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##### Activity
Mitko Georgiev
@mitkoge
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

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
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.
ok thanks, good to know
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?
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?

Sheehan Olver
@dlfivefifty

Not an easy way.

I take that back. I think you can always do ApproxFun.SpaceOperator(op, newdomainspace, newrangespace) change the spaces of an operator. So just write Q with the "right" spaces