Exercises — Lyapunov stability — Lyapunov function, positive definiteness

(a) PD. $V(0)=0$; if $\mathbf{x}\ne 0$ then at least one coordinate is nonzero and both terms are $\ge 0$ with the sum strictly $>0$. A genuine bowl. (b) PSD only. $V=0$ along the entire $x_2$-axis ($x_1=0,x_2$ arbitrary), not just the origin. Non-negative but not strictly positive off the origin — a flat channel. (c) Neither. $V(0,1)=-1<0$. It is a saddle, not a bowl. (d) PSD only. $V=0$ along the whole line $x_1=x_2$. Non-negative but zero off the origin.

$\dot{V}=-x_2^2\le 0$ is negative semi-definite ($\dot{V}=0$ whenever $x_2=0$, e.g. all along the $x_1$-axis). The basic theorem gives stability only, not asymptotic stability. Upgrading to asymptotic stability needs LaSalle's Invariance Principle.

The leading principal minors are the determinants of the top-left $k\times k$ blocks.

• $1\times 1$: ${\Delta }_1=2>0$. ✓
• $2\times 2$: ${\Delta }_2=\det P=(2)(2)-(1)(1)=3>0$. ✓

All leading principal minors positive ⟹ by Sylvester's criterion $P$ is PD, so $V$ is PD. (Cross-check: eigenvalues are $2\pm 1=\{1,3\}$, both $>0$.)

$\nabla V=(x_1,x_2)$, and $\mathbf{f}=(-x_1^3,-x_2)$. Dot them: $\dot{V}=x_1(-x_1^3)+x_2(-x_2)=-x_1^4-x_2^2.$ For every $\mathbf{x}\ne 0$ this is strictly negative (one of $x_1^4,x_2^2$ is positive). $V$ PD and $\dot{V}<0$ ⟹ asymptotically stable at the origin. It is even globally asymptotically stable: $V=\frac{1}{2}(x_1^2+x_2^2)\to \infty$ as $\Vert \mathbf{x}\Vert \to \infty$, so $V$ is radially unbounded (see the definition at the top), which upgrades the local conclusion to a global one.

$\dot{V}=a{x}_1{\dot{x}}_1+x_2{\dot{x}}_2=a{x}_1{x}_2+x_2(-9x_1)=(a-9)x_1{x}_2$. For $\dot{V}\equiv 0$ (identically zero for all $x_1,x_2$) the coefficient $(a-9)$ must vanish, so $a=9$. Check the constraint: the problem restricts us to $a>0$, and indeed $9>0$, so this value is admissible — the required $a$ lives inside the allowed range. This gives the true energy $V=\frac{1}{2}(9x_1^2+x_2^2)$, conserved. $\dot{V}=0\le 0$ ⟹ stable (orbits are closed ellipses), but not asymptotically stable — with no damping it never settles.

$\dot{V}=x_1(-x_1+2x_1^2{x}_2)+x_2(-x_2)=-x_1^2+2x_1^3{x}_2-x_2^2.$ Why regroup? The raw expression mixes a clearly-negative part ($-x_1^2$, $-x_2^2$) with an ambiguous middle term $2x_1^3{x}_2$ whose sign flips with the quadrant. To decide when the whole thing is negative we want to factor the ambiguous term into a piece already attached to $-x_1^2$, so the uncertainty collapses into a single yes/no condition. Pull $-x_1^2$ out of the two $x_1$-terms: $\dot{V}=-x_1^2(1-2x_1{x}_2)-x_2^2.$ Reading off the sign carefully. The second term $-x_2^2$ is $\le 0$ always, and it is strictly negative whenever $x_2\ne 0$. The first term $-x_1^2(1-2x_1{x}_2)$ is $\le 0$ as soon as the bracket $1-2x_1{x}_2\ge 0$, i.e. $x_1{x}_2\le \frac{1}{2}$. So on the closed region $\{x_1{x}_2\le \frac{1}{2}\}$ we already have $\dot{V}\le 0$, and in fact $\dot{V}<0$ for every $\mathbf{x}\ne 0$ there: if $x_2\ne 0$ the $-x_2^2$ term makes it strict; if $x_2=0$ (but $\mathbf{x}\ne 0$, so $x_1\ne 0$) the bracket is $1-0=1>0$ and $-x_1^2\cdot 1<0$ makes it strict. In particular, on the boundary $x_1{x}_2=\frac{1}{2}$ the bracket is exactly $0$, killing the first term, but $-x_2^2<0$ (there $x_2\ne 0$) still gives $\dot{V}<0$. So the certified set can be taken as the closed region $\{x_1{x}_2\le \frac{1}{2}\}$ — no need for the strict inequality — and this is a guaranteed piece of the region of attraction.

