Nontrivial solutions for a Robin problem with a nonlinear term asymptotically linear at and superlinear at +

Transcript

1 Revista Colombiana de Matemáticas Volumen 4(008), páginas 0-6 Nontrivial solutions for a Robin problem with a nonlinear term asymptotically linear at and superlinear at Soluciones no triviales para un problema de Robin con un término no lineal asintótico en y superlineal en Rafael A. Castro T.,a Universidad Industrial de Santander, Bucaramanga, Colombia Abstract. In this paper we study the existence of solutions for a Robin problem, with a nonlinear term with subcritical growth respect to a variable. Key words and phrases. Robin problems, weak solutions, Sobolev spaces, functional, Palais-Smale condition, critical points. 000 Mathematics Subject Classification. 35J67, 35J5, 35J60. Resumen. En este artículo estudiamos la existencia de soluciones de un problema de Robin, con término no lineal con crecimiento subcrítico respecto a una variable. Palabras y frases clave. Problemas de Robin, soluciones débiles, espacios de Sobolev, funcionales, condición de Palais-Smale, puntos críticos.. Introduction In this paper, we study the existence of nontrivial solutions of the following problem with the real parameter α 0: u H (, ), (P) u = f(x, u(x)), in, γ u αγ 0 u = 0, on. a Supported by CAPES (Brazil) and DIF de Ciencias (UIS) Código CB00. 0

2 0 RAFAEL A. CASTRO T. Here is the Laplace operator, is a bounded domain in R n (n ) simply connected and with smooth boundary. The case α = 0 was studied by Arcoya and Villegas in []. The function f : R R, satisfies the following conditions: f 0 ) The function f is continuous. f ) f(x, s) c ( s σ ), x and s R, where the exponent σ is a constant such that f ) There exists λ > 0 such that < σ < n n if n 3, < σ < if n =. [f(x, s) λs] = 0, uniformly in x. s f 3 ) There exist s 0 > 0 and θ ( 0, ) such that 0 < F (x, s) θsf(x, s), x, s s 0, where F (x, s) = s 0 f(x, t)dt is a primitive of f. The boundary condition γ u αγ 0 u = 0 involves the trace operators: γ 0 : H () H / ( ) and γ : H (, ) H / ( ), where H (, ) = { u H () : u L () } with the norm u H (, ) = ( u H () u L ()) /, for each u H (, ), γ u H / ( ) and γ 0 u H / ( ). Identifying the element γ 0 u with the functional γ0 u H/ ( ) defined by γ0 u, w = (γ 0 u)wds, w H / ( ), the boundary condition makes sense in H / ( ). The mathematical difficulties that arise by involving this type of boundary conditions are in the Condition of Palais -Smale.. Preinary results To get the results of existence Theorems 4. and 4. we will use the following Theorem. Theorem. (Theorem of Silva, E. A.). Let X = X X be a real Banach space, with dim(x ) <. If Φ C (X, R) satisfies the Palais-Smale condition and the following conditions: I) Φ(u) 0, u X. II) There exists ρ 0 > 0 such that Φ(u) 0, u B ρ0 (0) X. Volumen 4, Número, Año 008

3 NONTRIVIAL SOLUTIONS FOR A ROBIN PROBLEM 03 III) There exist e X {0} and a constant M such that Φ(v te) M, v X and t > 0. Then Φ has at least a critical point different from zero. Proof. See [6, Lemma.3, p. 460]. We use the decomposition of H () as orthogonal sum of two subspaces established in [3]. We denote the sequence of eigenvalues of the problem { u = µu, in, (.) γ u αγ 0 u = 0, on, in the case α < 0 with {µ j } j=, where µ = inf u 0 u H () u α (γ 0u) ds u < 0. (.) With X we denote the space associated to the first eigenvalue µ, and with X = X the orthogonal complement of X respect to the inner product defined by (u, v) k = u v α (γ 0 u)(γ 0 v)ds k uv, u, v H (), (.3) where k is a positive constant suitably selected in [3]. Then and H () = X X, (.4) ϕ α (γ 0 ϕ) ds = µ ϕ, ϕ X. (.5) In the case α > 0, the constant k in (.3) is positive and arbitrary. We denote with {β j } j= the eigenvalues of Problem (.), in particular, we have β = u α (γ 0u) ds > 0. (.6) u inf u 0 u H () With Y we denote the space associated to β and Y = Y the orthogonal complement of Y with respect to the inner product defined by the formula (.3). Then and H () = Y Y, (.7) ϕ α (γ 0 ϕ) ds = β ϕ, ϕ Y. (.8) Revista Colombiana de Matemáticas

