Solving symbolically for variables that are equal to several equations

Hello,

I'm trying to solve for any vaiable in a series of equations that are all equal to one another such as in the equation bellow.

In it, I know that "v" has two possible solutions, and I'd like to find a way to have MATLAB recognize this and give me both possible solutions as an array. So far, I have gotten my test program to give me one solution at a time, but not both.

clear all
close all
clc

syms xi v r a mu


spec_eng1 = [xi == (v^2)/2-mu/r, xi == -mu/(2*a)]
solve(spec_eng1, v)
% ans =

% Empty sym: 0-by-1

sol = eng.feval_internal('solve', eqns, vars, solveOptions);
spec_eng2 = xi == (v^2)/2-mu/r == -mu/(2*a)
solve(spec_eng2, v)
% Error using mupadengine/feval_internal
% Invalid argument.
% 
% Error in sym/solve (line 293)
% sol = eng.feval_internal('solve', eqns, vars, solveOptions);


% These outputs the solutions, but MATLAB doesn't know their related to
% one another

spec_eng3 = xi == (v^2)/2-mu/r
solve(spec_eng3, v)
% ans =
%
%  (2^(1/2)*(mu + r*xi)^(1/2))/r^(1/2)
% -(2^(1/2)*(mu + r*xi)^(1/2))/r^(1/2)


spec_eng4 = -mu/(2*a) == (v^2)/2-mu/r
solve(spec_eng4, v)
% ans =
% 
%  (mu^(1/2)*(2*a - r)^(1/2))/(a^(1/2)*r^(1/2))
% -(mu^(1/2)*(2*a - r)^(1/2))/(a^(1/2)*r^(1/2))

I don't really know what else to say. Any help with this would be appreicated. Thanks

Answer 1

[Torsten]

syms xi v r a mu
spec_eng1 = [xi == (v^2)/2-mu/r];
solve(spec_eng1, v)

ans = 

spec_eng2 = (v^2)/2-mu/r == -mu/(2*a);
solve(spec_eng2, v)

ans = 

Answer 2

[Paul]

Hmm. Don't know why solve doesn't find a solution,

syms xi v r a mu
spec_eng1 = [xi == (v^2)/2-mu/r, xi == -mu/(2*a)]

spec_eng1 = 

solve(spec_eng1,v,'ReturnConditions',true)

ans = struct with fields:
             v: [0×1 sym]
    parameters: [1×0 sym]
    conditions: [0×1 sym]

Here's a workaround that for this simple case returns both solutions in an array.

vsol = solve(subs(spec_eng1(1),xi,solve(spec_eng1(2),xi)),v)

vsol = 

Or reverse the order

vsol = solve(subs(spec_eng1(2),xi,solve(spec_eng1(1),xi)),v)

vsol = 

[Walter Roberson]

You cannot successfully solve() N equality equations for fewer than N variables. If some of the equations are inequalities then there is a chance it would work, but typically it would fail anyhow.

[Paul]

So if we know we want xi to be eliminated from the expression for v, we'd force that by adding to the solved-for variables, I suppose.

syms xi v r a mu
spec_eng1 = [xi == (v^2)/2-mu/r, xi == -mu/(2*a)]

spec_eng1 = 

sol = solve(spec_eng1,[v xi])

sol = struct with fields:
     v: [2×1 sym]
    xi: [2×1 sym]

sol.v

ans = 

sol.xi

ans = 

[Walter Roberson]

Yes, exactly,

In some cases you can do, for example,

syms xi v r a mu
spec_eng1 = [xi == (v^2)/2-mu/r, xi == -mu/(2*a)]

spec_eng1 = 

arrayfun(@(X) isolate(X, mu), spec_eng1)

ans = 

but not in the case of v because the second expression does not contain v