GOOD RADON MEASURE FOR ANISOTROPIC PROBLEMS WITH VARIABLE EXPONENT

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1 Electronic Journal of Differential Equations, Vol , No. 221, pp ISSN: URL: or GOOD RADON MEASURE FOR ANISOTROPIC PROBLEMS WITH VARIABLE EXPONENT IBRAHIME KONATÉ, STANISLAS OUARO Abstract. We study nonlinear anisotropic problems with bounded Radon diffuse measure and variable exponent. We prove the existence and uniqueness of entropy solution. 1. Introduction We consider the anisotropic elliptic Dirichlet boundary-value problem bu = µ in u = 0 on, 1.1 where is an open bounded domain of R N N 3, with smooth boundary, b : R R is a continuous, surjective and non-decreasing function, with b0 = 0 and µ is a bounded Radon diffuse measure this is µ does not charge the sets of zero p m -capacity such that µ > 0. All papers concerned by problems like 1.1 considered particular cases of data b and measure µ. Indeed, in [13], the authors considered b 0, which permit them to exploit minimization technics to prove the existence of weak solutions and mini-max theory to prove that weak solutions are multiple. Using the same methods, Koné et al. see [10] studied the problem = µ in 1.2 u = 0 on, where µ is a bounded Radon measure. Note that 1.2 is a particular case of 1.1, where b 0. Koné et al. also studied problem 1.2 when µ L see [9] and Ouaro, when µ L 1 see [15]. Ibrango and Ouaro studied the following problem see [8] + bu = µ u = 0 on, in 2010 Mathematics Subject Classification. 36J60, 35J65, 35J20, 35J25. Key words and phrases. Generalized Lebesgue-Sobolev spaces; anisotropic Sobolev space; weak solution; entropy solution; Dirichlet boundary condition; bounded Radon diffuse measure; Marcinkiewicz space. c 2016 Texas State University. Submitted March 28, Published August 17,

2 2 I. KONATÉ, S. OUARO EJDE-2016/221 where µ L 1. In [8], the authors used the technic of monotone operators in Banach spaces and approximation methods to prove the existence and uniqueness of entropy solution of problem 1.3. In this paper, by using the same technics as in [8], we extend their result, taking into account a measure µ which is zero on the subsets of zero p -capacity i.e., the capacity defined starting from W 1,p 0. In order to do that, we use a decomposition theorem proved by Nyanquini et al. in [14]: every bounded Radon measure that is zero on the sets of zero p -capacity can be split in the sum of an element in W 1,p the dual space of W 1,p 0, and a function in L 1, and conversely, every bounded measure in L 1 + W 1,p is zero on the sets of zero p -capacity. Using the decomposition of measures result of Nyanquini et al. see [14], we prove that there exists a unique entropy solution of 1.1. The proof of our result will strongly rely on the structure of the measure µ, that is, µ belongs to L 1 + W 1,p. Note that, since b is not necessarily invertible, then, the uniqueness of the entropy solution is proved in terms of bu which is clearly equivalent to the uniqueness of u if and only if b is invertible. Note that a good Radon measure for the problem 1.1 is a Radon measure for which, the entropy solution of problem 1.1 is unique. Many papers are related to problems involving variable exponents due to their applications to elastic mechanics, electrorheological fluids or image restoration. We denote by M b the space of bounded Radon measures in, equiped with its standard norm Mb. Note that, if µ belongs to M b, then µ the total variation of µ is a bounded positive measure on. Given µ M b, we say that µ is diffuse with respect to the capacity W 1,p 0 p -capacity for short if µa = 0, for every set A such that Cap p A, = 0. For A, we denote S p A := {u W 1,p 0 C 0 : u = 1 on A, u 0 on }. The p -capacity of every subset A with respect to is defined by Cap p A, := inf u S p A { } u px. In the case S p A =, we set Cap p A, = +. The set of bounded Radon diffuse measure in the variable exponent setting is denoted by M p b. We recall the decomposition result of bounded Radon diffuse measure proved by Nyanquini et al see [14]. Theorem 1.1. Let p : 1, + be a continuous function and µ M b. Then, µ M p b if and only if µ L 1 + W 1,p. Recall that, in this paper, we assume that µ M pm b, where p m is to be defined later. Remark 1.2. We do not have uniqueness of entropy solution if the measure µ does not belong to the space M pm b see Proof of uniqueness. Therefore, M pm b is the set of good Radon measure for problem 1.1. The remaining part of this article is organized as follows: In Section 2, we introduce some preliminary results. In Section 3, we study the existence and uniqueness of entropy solution. We refer to [3, 4, 7] as papers dealing with measures including the case of variable exponents.