4 04 RAFAEL A. CASTRO T. 3. Condition of Palais-Smale Following Arcoya - Villegas [], Figueiredo [4], and using theorems 3., 3. and 3.3 of [3] we establish the conditions under which the functional Φ(u) = u α (γ 0 u) ds F (x, u), u H (), satisfies the Palais-Smale condition. We prove the cases α > 0 and α < 0. The condition of Palais-Smale (P.S) affirms: any sequence {u n } n= in H () such that Φ(u n ) c and Φ (u n ) = 0 in H (), contains a convergent subsequence in the norm of H (). In virtue of the density of C ( ) in H () and by the continuity of the operator γ 0 : H () L ( ), we have the following lemma: Lemma 3.. Let us suppose R n bounded with boundary of class C. If u H (), u (x) = max{u(x), 0} and u (x) = max{u(x), 0} then x x γ 0 ( u ) γ 0 ( u ) ds = 0. (3.) Proof. Let {u n } n= be a sequence in C ( ) such that u n u in H () then u n u and u n u in H (), see []. By the continuity of the operator γ 0 : H () L ( ) we have γ 0 (u n ) γ 0 (u ) and γ 0 (u n ) γ 0 (u ) in L ( ) then: ( γ 0 u ) ( γ 0 u ) ( ) ( ) ds = γ 0 u n γ0 u n ds = u n u n ds = 0. From (3.) we have: (γ 0 u) ( γ 0 u ) ( ds = γ0 u ) ds, (3.) (γ 0 u) ( γ 0 u ) ( ds = γ0 u ) ds. (3.3) Lemma 3. (Condition of Palais-Smale). If α < 0 we suppose (f 0 ), (f ), (f ) and (f 3 ). In the case α > 0, moreover, we also suppose the following conditions S ) The number λ of condition (f ) is not an eigenvalue of the operator with boundary condition γ u αγ 0 u = 0. S ) The numbers σ and θ of the conditions (f ) and (f 3 ) are such that σθ if n 3 and n σθ < if n =. Volumen 4, Número, Año 008

5 NONTRIVIAL SOLUTIONS FOR A ROBIN PROBLEM 05 Then u H () the functional Φ(u) = u α (γ 0 u) ds satisfies the condition of Palais-Smale (P.S.). F (x, u), Proof. Let {u n } n= be a sequence in H () such that Φ(u n ) = u n α (γ 0 u n ) ds F (x, u n ) C, (3.4) and v H () Φ (u n ), v = u n v α (γ 0 u n )(γ 0 v)ds f(x, u n )v ε n v, (3.5) for some constant C > 0 and ε n 0. To show that {u n } n= has a convergent subsequence it is enough to prove that {u n } n= is bounded. Case α < 0. We argue by contradiction. Let us consider a subsequence of {u n } n=, which we denote in the same way, such that u n =. Let z n = un u. Then there exists a subsequence of {z n n} which we denote in the same way, such that z n z 0 weakly in H (), z 0 H (), z n z 0 in L (), γ 0 z n γ 0 z 0 in L ( ), z n (x) z 0 (x) a.e. x, z n (x) q(x) a.e. x, q L (). (3.6) Dividing the terms of (3.5) by and taking the it v H () we obtain f(x, u n ) v = z 0 v α (γ 0 z 0 )(γ 0 v)ds. (3.7) From (3.7) with v = in, we get f(x, u n ) We obtain the desired contradiction in three steps. = α γ 0 z 0 ds <. (3.8) Revista Colombiana de Matemáticas

6 06 RAFAEL A. CASTRO T. First step. We shall prove z 0 (x) = 0 a.e. x, and γ 0 z 0 (x) = 0 a.e. x. (3.9) First we prove z 0 (x) 0 a.e. x, and γ 0 z 0 (x) 0 a.e. x. (3.0) Let = {x : z 0 (x) > 0} and be the measure of Lebesgue of. Choosing v = z 0 in (3.7) we obtain f(x, u n ) u n z 0 = z 0 ( α γ0 z ) 0 ds <. (3.) Using conditions (f 3 ) and (f ), for x we obtain f(x, u n (x))z 0 (x) (λq(x) K )z 0 (x). (3.) Indeed, condition (f 3 ) implies the existence of a constant c > 0 such that Then we can choose s > s 0 such that f(x, s) cs θ, s s 0. (3.3) f(x, s) λs, s s. (3.4) On the other hand, by (f ), for ε > 0 there is s < 0 such that f(x, s) λs ε, s s and x, (3.5) by the continuity of the function f there exists a constant K such that f(x, s) λs K, s (, s ] and x. (3.6) From (3.4) and (3.6) we get Now, using (3.7) with x we have f(x, s) λs K s R, x. (3.7) f(x, u n (x))z 0 (x) (λu n(x) K ) z 0 (x) (λz n (x) K )z 0 (x) (λq(x) K )z 0 (x). From (3.6) we have u n (x) = for a.e. x and using (3.3) the superlinearity of f in we have for a. e. x f(x, u n )z 0 (x) f(x, u n ) = u n (x) z n(x)z 0 (x) =. Volumen 4, Número, Año 008

