Borrar filtros
Borrar filtros

Solving a System of ODEs

1 visualización (últimos 30 días)
Tom Keaton
Tom Keaton el 20 de Jun. de 2018
Comentada: Star Strider el 19 de Jul. de 2018
My Input:
syms x(t) y(t) z(t)
%Units of kg and seconds
q = 1.60217662*(10^-19);
B = 2;
m_e = 1.60217662*(10*-31);
a = (q*B)/m_e;
%DiffiQ to solve
ode1 = diff(x,2) == a*diff(y);
ode2 = diff(y,2) == -a*diff(x);
ode3 = diff(z,2) == 0;
odes = [ode1; ode2; ode3];
%Solutions
S = dsolve(odes);
xSol(t) = S.x;
ySol(t) = S.y;
zSol(t) = S.z;
[xSol(t), ySol(t), zSol(t)] = dsolve(odes);
%Initial Conditions
condx1 = x(0) == 1;
condy1 = y(0) == 1;
condz1 = z(0) == 0;
%{
condvx1 = diff(x) == 1;
condvy1 = diff(y) == 2;
condvz1 = diff(z) == 2;
%}
%Plot Trajectory
t = 0:pi/50:8*pi;
plot3(ode1,ode2,ode3,'r','LineWidth',3)
My Output: Nothing. The program runs but nothing happens. I have to force close MatLab so that the script stops. I left it running for an hour to see but I am starting to think that there is another issue. Any advice is appreciated!
PS - Using Matlab R2018a and yes I purposefully commented out a section of initial conditions.

Respuesta aceptada

Star Strider
Star Strider el 20 de Jun. de 2018
Try this:
...
%Initial Conditions
condx1 = x(0) == 1;
condy1 = y(0) == 1;
condz1 = z(0) == 0;
%Solutions
S = dsolve(odes, condx1, condy1, condz1);
xSol(t) = S.x;
ySol(t) = S.y;
zSol(t) = S.z;
Sx = matlabFunction(xSol);
Sy = matlabFunction(ySol);
Sz = matlabFunction(zSol);
Then evaluate the anonymous functions in terms of the variables and constants. Note that you will have to either evaluate ‘Sz’ first, then pass the result as ‘z’ or pass it as an evaluated function ‘Sz(t,Cz)’ as the ‘z’ argument.
  22 comentarios
Tom Keaton
Tom Keaton el 19 de Jul. de 2018
Editada: Tom Keaton el 19 de Jul. de 2018
I see. The past few days I have been messing around with these and trying to implement them in other ways. I will keep messing around with them, especially since the equations I have right now are only simplified versions of what I really am trying to do and I can only apply this separation trick to these. I will close this thread and accept the answer since the original question has been answered. I will be opening up more threads in the near future as I continue developing this simulation. Thank you again for all the help and patience!
Star Strider
Star Strider el 19 de Jul. de 2018
As always, my pleasure!

Iniciar sesión para comentar.

Más respuestas (0)

Community Treasure Hunt

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

Start Hunting!

Translated by