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I am having problems finding the roots of the following non linear discontinuous equation

Asked by David Campbell on 17 Jul 2013
Accepted Answer by Matt J

Below is the code I am using, and the error message I recieve:

function y = firstorder(x)

mat = 'mnfepas';

props = material_data(mat);

S = props(1,3);

L = props(1,4);

J = props(1,5);

Tc = props(1,6);

Ns = props(1,8);

g = props(10);

T = 200;

B = 1;

muB = 9.27e-24; % units: Am^2 or J/T

kB = 1.3807e-23; % units: J/K

nu = 1.75;

y = (1/T)*(3*Tc*J*((0.5/J)*((2*J+1)*coth((J+0.5)*x/J)-coth(0.5*x/J)))/(J+1) + g*muB*J*B/kB + (9/5)*(((2*J+1)^4)-1)*Tc*nu*(((0.5/J)*((2*J+1)*coth((J+0.5)*x/J)-coth(0.5*x/J)))^3)/((2*J+2)^4))-x;


Then I use the following:

x0 = 3;

x = fzero(firstorder,x0);

And then I get the following error message:

Error using firstorder (line 21) Not enough input arguments.

Error in nonlinear_brillouin_solver (line 2) x = fzero(firstorder,x0);

the function y is discontinuous at zero due to the coth, and at some values of T the function will cross the x axis multiple times, so I should get multiple roots... i think

Any help would be greatly appreciated



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2 Answers

Answer by Matt J
on 17 Jul 2013
 Accepted answer

Should be

   x = fzero(@firstorder,x0);


Thank you, does fzero work on discontinuous functions? My function has a discontinuity at x = 0, and also can it return more than one x value?

Matt J
on 17 Jul 2013

No fzero cannot return more than 1 value. The presence of discontinuities can definitely confuse fzero, as in the following example. You should therefore specify a search interval where the function is continuous.

>> fzero(@(x) 1/x,[-2,2])
ans =

Answer by Youssef Khmou
on 17 Jul 2013

hi David,

if you have numerical values of the coefficients, you can use the function root instead as in this example :

6 x^3 - 4 x^2 + 10 x + 6 =0

 R=roots([6 -4 10 6])


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