7 NONTRIVIAL SOLUTIONS FOR A ROBIN PROBLEM 07 If > 0, by the Fatou s Lemma, we get f(x, u n (x)) = z 0 (x) u n f(x, u n ) u n z 0(x), then f(x, u n (x)) z 0 (x) =, u n in contradiction with (3.). Hence = 0 and z 0 (x) 0 a.e. x. If y, then γ 0 z 0 (y) = z 0 (x)dx 0. r 0 B(y, r) ) B(y,r) See [5, p. 43]. Below we prove that z 0 (x)dx = 0 = γ 0 z 0 (s)ds. (3.8) Let v = u n in (3.5) and subtracting this identity from (3.4), we obtain { } f(x, un ) u n F (x, u n ) ε n u n C. (3.9) Dividing this inequality by and passing to the it, we get f(x,u n) u n F (x, u n ) dx = 0. (3.0) On the other hand, given ε > 0, conditions (f 0 ) and (f ) imply the existence of a constant k ε > 0 such that f(x, s)s F (x, s) ε s k ε, s s. (3.) Using (3.) we have u n(x) s f(x,u n) u n F (x, u n ) ε u n K ε and, since ε is arbitrary, u n(x) s εc K ε f(x,u n) u n F (x, u n ) = 0. (3.) Revista Colombiana de Matemáticas

8 08 RAFAEL A. CASTRO T. The identities (3.0) and (3.) show that u n(x)>s Using (3.6) and condition (f 3 ), we obtain where u n(x)>s f(x,u n) u n F (x, u n ) ( ) { θ s ( ) { θ s f(x,u n) u n F (x, u n ) = 0. (3.3) ( ) θ s f(x, u n (x)) u n(x)>s u n } f(x, u n ) f(x, u n ) f(x, u n ) u n(x) s u n λ χ n z n K { if un (x) s χ n (x) =, 0 otherwise. Using (3.8), (3.0) and getting the it we have ( ) { 0 θ s f(x, u n ) λ ( ) { } = θ s α γ 0 z 0 ds λ z 0 ( ) { = θ s α γ 0 z 0 ds λ Hence Then }, X n z n dx K } } z 0 0. ( ) { } θ s α γ 0 z 0 ds λ z 0 = 0. (3.4) γ 0 z 0 ds = 0 = z 0. Using (3.0) we have (3.9). Now, the it (3.7) is Second step. We shall prove now that f(x, u n ) v = 0, v H (). (3.5) sup f(x, u n (x)) z n 0. (3.6) Volumen 4, Número, Año 008

9 NONTRIVIAL SOLUTIONS FOR A ROBIN PROBLEM 09 We denote: I = I = I 3 = u n(x)<0 0 u n(x) s 0 f(x, u n (x)) z n, f(x, u n (x)) z n, u n(x)>s 0 f(x, u n ) z n. Let us prove that I = 0. (3.7) sf(x,s)λs From condition (f ), we have s s = 0, so, given ε > 0 by the continuity of f there exists a constant c ε > 0 such that f(x, s)s λs c ε ε s, s 0, (3.8) then u n<0 f(x, u n )u n c ε dx ε u n<0 c (c λ) u n. u n<0 u n λ u n u n<0 Dividing the last inequality by and getting the it yields (3.7), because z n z 0 in L () and z 0 (x) = 0 a.e. x. Let us see that If L = max { f(x, s) : (x, s) [0, s 0 ] } then f(x, u n ) 0 u n(x) s 0 I = 0. (3.9) z n f(x, u n ) 0 u n s 0 u n (x) Ls 0. Therefore, I = 0. To prove that I 3 = 0, first we see that sup I 3 0. (3.30) Revista Colombiana de Matemáticas

