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Liapunov stability test (G = [0 1; −0.5 1])

Short answer

Solve GᵀPG − P = −I using G=[0 1; −0.5 1]. The solution P=[2.2 −1.6; −1.6 4.8] has leading minors 2.2>0 and determinant 8>0, so it is positive definite and the system is asymptotically stable.

Check the stability of given system by Liapunov stability test: [x₁(k+1); x₂(k+1)]=[0 1; -0.5 1][x₁(k); x₂(k)].

Liapunov stability theorem (discrete time): the equilibrium x=0\mathbf{x} = \mathbf{0} of x(k+1)=Gx(k)\mathbf{x}(k+1) = \mathbf{G}\mathbf{x}(k) is asymptotically stable if and only if, for any positive-definite Hermitian matrix Q\mathbf{Q}, there is a positive-definite matrix P\mathbf{P} satisfying

GTPG−P=−Q\mathbf{G}^T\mathbf{P}\mathbf{G} - \mathbf{P} = -\mathbf{Q}

Then V(x)=xTPxV(\mathbf{x}) = \mathbf{x}^T\mathbf{P}\mathbf{x} is a Liapunov function, and ΔV=xT(GTPG−P)x=−xTQx<0\Delta V = \mathbf{x}^T(\mathbf{G}^T\mathbf{P}\mathbf{G}-\mathbf{P})\mathbf{x} = -\mathbf{x}^T\mathbf{Q}\mathbf{x} < 0.

Choose Q=I\mathbf{Q} = \mathbf{I} and P=[p11p12p12p22]\mathbf{P} = \begin{bmatrix} p_{11} & p_{12} \\ p_{12} & p_{22} \end{bmatrix}.

Forming the equations

G=[01−0.51],GT=[0−0.511]\mathbf{G} = \begin{bmatrix} 0 & 1 \\ -0.5 & 1 \end{bmatrix},\qquad \mathbf{G}^T = \begin{bmatrix} 0 & -0.5 \\ 1 & 1 \end{bmatrix} PG=[−0.5p12p11+p12−0.5p22p12+p22]\mathbf{P}\mathbf{G} = \begin{bmatrix} -0.5p_{12} & p_{11}+p_{12} \\ -0.5p_{22} & p_{12}+p_{22} \end{bmatrix} GTPG=[0.25p22−0.5(p12+p22)−0.5(p12+p22)p11+2p12+p22]\mathbf{G}^T\mathbf{P}\mathbf{G} = \begin{bmatrix} 0.25p_{22} & -0.5(p_{12}+p_{22}) \\ -0.5(p_{12}+p_{22}) & p_{11}+2p_{12}+p_{22} \end{bmatrix}

Setting GTPG−P=−I\mathbf{G}^T\mathbf{P}\mathbf{G} - \mathbf{P} = -\mathbf{I} element by element:

(1,1):0.25p22−p11=−1(1,2):−0.5p12−0.5p22−p12=0  ⇒  p22=−3p12(2,2):p11+2p12+p22−p22=−1  ⇒  p11=−1−2p12\begin{aligned} (1,1):&\quad 0.25p_{22} - p_{11} = -1\\ (1,2):&\quad -0.5p_{12} - 0.5p_{22} - p_{12} = 0 \;\Rightarrow\; p_{22} = -3p_{12}\\ (2,2):&\quad p_{11} + 2p_{12} + p_{22} - p_{22} = -1 \;\Rightarrow\; p_{11} = -1 - 2p_{12} \end{aligned}

Solving

Substitute into (1,1): 0.25(−3p12)−(−1−2p12)=−10.25(-3p_{12}) - (-1-2p_{12}) = -1, so 1.25p12=−21.25p_{12} = -2 and p12=−1.6p_{12} = -1.6. Then

p22=4.8,p11=−1+3.2=2.2p_{22} = 4.8,\qquad p_{11} = -1 + 3.2 = 2.2 P=[2.2−1.6−1.64.8]\mathbf{P} = \begin{bmatrix} 2.2 & -1.6 \\ -1.6 & 4.8 \end{bmatrix}

Test for positive definiteness (Sylvester's criterion)

  • Δ1=p11=2.2>0\Delta_1 = p_{11} = 2.2 > 0
  • Δ2=∣P∣=2.2×4.8−(1.6)2=10.56−2.56=8>0\Delta_2 = |\mathbf{P}| = 2.2\times4.8 - (1.6)^2 = 10.56 - 2.56 = 8 > 0

P\mathbf{P} is positive definite (its eigenvalues are 1.44 and 5.56).

Conclusion

The origin is asymptotically stable in the large, and V(x)=2.2x12−3.2x1x2+4.8x22V(\mathbf{x}) = 2.2x_1^2 - 3.2x_1x_2 + 4.8x_2^2 is a Liapunov function.

Cross-check: ∣zI−G∣=z2−z+0.5=0|z\mathbf{I}-\mathbf{G}| = z^2 - z + 0.5 = 0 gives z=0.5±j0.5z = 0.5 \pm j0.5, with ∣z∣=0.707<1|z| = 0.707 < 1. Both eigenvalues are inside the unit circle, which agrees.

✓ Answer checked by Claude Opus 5.5 (high effort)

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