Hyperfunction method for numerical integration and Fredholm integral equations of the second kind

1 / 23
Hyperfunction Method for Numerical Integration
and Fredholm Integral Equations of the Second Kind
Hidenori Ogata
The University of Electro-Communications, Japan
13 July, 2017

Aim of this study
2 / 23
Hyperfunction theory (M. Sato, 1958)✓ ✏
• A theory of generalized functions based on complex function theory.
• A “hyperfunction” is expressed in terms of complex analytic functions.
hyperfunctions
= functions with singularities
pole
discontinuity
delta impluse, ...
←−
complex analytic function
easy to treat
numerically
✒ ✑
In this talk, we propose hyperfunction methods for
• numerical integration
• Fredholm integral equations of the second kind.

Contents
3 / 23
1. Hyperfunction thoery
2. Hyperfunction method for numerical integration
3. Hyperfunction method for Fredholm integral equations
4. Summary

Contents
4 / 23
4. Summary

1. Hyperfunction theory
5 / 23
Hyperfunction theory (M. Sato, 1958)✓ ✏
• hyperfunction on an interval I
. . . the difference between the values of a complex analytic funtion F(z) on I
f(x) = [F(z)] ≡ F(x + i0) − F(x − i0).
F(z) : deﬁning function of the hyperfunction f(x)
analytic in D I, where D is a complex neighborhood of I
✒ ✑
D
I
F(z)
=Re z
m z

1. Hyperfunctions: examples
6 / 23
Dirac’s delta function
δ(x) = −
1
2πi
1
x + i0
−
1
x − i0
.

1. Hyperfunctions: examples
6 / 23
Dirac’s delta function
δ(x) = −
1
2πi
1
x + i0
−
1
x − i0
.
O
D
a b
C
+ǫ
−ǫ
Suppose that φ(z) is analytic in D. By Cauchy’s integral formula,
φ(0) =
b
a
φ(x)δ(x)dx = −
1
2πi
b
a
φ(x)
1
x + i0
−
1
x − i0
dx.

1. Hyperfunction: examples
7 / 23
Heaviside step function
H(x) =
1 ( x > 0 )
0 ( x < 0 )
= F(x + i0) − F(x − i0), F(z) = −
1
2πi
log(−z).
-0.4 -0.2 0 0.2 0.4 0.6 0.8 1Re z
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
Im z
-1
-0.5
0
0.5
1
Re F(z)
The real part of F(z) = −
1
2πi
log(−z).

1. Hyperfunction theory: integral
8 / 23
integral of a hyperfunction✓ ✏
f(x) = F(x + i0) − F(x − i0) : hyperfunction on an interval I
I
f(x)dx ≡ −
C
F(z)dz,
C : closed path encircling I in the positive sense and included in D
(F(z) is analytic in D I)
✒ ✑
D
C
I

1. Hyperfunction theory: integral
8 / 23
integral of a hyperfunction✓ ✏
f(x) = F(x + i0) − F(x − i0) : hyperfunction on an interval I
I
f(x)dx ≡ −
C
F(z)dz,
C : closed path encircling I in the positive sense and included in D
(F(z) is analytic in D I)
✒ ✑
D
C
I
I
f(x)dx =
I
[F(x + i0) − F(x − i0)] dx.

Contents
9 / 23
4. Summary

10 / 23
We consider an integral of the form
I
f(x)w(x)dx,
f(x) : analytic in D (I ⊂ D ⊂ C, )
w(x) : weight function.
D
I

10 / 23
I
f(x)w(x)dx,
D
I
We can regard the integrand as a hyperfunction.
✓ ✏
f(x)w(x)χI(x) = −
1
2πi
{f(x + i0)Ψ(x + i0) − f(x − i0)Ψ(x − i0)}
with χI(x) =
1 (x ∈ I)
0 (x ∈ I)
, Ψ(z) =
I
w(x)
z − x
dx.
✒ ✑

10 / 23
I
f(x)w(x)dx,
D
C : z = ϕ(u)
I
We can regard the integrand as a hyperfunction.
✓ ✏
I
f(x)w(x)dx =
1
2πi C
f(z)Ψ(z)dz
=
1
2πi
τperiod
0
f(ϕ(τ))Ψ(ϕ(τ))ϕ′
(τ)dτ,
C : z = ϕ(τ) ( 0 ≦ τ ≦ τperiod ) periodic function (of period τperiod)
✒ ✑
Approximating the complex integral by the trapezoidal rule, we have ...