10 0 RAFAEL A. CASTRO T. From (3.9) and (3.) we have { F (x, u n ) } u n>s 0 f(x, u n)u n u n s 0 f(x, u n)u n F (x, u n ) c ε n u n (ε u n k ε ) c ε n u n s 0 u n (ε u n k ε ) c ε n u n cε c ε n u n. On the other hand, condition (f 3 ) implies ( ) θ f(x, u n )u n u n>s 0 u n>s 0 { } f(x, u n)u n F (x, u n ). So, ( ) ( θ f(x, u n )u n c u n>s 0 cε ε n ). Dividing by, we obtain then sup I 3 0. Hence u n>s 0 f(x, u n ) z n c ( sup Third step. Finally we prove cε ε n ), f(x, u n ) z n = sup{i I I 3 } 0. sup f(x, u n ) z n =, (3.3) which contradicts (3.6). From (3.5) with v = z n and dividing by, we get ε n zn α (γ 0 z n ) f(x, u n ) ds z n ε n. By taking superior it we obtain (3.3). Volumen 4, Número, Año 008

11 NONTRIVIAL SOLUTIONS FOR A ROBIN PROBLEM Case α > 0. In this case we have that (u, v) = u v α (γ 0 u)(γ 0 v)ds, u, v H (), defines an inner product in H () and the norm u = (u, u) is equivalent to the usual norm of H (). As a matter of fact, from (.6) we get u β u, then u ( β ) u = d u, u H (). On the other hand, the inequality γ 0 u L ( ) c u implies, Then u ( αc ) u = d u, u H (). (d ) / u u d u, u H (). Henceforth, we denote the constants with the same letter c and expressions of the form cε n with ε n. Using the inner product previously defined and its associated norm, the inequalities (3.4) and (3.5) take the form Φ(u n ) = u n F (x, u n ) C, (3.3) Φ (u n ), v = (u n, v) f(x, u n )v ε n v d ε n v = ε n v, (3.33) where, ε n = 0 and v H (). Next we shall prove that the sequence {u n } n= is bounded. With this purpose first we establish the inequality u n c c and second, we prove that u n is bounded. The desired result will follow from the equality u = u u, u H (). First step. We shall prove F (x, u n )dx c ε n c u n L. (3.34) u n s 0 From (f 3 ) we have u n s 0 F (x, u n ) ( ) θ {f(x, u n )u n F (x, u n )} dx. (3.35) u n s 0 From (3.3) and (3.33) we get {f(x, u n )u n F (x, u n )}dx c ε n. (3.36) Revista Colombiana de Matemáticas

12 RAFAEL A. CASTRO T. Hence {f(x, u n )u n F (x, u n )}dx c ε n u n s 0 F (x, u n ) f(x, u n )u n. (3.37) u n<s 0 Conditions (f 0 ) and (f ) imply F (x, s) f(x, s)s c c s, s < 0, x, (3.38) from (3.37) and (3.38) we get {f(x, u n )u n F (x, u n )} dx c u n s 0 Now, from (3.35) and (3.39), we obtain (3.34). ε n c u n L. (3.39) Second step. We shall prove now that u n F (x, u n ) c ε n u n c u n L. (3.40) u n<0 Making v(x) = u n (x) in (3.33) we have u n f(x, u n )u n ε n u n. (3.4) u n<0 From (3.38) and (3.4) we obtain u n F (x, u n ) = u u n<0 n f(x, u n )u n u n<0 f(x, u n )u n F (x, u n ) u n<0 u n<0 u n f(x, u n )u n u n<0 {f(x, u n )u n F (x, u n )} dx u n<0 ε n u n f(x, u n )u n F (x, u n ) dx u n<0 c ε n u n c u n L. Volumen 4, Número, Año 008

13 NONTRIVIAL SOLUTIONS FOR A ROBIN PROBLEM 3 Third step. Next we shall verify the inequality u n c c u n. (3.4) From (3.3) we have u n F (x, u n ) F (x, u n ) u n u n 0 u n<0 u n u n F (x, u n ) = u n F (x, u n ) c, and with (3.40) we get u n u n 0 F (x, u n ) c u n<0 c c u n. F (x, u n ) u n Then the above inequality and (3.34) give u n c c u n F (x, u n ) u n 0 c c u n F (x, u n ) 0 u n(x) s 0 F (x, u n ) u n>s 0 c c u n ε n u n. Therefore u n c c u n ε n u n, since ε n = 0 this inequality yields (3.4). Fourth step. We consider the following exhaustive cases: i) There exists a constant c such that u n c, or ii) u n =, passing to a subsequence if it would be necessary. In case i), using (3.4) we have u n c, n N and, from the equality u = u u u H (), we conclude that (u n ) n= is bounded. Next let us prove that case ii) can not occur. First, from (3.8) and (3.4) we get u n λ ( ) u n c c u n. (3.43) If w n = u n then there exists w u n 0 H () and a subsequence from {w n } n= that we denote in the same way, such that it converges to w o weakly in H () Revista Colombiana de Matemáticas

