I don't know what you are trying to say, but did you try running the two programs with the same relative and absolute tolerances? Do the results look more similar now?
On 12 June 2016 at 02:01, <jmarcellopere...@ufpi.edu.br> wrote: > Hi daniel > in fact, there is a relationship with the tolerance between the methods. > > Em sábado, 11 de junho de 2016 11:51:53 UTC-3, Daniel Carrera escreveu: >> >> Probably because the Julia module has different default values for the >> absolute and relative tolerance. For Julia, the defaults are reltol=1e-5 >> and abstol=1e-8 but in Matlab the defaults are RelTol=1e-3 and AbsTol=1e-6. >> >> https://github.com/JuliaLang/ODE.jl >> http://se.mathworks.com/help/matlab/ref/odeset.html >> >> That would also explain why the Julia program took so much longer. Try >> again with the same tolerance and see if the results are more similar. Oh, >> and like Erik said, you probably meant to have parenthesis around "1/8". If >> the results still look difference, look for more options that might have >> different default values. >> >> Cheers, >> Daniel. >> >> >> >> On Saturday, 11 June 2016 04:29:47 UTC+2, jmarcell...@ufpi.edu.br wrote: >>> >>> this is the test for equation differential using runge-kutta45: f(x,y)= >>> (-5*x - y/5)^1/8 + 10 >>> >>> >>> <https://lh3.googleusercontent.com/-D6U4d5pN0pw/V1t3MaV3JpI/AAAAAAAAACU/-E_ceVhrTxkT3SAgmLUy5nHDhRJTIikSgCLcB/s1600/rk45.png> >>> >>> why the numerical result is different? I used : >>> >>> function Rk_JL() >>> f(x,y)= (-5*x - y/5)^1/8 + 10 >>> tspan = 0:0.001:n >>> y0 = [0.0, 1.0] >>> return ODE.ode45(f, y0,tspan);end >>> >>> >>> and >>> >>> >>> function [X1,Y1] = RK_M() >>> f = @(x,y) (-5*x - y/5)^1/8 + 10; >>> tspan = 0:0.001:n; >>> y0 = 1 >>> [X1,Y1]= ode45(f,tspan,1);end >>> >>>
