Exercises — Lyapunov stability — Lyapunov function, positive definiteness

(a) PD. $V(0)=0$; agar $\mathbf{x}\ne 0$ toh kam se kam ek coordinate nonzero hogi aur dono terms $\ge 0$ honge aur unka sum strictly $>0$ hoga. Ek genuine bowl. (b) PSD only. $V=0$ poori $x_2$-axis ke saath ($x_1=0,x_2$ arbitrary), sirf origin par nahi. Non-negative hai lekin origin ke bahar strictly positive nahi — ek flat channel. (c) Neither. $V(0,1)=-1<0$. Yeh ek saddle hai, bowl nahi. (d) PSD only. $V=0$ poori line $x_1=x_2$ ke saath. Non-negative hai lekin origin ke bahar zero hai.

$\dot{V}=-x_2^2\le 0$ negative semi-definite hai ($\dot{V}=0$ jab bhi $x_2=0$, jaise $x_1$-axis ke poore length par). Basic theorem sirf stability deta hai, asymptotic stability nahi. Asymptotic stability tak upgrade karne ke liye LaSalle's Invariance Principle chahiye.

Leading principal minors top-left $k\times k$ blocks ke determinants hain.

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

Saare leading principal minors positive hain ⟹ Sylvester's criterion se $P$ PD hai, toh $V$ PD hai. (Cross-check: eigenvalues $2\pm 1=\{1,3\}$ hain, dono $>0$.)

$\nabla V=(x_1,x_2)$, aur $\mathbf{f}=(-x_1^3,-x_2)$. Inhe dot karo: $\dot{V}=x_1(-x_1^3)+x_2(-x_2)=-x_1^4-x_2^2.$ Har $\mathbf{x}\ne 0$ ke liye yeh strictly negative hai ($x_1^4,x_2^2$ mein se ek positive hoga). $V$ PD aur $\dot{V}<0$ ⟹ origin par asymptotically stable. Yeh globally bhi asymptotically stable hai: $V=\frac{1}{2}(x_1^2+x_2^2)\to \infty$ jaise $\Vert \mathbf{x}\Vert \to \infty$, toh $V$ radially unbounded hai (page ke upar definition dekho), jo local conclusion ko global mein upgrade karta hai.

$\dot{V}=a{x}_1{\dot{x}}_1+x_2{\dot{x}}_2=a{x}_1{x}_2+x_2(-9x_1)=(a-9)x_1{x}_2$. $\dot{V}\equiv 0$ (saare $x_1,x_2$ ke liye identically zero) ke liye coefficient $(a-9)$ zero hona chahiye, toh $a=9$. Constraint check: problem $a>0$ restrict karta hai, aur indeed $9>0$ hai, toh yeh value allowed range ke andar hai — required $a$ allowed range mein hai. Yeh true energy $V=\frac{1}{2}(9x_1^2+x_2^2)$ deta hai, conserved. $\dot{V}=0\le 0$ ⟹ stable (orbits closed ellipses hain), lekin asymptotically stable nahi — bina damping ke yeh kabhi settle nahi hoga.

