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Bounding functions methods for fully nonlinear boundary value problems
Nonlinear Analysis 64 (2006) 696 – 705 www.elsevier.com/locate/na
Bounding functions methods for fully nonlinear boundary value problems夡 Guangwa Wang∗ , Mingru Zhou, Li Sun Department of Mathematics, Xuzhou Normal University, Xuzhou 221116, PR China Received 19 March 2005; accepted 20 June 2005
Abstract Using the bounding functions method and the theory of topological degree, this paper presents the existence criterion of solution for third-order BVP with nonlinear boundary conditions and extends the existing results. 䉷 2005 Elsevier Ltd. All rights reserved. MSC: 34B Keywords: Third-order differential equation; Boundary value problem; Existence of solutions; Differential inequality; Bounding function pair; Modified function
1. Introduction Considerable attention has been given to the study of the existence of solutions for boundary value problems (BVPs) of nonlinear equations (cf. [1–14]). But, there are no issued results on the study of the existence of solutions for the general nonlinear system with the general nonlinear boundary conditions. In this paper, we extend the thoughts and
夡 The
NSF(04XLB03) of Xuzhou Normal University.
∗ Corresponding author. Tel.: 86 516 3403006.
E-mail addresses: [email protected] (G. Wang), [email protected] (M. Zhou), [email protected] (L. Sun). 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.06.027
G. Wang et al. / Nonlinear Analysis 64 (2006) 696 – 705
697
methods of Fabry and Habets [7] to study the BVP for third-order nonlinear system
y = f (t, y, y , y ), Pi (y(a), y (a), y (a)) = 0, P3 (y(b), y (b), y (b)) = 0,
i = 1, 2,
(1.1)
where t ∈ I = [a, b], f (t, 0 , 1 , 2 ) ∈ C(I × R 3 , R), Pi (0 , 1 , 2 ) ∈ C(R 3 , R), i = 1, 2, 3. Our method is not only modifying the nonlinear function in the original equations, but also transforming the original nonlinear boundary conditions into some new boundary conditions which are easy to discuss. Thus, we get the new BVP which will be discussed in the first place, then the judgement of the existence of solutions for the original BVP will be attained naturally. This technique dealing with the nonlinear problem is simpler and clearer compared with the method of shooting. As far as we know, the method has scarcely been used in the available reference materials. The paper is organized as follows. In Section 2, we give some basic concepts and the preparative theorem. In Section 3, we present and prove the main result. In Section 4, a more generalized BVP is studied. Finally, in Section 5, we use the results to solve an example. 2. Preparative theorem 2.1. Basic concepts First of all, we define a function r if x < r; (r, x, s) ≡ x if r x s; s if s < x, where r, x, s ∈ R, r s. Definition 1. Assume that (t), (t) ∈ C 3 (I, R). The pair of functions ((t), (t)) is called a bounding function pair (or simply, a bounding pair) of BVP(1.1), in case for all u(t) ∈ C 3 (I, R), (i) (j ) (t)(j ) (t), t ∈ I, j = 0, 1; (ii) (t)f (t, u(t), (t), (t)), (t)f (t, u(t), (t), (t)), where u(t) = ((t), u(t), (t)); (iii) P1 ((a), (a), u (a))0 P1 ((a), (a), u (a)), where u (a) = (−N, u (a), N ), N is some positive constant; P2 (u(a), (a), (a))0 P2 (u(a), (a), (a)), where u(a) = ((a), u(a), (a)); P3 (u(b), (b), (b))0 P3 (u(b), (b), (b)), where u(b) = ((b), u(b), (b)).
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Definition 2. A continuous function f (t, 0 , 1 , 2 ) is said to satisfy a Nagumo condition with respect to variable 2 on the set D = {(t, 0 , 1 , 2 ) | t ∈ I ; |j |r, j = 0, 1, r is a positive constant; 2 ∈ R} in case there exists function (t) ∈ C([0, +∞), (0, +∞)) such that |f (t, 0 , 1 , 2 )|(|2 |) and
+∞ 0
s ds = +∞. (s)
2.2. The modified problem Assume that there are two functions (t), (t) ∈ C 3 (I, R) satisfying (j ) (t)(j ) (t),
j = 0, 1.
We define function f (t, y, y , y ) ≡ f (t, y, y , y ) + h(y ), where y (j ) (t) = ((j ) (t), y (j ) (t), (j ) (t)),
j = 0, 1
and y (t) = (N, y (t), N ). N satisfies ⎧ 2M (t)|, | (t)| , ⎪ N > max , | ⎨ t∈I b−a N s ds ⎪ ⎩ 2M/(b−a) (s) > 2M,
(2.1)
in which M > maxt∈I {| (t)|, | (t)|}. h(y ) is continuous, bounded and < 0 if y < , h(y ) = 0 if y , > 0 if y > . Such function h(·) is easy to obtain, for example, let h(y ) ≡
y − y . 1 + |y − y |
In addition, we define P i (y(t), y (t), y (t)) ≡ ((i−1) (t), y (i−1) (t) − Pi (y(t), y (t), y (t)), (i−1) (t)), i = 1, 2, P 3 (y(t), y (t), y (t)) ≡ ( (t), y (t) − P3 (y(t), y (t), y (t)), (t)).
