Solve an equation that has data arrays as constants

Hello,

I want to solve the following equation:

1/c0 c^2 - c +X*t / c0 = 0

where c0 is a given constant but X and t are constant arrays: they are data sets where X varies with time, t. The goal is to solve for c, which too will be a data array that varies with time, t.

I tried:

func = @(c) 1/c0.*c^2 - c + X.*t/c0;

c = fzero(func,1)

But got errors likely because X and t are arrays and not single values. Any other ideas? Time sensitive, thank you!!!

Answer 1

[Walter Roberson]

You have c^2 and say that c will be an array.

Could you confirm that you c^2 meaning c*c meaning algebraic matrix multiplication (inner product) of c with itself?

If you do then all of the values have to be searched for simultaneously, which makes it a multivariate problem, but fzero() can only be used for univariate situations. fsolve() can be used for multivariate cases.

Also is X*t intended to be algebraic matrix multiplication like in your problem definition, or is it X.*t like in your code?

XTc0 = X.*t./c0;
c = arrayfun(@(xtc0) fsolve(@(c) 1/c0.*c.^2 - c + xtc0), XTc0)

[Walter Roberson]

If it should be c.^2 instead of c^2 then

syms xtc0
syms c0 c positive
eqn = 1/c0.*c.^2 - c + xtc0

eqn = 

solve(eqn, c)

Warning: Solutions are only valid under certain conditions. To include parameters and conditions in the solution, specify the 'ReturnConditions' value as 'true'.

ans = 

and you would not have to use fsolve at all.

[Diego Torres-Siclait]

Thank you for the fast reply! Yes sorry it should be each corresponding c value multiplied with itself so c.^2 is correct. Im having trouble following why you use 'syms xtc0' if X and t are given data arrays. Defining 'XTc0 = X.*t./c0' made more sense to me. Thank you!!

Also yes its X.*t since each X individual value is multiplying with each respective t value creating data points (both arrays are of same length).

[Walter Roberson]

XTc0 = X.*t./c0;

can be computed in advance, since it does not depend upon the current c value.

Because of that, your problem comes down to solving

1/c0.*c.^2 - c + xtc0

repeatedly where xtc0 iterates over all values in the XTc0 matrix.

But for any particular value, that is just a quadratic equation, with general form as given by "ans =" above. And that in turn means that you can compute all of the solutions without using arrayfun or fsolve: just program the first of the two formulas and replace xtc0 with X.*t./c0 :

c_plus = sqrt(c0) .* (sqrt(c0) + sqrt(c0 - 4 * X.*t./c0))/2;

should now be the entire set of solutions.

If the above is real-valued then there would also be another set of real-valued solutions

c_minus = sqrt(c0) .* (sqrt(c0) - sqrt(c0 - 4 * X.*t./c0))/2;

which might or might not be positive, depending on the values involved.

[Diego Torres-Siclait]

great, thank you Walter!