Solution of the equation

Question

Dmitry 2023-1-17

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1895900-solution-of-the-equation

评论： Dmitry 2023-1-26

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.

My code:

%% initial conditions

global d k0 h_bar ksi m E;

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;

%t = T./Tc;

t = 0.1:0.1:2;

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

D = 10^(-8); % толщина пленки

ksi = 10^(-9);

%d = D/ksi;

d = 1000;

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;

%% calculation

F = f_calc(t,nD);

plot(t,F)

grid on

function F = f_calc(t,nD)

global d k0 h_bar ksi m;

F = zeros(1,numel(t));

for i = 1:numel(t)

for k = 0:nD(i)

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

end

F = -F;

%F = -(1/d).*F;

%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

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Walter Roberson 2023-1-18

What problem? You showed an equation, you showed code. Are you getting an error message? If not then at what point in the code do you start seeing unexpected results? As outsiders how would we know if the code was producing correct answers or not?

Dmitry 2023-1-18

Walter Roberson, I don't understand how to take into account in this problem that the upper limit of the sum depends on t

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Answer 1

Torsten 2023-1-18

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1895900-solution-of-the-equation#answer_1151215

编辑：Torsten 2023-1-18

在 MATLAB Online 中打开

This code doesn't work since there does not seem to be a solution for some or all of the d values you supplied.

I suggest you fix a value for d first and plot f_calc(t) for a number of t values to see if the curve passes the t-axis somewhere.

%% 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

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Torsten 2023-1-22

在 MATLAB Online 中打开

As you can see, there seems to be no t value in the range you specified such that F(t,d(i)) = 6.81 for all d values.

%% initial conditions

global k0 h_bar ksi m E C_2

Ef = 2.77*10^3;

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

m = 9.1093837*10^(-31);

Tc = 1.2;

ksi = 10^(-9);

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;

t = linspace(0.1, 2);

d = linspace(10, 1000);

%[TT,DD] = meshgrid(t,d);

%contour(TT, DD, @f_calc);

for i=1:numel(d)

for j = 1:numel(t)

F(i,j) = f_calc(t(j),d(i));

end

plot(t,F)

function F = f_calc(t, d)

global 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, d)-f_arg_1(k,t, d))./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 arg_1 = f_arg_1(n,t,d)

global k0

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

end

function arg_2 = f_arg_2(n,t, d)

global k0

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

end

Torsten 2023-1-22

编辑：Torsten 2023-1-22

I've got a graph that never crosses F - C_2 = 0 (which would mean that there exists a t value or which F(t,d(i)) - C_2 = 0).

Simply spoken: The graphs must somewhere cross y=0 for that your problem has a solution.

It's the same as if you graph f(x) = x^2 - 2 to see where the zeros of x^2 - 2 = 0 are located.

Dmitry 2023-1-26

Thank you so much!

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Solution of the equation

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