... modulus Then, the assertions of the theorem follow from Remark 4.1 We are now in the position to give the main results related to the stability problem Theorem 4.3 Let k = 2ν − The k-step methods ... problem, we can choose to fix some of the (k − 1) additional conditions at the beginning of the interval of integration and the remaining at the end Later on, we will refer to the discrete problem ... BVMs with (ν,k − ν)-boundary conditions, where ν is defined according to (2.18) Remark 4.1 From the statement (3) in Lemma 2.4 we have that ρk (z) is of type (r,k − 2r,r), with r a nonnegative integer...
... Krein-Rutman theorem, see 15, Theorem 19.3 , we may get the desired results Remark 3.4 Theorem 3.2 is a partial generalization of Lemma 3.1 It is enough to prove that the condition i on f in Theorem 3.2 ... nonresonance case, Problem 1.1 , 1.2 has a unique Green’s function In the resonance case, Problem 1.1 , 1.2 has no Green’s function any more Remark 3.10 It is worth remarking that Cabada and ... radius of B Define cK B {λ ∈ 0, ∞ : ∃ x ∈ K with x 1, x λBx} 2.20 Lemma 2.5 see 14, Lemma 2.1 Assume that K − K; i K has a nonempty interior and E ii A : 0, ∞ × K → E is K-completely continuous...
... theory [4, Theorems 1.27 and 1.40] However, as pointed out by Dancer [5, 6] and L´pez-G´mez [7], the proofs of these o o theorems contain gaps, the original statement of Theorem 1.40 of [4] is ... solutions emanating from u = in wide classes of boundary value problems for equations and systems [1, 2, 8, 9] Fortunately, L´pez-G´mez gave a corrected o o version of unilateral bifurcation theorem ... the bifurcation theorem of L´pez-G´mez [7, Theorem 6.4.3], we shall establish the following: o o Theorem 1.1 Suppose that f (t, u) satisfies (H1 ), (H2 ), and (H3 ), then problem (1.4) possesses...
... problem 2.4 - 2.5 , it is easy to verify that y t J u t − ω is a positive solution of problem 1.1 for more details, see 10 Consequently, we will concentrate our study on problem 2.4 - 2.5 Lemma ... theorem in cones due to Krasnoselskii 14 to study problem 1.1 It was proved that 1.1 has at least two positive solutions for the positone case and has at least one positive solution for the semipositone ... Value Problems of multiple positive solutions was obtained by using the fixed-point index theory Recently, instead of Schauder fixed-point theorem and fixed-point index theory, Chu and Zhou 10 employed...
... Theorem 1.2 guarantees that (1.1) has a solution u ∈ C(N + , R) with u(k) ≥ α(k) > for k ∈ N 210 Discrete initial and boundary value problems Example 2.2 Consider the boundary value problem ∆2 ... j =k Thus (2.3) holds Theorem 2.1 guarantees that (2.23) has a solution u ∈ C(N + , R) with u(k) > for k ∈ N Next we present a result for initial value problems Theorem 2.3 Let n0 ∈ {1,2, } be ... method for singular discrete initial value problems, Demonstratio Math 37 (2004), no 1, 115–122 R P Agarwal, H L¨ , and D O’Regan, Existence theorems for the one-dimensional singular u p-Laplacian...
... Boundary Value Problems Remark 2.4 The advantage of representing a solution of the initial problem 1.1 , 1.2 as a solution of 2.6 is that, although 2.6 remains to be a neutral system, its right-hand ... proof of the following theorem is similar to the proofs of Theorems 2.5 and 2.6 and its statement in the case of p exactly coincides with the statements of these theorems Therefore, we will restrict ... zero solution of system 1.1 is exponentially stable in the metric C0 10 Boundary Value Problems Theorem 2.6 Let the matrix D be nonsingular and D < Let the assumptions of Theorem 2.5 with γ < 2/τ...
