MATLAB: Solving two equations with 4 unknowns

equationsno apparent urgencyurgent

Hello all,

Can you please tell me how can I solve 4 equations with 2 unknowns in MATLAB?

Best Answer

Edited: Walk before you run:

syms x y
results = solve('3*x+y=9','x-7*y=-3',x,y)
results.x
results.y

look at

doc solve

to learn more.

Related Solutions

MATLAB: Solving a set of equations numerically where solutions are complex numbers

Not all nasty looking, high order nonlinear systems of equations have solutions. vpasolve is a capable numerical solver, but it has limits. One issue is that when working in high precision, things take longer to do. All computations are slower. It will probably happen that vpasolve will return a high precision solution eventually.

So I tried posing this as a problem for fsolve. Working in double precision is far faster. fsolve can sometimes have issues with complex valued functions. But it often does seem to work.

Here is your code, pasted into a function to be used for fsolve.

function eqs = funs(S)
S11 = S(1);
S22 = S(2);
S33 = S(3);
S12 = S(4);
S13 = S(5);
S23 = S(6);
A11= 0.2346 + 0.1345i;
A21= 0.8550 - 0.4363i;
A22= -0.0412 + 0.2931i;
B11= 0.2346 + 0.1345i;
B21= 0.8550 - 0.4363i;
B22= -0.0412 + 0.2931i;
D11= 0.2346 + 0.1345i;
D21= 0.8550 - 0.4363i;
D22= -0.0412 + 0.2931i;
c11= 0.169033453278504 + 0.127942837737115i;
c22= 0.169951074114244 + 0.148423165394202i;
c33= 0.168939745364374 + 0.142538799523638i;
c21= -0.00172796568379862 + 0.201091742020021i;
c31= -0.00152297772077853 + 0.199585730377658i;
c32= -0.00523160858486265 + 0.199472640309087i;
eqs = [(A11.*A22.*S11 - A21.^2.*S11 - A11 + A11.*B22.*S22 + A11.*D22.*S33 - A21.^2.*B22.*S12.^2 - A21.^2.*D22.*S13.^2 + A11.*A22.*B22.*S12.^2 + A11.*A22.*D22.*S13.^2 + A11.*B22.*D22.*S23.^2 + A21.^2.*B22.*S11.*S22 + A21.^2.*D22.*S11.*S33 + A21.^2.*B22.*D22.*S11.*S23.^2 + A21.^2.*B22.*D22.*S13.^2.*S22 + A21.^2.*B22.*D22.*S12.^2.*S33 - A11.*A22.*B22.*S11.*S22 - A11.*A22.*D22.*S11.*S33 - A11.*B22.*D22.*S22.*S33 - A11.*A22.*B22.*D22.*S11.*S23.^2 - A11.*A22.*B22.*D22.*S13.^2.*S22 - A11.*A22.*B22.*D22.*S12.^2.*S33 - 2.*A21.^2.*B22.*D22.*S12.*S13.*S23 - A21.^2.*B22.*D22.*S11.*S22.*S33 + 2.*A11.*A22.*B22.*D22.*S12.*S13.*S23 + A11.*A22.*B22.*D22.*S11.*S22.*S33)./(A22.*S11 + B22.*S22 + D22.*S33 + A22.*B22.*S12.^2 + A22.*D22.*S13.^2 + B22.*D22.*S23.^2 - A22.*B22.*S11.*S22 - A22.*D22.*S11.*S33 - B22.*D22.*S22.*S33 - A22.*B22.*D22.*S11.*S23.^2 - A22.*B22.*D22.*S13.^2.*S22 - A22.*B22.*D22.*S12.^2.*S33 + 2.*A22.*B22.*D22.*S12.*S13.*S23 + A22.*B22.*D22.*S11.*S22.*S33 - 1)-c11;
  -(A21.*B21.*(S12 + D22.*S13.*S23 - D22.*S12.*S33))./(A22.*S11 + B22.*S22 + D22.*S33 + A22.*B22.*S12.^2 + A22.*D22.*S13.^2 + B22.*D22.*S23.^2 - A22.*B22.*S11.*S22 - A22.*D22.*S11.*S33 - B22.*D22.*S22.*S33 - A22.*B22.*D22.*S11.*S23.^2 - A22.*B22.*D22.*S13.^2.*S22 - A22.*B22.*D22.*S12.^2.*S33 + 2.*A22.*B22.*D22.*S12.*S13.*S23 + A22.*B22.*D22.*S11.*S22.*S33 - 1)-c21;