3 EJDE-2016/221 GOOD RADON MEASURE 3 2. Preliminaries We study problem 1.1 under the following assumptions on the data. Let be a bounded domain in R N N 3 with smooth boundary domain and p = p 1,..., p N such that for any i = 1,..., N, p i : R is a continuous function with 1 < p i := ess inf x p i x p + i := ess sup x p i x <. 2.1 For i = 1,..., N, let : R R be a Carathéodory function satisfying: there exists a positive constant C 1 such that x, ξ C 1 j i x + ξ pix1, 2.2 for almost every x and for every ξ R, where j i is a non-negative function in L p i 1, with p + 1 ix p i x = 1; for ξ, η R with ξ η and for every x, there exists a positive constant C 2 such that { C 2 ξ η pix if ξ η 1 x, ξ x, ηξ η 2.3 C 2 ξ η p i if ξ η < 1 there exists a positive constant C 3 such that x, ξ.ξ C 3 ξ pix, 2.4 for ξ R and almost every x. The hypotheses on are classical in the study of nonlinear problems see [12]. Throughout this paper, we assume that and pn 1 Np 1 < pn 1 p i < N p, p + i p i 1 p < p N i pn 1, p i > 1, 2.6 where N p = N 1. p i A prototype example that is covered by our assumption is the following anisotropic p -harmonic problem: set x, ξ = ξ pix2 ξ, where p i x 2 for i = 1,..., N. Then, we obtain the problem u pix2 u bu = µ in u = 0 on, 2.7 which, in the particular case where p i = p for any i = 1,..., N, is the p-laplace equation. We also recall in this section some definitions and basic properties of anisotropic Lebesgue and Sobolev spaces. We refer to [17, 18] for details and related properties. Set C + = { p C : min px > 1 a.e. x } x

4 4 I. KONATÉ, S. OUARO EJDE-2016/221 and denote by p M x := maxp 1 x,..., p N x and p m x := minp 1 x,..., p N x. For any p C +, the variable exponent Lebesgue space is defined by { L p := u : u is a measurable real valued function such that } u px <, endowed with the so-called Luxembourg norm u p := inf { λ > 0 : ux λ px 1 }. The p -modular of the L p space is the mapping ρ p : L p R defined by ρ p u := u px. For any u L p, the following inequality see [5, 6] will be used later min{ u p p ; u p+ p } ρ p u max{ u p p ; u p+ p }. 2.8 For any u L p and v L q, with 1 px + 1 qx = 1 in, we have the Hölder type inequality see [11] uv 1 p + 1 q u p v q. 2.9 If is bounded and p, q C + such that px qx for any x, then the embedding L p L q is continuous see [11, Theorem 2.8]. Herein, we need the anisotropic Sobolev space W 1, p 0 := { u W 1,1 u 0 : L pi, i = 1,..., N }, which is a separable and reflexive Banach space see [11] under the norm u p = u pi. We introduce the numbers and define P = N N 1 p i q = Np 1 N 1, q = Np 1 N p = Nq N q, 1, P + = max{p 1,..., p N }, P, = max{p +, P }. Remark 2.1. Since µ M pm b, the Theorem 1.1 implies that there exist f L 1 and F L p m N such that where 1 p + 1 mx p m x = 1 for all x. µ = f div F, 2.10 We have the following embedding result see [13, Theorem 1].

