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.

2025 4 7
2 个回答
更新时间：2025 4 9
9 次查看（30 天）

0 个投票

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

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回答（2 个）

Matt J
Matt J 2025-4-7
编辑：Matt J 2025-4-7

0 个投票

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.

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Arunkarthiheyan
Arunkarthiheyan 2025-4-8
Efu(1) and Efe(1) are the unknowns.. The values obtained from fsolve will be stored in Efu(1) and Efe(1).
Torsten
Torsten 2025-4-8
编辑：Torsten 2025-4-8
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
Arunkarthiheyan 2025-4-9
@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
Torsten 2025-4-9
编辑：Torsten 2025-4-9
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;

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Star Strider
Star Strider 2025-4-7

0 个投票

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.
.

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Matt J
Matt J 2025-4-7
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
Star Strider
Star Strider 2025-4-9
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.

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