  -(A21.*D21.*(S13 + B22.*S12.*S23 - B22.*S13.*S22))./(A22.*S11 + B22.*S22 + D22.*S33 + A22.*B22.*S12.^2 + A22.*D22.*S13.^2 + B22.*D22.*S23.^2 - A22.*B22.*S11.*S22 - A22.*D22.*S11.*S33 - B22.*D22.*S22.*S33 - A22.*B22.*D22.*S11.*S23.^2 - A22.*B22.*D22.*S13.^2.*S22 - A22.*B22.*D22.*S12.^2.*S33 + 2.*A22.*B22.*D22.*S12.*S13.*S23 + A22.*B22.*D22.*S11.*S22.*S33 - 1)-c31;
   (A22.*B11.*S11 - B21.^2.*S22 - B11 + B11.*B22.*S22 + B11.*D22.*S33 - A22.*B21.^2.*S12.^2 - B21.^2.*D22.*S23.^2 + A22.*B11.*B22.*S12.^2 + A22.*B11.*D22.*S13.^2 + B11.*B22.*D22.*S23.^2 + A22.*B21.^2.*S11.*S22 + B21.^2.*D22.*S22.*S33 + A22.*B21.^2.*D22.*S11.*S23.^2 + A22.*B21.^2.*D22.*S13.^2.*S22 + A22.*B21.^2.*D22.*S12.^2.*S33 - A22.*B11.*B22.*S11.*S22 - A22.*B11.*D22.*S11.*S33 - B11.*B22.*D22.*S22.*S33 - A22.*B11.*B22.*D22.*S11.*S23.^2 - A22.*B11.*B22.*D22.*S13.^2.*S22 - A22.*B11.*B22.*D22.*S12.^2.*S33 - 2.*A22.*B21.^2.*D22.*S12.*S13.*S23 - A22.*B21.^2.*D22.*S11.*S22.*S33 + 2.*A22.*B11.*B22.*D22.*S12.*S13.*S23 + A22.*B11.*B22.*D22.*S11.*S22.*S33)./(A22.*S11 + B22.*S22 + D22.*S33 + A22.*B22.*S12.^2 + A22.*D22.*S13.^2 + B22.*D22.*S23.^2 - A22.*B22.*S11.*S22 - A22.*D22.*S11.*S33 - B22.*D22.*S22.*S33 - A22.*B22.*D22.*S11.*S23.^2 - A22.*B22.*D22.*S13.^2.*S22 - A22.*B22.*D22.*S12.^2.*S33 + 2.*A22.*B22.*D22.*S12.*S13.*S23 + A22.*B22.*D22.*S11.*S22.*S33 - 1)-c22;
  -(B21.*D21.*(S23 + A22.*S12.*S13 - A22.*S11.*S23))./(A22.*S11 + B22.*S22 + D22.*S33 + A22.*B22.*S12.^2 + A22.*D22.*S13.^2 + B22.*D22.*S23.^2 - A22.*B22.*S11.*S22 - A22.*D22.*S11.*S33 - B22.*D22.*S22.*S33 - A22.*B22.*D22.*S11.*S23.^2 - A22.*B22.*D22.*S13.^2.*S22 - A22.*B22.*D22.*S12.^2.*S33 + 2.*A22.*B22.*D22.*S12.*S13.*S23 + A22.*B22.*D22.*S11.*S22.*S33 - 1)-c32;
  (A22.*D11.*S11 - D21.^2.*S33 - D11 + B22.*D11.*S22 + D11.*D22.*S33 - A22.*D21.^2.*S13.^2 - B22.*D21.^2.*S23.^2 + A22.*B22.*D11.*S12.^2 + A22.*D11.*D22.*S13.^2 + B22.*D11.*D22.*S23.^2 + A22.*D21.^2.*S11.*S33 + B22.*D21.^2.*S22.*S33 + A22.*B22.*D21.^2.*S11.*S23.^2 + A22.*B22.*D21.^2.*S13.^2.*S22 + A22.*B22.*D21.^2.*S12.^2.*S33 - A22.*B22.*D11.*S11.*S22 - A22.*D11.*D22.*S11.*S33 - B22.*D11.*D22.*S22.*S33 - A22.*B22.*D11.*D22.*S11.*S23.^2 - A22.*B22.*D11.*D22.*S13.^2.*S22 - A22.*B22.*D11.*D22.*S12.^2.*S33 - 2.*A22.*B22.*D21.^2.*S12.*S13.*S23 - A22.*B22.*D21.^2.*S11.*S22.*S33 + 2.*A22.*B22.*D11.*D22.*S12.*S13.*S23 + A22.*B22.*D11.*D22.*S11.*S22.*S33)./(A22.*S11 + B22.*S22 + D22.*S33 + A22.*B22.*S12.^2 + A22.*D22.*S13.^2 + B22.*D22.*S23.^2 - A22.*B22.*S11.*S22 - A22.*D22.*S11.*S33 - B22.*D22.*S22.*S33 - A22.*B22.*D22.*S11.*S23.^2 - A22.*B22.*D22.*S13.^2.*S22 - A22.*B22.*D22.*S12.^2.*S33 + 2.*A22.*B22.*D22.*S12.*S13.*S23 + A22.*B22.*D22.*S11.*S22.*S33 - 1)-c33];

