equations with difference kernel k •••• t k •••• t

... order to obtain the ultradiscrete limit, we start with an equation for x, introduce X through x = eX/ , and then appropriately take the limit → Clearly, the substitution x = eX/ requires x to be ... reverse transcription inhibitors) The idea behind this strategy is that it should take much longer for the virus to evolve to a form resistant simultaneously to more than one toxic factor This rationale ... in detail the differential equations obtained We start by putting x = T, y = V , and z = I, where is a parameter which goes to at the continuous limit and is related to the time step through t =...

Ngày tải lên: 23/06/2014, 00:20

Ngày tải lên: 27/12/2013, 15:28

Ngày tải lên: 27/12/2013, 15:28

... such that x∗ = and x∗ L0 fk k k nk nk−1 +1 = L0 fk lp (nk−1 +1,nk ;Kq ) We define a sequence of causal operators Ak ∈ L(lp (0, ∞; Kq), lp(0, ∞; Ks)) by Ak h = fk+1 L0 fk lp (nk−1 +1,nk ;Kq ) The ... y(·) is a solution of the system with the initial condition y(0) = Put h(n) := En (Ag)(n) It is easy to see that h(n) has a compact support Substituting into the first one we obtain y(n + 1) = Bn ... Hence, it follows that y(·) lp(0,∞;Kd ) < ∞ That contradicts to y(·) ∈ lp (0, ∞; Kd) This means that (4) is not globally stable Summing up we obtain Theorem 3.4 For lp −stability, the complex stability...

Ngày tải lên: 22/03/2014, 11:20

... holds Thus we can deduce the conclusion in terms of Theorem 3.1 Remark 3.3 If Hk(x) = x, then Equation becomes difference equations with distributed delays without impulses in vector form ∞ k g ... H1mk (x1 (mk ), x2 (mk )), x2 (mk + 1) = H2mk (x1 (mk ), x2 (mk )) (26) and m1 = 4, mk = mk-1 + k for k = 2, 3, One can check that all the properties given in (H) are satisfied provided that

Ngày tải lên: 21/06/2014, 03:20

... we get Δwk − wk − A k k k2 k k wk 1 k by wk 3.11 0, that is, Δwk wk − 2 k k wk k k k Gw k 0, 3.12 Advances in Difference Equations where G is defined by Gw k k k2 k k wk 1 k wk k wk k − ... results k zk zk c1 k yk c1 yk c1 k yk /yk yk /xk c2 k xk /xk k xk ∼ c1 yk /xk c2 xk c2 k xk c2 xk 3.3 Now we prove the main statement Necessity Let y ∈ NRV be a solution of 1.1 positive for k ≥ ... lim k k/yk xk lim k kzk k 1 j a lim k 1/ yj yj k lim 1 Δzk zk k zk k Δzk 1/ yk yk 1 − 3.39 10 Advances in Difference Equations Hence, − xk ∼ k/ yk k1 − /Lk , that is, xk ∼ k1 − Lk , where Lk 1/ − Lk...

Ngày tải lên: 21/06/2014, 11:20

... But it is difficult to find the analytical solution to 1.4 - 1.5 , which makes numerical solution important To authors’ knowledge, the finite difference method to dissipative SRLWEs with damping term ... convergent Figures and show that the height of wave crest is more and more low with time elapsing due to the effect of damping and dissipativeness It simulates that the continue energy E t of problem ... t is decreasing with respect to t, which implies that E t u 2 ux 2 ρ 2 ≤ E , L L L t ∈ 0, T Then, it indicates that u L2 ≤ C, ux L2 ≤ C, and ρ L2 ≤ C It is followed from Sobolev inequality that...

Ngày tải lên: 21/06/2014, 11:20

... ΔψN−1 k − k1 2 k2 1 ψN−1 k − k1 2 ΔϕN−1 k k12 k2 2 ϕN−1 k k11 k2 2 ψN−1 k − k1 2 k2 1 ψN−1 k − k1 2 ΔϕN−1 k k12 k2 2 ϕN−1 k − k1 2 ΔϕN−1 k 3.13 ψN−1 k Hence, f k k22 ϕN−1 k − k1 2 ΔϕN−1 k Noting ... k − k1 2 ΔϕN−1 k By the definition of f λ , 3.11 , and det K k12 f k k12 k2 2 ϕN−1 k ψN−1 k k12 3.12 1, k1 2 k1 1 − k1 2 ΔψN−1 k − k1 2 k2 1 − k2 2 ψN−1 k − k1 2 ΔϕN−1 k k12 k2 2 ϕN−1 k k11 k1 2 ... that is, ϕN−1 k k11 − k1 2 , ΔϕN−1 k k21 − k2 2 , k1 2 , ΔψN−1 k k22 ψN−1 k 3.40 Advances in Difference Equations 15 Differentiating f λ with respect to λ two times, we get f k k22 ϕN−1 k k11...