14 4 RAFAEL A. CASTRO T. and strongly in L (). Let us see that w 0 0. Dividing (3.43) by u n we obtain λ wn c u n c u. n Taking it when n we get w0 = λ, therefore w 0 0. Let us see that λ is an eigenvalue and w 0 its eigenfunction. First we prove ( u, v ) n λ u n v ( c ε n u n c u n σ ) v. (3.44) L pσ From (3.33) we get ( u, v ) n λ u n v ( u n, v ) λ u n v f (x, u n ) v ( u n, v ) λ u n v ( u n, v) λ u n v f (x, u n ) v = (u n, v) f (x, u n ) v ε n v. Then ( u n, v ) λ u n v ε n v (u n, v) λ u n v f(x, u n )v ε n v u n v λ u n v f(x, u n )v. (3.45) Next we estimate λ u n v f(x, u n)v. Conditions (f0 ) and (f ) imply f(x, s) λs c, s 0, x. (3.46) Volumen 4, Número, Año 008

15 NONTRIVIAL SOLUTIONS FOR A ROBIN PROBLEM 5 using (3.46), we obtain λ u n v f(x, u n )v {f(x, u n ) λu n }v u n<0 f(x, u n )v u n 0 ( ) χ un f(x, u n ) λu n v f x, u n v ( ) c χ un v f x, u n v ( ) c v f x, u n v c v c v c u n σ v c v c u σ n v L pσ L q, where the function χ un is defined by { if un (x) < 0, χ un (x) = 0 if u n (x) 0, and p = n n, q = n n for n 3, and we take < p < /σθ as long as σθ < for n =. Then we obtain (3.44). Now, dividing (3.44) by u n, we get (w n, v) λ ( ) w n v c ε n u n u c σ L pσ n u v. (3.47) n It is evident that Let us prove that cε n u n u = 0. From (3.4) we have n = 0. u n u n σ L pσ u n = 0. (3.48) Conditions (f 0 ) and (f 3 ) imply the existence of positive constants K and c such that F (x, s) θks /θ c, for s > 0. (3.49) Then (3.34) and (3.49) give u n /θ c εn u n c u n. (3.50) Dividing (3.50) by u n /σθ we have u n /σθ u n /θ c u n /σθ u n ε n u n /σθ c u n σθ. (3.5) Revista Colombiana de Matemáticas

16 6 RAFAEL A. CASTRO T. From (S ) we have σθ = δ for some δ > 0. Hence /θ u n u n /σ = 0. (3.5) From (S ) and the choice of p in the case n = we have that < pσ θ, then pσ /pσ 0 u n u n /σ /θ θ u n u n /σ = 0. Therefore Then, the it in (3.47) yields (w n, v) λ u n u n /σ pσ /p w n v = (w 0, v) λ = 0. w 0 v = 0. Hence (w 0, v) = λ w 0 v, v H (), so, λ is an eigenvalue of, with boundary condition γ u αγ 0 u = 0. But this contradicts hypothesis (S ). Hence, u n cannot tend to when n. 4. Results of existence In this section, we establish the existence of solutions of Problem (P). Theorem 4.. Suppose n, α < 0, (f 0 ), (f ), (f ), (f 3 ), and let µ, µ be the first and the second eigenvalues of with the boundary condition of the problem (P ) { u = f(x, u(x)), in, γ u αγ 0 u = 0, on, such that f 4 ) f(x,s) s µ, s R {0}, x. f 5 ) There exist ε 0 > 0 and p > 0 such that µ < µ p < µ, and f(x, s) s µ p, s (ε 0, ε 0 ) {0}, x. Volumen 4, Número, Año 008

