Mathematical Notes, vol. 74, no. 5, 2003, pp. 637–646. Translated from Matematicheskie Zametki, vol. 74, no. 5, 2003, pp. 676–685. c Original Russian Text Copyright 2003 by A. A. Kovalevskii.
Integrability of Solutions of Nonlinear Elliptic Equations with Right-Hand Sides from Logarithmic Classes A. A. Kovalevskii Received June 22, 2001; in final form, April 8, 2002
Abstract—We establish the existence of a weak solution to the Dirichlet problem belonging to a Sobolev space for nonlinear elliptic equations of second order with right-hand sides from a wide class of functions defined in terms of the logarithmic function. Key words: nonlinear elliptic equation, Dirichlet problem, Sobolev space, integrability of func-
tions, Laplace operator.
1. INTRODUCTION AND STATEMENT OF THE MAIN RESULT Suppose that Ω is a bounded open set in Rn ( n ≥ 2) and p ∈ (1, n) . Suppose that c1 , c2 are positive constants g is a nonnegative function from Lp/(p−1) (Ω) , and, for any i ∈ {1, . . . , n} , ai is the Carath´eodory function on Ω × Rn . We assume that for almost all x ∈ Ω and all ξ ∈ Rn the following inequalities hold: n i=1 n
|ai (x, ξ)| ≤ c1 |ξ|p−1 + g(x),
(1)
ai (x, ξ)ξi ≥ c2 |ξ|p .
(2)
i=1
In addition, we assume that for almost all x ∈ Ω and all ξ , ξ ∈ Rn , inequality is valid: n [ai (x, ξ) − ai (x, ξ )](ξi − ξi ) > 0.
ξ = ξ , the following (3)
i=1
Consider the following Dirichlet problem: n ∂ ai (x, ∇u) = f − ∂xi i=1
u=0
in
Ω,
(4)
on ∂Ω.
The solvability and the properties of the solutions of problem (4) with f from L1 (Ω) or from the class of bounded measures were studied in a large number of papers (see, for example, [1–4]). Further, we assume that f ∈ L1 (Ω) . 0001-4346/2003/7456-0637 $25.00
c 2003 Plenum Publishing Corporation
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A. A. KOVALEVSKII ◦
Definition 1. By a weak solution of problem (4) we mean a function u ∈ W 1,1 (Ω) such that the following conditions are satisfied: (1) for any i ∈ {1, . . . , n} , ai (x, ∇u) ∈ L1 (Ω) ; (2) for any function ϕ ∈ C0∞ (Ω) , n Ω
ai (x, ∇u)Di ϕ dx =
i=1
f ϕ dx. Ω
With regard to this definition, see, for example, [2]. Set n(p − 1) . r= n−1 In [2], it was shown that if p > 2 − 1/n , then there exists a weak solution of problem (4) ◦
belonging to the space W 1,λ (Ω) for any λ ∈ [1, r) . Moreover, the exponent λ = r is generally unattainable. However, as was established in [2], if p>2−
1 n
f ln(1 + |f |) ∈ L1 (Ω),
and
◦
then there exists a weak solution of problem (4) belonging to W 1,r (Ω) . Recently, the author [5] ◦
established the existence of a weak solution of problem (4) belonging to W 1,r (Ω) under the conditions p ≥ 2 − 1/n and (5) f [ln(1 + |f |)]σ ∈ L1 (Ω), where σ ∈ ((n − 1)/n, 1) . It turns out that this result can be strengthened as well. Namely, for the assertion on the existence of a weak solution of the problem under consideration belonging ◦
to W 1,r (Ω) to be valid, it suffices to replace the factor beside f in condition (5) by the product of an arbitrary finite number of successive superpositions raised to the power (n − 1)/n of the logarithmic function of |f | and of the next such superposition to the power σ > (n − 1)/n . The proof of this fact is the goal of the present paper. Let us state the main result of this paper exactly. We define the sequence of numbers sj as follows: s1 = 1,
sj = esj−1 ,
j = 2, 3, . . . .
