assume can't constrain vpasolve (2024)

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Faraz j on 4 Feb 2024

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Edited: Torsten on 6 Feb 2024

Accepted Answer: Torsten

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Hello,

I am attempting to solve a system of nonlinear equations, but unfortunately, vpasolve find answer outside the range I specified and also ignores the constrain condition using "assume".

Here is my code:

% ------------ Parameters --------------------

M = 2;

N = 4;

a = 22.3702/25.4;

beta_10 = 3.41;

k0 = 4.9348;

K2 = 493.2753;

V = [0.7509 1.0000 1.0000 0.7509];

Zb = [-13.0203-6.9290i -19.0747-10.8648i -38.1365-9.9235i -46.2847-14.8522i];

X = [0.1672 -0.2434 0.1502 -0.0871 ; 0.0871 -0.1502 0.2434 -0.1672];

% ------------ Required Functions --------------------

syms g(x) h1(y) h2(y) v(x)

g(x) = 1.177.*sin((pi.*x)./a).^2;

h1(y) = 1-275.*(y-1).^2;

h2(y) = -14.*(y-1);

h(y) = h1(y) + 1i*h2(y);

v(x) = 1.517+1.833.*x.^2;

% ------------ Equation Functions --------------------

x = sym('x',[M N]);

y = sym('y',[M N]);

E1 = []; E2 = []; E3 = []; eq3 = 0; eq4 = 0;

j = 1;

for i = 1:N

Fn = ((cos((beta_10/k0)*y(j,i)*v(x(j,i)))-cos(y(j,i)*v(x(j,i))))/sin(y(j,i)*v(x(j,i))))*(sin(pi*x(j,i))/a);

Ya_G0 = (K2*Fn^2) / (((K2*Fn^2)/(g(x(j,i))*h(y(j,i))))+(Zb(j,i)));

E1 = [E1; imag((K2*Fn^2)/(g(x(j,i))*h(y(j,i)))) == -1*imag(Zb(j,i))];

if i==1 && j==1

Fn_1 = Fn;

Ya_G0_1 = Ya_G0;

else

if mod(i-1,2) == 1

sign = -1;

else

sign = +1;

end

E2 = [E2; (Ya_G0/(Fn)) == (-1)^(j-1)*sign * abs(V(j,i)/V(1,1)) * (sin(y(j,i)*v(x(j,i)))/(sin(y(1,1)*v(x(1,1))))) * (Ya_G0_1/(Fn_1)) ];

end

eq3 = eq3 + Ya_G0;

end

E3 = [E3; eq3 == 2];

E = [E1.',E2.',E3.'];

assume(x,"real"); assumeAlso(x<0.48*a);

assume(y,"positive"); assumeAlso(y,"real");

X_T = X.';

X_trial = X_T(:).';

sol_initial = [X_trial(1:M*N/2),ones(1,M*N/2)];

% sol_range = [all_x_range(1:M*N/2,:);repmat([0.96 1.04],M*N/2,1)];

% sol_initial = sol_range;

x_T = x.'; x_unknown_vector = x_T(:).';

y_T = y.'; y_unknown_vector = y_T(:).';

unknown = [x_unknown_vector(1:M*N/2),y_unknown_vector(1:M*N/2)];

S = vpasolve(E,unknown,sol_initial);

MyFieldNames = fieldnames(S);

MyValuesX = zeros(1,M*N/2); MyValuesY = zeros(1,M*N/2);

for i=1:M*N/2

MyValuesX(1,i) = getfield(S,MyFieldNames{i});

MyValuesY(1,i) = getfield(S,MyFieldNames{i+N/2});

end

round(MyValuesY,3)

ans = 1×4

0.1500 -0.0870 1.0050 1.0070

I also used the range for "vpasolver" as commented "sol_range" in the above code, but I didn't get the desired result either.

The solution to "y" should be a number around 1.

Thanks for your help in advance.

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Moved: Torsten on 4 Feb 2024

vpasolve is a numerical solver - symbolic commands like "assume" are not respected. But you can set ranges for the variables in "vpasolve" by setting "init_param" to a numerical search interval.

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Faraz j on 4 Feb 2024

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I defined the range as

all_x_range = repmat([0 a/2;-a/2 0],M*N/2,1);

sol_range = [all_x_range(1:M*N/2,:);repmat([0.96 1.04],M*N/2,1)];

but still give me a solution outside of this range !!!!

Torsten on 4 Feb 2024

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Most probably, no solution for y in the specified range can be found.

