c Pleiades Publishing, Ltd., 2008. ISSN 0001-4346, Mathematical Notes, 2008, Vol. 83, No. 1, pp. 3–13. c R. A. Amanov, F. I. Mamedov, 2008, published in Matematicheskie Zametki, 2008, Vol. 83, No. 1, pp. 3–13. Original Russian Text
Regularity of the Solutions of Degenerate Elliptic Equations in Divergent Form R. A. Amanov1 and F. I. Mamedov2* 1
Institute of Mathematics and Mechanics, National Academy of Sciences of Azerbaijan 2 Dicle University, Diyarbakir, Turkey Received July 14, 2006
Abstract—A priori estimates of the solution to the Dirichlet problem and of its first derivatives in terms of weighted Lebesgue norms are obtained for linear and quasilinear equations with degeneracy from Ap Muckenhoupt classes. DOI: 10.1134/S000143460801001X Key words: elliptic equation of divergence form, Dirichlet problem, Lipschitz condition, Lebesgue norm, Lebesgue measure, Holder’s ¨ inequality.
1. INTRODUCTION Suppose that D is an arbitrary domain in Rn , n ≥ 1, and L is an operator of the form L=
n ∂ (Ai (x, u, ux )), ∂xi
(1)
i=1
where x = (x1 , x2 , . . . , xn ), ux = (∂u/∂x1 , ∂u/∂x2 , . . . , ∂u/∂xn ), and the Ai (x, ξ, η) are measurable functions such that, for all ξ ∈ R1 , x, η ∈ Rn , the following inequality holds: n
Ai (x, ξ, η)ηi ≥ ω|η|p ,
1 ≤ p < ∞.
(2)
i=1
Here and elsewhere, v, ω are measurable almost everywhere bounded and positive functions. We study generalized solutions of the Dirichlet problem n ∂fi in D, Lu = f0 + ∂xi
(3)
i=1
u|∂D = 0,
(4)
and prove a number of theorems estimating the solutions of this problem via the right-hand side of the equation. The questions examined in this paper in the unweighted case (v ≡ ω ≡ 1) were studied by Stampacchia [1], Weinberger [2], Chicco [3], and Mazya [4] for linear equations (the case p = 2) and by Ladyzhenskaya and Uraltseva [5, p. 340], Talenti [6], Drabek, Kufner, and Nicolosi [7, pp. 49, 111], etc. for quasilinear equations. For linear equations with summable lower coefficients, an L∞ -estimate of the solution was obtained in [8] under a smallness condition for the measure of the domain D. In Mat. Zametki [Math. Notes] referred to above, the equation does not degenerate (ω, v ≡ 1). The results of the present paper are related to the description of conditions on the functions ω, v1 , v2 , and the exponents 1 ≤ p1 , p2 < ∞ for which one has inequalities between the Lp1 (D, v1 )- and L∞ (D)norms of the solution (and its derivatives) and the Lp2 (D, v2 )-norm of the functions f0 , f1 , . . . , fn . For *
E-mail:
[email protected].
