bartekltg <bartekltg@gmail.com> wrote in message <kqp8ic$864$1@node2.news.atman.pl>...
> W dniu 2013-06-30 10:32, William pisze:
> > bartekltg <bartekltg@gmail.com> wrote in message
> > <kqnu6a$k9n$1@node1.news.atman.pl>...
> >> W dniu 2013-06-28 22:38, William pisze:
> >> > Hi,
> >> >
> >> > I've been working on a program using ODE's and I've been using ODE45 to
> >> > get numerical solutions for differential equations. I've been getting
> >> > incorrect results e.g. when I used
> >> > y'' = -y, I get a period that is ~ 2.8 instead of pi. This also
> >> happened
> >> > with various other functions that I've been using.
> >> >
> >> > Is there a solution to this problem?
> >>
> >> I'm great wizard from quadruple-precision-in-matlab-land.
> >> My crystal ball told me you have done something wrong[1]. Sadly,
> >> this crystal ball is not HD ready, so I can't read you code:(
> >> Maybe you can post it? :)
> >>
> >> > Thanks!
> >>
> >> [1]:
> >> % solution of y'' = -y; y(1,:) is y, y(1,:) is velocity (y')
> >>
> >> [t,y] = ode45 ( @(t,Y) [Y(2);-Y(1)],[0,10],[0,1]);
> >>
> >> % of course ode45 solves first order equation, so we wrote
> >> % y'' = -y as system {y'=u, u'=-y}
> >>
> >> plot(t,y(:,1))
> >> grid
> >> %looks nice
> >>
> >> fzero(@(x)interp1(t,y(:,1),x),pi)
> >>
> >> % ans = 3.14134838301112
> >> % meh, but significantly better than 2.8
> >>
> >> options = odeset ('RelTol',1e-10,'AbsTol',1e-10); % quite useful:)
> >> [t,y] = ode45 ( @(t,x) [x(2);-x(1)],[0,10],[0,1],options );
> >>
> >> root = fzero(@(x)interp1(t,y(:,1),x),pi)
> >> (root-pi)/pi
> >>
> >> % root = 3.14159265459964
> >> % ans = 3.21445091288877e-010
> >> % :-)
> >>
> >>
> >
> > So my code looks something like this
> >
> > [t,phi] = ode45(@diffeq,0:T/100:T,[1,0]);
> > %T is the period of the diff eq.
>
> How do you now, how big is T?
> I think five (approximate) periods will be better.
>
> BTW, tspan other than [tstart, tend] give you more points,
> but doesn't increase accuracy.
>
> > then
> >
> > function phiprime = diffeq(t,phi)
> > m =1;
> > g=0;
> > l=1;
> > H = 0;
> > phiprime=[phi(2); -3*H.*phi(2)-m^2*phi(1) - g*phi(1).^2 - l*phi(1).^3];
> >
> > where i just let phi(1) = y, phi(2) = y'
>
> > I've tried this code with g, l, H = 0 which becomes y'' = -y
>
> And it works!
>
> %l=0,h=0,g=0
> T=10;
> [t,phi] = ode45(@diffeq,0:T/100:T,[1,0]);
> plot(t,phi(:,1))
> root1 = fzero(@(x)interp1(t,phi(:,1),x),1*pi/2);
> root2 = fzero(@(x)interp1(t,phi(:,1),x),3*pi/2);
> [root1,root2]-[pi/2,3*pi/2]
> %ans = -0.000118529835060111 -0.000454050586505161
>
> Could you paste the code that generates wrong answer?
>
> > What exactly does the reltol, abstol do also.
>
> help odeset
> ;p
>
> ode45 (and all matlab solvers) is 'adaptative'.
> In every iteration compute estimated error, and,
> if it is too big, decrease step size (?t).
>
> Absolute tolerance tell us, how big error
> can be, and relative tolerance - how big compedred
> to value.
> error <= max(RelTol*|y| , AbsTol).
>
>
>
>
>
>
So I tried your code out on my matlab and for some reason, i got [root1,root2]-[pi/2,3*pi/2] gave me -0.1647 -0.4933 as opposed to what you got.
