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, [email protected] 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 >> >>
