 53 are Cauchy sequences in W 1p (Rds ), W 1p (Rdm ), and W 1p (Rdt ), respectively. Let the limits of these sequences be uˆ ∈ W 1p (Rds ), u¯ ∈ W 1p (Rdm ), and uˇ ∈ W 1p (Rdt ). Then they satisfy Eq. 46. Moreover, they are actually in W 2p (Rds ), W 2p (Rdm ), and W 2p (Rdt ), respectively. Notice that u¯ ∈ W 1p (Rdm ) and by the , i = 1, 2, definitions of R(i,±) λ ¯ (1) u¯ ; ϕ1 ) ∈ W 2p (Rds ), R(1,+) ( f, η1 f + L ¯ (1) u¯ ; ϕ1 ) ∈ W 2p (Rdsc ), ( f, η1 f + L R(1,−) λ λ ¯ (2) u, ¯ (2) u, ¯ f ; ϕ2 ) ∈ W 2p (Rdtc ), R(2,+) ¯ f ; ϕ2 ) ∈ W 2p (Rdt ). R(2,−) (η2 f + L (η2 f + L λ λ These observations and the equalities in (46) prove that uˆ ∈ W 2p (Rds ), u¯ ∈ W 2p (Rdm ), and uˇ ∈ W 2p (Rdt ).
212
Finally, the estimate (50) follows easily. The lemma is proved.
D. Kim
By combining the above lemmas we obtain the following theorem. Theorem 4.4 Take λ3 which is bigger one between λ3 s in Lemma 4.2 and 4.3. Then 1−1/ p (Rd−1 ), i = 1, 2, there exists a unique for any λ > λ3 , f ∈ L p (Rd ), ϕi ∈ W p ⎧ d ⎪ ⎨ uˆ in Rs u = u¯ in Rdm ⎪ ⎩ uˇ in Rdt satisfying uˆ ∈ W 2p (Rds ), u¯ ∈ W 2p (Rdm ), uˇ ∈ W 2p (Rdt ), and Eq. 42. Remark 4.5 Since we have the estimate as in (50), by using the same reasoning as in this section we can extend the result inductively to the case where the coefficients of second order terms are discontinuous at finitely many parallel hyper-planes. Acknowledgement I wish to thank N.V. Krylov for suggesting this problem and providing valuable comments on this manuscript.
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