17 NONTRIVIAL SOLUTIONS FOR A ROBIN PROBLEM 7 Then Problem (P ) has at least one nontrivial solution. Proof. We prove the conditions of Theorem.. The functional Φ associated to the problem (P ) is defined by Φ(u) = u α (γ 0 u) ds which satisfies the Palais-Smale condition by Lemma (3.). Using decomposition (.4), H () = X X, we have F (x, u), I) Φ(u) 0, u X. Indeed, condition (f 4 ) implies that F (x, s) µ s, s R, x. Then for each u X Φ(u) = u α (γ 0 u) ds F (x, u) = µ u F (x, u) (by (.5)) µ u µ u = 0. II) There exists ρ 0 > 0 such that Φ(u) 0, u Bρ 0 (0) X. Condition (f 5 ) implies F (x, s) (µ p) s, for s < ε 0, and x. (4.) On the other hand, for s ε 0, condition (f ) implies the existence of a positive constant m 0 such that f(x, s) m 0 s σ, for s ε 0, and x. (4.) Now, (4.) and (4.) implies F (x, s) { (µ p) s, if s < ε 0, m s σ, if s ε 0, (4.3) for any constant m and x. Using (4.3), the variational characterization of µ, the Sobolev Imbedding Theorem, and the norm u k = (u, u) k, where the inner product (u, v) k is Revista Colombiana de Matemáticas

18 8 RAFAEL A. CASTRO T. defined in (.3), we get for u X, Φ(u) = u k k u F (x, u) u k k u k u u <ε 0 u ε 0 (µ p) u m u σ u <ε 0 u ε 0 = u k (µ k p) u k u m u <ε 0 u ε 0 u k (µ k p) u c u σ m u ε 0 (where c = k/ε σ 0 ) = u k (µ k p) u m (m = c m) u k (µ k p) (µ k) u k m ( ) pδ u m µ k = m 4 u m 3 u σ. u σ u ε 0 u σ u σ u ε 0 u σ u ε 0 u σ So, Φ(u) u (m 4 u m 3 u σ ), u X. (4.4) Recalling that σ > by condition f, the function d : [0, ) R defined by ( ) d(ρ) = m 4 ρ m 3 ρ σ achieves its global maximum in ρ 0 = m4 /(σ). m 3σ Then ( Φ(u) ρ 0 d(ρ 0 ) = ρ 0 ) m 4 > 0, u B ρ0 (0) X. σ III) There exists e X {0} and a constant M such that Φ(v te) M, v X and t > 0. If n the space H () is not contained in L (). Let e X be a function which is unbounded from above, and λ the number given by λ = e α (γ 0e) ds. (4.5) e Then, µ λ and µ < λ. The value of λ does not change by substituting e by te, then we suppose that e satisfies the condition (λ λ) e < δ (µ λ), (4.6) Volumen 4, Número, Año 008

19 NONTRIVIAL SOLUTIONS FOR A ROBIN PROBLEM 9 where λ is the positive constant of condition (f ), δ = δ µ and δ k is a positive constant such that δ v L v k, v X where v L = sup x v(x). Moreover δ satisfies δ v L v, v X. (4.7) If v X we get 0 = (v, e) k = v e k ve, = (µ k) ve α (γ 0 v)(γ 0 e)ds where µ k > 0, then ve = 0 and we obtain v e α (γ 0 v)(γ 0 e)ds = 0. (4.8) From (f 3 ) there exist m 5 > 0 and s s 0 such that F (x, s) λ s m 5 s /θ, s s and x. (4.9) Conditions (f 0 ) and (f ) imply the existence of a positive constant m 6 > 0 such that F (x, s) λ s m 6 s, s s and x. (4.0) Now, if v X and t > 0 then (4.5), (4.8), (4.9) and (4.0) yield Φ(v te) = (v te) α (γ 0 (v te)) F (x, v te) = µ v t λ e F (x, v te) = µ v t λ e F (x, v te) v(x)te(x) s F (x, v te) v(x)te(x)>s µ v t λ e λ (v te) m 6 v te m 5 (v te) v(x)te(x) s /θ v(x)te(x)>s (µ λ) m 6 t v t (λ λ) e m 6 v e m 5 v(x)te(x)>s (v te) /θ. Revista Colombiana de Matemáticas

