EXISTENCE AND UNIQUENESS OF SOLUTIONS FOR NONLOCAL p-laplacian PROBLEMS

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1 Electronic Journal of ifferential Equations, Vol (2016), No. 274, ISSN: URL: htt://ejde.math.txstate.edu or htt://ejde.math.unt.edu EXISTENCE AN UNIQUENESS OF SOLUTIONS FOR NONLOCAL -LAPLACIAN PROBLEMS BEHROUZ EMAMIZAEH, AMIN FARJUIAN Abstract. We study the existence and uniqueness of ositive solutions to a class of nonlocal boundary-value roblems involving the -Lalacian. Our main tools are a variant of the Schaefer s fixed oint theorem, an inequality which suitably handles the -Lalacian oerator, and a Sobolev embedding which is alicable to the bounded domain. 1. Introduction We study the boundary-value roblem M( u ) u = f(x, u) in, u = 0 in which denotes the -Lalacian on. u = ( u 2 u) (1.1) and denotes the norm in W 1, 0 (), u = ( u dx ) 1/. As for the functions M : [0, ) [0, ) and f : R R, we shall refer to the following assumtions: (A1) M is continuous, and M(t) m 0 > 0, where m 0 is a constant. Moreover, the function: ξ(t) := M(t) 1 1 t 1/ is invertible, and henceforth we let q = 1. (A2) Let ˆM(t) := M(t ). Then for a constant κ (to be defined by (2.4)), the function ˆM is uniformly Hölder continuous with exonent 1 in the interval [0, κ). In other words L := su ˆM(t 1 ) ˆM(t 2 ) t 1,t 2 [0,κ), t 1 t 2 t 1 t 2 1 < The rincial eigenvalue of with irichlet boundary conditions on is defined as Λ := u dx u dx. (1.2) inf u W 1, 0 (), u Mathematics Subject Classification. 3505, 35J20, 35J92. Key words and hrases. Nonlocal roblems; existence; uniqueness; Schaefer s fixed oint theorem. c 2016 Texas State University. Submitted June 12, Published October 12,

2 2 B. EMAMIZAEH, A. FARJUIAN EJE-2016/274 The eigenvalue Λ is ositive, isolated, and simle [11]. We imose the following minimum condition on Λ: (A3) ˆm := m 0 a/λ > 0, in which m 0 comes from (A1), the constant a is introduced in (A5), and Λ is the rincial eigenvalue from (1.2). (A4) The function f is a Carathéodory function, and for some r (1, 1) f(x, s) A(x) s r + B(x), x, s R, (1.3) in which: A L () is a non-negative function B L 1+1/r () { n n if 1 < < n = if n. (A5) For some ositive constants a and b, sf(x, s) a s + b s, a.e. x, s R. (A6) f(x, s) 0 a.e. x, for all s 0 and f(x, 0) > 0, a.e. x. (A7) For a ositive constant A, (f(x, u) f(x, v))(u v) A u v 2, x, u, v R Remark 1.1. Note that when 2, the condition (A2) is satisfied when M is a constant function, hence the boundary value roblem (1.1) is no longer a nonlocal. Whence, even though the arguments to follow will hold for 2 but it is the case 1 < < 2 which is of interest. Remark 1.2. Let us mention that a function M that satisfies the conditions (A1) and (A2) (these are the main conditions on M), for the case (1, 2) is M(t) = m 0 + t β, where β 1 q. On the other hand, any function f(x, s) which is bounded and sf(x, s) is uniformly bounded in x satisfies (A4) (A7). The main results of this article are the following theorems. Theorem 1.3. Under assumtions (A1) (A7), the boundary value roblem (1.1) has a ositive solution. Theorem 1.4. Suose the conditions (A1) (A7) are satisfied. Then (1.1) has a unique ositive solution, rovided that L is sufficiently small and m 0 is sufficiently large. The secial case of roblem (1.1) when = 2 has been considered in [1], and [14]. In the former, the authors imose conditions on the functions M and f so that it is ossible to settle the issue of existence of solutions via the Mountain Pass Theorem. However, in the latter the authors use a different set of conditions, and aly the Galerkin method to obtain their results (see also [2]). Our aer is motivated by [14]. For the result of Theorem 1.3 regarding the existence of ositive solutions, we aly a variant of the Schaefer s fixed oint theorem couled with a well known maximum rincile. For the uniqueness result of Theorem 1.4, we use the ideas of [14]. In roving both existence and uniqueness of solutions we shall use an inequality which is articularly useful in dealing with the -Lalace oerator. See inequality (2.6) in Lemma 2.5.

