JuliaApproximation/ApproxFun.jl

Mitko Georgiev

@mitkoge

I tried tzhe following

using ApproxFun
sp1= Chebyshev(0.0..50.0) ∪ Chebyshev(50.0..60.0)
x1=Fun(sp1); g1 = Fun(exp(-x1^2),sp1)
s1= sum(g1) #NaN
sp2= Chebyshev(0.0..5.0) ∪ Chebyshev(5.0..60.0)
x2=Fun(sp2); g2 = Fun(exp(-x2^2),sp2)
s2= sum(g2) #0.8862269254527582

i expected s1~= s2. but it seems it is not.

Mikael Slevinsky

@MikaelSlevinsky

I pushed a fix for this on this branch JuliaApproximation/ApproxFun.jl@07f50d0. If the tests pass, then it could be merged.

Mitko Georgiev

@mitkoge

Wow i am impressed by your reactivity! Fixed, pushed and merged in a day! Thank you!

Mitko Georgiev

@mitkoge

May i ask another question?
I try to reproduce an example from this chat given by @cortner

using ApproxFun
dom = Interval(0.001, 1) * PeriodicInterval(-pi, pi)
space = Space(dom)
Dr = Derivative(space, [1,0])
Dθ = Derivative(space, [0,1])
Mr = Multiplication(Fun( (r, θ) -> r, space ), space)
rDr = Mr * Dr
L = rDr * rDr + Dθ * Dθ

but PeriodicInterval(-pi, pi) is no more recognized.
What alternative can be used? Circle?

Sheehan Olver

@dlfivefifty

It’s been renamed PeriodicSegment as it’s really an oriented line segment

Mitko Georgiev

@mitkoge

Thank you,
i tried

julia> dom = Interval(0.001, 1) * PeriodicSegment(-pi, pi)
ERROR: MethodError: no method matching *(::Interval{:closed,:closed,Float64}, ::PeriodicSegment{Floa
t64})

and

julia> dom = Segment(0.001, 1) * PeriodicSegment(-pi, pi)
ERROR: MethodError: no method matching *(::Segment{Float64}, ::PeriodicSegment{Float64})

Sheehan Olver

@dlfivefifty

Ah and * has changed to imes

imes

\times

Mitko Georgiev

@mitkoge

thank you.

using ApproxFun,LinearAlgebra
dm = Segment(0.001,1) × PeriodicSegment(-pi,pi)
sp = Space(dm)
Dr = Derivative(sp, [1,0]); Dθ = Derivative(sp, [0,1])
Mr = Multiplication(Fun( (r, θ) -> r, sp ), sp)
rDr = Mr * Dr
Lr = rDr * rDr; L = Lr + Dθ * Dθ

i modified it and now can play with.

Simon Etter

@ettersi

Not sure whether this is the right place to ask since technically my question concerns SingularIntegralEquations.jl, but that package seems to be related enough to ApproxFun.jl that I'll go ahead anyway.

In the code of SingularIntegralEquations it says

# stieltjesintegral is an indefinite integral of stieltjes
# normalized so that there is no constant term

Can someone elaborate on that normalisation condition?

Sheehan Olver

@dlfivefifty

Sure, fine to ask here. This normalisation condition comes from the fact that Jacobi polynomials have nice integration formulae coming from the weighted differentiation d/dx[(1-x)^a *(1+x)^b P_n^(a,b)] = C (1-x)^{a-1} (1+x)^{b-1} P_{n+1}^(a-1,b-1). Roughly speaking, for functions that vanish at ±1, derivatives and Stieltjes transforms commute, so provided a,b > 0 we can use this to write the integral of a Stieltjes transforms in terms of the integral of the integrand.

Note that the n = 0 case is special, and dictates the normalisation constant (as the construction for other n will automatically decay something like O(z^(-n)) )

For the cases actually implemented, I think its normalized to go to zero at ∞.

Sorry, that's not right.

You always have logarithmic growth (unless the integrand integrates to zero).

Sheehan Olver

@dlfivefifty

On -1..1 it just vanishes apart from the log term:

julia> x = Fun();

julia> f = exp(x)/sqrt(1-x^2);

julia> z = 10_000; stieltjesintegral(f,z) - log(z) * sum(f)
-0.00017756097918919522

julia> f = exp(x) * sqrt(1-x^2);

julia> z = 10_000; stieltjesintegral(f,z) - log(z) * sum(f)
-4.264886882765495e-5

Oh, wait, that's true for other intervals too:


julia> x = Fun(1..2);

julia> f = exp(x) * sqrt((x-1)*(2-x));

julia> z = 100_000; stieltjesintegral(f,z) - log(z) * sum(f)
-2.8356239955229512e-5

Sorry for making it seem more complicated than it actually it is. It's really simple: it's normalized so that stieltjes(f,z) = log(z) *sum(f) + o(1).

Which happens to be consistent with logkernel (whose normalization is imposed by the actual definition):

julia> logkernel(f, z) -log(abs(z))/π * sum(f)
-9.026071512430178e-6

Simon Etter

@ettersi

If I understand correctly, I can summarise what you said as: stieltjes(f,z) = integral log(z - w) f(w) dw + C with Csuch that stieltjes(f,z) = log(z) *sum(f) + o(1) for z to Inf. Correct?

Sheehan Olver

@dlfivefifty

Yes

Simon Etter

@ettersi

Great, thanks a lot!

