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Some new nonlinear Volterra–Fredholm-type discrete inequalities and their applications
Journal of Computational and Applied Mathematics 216 (2008) 451 – 466 www.elsevier.com/locate/cam
Some new nonlinear Volterra–Fredholm-type discrete inequalities and their applications Qing-Hua Ma Faculty of Information Science and Technology, Guangdong University of Foreign Studies, Guangzhou 510420, PR China Received 22 February 2006; received in revised form 26 March 2007
Abstract Some new explicit bounds on solutions to a class of new nonlinear Volterra–Fredholm-type discrete inequalities are established, which can be used as effective tools in the study of certain sum–difference equations. Application examples are also indicated. © 2007 Elsevier B.V. All rights reserved. MSC: 26D15; 26D20; 39A10 Keywords: Nonlinear discrete inequalities; Volterra–Fredholm-type; Gronwall–Bellman–Ou-Iang-type; Explicit bounds; Sum–difference equations
1. Introduction In the study of ordinary differential equations, integral equations and difference equations, one often deal with certain integral inequalities. The Gronwall–Bellman inequality [9,2] and its various linear and nonlinear generalizations are crucial in the discussion of the existence, uniqueness, continuation, boundedness, oscillation and stability and other qualitative properties of solutions of differential and integral equations. The literature on such inequalities and their applications is vast; see [1,19,22] and the references therein. A specific branch of this type integral inequalities is originated by Ou-Iang. In his study of boundedness of solutions to linear second order differential equations, Ou-Iang [20] established and used the following nonlinear integral inequality, which is now known as Ou-Iang’s inequality in the literature. Theorem A (Ou-Iang [20]). Let u and f be real-valued, nonnegative, and continuous functions defined on R+ = [0, +∞) and let c 0 be a real constant. Then the nonlinear integral inequality t u2 (t)c2 + 2 f (s)u(s) ds, t ∈ R+ 0
implies
t
u(t)c + 0
f (s) ds,
t ∈ R+ .
E-mail address: [email protected]. 0377-0427/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.cam.2007.05.021
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While Ou-Iang’s inequality is having a neat form and is interesting in its own right as an integral inequality, its importance lies equality heavily on its many beautiful applications in differential and integral equations (see, e.g., [22]). Since this, over the years, many generalizations of Ou-Iang’s inequality to various situations have been established; see, for example, [3–7], [11–18], [21–25] and the references cited therein. Among various generalizations of Ou-Iang’s inequality, discrete analogue is also an interesting direction. The point is, similar to the noteworthy contributions of the continuous version of the inequality to the study of differential equations, one naturally expects that discrete versions of the inequality should also play an important role in the study of difference equations. In this respect, fewer results have been established. Recent results in this direction include the works of Pachpatte [21], Pang–Agarwal [23], Ma [13], Meng–Li [17], Cheung [3], Cheung–Ren [7], and Ma–Cheung [14]. The aim of the present paper is to give some explicit bounds to some new nonlinear discrete inequalities involving two-variable functions which, on the one hand, generalize Ou-Iang’s inequality to Volterra–Fredholm form at the first time to literatures as we know and, on the other hand, give a handy and effective tool for the study of quantitative properties of solutions of sum–difference equations. We illustrate the usefulness of these inequalities by applying them to study the boundedness, uniqueness, and continuous dependence of the solutions of certain Volterra–Fredholm-type sum–difference equations.
2. Nonlinear discrete inequalities Throughout this paper, I := [m0 , M) ∩ Z and J := [n0 , N ) ∩ Z are two fixed lattices of integral points in R, where m0 , n0 ∈ Z, M, N ∈ Z ∪ {∞}. Let := I × J ⊂ Z 2 , R+ := [0, ∞), R1 := [1, ∞) and for any (s, t) ∈ , the sub-lattice [m0 , s] × [n0 , t] ∩ of will be denoted as (s,t) . If U is a lattice in Z (resp.Z 2 ), the collection of all R-valued functions on U is denoted by F(U ), that of all R+ valued functions by F+ (U ), and that of all R1 -valued functions by F1 (U ). For the sake of convenience, we extend the domain of definition of each function in F(U ) and F+ (U ) trivially to the ambient space Z (resp.Z 2 ). So, for example, a function in F(U ) is regarded as a function defined on Z (resp.Z 2 ) with support in U. As usual, the collection of all continuous functions and all i-times continuously differentiable functions of a topological space X into a topological space Y will be denoted by C(X, Y ) and C i (X, Y ), respectively. If U is a lattice in Z 2 , the partial difference operators 1 and 2 on u ∈ F(Z 2 ) or F+ (Z 2 ) are defined as 1 u(m, n) = u(m + 1, n) − u(m, n), (m, n) ∈ U , 2 u(m, n) = u(m, n + 1) − u(m, n), (m, n) ∈ U . For w ∈ C(R+ , R+ ), the function G1 is defined as G1 (v) =
v
ds , w(s)
v0
v v0 > 0.
Theorem 2.1. Suppose that u and a ∈ F+ (), k 0 constant and w ∈ C(R+ , R+) is nondecreasing with w(r) > 0 for r > 0; G1 (∞) =
∞ v0
ds =∞ w(s)
and H1 (t) = G1 (2t − k) − G1 (t) is strictly increasing for t k.
(2.1)
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If u(m, n) satisfies m−1 n−1
u(m, n)k +
a(s, t)w(u(s, t)) +
s=m0 t=n0
M−1 N−1
a(s, t)w(u(s, t))
(2.2)
s=m0 t=n0
for (m, n) ∈ , then
u(m, n)G−1 G1 H1−1 1
M−1 N−1
a(s, t)
+
s=m0 t=n0
m−1 n−1
a(s, t)
(2.3)
s=m0 t=n0
−1 for (m, n) ∈ , where G−1 1 and H1 are inverse functions of G1 and H1 , respectively.
Proof. Let k > 0 and define z(m, n) = k +
m−1 n−1
a(s, t)w(u(s, t)) +
M−1 N−1
a(s, t)w(u(s, t))
s=m0 t=n0
s=m0 t=n0
then u(m, n)z(m, n) z(m0 , n) = k +
(m, n) ∈ ,
M−1 N−1
(2.4)
a(s, t)w(u(s, t))
s=m0 t=n0
and 1 z(m, n) =
n−1
a(m, t)w(u(m, t))
t=n0
n−1
a(m, t)w(z(m, t))
t=n0
w(z(m, n))
n−1
a(m, t).
t=n0
Therefore, by the Mean-Value Theorem for integrals, for each (m, n) ∈ , there exists : z(m, n) z(m + 1, n) such that 1 G1 (z(m, n)) = G1 (z(m + 1, n)) − G1 (z(m, n)) z(m+1,n) ds = w(s) z(m,n) =
1 1 z(m, n). w()
Since w is nondecreasing, w()w(z(m, n)); hence 1 G1 (z(m, n))
1 1 z(m, n) w(z(m, n)) n−1 t=n0
a(m, t)
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for all (m, n) ∈ . Therefore, m−1
1 G1 (z(s, n))
s=m0
m−1 n−1
a(s, t)
s=m0 t=n0
On the other hand, it is easy to check that m−1
1 G1 (z(s, n)) = G1 (z(m, n)) − G1 (z(m0 , n)),
s=m0
thus m−1 n−1
G1 (z(m, n))G1 (z(m0 , n)) +
a(s, t)
s=m0 t=n0
Since G−1 1 is increasing, the last inequality yields m−1 n−1 −1 z(m, n)G1 G1 (z(m0 , n)) + a(s, t)
(2.5)
s=m0 t=n0
for all (m, n) ∈ . From the last inequality, we observe that 2z(m0 , n) − k = k + 2
M−1 N−1
a(s, t)w(u(s, t))
s=m0 t=n0
= z(M, N) G−1 1
G1 (z(m0 , N )) +
a(s, t)
s=m0 t=n0
= G−1 1
M−1 N−1
G1 (z(m0 , n)) +
M−1 N−1
a(s, t)
s=m0 t=n0
or G1 (2z(m0 , n) − k) − G1 (z(m0 , n))
M−1 N−1
a(s, t).
(2.6)
s=m0 t=n0
Since H1 (t) = G1 (2t − k) − G1 (t) is increasing for t > k, H1 (t) has inverse function H1−1 (t) and then from the last inequality we get M−1 N−1 −1 a(s, t) . (2.7) z(m0 , n)H1 s=m0 t=n0
Substituting (2.7) into (2.5) and combining with (2.4) we obtain the desired inequality (2.3). If k = 0, we carry out the above procedure with ε > 0 instead of k and subsequently let ε → 0. Theorem 2.1 . Let u(m, n), a(m, n), w(u), G1 (u) and k be as in Theorem 2.1. If u(m, n) satisfies (2.2) for (m, n) ∈ , and
1 (t) = G1 (2t − k) − G1 (t) − H
M−1 N−1 s=m0 t=n0
a(s, t)
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1 (t) = 0 has a solution c1 for t k, then is increasing and H m−1 n−1 −1 a(s, t) u(m, n)G1 G1 (c1 ) +
(2.8)
s=m0 n=n0
for (m, n) ∈ , where G1 and G−1 1 are defined as in Theorem 2.1. Proof. By the same steps from (2.4) to (2.7) in the proofs of Theorem 2.1, we have u(m, n)z(m, n), z(m, n)G−1 1
(2.9)
G1 (z(m0 , n)) +
m−1 n−1
a(s, t)
(2.10)
s=m0 t=n0
and M−1 N−1
G1 (2z(m0 , n) − k) − G1 (z(m0 , n))
a(s, t)
(2.11)
s=m0 t=n0
for (m, n) ∈ . From the assumption of Theorem 2.1 and (2.11), we have
1 (z(m0 , n))0 = H
1 (c1 ). H
1 is increasing, H
1 has an inverse function H
−1 ; hence, from the last inequality we get Since H 1 z(m0 , n)c1 . Substituting the last inequality into (2.10) and combining with (2.9) we get the desired inequality (2.8).
