Use symbolic variable for lyapunov function

Question

Kashish Pilyal 2022-3-8

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编辑： Sam Chak 2022-3-9

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.

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Sam Chak 2022-3-9

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Hi @Kashish Pilyal

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:

Kashish Pilyal 2022-3-9

@Sam Chak Thank you for the help.

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Answer 1

Sam Chak 2022-3-9

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在 MATLAB Online 中打开

Hi @Kashish Pilyal

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

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Sam Chak 2022-3-9

编辑：Sam Chak 2022-3-9

Hi @Torsten

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 2022-3-9

Ah, I didn't know this.

Thank you for the info.

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Answer 2

Torsten 2022-3-9

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编辑：Torsten 2022-3-9

在 MATLAB Online 中打开

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)

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Kashish Pilyal 2022-3-9

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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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