need suggestion to calculate alpha value in semi-infinite experimental results

I wanted to solve for alpha by using experimental data for different time, the1, and the2.

Please suggest how to go ahead.

alpha=sym('alpha',[1 1467]);

eqn = (2*5761.11)*time.^0.5*(alpha).^0.5./(the1*pi^0.5)*exp((-x1^2)./(4*time.*alpha))-(5761.11*x1)./the1*(1-(2/sqrt(pi))*(x1./(2*sqrt(time.*alpha))-((x1./(2*sqrt(time.*alpha)))^3)/3))== (2*5761.11)*time.^0.5*(alpha).^0.5./(the2*pi^0.5)*exp((-x2^2)./(4*time.*alpha))-(5761.11*x2)./the2*(1-(2/sqrt(pi))*(x2./(2*sqrt(time.*alpha))-((x2./(2*sqrt(time.*alpha)))^3)/3));

Alpha = solve(eqn,alpha);

[Rik]

Your code doesn't run for scalar values of the variables you indicated. x1 is also undefined. Once your code runs for single values, it should be trivial to extend it to multiple values by making a function.

time=1;the1=2;the2=3;

x1=4;

alpha=sym('alpha',[1 1467]);
eqn = (2*5761.11)*time.^0.5*(alpha).^0.5./(the1*pi^0.5)*exp((-x1^2)./(4*time.*alpha))-(5761.11*x1)./the1*(1-(2/sqrt(pi))*(x1./(2*sqrt(time.*alpha))-((x1./(2*sqrt(time.*alpha)))^3)/3))== (2*5761.11)*time.^0.5*(alpha).^0.5./(the2*pi^0.5)*exp((-x2^2)./(4*time.*alpha))-(5761.11*x2)./the2*(1-(2/sqrt(pi))*(x2./(2*sqrt(time.*alpha))-((x2./(2*sqrt(time.*alpha)))^3)/3));

Error using *
Dimensions do not match.

Alpha = solve(eqn,alpha)

[Walter Roberson]

time is a vector the same size as alpha? Are the1 and the2 scalar or vector?

[AKSHAY SHIRSAT]

Yes. time, alpha, the1, the2 are all of the same size.

[AKSHAY SHIRSAT]

@Rik, I have solved for a single time, the1, the2, alpha.

I wanted to extend the same for all my experimental results.

Can you please suggest how to go ahead for multiple values?

[Rik]

If you post your code the edit will probably be fairly simple. As you can see, your current code actually does not work. At least it doesn't on R2022a, and since you didn't specify your release anywhere, you implied you are using a sufficiently new release that the Matlab version can't cause any issues.

[AKSHAY SHIRSAT]

@Rik, I have attached my code which is running.

6000 (time), 79.72195(the1), 76.38214(the2) - I need to be replaced with multiple values to get the corresponding values of alpha.

system x

eqn = (2*5761.11)*6000^0.5*(x)^0.5/(79.72195*pi^0.5)*exp((-0.01^2)/(4*x*6000))-(5761.11*0.01)/79.38214*(1-(2/sqrt(pi))*(0.01/(2*sqrt(x*6000))-((0.01/(2*sqrt(x*6000)))^3)/3))== (2*5761.11)*6000^0.5*(x)^0.5/(76.38214*pi^0.5)*exp((-0.03^2)/(4*x*6000))-(5761.11*0.03)/76.38214*(1-(2/sqrt(pi))*(0.03/(2*sqrt(x*6000))-((0.03/(2*sqrt(x*6000)))^3)/3));

S = solve(eqn,x)

[Rik]

Do you have some function or class that is called system? Because the normal system command calls the command line of the OS.

