How to calculate the difference error between Runge-Kutta Order 4 Method and euler method?

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cindyawati cindyawati
评论： Sam Chak 2024-3-4
Ran in:
I want to calculate the difference error between Runge-Kutta order 4 method and Euler method. Because I know the Runge-Kutta order 4 method more than precision depend on euler. So, Can I calculate the difference error? This is my Runge-Kutta code. Thanks
tstart = 0;
tend = 180;
dt = 0.01;
T = (tstart:dt:tend).';
Y0 = [10 0 0 0 0 0 0 0];
f = @myode;
Y = fRK4(f,T,Y0);
M1 = Y(:,1);
M2 = Y(:,2);
M3 = Y(:,3);
M4 = Y(:,4);
M5 = Y(:,5);
M6 = Y(:,6);
O = Y(:,7);
P = Y(:,8);
figure
%subplot(3,1,1)
%hold on
plot(T,M1,'r','Linewidth',2)
xlabel('Times (days)')
ylabel('M1 (gr/ml)')
figure
%subplot(3,1,2)
%hold on
plot(T,M2,'b','Linewidth',2)
xlabel('Times (days)')
ylabel('M2 (gr/ml)')
figure
%subplot(3,1,3)
%hold on
plot(T,M3,'g','Linewidth',2)
xlabel('Times (days)')
ylabel('M3 (gr/ml)')
figure
%subplot(3,1,3)
%hold on
plot(T,M4,'b','Linewidth',5)
xlabel('Times (days)')
ylabel('M4 (gr/ml)')
figure
%subplot(3,1,3)
%hold on
plot(T,M5,'r','Linewidth',5)
xlabel('Times (days)')
ylabel('M5 (gr/ml)')
figure
%subplot(3,1,3)
%hold on
plot(T,M6,'g','Linewidth',5)
xlabel('times (days)')
ylabel('M6 (gr/ml)')
figure
%subplot(2,1,1)
%hold on
plot(T,O,'k','Linewidth',2)
xlabel('Times (days)')
ylabel('O (gr/ml)')
figure
%subplot(2,1,2)
%hold on
plot(T,P,'m','Linewidth',2)
xlabel('Times (days)')
ylabel('P (gr/ml)')
function Y = fRK4(f,T,Y0)
N = numel(T);
n = numel(Y0);
Y = zeros(N,n);
Y(1,:) = Y0;
for j = 2:N
t = T(j-1);
y = Y(j-1,:);
h = T(j) - T(j-1);
k0 = f(t,y);
k1 = f(t+0.5*h,y+k0*0.5*h);
k2 = f(t+0.5*h,y+k1*0.5*h);
k3 = f(t+h,y+k2*h);
Y(j,:) = y + h/6*(k0+2*k1+2*k2+k3);
end
end
function CM1 = myode (~,MM)
M1 = MM(1);
M2 = MM(2);
M3 = MM(3);
M4 = MM(4);
M5 = MM(5);
M6 = MM(6);
O = MM(7);
P = MM(8);
delta=50;
gamma=75;
K1=10^-4;
K2=5*10^-4;
K3=10^-3;
K4=5*10^-3;
K5=10^-2;
K6=5*10^-2;
Ko=0.1;
n=6;
Oa=10;
Pa=100;
mu_1=10^-3;
mu_2=10^-3;
mu_3=10^-3;
mu_4=10^-3;
mu_5=10^-3;
mu_6=10^-3;
mu_o=10^-4;
mu_p= 10^-5;
sumter=K2*M2+K3*M3+K4*M4+K5*M5;
CM1= zeros(1,8);
CM1(1) = (delta*M1*(1-(M1/gamma))-2*K1*M1*M1-M1*sumter-((Oa-n)*K6*M1*M6)-((Pa-Oa)*Ko*M1*O)-(mu_1*M1));
CM1(2) = (K1*M1*M1)-(K2*M1*M2)-(mu_2*M2);
CM1(3) = (K2*M1*M2)-(K3*M1*M3)-(mu_3*M3);
CM1(4) = (K3*M1*M3)-(K4*M1*M4)-(mu_4*M4);
CM1(5) = (K4*M1*M4)-(K5*M1*M5)-(mu_5*M5);
CM1(6) = (K5*M1*M5)-(K6*M1*M6)-(mu_6*M6);
CM1(7) = (K6*M1*M6)-(Ko*M1*O)-(mu_o*O);
CM1(8) = (Ko*M1*O)-(mu_p*P);
end