5 EJDE-2016/221 GOOD RADON MEASURE 5 Theorem 2.2. Assume that R N N 3 is a bounded domain with smooth boundary. Assume also that the relation 2.6 is fillfuled. For any q C verifying 1 < qx < P, for any x, the embedding W 1, p 0 L q is continuous and compact. The following result is due to Troisi see [20]. Theorem 2.3. Let p 1,..., p N [1, + ; g W 1,p1,...,p N and { p if p < N q = [1, + if p N. Then, there exists a constant C 4 > 0 depending on N, p 1,..., p N if p < N and also on q and meas if p N such that g L q C 4 N g 1/N L p i In this paper, we will use the Marcinkiewicz space M q 1 < q < + as the set of measurable function g : R for which the distribution satisfies an estimate of the form λ g k = meas{x : gx > k}, k λ g k Ck q, for some finite constant C > We will use the following pseudo norm in M q g M q := inf{c > 0 : λ g k Ck q, k > 0} Finally, we use throughout the paper, the truncation function T k, k > 0 by T k s = max{k, min{k; s}} It is clear that lim k T k s = s and T k s = min{ s ; k}. We define T 1, p 0 as the set of measurable functions u : R such that T k u W 1, p 0. In the sequel, we denote W 1, p 0 = E to simplify notation. 3. Existence and uniqueness result Definition 3.1. A measurable function u T 1, p 0 is an entropy solution of 1.1 if bu L 1 and T k u v + but k u v T k u vdµ, 3.1 for all v E L and for every k > 0. The existence result is as follows. Theorem 3.2. Assume and 2.10 hold. Then, there exists at least one entropy solution of problem 1.1.

6 6 I. KONATÉ, S. OUARO EJDE-2016/221 Proof. The proof is done in three steps. Step 1: Approximate problem. We consider the problem n + T n bu n = µ n in u n = 0 on, 3.2 where f n = T n f L and µ n = f n div F. Note that f n f in L 1 as n +, and f n 1 = f n f = f Definition 3.3. A measurable function u n E is a weak solution for 3.2 if for every v E. n v + T n bu n v = f n v + F. v, 3.4 Lemma 3.4. There exists at least one weak solution u n for problem 3.2 and bu n µ n. Proof. We define the operators A n and B n as follows. A n u, v = Au, v + T n buv u, v E, 3.5 where u v Au, v = x,, 3.6 x i B n, v = vdµ n. 3.7 The operator A n is onto see [8, Lemma 3.1] and [19, Corollary 2.2]. Therefore, for B n E, we can deduce the existence of a function u n E such that A n u n, v = B n, v. Now, we show that bu n µ n. Indeed, let us denote by H ɛ = min { s + ɛ ; 1, sign + 0 s = 1 if s > 0 0 if s 0. If γ is a maximal monotone operator defined on R, we denote by γ 0 the main section of γ; i.e., minimal absolute value of γs if γs γ 0 s = + if [s, + Dγ = if, s] Dγ =. We remark that, as ɛ approaches 0, H ɛ s apaprocahes sign + 0 s. We take v = H ɛ u n M as test function in 3.4, for the weak solution u n, where M > 0 a

7 EJDE-2016/221 GOOD RADON MEASURE 7 constant to be chosen later, to obtain = By 2.4 we have u n x, H ɛ u n M + T n bu n H ɛ u n M H ɛ u n Mdµ n. = 1 ɛ = 1 ɛ x, u n H ɛ u n M { unm+ ɛ <1} {0<u nm<ɛ} x, u n u n M + x, u n u n Then, 3.8 gives T n bu n H ɛ u n M which is equivalent to T n bu n T n bm H ɛ u n M H ɛ u n Mdµ n, µ n T n bm H ɛ u n M. We now let ɛ approach 0 in the inequality above to obtain + T n bu n T n bm µ n T n bm sign + 0 u n M. 3.9 Choosing M = b 1 0 µ n in the above inequality since b is surjective, we obtain + T n bu n T n µ n For any n µ n, we have µ n T n µ n = Then, 3.10 gives µ n T n µ n sign + 0 u n b 1 0 µ n. sign + 0 u n b 1 0 µ n µ n µ n sign + 0 u n b 1 0 µ n 0. T n bu n µ n Hence, for all n > µ n, we have T n bu n µ n = 0 a.e. in, which is equivalent to T n bu n µ n, for all n > µ n. 3.11

8 8 I. KONATÉ, S. OUARO EJDE-2016/221 Let us remark that as u n is a weak solution of 3.4, then u n is a weak solution of the problem ã i n + T n bu n = µ n in 3.12 u n = 0 on, where ã i x, ξ = x, ξ, bs = bs and µ n = µ n. From 3.11 we deduce that T n bu n µ n, for all n > µ n. Therefore, T n bu n µ n, for all n > µ n It follows from 3.11 and 3.13 that for all n > µ n, T n bu n µ n which implies bu n µ n a.e. in. We now consider the problem n + bu n = µ n in u n = 0 on, 3.14 where f n = T n f L and µ n L. It follows from Lemma 3.4 that there exists u n E with bu n L such that n v + bu n v = f n v + F. v, 3.15 for every v E. Our aim is to prove that these approximated solutions u n tend, as n approaches infinity, to a measurable function u which is an entropy solution of the problem 1.1. To start with, we establish some a priori estimates. Step 2: A priori estimates. Lemma 3.5. There exists a positive constant C 5 which does not depend on n, such that u n p i C5 k + 1, 3.16 for every k > 0. { u n k} Proof. We take v = T k u n as test function in 3.15 to obtain n un + bu n T k u n { u n k} = f n T k u n + F T k u n. Using relation 2.4 and the fact that bu nt k u n 0, we obtain u n pix N c 3 { u n k} f n T k u n + F T k u n. 3.17

