[PPT] - Greens Functions for Stieltjes Boundary Problems Markus Rosenkranz PowerPoint Presentation

SLIDE 1

Green’s Functions for Stieltjes Boundary Problems

Markus Rosenkranz Nitin Serwa

School of Mathematics, Statistics & Act. Sci. University of Kent

ISSAC, 2015

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 1 / 21

SLIDE 2

Outline

1

Motivation

2

Introduction

3

Stieltjes Boundary Conditions

4

Equitable Integro-Differential Operators

5

Extracting Green’s Function

6

Examples

7

Conclusion

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 2 / 21

SLIDE 3

Example of four point Boundary Problem.

Given f ∈ C ∞[a, b], find u ∈ C ∞[a, b] such that −u′′ = f , u(0) + u(1/3) = u(1) + u(2/3) = 0.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 3 / 21

SLIDE 4

Example of four point Boundary Problem.

Given f ∈ C ∞[a, b], find u ∈ C ∞[a, b] such that −u′′ = f , u(0) + u(1/3) = u(1) + u(2/3) = 0. Represented by pair

−D2, [⌊0⌋ + ⌊1/3⌋, ⌊1⌋ + ⌊2/3⌋]
.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 3 / 21

SLIDE 5

Example of four point Boundary Problem.

Given f ∈ C ∞[a, b], find u ∈ C ∞[a, b] such that −u′′ = f , u(0) + u(1/3) = u(1) + u(2/3) = 0. Represented by pair

−D2, [⌊0⌋ + ⌊1/3⌋, ⌊1⌋ + ⌊2/3⌋]
.

Evaluation functionals: ⌊ ⌋

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 3 / 21

SLIDE 6

Green’s Operator

Similarly: Regular boundary problem (T, B) for LODE. Meaning ∃ unique solution u ∈ C ∞[a, b] for all f ∈ C ∞[a, b]: Tu = f , β(u) = 0 (β ∈ B).

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 4 / 21

SLIDE 7

Green’s Operator

Similarly: Regular boundary problem (T, B) for LODE. Meaning ∃ unique solution u ∈ C ∞[a, b] for all f ∈ C ∞[a, b]: Tu = f , β(u) = 0 (β ∈ B). Algorithm to compute Green’s operator G : f → u.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 4 / 21

SLIDE 8

Green’s Operator

Similarly: Regular boundary problem (T, B) for LODE. Meaning ∃ unique solution u ∈ C ∞[a, b] for all f ∈ C ∞[a, b]: Tu = f , β(u) = 0 (β ∈ B). Algorithm to compute Green’s operator G : f → u. However, sometimes we want Green’s functions. Why?

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 4 / 21

SLIDE 9

Green’s Operator

Similarly: Regular boundary problem (T, B) for LODE. Meaning ∃ unique solution u ∈ C ∞[a, b] for all f ∈ C ∞[a, b]: Tu = f , β(u) = 0 (β ∈ B). Algorithm to compute Green’s operator G : f → u. However, sometimes we want Green’s functions. Why?

Nice intuition (see below).
Standard form for solutions.
Useful for communicating with engineers.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 4 / 21

SLIDE 10

Green’s Operator

Similarly: Regular boundary problem (T, B) for LODE. Meaning ∃ unique solution u ∈ C ∞[a, b] for all f ∈ C ∞[a, b]: Tu = f , β(u) = 0 (β ∈ B). Algorithm to compute Green’s operator G : f → u. However, sometimes we want Green’s functions. Why?

Nice intuition (see below).
Standard form for solutions.
Useful for communicating with engineers.

How to extract Green’s function from Green’s operators?

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 4 / 21

SLIDE 11

Intuition behind Green’s Function.

Consider again general boundary problem for LODE Tu = f , β(u) = 0 (β ∈ B). r r r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 5 / 21

SLIDE 12

Intuition behind Green’s Function.

