Existence of Solutions for a Class of Third Order Quasilinear Ordinary Differential Equations with Nonlinear Boundary Value Problems

Advances in Dynamical Systems and Applications ISSN 973-5321, Volume 3, Number 2, pp. 35 321 (28) http://campus.mst.edu/adsa Existence of Solutions for a Class of Third Order Quasilinear Ordinary Differential Equations with Nonlinear Boundary Value Problems Jianqing Zhao and Zuodong Yang Nanjing Normal University School of Mathematics and Computer Science Jiangsu Nanjing 2197, China Abstract In this paper, we are concerned with the third-order quasilinear ordinary differential equation (Φ p (u )) = f(t, u, u, u ), < t < 1, with the nonlinear boundary conditions or u() =, g(u (), u ()) = A, h(u (1), u (1)) = B u() = C, L(u (), u (1)) =, R(u (), u (1), u (), u (1)) =, where A, B, C R, Φ p (u) = u p 2 u(p > 1), f : [, 1] R 3 R is continuous, g, h, L : R 2 R, R : R 4 R are continuous, which occurs in the study of the p-laplace equation, generalized reaction-diffusion theory, non-newtonian fluid theory, and the turbulent flow of a gas in porous medium. Existence results are obtained by using a Nagumo condition and upper and lower solutions methods. AMS Subject Classifications: 34B4. Keywords: Third order quasilinear ordinary differential equations, nonlinear boundary value problem, upper and lower solutions, Nagumo condition. Received June 16, 27; Accepted October 13, 27 Communicated by Meng Fan Second address: Math Dept, Lianyungang Teachers College, Jiangsu Lianyungang 222, China Second address: School of Zhongbei, Nanjing Normal University, Jiangsu Nanjing 2146, China

36 J. Zhao and Z. Yang 1 Introduction This paper deals with the existence of solutions for the nonlinear boundary value problem (Φ p (u )) = f(t, u, u, u ), t I = [, 1] (1.1) satisfying the conditions or u() =, g(u (), u ()) = A, h(u (1), u (1)) = B (1.2) u() = C, L(u (), u (1)) =, R(u (), u (1), u (), u (1)) =, (1.3) where A, B, C R, f : [, 1] R 3 R is continuous, g, h, L : R 2 R, R : R 4 R are continuous, and Φ p (s) = s p 2 s, p > 1. Equations of the above form are mathematical models occurring in studies of the p-laplace equation, generalized reactiondiffusion theory [8], non-newtonian fluid theory, and the turbulent flow of a gas in porous medium [6]. In the non-newtonian fluid theory, the quantity p is characteristic of the medium. Media with p > 2 are called dilatant fluids and those with p < 2 are called pseudoplastics. If p = 2, then they are Newtonian fluids. The equation (Φ p (u )) = f(t, u, u ), t I = [, 1] (1.4) with different boundary conditions has been studied by many authors (see [1 3, 11, 12, 14, 17 2, 22, 23] and references therein). Third-order boundary value problems u = f(t, u, u, u ), t I = [, 1], (1.5) have been discussed in many papers in recent years, for instance, see [4, 5, 9, 13, 15, 16, 26] and references therein. However, it seems that little is known about the results for problem (1.1) (1.2) or (1.1) (1.3). Our results are motivated by the papers [1,2,4,5,7,9, 1, 13, 14, 16, 26] which studied linear, periodic and Neumann with nonlinear boundary conditions for equation (1.4) and (1.5). In this paper, we obtain some new results for the problem (1.1) and (1.2) or (1.1) and (1.3), which are complementary and extensible to the cases considered by the authors in [4, 5, 9, 13, 15, 16, 26]. 2 Preliminaries In this section, we present results for second order Volterra type integro-differential equations, which will help to prove our main results. First, let us consider the boundary value problem (Φ p (u )) = f(t, u, T u, u ), (2.1)

