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Basic results on hybrid differential equations
Nonlinear Analysis: Hybrid Systems 4 (2010) 414–424
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Nonlinear Analysis: Hybrid Systems journal homepage: www.elsevier.com/locate/nahs
Basic results on hybrid differential equations Bapurao C. Dhage a,∗ , V. Lakshmikantham b a
Kasubai, Gurukul Colony, Ahmedpur-413 515, Dist: Latur Maharashtra, India
b
Department of Applied Mathematics, Florida Institute of Technology, Melbourne, FL, USA
article
info
Article history: Received 30 September 2009 Accepted 7 October 2009 Keywords: Quadratic differential equation Existence theorem Differential inequalities Comparison principle
abstract In this paper, an existence theorem for hybrid nonlinear differential equations is proved under mixed Lipschitz and Carathéodory conditions. Some fundamental differential inequalities are also established which are utilized to prove the existence of extremal solutions and a comparison result. © 2009 Elsevier Ltd. All rights reserved.
1. Introduction In recent years, quadratic perturbations of nonlinear differential equations have attracted much attention. We call such differential equations hybrid differential equations. The details of different types of perturbations for a nonlinear differential and integral equations are given in Dhage and Lakshmikantham [1]. The existence theory for such hybrid equations can be developed using hybrid fixed point theory. See for example, Dhage [2–4]. The theory of differential inequalities for hybrid differential equations is crucial in the qualitative study of nonlinear differential equations. It is known that the differential inequalities play a significant role in the study of extremal solutions for nonlinear differential equations via the method of upper and lower solutions (see Heikkiä and Lakshmikantham [6]). In the present paper, we establish the existence and uniqueness results and some fundamental differential inequalities for hybrid differential equations initiating the study of theory of such systems. We then prove utilizing the theory of inequalities, its existence of extremal solutions and a comparison result. 2. Hybrid differential equation Let R be the real line and J = [t0 , t0 + a) be a bounded interval in R for some t0 , a ∈ R with a > 0. Let C (J × R, R) denote the class of continuous functions f : J × R → R and let C (J × R, R) denote the class of functions g : J × R → R such that (i) the map t 7→ g (t , x) is measurable for each x ∈ R, and (ii) the map x 7→ g (t , x) is continuous for each t ∈ J. The class C (J × R, R) is called the Carathéodory class of functions on J × R which are Lebesgue integrable when bounded by a Lebesgue integrable function on J. The differential equation in question is the following first order hybrid differential equation (in short HDE), d
x(t )
dt f (t , x(t )) x(t0 ) = x0 ∈ R
= g (t , x(t )) a.e. t ∈ J
where, f ∈ C (J × R, R \ {0}) and g ∈ C (J × R, R).
∗
Corresponding author. E-mail addresses: [email protected] (B.C. Dhage), [email protected] (V. Lakshmikantham).
1751-570X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.nahs.2009.10.005
(2.1)
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415
By a solution of the HDE (2.1) we mean a function x ∈ AC (J , R) such that (i) the function t 7→ f (tx,x) is absolutely continuous for each x ∈ R, and (ii) x satisfies the equations in (2.1), where AC (J , R) is the space of absolutely continuous real-valued functions defined on J. The theory of strict and nonstrict differential inequalities related to the ODEs is available in the literature (see Lakshmikantham and Leela [7] and the references therein). It is known that differential inequalities are useful for proving the existence of extremal solutions of the ODEs defined on J. 3. Existence result In this section, we prove the existence results for the HDE (2.1) on the closed and bounded interval J = [t0 , t0 + a] under mixed Lipschitz and Carathéodory conditions on the nonlinearities involved in it. We place the HDE (2.1) in the space C (J , R) of continuous real-valued functions defined on J. Define a supremum norm k · k in C (J , R) defined by
kxk = sup |x(t )| t ∈J
and a multiplication in C (J , R) by
(xy)(t ) = x(t )y(t ) for x, y ∈ C (J , R). Clearly C (J , R) is a Banach algebra with respect to above norm and multiplication in it. By L1 (J , R) denote the space of Lebesgue integrable real-valued functions on J equipped with the norm k · kL1 defined by
Z
t0 + a
kxkL1 =
|x(s)|ds.
t0
We prove the existence of solution for the HDE (2.1) via a fixed point theorem in Banach algebra due to Dhage [5]. Theorem 3.1 (Dhage [5]). Let S be a non-empty, closed convex and bounded subset of the Banach algebra X and let A : X → X and B : S → X be two operators such that (a) A is D -Lipschitz with D -function ψ , (b) B is completely continuous, (c) x = Ax By H⇒ x ∈ S for all y ∈ S, and (d) M ψ(r ) < r,where M = kB(S )k = sup{kBxk : x ∈ S }. Then the operator equation Ax Bx = x has a solution in S. We consider the following hypotheses in what follows.
(A0 ) The function x 7→ f (tx,x) is increasing in R almost everywhere for t ∈ J. (A1 ) There exists a constant L > 0 such that |f (t , x) − f (t , y)| ≤ L|x − y| for all t ∈ J and x, y ∈ R. (A2 ) There exists a function h ∈ L1 (J , R) such that
|g (t , x)| ≤ h(t ) a.e. t ∈ J for all x ∈ R. The following lemma is useful in what follows. Lemma 3.1. Assume that hypothesis (A0 ) holds. Then for any h ∈ L1 (J , R+ ), the function x ∈ AC (J , R+ ) is a solution of the HDE d dt
x( t ) f (t , x(t ))
= h( t )
a.e. t ∈ J
(3.1)
and x(0) = x0
(3.2)
if and only if x satisfies the hybrid integral equation (HIE) x(t ) = f (t , x(t ))
x0 f (t0 , x0 )
Z
t
h(s)ds ,
+
t ∈ J.
(3.3)
t0 x(t )
Proof. Let h ∈ L1 (J , R+ ). Assume first that x is a solution of the HDE (3.1)–(3.2). By definition, f (t ,x(t )) is absolutely h i continuous, and so, almost everywhere differentiable, whence to (3.1) from t0 to t, we obtain the HIE (3.3) on J.
d dt
x(t ) f (t ,x(t ))
is Lebesgue integrable on J. Applying integration
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Conversely, assume that x satisfies the QHIE (3.3). Then by direct differentiation we obtain (3.1). Again, substituting t = t0 in (3.3) yields x(t0 ) f (t0 , x(t0 ))
=
x0 f (t0 , x0 )
.
Since the mapping x 7→ f (tx,x) is increasing in R almost everywhere for t ∈ J, the mapping x 7→ whence x(t0 ) = x0 . Hence the proof of the lemma is complete.
x f (t0 ,x)
is injective in R,
Now we are in a position to prove the following existence theorem for HDE (2.1). Theorem 3.2. Assume that hypotheses (A0 )–(A2 ) hold. Further, if
x0 L f (t , x 0
0
+ khkL1 < 1 )
(3.4)
then the HDE (2.1) has a solution defined on J. Proof. Set X = C (J , R) and define a subset S of X defined by S = {x ∈ X | kxk ≤ N } where, N =
(3.5)
x F0 f (t 0,x ) +khk 1 L 0 0 x 1−L f (t 0,x ) +khk 1 L
and F0 = supt ∈J |f (t , 0)|.
0 0
Clearly S is a closed, convex and bounded subset of the Banach algebra X . Now, using hypotheses (A0 ) and (A2 ) it can be shown by an application of Lemma 3.1 that the HDE (2.1) is equivalent to the nonlinear HIE x(t ) = f (t , x(t ))
t
Z
x0
g (s, x(s))ds
+
f (t0 , x0 )
(3.6)
t0
for t ∈ J. Define two operators A : X → X and B : S → X by Ax(t ) = f (t , x(t )),
t ∈J
(3.7)
and Bx(t ) =
t
Z
x0 f (t0 , x0 )
g (s, x(s))ds,
+
t ∈ J.
(3.8)
t0
Then the HDIE (3.6) is transformed into the operator equation as Ax(t ) Bx(t ) = x(t ),
t ∈ J.
