Solving a nonlinear equation

Hi. I have to solve a nonlinear equation over a range of values of a given parameter.

clc; clear;close all;

    f = 0.35+0.1;
    x0=f^2
    r = 0.3:0.3:1.5;
    zeta = 0.25;
    eps = 1/6;
    eqn= @(A)(r.^4.*A.^2)-(2.*r.^2.*A.^2)-(2.*r.^2.*((3.*eps.*A.^4)./4))+(3.*((3.*eps.*A.^4)./4))+(((9.*eps.^2.*A.^6)./16))+(4.*zeta.*r.^2.*A.^2);
    
    
    Ans=fzero(eqn,x0);

I want the answer to give me a vector for A at the given value for r, but I can't get the code to work.

Answer 1

[Abolfazl Chaman Motlagh]

the eqn as you defined is a vector-value multi-variable function. fzero takes a scalar-value single-variable function.

so maybe you have a Writing error for eqn that make it vector value.

f = 0.35+0.1;
x0=f^2;
r = 0.3:0.3:1.5;
zeta = 0.25;
eps = 1/6;
eqn= @(A)(r.^4.*A.^2)-(2.*r.^2.*A.^2)-(2.*r.^2.*((3.*eps.*A.^4)./4))+(3.*((3.*eps.*A.^4)./4))+(((9.*eps.^2.*A.^6)./16))+(4.*zeta.*r.^2.*A.^2);
eqn(rand(1,5))

ans = 1×5

-0.0047 -0.0155 -0.0149 0.0471 0.1364

if you really want to solve the above system of equations you can use minimization functions like fminunc. or use solve for algebraic equations. ( you should change eqn for both of them)

for using fminunc the cost function should be a scalar value function. one easy method to change your eqn is taking norm from function:

eqn_= @(A)norm((r.^4.*A.^2)-(2.*r.^2.*A.^2)-(2.*r.^2.*((3.*eps.*A.^4)./4))+(3.*((3.*eps.*A.^4)./4))+(((9.*eps.^2.*A.^6)./16))+(4.*zeta.*r.^2.*A.^2));
x = fminunc(eqn_,ones(1,5));

Local minimum possible.

fminunc stopped because it cannot decrease the objective function
along the current search direction.

x

x = 1×5

0.4796 0.8806 0.9109 -0.0000 -0.0000

Y=eqn(x)

Y = 1×5

1.0e+-8 * -0.0519 -0.1379 -0.0523 0.0632 0.0381

norm(eqn(x))

ans = 1.7290e-09

[John D'Errico]

Is there a valid reason why you posted the same answer twice? My guess is the site was moving slow, so you reposted it. I deleted the first of your identical answers.