họat động 1 tìm hiểu nội dung bài học
ONE PARAMETER FAMILY OF LINEAR DIFFERENCE EQUATIONS AND THE STABILITY PROBLEM FOR THE NUMERICAL pdf
... [(k + 1) !]2 z +1 (z − 1) k +1 Lk +1 (2k + 2)! z 1 (3.6) By using (3.3) in the previous relation we get σk +1 (z) = (k + 1) ! 2k + z + z +1 k z +1 (z − 1) k +1 Lk L k 1 − (2k + 2)! k +1 z 1 z 1 k +1 z 1 (3.7) ... (2 .11 ) and by using (2 .13 ) Moreover, (2) and (3) are easy consequence of (2 .14 ) Concerning item (4), from (2 .14 ) we obtain ρk 1 (z) = zk 1 ρk z 1 −zk ρk z 1 = = zk 1 ρk 1 z 1 z 1 z 1 − (2 .17 ) ... 1 j =0 α(k 1) z j j (3 .13 ) 4 − k −2 α(k 1) z j + j k 1 j =0 α(k 1) z j j Then, from (2 .11 ) we deduce that 1 ρk +1 (z) = zρk (z) + ρk (z) − z2 ρk 1 (z) − 2zρk 1 (z) + ρk 1 (z) 2 4 − k −2 z+1...
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báo cáo hóa học:" Research Article Existence of Positive Solutions of Nonlinear Second-Order Periodic Boundary Value Problems" ppt
... vol 13 0, no 11 , pp 3325–3333, 2002 14 R Ma, “Existence of positive solutions of a fourth-order boundary value problem,” Applied Mathematics and Computation, vol 16 8, no 2, pp 12 19 12 31, 2005 15 ... 0, u 1, 1. 13 1 1 .14 and let v be the unique solution of the initial value problem v at v 0, v T 0, v T Let α: v −u T 1. 15 Boundary Value Problems Then the Green’s function K of 1. 1 , 1. 2 given ... Mathematics, vol 13 2, no 2, pp 3 41 356, 20 01 12 E N Dancer, “Global solution branches for positive mappings,” Archive for Rational Mechanics and Analysis, vol 52, pp 18 1 19 2, 19 73 13 M Zhang and...
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báo cáo hóa học:" Nodal solutions of second-order two-point boundary value problems" potx
... Proof of Theorems 1. 1 and 1. 2 We only prove Theorem 1. 1 since the proof of Theorem 1. 2 is similar ν It is clear that any solution of (2.4) of the form (1, u) yields a solution u of (1. 4) We shall ... manuscript 14 Acknowledgement The authors were very grateful to the anonymous referees for their valuable suggestions This study was supported by the NSFC (No 11 0 610 30, No 10 9 710 87) and NWNU-LKQN -10 - 21 ... reaction diffusion systems Diff Int Equ 7, 14 27 14 52 (19 84) 15 [10 ] Guo, D, Lakshmikantham, V: Nonlinear Problems in Abstract Cones Academic press, New York (19 88) [11 ] Anane, A, Chakrone, O, Monssa,...
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Báo cáo hóa học: " Research Article Multiplicity of Positive Periodic Solutions of Singular Semipositone Third-Order Boundary Value Problems" doc
... 0, R1 Choose n1 ∈ N0 such that 1/ n1 < R1 and let N1 < αn un t − t − ω > max L, ρ2 r {n1 , n1 ρH , 3 .14 1, } For n ∈ N1 , let L L ≤ max un t − t ρ2 ρ βn 3 .15 First we claim that βn > R1 for ... 99, no 1, pp 10 9 11 4, 19 87 K Lan and J R L Webb, “Positive solutions of semilinear differential equations with singularities,” Journal of Differential Equations, vol 14 8, no 2, pp 407–4 21, 19 98 J ... such that R1 1/ n F ξn y , Jy dy ≤ η 3.25 Boundary Value Problems On the other hand, by G3 we can choose n2 ∈ N1 large enough such that R1 R1 F ξn y , Jy dy ≥ 1/ n g0 Jy dy > η 3.26 1/ n for all...