Corollary 2.2. Let u(m, n), a(m, n) and k be as in Theorem 2.1. If u(m, n) satisfies u(m, n)k +
m−1 n−1
a(s, t)u(s, t) +
s=m0 t=n0
M−1 N−1
a(s, t)u(s, t)
(2.12)
s=m0 t=n0
for (m, n) ∈ , and (M, N ) = exp
M−1 N−1
a(s, t) < 2,
(2.13)
s=m0 t=n0
then u(m, n)
m−1 n−1 k exp a(s, t) 2 − (M, N ) s=m t=n 0
(2.14)
0
for (m, n) ∈ . Proof. In Theorem 2.1, by letting w(u) = u we obtain v v ds ds v , v v0 > 0, G1 (v) = = = ln v0 v0 w(s) v0 s H1 (t) = G1 (2t − k) − G1 (t) = ln
2t − k , t
t k,
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Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
and hence G−1 1 (v) = v0 exp v, H1−1 (t) =
k . 2 − exp t
From inequality (2.3), we obtain inequality (2.14).
Corollary 2.3. Let u(m, n), a(m, n) and k be as in Theorem 2.1, 0 < p < 1 be a constant. If u(m, n) satisfies u(m, n)k +
m−1 n−1
p
a(s, t)u (s, t) +
s=m0 t=n0
M−1 N−1
a(s, t)up (s, t)
(2.15)
s=m0 t=n0
for (m, n) ∈ , then u(m, n) (c11 )1−p + (1 − p)
m−1 n−1
1/(1−p) a(s, t)
(2.16)
s=m0 t=n0
for (m, n) ∈ , where c11 is the solution of equation
1 (t) = H
M−1 N−1 1 (2t − k)1−p − t 1−p − a(s, t) = 0 1−p s=m t=n 0
(2.17)
0
for t k. Proof. By Theorem 2.1 , we only need to prove that (2.17) has a solution c11 for t k. In fact, p 1/p p
(t) = (2 t) − (2t − k) > 0 H 1 [t (2t − k)]p
for t k,
1 (k) = − H
M−1 N−1
a(s, t) < 0
s=m0 t=n0
and 1−p
1 (t) = lim t lim H t→∞ t→∞ 1 − p
k 2− t
1−p
1 (t) = 0 has a unique solution c11 > k. so H
−1 −
M−1 N−1
a(s, t) = +∞,
s=m0 t=n0
Remark 2.1. Though (2.16) does not give an exact estimation to the solution of (2.15), it is enough to get the upper bound to the solution of (2.15) in many cases. Theorem 2.4. Suppose that u(m, n), a(m, n), w(u) and k are as in Theorem 2.1. Let (u) ∈ C 1 (R+ , R+ ) with (u) > 0 and (u) is increasing for u > 0, here (u) denotes the derivative of . If u(m, n) satisfies (u(m, n))k +
m−1 n−1
a(s, t) (u(s, t))w(u(s, t))
s=m0 t=n0
+
M−1 N−1 s=m0 t=n0
a(s, t) (u(s, t))w(u(s, t)),
(2.18)
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457
for (m, n) ∈ , and H2 (t) = G1 ◦ −1 (2t − k) − G1 ◦ −1 (t) is increasing for t k, then u(m, n)G−1 1
H2−1
G1
M−1 N−1
a(s, t)
+
s=m0 t=n0
m−1 n−1
a(s, t)
(2.19)
s=m0 t=n0
for (m, n) ∈ , where H2−1 is the inverse functions of H2 , G1 and G−1 1 are defined as in Theorem 2.1. Proof. Similar to the proof of Theorem 2.1, it suffices to consider the case k > 0. Denote by z2 (m, n) the right-hand side of (2.18). Then z2 > 0, u −1 (z2 ), and z2 is nondecreasing in each variable. Hence, for any (m, n) ∈ , 1 z2 (m, n) =
n−1
a(m, t) (u(m, t))w(u(m, t))
t=n0
n−1
a(m, t) (−1 (z2 (m, t)))w(−1 (z2 (m, t)))
t=n0
(−1 (z2 (m, n)))
n−1
a(m, t)w(−1 (z2 (m, t)))
t=n0
or n−1 1 z2 (m, n) a(m, t)w(−1 (z2 (m, t))). (−1 (z2 (m, n))) t=n 0
On the other hand, using the differential Mean-Value Theorem and the last inequality we have 1 [−1 (z2 (m, n))] = −1 (z2 (m + 1, n)) − −1 (z2 (m, n)) =
1 1 z2 (m, n) 1 z2 (m, n) −1 (−1 ()) ( (z2 (m, n))) n−1
a(m, t)w(−1 (z2 (m, t))).
(2.20)
t=n0
Keeping n fixed in (2.20) and setting m = s and then summing over s = m0 , m0 + 1, . . . , m − 1, we get −1
−1
(z2 (m, n))
(z2 (m0 , n)) +
m−1 n−1
a(s, t)w(−1 (z2 (m, t)))
(2.21)
s=m0 t=n0
for all (m, n) ∈ . Now by applying Theorem 2.1 in [3] to the function −1 (z2 (m, n)), we have −1
(z2 (m, n))G−1 1
G1
−1
(z2 (m0 , n)) +
m−1 n−1 s=m0 t=n0
a(s, t) .
(2.22)
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Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
Observing that 2z2 (m0 , n) − k = k + 2
M−1 N−1
a(s, t) (u(s, t))w(u(s, t)) = z2 (M, N ),
s=m0 t=n0
and then from (2.22) we have G1 ◦ −1 (2z2 (m0 , n) − k) = G1 ◦ −1 (z2 (M, N )) G1 ◦ −1 (z2 (m0 , N )) +
M−1 N−1
a(s, t)
s=m0 t=n0
= G1 ◦
−1
(z2 (m0 , n)) +
M−1 N−1
a(s, t)
s=m0 t=n0
or G1 ◦ −1 (2z2 (m0 , n) − k) − G1 ◦ −1 (z2 (m0 , n))
M−1 N−1
a(s, t).
(2.23)
s=m0 t=n0
Since H2 (t) = G1 ◦ −1 (2t − k) − G1 ◦ −1 (t) is increasing for t k, H2 (t) has an inverse function H2−1 and then from (2.23) we get z2 (m0 , n)H2−1
M−1 N−1
a(s, t) .
s=m0 t=n0
Substituting the last inequality into (2.22) and by the definition of z2 (m, n) we obtain the desired inequality (2.19).
By similar argument as in the proofs of Theorem 2.1 , we have the following result immediately. Theorem 2.4 . Let u(m, n), a(m, n) and k be as in Theorem 2.4. If u(m, n) satisfies (2.18) for (m, n) ∈ , and
2 (t) = G1 ◦ −1 (2t − k) − G1 ◦ −1 (t) − H
M−1 N−1
a(s, t)
s=m0 t=n0
2 (t) = 0 has a solution c2 k, then is increasing and H u(m, n)G−1 1
G1 (c2 ) +
m−1 n−1
a(s, t)
(2.24)
s=m0 t=n0
for (m, n) ∈ . When = up (p 1 is a constant) in Theorem 2.4, we have the following: Corollary 2.5. Let u(m, n), a(m, n) and k be as in Theorem 2.4, p 1 is a constant. If u(t) satisfies p
u (m, n)k +
m−1 n−1 s=m0 t=n0
a(s, t)u
p−1
(s, t)w(u(s, t)) +
M−1 N−1 s=m0 t=n0
a(s, t)up−1 (s, t)w(u(s, t)),
(2.25)
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459
for (m, n) ∈ , and H21 (t) = G1 ((2t − k)1/p ) − G1 (t 1/p ) is increasing for (m, n) ∈ , then
−1 G1 H21 u(m, n)G−1 1
M−1 N−1
a(s, t)
+
s=m0 t=n0
m−1 n−1
a(s, t)
(2.26)
s=m0 t=n0
for (m, n) ∈ . Corollary 2.6. Let u(m, n) ∈ F1 (U ), a(m, n) and k be as in Theorem 2.4. If u(t) satisfies up (m, n)k +
m−1 n−1
a(s, t)up (s, t)w(ln u(s, t))
s=m0 t=n0
+
M−1 N−1
a(s, t)up (s, t)w(ln u(s, t)),
(2.27)
s=m0 t=n0
for (m, n) ∈ , and
1 1 H22 (t) = G1 ln(2t − k) − G1 ln t p p is increasing for (m, n) ∈ , then
−1 G1 H22 u(m, n)G−1 1
M−1 N−1 s=m0 t=n0
a(s, t)
+
m−1 n−1
a(s, t)
(2.28)
s=m0 t=n0
for (m, n) ∈ . Proof. Taking v(m, n) = ln u(m, n), then (2.27) reduces to exp(pv(m, n))k +
m−1 n−1
a(s, t) exp(pv(s, t))w(v(s, t))
s=m0 t=n0
+
M−1 N−1
a(s, t) exp(pv(m, n))w(v(s, t)),
(2.29)
s=m0 t=n0
for (m, n) ∈ , which is a special case of inequality (2.18) when (v) = exp(pv). In this special case,
1 1 H2 (t) = H22 (t) = G1 ln(2t − k) − G1 ln t . p p By Theorem 2.4, we get the desired inequality (2.28) directly.