... “Existence and uniqueness theorems for three-point boundary value problems,” SIAM Journal on Mathematical Analysis, vol 20, no 3, pp 716–726, 1989 R P Agarwal, Boundary Value Problems for Higher Order ... problems,” Journal of Mathematical Analysis and Applications, vol 185, no 2, pp 302– 320, 1994 A Cabada, M R Grossinho, and F Minhos, “Extremal solutions for third-order nonlinear problems ´ ... problems,” Journal of Mathematical Analysis and Applications, vol 294, no 1, pp 104–112, 2004 Y Feng and S Liu, “Solvability of a third-order two-point boundary value problem,” Applied Mathematics...
... well-known Riemann function Proofs of the main theorems 5.1 Proof of Theorem 3.1 To prove Theorem 3.1 we need the next result from 20 −2 − 2α−1 −1 ς α , 4.30 I Gy˝ ri and L Horv´ th o a 11 Theorem A ... satisfies ≤ K sup h m z n z0 n ≥ , 5.3 m≥0 Now, we prove some lemmas Lemma 5.1 The hypotheses of Theorem 3.1 imply that the hypotheses of Theorem A are satisfied, and hence the solution z : z n n≥0 of ... hence the proof of Theorem 3.1 is complete 5.2 Proof of Theorem 3.3 Theorem 3.3 will be proved after some preparatory lemmas In the next lemma, we show that 3.7 can be transformed into an equation...
... and no solution for λ > λ∗ Remark 2.3 Our theorems generalize Theorems 1.1–1.4 and the main results in [9] In fact, Theorems 1.1–1.4 are corollaries of our theorems Moreover, the nonlinear term ... Theorem 2.1 cannot be obtained by Theorems 1.1–1.4 and the abstract results in [12] Remark 2.4 The nonlinear term f was assumed to be nondecreasing in Theorems 1.2 and 1.4, but in Theorem 2.2 ... (0,1), (2.5) β(1) − γβ(η) ≥ According to [13, Lemma 4], we have the following lemma Lemma 2.6 Assume that (H1 ) holds and τ ≥ Then the initial value problems u (t) = τa(t)u(t), u(α) = 0, ≤ α < t
... B(k−m− j) em = ΔeBk C + m Δψ( j − 1) j =−m+1 B(k−m− j) Δ em Δψ( j − 1) (3.5) j =−m+1 = Δ eBk C + m B(k−m− j) Δ em Δψ( j − 1) j =−m+1 We use formula (2.8): B(k Δx(k) = Bem −m) C + B(k−2m− j) Bem (3.6) ... m k B(k−2m− j) em k −m ω( j) = B j =1 B(k−2m− j) em ω( j) + f (k) (3.26) j =1 Since eB(−m) ≡ I and m k B(k−2m− j) em j =1 k −m ω( j) = B(k−2m− j) em j =1 k ω( j) + B(k−2m− j) em j =k−m+1 ω( j) ... eB((k+1)−m−(k+1)) ω(k + 1) + m k B(k−m− j) Δ em k −m ω( j) = B B(k−2m− j) em j =1 ω( j) + f (k) j =1 (3.24) Using formula (2.8) we get B(k−m− j) em B(k−2m− j) = Bem , (3.25) and the last relation becomes...
... the proof Using Theorem 3.1 and Lemma 3.2, we can show that (3.1) has an almost periodic solution Theorem 3.3 If the bounded solution {u(n)}n≥0 of (3.1) is ᐁ, then system (3.1) has an almost ... almost periodic in n uniformly for w ∈ Ω If Ω is the empty set and Y = X in Theorem 2.3, then {ξ(n)} is an almost periodic sequence Theorem 2.4 If f ∈ Ꮽᏼ(Z × Ω : Y ), then there exists a sequence ... difference systems,” Journal of Mathematical Analysis and Applications, vol 313, no 2, pp 678–688, 2006 [15] Y Song, “Almost periodic solutions of discrete Volterra equations,” Journal of Mathematical...
... s→∞ β + βsβ β β (2.4) The lemma follows Remark 2.2 If β = 1, we have F(s) f (s)( f (s))−2 = We not care of this special case because it has been discussed in [2] Lemma 2.3 Let Φ = Φ(δ) be defined ... we find F(s) f (s) f (s) −2 = + e −s − e −e − e −s −e s s (3.2) The lemma follows Lemma 3.2 Let f (t) and F(s) be as in Lemma 3.1 If ∞ Ψ(δ) 2F(s) −1/2 ds = δ (3.3) we have −Ψ (δ) = + O(1)e−Ψ(δ) ... of quasilinear elliptic problems, Zeitschrift f¨ r u Angewandte Mathematik und Physik 54 (2003), no 5, 731–738 [5] C Bandle and E Giarrusso, Boundary blow up for semilinear elliptic equations...