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CONSTRUCTION OF UPPER AND LOWER SOLUTIONS FOR SINGULAR DISCRETE INITIAL AND BOUNDARY VALUE PROBLEMS potx
... satisfies −∆2 u(k − 1) = u(k), k ∈ N = {1, 2, ,T }, u(0) = u(T + 1) = 0, (1. 11) H L¨ and D O’Regan 207 u then u(k) ≥ µ(k) u for k ∈ N + ; (1. 12) here T +1 k k , T +1 T µ(k) = (1. 13) Lemma 1. 5 [8] Let ... + 1, ,b} ⊂ N If u ∈ C(N + , R) satisfies ∆ ϕ p ∆u(k − 1) u(a − 1) ≥ 0, k ∈ [a,b], ≤ 0, (1. 14) u(b + 1) ≥ 0, then u(k) ≥ for k ∈ [a − 1, b + 1] = {a − 1, a, ,b + 1} ⊂ N + In Theorems 1. 2 and 1. 3 ... =0 k0 k0 ϕ 1 p k0 1 j =1 q(k) k= j +1 ϕp (2 .15 ) q(k) k= j Similarly, we sum (2 .11 ) from k0 to j (k0 ≤ j ≤ T + 1) to obtain j −ϕ p ∆u( j) = −ϕ p ∆u k0 − + h(M) q(k) for j ≥ k0 (2 .16 ) k=k0 Now...
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báo cáo hóa học:" Research Article Exponential Stability and Estimation of Solutions of Linear Differential Systems of Neutral Type with Constant Coefficients" pot
... , 1 0 .1 A 0 .1 1 2, τ , 1, B 0 .1 0 0 .1 , 4 .1 that is, the system x1 t ˙ 0.5x1 t − − x1 t ˙ x2 t ˙ 0.5x2 t − ˙ with initial conditions 1. 2 Set β G1 0 , 0.1x2 t 0.1x1 t − , 0.1x1 t − x2 t 0.1x2 ... Problems 17 that is, the system 0.1x1 t − − 3x1 t − 2x2 t ˙ x1 t ˙ 0.6 213 x2 t − , 4.8 0.1x2 t − ˙ x2 t ˙ with initial conditions 1. 2 Set β 0.5 0 .1 G1 0 .1 0 .1 , 0.6 213 x1 t − , 1x1 t 0 .1 and G2 0 .1 0 ... 5.0033 2.5058, 1. 9967 3. 414 2 1. 7099, 1. 9967 1. 1, B γ0 0 .1, 1 G1 , H 0 .14 45 5.0033 0 .1, D 0.5008, 1. 9967 0.5, DA B 0.0289, 4.5 0 .1, 0.0289 0.45, M 2.0266 Since γ0 < 2/τ ln 1/ D 1. 3863, all...
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Báo cáo hóa học: "Research Article Existence and Uniqueness of Solutions for Higher-Order Three-Point Boundary Value Problems" ppt
... x1 t1 − β n t1 n−2 ≥ f t1 , w0 t1 , x1 t1 , , wn−2 t1 , x1 n−2 x1 n−2 t1 − wn−2 t1 , x1 n 1 φ x1 t1 n−2 ≥ f t1 , β t1 , , β n 1 t1 n 1 , x1 t1 x1 t1 t1 − f t1 , β t1 , , β n 1 t1 n 1 ... n t1 λf t1 , w0 t1 , x t1 , , wn−2 t1 , x n−2 t1 x n−2 t1 − λwn−2 t1 , x n−2 t1 , x n 1 t1 φ x n 1 t1 λf t1 , w0 t1 , x t1 , , wn−3 t1 , x n−3 t1 3 .13 , β n−2 t1 , x n−2 t1 − λβ n−2 t1 φ ... n 1 t1 − β n−2 t1 φ x1 3.40 t1 − f t1 , β t1 , · · · , β n 1 t1 n−2 x1 Case t1 n 1 t1 − β n−2 t1 φ x1 t1 > a In this case, we have n−2 max x1 t∈ a,c t − ψn−2 t n−2 : x1 a − β n−2 a > 0, 3. 41 10...