Remark 2.2. Eqs. (2.25) and (2.27) are new interesting discrete Volterra–Fredholm–Ou-Iang-type and Volterra– Fredholm–Engler [8]–Haraux [10] type inequality of two-variable, respectively. Using Theorems 2.1 and 2.4, we can get more generalized results as follows.
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Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
−1 Theorem 2.7. Let u(m, n), a(m, n), w(t), G1 , G−1 1 , H1 , H1 and k be as in Theorem 2.1, b(m, n) ∈ F+ (). If u(m, n) satisfies m−1 n−1
u(m, n)k +
a(s, t)w(u(s, t)) +
s=m0 t=n0
for (m, n) ∈ , then
u(m, n)G−1 1
b(s, t)w(u(s, t))
(2.30)
s=m0 t=n0
G1
M−1 N−1
H1−1
M−1 N−1
∗
a (s, t)
+
s=m0 t=n0
m−1 n−1
∗
a (s, t)
(2.31)
s=m0 t=n0
for (m, n) ∈ , where a ∗ (m, n) ∈ F+ () such that both a(m, n) and b(m, n) are less than or equal to a ∗ (m, n). Proof. From (2.30) and the assumptions, we have m−1 n−1
u(m, n) k +
a ∗ (s, t)w(u(s, t)) +
s=m0 t=n0
M−1 N−1
a ∗ (s, t)w(u(s, t))
s=m0 t=n0
for (m, n) ∈ . An application of Theorem 2.1 to the last inequality yields (2.31).
Theorem 2.8. Let u(m, n), a(m, n), b(m, n), a ∗ (m, n) and k be as in Theorem 2.7; (u) be as in Theorem 2.4. wi ∈ C(R+ , R+ ) be nondecreasing functions with wi > 0 for u > 0, i = 1, 2. If u(m, n) satisfies (u(m, n))k +
n−1 m−1
a(s, t) (u(s, t))w1 (u(s, t))
s=m0 t=n0
+
M−1 N−1
b(s, t) (u(s, t))w2 (u(s, t))
(2.32)
s=m0 t=n0
for (m, n) ∈ , there is a function W (u) ∈ C(R+ , R+ ) that is nondecreasing such that both w1 and w2 are less than or equal to W, v +∞ ds ds G2 (v) = , v v0 > 0, G2 (+∞) = = +∞, W (s) v0 W (s) v0 and H3 (t) = G2 ◦ −1 (2t − k) − G2 ◦ −1 (t) is increasing for t k, then u(m, n)G−1 2
G2
H3−1
M−1 N−1
∗
a (s, t)
s=m0 t=n0
+
m−1 n−1
∗
a (s, t)
s=m0 t=n0
−1 for (m, n) ∈ , where G−1 2 and H3 are inverse functions of G2 and H3 , respectively.
Proof. From (2.32) and the assumptions, we can obtain (u(m, n))k +
m−1 n−1
a ∗ (s, t) (u(s, t))W (u(s, t))
s=m0 t=n0
+
M−1 N−1 s=m0 t=n0
a ∗ (s, t) (u(s, t))W (u(s, t)).
(2.33)
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Now, an application of Theorem 2.4 to the last inequality yields the desired inequality (2.33).
461
By the same argument as in the proofs of Theorem 2.1 , we have the following result immediately. Theorem 2.8 . Let u(m, n), a(m, n), b(m, n), a ∗ (m, n), wi (i =1, 2), W and k be as in Theorem 2.8. If u(m, n) satisfies (2.32), M−1 N−1
3 (t) = G2 ◦ −1 (2t − k) − G2 ◦ −1 (t) − H
a ∗ (s, t)
s=m0 t=n0
3 (t) = 0 has a solution c3 k, then is increasing and H u(m, n)G−1 2
G2 (c3 ) +
m−1 n−1
∗
a (s, t)
(2.34)
s=m0 t=n0
for (m, n) ∈ . Remark 2.3. In Theorems 2.7, 2.8 and 2.8 , we can choose function a ∗ (m, n)=a(m, n)+b(m, n) or max{a(m, n), b(m, n)} as well as in function W. By Theorem 2.8 , we can get the following interesting result. Corollary 2.9. Let u(m, n), a(m, n), b(m, n), a ∗ (m, n) and k be as in Theorem 2.8, p 1 and 0 < q < 1 be constants. If u(m, n) satisfies up (m, n)k +
m−1 n−1
a(s, t)up (s, t) +
s=m0 t=n0
M−1 N−1
b(s, t)up+q−1 (s, t)
(2.35)
s=m0 t=n0
for (m, n) ∈ , and exp
M−1 N−1
a ∗ (s, t) < 21/p ,
s=m0 t=n0
then
1/(1−q) m−1 n−1 1−q ∗ u(m, n) 1 + c˜3 exp (1 − q) a (s, t) − 1
(2.36)
s=m0 t=n0
for (m, n) ∈ , where c˜3 is the solution of
3 (t) = H
M−1 N−1 1 + (2t − k)(1−q)/p 1 ln − a ∗ (s, t) = 0 (1−q)/p 1−q 1+t s=m t=n 0
(2.37)
0
for t k. Proof. In Theorem 2.8 , by letting w1 (u) = u, w2 (u) = uq and W = w1 + w2 we obtain G2 (v) =
v v0
ds = w1 (s) + w2 (s)
v v0
ds 1 1 + v 1−q = , ln s + sq 1 − q 1 + v 1−q 0
v v0 > 0.
(2.38)
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Hence, G−1 2 (v) =
1−q
1 + v0
1/(1−q)
exp((1 − q)v) − 1
.
(2.39)
By computation, we have
3 (t) = H
M−1 N−1 1 + (2t − k)(1−q)/p 1 − a ∗ (s, t), ln (1−q)/p 1−q 1+t s=m t=n 0
(t) = H 3
k
+ 2t 1−(1−q)/p
[2t − k + (2t − k)
− (2t − k)
1−(1−q)/p
0
1−(1−q)/p
](t + t 1−(1−q)/p )
>0
(2.40)
for t k,
3 (k) = − H
M−1 N−1
a ∗ (s, t) < 0
(2.41)
s=m0 t=n0
and
3 (t) = lim lim H
t→+∞
t→+∞
= ln 2
M−1 N−1 1 1 + (2t − k)(1−q)/p − a ∗ (s, t) ln (1−q)/p 1−q 1+t s=m t=n 0
1/p
−
M−1 N−1
0
a ∗ (s, t) > 0.