G. Wang et al. / Nonlinear Analysis 64 (2006) 696 – 705
Then we consider the following modified problem y = f (t, y, y , y ), y (i−1) (a) = P i (y(a), y (a), y (a)), i = 1, 2, y (b) = P 3 (y(b), y (b), y (b)).
699
(2.2)
2.3. Preparative theorem Lemma 1. Assume that (A1) The function f (t, y, y , y ) in BVP (1.1) satisfies the Nagumo condition with respect to y (t) by Definition 2; (A2) BVP (1.1) has a bounding pair ((t), (t)) on the interval I by Definition 1, where N is defined by formula (2.1). Then BVP (2.2) has a solution y(t) ∈ C 3 (I, R) such that (i) (t)y (i) (t)(i) (t), |y (t)|N, t ∈ I ,
i = 0, 1;
where N is defined in f . The following three propositions will lead to the proof of Lemma 1. Proposition 1. The modified BVP (2.2) has a solution y(t) ∈ C 3 (I, R). Proof. Consider y = f (t, y, y , y ) ≡ g(t), y (i−1) (a) = P i (y(a), y (a), y (a)) ≡ gi (a), y (b) = P 3 (y(b), y (b), y (b)) ≡ g3 (b),
i = 1, 2,
(2.3)
where ∈ [0, 1]. From the representations of f , P i , i = 1, 2, 3, y (t), y (i−1) (a)(i = 1, 2) and y (b) all are bounded. Also, we may ensure that y (t), y (t), y(t) all are bounded functions in I. Let = {y(t) ∈ R|y (i) (t) < K, i = 0, 1, 2, K is some sufficiently large positive constant, ∀t ∈ I }, where · is the usual maximum norm · ∞ . Then is a bounded open set. BVP(2.3) can be equivalently written into the following integral equation t t2 t1 2 y(t) = c1 + c2 t + c3 t + g(s) ds dt1 dt2 ≡ T y,
a
a
b
(2.4)
where T is an integral operator with a parameter and (c1 , c2 , c3 ) is determined by the system of equations ⎧ ⎨ c1 + c2 a + c3 a 2 = g1 (a), c2 + c3 · 2a = g2 (a), b t ⎩ c2 + c3 · 2b = g3 (b) − a b1 g(s) ds dt1 . Let H (, y) = (I − T )(y), then H : [0, 1] × −→ R is continuous, where I is identical mapping. Let h (y) = H (, y), then 0∈h (j). In fact, ∀y ∈ j, y = K. Noticing that
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K is sufficiently large, we have h (y) = y − T y y − T y, = K − T y > 0, ∀ ∈ [0, 1]. Thus, 0 ∈ / h (j). By the homotopy invariance theorem of topological degree, deg(h , , 0) keeps a constant, in particular, deg(h1 , , 0) = deg(h0 , , 0). Noticing that 0 ∈ , by the normality of topological degree, we have that deg(h0 , , 0) = deg(I − T0 , , 0) = deg(I, , 0) = 1, and deg(h1 , , 0) = deg(I − T1 , , 0) = deg(I − T0 , , 0) = 1. Hence, by the solvability theorem of topological degree, it is clear that there exists some y(t) satisfying (2.4), then this proposition is proved. Proposition 2. Every solution y(t) of the modified BVP (2.2) satisfies (i) (t)y (i) (t)(i) (t),
t ∈ I i = 0, 1.
Proof. First, we show that (t)y (t) (t),
t ∈ I.
(2.5)
If (t)y (t) isn’t true, there exists some ∈ [a, b] such that max ( (t) − y (t)) = () − y () > 0. t∈I
Obviously, = a, b by the boundary conditions of BVP(2.2). Thus () − y () = 0, () − y () 0.
(2.6)
However, from the definition of (t) and the fact that y(t) is a solution of (2.2), we have () − y () f (, y(), (), ()) − f (, y(), y (), y ()) − h(y ()) = −h(y ()) > 0. This contradicts (2.6). Hence, (t)y (t),
t ∈ I.
A similar proof shows that y (t) (t),
t ∈ I.
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To sum up, (2.5) is true. By the mean value theorem, from (2.5) and (i−1) (a)y (i−1) (a)(i−1) (a), It is clear that Proposition 2 holds.
i = 1, 2.
Proposition 3. For the every solution y(t) of the modified BVP (2.2) holds |y (t)|N,
t ∈ I.
