Regularity of the solutions for a Robin problem and some applications


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1 Revista Colombiana de Matemáticas Volumen 42(2008)2, páginas Regularity of the solutions for a Robin problem and some applications Regularidad de las soluciones para un problema de Robin y algunas aplicaciones Rafael A. Castro T. 1,a 1 Universidad Industrial de Santander, Bucaramanga, Colombia Abstract. In this paper we study the regularity of the solutions for a Robin problem, with a nonlinear term with subcritical growth respect to a variable. We establish the Sobolev space H 1 () as the orthogonal sum of two subspaces, and we give the first step to demonstrate the existence of solutions of our problem. Key words and phrases. Robin problems, trace operators, variational formulation, weak solutions, Sobolev spaces, bootstrapping, Green formula, orthogonal sum of subspace Mathematics Subject Classification. 35B65, 35J25, 35J60, 35P99. Resumen. En este artículo estudiamos la regularidad de las soluciones de un problema de Robin, con término no lineal con crecimiento subcrítico respecto a una variable. Expresamos el espacio de Sobolev H 1 () como la suma de dos subespacios dando el primer paso para la demostración de existencia de soluciones de nuestro problema. Palabras y frases clave. Problemas de Robin, operador trazo, formulación variacional, soluciones débiles, espacios de Sobolev, argumento iterativo, formula de Green, suma ortogonal de subespacios. a Supported by CAPES (Brazil) and DIF de Ciencias (UIS) Código CB
2 128 RAFAEL A. CASTRO T. 1. Introduction We examine the regularity of the solutions to the problem u H 1 (, ), (P) u = f(x, u(x)), in, γ 1 u + αγ 0 u = 0, on, here is the Laplacian operator, is a bounded domain in R n (n 2) simply connected and with smooth boundary. The function f : R R, satisfies the conditions: f 0 ) The function f is continuous. f 1 ) There exists a constant c > 0 such that f(x, s) c (1 + s σ ), x and s R, where the exponent σ is a constant such that 1 < σ < n+2 n 2 if n 3, 1 < σ < if n = 2. The boundary condition γ 1 u + αγ 0 u = 0 involves the trace operators: γ 0 : H 1 () H 1/2 ( ), and γ 1 : H 1 (, ) H 1/2 ( ), where H 1 (, ) = {u H 1 () : u L 2 ()} with the norm ( 1/2 u H 1 (, ) = u 2 H 1 () + u 2 L ()), 2 for each u H 1 (, ), γ 1 u H 1/2 ( ), and γ 0 u H 1/2 ( ). Identifying the element γ 0 u with the functional γ0u H 1/2 ( ) defined by γ0 u, w = (γ 0 u)wds, w H 1/2 ( ), the boundary condition makes sense in H 1/2 ( ) (Theorem 2.3), and it appears in the functional of the problem (P), see Theorems 3.1, 3.2 and 3.3 in section 3. This work was motivated by ArcoyaVillegas [1], who proved the existence of nontrivial solution of the nonlinear Neumann s problem u = f(x, u), in, u η = 0, on. We study the existence of solutions of the problem (P) in [4]. We write this paper to separate the regularity conditions of the additional existence conditions. Volumen 42, Número 2, Año 2008
