EquationProblem

System of nonlinear equations

Description

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

Tip

For the full workflow, see Problem-Based Workflow for Solving Equations.

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 the EquationProblem 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

Equations — Problem equations

[] (default) | OptimizationEquality array | structure with OptimizationEquality arrays as fields

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

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

Description — Problem label

'' (default) | string | character vector

Problem label, specified as a string or character vector. The software does not use Description 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

Variables — Optimization variables in object

structure of OptimizationVariable objects

This property is read-only.

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

Data Types: struct

Object Functions

evaluate	Evaluate optimization expression or objectives and constraints in problem
issatisfied	Constraint satisfaction of an optimization problem at a set of points
optimoptions	Create optimization options
prob2struct	Convert optimization problem or equation problem to solver form
show	Display information about optimization object
solve	Solve optimization problem or equation problem
solvers	Determine default and valid solvers for optimization problem or equation problem
varindex	Map problem variables to solver-based variable index
write	Save optimization object description

Examples

Solve Nonlinear System of Equations, Problem-Based

To solve the nonlinear system of equations

$$\begin{array}{l}\exp (-\exp (-(x_1+x_2)))=x_2(1+x_1^2)\\ x_1\cos (x_2)+x_2\sin (x_1)=\frac{1}{2}\end{array}$$

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

x = optimvar('x',2);

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.

show(prob)

  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: [2x1 double]

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

exitflag = 
    EquationSolved

View the solution point.

disp(sol.x)

    0.3532
    0.6061

Unsupported Functions Require fcn2optimexpr

If your equation functions are not composed of elementary functions, you must convert the functions to optimization expressions using fcn2optimexpr. For the present example:

ls1 = fcn2optimexpr(@(x)exp(-exp(-(x(1)+x(2)))),x);
eq1 = ls1 == x(2)*(1 + x(1)^2);
ls2 = fcn2optimexpr(@(x)x(1)*cos(x(2))+x(2)*sin(x(1)),x);
eq2 = ls2 == 1/2;

See Supported Operations for Optimization Variables and Expressions and Convert Nonlinear Function to Optimization Expression.

Evaluate Expressions in Equation

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 = 
  1x2 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 = 1x2 logical array

   0   1

details = 
  1x2 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

R2024b: EquationProblem supports evaluate and issatisfied

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:

Constraint Type	Value
L <= R	L – R
L >= R	R – L
L == R	abs(L – R)

For details, see the evaluate and issatisfied reference pages.