(3.9)
We shall show that the operators A and B satisfy all the conditions of Theorem 3.1. First, we show that A is a Lipschitz operator on X with the Lipschitz constant L. Let x, y ∈ X . Then, by hypothesis (A1 ),
|Ax(t ) − Ay(t )| = |f (t , x(t )) − f (t , y(t ))| ≤ L|x(t ) − y(t )| ≤ kx − yk for all t ∈ J. Taking supremum over t, we obtain
kAx − Ayk ≤ Lkx − yk for all x, y ∈ X . Next, we show that B is a compact and continuous operator on S into X . First we show that B is continuous on S. Let {xn } be a sequence in S converging to a point x ∈ S. Then by Lebesgue dominated convergence theorem, lim Bxn (t ) = lim
n→∞
n→∞
= = =
Z
x0
f (t0 , x0 ) x0 f (t0 , x0 ) x0 f (t0 , x0 )
Z n→∞
t
g (s, xn (s))ds t0
Z th t0
Z
t0
+ lim +
g (s, xn (s))ds
+
f (t0 , x0 )
x0
t
i
lim g (s, xn (s)) ds
n→∞
t
g (s, x(s))ds
+ t0
= Bx(t ) for all t ∈ J. This shows that B is a continuous operator on S.
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Next we show that B is compact operator on S. It is enough to show that B(S ) is a uniformly bounded and equi-continuous set in X . Let x ∈ S be arbitrary. Then by hypothesis (A2 ),
Z t x0 + |Bx(t )| ≤ |g (s, x(s))|ds f (t0 , x0 ) t0 Z t x0 + h(s)ds ≤ f (t0 , x0 ) t0 x0 + khkL1 ≤ f (t0 , x0 ) for all t ∈ J. Taking supremum over t,
x0 kBxk ≤ f (t , x 0
0
+ khkL1 )
for all x ∈ S. This shows that B is uniformly bounded on S. Again, let t1 , t2 ∈ J. Then for any x ∈ S, one has
Z t Z t2 1 |Bx(t1 ) − Bx(t2 )| = g (s, x(s))ds − g (s, x(s))ds 0 0 Z t 1 ≤ |g (s, x(s))|ds t2
≤ |p(t1 ) − p(t2 )| Rt
where p(t ) = t h(s)ds. Since the function p is continuous on compact J, it is uniformly continuous. Hence, for > 0, there 0 exists a δ > 0 such that
|t1 − t2 | < δ H⇒ |Bx(t1 ) − Bx(t2 )| < for all t1 , t2 ∈ J and for all x ∈ S. This shows that B(S ) is an equi-continuous set in X . Now the set B(S ) is uniformly bounded and equi-continuous set in X , so it is compact by Arzelá–Ascoli theorem. As a result, B is a continuous and compact operator on S. Next, we show that hypothesis (c) of Theorem 3.1 is satisfied. Let x ∈ X and y ∈ S be arbitrary such that x = Ax By. Then, by assumption (A1 ), we have
|x(t )| = |Ax(t )| |By(t )| Z t x0 ≤ |f (t , x(t ))| + g (s, y(s))ds f (t0 , x0 ) t0 Z t x0 + | ≤ |f (t , x(t )) − f (t , 0)| + |f (t , 0)| g ( s , y ( s ))| ds f (t0 , x0 ) t0 Z t x0 + h(s)ds ≤ [L |x(t )| + F0 ] f (t0 , x0 ) t0 x F0 f (t 0,x ) + khkL1 0 0 . ≤ x0 1 − L f (t ,x ) + khkL1 0 0 Taking supremum over t,
x0 f (t0 ,x0 ) + khkL1 = N. kx k ≤ x 1 − L f (t 0,x ) + khkL1 0 0 F0
This shows that hypothesis (c) of Theorem 3.1 is satisfied. Finally, we have
x0 + khkL1 M = kB(S )k = sup{kBxk : x ∈ S } ≤ f (t0 , x0 ) and so,
x0 α M ≤ L f (t , x 0
0
+ khkL1 < 1. )
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Thus, all the conditions of Theorem 3.1 are satisfied and hence the operator equation Ax Bx = x has a solution in S. As a result, the HDE (2.1) has a solution defined on J. This completes the proof. 4. Hybrid differential inequalities We begin by proving a result relative to strict inequalities for the HDE (2.1). Theorem 4.1. Assume that hypothesis (A0 ) holds. Suppose that there exist functions y, z ∈ AC (J , R) such that d
y(t )
f (t , y(t ))
dt
≤ g (t , y(t )) a.e. t ∈ J
(4.1)
≥ g (t , z (t )) a.e. t ∈ J ,
(4.2)
and d dt
z (t )
f (t , z (t ))
one of the inequalities being strict. Then, y(t0 ) < z (t0 )
(4.3)
implies y(t ) < z (t )
(4.4)
for all t ∈ J. Proof. Suppose that (4.1) is strict inequality. Assume that the claim is false. Then there exists a t1 ∈ J, t1 > t0 such that y(t1 ) = z (t2 ) and y(t ) < z (t ) for t0 ≤ t < t1 . Define Y (t ) =
y(t ) f (t , y(t ))
and Z (t ) =
z (t ) f (t , z (t ))
for all t ∈ J. Then we have Y (t1 ) = Z (t1 )
(4.5)
and by virtue of hypothesis (A0 ), we get Y (t ) < Z (t )
(4.6)
for all t < t1 . From (4.5) it follows that Y (t1 + h) − Y (t1 ) h
>
Z (t1 + h) − Z (t1 ) h
for small h < 0. The above inequality implies that Y 0 (t1 ) ≥ Z 0 (t1 ) which implies because of (A0 ), g (t1 , y(t1 )) > g (t1 , z (t1 )). This is a contradiction and hence completes the proof.
Our next result is concerned with the nonstrict differential inequality which needs a type of one sided Lipschitz condition.
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Theorem 4.2. Assume that hypothesis (A0 ) holds and there exists a real number L > 0 such that g (t , x1 ) − g (t , x2 ) ≤ L
x1 f ( t , x1 )
−
x2 f (t , x2 )
a.e. t ∈ J
(4.7)
for all x1 , x2 ∈ R with x1 ≥ x2 . Suppose that there exist functions y, z ∈ AC (J , R) such that inequalities (4.1) and (4.2) are satisfied. Then the nonstrict inequality (4.4) holds on J. Proof. Let > 0 and let a real number L > 0 be given. Set z f (t , z )
=
z f (t , z )
+ e2Lt
(4.8)
so that z f (t , z )
>
z f (t , z )
H⇒ z > z .
Let Z = f (tz,z ) so that Z = f (tz,z ) for t ∈ J. Then, by inequality (4.2), Z0 = Z 0 + 2L e2Lt ≥ g (t , z ) + 2L e2Lt .
(4.9)
Since g (t , z ) − g (t , z ) ≤ L
z f (t , z )
−
z
f (t , z )
for all t ∈ J, one has Z0 (t ) ≥ g (t , z (t )) − L e2Lt + 2L e2Lt > g (t , z (t )), or d
z (t )
f (t , z (t ))
dt
> g (t , z (t ))
(4.10)
for all t ∈ J. Also, we have z (t0 ) > z (t0 ) > y(t0 ). Hence by an application of Theorem 4.1 with z = z yields that y(t ) < z (t )
(4.11)
for all t ∈ J. Taking the limit as → 0, we get y(t ) ≤ z (t ) for all t ∈ J.
Remark 4.1. The conclusion of Theorems 4.1 and 4.2 also remains true if we replace the derivatives in the inequalities (4.1) x(t ) and (4.2) by Dini-derivative D− of the function f (t ,x(t )) on the bounded interval J. 5. Existence of maximal and minimal solutions In this section, we shall prove the existence of maximal and minimal solutions for the HDE (2.1) on J = [t0 , t0 + a]. We need the following definition in what follows. Definition 5.1. A solution r of the HDE (2.1) is said to be maximal if for any other solution x to the HDE (2.1) one has x(t ) ≤ r (t ), for all t ∈ J. Again, a solution ρ of the HDE (2.1) is said to be minimal if ρ(t ) ≤ x(t ), for all t ∈ J, where x is any solution of the HDE (2.1) on J . We discuss the case of maximal solution only, as the case of minimal solution is similar and can be obtained with the same arguments with appropriate modifications. Given an arbitrary small real number > 0, consider the following initial value problem of HDE, d
x( t )
= g (t , x(t )) + a.e. t ∈ J
dt f (t , x(t )) x(t0 ) = x0 + ∈ R
where, f ∈ C (J × R, R \ {0}) and g ∈ C (J × R, R).