3
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AMANOV, MAMEDOV
linear equations without degeneracy such estimates were applied [4] to the proof of the solvability of the Dirichlet problem in the space Wq1 (D), 1 < q < 2, for the case in which the right-hand side does not possess sufficient summability (fi ≡ 0, f0 ∈ Lp (D), 1 < p < 2n/(n + 2)). Denote by Lp (D, v) the space of measurable functions in the domain D with finite norm 1 ≤ p < ∞; f Lp (D,ω) = ( |f |p ω dx)1/p , D
set Lp (D) = Lp (D, 1). By p denote the number conjugate to p ∈ (1, ∞), i.e., 1/p + 1/p = 1 (p = 1 n for p = ∞, p = ∞ for p = 1). For a measurable set E ⊂ R and an integrable functions v on E, denote v(E) = E v dx. The expression 1/p 1−p ω dx D
for p = 1 is regarded as ess supx∈D ω −1 (x). For 1 < p < ∞, the condition ω ∈ Ap implies p−1 1−p ω dy ω dy ≤ Cp ρnp Qx ρ
Qx ρ
for any ball Qxρ = {y ∈ Rn : |y − x| < ρ}, where Cp is a constant independent of x, ρ. It is well known [9] that C0∞ (D) is dense in Lp (D, ω) if ω ∈ Ap . Suppose that Lip(D) is a function space satisfying the Lipschitz condition in the domain D and Lip0 (D) is the subspace of functions ¯ compactly supported in the domain D. By W 1,p (D, ω) (W 1,p (D, ω)) denote the closure from Lip(D) 0 Lip(D) (Lip0 (D)) in the norm n ∂f 1,p . f ; W (D, ω) = f Lp (D,ω) + ∂xi p L (D,ω) i=1
1,p (D, ω) (W 1,p (D, ω)) if A sequence {fj }∞ 0 j=1 is said to approximate f ∈ W
fj ∈ Lip(D) (fj ∈ Lip0 (D))
and
fj − f ; W 1,p (D, ω) → 0 as j → ∞.
Suppose that a ∈ R1 , E ⊂ D, and u ∈ Lpω (D); then u(x) ≥ a (u(x) ≤ a) on the set E if mes{E ∩ u(x) < a} = 0
(mes{E ∩ u(x) > a} = 0)
1 (D) if u (x) ≥ a (u (x) ≤ a) on E for some ¯ in the sense of Wpω and also u(x) ≥ a (u(x) ≤ a) on E ⊂ D j j 1 ¯ in the sense of W 1 (D) if approximating sequence {uj }. For z, u ∈ Wpω (D), z ≥ u (z ≤ u) on E ⊂ D pω zj ≥ uj (zj ≤ uj ) on E for the corresponding approximating sequences.
2. AUXILIARY RESULTS Lemma 1 (see also [5, p. 75] and [10, p. 81 (Russian transl.)]). Suppose that p ≥ 1, k ∈ R1 , 1 (D), D ⊂ Rn is an arbitrary domain, and {u} = max{u(x), k}. Suppose that {u } u ∈ Wpω j k 1 approximates u; then {u}k ∈ Wpω (D) and 1 (D) → 0 {uj }k − {u}k Wpω
as j → ∞.
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Proof. From {uj }, let us choose a subsequence, preserving the previous notation, for it and suppose that uj → u, ∇uj → ∇u a.e. in D. Obviously, ¯ {uj }k ∈ Lip(D),
{u}k = k + χu>k (u − k),
{uj }k = k + χuj >k (uj − k),
whence {uj }k − {u}k = (uj − u)χuj >k + (u − k)(χuj >k − χu>k ); then, by Minkowski’s inequality, {uj }k − {u}k Lpω (D) ≤ (uj − u)χuj >k Lpω (D) + (u − k)(χuj >k − χu>k )Lpω (D) → 0 as j → ∞, because uj − uLpω (D) → 0, while the second summand tends to zero by the Lebesgue dominated convergence theorem (and the relation χuj >k → χu>k a.e. in D for characteristic functions). Further, ∇{uj }k = ∇uj χuj >k ,
∇{u}k = ∇uχu>k
a.e. in D; therefore, ∇{uj }k − ∇{u}k = (∇uj − ∇u)χuj >k + ∇u(χuj >k − χu>k ), whence, by Lebesgue’s theorem and ∇uj − ∇uLpω (D) → 0, we have 1/p p p |χuj >k − χu>k | |∇u| ω dx → 0. ∇{uj }k − ∇{u}k Lpω (D) ≤ ∇uj − ∇uLpω (D) + D
Suppose that {uj } approximates u, and, for some subsequence of {uj } (for which we preserve the previous notation) the following relation holds: lim {uj }k − {u}k Wpω 1 (D) = δ > 0.
