Solve non-linear equations system parametrically, Why are the answers empty?

2 次查看（过去 30 天）
Arash Pourkazemi
Arash Pourkazemi 2021-2-26
评论： Walter Roberson 2021-2-28
clear all;
clc;
syms X1 X2 X3 X4 X5 X6 X7 X8 X9 h k1 Lz1 k21 Lz2 k32 Lz3 a1 a2 a3 b1 b2 b3 w1 w2 w3 r1 r2 r3;
e0 = X1-1 == 0;
e1 = h*X1-k1*X7*X4==0;
e2 = X1*cos(X7*Lz1)+X4*sin(X7*Lz1)-X2==0;
e3 = -X1*X6*sin(X7*Lz1)+X4*X6*cos(X7*Lz1)-X8*X5*k21==0;
e4 = X2*cos(X8*Lz2)+X5*sin(X8*Lz2)-X3==0;
e5 = -X2*X8*sin(X8*Lz2)+X5*X8*cos(X8*Lz2)-X9*X6*k32==0;
e6 = X3*cos(X9*Lz3)+X6*sin(X9*Lz3)==0;
e7 = (a1/a2)*(b1^2+r1^2+X7^2+w1^2)-b2^2-r2^2-w2^2-X8^2 ==0;
e8 = (a1/a3)*(b1^2+r1^2+X7^2+w1^2)-b3^2-r3^2-w3^2-X9^2 ==0;
sol = solve([e0,e1,e2,e3,e4,e5,e6,e7,e8],[X1 X2 X3 X4 X5 X6 X7 X8 X9]);
Why are the answers empty?
X1: [0×1 sym]
X2: [0×1 sym]
X3: [0×1 sym]
X4: [0×1 sym]
X5: [0×1 sym]
X6: [0×1 sym]
X7: [0×1 sym]
X8: [0×1 sym]
X9: [0×1 sym]

请先登录，再回答此问题。

回答（3 个）

darova
darova 2021-2-26
You didn't assign values to these variables
% h k1 Lz1 k21 Lz2 k32 Lz3 a1 a2 a3 b1 b2 b3 w1 w2 w3 r1 r2 r3;
  5 个评论
显示 3更早的评论

