Comp. Appl. Math. DOI 10.1007/s40314-013-0067-x
Nonlinear second-order q-difference equations with three-point boundary conditions Phollakrit Thiramanus · Jessada Tariboon
Received: 23 November 2012 / Revised: 4 April 2013 / Accepted: 7 July 2013 © SBMAC - Sociedade Brasileira de Matemática Aplicada e Computacional 2013
Abstract We study sufficient conditions for the existence of solutions for three-point boundary value problems of nonlinear second-order q-difference equations. Boundary conditions of this paper are concerned with q-derivatives and q-integrals with different q-numbers. Our results are based on some standard fixed point theorems and Leray–Schauder degree theory. Some examples are also discussed to illustrate our results. Keywords problem
q-derivative · q-integral · q-difference equation · Three-point boundary value
Mathematics Subject Classification (2010)
39A13
1 Introduction In this paper, we study the existence of solutions for nonlinear second-order q-difference equation with three-point boundary conditions 2 Dq x(t) = f (t, x(t)), t ∈ J = [0, ηT ], (1.1) x(T ) = α 0 x(s)dr s, x(0) = β D p x(η), where 0 < p, q, r < 1, f ∈ C([0, T ] × R, R), 0 < η < T and α, β are given constants. The theory of q-difference calculus or quantum calculus was established in the beginning of the 20th century by the pioneer works of Jackson (1910), Carmichael (1912), Mason
Communicated by Eduardo Souza de Cursi. P. Thiramanus (B) · J. Tariboon Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok 10800, Thailand e-mail:
[email protected] J. Tariboon e-mail:
[email protected]
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(1915), Adams (1929), Trjitzinsky (1933), etc. For the last few decades, this theory has also evolved in many area of research and applications, see Ernst (1999), Finkelstein (1995), Finkelstein (1996), Floreanini and Vinet (1993), Floreanini and Vinet (1994), Floreanini and Vinet (1996), Freund and Zabrodin (1995), Gasper and Rahman (1990), Han and Zeng (1999). For some recent papers we refer the reader to Kac and Cheung (2002), Bangerezako (2004), Bangerezako (2005), Bochner and Unal (2005), Bohner and Sh (2010), Dobrogowska and Odzijewicz (2006), Gasper and Rahman (2007), Ismail and Simeonov (2009), El-Shahed and Hassan (2010), Ernst (2012) and references therein. However, the study of boundary value problems for nonlinear q-difference equations is still in the initial stage and it has become a significant step to develop this theory. Ahmad (2011) gave some excellent results for existence of solutions for nonlinear boundary value problem of third-order q-difference equation
Dq3 u(t) = f (t, u(t)), 0 ≤ t ≤ 1, u(0) = 0, Dq u(0) = 0, u(1) = 0.
(1.2)
The existence results for problem (1.2) were obtained by using Leray–Schauder degree theory and standard fixed point theorems. Moreover, he showed that if q → 1 then his results corresponded to the classical results. Ahmad et al. (2012a) investigated the existence of solutions for nonlinear second-order q-difference equation with non-separated boundary conditions
Dq2 u(t) = f (t, u(t)), t ∈ [0, T ], u(0) = ηu(T ), Dq u(0) = ηDq u(T ).
(1.3)
Using the Leray–Schauder nonlinear alternative and some standard fixed point theorems, the existence and uniqueness theorems of the problem (1.3) were proved. For some very recent results on nonlocal boundary value problems of nonlinear q-difference equations and inclusions, see Ahmad et al. (2012b), Ahmad and Nieto (2012), Ahmad and Ntouyas (2011). The aim of the present paper is to prove some existence and uniqueness results for the boundary value problem (1.1). Our results are based on Banach’s contraction mapping principle, Krasnoselskii’s fixed point theorem and Leray–Schauder degree theory. Since the problem (1.1) has different values of the q-numbers of q-derivative and q-integral, the existence results of such problem are also new. 2 Preliminaries Let us recall some basic concepts of q-calculus (Kac and Cheung 2002). For 0 < q < 1, we define the q-derivative of a real valued function x as Dq x(t) :=
x(t) − x(qt) , (1 − q)t
and Dq x(0) = limt→0 Dq x(t) provided Dx(0) exists, where D = d/dt. The higher order q-derivatives are given by Dq0 x(t) = x(t),
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Dqn x(t) = Dq Dqn−1 x(t),
n ∈ N.
