Why Dsolve considers my equations' variables constants!?

1 visualización (últimos 30 días)
Ibrahim Bakry
Ibrahim Bakry el 11 de Feb. de 2024
Editada: Walter Roberson el 11 de Feb. de 2024
clear all
Nx = 2; % number of state equations
Nu = 1; % number of control parameters
% x = sym('x%d', [1 Nx]);
% p = sym('p%d', [1 Nx]);
syms u t
syms x p [1 Nx]
% State equations
Dx(1) = x(2);
Dx(2) = -x(2) + u(1);
% Cost function inside the integral
g = 0.5*u(1)^2;
% Hamiltonian
H = g;
for i = 1 : Nx
H = H + p(i)*Dx(i);
end
% Costate equations
for i = 1 : Nx
Dp(i) = -diff(H,x(i));
end
% solve for control u
du = diff(H,u);
sol_u = solve(du,u);
% Substitute u to state equations
for i = 1 : Nx
Dx(i) = subs(Dx(i),u,sol_u);
end
% convert symbolic objects to strings for using 'dsolve'
% need ti automate this solving
clear x p
syms x(t) p(t) [1 Nx]
eqx = diff(x,t)==Dx
eqx(t) = 
eqp = diff(p,t)==Dp
eqp(t) = 
eq = [eqx,eqp];
sol_h = dsolve(eqx,eqp);
  1 comentario
Ibrahim Bakry
Ibrahim Bakry el 11 de Feb. de 2024
these are the final diff. equations:
eq = [diff(x1(t), t) == x2, diff(x2(t), t) == - p2 - x2, diff(p1(t), t) == 0, diff(p2(t), t) == p2 - p1]
the solution of dsolve is:
sol_h =
struct with fields:
x2: C2 - t*(p2 + x2)
x1: C1 + t*x2
p1: C3
p2: C4 - t*(p1 - p2)
But the answer should be like:
x2: (C3*exp(t))/2 - C4 + (C2*exp(-t))/2
x1: (C3*exp(t))/2 - C4*t - C1 - (C2*exp(-t))/2
p1: C4
p2: C4 - C3*exp(t)
I got the first solution by dsolve(eq) in my script, the second solution i got by dsolve(diff(x1(t), t) == x2, diff(x2(t), t) == - p2 - x2, diff(p1(t), t) == 0, diff(p2(t), t) == p2 - p1) in the workspace
Where is the ptoblem?

Iniciar sesión para comentar.

Respuesta aceptada

Torsten
Torsten el 11 de Feb. de 2024
syms t x1(t) x2(t) p1(t) p2(t)
eq = [diff(x1, t) == x2, diff(x2, t) == - p2 - x2, diff(p1, t) == 0, diff(p2, t) == p2 - p1]
eq(t) = 
dsolve(eq)
ans = struct with fields:
x2: (C3*exp(t))/2 - C4 + (C2*exp(-t))/2 x1: (C3*exp(t))/2 - C4*t - C1 - (C2*exp(-t))/2 p1: C4 p2: C4 - C3*exp(t)
  3 comentarios
Torsten
Torsten el 11 de Feb. de 2024
Editada: Walter Roberson el 11 de Feb. de 2024
You cannot create arrays of symbolic functions in MATLAB, and this would be necessary if you wanted to proceed as you try in your code:
Ibrahim Bakry
Ibrahim Bakry el 11 de Feb. de 2024
Thank you, this was helpful

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Symbolic Math Toolbox en Help Center y File Exchange.

Productos


Versión

R2023b

Community Treasure Hunt

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

Start Hunting!

Translated by