Consider again general boundary problem for LODE Tu = f , β(u) = 0 (β ∈ B). Green’s function gξ is solution for f = δξ. r r r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 5 / 21

SLIDE 13

Intuition behind Green’s Function.

Consider again general boundary problem for LODE Tu = f , β(u) = 0 (β ∈ B). Green’s function gξ is solution for f = δξ. Note that g(x, ξ) = gξ(x) and “δ(x, ξ) = δξ(x)”. r r r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 5 / 21

SLIDE 14

Intuition behind Green’s Function.

Consider again general boundary problem for LODE Tu = f , β(u) = 0 (β ∈ B). Green’s function gξ is solution for f = δξ. Note that g(x, ξ) = gξ(x) and “δ(x, ξ) = δξ(x)”. u = r gξ f (ξ) dξ = ⇒ Tu = r Tgξ f (ξ) dξ = r δξf (ξ) dξ = f β(u) = r β(gξ) f (ξ) dξ = 0

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 5 / 21

SLIDE 15

Green’s Functions for Stieltjes Boundary Problems

Extraction of Green’s functions: G g(x, ξ)

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 6 / 21

SLIDE 16

Green’s Functions for Stieltjes Boundary Problems

Extraction of Green’s functions: G g(x, ξ) Well known for “classical case” [3].

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 6 / 21

SLIDE 17

Green’s Functions for Stieltjes Boundary Problems

Extraction of Green’s functions: G g(x, ξ) Well known for “classical case” [3]. Sometimes one needs Stieltjes boundary conditions:

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 6 / 21

SLIDE 18

Green’s Functions for Stieltjes Boundary Problems

Extraction of Green’s functions: G g(x, ξ) Well known for “classical case” [3]. Sometimes one needs Stieltjes boundary conditions:

More than two evaluation points → multipoint BP.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 6 / 21

SLIDE 19

Green’s Functions for Stieltjes Boundary Problems

Extraction of Green’s functions: G g(x, ξ) Well known for “classical case” [3]. Sometimes one needs Stieltjes boundary conditions:

More than two evaluation points → multipoint BP. Derivatives of arbitrary order → ill-posed BP.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 6 / 21

SLIDE 20

Green’s Functions for Stieltjes Boundary Problems

Extraction of Green’s functions: G g(x, ξ) Well known for “classical case” [3]. Sometimes one needs Stieltjes boundary conditions:

More than two evaluation points → multipoint BP. Derivatives of arbitrary order → ill-posed BP. Global terms in the form of definite integrals → nonlocal BP.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 6 / 21

SLIDE 21

Green’s Functions for Stieltjes Boundary Problems

Extraction of Green’s functions: G g(x, ξ) Well known for “classical case” [3]. Sometimes one needs Stieltjes boundary conditions:

More than two evaluation points → multipoint BP. Derivatives of arbitrary order → ill-posed BP. Global terms in the form of definite integrals → nonlocal BP.

Are there Green’s functions g(x, ξ) for such Stieltjes BPs?

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 6 / 21

SLIDE 22

Green’s Functions for Stieltjes Boundary Problems

Extraction of Green’s functions: G g(x, ξ) Well known for “classical case” [3]. Sometimes one needs Stieltjes boundary conditions:

More than two evaluation points → multipoint BP. Derivatives of arbitrary order → ill-posed BP. Global terms in the form of definite integrals → nonlocal BP.

Are there Green’s functions g(x, ξ) for such Stieltjes BPs? Can we extract it from G?