3 REGULARITY OF THE SOLUTIONS FOR A ROBIN PROBLEM Preliminary results In this section, we introduce some notations and known theorems to familiarize the reader with the boundary condition γ 1 u+αγ 0 u = 0. We also remember that is an isomorphism between some spaces, properly enunciated in Theorem 2.5. Using the Trace Theorem, Theorem 2.5 and the bootstrapping we prove that the weak solutions of Problem (P) belong to the space H 2 (). (Theorem 3.2). The norm u 2 = u 2 + u 2, (2.1) is the usual norm of the Sobolev space H 1 (). Theorem 2.1. Suppose bounded with boundary of class C 1. Then, there exists a continuous and linear function γ 0 : H 1 () L 2 ( ) such that: a) γ 0 u = u, if u W 1,2 () C ( ), b) γ 0 u L 2 ( ) c 1 u, u H 1 (), here the constant c 1 depends on. The value γ 0 u is called trace of u on. Proof. See [7, Theorem 1, p. 258]. Theorem 2.2 (Trace Theorem). Let R n be open, bounded, and with smooth boundary. Then the trace operators γ j can be extended to continuous linear operators, mapping H m () onto H m j 1/2 ( ), for 0 j m 1. Moreover, the operator γ = (γ 0, γ 1,..., γ j,..., γ m 1) maps H m () onto the product space m 1 j=0 Hm j 1/2 ( ). Finally, the space H0 m () is the Kernel of the operator γ. Proof. See [2, Theorem 3.1, p. 189] or [5, Theorem 5, p. 905]. Theorem 2.3 (Green Formula). There exists a unique operator γ 1 mapping H 1 (, ) into H 1/2 ( ) such that Green Formula holds: γ 1 u, γ 0 v = v u + u v, u H 1 (, ), v H 1 (). (2.2) Remark. The space C ( ) is dense in H 1 (, ). Proof. See [2, Theorem 2.1, p. 174]. Theorem 2.4 (Compactness of the trace operator). Let R n be an open set, bounded, and with boundary C 0,1, m 1 an integer and 1 p < +. Then the next claims are true : a) If mp < n and 1 q < p(n 1) n mp, then there exists a unique mapping γ 0 : W m,p () L q ( ), linear and continuous, such that if u C ( ) then γ 0 u = u. If p > 1 then γ 0 is compact. b) If mp = n, then the affirmation (a) is valid for every q 1. Revista Colombiana de Matemáticas
4 130 RAFAEL A. CASTRO T. c) If mp > n, the trace γ 0 u of u W m,p () C ( ) is the classic restriction u. Proof. See [10, Theorem 6.2 p. 107], [8, Theorem ] or [3]. For the reader s convenience, we recall some notation and definitions of normality and covering. If x = (x 1, x 2,..., x n ) R n and α = (α 1,..., α n ) an n tuple of nonnegative integers α j, then x α denotes the monomial x α1 1 xα2 α = n j=1 α j. Similarly, if D j = x j is the differential operator of order α xαn n, which has degree for 1 j n, then D α = D α Dαn n Let A be a partial differential linear operator of order 2m, defined by Au = ( 1) k D k ( a kh D h u ), (2.3) k, h m where a kh C ( ), with real values. Let B 0,..., B m 1 be m boundary operators defined by B j u = b jh D h u, j = 0, 1,..., m 1, (2.4) h m j where 0 m j 2m 1 and the functions b jh are infinitely differential functions over. Definition 2.1. We say that the set {B j } m 1 j=0 is normal according to Aronszajn and Milgram, if for j i, m j m i, x, and normal vector η 0 to in x, then Q j (x, η) = b jh η h 0, j = 0, 1,..., m 1. h =m j Definition 2.2. The set {B j } m 1 j=0 covers the operator A defined in (2.3), if for each x and every pair of real nonnull vectors ξ, ξ tangent and normal to in x respectively, then the polynomial in the variable τ, P (x, ξ + τξ ) = a kh (ξ + τξ ) k+h, k = h =m j has m roots: τ 1 (ξ, ξ ),, τ m (ξ, ξ ), with positive imaginary parts, and the m polynomials in τ Q j (τ) := Q j (x, ξ + τξ ) = b j,h (ξ + τξ ) h, h =m j are linearly independent module the polynomial m i=1 (τ τ i(ξ, ξ )). Definition 2.3. We say that the set of operators satisfies condition (N R) if: {A, B 0, B 1,, B m 1 }, Volumen 42, Número 2, Año 2008