Proof. Suppose that there exist some ∈ [a, b] such that |y ( )| > N . We can assume that y ( ) > N, because the discussion for the case y( ) <−N is analogous. There exists ∈ (a, b) such that y () =
2M y (b) − y (a) < N. b−a b−a
Hence, there exists some subinterval [c, d] (or [d, c]) ⊂ [a, b] such that y (c) =
2M , b−a
y (d) = N ,
and 2M < y (t) < N, b−a
∀t ∈ [c, d](or [d, c]).
From condition (A2),
d
y (s)y (s)
d
y (s) ds ds
(y (s)) c c = |y (d) − y (c)| 2M. On the other hand, from (2.1) we know that
d
y (s)y (s)
N s ds
ds =
(y (s)) c 2R/(b−a) (s) N s ds = 2R/(b−a) (s) > 2M. This inequality contradicts the above one and Proposition 3 holds. The proof of Lemma 1 is now a simple consequence of Propositions 1, 2, 3.
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3. Main theorem Now, the main result of this paper is given in the following theorem. Theorem 1. Assume that the conditions (A1)–(A2) in Lemma 1 hold and add to (A3) The function P1 (0 , 1 , 2 ) is decreasing in 0 and increasing in 1 , while P2 decreasing in 2 and P3 increasing in 2 . Then BVP (1.1) has a solution y(t) ∈ C 3 (I, R) such that (i) (t)y (i) (t)(i) (t), |y (t)|N, t ∈ I ,
i = 0, 1;
where N is defined in f . Proof. From Lemma 1 and the definition of f , the solution y(t) of the modified BVP (2.2) satisfies Eq. (1.1). As soon as it is proved that y(t) satisfies the boundary conditions of (1.1) under condition (A3), we may say that y(t) just is a solution of (1.1). First, we prove P1 (y(a), y (a), y (a)) = 0.
(3.1)
Case 1: Suppose that (a)y(a) − P1 (y(a), y (a), y (a))(a). Then y(a) = P 1 (y(a), y (a), y (a)), = y(a) − P1 (y(a), y (a), y (a)). Thus P1 (y(a), y (a), y (a)) = 0. Case 2: Suppose that (a) > y(a) − P1 (y(a), y (a), y (a)). Then y(a) = P 1 (y(a), y (a), y (a)), = (a). Hence P1 (y(a), y (a), y (a)) > 0. From Proposition 2 and condition (A3), P1 ((a), (a), y (a)) > 0. It is easy to see that the last inequality contradicts (iii) of Definition 1. Therefore, case 2 is not true.
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Case 3: Suppose that y(a) − P1 (y(a), y (a), y (a)) > (a). Then P1 ((a), (a), y (a))P1 (y(a), y (a), y (a)) < 0. Obviously, the last inequality also contradicts (iii) of Definition 1. Therefore, this case cannot hold. To sum up, (3.1) holds. A similar proof shows that P2 (y(a), y (a), y (a)) = 0, The proof is completed.
P3 (y(b), y (b), y (b)) = 0.
4. A generalized problem Now, we consider the following boundary value problem with more general boundary conditions y = f (t, y, y , y ), (4.1) Pi (y(a), y (a), y (a), y(b), y (b), y (b)) = 0, where t ∈ I, i = 1, 2, 3, y ∈ R, f and Pi are continuous functions. Similarly to Definition 1, we give Definition 3. Assume (t), (t) ∈ C 3 (I, R), The pair of functions ((t), (t)) is called a bounding function pair of BVP(4.1), in case that for all u(t) ∈ C 3 (I, R), (i) Same as (i) of Definition 1; (ii) Same as (ii) of Definition 1; (iii) P1 ((a), (a), u (a), u(b), u (b), u (b))0 P1 ((a), (a), u (a), u(b), u (b), u (b)), P2 (u(a), (a), (a), u(b), u (b), u (b))0 P2 (u(a), (a), (a), u(b), u (b), u (b)), P3 (u(a), u (a), u (a), u(b), (b), (b))0 P3 (u(a), u (a), u (a), u(b), (b), (b)), where u(j ) (s)=((j ) (s), u(j ) (s), (j ) (s)), u (s)=(−N, u (s), N ), j = 0, 1, N is some positive constant. For BVP(4.1), we have the following existence theorem: Theorem 2. Assume that (A1) Same as (A1) of Theorem 1; (A2) BVP (4.1) has a bounding function pair ((t), (t)) in the interval I by Definition 3; (A3) The function P1 (0 , 1 , 2 , 0 , 1 , 2 ) is decreasing in 0 and increasing in 1 , while P2 decreasing in 2 and P3 increasing in 2 .
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G. Wang et al. / Nonlinear Analysis 64 (2006) 696 – 705
Then the conclusion of Theorem 1 still holds for BVP (4.1). Proof. Consider the modified problem y = f (t, y, y , y ), y (i−1) (a) = P i (a), y (b) = P 3 (b).
