I am getting the error as " Not enough input arguments " and " Failure in initial objective function evaluation. FSOLVE cannot continue". Please suggest me possible solutions.

Efu(1)=0;
Efe(1)=0;
landau_levels=@(EF, m, B)  ( (1 / (2 * B)) * ((EF^2 / m^2) - 1) );
amuu = 5.0;
amud = 5.0;
amue = 0.511;
amus = 150.0;
hbarc = 197.3271;
fscon = 137.036;
Bcs = ( 3 * (fscon^0.5) * (amus^2) ) / (hbarc^1.5);  
Bcu = ( 1.5 * (fscon^0.5) * (amuu^2) ) / (hbarc^1.5); 
Bcd = ( 3 * (fscon^0.5) * (amud^2) )/ (hbarc^1.5); 
Bce = ( (fscon^0.5) * (amue^2) )/ (hbarc^1.5);
Bc = (5e17 * 1.95e-14);
Bds = Bc / Bcs;
Bdu = Bc / Bcu;
Bdd = Bc / Bcd;
Bde = Bc / Bce;
Efs(1)=400;
Efu_ini= 300;
Efe_ini= 20;
Efd=Efs(1);
nu_u= landau_levels(Efu_ini,amuu,Bdu);
nu_e= landau_levels(Efe_ini,amue,Bde);
nu_d= landau_levels(Efd(1),amud,Bdd);
nu_s= landau_levels(Efs(1),amus,Bds);    
pF_u= sqrt( max( (Efu_ini^2 - amuu^2 * (1 + 2 * nu_u * Bdu)), 0) );
pF_e= sqrt( max( (Efe_ini^2 - amue^2 * (1 + 2 * nu_e * Bde)), 0) );
pF_d= sqrt( max( (Efd(1)^2 - amud^2 * (1 + 2 * nu_d * Bdd)), 0) );
pF_s= sqrt( max( (Efs(1)^2 - amus^2 * (1 + 2 * nu_s * Bds)), 0) );
n_u = (2 / (3 * pi^2)) * pF_u^3;
n_e = (2 / (3 * pi^2)) * pF_e^3;
n_d = (2 / (3 * pi^2)) * pF_d^3;
n_s = (2 / (3 * pi^2)) * pF_s^3;

fun= @(n_s, n_u, n_d, n_e, Efu_ini, Efe_ini, Efd) root2d(n_s, n_u, n_d, n_e, Efu_ini, Efe_ini, Efd);
x0=[300,20];
x= fsolve( fun,x0)
function F = root2d(n_s, n_u, n_d, n_e , Efu_ini ,Efe_ini ,Efd)
eq1=  (2 / 3) * n_u - (1 / 3) * (n_d + n_s) - n_e ;
eq2= Efu_ini + Efe_ini - Efd ;
end

Answer 1

[Matt J]

It is not clear from your code which variables are meant to be the 2 unknowns, and which are constants. In any case, your fun needs to receive the unknowns as a vector, not as separate arguments.

[Arunkarthiheyan]

Efu(1) and Efe(1) are the unknowns.. The values obtained from fsolve will be stored in Efu(1) and Efe(1).

[Torsten]

I named the unknowns Efu and Efe in function root2d. Now we don't know where these unknowns appear in your subsequent code (at the moment, they appear nowhere). You will have to replace them therein whereever necessary.

Efu_ini= 300;
Efe_ini= 20;
x0 = [Efu_ini,Efe_ini];
x = fsolve(@(x)root2d(x,Efu_ini,Efe_ini),x0)

function res = root2d(x,Efu_ini,Efe_ini)
  Efu = x(1);
  Efe = x(2);
  landau_levels=@(EF, m, B)  ( (1 / (2 * B)) * ((EF^2 / m^2) - 1) );
  amuu = 5.0;
  amud = 5.0;
  amue = 0.511;
  amus = 150.0;
  hbarc = 197.3271;
  fscon = 137.036;
  Bcs = ( 3 * (fscon^0.5) * (amus^2) ) / (hbarc^1.5);  
  Bcu = ( 1.5 * (fscon^0.5) * (amuu^2) ) / (hbarc^1.5); 
  Bcd = ( 3 * (fscon^0.5) * (amud^2) )/ (hbarc^1.5); 
  Bce = ( (fscon^0.5) * (amue^2) )/ (hbarc^1.5);
  Bc = (5e17 * 1.95e-14);
  Bds = Bc / Bcs;
  Bdu = Bc / Bcu;
  Bdd = Bc / Bcd;
  Bde = Bc / Bce;
  Efs(1)=400;
  Efd=Efs(1);
  nu_u= landau_levels(Efu_ini,amuu,Bdu);
  nu_e= landau_levels(Efe_ini,amue,Bde);
  nu_d= landau_levels(Efd(1),amud,Bdd);
  nu_s= landau_levels(Efs(1),amus,Bds);    
  pF_u= sqrt( max( (Efu_ini^2 - amuu^2 * (1 + 2 * nu_u * Bdu)), 0) );
  pF_e= sqrt( max( (Efe_ini^2 - amue^2 * (1 + 2 * nu_e * Bde)), 0) );
  pF_d= sqrt( max( (Efd(1)^2 - amud^2 * (1 + 2 * nu_d * Bdd)), 0) );
  pF_s= sqrt( max( (Efs(1)^2 - amus^2 * (1 + 2 * nu_s * Bds)), 0) );
  n_u = (2 / (3 * pi^2)) * pF_u^3;
  n_e = (2 / (3 * pi^2)) * pF_e^3;
  n_d = (2 / (3 * pi^2)) * pF_d^3;
  n_s = (2 / (3 * pi^2)) * pF_s^3;