j→∞
Repeating the previous arguments, we find that the subsequence itself possesses a subsequence for which (preserving the old notation) we can write lim {uj }k − {u}k Wpω 1 (D) = 0.
j→∞
1 (D) for any approximating The resulting contradiction proves that {uj }k → {u}k strongly in Wpω sequence, as was to be proved. 1 (D), u(x) ≤ k on ∂D in the sense of W 1 (D), then Using Lemma 1, we find that if u ∈ Wpω pω ◦
1 (D). Using the same arguments as in Lemma 1, we can show that if uk = {u}k − k belongs to Wpω 1 (D) and z ≥ u on ∂D in the sense of W 1 (D), then (u − z) = max{u − z, 0} belongs to z, u ∈ Wpω + pω ◦
1 (D). Wpω
The condition v ∈ A∞ implies the existence of numbers C, δ > 0 such that, for any ball Qxρ and its measurable subset E, the following inequality holds: mesn E δ v(E) ≤C , v(Qxρ ) mesn Qxρ where mesn E is the Lebesgue measure of the set E in Rn . The condition v ∈ RD implies the existence of numbers µ, δ ∈ (0, 1) such that, for any ball Qxρ , the following inequality holds: v dy ≤ µ v dy. Qx δρ
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The condition (v, ω) ∈ Aps , 1 ≤ p, s < ∞, implies that, for any ball Qxρ , the following inequalities hold: 1/s 1/p 1−n 1−p ρ v dy ω dy ≤ Cp,s < ∞, Qx ρ
Qx ρ
where the number Cp,s > 0 is independent of x and ρ. By the solution of the equation n ∂fi , fi ∈ Lp (D, ω 1−p ), Lu = f0 + ∂xi
(5)
i = 1, . . . , n,
i=1
in the domain D we mean a function u ∈ W 1,p (D, ω) such that, for all ϕ ∈ W01,p (D, ω), the following identity holds: n n ∂ϕ ∂ϕ Ai (x, u, ux ) dx = fi − f0 ϕ dx. (6) ∂xi ∂xi D D i=1
i=1
Lemma 2. Suppose that D ⊂ Rn is an arbitrary domain. Suppose that u ∈ W01,p (D, ω) is a solution of the equation n ∂fi (7) Lu = ∂xi i=1 for f p · ω 1−p ∈ L1,loc (D), where 1 < p < ∞, f = ni=1 |fi |, and condition (2) holds. Then for the inequality s/p q p 1−p (p(q−s))/s |u(x)| v(x) dx ≤ C f ω |u(x)| dx , (8) D
D
where C > 0 is a constant independent of u, f , ω, v, and D, to be valid, it suffices that the following conditions hold: v ∈ RD (v ∈ A∞ ) and (v, ω) ∈ Aps for q > 2, q > s > q − 1, s > p (s = p). ¯ . In (6), replace the test function ϕ by the Proof. Denote D = {x ∈ D : u(x) > 0}, D = D \ D function uk (x) = max(u(x) − k, 0), where k ∈ [0, ess supD u]. By Lemma 1, we have uk ∈ W01,p (D, ω). Then n p ω|∇uk | dx ≤ |fi | · |uxi | dx, Dk
i=1
Dk
¨ inequality, we obtain where Dk = {x ∈ D : u(x) > k}. Hence, applying Holder’s ω|∇uk |p dx ≤ f p · ω 1−p dx. Dk
(9)
Dk
Taking into account the inclusion uk ∈ W01,p (D, ω), [13, Theorem 5] ([14]), and the conditions v ∈ RD (v ∈ A∞ ) and (v, ω) ∈ Aps for s > p (s = p), we obtain the inequality 1/s 1/s s p vuk dx ≤ C0 Cp,s |∇uk | ω dx , (10) Dk
Dk
¨ inequality and inequalities (9) where C0 > 0 is independent of v, ω, u, and Dk . Further, using Holder’s and (10), we find 1/s 1/s s vuk dx ≤ vuk dx v dx Dk
Dk
Dk
1/s
≤ C0 Cp,s
Dk
1/p
f p · ω 1−p dx
v dx
.
