Hiii all, i am trying to minimize my problem n want to get the plot...program is getting minimized but plot is not coming... m sending program also n the expected plot image also....plzzz plzzz plzzz do help

clear ; clc
syms beeta n 

alpha=1;
t=0.2;
w0=1;
N=3;
p=1;

M=@(beeta)(1+(beeta)^2+((beeta))./4+((beeta))./36)+...
    2*((beeta+((beeta)^3)./2+((beeta)^5)./12+((beeta)^7)./144)+...
    (((beeta)^2)./2+((beeta)^4)./6+((beeta)^6)./48+((beeta)^8)./720)+...
    (((beeta)^3)./6+((beeta)^5)./24+((beeta)^7)./240+((beeta)^9)./4320));

j=@(beeta,eta,alphas,h,delta) w0*(alpha*(2-eta)+2*sqrt(alpha)*(1-eta)*(h+M(beeta)*delta*exp(2*alphas)));
teff=@(beeta,eta,alphas) t*exp(alpha*eta^2*exp(-4*alphas)*...
    (1+(2*beeta)^2+((2*beeta)^4)./4+((2*beeta)^6)./36)-...
    ((-1)*((2*beeta)+((2*beeta)^3)./2+((2*beeta)^5)./12+((2*beeta)^7)./144)));

for u=0:0.2:4
    ueff=@(eta) u-2*alpha*w0*eta*(2-eta);
      energy = @(beeta,eta,alphas,h,delta) -j(beeta,eta,alphas,h,delta) +w0*...
            (h^2+(.5*((1+(2*beeta)^2+((2*beeta)^4)./4+((2*beeta)^6)./36)*cosh(4*alpha)-1)))+...
             (.25*(ueff(eta)-abs(ueff(eta))))+...
             (delta*w0.*M(beeta)*exp(2*alphas))*(2*h+delta*exp(2*alphas))-...
             4*teff(beeta,eta,alphas).*quadgk(@(k)besselj(0,k).*besselj(1,k) ./...
             (k.*(1+exp(k*0.5*abs(ueff(eta))/(teff(beeta,eta,alphas))))),0,inf);

      engy=@(v) energy(v(1),v(2),v(3),v(4),v(5));

   options = optimoptions('fminunc','Algorithm','quasi-newton');   
   xstart=[0.001,0.002,0.001,0.001,0.002];

     [xmin,ymin]=fminunc(engy,xstart,options);
     fprintf('minimum Enrgy = %f \n',ymin) ;

      x(p)= u;
     y(p)= ymin;
  p=p+1 ;
  end
  hold all
  plot(x,y)

[Walter Roberson]

My investigations so far suggest that the minima is -infinity for some combinations. I am checking further to see if I can work out what the circumstances are.

[ZAHID MALIK]

okkk .whatever conclusion you will reach just post here plzz

Answer 1

[Walter Roberson]

For the first u (at least), energy(0,-10^100,0,0,0) is -inf, so the search is unbounded.

[ZAHID MALIK]

what shall i do then??? should i decrease u value?? how to handle with unbounded problem???

[Walter Roberson]

If you let

beeta = 0
alphas = 0
eta = 1 +/- (1/2)*sqrt(4-2*u)

then nearly everything falls out of the integral part of the expression, leaving int((1/2)*BesselJ(0, k)*BesselJ(1, k)/k, k, 0, Inf) which has the solution 1/Pi . That makes the overall energy expression,

Delta^2 - (2^(1/2)*(2 - u)^(1/2))/2 + H^2 - (4*exp(2 - 2^(1/2)*(2 - u)^(1/2) - u/2))/(5*pi) + 2*Delta*H - 2^(1/2)*Delta*(2 - u)^(1/2) - 2^(1/2)*H*(2 - u)^(1/2) + 53454369307773/4398046511104

This goes complex for u > 2 (which just means that particular substituted eta should not be done because eta would not be real-valued). But as u goes to -infinity the limit goes to -infinity. Therefore if you were thinking of allowing u to be unbounded instead of in the range 0 to 4, then your minimum would be -infinity.

The combination alphas = 0, delta = 1, u = 0, h = 2, eta = 0, goes to -infinity as beetas approaches either +infinity or -infinity

There are probably other combinations that go to -infinity. For example the same behaviour for beetas happens with alphas = 1 as well as alphas = 0 in the above combination.

At this point you should probably re-check whether you have put in the proper equations, and re-check whether there are any constraints on the variables. You have used fminunc which is for unconstrained optimization.