im trying to solve a system of 4 nonlinear equations with 4 unknowns

clc
clear all

f= @(x) [x(1)+.17*x(1)-x(2)-x(3)-x(4); (((x(3)^(1/2))*x(4))/x(2))-.465797; 2*x(1)-2*x(2)-2*x(3)-x(4); 2*(.17*x(1))-x(2)-x(4)]

s=fsolve(f, [2,2,1,1])

it says it stops prematurely.

im also not sure how to plot these equations to see where my geusses should be

Answer 1

[Torsten]

Equations (1), (3) and (4) give x(2) = 0. Thus it's impossible to satisfy equation (2) where x(2) is in the denominator.

syms x1 x2 x3 x4
f1 = x1+.17*x1-x2-x3-x4 == 0

f1 = 

f2 = 2*x1-2*x2-2*x3-x4 == 0

f2 = 

f3 = 2*0.17*x1-x2-x4 == 0

f3 = 

sol = solve([f1,f2,f3],[x1 x2 x4])

sol = struct with fields:
    x1: (100*x3)/83
    x2: 0
    x4: (34*x3)/83

Answer 2

[John D'Errico]

x = sym('x',[1,4])

x = 

eq1 = x(1)+.17*x(1)-x(2)-x(3)-x(4) == 0

eq1 = 

eq2 = (((x(3)^(1/2))*x(4))/x(2))-.465797 == 0

eq2 = 

eq3 = 2*x(1)-2*x(2)-2*x(3)-x(4) == 0

eq3 = 

eq4 = 2*(.17*x(1))-x(2)-x(4) == 0

eq4 = 

So you have 3 linear equations, and one nonlinear. Does a solution exist? We might ask solve, since this is a pretty simple system.

solve(eq1,eq2,eq3,eq4)

ans = struct with fields:
    x1: [0×1 sym]
    x2: [0×1 sym]
    x3: [0×1 sym]
    x4: [0×1 sym]

So solve found no solution.

vpasolve(eq1,eq2,eq3,eq4)

ans = struct with fields:
    x1: [0×1 sym]
    x2: [0×1 sym]
    x3: [0×1 sym]
    x4: [0×1 sym]

Even vpasolve fails to find a solution.

Equations 1, 3, and 4 are called a homogeneous linear system, although we only have 3 equations in those 4 unknowns. I'll convert them to a matrix form, thus..

[A,b] = equationsToMatrix([eq1,eq3,eq4])

A = 

b = 

We can recover the original equations by a matrix multiply, thus:

A*x.' == b

ans = 

Clearly x1=x2=x3=x4=0 is a solution to that subset of equations. Does a non-trivial solution exist? Yes. It comes from the function null.

Anull = null(A)

Anull = 

So we have that x = Anull, or any multiple of that vector is a solution to the homogeneous linear part of your problem. Now, return to eq2. Do you see the problem?

eq2

eq2 = 

The ONLY solution to those three equations has x2 == 0. But in eq2, we divide by x2.

Therefore, NO solution can possibly exist for this system of equations.

[Alex Sha]

1: Approximate numerical solution

x1: 4.59039465606019E-13
x2: 1.70124596189725E-20
x3: 4.79191761335149E-13
x4: 1.14474596953164E-14

Feval:
F1 = 4.64369367161173E-14
F2 = 1.11022302462516E-15
F3 = -5.17520851784955E-14
F4 = 1.4462594159827E-13

2: making some transformation for 2nd equation:

(((x3^(1/2))*x4)/x2)-.465797=0

equivalence transformed into:

(((x3^(1/2))*x4))-.465797*x2=0

then the solution will be:

x1: 0
x2: 0
x3: 0
x4: 0