Nonlinear second-order q-difference equations
The q-integral of a function x defined in the interval J = [0, T ] is given by t x(s)dq s :=
∞
(1 − q)q n [t x(tq n ) − ax(q n a)],
n=0
a
for t ∈ J and a constant a ∈ J . In particular, for a = 0, we denote t Iq x(t) =
x(s)dq s =
∞
t (1 − q)q n x(tq n ),
n=0
0
provided the series converges. Observe that Dq Iq x(t) = x(t),
(2.1)
and if x is continuous at t = 0, then Iq Dq x(t) = x(t) − x(0). In q-calculus, the product rule and integration by parts formula are t
Dq (gh)(t) = (Dq g(t))h(t) + g(qt)Dq h(t), t t x(s)Dq g(s)dq s = [x(s)g(s)]0 − Dq x(s)g(qs)dq s.
0
(2.2) (2.3)
0
In addition, reversing the order of integration is t s
t t x(r )dq r dq s =
0
0
x(r )dq sdq r. 0 qr
In the limit q → 1 the q-calculus corresponds to classical calculus. Lemma 2.1 Let αη2 = (T + β(1 − αη))(1 + r ), 0 < p, q, r < 1 be given constants. Then the boundary value problem (1.1) is equivalent to the integral equation t x(t) =
β[(t (1 − αη) − T )(1 + r ) + αη2 ] η(1 − p)[αη2 − (T + β(1 − αη))(1 + r )] 0 ⎛ η ⎞ pη × ⎝ (η − qs) f (s, x(s))dq s + (qs − pη) f (s, x(s))dq s ⎠ (t − qs) f (s, x(s))dq s +
0
0
(t + β)(1 + r ) + 2 αη − (T + β(1 − αη))(1 + r ) ⎛ T ⎞ η y × ⎝ (T −qs) f (s, x(s))dq s −α (y −qs) f (s, x(s))dq sdr y ⎠ . 0
0
0
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(2.4)
P. Thiramanus, J. Tariboon
Proof For t ∈ J , taking double q-integral to the q-difference equation Dq2 x = f (t, x),
(2.5)
f (s, x(s))dq sdq ν + a1 t + a2 .
(2.6)
we get that t ν x(t) = 0
0
Changing the order of q-integration, Eq. (2.6) can be expressed as t t f (s, x(s))dq νdq s + a1 t + a2 .
(2.7)
(t − qs) f (s, x(s))dq s + a1 t + a2 .
(2.8)
x(t) = 0 qs
Alternatively, we have t x(t) = 0
For t ∈ J and t = 0, using p-derivative to (2.8), we obtain ⎤ ⎡ t t f (s, x(s))dq s − q s f (s, x(s))dq s ⎦ + a1 D p x(t) = D p ⎣t 0
=
1 (1 − p)t
0
t (t − qs) f (s, x(s))dq s + 0
1 (1 − p)t
pt (qs − pt) f (s, x(s))dq s + a1 . 0
For t = 0, we have D p x(0) = lim D p x(t) t→0
∞ t (1 − q) n q (1 − q n+1 ) f (tq n , x(tq n )) − p 2 f ( ptq n , x( ptq n )) + a1 t→0 (1 − p)
= lim = a1 .
n=0
Since x(0) = a2 and x(0) = β D p x(η), we have β − βa1 +a2 = (1− p)η
η 0
β (η−qs) f (s, x(s))dq s + (1− p)η
pη (qs − pη) f (s, x(s))dq s. 0
(2.9) For t ∈ J, r -integrating (2.8) from 0 to η, where η ∈ (0, T ), we get ⎛ ⎞ η η y x(s)dr s = ⎝ (y − qs) f (s, x(s))dq s + a1 y + a2 ⎠ dr y 0
0
0
η y (y − qs) f (s, x(s))dq sdr y + a1
= 0
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0
η2 + a2 η. 1+r
Nonlinear second-order q-difference equations
From T x(T ) = and condition x(T ) = α a1
η 0
αη2 T− 1+r
(T − qs) f (s, x(s))dq s + a1 T + a2 , 0
x(s)dr s, it follows that
η y + a2 (1 − αη) = α
(y − qs) f (s, x(s))dq sdr y 0
0
T (T − qs) f (s, x(s))dq s.
