Two-point boundary value problems,Green’s functions, and product integration

Geordie McBain and Steve Armfield

geordie.mcbain@aeromech.usyd.edu.au

School of Aerospace, Mechanical, & Mechatronic Engineering

The University of Sydney, AUSTRALIA

CTAC 2004 – p. 1

2-pt boundary value problems

We consider numerical solution of 2-pt BVPs.

A simple 2-pt BVP is the 1-D conduction equation:

− d2

dx2u(x) = f(x) , (−1 < x < 1)

u(±1) = 0

This governs:the (unknown) temperature u(x) in a slab, given(known) 1-D heat generation f(x) , andisothermal surfaces.

More generally we have a linear ODE on a linesegment, and linear BCs on u(x), u′(x), . . .

CTAC 2004 – p. 2

Applications

Assume operator & BCs simple, but force term f(x)messy and possibly only known via numericalapproximation.

Also many force terms for same operator & BCs.

Such problems arise in some techniques for solving:nonlinear 2-pt BVPs by iteration;quasilinear parabolic PDEs on line segment byexplicit time-discretization of nonlinear terms:

heat equation, Burgers equation;Navier–Stokes DNS for Couette, Poiseuille flows,following Fourier decomposition in twohomogeneous directions.

CTAC 2004 – p. 3

Green’s functions

Quite generally, the solution of a linear 2-pt BVP is

u(x) =

∫ 1

−1G(x, ξ)f(ξ) dξ .

This is called Green’s integral representation .G(x, ξ) is Green’s function .

Depends on operator and BCs, but not force f(x) .

The Green’s function can be difficult to find.‘Variation of Parameters’ sometimes effective.

We assume Green’s function has been found.True in many cases, including Navier–Stokes.

CTAC 2004 – p. 4

Green’s function example

For the conduction equation −u′′(x) = f(x) , u(±1) = 0 , thesolution is

u(x) =

∫ 1

−1G(x, ξ)f(ξ) dξ

where

G(x, ξ) =

{

12(1 − x)(1 + ξ) , ξ 6 x12(1 + x)(1 − ξ) , x 6 ξ ,

.

-1 -0.5 0 0.5 1

G(x

,ξ)

INTEGRATION VARIABLE, ξ

CTAC 2004 – p. 5

Evaluating Green’s representation

It may seem as though once the Green’s function hadbeen found, the problem is solved; however, the integralis still to be evaluated.

Gaussian quadrature is doesn’t achieve its usual highaccuracy, since the Green’s function is cusped at x = ξ .

The O(n2) matrix–vector multiplication is not competitivewith O(n) sparse schemes for the differential equation,like central finite differences with Thomas’s algorithm.

The failure of high-order quadrature schemes wasnoted by Atkinson (1976, ACM Trans. math. Software),who also proposed (but didn’t implement) a remedy:product integration.

CTAC 2004 – p. 6

Product integration

Quadrature rules approximate an integral:

∫ b

a

f(ξ) dξ.=

N∑

k=1

wkf(ξk) .

Product integration approximates a weighted integral:

∫ b

a

K(ξ)f(ξ) dξ.=

N∑

k=1

wkf(ξk) .

The weights wk now depend on the factor K(ξ) .

Worthwhile if either kernel K is badly behaved or manyfunctions f to be integrated, or, as here, both.

CTAC 2004 – p. 7

Outline of the presentation

Here, we develop Atkinson’s idea of evaluating Green’sintegral representation using product integration.

Implement, using Chebyshev polynomials and Fejér’sfirst quadrature rule.

Solve some example problems:a simple 2-pt BVP (conduction equation)an oscillatory 2-pt BVP (ditto, with oscillatory f )a quasilinear parabolic PDE (Burgers equation).

Compare performance against:2nd-order central finite differences; andorthogonal collocation (‘pseudospectral method’).

CTAC 2004 – p. 8

Developing a product integration rule

We want the weights wk and abscissae ξk in

∫ 1

−1K(ξ)f(ξ) dξ

.=

n∑

k=1

wkf(ξk) .

Say the function can be expanded as

f(x).=

n∑

j=1

f̃jφj(x) .