9 EJDE-2016/221 GOOD RADON MEASURE 9 Since we deduce that We have = f n T k u n + F T k u n = T k u n dµ n k µ Ck, { u n k} N c 3 { u n k} u n p i { u n k; un >1} { u n k} u n pix Ck u n p N i + u n pix + N. meas C C 3 k + N. meas due to relation 3.18 C k { u n k; un 1} u n p i with C 5 = max { C C 3 ; N. meas }. We also have the following lemma see [9], [10]. Lemma 3.6. For any k > 0, there exists some positive constants C 6 and C 7 such that: i u n M q C 6 ; ii un M p i q C 7, i = 1,..., N. p Step 3: Convergence. According to [2] see also [10], we have the following lemma. Lemma 3.7. For i = 1,..., N, as n +, we have n in L 1, a.e. x Proposition 3.8. Assume hold. If u n E is a weak solution of 3.2 then, the sequence u n n N is Cauchy in measure. In particular, there exists a measurable function u and a sub-sequence still denoted by u n such that u n u in measure. Proposition 3.9. Assume hold. If u n E is a weak solution of 3.2, then i for i = 1,..., N, un converges in measure to the weak partial gradient of u; ii for i = 1,..., N and k > 0, x, T k u n converges to x, T k u in L 1 strongly and in L p i weakly.

10 10 I. KONATÉ, S. OUARO EJDE-2016/221 We can now pass to the limit in 3.4. Let v W 1, p 0 L and k > 0; we choose T k u n v as test function in 3.15 to obtain = n T k u n v + f n T k u n v + F T k u n v. bu n T k u n v 3.20 For the first term of the right-hand side of 3.20, we have f n xt k u n v fxt k u v, 3.21 since f n converges strongly to f in L 1 and T k u n v converges weakly- to T k u v in L and a.e. in. For the second term of the right-hand side of 3.20, we have F T k u n v F T k u v, 3.22 since T k u n v T k u v in L pm N and F L p m N. For the first term of 3.20, we have see [2]: lim inf N n T k u n v n T k u v. For the second term of 3.20, we have bu n T k u n v = bu n bvt k u n v + bvt k u n v. The quantity bu n bvt k u n v is nonnegative and since for all s R, s bs is continuous, we obtain bu n bvt k u n v bu bvt k u v a.e. in. Then, it follows by Fatou s Lemma that lim inf bu n bvt k u n v bu bvt k u v n + We have bv L 1. Since T k u n v converges weakly -* to T k u v L and bv L 1, it follows that lim n + bvt k u n v = bvt k u n v From 3.21, 3.22, 3.23, 3.24 and 3.25, we pass to the limit in 3.20 to obtain T k u v + but k u v ft k u v + F T k u v. Then u is an entropy solution of 1.1.

11 EJDE-2016/221 GOOD RADON MEASURE 11 Theorem Assume hold and let u be an entropy solution of 1.1. Then, u is unique. The proof of the above theorem is done in two steps. Step 1: A priori estimates. Lemma Assume hold. Let u be an entropy solution of 1.1. Then u p k i µ, k > C 3 { u k} and there exists a positive constant C 8 such that bu 1 C 8 meas + µ Proof. Let us take v = 0 in the entropy inequality 3.1. By the fact that but ku 0 and using relations 2.3 and 2.4, we deduce As relation 3.1 gives T k u = By 3.28, we deduce or { u >k} Therefore, So, we obtain { u k} but k u { u k} u 0, ft k u + F T k u but k u + but k u k µ { u >k} but k u = k bu + k bu k µ. {u<k} {u>k} bu = { u >k} Since b is non-decreasing, we have { u >k} { u k} bu µ. bu + bu { u k} bu + µ. u k bk bu bk bu max{bk, bk }. Then, we have bu { u k} max{bk, bk } = max{bk, bk } meas. Consequently, there exists a constant C 8 = max{bk, bk } such that bu 1 C 8 meas + µ.