(11)
Dk
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REGULARITY OF THE SOLUTIONS OF ELLIPTIC EQUATIONS
Denote
7
λ(k) =
vuk dx; Dk
then
λ (k) = −
v dx. Dk
Suppose that k(λ) = inf{t > 0 : λ(t) < λ} (λ > 0) is the inverse function to λ(k). Inequality (11) implies 1/p p 1−p f ω dx (−λ (k))1/s . (12) λ(k) ≤ C0 Cp,s Dk
Hence C0 Cp,s (−λ (k))1/s 1≤ λ(k)
p
f ω
1−p
1/p dx
.
Dk
Integrating this inequality, we obtain 1/p k(x) dλ 1/s −1 p 1−p (λ(t)) f ω dx dt − k(t) ≤ C0 Cp,s dt 0 Dt or
k(t) ≤ C0 Cp,s
λ(0)
−1
(λ(s)) x
where 0 < t < λ(0). Noting that λ(0) k(λ)q−1 dλ = 0
dλ − ds
ess supD u
−1/s
p
f ω
1−p
1/p dx
dλ,
Ds
λ(k) dkq−1 =
0
D
(u(x))q v dx,
we apply Hardy’s inequality [15, p. 41]: q−1 ∞ ∞ q − 1 q−1 ∞ q−1 q−1 f (s) ds dx ≤ x f dx, q−2 0 x 0 where q > 2, f (s) ≥ 0. We have (q > 2) (u(x))q v(x) dx D
(q−1)/p λ(0) dλ (q−1)/s q − 1 q−1 (C0 Cp,s )q−1 f p ω 1−p dx dλ − q−2 dk 0 Dk(λ) (q−1)/p ess sup u D dλ (s−q+1)/s f p ω 1−p dx dλ; − = C1 dk Dk 0
≤
¨ hence, by Holder’s inequality (with exponents s/(s − q + 1) and s/(q − 1)), we obtain (s−q+1)/s ess sup u(x) λ(0) D dλ µs/(s−q+1) q−1 k(λ) dλ ≤ C1 dk − dk 0 0 ess sup u(x) s/p (q−1)/s D −µs/(q−1) p 1−p × k f ω dx dk , 0
where
q−1
C1 = (C0 Cp,s ) MATHEMATICAL NOTES
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Dk
q−1 q−2
No. 1
q−1
2008
,
µ=
(q − 1)(s − q + 1) . s
(13)
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AMANOV, MAMEDOV
Hence
λ(0)
(s−q+1)/s
λ(0)
k(λ)q−1 dλ ≤ C1
k(λ)q−1 dλ
0
0
ess supD u(x)
×
k
−µs/(q−1)
or
λ(0) q−1
k(λ)
dλ ≤ C1
0
1−p
p
1−p
f ω
0
s/p
p
dx
(q−1)/s dk ,
Dk
ess supD u(x)
k
−µs/(q−1)
f ω
0
s/p dx
dk.
Dk
Let us apply Minkowski’s inequality to the right-hand side of (s ≥ p) of the last inequality: (q−1)/s u(x) p/s (q−1)/p ess sup u(x) D q−1 p 1−p −µs/(q−1) k(λ) dλ ≤ C1 f ω dx k dk D
0
= C2
0 p
D
f ω
1−p
(q−1)/p
(1−(µs)/(q−1))p/s
(u(x))
dx
,
(14)
where C2 = C1
1 , (q − s)(q−1)/s
This yields the estimate
(u(x)) v(x) dx ≤ C3 q
D
D
The same inequality also holds for −u(x) in D : q (−u(x)) v(x) dx ≤ C3 D
where
C3 =
D
q−1 q−2
q > s. s/q
p
1−p
p
1−p
f ω
f ω s
(p(q−s))/s
(u(x))
dx
(15)
. s/p
(p(q−s))/s
(−u(x))
C0 Cp,s q−s
dx
,
(16)
s .