Ngày tải lên: 21/06/2014, 20:20

... |x2 t − | − 15 x2 t − x2 t − t −∞ x2 t − 15 e−6 t s |x1 s |ds sin x1 t − t |x1 t − | 15 −∞ , m / mk , x2 t − e−12 t s |x2 s |ds 4.1 with x1 tk α 1k x1 t − β 1k x2 t k k x2 tk β 2k x1 t k 4.2 α 2k ... t ≤ eγ t1 t0 · · · eγ ≤ κ0 zeγ t t0 κ0 ze− λ−γ tk−1 −tk−2 e−λ t t0 t t0 κ0 ze−λ t t0 σ0 · · · k 1 N eσ N eσ N, 3.35 t ∈ tk−1 , tk , k 1, 2, This implies that the conclusion of the theorem ... ≤ Rk x t k ≤ Rk γ0 · · · k 1 κ0 ze−λ tk t0 ≤ γ0 · · · k 1 k κ0 ze−λ tk t0 σ0 · · · k 1 N 3.28 σ0 · · · k 1 k N This, together with 3.27 , leads to x t ≤ γ0 · · · k 1 k κ0 ze−λ t t0 ...

Ngày tải lên: 21/06/2014, 20:20

... quotient in 2.11 k is negative This contradicts the required condition for fj ∈ MHSs is complete j, m, α and so the proof Distortion bounds and extreme points k In this section, we will obtain distortion ... starlike functions,” Bulletin of the Korean Mathematical Society, vol 37, no 2, pp 291–301, 2000 K Al-Shaqsi and M Darus, “On subclass of harmonic starlike functions with respect to K- symmetric ... ρ t ct Proof According to the definition of t , we can write −1 z t z ∞ ρ n t j k Further, since fj ,t z are in MHSs −1 n αΦn ct t ∞ n 4.5 1, 2, , ρ Then by 2.7 , we have n − αΦn t ρ t ct...

Ngày tải lên: 22/06/2014, 02:20

... 2.2 It is clear that the stability of the trivial solution of 2.2 is equivalent to the stability of the equilibrium point of 2.1 Together with nonlinear equation 2.2 we will consider and its linear ... nonlinearity of 2.2 is more than 1, then obtained stability conditions at the same time are 47–49 conditions for stability in the probability of the trivial solution of nonlinear equation 2.2 and therefore ... Computers in Simulation, vol 45, no 3-4, pp 269–277, 1998 40 L Shaikhet, “Stability of predator-prey model with aftereffect by stochastic perturbation,” Stability and Control: Theory and Applications,...

Ngày tải lên: 22/06/2014, 11:20

... is attractive if for any ϕ k a solution tends to zero limn → ∞ x n asymptotically stable if it is both stable and attractive One of the methods to establish stability of difference equations is ... contrast to many other stability tests, we consider the case when the sum of coefficients m ak n or some of its subsum is allowed to be in the interval 0, , not just k 0, We illustrate our results ... demonstrate that in the case when the arguments tend to infinity but the delays are not bounded and all other conditions of Theorem 2.18 are satisfied, this does not imply exponential stability The equation...

Ngày tải lên: 22/06/2014, 11:20

... located at the right, that is, at the final values For the jth-order asymptotic approximation, j ≥ 1, we agree that (l) yk ≡ 0, l < 0, ≤ k ≤ N, (2.3) to give a compact writing of the resulting equations ... asymptotic solution In this section, we develop a perturbation method to obtain asymptotic approximate solutions for the whole order This iterative method facilitates a considerable reduction ... zeroth-order asymptotic approximation, the resulting equations are given by (0) yk = k , k = 0,1, ,n − 1, (0) (0) (0) an ,k yk+n + an−1 ,k yk+n−1 + · · · + a0 ,k yk = fk , (0) y N k = β k , k =...

Ngày tải lên: 22/06/2014, 22:20

... introduction and some notations Section is devoted to the statement of the main results (and to some comments on them) The proofs of the main results will be given in Section Introduction and notations ... remain valid without the restriction that the root λ0 of the characteristic equation (2.8) satisfies λ0 ≤ Such a restriction is not a necessity in the non-neutral case (i.e., in Theorem 3.9, Corollary ... characteristic equation (2.9) to have at least one positive root (II) The characteristic equation (2.9) has no positive roots greater than or equal to a + (III) Let a > −1 and let there exist a...