Faraz j on 5 Feb 2024

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Edited: Faraz j on 5 Feb 2024

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This is the solution vpasolve should find !

MyValuesX = [0.167 -0.2434 0.1502 -0.0871];

MyValuesY = [1.005 1.007 1.0075 1.0114];

Torsten on 5 Feb 2024

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Edited: Torsten on 5 Feb 2024

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Ok, vpasolve failed.

Here is a numerical solution with "lsqnonlin":

% ------------ Parameters --------------------

M = 2;

a = 22.3702/25.4;

beta_10 = 3.41;

k0 = 4.9348;

K2 = 493.2753;

V = [0.7509 1.0000 1.0000 0.7509];

Zb = [-13.0203-6.9290i -19.0747-10.8648i -38.1365-9.9235i -46.2847-14.8522i];

X = [0.1672 -0.2434 0.1502 -0.0871 ; 0.0871 -0.1502 0.2434 -0.1672];

% ------------ Required Functions --------------------

syms g(x) h1(y) h2(y) v(x)

g(x) = 1.177.*sin((pi.*x)./a).^2;

h1(y) = 1-275.*(y-1).^2;

h2(y) = -14.*(y-1);

h(y) = h1(y) + 1i*h2(y);

v(x) = 1.517+1.833.*x.^2;

% ------------ Equation Functions --------------------

x = sym('x',[M N]);

y = sym('y',[M N]);

E1 = []; E2 = []; E3 = []; eq3 = 0; eq4 = 0;

j = 1;

for i = 1:N

Fn = ((cos((beta_10/k0)*y(j,i)*v(x(j,i)))-cos(y(j,i)*v(x(j,i))))/sin(y(j,i)*v(x(j,i))))*(sin(pi*x(j,i))/a);

Ya_G0 = (K2*Fn^2) / (((K2*Fn^2)/(g(x(j,i))*h(y(j,i))))+(Zb(j,i)));

E1 = [E1; imag((K2*Fn^2)/(g(x(j,i))*h(y(j,i)))) - (-1*imag(Zb(j,i)))];

if i==1 && j==1

Fn_1 = Fn;

Ya_G0_1 = Ya_G0;

else

if mod(i-1,2) == 1

sign = -1;

else

sign = +1;

end

tmp=(Ya_G0/(Fn)) -( (-1)^(j-1)*sign * abs(V(j,i)/V(1,1)) * (sin(y(j,i)*v(x(j,i)))/(sin(y(1,1)*v(x(1,1))))) * (Ya_G0_1/(Fn_1)) );

E2 = [E2; tmp ];

end

eq3 = eq3 + Ya_G0;

end

E3 = [E3; eq3 - 2];

E = [E1.',E2.',E3.'];

Enum = matlabFunction(E);

Enum = @(z)real(Enum(z(1),z(2),z(3),z(4),z(5),z(6),z(7),z(8)));

all_x_range = repmat([0 a/2;-a/2 0],M*N/2,1);

sol_range = [all_x_range(1:M*N/2,:);repmat([0.96 1.04],M*N/2,1)];

lb = sol_range(:,1);

ub = sol_range(:,2);

format long

solnum = lsqnonlin(Enum,(lb+ub)/2,lb,ub,optimset('TolX',1e-12,'TolFun',1e-12))

Local minimum possible.lsqnonlin stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.

solnum = 8×1

0.167210839303176 -0.243345930008853 0.150185145920223 -0.087101777478330 1.005182169079357 1.007061443401629 1.007465052260141 1.011445713909488

Enum(solnum)

ans = 1×8

1.0e-12 * -0.071942451995710 0.156319401867222 0.099475983006414 -0.110134124042816 0 -0.000888178419700 -0.001554312234475 0.000444089209850

Faraz j on 6 Feb 2024

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Edited: Faraz j on 6 Feb 2024

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Thanks a lot ! you're life saver

One more issue @Torsten! for different M and N, how can I modify this

Enum = @(z)real(Enum(z(1),z(2),z(3),z(4),z(5),z(6),z(7),z(8)));

Torsten on 6 Feb 2024

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Edited: Torsten on 6 Feb 2024

Open in MATLAB Online

Enum = matlabFunction(E,'Vars',{[x(1,1:N),y(1,1:N)]})

Enum = @(z)real(Enum(z.'));

instead of

Enum = matlabFunction(E);

Enum = @(z)real(Enum(z(1),z(2),z(3),z(4),z(5),z(6),z(7),z(8)));

This is at least correct for the equations you solved above since j was not increased and remained 1 (so I don't know why you created x and y as x(M,N) and y(M,N) and not as x(1,N) and y(1,N))

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Matt J on 4 Feb 2024

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Edited: Matt J on 4 Feb 2024

There doesn't appear to be any good reason for you to be using sym variables. Your code doesn't seem to make use of any other Symbolic Math Toolbox functions like diff or simplify.... Just re-implement using non-symbolic variables and solve with lsqnonlin which does allow you to impose bounds on the variables. Or, use matlabFunction to convert your symbolic functions into nonsymbolic ones.

