H-Matrix approximation for the operator exponential with applications

92 I. P. Gavrilyuk et al.
we obtain for integral (2.6)
T (t) ≡ exp(−tL) ≈ T N (t) ≡ exp N (−tL)
N∑
(2.17)
= h F (kh, t; L).
k=−N
Note that F satisfies (2.13) with α = t ã. The error analysis is given by the
following Theorem (see Appendix for the proof).
Theorem 2.4 Choose k>1, ã = a/k, h = 3√ 2πdk/((N +1) 2 a), b=
b(k) =γ 0 − (k − 1)/(4a) and the integration parabola Γ b(k) = {z =
ãη 2 + b(k) − iη : η ∈ (−∞, ∞)}. Then there holds
‖T (t) − T N (t)‖ ≡‖exp(−tL) − exp N (−tL)‖
(2.18)
≤ Mc √ [
π
2 √ k exp[−s(N +1) 2/3 ]
√
at(1 − exp(−s(N +1) 2/3 ))
]
+ k exp[−ts(N +1)2/3 ]
t(N +1) 1/3 3√ ,
2πdka 2
where
s = 3√ (2πd) 2 a/k ,
(2.19) c = M 1 e t[ad2 /k+d−b] , d =(1− √ 1 ) k
k 2a ,
|2 a k
M 1 = max
z − i|
z∈D d 1+ √ | a k z2 + b − iz|
and M is the constant from the inequality of the strong P-positiveness.
The exponential convergence of our quadrature rule allows to introduce
the following algorithm for the approximation of the operator exponent at a
given time value t.
Algorithm 2.5 1. Choose k>1, d =(1− √ 1
k
) k 2a , N and determine z p
√
(p = −N,...,N) by z p = a k (ph)2 + b − iph, where h = 3 2πdk
a (N +
1) −2/3 and b = γ 0 − k−1
4a .
2. Find the resolvents (z p I −L) −1 ,p= −N,...,N (note that it can be
done in parallel).
3. Find the approximation exp N (−tL) for the operator exponent exp(−tL)
in the form
(2.20) exp N (−tL) = h
2πi
N∑
p=−N
e −tzp [
2 a k ph − i ]
(z p I −L) −1 .