So.. maybe this is something wrong with my matlab in general?![]()
> W dniu 2013-06-30 10:32, William pisze:
> > bartekltg <bartekltg@gmail.com> wrote in message
> > <kqnu6a$k9n$1@node1.news.atman.pl>...
> >> W dniu 2013-06-28 22:38, William pisze:
> >> > Hi,
> >> >
> >> > I've been working on a program using ODE's and I've been using ODE45 to
> >> > get numerical solutions for differential equations. I've been getting
> >> > incorrect results e.g. when I used
> >> > y'' = -y, I get a period that is ~ 2.8 instead of pi. This also
> >> happened
> >> > with various other functions that I've been using.
> >> >
> >> > Is there a solution to this problem?
> >>
> >> I'm great wizard from quadruple-precision-in-matlab-land.
> >> My crystal ball told me you have done something wrong[1]. Sadly,
> >> this crystal ball is not HD ready, so I can't read you code:(
> >> Maybe you can post it? :)
> >>
> >> > Thanks!
> >>
> >> [1]:
> >> % solution of y'' = -y; y(1,:) is y, y(1,:) is velocity (y')
> >>
> >> [t,y] = ode45 ( @(t,Y) [Y(2);-Y(1)],[0,10],[0,1]);
> >>
> >> % of course ode45 solves first order equation, so we wrote
> >> % y'' = -y as system {y'=u, u'=-y}
> >>
> >> plot(t,y(:,1))
> >> grid
> >> %looks nice
> >>
> >> fzero(@(x)interp1(t,y(:,1),x),pi)
> >>
> >> % ans = 3.14134838301112
> >> % meh, but significantly better than 2.8
> >>
> >> options = odeset ('RelTol',1e-10,'AbsTol',1e-10); % quite useful:)
> >> [t,y] = ode45 ( @(t,x) [x(2);-x(1)],[0,10],[0,1],options );
> >>
> >> root = fzero(@(x)interp1(t,y(:,1),x),pi)
> >> (root-pi)/pi
> >>
> >> % root = 3.14159265459964
> >> % ans = 3.21445091288877e-010
> >> % :-)
> >>
> >>
> >
> > So my code looks something like this
> >
> > [t,phi] = ode45(@diffeq,0:T/100:T,[1,0]);
> > %T is the period of the diff eq.
>
> How do you now, how big is T?
> I think five (approximate) periods will be better.
>
> BTW, tspan other than [tstart, tend] give you more points,
> but doesn't increase accuracy.
>
> > then
> >
> > function phiprime = diffeq(t,phi)
> > m =1;
> > g=0;
> > l=1;
> > H = 0;
> > phiprime=[phi(2); -3*H.*phi(2)-m^2*phi(1) - g*phi(1).^2 - l*phi(1).^3];
> >
> > where i just let phi(1) = y, phi(2) = y'
>
> > I've tried this code with g, l, H = 0 which becomes y'' = -y
>
> And it works!
>
> %l=0,h=0,g=0
> T=10;
> [t,phi] = ode45(@diffeq,0:T/100:T,[1,0]);
> plot(t,phi(:,1))
> root1 = fzero(@(x)interp1(t,phi(:,1),x),1*pi/2);
> root2 = fzero(@(x)interp1(t,phi(:,1),x),3*pi/2);
> [root1,root2]-[pi/2,3*pi/2]
> %ans = -0.000118529835060111 -0.000454050586505161
>
> Could you paste the code that generates wrong answer?
>
> > What exactly does the reltol, abstol do also.
>
> help odeset
> ;p
>
> ode45 (and all matlab solvers) is 'adaptative'.
> In every iteration compute estimated error, and,
> if it is too big, decrease step size (?t).
>
> Absolute tolerance tell us, how big error
> can be, and relative tolerance - how big compedred
> to value.
> error <= max(RelTol*|y| , AbsTol).
>
>
>
>
>
>
So I tried your code out on my matlab and for some reason, i got [root1,root2]-[pi/2,3*pi/2] gave me -0.1647 -0.4933 as opposed to what you got.
So.. maybe this is something wrong with my matlab in general?