MATLAB: Symbolic equation system solving with a lot of simple equations

equation solvingMATLABsolvesymbolic equation solvingsyms

I have got a lot of very simple symbolic equations and i want to solve the equation system for the vars but for some reason i cannot succed in it. Here is an example:

syms u1 u2 I1 I2 I3 I4 C1 C2 R1 R2 a b e1 e2 e3 e4 e5 e6 e7 e8 e9 e10 e11 e12 e13 e14 e15 e16 f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13 f14 f15 f16 dx1 dx2 dx3 dx4 dx5 dx6 X1 X2 X3 X4 X5 X6
eqns = [e1==u1, e1==e2+e3+e4, f1==f2, f2==f3, f3==f4, e2==R1*f2, e3==dx1, f3==X1/I1, e4==a*f5, f4==e5/a, e5==e6+e7+e8, f5==f6, f6==f7, f7==f8, e6==R2*f6, e7==dx2, f7==X2/I2, e8==b*e9, f8==f9/b, e9==e10, e10==e11, f9==f10+f11, e10==X3/C1, f10==dx3, e11==e12+e13+e14, f11==f12, f12==f13, f13==f14, e12==u2, e13==dx4, f13==X4/I3, e14==e15, e15==e16, f14==f15+f16, e15==X5/C2, f15==dx5, e16==dx6, f16==X6/I4];
var = [dx1 dx2 dx3 dx4 dx5 dx6];
sol=solve(eqns,var)

As you can see i've got a lot of variables too and i want to eliminate all of the e. and f. -s (e1,e2,…e16,f1,f2,…,f16) and get is solved for dx.-es. I can solve it by hand so it shouldn't be an issue with matlab (this is the result: "Empty sym: 0-by-1"), possibly i just missed some important step.

Do you think it is even possible to solve the symbolic equations this way or should i try some different method?

Best Answer

People seem to think that shorthand that makes sense to them is always going ot make sense to a computer.

syms f1 f2 f3 f4

When use the following form, what does MATLAB see?

f1==f2==f3==f4
ans =
((f1 == f2) == f3) == f4

Can we "solve" that problem? What if we added one more equation, that f4==1? That should make the problem well posed.

solve(f1==f2==f3==f4,f4 == 1)
ans = 
  struct with fields:

    f1: [0×1 sym]
    f2: [0×1 sym]

No solution was found.

The problem is YOU understand a transitive shorthand like this: a=b=c=d, to imply that all of them are equal. MATLAB does not take it that way.

Instead, you need to set the problem up with separate equations.

S = solve(f1==f2, f2==f3, f3==f4, f4 == 1)
S = 
struct with fields:

    f1: [1×1 sym]
    f2: [1×1 sym]
    f3: [1×1 sym]
    f4: [1×1 sym]

And we see that S.f1 is what we expect.

S.f1
ans =
1

Related Solutions

MATLAB: Solve a system of nonlinear equations with conditions

The calculation had finished by the time I went to have another look.

The solutions were parameterized. I have attached a .mat with the values. (To get the conditions in the form I present them here, simplify() with 'steps', 100)

x1 -> z
x2 -> z1
x3 -> z2
x4 -> z3
y1 -> z4
y2 -> z5

Four independent solutions were created:

Set #1 (that is, x1 x2 x3 x4 y1 y2 are the tuples of values [z z1 z2 z3 z4 z5] such that the below conditions all hold)

0 <= z
0 <= z1
0 <= z2
0 <= z3
z4 == 0
0 <= z5
K22*epsilon2 + epsilon2*z5 == K22*beta12*f12*z + K22*beta22*f22*z1 + K22*beta32*f32*z2 + K22*beta42*f42*z3
s1 == z*(f12*z5 + r1*((eta11*z + eta12*z1 + eta13*z2 + eta14*z3)/N - 1))
s2 == z1*(f22*z5 + r2*((eta21*z + eta22*z1 + eta23*z2 + eta24*z3)/N - 1))
s3 == z2*(f32*z5 + r3*((eta31*z + eta32*z1 + eta33*z2 + eta34*z3)/N - 1))
s4 == z3*(f42*z5 + r4*((eta41*z + eta42*z1 + eta43*z2 + eta44*z3)/N - 1))

Set #2: (that is, x1 x2 x3 x4 y1 y2 are the tuples of values [z z1 z2 z3 z4 z5] such that the below conditions all hold)

0 <= z
0 <= z1
0 <= z2
0 <= z3
0 <= z4
z5 == 0
K11*epsilon1 + epsilon1*z4 == K11*beta11*f11*z + K11*beta21*f21*z1 + K11*beta31*f31*z2 + K11*beta41*f41*z3
s1 == z*(f11*z4 + r1*((eta11*z + eta12*z1 + eta13*z2 + eta14*z3)/N - 1))
s2 == z1*(f21*z4 + r2*((eta21*z + eta22*z1 + eta23*z2 + eta24*z3)/N - 1))
s3 == z2*(f31*z4 + r3*((eta31*z + eta32*z1 + eta33*z2 + eta34*z3)/N - 1))
s4 == z3*(f41*z4 + r4*((eta41*z + eta42*z1 + eta43*z2 + eta44*z3)/N - 1))

Set #3: (that is, x1 x2 x3 x4 y1 y2 are the tuples of values [z z1 z2 z3 z4 z5] such that the below conditions all hold)

