- 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!