On the minimax inequality for a special class of functionals

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1 ISSN , Engl, UK World Journal of Modelling Simulation Vol. 3 (2007) No. 3, On the minimax inequality for a secial class of functionals G. A. Afrouzi, S. Heidarkhani, S. H. Rasouli Deartment of Mathematics, Faculty of Basic Sciences, Mazaran University, Babolsar, Iran (Received January , Acceted Aril ) Abstract. In this aer, we establish an equivalent statement of minimax inequality for a secial class of functionals, also some conditions that imly minimax inequality are ointed out equivalent formulations are roved. Keywords: minimax inequality, critical oint, multilicity results, dirichlet roblem 1 Introduction Throughout the sequel, R N (N 1) is nonemty bounded oen set with a boundary of class C 1 > N. Given two Gâteaux differentiable functionals Φ Ψ on a real Banach sace X, the minimax inequality su (Φ(u) + λψ(u) + λρ) < su (Φ(u) + λψ(u) + λρ), ρ R, (1) lays a fundamental role for establishing the existence of at least three critical oints for the functional Φ(u) + λψ(u), as the theorem of B. Ricceri below ensures (see [4]). In recent years, many authors have studied multile solutions from several oint of view with different aroaches (see, for examle, [1 3]); for instance, in [2], the author roves multilicity results for the roblem { u + λf(x, u) = 0, (2) u(a) = u(b) = 0, which for each λ [0, + ], admits at least three solutions in W 1,2 0 ([a, b]) when f : [a, b] R R is a continuous function. In this aer some conditions that imly minimax inequality (1) are ointed out equivalent formulations are roved. Moreover, the aim of this aer is to establishes an equivalent statement of minimax inequality (1) for a secial class of functionals. 2 Main results In the sequel, X will denote the Sobolev sace W 1, 0 () with the norm ( u = u(x) dx) 1/, address: afrouzi@umz.ac.ir. Published by World Academic Press, World Academic Union

2 World Journal of Modelling Simulation, Vol. 3 (2007) No. 3, f : R R is a ositive Caratheodory function, a(x) C() is a ositive weight function, for each (x,y) R k = g(x, y) = su \{0} y 0 f(x, ξ)dξ max x u(x), u since > N, one has k <, (see [[5], formula (6b)]). Assume that there exists continuous function b(x) 1 on [a 1, a 2 ], (a 1, a 2 R) such that g(x, u(x)) k b(x) u(x) (3) for each u X. We define u I = ( ( u(x) + a(x) u(x) )dx ) 1/, such that there exist ositive suitable constants c 1 c 2 : c 1 u u I c 2 u (i.e., the above norms are equivalent). (4) We now introduce two secial functionals on the Sobolev sace X as follows Φ(u) = u I Ψ(u) = g(x, u(x))dx for every u X. Let r, ρ R, w X be such that 0 < r < 0 < ρ < Ψ(w), we ut A 1 (r, w) = r Ψ(w), A 2(ρ, w) = ρ Ψ(w) A 3(r, w) = ( 1 A 1(r, w)) 1/. Clearly, A 1 (r, w), A 2 (ρ, w) A 3 (r, w) are ositive. In this work, m() is Lebesgue measure on set. From (3) since b(x) 1 on [a 1, a 2 ], we have g(x, u(x))dx k u. (5) Now, we ut α 1 = {k u R + ; Φ(u) r}, α 2 = {k u R + ; m() t k u ( t + a(x) t ) r} for every t X Since α r = α 1 α 2. su u(x) k u x for every x for every u X, we have m() t k u ( t + a(x) t ) ( u(x) + a(x) u(x) )dx WJMS for subscrition: o@wjms.org.uk

3 222 G. A. Afrouzi & S. Heidarkhani & S. Rasouli: On the minimax inequality for a secial class of functionals for every u X, m() t k u ( t + a(x) t ) Φ(u) for every u X; therefore {k u R + ; Φ(u) r} {k u R + ; m() Hence α 1 α 2, α r 0. Now, the main result: Theorem 1. Assume that there exist r R, w X such that (i) 0 < r <, (ii) m() t A3 (r,w) α r ( t + a(x) t ) > r. su (Φ(u) + λψ(u) + λρ) < Proof. From (ii), we obtain A 3 (r, w) α r {l R + ; m() su t k u ( t + a(x) t ) r}. (Φ(u) + λψ(u) + λρ) t l ( t + a(x) t ) r}. Moreover {l R + ; m() t l ( t + a(x) t ) r} A 3 (r, w) α r. In fact, arguing by contradiction, we assume that there is l 1 R + such that m() ( t + a(x) t ) r t l 1 l 1 < A 3 (r, w) α r, so m() ( t + a(x) t ) m() t A 3 (r,w) α r ( t + a(x) t ) r t l 1 this is a contradiction. So Therefore, So, we have Using of (5), one has {l R + ; m() {k u R + ; m() t l ( t + a(x) t ) r} A 3 (r, w) α r. t k u ( t + a(x) t ) r} + α r > A 3 (r, w), {k u R + ; Φ(u) r} > A 3 (r, w). {k u R + ; Φ(u) r} > A 1 (r, w). WJMS for contribution: submit@wjms.org.uk

