Why Dsolve considers my equations' variables constants!?

1 Ansicht (letzte 30 Tage)
Ibrahim Bakry
Ibrahim Bakry am 11 Feb. 2024
Bearbeitet: Walter Roberson am 11 Feb. 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 Kommentar
Ibrahim Bakry
Ibrahim Bakry am 11 Feb. 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?

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Akzeptierte Antwort

Torsten
Torsten am 11 Feb. 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 Kommentare
Torsten
Torsten am 11 Feb. 2024
Bearbeitet: Walter Roberson am 11 Feb. 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 am 11 Feb. 2024
Thank you, this was helpful

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