Estimates (15) and (16) imply (8). Lemma 2 is proved. 3. MAIN RESULTS Theorem 1. Suppose that D ⊂ solution of the equation
Rn
is an arbitrary domain. Suppose that u ∈ W01,p (D, ω) is a
n ∂fi ∂xi i=1 for f r · v(ω 1−p /v)r/p ∈ L1 (D), r > p , where f = ni=1 |fi |, p > 1, and condition (2) holds. Then, for the inequality 1/q 1−p r/p 1/r(p−1) ω |u(x)|q v(x) dx ≤C vf r dx , v D D
Lu =
(17)
where C > 0 is independent of v, ω, f , u, and D, to be valid, it suffices that the following conditions hold: v ∈ RD (v ∈ A∞ ) and (v, ω) ∈ Aps for q > 2, q − 1 < s < q, s > p (s = p), 1 1 1 1 + − − = 0. p q s r(p − 1) MATHEMATICAL NOTES
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¨ Proof. Using estimate (8), Lemma 2, and Holder’s inequality (with exponents r/p and r/(r − p )), we can write 1/q s(r−p )/pr q (p(q−s)r)/s(r−p )) |u(x)| v(x) dx ≤ C3 v(x)|u(x)| dx D
D
×
f p ω 1−p v v D
r/p
p s/rp dx
;
hence, in view of condition (18), we obtain estimate (17). Theorem 1 is proved. Theorem 2. Suppose that D ⊂ Rn is a bounded domain, 1 < p < ∞, and condition (2) holds. Suppose that u ∈ W01,p (D, ω) is a solution of the equation Lu =
n ∂fi ∂xi i=1
for v · (ω 1−p /v)r/p · f r ∈ L1 (D), r > p , p > 1, where f =
n
i=1 |fi |.
Then, for the inequality 1/p−1/s−1/r(p−1) 1−p r/p 1/r(p−1) ω r v(x) dx v f dx , ess sup |u(x)| ≤ C v x∈D D D
(19)
where C > 0 is independent of v, ω, f , u, and D, to be valid, it suffices that the following conditions hold: v ∈ RD (v ∈ A∞ ) and (v, ω) ∈ Aps for s > p (s = p), 1 1 1 − − > 0. p s r(p − 1)
(20)
Proof. Suppose that λ(k), k(λ) are the same as in Lemma 2. Then inequality (12) implies 1/p p 1−p f ω dx (−λ (k))1/s . λ(k) ≤ C0 Cp,s Dk
¨ Applying Holder’s inequality (with exponents r/p and r/(r − p )) to the right-hand side and taking the inequality r > p into account, we obtain (r−p )/rp 1−p r/p p /rp ω r v dx v f dx (−λ (k))1/s ; λ(k) ≤ C0 Cp,s v Dk Dk hence, by the equality λ (k) = − Dk v dx, we have λ(k) ≤ C0 Cp,s (−λ (k))1/p+1/s−1/r(p−1) Using condition (20), the equalities λ(0) =
D
v
f p ω 1−p v
r/p
1/r(p−1) dx
.