4 World Journal of Modelling Simulation, Vol. 3 (2007) No. 3, { g(x, u(x))dx; Φ(u) r} > A 1 (r, w), or with Ψ = T, we have Now, we claim for each ρ satisfying one has {Ψ(u); Φ(u) r} > r Ψ(w) {Ψ(u); Φ(u) r} < r ( Ψ(w)) su{t (u); Φ(u) r} < r T (w). su{t (u); Φ(u) r} < ρ < r T (w), su (Φ(u) + λ(ρ T (u)) < su (Φ(u) + λ(ρ T (u)). From su{t (u); Φ(u) r} < ρ, we obtain r {Φ(u); T (u) ρ} ρ > 0. Moreover, from ρ < r T (w) ρ > 0, one has ρ < T (w) ρ T (w) < r. Hence, ρ < {Φ(u); T (u) ρ}, T (w) Now, there exists λ R such that {Φ(u); T (u) ρ} ρ > {Φ(u); T (u) ρ}. T (w) ρ λ > {Φ(u); T (u) ρ} T (w) ρ λ < {Φ(u); T (u) ρ}. ρ Namely + λ(ρ T (w)) < {Φ(u); T (u) ρ} λρ < {Φ(u); T (u) ρ}. So, by choose Φ(0) = T (0) = 0 thanks of 0 < ρ < T (w), we obtain with resect to one has (Φ(u) + λ(ρ T (u)) < {Φ(u); T (u) ρ}, (Φ(u) + λ(ρ T (u)) < (Φ(0) + λ(ρ T (0)) = λρ, su (Φ(u) + λ(ρ T (u)) < {Φ(u); T (u) ρ}. In other h, for {u X; T (u) ρ} we have su (Φ(u) + λ(ρ T (u)) = {Φ(u); T (u) ρ} λ 0 WJMS for subscrition: o@wjms.org.uk

5 224 G. A. Afrouzi & S. Heidarkhani & S. Rasouli: On the minimax inequality for a secial class of functionals since Ψ = T, we have su (Φ(u) + λψ(u) + λρ) < su Remark 1. If in theorem 1, A 3 (r, w) α r 0; the theorem holds again. Because, A 3 (r, w) α 1 α 2 α 1, by arguing as in the roof of theorem 1, the results holds. If instead of condition (ii) in theorem 1, we ut m() t A3 (r,w)( t +a(x) t ) > r, then the result holds. Because m() t A3 (r,w) α r ( t + a(x) t ) m() t A3 (r,w)( t + a(x) t ) > r. So, we have: Theorem 2. Assume that there exist r R, w X such that (i) 0 < r <, (ii) m() t A3 (r,w)( t + a(x) t ) > r. su (Φ(u) + λψ(u) + λρ) < su Now, If we give ρ = A 1 (r, w), then we have r = A 2 (ρ, w) A 3 (r, w) = following result: ρ. So, we have the Proosition 1. The following assertions are equivalent: (a) there are r R, w X such that (i) 0 < r <, (ii) m() t A3 (r,w)( t + a(x) t ) > r. (b) there are ρ R, w X such that (j) 0 < ρ < Ψ(w), q ρ ( t + a(x) t ) > A 2 (ρ, w). (jj) m() t Finally, by using of the theorem 2 roosition 1, we have: Theorem 3. Assume that there exist ρ R, w X such that (j) 0 < ρ < Ψ(w), q ρ ( t + a(x) t ) > A 2 (ρ, w). (jj) m() t su (Φ(u) + λψ(u) + λρ) < su References [1] R. I. Avery, J. Henderson. Three symmetric ositive solutions for a second-order boundary value roblem. Al. Math. lett, 2000, 13: 1 7. [2] P. Cito. Existence of three solutions for a non-autonomous two oint boundary value roblem. J. Math. Anal. Al, 2000, 252: [3] G. Henderson, H. B. Thomson. Existence of multile solutions for second order boundary value roblem. J. Diff.Eqs, 2000, 166: [4] R. Ricceri. On a three critical oints theorem. Arch. Math. (Basel), 2000, 75: [5] G. Talenti. Some inequalities of sobolev tye on two-dimensional sheres. General Inequalities, 1987, 5: In: W. Walter (Ed.). WJMS for contribution: submit@wjms.org.uk