Then the integral becomes

∫ 1

−1K(ξ)f(ξ) dξ

.=

n∑

j=1

{∫ 1

−1K(ξ)φj(ξ) dξ

}

f̃j

CTAC 2004 – p. 9

Expansion by collocation

Now say the expansion is done by collocation

f̃j =n∑

k=1

tjkf(ξk) .

This is a discrete transform.It dictates the abscissae ξj .

The integration becomes

∫ 1

−1K(ξ)f(ξ) dξ

.=

n∑

k=1

n∑

j=1

{∫ 1

−1K(ξ)φj(ξ) dξ

}

tjk

f(ξk)

≡n∑

k=1

wkf(ξk)

CTAC 2004 – p. 10

Apply to Green’s integral representation

The product integral rule for

u(x) =

∫ 1

−1G(x, ξ)f(ξ) dξ

is

u(x) =n∑

k=1

wk(x)f(ξk)

where

wk(x) =n∑

j=1

{∫ 1

−1G(x, ξ)φj(ξ) dξ

}

tjk

CTAC 2004 – p. 11

Evaluate at abscissae

At discrete abscissae x = xi ,

u(xi) =n∑

k=1

wk(xi)f(ξk)

or

ui =n∑

k=1

γikfk

where ui = u(xi) , fk = f(xk) , and

γik = wk(xi)

=n∑

j=1

{∫ 1

−1G(xi, ξ)φj(ξ) dξ

}

tjk

CTAC 2004 – p. 12

Discrete Green’s representation

Thus, what we have is a matrix–vector product

u(xi) =

∫ 1

−1G(x, ξ)f(ξ) dξ

.= ui =

n∑

i=1

γikfk ,

where

γik =n∑

j=1

gijtjk (matrix–matrix product)

gij =

∫ 1

−1G(xi, ξ)φj(ξ) dξ (‘moments’ of G(x, ξ))

and the tjk are the coefficients of the discrete transform.

CTAC 2004 – p. 13

Moments of the Green’s function

Assume the basis and abscissae are chosen such thatthe coefficients of the discrete transform is known.

Then what remains is to calculate the (modified)moments

gij =

∫ 1

−1G(xi, ξ)φj(ξ) dξ

Possible analytically in some cases.

In general, numerical integration may be required.

CTAC 2004 – p. 14

Splitting the moment integral

Need to be careful, asGreen’s function has discontinuous derivatives at ξ = xi .

-1 -0.5 0 0.5 1

G(x

,ξ)

INTEGRATION VARIABLE, ξ

Split integration domain at ξ = xi:

gij =

∫ xi

−1G(xi, ξ)φj(ξ) dξ +

∫ 1

xi

G(xi, ξ)φj(ξ) dξ .

and apply a quadrature rule to each of the two parts.

CTAC 2004 – p. 15

Specific choices

Specific choices need to be made for:the basis;the abscissae;the quadrature rule.

Here, we use:Chebyshev polynomials for the basis functions,φj(x) = Tj−1(x), j = 1, 2, . . . , n ;

‘classical’ Chebyshev abscissae, xk = cos (2k−1)π2n

;So expansion is the discrete cosine transform.

Fejérs first quadrature rule.Equivalent to Chebyshev expansion on classicalabscissae on the subdomains.

CTAC 2004 – p. 16

Example 1: conduction with absorption

−u′′(x) = f(x), u(±1) = 0

f(x) =

(

c

cos cx4

)2

c =√

2 cosc

4.= 0.1336

Exact solution: u(x) = 2 log ccos cx

4

− log 2

Linear version of a ‘solid fuel ignition’ or Bratu problem.

-1 -0.5 0 0.5 1

TE

MP

ER

AT

UR

E, u

COORDINATE, xCTAC 2004 – p. 17

Example 1: results

10-16

10-12

10-8

10-4

100

104

1 10 100 1000

MA

XIM

UM

AB

SO

LUT

E N

OD

AL

ER

RO

R

NUMBER OF ABSCISSAE, n

finite differenceorthogonal collocation

present methodGreen + GaussianGreen + midpoint

CTAC 2004 – p. 18

Example 2: oscillatory absorption

−u′′(x) = f(x), u(±1) = 0

f(x) = m2π2 sin mπx

Exact solution: u(x) = sin mπx .

