How can I Solve This Non Linear Parametrical System?

2018 11 15
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更新时间：2018 11 16
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Hi, I tried to solve this equation but Matlab answered this. How can I solve? Thank you
>> syms x y z p r
>> [xAns yAns zAns] = solve([(-1+((1-p)^2))*x + (1-p)*r*y + p^(2)*z, 2*p(1-p)*x +((1-p)*(1-r)+r*p-1)*y + (1-r)*p*z, p^(2)*x + 2*r*(1-r)*y + (((1-r)^2) - 1)*z],[x y z])
Error using sym/subsindex (line 825)
Invalid indexing or function definition. Indexing must follow MATLAB indexing. Function arguments must be symbolic
variables, and function body must be sym expression.
Error in sym/subsref (line 870)
R_tilde = builtin('subsref',L_tilde,Idx);

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Yuri Caridi
Yuri Caridi 2018-11-16
编辑：Walter Roberson 2018-11-16

0 个投票

Thank you very much but now the problem changed and Matlab answered me this
[xAns yAns zAns] = solve([(-1+((1-p)^2))*x + (1-p)*r*y + r^(2)*z, 2*p*(1-p)*x +((1-p)*(1-r)+r*p-1)*y + ((1-r)^(2))*r*z,( p^(2))*x + p*(1-r)*y + (((1-r)^2) - 1)*z,x+y+z-1],[x y z])
xAns =
Empty sym: 0-by-1
yAns =
Empty sym: 0-by-1
zAns =
Empty sym: 0-by-1

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Walter Roberson
Walter Roberson 2018-11-16
编辑：Walter Roberson 2018-11-16
syms x y z p r
eqn = [(-1+((1-p)^2))*x + (1-p)*r*y + r^(2)*z, 2*p*(1-p)*x +((1-p)*(1-r)+r*p-1)*y + ((1-r)^(2))*r*z,( p^(2))*x + p*(1-r)*y + (((1-r)^2) - 1)*z,x+y+z-1]);
sol = solve(eqn);
xAns = sol.x
yAns = sol.y
zAns = sol.z
You were having a problem because you were trying to solve 5 equations for 3 unknowns.
MATLAB finds 5 solutions each with specific numeric values. My calculation is that there are at least 5 families of solutions, one of which is specific numbers, but the other 4 of which involve at least one free parameter, such as p = 1, r = 1, x = 1/2-(1/2)*y, y = y, z = 1/2-(1/2)*y
Walter Roberson
Walter Roberson 2018-11-16
sol = solve(eqn, 'returnconditions', true);
will give you a set of parameterized solutions:
>> [sol.x,sol.y,sol.z,sol.p,sol.r]
ans =
[ 1 - z1, z1, 0, 0, 0]
[ 0, 1 - z1, z1, 0, 0]
[ 0, -1, 2, 3, -1]
[ 1 - z2 - z1, z1, z2, 0, 0]
[ z1, 1 - 2*z1, z1, 1, 1]
[ 1, 0, 0, 0, z1]
[ 0, 0, 1, z1, 0]
Yuri Caridi
Yuri Caridi 2018-11-16
Last thing, if I would have the solution with the 2 free parameters p and r?
Walter Roberson
Walter Roberson 2018-11-16
编辑：Walter Roberson 2018-11-16
If you look at the above list from the returnconditions version, you can see that the first 5 are all for specific p and r. The last two are for free parameters p and r. They can be read as:
(This is not a correct interpretation interpretation of the equations; see below. I leave it in for purposes of the conversation.)
x = r; y = 0; z = 0; p = 0; r free
and
x = 0; y = 0; z = p; p free; r = 0;
Yuri Caridi
Yuri Caridi 2018-11-16
编辑：Yuri Caridi 2018-11-16
yes, I meant, the perfect solution to my problem (a probability one) would be with both p and r parameters as p = t and r = s with t,s in (0,1)
Walter Roberson
Walter Roberson 2018-11-16
编辑：Walter Roberson 2018-11-16
(This is not a correct interpretation interpretation of the equations; see below. I leave it in for purposes of the conversation.)
x = s; y = 0; z = 0; p = 0; r = s;
x = 0; y = 0; z = t; p = t; r = 0;
are the two parameterized solutions. There are also solutions for some particular t and s, as indicated in the table above.
Yuri Caridi
Yuri Caridi 2018-11-16
编辑：Yuri Caridi 2018-11-16
Sorry, I can't explain myself.
Coul Matlab answer me the solution of this system, for example, in this way
x = combination of p and r
y = combination of p and r
z = combination of p and r
p = p
r = r
thank you
Walter Roberson
Walter Roberson 2018-11-16
Are you looking for code to filter the table
>> [sol.x,sol.y,sol.z,sol.p,sol.r]
ans =
[ 1 - z1, z1, 0, 0, 0]
[ 0, 1 - z1, z1, 0, 0]
[ 0, -1, 2, 3, -1]
[ 1 - z2 - z1, z1, z2, 0, 0]
[ z1, 1 - 2*z1, z1, 1, 1]
[ 1, 0, 0, 0, z1]
[ 0, 0, 1, z1, 0]
to automatically locate the entries in which p or q are not constant? In other words, code to isolate down to the last two of those rows?
There are two solutions, which I mis-stated above
x = 0*p + 0*r + 1; y = 0*p + 0*r + 0; z = 0*p + 0*r + 0; 0; r
x = 0*p + 0*r + 0; y = 0*p + 0*r + 0; z = 0*p + 0*r + 1; p; 0;
There are no solutions for arbitrary p and r.
You can solve your equations incrementally, one equation at a time, first one for x, substitute that result into the second, solve for y, and so on. The solution for y (second one) is already conditional, valid only for some combinations of p and r. The substituted third equation becomes
(r*z*(r + p*r^2 - r^2 - 1))/(p - 1)
which features z as a constant multiple, and so has a general solution of z = 0. This forces specific values for x and y, independent of p and q.
Yuri Caridi
Yuri Caridi 2018-11-16
Yes, I was searching for this. Now its all clear.
Thank you very much for the patience.

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