Now, for any j ∈ N , suppose that bj : [sj , +∞) → [0, +∞) is a function such that bj (s) = ln . . . ln ln s,
s ∈ [sj , +∞).
j
Theorem 1. Let p ≥ 2 − 1/n , and let the following condition be satisfied : there exists an m ∈ N and a σ > (n − 1)/n such that (n−1)/n
m bj (sj + |f |) [bm+1 (sm+1 + |f |)]σ ∈ L1 (Ω). (6) f j=1 ◦
Then there exists a weak solution of problem (4) belonging to W 1,r (Ω) . The proof of Theorem 1 is carried out by applying results from [3] and a new result (established in the next section) on the integrability of functions satisfying a certain family of integral inequalities (Theorem 2).
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2. INTEGRABILITY OF FUNCTIONS SATISFYING CERTAIN INTEGRAL INEQUALITIES First, note that if j ∈ N and s > sj , then bj (s) > 0 . Lemma 1. Suppose that u is a measurable function on Ω . Suppose that M > 0 , τ > 0 , ε > 1 , and m ∈ N . Suppose that for any k > sm+1 the following inequality is valid : meas{|u| ≥ k} ≤ M k
−τ
m
−1 bj (k) [bm+1 (k)]−ε .
(7)
j=1
Then u ∈ Lτ (Ω) and the following estimate holds: 2m+ε eτ M . |u|τ dx ≤ (esm+2 )τ meas Ω + ε−1 Ω
(8)
Proof. We restrict ourselves to the assumption that m ≥ 2 . The case m = 1 is treated likewise. Suppose that k0 is a positive integer such that sm+1 < k0 ≤ sm+1 + 1 . For any k ∈ N , k ≥ k0 , we set m−1 −1
bj (k) [bm (k)]−ε . βk = k j=1
Suppose that k ∈ N , k ≥ k0 . By (7), we have meas{|u| ≥ ek } ≤ M e−τ k βk .
Then
{ek ≤|u| 1 , we have ∞
βk ≤
k=k0
2m+ε [bm (k0 )]1−ε , ε−1
it follows from (9) that u ∈ Lτ (Ω) and the following estimate is valid: 2m+ε eτ M [bm (k0 )]1−ε . |u|τ dx ≤ eτ k0 meas Ω + ε − 1 Ω Hence, taking into account the choice of the number k0 , we obtain inequality (8). The lemma is proved. Set p∗ = np/(n − p) . It is well known (see, for example, [6]) that there exists a positive ◦
constant c0 depending only on n and p such that for any function u ∈ W 1,p (Ω) we have
p∗
Ω
1/p∗
|u| dx
≤ c0
Ω
p
1/p
|∇u| dx
Suppose that for any k > 0 Tk is a function on R such that s if |s| ≤ k, Tk (s) = k sign s if |s| > k.
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.
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A. A. KOVALEVSKII ◦
Lemma 2. Suppose that u is a function on Ω . Suppose that k > 0 and Tk (u) ∈ W 1,p (Ω) . Then the set {|u| ≥ k} is measurable and the following inequality is valid: p∗
meas{|u| ≥ k} ≤ (c0 /k)
p
Ω
p∗ /p
|∇Tk (u)| dx
.
(11)
Proof. The fact that the set {|u| ≥ k} is measurable follows from the relation {|u| ≥ k} = {|Tk (u)| = k} and the fact that the function Tk (u) is measurable, too. By the same relation, we have k
p∗
meas{|u| ≥ k} ≤
∗
Ω
|Tk (u)|p dx.
Combining this with inequality (10) applied to the function Tk (u) , we obtain (11). The lemma is proved. Lemma 3. Suppose that u is a function on Ω , and suppose that for any k > 0 the inclusion ◦
Tk (u) ∈ W 1,p (Ω) is valid. Then for all N , k, k1 > 0 the following inequality holds: p∗
meas{|∇TN (u)| ≥ k} ≤ (c0 /k1 )
Ω
p∗ /p
p
|∇Tk1 (u)| dx
+k
−p
Ω
|∇Tk1 (u)|p dx.