(21)
D
vu dx,
λ(ess sup u(x)) = 0, x∈D
and integrating inequality (21), we obtain estimate (19) for the function u(x) in D . A similar estimate holds for the function −u(x) in D . Theorem 2 is proved. MATHEMATICAL NOTES
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Lemma 3. Suppose that D ⊂ Rn is an arbitrary domain, 1 < p < ∞, and condition (2) holds. Suppose that u ∈ W01,p (D, ω) is a solution of the equation (22)
Lu = h s
1−s
∈ L1 (D), p < s < ∞ (s = p). for |h| v Then for the inequality (s−1)/(p−1) q s 1−s (p−1)(q−s)/(s−1) |u(x)| v(x) dx ≤ C |h| v |u(x)| dx , D
(23)
D
where C > 0 is independent of v, ω, u, h, and D, to be valid, it suffices that the following conditions hold: v ∈ RD (v ∈ A∞ ) and (v, ω) ∈ Aps for 2 < q < ∞, q − 1 < s < q. Proof. In (6), we replace the test function by ϕ = uk (x) for f0 = h, fi = 0, i = 1, . . . , n, obtaining ω|∇u|p dx ≤ |h|uk dx. Dk
Dk
¨ Hence, by Holder’s inequality, we have p/s p vusk dx ≤ Cp,s Dk
1/s
1/s
|h|s v 1−s dx
vusk dx
Dk
;
Dk
therefore,
1/s
vuk dx Dk
≤
v dx
p C0 Cp,s
s 1−s
|h| v
Dk
1/(p−1)s dx
.
Dk
Using the notation from Lemma 2, we can write
1/s
λ(k) ≤ (−λ (k))
p C0 Cp,s
s 1−s
|h| v
1/(p−1)s dx
.
(24)
Dk
Further, we repeat all the arguments related to Lemma 2. As a result, we obtain inequality (23), where p s q−1 s C0 Cp,s , C0 > 0, C= q−s q−2 is independent of v, ω, u, h, and D. Theorem 3. Suppose that D ⊂ Rn is an arbitrary domain, 1 < p < ∞, and condition (2) holds. Suppose that u ∈ W01,p (D, ω) is a solution of the equation Lu = h ∈ 1 < r < ∞. for Then for the inequality 1/q 1/r(p−1) q r 1−r |u(x)| v dx ≤C |h| v dx , |h|r v 1−r
L1 (D),
D
(25)
D
where C > 0 is independent of v, ω, u, h, and D, to be valid, it suffices that the following conditions hold: v ∈ RD (v ∈ A∞ ) and (v, ω) ∈ Aps for s > p (s = p), q − 1 < s < q, s > r , 1/r p + 1/p q = 1/s. ¨ Estimate (25) is a consequence of the assertion of Lemma 3 and Holder’s inequality (with exponents r/s and r/(r − s )). MATHEMATICAL NOTES
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Theorem 4. Suppose that D ⊂ Rn is a bounded domain, 1 < p < ∞, and condition (2) holds. Suppose that u ∈ W01,p (D, ω) is a solution of the equation Lu = h ∈ 1 < r < ∞. for Then, for the inequality 1/(p−1)s −1/s−1/r(p−1) 1/(p−1) ess sup |u(x)| ≤ C v(x) dx |h|r v 1−r dx , |h|r v 1−r
L1 (D),
x∈D
D
(26)
D
where C > 0 is independent of v, ω, u, h, and D, to be valid, it suffices that the following conditions hold: v ∈ RD (v ∈ A∞ ) and (v, ω) ∈ Aps for s > p (s = p), r > s , s (p
1 1 1 − − > 0. − 1) s r(p − 1)
¨ Estimate (26) is a consequence of estimate (24) and Holder’s inequality: r 1/r(p−1) |h| p 1/s +1/s (p−1)−1/r(p−1) v dx . λ(k) ≤ C0 Cp,s (−λ (k)) v D
(27)
Integrating the differential inequality (27) with respect to λ(k) and taking into account the relations vu dx, λ(ess sup u(x)) = 0, λ(0) = x∈D
D
we obtain estimate (26).