(5.1)
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An existence theorem for the HDE (5.1) can be stated as follows: Theorem 5.1. Assume that hypotheses (A0 )–(A2 ) hold. Suppose also that the inequality (3.1) holds. Then for every small number > 0, the HDE (5.1) has a solution defined on J. Proof. By hypothesis, since
L
+ khkL1 < 1, f (t0 , x0 ) x0
there exists an 0 > 0 such that
x0 + L + khkL1 + a < 1 f (t , x + ) 0
(5.2)
0
for all 0 < ≤ 0 . Now the rest of the proof is similar to Theorem 3.2.
Our main existence theorem for maximal solution for the HDE (2.1) is Theorem 5.2. Assume that hypotheses (A0 )–(A2 ) hold. Further, if condition (3.1) holds, then the HDE (2.1) has a maximal solution defined on J.
∞
Proof. Let n 0 be a decreasing sequence of positive real numbers such that limn→∞ n = 0, where 0 is a positive real number satisfying the inequality
x0 + 0 L f (t , x + 0
0
0
+ khkL1 + 0 a < 1. )
(5.3)
The number 0 exists in view of the inequality (3.1). Then for any solution u of the HDE (2.1), by Theorem 5.1, one has u(t ) < r (t , n )
(5.4)
for all t ∈ J and n ∈ N ∪ {0}, where r (t , n ) is a solution of the HDE, d
x(t )
= g (t , x(t )) + n
dt f (t , x(t )) x(t0 ) = x0 + n ∈ R
a.e. t ∈ J
(5.5)
defined on J. Since, by Theorems 4.1 and 4.2, {r (t , n )} is a decreasing sequence of positive real numbers, the limit r (t ) = lim r (t , n ) n→∞
(5.6)
exists. We show that the convergence in (5.6) is uniform on J. To finish, it is enough to prove that the sequence {r (t , n )} is equi-continuous in C (J , R). Let t1 , t2 ∈ J be arbitrary. Then,
Z t1 Z t1 x0 + n |r (t1 , n ) − r (t2 , n )| = f (t1 , r (t1 , n )) + g (s, r (s, n ))ds + n ds f (t0 , x0 + n ) t0 t0 Z t2 Z t2 x0 + n − f (t2 , r (t2 , n )) + g (s, r (s, n ))ds + n ds f (t0 , x0 + n ) t0 t0 Z t1 Z t1 x0 + n ≤ f (t1 , r (t1 , n )) + g (s, x(s))ds + n ds f (t0 , x0 + n ) t0 t0 Z t1 Z t1 x0 + n − f (t2 , r (t2 , n )) + g (s, r (s, n ))ds + n ds f (t0 , x0 + n ) t0 t0 Z t1 Z t2 x0 + n + f (t2 , r (t2 , n )) + g (s, r (s, n ))ds + n ds f (t0 , x0 + n ) t0 t0 Z t2 Z t2 x0 + n − f (t2 , r (t2 , n )) + g (s, r (s, n ))ds + n ds f (t0 , x0 + n ) t0 t0 x0 + n + khkL1 + n a ≤ |f (t1 , r (t1 , n )) − f (t2 , r (t2 , n ))| f (t0 , x0 + n ) + F |p(t1 ) − p(t2 )| + |t1 − t2 |n Rt where, F = sup(t ,x)∈J ×[−N ,N ] |f (t , x)| and p(t ) = t h(s)ds. 0
(5.7)
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Since f is continuous on compact set J × [−N , N ], it is uniformly continuous there. Hence,
|f (t1 , r (t1 , n )) − f (t2 , r (t2 , n ))| → 0 as t1 → t2 uniformly for all n ∈ N. Similarly, since the function p is continuous on compact set J, it is uniformly continuous and hence
|p(t1 ) − p(t2 )| → 0 as t1 → t2 . Therefore, from the above inequality (5.7), it follows that
|r (t1 , n ) − r (t1 , n )| → 0 as t1 → t2 uniformly for all n ∈ N. Therefore, r (t , n ) → r (t ) as
n→∞
for all t ∈ J. Next, we show that the function r (t ) is a solution of the HDE (2.1) defined on J. Now, since r (t , n ) is a solution of the HDE (5.5), we have r (t , n ) = f (t , r (t , n ))
x0 + n
Z
f (t0 , x0 + n )
t
g (s, r (s, n ))ds +
+
Z
t
n ds
(5.8)
t0
t0
for all t ∈ J. Taking the limit as n → ∞ in the above Eq. (5.8) yields r (t ) = f (t , r (t ))
x0 f (t0 , x0 )
t
Z
g (s, r (s))ds
+
t0
for t ∈ J. Thus, the function r is a solution of the HDE (2.1) on J. Finally, from the inequality (5.4) it follows that u( t ) ≤ r ( t ) for all t ∈ J. Hence the HDE (2.1) has a maximal solution on J. This completes the proof.
6. Comparison theorems The main problem of the differential inequalities is to estimate a bound for the solution set for the differential inequality related to the HDE (2.1). In this section we prove that the maximal and minimal solutions serve as bounds for the solutions of the related differential inequality to HDE (2.1) on J = [t0 , t0 + a]. Theorem 6.1. Assume that hypotheses (A0 )–(A2 ) hold. Suppose that condition (3.1) holds. Further, if there exists a function u ∈ AC (J , R) such that d
u( t )
dt f (t , u(t )) u(t0 ) ≤ x0 ,
≤ g (t , u(t )) a.e. t ∈ J
(6.1)
Then, u( t ) ≤ r ( t )
(6.2)
for all t ∈ J, where r is a maximal solution of the HDE (2.1) on J. Proof. Let > 0 be arbitrary small. Then, by Theorem 5.2, r (t , ) is a maximal solution of the HDE (5.1) and that the limit r (t ) = lim r (t , )
(6.3)
→0
is uniform on J and the function r is a maximal solution of the HDE (2.1) on J. Hence, we obtain d
r (t , )
dt f (t , r (t , )) r (t0 , ) = x0 + .
= g (t , r (t , )) +
a.e. t ∈ J
(6.4)
From the above inequality it follows that d
r (t , )
dt f (t , r (t , )) r (t0 , ) > x0 .
> g (t , r (t , ))
a.e. t ∈ J
(6.5)
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Now we apply Theorem 4.1 to the inequalities (6.1) and (6.5) and conclude that u(t ) < r (t , )
(6.6)
for all t ∈ J. This further in view of limit (6.3) implies that inequality (6.2) holds on J. This completes the proof.
Theorem 6.2. Assume that hypotheses (A0 )–(A2 ) hold. Suppose that condition (3.1) holds. Further, if there exists a function v ∈ AC (J , R) such that
v(t ) ≥ g (t , v(t )) dt f (t , v(t )) v(t0 ) ≥ x0 , d
a.e. t ∈ J
(6.7)
Then,
ρ(t ) ≤ v(t )
(6.8)
for all t ∈ J, where ρ is a minimal solution of the HDE (2.1) on J. Note that Theorem 6.1 is useful to prove the boundedness and uniqueness of the solutions for the HDE (2.1) on J. A result in this direction is Theorem 6.3. Assume that hypotheses (A0 )–(A2 ) hold. Suppose that there exists a function G : J × R+ → R+ such that
|g (t , x1 ) − g (t , x2 )| ≤ G t ,
− f (t , x1 ) f ( t , x2 ) x1
x2
a.e. t ∈ J
(6.9)
for all x1 , x2 ∈ R+ . If identically zero function is the only solution of the differential equation m0 (t ) = G(t , m(t ))
a.e. t ∈ J ,
m(t0 ) = 0,
(6.10)
then the HDIE (2.1) has a unique solution on J. Proof. By Theorem 3.1, the HDE (2.1) has a solution defined on J. Suppose that there are two solutions u1 and u2 of the HDE (2.1) existing on J. Define a function m : J → R+ by
u1 ( t ) u2 (t ) . m(t ) = − f (t , u1 (t )) f (t , u2 (t ))
(6.11)
As (|x(t )|)0 ≤ |x0 (t )| for t ∈ J, we have that
d m0 (t ) ≤ dt
u1 (t )
−
d
u2 (t )
f (t , u2 (t ))
f (t , u1 (t )) dt ≤ |g (t , u1 (t )) − g (t , u2 (t ))| u1 (t ) u2 ( t ) ≤ G t, − f (t , u1 (t )) f (t , u2 (t )) = G(t , m(t ))
for almost everywhere t ∈ J; and that m(t0 ) = 0. Now, we apply Theorem 6.1 with f (t , x) ≡ 1 to get that m(t ) = 0 for all t ∈ J. This gives u1 ( t ) f (t , u1 (t ))
=
u2 ( t ) f (t , u2 (t ))
for all t ∈ J. Finally, in view of hypothesis (A0 ) we conclude that u1 (t ) = u2 (t ) on J. This completes the proof.