−
(2.10)
0
Solving (2.9)–(2.10), we obtain 1+r a1 = αη2 − (T + β(1 − αη))(1 + r ) ⎡ ⎞ ⎛ η pη β(1 − αη) ⎝ (η − qs) f (s, x(s))dq s + (qs − pη) f (s, x(s))dq s ⎠ ×⎣ η(1 − p) 0
0
η y
T (T − qs) f (s, x(s))dq s − α
+ 0
0
⎤
(y − qs) f (s, x(s))dq sdr y ⎦ ,
0
1+r a2 = αη2 − (T + β(1 − αη))(1 + r ) ⎡ ⎞ ⎛ η pη 2 − T (1 + r )) β(αη ⎝ (η − qs) f (s, x(s))dq s + (qs − pη) f (s, x(s))dq s ⎠ ×⎣ η(1 − p)(1 + r )
0
+β
0
η y
T (T − qs) f (s, x(s))dq s − βα 0
0
⎤
(y − qs) f (s, x(s))dq sdr y ⎦ .
0
Substituting the values of a1 and a2 in (2.8), we obtain (2.4). This completes the proof.
In the following, for the sake of convenience, we set := αη2 − (T + β(1 − αη))(1 + r ) = 0, (1 + p)|β| |α|ηT (1 + r ) + |α|η2 1 := T2 + 1+q || (1 + r )|T + β||T 2 (1 + r + r 2 ) − αη3 | . + (1 + r + r 2 )||
(2.11)
(2.12)
Let C denote the Banach space of all continuous functions from [0, T ] to R endowed with the norm defined by x = sup{|x(t)|, t ∈ [0, T ]}.
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3 Main results Now we are in position to establish the main results. Our first result is based on Banach’s fixed point theorem. Theorem 3.1 Let f : [0, T ] × R → R be a jointly continuous function satisfying the conditions (H1 ) | f (t, x) − f (t, z)| ≤ L|x − z|, ∀t ∈ [0, T ], x, z ∈ R, (H2 ) := L < 1, where L is a Lipschitz constant and is given by (2.12). Then, the boundary value problem (1.1) has a unique solution. Proof We define the operator F : C → C by t (F x)(t) =
(t −qs) f (s, x(s))dq s + 0
⎛
β[(t (1 − αη) − T )(1 + r ) + αη2 ] η(1 − p)
⎞ pη × ⎝ (η − qs) f (s, x(s))dq s + (qs − pη) f (s, x(s))dq s ⎠ η 0
0
(t + β)(1 + r ) + ⎛ T ⎞ η y × ⎝ (T −qs) f (s, x(s))dq s −α (y −qs) f (s, x(s))dq sdr y ⎠ , (3.1) 0
0
0
for x ∈ C and t ∈ [0, T ] where is defined by (2.11). Setting supt∈[0,T ] | f (t, 0)| = M, we choose a constant R such that M R≥ , (3.2) 1−δ where δ satisfies ≤ δ < 1. Now we will show that F B R ⊂ B R , where B R = {x ∈ C : x ≤ R}. For any x ∈ B R , we get t β[(t (1 − αη) − T )(1 + r ) + αη2 ] (F x) = sup (t − qs) f (s, x(s))dq s + η(1 − p) t∈[0,T ] 0 ⎞ ⎛ η pη × ⎝ (η − qs) f (s, x(s))dq s + (qs − pη) f (s, x(s))dq s ⎠ 0
0
(t + β)(1 + r ) + ⎞ ⎛ T η y ⎠ ⎝ (T − qs) f (s, x(s))dq s − α (y − qs) f (s, x(s))dq sdr y × 0 0 0 t ≤ sup (t − qs)(| f (s, x(s)) − f (s, 0)| + | f (s, 0)|)dq s t∈[0,T ] 0