(2.42)
s=m0 t=n0
By (2.40)–(2.42) we obtain that (2.37) has a solution c˜3 > k. Now by (2.34), (2.38) and (2.39) we can get the desired (2.36). 3. Some applications In this section, we apply our results to study the boundedness, uniqueness, and continuous dependence of the solutions of certain Volterra–Fredholm sum–difference equations of the form up (m, n) = l(m, n) +
m−1 n−1
F (s, t, u(s, t)) +
s=m0 t=n0
M−1 N−1
G(s, t, u(s, t))
(3.1)
s=m0 t=n0
for (m, n) ∈ , where l ∈ F(), F, G ∈ F( × R), p 1 is a constant. Following theorem gives the bound on the solutions of Eq. (3.1). Theorem 3.1. Assume that the functions l, F and G in (3.1) satisfy the conditions |l(m, n)| k,
(3.2)
|F (m, n, v)|a(m, n)|v|p+q−1
(3.3)
|G(m, n, v)| b(m, n)|v|p+q−1 ,
(3.4)
and
where a(m, n) and b(m, n) are same as in Theorem 2.7, 0 < q < 1 is a constant, then all solutions of (3.1) satisfy u(m, n) (c11 )
1−q
+ (1 − q)
m−1 n−1 s=m0 t=n0
1/(1−q) ∗
a (s, t)
(3.5)
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463
for (m, n) ∈ , where c11 is the solution of equation
∗ (t) = H 2
M−1 N−1 1 a ∗ (s, t) = 0 (2t − k)(1−q)/p − t (1−q)/p − 1−q s=m t=n 0
(3.6)
0
for t k, where a ∗ (m, n) ∈ F+ () such that both a(m, n) and b(m, n) are less than or equal to a ∗ (m, n). Proof. Using the conditions (3.2)–(3.4) in (3.1) we have |u(m, n)|p k +
m−1 n−1
a ∗ (s, t)|u(s, t)|p+q−1 +
s=m0 t=n0
M−1 N−1
a ∗ (s, t)|u(s, t)|p+q−1
s=m0 t=n0
which is a special case of Theorem 2.4 when (u) = up and w(u) = uq . By Theorem 2.4 , we only need to prove
2 (t) = H
∗ (t) = G1 ◦ −1 (2t − k) − G1 ◦ −1 (t) − H 2
=
M−1 N−1
a ∗ (s, t)
s=m0 t=n0
M−1 N−1 1 a ∗ (s, t) (2t − k)(1−q)/p − t (1−q)/p − 1−q s=m t=n 0
0
∗ (t) = 0 has a solution c11 . In fact, taking r = (1 − q)/p, by computation we have is increasing and H 2 1 (21/(1−r) t)1−r − (2t − k)1−r d ∗ >0 H2 (t) = dt p [t (2t − k)]1−r for t k,
∗ (k) = − H 2
M−1 N−1
a ∗ (s, t) < 0
s=m0 t=n0
and
∗ (t) = lim lim H 2
t→+∞
t→+∞
tr 1−q
M−1 N−1 k r ∗ −1 − a (s, t) = +∞, 2− t s=m t=n
∗ (t) = 0 has a unique solution c11 > k. so H 2
0
0
Secondly, we consider the uniqueness of the solutions of (3.1). Theorem 3.2. Assume that the functions F and G in Eq. (3.1) satisfy the conditions |F (m, n, v) − F (m, n, v)|a(m, ¯ n)|v p − v¯ p |,
(3.7)
|G(m, n, v) − G(m, n, v)| ¯ b(m, n)|v p − v¯ p |,
(3.8)
for some a, b ∈ F+ (), and if (M, N ) = exp
M−1 N−1
a ∗ (s, t) < 2,
s=m0 t=n0
where a ∗ (m, n) is as in Theorem 3.1, then (3.1) has at most one positive solution on .
(3.9)
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Proof. Let u(m, n) and u(m, ¯ n) be two solutions of (3.1). By (3.1) and conditions (3.7) and (3.8), we have m−1 n−1
|up (m, n) − u¯ p (m, n)|
a ∗ (s, t)|up (s, t) − u¯ p (s, t)|
s=m0 t=n0
+
M−1 N−1
a ∗ (s, t)|up (s, t) − u¯ p (s, t)|.
s=m0 t=n0
An application of Corollary 2.2 to the function |up (m, n) − u¯ p (m, n)| yields that |up (m, n) − u¯ p (m, n)| 0 Hence, u(m, n) = u(m, ¯ n) on .
for all (m, n) ∈ .
Finally, we investigate the continuous dependence of the solutions of (3.1) on the functions F and G. For this, we consider the following variation of (3.1):
¯ n) + up (m, n) = l(m,
m−1 n−1
F¯ (s, t, u(s, t)) +
s=m0 t=n0
M−1 N−1
¯ t, u(s, t)) G(s,
(3.1)
s=m0 t=n0
¯ ∈ F( × R), p 1 is a constant as in (3.1). for (m, n) ∈ , where F¯ , G Theorem 3.3. Consider (3.1) and (3.1). If p p p p (i) |F (s, t, v1 ) − F (s, t, v2 )|a(s, t)|v1 − v2 | and |G(s, t, v1 ) − G(s, t, v2 )| b(s, t)|v1 − v2 |; ¯ (ii) |l(m, n) − l(m, n)|ε/2; N−1 ∗ (iii) (M, N ) = exp M−1 s=m0 t=n0 a (s, t) < 2; (iv) for all solutions u¯ of (3.1), m−1 n−1 s=m0 t=n0
ε |F (s, t, u) ¯ − F¯ (s, t, u)| ¯ 4
and M−1 N−1 s=m0 t=n0
ε ¯ t, u)| |G(s, t, u) ¯ − G(s, ¯ 4
for all (s, t) ∈ and v1 , v2 ∈ R, where ε > 0 is an arbitrary constant, a ∗ (m, n) is defined as in Theorem 3.1, then |up (s, t) − u¯ p (s, t)|
m−1 n−1 ε exp a ∗ (s, t) 2 − (M, N ) s=m t=n 0
(3.10)
0
for (m, n) ∈ . Hence, up depend continuously on F and G. In particular, if u does not change the sign, it depends continuously on F and G.
Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
465
Proof. Let u(m, n) and u(m, ¯ n) be solutions of (3.1) and (3.1), respectively. Then, u(m, n) satisfies (3.1) and u(m, ¯ n) satisfies (3.1). Hence, ¯ |up (m, n) − u¯ p (m, n)||l(m, n) − l(m, n)| +
m−1 n−1
|F (s, t, u(s, t)) − F¯ (s, t, u(s, ¯ t))|
s=m0 t=n0
+
M−1 N−1
¯ t, u(s, |G(s, t, u(s, t)) − G(s, ¯ t))|
s=m0 t=n0 m−1 n−1 ε + |F (s, t, u(s, t)) − F (s, t, u(s, ¯ t))| 2 s=m t=n 0
+
0
n−1 m−1
|F (s, t, u(s, ¯ t)) − F¯ (s, t, u(s, ¯ t))|
s=m0 t=n0
+
M−1 N−1
|G(s, t, u(s, t)) − G(s, t, u(s, ¯ t))|
s=m0 t=n0
+
M−1 N−1
¯ t, u(s, |G(s, t, u(s, ¯ t)) − G(s, ¯ t))|
s=m0 t=n0
ε +
m−1 n−1 s=m0 t=n0
ε +
m−1 n−1
M−1 N−1
a(s, t)|up (s, t) − u¯ p (s, t)| +
b(s, t)|up (s, t) − u¯ p (s, t)|
s=m0 t=n0
a ∗ (s, t)|up (s, t) − u¯ p (s, t)| +
s=m0 t=n0
M−1 N−1
a ∗ (s, t)|up (s, t) − u¯ p (s, t)|
s=m0 t=n0
by assumptions (i)–(iv) and the definition of a ∗ . Now, by applying Corollary 2.2 to the function |up (m, n) − u¯ p (m, n)|, we get |up (m, n) − u¯ p (m, n)|
m−1 n−1 ε a ∗ (s, t). exp 2 − (M, N ) s=m t=n 0
Evidently, if the function
m−1 n−1 s=m0
t=n0
0
a ∗ (s, t) is bounded on , then
|up (m, n) − u¯ p (m, n)|εK for some K > 0 and (m, n) ∈ . Hence, up depends continuously on F and G.
Acknowledgement The author is grateful to the referees for their very helpful and detailed comments in improving this paper. The research is partially supported by the NSF of Guangdong Province and the Research Group Grants Council of the Guangdong University of Foreign Studies (Project No. GW2006-TB-002) of China. References [1] D. Bainov, P. Simeonov, Integral Inequalities and Applications, Kluwer Academic Publishers, Dordrecht, 1992. [2] R. Bellman, The stability of solutions of linear differential equations, Duke Math. J. 10 (1943) 643–647.