Third Order Quasilinear Differential Equations 37 where T u(t) = φ(t) + u() = D, h(u(1), u (1)) =, (2.2) K(t, s)u(s)ds, function K(t, s) C([, 1] [, 1]), φ(t) C[, 1], K(t, s) on [, 1] [, 1], and D is a constant. Before introducing the main result of existence of solutions of problem (2.1) (2.2), we will give the following definition. Definition 2.1. We say that a function α C 1 [, 1] is a lower solution of equation (2.1) if Φ p (α ) C 1 (, 1) and satisfies (Φ p (α )) f(t, α, T α, α ), for t I = [, 1]. Analogously, we say that β C 1 [, 1] is an upper solution of equation (2.1) if Φ p (β ) C 1 (, 1) and satisfies (Φ p (β )) f(t, β, T β, β ), for t I. Definition 2.2. Let D be a subset of [, 1] R 3. We say that f(t, u, v, w) satisfies a Nagumo condition in D, if f(t, u, v, w) C([, 1] R 3, R) and given any a >, there exists a positive function Ψ(x) C[, ) : [, ) [a, + ) such that for arbitrary (t, u, v, w) D and such that f(t, u, v, w) Ψ( w ) Φ 1 p (u)/ψ(φ 1 p (u))du =. By a modification of the method given in [21, 24], we obtain the following lemma. Lemma 2.3. Let α and β be a lower and an upper solution, respectively, for equation (2.1) and such that α β in I. Assume that (i) f(t, u, v, w) is nonincreasing in v; (ii) f satisfies a Nagumo condition in Then the boundary value problem {(t, u, v, w) [, 1] R 3 : α(t) u β(t)}. (Φ p (u )) = f(t, u, T u, u ), u() = A, u(1) = B has at least one solution α(t) u(t) β(t) for all α() A β(), α(1) B β(1).

38 J. Zhao and Z. Yang Proof. Let u (t) = β(t). Then f(t, α(t), T u (t), α (t)) f(t, α, T α(t), α (t)) (Φ p (α )) f(t, β(t), T u (t), β (t)) (Φ p (β )), t I = [, 1]. From [21, Theorem 1] or [18, Theorem 2.2] or [19, Theorem 2.2], we have the problem (Φ p (u )) = f(t, u, T u, u ), u() = A, u(1) = B has at least one solution u 1 (t) such that α(t) u 1 (t) β(t) = u (t). We consider again the problem Clearly, and (Φ p (u )) = f(t, u, T u 1, u ), u() = A, u(1) = B. (2.3) f(t, α, T u 1, α ) (Φ p (α )), f(t, u 1, T u 1, u 1) f(t, u 1, T u, u 1) = (Φ p (u 1)). From Lemma 2.3, the above problem has at least one solution u 2 (t) such that α(t) u 2 (t) u 1 (t). By mathematical induction, we can construct a nonincreasing sequence {u n (t)} such that α(t) u n (t) u n 1 (t) u (t) = β(t). First let M > such that α(t) M, β(t) M, t [, 1] = I. Then u n (t) M. Next we shall prove u n(t) N. From condition (H 4 ), we obtain N p 1 (2M) p 1 Φ 1 p (u) du > 2M, (2.4) h(φ 1 p (u)) where N > max{2m, max t I { β (t), α (t) }}. If not, then there is t 2 [, 1] such that u (t 2 ) > N. We can suppose u (t 2 ) > N. The mean value theorem implies that there is ξ [, 1] such that u(1) u() = u (ξ). Thus u (ξ) 2M < N. Then there is an interval [c, d] [, 1] such that u (c) = 2M, u (d) = N, and if t [c, d], 2M u N. From (2.3), we have (Φ p (u )) h(u ). Then d Φ 1 c p (Φ(u ))(Φ(u )) h(φ 1 p (Φ p (u ))) ds d c u (Φ p (u )) ds h(u ) d c u ds 2M. Therefore, we have N p 1 (2M) p 1 Φ 1 p (u) du 2M. h(φ 1 p (u))

