EquationProblem - System of nonlinear equations - MATLAB (original) (raw)

System of nonlinear equations

Description

Specify a system of equations using optimization variables, and solve the system using solve.

Creation

Create an EquationProblem object by using the eqnproblem function. Add equations to the problem by creating OptimizationEquality objects and setting them as Equations properties of theEquationProblem object.

prob = eqnproblem; x = optimvar('x'); eqn = x^5 - x^4 + 3*x == 1/2; prob.Equations.eqn = eqn;

Warning

The problem-based approach does not support complex values in the following: an objective function, nonlinear equalities, and nonlinear inequalities. If a function calculation has a complex value, even as an intermediate value, the final result might be incorrect.

Properties

expand all

Problem equations, specified as an OptimizationEquality array or structure withOptimizationEquality arrays as fields.

Example: sum(x.^2,2) == 4

Problem label, specified as a string or character vector. The software does not useDescription for computation. Description is an arbitrary label that you can use for any reason. For example, you can share, archive, or present a model or problem, and store descriptive information about the model or problem in Description.

Example: "An iterative approach to the Traveling Salesman problem"

Data Types: char | string

This property is read-only.

Optimization variables in the object, specified as a structure of OptimizationVariable objects.

Data Types: struct

Object Functions

Examples

collapse all

To solve the nonlinear system of equations

exp(-exp(-(x1+x2)))=x2(1+x12)x1cos(x2)+x2sin(x1)=12

using the problem-based approach, first define x as a two-element optimization variable.

Create the first equation as an optimization equality expression.

eq1 = exp(-exp(-(x(1) + x(2)))) == x(2)*(1 + x(1)^2);

Similarly, create the second equation as an optimization equality expression.

eq2 = x(1)*cos(x(2)) + x(2)*sin(x(1)) == 1/2;

Create an equation problem, and place the equations in the problem.

prob = eqnproblem; prob.Equations.eq1 = eq1; prob.Equations.eq2 = eq2;

Review the problem.

EquationProblem :

Solve for:
   x


eq1:
   exp((-exp((-(x(1) + x(2)))))) == (x(2) .* (1 + x(1).^2))

eq2:
   ((x(1) .* cos(x(2))) + (x(2) .* sin(x(1)))) == 0.5

Solve the problem starting from the point [0,0]. For the problem-based approach, specify the initial point as a structure, with the variable names as the fields of the structure. For this problem, there is only one variable, x.

x0.x = [0 0]; [sol,fval,exitflag] = solve(prob,x0)

Solving problem using fsolve.

Equation solved.

fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.

sol = struct with fields: x: [2×1 double]

fval = struct with fields: eq1: -2.4070e-07 eq2: -3.8255e-08

exitflag = EquationSolved

View the solution point.

Create a set of equations in two optimization variables.

x = optimvar("x"); y = optimvar("y"); prob = eqnproblem; prob.Equations.eq1 = x^2 + y^2/4 == 2; prob.Equations.eq2 = x^2/4 + 2*y^2 == 2;

Solve the system of equation starting from x=1,y=1/2.

x0.x = 1; x0.y = 1/2; sol = solve(prob,x0)

Solving problem using fsolve.

Equation solved.

fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.

sol = struct with fields: x: 1.3440 y: 0.8799

Evaluate the equations at the points x0 and sol.

vars = optimvalues(prob,x=[x0.x sol.x],y=[x0.y sol.y]); vals = evaluate(prob,vars)

vals = 1×2 OptimizationValues vector with properties:

Variables properties: x: [1 1.3440] y: [0.5000 0.8799]

Equation properties: eq1: [0.9375 8.4322e-10] eq2: [1.2500 6.7431e-09]

The first point, x0, has nonzero values for both equations eq1 and eq2. The second point, sol, has nearly zero values of these equations, as expected for a solution.

Find the degree of equation satisfaction using issatisfied.

[satisfied details] = issatisfied(prob,vars)

satisfied = 1×2 logical array

0 1

details = 1×2 OptimizationValues vector with properties:

Variables properties: x: [1 1.3440] y: [0.5000 0.8799]

Equation properties: eq1: [0 1] eq2: [0 1]

The first point, x0, is not a solution, and satisfied is 0 for that point. The second point, sol, is a solution, and satisfied is 1 for that point. The equation properties show that neither equation is satisfied at the first point, and both are satisfied at the second point.

Version History

Introduced in R2019b

expand all

You can now evaluate optimization expressions and constraints using evaluate and issatisfied for EquationProblem objects.

The value of a constraint depends on the constraint type. An equation is equivalent to an == constraint. For expressions L and R:

For details, see the evaluate andissatisfied reference pages.