(4.2)
where i = 1, 2, the modified function f (t, y, y , y ) is defined as BVP(2.2), and P i (t) ≡ P i (y(t), y (t), y (t), y(b + a − t), y (b + a − t), y (b + a − t)) ≡ ((i−1) (t), y (i−1) (t) − Pi (y(t), y (t), y (t), y(b + a − t), y (b + a − t), y (b + a − t)), (i−1) (t)),
i = 1, 2,
P 3 (t) ≡ P 3 (y(b + a − t), y (b + a − t), y (b + a − t), y(t), y (t), y (t)) ≡ ( (t), y (t) − P3 (y(b + a − t), y (b + a − t), y (b + a − t), y(t), y (t), y (t)), (t)). Using the same argument as the proof of Lemma 1, it follows that under the conditions (A1) and (A2) , BVP (4.2) has a solution y(t) satisfying the two inequalities in the conclusions of Lemma 1. Furthermore, in an analogous way to the proof of Theorem 1, it follows that the solution y(t) of BVP (4.2) just is a solution of BVP(4.1). Consequently, the proof of Theorem 2 is completed. The details of the proof will be omitted. 5. An example In this section, we exhibit an example where the existence result of Theorem 1 and Theorem 2 are applied. With this example we try to illustrate the kind of problems that we can study with these techniques and show that a bounding pair according to Definition 1 or Definition 3 can exist naturally. Example. Consider the boundary value problem ⎧ y = (t − y)2 + t (1 + t)2 y + yy sin(y ), ⎪ ⎪ ⎨ 4y(1) − 1 (y (1))3 + ky(2) = A, 8 (1) − 1 y (1) + k (y (2))2 = B, 5y ⎪ ⎪ 8 2 ⎩ ky(1) − y(2) + (y (2))2 + 4(y (2))3 = C.
t ∈ [1, 2], (5.1)
When k = 0, A ∈ [−5, 31 8 ], B ∈ [−9, 5], C ∈ [−12, −1], let (t) = −t 2 ,
(t) = t.
It is easy to prove that ((t), (t)) is a bounding pair of BVP(5.1) and all assumptions of Theorem 1 are fulfilled. Hence BVP (5.1) has at least one solution y(t), which satisfies −t 2 y(t) t,
−2t y (t)1,
t ∈ [1, 2].
G. Wang et al. / Nonlinear Analysis 64 (2006) 696 – 705
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When k = 1, A ∈ [−3, − 18 ], B ∈ [−7, 41 8 ], C ∈ [−11, −2], let (t) = −t 2 ,
(t) = t.
It is easy to prove that ((t), (t)) is a bounding pair of BVP(5.1) and all assumptions of Theorem 2 are fulfilled. Hence BVP(5.1) has at least one solution y(t), which satisfies −t 2 y(t) t,
−2t y (t)1,
t ∈ [1, 2].
Remark. Theorem 1 and Theorem 2 include or improve the results in Refs. [1–14] since our equation and boundary conditions are fully nonlinear. Obviously, the results in the references all are not available to our Example. References [1] A. Cabada, M.d.R. Grossinho, F. Minhos, On the solvability of some discontinuous third order nonlinear differential equations with two point boundary conditions, J. Math. Anal. Appl. 285 (2003) 174–190. [2] K.W. Chang, F.A. Howes, Nonlinear Singular Perturbation Phenomena: Theory and Application, Springer, Berlin, New York, 1984. [3] M.A. O’Donnell, Semi-linear systems of boundary value problems, SIAM J. Math. Anal. 15 (2) (1984) 316–332. [4] Z. Du, W. Ge, X. Lin, Existence of solutions for a class of third-order nonlinear boundary value problems, J. Math. Anal. Appl. 294 (2004) 104–112. [5] J. Ehme, P.W. Eloe, J. Henderson, Upper and lower solution methods for fully nonlinear boundary value problems, J. Differential Equations 180 (2002) 51–64. [6] L.H. Erbe, Existence of solutions to boundary value problems for second order differential equations, Nonlinear Anal. Theory, Methods Applicat. 6 (11) (1982) 1155–1163. [7] Ch. Fabry, P. Habets, Upper and lower solutions for second-order boundary value problems with nonlinear boundary conditions, Nonlinear Anal. Theory, Methods Applicat. 10 (10) (1986) 985–1007. [8] S. Felix, Nonlinear fourth order two-point boundary value problems, Rocky Mount. J. Math. 25 (2) (1995) 757–781. [9] G. Wang, M. Zhou, L. Sun, Existence of solutions of Robin problems for the system of nth order differential equations, J. Nanjing University Math. Biquarterly 19 (1) (2002) 68–79. [10] F.A. Howes, Differential inequalities and applications to nonlinear singular perturbation problems, J. Differential Equations 20 (1976) 133–149. [11] W.G. Kelly, Some existence theorems for nth-order boundary value problems, J. Differential Equations 18 (1975) 158–169. [12] M. Pei, Nonlinear two-point boundary value problems for nth order nonlinear differential equations, CHN. Math. Sinica 43 (5) (2000) 921–930. [13] X. Yang, The method of lower and upper solutions for systems of boundary value problems, Appl. Math. Comput. 144 (2003) 169–172. [14] X. Yang, Upper and lower solutions for periodic problems, Appl. Math. Comput. 137 (2003) 413–422.