Third Order Quasilinear Differential Equations 39 This is a contradiction in light of (2.4). Thus we have u (t) N. On the other hand, u n (t) satisfies the equation, so {(Φ p (u n)) } is uniformly bounded on I. Therefore, {u n (t)}, {Φ p (u n)} are uniformly bounded and equicontinuous. By Arzela Ascoli s theorem, there exists a subsequence {u nk } satisfying lim k Φ p (u n k ) = v. Thus, we obtain and hence u nk = A + lim k u n k = Φ 1 p (v) u n k (s)ds A + Φ 1 p (v)ds = u (k ). Then there exist u C 1 (I) such that lim u nk (t) = u. From the dominated convergence theorem, we conclude that u is a solution of problem (2.3). Lemma 2.4. Let α and β be a lower and an upper solution, respectively, for equation (2.1) and such that α β in I. Assume that conditions (i) (ii) of Lemma 2.3 are satisfied, and h(u, v) is nondecreasing in v and continuous on R 2, and Then the boundary value problem h(α(1), α (1)) B h(β(1), β (1)). (Φ p (u )) = f(t, u, T u, u ), u() = D, h(u(1), u (1)) = B (2.5) has at least one solution α(t) u(t) β(t) for all α() D β(). Proof. First, we assume α(1) = β(1). Thus, by α(t) β(t), we get α (1) β (1). On the other hand, it is clear that α (1) β (1) from h(α(1), α (1)) h(β(1), β (1)), which means α (1) = β (1). Then the problem (Φ p (u )) = f(t, u, T u, u ), u() = D, u(1) = α(1) has at least one solution α(t) u(t) β(t), which is a solution of problem (2.5). Next, we consider that α(1) < β(1). Applying Lemma 2.3, we know that the problem (Φ p (u )) = f(t, u, T u, u ), u() = D, u(1) = α(1) has at least one solution α (t), and α(t) α (t) β(t). It follows that α (1) α (1). From the assumptions on h, we see that h(α (1), α (1)) h(α(1), α (1)) B. (2.6) If equality in (2.6) holds, then α (t) is a solution of problem (2.5). Thus the proof is complete. Otherwise, we consider the boundary value problem (Φ p (u )) = f(t, u, T u, u ), u() = D, u(1) = β(1).

31 J. Zhao and Z. Yang Clearly, the same reasoning gets to a solution β (t) and such that α (t) β (t) β(t), t 1, h(β (1), β (1)) h(β(1), β (1)) B. (2.7) Consequently, if equality in (2.7) holds, then the proof is completed. Otherwise, we choose d 1 = (β (1) + α (1))/2, and we consider the problem (Φ p (u )) = f(t, u, T u, u ), u() = D, u(1) = d 1. Applying Lemma 2.3, we obtain a solution u 1 (t) from the above problem, and α u 1 (t) β. If h(u(1), u (1)) = B, then the proof is completed. If h(u(1), u (1)) > B, then let α 1 (t) = α (t), β 1 (t) = u 1 (t); if h(u(1), u (1)) < B, then let α 1 (t) = u 1 (t), β 1 (t) = β (t). Hence β 1 (1) α 1 (1) = 1 2 [β (1) α (1)]. Assume, by the induction method, that we have obtained α n (t), β n (t) (n = 1, 2,, m), which satisfy α n 1 (t) α n (t) β n (t) β n 1 (t), t 1, (2.8) β n (1) α n (1) = 1 2 [β n 1(1) α n 1 (1)]. (2.9) Then, choose d m+1 = 1 2 [β m(1) + α m (1)], and we consider the problem (Φ p (u )) = f(t, u, T u, u ), u() = D, u(1) = d m+1. Consequently, by the same method used to obtain α 1 (t) and β 1 (t), we have α m+1 (t) and β m+1 (t), which satisfy α m (t) α m+1 (t) β m+1 (t) β m (t), t 1, β m+1 (1) α m+1 (1) = 1 2 [β m(1) α m (1)]. Hence, we have the relations (2.8) and (2.9) for every n. In view of how we picked α n (t) and β n (t), it is easy to see that From (2.1), we obtain that h(α n (1), α n(1)) < B, h(β n (1), β n(1)) > B. (2.1) β n (1) α n (1) = 1 2 n [β (1) α (1)]. (2.11) Also, the Nagumo condition shows {α n (t)}, {β n (t)}, {α n(t)}, {β n(t)} are equicontinuous and uniformly bounded in t 1. Therefore, applying the Arzela Ascoli theorem to the sequences {α n (t)} and {β n (t)}, there exist two subsequences {β nj (t)}

