How to solve 4th order Runge-Kutta for different initial conditions?

6 次查看（过去 30 天）
Kenny Gee
Kenny Gee 2022-4-22
评论： Kenny Gee 2022-4-24
I have a code that solves the 2 populations with 1 initial conditions and just plot that. How do I make 2D plots for different initial conditions?
%dx/dt = 5*x+1*x*y
%dy/dt = -2*y+1*x*y
clear
clc
f=@(t,z) [5*z(1)+1*z(1)*z(2);
-2*z(2)+1*z(1)*z(2)];
% Initial Conditions
t(1)=0;
z(:,1)=[1,1];
% Step size
s=1;
e=10;
N=e/s;
% Update Loop
for i=1:N
% Update time
t(i+1)=t(i)+s;
% Update for y
k1=f(t(i) ,z(:,i));
k2=f(t(i)+s/2,z(:,i)+s/2*k1);
k3=f(t(i)+s/2,z(:,i)+s/2*k2);
k4=f(t(i)+s ,z(:,i)+s *k3);
z(:,i+1)=z(:,i)+h/6*(k1 + 2*k2 + 2*k3 + k4);
end

请先登录，再回答此问题。

采纳的回答

Alan Stevens
Alan Stevens 2022-4-22
Ran in:
Here's a rather crude method (together with some corrections). You should be able to turn this into a much more elegant version:
%dx/dt = 5*x+1*x*y
%dy/dt = -2*y+1*x*y
f=@(t,z) [5*z(1)+1*z(1)*z(2);
-2*z(2)+1*z(1)*z(2)];
% Example initial conditions
x0 = [1, 0.5, 0.1];
y0 = [-1, -0.5, -0.1];
% Loop through different initial conditions
for j = 1:numel(x0)
% Initial Conditions
z(:,1)=[x0(j),y0(j)]; %%% Make y(0) negative
% Step size
s=0.1; %%% Need much smaller step size
e=10;
N=e/s;
t = zeros(1,N);
% Update Loop
for i=1:N
% Update time
t(i+1)=t(i)+s;
% Update for y
k1=f(t(i) ,z(:,i));
k2=f(t(i)+s/2,z(:,i)+s/2*k1);
k3=f(t(i)+s/2,z(:,i)+s/2*k2);
k4=f(t(i)+s ,z(:,i)+s *k3);
z(:,i+1)=z(:,i)+s/6*(k1 + 2*k2 + 2*k3 + k4); %%% s not h
end
x = z(1,:); y = z(2,:);
figure
subplot(2,1,1)
plot(t,x),grid
xlabel('t'),ylabel('x')
subplot(2,1,2)
plot(t,y),grid
xlabel('t'),ylabel('y')
end
  3 个评论
显示 1更早的评论
Alan Stevens
Alan Stevens 2022-4-24
Ran in:
Like this?
%dx/dt = 5*x+1*x*y
%dy/dt = -2*y+1*x*y
f=@(t,z) [5*z(1)+1*z(1)*z(2);
-2*z(2)+1*z(1)*z(2)];
% Step size
s=0.1;
e=10;
N=e/s;
% Example initial conditions
x0 = [1, 0.5, 0.1];
y0 = [-1, -0.5, -0.1];
t = zeros(1,N);
x = zeros(numel(x0),N); %%%%%%%%%%
y = zeros(numel(y0),N); %%%%%%%%%%
% Loop through different initial conditions
for j = 1:numel(x0)
z(1,1) = x0(j);
z(2,1) = y0(j);
% Update Loop
for i=1:N-1
% Update time
t(i+1)=t(i)+s;
% Update for y
k1=f(t(i) ,z(:,i));
k2=f(t(i)+s/2,z(:,i)+s/2*k1);
k3=f(t(i)+s/2,z(:,i)+s/2*k2);
k4=f(t(i)+s ,z(:,i)+s *k3);
z(:,i+1)=z(:,i)+s/6*(k1 + 2*k2 + 2*k3 + k4); %%% s not h
end
x(j,:) = z(1,:); y(j,:) = z(2,:); %%%%%%%%%%%%%
end
lbl1 = ['(' num2str(x0(1)) ',' num2str(y0(1)) ')'];
lbl2 = ['(' num2str(x0(2)) ',' num2str(y0(2)) ')'];
lbl3 = ['(' num2str(x0(3)) ',' num2str(y0(3)) ')'];
figure
plot(t,x),grid
xlabel('t'),ylabel('x')
legend(lbl1,lbl2,lbl3)
figure
plot(t,y),grid
xlabel('t'),ylabel('y')
legend(lbl1,lbl2,lbl3)

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

标签

产品


版本

R2022a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by