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How to find error in Runge Kutta 4th Order

2 次查看（过去 30 天）
Ana Carla Díaz Aguirre
编辑： Jan 2021-5-28
function RungeKutta4=RK4 (f,a,b,n,t0,y0)
h=(b-a)/n;
%n=(b-a)/h;
t=a:h:b;
k1=zeros(1,n+1);
k2=k1;
k3=k2;
k4=k3;
y=k4;
y(1,1)=y0;
RK4Y=k4;
for i=1:length(t)
k1(1,i)=h*feval('f',t(i),y(i));
k2(1,i)=h*feval('f',t(i)+h/2,y(i)+k1(i)/2);
k3(1,i)=h*feval('f',t(i)+h/2,y(i)+k2(i)/2);
k4(1,i)=h*feval('f',t(i)+h,y(i)+k3(i));
RK4Y(1,i)=y(i)+1/6*(k1(i)+2*k2(i)+2*k3(i)+k4(i));%y1
y(1,i+1)=RK4Y(i);
end
RK4Y
RungeKutta4Y=RK4Y(n)
plot(t,RK4Y,'ro');hold on;
g=350*(1-exp((-0.2*t)/(7-5*exp(-0.2*t))));
plot(t,g);hold off;
end

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回答（1 个）

Jan
Jan 2021-5-28
编辑：Jan 2021-5-28
You did not mention, which problems occur. So before the problem can be solved, the readers have to guess, what the problem is.
You do not want to evaluate the character 'f', but the function handle stored in the variable f :
% Replace:
feval('f', ...
% by
feval(f, ...
There is no need to store the intermediate values of k1 to k4.
Collecting the trajectory in the variable y is enough. There is no need to store in in RK4Y again.
Storing the result in y(i+1) creates a final value at y(length(t) + 1). So y is longer than t. Solution: Run the loop until length(t)-1.

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