5 REGULARITY OF THE SOLUTIONS FOR A ROBIN PROBLEM 131 a) The set of boundary operators B = {B j : j = 0, 1,, m 1} is normal according to Aronszajn and Milgram, and b) The set B covers operator A. Proposition 2.1. The operators n 2 =, and the boundary operator B 0 u = x 2 j=1 j n j=1 η j (x) u x j, where η(x) = (η 1,, η n ) is a unit normal vector field over and external to, satisfies Condition (N R). Proof. a) Let x and v(x) = (v 1,, v N ) 0 a normal vector field to, N Q 0 (x, η) = η j (x) v j (x) = η v 0, j=1 therefore, the operator {B 0 } is normal according to Aronszajn and Milgram. b) Now, the operator B 0 covers the operator. In fact, let ξ = (ξ 1,, ξ n ) and ξ = (ξ 1,, ξ n ) be tangent and normal vectors respectively to at the point x. By the ellipticity of the operator we have P (x, ξ + τξ ) = ξ + τξ 2 = ξ 2 τ 2 ξ 2, then P (x, ξ + τξ ) has 2 roots: τ 1 = ξ ξ i, and τ 2 = τ 1. On the other hand, the polynomial in τ N Q 0 (x, ξ + τξ ) = n j (x)(ξ j + τξ j ) = (η(x) ξ )τ = η ξ τ, j=1 is not multiple of the polynomial τ τ 1. Let 1 < p <, r 0 and η(x) = (η 1,, η n ) be a real vector normal field over and external to. We define : n N = u W 2,p () : u = 0, B 0 u = η j (x) u = 0 x j=1 j, { } N = u W 2,p () : ( ) u = 0V, B0u = 0, where p = p p 1 and (( ), B0) is the formal adjoint operator of (, B 0 ). It is easy to see that (, B 0 ) is selfadjoint. Spaces N and N are subspaces of C ( ), and thus they neither depend on p nor on p. If is simply connected then N = N is the space of the constants. In this work, the set is simply connected. We will consider the sets: W 2+r,p B 0 () = { u W 2+r,p () : B 0 u = 0 }, and {W r,p (); N } = { u W r,p () : u(x)dx = 0}. Revista Colombiana de Matemáticas
6 132 RAFAEL A. CASTRO T. Theorem 2.5. The operator is an isomorphism on {W r,p (); N } of ()/N for all real r 0 and 1 < p < +. W 2+r,p B 0 Proof. See [9, Theorem 3.1]. 3. Regularity results In Theorem 3.1 we establish the derivative of the functional Φ : H 1 () R, corresponding to our problem (P). In Theorem 3.2 we show that the critical points of Φ belong to the space H 2 (). Finally in Theorem 3.3 we obtain the variational formulation of the problem (P). Thus Theorem 3.2 is a regularity result for weak solutions of the problem (P). Theorem 3.1. The functional Φ : H 1 () R, defined by Φ(u) = 1 u 2 + α (γ 0 u) 2 ds F (x, u), 2 2 where α R, F (x, s) = s 0 f(x, t) dt belongs to class C1 (H 1 (), R) and its derivative is Φ (u), v = u v + α (γ 0 u)(γ 0 v)ds f(x, u)v, u, v H 1 (). (3.1) Proof. If Φ 1 (u) = 1 2 Φ 1 (u + tv) Φ 1 (u) = t (γ 0 u) 2 ds, {γ 0 u.γ 0 v + t2 } (γ 0v) 2 ds, then, Φ Φ 1 (u + tv) Φ 1 (u) 1 (u), v = lim = (γ 0 u)(γ 0 v)ds. t 0 t Using Theorem 2.1, the Gateaux derivative of Φ 1 is continuous; indeed, Φ 1(u), v γ 0 u L 2 ( ) γ 0 v L 2 ( ) C u v. Hence, Φ 1 C 1 ((), R). It is well known that the functional Φ 2 (u) = 1 u 2 F (x, u), 2 belongs to the class C 1 (H 1 (), R) whose derivative is Φ 2 (u), v = u. v f(x, u)v. Then we obtain (3.1). Volumen 42, Número 2, Año 2008
7 REGULARITY OF THE SOLUTIONS FOR A ROBIN PROBLEM 133 Theorem 3.2. Let R n, n 2 be a bounded and simply connected domain with smooth boundary, f : R R a function satisfying (f 0 ) and (f 1 ), and α a real number. If u H 1 () satisfies Φ (u), v = u v + α (γ 0 u)(γ 0 v)ds f(x, u)v = 0, (3.2) for all v H 1 (), then u H 2 (). Proof. Since γ 1 : H 2 () H 1/2 ( ) is onto (Trace Theorem), and αγ 0 u H 1/2 ( ), there exists w H 2 () such that w η = αγ 0u. (3.3) From (3.2), (3.3) and the Green formula we obtain h v = gv, v H 1 (), (3.4) where (i) First we prove that h = u w, and g(x) = f(x, u) + w(x). (3.5) g { W 0,p1 (); N }, where 1 < p 1 2. (3.6) Using Condition (f 1 ) in the cases: a) If n = 2, then b) If n > 2, taking into account that If p 1 = 2 σ then 1 < 2n Consequently f L 2 (), and so g L 2 (). (3.7) 1 < σ < n + 2 n 2, and 2 = 2n n 2. n + 2 < p 1 < 2n n 2. (3.8) ( f(x, u) p1 C 1 + u 2 ) < +, (3.9) thus, f L p1 (), then g L p1 () for 1 < p 1 2. From (3.4) with v = 1 over, we have Thus, g { W 0,p1 (); N }. (ii) Now, we will prove that g(x)dx = 0. u W 2,p1 (). (3.10) Revista Colombiana de Matemáticas