7. Existence of extremal solutions in vector segment Sometimes it is desirable to have knowledge of the existence of extremal positive solutions for the HDE (2.1) on J. In this section we shall prove the existence of maximal and minimal positive solutions for the HDE (2.1) between the given upper and lower solutions on J = [t0 , t0 + a]. We use a hybrid fixed point theorem of Dhage [3] in ordered Banach spaces for establishing our results. We need the following preliminaries in what follows. A non-empty closed set K in a Banach algebra X is called a cone with vertex 0, if (i) K +K ⊆ K , (ii) λK ⊆ K for λ ∈ R, λ ≥ 0 and (iii) {−K } ∩ K = 0, where 0 is the zero element of X . A cone K is called to be positive if (iv) K ◦ K ⊆ K , where ‘‘◦’’ is a multiplication composition in X . We introduce an order relation ≤ in X as follows. Let x, y ∈ X . Then x ≤ y if and only if y − x ∈ K . A cone K is called to be normal if the norm k·k is semi-monotone increasing on K , that is, there is a constant N > 0 such that kxk ≤ N kyk for all x, y ∈ K with x ≤ y. It is known that if the cone K is normal in X , then every order-bounded set in X is norm-bounded. The details of cones and their properties appear in Heikkilä and Lakshmikantham [6].
B.C. Dhage, V. Lakshmikantham / Nonlinear Analysis: Hybrid Systems 4 (2010) 414–424
423
Lemma 7.1 (Dhage [3]). Let K be a positive cone in a real Banach algebra X and let u1 , u2 , v1 , v2 ∈ K be such that u1 ≤ v1 and u2 ≤ v2 . Then u1 u2 ≤ v1 v2 . For any a, b ∈ X , a ≤ b, the order interval [a, b] is a set in X given by
[a, b] = {x ∈ X : a ≤ x ≤ b}. Definition 7.1. A mapping Q : [a, b] → X is said to be nondecreasing or monotone increasing if x ≤ y implies Qx ≤ Qy for all x, y ∈ [a, b]. We use the following fixed point theorems of Dhage [4] for proving the existence of extremal solutions for the BVP (2.1) under certain monotonicity conditions. Theorem 7.1 (Dhage [4]). Let K be a cone in a Banach algebra X and let a, b ∈ X be such that a ≤ b. Suppose that A, B : [a, b] → K are two nondecreasing operators such that (a) A is a Lipschitz with the Lipschitz constant α , (b) B is completely continuous, and (c) Ax Bx ∈ [a, b] for each x ∈ [a, b]. Further, if the cone K is positive and normal, then the operator equation Ax Bx = x has a least and a greatest positive solution in [a, b], whenever α M < 1, where M = kB([a, b])k := sup{kBxk : x ∈ [a, b]}. We equip the space C (J , R) with the order relation ≤ with the help of cone K defined by K = {x ∈ C (J , R) : x(t ) ≥ 0, ∀ t ∈ J }.
(7.1)
It is well known that the cone K is positive and normal in C (J , R). We need the following definitions in what follows. Definition 7.2. A function a ∈ AC (J , R) is called a lower solution of the HDE (2.1) defined on J if it satisfies (4.1). Similarly, a function b ∈ AC (J , R) is called an upper solution of the HDE (2.1) defined on J if it satisfies (4.2). A solution to the HDE (2.1) is a lower as well as an upper solution for the HDE (2.1) defined on J and vice versa. We consider the following set of assumptions:
(B0 ) (B1 ) (B2 ) (B3 ) (B4 )
f : J × R → R+ − {0}, g : J × R → R+ . The HDE (2.1) has a lower solution a and an upper solution b defined on J with a ≤ b. The function x 7→ f (tx,x) is increasing in the interval mint ∈J a(t ), maxt ∈J b(t ) almost everywhere for t ∈ J. The functions f (t , x) and g (t , x) are nondecreasing in x almost everywhere for t ∈ J . There exists a function k ∈ L1 (J , R+ ) such that g (t , b(t )) ≤ k(t ).
We remark that hypothesis (B4 ) holds in particular if f is continuous and g is L1 -Carathéodory on J × R. Theorem 7.2. Suppose that assumptions (A1 ) and (B0 ) through (B4 ) hold. Further if
x0 L f (t , x 0
0
+ kkkL1 < 1, )
(7.2)
then the HDE (2.1) has a minimal and a maximal positive solution defined on J. Proof. Now, the HDE (2.1) is equivalent to integral equation (3.6) defined on J. Let X = C (J , R). Define two operators A and B on X by (3.7) and (3.8) respectively. Then the integral equation (3.6) is transformed into an operator equation Ax(t ) Bx(t ) = x(t ) in the Banach algebra X . Notice that hypothesis (B0 ) implies A, B : [a, b] → K . Since the cone K in X is normal, [a, b] is a norm-bounded set in X . Now it is shown, as in the proof of Theorem 3.2, that A is a Lipschitz with the Lipschitz constant L and B is completely continuous operator on [a, b]. Again, hypothesis (B3 ) implies that A and B are nondecreasing on [a, b]. To see this, let x, y ∈ [a, b] be such that x ≤ y. Then, by hypothesis (B3 ), Ax(t ) = f (t , x(t )) ≤ f (t , y(t )) = Ay(t ) for all t ∈ J. Similarly, we have Bx(t ) =
f (t0 , x0 )
≤
x0 x0
f (t0 , x0 ) = By(t )
Z
t
g (s, x(s))ds
+
t0
Z
t
g (s, x(s))ds
+ t0
424
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for all t ∈ J. So A and B are nondecreasing operators on [a, b]. Again, Lemma 7.1 and hypothesis (B3 ) together imply that a(t ) ≤ [f (t , a(t ))]
≤ [f (t , x(t ))] ≤ [f (t , b(t ))]
x0 f (t0 , x0 )
x0 f (t0 , x0 )
x0 f (t0 , x0 )
t
Z
g (s, x(s))ds
+
t0 t
Z
g (s, x(s))ds
+
t0
Z
t
g (s, x(s))ds
+
t0
≤ b(t ), for all t ∈ J and x ∈ [a, b]. As a result a(t ) ≤ Ax(t )Bx(t ) ≤ b(t ), for all t ∈ J and x ∈ [a, b]. Hence, Ax Bx ∈ [a, b] for all x ∈ [a, b]. Again,
x0 + kkkL1 M = kB([a, b])k = sup{kBxk : x ∈ [a, b]} ≤ f (t0 , x0 ) and so,
x0 α M ≤ L f (t , x 0
0
+ kkkL1 < 1. )
Now, we apply Theorem 7.1 to the operator equation Ax Bx = x to yield that the HDE (2.1) has a minimal and a maximal positive solution in [a, b] defined on J. This completes the proof. References [1] B.C. Dhage, V. Lakshmikantham, Quadratic perturbations of periodic boundary value problems of second order ordinary differential equations (submitted for publication). [2] B.C. Dhage, On α -condensing mappings in Banach algebras, Math. Student 63 (1994) 146–152. [3] B.C. Dhage, A nonlinear alternative in Banach algebras with applications to functional differential equations, Nonlinear Funct. Anal. Appl. 8 (2004) 563–575. [4] B.C. Dhage, Fixed point theorems in ordered Banach algebras and applications, Panamer. Math. J. 9 (4) (1999) 93–102. [5] B.C. Dhage, A fixed point theorem in Banach algebras with applications to functional integral equations, Kyungpook Math. J. 44 (2004) 145–155. [6] S. Heikkilä, V. Lakshmikantham, Monotone Iterative Technique for Nonlinear Discontinues Differential Equations, Marcel Dekker Inc, New York, 1994. [7] V. Lakshmikantham, S. Leela, Differential and Integral Inequalities, Academic Press, New York, 1969.