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Nonlinear second-order q-difference equations
β[(t (1 − αη) − T )(1 + r ) + αη2 ] η(1 − p) ⎛ η × ⎝ (η − qs)(| f (s, x(s)) − f (s, 0)| + | f (s, 0)|)dq s +
0
⎞ pη + (qs − pη)(| f (s, x(s)) − f (s, 0)| + | f (s, 0)|)dq s ⎠ 0
(t + β)(1 + r ) + ⎛ T × ⎝ (T − qs)(| f (s, x(s)) − f (s, 0)| + | f (s, 0)|)dq s 0
η y −α 0
0
⎞ ⎠ (y − qs)(| f (s, x(s)) − f (s, 0)| + | f (s, 0)|)dq sdr y
t ≤ sup (t − qs)(L|x(s)| + | f (s, 0)|)dq s t∈[0,T ] 0
β[(t (1 − αη) − T )(1 + r ) + αη2 ] η(1 − p) ⎞ ⎛ η pη × ⎝ (η−qs)(L|x(s)|+| f (s, 0)|)dq s + (qs − pη)(L|x(s)|+| f (s, 0)|)dq s ⎠ +
0
0
⎛ T (t + β)(1 + r ) ⎝ + (T − qs)(L|x(s)| + | f (s, 0)|)dq s 0 ⎞ η y (y − qs)(L|x(s)| + | f (s, 0)|)dq sdr y ⎠ −α 0 0 t ≤ (L x + M) sup (t − qs)dq s t∈[0,T ] 0 ⎞ ⎛ η pη β[(t (1 − αη) − T )(1 + r ) + αη2 ] ⎝ (η − qs)dq s + (qs − pη)dq s ⎠ + η(1 − p) 0
0
⎞ ⎛ T η y (t + β)(1 + r ) ⎝ + (T − qs)dq s − α (y − qs)dq sdr y ⎠ 0 0 0 2 (1 + p)|β| |t (1 − αη) − T |(1 + r ) + |α|η2 t ≤ (L x + M) sup + (1 + q)|| t∈[0,T ] 1 + q
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(1 + r )|t + β||T 2 (1 + r + r 2 ) − αη3 | (1 + q)(1 + r + r 2 )|| ≤ (L R + M) ≤ ( + 1 − δ)R ≤ R.
+
Next, for x, z ∈ C and for each t ∈ [0, T ], we obtain (F x) − (F z) = sup |(F x)(t) − (F z)(t)| t∈[0,T ]
t ≤ sup (t − qs)[ f (s, x(s)) − f (s, z(s))]dq s t∈[0,T ] 0
β[(t (1 − αη) − T )(1 + r ) + αη2 ] + η(1 − p) ⎛ η × ⎝ (η − qs)[ f (s, x(s)) − f (s, z(s))]dq s 0
⎞ pη + (qs − pη)[ f (s, x(s)) − f (s, z(s))]dq s ⎠ 0
⎛ T (t + β)(1 + r ) ⎝ (T − qs)[ f (s, x(s)) − f (s, z(s))]dq s + 0 ⎞ η y ⎠ (y − qs)[ f (s, x(s)) − f (s, z(s))]dq sdr y −α 0 0 t ≤ L sup (t − qs)dq s t∈[0,T ] 0 ⎞ ⎛ η pη β[(t (1−αη)−T )(1+r )+αη2 ] ⎝ (η−qs)dq s + (qs − pη)dq s ⎠ + η(1− p) 0
0
⎛ T ⎞ η y (t +β)(1+r ) ⎝ + (T −qs)dq s − α (y −qs)dq sdr y ⎠ x −z 0 0 0 2 (1 + p)|β| |t (1 − αη) − T |(1 + r ) + |α|η2 t ≤ L sup + (1 + q)|| t∈[0,T ] 1 + q 2 2 (1 + r )|t + β||T (1 + r + r ) − αη3 | x − z + (1 + q)(1 + r + r 2 )|| ≤ x − z . As < 1, F is a contraction. Thus, by Banach’s contraction mapping principle, the problem (1.1) has a unique solution. This completes the proof. Our second result is based on the following Krasnoselskii’s fixed point theorem (Krasnoselskii 1955).
