Compare results of different step size using Euler's method

Question

Lucky 2019-10-30

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/488286-compare-results-of-different-step-size-using-euler-s-method

评论： Jan Smid 2021-3-15

x1=zeros(10000,1);
x2=zeros(10000,1);
x3=zeros(10000,1);
Time=zeros(10000,1);
for i=1:10001
    t=(i-1)*0.0005;
    Time(i)=t;
    if t==0
        x1(i)=0;
        x2(i)=0;
        x3(i)=2;
    else
        x1(i)=x1(i-1)+x2(i-1)*0.0005;
        x3(i-1)=2-9*sin(x1(i-1))-x2(i-1);
        x2(i)=x2(i-1)+x3(i-1)*0.0005;
end
Xa=x1*180/pi;
plot(Time,Xa,'red')

I can get the plot for one time interval but I want to get it for different time intervals such as at 0.001, 0.01,0.1 and also compare the results on the same plot. how do i do this? please help??

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

ME 2019-10-30

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/488286-compare-results-of-different-step-size-using-euler-s-method#answer_398962

在 MATLAB Online 中打开

I have made a few tweaks to your code and it will now allow you to alter one parameter value (the time step size) and produce a new approximation that will be added to your previous plot.

clear all
h = 0.00001;
x1=zeros(round(5/h),1);
x2=zeros(round(5/h),1);
x3=zeros(round(5/h),1);
Time=zeros(round(5/h),1);
for i=1:(5/h)+1
    t=(i-1)*h;
    Time(i)=t;
    if t==0
        x1(i)=0;
        x2(i)=0;
        x3(i)=2;
    else
        x1(i)=x1(i-1)+x2(i-1)*h;
        x3(i-1)=2-9*sin(x1(i-1))-x2(i-1);
        x2(i)=x2(i-1)+x3(i-1)*h;
    end
end
Xa=x1*180/pi;
hold on
plot(Time,Xa)

Here h is your time step and I have just modified everything else to produce the correct length arrays for storing the results. I also took out the 'red' from your plotting command so that each approximation will automatically plot in a different colour.

2 个评论
显示无隐藏无

Lucky 2019-10-30

Thanks a lot.. This code works fine... But how do I change the value of h in it.

e.g h = 0.00001;0.005;0.01;

I know how to plot them together but how do I put in different values of h together in one code.

Thanks

ME 2019-10-30

编辑：ME 2019-10-30

在 MATLAB Online 中打开

I guess you'd have to adjust this to something like:

h = sort([0.00001 0.005 0.01],'descend');
for step = 1:numel(h)
    x1=zeros(round(5/h(step)),1);
    x2=zeros(round(5/h(step)),1);
    x3=zeros(round(5/h(step)),1);
    Time=zeros(round(5/h(step)),1);
    for i=1:(5/h(step))+1
        t=(i-1)*h(step);
        Time(i)=t;
        if t==0
            x1(i)=0;
            x2(i)=0;
            x3(i)=2;
        else
            x1(i)=x1(i-1)+x2(i-1)*h(step);
            x3(i-1)=2-9*sin(x1(i-1))-x2(i-1);
            x2(i)=x2(i-1)+x3(i-1)*h(step);
        end
    end
    Xa=x1*180/pi;
    hold on
    plot(Time,Xa)
end

Here the sort is to make sure that you start with the largest step size, otherwise you'll run into issues when you come to plotting.

请先登录，再进行评论。

Answer 2

Lucky 2019-10-30

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/488286-compare-results-of-different-step-size-using-euler-s-method#answer_398986

在 MATLAB Online 中打开

Thanks a lot again, but it still doesn't solve my problem.

I need all the plots on a single figure.

clear all
h = sort([0.00001 0.005 0.01 1],'descend');
for step = 1:numel(h)
    x1=zeros(round(5/h(step)),1);
    x2=zeros(round(5/h(step)),1);
    x3=zeros(round(5/h(step)),1);
    Time=zeros(round(5/h(step)),1);
    for i=1:(5/h(step))+1
        t=(i-1)*h(step);
        Time(i)=t;
        if t==0
            x1(i)=0;
            x2(i)=0;
            x3(i)=2;
        else
            x1(i)=x1(i-1)+x2(i-1)*h(step);
            x3(i-1)=2-9*sin(x1(i-1))-x2(i-1);
            x2(i)=x2(i-1)+x3(i-1)*h(step);
        end
    end
end
    Xa=x1*180/pi;
    figure(1)
    subplot(2,1,1)
    plot(Time,Xa)
    hold on
    plot(Time,Xa)
    hold off

3 个评论
显示 1更早的评论隐藏 1更早的评论

Lucky 2019-10-30

This works well... Thanks a lot...

Greatly appreciate your help

Jan Smid 2021-3-15

Hello, may I ask what is the reasoning behind the "5/h(step)" while predefining the correct length arrays? It's not clear to me what exactly does this part of the code do.

请先登录，再进行评论。