(12)
Proof. Suppose that N , k, k1 > 0 . Set E = {|u| < k1 , |∇TN (u)| ≥ k}. Obviously, meas{|∇TN (u)| ≥ k} ≤ meas{|u| ≥ k1 } + meas E.
(13)
Since k ≤ |∇TN (u)| on E , ∇TN (u) = ∇Tk1 (u)
a.e. on {|u| ≤ N } ∩ {|u| ≤ k1 },
∇TN (u) = 0
a.e. on {|u| > N },
we obtain p
k meas E ≤
E
p
|∇TN (u)| dx ≤
Ω
|∇Tk1 (u)|p dx.
Hence from (13) and Lemma 2 we obtain (12). The lemma is proved. Lemma 4. Suppose that u is a function on Ω , λ ≥ 1 , and suppose that for any N ∈ N the inclusion TN (u) ∈ W 1,λ (Ω) is valid. Suppose that the sequence {TN (u)} is bounded in W 1,λ (Ω) . Then and TN (u) → u u ∈ W 1,λ (Ω) strongly in W 1,λ (Ω) .
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Proof. Since the sequence {TN (u)} is bounded in Lλ (Ω) and is pointwise convergent to u in Ω , using Fatou’s lemma and Lebesgue’s theorem on the passage to the limit under the integral sign, we see that u ∈ Lλ (Ω) and strongly in Lλ (Ω).
TN (u) → u
(14)
Further, suppose that i ∈ {1, . . . , n} . Since for arbitrary k, k1 ∈ N , k < k1 , we have a.e. on {|u| ≤ k},
Di Tk (u) = Di Tk1 (u)
we can state the following: there exists a function vi : Ω → R such that for any N ∈ N vi = Di TN (u)
a.e. on {|u| ≤ N }.
(15)
a.e. on Ω.
(16)
This yields Di TN (u) → vi
In addition, it follows from (15) and the definition of the function TN that for any N ∈ N the following inequality holds: a.e. on Ω. (17) |Di TN (u)| ≤ |vi | Since the sequence {Di TN (u)} is bounded in Lλ (Ω) (16), in view of Fatou’s lemma, we have vi ∈ Lλ (Ω) . Then, using (16), (17), and Lebesgue’s lemma on the passage to the limit under the integral sign, we find that strongly in Lλ (Ω).
Di TN (u) → vi
(18)
It follows from (14) and (18) that there exists a generalized derivative Di u where Di u = vi a.e. on Ω . Now it is obvious that Di u ∈ Lλ (Ω) and Di TN (u) → Di u strongly in Lλ (Ω) . Since this is valid for any i ∈ {1, . . . , n} , taking (14) into account, we find that u ∈ W 1,λ (Ω)
TN (u) → u
and
strongly in W 1,λ (Ω) . The lemma is proved. Theorem 2. Suppose that Φ is a nonnegative measurable function on Ω , m ∈ N , σ > (n−1)/n , and suppose that (n−1)/n
m bj (sj + Φ) [bm+1 (sm+1 + Φ)]σ ∈ L1 (Ω). Φ j=1 ◦
Suppose that p ≥ 2 − 1/n , u is a function on Ω , and the inclusion Tk (u) ∈ W 1,p (Ω) is valid for any k > 0 as well as the inequality Ω
p
|∇Tk (u)| dx ≤
◦
Then u ∈ W 1,r (Ω) . MATHEMATICAL NOTES
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Ω
Φ |Tk (u)| dx.