Bounding functions methods for fully nonlinear boundary value problems夡 Guangwa Wang∗ , Mingru Zhou, Li Sun Department of Mathematics, Xuzhou Normal University, Xuzhou 221116, PR China Received 19 March 2005; accepted 20 June 2005
Abstract Using the bounding functions method and the theory of topological degree, this paper presents the existence criterion of solution for third-order BVP with nonlinear boundary conditions and extends the existing results. 䉷 2005 Elsevier Ltd. All rights reserved. MSC: 34B Keywords: Third-order differential equation; Boundary value problem; Existence of solutions; Differential inequality; Bounding function pair; Modified function
1. Introduction Considerable attention has been given to the study of the existence of solutions for boundary value problems (BVPs) of nonlinear equations (cf. [1–14]). But, there are no issued results on the study of the existence of solutions for the general nonlinear system with the general nonlinear boundary conditions. In this paper, we extend the thoughts and
夡 The
NSF(04XLB03) of Xuzhou Normal University.
∗ Corresponding author. Tel.: 86 516 3403006.
E-mail addresses: [email protected] (G. Wang), [email protected] (M. Zhou), [email protected] (L. Sun). 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.06.027
G. Wang et al. / Nonlinear Analysis 64 (2006) 696 – 705
697
methods of Fabry and Habets [7] to study the BVP for third-order nonlinear system
y = f (t, y, y , y ), Pi (y(a), y (a), y (a)) = 0, P3 (y(b), y (b), y (b)) = 0,
i = 1, 2,
(1.1)
where t ∈ I = [a, b], f (t, 0 , 1 , 2 ) ∈ C(I × R 3 , R), Pi (0 , 1 , 2 ) ∈ C(R 3 , R), i = 1, 2, 3. Our method is not only modifying the nonlinear function in the original equations, but also transforming the original nonlinear boundary conditions into some new boundary conditions which are easy to discuss. Thus, we get the new BVP which will be discussed in the first place, then the judgement of the existence of solutions for the original BVP will be attained naturally. This technique dealing with the nonlinear problem is simpler and clearer compared with the method of shooting. As far as we know, the method has scarcely been used in the available reference materials. The paper is organized as follows. In Section 2, we give some basic concepts and the preparative theorem. In Section 3, we present and prove the main result. In Section 4, a more generalized BVP is studied. Finally, in Section 5, we use the results to solve an example. 2. Preparative theorem 2.1. Basic concepts First of all, we define a function r if x < r; (r, x, s) ≡ x if r x s; s if s < x, where r, x, s ∈ R, r s. Definition 1. Assume that (t), (t) ∈ C 3 (I, R). The pair of functions ((t), (t)) is called a bounding function pair (or simply, a bounding pair) of BVP(1.1), in case for all u(t) ∈ C 3 (I, R), (i) (j ) (t)(j ) (t), t ∈ I, j = 0, 1; (ii) (t)f (t, u(t), (t), (t)), (t)f (t, u(t), (t), (t)), where u(t) = ((t), u(t), (t)); (iii) P1 ((a), (a), u (a))0 P1 ((a), (a), u (a)), where u (a) = (−N, u (a), N ), N is some positive constant; P2 (u(a), (a), (a))0 P2 (u(a), (a), (a)), where u(a) = ((a), u(a), (a)); P3 (u(b), (b), (b))0 P3 (u(b), (b), (b)), where u(b) = ((b), u(b), (b)).
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Definition 2. A continuous function f (t, 0 , 1 , 2 ) is said to satisfy a Nagumo condition with respect to variable 2 on the set D = {(t, 0 , 1 , 2 ) | t ∈ I ; |j |r, j = 0, 1, r is a positive constant; 2 ∈ R} in case there exists function (t) ∈ C([0, +∞), (0, +∞)) such that |f (t, 0 , 1 , 2 )|(|2 |) and
+∞ 0
s ds = +∞. (s)
2.2. The modified problem Assume that there are two functions (t), (t) ∈ C 3 (I, R) satisfying (j ) (t)(j ) (t),
j = 0, 1.
We define function f (t, y, y , y ) ≡ f (t, y, y , y ) + h(y ), where y (j ) (t) = ((j ) (t), y (j ) (t), (j ) (t)),
j = 0, 1
and y (t) = (N, y (t), N ). N satisfies ⎧ 2M (t)|, | (t)| , ⎪ N > max , | ⎨ t∈I b−a N s ds ⎪ ⎩ 2M/(b−a) (s) > 2M,
(2.1)
in which M > maxt∈I {| (t)|, | (t)|}. h(y ) is continuous, bounded and < 0 if y < , h(y ) = 0 if y , > 0 if y > . Such function h(·) is easy to obtain, for example, let h(y ) ≡
y − y . 1 + |y − y |
In addition, we define P i (y(t), y (t), y (t)) ≡ ((i−1) (t), y (i−1) (t) − Pi (y(t), y (t), y (t)), (i−1) (t)), i = 1, 2, P 3 (y(t), y (t), y (t)) ≡ ( (t), y (t) − P3 (y(t), y (t), y (t)), (t)).
