# Solving symbolically for variables that are equal to several equations

4 vues (au cours des 30 derniers jours)
Andrew Matteson le 13 Sep 2023
Commenté : Walter Roberson le 14 Sep 2023
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
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### Réponses (2)

Torsten le 13 Sep 2023
Déplacé(e) : Torsten le 13 Sep 2023
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 =
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Paul le 13 Sep 2023
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 =
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Paul le 14 Sep 2023
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 le 14 Sep 2023
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

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