(19)
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A. A. KOVALEVSKII
Proof. By ci , i = 3, 4, . . . , we denote positive constants depending only on n , p , m , σ , meas Ω , and the norm on L1 (Ω) of the function (n−1)/n
m bj (sj + Φ) [bm+1 (sm+1 + Φ)]σ . Φ j=1
Define the numbers tj as follows: tj = etj−1 ,
t1 = 4, Set γ=
p−1 , 2p
j = 2, 3, . . . , m + 1. 2
γ1 = max(e1/γ , tm+1 ).
◦
Choose k ≥ γ1 . We have Tk (u) ∈ W 1,p (Ω) , and, by (19), the following inequality is valid: Ω
|∇Tk (u)|p dx ≤ k γ
Ω
|Tk (u)| dx + k
Φ dx.
(20)
{Φ>kγ }
Using (10) and H¨ older’s and Young’s inequalities, we obtain k
γ
Ω
(p∗ −1)/p∗
γ
p
1/p
|Tk (u)| dx ≤ c0 k (meas Ω) |∇Tk (u)| dx Ω 1 p−1 c3 k 1/2 . ≤ |∇Tk (u)|p dx + p Ω p
Combining this with (20), we see that
p
Ω
|∇Tk (u)| dx ≤ c3 k
1/2
+ 3k
Φ dx.
(21)
{Φ>kγ }
Let us estimate the integral on the right-hand side of (21). First, we note that, by the inequality k ≥ tm+1 and the definition of the numbers tj , for any j ∈ {1, . . . , m} we have bj (k) ≥ 4.
(22)
Obviously, we have the following property: if s > k γ ,
then b1 (s1 + s) > γ b1 (k).
(23)
Using (22), (23), and the inequality ln k ≥ 1/γ 2 , we establish that if s > k γ , then for any j ∈ {2, . . . , m + 1} the following inequality is valid: bj (sj + s) > bj (k)/2. This fact and property (23) allow us to conclude that
m
(n−1)/n (n−1)/n m bj (sj + Φ) [bm+1 (sm+1 + Φ)]σ > γ m+σ bj (k) [bm+1 (k)]σ
j=1
j=1
on the set {Φ > k γ }. MATHEMATICAL NOTES
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Therefore,
{Φ>kγ }
Φ dx ≤ c4
m
−(n−1)/n bj (k) [bm+1 (k)]−σ .
643
(24)
j=1
Note that for arbitrary λ, s > 0 the following inequality holds: λ ln s < sλ . This yields
m
(25)
(n−1)/n bj (k) [bm+1 (k)]σ < [2(m + σ)]m+σ k 1/2 .
(26)
j=1
Inequalities (21), (24), and (26) imply that for any k ≥ γ1 the following inequality holds:
p
Ω
|∇Tk (u)| dx ≤ c5 k
m
−(n−1)/n bj (k) [bm+1 (k)]−σ .
(27)
j=1
By Lemma 2 and (27), for any k > sm+1 we have meas{|u| ≥ k} ≤ c6 k
−n(p−1)/(n−p)
m
−(n−1)/(n−p) bj (k) [bm+1 (k)]−σn/(n−p) .
j=1
Hence, taking the inequality σ > (n − 1)/n into account and using Lemma 1, we find that u ∈ Ln(p−1)/(n−p) (Ω) . Further, set γ2 =
(n−1)/(n−p) γ1
m
(n−1)/n(n−p) bj (γ1 ) [bm+1 (γ1 )]σ/(n−p) ,
j=1
and choose k > γ2 . Suppose that ψ is a function on (sm+1 , +∞) such that for any s ∈ (sm+1 , +∞) n−1
ψ(s) = s
m
(n−1)/n bj (s) [bm+1 (s)]σ .
j=1
Since the function ψ is continuous, ψ(s) → +∞ as s → +∞ and ψ(γ1 ) < k n−p , there exists a k1 > γ1 such that (28) ψ(k1 ) = k n−p . It follows that −n(p−1)/(n−p)
k1
m
−(n−1)/(n−p) bj (k1 ) [bm+1 (k1 )]−σn/(n−p)
j=1
=k
−p
k1
m
−(n−1)/n −1 m −σ −r bj (k1 ) [bm+1 (k1 )] = k bj (k1 ) [bm+1 (k1 )]−σn/(n−1) .