Theorem 5. Suppose that D ⊂ Rn is an arbitrary domain and u ∈ W 1,r (D, v(ω/v)r /2 ) is a solution of the equation n ∂u ∂ aij (x) =h (28) Lu = ∂xi ∂xj i,j=1
subject to the condition n
aij (x)ξi ξj ≥ ω(x)|ξ|2
(29)
i,j=1
for all x ∈ D, for all ξ ∈ Rn , and |h|q · v 1−q ∈ L1 (D) for 2 < q < ∞, 2 < r < ∞. Then, for the inequality 1−p r /2 1/r 1/q ω r q 1−q |∇u| v dx ≤C |h| v dx , v D D
(30)
where C > 0 is independent of v, ω, u, h, and D, to be valid, it suffices that the following conditions hold: v ∈ RD (v ∈ A∞ ) and (v, ω) ∈ A2s , β = v(1/ωv)r/2 ∈ Ar for s > 2 (s = 2), q < s < q − 1, 1 1 1 1 + − − = 0. 2 q s r
(31)
Proof. Let us follow the relevant method given in [10, p. 84 (Russian transl.), Sec. 5]. Suppose that ψi ∈ C0∞ (D) and zi ∈ W01,p (D, ω) is a solution of the equation L zi = MATHEMATICAL NOTES
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i = 1, . . . , n.
(32)
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AMANOV, MAMEDOV
Then
D
uψxi i dx =
L uzi dx =
uL zi dx = D
hzi dx.
D
D
Using the estimate from Theorem 1 (p = 2, L = L ), we can write 1/q 1/q i i q q 1−q uxi ψ dx = uψxi dx ≤ |zi | v dx |h| v dx − D
D
≤
v|ψ |
i r
D
D
1 ωv
D
1/r
r/2
q 1−q
|h| v
dx
1/q dx
(33)
.
D
Noting the duality of the weighted Lebesgue spaces from inequality (33) , we obtain estimate (30). Note that the condition β ∈ Ar is needed for the validity of the equality C¯0∞ = Lr (D, β). Theorem 5 is proved.
Theorem 6. Suppose that D ⊂ Rn is an arbitrary domain and u ∈ W01,p (D, ω) ∩ Lr (D, v) is a solution of Eq. (28) subject to the conditions (29) and |h|q v 1−q ∈ L1 (D) for 2 < q < ∞. Then for the inequality
r
1/r
|u(x)| v dx
q 1−q
≤C
D
|h| v
1/q dx
(34)
,
D
where C > 0 is independent of v, ω, u, h, and D, to be valid, it suffices that the following conditions hold: v ∈ Ar and (v, ω) ∈ A2s , for 2 < s < ∞, q − 1 < s < q, r > s , 1/(2r ) = 1/s − 1/(2q). Proof. The proof is similar to that of the previous theorem. Suppose that ψ ∈ C0∞ (D), z ∈ W01,p (D, ω) is a solution of the equation L z = ψ. (For a discussion of the question of the existence of a solution to this problem, see, for example, [11].) Then 1/q 1/q q q 1−q uψ dx = uL z dx = L uz dx = hz dx ≤ |z| v dx |h| v dx . D
D
D
D
D
D
Using estimate (26) from the assertion of Theorem 4, we obtain the following upper bound for the right-hand side of the inequality given above: 1/q |ψ|r v 1−r dx |h|q v 1−q dx . C D
D
Hence, because of the duality of the weighted Lebesgue spaces, we obtain (34).