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Nonlinear second-order q-difference equations
Theorem 3.2 Let K be a closed convex and nonempty subset of a Banach space X . Suppose that A and B map K into X such that (i) Ax + By ∈ K whenever x, y ∈ K , (ii) A is compact and continuous, (iii) B is a contraction mapping. Then there exists z ∈ K such that z = Az + Bz. Theorem 3.3 Suppose that (H1 ) and (H2 ) hold. In addition, assume that a continuous function f : [0, T ] × R → R satisfies (H3 )| f (t, x)| ≤ μ(t), ∀(t, x) ∈ [0, T ] × R, and μ ∈ C([0, T ], R+ ). If 1 1+q +
(1 + p)|β| |α|ηT (1 + r ) + |α|η2 ||
(1 + r )|T + β||T 2 (1 + r + r 2 ) − αη3 | (1 + r + r 2 )||
< 1,
(3.3)
then the boundary value problem (1.1) has at least one solution on [0, T ]. Proof Let us set maxt∈J |μ(t)| = μ and let us choose a constant R ≥ μ , where is given by (2.12). We define a suitable ball B R = {u ∈ C : u ≤ R} and the operators (F1 x) and (F2 x) on the set B R by t (F1 x)(t) =
(t − qs) f (s, x(s))dq s, 0
(F2 x)(t) =
β[(t (1 − αη) − T )(1 + r ) + αη2 ] η(1 − p) ⎞ ⎛ η pη × ⎝ (η − qs) f (s, x(s))dq s + (qs − pη) f (s, x(s))dq s ⎠ 0
0
(t + β)(1 + r ) + ⎛ T ⎞ η y × ⎝ (T − qs) f (s, x(s))dq s − α (y − qs) f (s, x(s))dq sdr y ⎠ . 0
For any x, z ∈ B R , we get
0
0
(1 + p)|β| |α|ηT (1 + r ) + |α|η2 T + || (1 + r )|T + β||T 2 (1 + r + r 2 ) − αη3 | + (1 + r + r 2 )|| = μ ≤ R,
μ (F1 x) + (F2 z) ≤ 1+q
2
which implies that (F1 x)+(F2 z) ∈ B R . From (H1 ) and (H2 ), it follows that F2 is a contraction mapping. The assumption (H3 ) together with the continuity of f implies that the operator
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F1 is continuous and uniformly bounded on B R . Now, we set sup(t,x)∈[0,T ]×B R | f (t, x)| = f max < ∞. Then for t1 , t2 ∈ [0, T ] with t2 < t1 and x ∈ B R , we have |(F1 x)(t1 ) − (F1 x)(t2 )| t1 t2 = (t1 − qs) f (s, x(s))dq s − (t2 − qs) f (s, x(s))dq s 0
0
t2 t1 = (t1 − t2 ) f (s, x(s))dq s + (t1 − qs) f (s, x(s))dq s, t2
0
⎛ t ⎞ 2 t1 ≤ f max ⎝ (t1 − t2 )dq s + (t1 − qs)dq s ⎠ . t2
0
As t1 − t2 → 0 the right-hand side of the above inequality which is independent of x and tends to zero. This implies that F1 is equicontinuous. Since F1 is uniformly bounded and equicontinuous on B R , F1 is relatively compact on B R . Hence, by the Arzelá–Ascoli Theorem, F1 is compact on B R . Therefore, all the assumptions of Theorem 3.2 are satisfied and the conclusion of Theorem 3.2 implies that the boundary value problem (1.1) has at least one solution on [0, T ]. This completes the proof. Our third result is based on Leray–Schauder degree theory. Theorem 3.4 Let f : [0, T ] × R → R. Suppose that there exist constants 0 ≤ κ < −1 where are given by (2.12) and M > 0 such that | f (t, x)| ≤ κ|x| + M for all t ∈ [0, T ], x ∈ R. Then the boundary value problem (1.1) has at least one solution on [0, T ]. Proof In view of fixed point problem, we define an operator F : C → C by x = F x,
(3.4)
where t (F x)(t) =
β[(t (1 − αη) − T )(1 + r ) + αη2 ] η(1 − p) ⎞ ⎛ η pη × ⎝ (η − qs) f (s, x(s))dq s + (qs − pη) f (s, x(s))dq s ⎠ (t − qs) f (s, x(s))dq s +
0
0
0
(t + β)(1 + r ) + ⎛ T ⎞ η y × ⎝ (T − qs) f (s, x(s))dq s − α (y − qs) f (s, x(s))dq sdr y ⎠ , 0
0
0
and is defined by (2.11). From (3.4), we just need to prove the existence of at least one solution x ∈ C satisfying (3.4). Therefore, it is sufficient to show that F : B r → C [0, T ] satisfies x = λF x,
∀x ∈ ∂ Br ,
∀λ ∈ [0, 1],
where Br = {x ∈ C : maxt∈[0,T ] |x(t)| < r, r > 0}. Now, we define
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(3.5)
Nonlinear second-order q-difference equations
H (λ, x) = λF x,
x ∈ C,
λ ∈ [0, 1].