11 / 23
Hyperfunction method✓ ✏
I
f(x)w(x)dx ≃
h
2πi
N−1
k=0
f(ϕ(kh))Ψ(ϕ(kh))ϕ′
(kh),
with Ψ(z) =
b
a
w(x)
z − x
dx and h =
τperiod
N
.
✒ ✑
D
C : z = ϕ(τ), 0 ≦ τ ≦ τperiod
I

11 / 23
I
f(x)w(x)dx ≃
h
2πi
N−1
k=0
(kh),
with Ψ(z) =
b
a
w(x)
z − x
dx and h =
τperiod
N
.
✒ ✑
Ψ(z) for some typical weight functions w(x)
I w(x) Ψ(z)
(a, b) 1 log
z − a
z − b
∗
(0, 1) xα−1
(1 − x)β−1
B(α, β)z−1
F(α, 1; α + β; z−1
)∗∗
( α, β > 0 )
∗ log z is the branch s.t. −π ≦ arg z < π.
∗∗ F(α, 1; α + β; z−1
) can be easily evaluated using a continued fraction.

11 / 23
I
f(x)w(x)dx ≃
h
2πi
N−1
k=0
(kh),
with Ψ(z) =
b
a
w(x)
z − x
dx and h =
τperiod
N
.
✒ ✑
If f(z) is real-valued on R, we can reduce the number of sampling points N by half
using the reﬂection principle.

2. Numerical integration: example
13 / 23
✓ ✏
1
0
ex
xα−1
(1 − x)β−1
dx = B(α, β)F(α; α + β; 1) ( α, β > 0 ).
✒ ✑
We computed this integral by
• hyperfunction method (with N reduction),
• DE formula (efﬁcient for integrals with end-point singularities)
• Gauss-Jacobi formula
• C++ program, double precision
• complex integral path for the hyperfunction method (an ellipse)
z = ϕ(τ) =
1
2
+
1
4
ρ +
1
ρ
cos τ +
i
4
ρ −
1
ρ
sin τ ( ρ = 10 )
= 0.5 + 2.575 cos τ + i2.425 sin τ.

14 / 23
-16
-14
-12
-10
-8
-6
-4
-2
0
0 10 20 30 40 50 60
log10(error)
N
hyperfunction
hyperfunction
Gauss-Jacobi
Gauss-Jacobi
DE
DE
-16
-14
-12
-10
-8
-6
-4
-2
0
0 20 40 60 80 100 120
log10(error)
N
hyperfunction
hyperfunction
Gauss-Jacobi
DE
DE
α = β = 0.5 α = β = 10−4
(very strong singularities)
The errors of the hyperfunction method, Gauss-Jacobi formula and the DE formula
hyperfunction Gauss-Jacobi DE
α = β = 0.5 O(0.025N
) O((8.2 × 10−4
)N
) O(0.36N
)
α = β = 10−4
O(0.029N
) — O(0.70N
)

14 / 23
-16
-14
-12
-10
-8
-6
-4
-2
0
0 10 20 30 40 50 60
log10(error)
N
hyperfunction
hyperfunction
Gauss-Jacobi
Gauss-Jacobi
DE
DE
-16
-14
-12
-10
-8
-6
-4
-2
0
0 20 40 60 80 100 120
log10(error)
N
hyperfunction
hyperfunction
Gauss-Jacobi
DE
DE
α = β = 0.5 α = β = 10−4
(very strong singularities)
The hyperfunction method converges geometricaly,
and its performance is not affected by the end-point singularities.

Contents
15 / 23
4. Summary

3. Hyperfunction method for integral equations
16 / 23
Fredholm integral equation for unknown u(x)✓ ✏
λu(x) −
b
a
K(x, ξ)u(ξ)w(ξ)dξ = g(x),
w(ξ) : weight function, K(x, ξ), g(x), λ(= 0) : given.
✒ ✑
We apply the hyperfunction method to this integral equation.

17 / 23
λu(x) −
b
a
K(x, ξ)u(ξ)w(ξ)dξ = g(x).
(Assumption)
• g(z) : analytic in D except for
a ﬁnite number of poles at a1, . . . , aK
• K(z, ζ) : analytic function in D w.r.t. z and ζ D
a b
ak

17 / 23
λu(x) −
b
a
K(x, ξ)u(ξ)w(ξ)dξ = g(x).
(Assumption)
• g(z) : analytic in D except for
a ﬁnite number of poles at a1, . . . , aK
• K(z, ζ) : analytic function in D w.r.t. z and ζ D
a b
ak
ua(z) ≡ u(z) − λ−1
g(z) is analytic in D.
ua(x) satisﬁes the integral equation
✓ ✏
λua(x) −
b
a
K(x, ξ)ua(ξ)w(ξ)dξ =
1
λ
b
a
K(x, ξ)g(ξ)w(ξ)dξ.
✒ ✑
1. We discretize the integral equation for ua(x) by the hyperfunction method.
2. We solve the discretized equation by the collocation method.