G. Wang et al. / Nonlinear Analysis 64 (2006) 696 – 705
Then we consider the following modified problem y = f (t, y, y , y ), y (i−1) (a) = P i (y(a), y (a), y (a)), i = 1, 2, y (b) = P 3 (y(b), y (b), y (b)).
699
(2.2)
2.3. Preparative theorem Lemma 1. Assume that (A1) The function f (t, y, y , y ) in BVP (1.1) satisfies the Nagumo condition with respect to y (t) by Definition 2; (A2) BVP (1.1) has a bounding pair ((t), (t)) on the interval I by Definition 1, where N is defined by formula (2.1). Then BVP (2.2) has a solution y(t) ∈ C 3 (I, R) such that (i) (t)y (i) (t)(i) (t), |y (t)|N, t ∈ I ,
i = 0, 1;
where N is defined in f . The following three propositions will lead to the proof of Lemma 1. Proposition 1. The modified BVP (2.2) has a solution y(t) ∈ C 3 (I, R). Proof. Consider y = f (t, y, y , y ) ≡ g(t), y (i−1) (a) = P i (y(a), y (a), y (a)) ≡ gi (a), y (b) = P 3 (y(b), y (b), y (b)) ≡ g3 (b),
i = 1, 2,
(2.3)
where ∈ [0, 1]. From the representations of f , P i , i = 1, 2, 3, y (t), y (i−1) (a)(i = 1, 2) and y (b) all are bounded. Also, we may ensure that y (t), y (t), y(t) all are bounded functions in I. Let = {y(t) ∈ R|y (i) (t) < K, i = 0, 1, 2, K is some sufficiently large positive constant, ∀t ∈ I }, where · is the usual maximum norm · ∞ . Then is a bounded open set. BVP(2.3) can be equivalently written into the following integral equation t t2 t1 2 y(t) = c1 + c2 t + c3 t + g(s) ds dt1 dt2 ≡ T y,
a
a
b
(2.4)
where T is an integral operator with a parameter and (c1 , c2 , c3 ) is determined by the system of equations ⎧ ⎨ c1 + c2 a + c3 a 2 = g1 (a), c2 + c3 · 2a = g2 (a), b t ⎩ c2 + c3 · 2b = g3 (b) − a b1 g(s) ds dt1 . Let H (, y) = (I − T )(y), then H : [0, 1] × −→ R is continuous, where I is identical mapping. Let h (y) = H (, y), then 0∈h (j). In fact, ∀y ∈ j, y = K. Noticing that
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K is sufficiently large, we have h (y) = y − T y y − T y, = K − T y > 0, ∀ ∈ [0, 1]. Thus, 0 ∈ / h (j). By the homotopy invariance theorem of topological degree, deg(h , , 0) keeps a constant, in particular, deg(h1 , , 0) = deg(h0 , , 0). Noticing that 0 ∈ , by the normality of topological degree, we have that deg(h0 , , 0) = deg(I − T0 , , 0) = deg(I, , 0) = 1, and deg(h1 , , 0) = deg(I − T1 , , 0) = deg(I − T0 , , 0) = 1. Hence, by the solvability theorem of topological degree, it is clear that there exists some y(t) satisfying (2.4), then this proposition is proved. Proposition 2. Every solution y(t) of the modified BVP (2.2) satisfies (i) (t)y (i) (t)(i) (t),
t ∈ I i = 0, 1.
Proof. First, we show that (t)y (t) (t),
t ∈ I.
(2.5)
If (t)y (t) isn’t true, there exists some ∈ [a, b] such that max ( (t) − y (t)) = () − y () > 0. t∈I
Obviously, = a, b by the boundary conditions of BVP(2.2). Thus () − y () = 0, () − y () 0.
(2.6)
However, from the definition of (t) and the fact that y(t) is a solution of (2.2), we have () − y () f (, y(), (), ()) − f (, y(), y (), y ()) − h(y ()) = −h(y ()) > 0. This contradicts (2.6). Hence, (t)y (t),
t ∈ I.
A similar proof shows that y (t) (t),
t ∈ I.
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To sum up, (2.5) is true. By the mean value theorem, from (2.5) and (i−1) (a)y (i−1) (a)(i−1) (a), It is clear that Proposition 2 holds.
i = 1, 2.
Proposition 3. For the every solution y(t) of the modified BVP (2.2) holds |y (t)|N,
t ∈ I.