  eq1=  (2 / 3) * n_u - (1 / 3) * (n_d + n_s) - n_e ;
  eq2= Efu_ini + Efe_ini - Efd ;
  res = [eq1,eq2];
end

[Arunkarthiheyan]

Thank you @Torsten.. It works fine now.

[Arunkarthiheyan]

@Torsten I am getting the values of x as the initial values 300 and 20.

The logical error says the following

"No solution found.

fsolve stopped because the problem appears regular as measured by the gradient,

but the vector of function values is not near zero as measured by the

value of the function tolerance."

Can you help me fix this?

[Torsten]

Which variables are the unknowns (I named them Efu and Efe) in this part of the code where the equations to be solved are deduced ? If you don't know what I mean: can you write down the equations you are trying to solve in a mathematical way and mark the two unknowns ?

  landau_levels=@(EF, m, B)  ( (1 / (2 * B)) * ((EF^2 / m^2) - 1) );
  amuu = 5.0;
  amud = 5.0;
  amue = 0.511;
  amus = 150.0;
  hbarc = 197.3271;
  fscon = 137.036;
  Bcs = ( 3 * (fscon^0.5) * (amus^2) ) / (hbarc^1.5);  
  Bcu = ( 1.5 * (fscon^0.5) * (amuu^2) ) / (hbarc^1.5); 
  Bcd = ( 3 * (fscon^0.5) * (amud^2) )/ (hbarc^1.5); 
  Bce = ( (fscon^0.5) * (amue^2) )/ (hbarc^1.5);
  Bc = (5e17 * 1.95e-14);
  Bds = Bc / Bcs;
  Bdu = Bc / Bcu;
  Bdd = Bc / Bcd;
  Bde = Bc / Bce;
  Efs(1)=400;
  Efd=Efs(1);
  nu_u= landau_levels(Efu_ini,amuu,Bdu);
  nu_e= landau_levels(Efe_ini,amue,Bde);
  nu_d= landau_levels(Efd(1),amud,Bdd);
  nu_s= landau_levels(Efs(1),amus,Bds);    
  pF_u= sqrt( max( (Efu_ini^2 - amuu^2 * (1 + 2 * nu_u * Bdu)), 0) );
  pF_e= sqrt( max( (Efe_ini^2 - amue^2 * (1 + 2 * nu_e * Bde)), 0) );
  pF_d= sqrt( max( (Efd(1)^2 - amud^2 * (1 + 2 * nu_d * Bdd)), 0) );
  pF_s= sqrt( max( (Efs(1)^2 - amus^2 * (1 + 2 * nu_s * Bds)), 0) );
  n_u = (2 / (3 * pi^2)) * pF_u^3;
  n_e = (2 / (3 * pi^2)) * pF_e^3;
  n_d = (2 / (3 * pi^2)) * pF_d^3;
  n_s = (2 / (3 * pi^2)) * pF_s^3;

Answer 2

[Star Strider]

Note that ‘F’ is the output of ‘root2d’, however ‘F’ is nowhere defined as a calculation result in that code:

function F = root2d(n_s, n_u, n_d, n_e , Efu_ini ,Efe_ini ,Efd)

eq1= (2 / 3) * n_u - (1 / 3) * (n_d + n_s) - n_e ;

eq2= Efu_ini + Efe_ini - Efd ;

end

That might be something to consider fixing.

.

[Matt J]

Must not be a scalar, I think you mean. It should be,

function F = root2d(_____)

    eq1=  (2 / 3) * n_u - (1 / 3) * (n_d + n_s) - n_e ;
    eq2= Efu_ini + Efe_ini - Efd ;
    
    F=[eq1;eq2];

end

[Arunkarthiheyan]

Thank you for the suggestions @Star Strider @Matt J

[Star Strider]

The fsolve function is a root-finder, that is it finds the values of the parameters where the function crosses or equals zero. With your function, fsolve finds a minimum, however it may not be able to find a root.

You might be able to use either contour or fimplicit to see graphically if it has any roots.