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1530 JOURNAL OF COMPUTERS, VOL. 8, NO. 6, JUNE 2013 The derivative and the forward jump operator are related by the useful formula σ f f μ f Δ σ = + , where f : = f σ. We will also make use of the following product and quotient rules for the derivative of the product f g and the quotient f g( gg σ ≠ 0) of two differentiable functions f and g : Δ ⎛ f ⎞ f g− f g ( f g) Δ = f Δ g+ f σ g Δ , and ⎜ ⎟ = ⎝ g ⎠ gg σ Δ Δ . (6) By using the product rule, the derivative of f () t = ( t − α) m for m ∈ andα ∈ T can be calculated as m−1 Δ v m−v−1 () = ( σ() −α)( −α) . v= 0 f t ∑ t t (7) For a, b∈ T and a differentiable function f , the Cauchy integral of f Δ is defined by b Δ ∫a f () t Δ t = f( b) − f( a) . The integration by parts formula follows from (6) and reads b Δ b b σ Δ ∫ f () tgt () Δ t= ftgt () ()| −∫ f tg() tΔt. a To prove our main results, we will use the formula 1 1 ( x γ Δ ( t)) [ hx σ (1 h) x] γ − Δ = γ + − dhx ( t) 0 a a ∫ . (8) which is a simple consequence of Keller′s chain rule [2]. Also, we need the following lemma [5]. Lemma 1 Assume A and B are nonnegtive constants, λ > 1, then λ − − ≤( − 1) . 1 AB λ A λ λ B λ The reader is referred to [2] for more detailed and extensive developments in calculus on time scales. III. MAIN RESULTS First, we state the oscillation criteria for (4). Set yt () = xt () + ctx () ( τ ()). t (9) Theorem 1 Assume that (H1) - (H5) hold, Furthermore, suppose that there exists a positive Δ−differentiable function g() t such that for all sufficiently large T ∗ , and ∗ for all δ (T) > T , we have t ∗ limsup ∫ ( β (, sT ) gsQs () () − t→∞ T α( s)(( g ( s)) ) ( γ + 1) g ( s) Δ γ + 1 + γ+ 1 γ ) Δ s =∞. Then every solution of (4) is oscillatory on [ t , ∞) 0 T . (10) proof Suppose (4) has a nonoscillatory solution x (t). without loss of generality, there exists some t 1 ≥ t 0 , sufficiently large such that xt () > 0, x( τ ( t)) > 0, x( δ ( t)) > 0 for all t ≥ t . Hence In the view of (9), by (H1) we 1 get yt () > 0. from (4) and by (H2) - (H5), we have that Δ ( y )) γ γ α Δ ≤ −( q() t − p()) t x ( δ()) t < 0, and using the same proof of Theorem 1 [4], there exists t ≥ t such that for all t ≥ t , we have 2 1 2 Δ ⎧yt () > 0, y () t > 0, (11) ⎨ Δ ( ( y ) γ Δ γ ⎩ α ) ≤ −( q( t) − p( t))(1 − c( δ( t))) y ( δ( t)) < 0. By the definition of Qt (), we get Δ ( ( y ) γ γ α ) Δ ≤ − Q( t) y ( δ( t)) < 0. (12) Make the generalized Riccati substitution Δ γ α()( t y ()) t wt () = gt () . (13) γ y () t By the product and quotient rules, we have for all t ≥ t2 Δ γ Δ Δ gt ()( α()( t y())) t ⎛ gt () ⎞ Δ γ w () t = ( ()( t y ())) t γ +⎜ γ ⎟ α y () t ⎝ y () t ⎠ Δ γ Δ gt ()( α()( t y())) t = + γ y () t Δ γ Δ g () t g()( t y ()) t Δ γ σ ( − )( α( t)( y ( t)) ) . γσ γ γσ y () t y () t y () t From (12) - (14), we obtain Δ Δ ⎛ y( δ ( t)) ⎞ g ( t) σ w () t ≤− g() t Q() t ⎜ ⎟ + w () t σ ⎝ yt () ⎠ g () t σ γ Δ gtw () ()( t y()) t − σ γ . g () t y () t γ Δ σ (14) (15) First consider the case when δ () t ≥ t . For all large t, Δ from y () t > 0, we have which implies that y( δ ( t)) ≥ 1 , yt () Δ σ γ Δ Δ g () t σ g() t w ()( t y ()) t w () t ≤− g() t Q() t + w () t − . (16) σ σ γ g () t g () t y () t Next consider the case when δ () t ≤ t, for all large t. By using α( y Δ ) γ is strictly decreasing on [ t 2 , ∞ ) , we can choose t ≥ t such that δ () t ≥ t , for t ≥ t . Then we 3 2 2 3 obtain © 2013 ACADEMY PUBLISHER