Contents lists available at ScienceDirect
Nonlinear Analysis: Hybrid Systems journal homepage: www.elsevier.com/locate/nahs
Basic results on hybrid differential equations Bapurao C. Dhage a,∗ , V. Lakshmikantham b a
Kasubai, Gurukul Colony, Ahmedpur-413 515, Dist: Latur Maharashtra, India
b
Department of Applied Mathematics, Florida Institute of Technology, Melbourne, FL, USA
article
info
Article history: Received 30 September 2009 Accepted 7 October 2009 Keywords: Quadratic differential equation Existence theorem Differential inequalities Comparison principle
abstract In this paper, an existence theorem for hybrid nonlinear differential equations is proved under mixed Lipschitz and Carathéodory conditions. Some fundamental differential inequalities are also established which are utilized to prove the existence of extremal solutions and a comparison result. © 2009 Elsevier Ltd. All rights reserved.
1. Introduction In recent years, quadratic perturbations of nonlinear differential equations have attracted much attention. We call such differential equations hybrid differential equations. The details of different types of perturbations for a nonlinear differential and integral equations are given in Dhage and Lakshmikantham [1]. The existence theory for such hybrid equations can be developed using hybrid fixed point theory. See for example, Dhage [2–4]. The theory of differential inequalities for hybrid differential equations is crucial in the qualitative study of nonlinear differential equations. It is known that the differential inequalities play a significant role in the study of extremal solutions for nonlinear differential equations via the method of upper and lower solutions (see Heikkiä and Lakshmikantham [6]). In the present paper, we establish the existence and uniqueness results and some fundamental differential inequalities for hybrid differential equations initiating the study of theory of such systems. We then prove utilizing the theory of inequalities, its existence of extremal solutions and a comparison result. 2. Hybrid differential equation Let R be the real line and J = [t0 , t0 + a) be a bounded interval in R for some t0 , a ∈ R with a > 0. Let C (J × R, R) denote the class of continuous functions f : J × R → R and let C (J × R, R) denote the class of functions g : J × R → R such that (i) the map t 7→ g (t , x) is measurable for each x ∈ R, and (ii) the map x 7→ g (t , x) is continuous for each t ∈ J. The class C (J × R, R) is called the Carathéodory class of functions on J × R which are Lebesgue integrable when bounded by a Lebesgue integrable function on J. The differential equation in question is the following first order hybrid differential equation (in short HDE), d
x(t )
dt f (t , x(t )) x(t0 ) = x0 ∈ R
= g (t , x(t )) a.e. t ∈ J
where, f ∈ C (J × R, R \ {0}) and g ∈ C (J × R, R).
∗
Corresponding author. E-mail addresses: [email protected] (B.C. Dhage), [email protected] (V. Lakshmikantham).
1751-570X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.nahs.2009.10.005
(2.1)
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415
By a solution of the HDE (2.1) we mean a function x ∈ AC (J , R) such that (i) the function t 7→ f (tx,x) is absolutely continuous for each x ∈ R, and (ii) x satisfies the equations in (2.1), where AC (J , R) is the space of absolutely continuous real-valued functions defined on J. The theory of strict and nonstrict differential inequalities related to the ODEs is available in the literature (see Lakshmikantham and Leela [7] and the references therein). It is known that differential inequalities are useful for proving the existence of extremal solutions of the ODEs defined on J. 3. Existence result In this section, we prove the existence results for the HDE (2.1) on the closed and bounded interval J = [t0 , t0 + a] under mixed Lipschitz and Carathéodory conditions on the nonlinearities involved in it. We place the HDE (2.1) in the space C (J , R) of continuous real-valued functions defined on J. Define a supremum norm k · k in C (J , R) defined by
kxk = sup |x(t )| t ∈J
and a multiplication in C (J , R) by
(xy)(t ) = x(t )y(t ) for x, y ∈ C (J , R). Clearly C (J , R) is a Banach algebra with respect to above norm and multiplication in it. By L1 (J , R) denote the space of Lebesgue integrable real-valued functions on J equipped with the norm k · kL1 defined by
Z
t0 + a
kxkL1 =
|x(s)|ds.
t0
We prove the existence of solution for the HDE (2.1) via a fixed point theorem in Banach algebra due to Dhage [5]. Theorem 3.1 (Dhage [5]). Let S be a non-empty, closed convex and bounded subset of the Banach algebra X and let A : X → X and B : S → X be two operators such that (a) A is D -Lipschitz with D -function ψ , (b) B is completely continuous, (c) x = Ax By H⇒ x ∈ S for all y ∈ S, and (d) M ψ(r ) < r,where M = kB(S )k = sup{kBxk : x ∈ S }. Then the operator equation Ax Bx = x has a solution in S. We consider the following hypotheses in what follows.
(A0 ) The function x 7→ f (tx,x) is increasing in R almost everywhere for t ∈ J. (A1 ) There exists a constant L > 0 such that |f (t , x) − f (t , y)| ≤ L|x − y| for all t ∈ J and x, y ∈ R. (A2 ) There exists a function h ∈ L1 (J , R) such that
|g (t , x)| ≤ h(t ) a.e. t ∈ J for all x ∈ R. The following lemma is useful in what follows. Lemma 3.1. Assume that hypothesis (A0 ) holds. Then for any h ∈ L1 (J , R+ ), the function x ∈ AC (J , R+ ) is a solution of the HDE d dt
x( t ) f (t , x(t ))
= h( t )
a.e. t ∈ J
(3.1)
and x(0) = x0
(3.2)
if and only if x satisfies the hybrid integral equation (HIE) x(t ) = f (t , x(t ))
x0 f (t0 , x0 )
Z
t
h(s)ds ,
+
t ∈ J.
(3.3)
t0 x(t )
Proof. Let h ∈ L1 (J , R+ ). Assume first that x is a solution of the HDE (3.1)–(3.2). By definition, f (t ,x(t )) is absolutely h i continuous, and so, almost everywhere differentiable, whence to (3.1) from t0 to t, we obtain the HIE (3.3) on J.
d dt
x(t ) f (t ,x(t ))
is Lebesgue integrable on J. Applying integration
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Conversely, assume that x satisfies the QHIE (3.3). Then by direct differentiation we obtain (3.1). Again, substituting t = t0 in (3.3) yields x(t0 ) f (t0 , x(t0 ))
=
x0 f (t0 , x0 )
.
Since the mapping x 7→ f (tx,x) is increasing in R almost everywhere for t ∈ J, the mapping x 7→ whence x(t0 ) = x0 . Hence the proof of the lemma is complete.
x f (t0 ,x)
is injective in R,
Now we are in a position to prove the following existence theorem for HDE (2.1). Theorem 3.2. Assume that hypotheses (A0 )–(A2 ) hold. Further, if
x0 L f (t , x 0
0
+ khkL1 < 1 )
(3.4)
then the HDE (2.1) has a solution defined on J. Proof. Set X = C (J , R) and define a subset S of X defined by S = {x ∈ X | kxk ≤ N } where, N =
(3.5)
x F0 f (t 0,x ) +khk 1 L 0 0 x 1−L f (t 0,x ) +khk 1 L
and F0 = supt ∈J |f (t , 0)|.
0 0
Clearly S is a closed, convex and bounded subset of the Banach algebra X . Now, using hypotheses (A0 ) and (A2 ) it can be shown by an application of Lemma 3.1 that the HDE (2.1) is equivalent to the nonlinear HIE x(t ) = f (t , x(t ))
t
Z
x0
g (s, x(s))ds
+
f (t0 , x0 )
(3.6)
t0
for t ∈ J. Define two operators A : X → X and B : S → X by Ax(t ) = f (t , x(t )),
t ∈J
(3.7)
and Bx(t ) =
t
Z
x0 f (t0 , x0 )
g (s, x(s))ds,
+
t ∈ J.
(3.8)
t0
Then the HDIE (3.6) is transformed into the operator equation as Ax(t ) Bx(t ) = x(t ),
t ∈ J.