Third Order Quasilinear Differential Equations 311 and {α ni (t)} such that as j, β nj (t) u (t), β n j (t) u (t), uniformly on [, 1], and α ni (t) u (t), α n i (t) u (t), uniformly on [, 1] as i. Therefore u (t) and u (t) satisfy (2.1), and we have From (2.1) and (2.7), it is obvious that h(u (1), u (1)) B, h(u (1), u (1)) B. (2.12) u (t) u (t), t 1. (2.13) On the other hand, by (2.11), we will show that u (1) = u (1). Thus, we have from (2.13) u (1) u (1). (2.14) From (2.12) and (2.14), we get By (2.15), it is easy to show the relations Hence, we complete the proof. B h(u (1), u (1)) h(u (1), u (1)) B. (2.15) B = h(u (1), u (1)) = h(u (1), u (1)). As in the proof of Lemma 2.4, we can show the following. Lemma 2.5. Let α and β be a lower and an upper solution, respectively, for equation (2.1) and such that α β in I. Assume that conditions (i) (ii) of Lemma 2.3 are satisfied, g(u, v) is nonincreasing in v and continuous on R 2, and Then the boundary value problem g(β(), β ()) A g(α(), α ()). (Φ p (u )) = f(t, u, T u, u ), u(1) = F, g(u(), u ()) = A has at least one solution α(t) u(t) β(t) for all α(1) F β(1). As in the proofs of Lemma 2.4 and Lemma 2.5, we can show the following. Lemma 2.6. Let α and β be a lower and an upper solution, respectively, for equation (2.1) and such that α β in I. Assume that conditions (i) (ii) of Lemma 2.3 are satisfied, g(u, v), h(u, v) are nonincreasing in v and nondecreasing in u and continuous on R 2, and { g(β(), β ()) A g(α(), α ()), Then the boundary value problem h(α(1), α (1)) B h(β(1), β (1)). (Φ p (u )) = f(t, u, T u, u ), g(u(), u ()) = A, h(u(1), u (1)) = B has at least one solution α(t) u(t) β(t).

312 J. Zhao and Z. Yang Secondly, we consider the boundary value problem consisting of (2.1) and L(u(), u(1)) =, R(u(), u(1), u (), u (1)) =, (2.16) which contains the boundary conditions Lemma 2.7. Assume that u() = u(1), u () = u (1). (i) Conditions (i) (ii) of Lemma 2.3 are satisfied; (ii) L(ξ, η) C(R 2 ) and there exist nondecreasing functions γ(s) and θ(s), which are continuous on R, such that L(γ(s), θ(s)) for s R; (iii) R(ξ, η, u, v) C(R 4 ) is nondecreasing with respect to u, and is nonincreasing with respect to v; (iv) let α and β be a lower and an upper solution, respectively, for equation (2.1) and satisfy the following hypotheses R(β(), β(1), β (), β (1)) R(α(), α(1), α (), α (1)), and there exist s 1 and s 2 (s 1 s 2 ) such that γ(s 1 ) = α(), θ(s 1 ) = α(1); γ(s 2 ) = β(), θ(s 2 ) = β(1). Then the boundary value problem (2.1), (2.16) has a solution u(t) such that α(t) u(t) β(t) in I. Proof. By the monotonicity of γ(s) and θ(s), for arbitrary s (s 1 s s 2 ) we have α() γ(s) β(), α(1) θ(s) β(1). Hence, by Lemma 2.3, the problem (Φ p (u )) = f(t, u, T u, u ), u(, s) = γ(s), u(1, s) = θ(s) has at least one solution α(t) u(t, s) β(t). If s = s 1, then u(, s 1 ) = α(), u(1, s 1 ) = α(1). Since u (, s 1 ) α () and u (1, s 1 ) α (1), Similarly if s = s 2, then we have R(u(, s 1 ), u(1, s 1 ), u (, s 1 ), u (1, s 1 )) = R(α(), α(1), u (, s 1 ), u (1, s 1 )) R(α(), α(1), α (), α (1)). (2.17) u(, s 2 ) = β(), u(1, s 2 ) = β(1); u (, s 2 ) β (), u (1, s 2 ) β (1).