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[3] W.S. Cheung, Some discrete nonlinear inequalities and applications to boundary value problems for difference equations, J. Differential Equation Appl. 10 (2004) 213–223. [4] W.S. Cheung, Some new nonlinear inequalities and applications to boundary value problems, Nonlinear Anal. TMA 64 (2006) 2112–2128. [5] W.S. Cheung, Q.H. Ma, Nonlinear retarded integral inequalities for functions in two variables, J. Concrete Appl. Math. 2 (2004) 119–134. [6] W.S. Cheung, Q.H. Ma, On certain new Gronwall–Ou-Iang type integral inequalities in two variables and their applications, J. Inequality Appl. 8 (2005) 347–361. [7] W.S. Cheung, J.L. Ren, Discrete nonlinear inequalities and applications to boundary value problems, J. Math. Anal. Appl. 319 (2006) 708–724. [8] H. Engler, Global regular solutions for the dynamic antiplane shear problem in nonlinear viscoelasticity, Math. Z. 202 (1989) 251–259. [9] T.H. Gronwall, Note on the derivatives with respect to a parameter of solutions of a system of differential equations, Ann. Math. 20 (1919) 292–296. [10] A. Haraux, Nonlinear Evolution Equations. Global Behavior of Solutions, Lecture Note in Mathematics, Vol. 847, Springer, Berlin, New York, 1981. [11] W.N. Li, M.A. Han, F.W. Meng, Some new delay integral inequalities and their applications, J. Comput. Appl. Math. 180 (2005) 191–200. [12] O. Lipovan, A retarded integral inequality and its applications, J. Math. Anal. Appl. 285 (2003) 436–443. [13] Q.H. Ma, N-independent-variable discrete inequalities of Gronwall–Ou-Iang type, Ann. Differential Equations 16 (2000) 813–820. [14] Q.H. Ma, W.S. Cheung, Some new nonlinear difference inequalities and their applications, J. Comupt. Appl. Math. 202 (2007) 339–351. [15] Q.H. Ma, L. Debnath, A more generalized Gronwal-like integral inequality with applications, Internat. J. Math. Math. Sci. 33 (2003) 927–934. [16] Q.H. Ma, E.H. Yang, Some new nonlinear delay integral inequalities, J. Math. Anal. Appl. 252 (2000) 864–878. [17] F.W. Meng, W.N. Li, On some new nonlinear discrete inequalities and their applications, J. Comupt. Appl. Math. 158 (2003) 407–417. [18] F.W. Meng, W.N. Li, On some new integral inequalities and their applications, Appl. Math. Comput. 148 (2004) 381–392. [19] D.S. Mitrinovi´c, J.K. Peˇcari´c, A.M. Fink, Inequalities Involving Functions and their Integrals and Derivatives, Dordrecht, Boston, London, Kluwer Academic Publishers, 1991. [20] L. Ou-Iang, The boundedness of solutions of linear differential equations y + A(t)y = 0, Shuxue Jinzhan 3 (1957) 409–418. [21] B.G. Pachpatte, On some new inequalities related to certain inequality arising in the theory of differential equations, J. Math. Anal. Appl. 189 (1995) 128–144. [22] B.G. Pachpatte, Inequalities for Differential and Integral Equations, Academic Press, New York, 1998. [23] P.Y.H. Pang, R.P. Agarwal, On an integral inequality and discrete analogue, J. Math. Anal. Appl. 194 (1995) 569–577. [24] E.H. Yang, On some nonlinear integral and discrete inequalities related to Ou-Iang’s inequality, Acta Math. Sinica, English Ser. 14 (1998) 353–360. [25] E.H. Yang, A new integral inequality with power nonlinear and its discrete analogue, Acta Math. Appl. Sinica 17 (2001) 233–239.
Some new nonlinear Volterra–Fredholm-type discrete inequalities and their applications Qing-Hua Ma Faculty of Information Science and Technology, Guangdong University of Foreign Studies, Guangzhou 510420, PR China Received 22 February 2006; received in revised form 26 March 2007
Abstract Some new explicit bounds on solutions to a class of new nonlinear Volterra–Fredholm-type discrete inequalities are established, which can be used as effective tools in the study of certain sum–difference equations. Application examples are also indicated. © 2007 Elsevier B.V. All rights reserved. MSC: 26D15; 26D20; 39A10 Keywords: Nonlinear discrete inequalities; Volterra–Fredholm-type; Gronwall–Bellman–Ou-Iang-type; Explicit bounds; Sum–difference equations
1. Introduction In the study of ordinary differential equations, integral equations and difference equations, one often deal with certain integral inequalities. The Gronwall–Bellman inequality [9,2] and its various linear and nonlinear generalizations are crucial in the discussion of the existence, uniqueness, continuation, boundedness, oscillation and stability and other qualitative properties of solutions of differential and integral equations. The literature on such inequalities and their applications is vast; see [1,19,22] and the references therein. A specific branch of this type integral inequalities is originated by Ou-Iang. In his study of boundedness of solutions to linear second order differential equations, Ou-Iang [20] established and used the following nonlinear integral inequality, which is now known as Ou-Iang’s inequality in the literature. Theorem A (Ou-Iang [20]). Let u and f be real-valued, nonnegative, and continuous functions defined on R+ = [0, +∞) and let c 0 be a real constant. Then the nonlinear integral inequality t u2 (t)c2 + 2 f (s)u(s) ds, t ∈ R+ 0
implies
t
u(t)c + 0
f (s) ds,
t ∈ R+ .
E-mail address: [email protected]. 0377-0427/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.cam.2007.05.021
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Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
While Ou-Iang’s inequality is having a neat form and is interesting in its own right as an integral inequality, its importance lies equality heavily on its many beautiful applications in differential and integral equations (see, e.g., [22]). Since this, over the years, many generalizations of Ou-Iang’s inequality to various situations have been established; see, for example, [3–7], [11–18], [21–25] and the references cited therein. Among various generalizations of Ou-Iang’s inequality, discrete analogue is also an interesting direction. The point is, similar to the noteworthy contributions of the continuous version of the inequality to the study of differential equations, one naturally expects that discrete versions of the inequality should also play an important role in the study of difference equations. In this respect, fewer results have been established. Recent results in this direction include the works of Pachpatte [21], Pang–Agarwal [23], Ma [13], Meng–Li [17], Cheung [3], Cheung–Ren [7], and Ma–Cheung [14]. The aim of the present paper is to give some explicit bounds to some new nonlinear discrete inequalities involving two-variable functions which, on the one hand, generalize Ou-Iang’s inequality to Volterra–Fredholm form at the first time to literatures as we know and, on the other hand, give a handy and effective tool for the study of quantitative properties of solutions of sum–difference equations. We illustrate the usefulness of these inequalities by applying them to study the boundedness, uniqueness, and continuous dependence of the solutions of certain Volterra–Fredholm-type sum–difference equations.
2. Nonlinear discrete inequalities Throughout this paper, I := [m0 , M) ∩ Z and J := [n0 , N ) ∩ Z are two fixed lattices of integral points in R, where m0 , n0 ∈ Z, M, N ∈ Z ∪ {∞}. Let := I × J ⊂ Z 2 , R+ := [0, ∞), R1 := [1, ∞) and for any (s, t) ∈ , the sub-lattice [m0 , s] × [n0 , t] ∩ of will be denoted as (s,t) . If U is a lattice in Z (resp.Z 2 ), the collection of all R-valued functions on U is denoted by F(U ), that of all R+ valued functions by F+ (U ), and that of all R1 -valued functions by F1 (U ). For the sake of convenience, we extend the domain of definition of each function in F(U ) and F+ (U ) trivially to the ambient space Z (resp.Z 2 ). So, for example, a function in F(U ) is regarded as a function defined on Z (resp.Z 2 ) with support in U. As usual, the collection of all continuous functions and all i-times continuously differentiable functions of a topological space X into a topological space Y will be denoted by C(X, Y ) and C i (X, Y ), respectively. If U is a lattice in Z 2 , the partial difference operators 1 and 2 on u ∈ F(Z 2 ) or F+ (Z 2 ) are defined as 1 u(m, n) = u(m + 1, n) − u(m, n), (m, n) ∈ U , 2 u(m, n) = u(m, n + 1) − u(m, n), (m, n) ∈ U . For w ∈ C(R+ , R+ ), the function G1 is defined as G1 (v) =
v
ds , w(s)
v0
v v0 > 0.
Theorem 2.1. Suppose that u and a ∈ F+ (), k 0 constant and w ∈ C(R+ , R+) is nondecreasing with w(r) > 0 for r > 0; G1 (∞) =
∞ v0
ds =∞ w(s)
and H1 (t) = G1 (2t − k) − G1 (t) is strictly increasing for t k.
(2.1)
Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
453
If u(m, n) satisfies m−1 n−1
u(m, n)k +
a(s, t)w(u(s, t)) +
s=m0 t=n0
M−1 N−1
a(s, t)w(u(s, t))
(2.2)
s=m0 t=n0
for (m, n) ∈ , then
u(m, n)G−1 G1 H1−1 1
M−1 N−1
a(s, t)
+
s=m0 t=n0
m−1 n−1
a(s, t)
(2.3)
s=m0 t=n0
−1 for (m, n) ∈ , where G−1 1 and H1 are inverse functions of G1 and H1 , respectively.
Proof. Let k > 0 and define z(m, n) = k +
m−1 n−1
a(s, t)w(u(s, t)) +
M−1 N−1
a(s, t)w(u(s, t))
s=m0 t=n0
s=m0 t=n0
then u(m, n)z(m, n) z(m0 , n) = k +
(m, n) ∈ ,
M−1 N−1
(2.4)
a(s, t)w(u(s, t))
s=m0 t=n0
and 1 z(m, n) =
n−1
a(m, t)w(u(m, t))
t=n0
n−1
a(m, t)w(z(m, t))
t=n0
w(z(m, n))
n−1
a(m, t).
t=n0
Therefore, by the Mean-Value Theorem for integrals, for each (m, n) ∈ , there exists : z(m, n) z(m + 1, n) such that 1 G1 (z(m, n)) = G1 (z(m + 1, n)) − G1 (z(m, n)) z(m+1,n) ds = w(s) z(m,n) =
1 1 z(m, n). w()
Since w is nondecreasing, w()w(z(m, n)); hence 1 G1 (z(m, n))
1 1 z(m, n) w(z(m, n)) n−1 t=n0
a(m, t)
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Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
for all (m, n) ∈ . Therefore, m−1
1 G1 (z(s, n))
s=m0
m−1 n−1
a(s, t)
s=m0 t=n0
On the other hand, it is easy to check that m−1
1 G1 (z(s, n)) = G1 (z(m, n)) − G1 (z(m0 , n)),
s=m0
thus m−1 n−1
G1 (z(m, n))G1 (z(m0 , n)) +
a(s, t)
s=m0 t=n0
Since G−1 1 is increasing, the last inequality yields m−1 n−1 −1 z(m, n)G1 G1 (z(m0 , n)) + a(s, t)
(2.5)
s=m0 t=n0
for all (m, n) ∈ . From the last inequality, we observe that 2z(m0 , n) − k = k + 2
M−1 N−1
a(s, t)w(u(s, t))
s=m0 t=n0
= z(M, N) G−1 1
G1 (z(m0 , N )) +
a(s, t)
s=m0 t=n0
= G−1 1
M−1 N−1
G1 (z(m0 , n)) +
M−1 N−1
a(s, t)
s=m0 t=n0
or G1 (2z(m0 , n) − k) − G1 (z(m0 , n))
M−1 N−1
a(s, t).
