Finding a variable in a very complicated equation

There are two equations:

y=P*exp(-alfa*x)*cos(alfa*x)/(2*alfa^3*EI)

alfa=(k/(4*EI))^(1/4)

x, y, P, EI are known, I need to find k using these equations, accuracy of 0,0001 is enough

I think that the only way to find k is to use the trial and error (iteration) method, unfortunately I can't manage to write a good script to calculate that

Thank You for all Your help in advance

Answer 1

[Star Strider]

It would be nice if the constants were supplied. Using random numbers for them —

x = rand;
y = rand; 
P = rand;
EI = rand;
alfa = @(k,EI) (k./(4*EI)).^(1/4);
yfcn = @(x,y,P,EI,k) P.*exp(-alfa(k,EI).*x).*cos(alfa(k,EI).*x)./(2*alfa(k,EI).^3.*EI) - y;
[k_est,fv] = fsolve(@(k)yfcn(x,y,P,EI,k), 1)

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.

k_est = 2.1218

fv = 4.1983e-06

.

[Nikodem Podlaszewski]

Thanks for very much for your answer

The values are:

E=210000 % [N/mm^2]
I=25*3^3/12 % [mm^4]
EI=E*I % [Nmm^2]
P=19.62 % [N]

x [mm] and y [mm] are formed into 60 pairs and create a matrix

a part of the matrix is

x   y
400 -0.01
300 -0.09
200 -0.11
100 0.14
0   0.87
400 -0.02
300 -0.09
200 -0.10
100 0.12
0   0.80 

Could you modify the script to calculate the value of k for each of the pairs? Thank you in advance

[Star Strider]

Sure!

xy = [400 -0.01
      300 -0.09
      200 -0.11
      100  0.14
        0  0.87
      400 -0.02
      300 -0.09
      200 -0.10
      100  0.12
        0  0.80];

E=210000; % [N/mm^2]
I=25*3^3/12; % [mm^4]
EI=E*I; % [Nmm^2]
P=19.62; % [N]
alfa = @(k,EI) (k./(4*EI)).^(1/4);
yfcn = @(x,y,P,EI,k) P.*exp(-alfa(k,EI).*x).*cos(alfa(k,EI).*x)./(2*alfa(k,EI).^3.*EI) - y;
for kk = 1:size(xy,1)
    [k_est(kk,:),fv(kk,:)] = fsolve(@(k)yfcn(xy(kk,1),xy(kk,2),P,EI,k), 1);
end

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.


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.


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.


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.


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.


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.


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.


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.


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.


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.

Out = array2table([xy k_est, fv], 'VariableNames',{'x','y','k_est','fv'})

Out = 10×4 table
     x       y       k_est         fv     
    ___    _____    _______    ___________

    400    -0.01    0.47147    -5.6502e-06
    300    -0.09    0.23986    -3.4322e-08
    200    -0.11    0.42319       0.062807
    100     0.14    0.53096     1.0563e-06
      0     0.87     0.4441     7.7512e-08
    400    -0.02    0.36061    -5.5235e-06
    300    -0.09    0.23986    -3.4322e-08
    200     -0.1    0.42319       0.052807
    100     0.12    0.59015     1.3671e-07
      0      0.8    0.49665     3.0442e-11

figure
stem3(Out.x, Out.y, Out.k_est, 'filled')
grid on
xlabel('x')
ylabel('y')
zlabel('k_{est}')

It would have been nice to have all that information at the outset.

[Nikodem Podlaszewski]

Thank you very much! That's exactly what I needed. Could you also tell me what fv means? And what does @ do, for example in @(k, EI)?

[Star Strider]

As always, my pleasure!

The ‘fv’ variable is the function value of ‘yfcn’ at the convergence. The fsolve function is a root-finder, so ‘fv’ should be close to 0. The ‘fv’ value tells how close it actually is. Also, it is optional, however I usually get it so I can determine how close fsolve got to solving it.

The ‘@’ sign denotes a function handle that MATLAB uses to define functions or to refer to them. See the documentation on What Is a Function Handle? for more information.

[Nikodem Podlaszewski]

Also, in your outcome appeared lines such as

"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."

What does it mean? In the Out 10x4 table there are solutions for all pairs so I don't get why there is no solution found.

[Star Strider]

The fsolve function is a root (zero) finder, so if it cannot find a solution (zero-crossing) within its accepted tolerances (the function value at convergence is greater than the tolerance, so greater than zero), it reports that as a minimum, however not a solution.

It is quite possible that the function has no real solution for those specific values. If you give it a complex initial estimate, perhaps 1+1i instead of 1, fsolve will search the complex space in that region for a complex solution.

[Nikodem Podlaszewski]

I have an another problem. After inputting the code to MatLab and hitting "Run and Time" the results don't show anywhere. Instead, I see Profiler and a warning that says "The variable "k_est" appears to change size on every loop iteration (within a script). Consider preallocating for speed" The same appears for the variable "fv". I've read a lot about it and it appears to be just warnings but I can't see the results (nothing is being showed in Command Windows nor the output Array2table. Could you help me to get the results from that?

[Star Strider]

Apparently, there is more to your code than what you posted.

If you are using a for loop, it is relatively straightforward to preallocate ‘k_est’ and ‘fv’ —

x = rand;
y = rand; 
P = rand;
EI = rand;
alfa = @(k,EI) (k./(4*EI)).^(1/4);
yfcn = @(x,y,P,EI,k) P.*exp(-alfa(k,EI).*x).*cos(alfa(k,EI).*x)./(2*alfa(k,EI).^3.*EI) - y;
num_iter = 10;                                                      % Use The Appropriate Value
k_est = zeros(num_iter,1);                                          % Preallocate Vector
fv = zeros(num_iter,1);                                             % Preallocate Vector
for k1 = 1:num_iter
    [k_est(k1),fv(k1)] = fsolve(@(k)yfcn(x,y,P,EI,k), 10);
end

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.


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.


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.


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.


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.


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.


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.


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.


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.


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.

Output = table(k_est,fv)

Output = 10×2 table
     k_est          fv    
    ________    __________

    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08
    0.077417    1.1087e-08

If you are using a while loop, preallocate the vectors to be greater than what you will likely require, then after the looop finishes (since you will have used a counter to create the ‘k1’ values) re-define —

k_est = k_est(1:k1);
fv = fv(1:k1);

This discards the rest of the vector, saving memory.