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 6 / 21

SLIDE 23

Integro-Differential Operators

Recall algebraic setting for boundary problems [3]: r

r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 7 / 21

SLIDE 24

Integro-Differential Operators

Recall algebraic setting for boundary problems [3]: Ordinary integro-differential K-algebra (F, ∂, r ).

r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 7 / 21

SLIDE 25

Integro-Differential Operators

Recall algebraic setting for boundary problems [3]: Ordinary integro-differential K-algebra (F, ∂, r ). Characters on F: multiplicative linear functionals.

r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 7 / 21

SLIDE 26

Integro-Differential Operators

Recall algebraic setting for boundary problems [3]: Ordinary integro-differential K-algebra (F, ∂, r ). Characters on F: multiplicative linear functionals. Ring of integro-differential operators over F with characters Φ.

r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 7 / 21

SLIDE 27

Integro-Differential Operators

Recall algebraic setting for boundary problems [3]: Ordinary integro-differential K-algebra (F, ∂, r ). Characters on F: multiplicative linear functionals. Ring of integro-differential operators over F with characters Φ.

Standard integro-differential operator ring FΦ[∂, r ] r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 7 / 21

SLIDE 28

Integro-Differential Operators

Recall algebraic setting for boundary problems [3]: Ordinary integro-differential K-algebra (F, ∂, r ). Characters on F: multiplicative linear functionals. Ring of integro-differential operators over F with characters Φ.

Standard integro-differential operator ring FΦ[∂, r ] Equitable operator ring F[∂, r

Φ].

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 7 / 21

SLIDE 29

Integro-Differential Operators

Recall algebraic setting for boundary problems [3]: Ordinary integro-differential K-algebra (F, ∂, r ). Characters on F: multiplicative linear functionals. Ring of integro-differential operators over F with characters Φ.

Standard integro-differential operator ring FΦ[∂, r ] Equitable operator ring F[∂, r

Φ].

Isomorphic rings (alternative normal forms).

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 7 / 21

SLIDE 30

Integro-Differential Operators

Recall algebraic setting for boundary problems [3]: Ordinary integro-differential K-algebra (F, ∂, r ). Characters on F: multiplicative linear functionals. Ring of integro-differential operators over F with characters Φ.

Standard integro-differential operator ring FΦ[∂, r ] Equitable operator ring F[∂, r

Φ].

Isomorphic rings (alternative normal forms).

Later on will have F = C ∞(R) again.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 7 / 21

SLIDE 31

Stieltjes Boundary Conditions

Definition

The elements of right ideal |Φ) = Φ · FΦ[∂, r ] are called Stieltjes boundary conditions. r r r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 8 / 21

SLIDE 32

Stieltjes Boundary Conditions

Definition

The elements of right ideal |Φ) = Φ · FΦ[∂, r ] are called Stieltjes boundary conditions. Normal forms: β(u) = α1(uk1) + · · · + αr(ukr ) + r γ1

β1 + · · · +

r γs

βs

r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 8 / 21

SLIDE 33

Stieltjes Boundary Conditions

Definition

The elements of right ideal |Φ) = Φ · FΦ[∂, r ] are called Stieltjes boundary conditions. Normal forms: β(u) = α1(uk1) + · · · + αr(ukr ) + r γ1

β1 + · · · +

r γs

βs

Boundary operators: Two-sided ideal (Φ) r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 8 / 21

SLIDE 34

Stieltjes Boundary Conditions

Definition

The elements of right ideal |Φ) = Φ · FΦ[∂, r ] are called Stieltjes boundary conditions. Normal forms: β(u) = α1(uk1) + · · · + αr(ukr ) + r γ1

β1 + · · · +

r γs

βs

Boundary operators: Two-sided ideal (Φ) = left F-module generated by |Φ) r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 8 / 21

SLIDE 35

Stieltjes Boundary Conditions

Definition

The elements of right ideal |Φ) = Φ · FΦ[∂, r ] are called Stieltjes boundary conditions. Normal forms: β(u) = α1(uk1) + · · · + αr(ukr ) + r γ1

β1 + · · · +

r γs

βs

Boundary operators: Two-sided ideal (Φ) = left F-module generated by |Φ) Standard decomposition: FΦ[∂, r ] = F[∂] ∔ F[ r ] ∔ (Φ)