After saving this function on my search path, I used fsolve, as follows:

format long g
Sfinal = fsolve(@funs,(1+i)*ones(1,6))
Equation solved.
fsolve completed because the vector of function values is near zero
as measured by the default value of the function tolerance, and
the problem appears regular as measured by the gradient.
<stopping criteria details>
Sfinal =
Columns 1 through 3
      -0.0576463452402932 -    0.0753664180207837i        -0.0744682052419572 -    0.0630375797716533i        -0.0698118920703255 -    0.0672129482711116i
Columns 4 through 6
       -0.185173231731065 +     0.109260636702373i         -0.183468124937688 +      0.10862578397362i         -0.186854054845906 +     0.104967752613526i

As a test that this is a solution in double precision, evaluate funs at that point.

funs(Sfinal)
ans =
       2.04330996567137e-12 +  8.08547673258886e-13i
       2.06829562157673e-12 +  8.28642710004601e-13i
       2.07329856408145e-12 +  8.41465785939022e-13i
       2.07303618715571e-12 +   7.6394446324457e-13i
       2.08355208086708e-12 +  8.14071032806396e-13i
       2.05693795329864e-12 +  7.88147325181399e-13i

Note that when calling fsolve, it was necessary to use a complex start point. Even then, I have seen cases where fsolve fails for complex valued problems, though I have not yet pinned down why it may fail. It did seem to work here though.

MATLAB: Solve system of large linear symbolic equations

On my system, Maple ran out of memory trying to solve() that equation. There is a possibility that a system with more memory might be able to arrive at an answer, given enough time.

Solving the entire system of 12 equations when they are each this large, would take a lot of time and memory.

I would recommend following John's advice to simplify the system.

Best Answer

Related Solutions

MATLAB: Solving a set of equations numerically where solutions are complex numbers

MATLAB: Solve system of large linear symbolic equations

Related Question