N ~= 0
0 <= z
0 <= z1
0 <= z2
0 <= z3
z4 == 0
z5 == 0
N*s1 == r1*z*(eta11*z - N + eta12*z1 + eta13*z2 + eta14*z3)
N*s2 == r2*z1*(eta21*z - N + eta22*z1 + eta23*z2 + eta24*z3)
N*s3 == r3*z2*(eta31*z - N + eta32*z1 + eta33*z2 + eta34*z3)
N*s4 == r4*z3*(eta41*z - N + eta42*z1 + eta43*z2 + eta44*z3)

Set #4: (that is, x1 x2 x3 x4 y1 y2 are the tuples of values [z z1 z2 z3 z4 z5] such that the below conditions all hold)

0 <= z
0 <= z1
0 <= z2
0 <= z3
0 <= z4
0 <= z5
K11*epsilon1 + epsilon1*z4 + K11*g12*z5 + epsilon1*omega2*z5 == K11*beta11*f11*z + K11*beta21*f21*z1 + K11*beta31*f31*z2 + K11*beta41*f41*z3
K22*epsilon2 + epsilon2*z5 + epsilon2*omega1*z4 + K22*g12*gamma12*z4 == K22*beta12*f12*z + K22*beta22*f22*z1 + K22*beta32*f32*z2 + K22*beta42*f42*z3
s1 == z*(f11*z4 + f12*z5 + r1*((eta11*z + eta12*z1 + eta13*z2 + eta14*z3)/N - 1))
s2 == z1*(f21*z4 + f22*z5 + r2*((eta21*z + eta22*z1 + eta23*z2 + eta24*z3)/N - 1))
s3 == z2*(f31*z4 + f32*z5 + r3*((eta31*z + eta32*z1 + eta33*z2 + eta34*z3)/N - 1))
s4 == z3*(f41*z4 + f42*z5 + r4*((eta41*z + eta42*z1 + eta43*z2 + eta44*z3)/N - 1))

Code:

syms N x1 x2 x3 x4 y1 y2 real
assume(N >= 0 & x1 >= 0 & x2 >= 0 & x3 >= 0 & x4 >= 0 & y1 >= 0 & y2 >= 0)
syms r1 r2 r3 r4 positive
syms s1 s2 s3 s4 positive
syms epsilon1 epsilon2 omega1 positive 
syms omega2 K11 K22 positive
syms beta11 beta21 beta31 beta41 positive
syms beta12 beta22 beta32 beta42 gamma12 positive
syms eta11 eta12 eta13 eta14 positive
syms eta21 eta22 eta23 eta24 positive
syms eta31 eta32 eta33 eta34 positive
syms eta41 eta42 eta43 eta44 positive
syms R c1 c2 c3 c4 positive
syms f11 f12 f21 f22 f31 f32 f41 f42 g12 positive
F1=R-c1*x1-c2*x2-c3*x3-c4*x4;
F2=x1*(r1*(1-(eta11*x1+eta12*x2+eta13*x3+eta14*x4)/N)-f11*y1-f12*y2)+s1==0;
F3=x2*(r2*(1-(eta21*x1+eta22*x2+eta23*x3+eta24*x4)/N)-f21*y1-f22*y2)+s2==0;
F4=x3*(r3*(1-(eta31*x1+eta32*x2+eta33*x3+eta34*x4)/N)-f31*y1-f32*y2)+s3==0;
F5=x4*(r4*(1-(eta41*x1+eta42*x2+eta43*x3+eta44*x4)/N)-f41*y1-f42*y2)+s4==0;      
F6=y1*(-epsilon1*(1+(y1+omega2*y2)/K11)-g12*y2+beta11*f11*x1+beta21*f21*x2+beta31*f31*x3+beta41*f41*x4)==0;
       
F7=y2*(-epsilon2*(1+(omega1*y1+y2)/K22)-gamma12*g12*y1+beta12*f12*x1+beta22*f22*x2 +beta32*f32*x3+beta42*f42*x4)==0 ;           
eqns=[F2,F3,F4,F5,F6,F7];
S = solve(eqns,[x1,x2,x3,x4,y1,y2], 'returnconditions', true);

MATLAB: This functions doesn’t work on matlab coder

First of all, you have

if Errx1 < Err && Errx2 , Err;

which is the same as

if (Errx1 < Err) && (Errx2 ~= 0)
  Err;

You probably wanted

 if Errx1 < Err && Errx2 < Err

Secondly, you have the structure

for j=1:iter
  ...
  if Errx1 < Err && Errx2 < Err
     xsol1 = x1;
     break;
  end
end

Consider what happens if Errx1 < Err && Errx2 < Err does not become true before the end of the "for j" loop: xsol1 will not get assigned to in that situation.

When you use MATLAB Coder, or MATLAB Function Block (in Simulink) you should get into the habit of always assigning some value to your output variables as one of the first things you do. That will have three benefits: (1) It will tell the code generation what size and datatype the variable is expected to be; (2) If you assign the result of a function call to the variable, it will tell the code generation that it needs to copy back from the MxArray format into the direct C/C++ variables used for code generation; and (3) If you accidentally have any paths in which you did not assign a value to the variable, it gives the variable a well defined value to return.

Best Answer

Related Solutions

MATLAB: Solve a system of nonlinear equations with conditions

MATLAB: This functions doesn’t work on matlab coder

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