Proof. Suppose that there exist some ∈ [a, b] such that |y ( )| > N . We can assume that y ( ) > N, because the discussion for the case y( ) <−N is analogous. There exists ∈ (a, b) such that y () =
2M y (b) − y (a) < N. b−a b−a
Hence, there exists some subinterval [c, d] (or [d, c]) ⊂ [a, b] such that y (c) =
2M , b−a
y (d) = N ,
and 2M < y (t) < N, b−a
∀t ∈ [c, d](or [d, c]).
From condition (A2),
d
y (s)y (s)
d
y (s) ds ds
(y (s)) c c = |y (d) − y (c)| 2M. On the other hand, from (2.1) we know that
d
y (s)y (s)
N s ds
ds =
(y (s)) c 2R/(b−a) (s) N s ds = 2R/(b−a) (s) > 2M. This inequality contradicts the above one and Proposition 3 holds. The proof of Lemma 1 is now a simple consequence of Propositions 1, 2, 3.
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3. Main theorem Now, the main result of this paper is given in the following theorem. Theorem 1. Assume that the conditions (A1)–(A2) in Lemma 1 hold and add to (A3) The function P1 (0 , 1 , 2 ) is decreasing in 0 and increasing in 1 , while P2 decreasing in 2 and P3 increasing in 2 . Then BVP (1.1) has a solution y(t) ∈ C 3 (I, R) such that (i) (t)y (i) (t)(i) (t), |y (t)|N, t ∈ I ,
i = 0, 1;
where N is defined in f . Proof. From Lemma 1 and the definition of f , the solution y(t) of the modified BVP (2.2) satisfies Eq. (1.1). As soon as it is proved that y(t) satisfies the boundary conditions of (1.1) under condition (A3), we may say that y(t) just is a solution of (1.1). First, we prove P1 (y(a), y (a), y (a)) = 0.
(3.1)
Case 1: Suppose that (a)y(a) − P1 (y(a), y (a), y (a))(a). Then y(a) = P 1 (y(a), y (a), y (a)), = y(a) − P1 (y(a), y (a), y (a)). Thus P1 (y(a), y (a), y (a)) = 0. Case 2: Suppose that (a) > y(a) − P1 (y(a), y (a), y (a)). Then y(a) = P 1 (y(a), y (a), y (a)), = (a). Hence P1 (y(a), y (a), y (a)) > 0. From Proposition 2 and condition (A3), P1 ((a), (a), y (a)) > 0. It is easy to see that the last inequality contradicts (iii) of Definition 1. Therefore, case 2 is not true.
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Case 3: Suppose that y(a) − P1 (y(a), y (a), y (a)) > (a). Then P1 ((a), (a), y (a))P1 (y(a), y (a), y (a)) < 0. Obviously, the last inequality also contradicts (iii) of Definition 1. Therefore, this case cannot hold. To sum up, (3.1) holds. A similar proof shows that P2 (y(a), y (a), y (a)) = 0, The proof is completed.
P3 (y(b), y (b), y (b)) = 0.
4. A generalized problem Now, we consider the following boundary value problem with more general boundary conditions y = f (t, y, y , y ), (4.1) Pi (y(a), y (a), y (a), y(b), y (b), y (b)) = 0, where t ∈ I, i = 1, 2, 3, y ∈ R, f and Pi are continuous functions. Similarly to Definition 1, we give Definition 3. Assume (t), (t) ∈ C 3 (I, R), The pair of functions ((t), (t)) is called a bounding function pair of BVP(4.1), in case that for all u(t) ∈ C 3 (I, R), (i) Same as (i) of Definition 1; (ii) Same as (ii) of Definition 1; (iii) P1 ((a), (a), u (a), u(b), u (b), u (b))0 P1 ((a), (a), u (a), u(b), u (b), u (b)), P2 (u(a), (a), (a), u(b), u (b), u (b))0 P2 (u(a), (a), (a), u(b), u (b), u (b)), P3 (u(a), u (a), u (a), u(b), (b), (b))0 P3 (u(a), u (a), u (a), u(b), (b), (b)), where u(j ) (s)=((j ) (s), u(j ) (s), (j ) (s)), u (s)=(−N, u (s), N ), j = 0, 1, N is some positive constant. For BVP(4.1), we have the following existence theorem: Theorem 2. Assume that (A1) Same as (A1) of Theorem 1; (A2) BVP (4.1) has a bounding function pair ((t), (t)) in the interval I by Definition 3; (A3) The function P1 (0 , 1 , 2 , 0 , 1 , 2 ) is decreasing in 0 and increasing in 1 , while P2 decreasing in 2 and P3 increasing in 2 .
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Then the conclusion of Theorem 1 still holds for BVP (4.1). Proof. Consider the modified problem y = f (t, y, y , y ), y (i−1) (a) = P i (a), y (b) = P 3 (b).