(2.6)
s=m0 t=n0
Since H1 (t) = G1 (2t − k) − G1 (t) is increasing for t > k, H1 (t) has inverse function H1−1 (t) and then from the last inequality we get M−1 N−1 −1 a(s, t) . (2.7) z(m0 , n)H1 s=m0 t=n0
Substituting (2.7) into (2.5) and combining with (2.4) we obtain the desired inequality (2.3). If k = 0, we carry out the above procedure with ε > 0 instead of k and subsequently let ε → 0. Theorem 2.1 . Let u(m, n), a(m, n), w(u), G1 (u) and k be as in Theorem 2.1. If u(m, n) satisfies (2.2) for (m, n) ∈ , and
1 (t) = G1 (2t − k) − G1 (t) − H
M−1 N−1 s=m0 t=n0
a(s, t)
Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
455
1 (t) = 0 has a solution c1 for t k, then is increasing and H m−1 n−1 −1 a(s, t) u(m, n)G1 G1 (c1 ) +
(2.8)
s=m0 n=n0
for (m, n) ∈ , where G1 and G−1 1 are defined as in Theorem 2.1. Proof. By the same steps from (2.4) to (2.7) in the proofs of Theorem 2.1, we have u(m, n)z(m, n), z(m, n)G−1 1
(2.9)
G1 (z(m0 , n)) +
m−1 n−1
a(s, t)
(2.10)
s=m0 t=n0
and M−1 N−1
G1 (2z(m0 , n) − k) − G1 (z(m0 , n))
a(s, t)
(2.11)
s=m0 t=n0
for (m, n) ∈ . From the assumption of Theorem 2.1 and (2.11), we have
1 (z(m0 , n))0 = H
1 (c1 ). H
1 is increasing, H
1 has an inverse function H
−1 ; hence, from the last inequality we get Since H 1 z(m0 , n)c1 . Substituting the last inequality into (2.10) and combining with (2.9) we get the desired inequality (2.8).
Corollary 2.2. Let u(m, n), a(m, n) and k be as in Theorem 2.1. If u(m, n) satisfies u(m, n)k +
m−1 n−1
a(s, t)u(s, t) +
s=m0 t=n0
M−1 N−1
a(s, t)u(s, t)
(2.12)
s=m0 t=n0
for (m, n) ∈ , and (M, N ) = exp
M−1 N−1
a(s, t) < 2,
(2.13)
s=m0 t=n0
then u(m, n)
m−1 n−1 k exp a(s, t) 2 − (M, N ) s=m t=n 0
(2.14)
0
for (m, n) ∈ . Proof. In Theorem 2.1, by letting w(u) = u we obtain v v ds ds v , v v0 > 0, G1 (v) = = = ln v0 v0 w(s) v0 s H1 (t) = G1 (2t − k) − G1 (t) = ln
2t − k , t
t k,
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and hence G−1 1 (v) = v0 exp v, H1−1 (t) =
k . 2 − exp t
From inequality (2.3), we obtain inequality (2.14).
Corollary 2.3. Let u(m, n), a(m, n) and k be as in Theorem 2.1, 0 < p < 1 be a constant. If u(m, n) satisfies u(m, n)k +
m−1 n−1
p
a(s, t)u (s, t) +
s=m0 t=n0
M−1 N−1
a(s, t)up (s, t)
(2.15)
s=m0 t=n0
for (m, n) ∈ , then u(m, n) (c11 )1−p + (1 − p)
m−1 n−1
1/(1−p) a(s, t)
(2.16)
s=m0 t=n0
for (m, n) ∈ , where c11 is the solution of equation
1 (t) = H
M−1 N−1 1 (2t − k)1−p − t 1−p − a(s, t) = 0 1−p s=m t=n 0
(2.17)
0
for t k. Proof. By Theorem 2.1 , we only need to prove that (2.17) has a solution c11 for t k. In fact, p 1/p p
(t) = (2 t) − (2t − k) > 0 H 1 [t (2t − k)]p
for t k,
1 (k) = − H
M−1 N−1
a(s, t) < 0
s=m0 t=n0
and 1−p
1 (t) = lim t lim H t→∞ t→∞ 1 − p
k 2− t
1−p
1 (t) = 0 has a unique solution c11 > k. so H
−1 −
M−1 N−1
a(s, t) = +∞,
s=m0 t=n0
Remark 2.1. Though (2.16) does not give an exact estimation to the solution of (2.15), it is enough to get the upper bound to the solution of (2.15) in many cases. Theorem 2.4. Suppose that u(m, n), a(m, n), w(u) and k are as in Theorem 2.1. Let (u) ∈ C 1 (R+ , R+ ) with (u) > 0 and (u) is increasing for u > 0, here (u) denotes the derivative of . If u(m, n) satisfies (u(m, n))k +
m−1 n−1
a(s, t) (u(s, t))w(u(s, t))
s=m0 t=n0
+
M−1 N−1 s=m0 t=n0
a(s, t) (u(s, t))w(u(s, t)),
(2.18)
Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
457
for (m, n) ∈ , and H2 (t) = G1 ◦ −1 (2t − k) − G1 ◦ −1 (t) is increasing for t k, then u(m, n)G−1 1
H2−1
G1
M−1 N−1
a(s, t)
+
s=m0 t=n0
m−1 n−1
a(s, t)
(2.19)
s=m0 t=n0
for (m, n) ∈ , where H2−1 is the inverse functions of H2 , G1 and G−1 1 are defined as in Theorem 2.1. Proof. Similar to the proof of Theorem 2.1, it suffices to consider the case k > 0. Denote by z2 (m, n) the right-hand side of (2.18). Then z2 > 0, u −1 (z2 ), and z2 is nondecreasing in each variable. Hence, for any (m, n) ∈ , 1 z2 (m, n) =
n−1
a(m, t) (u(m, t))w(u(m, t))
t=n0
n−1
a(m, t) (−1 (z2 (m, t)))w(−1 (z2 (m, t)))
t=n0
(−1 (z2 (m, n)))
n−1
a(m, t)w(−1 (z2 (m, t)))
t=n0
or n−1 1 z2 (m, n) a(m, t)w(−1 (z2 (m, t))). (−1 (z2 (m, n))) t=n 0
On the other hand, using the differential Mean-Value Theorem and the last inequality we have 1 [−1 (z2 (m, n))] = −1 (z2 (m + 1, n)) − −1 (z2 (m, n)) =
1 1 z2 (m, n) 1 z2 (m, n) −1 (−1 ()) ( (z2 (m, n))) n−1
a(m, t)w(−1 (z2 (m, t))).
(2.20)
t=n0
Keeping n fixed in (2.20) and setting m = s and then summing over s = m0 , m0 + 1, . . . , m − 1, we get −1
−1
(z2 (m, n))
(z2 (m0 , n)) +
m−1 n−1
a(s, t)w(−1 (z2 (m, t)))
(2.21)
s=m0 t=n0
for all (m, n) ∈ . Now by applying Theorem 2.1 in [3] to the function −1 (z2 (m, n)), we have −1
(z2 (m, n))G−1 1
G1
−1
(z2 (m0 , n)) +
m−1 n−1 s=m0 t=n0
a(s, t) .
(2.22)
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Observing that 2z2 (m0 , n) − k = k + 2
M−1 N−1
a(s, t) (u(s, t))w(u(s, t)) = z2 (M, N ),
s=m0 t=n0
and then from (2.22) we have G1 ◦ −1 (2z2 (m0 , n) − k) = G1 ◦ −1 (z2 (M, N )) G1 ◦ −1 (z2 (m0 , N )) +
M−1 N−1
a(s, t)
s=m0 t=n0
= G1 ◦
−1
(z2 (m0 , n)) +
M−1 N−1
a(s, t)
s=m0 t=n0
or G1 ◦ −1 (2z2 (m0 , n) − k) − G1 ◦ −1 (z2 (m0 , n))
M−1 N−1
a(s, t).
(2.23)
s=m0 t=n0
Since H2 (t) = G1 ◦ −1 (2t − k) − G1 ◦ −1 (t) is increasing for t k, H2 (t) has an inverse function H2−1 and then from (2.23) we get z2 (m0 , n)H2−1
M−1 N−1
a(s, t) .
s=m0 t=n0
Substituting the last inequality into (2.22) and by the definition of z2 (m, n) we obtain the desired inequality (2.19).
