On the minimax inequality and its application to existence of three solutions for elliptic equations with Dirichlet boundary condition

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1 ISSN England UK World Journal of Modelling and Simulation Vol. 3 (2007) No On the minimax inequality and its alication to existence of three solutions for ellitic equations with Dirichlet boundary condition G. A. Afrouzi S. Heidarkhani Deartment of Mathematics Faculty of Basic Sciences Mazandaran University Babolsar (Received August Acceted December ) Abstract. In this aer we establish an equivalent statement of minimax inequality for a secial class of functionals. As an alication a result for the existence of three solutions to the Dirichlet roblem u + λf(x u) = a(x) u 2 u in on where u =div( u 2 u) is the -Lalacian oerator R N (N 1) is non-emty bounded oen set with smooth boundary > N λ > 0 f : R R is a ositive continuous function and ositive function a(x) C() is emhasized. Keywords: minimax inequality critical oint three solutions multilicity results Dirichlet roblem 1 Introduction Throughout the sequel R N (N 1) is nonemty bounded oen set with smooth boundary and > N. Given two Gâteaux differentiable functionals Φ and Ψ on a real Banach sace X the minimax inequality su inf (Φ(u) + λ(ρ Ψ(u))) < inf su (Φ(u) + λ(ρ Ψ(u))) ρ R (1) lays a fundamental role for establishing the existence of at least three critical oints for the functional Φ(u) λψ(u). The main result of this aer (Theorem 2) establishes an equivalent statement of minimax inequality (1) for a secial class of functionals while its consequences (Theorem 3 and Theorem 5) guarantee some conditions so minimax inequality holds. Finally we aly Theorem 1 to ellitic equations by using an immediate consequence of Theorem 2 and we consider the boundary value roblem u + λf(x u) = a(x) u 2 u in (2) on where u =div( u 2 u) is the -Lalacian oerator λ > 0 f : R R is a ositive continuous function and ositive function a(x) C() and we establish some conditions on f so roblem (2) admits at least three weak solutions. We say u is a weak solution to (2) if u W 1 0 () and u(x) 2 u(x) v(x)dx λ f(x u(x))v(x)dx = a(x) u(x) 2 u(x)v(x)dx address: afrouzi@umz.ac.ir. Published by World Academic Press World Academic Union

2 84 G. Afrouzi & S. Heidarkhani: On the minimax inequality and its alication for every v W 1 0 (). Also by a similar arguments as in the roblem (2) we will have the existence of at least three weak solutions for the roblem u + λh 1 (x)h 2 (u) = a(x) u 2 u in (3) on where h 1 C() and h 2 C(R) are two ositive functions and for the roblem u + λf(u) = a(x) u 2 u in on (4) where f : R R is a ositive continuous function. In recent years many authors have studied multile solutions from several oints of view and with different aroaches and we refer to [1]-[6] and the references therein for more details for instance in their interesting aer [3] the authors studied roblem u + λf(u) = 0 (5) u(0) = u(1) = 0 (indeendent of λ in the case) where f : R R is a continuous function and λ is a real arameter by using a multile fixed-oint theorem to obtain three symmetric ositive solutions under growth conditions on f and in [4] the author roves multilicity results for the roblem (5) which for each λ [0 + ] admits at least three solutions in W 12 0 ([0 1]) where f is a continuous function. Also in [6] the authors established the existence of three ositive solutions for