(3.9)
We shall show that the operators A and B satisfy all the conditions of Theorem 3.1. First, we show that A is a Lipschitz operator on X with the Lipschitz constant L. Let x, y ∈ X . Then, by hypothesis (A1 ),
|Ax(t ) − Ay(t )| = |f (t , x(t )) − f (t , y(t ))| ≤ L|x(t ) − y(t )| ≤ kx − yk for all t ∈ J. Taking supremum over t, we obtain
kAx − Ayk ≤ Lkx − yk for all x, y ∈ X . Next, we show that B is a compact and continuous operator on S into X . First we show that B is continuous on S. Let {xn } be a sequence in S converging to a point x ∈ S. Then by Lebesgue dominated convergence theorem, lim Bxn (t ) = lim
n→∞
n→∞
= = =
Z
x0
f (t0 , x0 ) x0 f (t0 , x0 ) x0 f (t0 , x0 )
Z n→∞
t
g (s, xn (s))ds t0
Z th t0
Z
t0
+ lim +
g (s, xn (s))ds
+
f (t0 , x0 )
x0
t
i
lim g (s, xn (s)) ds
n→∞
t
g (s, x(s))ds
+ t0
= Bx(t ) for all t ∈ J. This shows that B is a continuous operator on S.
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Next we show that B is compact operator on S. It is enough to show that B(S ) is a uniformly bounded and equi-continuous set in X . Let x ∈ S be arbitrary. Then by hypothesis (A2 ),
Z t x0 + |Bx(t )| ≤ |g (s, x(s))|ds f (t0 , x0 ) t0 Z t x0 + h(s)ds ≤ f (t0 , x0 ) t0 x0 + khkL1 ≤ f (t0 , x0 ) for all t ∈ J. Taking supremum over t,
x0 kBxk ≤ f (t , x 0
0
+ khkL1 )
for all x ∈ S. This shows that B is uniformly bounded on S. Again, let t1 , t2 ∈ J. Then for any x ∈ S, one has
Z t Z t2 1 |Bx(t1 ) − Bx(t2 )| = g (s, x(s))ds − g (s, x(s))ds 0 0 Z t 1 ≤ |g (s, x(s))|ds t2
≤ |p(t1 ) − p(t2 )| Rt
where p(t ) = t h(s)ds. Since the function p is continuous on compact J, it is uniformly continuous. Hence, for > 0, there 0 exists a δ > 0 such that
|t1 − t2 | < δ H⇒ |Bx(t1 ) − Bx(t2 )| < for all t1 , t2 ∈ J and for all x ∈ S. This shows that B(S ) is an equi-continuous set in X . Now the set B(S ) is uniformly bounded and equi-continuous set in X , so it is compact by Arzelá–Ascoli theorem. As a result, B is a continuous and compact operator on S. Next, we show that hypothesis (c) of Theorem 3.1 is satisfied. Let x ∈ X and y ∈ S be arbitrary such that x = Ax By. Then, by assumption (A1 ), we have
|x(t )| = |Ax(t )| |By(t )| Z t x0 ≤ |f (t , x(t ))| + g (s, y(s))ds f (t0 , x0 ) t0 Z t x0 + | ≤ |f (t , x(t )) − f (t , 0)| + |f (t , 0)| g ( s , y ( s ))| ds f (t0 , x0 ) t0 Z t x0 + h(s)ds ≤ [L |x(t )| + F0 ] f (t0 , x0 ) t0 x F0 f (t 0,x ) + khkL1 0 0 . ≤ x0 1 − L f (t ,x ) + khkL1 0 0 Taking supremum over t,
x0 f (t0 ,x0 ) + khkL1 = N. kx k ≤ x 1 − L f (t 0,x ) + khkL1 0 0 F0
This shows that hypothesis (c) of Theorem 3.1 is satisfied. Finally, we have
x0 + khkL1 M = kB(S )k = sup{kBxk : x ∈ S } ≤ f (t0 , x0 ) and so,
x0 α M ≤ L f (t , x 0
0
+ khkL1 < 1. )
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Thus, all the conditions of Theorem 3.1 are satisfied and hence the operator equation Ax Bx = x has a solution in S. As a result, the HDE (2.1) has a solution defined on J. This completes the proof. 4. Hybrid differential inequalities We begin by proving a result relative to strict inequalities for the HDE (2.1). Theorem 4.1. Assume that hypothesis (A0 ) holds. Suppose that there exist functions y, z ∈ AC (J , R) such that d
y(t )
f (t , y(t ))
dt
≤ g (t , y(t )) a.e. t ∈ J
(4.1)
≥ g (t , z (t )) a.e. t ∈ J ,
(4.2)
and d dt
z (t )
f (t , z (t ))
one of the inequalities being strict. Then, y(t0 ) < z (t0 )
(4.3)
implies y(t ) < z (t )
(4.4)
for all t ∈ J. Proof. Suppose that (4.1) is strict inequality. Assume that the claim is false. Then there exists a t1 ∈ J, t1 > t0 such that y(t1 ) = z (t2 ) and y(t ) < z (t ) for t0 ≤ t < t1 . Define Y (t ) =
y(t ) f (t , y(t ))
and Z (t ) =
z (t ) f (t , z (t ))
for all t ∈ J. Then we have Y (t1 ) = Z (t1 )
(4.5)
and by virtue of hypothesis (A0 ), we get Y (t ) < Z (t )
(4.6)
for all t < t1 . From (4.5) it follows that Y (t1 + h) − Y (t1 ) h
>
Z (t1 + h) − Z (t1 ) h
for small h < 0. The above inequality implies that Y 0 (t1 ) ≥ Z 0 (t1 ) which implies because of (A0 ), g (t1 , y(t1 )) > g (t1 , z (t1 )). This is a contradiction and hence completes the proof.
Our next result is concerned with the nonstrict differential inequality which needs a type of one sided Lipschitz condition.
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419
Theorem 4.2. Assume that hypothesis (A0 ) holds and there exists a real number L > 0 such that g (t , x1 ) − g (t , x2 ) ≤ L
x1 f ( t , x1 )
−
x2 f (t , x2 )
a.e. t ∈ J
(4.7)
for all x1 , x2 ∈ R with x1 ≥ x2 . Suppose that there exist functions y, z ∈ AC (J , R) such that inequalities (4.1) and (4.2) are satisfied. Then the nonstrict inequality (4.4) holds on J. Proof. Let > 0 and let a real number L > 0 be given. Set z f (t , z )
=
z f (t , z )
+ e2Lt
(4.8)
so that z f (t , z )
>
z f (t , z )
H⇒ z > z .
Let Z = f (tz,z ) so that Z = f (tz,z ) for t ∈ J. Then, by inequality (4.2), Z0 = Z 0 + 2L e2Lt ≥ g (t , z ) + 2L e2Lt .
(4.9)
Since g (t , z ) − g (t , z ) ≤ L
z f (t , z )
−
z
f (t , z )
for all t ∈ J, one has Z0 (t ) ≥ g (t , z (t )) − L e2Lt + 2L e2Lt > g (t , z (t )), or d
z (t )
f (t , z (t ))
dt
> g (t , z (t ))
(4.10)
for all t ∈ J. Also, we have z (t0 ) > z (t0 ) > y(t0 ). Hence by an application of Theorem 4.1 with z = z yields that y(t ) < z (t )
(4.11)
for all t ∈ J. Taking the limit as → 0, we get y(t ) ≤ z (t ) for all t ∈ J.
Remark 4.1. The conclusion of Theorems 4.1 and 4.2 also remains true if we replace the derivatives in the inequalities (4.1) x(t ) and (4.2) by Dini-derivative D− of the function f (t ,x(t )) on the bounded interval J. 5. Existence of maximal and minimal solutions In this section, we shall prove the existence of maximal and minimal solutions for the HDE (2.1) on J = [t0 , t0 + a]. We need the following definition in what follows. Definition 5.1. A solution r of the HDE (2.1) is said to be maximal if for any other solution x to the HDE (2.1) one has x(t ) ≤ r (t ), for all t ∈ J. Again, a solution ρ of the HDE (2.1) is said to be minimal if ρ(t ) ≤ x(t ), for all t ∈ J, where x is any solution of the HDE (2.1) on J . We discuss the case of maximal solution only, as the case of minimal solution is similar and can be obtained with the same arguments with appropriate modifications. Given an arbitrary small real number > 0, consider the following initial value problem of HDE, d
x( t )
= g (t , x(t )) + a.e. t ∈ J
dt f (t , x(t )) x(t0 ) = x0 + ∈ R
where, f ∈ C (J × R, R \ {0}) and g ∈ C (J × R, R).