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Faraz j on 5 Feb 2024

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Edited: Faraz j on 5 Feb 2024

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Thanks for your suggestion !

I wanted to try this approach, but building the function was complicated for me without knowing matlabFunction.

I am not good at this, I think. I tried this and got an error

fun = matlabFunction(E);

lsqnonlin(fun,sol_initial)

what am I doing wrong?

Matt J on 5 Feb 2024

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what am I doing wrong?

Who knows? You haven't shown copy/pastes of the errors.

Faraz j on 6 Feb 2024

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Thanks for your advice, @Torsten solved the issue !

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assume can't constrain vpasolve (2024)

FAQs

What does vpasolve mean in MATLAB? ›

An equation or a system of equations can have multiple solutions. To find these solutions numerically, use the function vpasolve . For polynomial equations, vpasolve returns all solutions. For nonpolynomial equations, vpasolve returns the first solution it finds.

Show Me More ›
What is the difference between solve and Vpasolve? ›

solve solves equations and inequalities that contain parameters. vpasolve does not solve inequalities, nor does it solve equations that contain parameters. solve can return parameterized solutions. vpasolve does not return parameterized solutions.

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What is the precision of Vpasolve in MATLAB? ›

By default, vpasolve finds the solutions to 32 significant digits. To change the number of significant digits, use the digits function.

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What is approximate solution in MATLAB? ›

approximate( solution ) generates either a Simulink® model containing a subsystem made up of the Lookup Table block, or a lookup table as a MATLAB® function, depending on the ApproximateSolutionType property of the FunctionApproximation.

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What is VPA used for in MATLAB? ›

Description. xVpa = vpa( x ) uses variable-precision arithmetic (arbitrary-precision floating-point numbers) to evaluate each element of the symbolic input x to at least d significant digits, where d is the value of the digits function. The default value of digits is 32.

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How to solve for a variable? ›

These are the general steps for solving an equation with variables:
  1. Step 1: Simplify both sides of the equation.
  2. Step 2: Move all of the parts containing the variable you are solving for to the same side of the equation.
  3. Step 3: Isolate the variable using inverse operations.
Dec 26, 2023

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How do I fix precision in MATLAB? ›

You can set a higher precision by using the digits function. Approximate a sum using the default precision of 32 digits. If at least one input is wrapped with vpa , all other inputs are converted to variable precision automatically. You must wrap all inner inputs with vpa , such as exp(vpa(200)) .

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How many decimal places is a VPA in MATLAB? ›

The default precision for vpa is 32 digits. Increase precision beyond 32 digits by using digits . Find pi using vpa , which uses the default 32 digits of precision. Confirm that the current precision is 32 by using digits .

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How do you set MATLAB to double precision? ›

Description. Y = double( X ) converts the values in X to double precision.

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What is the difference between exact solution and approximate solution? ›

My understanding is that "exact" translates to "produces a provably optimal solution" and "approximate" translates to "may or may not find an optimal solution but will not provide proof of optimality".

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An approximation is anything that is intentionally similar but not exactly equal to something else.

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An approximation is anything that is similar, but not exactly equal, to something else. A number can be approximated by rounding.

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dsolve sorts the dependent variables alphabetically, and then assigns the solutions for the variables to output variables or symbolic arrays.

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X = linsolve( A , B ) solves the matrix equation AX = B, where A is a symbolic matrix and B is a symbolic column vector. [ X , R ] = linsolve( A , B ) also returns the reciprocal of the condition number of A if A is a square matrix. Otherwise, linsolve returns the rank of A .

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How to use simplify in MATLAB? ›

S = simplify( expr ) performs algebraic simplification of expr . If expr is a symbolic vector or matrix, this function simplifies each element of expr . S = simplify( expr , Name,Value ) performs algebraic simplification of expr using additional options specified by one or more Name,Value pair arguments.

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