Use symbolic variable for lyapunov function

4 views (last 30 days)
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

Sign in to comment.

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
Torsten
Torsten on 9 Mar 2022
Ah, I didn't know this.
Thank you for the info.

Sign in to comment.

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.

Sign in to comment.

Products


Release

R2021a

Community Treasure Hunt

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

Start Hunting!

Translated by