Third Order Quasilinear Differential Equations 313 Hence Define the sets Ω 1, Ω 2 by R(u(, s 2 ), u(1, s 2 ), u (, s 2 ), u (1, s 2 )) = R(β(), β(1), u (, s 2 ), u (1, s 2 )) R(β(), β(1), β (), β (1)). (2.18) Ω 1 = {s : R(u(, s), u(1, s), u (, s), u (1, s)) >, s 1 s s 2 }, Ω 2 = {s : R(u(, s), u(1, s), u (, s), u (1, s)) <, s 1 s s 2 }. From (2.17) and (2.18), we know that Ω 1 Ω 2 To see this, let c n =. We claim that Ω 1 is closed. Ω 1 with lim c n = c. Then with u n = u cn, it follows that n R(u n (, s), u n (1, s), u n(, s), u n(1, s)) >, and there exists a subsequence of u n which converges uniformly on [, 1], to a solution u of (1.1) satisfying u (, s) = γ(s) and u (1, s) = θ(s) and R(u (, s), u (1, s), u (, s), u (1, s)). By assumption, equality cannot occur, so that R(u (, s), u (1, s), u (, s), u (1, s)) >, and thus c Ω 1. Therefore, Ω 1 is closed. Likewise, we may show that Ω 2 is closed. Thus there exists an s [s 1, s 2 ] such that The proof is complete. R(u(, s ), u(1, s ), u (, s ), u (1, s )) =. 3 Main Results In this section, we discuss the existence of solutions for the third order nonlinear boundary value problem (1.1) (1.2) or (1.1), (1.3). Definition 3.1. We say that a function α C 2 [, 1] is a lower solution of equation (1.1) if Φ p (α ) C 1 (, 1) and satisfies (Φ p (α )) f(t, α, α, α ), for t I. Analogously, we say that β C 2 [, 1] is an upper solution of equation (1.1) if Φ p (β ) C 1 [, 1] and satisfies (Φ p (β )) f(t, β, β, β ), for t I. By a modification of the method given in [15, 16], we have the following main theorem for (1.1) (1.2). Theorem 3.2. Assume that