8 134 RAFAEL A. CASTRO T. By virtue of Theorem 2.5, there exists a unique h W 2,p1 B 0 ()/N such that { h = g, in, h (3.11) η = 0, on. In the case n 3, we have 1 p = p 1 > 1 n+2 2n = 1 2, then p 1 < 2, and hence H 1 () L p 1 (). If n = 2, H 1 () L p 1 (). From (3.11) we get v h = vg, v H 1 (). (3.12) Integrating by parts, the first integral of (3.12) gives h v = gv, v H 1 (). (3.13) Subtracting (3.13) from (3.4) we obtain (h h) v = 0, v H 1 (). (3.14) Let us see that h W 1,2 (). Since h W 2,p1(), if p 1 = 2 then h W 1,2 (); if p 1 < 2, then we have the cases: n = 2 and n > 2. If n = 2, given that p 1 > 1 then 2 < 2p1 2 p 1 ; if n > 2, as p 1 > 2n np1 n+2, then 2 < n p 1. Therefore, in both cases W 2,p1 () W 1,2 (), then h W 1,2 (). Making v = h h in (3.14) we obtain v 2 dx = 0. By the Poincaré inequality we have that v(x) = v = 1 v dx, x. Then h(x) = h(x)+v, thus D α h = D α h for any index α, with α 2. Since h W 2,p1 (), then h W 2,p1 (), therefore u W 2,p1 (). (iii) Finally we will prove that u W 2,2 (). (3.15) If n = 2, by (3.7) and case (i), g { W 0,2 (); N }, and by case (ii), u W 2,2 (). If n > 2, we have the following cases: a) 2p 1 > n, b) 2p 1 = n, c) 2p 1 < n. In case a), W 2,p1 () C ( ) L (), thus u L q (), q [1, + ), and on the other hand, we have f(x, u) 2 ( c 1 + u 2σ ) < +, (3.16) then f L 2 (), g L 2 (), thus by cases (i) and (ii), u W 2,2 (). Volumen 42, Número 2, Año 2008
9 REGULARITY OF THE SOLUTIONS FOR A ROBIN PROBLEM 135 In case b), W 2,p1 () L q (), q [p 1, ), 2σ > 2 > p 1 and by (3.16), then f L 2 (). Similarly to the last case u W 2,2 (). In case c), we use bootstrapping. As W 2,p1 () L p 1 (), where p 1 = thus u L p 1 (). Let p2 = p 1 σ be, and f(x, u) p2 c ( ) 1 + u p 1 < +. np1 n 2p 1, Then, f L p2 (), g L p2 (), thus u W 2,p2 (). If p 2 < 2 and 2p 2 < n then u L p 2 (), p 2 = np2 n 2p 2. Considering p 3 = p 2 σ then f L p3, g L p3 (), thus u W 2,p3 (). Therefore, we obtain the sequence {p j }, such that 1 < p 1 < p 2 < < p s, where p s 2. In fact, since p 1 > 2n n+2, there exists ε > 0 such that p 1 = (1 + ε) and we obtain 2n n+2 p 2 = p 1 /σ (n 2)(1 + ε) p 1 2 = > 1 + ε, /σ n 2 4ε then p 2 p 1 > ε. If 1 < p 1 < p 2 < < p i 1 < p i, where j = 2, 3,, i, then p i+1 p i = p i p i 1 = p i(n 2p i 1 ) p i 1 (n 2p i ) > p i > 1 + ε. p i 1 p j p j 1 > 1 + ε for Hence {p j } is an strictly increasing sequence, therefore, there exists p s 2 and u L p s 1 (), then f L p s (), g L ps () and by case (ii) u W 2,ps (). Therefore u W 2,2 (). Theorem 3.3. Based on the conditions in Theorem 3.2 on, f and α, the problems (P) and (Q) are equivalent: i) u H 1 (, ), (P) ii) u = f(x, u(x)), x, iii) γ 1 u + αγ 0 u = 0, on, and (Q) { i) u H 1 (), ii) u v = f(x, u)v α (γ 0u)(γ 0 v)ds, v H 1 (). Proof. Let u be a solution to problem (P), then ( u)v = f(x, u(x))v, v H 1 (). Using the Green Formula (2.2) we obtain u v γ 1 u, γ 0 v = f(x, u)vdx, v H 1 (). Now, taking the boundary condition, u is a solution to problem (Q). Revista Colombiana de Matemáticas