12 12 I. KONATÉ, S. OUARO EJDE-2016/221 Lemma Assume hold true and µ M p b. If u is an entropy solution of 1.1, then there exists a constant D which depends on µ and such that D meas{ u > k}, k > minbk, bk and a constant D which depends on µ and such that meas { u > k } D Proof. Since b is non-decreasing, we have k 1 p M, k k > 0, u > k bu minbk, bk. For any k > 0, the relation 3.27 and the fact that bu minbk, bk give minbk, bk bu C 8 meas + µ. { u >k} Therefore, that is { u >k} minbk, bk meas u > k C 8 meas + µ = D; meas u > k For the proof of 3.30, we refer to [1]. D minbk, bk. We have the following two lemmas whose proofs can be found in [8]. Lemma Assume hold, and let f L 1. If u is an entropy solution of 1.1, then lim f χ { u >ht} = 0, h + where h > 0 and t > 0. Lemma Assume hold, and let f L 1. If u is an entropy solution of 1.1, then there exists a positive constant K such that ρ p i u pix1 χ Fh,k K, for i = 1,..., N, 3.31 where = {h < u h + k}, h > 0, k > 0. Step 2: Uniqueness of the entropy solution. Let h > 0 and u, v be two entropy solutions of 1.1. We write the entropy inequality corresponding to the solution u, with T h v as test function, and to the solution v, with T h u as test function. We obtain ft k u T h v + T k u T h v + F T k u T h v but k u T h v 3.32

13 EJDE-2016/221 GOOD RADON MEASURE 13 and x, v T k v T h u + ft k v T h u + F T k v T h u. Adding 3.32 and 3.33, we obtain [ N T k u T h v [ N + + Let us define x, v T k v T h u but k u T h v + bvt k v T h u ] F. T k u T h v bvt k v T h u fx[t k u T h v + T k v T h u]. ] F. T k v T h u E 1 = { u v k; v h}, E 2 = E 1 { u h}, E 3 = E 1 { u > h}, E 1 = { v u k; u h}, E 3 = E 1 { v > h} Assertion 1. E 3 and B = { u h sign 0 v k, v > h} F hk, 2k. Indeed, We decompose E 3 as E 3 = E 3 + E 3 where E+ 3 = { u v k, v h, u > h} and E3 = { u v k, v h, u < h}. In E+ 3, we have h v h and k u v k so that v k u v + k h + k. Since u > h and v + k h + k, we obtain h u h + k. Hence, E 3 +. In E3, we have h v h and k uv k so that vk u v+k h+k; since u < h and k h v k h k, we obtain h k u h so that h u h + k. Hence, E3. We split B as B = B + B = { u h k, v > h} { u + h k, v < h}. In B + B can be treated in the same way, we have k u h k and v > h so that h k u h + k. Hence B + F hk,2k. Assertion 2. On E 3 and on B we have according to Hölder inequality 1 F. u F p m p 1 m u p pm m, 3.35 E 3 E 3 E3 with 1 lim F p m p 1 m u p m p m = 0, h + E 3 E3 1 where + 1 p m p = 1. Indeed, m lim n + E 3 F p 1 m p m = 0,

14 14 I. KONATÉ, S. OUARO EJDE-2016/221 as F p m χ E3 belongs to L 1 and as E 3, then F p m χ E3 converges to zero as h +. Then, by Lebesgue dominated convergence theorem, lim F p m = 0. h + E 3 Now, it remains to prove that E 3 u p m is bounded with respect to h. We use the notation. I 1 = {i {1,..., N} : u 1}, I 2 = {i {1,..., N} : u > 1}. Then we have u pix = u pix + u pix i I 1 i I 2 u pix i I 2 u p m i I 2 u u p p m i I 1 u m N meas u p m x N meas i L p m C 9 u p m L p m N N meas, p m where we used Poincaré inequality. We deduce that u pix C 9 u p m N meas Choosing T h u as test fonction in 3.1, we obtain T k u T h u + but k u T h u ft k u T h u + F. T k u T h u According to the fact that T k u T h u = u on {h u h + k}, and zero elsewhere, and but ku T h u 0, we deduce from 3.37 that k { u h} u 2 f + C 3 C 9 p m 1 p m F C3 C 9 p m 2 1 p m u. 3.38