classes of nondecreasing -sublinear function f belonging to C 1 ([0 )) for a - Lalacian version of [3] i.e. the roblem u = λf(u) in (6) on where > 1 λ > 0 is a arameter and is a bounded domain in R N ; N 2 with of class C 2 and connected. In [2] using variational methods the authors ensure the existence of a sequence of arbitrarily small ositive solutions for roblem u + λf(x u) = 0 in (7) on when the function f has a suitable oscillating behaviour at zero. In articular in [1] we obtained the existence of an interval Λ [0 + ] and a ositive real number q such for each λ Λ the roblem (2) where R N (N 2) is non-emty bounded oen set with smooth boundary > N λ > 0 f : R R is a continuous function and ositive weight function a(x) C() admits at least three weak solutions in whose norms in W 1 0 () are less than q. We now recall the three critical oints theorem of B. Ricceri [7] by choosing h(λ) = λρ: Theorem 1. Let X be a searable and reflexive real Banach sace; Φ : X R a continuously Gâteaux differentiable and sequentially weakly lower semicontinuous functional whose Gâteaux derivative admits a continuous inverse on X ; T : X R a continuously Gâteaux differentiable functional whose Gâteaux derivative is comact. Assume lim u + (Φ(u) + λt (u)) = + for all λ [0 + ] and there exists ρ R such su inf (Φ(u) + λt (u) + λρ) < inf (Φ(u) + λt (u) + λρ). su Then there exists an oen interval Λ [0 + ] and a ositive real number q such for each λ Λ the equation Φ (u) + λt (u) = 0 has at least three solutions in X whose norms are less than q. WJMS for contribution: submit@wjms.org.uk

3 World Journal of Modelling and Simulation Vol. 3 (2007) No Main results In the sequel let f : R R be a ositive continuous function and g : R R be the function defined as follows g(x t) = t 1 0 f(x ξ)dξ for each (x t) R. X will denote the Sobolev sace W0 () with the norm u := ( u(x) dx ) 1/ Now we define u := ( ( u + a(x) u )dx ) 1/ such there exist ositive suitable constants c 1 and c 2 : c 1 u u c 2 u (8) (i.e. the above norms are equivalent). We now introduce two ositive secial functionals on the Sobolev sace X as follows Φ(u) := u every u X and Ψ(u) := g(x u(x))dx for every u X. Let ρ r R w X be such 0 < ρ < Ψ(w) and 0 < r < Φ(w). We ut for β 1 = β 1 (ρ w) := ρ Φ(w) Ψ(w) β 2 = β 2 (r w) := r Ψ(w) Φ(w) (9) (10) β 3 = β 3 (ρ w) := k c 1 m() 1 N 1 ( β 1 (ρ w)) 1/ (11) and σ = σ(c 1 u) = k c 1 m() 1 N 1 u (12) for every u X where k = k(n ) is a ositive constant and m() is the Lebesgue measure of the set. Clearly β 1 β 2 β 3 and σ are ositive. Now we ut δ 1 := infσ R + ; Ψ(u) ρ} δ 2 := infσ R + ; m() max x g(x σ) ρ} and Clearly δ 1 δ 2. δ ρ := δ 1 δ 2 (13) Taking into account for every u X one has su x u(x) km() 1 N 1 u for each u X namely su x u(x) k c 1 m() 1 N 1 u for each u X so Ψ(u) = g(x u(x))dx m() max x g(x σ). Namely Ψ(u) m() max x g(x σ); therefore σ R + ; Ψ(u) ρ} σ R + ; m() max x g(x σ) ρ }. So we have infσ R + ; Ψ(u) ρ} infσ R + ; m() max x g(x σ) ρ }. Hence δ ρ 0. The main result of this aer is the following theorem: Theorem 2. Assume there exist ρ R w X such (i) 0 < ρ < Ψ(w) (ii) m() max x g(x β 3 δ ρ ) < ρ; where β 3 is given by (11) and δ ρ by (13). Then su inf (Φ(u) + λ(ρ Ψ(u))) < inf su (Φ(u) + λ(ρ Ψ(u))). Proof. From (ii) we obtain β 3 δ ρ θ R + ; m() max x g(x θ) ρ}. Moreover infθ R + ; m() max x g(x θ) ρ} β 3 δ ρ ; in fact arguing by contradiction we assume there is ϑ R + such m() max x g(x ϑ) ρ and ϑ < β 3 δ ρ so m() max x g(x β 3 δ ρ ) WJMS for subscrition: info@wjms.org.uk

