Question

I m试图确定它在Runge-Kuttameth(“RK4”)上采取多少步骤,以达到普通差别方程式的确切解决办法的0.01%。 http://mathworld.wolfram.com/EulerForwardMethod.html Euler方法。两者都应导致对伐木地块的直线。我的欧勒解决办法似乎正确,但我正为朝鲜找到一个治愈的办法。他们是根据同一法典制定的,因此我完全混淆了这个问题。

EDIT:Sorry for去除wikipedia链接。请允许我保持不止一个环节,因为我是新用户。然而,这两种方法都与我的执行相类似,在瓦基佩西亚详细。

如果有人想解决我的问题,守则如下,图表载于。是的,这是一个家庭工作问题;由于我理解我的思想过程中的错误,我把这个问题说出来。

f = @(x,y) x+y; %this is the eqn (the part after the @(t,y)

这是我的《朝鲜法典》:

k1=@(x,y) h*f(x,y);
k2=@(x,y) h*f(x+1/2*h,y+1/2*k1(x,y));
k3=@(x,y) h*f(x+1/2*h,y+1/2*k2(x,y));
k4=@(x,y) h*f(x+h,y+k3(x,y));

clear y x exact i
x(1)=0;
y(1)=2;
xn=1;
x0=0;

exact=3*exp(xn)-xn-1; %exact solution at x=1
%# Evaluate RK4 with 1 step for x=0...1
N=1; %# number of steps
h=(xn-x0)/N; %# step size
i=1;
y(i+1)=y(i)+1/6*k1(x(i),y(i))+1/3*k2(x(i),y(i))+1/3*k3(x(i),y(i))+1/6*k4(x(i),y(i));
RK4(N)=y(i+1);  %# store result of RK4 in vector RK4(# of steps)
E_RK4(N)=-(RK4(N)-exact)/exact*100;%keep track of %error for each N
Nsteps_RK4(N)=N;


%# repeat for increasing N until error is less than 0.01%
while -(RK4(N)-exact)/exact > 0.0001
    i=1;
    N=N+1;
    h=(xn-x0)/N;
    for i=1:N
        y(i+1)=y(i)+1/6*k1(x(i),y(i))+1/3*k2(x(i),y(i))+1/3*k3(x(i),y(i))+1/6*k4(x(i),y(i));
        x(i+1)=x(i)+h;
    end
    RK4(N)=y(i+1);
    Nsteps_RK4(N)=N;
    E_RK4(N)=-(RK4(N)-exact)/exact*100; %# keep track of %error for each N
end

Nsteps_RK4(end);

这是我的欧勒法典:

%# Evaluate Euler with 1 step for x=0...1
clear y x i
x(1)=0;
y(1)=2;
N=1; %# number of steps
h=(xn-x0)/N; %# step size
i=1;
y(i+1)= y(i)+h*f(x(i),y(i));
Euler(N)=y(i+1); %# store result of Euler in vector Euler(# of steps)
E_Euler(N)=-(Euler(N)-exact)/exact*100;%# keep track of %error for each N
Nsteps_Euler(N)=N;
%# repeat for increasing N until error is less than 0.01%
while -(Euler(N)-exact)/exact > 0.0001
    i=1;
    N=N+1;
    h=(xn-x0)/N;
    for i=1:N
        y(i+1)= y(i)+h*f(x(i),y(i));   
        x(i+1)=x(i)+h;
    end
    Euler(N)=y(i+1);
    Nsteps_Euler(N)=N;
    E_Euler(N)=-(Euler(N)-exact)/exact*100; %# keep track of %error for each N
end

Answer 1

见:。

Variables specified in the body of the expression [...] must have a value assigned to them at the time you construct an anonymous function that uses them. Upon construction, MATLAB captures the current value for each variable specified in the body of that function. The function will continue to associate this value with the variable even if the value should change in the workspace or go out of scope.

<代码>h载于k1.k4,即使您在<代码>上改动了>。

一种解决办法是,在匿名职能中添加<代码>h:

k1=@(x,y,h) h*f(x,y);

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