Theorem 7. Suppose that D ⊂ Rn is a bounded domain, u ∈ W01,r (D, v(ω/v)r /2 ) is a solution of Eq. (28) for 2 < r < ∞, and conditions (29) and |h| ∈ L1 (D) hold. Then for the inequality r /2 1/r 1/2−1/s−1/r ω r |∇u| v dx ≤C v dx |h| dx , v D D D
(35)
where C > 0 is independent of v, ω, u, h, and D, to be valid, it suffices that the following conditions hold: v ∈ RD and (v, ω) ∈ A2s , β = v(1/(ωv))r/2 ∈ Ar for 2 < s < ∞, 1/2 − 1/s − 1/r > 0. Proof. The proof of Theorem 7 is similar to that of Theorem 5; one only needs to use Theorem 2 instead of Theorem 1. MATHEMATICAL NOTES
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REGULARITY OF THE SOLUTIONS OF ELLIPTIC EQUATIONS
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Theorem 8. Suppose that D ⊂ Rn is a bounded domain and u ∈ W01,p (D, ω) ∩ Lr (D, v) is a solution of Eq. (28) subject to conditions (29), 1 ≤ r < ∞, and |h| ∈ L1 (D). Then for the inequality 1/r 1/2r −1/s r |u(x)| v dx ≤C v dx |h| dx , (36) D
D
D
where C > 0 is independent of v, ω, u, h, and D, to be valid, it suffices that the following conditions hold: v ∈ Ar and (v, ω) ∈ A2s , for 2 ≤ s < ∞, s < r < ∞, 1/2r − 1/s > 0. Proof. The proof of Theorem 8 is similar to that of Theorem 6. REFERENCES 1. G. Stampacchia, “Contributi alla regolarizzazione della soluzioni del problemi al contro per equazi del secondo ordine ellitici,” Ann. Scuola Norm. Sup. Pisa (3) 12, 223–245 (1958). 2. H. F. Weinberger, “Symmetrization in uniformly elliptic problems,” in Studies in Mathematical Analysis and Related Topics (Stanford Univ. Press, Stanford, Calif., 1962), pp. 424–428. 3. M. Chicco, “An a priori inequality concerning elliptic second-order partial differential equations of variational type,” Matematiche (Catania) 26 (1), 173–182 (1971). 4. V. G. Mazya, “On weak solutions of the Dirichlet and Neumann problems,” Trudy Moskov. Mat. Obshch. 20, 137–172 (1969). 5. O. A. Ladyzhenskaya and N. N. Uraltseva, Linear and Quasilinear Equations of Elliptic Type (Nauka, Moscow, 1973) [in Russian]. 6. G. Talenti, “Elliptic equations and rearrangements,” Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4) 3 (4) 697–718 (1976). 7. P. Drabek, A. Kufner, and F. Nicolosi, Nonlinear Elliptic Equations (Singular and Degenerated Case) (University of West Bohemia, Pilshen, 1996). 8. F. I. Mamedov, “Regularity of solutions of linear and quasilinear equations of elliptic type in divergence form,” the regularity of the solutions of the linear and quasilinear equations of elliptic type in the divergent form,” Mat. Zametki 53 (1), 50–58 (1993) [Math. Notes 53 (1), 68–82 (1993)]. ¨ 9. Kilpelainen, “Smooth approximation in weighted Sobolev spaces,” Comment. Math. Univ. Corolin. 38 (1), 29–35 (1997). 10. W. Littman, G. Stampacchia, and H. F. Weinberger, “Regular points for equations with discontinuous coefficients,” Ann. Scuola Norm. Sup. Pisa (3) 17, 43–77 (1963) [Mathematics, Collection of Translations, 9 (2), 72–97 (1965)]. 11. E. B. Fabes, C. E. Kenig, and R. P. Serapioni, “The local regularity of solutions of degenerate elliptic equations,” Comm. Partial Differential Equations 7 (1), 77–116 (1982). 12. E. B. Fabes, D. Jerison, and C. Kenig, “The Wiener test for degenerate elliptic equations,” Ann. Inst. Fourier (Grenoble) 32 (3), 151–182 (1982). 13. E. T. Sawyer and R. L. Wheeden, “Weighted inequalities for fractional integrals on Euclidean and homogeneous spaces,” Amer. J. Math. 114 (4), 813–874 (1992). 14. F. I. Mamedov, “On two-weighted Sobolev inequalities in unbounded domains,” Proc. Razmadze Math. Inst. 21, 117–123 (1999). 15. V. G. Mazya, Sobolev Spaces (Izd. Leningrad. Univ. Leningrad, 1985) [in Russian].
MATHEMATICAL NOTES
Vol. 83
No. 1
2008
