How to solve an equation with the variable on both RHS and LHS?

I have an equation with 2 variables, however, one variable is present on both the LHS and the RHS, on the RHS it is raised to a fractional power usually between 0 and 1, here, it is raised to 0.615.

I would like one variable as an equation of the other. I have tried to plot it, by saving values in seperate vectors, but since both vectors are of different sizes, the plot is unsuccessful.

I am trying to code this particular, equation where only Gf_bar and w_bar are variables, everything else is a constant. Ideally, I would like an equation of Gf_bar in terms of w_bar. I tried to plot Gf_bar against w_bar and fit a curve to obtain the equation.

As is seen, clearly, Gfbar and wbar are vectors but of different lengths. I would like to know how to solve this.

[Alan Stevens]

You should be able to use fzero to find Gfbar for value sof wbar.

In your definitions for c3 and c4 you raise wbar to the power n, though the original equation has them raised to the power 2.

It would be easier if you used the insert code button to show your code, rather than showing an image (it also doesn't tell us what value OF takes).

[Khanak Saxena]

Thank you for pointing out that wbar is raised to 2. OF is by user input, I have been taking 2.4 as its value. How much of the code inouts would you like to see? The entire code is too large to insert here I suppose. I also wanted to add that Gfbar_eq is declared as a symbol in the beginning og the code.

If i use fzero to find these values, will the size of the vector be the same?

[Khanak Saxena]

I would like to add that the system doesn't give me a solution in eqn_Gfbar at all despite the employment of the simplify function.

This is exactly the errors got, line 59 would be Gfbar=solve(eqn_Gfbar,Gfbar_eq,'Real',true,'IgnoreAnalyticConstraints',true);

and line 63 is

plot(wbar,Gfbar,'b')

[Alan Stevens]

There is almost certainly no symbolic solution here. You don't need a symbolic solution to find values of Gbar for various values of wbar. See below.

Answer 1

[Alan Stevens]

Like this for example (obviously, you will need to replace my arbitrary data with your actual data):

OF = 2.4;
wbar = linspace(0,1,100);
n = 0.615;
G0 = 0.1; 
Gbar = zeros(1,numel(wbar));
for k = 1:numel(wbar)
Gbar(k)= fzero(@Gbarfn,G0,[],wbar(k),OF,n);
end

plot(wbar,Gbar), grid
xlabel('wbar'),ylabel('Gbar')

function F = Gbarfn(G,wbar,OF,n)

        c1 = 1/(OF + 1)^n;
        c2 = 4 + 2*pi*wbar;
        c3 = 1 + 4*wbar +pi*wbar.^2;
        c4 = 1./(1 + 4*wbar + 4*wbar.^2);
        
        F = c1*( OF*c4 - G ).^n*c2./c3 - G;

end

[Khanak Saxena]

Thank you so much for this. A few questions for you. I understand that you created a function Gbarfn with G, wbar, OF and n as parameters and you call that function in the loop. Before the loop, youn made sure Gbar was the same size as wbar and had all zeros.

Inside the loop you sytematically save the value of the required Gbar for the corresponding wbar value. The only thing I do not understand is " Go,[ ] " in the fzero function and its purpose here. Will G0 be called in each loop and if it is being called in each loop, what purpose is it serving?

[Alan Stevens]

G0 is simply an initial guess for Gbar, required by fzero. Each time fzero is called it needs an initial guess for the parameter it is solving for. I've used the same initial guess for each Gbar, though that isn't essential.

The [] is a placeholder in case one wants to specify any non default options (like a tighter tolerance for example). See fzero documentation for details.

[Khanak Saxena]

Thank you so much