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Báo cáo hóa học: "Research Article Asymptotic Representation of the Solutions of Linear Volterra Difference Equations" pdf
... − i 1 λj < 1, i , j j 0, 1, 5.74 0, 5.75 i It follows from 5.74 , the mean value inequality and 5.75 that λ0 − 1 ≤ ∞ max iK i i max < max 1 , λ0 1 1 , λ0 1 1 , |λ0 | | 1 | i 1 1 − λ0 1 −2 ... Now, let m 1, 1 : l ∈ {1, } and K l / Then 1 and 4 .15 reduce to 1/ l A> l K l 1/ l A
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Báo cáo hóa học: "EXISTENCE AND MULTIPLICITY OF SOLUTIONS FOR SOME THREE-POINT NONLINEAR BOUNDARY VALUE PROBLEMS" pptx
... (t) ≤ t η q1,τ (s) pη,τ (1) a(s)x(s)ds pη,τ (1) + ≤ ≤ ≤ t pη,τ (1) q1,τ (t) + γ pη,τ (t) q1,τ (s) a(s)x(s)ds+ Lτ x (η) q1,τ (η) q1,τ (η) η q1,τ (s)a(s)x(s)ds+ Lτ x (η) q1,τ (η) (1 − t)/ (1 − η) +γ ... a(s)x(s)ds p0,τ (1) + p0,τ (η) p0,τ (1) − γ p0,τ (η) q1,τ (s) a(s)x(s)ds q1,τ (0) η p0,τ (1) p0,τ (1) − γ p0,τ (η) ≥ (2 .17 ) η p0,τ (η) q1,τ (s) a(s)x(s)ds q1,τ (0) η q1,τ (s)a(s)x(s)ds (2 .19 ) X Xian ... − xn (1) xn (η) (1 − γ) zλ (η) (1 − γ) ds ≥ ≥ ≥ g(s) g xn (1) g xn (1) g zλ (1) (2.33) By (2.32) and (2.33), we have F t n ¯ h Rλ ≥ λ 1+ g Rλ + 1 ¯ h Rλ g Rλ + 1 zλ (η) (1 − γ) g zλ (1) (2.34)...
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Báo cáo hóa học: "REPRESENTATION OF SOLUTIONS OF LINEAR DISCRETE SYSTEMS WITH CONSTANT COEFFICIENTS AND PURE DELAY" ppt
... + 1) ! (m + 1) + − ( j − 1) m ! − (m + 1) + − ( j − 1) m − j ! Bj j! j =1 + = B +1 ( + 1) ! (m + 1) − ( j − 1) m j B j 1 ( j − 1) ! j =1 B j 1 · = B+B j =2 +1 +1 (m + 1) − ( j − 1) m ! +B (m + 1) ... j − 1) m ! (m + 1) − ( j − 1) m − j ! (m + 1) − ( j − 1) m ! (m + 1) + − ( j − 1) m − j ! Bj j! j =1 × (m + 1) − ( j − 1) m ! (m + 1) − ( j − 1) m − j ! (m + 1) + − ( + − 1) m ! (m + 1) + − ( + − 1) m ... ( j − 1) m j (2. 21) Therefore ΔeBk = eB(k +1) − eBk = eB( m m m m +1 (m +1) +1) − eB (m +1) m j =1 (m + 1) + − ( j − 1) m − Bj · j j =1 Bj j! j =1 (m + 1) + − ( j − 1) m ! − (m + 1) + − ( j − 1) m −...
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Báo cáo hóa học: "Research Article Periodic and Almost Periodic Solutions of Functional Difference Equations with Finite Delay" pdf
... 37, no 3, pp 4 01 413 , 19 94 [7] S Elaydi and I Gy¨ ri, “Asymptotic theory for delay difference equations,” Journal of Difference o Equations and Applications, vol 1, no 2, pp 99 11 6, 19 95 [8] S Elaydi, ... USA, 19 91 [10 ] Y Song and C T H Baker, “Perturbation theory for discrete Volterra equations,” Journal of Difference Equations and Applications, vol 9, no 10 , pp 969–987, 2003 Yihong Song 15 [11 ] ... bounded solution of (1. 1) In Section 3, we discuss the existence of periodic solutions of (1. 1) In Section 4, we discuss the existence of almost periodic solutions of (1. 1) Preliminaries We formalize...