15 EJDE-2016/221 GOOD RADON MEASURE 15 Using 2.4 in the left hand side of 3.38, Young inequality in the right hand side of 3.38 and setting we obtain N C 3 k { u h} 2 C 10 = C 3 C 9 p m p m p m p m 1 p m u pix f + C 10 F p m + C C 9 u p m. 2 From 3.36 and 3.39, we obtain C 3 C 9 u p m F h,k k f + C 10 F p m + C 3C 9 u p m + N meas. { u h} 2 Therefore, C 3 C 9 u p m 2 F h,k k f + C 10 F p m + N meas. { u h}, 3.40 Since E 3, we deduce from 3.40 that E 3 u p m is bounded. Since B F hk,2k, reasoning as before, we obtain 1 F. u F p m p 1 m u p m p m, B with We have = B B 1 lim F p m p 1 m u p m p m = 0. h + B B { ut h v k} { ut h v k} + T k u T h v F T k u T h v { ut h v k} { v h} { ut h v k} { v >h} { ut k v k} { v h} { ut k v k} { v >h} T k u T h v T k u T h v F T k u T h v F T k u T h v

16 16 I. KONATÉ, S. OUARO EJDE-2016/221 = + { uv k} { v h} { uh sign 0 v k} { v h} { uv k} { v h} { uh sign 0 v k} { v >h} E 1 u v F u v u v u F T k u h sign 0 v F u v E 1 = u v + E 2 E 3 F u v F u v E 2 E 3 Then, we obtain { ut h v k} { ut k v k} E 2 { uh sign 0 v k} { v >h} F u T k u T h v F. T k u T h v u v + F u v + E 2 We deduce from 3.35 that { ut h v k} { ut k v k} E 2 u v { uh sign 0 v k} { v >h} E 3 F v F u E 3 E 3 T k u T h v F T k u T h v u v [ N ] v F v E 3 E 3 1 F u v F p m p m E 2 E 3 E3 F u. u v F u. B 1 u p m p m

17 EJDE-2016/221 GOOD RADON MEASURE 17 1 F p m p 1 m u p m p m B B According to 2.2 and the Hölder type inequality, we have where E 3 C 1 C 1 [ N N N ] u v F. v E 3 E 3 + F p E 3 j i x + u pix1 v + F v E 3 j i x + u 1 m pix1 p i,{h< u <h+k} p m E3 1 v p m p m, u pix1 p i,{h< u <h+k} = u For i = 1,..., N, the quantity v p i,{hk< u <h} pix1 x L p. i {h< u <h+k} i j i x + u pix1 p i,{h< u <h+k} according to 2.8 and Lemma Thanks to Lemma 3.13 and Assertion 2, the quantity v E 3 F p 1 m p m E3 1 v p m p m is finite p i,{hk< u <h} and converge to zero as h approaches infinity. Consequently, the second term at the right-hand side of 3.41 converges to zero as h approaches infinity. Therefore, { ut h v k} { ut h v k} I h + E 2 T k u T h v F. T k u T h v u v E 2 F. u v, with lim h + I h = 0. We adopt the same process replacing respectively E 1, E 3 by E 1 and E 3 to treat the second term of 3.34, which give { vt h u k} { vt h u k} J h + with lim h + J h = 0. E 2 x, v T k v T h u F. T k v T h u x, v v u E 2 F. v u,

18 18 I. KONATÉ, S. OUARO EJDE-2016/221 The other two terms in the left-hand side of 3.34 are denoted by K h = but k u T h v + bvt k v T h u. We have but k u T h v but k u v a.e. in as h +, but k u T h v k bu L 1. Then, by the Lebesgue dominated convergence theorem, we obtain but k u T h v = but k u v. lim h + In the same way, we obtain bvt k v T h u = lim h + but k v u. Then lim K h = bu bvt k u v. h + Now, let us consider the integral of the right-hand side of We have lim fx T k u T h v + T k v T h u = 0 a.e. in, h + fx T k u T h v + T k v T h u 2k f L 1. By the Lebesgue dominated convergence theorem, we obtain fx T k u T h v + T k v T h u = 0. lim h + After passing to the limit as h tends to infinity in 3.34, we obtain x, v u v { uv k} + bu bvt k u v 0. Since b and x, are monotone, we have x, v u v = 0, { uv k} 3.42 bu bvt k u v = We deduce from 3.43 that bu bv 1 k T ku v = lim k 0 We deduce from 2.3 and 3.42 that bu bv = u v = 0 a.e. in, for i = 1,..., N. Therefore, there exists c R such that u v = c a.e. in and using 3.44 we obtain bu = bv.

