Solve: System of symbolic linear equations

Hello all,

Although this question has been asked before but I could not find a relevant answer to the problem.

I am trying to solve a system of linear equations but 'solve' does not give me a solution or it takes too long to find one.

clear all;close all;

syms v2 v3 v4 v5 v6 real positive
syms r4 r5 r6 r7 c4 c6 omega real positive
syms Av1 Av2 real positive
syms s %Laplace domain 's'. Cannot define as complex?

zc4=1/(s*c4)
zc6=1/(s*c6)

eqn1 = ((v3-v2)/r4) + ((v3-v4)/zc4) + ((v3-v6)/r5) == 0 

eqn2 = ((v6-v3)/r5) + ((v6-v5)/zc6) + ((v6-v5)/r7) == 0 

eqn3 = ((v4-v3)/zc4) + ((v4-v5)/r6) == 0 

eqn4 = ((v5-v4)/r6) + ((v5-v6)/r7) + ((v5-v6)/zc6) == 0 

eqn5 = v4 == -Av1*v3

eqn6 = v5 == -v6/Av2

eqns = [eqn1 eqn2 eqn3 eqn4 eqn5 eqn6];
vars = [ v3 v4 v5  v6 Av1 Av2 ];
S = solve(eqns,vars,'ReturnConditions',true,'Real',true)
v6_soln=S.v6
%vars= [ v3 v4 v5 ]
%S = eliminate(eqns,vars) % takes too long.

With syms defined as real and positive, solve does not find a solution. With syms defined as 'real' only solve takes too long to come to a solution, perhaps because it evaluates all possible solutions. I have also tried 'eiliminate' function. The system of equations will have multiple solutions if no conditions are assumed for syms. I'd like solve to return real solutions only.

Any help will be appreciated.

[Dyuman Joshi]

You can get a solution this way -

syms v2 v3 v4 v5 v6
syms r4 r5 r6 r7 c4 c6
syms Av1 Av2
syms s %Laplace domain 's'. Cannot define as complex?

All numbers are complex numbers.

zc4=1/(s*c4);
zc6=1/(s*c6);

eqn1 = ((v3-v2)/r4) + ((v3-v4)/zc4) + ((v3-v6)/r5) == 0 ;

eqn2 = ((v6-v3)/r5) + ((v6-v5)/zc6) + ((v6-v5)/r7) == 0 ;

eqn3 = ((v4-v3)/zc4) + ((v4-v5)/r6) == 0 ;

eqn4 = ((v5-v4)/r6) + ((v5-v6)/r7) + ((v5-v6)/zc6) == 0 ;

eqn5 = v4 == -Av1*v3;

eqn6 = v5 == -v6/Av2;

eqns = [eqn1; eqn2; eqn3; eqn4; eqn5; eqn6];
vars = [ v3; v4; v5; v6; Av1; Av2];
S = solve(eqns,vars,'real',true)

S = struct with fields:
     v3: v2
     v4: v2
     v5: v2
     v6: v2
    Av1: -1
    Av2: -1

v6_soln=S.v6

v6_soln = 

%vars= [ v3 v4 v5 ]
%S = eliminate(eqns,vars) % takes too long.

It's not clear to me, why you are eliminating variables from the equations?

Do you only want solution in terms of v6, Av1 and Av2? Then specify those as variables only, to solve for.

[RAN]

Hi,

Yes my aim is to find a solution for v6 in terms of v2 i.e v6=f(v2,av1,av2...), by elminating v3 v4 and v5.

With your modification I run into the same problem that solve takes too long (longer than 30 mins)

Did solve find a solution for you?

[Dyuman Joshi]

Sorry, I forgot to press the run button. Please refer to my above comment again, I have edited it. Yes, it did find a solution.

30 mins is quite sometime. What are system specifications? And which version of MATLAB are you using?

[RAN]

I'd like the solution for v6=f(v2,av1,av2...) by eliminating v3,v4,v5. In your case the solution returned is not relevant.

Also when I mentioned that solve is taking a long time, I meant that I have kept it running for 30 mins, but it never converged to a solution.

[Dyuman Joshi]

What is the correct solution then? If you have a reference solution, then do mention it.

"Yes my aim is to get a solution as ito find a solution for v6 in terms of v2 i.e v6=f(v2,av1,av2...)"

This is unclear. Which variables, exactly, should v6 be a function of?

[RAN]

"Yes my aim is to get a solution as ito find a solution for v6 in terms of v2 i.e v6=f(v2,av1,av2...)" I have edited that statement due to typos.

v6=f(v2 r4 r5 r6 r7 c4 c6 av1 av2)

The solution I am trying to get to is v6=(v2)*((r5*r7)/(c1*c6*r5*r6*r7*s^2+c1*r5*r6*s+r7) ). This solution is obtained after putting limits on av1-->infinity, av2--> infinity which is the next step.