3 EJE-2016/274 EXISTENCE AN UNIQUENESS OF SOLUTIONS 3 Nonlocal roblems have been used in modeling various hysical henomena, and the roblem (1.1) which we have considered in this note is related to the steady state version of the Kirchhoff equation [12] ( ) u tt M u 2 dx u = f(x, t), (1.4) where the coefficient of the diffusion term deends on the unknown function u(x, t) globally. It was the aer [13] by Lions that introduced an abstract setting for (1.4). Other relevant work are [3, 7, 9]. Some nonlocal roblems in statistical mechanics are studied in [4, 5]. 2. Preliminaries This section contains the basic material that we need for roving Theorems 1.3 and 1.4. We begin with the following definition. efinition 2.1. We say that u W 1, 0 () is a solution of (1.1) if the following integral equation is satisfied: M( u ) u 2 u v dx vf(x, u) dx = 0, v W 1, 0 (). (2.1) The convergence of the second integral in (2.1) follows from the following general result regarding Nemytskii maings. Lemma 2.2. Let g : R R be a Carathéodory function and suose that there is a constant c > 0, a function l(x) L γ () (where 1 γ ) and τ > 0 such that g(x, s) c s τ + l(x), x, s R. Then N g : L γτ () L γ () defined by N g (u)(x) = g(x, u(x)) is continuous and bounded, i.e. it mas bounded sets into bounded sets. For a roof of the above lemma, see [10, Theorem 2.3]. Let us review some basic facts regarding the roblem u = h(x) u = 0 on, in (2.2) where h(x) L 1+1/r (). It is well known, see for examle [15], that (2.2) has a unique solution u W 1, 0 () which is the unique minimizer of the strictly convex functional Φ(w) = 1 w dx hw dx relative to w W 1, 0 (). Therefore, the inverse maing ( ) 1 : L 1+1/r () W 1, 0 () which takes every h L 1+1/r () to the unique solution of (2.2) is well-defined. It is straightforward to verify that ( ) 1 (ηh) = η 1/( 1) ( ) 1 (h), h L 1+1/r (), η > 0 and that the following inequality holds: ( ) 1 (h) C h 1/( 1) 1+1/r, h L1+1/r () (2.3)

4 4 B. EMAMIZAEH, A. FARJUIAN EJE-2016/274 where C is a ositive constant. Henceforth we shall use C as a generic symbol for the several constants which aear in various laces in this document, whose values could be different. Lemma 2.3. Assume that u W 1, 0 () is a solution of (1.1), and that (A1) (A6) hold. Then ( b 1/q ) q/ u =: κ, (2.4) ˆmΛ 1/ where q =. Here denotes the n-dimensional Lebesgue measure of. 1 Proof. Setting v = u in (2.1), assumtion (A1) imlies m 0 u M( u ) u = f(x, u)u dx a u dx + b u dx (by assumtion (A5)) a u Λ + b 1/q u (by (1.2) and Hölder) a u Λ + b 1/q u (again by (1.2)) Λ 1/ From these inequalities, we infer that ˆm u b 1/q u Λ 1/, which in turn imlies the desired estimate (2.4). (2.5) We also need the following variant of the Schaefer s fixed oint theorem, see for examle [16], but we include the roof for the convenience of the reader. Lemma 2.4. Let X be a Banach sace and assume that: (a) P X is a non-emty, closed, and convex subset of X. (b) T : P P is a strongly continuous maing, i.e. T is continuous and for every bounded sequence (u n ) P, the image (T u n ) has a strongly convergent subsequence. (c) The set S = {x P x = λt x, for some λ [0, 1]} is bounded. Then T has a fixed oint, i.e. there exists x P such that T x = x. Proof. Consider the orthogonal rojection P : X P of X on P. This rojection satisfies: x X : P x x = inf x m, m P The maing T P : X P X is clearly strongly continuous. efine S = {x X : x = λ(t P )x, for some λ [0, 1]}. Hence, S S and S is bounded. Now we can invoke the classical Schaefer s fixed oint theorem, alied to T P, and deduce that T P has a fixed oint, say x 0 X. Thus: x 0 = T (P x 0 ) x 0 range(t ) x 0 P x 0 = P x 0 Thus, x 0 is a desired fixed oint of T. x 0 = T x 0