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Báo cáo hóa học: "SECOND-ORDER ESTIMATES FOR BOUNDARY BLOWUP SOLUTIONS OF SPECIAL ELLIPTIC EQUATIONS CLAUDIA ANEDDA, ANNA BUTTU, AND GIOVANNI " pot
... Φ(δ) 1 β δxi + α δ Φ(δ) 1 2β δx i (2 .16 ) Using (1. 7) we find Δw = Φ (δ) − Φ (δ)H + β 1 ΔHδ Φ(δ) + β 1 H δ Φ(δ) + α δ Φ(δ) 1 2β 1 β 1 β + 2β 1 ∇H · ∇δ δ Φ(δ) − β 1 H δ Φ(δ) − α δ Φ(δ) 1 2β H 1 β 1 β ... et s2β 1 − sβ t β 1 dt = lim s→∞ s→∞ e sβ s tβ 0e (2β − 1) sβ 1 − βt β 1 dt e sβ β β 1 s β (β − 1) es s + et (2β − 1) (β − 1) sβ−2 dt = lim s→∞ βesβ sβ 1 β lim s β β 1 et dt = + (2β − 1) (β − 1) lim ... Communications in Applied Analysis (2000), no 1, 12 1 13 1 [10 ] L Bieberbach, Δu = eu und die automorphen Funktionen, Mathematische Annalen 77 (19 16), no 2, 17 3– 212 [11 ] M del Pino and R Letelier, The influence...
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ON THE SOLVABILITY OF INITIAL-VALUE PROBLEMS FOR NONLINEAR IMPLICIT DIFFERENCE EQUATIONS PHAM KY ANH pdf
... = Qn G 1 Bn Qn 1, n Qn,n 1 xn = Qn,n 1 xn = Vn Vn 1 Qn 1 xn =Vn Vn 1 , n n (2.5) is reduced to − = Vn 1 Vn Qn G 1 qn − Qn G 1 Bn un n n (2.7) xn = un + = I − Qn 1, n G 1 Bn un + Qn 1, n G 1 qn ... xn +1 + Pn G 1 Bn xn = Pn G 1 qn , n n 1 1 Qn Gn Bn xn = Qn Gn qn (2.4) (2.5) Further, denoting un = Pn 1 xn , = Qn 1 xn (n 0) and observing that Pn G 1 Bn Qn 1 xn = n Pn G 1 Bn Qn 1, n Qn,n 1 ... Qn,n 1 xn = Pn Qn Qn,n 1 xn = 0, we find Pn G 1 Bn xn = n n Pn G 1 Bn un Thus, (2.4) becomes an ordinary difference equation n un +1 + Pn G 1 Bn un = Pn G 1 qn n n (2.6) 1 1 Since Qn G 1 Bn Qn−1...
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NONOSCILLATORY HALF-LINEAR DIFFERENCE EQUATIONS AND RECESSIVE SOLUTIONS ˇ ´ MARIELLA CECCHI, pdf
... Φ∗ zn [1] [1] ≤ H xn − zn (3 .13 ) Summing (1. 1) from n to ∞, n ≥ N, we obtain ∞ [1] xn = c + bk Φ xk +1 , k=n ∞ [1] zn = c + bk Φ zk +1 (3 .14 ) k=n Thus from (3 .12 ) and (3 .13 ), we have [1] [1] Φ∗ ... Ak +1 p−2 k=n ∞ ≤ HM bk uk +1 Ak +1 ∞ uj Φ∗ a j j =k +1 p−2 k=n ∞ = HM ∞ Φ∗ a j j =k +1 (3 .17 ) bk Φ Ak +1 uk +1 k=n Taking into account (1. 10), we can apply Lemma 2.3 and we obtain un ≡ for n ≥ N [1] ... [1] Φ∗ xn − Φ∗ zn ∞ bk Φ xk +1 − Φ zk +1 ≤H k=n ∞ ≤ HM bk Ak +1 k=n p−2 ∞ Φ∗ a j j =k +1 Φ∗ z [1] − Φ∗ x [1] j j (3 .15 ) Putting un = sup [1] [1] Φ∗ xk − Φ∗ zk :k≥n , (3 .16 ) 200 Recessive solutions...