10 136 RAFAEL A. CASTRO T. thus If u is a solution to the problem (Q) then u. v = f(x, u)v α (γ 0 u)(γ 0 v)ds, v H 1 (), (3.17) u. v = f(x, u)v, v C c (). (3.18) By Theorem 3.2 u H 2 (), and using the Green formula, we obtain v u = f(x, u)v, v Cc (), then u = f(x, u(x)). Changing f for u in (3.17) and using the Green formula, we obtain γ 1 u + αγ 0 u = Applications The purpose of this section is to give the first step for the proof of the existence of solutions of our the problem (P) (see [4]), which consists in expressing the Sobolev space H 1 () as the orthogonal sum of two subspaces. To achieve this goal we consider the problem (P) with f(x, u(x)) = µu(x) where µ is a parameter. Making use of spectral analysis of compact symmetric operator, we obtain the existence of eigenvalues and by virtue of theorems 3.2 and 3.3, we establish some properties of the first eigenvalue and its associated eigenfunctions Existence of eigenvalues. Theorem 4.1. Let α be a real parameter, α 0. The eigenvalue problem { u = µu, in, (4.1.1) γ 1 u + αγ 0 u = 0, on, has an increasing sequence of eigenvalues. In the case α < 0 the first eigenvalue is negative, and in the case α > 0 the first eigenvalue is positive. Proof. 1) Case α < 0. In this case, to establish the existence of the eigenvalues of the problem (4.1.1) first we introduce an scalar product (.,.) k in H 1 (). To make this, first we have that given ε > 0, there exists a constant C(ε) > 0 such that α (γ 0 u) 2 ds αε u 2 + αc(ε) u 2, u H 1 (). (4.1.2) Choosing k and ε such that 0 < ε < 1 α and k > αc(ε), (4.1.3) Volumen 42, Número 2, Año 2008
11 REGULARITY OF THE SOLUTIONS FOR A ROBIN PROBLEM 137 we define the symmetric bilinear form in H 1 () in this way for u, v H 1 (): a k [u, v] = u v + α (γ 0 u)(γ 0 v)ds + k uv. (4.1.4) This defines an scalar product and we denote it by (.,.) k. In fact, the form a k [.,.] is coercive: using (4.1.2) we have a k [u, u] δ u 2, u H 1 (), with δ = min{1+αε, k+αc(ε)}. On the other hand, using the inequality γ 0 u L 2 ( ) c 1 u, u H 1 (), and the Schwarz inequality we have a k [u, v] c 3 u v, where c 3 = 1 + α c k, i.e. a k [.,.] is continuous. We define for each u H 1 () a norm u k = (u, u) k. Then δ u 2 u 2 k c 3 u 2, u H 1 (), (4.1.5) i.e., the norm k is equivalent to the usual norm of H 1 (). With our k chosen in (4.1.3), the problem (4.1.1) is equivalent to { u + ku = (µ + k)u, in, γ 1 u + αγ 0 u = 0, on. (4.1.6) We prove that the eigenvalues µ + k are positive. Let u be a weak solution of (4.1.6), then (u, v) k = (µ + k) uv, v H 1 (). (4.1.7) If moreover, u 0 and v = u in (4.1.7) we get that µ + k > 0. For fixed u H 1 (), the map v uv is a bounded linear functional in H 1 (), uv u v u v 1 δ u k v k, v H 1 (). (4.1.8) So, by the RieszFrechet Representation Theorem, there is an element in H 1 () that we denote by T u, such that uv = (T u, v) k, v H 1 (). (4.1.9) Thus (4.1.9) defines the operator T : H 1 () H 1 () that is linear and symmetric: (T u, v) k = (u, T v) k, u, v H 1 (). By virtue of (4.1.8) and (4.1.9) we have T u k 1 δ u k, u H 1 (), i.e. T is bounded. Now, we are going to prove that the operator T is compact. Let {u n } be a bounded sequence in H 1 (). By the theorem on compact imbedding of H 1 () in L 2 () there exists a subsequence {u nj } and a element u in H 1 () such that u nj u in H 1 (), and u nj u in L 2 (), when j. Revista Colombiana de Matemáticas