3. Integral equations: Collocation equation
18 / 23
h
2πi
N
k=1
λ
ϕ(kh) − zi
− K(zi, ϕ(kh))Ψ(ϕ(kh)) ϕ′
(ϕ(kh))ua(ϕ(kh))
=
1
2πiλ C
K(zi, ζ)g(ζ)Ψ(ζ)dζ−
1
λ
N
k=1
Res(K(zi, ·)Ψg, ak) (i = 1, . . . , N),
where
C : z = ϕ(τ) ( 0 ≦ τ ≦ τperiod ) closed path encircling [a, b],
periodic function (period τperiod)
z1, . . . , zN : the collocation points inside C, h = τperiod/N.
The collocation equation
... a system of linear equations for ua(ϕ(kh))
( k = 1, . . . , N ). a b
ak
C : z = ϕ(τ)
D
zi

3. Integral equations: Collocation equation
18 / 23
h
2πi
N
k=1
λ
ϕ(kh) − zi
− K(zi, ϕ(kh))Ψ(ϕ(kh)) ϕ′
(ϕ(kh))ua(ϕ(kh))
=
1
2πiλ C
K(zi, ζ)g(ζ)Ψ(ζ)dζ−
1
λ
N
k=1
Res(K(zi, ·)Ψg, ak) (i = 1, . . . , N),
where
C : z = ϕ(τ) ( 0 ≦ τ ≦ τperiod ) closed path encircling [a, b],
periodic function (period τperiod)
z1, . . . , zN : the collocation points inside C, h = τperiod/N.
The approximate solution u(z) is given by
u(z) =
1
2πi C
ua(ζ)
ζ − z
dζ + g(z)
≃
h
2πi
N
j=1
ua(ϕ(kh))
ϕ(kh) − z
ϕ′
(kh) + g(z).
a b
ak
C : z = ϕ(τ)
D
zi

3. Integral equations: example
19 / 23
✓ ✏
u(x) +
1
0
(x − ξ)u(ξ)ξα−1
(1 − ξ)β−1
dξ = g(x),
g(x) =
1
1 + x2
+ B(α, β) Re{F(α, 1; α + β; i)}x
− B(α + 1, β) Re{F(α + 1, 1; α + β + 1; i)} ( α = β = 0.5, 10−4
).
✒ ✑
We solved the integral equation by the hyperfunction method, DE-Nystr¨om method and
Gauss-Jacobi-Nystr¨om method.
• complex integral path
C : z = ϕ(τ) =
1
2
+
1
4
ρ +
1
ρ
cos τ +
i
4
ρ −
1
ρ
sin τ ( ρ = 200 )
• collocation points zi = ϕcol
2π(i − 1)
N
( i = 1, . . . , N )
ϕc(τ) =
1
2
+
1
4
ρc +
1
ρc
cos τ +
i
4
ρc −
1
ρc
sin τ ( 1 < ρc < ρ ).

3. Integral equations: example (α = β = 0.5)
20 / 23
-60
-50
-40
-30
-20
-10
0
0 20 40 60 80 100
log10(error)
N
rhoc=1.2
rhoc=2.0
rhoc=4.0
rhoc=6.0
rhoc=8.0
DE
Gauss-Jacobi
0
20
40
60
80
100
0 20 40 60 80 100
log10(cond)
N
rhoc=1.2
rhoc=2.0
rhoc=4.0
rhoc=6.0
rhoc=8.0
DE
Gauss-Jacobi
error ǫN condition number κN of
the collocation equation
(rhoc = ρc)
ρc/ρ 0.006 0.01 0.02 0.03 0.04
ǫN O(0.0058N
) O(0.010N
) O(0.020N
) O(0.030N
) O(0.040N
)
κN O(160N
) O(97N
) O(48N
) O(32N
) O(23N
)