What the figure shows (caption). Figure L3.1 — the $(x_1,x_2)$-plane. Green shaded area: every point with $x_1{x}_2\le \frac{1}{2}$, the certified region where $\dot{V}<0$. Red curves: the boundary hyperbola $x_1{x}_2=\frac{1}{2}$ in quadrants I and III, beyond which the certificate stops guaranteeing decrease. The whole 2nd and 4th quadrants ($x_1{x}_2<0$) lie safely inside; the origin sits in the interior, as it must for an equilibrium.

$\dot{V}=4x_1{\dot{x}}_1+x_2{\dot{x}}_2=4x_1{x}_2+x_2(-4x_1-x_2)=-x_2^2\le 0$. The spring terms $4x_1{x}_2$ cancel; only the damper survives. $\dot{V}=0⟺x_2=0$. LaSalle: find the largest invariant set inside $\{x_2=0\}$. On that set ${\dot{x}}_2=-4x_1-0=-4x_1$; to stay on $x_2=0$ we need ${\dot{x}}_2=0$, forcing $x_1=0$. The only invariant subset is $\{0\}$, so asymptotically stable. Physics agrees: friction eventually kills the motion.

$\dot{V}=x_1{x}_2+x_2(-x_1)=0$: energy is conserved (undamped rotation, closed circular orbits). Correct conclusion: stable, not asymptotically stable (it is a centre). Declaring instability is wrong — orbits stay bounded on circles. To prove instability you would need a $V$ satisfying Chetaev's Instability Theorem; a mere failure to get $\dot{V}<0$ proves nothing.

First form $M=A^{\top }P+PA$ entry by entry, with $A^{\top }=\left[\begin{array}{cc}0& -1\\ 1& -1\end{array}\right]$. Multiplying out: $A^{\top }P=\left[\begin{array}{cc}0& -1\\ 1& -1\end{array}\right]\left[\begin{array}{cc}{p}_{11}& p_{12}\\ p_{12}& p_{22}\end{array}\right]=\left[\begin{array}{cc}-p_{12}& -p_{22}\\ p_{11}-p_{12}& p_{12}-p_{22}\end{array}\right],$ $PA=\left[\begin{array}{cc}{p}_{11}& p_{12}\\ p_{12}& p_{22}\end{array}\right]\left[\begin{array}{cc}0& 1\\ -1& -1\end{array}\right]=\left[\begin{array}{cc}-p_{12}& p_{11}-p_{12}\\ -p_{22}& p_{12}-p_{22}\end{array}\right].$ Add them ($M=A^{\top }P+PA$, symmetric, so three independent entries) and set $M=-I$:

• $(1,1)$: $-p_{12}+(-p_{12})=-2p_{12}=-1\Rightarrow p_{12}=\frac{1}{2}$.
• $(1,2)$: $-p_{22}+(p_{11}-p_{12})=p_{11}-p_{12}-p_{22}=0$. Why this relation? The $(1,2)$ entry of $A^{\top }P$ is $-p_{22}$ and of $PA$ is $p_{11}-p_{12}$; adding gives $p_{11}-p_{12}-p_{22}$, and the right side $-I$ has a $0$ off-diagonal — hence $p_{11}-p_{12}-p_{22}=0$, i.e. $p_{11}=p_{12}+p_{22}$.
• $(2,2)$: $(p_{12}-p_{22})+(p_{12}-p_{22})=2p_{12}-2p_{22}=-1\Rightarrow p_{22}=\frac{1+2p_{12}}{2}=1$.

Then $p_{11}=p_{12}+p_{22}=\frac{1}{2}+1=\frac{3}{2}$: $P=\left[\begin{array}{cc}\frac{3}{2}& \frac{1}{2}\\ \frac{1}{2}& 1\end{array}\right].$ PD check (Sylvester): ${\Delta }_1=\frac{3}{2}>0$; ${\Delta }_2=\det P=\frac{3}{2}\cdot 1-\frac{1}{4}=\frac{5}{4}>0$. ✓ So $P$ PD, and since $A$'s eigenvalues have negative real part (they are $-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$), the origin is asymptotically stable — see Lyapunov Equation A^TP+PA=-Q.

$\dot{V}=x\dot{x}=x(x^2+u)$. We want this $<0$ for $x\ne 0$. Cancel the destabilising $x^2$ and add damping: pick $u=-x^2-kx,k>0.$ Then $\dot{x}=x^2+(-x^2-kx)=-kx$, and $\dot{V}=x(-kx)=-k{x}^2<0$ for $x\ne 0$. Origin is (globally) asymptotically stable. This "cancel + stabilise" pattern is the seed of Backstepping Control.

$\dot{V}=x\dot{x}=x(\sin x+u)=x(\sin x-\sin x-k\mathrm{sgn}(x))=-kx\mathrm{sgn}(x)$. Using the identity $x\mathrm{sgn}(x)=|x|$ from the $\mathrm{sgn}$ definition, $\dot{V}=-k|x|<0$ for $x\ne 0$ (and $\dot{V}=0$ exactly at $x=0$, where both sides vanish because $\mathrm{sgn}(0)=0$). The discontinuous $-k\mathrm{sgn}(x)$ term is the hallmark of Sliding Mode Control: it drives $|x|$ down at a constant rate $k$, giving finite-time convergence — used in robust attitude control.