12 138 RAFAEL A. CASTRO T. By (4.1.9) and the definition of the operator T, we have T unj T u 2 k = (u nj u)t (u nj u) Then u nj u L2 () T (u nj u) L2 () u nj u L 2 () T (u nj u) 1 δ u nj u L 2 () T (u nj ) T (u) k. lim j T unj T u k = 0. Let us see that the eigenvalues of problem (4.1.6) are the inverse eigenvalues of the operator T. In fact, if u is a weak solution of (4.1.6), then by (4.1.7) and (4.1.9) we obtain ( ) 1 T u = u. (4.1.10) µ + k Reciprocally, if u satisfies (4.1.10) then u verifies (4.1.7) by (4.1.9). Our next step is to establish the existence of eigenvalues of operator T. Since, 1 µ 1 + k := sup {(T u, u) k : u k = 1} > 0, (4.1.11) then by [6, Lemma 1.1, p. 36], there exists ϕ 1 H 1 () with ϕ 1 k = 1 such that (T ϕ 1, ϕ 1 ) k = 1 ( ) 1 µ 1 + k, T ϕ 1 = ϕ 1. (4.1.12) µ 1 + k On the other hand, we have hence 1 µ 1 + k = sup {(T u, u) k : u k = 1} ( = sup T u ) u, u 0 u k u k k = sup u2 u 0 u 2, k (4.1.13) Then µ 1 + k = inf u 0 u H 1 () µ 1 = inf u 0 u H 1 () u 2 k. (4.1.14) u2 u 2 + α (γ 0u) 2 ds u2, (4.1.15) Volumen 42, Número 2, Año 2008
13 REGULARITY OF THE SOLUTIONS FOR A ROBIN PROBLEM 139 in particular, for the constant function ϕ(x) = b, x, b 0, we have µ 1 ϕ 2 + α (γ 0ϕ) 2 ds = αb2 n 1 ϕ2 b 2 < 0, where n 1 denotes the (n 1)dimensional Lebesgue s measure and the Lebesgue s measure ndimensional of. Let ϕ 1,, ϕ j 1 be the j 1 eigenfunctions that correspond to the j 1 eigenvalues of T. We define 1 µ j + k := sup{(t u, u) k : u k = 1, u ϕ 1,, ϕ j 1 } > 0, (4.1.16) then by [6, Proposition 1.3, p. 37], there exists ϕ j H 1 () with ϕ j k = 1 such that (T ϕ j, ϕ j ) k = 1 µ j + k, T ϕ j = 1 µ j + k ϕ j. (4.1.17) Thus, the eigenvalues of Problem (4.1.6) are positive and form the sequence 0 < µ 1 + k µ 2 + k, then the eigenvalue problem (4.1.1) has a sequence k < µ 1 µ 2 µ 3. (4.1.18) 2) Case α > 0. In this case, the bilinear form a k [u, v] = u v + k uv + α (γ 0 u)(γ 0 v)ds, u, v H 1 (), is coercive for every value of the constant k > 0. In fact a k [u, u] u 2 + k u 2 m u 2, where m = min{1, k}. On the other hand, a k [.,.] is continuous, hence a k [u, v] m 2 u v, where m 2 = 1 + k + αc 2 1. This implies m u 2 u 2 k m 2 u 2, u H 1 (), (4.1.19) i.e., the norm. k is equivalent to the usual norm of H 1 (). The procedure to determine the existence of the eigenvalues of Problem (4.1.1) with α > 0 continues in analogous form to the case α < 0. Considering the eigenvalue problem (4.1.6) with an arbitrary positive constant k, we find that it has a sequence of eigenvalues β 1 + k β 2 + k β 3 + k. Hence inf u 0 u H 1 () β 1 β 2 β 3 (4.1.20) are the eigenvalues of problem (4.1.1). The first eigenvalue β 1 is positive. In fact, by virtue of (4.1.15) we have β 1 = u 2 + α (γ 0u) 2 ds 0, (4.1.21) u2 Revista Colombiana de Matemáticas
14 140 RAFAEL A. CASTRO T. if β 1 = 0, let ϕ be an eigenfunction associated to β 1. Then 0 = ϕ 2 + α (γ 0ϕ) 2 ds, ϕ2 thus γ 0 ϕ = 0, ϕ H 1 0 () and ϕ = 0. Then ϕ 0 which is absurd because ϕ k = The first eigenvalue and its associated eigenfunctions. Theorem 4.2. If ϕ is an eigenfunction of problem (4.1.1) associated to the first eigenvalue λ 1, then ϕ C ( ). Proof. By Theorem 3.3 ϕ verifies ϕ v + α γ 0 ϕ γ 0 v = µ 1 ϕv, v H 1 (). (4.1.22) Using Theorem 3.2 with f(x, ϕ(x)) = λ 1 ϕ(x) we have ϕ H 2 (), γ 0 ϕ H 2 1/2 ( ). Then there exists w 1 H 3 () (Trace Theorem) such that γ 1 w 1 = αγ 0 ϕ. By the Green formula (2.2) we have ϕ. v = {λ 1 ϕ + w 1 }v + w 1 