Solution of the equation

We have such a function:

We set the array d = [10:10:10000], d(i) is substituted in and equated to the constant C_2, i.e. we get the equation F(d(i), t)=C_2 (C_2 = 6.81), and we need to find this 't' for the corresponding d(i), that is, as a result, we should have a graph d(t). I think that most likely it will be a discontinuous fast-oscillating function. In essence, our task is reduced to solving the equation F(t, d) = C_2

My code:

%% initial conditions

global d k0 h_bar ksi m E C_2

Ef = 2.77*10^3;

Kb = physconst('boltzmann'); % 1.38*10^(-23)

T = 0.12:0.24:6.4;

m = 9.1093837*10^(-31);

Tc = 1.2;

D = 10^(-8); % толщина пленки

ksi = 10^(-9);

dd = [10:10:10000];

E = Ef/(pi*Kb*Tc);

h_bar = (1.0545726*10^(-34));

k0 = (ksi/h_bar)*sqrt(2.*m.*pi.*Kb.*Tc);

C_2 = 6.81;

t0 = 1.0;

for i = 1:numel(dd)

d = dd(i);

t(i) = fsolve(@f_calc,t0);

t0 = t(i);

end

plot(t,F)

function F = f_calc(t)

global d k0 h_bar ksi m C_2

nD = floor(375/(2*pi*t*1.2) - 0.5);

F = 0;

for k = 0:nD

F = F + 1/(2*k+1).*(k0.*real(((f_p1(k,t)-f_p2(k,t))./2))+(f_arg_2(k,t)-f_arg_1(k,t))./d);

end

F = F - C_2;

end

function p1 = f_p1(n,t)

p1 = ((1+1i)./sqrt(2)).*sqrt(t.*(2.*n+1));

end

function p2 = f_p2(n,t)

global E

p2 = sqrt(3601+1i.*t.*(2.*n+1));

end

function n_lg = f_lg(n,t)

global d k0

arg_of_lg = (1+exp(-1i*d*k0.*f_p1(n,t)))/(1+exp(-1i*d*k0.*f_p2(n,t)));

n_lg = log(abs(arg_of_lg));

end

function arg_1 = f_arg_1(n,t)

global d k0

arg_1 = angle(1+exp(-1i*d*k0.*f_p1(n,t)));

end

function arg_2 = f_arg_2(n,t)

global d k0

arg_2 = angle(1+exp(-1i*d*k0.*f_p2(n,t)));

end

[Torsten]

Why do you open a new question ?

Did you fix a value d and plot f(t) = F(t, d) - C_2 to see whether a root t of your equation exists at all ?