19 EJDE-2016/221 GOOD RADON MEASURE 19 References [1] B. K. Bonzi, S. Ouaro; Entropy solution for a doubly nonlinear elliptic problem with variable exponent, J. Math. Anal. Appl , [2] B. K. Bonzi, S. Ouaro, F. D. Y. Zongo; Entropy solution for nonlinear elliptic anisotropic homogeneous Neumann Problem, Int. J. Differ. Equ., Article ID [3] H. Brezis, W. Strauss; Semi-linear second-order elliptic equations in L 1, J. Math. Soc. Japan , [4] M. Colturato and M. Degiovanni; A Fredholm alternative for quasilinear elliptic equations with right-hand side measure, Adv. Nonlinear Anal , no. 2, [5] X. Fan; Anisotropic variable exponent Sobolev spaces and p -Laplacien equations, Complex Var. Elliptic Equ , [6] X. Fan, D. Zhao; On the spaces L px and W 1,px, J. Math. Anal. Appl , [7] Y. Fu, Y. Shan; On the removability of isolated singular points for elliptic equations involving variable exponent, Adv. Nonlinear Anal , no. 2, [8] I. Ibrango, S. Ouaro; Entropy solutions for anisotropic nonlinear Dirichlet problems, Ann. Univ. Craiova Ser. Mat. Inform , [9] B. Koné, S. Ouaro, S. Traoré; Weak solutions for anisotropic nonlinear elliptic equations with variable exponents, Electron. J. Diff. Equ , [10] B. Koné, S. Ouaro, S. Soma; Weak solutions for anisotropic nonlinear elliptic problem with variable exponent and measure data, Int. J. Evol. Equ. Vol. 5, Issue 2011, [11] O. Kovacik, Z. J. Rakosnik; On spaces L px and W 1,px, Czech. Math. J no 1, [12] J. Leray, J.-L. Lions; Quelques resultats de Visik sur les problèmes elliptiques non linéaires par les méthodes de Minty et Browder, Bull. Soc. Math. France , [13] M. Mihăilescu, P. Pucci, V. Rădulescu; Eigenvalue problems for anisotropic quasi-linear elliptic equations with variable exponent, J. Math. Anal. Appl , no. 1, [14] I. Nyanquini, S. Ouaro, S. Safimba; Entropy solution to nonlinear multivalued elliptic problem with variable exponents and measure data, Ann. Univ. Craiova Ser. Mat. Inform , no. 2, [15] S. Ouaro; Well-posedness results for anisotropic nonlinear elliptic equations with variable exponent and L 1 -data, Cubo J , [16] S. Ouaro, A. Ouedraogo, S. Soma; Multivalued problem with Robin boundary condition involving diffuse measure data and variable exponent, Adv. Nonlinear Anal , no. 4, [17] V. Rădulescu; Nonlinear elliptic equations with variable exponent: old and new, Nonlinear Anal , [18] V. Rădulescu, D. Repovš; Partial Differential Equations with Variable Exponents. Variational Methods and Qualitative Analysis, Monographs and Research Notes in Mathematics, CRC Press, Boca Raton, FL, [19] R. E. Showalter; Monotone operators in Banach space and nonlinear partial differential equations, Mathematical Surveys and Monographs, vol , American Mathematical Society. [20] M. Troisi; Teoremi di inclusione per spazi di Sobolev non isotropi, Ricerche di Matematica , Ibrahime Konaté Laboratoire de Mathématiques et Informatique LAMI, UFR. Sciences Exactes et appliquées, Université de Ouagadougou, 03 BP 7021 Ouaga 03, Ouagadougou, Burkina Faso address: ibrakonat@yahoo.fr Stanislas Ouaro Laboratoire de Mathématiques et Informatique LAMI, UFR. Sciences Exactes et Appliquées, Université de Ouagadougou, 03 BP 7021 Ouaga 03, Ouagadougou, Burkina Faso address: souaro@univ-ouaga.bf, ouaro@yahoo.fr