(5.1)
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B.C. Dhage, V. Lakshmikantham / Nonlinear Analysis: Hybrid Systems 4 (2010) 414–424
An existence theorem for the HDE (5.1) can be stated as follows: Theorem 5.1. Assume that hypotheses (A0 )–(A2 ) hold. Suppose also that the inequality (3.1) holds. Then for every small number > 0, the HDE (5.1) has a solution defined on J. Proof. By hypothesis, since
L
+ khkL1 < 1, f (t0 , x0 ) x0
there exists an 0 > 0 such that
x0 + L + khkL1 + a < 1 f (t , x + ) 0
(5.2)
0
for all 0 < ≤ 0 . Now the rest of the proof is similar to Theorem 3.2.
Our main existence theorem for maximal solution for the HDE (2.1) is Theorem 5.2. Assume that hypotheses (A0 )–(A2 ) hold. Further, if condition (3.1) holds, then the HDE (2.1) has a maximal solution defined on J.
∞
Proof. Let n 0 be a decreasing sequence of positive real numbers such that limn→∞ n = 0, where 0 is a positive real number satisfying the inequality
x0 + 0 L f (t , x + 0
0
0
+ khkL1 + 0 a < 1. )
(5.3)
The number 0 exists in view of the inequality (3.1). Then for any solution u of the HDE (2.1), by Theorem 5.1, one has u(t ) < r (t , n )
(5.4)
for all t ∈ J and n ∈ N ∪ {0}, where r (t , n ) is a solution of the HDE, d
x(t )
= g (t , x(t )) + n
dt f (t , x(t )) x(t0 ) = x0 + n ∈ R
a.e. t ∈ J
(5.5)
defined on J. Since, by Theorems 4.1 and 4.2, {r (t , n )} is a decreasing sequence of positive real numbers, the limit r (t ) = lim r (t , n ) n→∞
(5.6)
exists. We show that the convergence in (5.6) is uniform on J. To finish, it is enough to prove that the sequence {r (t , n )} is equi-continuous in C (J , R). Let t1 , t2 ∈ J be arbitrary. Then,
Z t1 Z t1 x0 + n |r (t1 , n ) − r (t2 , n )| = f (t1 , r (t1 , n )) + g (s, r (s, n ))ds + n ds f (t0 , x0 + n ) t0 t0 Z t2 Z t2 x0 + n − f (t2 , r (t2 , n )) + g (s, r (s, n ))ds + n ds f (t0 , x0 + n ) t0 t0 Z t1 Z t1 x0 + n ≤ f (t1 , r (t1 , n )) + g (s, x(s))ds + n ds f (t0 , x0 + n ) t0 t0 Z t1 Z t1 x0 + n − f (t2 , r (t2 , n )) + g (s, r (s, n ))ds + n ds f (t0 , x0 + n ) t0 t0 Z t1 Z t2 x0 + n + f (t2 , r (t2 , n )) + g (s, r (s, n ))ds + n ds f (t0 , x0 + n ) t0 t0 Z t2 Z t2 x0 + n − f (t2 , r (t2 , n )) + g (s, r (s, n ))ds + n ds f (t0 , x0 + n ) t0 t0 x0 + n + khkL1 + n a ≤ |f (t1 , r (t1 , n )) − f (t2 , r (t2 , n ))| f (t0 , x0 + n ) + F |p(t1 ) − p(t2 )| + |t1 − t2 |n Rt where, F = sup(t ,x)∈J ×[−N ,N ] |f (t , x)| and p(t ) = t h(s)ds. 0
(5.7)
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Since f is continuous on compact set J × [−N , N ], it is uniformly continuous there. Hence,
|f (t1 , r (t1 , n )) − f (t2 , r (t2 , n ))| → 0 as t1 → t2 uniformly for all n ∈ N. Similarly, since the function p is continuous on compact set J, it is uniformly continuous and hence
|p(t1 ) − p(t2 )| → 0 as t1 → t2 . Therefore, from the above inequality (5.7), it follows that
|r (t1 , n ) − r (t1 , n )| → 0 as t1 → t2 uniformly for all n ∈ N. Therefore, r (t , n ) → r (t ) as
n→∞
for all t ∈ J. Next, we show that the function r (t ) is a solution of the HDE (2.1) defined on J. Now, since r (t , n ) is a solution of the HDE (5.5), we have r (t , n ) = f (t , r (t , n ))
x0 + n
Z
f (t0 , x0 + n )
t
g (s, r (s, n ))ds +
+
Z
t
n ds
(5.8)
t0
t0
for all t ∈ J. Taking the limit as n → ∞ in the above Eq. (5.8) yields r (t ) = f (t , r (t ))
x0 f (t0 , x0 )
t
Z
g (s, r (s))ds
+
t0
for t ∈ J. Thus, the function r is a solution of the HDE (2.1) on J. Finally, from the inequality (5.4) it follows that u( t ) ≤ r ( t ) for all t ∈ J. Hence the HDE (2.1) has a maximal solution on J. This completes the proof.
6. Comparison theorems The main problem of the differential inequalities is to estimate a bound for the solution set for the differential inequality related to the HDE (2.1). In this section we prove that the maximal and minimal solutions serve as bounds for the solutions of the related differential inequality to HDE (2.1) on J = [t0 , t0 + a]. Theorem 6.1. Assume that hypotheses (A0 )–(A2 ) hold. Suppose that condition (3.1) holds. Further, if there exists a function u ∈ AC (J , R) such that d
u( t )
dt f (t , u(t )) u(t0 ) ≤ x0 ,
≤ g (t , u(t )) a.e. t ∈ J
(6.1)
Then, u( t ) ≤ r ( t )
(6.2)
for all t ∈ J, where r is a maximal solution of the HDE (2.1) on J. Proof. Let > 0 be arbitrary small. Then, by Theorem 5.2, r (t , ) is a maximal solution of the HDE (5.1) and that the limit r (t ) = lim r (t , )
(6.3)
→0
is uniform on J and the function r is a maximal solution of the HDE (2.1) on J. Hence, we obtain d
r (t , )
dt f (t , r (t , )) r (t0 , ) = x0 + .
= g (t , r (t , )) +
a.e. t ∈ J
(6.4)
From the above inequality it follows that d
r (t , )
dt f (t , r (t , )) r (t0 , ) > x0 .
> g (t , r (t , ))
a.e. t ∈ J
(6.5)
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Now we apply Theorem 4.1 to the inequalities (6.1) and (6.5) and conclude that u(t ) < r (t , )
(6.6)
for all t ∈ J. This further in view of limit (6.3) implies that inequality (6.2) holds on J. This completes the proof.
Theorem 6.2. Assume that hypotheses (A0 )–(A2 ) hold. Suppose that condition (3.1) holds. Further, if there exists a function v ∈ AC (J , R) such that
v(t ) ≥ g (t , v(t )) dt f (t , v(t )) v(t0 ) ≥ x0 , d
a.e. t ∈ J
(6.7)
Then,
ρ(t ) ≤ v(t )
(6.8)
for all t ∈ J, where ρ is a minimal solution of the HDE (2.1) on J. Note that Theorem 6.1 is useful to prove the boundedness and uniqueness of the solutions for the HDE (2.1) on J. A result in this direction is Theorem 6.3. Assume that hypotheses (A0 )–(A2 ) hold. Suppose that there exists a function G : J × R+ → R+ such that
|g (t , x1 ) − g (t , x2 )| ≤ G t ,
− f (t , x1 ) f ( t , x2 ) x1
x2
a.e. t ∈ J
(6.9)
for all x1 , x2 ∈ R+ . If identically zero function is the only solution of the differential equation m0 (t ) = G(t , m(t ))
a.e. t ∈ J ,
m(t0 ) = 0,
(6.10)
then the HDIE (2.1) has a unique solution on J. Proof. By Theorem 3.1, the HDE (2.1) has a solution defined on J. Suppose that there are two solutions u1 and u2 of the HDE (2.1) existing on J. Define a function m : J → R+ by
u1 ( t ) u2 (t ) . m(t ) = − f (t , u1 (t )) f (t , u2 (t ))
(6.11)
As (|x(t )|)0 ≤ |x0 (t )| for t ∈ J, we have that
d m0 (t ) ≤ dt
u1 (t )
−
d
u2 (t )
f (t , u2 (t ))
f (t , u1 (t )) dt ≤ |g (t , u1 (t )) − g (t , u2 (t ))| u1 (t ) u2 ( t ) ≤ G t, − f (t , u1 (t )) f (t , u2 (t )) = G(t , m(t ))
for almost everywhere t ∈ J; and that m(t0 ) = 0. Now, we apply Theorem 6.1 with f (t , x) ≡ 1 to get that m(t ) = 0 for all t ∈ J. This gives u1 ( t ) f (t , u1 (t ))
=
u2 ( t ) f (t , u2 (t ))
for all t ∈ J. Finally, in view of hypothesis (A0 ) we conclude that u1 (t ) = u2 (t ) on J. This completes the proof.
