[julia-users] A Julia code slower than Matlab, can it be faster?

[博陈]

This is my first julia code, I am happy it did the right thing, but 
compared with the Matlab code that did the same thing, it runs so slowly. 
The Matlab code takes about 90s, but the julia below takes 130s.

Potential(x::Float64, y::Float64) = -2./sqrt(x^2+1.) -2./sqrt(y^2+1.)  + 
1./sqrt((x-y)^2+1.)
Ko(x::Float64,y::Float64) = x^2+y^2

function ground()
    R = 320.
    Ks = 2^11
    x = linspace(-R, R, Ks+1)
    dx = 2R/Ks
    dt = 0.1
    pm = 2π/2dx
    px = circshift( linspace(-pm, pm, Ks+1),  round(Int64, Ks/2) )
    FFTW.set_num_threads(8)

    V = Float64[Potential(ix, iy)  for ix in x, iy in x]
    ko = Float64[e^(-dt/4* (Ko(ipx,ipy))) for ipx in px, ipy in px]
    ko2 = ko.^2
    vo = Float64[e^(-dt* (Potential(ix,iy))) for ix in x, iy in x]
    ϕ = Array(Complex128,(Ks+1,Ks+1))
    Pl = plan_fft!(ϕ; flags=FFTW.MEASURE)
    ϕ = V.*complex(1.)

    normphi = sqrt(sumabs2(ϕ))*dx
    ϕ /= normphi

    ϕ = Pl*ϕ
    ϕ .*= ko
    ϕ = Pl\ϕ

    Δϕ = 1.
    nstep = 0
    print("start loop")
     while Δϕ > 1.e-15
        ϕ′ = ϕ
        ϕ .*= vo

        normphi = sqrt(sumabs2(ϕ))*dx
        ϕ /= normphi

        ϕ = Pl*ϕ
        ϕ .*= ko2
        ϕ = Pl\ϕ
# if check  Δϕ for every step, 35s is needed.
        if nstep>500
            Δϕ = maxabs(ϕ-ϕ′)
        end
        nstep += 1
        if mod(nstep,200)==0
            print(nstep,"  ")
            print("Δϕ=",Δϕ,"   ")
        end

    end
    ϕ .*= vo

    ϕ = Pl*ϕ
    ϕ .*= ko
    ϕ = Pl\ϕ

    normphi = sqrt(sumabs2(ϕ))*dx
    ϕ /= normphi
end