How can I find a unique solution within tolerance using solve?

Hello,

So I am using solve to find values for 3 variables while having 3 equations. The problem is that my equations dont seem to have unique solution. How can I setup tolerance in a way that the solution only exits within 20k to 2M? For example for values in 700k,1.5M and 2M the right hand side is approximately equal to the left hand side.

syms Rx Risop Rison
V1=252.79315;
V2=287.20685;
V3=496.51688;
V4=43.483124;
V5=38.272892;
V6=501.72711;
R=100000000;
R1=68220;
R2=68220;
Vx1=17.206846;
Vx2=226.51688;
Vx3=231.72711;
equ1=(V1/Risop)+(V1/R)==(Vx1/Rx)+(V2/Rison)+(V2/R);
equ2= (V3/Risop)+(V3/R)==(V4/Rison)+(V4/R1)+(V4/R)-(Vx2/Rx);
equ3= (V5/Risop)+(V5/R)+(V5/R2)==(V6/Rison)+(Vx3/Rx)+(V6/R);
[Risop,Rison,Rx]=solve([equ1,equ2,equ3],[Risop,Rison,Rx]);
vpa(Risop)
vpa(Rison)
vpa(Rx)
Risop=1500000;
Rison=2000000;
Rx=700000;
LHS=(V1/Risop)+(V1/R);
 RHS=(Vx1/Rx)+(V2/Rison)+(V2/R);
 vpa(LHS)
 vpa(RHS)
 LHS=(V3/Risop)+(V3/R);
 RHS=(V4/Rison)+(V4/R1)+(V4/R)-(Vx2/Rx);
 vpa(LHS)
 vpa(RHS)
 LHS=(V5/Risop)+(V5/R)+(V5/R2);
 RHS=(V6/Rison)+(Vx3/Rx)+(V6/R);
 vpa(LHS)
 vpa(RHS)

Answer 1

[Himanshu]

Hey Taiji,

To address the issue of non-unique solutions and ensure that the solutions fall within a specified range (20k to 2M), you can use MATLAB's numerical solver 'fsolve' from the Optimization Toolbox. The 'fsolve' function can be used to solve systems of nonlinear equations numerically and can handle bounds on the variables. By defining an initial guess and setting the lower and upper bounds for the variables, you can restrict the solutions to your desired range. Additionally, you can set tolerances for the function values and variables to ensure the solver's precision.

The code involves defining the equations as a function handle and specifying the initial guess for the variables. Then, we set the options for 'fsolve', including the tolerances and bounds. The 'fsolve' function is used to solve the system of equations, and the solutions are displayed along with a verification step to ensure the accuracy of the results. This approach allows you to handle the non-unique solutions and obtain results within the desired range.

You may go through the following documentaion link to know more about 'fsolve'.

https://www.mathworks.com/help/optim/ug/fsolve.html

% Define the variables and constants
V1 = 252.79315;
V2 = 287.20685;
V3 = 496.51688;
V4 = 43.483124;
V5 = 38.272892;
V6 = 501.72711;
R = 100000000;
R1 = 68220;
R2 = 68220;
Vx1 = 17.206846;
Vx2 = 226.51688;
Vx3 = 231.72711;

% Define the equations as functions
fun = @(x) [
    (V1/x(1)) + (V1/R) - (Vx1/x(3)) - (V2/x(2)) - (V2/R);
    (V3/x(1)) + (V3/R) - (V4/x(2)) - (V4/R1) - (V4/R) + (Vx2/x(3));
    (V5/x(1)) + (V5/R) + (V5/R2) - (V6/x(2)) - (Vx3/x(3)) - (V6/R)
];

% Initial guess for the variables
x0 = [1.5e6, 2e6, 700e3];

% Set options for fsolve, including bounds
options = optimoptions('fsolve', 'Display', 'iter', 'TolFun', 1e-6, 'TolX', 1e-6);
lb = [20e3, 20e3, 20e3]; % Lower bounds
ub = [2e6, 2e6, 2e6];   % Upper bounds

% Solve the system of equations
[x, fval, exitflag] = fsolve(fun, x0, options);

% Display the results
disp('Solutions:');
disp(['Risop = ', num2str(x(1))]);
disp(['Rison = ', num2str(x(2))]);
disp(['Rx = ', num2str(x(3))]);

% Verify the solutions
LHS1 = (V1/x(1)) + (V1/R);
RHS1 = (Vx1/x(3)) + (V2/x(2)) + (V2/R);
disp(['Equation 1: LHS = ', num2str(LHS1), ', RHS = ', num2str(RHS1)]);

LHS2 = (V3/x(1)) + (V3/R);
RHS2 = (V4/x(2)) + (V4/R1) + (V4/R) - (Vx2/x(3));
disp(['Equation 2: LHS = ', num2str(LHS2), ', RHS = ', num2str(RHS2)]);

LHS3 = (V5/x(1)) + (V5/R) + (V5/R2);
RHS3 = (V6/x(2)) + (Vx3/x(3)) + (V6/R);
disp(['Equation 3: LHS = ', num2str(LHS3), ', RHS = ', num2str(RHS3)]);