20 0 RAFAEL A. CASTRO T. From (4.7) we have Φ(v te) δ (µ λ) v L m 6 v L t (λ λ) e m 6 t e m 5 (v te) v(x)te(x)>s /θ. (4.) Observing (4.) we have the following cases: Case. If λ > λ, then Φ(v te) δ (µ λ) v L m 6 v L t (λ λ) e m 6 t e, where the coefficients of v L and t are negative, therefore there exists a constant M > 0 such that Φ(v te) M, v X and t > 0. Case. If 0 < λ λ, and v 0 = min { v(x) : x } then v 0 t s or v 0 t > s. Let t s v 0. If v 0 = 0, from (4.) we have Φ(v te) δ (µ λ) v L m 6 v L s (λ λ) e m 6 s e. Since the coefficient of v L is negative, there is M > 0 such that Φ(vte) M. If v 0 0 then v 0 v L and from (4.) we have, Φ(v te) δ (µ λ) v L m 6 v L (s v 0 ) (λ λ) e (s v 0 )m 6 Using the inequality and calling we obtain e. (s v 0 ) ( s v 0 ), (4.) [ δ Φ(v te) (µ λ) (λ λ) c v L s (λ λ) ) c = m 6 ( e, (4.3) e ] v L e s m 6 e. Volumen 4, Número, Año 008

21 NONTRIVIAL SOLUTIONS FOR A ROBIN PROBLEM The coefficient of v L is negative, therefore there exists M 3 > 0 such that Φ(v te) M 3. In the case t > s v 0, let = {x : e(x) > } then > 0. Since the function e is not bounded from above, and {x : v(x)te(x) > s } then (4.) yields Φ(v te) δ (µ λ) v L m 6 v L t (λ λ) e m 6 t e m 5 (v 0 t) /θ. (4.4) Setting v 0 t = s we have Φ(v te) δ (µ λ) v L m 6 v L (s v 0) (λ λ) e m 6 (s v 0 ) e m 5 s /θ. From (s v 0 ) s v L and (4.3) we obtain ( δ ) Φ(v te) (µ λ) (λ λ) e v L c v L m 6 s e s (λ λ) e m 5 s /θ. Since the coefficients of v L and s/θ are negative, then there exists M 4 > 0 such that Φ(v te) M 4. If M = max{m, M 3, M 4 }, then Φ(v te) M v X, and t > 0. In the following theorem we consider the case α > 0, and we use the following condition (f ): the number λ of condition (f ) is such that λ > β, and λ β j, for j =, 3,, (λ is not an eigenvalue). Theorem 4.. Suppose: n, α > 0, (f 0 ), (f ), (f 3 ), (f ), (S ), and the conditions: (f 4 ) f(x,s) s β, s R {0}, x, (f 5 ) there exist ε 0 > 0 and β (β, β ) such that f(x, s) s Then the problem { (P ) has at least a nontrivial solution. β s (ε 0, ε 0 ) {0} x. u = f(x, u(x)), in, γ u αγ 0 u = 0, on, Revista Colombiana de Matemáticas

22 RAFAEL A. CASTRO T. Proof. To prove the conditions of Theorem. we use the decomposition (.7), H () = Y Y. The functional Φ associated to problem (P ) is Φ(u) = u α (γ 0 u) ds F (x, u), which satisfies the Palais-Smale condition by Lemma (3.). I) Φ(u) 0, u Y. Let u Y, condition (f4 ) and the equality (.8), yields Φ(u) = u α (γ 0 u) ds F (x, u) β u β u = 0. II) There exists ρ 0 > 0 such that Φ(u) 0 u B ρ0 (0) Y. Condition (f 5 ) implies F (x, s) β s, s ε 0, x. (4.5) On the other hand, for s ε 0 and x condition (f ) implies the existence of a positive constant m 0 such that f(x, s) m 0 s σ and its integrals yield F (x, s) m s σ, s ε 0 and x, (4.6) for any constant m > 0. From (4.5) and (4.6), we have F (x, s) β s m s σ, s R, x. (4.7) If u Y then Φ(u) = u α (γ 0 u) ds F (x, u) = u F (x, u) u β u m u σ u β u β mc u σ ) ( ββ d u mc u σ. Since σ >, there exist ρ 0 > 0 and a > 0 such that Φ(u) a > 0, u Bρ 0 (0) Y. III) There exist a function e Y {0} and a constant M > 0 such that Φ(v te) M, v Y and t > 0. Volumen 4, Número, Año 008