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 8 / 21

SLIDE 36

Equitable Operators

Equitable operator ring used for extracting Green’s function via ι: FΦ[∂, r ] → F[∂, r

Φ],

the translation isomorphism defined as follows: r r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 9 / 21

SLIDE 37

Equitable Operators

Equitable operator ring used for extracting Green’s function via ι: FΦ[∂, r ] → F[∂, r

Φ],

the translation isomorphism defined as follows: Fixes F and ∂. r r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 9 / 21

SLIDE 38

Equitable Operators

Equitable operator ring used for extracting Green’s function via ι: FΦ[∂, r ] → F[∂, r

Φ],

the translation isomorphism defined as follows: Fixes F and ∂. Translate characters by ι(ϕ) = id − r

ϕ∂.

r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 9 / 21

SLIDE 39

Equitable Operators

Equitable operator ring used for extracting Green’s function via ι: FΦ[∂, r ] → F[∂, r

Φ],

the translation isomorphism defined as follows: Fixes F and ∂. Translate characters by ι(ϕ) = id − r

ϕ∂.

Translate back integrals by ι−1( r

ϕ) = (id −ϕ)

r .

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 9 / 21

SLIDE 40

Extracting Green’s Function

Interval J ⊂ R containing all evaluation points.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 10 / 21

SLIDE 41

Extracting Green’s Function

Interval J ⊂ R containing all evaluation points.

Theorem

The Green’s function of any regular Stieltjes boundary problem with m evaluations α1, . . . , αm has the form g(x, ξ) = ˜ g(x, ξ) + ˆ g(x, ξ), where the functional part ˜ g ∈ C(J2) is defined by the 2(m − 1) case branches ξ ∈ [αi, αi+1] x ≤ ξ, ξ ∈ [αi, αi+1] ξ ≤ x, while the distributional part ˆ g(x, ξ) is an F-linear combination of the δ(ξ − αi) and their derivatives.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 10 / 21

SLIDE 42

Illustration of Proof

Consider an example: Green’s operator G = x r − r x + x⌊1⌋ r x − x⌊1⌋ r + ex⌊−1⌋∂ ∈ FΦ[∂, r ]. r r r r r r r r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 11 / 21

SLIDE 43

Illustration of Proof

Consider an example: Green’s operator G = x r − r x + x⌊1⌋ r x − x⌊1⌋ r + ex⌊−1⌋∂ ∈ FΦ[∂, r ]. Translate to equitable ring by ⌊α⌋ r → r

0 −

r

α.

r r r r r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 11 / 21

SLIDE 44

Illustration of Proof

Consider an example: Green’s operator G = x r − r x + x⌊1⌋ r x − x⌊1⌋ r + ex⌊−1⌋∂ ∈ FΦ[∂, r ]. Translate to equitable ring by ⌊α⌋ r → r

0 −

r

α.

Yields G = x r

0 x − x

r

1 x −

r

0 x + x

r

1 + ex⌊−1⌋∂ ∈ F[∂,

r

Φ].

r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 11 / 21

SLIDE 45

Illustration of Proof

Consider an example: Green’s operator G = x r − r x + x⌊1⌋ r x − x⌊1⌋ r + ex⌊−1⌋∂ ∈ FΦ[∂, r ]. Translate to equitable ring by ⌊α⌋ r → r

0 −

r

α.

Yields G = x r

0 x − x

r

1 x −

r

0 x + x

r

1 + ex⌊−1⌋∂ ∈ F[∂,

r

Φ].

Extract ˜ g by f r

αg = f (x) g(ξ) [α ≤ ξ][ξ ≤ x]

− f (x) g(ξ) [ξ ≤ α][x ≤ ξ].

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 11 / 21

SLIDE 46

Illustration of Proof

Consider an example: Green’s operator G = x r − r x + x⌊1⌋ r x − x⌊1⌋ r + ex⌊−1⌋∂ ∈ FΦ[∂, r ]. Translate to equitable ring by ⌊α⌋ r → r

0 −

r

α.