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采纳的回答

Aquatris
Aquatris 2024-3-4
编辑：Aquatris 2024-3-4
Ran in:
You can add a 'eul' function for euler integration and take the norm of the difference between the resulting outputs.
clear all ,clc
tstart = 0;
tend = 180;
dt = 0.001;
T = (tstart:dt:tend).';
Y0 = [10 0 0 0 0 0 0 0];
f = @myode;
Y = fRK4(f,T,Y0); % RK4 integration
Y2 = eul(f,T,Y0); % euler integration
norm((Y-Y2)./max(Y))
ans = 0.0280
%% here is the euler integration
function Y = eul(f,T,Y0)
N = numel(T);
n = numel(Y0);
Y = zeros(N,n);
Y(1,:) = Y0;
for j = 2:N
t = T(j-1);
y = Y(j-1,:);
h = T(j) - T(j-1);
Y(j,:) = y + f(t,y)*h; % y(n+1) = y(n) + dy*dt
end
end
% here is your RK4
function Y = fRK4(f,T,Y0)
N = numel(T);
n = numel(Y0);
Y = zeros(N,n);
Y(1,:) = Y0;
for j = 2:N
t = T(j-1);
y = Y(j-1,:);
h = T(j) - T(j-1);
k0 = f(t,y);
k1 = f(t+0.5*h,y+k0*0.5*h);
k2 = f(t+0.5*h,y+k1*0.5*h);
k3 = f(t+h,y+k2*h);
Y(j,:) = y + h/6*(k0+2*k1+2*k2+k3);
end
end
function CM1 = myode (~,MM)
M1 = MM(1);
M2 = MM(2);
M3 = MM(3);
M4 = MM(4);
M5 = MM(5);
M6 = MM(6);
O = MM(7);
P = MM(8);
delta=50;
gamma=75;
K1=10^-4;
K2=5*10^-4;
K3=10^-3;
K4=5*10^-3;
K5=10^-2;
K6=5*10^-2;
Ko=0.1;
n=6;
Oa=10;
Pa=100;
mu_1=10^-3;
mu_2=10^-3;
mu_3=10^-3;
mu_4=10^-3;
mu_5=10^-3;
mu_6=10^-3;
mu_o=10^-4;
mu_p= 10^-5;
sumter=K2*M2+K3*M3+K4*M4+K5*M5;
CM1= zeros(1,8);
CM1(1) = (delta*M1*(1-(M1/gamma))-2*K1*M1*M1-M1*sumter-((Oa-n)*K6*M1*M6)-((Pa-Oa)*Ko*M1*O)-(mu_1*M1));
CM1(2) = (K1*M1*M1)-(K2*M1*M2)-(mu_2*M2);
CM1(3) = (K2*M1*M2)-(K3*M1*M3)-(mu_3*M3);
CM1(4) = (K3*M1*M3)-(K4*M1*M4)-(mu_4*M4);
CM1(5) = (K4*M1*M4)-(K5*M1*M5)-(mu_5*M5);
CM1(6) = (K5*M1*M5)-(K6*M1*M6)-(mu_6*M6);
CM1(7) = (K6*M1*M6)-(Ko*M1*O)-(mu_o*O);
CM1(8) = (Ko*M1*O)-(mu_p*P);
end
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Sam Chak
Sam Chak 2024-3-4
Ran in:
Hi @cindyawati cindyawati
If the integration step size is chosen to be extremely small, the error within the finite tspan could become negligible. In this case, the computed error is 10 times smaller. Based on this finding, what conclusions can you draw?
tstart = 0;
tend = 180;
f = @myode;
Y0 = [10 0 0 0 0 0 0 0];
dt = 0.00001;
T = (tstart:dt:tend).';
Y = fRK4(f,T,Y0); % RK4 integration
Y2 = eul(f,T,Y0); % euler integration
error = norm((Y-Y2)./max(Y))
error = 0.0028

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