v. Making h = ϕ w 1 and G = λ 1 ϕ + w 1, we obtain { h v = Gv, v H 1 (), γ 1 (h) = 0, (4.1.23) thus, G { H 1 (); N }. By Theorem 2.5 there exists h H 3 () such that { h = G in, (4.1.24) h n = 0 on. Multiplying the first equality in (4.1.24) by v and integrating we obtain h. v = Gv, v H 1 (). (4.1.25) Subtracting (4.1.25) from the first equation in (4.1.23) we have (h h) v = 0, v H 1 (). Making v = h h we obtain (h h) 2 ( = 0, then h h ) = 0, h h = M for some constant M, then ϕ = w 1 + h + M H 3 (). Since ϕ H 3 () then γ 0 ϕ H 3 1/2 ( ). Using again Trace Theorem there exists w 2 H 4 () such that γ 1 w 2 = αγ 0 ϕ and proceeding as in the above case, we conclude that ϕ H 4 (). Continuing on this way we obtain that ϕ H m () for any integer m 0. Thus, for D β ϕ there exists an integer m β such that 2m β > n and Volumen 42, Número 2, Año 2008
15 REGULARITY OF THE SOLUTIONS FOR A ROBIN PROBLEM 141 D β ϕ H m β (). Then by the inclusion of Sobolev and smoothness of the boundary, D β ϕ C ( ). Theorem 4.3. The first eigenvalue of problem (4.1.1) is simple. Proof. i) We claim that if ϕ is an eigenfunction associated to the first eigenvalue λ 1, then ϕ does not change sign. Looking for by contradiction, suppose that ϕ changes sign. Then ϕ verifies the equality (ϕ, ϕ) k = (λ 1 + k) ϕ 2. (4.1.26) Writing the function ϕ in the form ϕ(x) = ϕ + (x) + ϕ (x), where ϕ + (x) = max x {ϕ(x), 0} and ϕ (x) = min x {ϕ(x), 0}, the left member of (4.1.26) can be written (ϕ, ϕ) k = ( ϕ +, ϕ +) k + ( ϕ, ϕ ) k, because ϕ + and ϕ are orthogonals, i.e. ( ϕ +, ϕ ) k = ϕ + ϕ + k = 0. ϕ + ϕ + α γ 0 ϕ + γ 0 ϕ ds Indeed, we have ϕ+ ϕ = 0 and ϕ+ ϕ = 0. Furthermore, by Theorem 4.2, ϕ C ( ). Then γ 0 ϕ + and γ 0 ϕ are restrictions of ϕ + and ϕ on, so γ 0 ϕ + γ 0 ϕ ds = ϕ + ϕ ds = 0. Let b 1 = (ϕ +, ϕ + ) k and b 2 = (ϕ, ϕ ) k, then b 1 > 0 and b 2 > 0. On the other hand, ϕ 2 ( = ϕ + ) 2 ( + ϕ ) 2 = a1 + a 2. Then using the above notation we can write (4.1.26) in the following way: b 1 + b 2 = (λ 1 + k)(a 1 + a 2 ). (4.1.27) From (4.1.27) and (4.1.14) we have 1 λ 1 + k = a 1 + a 2 = sup u2 b 1 + b 2 u 0 u 2. (4.1.28) k For the numbers a1 = a2 b 2. So a 1 b 1 b 1, a2 b 2, and a1+a2 b 1+b 2, we have the unique possibility a1+a2 b 1+b 2 = 1 λ 1 + k = (ϕ+ ) 2 ϕ + 2 = (ϕ ) 2 k ϕ 2. (4.1.29) k Using (4.1.13), the previous equalities imply that ϕ + and ϕ are eigenfunctions corresponding to λ 1. Then v H 1 () we have λ 1 ϕ + v = ϕ + v + α γ 0 ϕ + γ 0 v, (4.1.30) Revista Colombiana de Matemáticas
16 142 RAFAEL A. CASTRO T. By Theorem 4.2, we have that ϕ + C ( ). On the other side we observe that: ϕ + + kϕ + = (λ 1 + k)ϕ + 0, in. Since ϕ changes sign in, there exists x 0 inside of such that ϕ + (x 0 ) = 0 which is its minimum value in. Then by the Strong Maximum Principle ϕ + is constant in, thus ϕ + = 0 in, which is absurd. Therefore ϕ can not change sign in. ii) Next we prove that the geometric multiplicity of λ 1 is one. Let ϕ 1 and ϕ 2 be two eigenfunctions associate to the eigenvalue λ 1. By step i), for each t R the eigenfunction ϕ 1 + tϕ 2 has definite sign in, so the sets A = {t R : ϕ 1 + tϕ 2 0, in } and B = {t R : ϕ 1 + tϕ 2 0, in } are non empty, closed and A B = R. Since the set of real number R is connected, there exists t A B, t 0 such as ϕ 1 + tϕ 2 = 0, i.e. ϕ 1 and ϕ 2 are linearly dependent. Therefore, the eigenspace associate to the eigenvalue λ 1 is generated by a single eigenfunction, that we denote by ϕ 1. iii) The algebraic