Runga-Kutta method in the matrix form
1 view (last 30 days)
Show older comments
Hi,
I have a set of differential eqns as: dV/dt = A*V, here, V = [V1 V2 V3] and A= [a11 a12 a13; a21 a22 a23; a31 a32 a33]. Value of A is so complex that writing down the eqns. in dV1/dt = ... dV2/dt = ... dV3/dt = ... is not convenient. So I need to solve this using Runga-Kutta method in the matrix form. I am trying the following:
V=zeros(length(t),3);
V(1,:)=[u0x, u0y, u0z];
Fx = @(t, V) A*V ;
for i=1: length(t)-1
K1 = ...
K2 = ...
K3 = ...
K4 =...
V(i+1) = V(i) + (1/6)*(K1+2*K2+2*K3+K4)*h;
end
end
However, this gives the error - Dimensions of matrices being concatenated are not consistent.
Could you please suggest how do I move ahead?
0 Comments
Answers (1)
Torsten
on 31 Aug 2023
Moved: Torsten
on 31 Aug 2023
Supply your full code so that we can test it, not only dots for missing parts.
It should be obvious that
V(i+1) = V(i) + (1/6)*(K1+2*K2+2*K3+K4)*h;
must throw an error because your solution V is not a scalar, but a 3x1 vector.
2 Comments
Torsten
on 31 Aug 2023
clear;
close all;
clc;
t0 = 0; %seconds
tf =20; %seconds
tspan=[t0 tf] ;
h=0.01;
x0=0;
y0=0;
z0=-0.5;
% Initial field velocity @t=0, x=0, z=0
u0x=exp(z0)*cos(x0-t0);
u0y=0;
u0z=exp(z0)*sin(x0-t0);
t = tspan(1):h:tspan(2);
VX=zeros(length(t),3).';
VX(:,1)=[u0x, u0y, u0z].';
Fx = @(t, VX) [1 0 1; 1 1 0; 1 0 1]*VX;
for i=1: length(t)-1
K1 = Fx(t(i),VX(:,i));
K2 = Fx(t(i)+0.5*h, VX(:,i)+0.5*h*K1);
K3 = Fx(t(i)+0.5*h, VX(:,i)+0.5*h*K2);
K4 = Fx(t(i)+h, VX(:,i)+h*K3);
VX(:,i+1) = VX(:,i) + (1/6)*(K1+2*K2+2*K3+K4)*h;
end
plot(t,VX,'LineWidth',2)
xlabel('$t$','FontSize',20,'FontWeight','bold', 'Interpreter','latex');
ylabel('$vx$', 'FontSize',20,'FontWeight','bold', 'Interpreter','latex');
set(gca,'FontSize',15);
See Also
Categories
Find more on Numerical Integration and Differential Equations in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!