Question bank — Stability of equilibria — stable, unstable, saddle, spiral, centre

False ::: $\Delta <0$ hone par do eigenvalues ke signs opposite hote hain, isliye ek hamesha grow karta hai — yeh ek saddle hai, jo unstable hota hai. Stability ke liye $\tau <0$ aur $\Delta >0$ dono saath chahiye.

False ::: $\tau <0$ ka sirf matlab hai ki eigenvalues ka sum negative hai. Agar $\Delta <0$ hai toh phir bhi saddle milega (e.g. ${\lambda }_1=-3,{\lambda }_2=+1$ se $\tau =-2<0$ milta hai lekin ek grow karta hai). Saath mein $\Delta >0$ bhi chahiye.

False ::: Centre sirf neutrally (marginally) stable hota hai. Nudge ki gayi trajectory ek nayi closed orbit par settle ho jaati hai aur fixed point par kabhi wapas nahi aati, isliye woh "converges as $t\to \infty$" ki test fail kar deti hai jo asymptotic stability ke liye zaroor hai. Dekho Lyapunov stability.

Mostly false ::: Complex eigenvalues ka matlab rotation hai, lekin agar real part exactly zero ho ($\tau =0$) toh centre milta hai (closed loops), spiral nahi. Sirf $\mathrm{Re}(\lambda )\ne 0$ se inward/outward spiral milta hai.

True, ek caveat ke saath ::: Dono real aur negative $\Rightarrow$ har mode decay karta hai $\Rightarrow$ stable node. Lekin "dono negative" ke liye pehle se $\Delta ={\lambda }_1{\lambda }_2>0$, $\tau ={\lambda }_1+{\lambda }_2<0$, aur $D\ge 0$ chahiye. $D\ge 0$ kyun? Kyunki eigenvalues $\frac{\tau \pm \sqrt{D}}{2}$ hote hain, aur agar $D<0$ ho toh woh square root imaginary hoga — roots ek complex pair honge, do real negatives nahi. Real eigenvalues ke liye non-negative discriminant zaroor chahiye.

False ::: Yeh sirf hyperbolic points par reliable hai, jahan kisi eigenvalue ka real part zero na ho (Linearisation and Hartman–Grobman theorem). Borderline centre par ($\tau =0$) discarded nonlinear terms ise ek slow spiral mein badal sakte hain.

$t\to -t$) ek stable node ko unstable node mein badal deta hai. True ::: Time reverse karne se har eigenvalue ka sign flip hota hai ($\lambda \to -\lambda$), isliye $\tau \to -\tau$ jabki $\Delta$ (ek product) unchanged rehta hai. Ek stable node ($\tau <0$) unstable node ($\tau >0$) ban jaata hai; saddle saddle hi rehta hai.

$D=0$ par hote hain. False ::: Spiral/centre region $D<0$ hai; star aur degenerate nodes exactly $D={\tau }^2-4\Delta =0$ par rehte hain. Parabola Trace–determinant plane mein real-eigenvalue nodes aur complex-eigenvalue spirals ke beech ki borderline hai.

$J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ mein $\tau =0,\Delta =1$ hai, isliye yeh ek stable centre hai." Error ::: Ek centre asymptotic sense mein stable nahi hota — ise sirf centre (neutrally stable) kaho. "Stable centre" kehna galat hai kyunki isse lagta hai trajectories origin par wapas aati hain.

$(\pi ,0)$ mein $\Delta =1>0$ hai, isliye yeh stable node hai." Error ::: Dobara compute karo: $-\cos \pi =+1$, jisse $J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ milta hai aur $\Delta =(0)(0)-(1)(1)=-1<0$ — yeh ek saddle hai. $\Delta$ ke sign mein galti ne ek unstable equilibrium ko fake stable mein badal diya.

Error ::: Jacobian matrix ko har equilibrium par alag se evaluate karna padta hai — partial derivatives position par depend karte hain (e.g. $-\cos x$, $x=0$ aur $x=\pi$ par alag hota hai), isliye ek hi system centre aur saddle dono host kar sakta hai.

$D<0$ ka matlab hai eigenvalues negative hain, isliye point stable hai." Error ::: $D<0$ eigenvalues ko complex banata hai, "negative" nahi — $D$ ek square root ke andar hai, woh khud $\lambda$ nahi hai. Resulting spiral ki stability $\mathrm{Re}(\lambda )=\tau /2$ se decide hoti hai, yaani $\tau$ ke sign se.

$D<0$ ke saath $\tau >0$ bhi theek hai — spirals bounded hote hain, isliye yeh blow up nahi kar sakta." Error ::: $D<0,\tau >0$ ek unstable spiral hai: $\mathrm{Re}(\lambda )=\tau /2>0$ ka matlab hai ki amplitude rotate karte hue badhta hai, isliye trajectories bahar ki taraf spiral karti hain aur bhaag jaati hain. Rotation akele boundedness guarantee nahi karta.

$\det (J-\lambda I)={\lambda }^2-\tau \lambda +\Delta$ hai, eigenvalues $\tau$ aur $\Delta$ hain." Error ::: $\tau$ aur $\Delta$ eigenvalues ka sum aur product hain (Vieta), eigenvalues khud nahi. Roots hain $\lambda =\frac{\tau \pm \sqrt{{\tau }^2-4\Delta }}{2}$.