multiplicity of λ 1 is one, (λ 1 is simple). Let N(λ 1 I + ) = {u : λ 1 u + u = 0 and γ 1 u + αγ 0 u = 0}, N(λ 1 I + ) 2 = {u : (λ 1 I + )(λ 1 u + u) = 0 and γ 1 u + αγ 0 u = 0}. It is clear that N(λ 1 I + ) N(λ 1 I + ) 2. Next we show that N(λ 1 I + ) 2 N(λ 1 I + ). Indeed, let ϕ N(λ 1 I + ) 2 be, then λ 1 ϕ + ϕ N(λ 1 I + ), λ 1 ϕ + ϕ = tϕ 1 for some real t. By the other side, we have t(ϕ 1, ϕ 1 ) k = (tϕ 1, ϕ 1 ) k = (λ 1 ϕ + ϕ, ϕ 1 ) k = λ 1 (ϕ, ϕ 1 ) k + ( ϕ, ϕ 1 ) k = λ 1 (ϕ, ϕ 1 ) k + (λ 1 + k) ϕ 1 ϕ, (by (4.1.7)) = (λ 1 λ 1 )(ϕ, ϕ 1 ) k = 0. Hence t(ϕ 1, ϕ 1 ) k = 0, then t = 0. Thus λ 1 ϕ+ ϕ = 0 and ϕ N(λ 1 I + ) Orthogonal sum. Finally we express the Sobolev space H 1 () as the orthogonal sum of two subspaces, where an addend has finite dimension. In the case α < 0, from (4.1.18) and Theorem (4.3), the sequence of eigenvalues has the form k < µ 1 < µ 2 µ 3 (4.2.1) Let X 1 be the space associated to the first eigenvalue µ 1 and X 2 = X 1 the orthogonal complement of X 1 with respect to the inner product of (, ) k defined in (4.1.4). Then we have H 1 () = X 1 X 2, (4.2.2) Volumen 42, Número 2, Año 2008
17 REGULARITY OF THE SOLUTIONS FOR A ROBIN PROBLEM 143 and ϕ 2 + α (γ 0 ϕ) 2 ds = λ 1 ϕ 2, ϕ X 1. (4.2.3) In the case α > 0, from (4.1.20) and Theorem (4.3), the sequence of eigenvalues has the form β 1 < β 2 β 3 (4.2.4) Let Y 1 be the space associate to β 1 and Y 2 = Y 1 the orthogonal complement of Y 1 with respect to the inner product (, ) k, but here k > 0 is an arbitrary constant. Then H 1 () = Y 1 Y 2, (4.2.5) and ϕ 2 + α (γ 0 ϕ) 2 ds = β 1 ϕ 2, ϕ Y 1. (4.2.6) by virtue of Theorem 4.3, the subspaces X 1 and Y 1 have dimension one. Acknowledgement. The author is deeply thankful to Helena J. Nussenzveig Lopes, and to Djairo G. De Figueiredo for their valuable comments. References [1] Arcoya, D., and Villegas, S. Nontrivial solutions for a Neumann problem with a nonlinear term asymptotically linear at and superlinear at +. Math.Z 219, 4 (1995), [2] Aubin, J. P. Approximation of elliptic boundary value problems. WileyInterscience, New York, [3] Baiocchi, C., and Capelo, A. Variational and quasivariational inequalities, aplications to freeboundary problems. John Wiley and Sons, New York, [4] Castro, R. A. Nontrivial solutions for a Robin problem with a nonlinear term asymptotically linear at and superlinear at. Revista Colombiana de Matemáticas (2008). Este volumen. [5] Dautray, R., and Lions, J. L. Analyse mathématique et calcul numérique; pour les sciences et les techniques. Masson, Paris, Vol. 3. [6] de Figueiredo, D. G. Positive solutions of semilinear elliptic problems, vol. 957 of Lecture Notes in Mathematics. SpringerVerlag, Berlin, Heidelberg, New York, [7] Evans, L. C. Partial differential equations, vol. 19 of Graduate Studies in Mathematics. American Mathematical Society, Providence, [8] Kufner, A., John, O., and Fučik, S. Function spaces. Academia, Prague, [9] Lions, J. L., and Magenes, E. Problèmes aux limites non homogènes (vi). Journal D Analyse Mathématique XI (1963), [10] Nečas, J. Les méthodes directes en théorie des équations elliptiques. Masson, Prague, Paris/Academia. (Recibido en julio de Aceptado en agosto de 2008) Revista Colombiana de Matemáticas
18 144 RAFAEL A. CASTRO T. Escuela de Matemáticas Universidad Industrial de Santander Apartado Aéreo 678 Bucaramanga, Colombia Volumen 42, Número 2, Año 2008
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