j=1
(29)
j=1
In addition, relation (28) implies that k1 < k and k n−p < k1n+m+σ . Using the last inequality, we prove that for any j ∈ {1, . . . , m + 1} bj (k) < MATHEMATICAL NOTES
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A. A. KOVALEVSKII
It follows from Lemma 3 and relations (27), (29), and (30) that for any N ∈ N meas{|∇TN (u)| ≥ k} ≤ c7 k
−r
m
−1 bj (k) [bm+1 (k)]−σn/(n−1) .
j=1
The result obtained above, the inequality σ > (n − 1)/n , and Lemma 1 imply that for any N ∈ N |∇TN (u)|r dx ≤ c8 . (31) Ω
By the condition p ≥ 2 − 1/n , we have r ≥ 1 , and since p < n , it follows that r < p . Then for ◦
any N ∈ N we have TN (u) ∈ W 1,r (Ω) . Hence, since the function u belongs to the space Lr (Ω) , ◦
in view of (31) and Lemmas 4, it follows that u ∈ W 1,r (Ω) . Thus, the theorem is proved. 3. PROOF OF THEOREM 1 Suppose that the assumptions of Theorem 1 are satisfied. By (1)–(3) and Theorem 6.1 from [3], ◦
there exists a function u on Ω such that for any k > 0 the inclusion Tk (u) ∈ W 1,p (Ω) is valid ◦
and for all ϕ ∈ W 1,p (Ω) ∩ L∞ (Ω) , k > 0 , and k1 > k + ϕL∞ (Ω) the following inequality holds: {|u−ϕ| 0 1 |∇Tk (u)|p dx ≤ |f | |Tk (u)| dx. c2 Ω Ω ◦
Now, invoking (6) and using Theorem 2, we find that u ∈ W 1,r (Ω) . Combining this with (1), we see that the inclusion ai (x, ∇u) ∈ L1 (Ω) is valid for any i ∈ {1, . . . , n} . In addition, by (32) and Corollary 4.3 from [3], for any function ϕ ∈ C0∞ (Ω) we have n Ω
ai (x, ∇u)Di ϕ dx =
i=1
f ϕ dx. Ω
Thus, the function u is a weak solution of problem (4). The theorem is proved. 4. CONCLUDING REMARKS Condition (6) on the right-hand side of the equation in problem (4) is only a sufficient condition ◦
for the existence of a weak solution belonging to W 1,r (Ω) . Indeed, using the Laplace operator as an example (this corresponds to the case p = 2), we can show that there may exist a weak ◦
solution of problem (4) belonging to the space W 1,r (Ω) also in the case of a weaker constraint on the right-hand side of the equation, as compared to condition (6). Suppose that Ω = {x ∈ Rn : |x| < 1} , and suppose that g1 is a function of class C 2 ((0, +∞)) such that g1 = 0 on [1/2, +∞) and for any s ∈ (0, e−e ) we have g1 (s) =
1 ln s
−(n−1)/n
−(2n−1)/n 1 ln ln . s MATHEMATICAL NOTES
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Further, suppose that u1 and f1 are functions on Ω such that for any x ∈ Ω \ {0} we have u1 (x) = |x|2−n g1 (|x|),
f1 (x) = (n − 3)|x|1−n g1 (|x|) − |x|2−n g1 (|x|).
