Iterative Process to find a variable in which two equations are equal and opposite

I am struggling to write a code that will find the multiple values of a variable in two equations where the equations will be equal in magnitude but opposite in sign. There can be a miniscule tolerance.

Here is what I have so far, it's definitely not functional, but I'm not sure what to do here.

% Equation
Eo=1;
Uo=1;
Fc=1.03;
Wc=2*pi*Fc;
b=0.3;
Ed=100;
Ud=1;
h = input('Enter initial value h: ');
A = sqrt(Eo/Uo)*cot(Wc*sqrt(Uo*Eo)*(b-h));
B = sqrt(Ed/Ud)*cot(Wc*sqrt(Ud/Ed)*h);
Error = A-B;
iter = 1;
while Error > 1e-3 && iter < 1000 %To avoid infinite loops
    h = h+0.0001
    A = sqrt(Eo/Uo)*cot(Wc*sqrt(Uo*Eo)*(b-h));
    B = sqrt(Ed/Ud)*cot(Wc*sqrt(Ud/Ed)*h);
    Error = A-B;
    iter = iter + 1;
end
if iter == 1000
    warning('Max Iterations reached')
end

Answer 1

[Torsten]

Here is at least one solution for h:

% Equation
Eo=1;
Uo=1;
Fc=1.03;
Wc=2*pi*Fc;
b=0.3;
Ed=100;
Ud=1;
h0 = 10;
fun = @(h)sqrt(Eo/Uo)*cot(Wc*sqrt(Uo*Eo)*(b-h))+sqrt(Ed/Ud)*cot(Wc*sqrt(Ud/Ed)*h);
h = fsolve(fun,h0)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

h = 10.9966

sqrt(Eo/Uo)*cot(Wc*sqrt(Uo*Eo)*(b-h))

ans = -9.0826

sqrt(Ed/Ud)*cot(Wc*sqrt(Ud/Ed)*h)

ans = 9.0826

Answer 2

[Les Beckham]

I would use fminsearch for this. See below.

Note that I am using abs(A - B) for the Error term.

% minimize error between two equations by adjusting the h parameter

minimizeError % call the main minimization function

 
 Iteration   Func-count         f(x)         Procedure
     0            1           154.02         
     1            2          146.616         initial simplex
     2            4          133.725         expand
     3            6          113.617         expand
     4            8           86.965         expand
     5           10          56.7086         expand
     6           12          37.4199         expand
     7           14           16.459         expand
     8           16          9.35441         expand
     9           18          2.62892         reflect
    10           20          2.04807         contract outside
    11           22         0.522464         contract inside
    12           25         0.522464         shrink
    13           27         0.522464         contract outside
    14           29         0.189874         contract inside
    15           31         0.131747         contract inside
    16           33        0.0361019         contract inside
    17           35        0.0361019         contract inside
    18           37       0.00443706         contract inside
    19           39       0.00443706         contract inside
    20           41       0.00443706         contract inside
    21           43      0.000679668         contract inside
    22           45      0.000679668         contract inside
    23           47      0.000598173         contract inside
 
Optimization terminated:
 the current x satisfies the termination criteria using OPTIONS.TolX of 1.000000e-04 
 and F(X) satisfies the convergence criteria using OPTIONS.TolFun of 1.000000e-04 


Error is minimized with h = 2.161133

function minimizeError()
% Equation
Eo=1;
Uo=1;
Fc=1.03;
Wc=2*pi*Fc;
b=0.3;
Ed=100;
Ud=1;
h = 0.1; %input('Enter initial value h: ');

opt = optimset('Display', 'iter');
h = fminsearch(@costfun, h, opt);

fprintf('\nError is minimized with h = %f', h);

function Error = costfun(h) % define the cost function for the minimization
    A = sqrt(Eo/Uo)*cot(Wc*sqrt(Uo*Eo)*(b-h));
    B = sqrt(Ed/Ud)*cot(Wc*sqrt(Ud/Ed)*h);
    Error = abs(A - B);
end

end

Answer 3

[John D'Errico]

h is the unknown. The problem becomes simple, but even so, there are some serious issues you need to understand.

Eo=1;
Uo=1;
Fc=1.03;
Wc=2*pi*Fc;
b=0.3;
Ed=100;
Ud=1;

A_h = @(h) sqrt(Eo/Uo)*cot(Wc*sqrt(Uo*Eo)*(b-h));
B_h = @(h) sqrt(Ed/Ud)*cot(Wc*sqrt(Ud/Ed)*h);

You want to solve for h, such that the two are the same value. However, you STATE that you are looking for a solution where the two have opposite signs. I think you really meant that the difference is zero, so the two are EXACTLY equal, not opposite in sign. That conclusion is based on your search for a zero of A-B.

First, we wil plot the difference.

A_B = @(h) A_h(h) - B_h(h);
fplot(A_B)
yline(0,'k')

And here we should see there will be infinitely many solutions. i assume you want the principal solution, so the first positive one?

fplot(A_B,[0,1])
yline(0,'k')
ylim([-100,100])
grid on

It looks like there is a zero crossing near 0.3, but the slope of the curve there will be extremely high, and that solution happens within a hair's breath of a singularity. That means I cannot easily give fzero a bracket. A good starting value, based on the information available here form the plot will be sufficient.

[hsol,fval,exitflag] = fzero(A_B,.29)

hsol = 0.2970

fval = 3.0695e-12

exitflag = 1

There will of course be infinitely many other solutions. Remember the importance of plotting the curve BEFORE you just throw a solver at it.