By similar argument as in the proofs of Theorem 2.1 , we have the following result immediately. Theorem 2.4 . Let u(m, n), a(m, n) and k be as in Theorem 2.4. If u(m, n) satisfies (2.18) for (m, n) ∈ , and
2 (t) = G1 ◦ −1 (2t − k) − G1 ◦ −1 (t) − H
M−1 N−1
a(s, t)
s=m0 t=n0
2 (t) = 0 has a solution c2 k, then is increasing and H u(m, n)G−1 1
G1 (c2 ) +
m−1 n−1
a(s, t)
(2.24)
s=m0 t=n0
for (m, n) ∈ . When = up (p 1 is a constant) in Theorem 2.4, we have the following: Corollary 2.5. Let u(m, n), a(m, n) and k be as in Theorem 2.4, p 1 is a constant. If u(t) satisfies p
u (m, n)k +
m−1 n−1 s=m0 t=n0
a(s, t)u
p−1
(s, t)w(u(s, t)) +
M−1 N−1 s=m0 t=n0
a(s, t)up−1 (s, t)w(u(s, t)),
(2.25)
Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
459
for (m, n) ∈ , and H21 (t) = G1 ((2t − k)1/p ) − G1 (t 1/p ) is increasing for (m, n) ∈ , then
−1 G1 H21 u(m, n)G−1 1
M−1 N−1
a(s, t)
+
s=m0 t=n0
m−1 n−1
a(s, t)
(2.26)
s=m0 t=n0
for (m, n) ∈ . Corollary 2.6. Let u(m, n) ∈ F1 (U ), a(m, n) and k be as in Theorem 2.4. If u(t) satisfies up (m, n)k +
m−1 n−1
a(s, t)up (s, t)w(ln u(s, t))
s=m0 t=n0
+
M−1 N−1
a(s, t)up (s, t)w(ln u(s, t)),
(2.27)
s=m0 t=n0
for (m, n) ∈ , and
1 1 H22 (t) = G1 ln(2t − k) − G1 ln t p p is increasing for (m, n) ∈ , then
−1 G1 H22 u(m, n)G−1 1
M−1 N−1 s=m0 t=n0
a(s, t)
+
m−1 n−1
a(s, t)
(2.28)
s=m0 t=n0
for (m, n) ∈ . Proof. Taking v(m, n) = ln u(m, n), then (2.27) reduces to exp(pv(m, n))k +
m−1 n−1
a(s, t) exp(pv(s, t))w(v(s, t))
s=m0 t=n0
+
M−1 N−1
a(s, t) exp(pv(m, n))w(v(s, t)),
(2.29)
s=m0 t=n0
for (m, n) ∈ , which is a special case of inequality (2.18) when (v) = exp(pv). In this special case,
1 1 H2 (t) = H22 (t) = G1 ln(2t − k) − G1 ln t . p p By Theorem 2.4, we get the desired inequality (2.28) directly.
Remark 2.2. Eqs. (2.25) and (2.27) are new interesting discrete Volterra–Fredholm–Ou-Iang-type and Volterra– Fredholm–Engler [8]–Haraux [10] type inequality of two-variable, respectively. Using Theorems 2.1 and 2.4, we can get more generalized results as follows.
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Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
−1 Theorem 2.7. Let u(m, n), a(m, n), w(t), G1 , G−1 1 , H1 , H1 and k be as in Theorem 2.1, b(m, n) ∈ F+ (). If u(m, n) satisfies m−1 n−1
u(m, n)k +
a(s, t)w(u(s, t)) +
s=m0 t=n0
for (m, n) ∈ , then
u(m, n)G−1 1
b(s, t)w(u(s, t))
(2.30)
s=m0 t=n0
G1
M−1 N−1
H1−1
M−1 N−1
∗
a (s, t)
+
s=m0 t=n0
m−1 n−1
∗
a (s, t)
(2.31)
s=m0 t=n0
for (m, n) ∈ , where a ∗ (m, n) ∈ F+ () such that both a(m, n) and b(m, n) are less than or equal to a ∗ (m, n). Proof. From (2.30) and the assumptions, we have m−1 n−1
u(m, n) k +
a ∗ (s, t)w(u(s, t)) +
s=m0 t=n0
M−1 N−1
a ∗ (s, t)w(u(s, t))
s=m0 t=n0
for (m, n) ∈ . An application of Theorem 2.1 to the last inequality yields (2.31).
Theorem 2.8. Let u(m, n), a(m, n), b(m, n), a ∗ (m, n) and k be as in Theorem 2.7; (u) be as in Theorem 2.4. wi ∈ C(R+ , R+ ) be nondecreasing functions with wi > 0 for u > 0, i = 1, 2. If u(m, n) satisfies (u(m, n))k +
n−1 m−1
a(s, t) (u(s, t))w1 (u(s, t))
s=m0 t=n0
+
M−1 N−1
b(s, t) (u(s, t))w2 (u(s, t))
(2.32)
s=m0 t=n0
for (m, n) ∈ , there is a function W (u) ∈ C(R+ , R+ ) that is nondecreasing such that both w1 and w2 are less than or equal to W, v +∞ ds ds G2 (v) = , v v0 > 0, G2 (+∞) = = +∞, W (s) v0 W (s) v0 and H3 (t) = G2 ◦ −1 (2t − k) − G2 ◦ −1 (t) is increasing for t k, then u(m, n)G−1 2
G2
H3−1
M−1 N−1
∗
a (s, t)
s=m0 t=n0
+
m−1 n−1
∗
a (s, t)
s=m0 t=n0
−1 for (m, n) ∈ , where G−1 2 and H3 are inverse functions of G2 and H3 , respectively.
Proof. From (2.32) and the assumptions, we can obtain (u(m, n))k +
m−1 n−1
a ∗ (s, t) (u(s, t))W (u(s, t))
s=m0 t=n0
+
M−1 N−1 s=m0 t=n0
a ∗ (s, t) (u(s, t))W (u(s, t)).
(2.33)
Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
Now, an application of Theorem 2.4 to the last inequality yields the desired inequality (2.33).
461
By the same argument as in the proofs of Theorem 2.1 , we have the following result immediately. Theorem 2.8 . Let u(m, n), a(m, n), b(m, n), a ∗ (m, n), wi (i =1, 2), W and k be as in Theorem 2.8. If u(m, n) satisfies (2.32), M−1 N−1
3 (t) = G2 ◦ −1 (2t − k) − G2 ◦ −1 (t) − H
a ∗ (s, t)
s=m0 t=n0
3 (t) = 0 has a solution c3 k, then is increasing and H u(m, n)G−1 2
G2 (c3 ) +
m−1 n−1
∗
a (s, t)
(2.34)
s=m0 t=n0
for (m, n) ∈ . Remark 2.3. In Theorems 2.7, 2.8 and 2.8 , we can choose function a ∗ (m, n)=a(m, n)+b(m, n) or max{a(m, n), b(m, n)} as well as in function W. By Theorem 2.8 , we can get the following interesting result. Corollary 2.9. Let u(m, n), a(m, n), b(m, n), a ∗ (m, n) and k be as in Theorem 2.8, p 1 and 0 < q < 1 be constants. If u(m, n) satisfies up (m, n)k +
m−1 n−1
a(s, t)up (s, t) +
s=m0 t=n0
M−1 N−1
b(s, t)up+q−1 (s, t)
(2.35)
s=m0 t=n0
for (m, n) ∈ , and exp
M−1 N−1
a ∗ (s, t) < 21/p ,
s=m0 t=n0
then
1/(1−q) m−1 n−1 1−q ∗ u(m, n) 1 + c˜3 exp (1 − q) a (s, t) − 1
(2.36)
s=m0 t=n0
for (m, n) ∈ , where c˜3 is the solution of
3 (t) = H
M−1 N−1 1 + (2t − k)(1−q)/p 1 ln − a ∗ (s, t) = 0 (1−q)/p 1−q 1+t s=m t=n 0
(2.37)
0
for t k. Proof. In Theorem 2.8 , by letting w1 (u) = u, w2 (u) = uq and W = w1 + w2 we obtain G2 (v) =
v v0
ds = w1 (s) + w2 (s)
v v0
ds 1 1 + v 1−q = , ln s + sq 1 − q 1 + v 1−q 0
v v0 > 0.
(2.38)
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Hence, G−1 2 (v) =
1−q
1 + v0
1/(1−q)
exp((1 − q)v) − 1
.
(2.39)
By computation, we have
3 (t) = H
M−1 N−1 1 + (2t − k)(1−q)/p 1 − a ∗ (s, t), ln (1−q)/p 1−q 1+t s=m t=n 0
(t) = H 3
k
+ 2t 1−(1−q)/p
[2t − k + (2t − k)
− (2t − k)
1−(1−q)/p
0
1−(1−q)/p
](t + t 1−(1−q)/p )
>0
(2.40)
for t k,
3 (k) = − H
M−1 N−1
a ∗ (s, t) < 0
(2.41)
s=m0 t=n0
and
3 (t) = lim lim H
t→+∞
t→+∞
= ln 2
M−1 N−1 1 1 + (2t − k)(1−q)/p − a ∗ (s, t) ln (1−q)/p 1−q 1+t s=m t=n 0
1/p
−
M−1 N−1
0
a ∗ (s, t) > 0.