It is easy to verify that the function u1 is a weak solution of the problem −∆u = f1
in Ω ,
u = 0 on ∂Ω
◦
and, moreover, u1 ∈ W 1,r (Ω) ; for any σ ∈ (0, (n − 1)/n) we have f1 [ln(1 + |f1 |)](n−1)/n [ln ln(e + |f1 |)]σ ∈ L1 (Ω), but
/ L1 (Ω) f1 [ln(1 + |f1 |)](n−1)/n [ln ln(e + |f1 |)](n−1)/n ∈
and, therefore, the function f1 does not satisfy condition (6). In the general case treated in this paper, the loss of accuracy in the condition on the right-hand ◦
side of the equation ensuring the existence of a weak solution from W 1,r (Ω) to problem (4) is due to the fact that, when estimating the integrals of the function Φ over the sets {Φ > kγ } in (the main) Theorem 2 (see (24)), the integrals of the function
(n−1)/n m Φ bj (sj + Φ) [bm+1 (sm+1 + Φ)]σ j=1
over the same sets were estimated from above by the same constant, while the sequence of these integrals tends to zero as k → ∞ . However, for a function Φ of general position it seems impossible to give the definitive estimate of the accuracy of this convergence to zero depending on k and thus to use the additional factor, which converges to zero as k → ∞ , on the right-hand side of the inequality whose crude form is inequality (24). This justifies the estimate of the integrals of the function (33) by only a constant, although this, as has already been stated, leads to some loss in the accuracy of the condition ensuring the existence of a weak solution of the problem belonging ◦
to the space W 1,r (Ω) . Finally, we present an example in which the right-hand side of the equation possesses “logarithmic” integrability, but the weak solution of the corresponding Dirichlet problem does not belong ◦
to W 1,r (Ω) . Suppose that Ω = {x ∈ Rn : |x| < 1} , and suppose that g2 is a function of class C 2 ((0, +∞)) such that g2 = 0 on [1/2, +∞) and for any s ∈ (0, e−1 ) we have
−(n−1)/n 1 g2 (s) = ln . s Further, suppose that u2 and f2 are functions on Ω such that for any x ∈ Ω \ {0} the following expressions are valid: u2 (x) = |x|2−n g2 (|x|),
f2 (x) = (n − 3)|x|1−n g2 (|x|) − |x|2−n g2 (|x|).
Then the function u2 is a weak solution of the problem −∆u = f2
in Ω ,
u = 0 on ∂Ω .
◦
◦
/ Lr (Ω) ; therefore, u2 ∈ / W 1,r (Ω) . Moreover, for any λ ∈ [1, r) we have u2 ∈ W 1,λ (Ω) , but |∇u2 | ∈ In addition, for any σ ∈ (0, (n − 1)/n) we have f2 [ln(1 + |f2 |)]σ ∈ L1 (Ω), but
f2 [ln(1 + |f2 |)](n−1)/n ∈ / L1 (Ω). MATHEMATICAL NOTES
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REFERENCES 1. L. Boccardo and T. Gallou¨et, “Nonlinear elliptic and parabolic equations involving measure data,” J. Funct. Anal., 87 (1989), 149–169. 2. L. Boccardo and T. Gallou¨et, “Nonlinear elliptic equations with right-hand side measures,” Comm. Partial Differential Equations, 17 (1992), 641–655. 3. Ph. B´enilan, L. Boccardo, and T. Gallou¨et, R. Gariepy and M. Pierre, J. L. Vazquez, “An L1 -theory of existence and uniqueness of solutions of nonlinear elliptic equations,” Ann. Scuola Norm. Sup. Pisa Cl. Sci., 22 (1995), 241–273. 4. L. Boccardo and T. Gallou¨et, “Summability of the solutions of nonlinear elliptic equations with righthand side measures,” J. Convex Analysis, 3 (1996), 361–365. 5. A. A. Kovalevskii, “Integrability of solutions of nonlinear elliptic equations with right-hand sides from classes close to L1 ,” Mat. Zametki [Math. Notes], 70 (2001), no. 3, 375–385. 6. D. Gilbarg and N. S. Trudinger, Elliptic Partial Differential Equations of Second Order, SpringerVerlag, Berlin, 1983. Institute of Applied Mathematics and Mechanics, National Academy of Sciences of Ukraine E-mail :
[email protected]
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