3. Integral equations: example (α = β = 0.5)
20 / 23
-60
-50
-40
-30
-20
-10
0
0 20 40 60 80 100
log10(error)
N
rhoc=1.2
rhoc=2.0
rhoc=4.0
rhoc=6.0
rhoc=8.0
DE
Gauss-Jacobi
0
20
40
60
80
100
0 20 40 60 80 100
log10(cond)
N
rhoc=1.2
rhoc=2.0
rhoc=4.0
rhoc=6.0
rhoc=8.0
DE
Gauss-Jacobi
(rhoc = ρc)
ρc/ρ 0.006 0.01 0.02 0.03 0.04
ǫN O(0.0058N
) O(0.010N
) O(0.020N
) O(0.030N
) O(0.040N
)
κN O(160N
) O(97N
) O(48N
) O(32N
) O(23N
)
• error ǫN = O[(ρc/ρ)N
], cond. number κN = O[(ρ/ρc)N
].
• converges faster than the DE-Nystr¨om method.
• The linear system of the collocation equation is very ill-conditioned.

3. Integral equations: example (α = β = 10−4
)
21 / 23
-60
-50
-40
-30
-20
-10
0
0 20 40 60 80 100
log10(error)
N
rhoc=1.2
rhoc=2.0
rhoc=4.0
rhoc=6.0
rhoc=8.0
DE
Gauss-Jacobi
0
20
40
60
80
100
0 20 40 60 80 100
log10(cond)
N
rhoc=1.2
rhoc=2.0
rhoc=4.0
rhoc=6.0
rhoc=8.0
DE
Gauss-Jacobi
(rhoc = ρc)
ρcol/ρ 0.006 0.01 0.02 0.03 0.04
ǫN O(0.0058N
) O(0.010N
) O(0.020N
) O(0.030N
) O(0.040N
)
κN O(160N
) O(97N
) O(48N
) O(32N
) O(24N
)

3. Integral equations: example (α = β = 10−4
)
21 / 23
-60
-50
-40
-30
-20
-10
0
0 20 40 60 80 100
log10(error)
N
rhoc=1.2
rhoc=2.0
rhoc=4.0
rhoc=6.0
rhoc=8.0
DE
Gauss-Jacobi
0
20
40
60
80
100
0 20 40 60 80 100
log10(cond)
N
rhoc=1.2
rhoc=2.0
rhoc=4.0
rhoc=6.0
rhoc=8.0
DE
Gauss-Jacobi
(rhoc = ρc)
ρcol/ρ 0.006 0.01 0.02 0.03 0.04
ǫN O(0.0058N
) O(0.010N
) O(0.020N
) O(0.030N
) O(0.040N
)
κN O(160N
) O(97N
) O(48N
) O(32N
) O(24N
)
• error ǫN = O[(ρcol/ρ)N
], cond. number κN = O[(ρ/ρcol)N
].
• The DE-Nystr¨om method does not work if the end-point singularities are
very strong.

Contents
22 / 23
4. Summary

4. Summary
23 / 23
• We applied hyperfunction theory to numerical integration and Fredholm integral
equations of the second kind.
◦ Hyperfunction theory: a generalized function theory where a “hyperfunction” is
expressed in terms of complex analytic functions.
◦ A hyperfunction integral is given by a complex loop integral, which is evaluated
numerically in the hyperfunction method.
• Hyperfunction method
◦ (Theoretical error estimate) geometric convergence
◦ (Numerical examples) efﬁciency for problems with strong end-point singularities
◦ Integral equation: The linear system of the collocation equation is
very ill-conditioned.
• Problems for future study
◦ Volterra integral equations.
◦ theoretical error estimate.

4. Summary
23 / 23
• We applied hyperfunction theory to numerical integration and Fredholm integral
equations of the second kind.
◦ Hyperfunction theory: a generalized function theory where a “hyperfunction” is
expressed in terms of complex analytic functions.
◦ A hyperfunction integral is given by a complex loop integral, which is evaluated
numerically in the hyperfunction method.
• Hyperfunction method
◦ (Theoretical error estimate) geometric convergence
◦ (Numerical examples) efﬁciency for problems with strong end-point singularities
◦ Integral equation: The linear system of the collocation equation is
very ill-conditioned.
• Problems for future study
◦ Volterra integral equations.
◦ theoretical error estimate.
Thank you!

Hyperfunction method for numerical integration and Fredholm integral equations of the second kind

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Hyperfunction method for numerical integration and Fredholm integral equations of the second kind

Similar to Hyperfunction method for numerical integration and Fredholm integral equations of the second kind (20)

Recently uploaded

Recently uploaded (20)

Hyperfunction method for numerical integration and Fredholm integral equations of the second kind