Yields G = x r

0 x − x

r

1 x −

r

0 x + x

r

1 + ex⌊−1⌋∂ ∈ F[∂,

r

Φ].

Extract ˜ g by f r

αg = f (x) g(ξ) [α ≤ ξ][ξ ≤ x]

− f (x) g(ξ) [ξ ≤ α][x ≤ ξ]. Extract ˆ g by f ⌊α⌋∂i = (−1)if (x) δ(i)(ξ − α).

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 11 / 21

SLIDE 47

Illustration of Proof

Consider an example: Green’s operator G = x r − r x + x⌊1⌋ r x − x⌊1⌋ r + ex⌊−1⌋∂ ∈ FΦ[∂, r ]. Translate to equitable ring by ⌊α⌋ r → r

0 −

r

α.

Yields G = x r

0 x − x

r

1 x −

r

0 x + x

r

1 + ex⌊−1⌋∂ ∈ F[∂,

r

Φ].

Extract ˜ g by f r

αg = f (x) g(ξ) [α ≤ ξ][ξ ≤ x]

− f (x) g(ξ) [ξ ≤ α][x ≤ ξ]. Extract ˆ g by f ⌊α⌋∂i = (−1)if (x) δ(i)(ξ − α). Resulting Green’s function:

˜ g(x, ξ) =

(x − 1)ξ

for 0 ≤ ξ ≤ x ≤ 1, x(ξ − 1) for 0 ≤ x ≤ ξ ≤ 1. ˆ g(x, ξ) = −exδ′(ξ + 1)

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 11 / 21

SLIDE 48

Four Point Boundary Problem

Back to our first example: −u′′ = f , u(0) + u(1/3) = u(1) + u(2/3) = 0 r r r r r r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 12 / 21

SLIDE 49

Four Point Boundary Problem

Back to our first example: −u′′ = f , u(0) + u(1/3) = u(1) + u(2/3) = 0 Green’s operator (computed using INTDIFFOP by A. Korporal): G = x r − r x + (−5/24 + x/4)⌊1/3⌋ r + (5/8 − 3x/4)⌊1/3⌋ r x + (1/8 − 3x/4)⌊1⌋ r x + (1/12 − x/2)⌊2/3⌋ r + (−1/8 + 3x/4)⌊2/3⌋ r x

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 12 / 21

SLIDE 50

Four Point Boundary Problem

Back to our first example: −u′′ = f , u(0) + u(1/3) = u(1) + u(2/3) = 0 Green’s operator (computed using INTDIFFOP by A. Korporal): G = x r − r x + (−5/24 + x/4)⌊1/3⌋ r + (5/8 − 3x/4)⌊1/3⌋ r x + (1/8 − 3x/4)⌊1⌋ r x + (1/12 − x/2)⌊2/3⌋ r + (−1/8 + 3x/4)⌊2/3⌋ r x Green’s function (computed using extra code) g(x, ξ) =       

(3/4)xξ − (5/8)ξ : 0 ≤ ξ ≤ 1/3, ξ ≤ x (3/4)xξ + (3/8)ξ − x : 0 ≤ ξ ≤ 1/3, x ≤ ξ (3/2)xξ − (5/4)ξ − (1/4)x + 5/24 : 1/3 ≤ ξ ≤ 2/3, ξ ≤ x (3/2)xξ − (1/4)ξ − (5/4)x + 5/24 : 1/3 ≤ ξ ≤ 2/3, x ≤ ξ (3/4)xξ − (9/8)ξ + (1/4)x + 1/8 : 2/3 ≤ ξ ≤ 1, ξ ≤ x (3/4)xξ − (1/8)ξ − (3/4)x + 1/8 : 2/3 ≤ ξ ≤ 1, x ≤ ξ

