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Quadratic spline collocation for Volterra integral equations
- D. Saveljeva
Quadratic spline collocation for Volterra integral equations D. - - PDF document
Quadratic spline collocation for Volterra integral equations D. - - PDF document
Quadratic spline collocation for Volterra integral equations D. Saveljeva University of Tartu, Estonia Consider Volterra integral equation t y ( t ) = 0 K ( t, s, y ( s )) ds + f ( t ) , t [0 , T ] , (1) with given functions f : [0 , T
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We have (1) ⇔ y = Ky + f with (Kv)(t) =
t
0 K(t, s, v(s))ds,
v ∈ C[0, T]. Then our method are u = PNKu + PNf, where the projections PN are determined by (PNf)(0) = f(0), (PNf)(τi) = f(τi), i = 1, . . . , N, (PNf)′′(tN−1 − 0) = (PNf)′′(tN−1 + 0), PN : C[0, T] → C[0, T].
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Let E be Banach spaces. Suppose we have an equation u = Ku + f, (2) where K ∈ K(E, E) and f ∈ E. Let it be given a sequence of approximating operators PN ∈ L(E, E), N = 1, 2, . . . . Consider also equations uN = PNKuN + PNf. (3) General convergence theorem. Suppose u = Ku
- nly if u = 0 and PNu → u for all u ∈ E in the
process N → ∞. Then
- 1. (2) has the unique solution u∗;
- 2. there is N0 such that for N ≥ N0, (3) has the
unique solution u∗
N;
- 3. u∗
N → u∗ as N → ∞;
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- 4. there are C1,
C2, C3 > 0 such that C1PNu∗−u∗ ≤ u∗
N −u∗ ≤ C2PNu∗−u∗ (4)
and u∗
N − PNu∗ ≤ C3K(PNu∗ − u∗).
(5) PNu → u ∀u ∈ E ⇒ (4) and (5); ⇓ PNK − K → 0 ⇒ (4); ⇓ PNK → K compactly ⇒ (4); ⇓ I − PNK → I − K stably or regularly ⇒ (4) and u∗
N − PNu∗ ≤ constPNK(PNu∗ − u∗).
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Results: c ∈ (0, 1) In the space C[0, T] for quasi-uniform mesh we get PN → I as N → ∞. In the space C1[0, T] for mesh with hN/hN−1 = O(1) it holds PN → I as hmax → 0. c = 1 In the space C[0, T] and C1[0, T] it holds PN = O(N). There is no compact convergence PNK → K, hence, there is no convergence PNK − K → 0 in both spaces. There is regular convergence I − PNK → I − K in the space C[0, T]. Conjecture: There is regular convergence in the space C1[0, T].
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|uN(τi)−y(τi)| = |uN(τi)−PNy(τi)| ≤ uN −PNy ≤ ≤ constPNK(PNy − y) Results in superconvergence: Theorem. Suppose that K and ∂K/∂s are con- tinuous on {(t, s)| 0 ≤ s ≤ t ≤ T} and y′′′ ∈ Lip 1. Then, for c = 1/2, it holds max
1≤i≤N |uN(ti−1 + h/2) − y(ti−1 + h/2)| = O(h4).
- Theorem. Suppose that K, ∂K/∂s, ∂K/∂t, ∂2K/∂t2
and ∂3K/∂t3 are continuous on {(t, s) : 0 ≤ s ≤ t ≤ T}. Suppose also the function t → K(t, t) is twice continuously differentiable on [0, T] and y′′′ ∈ Lip 1. Then, for c = 1, max
0≤i≤N |uN(ti) − y(ti)| = O(h4)
in the case of uniform mesh. Conjecture: There is superconvergence in the col- location points in case of c = O(h2).
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Numerical examples: We considered a test equation y(t) = λ
t
0 y(s)ds + f(t),
t ∈ [0, 1], (6) with function f(t) = 1 2((1 − λ) sin t + (1 + λ) cos t + (1 − λ)et), and the exact solution y(t) = (sin t + cos t + et)/2. Another test equation is y(t) =
t
0 (t − s)y(s)ds + sin t,
t ∈ [0, 1], with the exact solution y(t) = (2 sin t + et − e−t)/4. We calculated the error of the method u − y∞ approximately as max
1≤n≤N
max
0≤k≤10 |u(tn−1 + k
10h) − y(tn−1 + k 10h)|.
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Quadratic spline collocation for test equation (6): λ = −1, f(t) = sin t + et N 4 16 64 c = 1 1.13 · 10−3 51 2.21 · 10−5 61 3.63 · 10−7 63 2.54 · 10−2 13 1.87 · 10−3 16 1.21 · 10−4 16 c = 0.5 4.41 · 10−3 50 8.84 · 10−5 61 1.46 · 10−6 63 2.54 · 10−2 13 1.87 · 10−3 16 1.21 · 10−4 16 c = 0.1 7.71 · 10−3 48 1.62 · 10−4 60 2.70 · 10−6 63 7.11 · 10−2 12 5.81 · 10−3 15 3.85 · 10−4 16 λ = 2, f(t) = (3 cos t − sin t − et)/2 N 4 16 64 c = 1 1.88 · 10−3 47 4.02 · 10−5 59 6.82 · 10−7 66 2.64 · 10−2 10 2.52 · 10−3 14 1.74 · 10−4 16 c = 0.7 2.15 · 10−3 44 4.94 · 10−5 59 8.43 · 10−7 62 4.10 · 10−2 12 3.40 · 10−3 15 2.27 · 10−4 16 c = 0.1 7.24 · 10−3 46 1.56 · 10−4 60 2.62 · 10−6 63 6.99 · 10−2 12 5.79 · 10−3 15 3.85 · 10−4 16 Numerical results for y(t) =
t
- (t − s)y(s)ds + sin t:
N 8 32 128 c = 1 4.81 · 10−5 52 9.20 · 10−7 61 1.51 · 10−8 63 c = 0.5 1.95 · 10−4 53 3.71 · 10−6 61 6.07 · 10−8 63 c = 0.1 3.56 · 10−4 52 6.84 · 10−6 61 1.12 · 10−7 63
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