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Use symbolic variable for lyapunov function

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I am trying to find a value for a lyapunov function but I do not know the numeric values. When I run the lyapunov command, I get an error that only numeric arrays can be used. Is there a way for using only symbolic variable to get the answer.
  4 Comments
Sam Chak
Sam Chak on 9 Mar 2022
I have tested and verified the results symbolically that holds.
clear all; clc
syms a b c
A = sym('A', [3 3]); % state matrix
P = sym('P', [3 3]); % positive definite matrix
A = [sym('0') sym('1') sym('0');
-a -b sym('0');
sym('0') c -c];
P = [((a^3 + 2*a^2*b*c + 2*a^2*c^2 + a^2 + a*b^2 + a*b*c + a*c^2 + b^3*c + b^2*c^2)/(2*a*b*(c^2 + b*c + a))) (1/(2*a)) (-a/(2*(c^2 + b*c + a)));
(1/(2*a)) ((a^2 + 2*a*c^2 + b*a*c + a + c^2 + b*c)/(2*a*b*(c^2 + b*c + a))) (c/(2*(c^2 + b*c + a)));
(-a/(2*(c^2 + b*c + a))) (c/(2*(c^2 + b*c + a))) (1/(2*c))];
Q = sym(eye(3)); % identity matrix
L = A.'*P + P*A + Q; % Lyapunov equation
simplify(L)
Result:

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Accepted Answer

Sam Chak
Sam Chak on 9 Mar 2022
If you are writing for a journal paper or a thesis, the following explanation might be helpful.
Let , , and .
There are a few ways to solve this symbolically.
syms a b c p11 p12 p22 p23 p33 p31
eqns = [1 - 2*a*p12 == 0, - a*p22 - b*p12 + c*p31 + p11 == 0, 1 - 2*b*p22 + 2*c*p23 + 2*p12 == 0, - b*p23 - c*p23 + c*p33 + p31 == 0, 1 - 2*c*p33 == 0, - a*p23 - c*p31 == 0];
S = solve(eqns);
sol = [S.p11; S.p12; S.p22; S.p23; S.p33; S.p31]
Result:
The result has been verified numerically:
clear all; clc
A = [0 1 0; -1 -2 0; 0 1 -1]
Q = eye(3)
P = lyap(A', Q)
A'*P + P*A
  5 Comments
Sam Chak
Sam Chak on 9 Mar 2022
Edited: Sam Chak on 9 Mar 2022
My apologies for failing to inform that P has to be a symmetric matrix . Allow me to quote the theorem directly from Prof. Hassan Khalil's book, "Nonlinear Control":
Theorem: A matrix A is Hurwitz if and only if for every positive definite symmetric matrix Q, there exists a positive definite symmetric matrix P that satisfies the Lyapunov equation . Moreover, if A is Hurwitz, then P is the unique solution.
From the property of symmetry, we know that , , and .
I'm still learning and not good at expressing the control law and equations in the symbolic form in MATLAB. That's why I worked out the equations manually and then used MATLAB to solve the derived set of linear equations. Thanks for your original script in solving the symbolic equations.
Torsten
Torsten on 9 Mar 2022
Ah, I didn't know this.
Thank you for the info.

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More Answers (1)

Torsten
Torsten on 9 Mar 2022
Edited: Torsten on 9 Mar 2022
syms k_p k_d h
A = sym('A', [3 3]);
X = sym('X', [3 3]);
A = [sym('0') sym('1') sym('0');
-k_p -k_d sym('0');
sym('0') sym('1')/h sym('-1')/h];
Q = sym(eye(3));
N = sym(zeros(3));
B = A.'*X + X*A + Q;
F = solve(B==N)
  1 Comment
Kashish Pilyal
Kashish Pilyal on 9 Mar 2022
Thank you for the answer but I have tried this method too. The matrix F in this case comes out to be empty. It is 0 by 1 symbolic. I actually managed to get the answer now. I had to write all equations seperately like this
syms P [3 3]
X= (A'*P)+(P*A);
eqnA=X(3,3)==-1;
eqnB=X(3,2)==0;
eqnC=X(3,1)==0;
eqnD=X(2,3)==0;
eqnE=X(1,3)==0;
eqnF=X(1,1)==-1;
eqnG=X(2,1)==0;
eqnH=X(2,2)==-1;
eqnI=X(1,2)==0;
Then I used solve and got the values of each element although the lyapunov function equation A'P+PA=-Q (in my case I) does not hold properly still. In other words the lyapunov equation does not give the identity matrix as output.

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