Mitko Georgiev

@mitkoge

Hi i was playing with modified example from Quantum states.jl

Using  ApproxFun
fc1 = Fun(1, 0..1); fc2 = Fun(2, 1..2); fc3 = Fun(3, 2..3)

fc= fc1+fc2+fc3
D = Derivative(); sp = space(fc); x= Fun(sp);
B = [Dirichlet(sp); continuity(sp,0:1)]
λ, v = ApproxFun.eigs(B, -D^2 + fc, 500,tolerance=1E-10)  
#returns result

λ, v = ApproxFun.eigs(B, -D^2 - 1/x*D + fc, 500,tolerance=1E-10)
# says Not implemented  but if fc=fc1+fc2 it manages to return result

Is 1/x*D kind of a problematic composition?

Sheehan Olver

@dlfivefifty

Yes. It would be easy to work around by multiplying by x and updating the generalised eigenvalue call appropriately in the implementation of eigs.

Mitko Georgiev

@mitkoge

Thanks,
you mean calling like this

λ, v = ApproxFun.eigs_overx(B, -x*D^2 - D + x*fc, 500,tolerance=1E-10)

and somehow divide C1 to x before calling to

    λ,V = eigen(A1,C1)

inside the copy of eigs?

Sheehan Olver

@dlfivefifty

No multiply by x: instead of solving A*v = λ*C*v we solve x * A * v = λ * x * C* v. That is, in line 34 change to C = Conversion(ds,rangespace(A)); C = x * C should be enough.

Note eigs is quite stale and needs to be rewritten, but that's low priority for me. I'll accept any PRs.

Mitko Georgiev

@mitkoge

Thanks for the great support,

Mitko Georgiev

@mitkoge

i followed you and copy paste defined

    function eigs(Bcs_in::Operator,A_in::Operator,fmul::Fun,n::Integer;tolerance::Float64=100eps())

with the correction

        C = Conversion(ds,rangespace(A)); C = fmul * C

now have only to check if it works correct. :-)

λ, v = ApproxFun.eigs(B, L, x, 500,tolerance=1E-10)

Mitko Georgiev

@mitkoge

:point_up: 22. November 2018 12:35
I intendet to help with replacing it in the docs but as i see it is already fixed.
Unfortunately it might be not at http://juliaapproximation.github.io/ApproxFun.jl/latest

Ashton Bradley

@AshtonSBradley

is it possible to solve a nonlinear PDE in approxfun? for example I would like to add a simple nonlinear term to this problem
https://github.com/JuliaApproximation/ApproxFun.jl#solving-partial-differential-equations
, presumably switching to newton solver, but not sure how to inform it about boundary conditions etc in 2D

Sheehan Olver

@dlfivefifty

Not yet: I want to make the PDE solver faster before trying to implement that sort of thing. For now you can try doing Newton iteration by hand.

Ashton Bradley

@AshtonSBradley

ok thanks, good to know

Ashton Bradley

@AshtonSBradley

for the basic procedure of setting up for manual iteration, can I take it that this https://github.com/JuliaApproximation/ApproxFunExamples/blob/master/ODEs/Blasius%20and%20Falkner-Skan.jl
is a reasonable starting point, that I just need to generalise to 2D?

Ashton Bradley

@AshtonSBradley

Ok, maybe ignore that

Sheehan Olver

@dlfivefifty

It’s pretty simple: just differentiate wrt u but treating d/dx and d/dy as constants.

To get the Jacobian

Mitko Georgiev

@mitkoge

excuse if my question is too naive but i try this

x= Fun(0..1); sx=space(x);
a=Fun(0..pi); sa=space(a);
fa= DefiniteIntegral(x*a,sx)
# i expect fa= a here

but i get MethodError. Are integrals of two arguments possible?

Mitko Georgiev

@mitkoge

or may be like this

x= Fun(0..1); sx=space(x)
a=Fun(0..pi); sa=space(a)
fm= x*a; sm= space(fm)
fI=  DefiniteIntegral(sx)
(fI*fm)(0.2)
#expect 0.2

Sheehan Olver

@dlfivefifty

Are you expecting it to be a 2D function? And do you actually expect the result to be 0.1 (that is, a/2 since you integrate out x)?

That's possible via:

x,a = Fun((0..1) × (0..π))
Q = DefiniteIntegral(0..1) ⊗ I
(Q*(x*a))(0.0,0.2)

Mitko Georgiev

@mitkoge

Wow exciting how elegant is this! Thank you!
As you can see i am exploring it the dummy way so please bear with me.

Mitko Georgiev

@mitkoge

let Q2= Q (x a) . Q2 is still 2D Fun((x,a)-> smth). Is there a way to curry it to 1D function Q1 that is Fun(a-> smth)?

Sheehan Olver

@dlfivefifty

Not an easy way. What we actually want is DefiniteIntegral(S, [1,0]) that wraps the kronecker product. It’s not too hard to add this but I don’t have time right now

Mitko Georgiev

@mitkoge

to practice your guiding i wrote

function Hankel(fx::Fun, spx,spa::Space)
    spxa= spx ⊗ spa
    fx2= Fun((x,a)-> fx(x), spxa)
    fj= Fun((x,a) -> besselj0(x*a), spxa)
    fa2= (DefiniteIntegral(dmx1) ⊗ I)*(fx2*fj*x)
    fa= Fun(a->fa2(0,a), spa)
end

from the definition. It seems to work. Thank you!
Is this the style ApproxFun is expected to be used?

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