(2.42)
s=m0 t=n0
By (2.40)–(2.42) we obtain that (2.37) has a solution c˜3 > k. Now by (2.34), (2.38) and (2.39) we can get the desired (2.36). 3. Some applications In this section, we apply our results to study the boundedness, uniqueness, and continuous dependence of the solutions of certain Volterra–Fredholm sum–difference equations of the form up (m, n) = l(m, n) +
m−1 n−1
F (s, t, u(s, t)) +
s=m0 t=n0
M−1 N−1
G(s, t, u(s, t))
(3.1)
s=m0 t=n0
for (m, n) ∈ , where l ∈ F(), F, G ∈ F( × R), p 1 is a constant. Following theorem gives the bound on the solutions of Eq. (3.1). Theorem 3.1. Assume that the functions l, F and G in (3.1) satisfy the conditions |l(m, n)| k,
(3.2)
|F (m, n, v)|a(m, n)|v|p+q−1
(3.3)
|G(m, n, v)| b(m, n)|v|p+q−1 ,
(3.4)
and
where a(m, n) and b(m, n) are same as in Theorem 2.7, 0 < q < 1 is a constant, then all solutions of (3.1) satisfy u(m, n) (c11 )
1−q
+ (1 − q)
m−1 n−1 s=m0 t=n0
1/(1−q) ∗
a (s, t)
(3.5)
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463
for (m, n) ∈ , where c11 is the solution of equation
∗ (t) = H 2
M−1 N−1 1 a ∗ (s, t) = 0 (2t − k)(1−q)/p − t (1−q)/p − 1−q s=m t=n 0
(3.6)
0
for t k, where a ∗ (m, n) ∈ F+ () such that both a(m, n) and b(m, n) are less than or equal to a ∗ (m, n). Proof. Using the conditions (3.2)–(3.4) in (3.1) we have |u(m, n)|p k +
m−1 n−1
a ∗ (s, t)|u(s, t)|p+q−1 +
s=m0 t=n0
M−1 N−1
a ∗ (s, t)|u(s, t)|p+q−1
s=m0 t=n0
which is a special case of Theorem 2.4 when (u) = up and w(u) = uq . By Theorem 2.4 , we only need to prove
2 (t) = H
∗ (t) = G1 ◦ −1 (2t − k) − G1 ◦ −1 (t) − H 2
=
M−1 N−1
a ∗ (s, t)
s=m0 t=n0
M−1 N−1 1 a ∗ (s, t) (2t − k)(1−q)/p − t (1−q)/p − 1−q s=m t=n 0
0
∗ (t) = 0 has a solution c11 . In fact, taking r = (1 − q)/p, by computation we have is increasing and H 2 1 (21/(1−r) t)1−r − (2t − k)1−r d ∗ >0 H2 (t) = dt p [t (2t − k)]1−r for t k,
∗ (k) = − H 2
M−1 N−1
a ∗ (s, t) < 0
s=m0 t=n0
and
∗ (t) = lim lim H 2
t→+∞
t→+∞
tr 1−q
M−1 N−1 k r ∗ −1 − a (s, t) = +∞, 2− t s=m t=n
∗ (t) = 0 has a unique solution c11 > k. so H 2
0
0
Secondly, we consider the uniqueness of the solutions of (3.1). Theorem 3.2. Assume that the functions F and G in Eq. (3.1) satisfy the conditions |F (m, n, v) − F (m, n, v)|a(m, ¯ n)|v p − v¯ p |,
(3.7)
|G(m, n, v) − G(m, n, v)| ¯ b(m, n)|v p − v¯ p |,
(3.8)
for some a, b ∈ F+ (), and if (M, N ) = exp
M−1 N−1
a ∗ (s, t) < 2,
s=m0 t=n0
where a ∗ (m, n) is as in Theorem 3.1, then (3.1) has at most one positive solution on .
(3.9)
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Proof. Let u(m, n) and u(m, ¯ n) be two solutions of (3.1). By (3.1) and conditions (3.7) and (3.8), we have m−1 n−1
|up (m, n) − u¯ p (m, n)|
a ∗ (s, t)|up (s, t) − u¯ p (s, t)|
s=m0 t=n0
+
M−1 N−1
a ∗ (s, t)|up (s, t) − u¯ p (s, t)|.
s=m0 t=n0
An application of Corollary 2.2 to the function |up (m, n) − u¯ p (m, n)| yields that |up (m, n) − u¯ p (m, n)| 0 Hence, u(m, n) = u(m, ¯ n) on .
for all (m, n) ∈ .
Finally, we investigate the continuous dependence of the solutions of (3.1) on the functions F and G. For this, we consider the following variation of (3.1):
¯ n) + up (m, n) = l(m,
m−1 n−1
F¯ (s, t, u(s, t)) +
s=m0 t=n0
M−1 N−1
¯ t, u(s, t)) G(s,
(3.1)
s=m0 t=n0
¯ ∈ F( × R), p 1 is a constant as in (3.1). for (m, n) ∈ , where F¯ , G Theorem 3.3. Consider (3.1) and (3.1). If p p p p (i) |F (s, t, v1 ) − F (s, t, v2 )|a(s, t)|v1 − v2 | and |G(s, t, v1 ) − G(s, t, v2 )| b(s, t)|v1 − v2 |; ¯ (ii) |l(m, n) − l(m, n)|ε/2; N−1 ∗ (iii) (M, N ) = exp M−1 s=m0 t=n0 a (s, t) < 2; (iv) for all solutions u¯ of (3.1), m−1 n−1 s=m0 t=n0
ε |F (s, t, u) ¯ − F¯ (s, t, u)| ¯ 4
and M−1 N−1 s=m0 t=n0
ε ¯ t, u)| |G(s, t, u) ¯ − G(s, ¯ 4
for all (s, t) ∈ and v1 , v2 ∈ R, where ε > 0 is an arbitrary constant, a ∗ (m, n) is defined as in Theorem 3.1, then |up (s, t) − u¯ p (s, t)|
m−1 n−1 ε exp a ∗ (s, t) 2 − (M, N ) s=m t=n 0
(3.10)
0
for (m, n) ∈ . Hence, up depend continuously on F and G. In particular, if u does not change the sign, it depends continuously on F and G.
Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
465
Proof. Let u(m, n) and u(m, ¯ n) be solutions of (3.1) and (3.1), respectively. Then, u(m, n) satisfies (3.1) and u(m, ¯ n) satisfies (3.1). Hence, ¯ |up (m, n) − u¯ p (m, n)||l(m, n) − l(m, n)| +
m−1 n−1
|F (s, t, u(s, t)) − F¯ (s, t, u(s, ¯ t))|
s=m0 t=n0
+
M−1 N−1
¯ t, u(s, |G(s, t, u(s, t)) − G(s, ¯ t))|
s=m0 t=n0 m−1 n−1 ε + |F (s, t, u(s, t)) − F (s, t, u(s, ¯ t))| 2 s=m t=n 0
+
0
n−1 m−1
|F (s, t, u(s, ¯ t)) − F¯ (s, t, u(s, ¯ t))|
s=m0 t=n0
+
M−1 N−1
|G(s, t, u(s, t)) − G(s, t, u(s, ¯ t))|
s=m0 t=n0
+
M−1 N−1
¯ t, u(s, |G(s, t, u(s, ¯ t)) − G(s, ¯ t))|
s=m0 t=n0
ε +
m−1 n−1 s=m0 t=n0
ε +
m−1 n−1
M−1 N−1
a(s, t)|up (s, t) − u¯ p (s, t)| +
b(s, t)|up (s, t) − u¯ p (s, t)|
s=m0 t=n0
a ∗ (s, t)|up (s, t) − u¯ p (s, t)| +
s=m0 t=n0
M−1 N−1
a ∗ (s, t)|up (s, t) − u¯ p (s, t)|
s=m0 t=n0
by assumptions (i)–(iv) and the definition of a ∗ . Now, by applying Corollary 2.2 to the function |up (m, n) − u¯ p (m, n)|, we get |up (m, n) − u¯ p (m, n)|
m−1 n−1 ε a ∗ (s, t). exp 2 − (M, N ) s=m t=n 0
Evidently, if the function
m−1 n−1 s=m0
t=n0
0
a ∗ (s, t) is bounded on , then
|up (m, n) − u¯ p (m, n)|εK for some K > 0 and (m, n) ∈ . Hence, up depends continuously on F and G.
Acknowledgement The author is grateful to the referees for their very helpful and detailed comments in improving this paper. The research is partially supported by the NSF of Guangdong Province and the Research Group Grants Council of the Guangdong University of Foreign Studies (Project No. GW2006-TB-002) of China. References [1] D. Bainov, P. Simeonov, Integral Inequalities and Applications, Kluwer Academic Publishers, Dordrecht, 1992. [2] R. Bellman, The stability of solutions of linear differential equations, Duke Math. J. 10 (1943) 643–647.
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Q.-H. Ma / Journal of Computational and Applied Mathematics 216 (2008) 451 – 466
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