Error trying to solve 2 Second ODE

1 Ansicht (letzte 30 Tage)
Scott Angus
Scott Angus am 15 Nov. 2019
Kommentiert: Scott Angus am 15 Nov. 2019
Hi,
I have a function that solves a differential equation using the RK method:
function [tvec, xvec] = rksolve(f, t0, tf, x0, dt)
%Solve f with initial conditions x0 between t0 -> tf
%with increments dt
%Set initial conditions
t = t0;
x = x0;
%Answer output
tvec = t;
xvec = x;
while t < tf
%RK method of solving
k1 = f(x, t);
k2 = f(x+0.5*k1*dt, t+0.5*dt);
k3 = f(x+0.5*k2*dt, t+0.5*dt);
k4 = f(x+k3*dt, t+dt);
x = x + ((k1 + 2*k2 + 2*k3 +k4)*dt)/6;
t = t+dt;
%Store outputs
tvec = [tvec t];
xvec = [xvec x];
end
But when I try to run it with the equations defined in func.m, it comes up with a horzcat error in the line:
xvec = [xvec x];
func.m:
function a = coords(x, t);
%Initial Conditions
xpos = x(1);
vx = x(2);
ypos = x(3);
vy = x(4);
a = [(-xpos - 2*xpos*ypos); (-ypos - (xpos^2) +ypos^2)];
%dxdt = vx
%dvxdt = -x -2*x*y;
%dydt = vy;
%dvydt = -y -(x^2) + y^2;
Any help would be greatly appreciated!
  1 Kommentar
darova
darova am 15 Nov. 2019
What about preallocation?
n = round((tf-t0)/dt+1);
tvec1 = zeros(1,n);
i = 1;
while t < tvec
%% ...
tvec1(i) = t;
i = i + 1;
end
tvec = tvec1;

Melden Sie sich an, um zu kommentieren.

Antworten (1)

James Tursa
James Tursa am 15 Nov. 2019
Bearbeitet: James Tursa am 15 Nov. 2019
Just looking at func.m, it appears you pass in a 4-element x vector, but you only return a 2-element a vector. You need to return the derivative for all the states. So something like this instead based on your comments:
a = [ vx; (-xpos - 2*xpos*ypos); vy; (-ypos - (xpos^2) +ypos^2) ];
  2 Kommentare
Scott Angus
Scott Angus am 15 Nov. 2019
Thanks for your response. I tried changing the output of func.m to
a = [ vx; (-xpos - 2*xpos*ypos); vy; (-ypos - (xpos^2) +ypos^2) ];
but still come up with an error:
>> [ts, xmat] = rksolve(@func, 0, 200, [0, 0.3, 0, 0], 0.2)
Error using horzcat
Dimensions of arrays being concatenated are not consistent.
Error in rksolve (line 25)
xvec = [xvec x];
I'm not sure why because the x and xvec dimensions should be the same?
Scott Angus
Scott Angus am 15 Nov. 2019
I found the issue. I was passing the function rksolve with a horizontal vector for x0 instead of a vertical vector. Adding semicolons between the components of the vector when I passed it solved the problem.

Melden Sie sich an, um zu kommentieren.

Community Treasure Hunt

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

Start Hunting!

Translated by