5 EJE-2016/274 EXISTENCE AN UNIQUENESS OF SOLUTIONS 5 We shall also need the following result, see for examle [8] and [6] Lemma 2.5. For any vectors X, Y R n, the following inequalities hold: X Y if 2 C X 2 X Y 2 Y, X Y X Y 2 ( X + Y ) if 1 2, 2 (2.6) in which, denotes the usual dot roduct in R n, and C is a constant deending on. The following lemma is elementary, so we omit its roof. Lemma 2.6. Let t n R t and fn L f. Then t n f n L tf. L Lemma 2.7. If f 1+1/r n f, then ( ) 1 (f n ) W 1, 0 ( ) 1 (f). Proof. Set v n = ( ) 1 (f n ) and v = ( ) 1 (f). Thus: and v n = f n in v n = 0 on v = f v = 0 in on The derivation of the following equation is then straightforward. ( v n 2 v n v 2 v) ( v n v) dx = (f n f)(v n v) dx. (2.7) Alying the Hölder s inequality and the embedding W 1, 0 () L 1+1/r (), a bound on the integral on the right hand side of (2.7) is obtained as follows: (f n f)(v n v) dx f n f 1+1/r v n v r+1 (2.8) C f n f 1+1/r v n v. Now we consider two cases: Case 2. From Lemma 2.5 (setting X = v n, Y = v), (2.7) and (2.8), we obtain v n v C f n f 1+1/r v n v, hence v n v C f n f 1/( 1) 1+1/r. Thus, v n v in W 1, 0 (). Case 1 2. This case requires more work. We begin with the observation v n v v n v = ( v n + v ) (2 ) 2 ( ( v + v ) (2 ) 2 dx v n v 2 ) /2( ( v n + v ) 2 dx ( v n + v ) dx ) (2 )/2 (2.9)

6 6 B. EMAMIZAEH, A. FARJUIAN EJE-2016/274 This follows from Hölder s inequality which is alicable since 2 inequality (2.3), we obtain v n C f n 1/( 1) 1+1/r (f n ) is bounded in L 1+1/r (), we infer that max( v n, v ) C, for all n N. Thus, from (2.9) we obtain ( v n v C 1. Alying and v C f 1/( 1) 1+1/r. Since v n v 2 ( v n + v ) 2 ) /2 (2.10) Now, by setting X = v n and Y = v in Lemma 2.5, together with (2.7), (2.8), and (2.10) we find that v n v = v n C f n f 1+1/r v n v, This imlies that v n v C f n f 1/( 1) 1+1/r. So, v n v in W 1, 0 (), as desired. The roof is comlete. 3. Proofs of main theorems To rove Theorem 1.3 we shall aly Lemma 2.4. To this end, we set P = L r+1 + (). Note that by Lemma 2.2 we have u P : N f (u) L 1+1/r (). From assumtion (A6) we infer that N f (u) is non-negative. For every u P, we define: T u = t 1/ v v, in which v := ( ) 1 (N f (u)) and t := ξ 1 ( v ). Observe that w = T u satisfies M( w ) w = f(x, u) w = 0 on. in (3.1) Since T u W 1, 0 (), the embedding W 1, 0 () L r+1 () imlies T u L r+1 (). Thus, by alying a classical maximum rincile (see for examle [17]) to (3.1), we deduce that w is ositive, i.e. T u L r+1 + (). The above discussion ensures that the maing T : P P is well defined. Note that if u is a fixed oint of T, then u will be a solution of (1.1). The existence of such a fixed oint will confirm the assertion of Theorem Proof of Theorem 1.3. We just need to verify that the maing T satisfies the hyotheses of Lemma 2.4. Continuity. Let (u n ) P be a sequence such that u n u in L r+1 (). Note that since P is closed then u must be non-negative. We need to show that T u n T u in L r+1 (). In view of the embedding W 1, 0 () L r+1 (), it suffices to show T u n T u in W 1, 0 (). To this end, we first recall Lemma 2.2 which ensures that N f (u n ) N f (u) in L 1+1/r (). Whence, by Lemma 2.7: ( ) 1 (N f (u n )) ( ) 1 (N f (u)) in W 1, 0 (). By the continuity of the norm we also have On the other hand, ( ) 1 (N f (u n )) ( ) 1 (N f (u)). T u n = t 1/ ( ) 1 (N f (u n )) n ( ) 1 (N f (u n )),