[John D'Errico]

Your system of equations is NOT linear. It is NONLINEAR.

eqn5 = v4 == -Av1*v3

Do you see that BOTH Av1 and v3 are unknowns? The fact that you have a product of unknowns makes it nonlinear. As well, the system is homogeneous, so all zeros solves the problem. It also means that USUALLY, only if the system is rank deficient will a non-trivial solution exist. That your problem is nonlinear makes it more difficult.

And trying to enforce solutions as positive, means the problem gets far more complicated. I'm not at all surprised that solve fails there.

Next, while you talk about converging to a solution, that statement is meaningless, since convergence cannot apply to a problem with symbolic parameters.

So, that you have a nonlinear, homogenous, constrained system of equations, I am not at all suprised that solve has some issues.

[RAN]

Hi,

Please can you elaborate further on known and unknown variables? In my mind Av1 and Av2 are defined the same like r4 r5 r6 r7 c4 c6. The equation I am trying to get at has v6 as the dependent variable and v2 as independent. Maybe I am structring the problem wrong? Please can you let me know what changes would be needed?

Also I agree converging to a solution in terms of symbolic equation is meaningless. I meant finding a symbolic solution.

[John D'Errico]

As far as solve is concerned, you tell it this:

vars = [ v3; v4; v5; v6; Av1; Av2];
S = solve(eqns,vars,'real',true)

So what are unknowns? EVERYTHING in vars. both Av1 and v3 are in vars. This means your problem is nonlinear. It also makes the problem more difficult, of course.

That later on, you decide that some of those variables are not really, truly, absolutely unknowns in your mind, that is irrelevant. All that matters is what you told solve.

[Torsten]

Why do you expect there could be a solution for v6 as a function of v2 ? This could only be true if you had one equation less than unknowns, but you have one equation more than unknowns.

[John D'Errico]

You have said that you want to solve for v6, in terms of v2.

The thing is solve is a computer program. Code sometimes is seemingly intelligent, yet usually is pretty stupid. That is definitely the case of solve.

syms v2 v3 v4 v5 v6 real positive
syms r4 r5 r6 r7 c4 c6 omega real positive
syms Av1 Av2 real positive
syms s %Laplace domain 's'
zc4=1/(s*c4);
zc6=1/(s*c6);
eqn1 = ((v3-v2)/r4) + ((v3-v4)/zc4) + ((v3-v6)/r5) == 0 ;
eqn2 = ((v6-v3)/r5) + ((v6-v5)/zc6) + ((v6-v5)/r7) == 0 ;
eqn3 = ((v4-v3)/zc4) + ((v4-v5)/r6) == 0 ;
eqn4 = ((v5-v4)/r6) + ((v5-v6)/r7) + ((v5-v6)/zc6) == 0 ;
eqn5 = v4 == -Av1*v3;
eqn6 = v5 == -v6/Av2;

The obvious is to eliminate v4 and v5. That part is simple, since we can those last two equations.

eqn1 = eliminate([eqn1,eqn5,eqn6],[v4,v5])

eqn1 = 

eqn2 = eliminate([eqn2,eqn5,eqn6],[v4,v5])

eqn2 = 

eqn3 = eliminate([eqn3,eqn5,eqn6],[v4,v5])

eqn3 = 

eqn4 = eliminate([eqn4,eqn5,eqn6],[v4,v5])

eqn4 = 

Next, isolate v6. What can we learn from that?

eqn1 = isolate(eqn1 == 0,v6)

eqn1 = 

eqn2 = isolate(eqn2 == 0,v6)

eqn2 = 

eqn3 = isolate(eqn3 == 0,v6)

eqn3 = 

eqn4 = isolate(eqn4 == 0,v6)

eqn4 = 

At least, what I see is something that is going to be difficult to now reduce the problem too much further. We want next to emininate the unknowns Av1, Av2, and v3, I think.

[RAN]

@John D'Errico Ok now I understand. I did end up finding a solution by substituting v4=-v3*Av1 and v5=-v6/Av2 in the first four eqns i.e eqn1,eqn2, eqn3, eqn4. Then I used 'solve' to solve for v3 and v6 but seperately. i.e [eqn1 eqn2],[eqn1 eqn3],[eqn1 eqn4] and the last combination gives me the correct solution. This also makes sense since eqn2 and eqn3 are redundant. But I was hoping that 'solve' function gives me four solutions for me to choose from instead of solving them in steps.

@Torsten The structure of the problem i have formulated is wrong. Yes I would need one equation less.