Ngày tải lên: 22/06/2014, 22:20

... + d Then BVP (1.1) has at least two positive solutions x1 and x2 such that a < max x1 (t) < b < max x2 (t) < c t ∈[0,h] t ∈[0,h] (2.16) The proof is omitted since it is similar to that of Theorem ... 2 Positive solutions of difference equations Recently, the existence of positive solutions of finite difference equations with different boundary value conditions is investigated in [1–5] ... at least two positive solutions The proof is complete 8 Positive solutions of difference equations Theorem 2.5 Let the following conditions be satisfied: (F) f0 < k3 p−1 ; (G) there exists a p2...

Ngày tải lên: 22/06/2014, 22:20

... from then the theory of functional difference equations with infinite delay has drawn the attention of several authors (see [6–12, 16–18]) The fact that the state space for functional difference equations ... C is a suitable constant Remark 4.5 Let (α, K, P(n)) be the data of an exponential dichotomy Now we turn our attention to the following question: how one can constructing other projectors P(n) ... f (0) = This paper deals with maximal regularity for functional difference equation with infinite delay under the hypothesis that the solution operator (1.1) has an exponential dichotomy The problem...

Ngày tải lên: 22/06/2014, 22:20

... integer lk such that lk ω ≤ nk ≤ (lk + 1)ω Set nk = lk ω + k Then ≤ k < ω for all k ≥ Since { k }k 1 is bounded set, we can assume that, taking a subsequence if necessary, k = j∗ for all k ≥ 1, ... almost periodic solutions of nonlinear functional difference equations with finite delay via uniform stability properties of a bounded solution This motivates us to investigate almost periodic solutions ... set uk (n) = u(n + nk ) Notice that un+nk ( j) = u(n + nk + j) = uk (n + j) = uk ( j) and n hence, un+nk = uk Thus, n uk (n + 1) = u n + nk + = F n + nk ,un+nk = F n + nk ,uk = F n + j∗ ,uk...

Ngày tải lên: 22/06/2014, 22:20

... integrating (1.14) from t = t0 to t = T, by virtue of (1.15), we have T t0 E V t + h0 − V (t) dt ≤ −c2 T T E x (t) dt + t0 t0 γ (t) dt (1.28) γ (t) dt (1.29) or T+ h0 T EV (t) dt − t0 +h0 t0 T EV (t) dt ... m=0 then EB (t) DAX (t)  k = Eb (t)    k am x (t − m) dl ,k+ 1 x (t − k + l) + dk+1 ,k+ 1 m=0 l=1 k 1 = Eb (t)   am dk+1 ,k+ 1 + dk−m ,k+ 1 x (t − m) + ak dk+1 ,k+ 1 x (t − k)  (3.13) m=0 k Qkm Eb (t) x (t − ... Eb (t) x (t − m), = dk+1 ,k+ 1 m=0 where Qkm = am + dk−m ,k+ 1 , dk+1 ,k+ 1 m = 0, ,k − 1, Qkk = ak (3.14) Note that k k Qkm Eb (t) x (t − m) = m=0 k Qkm Eη (t + 1)x (t − m) + EF2 (t) m=0 Qkm x (t − m), (3.15)...

Ngày tải lên: 23/06/2014, 00:20

... (2.4) Set y (t) = λ t/ σ x (t) for t ≥ −τ (2.5) Then, by taking into account the fact that τ j = m j σ ( j = 1, ,k) and using the hypothesis that λ0 is a (positive) root of the characteristic equation ... j λ0 k −m j y (t − σ) − j =1 b j λ0 j =1 So, the fact that x satisfies (1.1) for t ≥ is equivalent to the fact that y satisfies k y (t) − y (t − σ) = − −m j b j λ0 + λ t/ σ f (t) y (t − σ) − y t − τ ... difference equations (with discrete variable), see Driver et al [3] and Pituk [13, 14] Motivated by the results in [4] as well as by those in the above-mentioned papers, we here make a first attempt to...

Ngày tải lên: 23/06/2014, 00:20

... such that x∗ = and x∗ L0 fk k k nk nk−1 +1 = L0 fk lp (nk−1 +1,nk ;Kq ) We define a sequence of causal operators Ak ∈ L(lp (0, ∞; Kq), lp(0, ∞; Ks)) by Ak h = fk+1 L0 fk lp (nk−1 +1,nk ;Kq ) The ... y(·) is a solution of the system with the initial condition y(0) = Put h(n) := En (Ag)(n) It is easy to see that h(n) has a compact support Substituting into the first one we obtain y(n + 1) = Bn ... Hence, it follows that y(·) lp(0,∞;Kd ) < ∞ That contradicts to y(·) ∈ lp (0, ∞; Kd) This means that (4) is not globally stable Summing up we obtain Theorem 3.4 For lp −stability, the complex stability...

Ngày tải lên: 21/07/2014, 16:21