Numerical Methods Contents - SAM

⇒
if y ∈ C 2 ([0, T]), then
y(t k+1 ) − y(t k ) = ẏ(t k )h k + 1 2ÿ(ξ k)h 2 k for some t k ≤ ξ k ≤ t k+1
= f(y(t k ))h k + 1 2ÿ(ξ k)h 2 k ,
Total error arises from accumulation of one-step errors!
since t ↦→ y(t) solves the ODE, which implies
the one-step error from (11.3.4)
τ(h k ,y(t k ))=Ψ h ky(t k ) − y(t k + h k )
ẏ(t k ) = f(y(t k )). This leads to an expression for
(11.3.2)
= y(t k ) + h k f(y(t k )) − y(t k ) − f(y(t k ))h k + 2ÿ(ξ 1 k)h 2 (11.3.6)
k
= 1 2ÿ(ξ k)h 2 k .
error bound = O(h), h := max h l (➤
l
1st-order algebraic convergence)
Error bound grows exponentially with the length T of the integration interval.
Most commonly used single step methods display algebraic convergence of integer order with respect
to the meshwidth h := max k h k . This offers a criterion for gauging their quality.
Sloppily speaking, we observe τ(h k ,y(t k )) = O(h 2 k ) uniformly for h k → 0.
The sequence (y k ) k generated by a
➂ Estimate for the propagated error from (11.3.3)
∥
∥Ψ h ky k − Ψ h ky(t k ) ∥ = ‖y k + h k f(y k ) − y(t k ) − h k f(y(t k ))‖
(11.3.1)
≤ (1 + Lh k ) ‖y k − y(t k )‖ .
(11.3.7)
Ôº ½½º¿
single step method (→ Def. 11.2.1) of order (of consistency) p ∈ N
for ẏ = f(t,y) on a mesh M := {t 0 < t 1 < · · · < t N = T } satisfies
max
k ‖y k − y(t k )‖ ≤ Ch p for h := max
k=1,...,N |t k − t k−1 | → 0 ,
with C > 0 independent of M, provided that f is sufficiently smooth.
Ôº ½½º
➂ Recursion for error norms ǫ k := ‖e k ‖ by △-inequality:
11.4 Runge-Kutta methods
ǫ k+1 ≤ (1 + h k L)ǫ k + ρ k , ρ k := 2 1h2 k max ‖ÿ(τ)‖ . (11.3.8)
t k ≤τ≤t k+1
Taking into account ǫ 0 = 0 this leads to
Use the elementary estimate
ǫ k ≤
k∑ l−1 ∏
(1 + Lh j ) ρ l , k = 1, ...,N . (11.3.9)
l=1 j=1
(1 + Lh j ) ≤ exp(Lh j ) (by convexity of exponential function):
So far we only know first order methods, the explicit and implicit Euler method (11.2.1) and (11.2.4),
respectively.
Now we will build a class of methods that achieve orders > 1. The starting point is a simple integral
equation satisfied by solutions of initial value problems:
Note:
(11.3.9) ⇒ ǫ k ≤
l−1 ∑
h j ≤ T for final time T
j=1
k∑ l−1 ∏
exp(Lh j ) · ρ l =
l=1 j=1
k∑
l=1
exp(L ∑ l−1
j=1 h j)ρ l .
IVP:
∫
ẏ(t) = f(t,y(t)) ,
t1
⇒ y(t 1 ) = y 0 + f(τ,y(τ)) dτ
y(t 0 ) = y 0 t 0
k∑
ρ
k∑
ǫ k ≤ exp(LT) ρ l ≤ exp(LT) max k
h
k h l
l=1
k
l=1
≤ T exp(LT) max h l · max ‖ÿ(τ)‖ .
l=1,...,k t 0 ≤τ≤t k
‖y k − y(t k )‖ ≤ T exp(LT) max h l · max ‖ÿ(τ)‖ .
l=1,...,k t 0 ≤τ≤t k
Ôº ½½º¿
Idea: approximate integral by means of s-point quadrature formula (→ Sect. 10.1,
defined on reference interval [0, 1]) with nodes c 1 ,...,c s , weights b 1 ,...,b s .
s∑
y(t 1 ) ≈ y 1 = y 0 + h b i f(t 0 + c i h, y(t 0 + c i h) ) , h := t 1 − t 0 . (11.4.1)
i=1
Obtain these values by bootstrapping
Ôº¼ ½½º