As shown in Theorems 3.1 and 3.3, we have that an operator F is continuous, uniformly bounded and equicontinuous. Then, by the Arzelá–Ascoli Theorem, a continuous map h λ defined by h λ (x) = x − H (λ, x) = x − λF x is completely continuous. If (3.5) holds, then the following Leray–Schauder degrees are well defined. From the homotopy invariance of topological degree, it follows that deg(h λ , Br , 0) = deg(I − λF, Br , 0) = deg(h 1 , Br , 0) = deg(h 0 , Br , 0) = deg(I, Br , 0) = 1 = 0, 0 ∈ Br , where I denotes the unit operator. By the nonzero property of Leray–Schauder degree, h 1 (x) = x − F x = 0 for at least one x ∈ Br . Next, we prove (3.5). Assume that x = λF x for some λ ∈ [0, 1], we get x = λ(F x) t β[(t (1 − αη) − T )(1 + r ) + αη2 ] ≤ sup (t − qs) f (s, x(s))dq s + η(1 − p) t∈[0,T ] 0 ⎞ ⎛ η pη × ⎝ (η − qs) f (s, x(s))dq s + (qs − pη) f (s, x(s))dq s ⎠ 0
0
(t + β)(1 + r ) + ⎛ T ⎞ η y × ⎝ (T − qs) f (s, x(s))dq s − α (y − qs) f (s, x(s))dq sdr y ⎠ 0
0
0
t ≤ (κ x + M) sup | t∈[0,T ]
(t − qs)dq s 0
⎛ η ⎞ pη β[(t (1 − αη) − T )(1 + r ) + αη2 ] ⎝ × (η − qs)dq s + (qs − pη)dq s ⎠ + η(1 − p) 0
0
⎞ ⎛ T η y (t + β)(1 + r ) ⎝ ⎠ (T − qs)dq s − α (y − qs)dq sdr y + 0 0 0 (1 + p)|β| |t (1 − αη) − T |(1 + r ) + |α|η2 (κ x + M) 2 ≤ sup t + 1+q || t∈[0,T ] 2 2 3 (1 + r )|t + β||T (1 + r + r ) − αη | + (1 + r + r 2 )|| = (κ x + M). Computing directly for x , we obtain x ≤ Set r =
M 1−κ
M . 1 − κ
+ 1, then (3.5) holds. This completes the proof.
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4 Some examples In this section, in order to illustrate our results, we consider some examples. Example 4.1 Consider the following three-point boundary value problem D 21 x(t) = 3
1 x · , (t + 4)2 x + 1
3 x(0) = D 2 x 4 3
5 , 2
t ∈ [0, 3].
(4.1)
5
4 x(3) = 5
2
x(s)d 1 s. 4
(4.2)
0
Set q = 1/3, T = 3, η = 5/2, β = 3/4, α = 4/5, p = 2/3, r = 1/4, f (t, x) = (1/(t + 4)2 )( x /( x + 1)). Since | f (t, x) − f (t, z)| ≤ (1/16) x − z , then, (H1 ) is satisfied with = αη2 − (T + β(1 − αη))(1 + r ) = 35/16 = 0, (1+ p)|β| |α|ηT (1 + r )+|α|η2 T2 (1 + r )|T + β||T 2 (1 + r + r 2 ) − αη3 | = + + 1+q (1+q)|| (1 + q)(1 + r + r 2 )|| = 1269/98, and L = 1/16. Hence L ≈ 0.80931 < 1, and therefore from Theorem 3.1, the boundary value problem (4.1)–(4.2) has a unique solution on [0,3]. Example 4.2 Consider the following three-point boundary value problem D 21 x(t) = 4
sin(2π x) x + , 6π x + 1
1 x(0) = D 1 x 3 3
3 , 4
t ∈ [0, 1].
(4.3)
x(s)d 2 s.
(4.4)
3
4 x(1) = 4
5
0
Set q = 1/4, T = 1, η = 3/4, β = 1/3, α = 4, p = 1/3, r = 2/5. Here, | f (t, x)| = | sin(2π x)/6π + x /( x + 1)| ≤ ( x /3) + 1. So, M = 1, and = αη2 − (T + β(1 − αη))(1 + r ) = 107/60 = 0, (1 + p)|β| |α|ηT (1+r ) + |α|η2 T2 (1+r )|T + β||T 2 (1 + r + r 2 ) − αη3 | = + + 1+q (1 + q)|| (1 + q)(1 + r + r 2 )|| = 14984/6955, 1 1 ≈ 0.46416. κ= < 3 Hence, by Theorem 3.4, the boundary value problem (4.3)–(4.4) has at least one solution on [0,1]. Acknowledgments Thailand.
This research is supported by King Mongkut’s University of Technology North Bangkok,
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