314 J. Zhao and Z. Yang (i) There exists an upper solution β and a lower solution α of the equation (1.1) on I = [, 1], respectively, such that α(t) β(t), α (t) β (t), α() = β() =, t 1; (ii) f(t, u, u, u ) is nonincreasing in u and continuous on [, 1] R 3 ; (iii) f(t, u, u, u ) satisfies a Nagumo condition in {(t, u, v, w) [, 1] R 3 : α(t) u α(t), α (t) u (t) β (t)}; (iv) g(u, v), h(u, v) are continuous on R 2, g(u, v) is decreasing in v and h(u, v) is increasing in v, and { g(β (), β ()) A g(α (), α ()), h(α (1), α (1)) B h(β (1), β (1)). Then the boundary value problem (1.1) (1.2) has a solution u(t) such that α(t) u(t) β(t) for t I. Proof. Set u = z. Then we have u(t) = z(s)ds. Thus, the boundary value problem (1.1) (1.2) can be written as a boundary value problem for the second order integrodifferential equation of Volterra type as (Φ p (z )) = f ( t, z(s)ds, z, z ), (3.1) g(z(), z ()) = A, h(z(1), z (1)) = B. (3.2) In order to employ Lemma 2.5, we construct the lower and upper solutions for the boundary value problem (3.1) (3.2) by using α(t), β(t) and hypotheses (i) (iv). We set Then, it is easy to show that because of (i). Note that α(t) = β(s)ds. Now, using (iv) and the monotonicity of f from (ii), we obtain α(t) = α (t), β(t) = β (t). α(s)ds, β(t) = (Φ p (α )) f α(t) β(t) ( t, ) α(s)ds, α(t), α (t),

Third Order Quasilinear Differential Equations 315 (Φ p (β )) f ( t, ) β(s)ds, β(t), β (t), g(β(), β ()) A g(α(), α ()), h(α(1), α (1)) B h(β(1), β (1)). Thus, we see that the functions α and β are the lower and the upper solutions, respectively for the boundary value problem (3.1) (3.2). Hence, by Lemma 2.6, we have α(t) z(t) β(t), t 1, where z(t) is a solution of the boundary value problem (2.17) (2.18). Finally, from the relation z(t) = u (t), we can recover u(t) = z(s)ds. Next, we give the following main theorem for (1.1), (1.3). Theorem 3.3. Assume that (i) There exists an upper solution β and a lower solution α of the equation (1.1) on I = [, 1], respectively, such that α(t) β(t), α (t) β (t), α() = β() =, t 1; (ii) f(t, u, u, u ) C([, 1] R 3 ) and nonincreasing with respect to u; (iii) conditions (ii) (iii) of Lemma 2.7 hold; (iv) f satisfies a Nagumo condition in {(t, u, v, w) [, 1] R 3 : α(t) u α(t), α (t) v(t) β (t)}; (v) R satisfies R(β (), β (1), β (), β (1)) R(α (), α (1), α (), α (1)). In addition, if there exist s 1 and s 2 (s 1 s 2 ) with r(s 1 ) = α (), θ(s 1 ) = α (1), r(s 2 ) = β (), θ(s 2 ) = β (1), then the boundary value problem (1.1), (1.3) has a solution u(t) such that α(t) u(t) β(t).

316 J. Zhao and Z. Yang Proof. Set u = z. Then we have u(t) = z(s)ds. Thus, the boundary value problem (1.1), (1.3) can be written as a boundary value problem for the second order integrodifferential equation of Volterra type as (Φ p (z )) = f ( t, z(s)ds, z, z ), (3.3) L(z(), z ()) = A, R(z(), z (), z(1), z (1)) = B. (3.4) In order to employ Lemma 2.7, we construct the lower and upper solutions for the boundary value problem (3.1) (3.2) by using α(t), β(t) and hypotheses (i) (v). We set where α(t) = α(t) + δ 1, β(t) = β(t) δ 2, δ 1 = C α(), δ 2 = β() C. Then, it is easy to show that α() = C = β(). Moreover, if we write α (t) = α (t), β (t) = β (t), it is easy to show that α (t) β (t), for t 1. Note that α(t) = C + (Φ p (α (t))) f (Φ p (β (t))) f α (s)ds, β(t) = C + ( t, α (t), C + ( t, β (t), C + β (s)ds. We have ) α (s)ds, α (t), ) β (s)ds, β (t), R(β (), β (1), β (), β (1)) R(α (), α (1), α (), α (1)), and there exist s 1, s 2 (s 1 s 2 ), such that r(s 1 ) = α (), θ(s 1 ) = α (1), r(s 2 ) = β (), θ(s 2 ) = α (1). Consequently, by Lemma 2.7, we obtain a solution z(t) of (3.3) (3.4), such that α (t) z(t) β (t). From the relation u (t) = z(t), we can recover u(t) = C + So, the proof is completed. z(s)ds, α(t) u(t) β(t) ( t 1).