4 86 G. Afrouzi & S. Heidarkhani: On the minimax inequality and its alication m() max x g(x ϑ) ρ and this is a contradiction. So infθ R + ; m() max x g(x θ) ρ} > β 3 δ ρ. Therefore infσ R + ; m() max x g(x σ) ρ} > β 3 δ ρ ; namely β 3 < δ 1. So we have inf u } R + ; Ψ(u) ρ > β 1 namely inf u Ψ 1 ([ρ+ ]) Φ(u) > ρ Φ(w) Ψ(w) and taking in to account (i) holds one has inf u Ψ 1 ([ρ+ ]) Φ(u) ρ > Φ(w) inf u Ψ 1 ([ρ+ ]) Φ(u) Ψ(w) ρ. Now let λ R and taking into account the revious inequality one has either λ > Φ(w) inf u Ψ 1 ([ρ+ ]) Φ(u) Ψ(w) ρ or λ < inf u Ψ 1 ([ρ+ ]) Φ(u) ρ. Namely inf u Ψ 1 ([ρ+ ]) Φ(u) > Φ(w) + λ(ρ Ψ(w)) or λρ < inf u Ψ 1 ([ρ+ ]) Φ(u). Therefore thanks to the 0 < ρ < Ψ(w) we obtain inf u X (Φ(u) + λ(ρ Ψ(u))) < inf u Ψ 1 ([ρ+ ]) Φ(u) and then one has su λ 0 inf u X (Φ(u) + λ(ρ Ψ(u))) < inf u Ψ 1 ([ρ+ ]) Φ(u). Therefore thanks to the inf u X su λ 0 (Φ(u) + λ(ρ Ψ(u))) = inf u Ψ 1 ([ρ+ ]) Φ(u) we have the su λ 0 inf u X (Φ(u) + λ(ρ Ψ(u))) < inf u X su λ 0 (Φ(u) + λ(ρ Ψ(u))). Remark 1. su λ 0 inf u X (Φ(u)+λ(ρ Ψ(u))) is well define because λ inf u X (Φ(u)+λ(ρ Ψ(u))) is uer semicontinuous in [0 + ] and tends to as λ +. Remark 2. If in Theorem 2 β 3 δ ρ 0; the Theorem holds again. Because β 3 δ 1 δ 2 δ 1. Arguing as before roof Theorem 2 result holds. If instead of condition (ii) in Theorem 2 we ut m() max x g(x β 3 ) < ρ then the result holds because m() max x g(x β 3 δ ρ ) m() max x g(x β 3 ) < ρ. So we have the following result: Theorem 3. Assume there exist ρ R w X such (i) 0 < ρ < Ψ(w) (ii) m() max x g(x β 3 ) < ρ. where β 3 is given by (11). Then su inf (Φ(u) + λ(ρ Ψ(u))) < inf Now we oint out the following result: Proosition 1. The following assertions are equivalent: (a) there are ρ R w X such (i) 0 < ρ < Ψ(w) (ii) m() max x g(x β 3 ) < ρ; where β 3 is given by (11). (b) there are r R w X such (j) 0 < r < Φ(w) su (Φ(u) + λ(ρ Ψ(u))). (jj) m() max x g(x k c 1 m() 1 N 1 r) < β 2 ; where β 2 is given by (10). Proof. (a) (b). First we note 0 < Φ(w) because if 0 Φ(w) from (i) one has ρ Φ(w) Ψ(w) namely β 3 k c 1 m() 1 N 1 w. Hence taking into account (ii) one has Φ(w) Ψ(w) m() max g(x k m() 1 N 1 x c 1 w ) m() max g(x β 3 ) < ρ x and is in contradiction to (i). We now ut β 1 = r. We obtain ρ = β 2 and β 3 = k c 1 m() 1 N 1 Therefore from (i) and (ii) one has 0 < r < Φ(w) and r. m() max g(x k m() 1 N 1 r) < β 2. x c 1 (b) (a). First we note 0 < Ψ(w) because if 0 Ψ(w) from (j) one has r Ψ(w) Φ(w) 0; namely β 2 0. Hence from (jj) one has WJMS for contribution: submit@wjms.org.uk

5 World Journal of Modelling and Simulation Vol. 3 (2007) No = Ψ(0) m() max g(x k m() 1 N 1 r) < 0 x c 1 and this is a contradiction. We now ut β 2 = ρ. We obtain r = β 1 and k c 1 m() 1 N 1 from (j) and (jj) we have the conclusion. r = β 3. Therefore The following Theorem is another consequence of Theorem 2. Theorem 4. Assume there exist r R w X such (j) 0 < r < Φ(w) (jj) m() max x g(x k c 1 m() 1 N 1 r) < β 2 where β 2 is given by (10). Then there exists ρ R such su inf (Φ(u) + λ(ρ Ψ(u))) < inf Proof. It follows from Theorem 3 and Proosition 1. su (Φ(u) + λ(ρ Ψ(u))). Finally we interested in ensuring the existence of at least three weak solutions for the Dirichlet roblem (2). Now we have the following result: Theorem 5. Assume there exist ρ R b 1 L 1 () w X and a ositive constant γ with γ < such (i) 0 < ρ < g(x w(x))dx (ii) m() max x g(x β 3 ) < ρ. (iii) g(x t) b 1 (x)(1 + t γ ) almost everywhere in and for each t R. where β 3 is given by (11). Then there exists an oen interval Λ [0 + ] and a ositive real number q such for each λ Λ roblem (2) admits at least three solutions in X whose norms are less than q. Proof. For each u X we ut Φ(u) = u T (u) = g(x u(x))dx and J(u) = Φ(u) + λt (u). In articular for each u v X one has Φ (u)(v) = ( u(x) 2 u(x) v(x) + a(x) u(x) 2 u(x)v(x))dx and T (u)(v) = f(x u(x))v(x)dx. It is well known the critical oints of J are the weak solutions of (2) and our goal is to rove Φ and T satisfy the assumtions of Theorem 1. Clearly Φ is a continuously Gâteaux differentiable and sequentially weakly lower semi continuous functional whose Gâteaux derivative admits a continuous inverse on X and T is a continuously Gâteaux differentiable functional whose Gâteaux derivative is comact. Thanks to (iii) for each λ > 0 one has lim u + (Φ(u) + λt (u)) = + for all λ [0 + ]. Furthermore thanks to Theorem 3 from (i) and (ii) we have su inf (Φ(u) + λt (u) + λρ) < inf su (Φ(u) + λt (u) + λρ). Therefore we can aly Theorem 1. It follows there exists an oen interval Λ [0 + ] and a ositive real number q such for each λ Λ roblem (2) admits at least three solutions in X whose norms are less than q. We also have the following existence result: WJMS for subscrition: info@wjms.org.uk

6 88 G. Afrouzi & S. Heidarkhani: On the minimax inequality and its alication Theorem 6. Assume there exist r R b 2 L 1 () w X and a ositive constant γ with γ < such (j) 0 < r < w (jj) m() max x g(x k c 1 m() 1 N 1 r) < β 2 ; (jjj) g(x t) b 2 (x)(1 + t γ ) almost everywhere in and for each t R. where β 2 is given by (10). Then there exists an oen interval Λ [0 + ] and a ositive real number q such for each λ Λ roblem (2) admits at least three solutions in X whose norms are less than q. Proof. It follows from Theorem 4 and Theorem 5. Let h 1 C() and h 2 C(R) be two ositive functions. Put f(x t) = h 1 (x)h 2 (t) for each (x t) R H(t) = t 0 h 2(ξ)dξ for all t R and b 3 (x) = b 1(x) h 1 (x) for almost every x. Then with use the Theorem 5 we have the following result: Theorem 7. Assume there exist ρ R b 3 L 1 () w X and a ositive constant γ with γ < such (i ) 0 < ρ < (h 1(x)H(w(x)))dx (ii ) m() max x h 1 (x) < ρ H(β 3 ). (iii ) H(t) b 3 (x)(1 + t γ ) almost everywhere in and for each t R. where β 3 is given by (11). Then there exists an oen interval Λ [0 + ] and a ositive real number q such for each λ Λ roblem (3) admits at least three solutions in X whose norms are less than q. Put b 4 (x) = b 2(x) h 1 (x) existence result: for almost every x. Then with use the Theorem 6 we have the following Theorem 8. Assume there exist r R b 4 L 1 () w X and a ositive constant γ with γ < such (j ) 0 < r < w (jj ) m() max x h 1 (x) < β 2 H( k c 1 m() 1 N 1 ; r) (jjj ) H(t) b 4 (x)(1 + t γ ) almost everywhere in and for each t R. where β 2 is given by (10). Then there exists an oen interval Λ [0 + ] and a ositive real number q such for each λ Λ roblem (3) admits at least three solutions in X whose norms are less than q. We now want to oint out two simle consequences of Theorem 5 and Theorem 6 resectively. Let f : R R be a ositive continuous function. Put g(t) = t 0 f(ξ)dξ for each t R. So we have the following results: Theorem 9. Assume there exist ρ R w X and two ositive constants γ and η 1 with γ < such (i ) 0 < ρ < g(w(x))dx (ii ) m()g(β 3 ) < ρ. (iii ) g(t) η 1 (1 + t γ ) for each t R. where β 3 is given by (11). Then there exists an oen interval Λ [0 + ] and a ositive real number q such for each λ Λ roblem (4) admits at least three solutions in X whose norms are less than q. Theorem 10. Assume there exist r R w X and two ositive constants γ and η 2 with γ < such (j ) 0 < r < w (jj ) m()g( k c 1 m() 1 N 1 r) < β 2 ; (jjj ) g(t) η 2 (1 + t γ ) for each t R. where β 2 is given by (10). Then there exists an oen interval Λ [0 + ) and a ositive real number q such for each λ Λ roblem (4) admits at least three solutions in X whose norms are less than q. WJMS for contribution: submit@wjms.org.uk

7 World Journal of Modelling and Simulation Vol. 3 (2007) No Now in the case N = 1 and = 2 we resent two simle consequence of Theorem 9 and Theorem 10 resectively.. For simlicity we fix = [0 1] and consider a ositive continuous function f : R R. Moreover ut g(t) = t 0 f(ξ)dξ for all t R. Taking into account in this situation k = 1 2 and β 3 = 1 β 1 c 1 2 we have the following results: Theorem 11. Assume there exist ρ R w X and two ositive constants γ and η 3 with γ < 2 such (i ) 0 < ρ < 1 0 g(w(x))dx (ii ) g( 1 β 1 c 1 2 ) < ρ. (iii ) g(t) η 3 (1 + t γ ) for each t R. where β 1 is given by (9). Then there exists an oen interval Λ [0 + ) and a ositive real number q such for each λ Λ roblem u (x) + λf(u(x)) = a(x)u(x) x [0 1] (14) u(0) = u(1) = 0 admits at least three solutions in W 12 0 ([0 1]) whose norms are less than q. Theorem 12. Assume there exist r R w X and two ositive constants γ and η 4 with γ < 2 such (j ) 0 < r < w 2 2 (jj ) g( 1 r c 1 2 ) < β 2; (jjj ) g(t) η 4 (1 + t γ ) for each t R. where β 2 is given by (10). Then there exists an oen interval Λ [0 + ) and a ositive real number q such for each λ Λ roblem (14) admits at least three solutions in W 12 0 ([0 1]) whose norms are less than q. Examle 1. Consider the roblem u + λ(e u u 2 (3 + u)) = e x u u(0) = u(1) = 0. Then there exists an oen interval Λ [0 + ) and a ositive real number q such for each λ Λ roblem (15) admits at least three solutions in W 12 0 ([0 1]) whose norms are less than q. In fact by choosing ρ = 1 4 and w(x) = x so β 1(ρ w) = e e all assumtions of Theorem 11 are satisfied with γ = 1 c 1 is ositive constant such the inequality (8) hold for m(x) = e x and η 3 sufficiently large also with choose r = 1 2 and w(x) = x so β 2(r w) = 6 2e e 1 all assumtions of Theorem 12 are satisfied with γ = 1 c 1 is ositive constant such the inequality (8) hold for m(x) = e x and η 4 sufficiently large. (15) References [1] G. A. Afrouzi S. Heidarkhani. Three solutions for a dirichlet boundary value roblem involving the -lalacian. Nonlinear Analysis. To aear. [2] G. Anello G. Cordaro. Positive infinitely many arbitrarily small solutions for the Dirichlet roblem involving the -Lalacian. Proc. Roy. Soc. Edinburgh Sect. A ( ). [3] R. I. Avery J. Henderson. Three symmetric ositive solutions for a second-order boundary value roblem. Al. Math. lett (1-7). [4] G. Bonanno. Existence of three solutions for a two oint boundary value roblem. Al. Math. Lett (53-57). [5] P. K T. Ouyang. Exact multilicity results for two classes of boundary value roblem. Diff. Integral Equations ( ). [6] M. R. Shivaji. Multile ositive solutions for classes of -Lalacian equations. Differential and Integral Equations (11-12): [7] B. Ricceri. On a three critical oints theorem. Arch. Math. (Basel) : WJMS for subscrition: info@wjms.org.uk