Direct: $\dot{V}=x_1(-x_1+x_2)+x_2(-x_1-x_2)=-x_1^2+x_1{x}_2-x_1{x}_2-x_2^2=-x_1^2-x_2^2<0$ for $\mathbf{x}\ne 0$. Asymptotically stable. Matrix view: $A=\left[\begin{array}{cc}-1& 1\\ -1& -1\end{array}\right]$, $P=\frac{1}{2}I$. Then $A^{\top }P+PA=\frac{1}{2}(A^{\top }+A)=\frac{1}{2}\left[\begin{array}{cc}-2& 0\\ 0& -2\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]=-I=-Q.$ $Q=I$ is PD, $P$ is PD ⟹ same conclusion. The cross-terms cancelled because the skew part of $A$ contributes nothing to $A^{\top }+A$.

What the figure shows (caption). Figure L5.1 — blue closed loops are level sets of the bowl $V=\frac{1}{2}(x_1^2+x_2^2)$ (circles, since $V$ is symmetric). Orange arrows are the flow field $\mathbf{f}=(-x_1+x_2,-x_1-x_2)$, the state's actual velocity at each point. Every orange arrow points from a higher circle toward a lower one (inward across the level sets) — the visual statement of $\dot{V}=-\Vert \mathbf{x}{\Vert }^2<0$. The red dot at the origin is the sink all arrows spiral into.

Diagonal genuinely fails. $\dot{V}=x_1(-x_1+4x_2)+x_2(-x_2)=-x_1^2+4x_1{x}_2-x_2^2$. This quadratic form has matrix $\left[\begin{array}{cc}-1& 2\\ 2& -1\end{array}\right]$, whose eigenvalues are $-1\pm 2=\{1,-3\}$ — one positive. Concretely at $(x_1,x_2)=(1,1)$: $\dot{V}=-1+4-1=+2>0$. So along that direction the "energy" of this bowl increases: the diagonal candidate is worthless here, and (crucially) this does not mean the system is unstable — we just chose a bad bowl.

Build the right bowl via the Lyapunov equation. $A=\left[\begin{array}{cc}-1& 4\\ 0& -1\end{array}\right]$, so $A^{\top }=\left[\begin{array}{cc}-1& 0\\ 4& -1\end{array}\right]$. Form $M=A^{\top }P+PA$ with symmetric $P$ and set $M=-I$. Working the three independent entries:

• $(1,1)$: the $(1,1)$ entry of $A^{\top }P$ is $-p_{11}$ and of $PA$ is $-p_{11}$, summing to $-2p_{11}=-1\Rightarrow p_{11}=\frac{1}{2}$.
• $(1,2)$: entry of $A^{\top }P$ is $-p_{12}$, of $PA$ is $4p_{11}-p_{12}$; sum $=4p_{11}-2p_{12}=0\Rightarrow p_{12}=2p_{11}=1$.
• $(2,2)$: entry of $A^{\top }P$ is $4p_{12}-p_{22}$, of $PA$ is $-p_{22}$; sum $=4p_{12}-2p_{22}=-1\Rightarrow p_{22}=\frac{1+4p_{12}}{2}=\frac{5}{2}$.

$P=\left[\begin{array}{cc}\frac{1}{2}& 1\\ 1& \frac{5}{2}\end{array}\right],{\Delta }_1=\frac{1}{2}>0,{\Delta }_2=\frac{1}{2}\cdot \frac{5}{2}-1=\frac{1}{4}>0.$ $P$ is PD, so $V={\mathbf{x}}^{\top }P\mathbf{x}$ is a valid Lyapunov function with $\dot{V}=-\Vert \mathbf{x}{\Vert }^2<0$ everywhere. The off-diagonal $p_{12}=1$ tilts the bowl so its level sets align with the sheared flow — exactly what the naive circular bowl failed to do.

Try $V=\frac{1}{2}(x_1^2+x_2^2)$. It is PD and radially unbounded ($V\to \infty$ as $\Vert \mathbf{x}\Vert \to \infty$; see the definition at the top of the page — this is what will let us claim global stability). $\dot{V}=x_1(-x_1^3+x_2)+x_2(-x_1-x_2^3)=-x_1^4+x_1{x}_2-x_1{x}_2-x_2^4=-x_1^4-x_2^4.$ The linear cross-terms $x_1{x}_2$ cancel exactly (skew coupling), leaving $\dot{V}=-x_1^4-x_2^4<0$ for all $\mathbf{x}\ne 0$. With $V$ radially unbounded and $\dot{V}$ negative definite globally ⟹ globally asymptotically stable. ∎

Given $V$ PD and $\dot{V}=-x_2^2$, the origin is ::: stable (semi-definite $\dot{V}$); need LaSalle or strict $\dot{V}<0$ for asymptotic stability.