$\dot{V}=x_1(-x_1+2x_1^2{x}_2)+x_2(-x_2)=-x_1^2+2x_1^3{x}_2-x_2^2.$ Regroup kyun karein? Raw expression clearly-negative part ($-x_1^2$, $-x_2^2$) ko ek ambiguous middle term $2x_1^3{x}_2$ ke saath mix karta hai jiska sign quadrant ke saath flip hota hai. Yeh decide karne ke liye ki poori cheez kab negative hai, hum ambiguous term ko already $-x_1^2$ se attached piece mein factor karna chahte hain, taaki uncertainty ek single yes/no condition mein collapse ho jaaye. $-x_1^2$ ko do $x_1$-terms se bahar nikalo: $\dot{V}=-x_1^2(1-2x_1{x}_2)-x_2^2.$ Sign carefully padhna. Second term $-x_2^2$ hamesha $\le 0$ hai, aur yeh strictly negative hai jab $x_2\ne 0$ ho. First term $-x_1^2(1-2x_1{x}_2)$ $\le 0$ hota hai jaise hi bracket $1-2x_1{x}_2\ge 0$, yaani $x_1{x}_2\le \frac{1}{2}$. Toh closed region $\{x_1{x}_2\le \frac{1}{2}\}$ par humein already $\dot{V}\le 0$ milta hai, aur actually $\dot{V}<0$ har $\mathbf{x}\ne 0$ ke liye wahan hai: agar $x_2\ne 0$ toh $-x_2^2$ term ise strict banata hai; agar $x_2=0$ (lekin $\mathbf{x}\ne 0$, toh $x_1\ne 0$) toh bracket $1-0=1>0$ hai aur $-x_1^2\cdot 1<0$ ise strict banata hai. Khaas taur par, boundary $x_1{x}_2=\frac{1}{2}$ par bracket exactly $0$ hai, first term kill ho jaata hai, lekin $-x_2^2<0$ (wahan $x_2\ne 0$) phir bhi $\dot{V}<0$ deta hai. Toh certified set ko closed region $\{x_1{x}_2\le \frac{1}{2}\}$ liya ja sakta hai — strict inequality ki zaroorat nahi — aur yeh region of attraction ka ek guaranteed hissa hai.

Figure kya dikhata hai (caption). Figure L3.1 — $(x_1,x_2)$-plane. Green shaded area: har point $x_1{x}_2\le \frac{1}{2}$ ke saath, certified region jahan $\dot{V}<0$ hai. Red curves: boundary hyperbola $x_1{x}_2=\frac{1}{2}$ quadrants I aur III mein, jiske baad certificate decrease guarantee karna band kar deta hai. Poore 2nd aur 4th quadrants ($x_1{x}_2<0$) safely andar hain; origin interior mein baitha hai, jaisa ek equilibrium ke liye hona chahiye.

$\dot{V}=4x_1{\dot{x}}_1+x_2{\dot{x}}_2=4x_1{x}_2+x_2(-4x_1-x_2)=-x_2^2\le 0$. Spring terms $4x_1{x}_2$ cancel ho jaate hain; sirf damper bachta hai. $\dot{V}=0⟺x_2=0$. LaSalle: $\{x_2=0\}$ ke andar sabse bada invariant set find karo. Us set par ${\dot{x}}_2=-4x_1-0=-4x_1$; $x_2=0$ par rehne ke liye ${\dot{x}}_2=0$ chahiye, jo $x_1=0$ force karta hai. Sirf invariant subset $\{0\}$ hai, toh asymptotically stable. Physics agree karta hai: friction aakhirkar motion khatam kar deta hai.

$\dot{V}=x_1{x}_2+x_2(-x_1)=0$: energy conserved hai (undamped rotation, closed circular orbits). Sahi conclusion: stable, asymptotically stable nahi (yeh ek centre hai). Instability declare karna galat hai — orbits circles par bounded rehti hain. Prove karne ke liye instability tumhe Chetaev's Instability Theorem satisfy karne wala ek $V$ chahiye hoga; $\dot{V}<0$ na milna kuch prove nahi karta.

Pehle $M=A^{\top }P+PA$ entry by entry form karo, $A^{\top }=\left[\begin{array}{cc}0& -1\\ 1& -1\end{array}\right]$ ke saath. Multiply out karke: $A^{\top }P=\left[\begin{array}{cc}0& -1\\ 1& -1\end{array}\right]\left[\begin{array}{cc}{p}_{11}& p_{12}\\ p_{12}& p_{22}\end{array}\right]=\left[\begin{array}{cc}-p_{12}& -p_{22}\\ p_{11}-p_{12}& p_{12}-p_{22}\end{array}\right],$ $PA=\left[\begin{array}{cc}{p}_{11}& p_{12}\\ p_{12}& p_{22}\end{array}\right]\left[\begin{array}{cc}0& 1\\ -1& -1\end{array}\right]=\left[\begin{array}{cc}-p_{12}& p_{11}-p_{12}\\ -p_{22}& p_{12}-p_{22}\end{array}\right].$ Inhe add karo ($M=A^{\top }P+PA$, symmetric, toh teen independent entries) aur $M=-I$ set karo:

• $(1,1)$: $-p_{12}+(-p_{12})=-2p_{12}=-1\Rightarrow p_{12}=\frac{1}{2}$.
• $(1,2)$: $-p_{22}+(p_{11}-p_{12})=p_{11}-p_{12}-p_{22}=0$. Yeh relation kyun? $A^{\top }P$ ka $(1,2)$ entry $-p_{22}$ hai aur $PA$ ka $p_{11}-p_{12}$; add karne par $p_{11}-p_{12}-p_{22}$ milta hai, aur right side $-I$ ka off-diagonal $0$ hai — isliye $p_{11}-p_{12}-p_{22}=0$, yaani $p_{11}=p_{12}+p_{22}$.
• $(2,2)$: $(p_{12}-p_{22})+(p_{12}-p_{22})=2p_{12}-2p_{22}=-1\Rightarrow p_{22}=\frac{1+2p_{12}}{2}=1$.

Phir $p_{11}=p_{12}+p_{22}=\frac{1}{2}+1=\frac{3}{2}$: $P=\left[\begin{array}{cc}\frac{3}{2}& \frac{1}{2}\\ \frac{1}{2}& 1\end{array}\right].$ PD check (Sylvester): ${\Delta }_1=\frac{3}{2}>0$; ${\Delta }_2=\det P=\frac{3}{2}\cdot 1-\frac{1}{4}=\frac{5}{4}>0$. ✓ Toh $P$ PD hai, aur kyunki $A$ ke eigenvalues ka real part negative hai (woh $-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$ hain), origin asymptotically stable hai — Lyapunov Equation A^TP+PA=-Q dekho.

$\dot{V}=x\dot{x}=x(x^2+u)$. Hum chahte hain yeh $<0$ ho $x\ne 0$ ke liye. Destabilising $x^2$ cancel karo aur damping add karo: choose karo $u=-x^2-kx,k>0.$ Phir $\dot{x}=x^2+(-x^2-kx)=-kx$, aur $\dot{V}=x(-kx)=-k{x}^2<0$ $x\ne 0$ ke liye. Origin (globally) asymptotically stable hai. Yeh "cancel + stabilise" pattern Backstepping Control ka seed hai.

$\dot{V}=x\dot{x}=x(\sin x+u)=x(\sin x-\sin x-k\mathrm{sgn}(x))=-kx\mathrm{sgn}(x)$. $\mathrm{sgn}$ definition se identity $x\mathrm{sgn}(x)=|x|$ use karke, $\dot{V}=-k|x|<0$ $x\ne 0$ ke liye (aur $\dot{V}=0$ exactly $x=0$ par, jahan dono sides vanish hote hain kyunki $\mathrm{sgn}(0)=0$). Discontinuous $-k\mathrm{sgn}(x)$ term Sliding Mode Control ki hallmark hai: yeh $|x|$ ko constant rate $k$ par drive karta hai neeche, finite-time convergence deta hai — robust attitude control mein use hota hai.

Direct: $\dot{V}=x_1(-x_1+x_2)+x_2(-x_1-x_2)=-x_1^2+x_1{x}_2-x_1{x}_2-x_2^2=-x_1^2-x_2^2<0$ $\mathbf{x}\ne 0$ ke liye. Asymptotically stable. Matrix view: $A=\left[\begin{array}{cc}-1& 1\\ -1& -1\end{array}\right]$, $P=\frac{1}{2}I$. Phir $A^{\top }P+PA=\frac{1}{2}(A^{\top }+A)=\frac{1}{2}\left[\begin{array}{cc}-2& 0\\ 0& -2\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]=-I=-Q.$ $Q=I$ PD hai, $P$ PD hai ⟹ same conclusion. Cross-terms cancel ho gaye kyunki $A$ ka skew part $A^{\top }+A$ mein kuch contribute nahi karta.