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 12 / 21

SLIDE 51

Graph of its Green’s Function

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 13 / 21

SLIDE 52

Nonclassical Boundary Problem

Example with three evaluations, nonlocal part and higher-order derivative: u′′ − u = f , u′′′(−1) − 1

0 u(ξ) ξ dξ = 0,

u′(−1) − u′′(1) + 1

−1 u(ξ) dξ = 0,

r r r r r r r r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 14 / 21

SLIDE 53

Nonclassical Boundary Problem

Example with three evaluations, nonlocal part and higher-order derivative: u′′ − u = f , u′′′(−1) − 1

0 u(ξ) ξ dξ = 0,

u′(−1) − u′′(1) + 1

−1 u(ξ) dξ = 0,

Green’s operator (with σ := 2(2e − 3)(e − 1) for brevity):

σG = σ/2 (exr e−x − e−xr ex) + 2(−ex+3 + ex+2 − ex+1 + e−x+2 − e−x+1)(⌊−1⌋∂ + ⌊1⌋ r x) + (e − 1)(−ex+2 − 2ex+1 + e−x+1)(⌊−1⌋ r + ⌊1⌋ r ) + (3ex+2 − ex+1 − 3e−x+1 + 3e−x)⌊1⌋ r ex + (2ex+2 − 3ex+1)(e−1⌊−1⌋ r e−x + e⌊−1⌋ r ex) + (−ex+3 − ex+2 + 2ex+1 + e−x+2 − e−x+1)⌊1⌋

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 14 / 21

SLIDE 54

Green’s Function for Nonclassical Problem

Green’s function

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 15 / 21

SLIDE 55

Green’s Function for Nonclassical Problem

Green’s function Distributional part

σ ˆ g(x, ξ) = (−ex+3 − ex+2 + 2ex+1 + e−x+2 − e−x+1) δ(ξ − 1) + 2 (−ex+3 + ex+2 − ex+1 + e−x+2 − e−x+1) δ′(ξ − 1)

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 15 / 21

SLIDE 56

Green’s Function for Nonclassical Problem

Green’s function Distributional part

σ ˆ g(x, ξ) = (−ex+3 − ex+2 + 2ex+1 + e−x+2 − e−x+1) δ(ξ − 1) + 2 (−ex+3 + ex+2 − ex+1 + e−x+2 − e−x+1) δ′(ξ − 1)

Functional part

˜ g(x, ξ) =                                                                −1 ≤ ξ ≤ 0 ξ ≤ x 3ex+2+ξ + 3ex−ξ − 2ex+1−ξ − 2e3+x+ξ +e3+x + e−x+1 + ex+2 − e−x+2 − 2ex+1 −1 ≤ ξ ≤ 0 x ≤ ξ −2ex+1 + 2e−x+2+ξ − 5e−x+1+ξ − 2ex+2−ξ −2e3+x+ξ + 3e−x+ξ + e−x+1 + ex+2 +e3+x + 3ex+1−ξ + 3ex+2+ξ − e−x+2 0 ≤ ξ ≤ 1 ξ ≤ x −2e3+x ξ − 2e−x+1ξ + 2ex+2ξ + 2e−x+2ξ −2ex+1ξ + 3ex+2+ξ + 3ex−ξ − 5ex+1−ξ +2e−x+1+ξ − ex+1+ξ − 2e−x+2+ξ + 2ex+2−ξ −e3+x − e−x+1 − ex+2 + e−x+2 + 2ex+1 0 ≤ ξ ≤ 1 x ≤ ξ −2e3+x ξ − 2e−x+1ξ + 2ex+2ξ + 2e−x+2ξ −2ex+1ξ + 3e−x+ξ + 3ex+2+ξ − e3+x −e−x+1 − ex+2 + e−x+2 + 2ex+1 −3e−x+1+ξ − ex+1+ξ Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 15 / 21

