Solving linear system of equations

I want to solve a system of three equations. I'm using this code to calculate the eigenvectors and eigenvalues, then write the equations for S1, S2 and S3:

%Parâmetros
x=0;
del=0;
al1=0.707;
oc=0.3;
Gamma1=1;
Gamma3=1;
g1=1/100;

%Matriz a ser diagonalizada
sig23e = [-(1i*(del+x)+Gamma1+g1) 1i*al1*oc    0                       ];
sig13e = [1i*al1*oc               -(1i*x+g1/2) -1i*oc                  ];
sig14e = [0                       -1i*oc       1i*(del-x)-Gamma1/2-g1/2];

A = [sig23e; sig13e; sig14e];

[V,D] = eig(A);

%Solução final
syms A1 A2 A3 sig13
S1 = V(:,1)*exp(D(1,1))
S2 = V(:,2)*exp(D(2,2))
S3 = V(:,3)*exp(D(3,3))

S = [S1,S2,S3];

eqn1 = S(1,:) == 0;
eqn2 = S(2,:) == 0;
eqn3 = S(3,:) == sig13;

sol = solve([eqn1,eqn2,eqn3],[A1,A2,A3])

The problem arises when calculating S1, S2, and S3, if I let them in the format above, the output are something like this:

S1 =

0.3700 - 0.0000i

-0.0000 - 0.0677i

0.0436 + 0.0000i

Which is fine. But when I do the following change to the last part to add the variables A1, A2 and A3:

syms A1 A2 A3 sig13
S1 = A1*V(:,1)*exp(D(1,1))
S2 = A2*V(:,2)*exp(D(2,2))
S3 = A3*V(:,3)*exp(D(3,3))

S = [S1,S2,S3];

eqn1 = S(1,:) == 0;
eqn2 = S(2,:) == 0;
eqn3 = S(3,:) == sig13;

sol = solve([eqn1,eqn2,eqn3],[A1,A2,A3])

The output becomes horrid and the solution using "sol.A1" gives me "Empty sym: 0-by-1"

[Torsten]

Maybe you could describe in your own words what you try to do. It's hard to deduce it from your code.

Do you try to solve some system of ordinary differential equation using matrix exponentials ?

[Luis]

Yeah, sure. I'm trying to solve this problem x' = A*x. Where x is a column vector of the variables, x' is the derivative of x respect to time and A is the coefficient matrix. The general solution for this problem is x = exp(λ*t). Therefore, the solution is now found by finding the eigenvector and eigenvalues of A. I do this on this part of the code:

[V,D] = eig(A);

Now, the general form of the solution is given by x = A1*S1 + A2*S2 + A3*S3, where the S's are the eigenvector times the exponential of the associated eigenvalue and A's are constants to be determined.

Therefore, the last part:

syms A1 A2 A3 sig13
S1 = A1*V(:,1)*exp(D(1,1))
S2 = A2*V(:,2)*exp(D(2,2))
S3 = A3*V(:,3)*exp(D(3,3))
S = [S1,S2,S3];
eqn1 = S(1,:) == 0;
eqn2 = S(2,:) == 0;
eqn3 = S(3,:) == sig13;
sol = solve([eqn1,eqn2,eqn3],[A1,A2,A3])

Was supposed to give the values of A1, A2 and A3 from eqn1, eqn2 and eqn3.

Answer 1

[Torsten]