7 EJE-2016/274 EXISTENCE AN UNIQUENESS OF SOLUTIONS 7 in which t n = ξ 1 ( ( ) 1 (N f (u n )) ). Since ξ is continuous, we obtain t n t := ξ 1 ( ( ) 1 (N f (u)) ). Now we aly Lemma 2.6 to conclude that T u n T u in W 1, 0 (), as desired. Comactness. Consider a bounded sequence (u n ) P. Setting w n = T u n, we will have M( w n ) w n = f(x, u n ) in (3.2) w n = 0 on. From (3.2) we obtain M( w n ) w n = An alication of Hölder s inequality then gives f(x, u n )w n dx. M( w n ) w n N f (u n ) 1+1/r w n r+1. (3.3) The inequality (3.3), the embedding W 1, 0 () L r+1 (), and the assumtion (A1) together lead to m 0 w n C N f (u n ) 1+1/r w n, Hence, we get w n C N f (u n ) /r. This, couled with the boundedness of the oerator N f (see Lemma 2.2), imlies that (w n ) is bounded in W 1, 0 (). So, there exists a subsequence (w nj ) (w n ) such that w nj w in W 1, 0 (), for some w W 1, 0 (). Since the embedding W 1, 0 () L r+1 () is comact, we deduce that w nj w in L r+1 (). This means that (T u nj ) is strongly convergent in L r+1 () and as a result T : P P is comact. Boundedness of S. The final ste is to rove the boundedness of the set S = {u P : u = λt u, for some λ [0, 1]}. To that end, let us fix a u S and assume that u = λt u for some λ [0, 1]. Thus, we must have u = λt 1/ ( ) 1 (N f (u)) ( ) 1 (N f (u)), where t = ξ 1 ( ( ) 1 (N f (u)) ). Since u = λt 1/ and assuming that λ 0, then t = u /λ and M( u λ ) = M(t). So, we obtain where q = 1 ( u ) M λ u = M(t)λ 1 t 1/q ( ) 1 f(x, u) (N f (u)) 1 = λ 1 f(x, u), 1,. Since u W0 (), from (3.4), (A1), and (A5) one gets ( u m 0 u ) M λ u = λ 1 f(x, u)u dx λ 1( ) a u dx + b u dx a u Λ + b 1/q u Λ 1/. (3.4) (3.5)

8 8 B. EMAMIZAEH, A. FARJUIAN EJE-2016/274 From (3.5) and (A3) we obtain u κ (which was defined in (2.4)). Note that in case λ = 0, this last inequality trivially holds. Finally, by invoking the embedding W 1, 0 () L r+1 (), we infer that u r+1 C. Whence, S is bounded, as desired. This comletes the roof Proof of Theorem 1.4. The existence of solutions is guaranteed by Theorem 1.3. We rove uniqueness by contradiction. Let us assume that u 1 and u 2 are two solutions of (1.1), satisfying M( u i ) u i = f(x, u i ) u i = 0 on, in for i = 1, 2. From (3.6) we obtain (M( u 1 ) u 1 2 u 1 M( u 2 ) u 2 2 u 2 ) w dx = (f(x, u 1 ) f(x, u 2 ))w dx. By rearranging terms, we obtain M( u 2 ) ( u 1 2 u 1 u 2 2 u 2 ) w dx = (M( u 2 ) M( u 1 )) u 1 2 u 1 w dx + (f(x, u 1 ) f(x, u 2 ))w dx L u2 u 1 1 u 1 1 w + A w, (3.6) (3.7) (3.8) where we have used (A2) and (A7) in the last inequality. Note that ( L u 2 u 1 1 u 1 1 w + A w Lκ 1 + A ) w. (3.9) Λ On the other hand, using similar arguments as in the roof of Lemma 2.7, we obtain the estimate ( u 1 2 u 1 u 2 2 u 2 ) w dx C w, (3.10) in which the constant C deends on κ if < 2, otherwise it does not. From (3.8), (3.9), and (3.10) we obtain ( m 0 C w Lκ 1 + A ) w. (3.11) Λ Since u 1 u 2, (3.11) imlies Now, if m 0 is large enough, and L is small enough as m 0 C AΛ Lκ 1. (3.12) m 0 > AC 1 Λ and L < m 0C AΛ κ 1. then we obtain the desired contradiction, and the roof is comlete.

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