If you insert your code as a 'line of code' instead of 'code example' you will be able to run your code. You need to make sure it runs in the online environment. Then you can be certain that we can reproduce your efforts.

time=6000;
the1=79.72195;
th2=76.38214;
system x

/bin/bash: x: command not found

ans = 127

eqn = (2*5761.11)*6000^0.5*(x)^0.5/(79.72195*pi^0.5)*exp((-0.01^2)/(4*x*6000))-(5761.11*0.01)/79.38214*(1-(2/sqrt(pi))*(0.01/(2*sqrt(x*6000))-((0.01/(2*sqrt(x*6000)))^3)/3))== (2*5761.11)*6000^0.5*(x)^0.5/(76.38214*pi^0.5)*exp((-0.03^2)/(4*x*6000))-(5761.11*0.03)/76.38214*(1-(2/sqrt(pi))*(0.03/(2*sqrt(x*6000))-((0.03/(2*sqrt(x*6000)))^3)/3));

Unrecognized function or variable 'x'.

S = solve(eqn,x)

[AKSHAY SHIRSAT]

syms x instead of system x.

syms x
eqn = (2*5761.11)*6000^0.5*(x)^0.5/(79.72195*pi^0.5)*exp((-0.01^2)/(4*x*6000))-(5761.11*0.01)/79.38214*(1-(2/sqrt(pi))*(0.01/(2*sqrt(x*6000))-((0.01/(2*sqrt(x*6000)))^3)/3))== (2*5761.11)*6000^0.5*(x)^0.5/(76.38214*pi^0.5)*exp((-0.03^2)/(4*x*6000))-(5761.11*0.03)/76.38214*(1-(2/sqrt(pi))*(0.03/(2*sqrt(x*6000))-((0.03/(2*sqrt(x*6000)))^3)/3));
S = solve(eqn,x)

Warning: Unable to solve symbolically. Returning a numeric solution using vpasolve.

S = 

0.000029352683458310990175159943542835

[Torsten]

And does it work ?

[Rik]

Now you have code that works, but no longer contains your variables. Why did you do that? Now you have to go in and change that 6000 to the value of time, instead of changing the value of time.

You need to write code that runs in the online environment (use the green arrow) and that depends on your variables. You're almost there.

Answer 1

[Sandeep]

Hi Akshay,

It is my understanding that you are expecting an approach to solve for alpha using experimental data for different times (time), the1, and the2 which can be semi-infinite in number. You can follow these steps as an approach for the solution,

1. Define the symbolic variable alpha using the sym function.
2. Set up the equation (eqn) that relates alpha to the experimental data. Make sure to replace the variables in the equation with their corresponding values.
3. Solve the equation (eqn) to find the values of alpha that satisfy the equation. The variable Alpha will contain the solutions for alpha based on the experimental data provided.

Please refer to the sample implementation given below,

function Alpha = solveAlpha(time_values, the1_values, the2_values)
    syms alpha

    Alpha = zeros(length(time_values), length(the1_values));  % Preallocate array to store results

    for i = 1:length(time_values)
        for j = 1:length(the1_values)
            eqn = (2 * 5761.11) * time_values(i)^0.5 * (alpha)^0.5 / (the1_values(j) * pi^0.5) * exp((-x1^2) / (4 * time_values(i) * alpha)) - (5761.11 * x1) / the1_values(j) * (1 - (2 / sqrt(pi)) * (x1 / (2 * sqrt(time_values(i) * alpha)) - ((x1 / (2 * sqrt(time_values(i) * alpha)))^3) / 3)) == (2 * 5761.11) * time_values(i)^0.5 * (alpha)^0.5 / (the2_values(j) * pi^0.5) * exp((-x2^2) / (4 * time_values(i) * alpha)) - (5761.11 * x2) / the2_values(j) * (1 - (2 / sqrt(pi)) * (x2 / (2 * sqrt(time_values(i) * alpha)) - ((x2 / (2 * sqrt(time_values(i) * alpha)))^3) / 3));
            solutions = solve(eqn, alpha);
            Alpha(i, j) = double(solutions);
        end
    end
end

Make sure to replace x1 and x2 with the actual values you have for the variables. The nested loops will iterate over each combination of time, the1, and the2 values, solve the equation, and store the corresponding values of alpha in the Alpha matrix.

Hope you find it helpful.

Sandeep BS