SLIDE 57

Graph of Functional Part of its Green’s Function

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 16 / 21

SLIDE 58

Summary

From Green’s operator to Green’s function:

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 17 / 21

SLIDE 59

Summary

From Green’s operator to Green’s function: Generalised Green’s functions for Stieltjes BPs.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 17 / 21

SLIDE 60

Summary

From Green’s operator to Green’s function: Generalised Green’s functions for Stieltjes BPs. Extraction algorithm from Green’s operators.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 17 / 21

SLIDE 61

Summary

From Green’s operator to Green’s function: Generalised Green’s functions for Stieltjes BPs. Extraction algorithm from Green’s operators. Form of Green’s functions: Decomposition g(x, ξ) = ˜ g(x, ξ) + ˆ g(x, ξ).

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 17 / 21

SLIDE 62

Summary

From Green’s operator to Green’s function: Generalised Green’s functions for Stieltjes BPs. Extraction algorithm from Green’s operators. Form of Green’s functions: Decomposition g(x, ξ) = ˜ g(x, ξ) + ˆ g(x, ξ). Functional part ˜ g(x, ξ) defined by 2(m − 1) case branches.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 17 / 21

SLIDE 63

Summary

From Green’s operator to Green’s function: Generalised Green’s functions for Stieltjes BPs. Extraction algorithm from Green’s operators. Form of Green’s functions: Decomposition g(x, ξ) = ˜ g(x, ξ) + ˆ g(x, ξ). Functional part ˜ g(x, ξ) defined by 2(m − 1) case branches. Distributional part ˆ g(x, ξ) is F-linear combination of δ(ξ − αi) and their derivatives.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 17 / 21

SLIDE 64

Future work

In this paper: F = C ∞(R). r

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 18 / 21

SLIDE 65

Future work

In this paper: F = C ∞(R). How about general (F, ∂, r )?

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 18 / 21

SLIDE 66

Future work

In this paper: F = C ∞(R). How about general (F, ∂, r )? Need algebraic structures where these Green’s functions “live”:

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 18 / 21

SLIDE 67

Future work

In this paper: F = C ∞(R). How about general (F, ∂, r )? Need algebraic structures where these Green’s functions “live”: Functional part g(x, ξ): Ring F ⊗ F is sufficient.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 18 / 21

SLIDE 68

Future work

In this paper: F = C ∞(R). How about general (F, ∂, r )? Need algebraic structures where these Green’s functions “live”: Functional part g(x, ξ): Ring F ⊗ F is sufficient. Distributional part ˆ g(x, ξ): Integro-differential module generated over F ⊗ F by “algebraic Diracs”.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 18 / 21

SLIDE 69

Thank you

SLIDE 70

For Further Reading I

A. Korporal, G. Regensburger, and M. Rosenkranz.

A Maple package for integro-differential operators and boundary problems, 2010. Also presented as a poster at ISSAC ’10.

M. Rosenkranz and G. Regensburger.

Integro-differential polynomials and operators. In D. Jeffrey, editor, ISSAC’08: Proceedings of the 2008 International Symposium on Symbolic and Algebraic Computation. ACM Press, 2008.

M. Rosenkranz and G. Regensburger.

Solving and factoring boundary problems for linear ordinary differential equations in differential algebras. Journal of Symbolic Computation, 43(8):515–544, 2008.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 20 / 21

SLIDE 71

For Further Reading II

M. Rosenkranz.

A new symbolic method for solving linear two-point boundary value problems on the level of operators.

J. Symbolic Comput., 39(2):171–199, 2005.
I. Stakgold and M. Holst.

Green’s functions and boundary value problems. Pure and Applied Mathematics (Hoboken). John Wiley & Sons, Inc., Hoboken, NJ, third edition, 2011.

Markus Rosenkranz, Nitin Serwa (Uni. of Kent) Green’s Functions ISSAC, 2015 21 / 21