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Tài liệu Solution of Linear Algebraic Equations part 1 docx
... coefficients, and b is the right-hand side written as a column vector, a 11 a 21 A= aM a12 a22 ··· aM a1N a2N aM N b1 b b= ··· bM (2.0.3) By convention, the first index on an element ... RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by Cambridge University Press.Programs Copyright (C) 19 88 -19 92 by Numerical Recipes Software Permission is ... right-hand sides, the b’s, are changed (§2 .1 §2 .10 ) • Calculation of the matrix A 1 which is the matrix inverse of a square matrix A, i.e., A · A 1 = A 1 · A = 1, where is the identity matrix (all...
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Tài liệu Solution of Linear Algebraic Equations part 2 ppt
... = b 11 b 21 b 31 b 41 b12 b22 b32 b42 x12 x22 x32 x42 x13 x23 x33 x43 b13 b23 b33 b43 0 0 y 11 y 21 y 31 y 41 0 0 y12 y22 y32 y42 y13 y23 y33 y43 y14 y24 ... unknown vector x: A·x= b A · C1 · C 1 ·x= b A · C1 · C2 · C 1 · C 1 · x = b (A · C1 · C2 · C3 · · ·) · · · C 1 · C 1 · C 1 ·x= b (1) · · · C 1 · C 1 · C 1 · x = b (2 .1. 7) Sample page from NUMERICAL ... 2 .1 Gauss-Jordan Elimination Elimination on Column-Augmented Matrices Consider the linear matrix equation a12 a22 a32 a42 a13 a23 a33 a43 a14 x 11 a24 x 21 · a34 x 31 a44 x41...
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Tài liệu Solution of Linear Algebraic Equations part 11 ppt
... a 11 a 21 a12 a22 · b 11 b 21 b12 b22 = c 11 c 21 c12 c22 (2 .11 .1) Eight, right? Here they are written explicitly: c 11 = a 11 × b 11 + a12 × b 21 c12 = a 11 × b12 + a12 × b22 c 21 = a 21 × b 11 + a22 × b 21 ... Strassen [1] The formulas are: Q1 ≡ (a 11 + a22 ) × (b 11 + b22 ) Q2 ≡ (a 21 + a22 ) × b 11 Q3 ≡ a 11 × (b12 − b22 ) Q4 ≡ a22 × (−b 11 + b 21 ) Q5 ≡ (a 11 + a12 ) × b22 Q6 ≡ (−a 11 + a 21 ) × (b 11 + b12 ) Q7 ... 2 .10 QR Decomposition 99 R = Qn 1 · · · Q1 · A (2 .10 .5) Since the Householder matrices are orthogonal, Q = (Qn 1 · · · Q1 ) 1 = Q1 · · · Qn 1 (2 .10 .6) In most applications...
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Tài liệu Solution of Linear Algebraic Equations part 3 pdf
... (2.3 .1) would look like this: 11 α 21 α 31 α 41 α22 α32 α42 0 α33 α43 11 · 0 α44 12 β22 0 13 β23 β33 14 β24 β34 β44 a 11 a = 21 a 31 a 41 a12 a22 a32 a42 a13 a23 a33 a43 a14 ... Thus, equation (2.3.4) can be solved by forward substitution as follows, y1 = b1 11 i 1 yi = αij yj bi − αii j =1 (2.3.6) i = 2, 3, , N while (2.3.5) can then be solved by backsubstitution ... xi = βij xj yi − i = N − 1, N − 2, , βii j=i +1 Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-5 21- 4 310 8-5) Copyright (C) 19 88 -19 92 by Cambridge University...
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Tài liệu Solution of Linear Algebraic Equations part 4 docx
... (equation 2.3 .11 ) are not stored at all.] In brief, Crout’s method fills in the combined matrix of α’s and β’s, 11 α 21 α 31 α 41 12 β22 α32 α42 13 β23 β33 α43 14 β24 β34 β44 (2.3 .14 ) by ... αii = 1, i = 1, , N (equation 2.3 .11 ) • For each j = 1, 2, 3, , N these two procedures: First, for i = 1, 2, , j, use (2.3.8), (2.3.9), and (2.3 .11 ) to solve for βij , namely i 1 βij ... be solved is u1 r1 b c1 · · · u2 r2 a2 b c2 · · · ··· · · · · = · · · (2.4 .1) uN 1 rN 1 · · · aN 1 bN 1 cN 1 ··· aN bN uN rN Sample...
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