... on Nonlinear Problems in Mathematical Analysis Pure and Applied Mathematics, Academic Press, New York, 1977 L C Loi, N H Du, and P K Anh, On linear implicit non-autonomous systems of difference ... Institute of Mathematics, Chinese Academy of Sciences References [1] [2] [3] [4] P K Anh and L C Loi, On multipoint boundary-value problems for linear implicit nonautonomous systems of difference ... following lemma Lemma 2.1 The matrix Gn := An + Bn Qn−1,n is nonsingular if and only if Sn ∩ KerAn−1 = {0}, (2.3) where, as in the DAE case, Sn := {ξ ∈ Rm : Bn ξ ∈ ImAn } The proof of Lemma 2.1...
... contradiction with (3.2) The second statement follows from Lemma 2.2(i) The following uniqueness result will play an important role in our later consideration Theorem 3.4 Assume (1.10) For any fixed ... (3.4) Proof The existence follows from Lemma 2.1 and the homogeneity property It remains to prove the uniqueness The argument is suggested by [12, Theorem 4.3] Without loss of generality, let ... different from zero, in view of Remark 3.7, also the limit limn nun is finite and different from zero for any recessive solution u of (4.9) Concluding remarks Theorems 4.2 and 4.3, and Example 4.4...
... an element aij denotes its row, the second index its column For most purposes you don’t need to know how a matrix is stored in a computer’s physical memory; you simply reference matrix elements ... America only),or send email to trade@cup.cam.ac.uk (outside North America) • Accumulated roundoff errors in the solution process can swamp the true solution This problem particularly emerges if N is ... be minimized, then the overdetermined linear problem reduces to a (usually) solvable linear problem, called the • Linear least-squares problem The reduced set of equations to be solved can be...
... resembles the identity matrix in any way!) Obviously we will run into trouble if we ever encounter a zero element on the (then current) diagonal when we are going to divide by the diagonal element ... pivot element, so we interchange rows, if needed, to put the pivot element on the diagonal The columns are not physically interchanged, only relabeled: indxc[i], the column of the ith pivot element, ... diagonal when we are going to divide by the diagonal element (The element that we divide by, incidentally, is called the pivot element or pivot.) Not so obvious, but true, is the fact that Gauss-Jordan...
... all elements in a column of the matrix situated below a chosen element Thus we arrange for the first Householder matrix Q1 to zero all elements in the first column of A below the first element Similarly ... in the first column of A below the first element Similarly Q2 zeroes all elements in the second column below the second element, and so on up to Qn−1 Thus 100 Chapter Solution of Linear Algebraic ... s=SIGN(1.0/sqrt(1.0+(fact*fact)),b); c=fact*s; } for (j=i;j
... matrices, L·U=A (2.3.1) where L is lower triangular (has elements only on the diagonal and below) and U is upper triangular (has elements only on the diagonal and above) For the case of a × matrix ... Gauss-Jordan However, the unit vectors are quite special in containing all zeros except for one element If this is taken into account, the right-side manipulations can be reduced to only N loop ... solution with any number of right-hand sides, and for matrix inversion For this reason we will not implement the method of Gaussian elimination as a routine CITED REFERENCES AND FURTHER READING: Ralston,...
... (pivot element), then all the divisions by that element en masse This is Crout’s method with partial pivoting Our implementation has one additional wrinkle: It initially finds the largest element ... informabig=0.0; tion for (j=1;j big) big=temp; if (big == 0.0) nrerror("Singular matrix in routine ludcmp"); No nonzero largest element vv[i]=1.0/big; Save the scaling ... Press) 2.4 Tridiagonal and Band Diagonal Systems of Equations The special case of a system of linear equations that is tridiagonal, that is, has nonzero elements only on the diagonal plus or minus...