Third Order Quasilinear Differential Equations 317 Finally, we consider the third-order two-point boundary value problem ( u p 2 u ) + f(t, u) =, t 1, (3.5) u() = u () = u (1) =. (3.6) Theorem 3.4. Assume that there exists an upper solution β and a lower solution α of the equation (3.5) on I = [, 1], respectively, such that α(t) β(t), α (t) β (t), α() = β() =, t 1. Then the boundary value problem (3.5) (3.6) has a solution u(t) such that α(t) u(t) β(t). Proof. Set u = z. Then we have u(t) = z(s)ds. Thus, the boundary value problem (3.5) (3.6) can be written as a boundary value problem for the second order integrodifferential equation of Volterra type as ( ) (Φ p (z )) = f t, z(s)ds, t 1, (3.7) z() = z(1) =. (3.8) In order to employ Lemma 2.3, we construct the lower and upper solutions for the boundary value problem (3.7) (3.8) by using α(t), β(t). We set Then, it is easy to show that Note that α(t) = α(s)ds, β(t) = α(t) = α (t), β(t) = β (t). α(t) β(t). (Φ p (α )) f (Φ p (β )) f β(s)ds. By using the assumptions, we obtain ( t, ( t, ) α(s)ds, ) β(s)ds. Consequently, by Lemma 2.3, we obtain a solution z(t) of (3.7) (3.8), such that α(t) z(t) β(t). From the relation u (t) = z(t), we can recover u(t) = C + So, the proof is completed. z(s)ds, α(t) u(t) β(t) ( t 1). Remark 3.5. When p = 2, related results for problem (3.5) (3.6) have been obtained in [25]. Thus, Theorem 3.4 extends some results in the literature [25].

318 J. Zhao and Z. Yang 4 Application The following example illustrates an application of the main results for this paper. Example 4.1. Consider the boundary value problem ( u p 2 u ) (t u) 2 t(4 + t 2 )u (u ) p sin(u ) =, t I = [, 1] (4.1) satisfying the conditions Let u() =, 5(u ()) 2 1 2 u () = 5, (u (1)) 2 + (u (1)) 3 = 1. (4.2) f(t, u, v, w) = (t u) 2 + t(4 + t 2 )v + v p sin w, g(v, w) = 5v 2 1 2 w, h(v, w) = v2 + w 3. f is continuous on [, 1] R 3 and nonincreasing in u when α(t) u(t) β(t), t [, 1]. g, h are continuous on R 2, g(v, w) is decreasing in w, h(v, w) is increasing in w. Furthermore, f satisfies a Nagumo condition in In fact Then {(t, u, v, w) [, 1] R 3 : t u t, 1 v 1}. f u = 2(t u) 4t 4, f v = t(4 + t 2 ) + 2v sin w 5 + 2 v 7, f w = v p cos w v p 1. f(t, u, v, w) f u u + f v v + f w w + f(t,,, ) 4 + 5 + 2 v + v p + 1 1 + 2 v + v p 13 = M. It is easy to prove that α(t) = t, β(t) = t are lower and upper solutions of BVP (4.1) (4.2), respectively. In fact Moreover, and ( α p 2 α ) (t α) 2 t(4 + t 2 )α (α ) p sin(α ) = (2t) 2 + t(4 + t 2 ) = t 3 + 4t 4t 2 = t(t 2 4t + 4) = t(t 2) 2, ( β p 2 β ) (t β) 2 t(4 + t 2 )β (β ) p sin(β ) = t(4 + t 2 ). α(t) β(t), α() = β() = g(β (), β ()) 5 g(α (), α ()), h(α (1), α (1)) 1 h(β (1), β (1)). From Theorem 3.2, the boundary value problem (4.1) (4.2) has a solution.

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