-1 -0.5 0 0.5 1

TE

MP

ER

AT

UR

E, u

COORDINATE, x

Oscillations mean a certain number of abscissae will berequired for resolution before asymptotic convergencebegins.

CTAC 2004 – p. 19

Example 2: results

10-14

10-12

10-10

10-8

10-6

10-4

10-2

100

102

104

1 10 100 1000

MA

XIM

UM

AB

SO

LUT

E N

OD

AL

ER

RO

R

NUMBER OF ABSCISSAE, n

finite differenceorthogonal collocation

present method

CTAC 2004 – p. 20

Quasilinear parabolic equations

Application of ‘time-splitting’ to parabolic equationsleads to a sequence of 2-pt BVPs with the same linearoperator and BCs but different RHSs.

explicit stepping for nonlinear or time-varying termsimplicit for steady linear terms

2nd-order Adams–Bashforth/Crank–Nicolson popular

Example: the heat equation u̇ − νu′′ = ϕ becomes(

1 − ν∆t

2

d2

dx2

)

uk =

(

1 +ν∆t

2

d2

dx2

)

uk−1

+∆t

2{3ϕk−1 − ϕk−2}

CTAC 2004 – p. 21

Resulting 2-pt BVP

Like u − λ2u′′ = f at each time-step.

The Green’s function is

G(x, ξ) =

{

sinh λ(1−x) sinh λ(1+ξ)λ sinh 2λ

, ξ 6 xsinh λ(1+x) sinh λ(1−ξ)

λ sinh 2λ, x 6 ξ

λ =√

2ν∆t

decreases with time-step.

CTAC 2004 – p. 22

Green’s function for u − λ2u′′= f

λ = 1

-1 -0.5 0 0.5 1

G(x

,ξ)

INTEGRATION VARIABLE, ξ

λ = 10

-1 -0.5 0 0.5 1

G(x

,ξ)

INTEGRATION VARIABLE, ξ

Narrows with decreasing time-step.CTAC 2004 – p. 23

Example 3: Burgers equation

Unsteady viscous 1-D Burgers equation:

u̇ − νu′′ = −uu′, u(±1, t) = 0, u(x, 0) = u0(x) .

Quasilinear heat equation with f(x, t) = uu′ .

Use ν = 1/100π and u0(x) = − sin πx (much studied).

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

-1 -0.5 0 0.5 1

u

x

t=1/πt=2/πt=3/π

A near-shock forms and then decays.

Compare peak slope with analytical value (−150.00516 ).CTAC 2004 – p. 24

Example 3: results

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100 1000

AB

SO

LUT

E R

ELA

TIV

E E

RR

OR

NUMBER OF ABSCISSAE, n

FDM, τ/dt=128FDM, 256FDM, 512

FDM, 1024present, 128present, 256present, 512

present, 1024

CTAC 2004 – p. 25

Computational time requirements

There is some competition between:sparse low-order direct (attacking ODE) methodsdense high-order Green’s function methods.

Present method requires matrix–vector product: O(n2) .

Finite difference method requires Thomas: O(n) .

Can more rapid convergence offset complexity?

Compare timings for the oscillatory 2-pt BVP.Excluding initialization time!

Measure CPU times for oscillatory BVP.104 – 106 runs after initialization.

CTAC 2004 – p. 26

Timings for oscillatory BVP

10-12

10-10

10-8

10-6

10-4

10-2

10-5 10-4 10-3 10-2

MA

XIM

UM

NO

DA

L E

RR

OR

CPU TIME (s)

finite differencepresent method

CTAC 2004 – p. 27

Conclusions

Given the Green’s function, we have developed a veryaccurate numerical scheme for 2-pt BVPs.

Can also be used for time-step of parabolic 1-D PDEs.

Utilizes product integration of Green’s integral:Chebyshev expansion of RHSFejér’s first quadrature rule on either side of cusp

Converges as rapidly as orthogonal collocation.

Doesn’t suffer collocation’s ill-conditioning for large n.

Operation count: O(n2) dense matrix–vector product.

Faster than O(n) sparse method for same accuracy.Excluding initialization time!

Suited to multiple-RHS problems; e.g. 1-D parabolic.CTAC 2004 – p. 28