请先登录，再进行评论。

Walter Roberson
Walter Roberson 2021-2-27
There are four solution sets, that look something like
[X1 == 1,
X2 == (cos(Lz1*h*sin(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))/cos(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))/k1)*sin(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))+cos(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))*sin(Lz1*h*sin(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))/cos(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^
2*Z^2*k1^2-Lz1^2*h^2))/k1))/sin(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2)),
X3 == cos(Lz2*RootOf((Lz1^2*a2*h^2+RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2)^2*a2*k1^2)*Z^2-Lz1^2*a1*b1^2*h^2*k1^2-Lz1^2*a1*h^2*k1^2*r1^2-Lz1^2*a1*h^2*k1^2*w1^2+Lz1^2*a2*b2^2*h^2*k1^2+Lz1^2*a2*h^2*k1^2*r2^2+Lz1^2*a2*h^2*k1^2*w2^2-RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2)^2*a1*h^2*k1^2)/cos(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))/k1)*(cos(Lz1*h*sin(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))/cos(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))/k1)*sin(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))+cos(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))*sin(Lz1*h*sin(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))/cos(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))/k1))/sin(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2)),
X4 == cos(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))/sin(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2)),
X5 == 0,
X6 == 0,
X7 == h*sin(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))/cos(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))/k1,
X8 == RootOf((Lz1^2*a2*h^2+RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2)^2*a2*k1^2)*Z^2-Lz1^2*a1*b1^2*h^2*k1^2-Lz1^2*a1*h^2*k1^2*r1^2-Lz1^2*a1*h^2*k1^2*w1^2+Lz1^2*a2*b2^2*h^2*k1^2+Lz1^2*a2*h^2*k1^2*r2^2+Lz1^2*a2*h^2*k1^2*w2^2-RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2)^2*a1*h^2*k1^2)/cos(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))/k1,
X9 == RootOf((Lz1^2*a3*h^2+RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2)^2*a3*k1^2)*Z^2-Lz1^2*a1*b1^2*h^2*k1^2-Lz1^2*a1*h^2*k1^2*r1^2-Lz1^2*a1*h^2*k1^2*w1^2+Lz1^2*a3*b3^2*h^2*k1^2+Lz1^2*a3*h^2*k1^2*r3^2+Lz1^2*a3*h^2*k1^2*w3^2-RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2)^2*a1*h^2*k1^2)/cos(RootOf(Lz1^2*cos(Z)^2*h^2+cos(Z)^2*Z^2*k1^2-Lz1^2*h^2))/k1]
Here, RootOf() an expression in Z indicates the set of values for Z such that the expression becomes 0 -- the roots of the equation.
However, the RootOf are nested here and it gets confusing as to which Z refers to which level.
These RootOf are of nonlinear equations, such as RootOf(Lz1^2*cos(Z)^2*h^2 + cos(Z)^2*Z^2*k1^2 - Lz1^2*h^2), which contains a constant*Z^2*cos(Z) sub-expression. You are not likely to find closed-form expressions for them.
In my tests in Maple, the first seven expressions could be solved for the first 7 variables, giving a family of four solutions. You could then do substitutions of family members one at a time into the remaining two equations, solve, and back-substitute to complete the family member. You probably cannot do the same thing in MATLAB, as MATLAB tends to just give up if it discovers it needs to take the root of a nonlinear expression.
Mahsa Babaee
Mahsa Babaee 2021-2-28
编辑：Mahsa Babaee 2021-2-28
Dear @Arash Pourkazemi
I am trying to solve exactly the same problem and I faced similar difficulties. I galad to have your contanct info for more exchanges.
  4 个评论
显示 2更早的评论
Walter Roberson
Walter Roberson 2021-2-28
I solved the equations in Maple.
In my tests in Maple, the first seven expressions could be solved for the first 7 variables, giving a family of four solutions. You could then do substitutions of family members one at a time into the remaining two equations, solve, and back-substitute to complete the family member. You probably cannot do the same thing in MATLAB, as MATLAB tends to just give up if it discovers it needs to take the root of a nonlinear expression.
Walter Roberson
Walter Roberson 2021-2-28
Ran in:
syms X1 X2 X3 X4 X5 X6 X7 X8 X9 a1 a2 a3 b1 b2 b3 h k1 k21 k32 Lz1 Lz2 Lz3 w1 w2 w3 r1 r2 r3;
e0 = X1-1 == 0;
e1 = h*X1-k1*X7*X4==0;
e2 = X1*cos(X7*Lz1)+X4*sin(X7*Lz1)-X2==0;
e3 = -X1*X6*sin(X7*Lz1)+X4*X6*cos(X7*Lz1)-X8*X5*k21==0;
e4 = X2*cos(X8*Lz2)+X5*sin(X8*Lz2)-X3==0;
e5 = -X2*X8*sin(X8*Lz2)+X5*X8*cos(X8*Lz2)-X9*X6*k32==0;
e6 = X3*cos(X9*Lz3)+X6*sin(X9*Lz3)==0;
e7 = (a1/a2)*(b1^2+r1^2+X7^2+w1^2)-b2^2-r2^2-w2^2-X8^2 ==0;
e8 = (a1/a3)*(b1^2+r1^2+X7^2+w1^2)-b3^2-r3^2-w3^2-X9^2 ==0;
sol027 = solve([e0, e2, e7], [X1, X7, X8])
sol027 = struct with fields:
X1: [4×1 sym] X7: [4×1 sym] X8: [4×1 sym]
sol027.X1
ans = 
sol027.X7
ans = 
sol027.X8
ans = 
eqn2a = (subs([e1; e3; e4; e5; e6; e8], [X1, X7, X8], [sol027.X1(1), sol027.X7(1), sol027.X8(1)]))
eqn2a = 
sol027a1 = solve(eqn2a(1:4),[X2,X3,X5,X9],'returnconditions', true)
sol027a1 = struct with fields:
X2: [1×1 sym] X3: [1×1 sym] X5: [1×1 sym] X9: [1×1 sym] parameters: [1×0 sym] conditions: [1×1 sym]
sol027a1.X2
ans = 
sol027a1.X5
ans = 
sol027a1.X9
ans = 
simplify(sol027a1.conditions)
ans = 
eqn3a = subs(eqn2a(5:end), [X2, X5, X9], [sol027a1.X2, sol027a1.X5, sol027a1.X9])
eqn3a = 
sol3a_4 = solve(eqn3a(1), X4)
Warning: Unable to find explicit solution. For options, see help.
sol3a_4 = Empty sym: 0-by-1
sol3a_6 = solve(eqn3a(1), X6)
Warning: Unable to find explicit solution. For options, see help.
sol3a_6 = Empty sym: 0-by-1
You can play around with the order to see if you can get one more variable solved for, but you are unlikely to get all variables solved for.
Also, in the step forming eqn2a you could choose 2nd, 3rd, or 4th solutions to the previous equations, as that could certainly make a difference in whether you can solve.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Symbolic Math Toolbox 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by