I think you mean

%Parâmetros
x=0;
del=0;
al1=0.707;
oc=0.3;
Gamma1=1;
Gamma3=1;
g1=1/100;
%Matriz a ser diagonalizada
sig23e = [-(1i*(del+x)+Gamma1+g1) 1i*al1*oc    0                       ];
sig13e = [1i*al1*oc               -(1i*x+g1/2) -1i*oc                  ];
sig14e = [0                       -1i*oc       1i*(del-x)-Gamma1/2-g1/2];
A = [sig23e; sig13e; sig14e];
[V,D] = eig(A);

syms t A1 A2 A3 sig13
% General solution of ODE system
S = V*[A1*exp(D(1,1)*t);A2*exp(D(2,2)*t);A3*exp(D(3,3)*t)];
% Determine A1, A2 and A3 such that x1(0) = 0, x2(0) = 0 and x3(0) = sig13
S0 = subs(S,t,0);
sol_coeff = solve(S0==[0;0;sig13],[A1 A2 A3]);
% Substitute the thus obtained values for A1, A2 and A3 in the general
% solution of the ODE system
sol = subs(S,[A1 A2 A3],[sol_coeff.A1,sol_coeff.A2,sol_coeff.A3]);
sol_to_plot = subs(sol(3),sig13,1);
figure(1)
hold on
fplot(real(sol_to_plot),[0 10])
fplot(imag(sol_to_plot),[0 10])
hold off
grid on

If you are allowed to use "dsolve" for the solution of x'=A*x, you could set

syms x1(t) x2(t) x3(t) sig13
x = [x1(t);x2(t);x3(t)];
eqns = diff(x,t) == A*x;
conds = [x1(0)==0,x2(0)==0,x3(0)==sig13];
sol = dsolve(eqns,conds,'MaxDegree',3);
sol_to_plot = subs(sol.x3,sig13,1);
figure(2)
hold on
fplot(real(sol_to_plot),[0 10])
fplot(imag(sol_to_plot),[0 10])
hold off
grid on

[Luis]

I see, there was also a miss on the code I was writing.

This was truly helpful. Thank you very much.

Answer 2

[Walter Roberson]

format long g

%Parâmetros
x=0;
del=0;
al1=0.707;
oc=0.3;
Gamma1=1;
Gamma3=1;
g1=1/100;

%Matriz a ser diagonalizada
sig23e = [-(1i*(del+x)+Gamma1+g1) 1i*al1*oc    0                       ];
sig13e = [1i*al1*oc               -(1i*x+g1/2) -1i*oc                  ];
sig14e = [0                       -1i*oc       1i*(del-x)-Gamma1/2-g1/2];

A = [sig23e; sig13e; sig14e];

[V,D] = eig(A);

%Solução final
syms A1 A2 A3 sig13
S1 = V(:,1)*exp(D(1,1))

S1 = 
          0.369975188379713 -  8.21509945358374e-17i
      -1.29955495866915e-17 -    0.0676920303661541i
         0.0435604819119288 +  2.39741692879728e-17i

S2 = V(:,2)*exp(D(2,2))

S2 = 
         0.0101550277323791 +     0.148630850284153i
          0.525554721308386 +     0.112301429960421i
         -0.260764335167304 -     0.445675984904963i

S3 = V(:,3)*exp(D(3,3))

S3 = 
        -0.0101550277323791 +     0.148630850284154i
          0.525554721308386 -     0.112301429960421i
          0.260764335167305 -     0.445675984904962i

S1, S2, and S3 are each 3 x 1 vectors.

S = [S1,S2,S3];

So S is a 3 x 3 vector

eqn1 = S(1,:) == 0;
eqn2 = S(2,:) == 0;
eqn3 = S(3,:) == sig13;

eqn1, eqn2, eqn3 are each 1 x 3 equations.

eqn1 and eq2 consists of comparing a set of 3 constants to 0. Unless the constants happen to be exactly 0, solve() involving eqn1 or eqn2 will not have a solution.

sol = solve([eqn1,eqn2,eqn3],[A1,A2,A3])

sol = struct with fields:
    A1: [0×1 sym]
    A2: [0×1 sym]
    A3: [0×1 sym]

[eqn1, eqn2, eqn3] is a 3 x 3 set of equations.

You are trying to solve() a 3 x 3 set of equations with respect to exactly 3 variables. The system is overdetermined and does not happen to have a solution.

symvar([eqn1, eqn2, eqn3])

ans = 

The equations do not have any A1, A2, or A3 at all.