(4.2)
where i = 1, 2, the modified function f (t, y, y , y ) is defined as BVP(2.2), and P i (t) ≡ P i (y(t), y (t), y (t), y(b + a − t), y (b + a − t), y (b + a − t)) ≡ ((i−1) (t), y (i−1) (t) − Pi (y(t), y (t), y (t), y(b + a − t), y (b + a − t), y (b + a − t)), (i−1) (t)),
i = 1, 2,
P 3 (t) ≡ P 3 (y(b + a − t), y (b + a − t), y (b + a − t), y(t), y (t), y (t)) ≡ ( (t), y (t) − P3 (y(b + a − t), y (b + a − t), y (b + a − t), y(t), y (t), y (t)), (t)). Using the same argument as the proof of Lemma 1, it follows that under the conditions (A1) and (A2) , BVP (4.2) has a solution y(t) satisfying the two inequalities in the conclusions of Lemma 1. Furthermore, in an analogous way to the proof of Theorem 1, it follows that the solution y(t) of BVP (4.2) just is a solution of BVP(4.1). Consequently, the proof of Theorem 2 is completed. The details of the proof will be omitted. 5. An example In this section, we exhibit an example where the existence result of Theorem 1 and Theorem 2 are applied. With this example we try to illustrate the kind of problems that we can study with these techniques and show that a bounding pair according to Definition 1 or Definition 3 can exist naturally. Example. Consider the boundary value problem ⎧ y = (t − y)2 + t (1 + t)2 y + yy sin(y ), ⎪ ⎪ ⎨ 4y(1) − 1 (y (1))3 + ky(2) = A, 8 (1) − 1 y (1) + k (y (2))2 = B, 5y ⎪ ⎪ 8 2 ⎩ ky(1) − y(2) + (y (2))2 + 4(y (2))3 = C.
t ∈ [1, 2], (5.1)
When k = 0, A ∈ [−5, 31 8 ], B ∈ [−9, 5], C ∈ [−12, −1], let (t) = −t 2 ,
(t) = t.
It is easy to prove that ((t), (t)) is a bounding pair of BVP(5.1) and all assumptions of Theorem 1 are fulfilled. Hence BVP (5.1) has at least one solution y(t), which satisfies −t 2 y(t) t,
−2t y (t)1,
t ∈ [1, 2].
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When k = 1, A ∈ [−3, − 18 ], B ∈ [−7, 41 8 ], C ∈ [−11, −2], let (t) = −t 2 ,
(t) = t.
It is easy to prove that ((t), (t)) is a bounding pair of BVP(5.1) and all assumptions of Theorem 2 are fulfilled. Hence BVP(5.1) has at least one solution y(t), which satisfies −t 2 y(t) t,
−2t y (t)1,
t ∈ [1, 2].
Remark. Theorem 1 and Theorem 2 include or improve the results in Refs. [1–14] since our equation and boundary conditions are fully nonlinear. Obviously, the results in the references all are not available to our Example. References [1] A. Cabada, M.d.R. Grossinho, F. Minhos, On the solvability of some discontinuous third order nonlinear differential equations with two point boundary conditions, J. Math. Anal. Appl. 285 (2003) 174–190. [2] K.W. Chang, F.A. Howes, Nonlinear Singular Perturbation Phenomena: Theory and Application, Springer, Berlin, New York, 1984. [3] M.A. O’Donnell, Semi-linear systems of boundary value problems, SIAM J. Math. Anal. 15 (2) (1984) 316–332. [4] Z. Du, W. Ge, X. Lin, Existence of solutions for a class of third-order nonlinear boundary value problems, J. Math. Anal. Appl. 294 (2004) 104–112. [5] J. Ehme, P.W. Eloe, J. Henderson, Upper and lower solution methods for fully nonlinear boundary value problems, J. Differential Equations 180 (2002) 51–64. [6] L.H. Erbe, Existence of solutions to boundary value problems for second order differential equations, Nonlinear Anal. Theory, Methods Applicat. 6 (11) (1982) 1155–1163. [7] Ch. Fabry, P. Habets, Upper and lower solutions for second-order boundary value problems with nonlinear boundary conditions, Nonlinear Anal. Theory, Methods Applicat. 10 (10) (1986) 985–1007. [8] S. Felix, Nonlinear fourth order two-point boundary value problems, Rocky Mount. J. Math. 25 (2) (1995) 757–781. [9] G. Wang, M. Zhou, L. Sun, Existence of solutions of Robin problems for the system of nth order differential equations, J. Nanjing University Math. Biquarterly 19 (1) (2002) 68–79. [10] F.A. Howes, Differential inequalities and applications to nonlinear singular perturbation problems, J. Differential Equations 20 (1976) 133–149. [11] W.G. Kelly, Some existence theorems for nth-order boundary value problems, J. Differential Equations 18 (1975) 158–169. [12] M. Pei, Nonlinear two-point boundary value problems for nth order nonlinear differential equations, CHN. Math. Sinica 43 (5) (2000) 921–930. [13] X. Yang, The method of lower and upper solutions for systems of boundary value problems, Appl. Math. Comput. 144 (2003) 169–172. [14] X. Yang, Upper and lower solutions for periodic problems, Appl. Math. Comput. 137 (2003) 413–422.