Error ::: Ek saddle mein ek stable direction hoti hai (woh eigenvector jahan $\lambda <0$) jiske along kuch khaas trajectories aati hain. Sirf woh trajectories jo iske saath perfectly aligned nahi hain, eventually unstable direction ke along bhaag jaati hain.

$\Delta$ ka sign kyun batata hai ki eigenvalues same sign share karte hain? Kyunki $\Delta ={\lambda }_1{\lambda }_2$ ::: Ek product tabhi positive hota hai jab dono factors ka sign same ho, aur negative jab opposite signs hon — woh akela number "same sign ya nahi" encode karta hai bina roots solve kiye.

Kyunki $\tau ={\lambda }_1+{\lambda }_2=2\mathrm{Re}(\lambda )$ complex pairs ke liye ::: Trace (twice) average real part hota hai, aur real part $e^{\lambda t}$ mein growth/decay exponent hota hai — isliye $\tau <0$ ka matlab hai average mode andar decay karta hai.

Kyunki fixed point ke paas flow almost linear hota hai ::: First-order Taylor expansion curvy velocity functions $f,g$ ko ek constant matrix $J$ mein badal deta hai jiske eigenvalues hum exactly read kar sakte hain, jo local behaviour ka ek proof deta hai na ki sirf ek picture.

Kyunki light damping $D=4{\beta }^2-4{\omega }_0^2<0$ rakhta hai ::: $\beta <{\omega }_0$ ke saath discriminant negative hota hai, jo complex eigenvalues force karta hai, isliye rotation (oscillation) aur decay saath hote hain — yahi exactly ek stable spiral hai. Dekho Damped harmonic oscillator.

$2\times 2$ real system mein exactly ek complex eigenvalue kyun nahi ho sakta? Kyunki characteristic polynomial ke real coefficients hain ::: Real polynomial ke complex roots conjugate pairs mein aate hain, isliye $\lambda$ aur $\bar{\lambda }$ hamesha saath aate hain — ya toh do complex eigenvalues milenge ya koi nahi.

$\pm i\omega$ closed loops kyun produce karta hai na ki slow inward drift? Kyunki real part zero hai ::: $e^{(0\pm i\omega )t}$ ki magnitude constant hoti hai aur sirf rotate karta hai, isliye trajectories na shrink karti hain na grow — geometrically woh hamesha ke liye ellipses trace karti hain, jo centre ki signature hai.

$(\tau ,\Delta )$ plane mein ek parabola kyun hai? Kyunki woh boundary exactly $D=0$ hai ::: ${\tau }^2-4\Delta =0$ set karne par $\Delta ={\tau }^2/4$ milta hai, jo ek upward parabola ki equation hai. Iske upar ($\Delta >{\tau }^2/4$) $D<0$ hai isliye spiral; neeche $D>0$ hai isliye real-eigenvalue nodes — upar wala sketch aur Trace–determinant plane dekho.

$D={\tau }^2-4\Delta =0$ par $\tau <0$ ke saath kya hota hai? Ek degenerate/star stable node ::: Eigenvalues coincide karte hain (repeated real $\lambda =\tau /2<0$). Saare modes phir bhi decay karte hain isliye yeh stable hai, lekin trajectories do alag directions ki jagah ek single ya shared direction ke along approach karti hain.

$\Delta >0,\tau >0,D<0$ mein kaun sa behaviour hota hai? Ek unstable spiral ::: Complex eigenvalues with positive real part $\tau /2>0$: trajectories rotate karti hain jabki unka amplitude badhta hai, fixed point se bahar ki taraf spiral karte hue. Yeh stable spiral ka mirror image hai ($\tau$ ka sign flip karo).

$\Delta =0$ (with $\tau \ne 0$) kya signify karta hai? Equilibria ki poori ek line ::: Ek eigenvalue zero hai, isliye fixed point non-isolated hai: ek neutral direction hai jiske along system move nahi karta, aur node/spiral mein classification break down ho jaati hai.

$\tau =0$ aur $\Delta =0$ hon toh kya? Ek doubly degenerate case ::: Dono eigenvalues zero hain; linearisation bilkul uninformative hai aur fate completely nonlinear terms se decide hoti hai — yahan linear picture par kabhi trust mat karo.

Kya nonlinear terms ise preserve karte hain ::: Kyunki $\tau =0$ non-hyperbolic hai, Hartman–Grobman apply nahi hota; actual system ek weak spiral ho sakta hai. Genuine centre confirm karne ka usual tarika ek conserved quantity (energy) hota hai.

Sirf measure-zero wale ::: Trajectories jo exactly stable eigenvector par launch ki gayi hain woh fixed point approach karti hain, lekin koi bhi real-world nudge us line se hata kar eventually nikal jaata hai — isliye saddle overall unstable maana jaata hai.

$\Delta >0$, $\tau >0$, aur $D=0$ ho, toh kaun sa type milta hai? Ek degenerate unstable node ::: Positive product, positive sum, zero discriminant se repeated positive eigenvalue milta hai — har mode grow karta hai, isliye yeh repel karta hai; "star" geometry ka sirf matlab hai ki do directions merge ho gayi hain.