23 NONTRIVIAL SOLUTIONS FOR A ROBIN PROBLEM 3 Let e Y be a function which is unbounded from above and λ the number defined by λ = e α (γ 0e) ds, (4.8) e then β λ and λ > λ or λ λ. We suppose that e satisfies the condition ( (λ λ) e < δ λ ), (4.9) β where δ > 0 is such that For v Y and k > 0 we have, (v, u) k = (v, u) k Making u = e we get 0 = (v, e) k = (v, e) k then and We also use δ v L v, v Y. (4.0) vu = (β k) vu, u H (). ve = (β k) ve, ve = 0, (4.) (v, e) = 0. (4.) F (x, s) λ s m 5 s /θ, s s, x and (4.3) F (x, s) λ s m 6 s, s s and x. (4.4) If v Y and t > 0 then using (4.8), (4.), (4.), (4.3) and (4.4), we obtain Φ(v te) = (v te) α (γ 0 (v te)) F (x, v te) = β v t λ e F (x, v te) v(x)te(x) s F (x, v te) v(x)te(x)>s Revista Colombiana de Matemáticas

24 4 RAFAEL A. CASTRO T. β v t λ e λ vte s (v te) m 6 v te vte s λ (v te) m 5 (v te) /θ vte>s vte>s (β λ) v t (λ λ) e m 6 v te m 5 (v te) /θ vte>s (β λ) v m 6 v t (λ λ) e tm 6 e m 5 (v te) vte>s /θ. Using β v = v, v Y and (4.0) we get Φ(v te) δ ( λβ ) v L m 6 v L (4.5) t (λ λ) e tm 6 Observing (4.5) we have the following cases: e m 5 vte>s (v te) /θ. Case λ > λ. In this case, the coefficients of v L and t in (4.5) are negative, therefore, there exists a constant M > 0 such that Φ(v te) M v Y and t > 0. Case 0 < λ λ. If v 0 = min { v(x) : x } then v 0 t s or v 0 t > s. Let t s v 0. If v 0 = 0 then from (4.5) we have, Φ(v te) δ ( λβ ) v L m 6 v L s (λ λ) e s m 6 then there exists M > 0 such that Φ(v te) M. If v 0 0 then v 0 v L v Y. From (4.5) we have ) Φ(v te) ( δ λβ v L m 6 v L (s v 0 ) (λ λ) e (s v 0 )m 6 e, (4.6) e. Volumen 4, Número, Año 008

25 NONTRIVIAL SOLUTIONS FOR A ROBIN PROBLEM 5 Using (4.) and (4.3), we obtain [ ) δ Φ(v te) ( λβ (λ λ) c v L s (λ λ) ] e v L e s m 6 e. From (4.9) there exists M 3 > 0 such that Φ(v te) M 3. In the case v 0 t > s, let = {x : e(x) > }. Clearly, > 0. From (4.5) we have ) Φ(v te) ( δ λβ v L m 6 v L t (λ λ) e tm 6 e m 5 (v 0 t) /θ. (4.7) Making v 0 t = s, using (4.) and (4.3), we obtain [ ) δ ] Φ(v te) ( λβ (λ λ) e v L c v L sm 6 e s (λ λ) e m 5 s /θ. and there exists M 4 > 0 such that Φ(vte) M 4. If M = max {M, M 3, M 4 } then Φ(v te) M, v Y and t > 0. Recalling that for n = the space H () is contained in L (), and the fact that the above proofs require an unbounded function in H (), we conclude that theorems 4. and 4. can not be applied to the case n =. Acknowledgement. The author is deeply thankful to Helena J. Nussenzveig Lopes, and to Djairo G. De Figueiredo for their valuable comments. References [] Arcoya, D., and Villegas, S. Nontrivial solutions for a Neumann problem with a nonlinear term asymptotically linear at and superlinear at. Math.-Z 9, 4 (995), [] Castro, R. A. Convergencia de dos sucesiones. Revista Integración-UIS 3, (005), [3] Castro, R. A. Regularity of the solutions for a Robin problem and some applications. Revista Colombiana de Matemáticas (008). Este volumen. [4] de Figueiredo, D. G. On superlinear elliptic problems with nonlinearities interacting only with higher eigenvalues. Rocky Mountain Journal of Mathematics 8, (988), [5] Evans, L. C., and Gariepy, R. F. Lectures notes on measure theory and fine properties of functions. University of Kentucky, Kentucky, 99. Department of Mathematics. Revista Colombiana de Matemáticas

26 6 RAFAEL A. CASTRO T. [6] Silva, E. A. Linking theorems and applications to semilinear elliptic problems at resonance. Nonlinear Analysis, Theory, Methods & Applications 6, 5 (99), (Recibido en junio de 007. Aceptado en agosto de 008) Escuela de Matemáticas Universidad Industrial de Santander Apartado Aéreo 678 Bucaramanga, Colombia rcastro@uis.edu.co Volumen 4, Número, Año 008