# Inverse Scattering-checkpoint

In [27]:
using ApproxFun, SingularIntegralEquations, RiemannHilbert, Plots, QuadGK, DualNumbers, ComplexPhasePortrait
import ApproxFunBase: SequenceSpace, BasisFunctional, ℓ⁰, SpaceOperator, piece, pieces, npieces
import ApproxFun: eigs
import ApproxFunOrthogonalPolynomials: Recurrence
import RiemannHilbert: RiemannDual, logpart, finitepart, istieltjes, LogNumber
import SingularIntegralEquations: ,
using RiemannHilbert.KdV

## The direct scattering transform

We first consider calculation of discrete spectrum of Schrödinger: $$\psi'' + V(x) \psi = \lambda \psi$$

In [196]:
V = x -> 1.2exp(-x^2)
d = -10..10 # computational domain
D = Derivative()

λ,Q = eigs(Dirichlet(d), D^2 + Fun(V,d), 500)
scatter(λ, zero.(λ); xlims=(-2,2))

Out[196]:

Associated with the continuous spectrum is the reflection coefficient. This can be calculated:

In [218]:
R = ReflectionCoefficient(V)
R(0.1)

Out[218]:
-0.9651837057640016 + 0.06801873361190361im

Behind the scenes, its solving an ODE for $\psi$ on (-∞,0] and $\phi_\pm$ on [0,∞), by truncating the domain:

In [214]:
M = 10
V = x -> 1.2exp(-x^2)
V₋,V₊ = Fun(V, (-M)..0), Fun(V, 0..M)

k = 1
@time ψ = [ivp(); D^2  + (V₋ + k^2)] \ [exp(-im*k*(-M)), -im*k*(exp(-im*k*(-M))), 0.0]

F = qr([rdirichlet(space(V₊)); rneumann(); D^2  + (V₊ + k^2)])
φ = F \ [exp(im*k*M), im*k*(exp(im*k*M)), 0.0]
φ = F \ [exp(-im*k*M), -im*k*(exp(-im*k*M)), 0.0]

plot(imag(ψ); legend=:bottomleft, label="\\psi", title="k = $k, imaginary part") plot!(imag(φ); label="\\phi\\_+") plot!(imag(φ); label="\\phi\\_-") 0.015890 seconds (161.25 k allocations: 6.126 MiB) Out[214]: In [215]: a,b = [φ(0) φ(0); φ'(0) φ'(0) ] \ [ψ(0); ψ'(0)] plot(imag(ψ); legend=:bottomleft, label="\\psi", title="k =$k, imaginary part")
plot!(imag(a*φ + b*φ); label="a\\phi\\_+ + b\\phi\\_-")

Out[215]:

In [220]:
q = Q[findmax(λ)[2]]
q = q/norm(q)
plot(x -> -V(x), -10,10; label="potential",  legend=:bottomright, ylims=(-2.5,2.5), linestyle=:dot)
plot!(-findmax(λ)[1]+abs2(q); label="eigenfunction")
plot!(abs2(ψ+a*φ + b*φ)+k; label="wave")

Out[220]:

## The inverse scattering transform

With $R$ we have han efficient way of calculating reflection coefficients. We now see how to solve the Riemann–Hilbert problem, but lets simplify things so that we only have continuous spectrum:

In [221]:
V = x -> 0.1exp(-x^2)
d = -10..10 # computational domain
D = Derivative()

λ,Q = eigs(Dirichlet(d), D^2 + Fun(V,d), 500)
scatter(λ, zero.(λ); xlims=(-2,2))

Out[221]:

The first step is to expand $R$ into a Chebyshev expansion. We use tFun which takes advantage of parallelisation:

In [222]:
@time ρ = tFun(R, -5.0..5, 300)
plot(ρ)

5.145820 seconds (73.69 M allocations: 2.392 GiB, 13.47% gc time)
Out[222]:

For any $t$ and $x$ we can construct the jump function, but we would need to deform for large $t$ and $x$. Lets keep life simple and take $t = x = 0$, in which case we get the following jump:

In [223]:
k = Fun(identity, space(ρ))
G = [1-abs2.(ρ) -conj.(ρ);
ρ           1.0];

We can now find the solution to $\Phi_+ = \Phi_- G$. RiemannHilbert.jl uses the transpose version ($\Phi_+ = G \Phi_-$) so we transpose twice:

In [224]:
@time Φ = transpose(rhsolve(transpose(G), 2*4*200)); # asymptotic to I
Φ = [1 1]*Φ; # asymptotic to [1 1]

0.615777 seconds (2.02 M allocations: 282.682 MiB)

It worked!

In [225]:
Φ(0.1+0.0im)  Φ(0.1-0.0im)*G(0.1)

Out[225]:
true

We can then recover $V$ from Φ. I'm not going to show that since there's a bug 😳

In [234]:
ψ(0.0)/a - (φ(0.0) + ρ(1)*φ(0.0))

Out[234]:
-1.1102230246251565e-16 + 1.1657341758564144e-15im

In [227]:
Φ(1.0-0.0im)

Out[227]:
1×2 Array{Complex{Float64},2}:
1.05043-0.372028im  0.942763+0.332064im

In [230]:
Φ(1.0+0.0im)

Out[230]:
1×2 Array{Complex{Float64},2}:
0.942763-0.332064im  1.05043+0.372028im

In [219]:
R(1)

Out[219]:
-0.07910491062180808 + 0.06606135311851252im

In [232]:
ψ(0.0)/a

Out[232]:
0.6642329847792622 + 0.3535499072338503im

In [231]:
(ϕ₋)(0.0)

Out[231]:
0.9732652639976271 + 0.027633169815842518im

In [233]:
ϕ₊(0.0)

Out[233]:
0.9732652639976271 - 0.027633169815842518im

In [ ]:
S = []

In [164]:
ψ(-10.0)exp(im*1*(-10.0))

Out[164]:
1.0000000000000007 + 5.551115123125783e-17im

In [146]:
Φ(1-0.0im)

Out[146]:
2×2 Array{Complex{Float64},2}:
1.03402-0.113411im  -0.0808899+0.20814im
0.0090288-0.286308im     1.00923+0.134907im

In [158]:
ϕ₊(0.0)

Out[158]:
0.7008654655943309 - 0.28472652028393397im

In [151]:
ϕ₋(1)

Out[151]:
0.5383978200907342 - 0.7947809510658824im

In [ ]: