# Use symbolic variable for lyapunov function

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Kashish Pilyal on 8 Mar 2022
Edited: Sam Chak on 9 Mar 2022
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.
Kashish Pilyal on 9 Mar 2022
@Sam Chak Thank you for the help.

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

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)
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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