JOURNAL OF COMPUTERS, VOL. 8, NO. 6, JUNE 2013 1531 and hence Δ γ 1 γ t ( α( sy ) ( s)) ) y() t − y( δ ()) t = ∫ Δs δ () t 1 γ α () s ≤ t y t Δs α () s Δ γ 1 γ t ( αδ ( ( ))( ( δ( ))) ) ∫ , δ () t 1 γ Δ γ 1 γ yt () ( αδ ( ())( t y ( δ()))) t t Δs ≤ 1+ ∫ . (17) δ () t 1 γ y( δ( t)) y( δ( t)) α ( s) Also, for t ≥ t , we can see that 3 ≥ 1 γ Δ γ δ () t y( δ( t)) > y( δ( t)) − y( t ) = s 2 ∫ Δ t2 1 γ Δs α () s 1 γ Δ γ δ () t ( αδ ( ( t))( y ( δ( t))) ) ∫ , t2 1 γ and therefore ( αδ ( ( ))( ( δ( ))) ) ( α( sy ) ( s)) ) α () s 1 γ Δ γ t y t δ () t s ≤ ⎜∫t2 1 γ y( δ( t)) α ( s) From (17) and the above inequality, we have ⎛ ⎝ Δ ⎞ ⎟ ⎠ −1 yt () t Δs δ () t Δs −1 ≤ ∫ ( ) t2 1 γ ∫ , (18) t2 1 γ y( δ( t)) α ( s) α ( s) therefore we get the desired inequality y( δ ( t)) ≥ ρ(, tt), for t ≥ t . (19) 2 3 yt () Using (19) in (15), when δ () t ≤ t, we get Δ Δ γ g () t σ w () t ≤− ρ (, t t ) g() t Q() t + w () t 2 σ g () t σ γ Δ gtw () ()( t y()) t − σ γ . g () t y () t From (16), (20) and the definition of β (, tt) , we have 2 By (8), we obtain Δ Δ g () t σ w () t ≤− β (, t t ) g() t Q() t + w () t 2 σ g () t σ γ Δ gtw () ()( t y()) t − σ γ . g () t y () t y t γ hy h y dh y t γ Δ 1 σ γ−1 Δ ( ( )) = ∫ [ + (1 − ) ] ( ) 0 σ γ−1 Δ ⎧γ( y ( t)) y ( t),0< γ ≤1, ≥ ⎨ ⎩γ yt y t γ ≥ γ −1 Δ ( ( )) ( ), 1. Since α( y Δ ) γ is strictly decreasing on[ t 2 , ∞ ), we get γ ( y ( t)) Δ ⎧γα t y t y t ⎪ α () t ≥ ⎨ ⎪ γα t y t y t ⎪ ⎩ α () t ( σ 1 γ ( )) ( σ γ−1 ( )) ( Δ σ ( )) 1 γ ( σ 1 γ γ−1 ( )) ( ( )) ( Δ σ ( )) 1 γ . ,0< γ ≤1, , γ ≥ 1. (20) (21) From the last inequality and (21), if 0< γ ≤ 1, we have Δ Δ g () t σ w () t ≤ − β (, t t ) g() t Q() t + w () t − 2 σ g () t σ 11 + γ σ γ gt ()( w()) t ⎛ y () t ⎞ 1 γ σ 1+ 1 γ ⎜ ⎟ α ()( t g ()) t ⎝ y() t ⎠ whereas if γ > 1 , we find that Δ Δ g () t σ w () t ≤− β (, t t ) g() t Q() t + w () t 2 σ g () t γ gt w t y t − α ()( t g ()) t y() t σ 11 + γ σ ()( ()) () 1 γ σ 1+ 1 γ . Δ And by using y () t > 0, we obtain that Δ Δ g () t + σ w () t ≤− β (, t t ) g() t Q() t + w () t 2 σ g () t γ gt () − α ()( t g ()) t 1 γ σ λ γ , σ λ ( w ( t)) , where λ : = ( γ + 1) γ . Define A ≥ 0 and B ≥ 0 by σ λ 1( γ +1) Δ λ γ gt ()( w()) t λ− 1 α ()( t g ()) t + A : = , B : = , 1 γ σ λ 1 λ α ( t)( g ( t)) λ( γg( t)) then using Lemma 1, we obtain (22) g Δ 1 () t () ()(( ())) () ( ()) t g Δ t γ + σ gt σ λ α + γ + w t − w t ≤ . σ 1γ σ λ γ+ 1 γ g () t α ()( t g ()) t ( γ + 1) g () t From the last inequality and (22), we have Δ γ + 1 Δ α()(( t g ())) t + w () t ≤ −β (, t t ) g() t Q() t . γ+ 1 γ 2 ( γ + 1) g ( t) Integrating both sides from t 3 to t, we get Δ γ + 1 t α( s)(( g ( s)) ) + ∫ [ β ( s, t ) g( s) Q( s) − ] Δs t3 2 γ+ 1 γ ( γ + 1) g ( s) ≤ wt ( ) −wt ( ) ≤ wt ( ), 3 3 which leads to a contradiction to (10). This completes the proof. Corollary 1 Assume that (H1) - (H5) hold, Furthermore, suppose that for all sufficiently large T ∗ , and for δ ( T ) > T ∗ , we have t ∗ α() s limsup ∫ ( sβ ( s, T ) Q( s) − ) Δ s =∞. t→∞ T γ+ 1 γ ( γ + 1) s Then every solution of (4) is oscillatory on [ t , ∞) 0 T . Corollary 2 Assume that (H1) - (H5) hold, Furthermore, suppose that for all sufficiently large T ∗ , and for δ ( T ) > T ∗ , we have t ∗ limsup ∫ β ( sT , ) Qs ( ) Δ s=∞. t→∞ T © 2013 ACADEMY PUBLISHER
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Journal of Computers ISSN 1796-203X
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Corn Moisture Measurement using a C
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Call for Papers and Special Issues
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(Contents Continued from Back Cover
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