2012 International Conference on Image, Vision and Computing (ICIVC 2012) IPCSIT vol. 50 (2012) © (2012) IACSIT Press, Singapore DOI: 10.7763/IPCSIT.2012.V50.46 A New Generalized Gronwall-Bellman Type Inequality Qinghua Feng+ School of Science, Shandong University of Technology, Zhangzhou Road 12, Zibo, Shandong, China, 255049 Abstract. In this paper, a new nonlinear integral inequality is established, which provide a handy tool for analyzing the global existence of solutions of differential and integral equations. Keywords: Integral inequality; Global existence; Integral equation; Differential equation 1. Introduction During the past decades, with the development of the theory of differential and integral equations, a lot of integral inequalities, for example [1-13], have been discovered, which play an important role in the research of boundedness, global existence, stability of solutions of differential and integral equations. In [12], Jiang proved the following theorem: Theorem A: R+ = [0, ∞) . Suppose that x (t ), f (t ), h(t ) ∈ C ( R+ , R+ ) . Then the following form of delay integral inequality: t x p (t ) ≤ C + ∫ [ f ( s ) x p ( s ) + h ( s ) x p (σ ( s ))]ds, t ∈ R+ 0 with the initial condition 1 x (t ) = φ (t ), t ∈ [α , 0], φ (σ (t )) ≤ ( ρ (t )) p , t ∈ R+ , σ (t ) ≤ 0 where p, q are constants, p > 0, q > 0, p ≠ q . σ (t) ∈C(R+ , R) , σ (t ) ≤ t, −∞ < α = inf{σ (t ), t ∈ R+ } ≤ 0,φ ∈C([α ,0], R+ ) , implies that t x (t ) ≤ exp( ∫ 0 t s p −q 1 f (s) p−q p−q h ( s ) exp( ∫ f (τ ) dτ ) ds ] p−q ds ) [C p + ∫ p p p 0 0 for t ∈ [0, t0 ] , where t0 is a positive number satisfying inf {C t∈[0, t0 ] p−q p t +∫ 0 s p−q p−q h ( s ) exp( ∫ f (τ ) dτ ) ds} > 0 . p p 0 In this paper, motivated by the above work, we will prove more general theorem and establish a new integral inequality. Also we will give one example so as to illustrate the validity of the present integral inequality. + Corresponding author. E-mail address: fqhua@sina.com 2. Main Results Theorem 2.1: Assume that x, a ∈ C ( R+ , R+ ) , a (t ) is non-decreasing. f , g , ∂ t f , ∂ t g ∈ C ( R+ × R+ , R+ ) . ω ∈ C ( R+ , R+ ) be nondecreasing with ω (u ) > 0 on (0, ∞) . If x (t ) satisfies the following delay integral inequality: t x p (t ) ≤ a(t ) + ∫ [ f ( s, t ) x p (σ 1 ( s)) + g ( s, t ) x q (σ 2 ( s))ω ( x(σ 3 ( s)))]ds, t ∈ R+ (1) 0 with the initial condition 1 x(t ) = φ (t ), t ∈ [α , 0], φ (σ i (t )) ≤ (a (t )) p , t ∈ R+ , σ i (t ) ≤ 0, i = 1, 2,3 (2) where p, q are constants, p > q > 0 , σi ∈C(R+ , R) , σi (t) ≤ t, −∞ < α = inf{min{σi (t), i = 1,2,3}, t ∈ R+} ≤ 0,φ ∈C([α,0], R+ ) , then for t ∈ R+ , x (t ) ≤ {Ω −1[Ω(exp( t p−q 1 p−q p−q p−q p−q F1 (t ))a p (t )) + exp( F1 (t )) ∫ F2 '( s ) exp( − F1 ( s )) ds ]} p−q p p p p 0 (3) where t t 0 r 0 F1 (t ) = ∫ f ( s , t ) ds, F2 (t ) = ∫ g ( s, t ) ds , Ω(r ) = ∫ 1 1 1 p −q ω (s ) (4) ds, Ω −1 is the inverse of Ω . Proof: We notice (3) holds for t = 0 obviously. Let the right side of (1) be ϕ p (t ) . Then x (t ) ≤ ϕ (t ) (5) When σ i (t ) ≥ 0 , we have x(σ i (t )) ≤ ϕ (σ i (t )) ≤ ϕ (t ) (6) When σ i (t ) ≤ 0 , we have 1 x (σ i (t )) = φ (σ i (t )) ≤ a p (t ) ≤ ϕ (t ) (7) So from (6), (7) we always have x (σ i (t )) ≤ ϕ (t ), i = 1, 2, 3 . Fix T > 0 . Then for t ∈ (0, T ] , we have t ϕ p (t ) ≤ a(T ) + ∫ [ f ( s, t )ϕ p (s) + g (s, t )ϕ q (s)ω (ϕ (s))]ds (8) 0 Let the right side of (8) be u p (t ) . Then ϕ (t ) ≤ u (t ), t ∈ (0, T ] (9) and u p −1 t ∂g ( s, t ) q 1 ∂f ( s, t ) p ϕ ( s)ω (ϕ (s))ds + g (t , t )ϕ q (t )ω (ϕ (t ))] (t )u '(t ) = [ ∫ ϕ ( s )ds + f (t , t )ϕ p (t ) + ∫ t ∂ p 0 ∂t 0 t t ≤ d ∫ f ( s, t )ds 0 dt t 1 p ϕ (t ) + p d ∫ g ( s, t )ω (u ( s))ds t ≤ 0 dt 1 q ϕ (t ) p t d ∫ f ( s, t )ds 1 p u (t ) + p 0 dt d ∫ g ( s, t )ω (u ( s ))ds 0 dt 1 q u (t ) p (10) Then t u '(t ) ≤ d ∫ f (s, t )ds 0 dt t 1 u(t ) + p d ∫ g(s, t)ω(u(s))ds 0 dt 1 1+q− p u (t ) p (11) Let v(t ) = u p − q (t ) . Then t v '(t ) ≤ d ∫ f (s, t )ds 0 dt t p−q v(t ) + p d ∫ g(s, t )ω(u(s))ds 0 dt t p−q ≤ p d ∫ f (s, t )ds 0 dt t p−q v(t ) + p d ∫ g(s, t )ds 0 dt 1 p−q ω(v p−q (t)) p that is, v '(t ) − 1 p−q p−q F1 '(t )v(t ) ≤ F2 '(t )ω(v p−q (t )) p p (1 2) Multiplying exp(− p − q F1 (t )) on both sides of (12), we have p d[exp(− p−q F1(t))v(t)] 1 p−q p−q p ≤ F2 '(t)ω(v p−q (t))exp(− F1(t)) dt p p (13) Integrating (13) from 0 to t , it follows exp(− t p −q p−q p−q F1(t))v(t) − v(0) ≤ ∫ F2 '(s)ω(v (s))exp(− F1 (s))ds p p p 0 1 p−q (14) 1 p−q p−q p −q F2 '(s)ω(v p−q (s))exp(− F1(s))ds}exp( F1(t)) p p p (15) p −q p Since v(0) = u p − q (0) = a (T ) , we have p−q t v(t) ≤ {a p (T ) + ∫ 0 and t p−q 1 1 p−q p−q u(t) ≤ exp( F1(t)) {a p (T ) + ∫ F2 '(s)ω(u(s))exp(− F1(s))ds}p−q p p p 0 (16) t p−q Let k(t) = a p (T ) + ∫ 0 1 p−q p−q F2 '(s)ω(v p−q (s))exp(− F1(s))ds . p p Then 1 1 u (t ) ≤ exp( F1 (t ))k p−q (t ) p (17) and k '(t) = 1 p−q p−q p−q 1 p−q F2 '(t)ω(u(t))exp(− F1(t)) ≤ F2 '(t)ω(exp( F1(T ))k p−q (t))exp(− F1(t)) p p p p p Then k '(t) p−q p−q ≤ F2 '(t)exp(− F1(t)) 1 1 p p p−q ω(exp( F1(T ))k (t)) p (18) Integrating (18) from 0 to t , it follows