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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
1. Hyperfunction thoery
2. Hyperfunction method for numerical integration
3. Hyperfunction method for Fredholm integral equations
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. 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
1. Hyperfunction thoery
2. Hyperfunction method for numerical integration
3. Hyperfunction method for Fredholm integral equations
4. Summary
2. Hyperfunction method for numerical integration
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
2. Hyperfunction method for numerical integration
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
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.
✒ ✑
2. Hyperfunction method for numerical integration
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
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 ...
2. Hyperfunction method for numerical integration
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
2. Hyperfunction method for numerical integration
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
.
✒ ✑
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.
2. Hyperfunction method for numerical integration
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
.
✒ ✑
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: theoretical error estimate
12 / 23
theoretical error estimate✓ ✏
If f(ϕ(w)) and ϕ(w) are analytic in | Im w| < d0,
|error| ≦
τperiod
π
max
Im w=±d
|f(ϕ(w))Ψ(ϕ(w))ϕ′
(w)|
×
exp(−(4πd/τperiod)N)
1 − exp(−(4πd/uperiod)N)
( 0 < ∀d < d0 ).
. . . geometric convergence.
✒ ✑
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.
2. Numerical integration: example
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
)
2. Numerical integration: example
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
1. Hyperfunction thoery
2. Hyperfunction method for numerical integration
3. Hyperfunction method for Fredholm integral equations
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.
3. Hyperfunction method for integral equations
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
3. Hyperfunction method for integral equations
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
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
)
• 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
error ǫN condition number κN of
the collocation equation
(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
error ǫN condition number κN of
the collocation equation
(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
1. Hyperfunction thoery
2. Hyperfunction method for numerical integration
3. Hyperfunction method for Fredholm integral equations
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!

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

  • 1. 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
  • 2. 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.
  • 3. Contents 3 / 23 1. Hyperfunction thoery 2. Hyperfunction method for numerical integration 3. Hyperfunction method for Fredholm integral equations 4. Summary
  • 4. Contents 4 / 23 1. Hyperfunction thoery 2. Hyperfunction method for numerical integration 3. Hyperfunction method for Fredholm integral equations 4. Summary
  • 5. 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
  • 6. 1. Hyperfunctions: examples 6 / 23 Dirac’s delta function δ(x) = − 1 2πi 1 x + i0 − 1 x − i0 .
  • 7. 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.
  • 8. 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.
  • 9. 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).
  • 10. 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
  • 11. 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.
  • 12. Contents 9 / 23 1. Hyperfunction thoery 2. Hyperfunction method for numerical integration 3. Hyperfunction method for Fredholm integral equations 4. Summary
  • 13. 2. Hyperfunction method for numerical integration 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
  • 14. 2. Hyperfunction method for numerical integration 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 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. ✒ ✑
  • 15. 2. Hyperfunction method for numerical integration 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 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 ...
  • 16. 2. Hyperfunction method for numerical integration 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
  • 17. 2. Hyperfunction method for numerical integration 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 . ✒ ✑ 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.
  • 18. 2. Hyperfunction method for numerical integration 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 . ✒ ✑ If f(z) is real-valued on R, we can reduce the number of sampling points N by half using the reflection principle.
  • 19. 2. Numerical integration: theoretical error estimate 12 / 23 theoretical error estimate✓ ✏ If f(ϕ(w)) and ϕ(w) are analytic in | Im w| < d0, |error| ≦ τperiod π max Im w=±d |f(ϕ(w))Ψ(ϕ(w))ϕ′ (w)| × exp(−(4πd/τperiod)N) 1 − exp(−(4πd/uperiod)N) ( 0 < ∀d < d0 ). . . . geometric convergence. ✒ ✑
  • 20. 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.
  • 21. 2. Numerical integration: example 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 )
  • 22. 2. Numerical integration: example 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.
  • 23. Contents 15 / 23 1. Hyperfunction thoery 2. Hyperfunction method for numerical integration 3. Hyperfunction method for Fredholm integral equations 4. Summary
  • 24. 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.
  • 25. 3. Hyperfunction method for integral equations 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
  • 26. 3. Hyperfunction method for integral equations 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.
  • 27. 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
  • 28. 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
  • 29. 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 ).
  • 30. 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 )
  • 31. 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 ) • 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.
  • 32. 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 error ǫN condition number κN of the collocation equation (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 )
  • 33. 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 error ǫN condition number κN of the collocation equation (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.
  • 34. Contents 22 / 23 1. Hyperfunction thoery 2. Hyperfunction method for numerical integration 3. Hyperfunction method for Fredholm integral equations 4. Summary
  • 35. 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.
  • 36. 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!