7. Existence of extremal solutions in vector segment Sometimes it is desirable to have knowledge of the existence of extremal positive solutions for the HDE (2.1) on J. In this section we shall prove the existence of maximal and minimal positive solutions for the HDE (2.1) between the given upper and lower solutions on J = [t0 , t0 + a]. We use a hybrid fixed point theorem of Dhage [3] in ordered Banach spaces for establishing our results. We need the following preliminaries in what follows. A non-empty closed set K in a Banach algebra X is called a cone with vertex 0, if (i) K +K ⊆ K , (ii) λK ⊆ K for λ ∈ R, λ ≥ 0 and (iii) {−K } ∩ K = 0, where 0 is the zero element of X . A cone K is called to be positive if (iv) K ◦ K ⊆ K , where ‘‘◦’’ is a multiplication composition in X . We introduce an order relation ≤ in X as follows. Let x, y ∈ X . Then x ≤ y if and only if y − x ∈ K . A cone K is called to be normal if the norm k·k is semi-monotone increasing on K , that is, there is a constant N > 0 such that kxk ≤ N kyk for all x, y ∈ K with x ≤ y. It is known that if the cone K is normal in X , then every order-bounded set in X is norm-bounded. The details of cones and their properties appear in Heikkilä and Lakshmikantham [6].
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Lemma 7.1 (Dhage [3]). Let K be a positive cone in a real Banach algebra X and let u1 , u2 , v1 , v2 ∈ K be such that u1 ≤ v1 and u2 ≤ v2 . Then u1 u2 ≤ v1 v2 . For any a, b ∈ X , a ≤ b, the order interval [a, b] is a set in X given by
[a, b] = {x ∈ X : a ≤ x ≤ b}. Definition 7.1. A mapping Q : [a, b] → X is said to be nondecreasing or monotone increasing if x ≤ y implies Qx ≤ Qy for all x, y ∈ [a, b]. We use the following fixed point theorems of Dhage [4] for proving the existence of extremal solutions for the BVP (2.1) under certain monotonicity conditions. Theorem 7.1 (Dhage [4]). Let K be a cone in a Banach algebra X and let a, b ∈ X be such that a ≤ b. Suppose that A, B : [a, b] → K are two nondecreasing operators such that (a) A is a Lipschitz with the Lipschitz constant α , (b) B is completely continuous, and (c) Ax Bx ∈ [a, b] for each x ∈ [a, b]. Further, if the cone K is positive and normal, then the operator equation Ax Bx = x has a least and a greatest positive solution in [a, b], whenever α M < 1, where M = kB([a, b])k := sup{kBxk : x ∈ [a, b]}. We equip the space C (J , R) with the order relation ≤ with the help of cone K defined by K = {x ∈ C (J , R) : x(t ) ≥ 0, ∀ t ∈ J }.
(7.1)
It is well known that the cone K is positive and normal in C (J , R). We need the following definitions in what follows. Definition 7.2. A function a ∈ AC (J , R) is called a lower solution of the HDE (2.1) defined on J if it satisfies (4.1). Similarly, a function b ∈ AC (J , R) is called an upper solution of the HDE (2.1) defined on J if it satisfies (4.2). A solution to the HDE (2.1) is a lower as well as an upper solution for the HDE (2.1) defined on J and vice versa. We consider the following set of assumptions:
(B0 ) (B1 ) (B2 ) (B3 ) (B4 )
f : J × R → R+ − {0}, g : J × R → R+ . The HDE (2.1) has a lower solution a and an upper solution b defined on J with a ≤ b. The function x 7→ f (tx,x) is increasing in the interval mint ∈J a(t ), maxt ∈J b(t ) almost everywhere for t ∈ J. The functions f (t , x) and g (t , x) are nondecreasing in x almost everywhere for t ∈ J . There exists a function k ∈ L1 (J , R+ ) such that g (t , b(t )) ≤ k(t ).
We remark that hypothesis (B4 ) holds in particular if f is continuous and g is L1 -Carathéodory on J × R. Theorem 7.2. Suppose that assumptions (A1 ) and (B0 ) through (B4 ) hold. Further if
x0 L f (t , x 0
0
+ kkkL1 < 1, )
(7.2)
then the HDE (2.1) has a minimal and a maximal positive solution defined on J. Proof. Now, the HDE (2.1) is equivalent to integral equation (3.6) defined on J. Let X = C (J , R). Define two operators A and B on X by (3.7) and (3.8) respectively. Then the integral equation (3.6) is transformed into an operator equation Ax(t ) Bx(t ) = x(t ) in the Banach algebra X . Notice that hypothesis (B0 ) implies A, B : [a, b] → K . Since the cone K in X is normal, [a, b] is a norm-bounded set in X . Now it is shown, as in the proof of Theorem 3.2, that A is a Lipschitz with the Lipschitz constant L and B is completely continuous operator on [a, b]. Again, hypothesis (B3 ) implies that A and B are nondecreasing on [a, b]. To see this, let x, y ∈ [a, b] be such that x ≤ y. Then, by hypothesis (B3 ), Ax(t ) = f (t , x(t )) ≤ f (t , y(t )) = Ay(t ) for all t ∈ J. Similarly, we have Bx(t ) =
f (t0 , x0 )
≤
x0 x0
f (t0 , x0 ) = By(t )
Z
t
g (s, x(s))ds
+
t0
Z
t
g (s, x(s))ds
+ t0
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B.C. Dhage, V. Lakshmikantham / Nonlinear Analysis: Hybrid Systems 4 (2010) 414–424
for all t ∈ J. So A and B are nondecreasing operators on [a, b]. Again, Lemma 7.1 and hypothesis (B3 ) together imply that a(t ) ≤ [f (t , a(t ))]
≤ [f (t , x(t ))] ≤ [f (t , b(t ))]
x0 f (t0 , x0 )
x0 f (t0 , x0 )
x0 f (t0 , x0 )
t
Z
g (s, x(s))ds
+
t0 t
Z
g (s, x(s))ds
+
t0
Z
t
g (s, x(s))ds
+
t0
≤ b(t ), for all t ∈ J and x ∈ [a, b]. As a result a(t ) ≤ Ax(t )Bx(t ) ≤ b(t ), for all t ∈ J and x ∈ [a, b]. Hence, Ax Bx ∈ [a, b] for all x ∈ [a, b]. Again,
x0 + kkkL1 M = kB([a, b])k = sup{kBxk : x ∈ [a, b]} ≤ f (t0 , x0 ) and so,
x0 α M ≤ L f (t , x 0
0
+ kkkL1 < 1. )
Now, we apply Theorem 7.1 to the operator equation Ax Bx = x to yield that the HDE (2.1) has a minimal and a maximal positive solution in [a, b] defined on J. This completes the proof. References [1] B.C. Dhage, V. Lakshmikantham, Quadratic perturbations of periodic boundary value problems of second order ordinary differential equations (submitted for publication). [2] B.C. Dhage, On α -condensing mappings in Banach algebras, Math. Student 63 (1994) 146–152. [3] B.C. Dhage, A nonlinear alternative in Banach algebras with applications to functional differential equations, Nonlinear Funct. Anal. Appl. 8 (2004) 563–575. [4] B.C. Dhage, Fixed point theorems in ordered Banach algebras and applications, Panamer. Math. J. 9 (4) (1999) 93–102. [5] B.C. Dhage, A fixed point theorem in Banach algebras with applications to functional integral equations, Kyungpook Math. J. 44 (2004) 145–155. [6] S. Heikkilä, V. Lakshmikantham, Monotone Iterative Technique for Nonlinear Discontinues Differential Equations, Marcel Dekker Inc, New York, 1994. [7] V. Lakshmikantham, S. Leela, Differential and Integral Inequalities, Academic Press, New York, 1969.