Ω[exp( t p−q p−q p−q p−q p −q p −q F1 (T ))k (t )] −Ω[exp( F1 (T ))a p (T )] ≤ exp( F1(T ))∫ F2 '(s)exp(− F1(s))ds p p p p p 0 (19) t p−q q− p p−q p−q p−q p−q F1(T ))Ω−1[Ω(exp( F1(T ))a p (T )) + exp( F1(T ))∫ F2 '(s)exp(− F1(s))ds] p p p p p 0 (20) Then k(t) ≤ exp( From (17) and (20), it follows u(t) ≤ {Ω−1[Ω(exp( t p−q 1 p −q p−q p−q p −q F1(T ))a p (T )) + exp( F1(T ))∫ F2 '(s)exp(− F1(s))ds]}p−q , t ∈(0,T ] p p p p 0 (21) Combining (5), (9), (21) we have x(t) ≤ {Ω−1[Ω(exp( t p−q 1 p −q p−q p−q p −q F1(T ))a p (T )) + exp( F1(T ))∫ F2 '(s)exp(− F1(s))ds]}p−q , t ∈(0,T ] p p p p 0 (22) Setting t = T and considering T ∈ R+ is arbitrary, we have completed the proof. Corallary 2.2: Assume that x, a, f , g ∈ C(R+ , R+ ), m ∈ C1 (R+ , R+ ) . If x, a, p, q,σi (t ),α,φ, ω are the same as in Theorem 2.1, and x(t) satisfies the following delay integral inequality: t x p (t ) ≤ a(t ) + m(t )∫ [ f (s) x p (σ1 (s)) + g (s) xq (σ 2 (s))ω( x(σ 3 (s)))]ds, t ∈ R+ (23) 0 with the initial condition (2), then for t ∈ R+ , t p−q p−q p−q p−q x(t ) ≤ {Ω −1[Ω(exp( F1 (t ))a (t )) + exp( F1 (t )) ∫ F2 '( s ) exp( − F1 ( s )) ds ]} p p p p 0 p −q p 1 p −q (24) where t t 0 0 (25) F1 (t ) = m (t ) ∫ f ( s ) ds , F2 (t ) = m (t ) ∫ g ( s ) ds 3. Application Example: We consider the following delay differential equation ( x p (t )) ' = F (t , x (σ 1 (t )), x(σ 2 (t )), x(σ 3 (t ))) (26) with the initial condition 1 x(t ) = φ (t ), t ∈ [α , 0],| φ (σ i (t )) |≤ A p , t ∈ R+ , σ i (t ) ≤ 0, i = 1, 2,3 where A>0 is a constants and A = x p (0) , F ∈ C( R+ × R3 , R) ,. σi (t) ∈C(R+ , R) , σi (t) ≤ t, −∞< α = inf{min{σi (t), i =1,2,3}, t ∈R+} ≤ 0, φ ∈C([α,0], R+ ) , Assume | F(t, x, y, z) |≤ f (t) | x | p +g(t) | x |q v(| z |) ∞ where f , g , v ∈ C ( R+ , R+ ) , v is nondecreasing, and v ( s ) ds = ∞ . p > q > 0 ∫ 1 Integrating (26) from 0 to t , it follows t x p (t ) − x p (0) = ∫ F ( s , x (σ 1 ( s )), x (σ 2 ( s )), x (σ 3 ( s ))) ds o So t | x p (t ) | − | x p (0) |≤| x p (t ) − x p (0) | ≤ | F ( s , x (σ 1 ( s )), x (σ 2 ( s )), x (σ 3 ( s ))) | ds ∫ o t ≤ ∫ [ f ( s ) | x (σ 1 ( s )) | p + g ( s ) | x (σ 2 ( s )) |q v (| x (σ 3 ( s )) |)]ds o Taking ω (u ) = v (u ) , from Theorem 2.1, we can reach the estimate | x(t ) |≤ {Ω −1[Ω(exp( t p−q 1 p−q p−q p−q p−q F1 (t )) | A | p ) + exp( F1 (t )) ∫ F2 '( s ) exp( − F1 ( s )) ds ]} p−q p p p p 0 where t t 0 0 F1 (t ) = ∫ f ( s ) ds, F2 (t ) = ∫ g ( s ) ds , which shows x(t ) does not blow up in finite time. 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