Figure kya dikhata hai (caption). Figure L5.1 — blue closed loops bowl $V=\frac{1}{2}(x_1^2+x_2^2)$ ke level sets hain (circles, kyunki $V$ symmetric hai). Orange arrows flow field $\mathbf{f}=(-x_1+x_2,-x_1-x_2)$ hain, har point par state ki actual velocity. Har orange arrow ek higher circle se lower circle ki taraf point karta hai (level sets ke across inward) — $\dot{V}=-\Vert \mathbf{x}{\Vert }^2<0$ ka visual statement. Origin par red dot woh sink hai jis mein saare arrows spiral karte hain.

Diagonal genuinely fail karta hai. $\dot{V}=x_1(-x_1+4x_2)+x_2(-x_2)=-x_1^2+4x_1{x}_2-x_2^2$. Is quadratic form ki matrix $\left[\begin{array}{cc}-1& 2\\ 2& -1\end{array}\right]$ hai, jiske eigenvalues $-1\pm 2=\{1,-3\}$ hain — ek positive. Concretely $(x_1,x_2)=(1,1)$ par: $\dot{V}=-1+4-1=+2>0$. Toh us direction mein is bowl ki "energy" increase karti hai: diagonal candidate yahan bekaar hai, aur (crucially) iska matlab system unstable nahi hai — humne bas ek bura bowl choose kiya.

Lyapunov equation se sahi bowl banao. $A=\left[\begin{array}{cc}-1& 4\\ 0& -1\end{array}\right]$, toh $A^{\top }=\left[\begin{array}{cc}-1& 0\\ 4& -1\end{array}\right]$. $M=A^{\top }P+PA$ form karo symmetric $P$ ke saath aur $M=-I$ set karo. Teen independent entries work karke:

• $(1,1)$: $A^{\top }P$ ka $(1,1)$ entry $-p_{11}$ hai aur $PA$ ka $-p_{11}$, sum $=-2p_{11}=-1\Rightarrow p_{11}=\frac{1}{2}$.
• $(1,2)$: $A^{\top }P$ ka entry $-p_{12}$ hai, $PA$ ka $4p_{11}-p_{12}$; sum $=4p_{11}-2p_{12}=0\Rightarrow p_{12}=2p_{11}=1$.
• $(2,2)$: $A^{\top }P$ ka entry $4p_{12}-p_{22}$ hai, $PA$ ka $-p_{22}$; sum $=4p_{12}-2p_{22}=-1\Rightarrow p_{22}=\frac{1+4p_{12}}{2}=\frac{5}{2}$.

$P=\left[\begin{array}{cc}\frac{1}{2}& 1\\ 1& \frac{5}{2}\end{array}\right],{\Delta }_1=\frac{1}{2}>0,{\Delta }_2=\frac{1}{2}\cdot \frac{5}{2}-1=\frac{1}{4}>0.$ $P$ PD hai, toh $V={\mathbf{x}}^{\top }P\mathbf{x}$ ek valid Lyapunov function hai $\dot{V}=-\Vert \mathbf{x}{\Vert }^2<0$ ke saath everywhere. Off-diagonal $p_{12}=1$ bowl ko tilt karta hai taaki uske level sets sheared flow ke saath align ho jaayein — exactly woh kaam jo naive circular bowl nahi kar paaya.

Try karo $V=\frac{1}{2}(x_1^2+x_2^2)$. Yeh PD aur radially unbounded hai ($V\to \infty$ jaise $\Vert \mathbf{x}\Vert \to \infty$; page ke upar definition dekho — yahi cheez humein global stability claim karne degi). $\dot{V}=x_1(-x_1^3+x_2)+x_2(-x_1-x_2^3)=-x_1^4+x_1{x}_2-x_1{x}_2-x_2^4=-x_1^4-x_2^4.$ Linear cross-terms $x_1{x}_2$ exactly cancel ho jaate hain (skew coupling), $\dot{V}=-x_1^4-x_2^4<0$ bachta hai saare $\mathbf{x}\ne 0$ ke liye. $V$ radially unbounded aur $\dot{V}$ globally negative definite ke saath ⟹ globally asymptotically stable. ∎

$V$ PD aur $\dot{V}=-x_2^2$ given hai, origin ::: stable hai (semi-definite $\dot{V}$); asymptotic stability ke liye LaSalle ya strict $\dot{V}<0$ chahiye.